Design of Machine Elements - V.M. Faires.pdf

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SECTION 1– DESIGN FOR SIMPLE STRESSES TENSION, COMPRESSION, SHEAR DESIGN PROBLEMS 1.

The link shown, made of AISI C1045 steel, as rolled, is subjected to a tensile load of 8000 lb. Let h = 1.5b . If the load is repeated but not reversed, determine the dimensions of the section with the design based on (a) ultimate strength, (b) yield strength. (c) If this link, which is 15 in. long., must not elongate more than 0.005 in., what should be the dimensions of the cross section?

Problems 1 – 3. Solution: For AISI C1045 steel, as rolled (Table AT 7) su = 96 ksi s y = 59 ksi E = 30 × 10 6 psi F A where F = 8000 lb A = bh but h = 1.5b therefore A = 1.5b 2 sd =

(a) Based on ultimate strength N = factor of safety = 6 for repeated but not reversed load (Table 1.1) s F sd = u = N A 96,000 8000 = 6 1.5b 2 5 b = 0.577 in say in . 8

Page 1 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES

h = 1.5b =

15 in 16

(b) Based on yield strength N = factor of safety = 3 for repeated but not reversed load (Table 1.1) s F sd = u = N A 59,000 8000 = 3 1.5b 2 9 in . b = 0.521 in say 16 27 h = 1.5b = in 32

(c) Elongation = δ =

FL AE

where, δ = 0.005 in F = 8000 lb E = 30× 10 6 psi L = 15 in A = 1.5b 2 then, FL δ= AE (8000)(15) 0.005 = 1.5b 2 30 × 106 3 b = 0.730 in say in . 4 1 h = 1.5b = 1 in 8

(

2.

)(

)

The same as 1 except that the material is malleable iron, ASTM A47-52, grade 35 018.

Solution: For malleable iron, ASTM A47-52, grade 35 018(Table AT 6) su = 55 ksi s y = 36.5 ksi E = 25× 10 6 psi

Page 2 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES F A where F = 8000 lb A = bh but h = 1.5b therefore A = 1.5b 2 sd =

(a) Based on ultimate strength N = factor of safety = 6 for repeated but not reversed load (Table 1.1) s F sd = u = N A 55,000 8000 = 6 1.5b 2 7 b = 0.763 in say in . 8 5 h = 1.5b = 1 in 16

(b) Based on yield strength N = factor of safety = 3 for repeated but not reversed load (Table 1.1) s F sd = u = N A 36,500 8000 = 3 1.5b 2 11 in . b = 0.622 in say 16 1 h = 1.5b = 1 in 32

(c) Elongation = δ = where, δ = 0.005 in F = 8000 lb E = 25× 10 6 psi L = 15 in A = 1.5b 2 then,

Page 3 of 131

FL AE



δ=

FL AE

0.005 =

(8000)(15)

(1.5b )(25 ×10 ) 2

b = 0.8 in say h = 1.5b = 1

3.

6

7 in . 8

5 in 16

The same as 1 except that the material is gray iron, ASTM 30.

Solution: For ASTM 30 (Table AT 6) su = 30 ksi , no s y E = 14.5 × 10 6 psi Note: since there is no s y for brittle materials. Solve only for (a) and (c) F A where F = 8000 lb A = bh but h = 1.5b therefore A = 1.5b 2 sd =

(a) Based on ultimate strength N = factor of safety = 7 ~ 8 say 7.5 (Table 1.1) s F sd = u = N A 30,000 8000 = 7 .5 1.5b 2 3 b = 1.1547 in say 1 in . 16 25 h = 1.5b = 1 in 32 FL (c) Elongation = δ = AE where, δ = 0.005 in F = 8000 lb

E = 14.5 × 10 6 psi

Page 4 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES L = 15 in A = 1.5b 2 then, FL δ= AE

0.005 =

(8000)(15)

(1.5b )(14.5 ×10 ) 2

6

b = 1.050 in say 1 h = 1.5b = 1

4.

1 in . 16

19 in 32

A piston rod, made of AISI 3140 steel, OQT 1000 F (Fig. AF 2), is subjected to a repeated, reversed load. The rod is for a 20-in. air compressor, where the maximum pressure is 125 psig. Compute the diameter of the rod using a design factor based on (a) ultimate strength, (b) yield strength.

Solution: From Fig. AF 2 for AISI 3140, OQT 1000 F su = 152.5 ksi s y = 132.5 ksi F = force =

π

(20)2 (125) = 39,270 lb = 39.27 kips

4 From Table 1.1, page 20 Nu = 8 Ny = 4

(a) Based on ultimate strength N F A= u su π 2 (8)(39.27 ) d = 4 152.5 5 d = 1.62 in say 1 in 8 (b) Based on yield strength NyF A= sy π 2 (4 )(39.27 ) d = 4 132.5

Page 5 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES 1 d = 1.23 in say 1 in 4

5.

A hollow, short compression member, of normalized cast steel (ASTM A27-58, 65 ksi), is to support a load of 1500 kips with a factor of safety of 8 based on the ultimate strength. Determine the outside and inside diameters if Do = 2 Di .

Solution: su = 65 ksi Nu = 8 F = 1500 kips A=

π

(D 4

2 o

)

− Di2 =

π

(4 D 4

2 i

)

− Di2 =

3πDi2 4

3πDi2 N u F (8)(1500) = A= = 4 65 su 7 Di = 8.85 in say 8 in 8 3  7 Do = 2 Di = 2 8  = 17 in 4  8 6.

A short compression member with Do = 2 Di is to support a dead load of 25 tons. The material is to be 4130 steel, WQT 1100 F. Calculate the outside and inside diameters on the basis of (a) yield strength, (b) ultimate strength.

Solution: From Table AT 7 for 4130, WQT 1100 F su = 127 ksi s y = 114 ksi From Table 1.1 page 20, for dead load N u = 3 ~ 4 , say 4 N y = 1.5 ~ 2 , say 2 Area, A =

π

(D 4

2 o

)

− Di2 =

π

(4 D 4

F = 25 tons = 50 kips

(a) Based on yield strength 3πDi2 N y F (2 )(50 ) A= = = 4 sy 114

Page 6 of 131

2 i

)

− Di2 =

3πDi2 4


SECTION 1– DESIGN FOR SIMPLE STRESSES 5 in 8 1 5 Do = 2 Di = 2  = 1 in 4 8 (b) Based on ultimate strength 3πDi2 N u F (4 )(50 ) = A= = 4 127 su 7 Di = 0.82 in say in 8 3 7 Do = 2 Di = 2  = 1 in 4 8

Di = 0.61 in say

7.

A round, steel tension member, 55 in. long, is subjected to a maximum load of 7000 lb. (a) What should be its diameter if the total elongation is not to exceed 0.030 in? (b) Choose a steel that would be suitable on the basis of yield strength if the load is gradually applied and repeated (not reversed).

Solution: (a) δ =

FL FL or A = δE AE

where, F = 7000 lb L = 55 in δ = 0.030 in E = 25× 10 6 psi π (7000)(55) A = d2 = 4 (0.030) 30 ×106 3 d = 0.74 in say in 4 (b) For gradually applied and repeated (not reversed) load Ny = 3

(

sy =

NyF A

=

)

(3)(7000) = 47,534 psi π (0.75)2 4

s y ≈ 48 ksi

say C1015 normalized condition ( s y = 48 ksi ) 8.

A centrifuge has a small bucket, weighing 0.332 lb. with contents, suspended on a manganese bronze pin (B138-A, ½ hard) at the end of a horizontal arm. If the pin is in double shear under the action of the centrifugal force, determine the diameter

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SECTION 1– DESIGN FOR SIMPLE STRESSES needed for 10,000 rpm of the arm. The center of gravity of the bucket is 12 in. from the axis of rotation. Solution: From Table AT 3, for B138-A, ½ hard sus = 48 ksi W F = ω 2r g where W = 0.332 lb g = 32.2 fps 2 2π n 2π (10,000 ) ω= = = 1047 rad sec 60 60 r = 12 in W 0.332 F = ω 2r = (1047 )2 (1) = 11,300 lb = 11.3 kips g 32.2 From Table 1.1, page 20 N = 3 ~ 4 , say 4 N F A= u su  π  (4 )(11.3) for double shear 2 d 2  = 48 4  25 in d = 0.774 in say 32

CHECK PROBLEMS 9.

The link shown is made of AISIC1020 annealed steel, with b =

3 in and 4

1 h = 1 in . (a) What force will cause breakage? (b) For a design factor of 4 based 2 on the ultimate strength, what is the maximum allowable load? (c) If N = 2.5 based on the yield strength, what is the allowable load?

Problem 9.

Page 8 of 131



Solution: For AISI C1020 annealed steel, from Table AT 7 su = 57 ksi s y = 42 ksi (a) F = su A  3  1  A = bh =  1  = 1.125 in 2  4  2  F = (57 )(1.125) = 64 kips s A (b) F = u Nu Nu = 4  3  1  A = bh =  1  = 1.125 in 2  4  2  (57 )(1.125) = 16 kips F= 4

(c) F =

sy A Ny

N y = 2 .5  3  1  A = bh =  1  = 1.125 in 2  4  2  (42)(1.125) = 18.9 kips F= 2

10.

A ¾-in.bolt, made of cold-finished B1113, has an effective stress area of 0.334 sq. in. and an effective grip length of 5 in. The bolt is to be loaded by tightening until the tensile stress is 80 % of the yield strength, as determined by measuring the total elongation. What should be the total elongation?

Solution: sL δ= E from Table AT 7 for cold-finished B1113 s y = 72 ksi then, s = 0.80s y = 0.8(72) = 57.6 ksi E = 30 ×106 psi = 30,000 ksi sL (57.6)(5) δ= = = 0.0096 in E 30,000

Page 9 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES 11.

A 4-lb. weight is attached by a 3/8-in. bolt to a rotating arm 14-in. from the center of rotation. The axis of the bolts is normal to the plane in which the centrifugal force acts and the bolt is in double shear. At what speed will the bolt shear in two if it is made of AISI B1113, cold finish?

Solution: From Table AT 7, sus = 62 ksi = 62,000 psi 2

 1  3 A = 2 (π )  = 0.2209 in 2  4 8 W F = ω 2 r = sus A g 4 ω 2 (14) = (62,000)(0.2209) 32.2 ω = 88.74 rad sec 2π n = 88.74 ω= 60 n = 847 rpm

12.

How many ¾-in. holes could be punched in one stroke in annealed steel plate of AISI C1040, 3/16-in. thick, by a force of 60 tons?

Solution: For AISI C1040, from Figure AF 1 su = 80 ksi sus = 0.75su = 0.75(80 ) ksi = 60 ksi A = π dt F = 60 tons = 120 kips n = number of holes F 120 n= = = 9 holes Asus (0.2209)(60 )

13.

What is the length of a bearing for a 4-in. shaft if the load on the bearing is 6400 lb. and the allowable bearing pressure is 200 psi of the projected area?

Solution: pDL = W where p = 200 psi D = 4 in W = 6400 lb

Page 10 of 131



(200)(4)L = 6400 L = 8 in

BENDING STRESSES DESIGN PROBLEMS 14.

A lever keyed to a shaft is L = 15 in long and has a rectangular cross section of h = 3t . A 2000-lb load is gradually applied and reversed at the end as shown; the material is AISI C1020, as rolled. Design for both ultimate and yield strengths. (a) What should be the dimensions of a section at a = 13 in ? (b) at b = 4 in ? (c) What should be the size where the load is applied?

Problem 14. Solution: For AISI C1020, as rolled, Table AT 7 su = 65 ksi s y = 49 ksi Design factors for gradually applied and reversed load Nu = 8 Ny = 4

th 3 , moment of inertial 12 but h = 3t h4 I= 36 I=

Moment Diagram (Load Upward)

Page 11 of 131



Based on ultimate strength s s= u Nu (a) s =

Mc Fac = I I

h 2 F = 2000 lbs = 2 kips (2)(13) h  65 2 = s= 4 8 h     36  h = 3.86 in h 3.86 t= = = 1.29 in 3 3 say 1 h = 4.5 in = 4 in 2 1 t = 1.5 in = 1 in 2 c=

(b) s =

Mc Fbc = I I

h 2 F = 2000 lbs = 2 kips (2)(4) h  65 2 = s= 4 8 h     36  h = 2.61 in h 2.61 t= = = 0.87 in 3 3 say h = 3 in t = 1 in c=

(c)

Page 12 of 131



3 − h 4.5 − 3 = 4 13 − 4 h = 2.33 in 1 − t 1.5 − 1 = 4 13 − 4 t = 0.78 in say 5 h = 2.625 in or h = 2 in 8

15.

A simple beam 54 in. long with a load of 4 kips at the center is made of cast steel, SAE 080. The cross section is rectangular (let h ≈ 3b ). (a) Determine the dimensions for N = 3 based on the yield strength. (b) Compute the maximum deflection for these dimensions. (c) What size may be used if the maximum deflection is not to exceed 0.03 in.?

Solution: For cast steel, SAE 080 (Table AT 6) s y = 40 ksi

E = 30×106 psi

Page 13 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES From Table AT 2 FL (4 )(54 ) = = 54 kips − in Max. M = 4 4 bh3 I= 12 but h = 3b h4 I= 36 sy

(a) s = c=

Ny

=

Mc I

h 2

(54) h 

40 2 = 3  h4     36  h = 4.18 in h 4.18 b= = = 1.39 in 3 3 1 h 4.5 1 say h = 4 in , b = = = 1.5 in = 1 in 2 3 3 2

(b) δ =

(c) δ =

0.03 =

FL3 = 48EI

(4000)(54)3 = 0.0384 in 3   ( )( ) 1 . 5 4 . 5 48(30 ×106 )  

12

 

FL3  h4  48E    36 

(4000)(54)3 (36)

48(30 ×106 )(h 4 ) h = 4.79 in h 4.79 b= = = 1.60 in 3 3 1 h 5.25 3 say h = 5.25 in = 5 in , b = = = 1.75 in = 1 in 4 3 3 4

Page 14 of 131



The same as 15, except that the beam is to have a circular cross section.

Solution: s Mc (a) s = y = Ny I

I=

πd4

64 d c= 2 d  M  32M 2 s=  4 = π d  π d3   64   40 32(54 ) = 3 πd3 d = 3.46 in 1 say d = 3 in 2

(b) δ =

I=

FL3 48EI

πd4

64 3 64 FL3 64(4000)(54 ) = = 0.0594 in δ= 48E (π d 4 ) 48(30 × 10 6 )(π )(3.5)4

(c) δ =

64 FL3 48E (π d 4 )

64(4000 )(54) 48(30 ×106 )(π )d 4 d = 4.15 in 1 say d = 4 in 4 3

0.03 =

17.

A simple beam, 48 in. long, with a static load of 6000 lb. at the center, is made of C1020 structural steel. (a) Basing your calculations on the ultimate strength, determine the dimensions of the rectangular cross section for h = 2b . (b) Determine the dimensions based on yield strength. (c) Determine the dimensions using the principle of “limit design.”

Page 15 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES Solution:

From Table AT 7 and Table 1.1 su = 65 ksi s y = 48 ksi

N u = 3 ~ 4 , say 4 N y = 1.5 ~ 2 , say 2 FL (6 )(48) = = 72 in − kips 4 4 Mc s= I h c= 2 bh3 I= 12 h but b = 2 4 h I= 24 h M  12 M 2 s = 4  = 3 h h 24 M=

(a) Based on ultimate strength s 12 M s= u = 3 Nu h 65 12(72 ) = 4 h3 h = 3.76 in

Page 16 of 131



b=

h 3.76 = = 1.88 in 2 2

h 3.75 3 7 = 1.875 in = 1 in say h = 3.75 in = 3 in , b = = 4 2 2 8

(b) Based on yield strength s y 12 M s= = 3 Ny h 48 12(72 ) = 2 h3 h = 3.30 in h 3.30 b= = = 1.65 in 2 2 h 3.5 1 3 say h = 3.5 in = 3 in , b = = = 1.75 in = 1 in 2 2 2 4 (c) Limit design (Eq. 1.6)

bh 2 4 h 2  h 2 72 = (48)   4 h = 2.29 in h 2.29 b= = = 1.145 in 2 2 1 h 2.5 1 say h = 2.5 in = 2 in , b = = = 1.25 in = 1 in 2 2 2 4 M = sy

18.

The bar shown is subjected to two vertical loads, F1 and F2 , of 3000 lb. each, that are L = 10 in apart and 3 in. ( a , d ) from the ends of the bar. The design factor is 4 based on the ultimate strength; h = 3b . Determine the dimensions h and b if the bar is made of (a) gray cast iron, SAE 111; (b) malleable cast iron, ASTM A4752, grade 35 018; (c) AISI C1040, as rolled (Fig. AF 1). Sketch the shear and moment diagrams approximately to scale.

Page 17 of 131



Problems18, 19. Solution: F1 = F2 = R1 = R2 = 3000 lb Moment Diagram

M = R1a = (3000)(3) = 9000 lbs − in = 9 kips − in N = factor of safety = 4 based on su bh3 12 h c= 2 h 3  h h4 3  I= = 12 36 I=

(a) For gray cast iron, SAE 111 su = 30 ksi , Table AT 6 h M  18M s Mc 2 s= u = = 4  = 3 N I h h     36  30 18(9 ) s= = 3 4 h h = 2.78 in h 2.78 b= = = 0.93 in 3 3 say h = 3.5 in , b = 1 in

(b) For malleable cast iron, ASTM A47-52, grade 35 018 Page 18 of 131



su = 55 ksi , Table AT 6 h M  18M s Mc 2 s= u = = 4  = 3 N I h h     36  55 18(9 ) s= = 3 4 h h = 2.28 in h 2.28 b= = = 0.76 in 3 3 1 3 say h = 2 in , b = in 4 4

(c) For AISI C1040, as rolled su = 90 ksi , Fig. AF 1 h M  18M s Mc 2 s= u = = 4  = 3 N I h h     36  90 18(9 ) s= = 3 4 h h = 1.93 in h 1.93 b= = = 0.64 in 3 3 7 5 say h = 1 in , b = in 8 8

19.

The same as 18, except that F1 acts up ( F2 acts down).

Solution:

[∑ M

A

=0

]

Page 19 of 131

R1 = R2 = 1875 lb


SECTION 1– DESIGN FOR SIMPLE STRESSES Shear Diagram

Moment Diagram

M = maximum moment = 5625 lb-in = 5.625 kips-in

(a) For gray cast iron su 18M = 3 N h 30 18(5.625) = 4 h3 h = 2.38 in h 2.38 b= = = 0.79 in 3 3 1 3 say h = 2 in , b = in 4 4 (b) For malleable cast iron s=

su 18M = 3 N h 55 18(5.625) = 4 h3 h = 1.95 in h 1.95 b= = = 0.65 in 3 3 7 5 say h = 1 in , b = in 8 8 s=

Page 20 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES (c) For AISI C1040, as rolled su 18M = 3 N h 90 18(5.625) = 4 h3 h = 1.65 in h 1.65 b= = = 0.55 in 3 3 1 1 say h = 1 in , b = in 2 2 s=

20.

The bar shown, supported at A and B , is subjected to a static load F of 2500 lb. at θ = 0 . Let d = 3 in , L = 10 in and h = 3b . Determine the dimensions of the section if the bar is made of (a) gray iron, SAE 110; (b) malleable cast iron, ASTM A47-52, grade 32 510; (c) AISI C1035 steel, as rolled. (d) For economic reasons, the pins at A, B, and C are to be the same size. What should be their diameter if the material is AISI C1035, as rolled, and the mounting is such that each is in double shear? Use the basic dimensions from (c) as needed. (e) What sectional dimensions would be used for the C1035 steel if the principle of “limit design” governs in (c)?

Problems 20, 21. Solution:

Page 21 of 131



[∑ M [∑ M

A

B

=0 =0

]

3RB = 13(2500)

]

RB = 10,833 lb 3RA = 10(2500 ) RA = 8333 lb

Shear Diagram

Moment Diagram

M = (2500)(10) = 25,000 lb − in = 25 kips − in h = 3b bh3 I= 12 h4 I= 36 h c= 2

h M  18M Mc 2 s= = 4  = 3 I h h     36 

(a) For gray cast iron, SAE 110 su = 20 ksi , Table AT 6 N = 5 ~ 6 , say 6 for cast iron, dead load s 18M s= u = 3 N h 20 18(25) = 6 h3

Page 22 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES h = 5.13 in h b = = 1.71 in 3 1 3 say h = 5 in , b = 1 in 4 4 (b) For malleable cast iron, ASTM A47-32 grade 32510 su = 52 ksi , s y = 34 ksi N = 3 ~ 4 , say 4 for ductile, dead load s 18M s= u = 3 N h 52 18(25) = 4 h3 h = 3.26 in h b = = 1.09 in 3 3 1 say h = 3 in , b = 1 in 4 4 (c) For AISI C1035, as rolled su = 85 ksi , s y = 55 ksi N = 4 , based on ultimate strength s 18M s= u = 3 N h 85 18(25) = 4 h3 h = 2.77 in h b = = 0.92 in 3 say h = 3 in , b = 1 in

(d) For AISI C1035, as rolled ssu = 64 ksi N = 4 , RB = 10.833 kips s R ss = su = B N A π   π A = 2 D 2  = D 2 4  2 64 10.833 ss = = π 2 4 D 2 D = 0.657 in

Page 23 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES 11 in 16 (e) Limit Design bh 2 M = sy 4 For AISI C1035 steel, s y = 55 ksi

say D =

b=

h 3

h 2  h 3 M = 25 = (55)   4 h = 1.76 in h b = = 0.59 in 3 7 5 say h = 1.875 in = 1 in , b = in 8 8

The same as 20, except that θ = 30o . Pin B takes all the horizontal thrust.

21.

Solution:

FV = F cos θ

[∑ M [∑ M

A

B

=0

=0

] ]

3RB = 13FV 3RB = 13(2500) cos 30 RB = 9382 lb 3RA = 10 FV 3RA = 10(2500 ) cos 30 RA = 7217 lb

Shear Diagram

Page 24 of 131



Moment Diagram

M = (2165)(10) = 21,650 lb − in = 21.65 kips − in 18M s= 3 h (a) For gray cast iron, SAE 110 su = 20 ksi , Table AT 6 N = 5 ~ 6 , say 6 for cast iron, dead load s 18M s= u = 3 N h 20 18(21.65) = 6 h3 h = 4.89 in h b = = 1.63 in 3 1 3 say h = 5 in , b = 1 in 4 4 (b) For malleable cast iron, ASTM A47-32 grade 32510 su = 52 ksi , s y = 34 ksi N = 3 ~ 4 , say 4 for ductile, dead load s 18M s= u = 3 N h 52 18(21.65) = 4 h3 h = 3.11 in h b = = 1.04 in 3 say h = 3 in , b = 1 in (c) For AISI C1035, as rolled su = 85 ksi , s y = 55 ksi N = 4 , based on ultimate strength

Page 25 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES su 18M = 3 N h 85 18(21.65) = 4 h3 h = 2.64 in h b = = 0.88 in 3 5 7 say h = 2 in , b = in 8 8 s=

(d) For AISI C1035, as rolled ssu = 64 ksi N = 4 , RBV = 9382 lb RBH = FH = F sin θ = 2500 sin 30 = 1250 lb 2 2 RB2 = RBV + RBH = (9382) + (1250) RB = 9465 lb s R ss = su = B N A π  π A = 2 D 2  = D 2 4  2 64 9.465 ss = = 4 π D2 2 D = 0.614 in 5 say D = in 8 (e) Limit Design bh 2 M = sy 4 For AISI C1035 steel, s y = 55 ksi 2

b=

h 3

h 2  h 3 M = 21.65 = (55)   4 h = 1.68 in h b = = 0.56 in 3 7 5 say h = 1.875 in = 1 in , b = in 8 8

Page 26 of 131

2



22.

A cast-iron beam, ASTM 50, as shown, is 30 in. long and supports two gradually applied, repeated loads (in phase), one of 2000 lb. at e = 10 in from the free end, and one of 1000 lb at the free end. (a) Determine the dimensions of the cross section if b = c ≈ 3a . (b) The same as (a) except that the top of the tee is below.

Problem 22. Solution: For cast iron, ASTM 50 su = 50 ksi , suc = 164 ksi For gradually applied, repeated load N = 7 ~ 8 , say 8 M = F1d + F2 (d + e ) where: F1 = 2000 lb F2 = 1000 lb d = 30 − 10 = 20 in d + e = 30 in M = (2000)(20) + (1000)(30) = 70,000 lb − in = 70 kips − in Mc I Solving for I , moment of inertia s=

(3a )(a ) a  + (3a )(a ) 5a  = [(3a )(a ) + (3a )(a )]y 2

y=

3a 2

Page 27 of 131

 2 



I=

(3a )(a )3 + (3a )(a )(a 2 ) + (a )(3a )3 + (3a )(a )(a 2 ) = 17 a 4 12

12

(a)

3a 2 5a cc = 2 Based on tension s Mct st = u = N I (70) 3a  50  2 = 8  17a 4     2  a = 1.255 in Based on compression s Mcc sc = uc = N I (70) 5a  164  2  = 8  17a 4     2  a = 1.001 in Therefore a = 1.255 in 1 Or say a = 1 in 4 And b = c = 3a = 3(1.25) = 3.75 in ct =

Page 28 of 131

2


SECTION 1– DESIGN FOR SIMPLE STRESSES 3 Or b = c = 3 in 4

(b) If the top of the tee is below

5a 2 3a cc = 2 17a 4 I= 2 M = 70 kips − in ct =

Based on tension s Mct st = u = N I (70) 5a  50  2  = 8  17 a 4     2  a = 1.488 in Based on compression s Mcc sc = uc = N I (70) 3a  164  2  = 8  17a 4     2  a = 0.845 in Therefore a = 1.488 in 1 Or say a = 1 in 2 1 And b = c = 3a = 4 in 2 CHECK PROBLEMS

Page 29 of 131



An I-beam is made of structural steel, AISI C1020, as rolled. It has a depth of 3 in. and is subjected to two loads; F1 and F2 = 2F1 ; F1 is 5 in. from one end and F2 is 5 in. from the other ends. The beam is 25 in. long; flange width is b = 2.509 in ; I x = 2.9 in 4 . Determine (a) the approximate values of the load to cause elastic failure, (b) the safe loads for a factor of safety of 3 based on the yield strength, (c) the safe load allowing for flange buckling (i1.24), (f) the maximum deflection caused by the safe loads.

Problems 23 – 25. Solution:

[∑ M [∑ F

V

A

=0

=0

]

]

5F1 + 20(2 F1 ) = 25RB RB = 1.8F1 F1 + 2 F1 = RA + RB RA = 3F1 − 1.8F1 = 1.2 F1

Shear Diagram

Moment Diagram

Page 30 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES M = 9F1 = maximum moment For AISI C1020, as rolled s y = 48 ksi Mc I d 3 where c = = = 1.5 in 2 2 (9 F1 )(1.5) s y = 48 = 2 .9 F1 = 10.31 kips F2 = 2 F1 = 20.62 kips

(a) s y =

sy

Mc N I 48 (9 F1 )(1.5) s= = 3 2 .9 F1 = 3.44 kips F2 = 2 F1 = 6.88 kips

(b) s =

(c)

=

L 25 = 9.96 < 15 (page 34) = b 2.509

sc = 20 ksi ( page 34, i1.24) Mc I (9 F1 )(1.5) 20 = 2 .9 F1 = 4.30 kips F2 = 2 F1 = 8.60 kips sc =

(d) For maximum deflection, by method of superposition, Table AT 2 3

ymax

Fb′  a(L + b′)  2 =  , a > b′ 3EIL  3

or 3

ymax

Fa  b′(L + a )  2 =  , b′ > a 3EIL  3

Page 31 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES ymax caused by F1 3

F a  b′(L + a1 )  2 ymax1 = 1 1  1  , b1′ > a1 3EIL  3 where E = 30,000 ksi a1 = 5 in b1′ = 20 in L = 25 in I = 2.9 in 4 3

ymax1

F1 (5)  20(25 + 5)  2 =  = 0.0022 F1 3(30,000)(2.9)(25)  3

ymax caused by F2 3

F b′  a (L + b2′ )  2 ymax 2 = 2 2  2  , a2 > b2′ 3EIL  3 where b2′ = 5 in a2 = 20 in 3

ymax 2

2 F1 (5)  20(25 + 5)  2 =  = 0.0043F1 3(30,000)(2.9)(25)  3

Total deflection = δ δ = ymax1 + ymax 2 = 0.022 F1 + 0.0043F1 = 0.0065 F1 Deflection caused by the safe loads in (a) δ a = 0.0065(10.31) = 0.067 in Deflection caused by the safe loads in (b) δ b = 0.0065(3.44) = 0.022 in Deflection caused by the safe loads in (c) δ c = 0.0065(4.30) = 0.028 in 24.

The same as 23, except that the material is aluminum alloy, 2024-T4, heat treated.

Solution: For aluminum alloy, 2024-T4, heat treated s y = 47 ksi (a) s y =

Mc I

Page 32 of 131



s y = 47 =

(9 F1 )(1.5)

2 .9 F1 = 10.10 kips F2 = 2 F1 = 20.20 kips

sy

Mc N I 47 (9 F1 )(1.5) s= = 3 2 .9 F1 = 3.36 kips F2 = 2 F1 = 6.72 kips

(b) s =

(c)

=

L 25 = 9.96 < 15 (page 34) = b 2.509

sc = 20 ksi ( page 34, i1.24) Mc I 9 F1 )(1.5) ( 20 = 2 .9 F1 = 4.30 kips F2 = 2 F1 = 8.60 kips sc =

(d) Total deflection = δ δ = ymax1 + ymax 2 = 0.022 F1 + 0.0043F1 = 0.0065 F1 Deflection caused by the safe loads in (a) δ a = 0.0065(10.10) = 0.066 in Deflection caused by the safe loads in (b) δ b = 0.0065(3.36) = 0.022 in Deflection caused by the safe loads in (c) δ c = 0.0065(4.30) = 0.028 in 25.

A light I-beam is 80 in. long, simply supported, and carries a static load at the midpoint. The cross section has a depth of d = 4 in , a flange width of b = 2.66 in , and I x = 6.0 in 4 (see figure). (a) What load will the beam support if it is made of C1020, as-rolled steel, and flange buckling (i1.24) is considered? (b) Consider the stress owing to the weight of the beam, which is 7.7 lb/ft, and decide whether or not the safe load should be less.

Page 33 of 131



Solution: (a) For C1020, as rolled, su = 65 ksi Consider flange buckling L 80 = = 30 b 2.66 L since 15 < < 40 b 22.5 22.5 sc = = = 15 ksi 2 2 30 ) ( L 1 +   1800 1 + 1800 b Mc s= I d 4 c = = = 2 in 2 2

From Table AT 2 FL F (80 ) = = 20 F M= 4 4 Mc s = sc = I 20 F )(2 ) ( 15 = 6 F = 2.25 kips , safe load (b) Considering stress owing to the weight of the beam wL2 (Table AT 2) 8 where w = 7.7 lb ft add’l M =

Page 34 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES wL2  7.7  (80) = = 513 lb − in = 0.513 kips − in  8  12  8 M = 20 F + 0.513 = total moment Mc s = sc = I (20 F + 0.513)(2) 15 = 6 F = 2.224 kips Therefore, the safe load should be less. 2

add’l M =

26.

What is the stress in a band-saw blade due to being bent around a 13 ¾-in. pulley? The blade thickness is 0.0265 in. (Additional stresses arise from the initial tension and forces of sawing.)

Solution:

t = 0.0265 = 0.01325 in 2 r = 13.75 + 0.01325 = 13.76325 in Using Eq. (1.4) page 11 (Text) Ec s= r where E = 30× 106 psi c=

(30 ×10 )(0.01325) = 28,881 psi s= 6

13.76325

27.

A cantilever beam of rectangular cross section is tapered so that the depth varies uniformly from 4 in. at the fixed end to 1 in. at the free end. The width is 2 in. and the length 30 in. What safe load, acting repeated with minor shock, may be applied to the free end? The material is AISI C1020, as rolled.

Solution: For AISI C1020, as rolled su = 65 ksi (Table AT 7) Designing based on ultimate strength, N = 6 , for repeated, minor shock load

Page 35 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES su 65 = = 10.8 ksi N 6 Loading Diagram s=

4 −1 h −1 = 30 x h = 0.10 x + 1 wh 3 I= 12 h c= 2 M = Fx

(Fx ) h 

Mc 3Fx  2  = 6 Fx = 3Fx = = 2 2 3 I 2h h  wh  (0.10 x + 1)2    12  Differentiating with respect to x then equate to zero to solve for x giving maximum stress.  (0.10 x + 1)2 (1) − 2( x )(0.10 x + 1)(0.10 )  ds = 3F  =0 dx (0.10 x + 1)4   0.10 x + 1 − 2(0.10 x ) = 0 x = 10 in h = 0.10(10) + 1 = 2 in s 3Fx s= u = 2 N h 3F (10 ) 10.8 = (2)2 F = 1.44 kips s=

TORSIONAL STRESSES DESIGN PROBLEMS

Page 36 of 131



A centrifugal pump is to be driven by a 15-hp electric motor at 1750 rpm. What should be the diameter of the pump shaft if it is made of AISI C1045 as rolled? Consider the load as gradually repeated.

Solution: For C1045 as rolled, s y = 59 ksi sus = 72 ksi Designing based on ultimate strength s s = us , N = 6 (Table 1.1) N 72 s= = 12 ksi 6 33,000hp 33,000(15) Torque, T = = = 45 ft − lb = 540 in − lb = 0.540 in − kips 2π n 2π (1750) For diameter, 16T s= π d3 16(0.540) 12 = π d3 d = 0.612 in 5 say d = in 8 29.

A shaft in torsion only is to transmit 2500 hp at 570 rpm with medium shocks. Its material is AISI 1137 steel, annealed. (a) What should be the diameter of a solid shaft? (b) If the shaft is hollow, Do = 2 Di , what size is required? (c) What is the weight per foot of length of each of these shafts? Which is the lighter? By what percentage? (d) Which shaft is the more rigid? Compute the torsional deflection of each for a length of 10 ft.

Solution: 33,000hp 33,000(2500) T= = = 23,036 ft − lb = 276 in − kips 2π n 2π (570 ) For AISI 1137, annealed s y = 50 ksi (Table AT 8) s ys = 0.6s y = 30 ksi

Designing based on yield strength N = 3 for medium shock, one direction

Page 37 of 131



Design stress s 30 s = ys = = 10 ksi N 3 (a) Let D = shaft diameter Tc J π D4 J= 32 D c= 2 16T s= π D3 16(276) 10 = π D3 D = 5.20 in 1 say D = 5 in 4 s=

(b) J =

[

π (Do4 − Di4 ) π (2 Di )4 − Di4 =

32 Do 2 Di c= = Di = 2 2 32T TDi s= = 4  15π Di  15π Di3    32  32(276) 10 = 15π Di3 Di = 2.66 in

32

] = 15π D

4 i

32

Do = 2 Di = 5.32 in say 5 Di = 2 in 8 1 Do = 5 in 4 (c) Density, ρ = 0.284 lb in3 (Table AT 7)

Page 38 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES For solid shaft w = weight per foot of length π  2 w = 12 ρ  D 2  = 3πρD 2 = 3π (0.284)(5.25) = 73.8 lb ft 4  For hollow shaft π  2 2 w = 12 ρ  (Do2 − Di2 ) = 3πρ (Do2 − Di2 ) = 3π (0.284) (5.25) − (2.625) = 55.3 lb ft 4 Therefore hollow shaft is lighter 73.8 − 55.3 Percentage lightness = (100%) = 33.5% 55.3

[

]

(d) Torsional Deflection TL JG where L = 10 ft = 120 in

θ=

G = 11.5 ×103 ksi For solid shaft, J =

θ=

π D4 32

(276)(120)  180  o = 0.039 rad = (0.039 )  = 2.2 π π  4 3    (5.25) (11.5 × 10 )  32 

For hollow shaft, J =

θ=

π (Do4 − Di4 ) 32

(276)(120)  180  o = 0.041 rad = (0.041)  = 2.4 π  4 4 3  π   [(5.25) − (2.625) ](11.5 × 10 )

 32  Therefore, solid shaft is more rigid, 2.2o < 2.4o

30.

The same as 29, except that the material is AISI 4340, OQT 1200 F.

Solution: 33,000hp 33,000(2500 ) T= = = 23,036 ft − lb = 276 in − kips 2π n 2π (570 ) For AISI 4340, OQT 1200 F s y = 130 ksi s ys = 0.6 s y = 0.6(130 ) = 78 ksi

Designing based on yield strength

Page 39 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES N = 3 for mild shock

Design stress s 78 = 26 ksi s = ys = N 3 (a) Let D = shaft diameter Tc J π D4 J= 32 D c= 2 16T s= π D3 16(276 ) 26 = π D3 D = 3.78 in 3 say D = 3 in 4 s=

(b) J =

[

π (Do4 − Di4 ) π (2 Di )4 − Di4 =

32 Do 2 Di c= = Di = 2 2 32T TDi s= = 4  15π Di  15π Di3    32  32(276 ) 26 = 15π Di3 Di = 1.93 in

32

] = 15π D

4 i

32

Do = 2 Di = 3.86 in say Di = 2 in Do = 4 in (c) Density, ρ = 0.284 lb in3 (Table AT 7)

Page 40 of 131



For solid shaft w = weight per foot of length π  2 w = 12 ρ  D 2  = 3πρD 2 = 3π (0.284)(3.75) = 37.6 lb ft 4  For hollow shaft π  2 2 w = 12 ρ  (Do2 − Di2 ) = 3πρ (Do2 − Di2 ) = 3π (0.284 ) (4 ) − (2 ) = 32.1 lb ft 4 Therefore hollow shaft is lighter 37.6 − 32.1 Percentage lightness = (100% ) = 17.1% 32.1

[

]

(d) Torsional Deflection TL JG where L = 10 ft = 120 in

θ=

G = 11.5 ×103 ksi For solid shaft, J =

π D4 32

(276)(120)  180  o = 0.148 rad = (0.148) θ=  = 8.48 π  4 3  π   (3.75) (11.5 ×10 )  32 

For hollow shaft, J =

θ=

π (Do4 − Di4 ) 32

(276)(120 )  180  o = 0.122 rad = (0.122 )  = 6.99 π π  4 4 3    [(4 ) − (2 ) ](11.5 × 10 )

 32  Therefore, hollow shaft is more rigid, 6.99o < 8.48o .

31.

A steel shaft is transmitting 40 hp at 500 rpm with minor shock. (a) What should be its diameter if the deflection is not to exceed 1o in 20 D ? (b) If deflection is primary what kind of steel would be satisfactory?

Solution: 33,000hp 33,000(40 ) (a) T = = = 420 ft − lb = 5.04 in − kips 2π n 2π (500 ) G = 11.5 ×103 ksi L = 20 D

Page 41 of 131



θ = 1o = θ= π

π 180

TL JG

rad

(5.04)(20 D )

=

 π D4    11.5 × 103 32   D = 1.72 in 3 say D = 1 in 4 180

(

(b) s =

)

16T 16(5.04 ) = = 4.8 ksi π D 3 π (1.75)3

Based on yield strength N =3 s ys = Ns = (3)(4.8) = 14.4 ksi s ys

14.4 = 24 ksi 0.6 0.6 Use C1117 normalized steel s y = 35 ksi sy =

32.

=

A square shaft of cold-finish AISI 1118 transmits a torsional moment of 1200 inlb. For medium shock, what should be its size?

Solution: For AISI 1118 cold-finish s y = 75 ksi s ys = 0.6 s y = 45 ksi N = 3 for medium shock s T s = ys = N Z′ where, h = b 2b 2 h 2b3 Z′ = = (Table AT 1) 9 9 T = 1200 in − lb = 1.2 in − kips 45 1.2(9 ) s= = 3 2b 3 b = h = 0.71 in 3 say b = h = in 4

Page 42 of 131



CHECK PROBLEMS 33.

A punch press is designed to exert a force sufficient to shear a 15/16-in. hole in a ½-in. steel plate, AISI C1020, as rolled. This force is exerted on the shaft at a radius of ¾-in. (a) Compute the torsional stress in the 3.5-in. shaft (bending neglected). (b) What will be the corresponding design factor if the shaft is made of cold-rolled AISI 1035 steel (Table AT 10)? Considering the shock loading that is characteristics of this machine, do you thick the design is safe enough?

Solution: For AISI C1020, as rolled sus = 49 ksi F = sus (π Dt ) 15 where D = in 16 1 t = in 2  15  1  F = 49(π )   = 72.2 kips  16  2  T = Fr 3 where r = in 4 3 T = (72.2 )  = 54.2 in − kips 4 16T π d3 where d = 3.5 in 16(54.2 ) s= = 6.44 ksi π (3.5)3

(a) s =

(b) For AISI 1035 steel, s us = 64 ksi for shock loading, traditional factor of safety, N = 10 ~ 15 Design factor , N = 34.

sus 64 = = 9.94 , the design is safe ( N ≈ 10 ) s 6.44

The same as 33, except that the shaft diameter is 2 ¾ in.

Solution: Page 43 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES d = 2.75 in 16T π d3 16(54.2 ) s= = 13.3 ksi π (2.75)3

(a) s =

(b) For AISI 1035 steel, s us = 64 ksi for shock loading, traditional factor of safety, N = 10 ~ 15 Design factor , N = 35.

sus 64 = = 4.8 , the design is not safe ( N < 10 ) s 13.3

A hollow annealed Monel propeller shaft has an external diameter of 13 ½ in. and an internal diameter of 6 ½ in.; it transmits 10,000 hp at 200 rpm. (a) Compute the torsional stress in the shaft (stress from bending and propeller thrust are not considered). (b) Compute the factor of safety. Does it look risky?

Solution: For Monel shaft, s us = 98 ksi (Table AT 3) N = 3 ~ 4 , for dead load, based on ultimate strength Tc J π Do4 − Di4 π (13.5)4 − (6.5)4 J= = = 3086 in 4 32 32 Do 13.5 c= = = 6.75 in 2 2 33,000hp 33,000(10,000 ) T= = = 262,606 ft − lb = 3152 in − kips 2π n 2π (200 ) (3152)(6.75) = 6.9 ksi s= 3086 (b) Factor of safety,

(a) s =

(

N=

)

[

sus 98 = = 14.2 , not risky s 6 .9

Page 44 of 131

]



STRESS ANALYSIS DESIGN PROBLEMS 36.

A hook is attached to a plate as shown and supports a static load of 12,000 lb. The material is to be AISI C1020, as rolled. (a) Set up strength equations for dimensions d , D , h , and t . Assume that the bending in the plate is negligible. (b) Determine the minimum permissible value of these dimensions. In estimating the strength of the nut, let D1 = 1.2d . (c) Choose standard fractional dimensions which you think would be satisfactory.

Problems 36 – 38. Solution: s = axial stress ss = shear stress (a) s=

4F F = 2 1 πd2 πd 4

Equation (1) d =

Page 45 of 131

4F πs



s=

F 1 π D 2 − D12 4

(

)

Equation (2) D = ss =

4F 4F 4F = = 2 2 2 2 2 π D − D1 π D − 1.44d 2 π D − (1.2d )

(

)

[

] (

4F + 1.44d 2 πs

F F = π D1h 1.2π dh

Equation (3) h = ss =

=

F 1.2π dss

F π Dt

Equation (4) t =

F π Dss

(b) Designing based on ultimate strength, Table AT 7, AISI C1020, as rolled su = 65 ksi sus = 49 ksi N = 3 ~ 4 say 4, design factor for static load s 65 s= u = = 16 ksi N 4 s 49 ss = us = = 12 ksi N 4 F = 12,000 lb = 12 kips

From Equation (1) 4F 4(12) d= = = 0.98 in πs π (16) From Equation (2) 4F 4(12) 2 D= + 1.44d 2 = + 1.44(0.98) = 1.53 in ( ) πs π 16 From Equation (3) F 12 h= = = 0.27 in 1.2π dss 1.2π (0.98)(12 ) From Equation (4) F 12 t= = = 0.21 in π Dss π (1.53)(12)

Page 46 of 131

)



(c) Standard fractional dimensions d = 1 in 1 D = 1 in 2 1 h = in 4 1 t = in 4

37.

The same as 36, except that a shock load of 4000 lb. is repeatedly applied.

Solution: (a) Same as 36. (b) N = 10 ~ 15 for shock load, based on ultimate strength say N = 15 , others the same. s 65 s= u = = 4 ksi N 15 s 49 ss = us = = 3 ksi N 15 F = 4000 lb = 4 kips

From Equation (1) 4F 4(4) d= = = 1.13 in πs π (4) From Equation (2) 4F 4(4) 2 D= + 1.44d 2 = + 1.44(1.13) = 1.76 in πs π (4) From Equation (3) F 4 h= = = 0.31 in 1.2π ds s 1.2π (1.13)(3) From Equation (4) F 4 t= = = 0.24 in π Dss π (1.76)(3)

Page 47 of 131



(c) Standard fractional dimensions 1 d = 1 in 8 3 D = 1 in 4 3 h = in 8 1 t = in 4

38.

The connection between the plate and hook, as shown, is to support a load F . Determine the value of dimensions D , h , and t in terms of d if the connection is to be as strong as the rod of diameter d . Assume that D1 = 1.2d , sus = 0.75su , and that bending in the plate is negligible.

Solution:

s=

F

1 πd2 4 1 F = π d 2s 4 1 s  (1) F = π d 2  u  4 N

Page 48 of 131



s=

F

F

=

1 1 π D 2 − D12 π D 2 − 1.44d 2 4 4 1 F = π D 2 − 1.44d 2 s 4 1 s  (2) F = π D 2 − 1.44d 2  u  4 N F F ss = = π D1h 1.2π dh F = 1.2π dhss

(

)

(

(

)

)

(

)

 0.75su  s  F = 1.2π dh us  = 1.2π dh   N  N  5s  (3) F = 0.9π dh u   N  F ss = π Dt F = π Dtss s   0.75su  F = π Dt  us  = π Dt   N  N  s  (4) F = 0.75π Dt  u  N

Equate (2) and (1) 1 s  1 s  F = π (D 2 − 1.44d 2 ) u  = π d 2  u  4 N 4 N 2 2 D = 2.44d D = 1.562d Equate (3) and (1) s  1 s  F = 0.9π dh u  = π d 2  u  N 4 N d h= = 0.278d 4(0.9) Equate (4) and (1) s  1 s  F = 0.75π Dt  u  = π d 2  u  N 4 N s  1 s  F = 0.75π (1.562d )(t ) u  = π d 2  u  N 4 N d t= = 0.214d 4(0.75)(1.562 )

Page 49 of 131



39.

(a) For the connection shown, set up strength equations representing the various methods by which it might fail. Neglect bending effects. (b) Design this connection for a load of 2500 lb. Both plates and rivets are of AISI C1020, as rolled. The load is repeated and reversed with mild shock. Make the connection equally strong on the basis of yield strengths in tension, shear, and compression.

Problems 39, 40 Solution: (a) ss =

F 1  5 π D 2  4 

4F 5π ss

Equation (1) D = s=

F t (b − 2 D )

Equation (2) b = s=

F + 2D ts

F 5 Dt

Equation (3) t =

F 5 Ds

(b) For AISI C1020, as rolled s y = 48 ksi (Table AT 7) s ys = 0.6 s y = 28 ksi N = 4 for repeated and reversed load (mild shock) based on yield strength 48 s= = 12 ksi 4 28 ss = = 7 ksi 4 From Equation (1)

Page 50 of 131



D=

4F 5π ss

where F = 2500 lb = 2.5 kips D=

4F 4(2.5) 5 in = = 0.30 in say 16 5π ss 5π (7 )

From Equation (3) F 5 2.5 t= in = = 0.13 in say 5 Ds 32 5 5 (12 )  16  From Equation (2) F 2.5 5 b = + 2D = + 2  = 1.96 in say 2 in ts  5   16   (12 )  32  40.

The same as 39, except that the material is 2024-T4, aluminum alloy.

Solution: (a) Same as 39. (b) ) For 2024-T4, aluminum alloy s y = 47 ksi (Table AT 3) s ys = 0.55s y = 25 ksi N = 4 for repeated and reversed load (mild shock) based on yield strength 47 s= = 12 ksi 4 25 ss = = 6 ksi 4 From Equation (1) 4F D= 5π ss where F = 2500 lb = 2.5 kips

D=

4F 4(2.5) 3 = = 0.33 in say in 8 5π ss 5π (6)

From Equation (3)

Page 51 of 131



t=

1 F 2.5 = = 0.11 in say in 5 Ds 8 3 5 (12 ) 8

From Equation (2) 1 F 2.5 3 b = + 2D = + 2  = 2.42 in say 2 in ts 2 1 8  (12 ) 8 41.

(a) For the connection shown, set up strength equations representing the various methods by which it might fail. (b) Design this connection for a load of 8000 lb. Use AISI C1015, as rolled, for the rivets, and AISI C1020, as rolled, for the plates. Let the load be repeatedly applied with minor shock in one direction and make the connection equally strong on the basis of ultimate strengths in tension, shear, and compression.

Problem 41. Solution: (a)

F sP = t (b − D )

3 F sP = 4 t (b − 2 D )

or

ssR =

Page 52 of 131

F 1  4 πD 2 (2 ) 4 

Equation (1)

Equation (2)



sR =

F 4 Dt

Equation (3)

(b) For AISI C1015, as rolled suR = 61 ksi , susR = 0.75suR = 45 ksi For AISI C1020, as rolled suP = 65 ksi N = 6 , based on ultimate strength s 65 s P = uP = = 10.8 ksi N 6 s 61 s R = uR = = 10.1 ksi N 6 s 45 ssR = usR = = 7.5 ksi N 6 F = 8000 lb = 8 kips Solving for D F ssR = 2π D 2

F 8 7 in = = 0.412 in say 2π ssR 2π (7.5) 16 Solving for t F sR = 4 Dt F 1 8 t= = = 0.453 in say in 4 Ds R 2 7 4 (10.1)  16  Solving for b F Using s P = t (b − D ) F 8 7 b= +D= + = 1.92 in say 2 in ts P 16 1  (10.8) 2 3 F 4 Using s P = t (b − 2 D ) D=

Page 53 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES 3F 3(8) 7 + 2D = + 2  = 1.99 in say 2 in 4ts P 16 1 4 (10.8)   2 Therefore b = 2 in 7 D = in 16 1 t = in 2 b=

42.

Give the strength equations for the connection shown, including that for the shear of the plate by the cotter.

Problems 42 – 44. Solution: Axial Stresses

s=

F 1 π D12 4

s=

Page 54 of 131

=

4F π D12

F (L − D2 )e

Equation (1)

Equation (2)



s=

s=

s=

F D2e

Equation (3)

F 1 π a 2 − D22 4

(

F 1 π D22 − D2 e 4

)

=

=

4F Equation (4) π a 2 − D22

(

4F Equation (5) π D − 4 D2e 2 2

Shear Stresses

Page 55 of 131

ss =

F 2eb

ss =

F 2(L − D2 + e )t

)

Equation (6)

Equation (7)



43.

ss =

F π at

Equation (8)

ss =

F π D1m

Equation (9)

ss =

F 2 D2 h

Equation (10)

A steel rod, as-rolled AISI C1035, is fastened to a 7/8-in., as-rolled C1020 plate by means of a cotter that is made of as-rolled C1020, in the manner shown. (a) Determine all dimensions of this joint if it is to withstand a reversed shock load F = 10 kips , basing the design on yield strengths. (b) If all fits are free-running fits, decide upon tolerances and allowances.

Solution: (See figure of Prob. 42) 7 t = in = 0.875 in , ssy = 0.6 s y 8 For steel rod, AISI C1035, as rolled s y1 = 55 ksi ssy1 = 33 ksi For plate and cotter, AISI C1020, as rolled s y2 = 48 ksi ssy2 = 28 ksi N = 5 ~ 7 based on yield strength say N = 7

From Equation (1) (Prob. 42) sy 4F s= 1 = N π D12 55 4(10 ) = 7 π D12 D1 = 1.27 in 1 say D1 = 1 in 4

Page 56 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES From Equation (9) ssy F ss = 1 = N π D1m 33 10 = 7  1 π 1  m  4 m = 0.54 in 9 say m = in 16 From Equation (3) sy F s= 1 = N D2 e 55 10 s= = 7 D2 e D2e = 1.273 From Equation (5) sy 4F s= 1 = 2 N π D2 − 4 D2e 55 4(10 ) = 2 7 π D2 − 4(1.273) D2 = 1.80 in 3 say D2 = 1 in 4 and D2e = 1.273  3 1 e = 1.273  4 e = 0.73 in 3 say e = in 4 By further adjustment 5 Say D2 = 2 in , e = in 8 From Equation (8) ssy F ss = 2 = N π at 28 10 = 7 π a (0.875) a = 0.91 in say a = 1 in

Page 57 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES From Equation (4) sy 4F s= 2 = 2 N π (a − D22 ) 48 4(10) = 7 π (a 2 − 2 2 ) a = 2.42 in 1 say a = 2 in 2 1 use a = 2 in 2 From Equation (7) ssy F ss = 2 = N 2(L − D2 + e )t 28 10 = 5 7  2 L − 2 + (0.875) 8  L = 2.80 in say L = 3 in From Equation (6) ssy F ss = 2 = N 2eb 28 10 = 7 5 2 b 8 b = 2 in From Equation (10) ssy F ss = 2 = N 2 D2 h 28 10 = 7 2(2)h 5 h = 0.625 in = in 8 Summary of Dimensions L = 3 in 5 h = in 8 b = 2 in 7 t = in 8

Page 58 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES 9 in 16 1 a = 2 in 2 1 D1 = 1 in 4 D2 = 2 in 5 e = in 8 m=

(b) Tolerances and allowances, No fit, tolerance = ± 0.010 in L = 3 ± 0.010 in h = 0.625 ± 0.010 in t = 0.875 ± 0.010 in m = 0.5625 ± 0.010 in a = 2.500 ± 0.010 in D1 = 1.25 ± 0.010 in For Free Running Fits (RC 7) Table 3.1 Female Male + 0.0030 − 0.0040 in b = 2.0 in b = 2.0 − 0.0000 − 0.0058 allowance = 0.0040 in + 0.0030 − 0.0040 in D2 = 2.0 D2 = 2.0 in − 0.0058 − 0.0000 allowance = 0.0040 in + 0.0016 − 0.0020 in e = 0.625 in e = 0.625 − 0.0000 − 0.0030 allowance = 0.0020 in

44.

A 1-in. ( D1 ) steel rod (as-rolled AISI C1035) is to be anchored to a 1-in. steel plate (as-rolled C1020) by means of a cotter (as rolled C1035) as shown. (a) Determine all the dimensions for this connection so that all parts have the same ultimate strength as the rod. The load F reverses direction. (b) Decide upon tolerances and allowances for loose-running fits.

Solution: (Refer to Prob. 42) (a) For AISI C1035, as rolled su1 = 85 ksi sus1 = 64 ksi For AISI C1020, as rolled

Page 59 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES su2 = 65 ksi sus2 = 48 ksi

Ultimate strength Use Equation (1) 1  1  2 Fu = su1  π D12  = (85) π (1) = 66.8 kips 4  4  Equation (9) Fu = sus1 π D1m 66.8 = (64)(π )(1)m m = 0.33 in 3 say m = in 8 From Equation (3) Fu = su1 D2e 66.8 = (85)D2 e D2e = 0.7859 From Equation (5) 1  Fu = su1  π D22 − D2 e  4  1  66.8 = (85) π D22 − 0.7859  4  D2 = 1.42 in 3 say D2 = 1 in 8 3   D2e = 1 e = 0.7859  8 e = 0.57 in 9 say e = in 16 From Equation (4) 1  Fu = su2  π (a 2 − D22 ) 4  2  1   3  66.8 = (65) π  a 2 −1    4    8   a = 1.79 in 3 say a = 1 in 4 From Equation (8)

Page 60 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES Fu = sus2 π at

66.8 = (48)(π )(a )(1) a = 0.44 in 1 say a = in 2 3 use a = 1 in 4 From Equation (2) Fu = su2 (L − D2 )e 3  9   66.8 = (65) L − 1   8  16   L = 3.20 in 1 say L = 3 in 4 From Equation (7) Fu = 2 sus2 (L − D2 − e )t 3 9  66.8 = 2(48) L − 1 − (1) 8 16   L = 1.51 in 1 say L = 1 in 2 1 use L = 3 in 4 From Equation (6) Fu = 2 sus1 eb  9 66.8 = 2(64 ) b  16  b = 0.93 in say b = 1 in From Equation (10) Fu = 2 sus1 D2 h  3 66.8 = 2(64 ) 1 h  8 h = 0.38 in 3 say h = in 8 Dimensions 1 L = 3 in 4

Page 61 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES 3 in 8 b = 1 in t = 1 in 3 m = in 8 3 a = 1 in 4 D1 = 1 in 3 D2 = 1 in 8 9 e = in 16 h=

(b) Tolerances and allowances, No fit, tolerance = ± 0.010 in L = 3.25 ± 0.010 in h = 0.375 ± 0.010 in t = 1.000 ± 0.010 in m = 0.375 ± 0.010 in a = 1.75 ± 0.010 in D1 = 1.000 ± 0.010 in For Loose Running Fits (RC 8) Table 3.1 Female Male + 0.0035 − 0.0045 b = 1.0 in b = 1.0 in − 0.0065 − 0.0000 allowance = 0.0045 in + 0.0040 − 0.0050 in D2 = 1.375 in D2 = 1.375 − 0.0000 − 0.0075 allowance = 0.0050 in + 0.0028 − 0.0035 in e = 0.5625 in e = 0.5625 − 0.0000 − 0.0051 allowance = 0.0035 in 45.

Give all the simple strength equations for the connection shown. (b) Determine the ratio of the dimensions a , b , c , d , m , and n to the dimension D so that the connection will be equally strong in tension, shear, and compression. Base the calculations on ultimate strengths and assume sus = 0.75su .

Page 62 of 131



Problems 45 – 47. Solution: (a) Neglecting bending 1  Equation (1): F = s π D 2  4   1  Equation (2): F = ss 2 π c 2    4 Equation (3): F = s(2bc ) Equation (4): F = s(ac ) Equation (5): F = s[2(d − c )b] Equation (6): F = ss (4mb ) Equation (7): F = ss (2nb ) Equation (8): F = s(d − c )a su s and ss = us N N Therefore ss = 0.75s Equate (2) and (1)  1   1  F = ss 2 π c 2  = s π D 2    4   4

(b) s =

1  1  0.75s c 2  = s D 2  2  4  c = 0.8165 D Equate (3) and (1) 1  F = s (2bc ) = s π D 2  4  1 2b(0.8165 D ) = π D 2 4 b = 0.4810 D

Page 63 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES Equate (4) and (1) 1  F = sac = s π D 2  4  1 a (0.8165 D ) = π D 2 4 a = 0.9619 D Equate (5) and (1) 1  F = s[2(d − c )b ] = s π D 2  4  1 2(d − 0.8165 D )(0.4810 ) = π D 2 4 d = 1.6329 D Equate (6) and (1) 1  F = ss (4mb ) = s π D 2  4  1 0.75(4m )(0.4810 D ) = π D 2 4 m = 0.5443D Equate (7) and (1) 1  F = ss (2nb ) = s π D 2  4  1 0.75(2n )(0.4810 D ) = π D 2 4 n = 1.0886 D Equate (8) and (1) 1  F = s (d − c )a = s π D 2  4  (1.6329 − D − 0.8165D )a = 1 π D 2 4 a = 0.9620 D Summary a = 0.9620 D b = 0.4810 D c = 0.8165 D d = 1.6329 D m = 0.5443D n = 1.0886 D 46.

The same as 45, except that the calculations are to be based on yield strengths. Let ssy = 0.6 s y .

Page 64 of 131


SECTION 1– DESIGN FOR SIMPLE STRESSES Solution: (Refer to Prob. 45) (a) Neglecting bending 1  Equation (1): F = s π D 2  4   1  Equation (2): F = ss 2 π c 2    4 Equation (3): F = s(2bc ) Equation (4): F = s(ac ) Equation (5): F = s[2(d − c )b] Equation (6): F = ss (4mb ) Equation (7): F = ss (2nb ) Equation (8): F = s(d − c )a

(b) s =

sy

and ss =

s sy

N N Therefore ss = 0.6s Equate (2) and (1)  1   1  F = ss 2 π c 2  = s π D 2    4   4 1  1  0. 6 s  c 2  = s  D 2  2  4  c = 0.9129 D Equate (3) and (1) 1  F = s (2bc ) = s π D 2  4  1 2b(0.9129 D ) = π D 2 4 b = 0.4302 D

Equate (4) and (1) 1  F = sac = s π D 2  4  1 a (0.9129 D ) = π D 2 4 a = 0.8603D Equate (5) and (1) 1  F = s[2(d − c )b ] = s π D 2  4 

Page 65 of 131


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS VARYING STRESSES – NO CONCENTRATION DESIGN PROBLEMS 141.

The maximum pressure of air in a 20-in. cylinder (double-acting air compressor) is 125 psig. What should be the diameter of the piston rod if it is made of AISI 3140, OQT at 1000 F, and if there are no stress raisers and no column action? Let N = 1.75 ; indefinite life desired. How does your answer compare with that obtained for 4?

Solution: For AISI 3140, OQT 1000 F su = 153 ksi s y = 134 ksi sn = 0.5su = 0.5(153) = 76.5 ksi For axial loading, with size factor sn = 0.5su = (0.8)(0.85)(76.5) = 52 ksi Soderberg line 1 sm sa = + N s y sn For double-acting π  2 Fmax = F = pA = (125) (20) = 39,270 lb = 39.27 kips 4 Fmin = − F = −39.27 kips sm = 0 4 F 4(39.27 ) 50 sa = = = 2 π d2 πd2 d  50   2 1 1 d = 0+   = N 1.75 52 d = 1.2972 in 5 say d = 1 in 16 comparative to Problem 4.

142.

A link as shown is to be made of AISI 2330, WQT 1000 F. The load F = 5 kips is repeated and reversed. For the time being, ignore stress concentrations. (a) If its surface is machined, what should be its diameter for N = 1.40 . (b) The same as (a), except that the surface is mirror polished. What would be the percentage saving in weight? (c) The same as (a), except that the surface is as forged.

Page 1 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS

Prob. 142 – 144 Solution: For AISI 2330, WQT 1000 F su = 105 ksi s y = 85 ksi sn = 0.5su = 0.5(105) = 52.5 ksi sm = 0 20 4F 4(5) sa = = = 2 2 πd πd π d2 Soderberg line 1 sm sa = + N s y sn 1 s = 0+ a N sn s sa = n N Size factor = 0.85 Factor for axial loading = 0.80

(a) Machined surface Surface factor = 0.85 (Fig. AF 5) sn = 0.5su = (0.80)(0.85)(0.85)(52.5) ksi = 30.345 ksi 20 30.345 sa = = 2 1.4 πD D = 0.542 in 9 say D = in 16 (b) Mirror polished surface Surface factor = 1.00 (Fig. AF 5) sn = 0.5su = (0.80)(0.85)(1.00)(52.5) ksi = 35.7 ksi 20 35.7 sa = = 2 πD 1.4

Page 2 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS D = 0.5 in 2

2

 9  1   −  16 2 Savings in weight =   2  (100% ) = 21% 9    16  (c) As forged surface Surface factor = 0.40 (Fig. AF 5) sn = 0.5su = (0.80)(0.85)(0.40 )(52.5) ksi = 14.28 ksi 20 14.28 sa = = 1.4 π D2 D = 0.79 in 3 say D = in 4

143.

The same as 142, except that, because of a corrosive environment, the link is made from cold-drawn silicon bronze B and the number of reversals of the load is expected to be less than 3 x 107.

Solution: For cold-drawn silicon bronze, Type B. sn = 30 ksi at 3 x 108 s y = 69 ksi su = 93.75 ksi 0.085

 3 × 108   = 36.5 ksi sn at 3 x 10 = (30) 7  10 3 ×   sn = (0.80)(0.85)(36.5) = 24.82 ksi 20 24.82 sa = = 2 1.4 πD D = 0.60 in 5 say D = in 8 7

144.

The same as 142, except that the link is made of aluminum alloy 2024-T4 with a minimum life of 107 cycles.

Solution: For AA 2024-T4 s y = 47 ksi su = 68 ksi sn = 20 ksi at 5 x108

Page 3 of 62



0.085

 5 × 108   sn at 10 (20) = 27.9 ksi 7   10  sn = (0.80)(0.85)(27.9) = 19 ksi 19 20 sa = = 2 π D 1.4 D = 0.685 in 11 say D = in 16 7

145.

A shaft supported as a simple beam, 18 in. long, is made of carburized AISI 3120 steel (Table AT 10). With the shaft rotating, a steady load of 2000 lb. is appliled midway between the bearings. The surfaces are ground. Indefinite life is desired with N = 1.6 based on endurance strength. What should be its diameter if there are no surface discontinuities?

Solution:

For AISI 3120 steel, carburized sn = 90 ksi s y = 100 ksi su = 141 ksi Size Factor = 0.85 Surface factor (ground) = 0.88 sn = (0.85)(0.88)(90) = 67.32 ksi sm = 0 32 M sa = π D3 FL (2000)(18) M= = = 9000 in − lb = 9.0 in − kips 4 4 Soderberg line 1 sm sa = + N s y sn

Page 4 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 s = 0+ a N sn s sa = n N 32(9) 67.32 = πD3 1 .6 D = 1.2964 in 1 say D = 1 in 4

146.

(a) A lever as shown with a rectangular section is to be designed for indefinite life and a reversed load of F = 900 lb . Find the dimensions of a section without discontinuity where b = 2.8t and L = 14 in . for a design factor of N = 2 . The material is AISI C1020, as rolled, with an as-forged surface. (b) compute the dimensions at a section where e = 4 in .

Problems 146, 147 Solution: For AISI C1020, as rolled su = 65 ksi s y = 48 ksi s n = 0.5su = 32.5 ksi Surface factor (as forged) = 0.55 (a) sm = 0 Mc sa = I 3 tb3 t (2.8t ) I= = = 1.8293t 4 12 12 b 2.8t c= = = 1.4t 2 2 M = FL = (900)(14) = 12,600 in − lb = 12.6 in − kips (12.6)(1.4t ) = 9.643 sa = 1.8293t 4 t3 sn = (0.85)(0.55)(32.5) = 15.20 ksi Page 5 of 62



Soderberg line 1 sm sa = + N s y sn 1 s = 0+ a N sn s sa = n N 9.643 15.20 = t3 2 t = 1.08 in b = 2.8t = 2.8(1.08) = 3.0 in 1 say t = 1 in , b = 3.0 in 16

(b) M = Fe = (900)(4 ) = 3,600 in − lb = 3.6 in − kips (3.6)(1.4t ) = 2.755 sa = 18293t 4 t3 2.755 15.20 = t3 2 t = 0.713 in b = 2.8t = 2.8(0.713) = 1.996 in 23 in , b = 2 in say t = 32

147.

The same as 146, except that the reversal of the load are not expected to exceed 105 (Table AT 10).

Solution: sn = 32.5 ksi 0.085

 106  sn at 10 = (32.5) 5  = 39.5 ksi  10  sn = (0.85)(0.55)(39.5) = 18.5 ksi 5

sn N 9.643 18.5 = t3 2

(a) sa =

Page 6 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS t = 1.014 in b = 2.8t = 2.8(1.014) = 2.839 in 13 say t = 1 in , b = 2 in 16 sn N 2.755 18.5 = t3 2

(b) sa =

t = 0.6678 in b = 2.8t = 2.8(0.6678) = 1.870 in 11 7 say t = in , b = 1 in 16 8

148.

A shaft is to be subjected to a maximum reversed torque of 15,000 in-lb. It is machined from AISI 3140 steel, OQT 1000 F (Fig. AF 2). What should be its diameter for N = 1.75 ?

Solution: For AISI 3140 steel, OQT 1000 F su = 152 ksi s y = 134 ksi sn = 0.5su = 76 ksi For machined surface, Surface factor = 0.78 Size factor = 0.85 sns = (0.6)(0.85)(0.78)(134) = 53.3 ksi s ys = 0.6s y = 0.6(134) = 80.4 ksi 1 sms sas = + N s ys sns sms = 0 16T sas = πD3 T = 15 in − kips 16(15) 240 sas = = πD3 πD3 1 s = 0 + as N sns s sas = ns N Page 7 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 240 53.3 = πD 3 1.75 D = 1.3587 in 3 say D = 1 in 8

149.

The same as 148, except that the shaft is hollow with the outside diameter twice the inside diameter.

Solution: Do = 2 Di 32 16TDo 16(15)(2 Di ) sas = = = 4 4 4 4 π (Do − Di ) π (2 Di ) − Di πDi3 s sas = ns N 32 53.3 = πDi3 1.75

[

]

Di = 0.694 in 11 3 say Di = in , Do = 1 in 16 8 150.

The link shown is machined from AISI 1035 steel, as rolled, and subjected to a repeated tensile load that varies from zero to 10 kips; h = 2b . (a) Determine these dimensions for N = 1.40 (Soderberg) at a section without stress concentration. (b) How much would these dimensions be decreased if the surfaces of the link were mirror polished?

Problems 150, 151, 158. Solution: For AISI 1035, steel as rolled su = 85 ksi s y = 55 ksi sn = 0.5su = 42.5 ksi

Page 8 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 (10 + 0 ) = 5 kips 2 1 Fa = (10 − 0 ) = 5 kips 2 F 5 10 sm = m = = 2 2 bh 1.5b 3b F 5 10 sa = a = = 2 2 bh 1.5b 3b (a) Soderberg line 1 sm sa = + N s y sn For machined surface, Factor = 0.88 Fm =

Size factor = 0.85 sn = (0.80)(0.85)(0.88)(42.5) = 25.4 ksi 10 1 10 = 2 + 2 1.40 3b (55) 3b (25.4 ) b = 0.5182 in 9 say b = in 16 27 h = 1.5b = in 32

(b) Mirror polished, Factor = 1.00 Size factor = 0.85 sn = (0.80)(0.85)(1.00)(42.5) = 28.9 ksi 10 1 10 = 2 + 2 1.40 3b (55) 3b (28.9 ) b = 0.4963 in 1 say b = in 2 3 h = 1.5b = in 4

151.

The same as 150, except that the link operates in brine solution. (Note: The corroding effect of the solution takes precedence over surface finish.)

Page 9 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Solution: Table AT 10, in brine, AISI 1035, sn = 24.6 ksi s y = 58 ksi sn = (0.80)(0.85)(24.6) = 16.73 ksi 10 1 10 = 2 + 2 1.40 3b (55) 3b (16.73) b = 0.60 in 5 say b = in 8 15 h = 1.5b = in 16

152.

The simple beam shown, 30-in. long ( = a + L + d ), is made of AISI C1022 steel, as rolled, left a forged. At a = 10 in , F1 = 3000 lb. is a dead load. At d = 10 in , F2 = 2400 lb. is repeated, reversed load. For N = 1.5 , indefinite life, and h = 3b , determine b and h . (Ignore stress concentration).

Problem 152, 153 Solution: For AISI C1022, as rolled su = 72 ksi s y = 52 ksi sn = 0.5su = 36 ksi For as forged surface Figure AF 5, factor = 0.52 Size factor = 0.85 sn = (0.85)(0.52)(36) = 16 ksi Loading:

Page 10 of 62



∑M

A

=0

10(3000) + 20(2400 ) = 30 R2 R2 = 2600 lb

∑F

V

=0

R1 + R2 = F1 + F2 R1 + 2600 = 3000 + 2400 R1 = 2800 lb Shear Diagram

M C1 = (2800 )(10 ) = 28,000 in − lb = 28 in − kips M D1 = (2600 )(10 ) = 26,000 in − lb = 26 in − kips

Then Loading

∑M

A

=0

10(3000) + 30 R2 = 20(2400 ) R2 = 600 lb

∑F

V

=0

R1 + F2 = F1 + R2 R1 + 2400 = 3000 + 600 R1 = 1200 lb

Page 11 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Shear Diagram

M C2 = (1200 )(10 ) = 12,000 in − lb = 12 in − kips M D2 = (600 )(10 ) = 6,000 in − lb = 6 in − kips

Then using M max = M C1 = 28 in − kips M min = M C2 = 12 in − kips 1 (M max + M min ) = 1 (28 + 12) = 20 in − kips 2 2 1 1 M a = (M max − M min ) = (28 − 12 ) = 8 in − kips 2 2 M c M c sm = m , sa = a I I 3 3 b(3b ) bh I= = = 2.25b 4 12 12 h c = = 1.5b 2 Mm Ma sm = , sa = 3 1.5b 1.5b3 1 sm sa = + N s y sn Mm =

 20   8      1  1.5b3   1.5b3  = + 1.5 52 16 b = 0.96 in say b = 1 in h = 3b = 3 in

153.

The same as 152, except that the cycles of F2 will not exceed 100,000 and all surfaces are machined.

Solution:

Page 12 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 0.085

 106  ( ) sn at 10 cycles = 36  5  = 43.8 ksi  10  su = 72 ksi Machined surface, factor = 0.90 sn = (0.85)(0.90)(43.8) = 33.5 ksi 5

 20   8      1  1.5b3   1.5b3  + = 1.5 52 33.5 b = 0.8543 in 7 say b = in 8 5 h = 3b = 2 in 8

154.

A round shaft, made of cold-finished AISI 1020 steel, is subjected to a variable torque whose maximum value is 6283 in-lb. For N = 1.5 on the Soderberg criterion, determine the diameter if (a) the torque is reversed, (b) the torque varies from zero to a maximum, (c) the torque varies from 3141 in-lb to maximum.

Solution: For AISI 1020, cold-finished su = 78 ksi s y = 66 ksi sn = 0.5su = 39 ksi size factor = 0.85 sns = (0.6)(0.85)(39) = 20 ksi s ys = 0.6s y = 0.6(66 ) = 40 ksi 1 sms sas = + N s ys sns (a) Reversed torque sms = 0 16T sas = πD3 T = 6283 in − lb 32 16(6283) 32,000 sas = psi = 3 ksi = 3 3 πD D D 1 sas = 0+ N sns

Page 13 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS  32   3 1 D  = 0+  1.5 20 D = 1.34 in 3 say D = 1 in 8

(b) Tmin = 0 , Tmax = 6283 in − lb 1 Tm = (6283) = 3141 in − lb 2 1 Ta = (6283) = 3141 in − lb 2 16 16(3141) 16,000 psi = 3 ksi = 3 3 πD D D 16 16(3141) 16,000 sas = psi = 3 ksi = 3 3 πD D D  16   16      1  D3   D 3  = + 1.5 40 20 D = 1.22 in 1 say D = 1 in 4 sms =

(c) Tmin = 3141 in − lb , Tmax = 6283 in − lb 1 Tm = (6283 + 3141) = 4712 in − lb 2 1 Ta = (6283 − 3141) = 1571 in − lb 2 24 16(4712 ) 24,000 psi = 3 ksi = 3 3 πD D D 8 16(1571) 8,000 sas = psi = 3 ksi = 3 3 πD D D  24   8      1  D3   D 3  = + 1.5 40 20 D = 1.145 in 5 say D = 1 in 32 sms =

Page 14 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS CHECK PROBLEMS 155.

A simple beam 2 ft. long is made of AISI C1045 steel, as rolled. The dimensions of the beam, which is set on edge, are 1 in. x 3 in. At the midpoint is a repeated, reversed load of 4000 lb. What is the factor of safety?

Solution: For AISI C1045, as rolled su = 96 ksi s y = 59 ksi sn = 0.5su = 0.5(96) = 48 ksi size factor = 0.85 sn = (0.85)(48) = 40.8 ksi 1 sm sa = + N s y sn sm = 0 6M sa = 2 bh h = 3 in b = 1 in FL (4000 )(24 ) M= = = 24,000 in − lb = 24 in − kips 4 4 6(24 ) sa = = 16 ksi (1)(3)2 1 16 = 0+ N 40.8 N = 2.55 156.

The same as 155, except that the material is normalized and tempered cast steel, SAE 080.

Solution: Table AT 6 sn′ = 35 ksi s y = 40 ksi sn = (0.85)(35) = 29.75 ksi 1 16 = 0+ N 29.75 N = 1.86 157.

A 1 ½-in. shaft is made of AISI 1045 steel, as rolled. For N = 2 , what repeated and reversed torque can the shaft sustain indefinitely?

Page 15 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Solution: For AISI 1045, as rolled su = 96 ksi s y = 59 ksi sn′ = 0.5su = 0.5(96) = 48 ksi sns = (0.6)(0.85)(48) = 24.48 ksi s ys = 0.6 s y = (0.6 )(59 ) = 35.4 ksi 1 sms sas = + N s ys sns sms = 0 1 s = 0 + as 2 24.48 sas = 12.24 ksi 16T sas = = 12.24 πD 3 T = 8 in − kips VARIABLE STRESSES WITH STRESS CONCENTRATIONS DESIGN PROBLEMS 158.

The load on the link shown (150) is a maximum of 10 kips, repeated and reversed. The link is forged from AISI C020, as rolled, and it has a ¼ in-hole drilled on the center line of the wide side. Let h = 2b and N = 1.5 . Determine b and h at the hole (no column action) (a) for indefinite life, (b) for 50,000 repetitions (no reversal) of the maximum load, (c) for indefinite life but with a ground and polished surface. In this case, compute the maximum stress.

Solution: For AISI C1020, as rolled su = 65 ksi s y = 48 ksi sn = 0.5su = 0.5(65) = 32.5 ksi For as forged surface Surface factor = 0.55 Size factor = 0.85 sn = (0.80)(0.85)(0.55)(32.5) = 12.2 ksi 1 sm K f sa = + N sy sn

Page 16 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Fig. AF 8, b h > 1 Assume K t = 3.5 d 1 Figure AF 7, r = = in = 0.125 in 2 8 a = 0.01 in 1 1 q= = = 0.926 a 0.01 1+ 1+ r 0.125 ( ) K f = q K t − 1 + 1 = 0.926(3.5 − 1) + 1 = 3.3 sm = 0 10 F = b(h − d ) b(2b − 0.25) K s 1 (a) = 0+ f a sn N 1 (3.3)(10) = 0+ 1.5 b(2b − 0.25)(12.2) 2 2b − 0.25b = 4.06 b 2 − 0.125b − 2.03 = 0 b = 1.489 in 1 say b = 1 in , h = 2b = 3 in 2 sa =

(b) For 50,000 repetitions or 50,000 cycles  106   sn = (12.2 ) 4   5 ×10  (log K ) 3

0.085

= 15.74 ksi (log 3.3 ) 3

f n (5 ×104 ) = = 2.0 log K 10log 3.3 10 f 1 K fl sa = N sn 1 (2.0)(10) = 1.5 b(2b − 0.25)(15.74) 2b 2 − 0.25b = 1.906 b 2 − 0.125b − 0.953 = 0 b = 1.04 in 1 1 say b = 1 in , h = 2b = 2 in 16 8

K fl =

(c) For indefinite life, ground and polished surface Surface factor = 0.90

Page 17 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS sn = (0.80)(0.85)(0.90)(32.5) = 20 ksi 1 K f sa = sn N 1 (3.3)(10) = 1.5 b(2b − 0.25)(20) b 2 − 0.125b − 1.2375 = 0 b = 1.18 in 3 3 say b = 1 in , h = 2b = 2 in 16 8 Kf F Maximum stress = b(h − d ) b h > 1 , d h = 0.25 2.375 = 0.105 Figure AF 8 K t = 3.5 K f = q (K t − 1) + 1 = 0.926(3.5 − 1) + 1 = 3.315 smax = 159.

(3.315)(10) = 13.14 ksi 1.1875(2.375 − 0.25) A connecting link as shown, except that there is a 1/8-in. radial hole drilled through it at the center section. It is machined from AISI 2330, WQT 1000 F, and it is subjected to a repeated, reversed axial load whose maximum value is 5 kips. For N = 1.5 , determine the diameter of the link at the hole (a) for indefinite life; (b) for a life of 105 repetitions (no column action). (c) In the link found in (a) what is the maximum tensile stress?

Problem 159 Solution: For AISI 2330, WQT 1000 F su = 135 ksi s y = 126 ksi sn = 0.5su = 0.5(135) = 67.5 ksi For machined surface, Fig. AF 7, surface factor = 0.80 Size factor = 0.85 sn = (0.80)(0.85)(0.80)(67.5) = 36.72 ksi

Page 18 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 sm K f sa = + sn N sy Fig. AF 8, b h > 1 Assume K t = 2.5 d 1 Figure AF 7, r = = in = 0.0625 in 2 16 a = 0.0025 in 1 1 q= = = 0.96 a 0.0025 1+ 1+ r 0.0625 K f = q(K t − 1) + 1 = 0.96(2.5 − 1) + 1 = 2.44 (a) Indefinite life, K f = 2.44 sn = 36.72 ksi sm = 0 F 4F 4(5) 20 sa = = = = 2 2 2 πD πD − 4 Dd  1  πD − 0.5 D πD 2 − 4 D  − Dd 4 8 K s 1 = 0+ f a sn N 1 (2.44)(20) = 1.5 36.72(πD 2 − 0.5D ) πD 2 − 0.5D = 2.00 D = 0.88 in 7 say D = in 8 (b) For a life of 105 repetitions or cycles  106  sn = (36.72 ) 5   10  (log K ) 3

0.085

= 44.66 ksi (log 2.4 ) 3

f n (105 ) K fl = log K f = 10log 2.44 10 1 K fl sa = N sn 1 (1.81)(20) = 1.5 44.66(πD 2 − 0.5 D ) πD 2 − 0.5D = 1.216 D = 0.71 in

Page 19 of 62

= 1.81



say D =

3 in 4

4K f F πD 2 − 0.5D 7 d 0.125 D = in , = = 0.14 8 D 0.875 Figure AF 8 K t = 2.6

(c) smax =

K f = q(K t − 1) + 1 = 0.96(2.6 − 1) + 1 = 2.54

smax =

160.

4(2.54)(5) 2

7 7 π   − 0.5  8 8

= 25.82 ksi

A machine part of uniform thickness t = b 2.5 is shaped as shown and machined all over from AISI C1020, as rolled. The design is for indefinite life for a load repeated from 1750 lb to 3500 lb. Let d = b . (a) For a design factor of 1.8 (Soderberg), what should be the dimensions of the part? (b) What is the maximum tensile stress in the part designed?

Problems 160, 161 Solution: For AISI C1020, as rolled su = 65 ksi s y = 48 ksi sn′ = 0.5su = 0.5(65) = 32.5 ksi For machined surface Surface factor = 0.90 Size factor = 0.85 sn = (0.80)(0.85)(0.90)(32.5) = 20 ksi 1 sm K f sa = + sn N sy (a) For flat plate with fillets Figure AF 9 b d r= = 3 3

Page 20 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS r 1 = = 0.333 d 3 h 2b = =2 d b K t = 1.65 a = 0.01 in 1 q= ≈ 1.0 a 1+ r K f ≈ K t = 1.65 Fm bt Fa sa = bt b t= 2 .5 1 Fm = (3500 + 1750 ) = 2625 lb 2 1 Fa = (3500 − 1750 ) = 875 lb 2 2625 6562.5 sm = = b2  b  b   2.5  875 2187.5 sa = =  b  b2 b   2.5  1 6562.5 (1.65)(2187.5) = + 1.8 48,000b 2 20,000b 2 b = 0.7556 in or b = 0.75 in b 0.75 t= = = 0.3 in 2 .5 2 .5 sm =

For flat plate with central hole Fig. AF 8, b h > 1 , d h = b 2b = 1 2 Assume K f ≈ K t = 2.9 Fm Fm F = = m (h − d )t (2b − b )t bt Fa Fa F sa = = = a (h − d )t (2b − b )t bt

sm =

Page 21 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 2625 6562.5 =  b  b2 b   2.5  875 2187.5 sa = =  b  b2 b   2.5  1 6562.5 (2.9)(2187.5) = + 1.8 48,000b 2 20,000b 2 b = 0.904 in 15 or b = 0.9375 in = in 16 b 3 t= = in 2 .5 8 15 d = b = in 16 15 3 15 use b = in , t = in , d = in 16 8 16 sm =

(b) smax = sm + K f sa d 15 = in 2 32 1 q= = 0.98 0.01 1+  15     32  K t = 2.9

r=

K f = q(K t − 1) + 1 = 0.98(2.9 − 1) + 1 = 2.86

Fm 6562.5 6562.5 = = = 7467 psi 2 bt b2  15     16  F 2187.5 2187.5 sa = a = = = 2489 psi 2 bt b2  15     16  smax = sm + K f sa = 7467 + (2.86 )(2489 ) = 14,586 psi sm =

162.

The beam shown has a circular cross section and supports a load F that varies from 1000 lb to 3000 lb; it is machined from AISI C1020 steel, as rolled. Determine the diameter D if r = 0.2 D and N = 2 ; indefinite life.

Page 22 of 62



Problems 162 – 164. Solution: For AISI C1020, as rolled su = 65 ksi s y = 48 ksi sn′ = 0.5su = 0.5(65) = 32.5 ksi For machined surface Surface factor = 0.90 Size factor = 0.85 sn = (0.85)(0.90 )(32.5) = 24.86 ksi

∑M

A

=0

12 F = 24 B F = 2B F B= 2 F A= B= 2 At discontinuity 6F M= = 3F 2 M max = 3(3000) in − lb = 9000 in − lb = 9 in − kips

M min = 3(1000) in − lb = 3000 in − lb = 3 in − kips 1 M m = (9 + 3) = 6 in − kips 2 1 M a = (9 − 3) = 3 in − kips 2 32 M s= πD3 Figure AF 12 D d =1.5d d =1.5 r d = 0.2d d = 0.2 K t = 1.42 assume K f ≈ K t = 1.42

Page 23 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 sm K f sa = + sn N sy 1 (32 )(6 ) (1.42 )(32 )(3) = + 2 48πD 3 24.86πD 3 D = 1.821 in 13 say D = 1 in 16 At maximum moment 12 F M= = 6F 2 M max = 6(3000) in − lb = 18000 in − lb = 18 in − kips

M min = 6(1000) in − lb = 6000 in − lb = 6 in − kips 1 M m = (18 + 6 ) = 12 in − kips 2 1 M a = (18 − 6 ) = 6 in − kips 2 32 M s= πD3 K f = 1.00 1 sm K f sa = + sn N sy 1 (32 )(12 ) (1.0 )(32 )(6 ) = + 2 48πD 3 24.86πD 3 D = 1.4368 in 13 Therefore use D = 1 in 16 164. The shaft shown is machined from C1040, OQT 1000 F (Fig. AF 1). It is subjected to a torque that varies from zero to 10,000 in-lb. ( F = 0 ). Let r = 0.2 D and N = 2 . Compute D . What is the maximum torsional stress in the shaft?

Solution:

For C1040, OQT 1000 F su = 104 ksi s y = 72 ksi

Page 24 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS sn′ = 0.5su = 0.5(104) = 52 ksi For machined surface Surface factor = 0.85 Size factor = 0.85 sns = (0.60)(0.85)(0.85)(52) = 22.5 ksi 1 Ta = Tm = (10,000) = 5000 in − lb = 5 in − kips 2 s ys = 0.6 s y = 0.6(72 ) = 43.2 ksi sms = sas =

16T πD3 K fs sas

1 sms = + N s ys sns Figure AF 12 D d =1.5d d =1.5 r d = 0.2d d = 0.2 K ts = 1.2 assume K fs ≈ K ts = 1.2

(16)(5) + (1.2)(16)(5) 1 = 2 43.2πD 3 22.5πD 3 D = 1.5734 in 9 say D = 1 in 16 smax = sm + K f sa smax =

165.

(16)(5)  9 π 1   16 

3

+

(1.2)(16)(5) = 14.686 ksi  9 π 1   16 

3

An axle (nonrotating) is to be machined from AISI 1144, OQT 1000 F, to the proportions shown, with a fillet radius r ≈ 0.25 D ; F varies from 400 lb to 1200 lb.; the supports are to the left of BB not shown. Let N = 2 (Soderberg line). (a) At the fillet, compute D and the maximum tensile stress. (b) Compute D at section BB. (c) Specify suitable dimensions keeping the given proportions, would a smaller diameter be permissible if the fillet were shot-peened?

Problems 165 – 167

Page 25 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Solution: For AISI 1144, OQT 1000 F su = 118 ksi s y = 83 ksi sn′ = 0.5su = 59 ksi For machined surface Surface factor = 0.83 Size factor = 0.85 sn = (0.85)(0.83)(59) = 41.62 ksi (a) At the fillet D d =1.5d d =1.5 r d = 0.25d d = 0.25 K t = 1.35 assume K f ≈ K t = 1.35 M = 6F M max = 6(1200) in − lb = 7200 in − lb = 7.2 in − kips M min = 6(400) in − lb = 2400 in − lb = 2.4 in − kips 1 M m = (7.2 + 2.4 ) = 4.8 in − kips 2 1 M a = (7.2 − 2.4 ) = 2.4 in − kips 2 32 M s= πD3 1 sm K f sa = + sn N sy 1 (32 )(4.8) (1.35)(32 )(2.4 ) = + 2 83πD 3 41.62πD 3 D = 1.4034 in 7 say D = 1 in 16

(b) At section BB, M = 30 F M max = 30(1200) in − lb = 36000 in − lb = 36 in − kips M min = 30(400) in − lb = 12000 in − lb = 12 in − kips 1 M m = (7.2 + 2.4 ) = 4.8 in − kips 2 1 M a = (7.2 − 2.4 ) = 2.4 in − kips 2

Page 26 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 32 M πD3 K f = 1 .0

s=

1 sm K f sa = + sn N sy

(32)(36) + (1.0)(32)(12) 1 = 2 83π (1.5 D )3 41.62π (1.5 D )3 D = 1.6335 in 11 say D = 1 in 16 (c) Specified dimension: D = 2 in , 1.5 D = 3 in A smaller diameter is permissible if the fillet were shot-peened because of increased fatigue strength. 166.

A pure torque varying from 5 in-kips to 15 in-kips is applied at section C. ( F = 0 ) of the machined shaft shown. The fillet radius r = D 8 and the torque passes through the profile keyway at C. The material is AISI 1050, OQT 1100 F, and N = 1.6 . (a) What should be the diameter? (b) If the fillet radius were increased to D 4 would it be reasonable to use a smaller D ?

Solution:

Tmax = 15 in − kips Tmin = 5 in − kips 1 Tm = (15 + 5) = 10 in − kips 2 1 Ta = (15 − 5) = 5 in − kips 2 For AISI 1050, OQT 1100 F su = 101 ksi s y = 58.5 ksi sn = 0.5su = 0.5(101) = 50.5 ksi Page 27 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS For machined surface Surface factor = 0.85 Size factor = 0.85 sns = (0.60)(0.85)(0.85)(50.5) = 21.9 ksi (a) At the fillet r d =r D = =1 8 D d =1.5 K ts = 1.3 assume K fs ≈ K ts = 1.3 At the key profile K fs = 1.6 use K fs = 1.6 s ys = 0.6 s y = 0.6(58.5) = 35.1 ksi

1 sms K fs sas = + N s ys sns

(16)(10) + (1.6)(16)(5) 1 = 1.6 35.1πD 3 21.9πD 3 D = 1.7433 in 3 say D = 1 in 4 (b) r = D 4 r D = 0.25 D d =1.5 Figure AF 12 K ts = 1.18 K fs ≈ K ts = 1.18 < 1.6 Therefore, smaller D is not reasonable. 170.

The beam shown is made of AISI C1020 steel, as rolled; e = 8 in . The load F is repeated from zero to a maximum of 1400 lb. Assume that the stress concentration at the point of application of F is not decisive. Determine the depth h and width t if h ≈ 4t ; N = 1.5 ± 0.1 for Soderberg line. Iteration is necessary because K f depends on the dimensions. Start by assuming a logical K f for a logical h (Fig. AF 11), with a final check of K f . Considerable estimation inevitable.

Page 28 of 62



Problem 170 Solution:

1 F 2 At the hole A= B=

F M = eB = (8)  = 4 F 2 M max = 4 F

M min = 0 1 M m = (4 F ) = 2 F = 2(1.4 ) = 2.8 in − kips 2 1 M a = (4 F ) = 2 F = 2(1.4 ) = 2.8 in − kips 2 Mc s= I (h − 2d )3 t I= 12 1 d = in = 0.5 in 2 1 11 c = 1 +   = 1.75 in 2 22 For AISI C1020, as rolled su = 65 ksi s y = 48 ksi sn = 0.5su = 0.5(65) = 32.5 ksi Size factor = 0.85 sn = (0.85)(32.5) = 27.62 ksi Fig. AF 7, c d = 1.75 0.5 = 3.5 > 0.5

Page 29 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Assume K t = 3.5  1  1  r =    = 0.25 in  2  2  a = 0.010 in 1 1 q= = = 0.962 a 0.010 1+ 1+ r 0.25 K f = q (K t − 1) + 1 = 0.962(3.5 − 1) + 1 = 3.4

1 sm K f sa = + sn N sy 1 12(2.8)(1.75) (3.4 )(12 )(2.8)(1.75) = + 3 1.5 48(h − 2d )3 t 27.62(h − 2d ) t

(h − 2d )3 t = 12.70 [h − 2(0.50)]3 t = 12.70 (4t − 1)3 t = 12.70 t = 0.8627 in 7 say t = in 8 h = 4t = 3.5 in 1 1 1 h > 1 + 1 + in 2 2 2 h > 3.5 in Figure AF 11, h d > 10 h = 10d = 10(0.50) = 5 in 1 d = 2 = 0.5 b 5 −1 1 2 2 Therefore K t = 3.5 , K f = 3.4

1 Use h = 5 in , t = 1 in 4

171.

Design a crank similar to that shown with a design factor of 1.6 ± 0.16 based on the modified Goodman line. The crank is to be forged with certain surfaces milled as shown and two ¼-in. holes. It is estimated that the material must be of the order of AISI 8630, WQT 1100 F. The length L = 17 in. , a = 5 in. , and the load varies form + 15 kips to –9 kips. (a) Compute the dimensions at section AB with h = 3b . Check the safety of the edges (forged surfaces). (Iteration involves; one could first make calculations for forged surfaces and then check safety at holes.) (b) Without redesigning but otherwise considering relevant factors ,

Page 30 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS quantitatively discuss actions that might be taken to reduce the size; holes must remain as located.

Problems 171-174. Solution: (a) AISI 8630, WQT 1100 F su = 96 ksi sn = 0.5su = 0.5(96) = 48 ksi Size factor = 0.85 As-forged surface (Fig. AF I) Surface factor = 0.4 sn = (0.85)(0.42)(48) = 17 ksi Milled surface (Machined) Surface factor = 0.85 sn = (0.85)(0.85)(48) = 34.68 ksi At AB, machined 1 sm K f sa = + sn N su Figure AF 11 1 b = in = 0.5 in 2 1 d = in = 0.25 in 4 d 0.25 = in = 0.5 b 0 .5 Assume K f = 3.50 q = 0.998 K f = q (K t − 1) + 1 = 0.998(3.5 − 1) + 1 = 3.495 Mc I 3 ( h − 2d ) b I= 12

s=

Page 31 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 1 1 h 1  1  1  1  − +    =  h − 1 +  = (4h − 4 + 1) = (4h − 3) 2 2  2  4  2  4 8 8 h = 3b 1 M  (4h − 3) 8 s=   3   1  h − 2 4  b    12 3 M (12b − 3) 2 s= (3b − 0.5)3 b 4.5M (4b − 1) s= (3b − 0.5)3 b M = F (L − a ) M max = (15)(17 − 5) = 180 in − kips

c=

M min = (− 9)(17 − 5) = 108 in − kips 1 M m =  (180 − 108) = 36 in − kips 2 1 M a =  (180 + 108) = 144 in − kips 2 1 sm K f sa = + sn N su 1 4.5(36 )(4b − 1) (3.495)(4.5)(144 )(4b − 1) = + 3 1.6 96(3b − 0.5)3 b 34.68(3b − 0.5) b (4b − 1) = 1 (3b − 0.5)3 b 107.2

(3b − 0.5)3 b = 107.2 (4b − 1) b = 2.6 in 5 say b = 2 in 8 7 h = 3b = 7 in 8 Checking at the edges (as forged) M max = (15)(17 ) = 255 in − kips M min = (− 9)(17 ) = −153 in − kips 1 M m =  (255 − 153) = 51 in − kips 2

Page 32 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 M a =  (255 + 153) = 204 in − kips 2 6M 6M 2 M s= 2 = 3 = 3 bh 9b 3b K f ≈ 1.0

1 sm K f sa = + sn N su 1 2(51) (1.0)(2)(204) = 3 + 3b3 (17 ) 1.6 3b (96) b = 2.373 in 3 say b = 2 in 8 5 3 since b = 2 in > 2 in , ∴ safe. 8 8 (c) Action: reduce number of repetitions of load. CHECK PROBLEMS 173.

For the crank shown, L = 15 in , a = 3 in , d = 4.5 in , b = 1.5 in . It is as forged from AISI 8630, WQT 1100 F, except for machined areas indicated. The load F varies from +5 kips to –3 kips. The crank has been designed without detailed attention to factors that affect its endurance strength. In section AB only, compute the factor of safety by the Soderberg criterion. Suppose it were desired to improve the margin of safety, with significant changes of dimensions prohibited, what various steps could be taken? What are your particular recommendations?

Solution:

For as forged surface sn = 17 ksi For machined surface sn = 34.68 ksi s n = 72 ksi In section AB, machined Page 33 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS M = F (L − a ) M max = (+ 5)(15 − 3) = 60 in − kips M min = (− 3)(15 − 3) = −36 in − kips 1 M m =  (60 − 36 ) = 12 in − kips 2 1 M a =  (60 + 36 ) = 48 in − kips 2 d = h = 4.5 in , b = 1.5 in h =3 b 4.5M (4b − 1) s= (3b − 0.5)3 b 4.5(12)[4(1.5) − 1] sm = = 2.8125 ksi [3(1.5) − 0.5]3 (1.5) 4.5(48)[4(1.5) − 1] sa = = 11.25 ksi [3(1.5) − 0.5]3 (1.5) 1 sm K f sa = + sn N sy K f = 3.495 from Problem 171. 1 2.8125 (3.495)(11.25) = + N 72 34.68 N = 0.85 < 1 , unsafe To increase the margin of safety 1. reduce the number of repetitions of loads 2. shot-peening 3. good surface roughness

Recommendation: No. 1, reducing the number of repetitions of loads. 175.

The link shown is made of AISI C1020, as rolled, machined all over. It is loaded 3 9 5 in tension by pins in the D = in holes in the ends; a = in , t = in , 8 16 16 1 h = 1 in . Considering sections at A, B, and C, determine the maximum safe 8 axial load for N = 2 and indefinite life (a) if it is repeated and reversed; (b) if it is repeated varying from zero to maximum; (c) if it is repeatedly varies or F = −W to F = 3W . (d) Using the results from (a) and (b), determine the ratio of the endurance strength for a repeated load to that for a reversed load (Soderberg line).

Page 34 of 62



Problems 175 - 178 Solution: For AISI C1020, as rolled su = 65 ksi s y = 48 ksi sn = 0.5su = 0.5(65) = 32.5 ksi Size factor = 0.85 For machined all over Surface factor = 0.90 sn = (0.85)(0.90)(0.80)(32.5) = 20 ksi 1 sm K f sa = + sn N sy at A, Figure AF 8 9 b = in 16 1 h = 1 in 8 3 d = D = in 8 5 t = in 16 3 d = 8 = 0.33 h 11 8 9 b 16 = = 0.5 1 h 1 8 K tA = 3.6 d 3 r = = in 2 16 a = 0.01 in

Page 35 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 = 0.95 a 0.01 1+ 1+ 3 r 16 ( ) k fA = q ktA − 1 + 1 = 0.95(3.6 − 1) + 1 = 3.47 q=

1

=

64 F F F = = (h − d )t 1 1 − 3  5  15  8 8  16  1 64 Fm 3.47(64)Fa = + 15(20) 2 15(48) 8 Fm + 1.48 Fa at A 1= 45 s=

At B Figure AF 9 9 d = a = in 16 1 h = 1 in 8 3 r = in 16 5 t = in 16 3 r 16 = = 0.33 9 d 16 1 1 h = 8 =2 9 d 16 K tB = 1.63 a = 0.01 in 1 1 q= = = 0.95 a 0.01 1+ 1+ 3 r 16 k fB = q (ktB − 1) + 1 = 0.95(1.63 − 1) + 1 = 1.6 s=

F F 256 F = = 9 5 dt    45     16  16 

Page 36 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 256 Fm 1.6(256)Fa = + 45(20) 2 45(48) 32 1= Fm + 0.455 Fa at B 135 at C, Figure AF 8,

b >1 h

1 in 8 9 h = a = in 16 1 d = 8 = 0.22 h 9 16 K tC = 3.5 d 1 r = = in 2 16 a = 0.01 in 1 1 q= = = 0.862 a 0.01 1+ 1+ 1 r 16 k fC = q (ktC − 1) + 1 = 0.862(3.5 − 1) + 1 = 3.2 D=

F 256 F F = = (h − d )t  9 − 1  5  35  16 8  16  1 256 Fm 3.2(256)Fa = + 35(20) 2 35(48) 32 1= Fm + 1.17 Fa at C 105 s=

Equations 8 Fm + 1.48 Fa 45 32 At B, 1 = Fm + 0.455 Fa 135 32 At C, 1 = Fm + 1.17 Fa 105

At A, 1 =

Page 37 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS (a) Repeated and reversed load Fm = 0 Fa = F use at A 8 1= Fm + 1.48 Fa 45 8 1 = (0 ) + 1.48 Fa 45 F = 0.676 kip (b) Fm = Fa = F 8 F + 1.48 F at A, 1 = 45 F = 0.603 kip 32 F + 0.455 F at B, 1 = 135 F = 1.480 kips 32 at C, 1 = F + 1.17 F 105 F = 0.678 kip use F = 0.603 kip (c) Fmin = −W , Fmax = 3W 1 Fm = (3W − W ) = W 2 1 Fa = (3W + W ) = 2W 2 8 at A, 1 = W + 1.48(2W ) 45 W = 0.319 kip 32 at B, 1 = W + 0.455(2W ) 135 W = 0.884 kip 32 at C, 1 = W + 1.17(2W ) 105 W = 0.378 kip use W = 0.319 kip Fmax = 0.957 kip

Page 38 of 62



(d) Ratio = 179.

F (b ) 0.603 = = 0.892 F (a ) 0.676

A steel rod shown, AISI 2320, hot rolled, has been machined to the following 3 1 dimensions: D = 1 in. , c = in. , e = in. A semicircular groove at the 4 8 1 1 midsection has r = in. ; for radial hole, a = in. An axial load of 5 kips is 8 4 repeated and reversed ( M = 0 ). Compute the factor of safety (Soderberg) and make a judgement on its suitability (consider statistical variations of endurance strength – i4.4). What steps may be taken to improve the design factor?

Problems 179-183 Solution: AISI 2320 hot-rolled (Table AT 10) su = 96 ksi s y = 51 ksi sn = 48 ksi Size factor = 0.85 Surface factor = 0.85 (machined) sn = (0.80)(0.85)(0.85)(48) = 27.74 ksi 1 sm K f sa = + sn N sy sm = 0 , reversed sa = s 1 K f sa = sn N s sa = n NK f at the fillet, Figure AF 12 1 r = e = in 8

Page 39 of 62



d =c=

3 in 4

D = 1 in 1 r 8 = = 0.17 d 3 4 D 1 = = 1.3 d 3 4 K t = 1.55 a = 0.010 in 1 1 q= = = 0.926 a 0.010 1+ 1+ 1 r 8 K f = q (K t − 1) + 1 = 0.926(1.55 − 1) + 1 = 1.51

sa = s =

N=

4(5) 3 π  4

2

= 11.32 ksi

sn 27.74 = = 1.62 sa K f (11.32 )(1.51)

At the groove, Figure AF 14 1  3 d = b = D − 2r = 1 in − 2 in  = in 8  4 D = 1 in 1 r = in 8 1 r 8 = = 0.17 d 3 4 D 1 = = 1.3 d 3 4 K t = 1.75 a = 0.010 in

Page 40 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 = 0.926 a 0.010 1+ 1+ 1 r 8 ( ) K f = q K t − 1 + 1 = 0.926(1.75 − 1) + 1 = 1.7

q=

1

=

4F 4(5) = = 11.32 ksi 2 2 πd 3 π  4 s 27.74 N= n = = 1.44 sa K f (11.32)(1.7 )

sa = s =

At the hole, Figure AF8 D = h = 1 in 1 d = a = in 4 1 D 4 = = 0.25 h 1 K t = 2.44 a = 0.010 in 1 1 q= = = 0.926 a 0.010 1+ 1+ 1 r 8 K f = q(K t − 1) + 1 = 0.926(2.44 − 1) + 1 = 2.33 sa = s =

F

πD 4

N=

2

− Dd

=

5

π (1)

2

1 − (1)  4 4

= 9.34 ksi

sn 27.74 = = 1.27 sa K f (9.34)(2.33)

Factor of safety is 1.27 From i4.4 s = 0.76 sn sn N= = 1.32 min > 1.27 0.76 sn Therefore, dimensions are not suitable. Steps to be taken: 1. Reduce number of cycle to failure Page 41 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 2. Good surface condition 3. Presetting 186.

A stock stud that supports a roller follower on a needle bearing for a cam is 5 7 3 made as shown, where a = in , b = in , c = in . The nature of the junction 8 16 4 of the diameters at B is not defined. Assume that the inside corner is sharp. The material of the stud is AISI 2317, OQT 1000 F. Estimate the safe, repeated load F for N = 2 . The radial capacity of the needle bearing is given as 1170 lb. at 2000 rpm for a 2500-hr life. See Fig. 20.9, p. 532, Text.

Problem 186 Solution: AISI 2317, OQT 1000 F su = 106 ksi s y = 71 ksi sn = 0.5su = 53 ksi Size factor = 0.85 sn = (0.85)(53) = 45 ksi Figure AF 12 5 d = a = in 8 3 D = c = in 4 r d ≈ 0 , sharp corner 3 D 4 = = 1.2 d 5 8 Assume K t = 2.7 K f ≈ K t = 2.7 s=

32 M π a3

7 M = Fb = F   = 0.4375 F  16 

Page 42 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 5 in = 0.625 in 8 32(0.4375)F s= = 18.25 F π (0.625)3 sm = sa = s = 18.25F 1 sm K f sa = + sn N sy

a=

1 18.25 F (2.7 )(18.25 F ) = + 2 71 45 F = 0.370 kip = 370 lb < less than radial capacity of the needle bearing. Ok.

187.

The link shown is made of AISI C1035 steel, as rolled, with the following 3 7 1 1 dimensions a = in. , b = in. , c = 1 in. , d = in. , L = 12 in. , r = in. The 8 8 2 16 axial load F varies from 3000 lb to 5000 lb and is applied by pins in the holes. (a) What are the factors of safety at points A, B, and C if the link is machined all over? What are the maximum stresses at these points?

Problems 187, 188 Solution: AISI C1035, as rolled su = 85 ksi s y = 55 ksi sn = 0.5su = 42.5 ksi size factor = 0.85 sn = (0.6)(0.85)(42.5) = 21.68 ksi 1 sm K f sa = + sn N sy 1 (5 + 3) = 4 kips 2 1 Fa = (5 − 3) = 1 kip 2

Fm =

Page 43 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS (a) at A, Figure AF 9 1 r = in 16 3 d = a = in 8 7 h = b = in 8 1 r 16 = = 0.17 3 d 8 7 h 8 = = 2.33 d 3 8 K t = 1.9 a = 0.010 in 1 1 q= = = 0.862 a 0.010 1+ 1+ 1 r 16 K f = q(K t − 1) + 1 = 0.862(1.9 − 1) + 1 = 1.78 s=

F ac

4 = 10.67 ksi  3  (1) 8 1 sa = = 2.67 ksi 3  (1) 8 1 10.67 (1.78)(2.67 ) = + N 55 21.68 N = 2.42 sm =

At B, same as A, K f = 1.78 F (b − a )c 4 sm = = 8 ksi  7 3  − (1) 8 8 s=

Page 44 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 = 2 ksi  7 3  − (1) 8 8 1 8 (1.78)(2 ) = + N 55 21.68 N = 3.23

sa =

At C, Figure AF 8 1 d = in 2 h = c = 1 in b h >1 1 d 2 = = 0.5 h 1 K t = 2. 2 a = 0.010 in d 1 r = = in = 0.25 in 2 4 1 1 q= = = 0.962 a 0.010 1+ 1+ r 0.25 K f = q(K t − 1) + 1 = 0.962(2.2 − 1) + 1 = 2.15 F (b − a )(c − d ) 4 sm = = 16 ksi 7 3   1   − 1 −   8 8  2  1 sm = = 4 ksi  7 3  1   − 1 −   8 8  2  1 16 (2.15)(4 ) = + N 55 21.68 N = 1.45 s=

(b) Maximum stresses at A s A = sm + K f sa = 10.67 + 1.78(2.67 ) = 15.42 ksi at B sB = sm + K f sa = 8 + 1.78(2 ) = 11.56 ksi

Page 45 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS at C sC = sm + K f sa = 16 + 2.15(4 ) = 24.6 ksi IMPACT PROBLEMS 189.

A wrought-iron bar is 1in. in diameter and 5 ft. long. (a) What will be the stress and elongation if the bar supports a static load of 5000 lb? Compute the stress and elongation if a 5000 lb. weight falls freely 0.05 in. and strikes a stop at the end of the bar. (b) The same as (a), except that the bar is aluminum alloy 3003H14.

Solution: D = 1 in. , L = 5 ft For wrought iron, E = 28× 106 psi (a) elongation F = 5000 lb FL (5000)(5)(12) = 0.01364 in = δ= π AE (1)2 28 ×106 4 Stress and elongation h = 0.05 in W = 5000 lb L = 5 ft = 60 in

(

)

1

W W  2hEA  2 s = + 1 +  A A LW  1

2  2 6 π  2 ( 0 . 05 ) 28 × 10 ( 1 )    5000 5000   4   = 24,741 psi 1 + s= + π 2 π 2 (60)(5000)  (1) (1)   4 4  sL (24,741)(60 ) = = 0.053 in δ= E 28 × 106

(

(b) Aluminum alloy 3003-H14 E = 10× 106 psi F = 5000 lb FL (5000)(5)(12) = 0.038 in = δ= AE π (1)2 10 ×106 4 Stress and elongation h = 0.05 in

(

Page 46 of 62

)

)


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS W = 5000 lb L = 5 ft = 60 in 1

W W  2hEA  2 s = + 1 +  A A LW  1

2  2 6 π  2 0 . 05 10 × 10 1 ( ) ( )    5000 5000   4   = 18,475 psi 1 + s= + π 2 π 2 (60)(5000)  (1) (1)   4 4  sL (18,475)(60 ) = = 0.111 in δ= E 10 × 106

(

190.

)

What should be the diameter of a rod 5 ft. long, made of an aluminum alloy 2024-T4, if it is to resist the impact of a weight of W = 500 lb dropped through a distance of 2 in.? The maximum computed stress is to be 20 ksi.

Solution: For aluminum alloy, 2024-T4 E = 10.6 × 106 psi W = 500 lb h = 2 in L = 5 ft = 60 in s = 20 ksi = 20,000 psi 1

W W  2hEA  2 s = + 1 +  A A LW 

(

)

1

5000 5000  2(2 ) 10.6 × 106 A  2 20,000 = + 1+ (60)(500)  A A  1

40 A = 1 + (1 + 1413 A)2 A=

πD 2 4

= 0.9332

D = 1.09 in , say D = 1

191.

1 in 16

A rock drill has the heads of the cylinder bolted on by 7/8-in. bolts somewhat as shown. The grip of the bolt is 4 in. (a) If the shank of the bolt is turned down to the minor diameter of the coarse-thread screw, 0.7387 in., what energy may each bolt absorb if the stress is not to exceed 25 ksi? (b) Short bolts used as described above sometimes fail under repeated shock loads. It was found in one instance that if long bolts, running from head to head, were used, service failures were eliminated. How much more energy will the bolt 21 in. long absorb for a stress of

Page 47 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 25 ksi. That the bolt 4 in. long? As before, let the bolt be turned down to the minor diameter. The effect of the threads on the strength is to be neglected.

Problem 191 Solution: 2 s2 U= ( AL ) = s AL 2E 2E 2 πD (a) A = 4 L = 4 in D = 0.7387 in E = 30×106 psi s = 25 ksi = 25,000 psi

(25,000)2  π (0.7387 )2 (4) 4 2 30 × 106

U=

(

)

= 17.86 in − lb

(b) L = 21 in

(25,000)2  π (0.7387 )2 (21)

4 = 93.75 in − lb 2 30 × 106 ∆U = 93.75 − 17.86 = 75.89 in − lb

U=

192.

(

)

As seen in the figure, an 8.05-lb body A moving down with a constant acceleration of 12 fps2, having started from rest at point C. If A is attached to a steel wire, W & M gage 8 (0.162 in. diameter) and if for some reason the sheave D is instantly stopped, what stress is induced in the wire?

Problems 192, 193 Solution: s 2 AL U= 2E 1 2 1 U = mv = m(2ah ) = mah = maL 2 2

Page 48 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS s 2 AL = maL 2E 2maE 2WaE s2 = = gA A W = 8.05 lb a = 12 fps 2 g = 32 fps 2 b = 12 fps 2 E = 30×106 psi A=

πD 2

4 8WaE 8(8.05)(12)(30 ×106 ) s2 = = πD 2 g π (0.162)2 (32) s = 93,741 psi

193.

The hoist A shown, weighing 5000 lb. and moving at a constant v = 4 fps is attached to a 2 in. wire rope that has a metal area of 1.6 sq. in. and a modulus E = 12×106 psi . When h = 100 ft , the sheave D is instantly stopped by a brake (since this is impossible, it represents the worst conceivable condition). Assuming that the stretching is elastic, compute the maximum stress in the rope.

Solution:

s 2 AL 2E 1 W 2 v U = mv 2 = 2 2g

U=

s 2 AL W 2 v = 2E 2g Wv 2 E gAL W = 5000 lb s2 =

Page 49 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS v = 4 fps

E = 12×106 psi A = 1.6 in 2 L = h = 100 ft g = 32 fps 2 s2 =

(5000)(4)2 (12 ×106 ) (32)(1.6)(100)

s = 13,693 psi

194.

A coarse-thread steel bolt, ¾ in. in diameter, with 2 in. of threaded and 3 in. of unthreaded shank, receives an impact caused by a falling 500-lb weight. The area at the root of the thread is 0.334 sq. in. and the effects of threads are to be neglected. (a) What amount of energy in in-lb. could be absorbed if the maximum calculated stress is 10 ksi? (b) From what distance h could the weight be dropped for this maximum stress? (c) How much energy could be absorbed at the same maximum stress if the unthreaded shank were turned down to the root diameter.

Solution: s 2 AL U= 2E (a) U = U1 + U 2 s12 A1L1 2E 2 s AL U2 = 2 2 2 2E A1 = 0.334 in 2 U1 =

A2 =

π

(0.75) = 0.442 in 2

4 s1 = 10,000 psi s A (10,000)(0.334 ) s2 = 1 1 = = 7556 psi A2 0.442 L1 = 2 in L2 = 3 in

E = 30 × 10 6 psi U1 =

(10,000)2 (0.334)(2) = 1.113 in − lb

U2 =

(7556)2 (0.442)(3) = 1.262 in − lb

2(30 ×106 )

2(30 ×106 )

Page 50 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS U = U1 + U 2 = 1.113 + 1.262 = 2.375 in − lb 1   W   2hEA  2  (b) s = 1 + 1 +  A  LW     1   2         W 2hE   s = 1 + 1 +  L L  A1   W  1 + 2        A1 A2      1   W  2hEA1 A2  2  s = 1 + 1 +  A1   W ( A2 L1 + A1L2 )     W = 500 lb

A1 = 0.334 in 2 A2 = 0.442 in 2 L1 = 2 in L2 = 3 in E = 30×106 psi s = 10,000 psi

(

)

1

 2h 30 ×106 (0.334 )(0.442 )  2 500 10,000 = 1 + 1 +  0.334  500[(0.442 )(2 ) + (0.334 )(3)] h = 0.0033 in

s 2 AL 2E A = 0.334 in 2 L = 5 in E = 30×106 psi s = 10,000 psi

(c) U =

2 10,000) (0.334)(5) ( U= = 2.783 in − lb

2(30 ×106 )

196.

A part of a machine that weighs 1000 lb. raised and lowered by 1 ½-in. steel rod that has Acme threads on one end (see i8.18 Text, for minor diameter). The length of the rod is 10 ft. and the upper 4 ft are threaded. As the part being

Page 51 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS lowered it sticks, then falls freely a distance of 1/8 in. (a) Compute the maximum stress in the rod. (b) What would be the maximum stress in the rod if the lower end had been turned down to the root diameter? Solution: 1   W   2hEA  2  s= 1 + 1 +  A  LW     1   2       W   2hE   s = 1 + 1 +  L L  A1   W  1 + 2        A1 A2      1   W  2hEA1 A2  2  s = 1 + 1 +  A1   W ( A2 L1 + A1L2 )     1 see i8.18 , D2 = 1 in , D1 = 1.25 in 2 2 π (1.25) A1 = = 1.227 in 2 4 π (1.5)2 A2 = = 1.767 in 2 4 L1 = 4 in L2 = 6 in 1 h = in = 0.125 in 8 W = 1000 lb E = 30×106 psi

(

)

1

 2(0.125) 30 × 106 (1.227 )(1.767 )  2 1000 s= 1 + 1 +  = 28,186 psi 1.227 1000[(1.767 )(4 ) + (1.227 )(6 )]   1   W   2hEA  2  (b) s = 1 + 1 +  A  LW     2 A = A1 = 1.227 in L = L1 + L2 = 10 in

Page 52 of 62



(

)

1

 2(0.125) 30 × 106 (1.227 )  2 1000 s= 1 + 1 +  = 25,552 psi 1.227 10(1000 )  

197.

A weight W of 50 lb is moving on a smooth horizontal surface with a velocity of 2 fps when it strikes head-on the end of a ¾-in. round steel rod, 6 ft. long. Compute the maximum stress in the rod. What design factor based on yield strength is indicated for AISI 1010, cold drawn?

Solution: 1

 2 2   Wv E  s=  g AL1 + We    o  W   W We = b 3 Wb = ρAL

ρ = 0.284 lb in3 π 3

2 2

  = 0.442 in 4 4 L = 6 ft = 72 in Wb = (0.284 )(0.442 )(72) = 9.038 lb 9.038 = 3.013 lb We = 3 W = 50 lb v = 2 fps A=

go = 32 fps 2

E = 30×106 psi L = 6 ft 1

2  2 6  (50 )(2) (30 ×10 )  = 8166 psi s=  (32)(0.442)(6)1 + 3.013    50    For AISI 1010, cold drawn s y = 55 ksi = 55,000 psi N=

s y 55,000 = = 6.74 s 8166

Page 53 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 199.

A rigid weight of 100 lb is dropped a distance of 25 in. upon the center of a 12 in., 50-lb. I-beam ( I x = 301.6 in 4 ) that is simply supported on supports 10 ft apart. Compute the maximum stress in the I-beam both with and without allowing for the beam’s weight.

Solution: Without beams weight y s = sst yst FL3 48EI F 48EI k= = 3 L y y=

1   W   2hk  2  y =δ = 1 + 1 +  k   W     6 E = 30×10 psi L = 10 ft = 120 in

I = 301.6 in 4 48(30 ×106 )(301.6) = 251,333 lb in (120)3 W = 100 lb h = 25 in k=

1    100   2(25)(251,333)  2  y= 1 + 1 +   = 0.1415 in 100  251,333    

WL3 (100)(120) yst = = = 0.0004 in 48EI 48(30 ×106 )(301.6) Mc sst = I WL (100 )(120) M= = = 3000 in − lb 4 4 h 12 c= = = 6 in 2 2 (3000)(6) = 59.68 psi sst = 301.6  0.1415  s = (59.68)  = 21,112 psi  0.0004  3

Page 54 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS with mass of beam 1

 2h  2  y = yst + yst 1 +  yst  h - correction factor =

1 W 1+ e W

17Wb 35 Wb = (50 lb ft )(10 ft ) = 500 lb 17(500 ) We = = 243 lb 35 1 h - correction factor = = 0.292 243 1+ 100 1     2(25)(0.292)  2  y = 0.00041 + 1 +   = 0.0764 in 0004 0 .    y  0.0764  s = sst = (59.68)  = 11,400 psi yst  0.0004  We =

201.

A 3000 lb. automobile (here considered rigid) strikes the midpoint of a guard rail that is an 8-in. 23-lb. I-beam, 40 ft. long; I = 64.2 in4 . Made of AISI C1020, as rolled, the I-beam is simply supported on rigid posts at its ends. (a) What level velocity of the automobile results in stressing the I-beam to the tensile yield strength? Compare results observed by including and neglecting the beam’s mass.

Solution: For AISI C1020, as rolled s y = 48 ksi = 48,000 psi Fδ Wv 2 = 2 2 go F 48 EI k= = 3 δ L Mc FLc s= = I 4I 4 Is F= Lc Fδ F 2 L3 16 I 2 s 2 L3 s 2 IL = = 2 2 = 2 96 EI L c (96 EI ) 6 Ec 2

Page 55 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS neglecting mass of beam Fδ s 2 IL Wv 2 = = 2 6 Ec 2 2 go 3Wv 2 Ec 2 2 g o IL W = 3000 lb go = 32 fps 2 h 8 c = = = 4 in 2 2 E = 30×106 psi I = 64.2 in4 L = 40 ft s = s y = 48 ksi = 48,000 psi s2 =

(

)

3Wv 2 Ec 2 3(3000 )v 2 30 × 106 (4 ) s = (48,000) = = 32(64.2)(40) 2 g o IL v = 6.62 fps Including mass of beam   2 2  3Wv Ec 1    s2 = 2 g o IL  1 + We     W  17Wb We = 35 Wb = (23 lb ft )(40 ft ) = 920 lb 17(920 ) We = = 447 lb 35 2

2

2

  2 2 2 2 6   × 3 Wv Ec 3 ( 3000 ) v ( 30 10 ) ( 4 ) 1 = s 2 = (48,000) =  447  2 go IL 32(64.2)(40) 1 +   3000  v = 7.10 fps 2

DATA LACKING – DESIGNER’S DECISIONS 202.

A simple beam is struck midway between supports by a 32.2-lb. weight that has fallen 20 in. The length of the beam is 12 ft. If the stress is not to exceed 20 ksi, what size I-beam should be used?

Solution:

Page 56 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1

 2h  2  y = yst + yst 1 +  yst  y s = sst yst h = 20 in s = 20,000 psi

WL3 yst = 48EI 1

y  96 EIh  2 = 1 + 1 +  yst WL3   with correction factor 1

   2   96 EIh 1  y   = 1 + 1 + 3 We  yst WL   1+    W  Mc WLd sst = = I 8I 17 wL We = 35 1   2        WLd   96 EIh  1    s= 1 + 1 + 8I   WL3  1 + 17 wL     35W        W = 32.2 lb h = 20 in L = 12 ft = 144 in

E = 30×106 psi

s=

(32.2)(144)d 8I

1   2       1  96 30 ×106 (I )(20)  1 + 1 +   (32.2)(144)3 1 + 17(w)(12)    35(32.2)      

(

)

1   579.6d   1   2  s=   1 + 1 + 599 I  I    1 + 0.181w     From The Engineer’s Manual By Ralph G. Hudson, S.B.

Page 57 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Use 3”, 5.7 lb, I = 2.5 in4 1

  2 579.6(3) 1 s= 1 + 1 + 599(2.5)   = 19,600 psi < 20,000 psi 2.5 1 + 0.181(5.7 )   

Therefore use 3-in depth, 5.7-lb I-beam ( I = 2.5 in4 ) 204.

A 10-in., 25.4-lb.., I-bean, AISI 1020, as rolled, is 10 ft. long and is simply supported at the ends shown. There is a static load of F1 = 10 kips , 4 ft from the left end, and a repeated reversed load of F2 = 10 kips , 3 ft from the right end. It is desired to make two attachments to the beam through holes as shown. No significant load is supported by these attachments, but the holes cause stress concentration. Will it be safe to make these attachments as planned? Determine the factor of safety at the point of maximum moment and at points of stress concentration.

Problem 204 Solution: Mass of beam negligible For AISI C1020, as rolled s y = 48 ksi su = 65 ksi

(∑ M

A

= 0)

4 F1 + (10 − 3)F2 = 10 B 1 B = (4 F1 + 7 F2 ) 10 (∑ M B = 0) 3F2 + (10 − 4)F1 = 10 A 1 A = (6 F1 + 3F2 ) 10 F1 = 10 kips F2 = −10 to 10 kips 1 Bmin = [4(10) + 7(− 10 )] = −3 kips 10

Page 58 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 [4(10) + 7(10)] = 11 kips 10 1 Amin = [6(10) + 3(− 10)] = 3 kips 10 1 Amax = [6(10) + 7(30)] = 9 kips 10 Figure AF 11, 1 e = 1 in , 2 1 d = in 4 1 1 c = 1 + 2  = 1.625 in 2 4 h = 10 in h 10 1 b = − e = − 1 = 3.5 in 2 2 2 d 0.25 = = 0.07 b 3 .5 e 1.50 = 6 > 0.5 = d 0.25 Use K t = 3.0 1 q= = 0.926 0.010 1+ 1 8 K f = q (K t − 1) + 1 = 0.926(3 − 1) + 1 = 2.85

Bmax =

sn = 0.5su = 0.5(65) = 32.5 ksi size factor = 0.85 sn = 0.85(32.5) = 27.6 ksi left hole, M = (2)A M max = 2(9) = 18 ft − kips M min = 2(3) = 6 ft − kips Mc s= I 1 M m = (18 + 6) = 12 ft − kips = 144 in − kips 2 1 M a = (18 − 6 ) = 6 ft − kips = 72 in − kips 2 c = 1.625 in I = 122.1 in 4 (Tables)

Page 59 of 62



sm = sa

(144)(1.625) = 1.92 ksi 122.1

(72 )(1.625) = 0.96 ksi =

122.1 1 sm K f sa = + sn N sy

1 1.92 (2.85)(0.96 ) = + N 48 27.6 N = 7 .2

right hole , M = (1.5)B M max = 1.5(11) = 16.5 ft − kips M min = 1.5(− 3) = −4.5 ft − kips Mc s= I 1 M m = (16.5 − 4.5) = 6 ft − kips = 72 in − kips 2 1 M a = (16.5 + 4.5) = 10.5 ft − kips = 126 in − kips 2 c = 1.625 in I = 122.1 in 4 (Tables) (72)(1.625) = 0.96 ksi sm = 122.1 (126)(1.625) = 1.68 ksi sa = 122.1 1 sm K f sa = + sn N sy 1 0.96 (2.85)(1.68) = + N 48 27.6 N = 5.67

at maximum moment, or at , F2 M max = 3(11) = 33 ft − kips M min = 3(− 3) = −9 ft − kips Mc s= I 1 M m = (33 − 9) = 12 ft − kips = 144 in − kips 2 1 M a = (33 + 9 ) = 21 ft − kips = 252 in − kips 2

Page 60 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 10 = 5 in 2 I = 122.1 in 4 (Tables) (144)(5) = 5.90 ksi sm = 122.1 ( 252)(5) sa = = 10.32 ksi 122.1 K f = 1 .0

c=

1 sm K f sa = + sn N sy 1 5.90 (1.0 )(10.32 ) = + N 48 27.6 N =2

Since the design factor at the holes is much larger than at the point of maximum moment, it is safe to make these attachment as planned. 205.

The runway of a crane consists of L = 20 ft. lengths of 15-in., 42.9-lb. I-beams, as shown, each section being supported at its ends; AISI C1020, as rolled. The wheels of the crane are 9 ft apart, and the maximum load expected is F = 10,000 lb on each wheel. Neglecting the weight of the beam, find the design factor (a) based on variable stresses for 105 cycles, (b) based on the ultimate strength. (Hint. Since the maximum moment will occur under the wheel, assume the wheels at some distance x from the point of support, and determine the dM reaction, R1 as a function of x ; = 0 gives position for a maximum bending dx moment.)

Problem 205. Solution: ∑ M R2 = 0

(

)

(L − x )F + (L − x − a )F = LR1 (2 L − 2 x − a )F R = 1

L x M = R1 x = (2 L − 2 x − a )F L

Page 61 of 62


SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS dM F = [(2 L − 2 x − a ) + x(− 2 )] = 0 dx L 2L − 2x − a − 2x = 0 1 a x = L−  2 2 2

a  a L−  F L−  2 2 2 L −  L − a  − a  F =  M max =   L  2  2L  L = 20 ft = 240 in a = 9 ft = 108 in F = 10,000 lb = 10 kips 2

108    240 −  (10 ) 2   M max = = 720.75 in − kips 2(240) For 15-in., 42.9 lb, I-beam I = 441.8 in4 15 c = = 7.5 in 2 Mc (720.75)(7.5) smax = = = 12.24 ksi I 441.8 For AISI C1029, as rolled su = 65 ksi

sn = 0.5su = 32.5 ksi size factor = 0.85 sn = 0.85(32.5) = 27.6 ksi (a) at 105 cycles 0.085

 106  sn = 27.6 5  = 34 ksi  10  s 34 N= n = =7 s 12.24

(b) N =

su 65 = = 5.31 s 12.24

- end -

Page 62 of 62


SECTION 3 – SCREW FASTENINGS SIMPLE TENSION INCLUDING TIGHTENING STRESSES DESIGN PROBLEMS

221.

A 5000-lb. gear box is provided with a steel (as rolled B1113) eyebolt for use in moving it. What size bolt should be used: (a) if UNC threads are used? (b) If UNF threads are used? (c) If the 8-thread series is used? Explain the basis of your choice of design factor.

Solution: B1113, as rolled s y = 45 ksi (Table AT-7) Fe = 5000 lb i5.6, sd =

sy 6

1

( As ) 2

3    D < in  4  

3 in 4 As ≈ 0.35 sq.in.

For D =

sd = sd =

sy 6 sy

1

(0.35) 2

10 use design factor = 10 45,000 psi 10 sd = 4500 psi sd =

As =

Fe 5000 = = 1.111 sq.in. sd 4500

Table AT 14 and Table 5.1

(a) UNC Threads 3 Use D = 1 in , As = 1.155 sq.in. 8

Page 1 of 42


SECTION 3 – SCREW FASTENINGS (b) UNF Threads 3 Use D = 1 in , As = 1.155 sq.in. 8

(c) 8-Thread Series 3 Use D = 1 in , As = 1.233 sq.in. 8

222.

A motor weighing 2 tons is lifted by a wrought-iron eye bolt which is screwed into the frame. Decide upon a design factor and determine the size of the eyebolt if (a) UNC threads are used, (b) UNF threads are used. Note: Fine threads are not recommended for brittle materials.

Solution: Table AT-7 Wrought iron, s y = 25 ksi Assume design factor = 10 sd =

sy

10 25,000 psi sd = 10 sd = 2500 psi As =

Fe 2(2000) = = 1.60 sq.in. 2500 sd

Table AT 17 (a) UNC Threads 3 Use D = 1 in , As = 1.90 sq.in. 4

(b) UNF Threads 1 Use D = 1 in , As = 1.581 sq.in. 2

Page 2 of 42


SECTION 3 – SCREW FASTENINGS 224.

A wall bracket, Fig. 8-13, Text, is loaded so that the two top bolts that fasten it to the wall are each subjected to a tensile load of 710 lb. The bolts are to be cold forged from AISI C1020 steel with UNC threads, Neglecting the effect of shearing stresses, determine the diameter of these bolts if they are well tightened.

Figure 8-13

Solution:

cold forged, AISI C1020 s y = 66 ksi (Table AT-7)

Fe = 710 lb s 3 3   Fe = y ( As ) 2  D < in  6 4   3 66,000 710 = ( As ) 2 6 3 As = 0.161 sq.in. , D < in 4

Table AT 14 , UNC Threads Use D =

9 in , As = 0.1820 sq.in. 16

Page 3 of 42


SECTION 3 – SCREW FASTENINGS 225.

A connection similar to Fig. 5.9, Text, is subjected to an external load Fe of 1250 lb. The bolt is made from cold-finished AISI B1113 steel with UNC threads. (a) Determine the diameter of the bolt if it is well tightened. (b) Compute the initial tension and corresponding approximate tightening torque if si = 0.85s y (i5.8). Figure 5.9

Solution: Cold-finished AISI B1113 Table A-7, s y = 72 ksi Fe = 1250 lb sy 3 (a) Fe = ( As ) 2 6 3 72,000 1250 = ( As ) 2 6 3 in 4 Table AT 14 , UNC Threads

As = 0.2214 sq.in. , D <

Use D =

5 in , As = 0.2260 sq.in. 8

(b) si = 0.85s y = 0.85(72,000) = 61,200 psi Initial Tension Fi = si As = (61,200)(0.2260) = 13,831 lb Tightening torque T = CDFi

Page 4 of 42


SECTION 3 – SCREW FASTENINGS 5 T = 0.2 DFi = 0.2 (13,831) = 1729 in − lb 8

The cylinder head of a 10 x 18 in. Freon compressor is attached by 10 stud bolts made of SAE Grade 5. The cylinder pressure is 200 psi. (a) What size bolts should be used? (b) What approximate tightening torque should be needed to induce a tightening stress si of 0.9 times the proof stress?

226.

Solution: Table 5.2 SAE Grade 5 Assume s y = 88 ksi π  2 200 (10) 4 = 1571 lb (a) Fe = 10 s 3 3 Fe = y ( As ) 2 , D < in 6 4 3 88,000 1571 = ( As ) 2 6 3 As = 0.2255 sq.in. , D < in 4 Table AT 14 , UNC Threads Use D =

5 in , As = 0.2260 sq.in. 8

(b) T = CDFi C = 0 .2 si = 0.9 s p s p = 85 ksi , (Table 5.2)

si = 0.9(85,000) = 76,500 psi Fi = si As = (76,500)(0.2260) = 17,289 lb Tightening torque 5 T = 0.2 DFi = 0.2 (17,289 ) = 2161 in − lb 8 227.

The American Steel Flange Standard specifies that 8 bolts are to be used on flanges for 4-in. pipe where the steam or water pressure is 1500 psi. It is also specified that, in calculating the bolt load, the outside diameter of the gasket, which is 6 3/16 in., should be used. Determine (a) the diameter of the UNC bolts if they are well-tightened and made of ASTM 354 BD (Table 5-2), (b) the approximate torque to tighten the nuts if the initial stress is 90 % of the proof

Page 5 of 42


SECTION 3 – SCREW FASTENINGS stress. The Standard specifies that 1 1/4 –in. bolts with 8 th./in. be used (these bolts are also subjected to bending). How does your answer compare? Solution: Table 5.2, ASTM 354 BD s p = 120 ksi s y = 125 ksi 2

 π  3  1500  6   4  16  = 5638 lb Fe = 8 (a) Fe =

sy

3

( As ) 2

6 3 125,000 5638 = ( As ) 2 6

, D<

3 in 4

3 in 4 Table AT 14 , UNC Threads

As = 0.4184 sq.in. , D <

Use D = D>

7 in , As = 0.4620 sq.in. 8

3 in 4

use sd = (1 − 0.85)s y = 0.15(125,000 ) = 18,750 psi Fe 5638 = = 0.3007 sq.in. sd 18,750 Table AT 14 , UNC Threads 3 Use D = in , As = 0.334 sq.in. 8 (b) T = 0.2 DFi As =

si = 0.9 s p

si = 0.9(120,000 ) = 108,000 psi Fi = si As = (108,000 )(0.3340) = 36,072 lb Tightening torque 3 T = 0.2 DFi = 0.2 (36,072 ) = 5411 in − lb 4

Page 6 of 42


SECTION 3 – SCREW FASTENINGS 1 D < 1 in as specified by the standard. 4

CHECK PROBLEMS 228.

A cap screw, ¾ in.-10-UNC-2, with a hexagonal head that is 9/16 in. thick, carries a tensile load of 3000 lb. If the material is AISI 1015, cold drawn, find the factor of safety based on ultimate strengths of (a) the threaded shank, (b) the head against being sheared off, and (c) the bearing surface under the head. (d) Is there any need to consider the strength of standard cap-screw heads in design?

Solution: For ¾ in. UNC, Table AT 14, As = 0.334 sq.in. Head:

1 A = 1 in. 8

For AISI 1015, cold drawn su = 77 ksi , sus = 58 ksi F 3000 = = 8982 psi As 0.334 s 77,000 N= u = = 8.57 8982 sd

(a) s =

(b) ss =

F πDt

9 in 16 3000 ss = = 2264 psi  3  9  π     4  16  s 58,000 N = us = = 25.6 2264 ss

t=

Page 7 of 42


SECTION 3 – SCREW FASTENINGS

(c) θ =

360o = 30 o 12

2

 1 1   1  A  A   1  8  tan 30 = 1.096 sq.in. Area = 6(2 )   tan θ  = 6(2 )   2  2  2   2  2      F 3000 sb = = = 4586 psi 2 Area − Ab π 3 1.096 −   4 4 s 77,000 N= u = = 16.8 4586 sb

(d)

No need to consider the strength of standard cap-screw heads since its factor of safety is very much higher than for the threaded shank.

229.

A bolt, 1 1/8 in.-7-UNC-2, is subjected to a tensile load of 10,000 lb. The head has a thickness of ¾ in. and the nut a thickness of 1 in. If the material is SAE grade 2 (Table 5.2), find the design factor as based on ultimate stresses (a) of the threaded shank, (b) of the head against being sheared off, and (c) of the bearing surface under the head. The bolt head is finished. (d) Is there any need to consider the strength of standard bolt heads in design?

Solution: 1 For SAE grade 2 (Table 5.2), D = 1 in 8 su = 55 ksi , sus = 0.75su For 1 1/8 in.-7-UNC-2 (Table AT 14) As = 0.763 sq.in.

A =1

11 in 16

F = 10,000 lb.

Page 8 of 42



F 10,000 = = 13,106 psi As 0.763 s 55,000 N= u = = 4.2 sd 13,106

(a) s =

(b) ss =

F πDt

3 in 4 10,000 ss = = 3773 psi  1  3  π 1    8  4  s 0.75(55,000 ) N = us = = 11 3773 ss

t=

(c) θ =

360o = 30 o 12

2

 11  1  1 A A 1       Area = 6(2 )   tan θ  = 6(2 )  16  tan 30 = 2.4661 sq.in.  2  2  2   2  2      F 10,000 sb = = = 6793 psi 2 Area − Ab π  1 2.4661 − 1  4  8 s 55,000 N= u = = 8.1 6793 sb

(d) No need to consider the strength of standard bolt head in design since its factor of safety is higher than for the threaded shank. 230.

An axial force is applied to a regular nut which of course tends to shear the threads on the screw. (a) What is the ratio of the force necessary to shear the threads (all threads initially in intimate contact) to the force necessary to pull the bolt in two? Use coarse threads, a 1 ½ -in. bolt, and assume that sus = 0.75su . The head thickness is 1 in. and the nut thickness is 1 5/16 in. (b) Is failure of the thread by shear likely in this bolt?

Page 9 of 42



Solution: 1 ½ - in. UNC As = 1.405 sq.in. (a) Fs = shear force = susπDt 1 D = 1 in. 2 5 t = 1 in. 16 sus = 0.75su  1  5  Fs = (0.75su )(π )1 1  = 4.6388su  2  16  F = su As = 1.405su 4.6388su = 3.3 Ratio = 1.405su

(b) Ratio > 1, failure by shear is not likely to occur. 231.

For bolted structural joints, specifications suggest that ½-in. bolts (highstrength material) be tightened to an initial tension of Fi = 12,500 lb . What should be the approximate tightening torque? How does your answer compare with T = 90 ft − lb ., which is the value in the specification?

Solution: 1 T = 0.2 DFi = 0.2 (12,500 ) = 1250 in − lb 2 T = 90 ft − lb = 1080 in − lb < 1250 in − lb o.k.

232.

One method of estimating the initial tensile stress in a tightened bolt is to turn the nut until it is snug, but with no significant stress in the bolt. Then the nut is turned through a predetermined angle that induces a certain unit strain corresponding to the desired stress. A ¾ - in. bolt of the type shown in Fig. 5.4, Text, is turned down until, for practical purposes, the diameter of the entire shank is the minor diameter. The material is AISI 4140, OQT 1200 oF. The grip is 5 in. and the effective strain length is estimated to be 5.3 in. If the initial tensile stress at the root diameter is to be about 75 % of the yield strength, through what angle should the nut be turned after it is just snug? The threads are UNC and the parts being bolted are assumed to be rigid.

Solution:

Page 10 of 42



For ¾ in., UNC Dr = 0.6273 in As = 0.334 sq.in. Th in. = 10 AISI 4140, OQT 1200 oF s y = 115 ksi

s = 0.75(115) = 86.25 ksi sL E L = 5.3 in

δ=

pitch, p =

1 in = 0.10 in 10

δ

(360o ) p sL 360o ) θ= ( pE (86,250)(5.3) (360o ) = 55o θ= (0.10)(30 ×106 ) θ=

233.

When both ends of a bolt are accessible for micrometer measurements, the total elongation δ caused by tightening can be determined by measuring lengths before and after tightening. In order to reduce this total elongation to unit elongation, thence to stress, the effective strain length for the bolt must be known. For a 1 ¼-in steel bolt, threaded for its full length, 8-thread series, the effective strain length has been found by experiment to be Le = 0.97G + 1.1 in. , where G is the grip (by W.A. McDonald, North Carolina State College). Let the bolt material be AISI 8742, OQT 1000 oF. (a) It is desired that the initial tensile stress be about 0.7 s y . What total elongation should be obtained for a grip length of 4.8 in.? (b) Investigate the approximate tightening torque for the specified condition. How could this torque be obtained?

Solution: 1 ¼ in., 8-thread series Table 5.1 Page 11 of 42


SECTION 3 – SCREW FASTENINGS Dr = 1.0966 in As = 1.000 sq.in. Th in. = 8 AISI 8742, OQT 1000 oF s y = 147 ksi (a) si = 0.70 s y si = 0.70(147 ) = 102.9 ksi = 102,900 psi si Le E Le = 0.97G + 1.1 in. G = 4.8 in Le = 0.97(4.8) + 1.1 in. = 5.756 in

δ=

δ=

si Le (102,900 )(5.756 ) = = 0.01975 in E 30 × 10 6

(b) θ =

δ p

=

64TL πDr4G

G = 11.5 ×106 psi 1 p = in = 0.125 in 8 0.01975 64T (5.756 ) = θ= 0.125 π (1.0966 )4 (11.5 × 106 ) T = 22,408 in − lb ELASTIC CONSIDERATIONS 235.

The member C shown is part of a swivel connection that is to be clamped by a 1-in. bolt D to the member B, which has large dimensions in the plane perpendicular to the paper. Both B and C are aluminum alloy 2024-T4, HT aged. The bolt is made of AISI C1113, cold-drawn steel; consider the unthreaded shank to be 2 in. long; it is well tightened with a torque of 250 ftlb.; UNC threads, unlubricated. (a) Estimate the initial tension by equation (5.2), assume elastic action, and compute the bolt elongation and the total deformation of B and C. Let the effective strain length be 2 in. (b) After tightening an external axial force Fe of 5000 lb. is applied to member C. Determine the total normal stresses in the bolt and in B and C. (c) Determine the load required to “open” the connection. Draw a diagram similar to Fig. 5.6, Text, locating points A, B, D and M.

Page 12 of 42



Prob. 235, 236 Solution: For aluminum alloy, 2024-T4 HT aged, E = 10.6 ×106 psi s y = 47 ksi For AISI C1113, cold-drawn steel, E = 30×106 psi s y = 72 ksi (a) T = 0.2 DFi D = 1 in. T = 250 ft − lb = 3000 in − lb Fi = 15,000 lb Deformations: L = 2 in. Table AT 14, 1-in. UNC Bolt, As = 0.66 sq.in. π 2 Ab = (1) = 0.785 sq.in. 4 Bolt: FL (15,000)(2) = 0.00127 in δi = i = Ab Ei (0.785)(30 × 10 6 ) Member B and C FL δc = i Ac Ec π π Ac = De2 − D 2 4 4 h De = (Nut or head width across flats) + 2 Table AT 14 1 A = 1 in 2

Page 13 of 42


SECTION 3 – SCREW FASTENINGS 1 2 1 De = 1 + = 2 in. 2 2 2 Ac = Ac =

π 4

π

De2 −

π 4

D2

[(2.5) − (1) ] = 4.1234 sq.in. 4 2

2

Fi L (15,000)(2) = 0.000686 in. = Ac Ec (4.1234 ) 10.6 × 10 6 (b) Fe = 5000 lb

δc =

(

)

 δb   kb  0.000686   = 5000  = Fe  ∆Fb = Fe    0.000686 + 0.00127   δb + δc   kb + k c  ∆Fb = 1754 lb Bolt: Ft = Fi + ∆Fb = 15,000 + 1754 = 16,754 lb F 16,754 sb = t = = 29,132 psi As 0.606 Member B and C  kc   Fc = Fi − Fe   kb + kc   δc   Fc = Fi − Fe  + δ δ c   b 0.00127   Fc = 15,000 − 5000  = 11,754 lb  0.00127 + 0.000686  F 11,754 sc = c = = 2851 psi Ac 4.1234 (c) Fo = opening load  δ + δc  0.00127 + 0.000686   = 15,000 Fo = Fi  i  = 23,102 lb 0.00127    δi  Fig. 5.6

Page 14 of 42



237.

A 1-in. steel bolt is used to clamp two aluminum (2014-T6, HT aged) plates together as shown by Fig. 5.9, Text. The aluminum plates have a total thickness of 2 in. and an equivalent diameter of 2 in. The bolt is heated to a temperature of 200 oF, the inserted in the aluminum plates, which are at 80 oF, and tightened so as to have a tensile tightening stress of 30 ksi in the unthreaded shank while steel at 200 oF. What is the tensile stress in the bolt after assembly has cooled to 80 oF? The deformations are elastic. Figure 5.9

Solution: For aluminum 2014-T6 E = 10.6 × 106 psi sb = 30,000 psi π  2 Fi = sb Ab = (30,000 ) (1) = 23,562 lb 4

Steel bolt. E b = 30 × 10 6 psi s L (30,000 )(2) δi = b = = 0.002 in. Eb 30 × 10 6 Fi L . Ac E c π π Ac = De2 − D 2 4 4 π 2 2 Ac = (2 ) − (1) = 2.3562 sq.in. 4

δc =

[

Page 15 of 42

]


SECTION 3 – SCREW FASTENINGS E c = 10.6 × 10 6 psi FL (23,562)(2) = 0.001887 in δc = i = Ac Ec (2.3562 ) 10.6 × 10 6 L′ = L − δ c = 2 − 0.001887 = 1.998113 in. ∆L = αL′∆t α = 0.000007 in. (in. − F ) for steel ∆L = (0.000007 )(1.998113)(80 − 200 ) = −0.001678 in. δ i′ = δ i + ∆L = 0.002 − 0.001678 = 0.000322 in. s′ L δ i′ = b Eb s′ (2 ) 0.000322 = b 6 30 × 10 sb′ = 4830 psi

(

238.

)

A 1 1/8-in. steel bolt A passes through a yellow brass (B36-8) tube B as shown. The length of the tube is 30 in. (virtually the unthreaded bolt length), the threads on the bolt are UNC, and the tube’s cross-sectional area is 2 sq. in. After the nut is snug it is tightened ¼ turn. (a) What normal stresses will be produced in the bolt and in the tube? Assume that washers, nut, and head are rigid. (b) What are the stresses if an axial load of 5 kips is now applied to the bolts end? Compute the bolt load that just results in a zero stress in the tube.

Prob. 238 Solution: For Yellow brass, B36-8, E = 15× 106 psi Steel bolt E = 30 × 106 psi Table AT 14, 1 1/8 in., UNC Dr = 0.9497 in. As = 0.763 sq.in. Th in = 7

Page 16 of 42


SECTION 3 – SCREW FASTENINGS L = 30 in.

θ=

δi

p 1 p = in. 7 1 θ = turn 4  1  1  1 δ i =    = in.  4  7  28 FL δi = i Ab Eb 1 Fi (30 ) = 28  π  1  2 6  1  (30 × 10 )  4  8  Fi = 35,500 lb Bolt: sb =

(a)

Fi 35,500 = = 46,527 psi As 0.763

Fi Ac Ac = 2 sq.in. F 35,500 sc = i = = 17,750 psi Ac 2

Tube: sc =

(b) Fe = 5000 lb kc =

Ac Ec (2 )(15 × 106 ) = = 1,000,000 lb in L 30 2

 π  1  6  1  (30 × 10 ) AE 4 8 kb = b b =    = 994,000 lb in L 30 Bolts:  kb   Fe Ft = Fi +   kb + k c  994,000   Ft = 35,500 +  (5000 ) = 38,000 lb  994,000 + 1,000,000  F 38,000 st = t = = 49,800 psi As 0.763 Page 17 of 42


SECTION 3 – SCREW FASTENINGS Tube:  kc   Fe Fc = Fi −  k + k  b c 1,000,000   Fc = 35,500 −  (5000 ) = 33,000 lb  994,000 + 1,000,000  F 33,000 sc = c = = 16,500 psi Ac 2 For zero stress in the tube k +k   994,000 + 1,000,000  Fo =  b c  Fi =  (35,500 ) = 70,787 lb 1,000,000    kc 

ENDURANCE STRENGTH DESIGN PROBLEMS 239.

As shown diagrammatically, a bearing is supported in a pillow block attached to an overhead beam by two cap screws, each of which, it may be assumed, carried half the total bearing load. This load acts vertically downward, varying from 0 to 1500 lb. The screws are to be made of AISI C1118, as rolled, and they are tightened to give an initial stress of about si = 0.5s y . The pillow block is made of class-20 cast iron. Assume that the effective length of screw is equal to the thickness t , as shown, and that the head and beam are rigid (overly conservative?). The equivalent diameter of the compression area may be taken as twice the bolt diameter. For a design factor of 1.75, determine the size of the screw: (a) from the Soderberg line, (b) from the modified Goodman line. (c) What size do you recommend using?

Problem 239 Solution: For AISI C1118, as rolled s y = 46 ksi su = 75 ksi si = 0.5s y

Page 18 of 42


SECTION 3 – SCREW FASTENINGS si = 0.5(46) = 23 ksi = 23,000 psi Fi = si As = 23 As kip = 23,000 As lb  kb   Fe ∆Fb =   kb + k c  AE kb = b b Lb Eb = 30 ×10 6 psi (steel)

Lb = t Ac Ec Lc For cast-iron class 20 Ec = 9.6 ×10 6 psi kc =

Lc = t π π Ac = De2 − D 2 4 4 π Ab = D 2 4 De = 2 D 3π 2 π π 2 D = 3 Ab Ac = (2 D ) − D 2 = 4 4 4 AE kb = b b Lb Ab (30 ×106 ) t 3 A (9.6 × 106 ) kc = b t 300 kb 30 ×106 = = 6 6 kb + kc 30 × 10 + 3(9.6 × 10 ) 588 ∆Fb1 = 0 kb =

 kb  300  1500   Fe =  ∆Fb 2 =    = 383 lb  588  2   kb + k c  1 1 Fm = Fi + (∆Fb 2 + ∆Fb1 ) = 23,000 As + (383 + 0 ) = 23,000 As + 192 lb 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (383 − 0 ) = 192 lb 2 2

Page 19 of 42


SECTION 3 – SCREW FASTENINGS sn = 0.5su = 0.5(75,000 psi ) = 37,500 psi For axial loading with size factor sn = (0.8)(0.85)(37,500 psi ) = 25,500 psi N = 1.75 Fm 23,000 As 192 192 = + = 23,000 + As As As As F 192 sa = a = As As

sm =

Table AT 12, K f = 1.8 (a) Soderberg line 1 sm K f sa = + sn N sy

  192  192   23,000 +  (1.8)  As  As  1   = + 25,500 1.75 46,000 As = 0.2482 sq.in. Table AT 14, UNC 3 Use D = in. , As = 0.334 sq.in. 4 (b) Modifies Goodman line 1 sm K f sa = + sn N su   192  192   23,000 +  (1.8)  As  As  1   = + 25,500 1.75 75,000 As = 0.0609 sq.in. Table AT 14, UNC 3 Use D = in. , As = 0.0775 sq.in. 8

Page 20 of 42



(c) Recommended, D = 240.

3 in. − UNC 4

A connection similar to Fig. 5.9, Text, is subjected to an external load that varied from 0 to 1250 lb. The bolt is cold forged from AISI B1113 steel; UNC threads.The aluminum parts C (3003 H14) have a total thickness of 1 ½ in. and an external diameter of 2 D . It is desired that the connection not open for an external load of 1.5Fe . Determine (a) the initial tensile load on the bolt, (b) the bolt diameter for N = 2 based on the Soderberg line. Fig. 5.9

Solution:  kc   lb (a) Fi = QFe  k k +  b c Q = 1 .5 AE kb = b b Lb π Ab = D 2 4 Eb = 30 ×10 6 psi 1 Lb = 1 in. 2 AE kc = c c Lc π π Ac = De2 − D 2 4 4 De = 2 D 3π 2 π π 2 Ac = (2 D ) − D 2 = D = 3 Ab 4 4 4 Ec = 10 × 10 6 psi (3003-H14 aluminum)

Page 21 of 42


SECTION 3 – SCREW FASTENINGS 1 Lc = 1 in. 2 A 30 ×106 kb = b  1 1   2 3 A 10 × 106 kc = b  1 1   2 kc 3 10 × 106 = = 0.5 kb + kc 30 ×10 6 + 3 10 × 106 Fe = 1250 lb

(

)

(

)

(

(

)

)

 kc   lb Fi = QFe   kb + kc  Fi = (1.5)(1250 )(0.5) = 937.5 lb (b) For AISI B1113 steel, cold forged su = 83 ksi s y = 72 ksi sn = 0.5su = 0.5(83) = 41.5 ksi = 41,500 psi For axial loading with size factor sn = (0.8)(0.85)(41,500 psi ) = 28,220 psi  kb   Fe ∆Fb =   kb + k c  ∆Fb1 = 0  kb    30 ×106  Fe =  (1250) = 625 lb ∆Fb 2 =  6 6   30 ×10 + 3 10 × 10   kb + k c 

(

)

1 (∆Fb 2 + ∆Fb1 ) = 937.5 + 1 (625 + 0) = 1250 lb 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (625 − 0 ) = 312.5 lb 2 2 F 1250 sm = m = As As Fm = Fi +

Page 22 of 42



sa =

Fa 312.5 = As As

Soderberg line, K f = 1.8 Table AT 12 1 sm K f sa = + sn N sy 1 1250 (1.8)(312.5) = + 2 72,000 As 28,220 As As = 0.07459 sq.in.

Table AT 14, UNC 3 Use D = in. , As = 0.0775 sq.in. 8 243.

This problem concerns the Freon compressor of 226: size, 10 x 18 in.; 10 studs, UNC; made of C1118, as rolled; 200 psi gas pressure. The initial tension in the bolts, assumed to be equally loaded, is such that a cylinder pressure of 300 psi is required for the joint to be on the opening. The bolted parts are cast steel and for the first calculations, it will be satisfactorily to assume the equivalent diameter of the compressed parts to be twice the bolt size. (a) For N = 2 on the Soderberg criterion, what bolt size is required? (b) Compute the torque required for the specified initial tension.

Solution:  π  (10) Fo = 300  = 2356 lb.  4  10  kc   Fi = Fo  + k k  b c AE kb = b b Lb π Ab = D 2 4 Eb = 30 ×10 6 psi 2

Lb = L AE kc = c c Lc π π Ac = De2 − D 2 4 4 De = 2 D

Page 23 of 42


SECTION 3 – SCREW FASTENINGS 3π 2 D = 3 Ab 4 4 4 Cast Steel, Ec = 30 ×10 6 psi Ac =

π

(2 D )2 − π

D2 =

Lc = L

(

)

Ab 30×10 6 L 3 A (30 × 10 6 ) kc = b = 3kb L  kc   3kb   = (2356 )  = 1767 lb Fi = Fo  k k k k + + 3 b   b c  b kb =

 kb   Fe (a) ∆Fb =  k k +  b c ∆Fb1 = 0  kb   kb  π  (10 )2 (200 )  Fe =    ∆Fb 2 =  = 393 lb k k k k + + 3 4 10   b   b c  b 1 (∆Fb 2 + ∆Fb1 ) = 1767 + 1 (393 + 0) = 1964 lb 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (393 − 0 ) = 196 lb 2 2 Fm 1964 sm = = As As F 196 sa = a = As As For C1118, as rolled su = 75 ksi Fm = Fi +

s y = 46 ksi

sn = 0.5su = 0.5(75) = 37.5 ksi = 37,500 psi For axial loading with size factor sn = (0.8)(0.85)(37,500 psi ) = 25,500 psi K f = 1.8 Table AT 12 1 sm K f sa = + sn N sy 1 1964 (1.8)(196) = + 2 46,000 As 25,500 As As = 0.1131 sq.in.

Page 24 of 42



Table AT 14, UNC 1 Use D = in. , As = 0.1419 sq.in. 2 (b) T = 0.2 DFi 1 T = 0.2 (1767 ) = 176.7 in − lb. 2

245.

A cast-iron (class 35) Diesel-engine cylinder head is held on 8 stud bolts with UNC threads. These bolts are made of AISI 3140 steel, OQT 1000 oF (Fig. AF2). Assume that the compressed material has an equivalent diameter twice the bolt size. The maximum cylinder pressure is 750 psi and the bore of the engine is 8 in. Let the initial bolt load be such that a cylinder pressure of 1500 psi brings the joint to the point of opening. For a design factor of 2, determine the bolt diameter (a) using the Soderberg equation, (b) using the Goodman equation. (c) What approximate torque will be required to induce the desired initial stress? (d) Determine the ratio of the initial stress to the yield strength. Considering the lessons of experience (i5.8), what initial stress would you recommend? Using this value, what factor of safety is computed from the Soderberg equation?

Solution:  π  (8) Fo = 1500  = 9425 lb. 4 8  kc   Fi = Fo  + k k  b c Ab Eb kb = Lb π Ab = D 2 4 Eb = 30 ×10 6 psi 2

Lb = L AE kc = c c Lc π π Ac = De2 − D 2 4 4 De = 2 D 3π 2 π π 2 Ac = (2 D ) − D 2 = D = 3 Ab 4 4 4 Page 25 of 42


SECTION 3 – SCREW FASTENINGS Ec = 14.5 × 10 6 psi , for cast-iron (class 35)

Lc = L Ab (30×10 6 ) L 3 A (14.5 × 10 6 ) kc = b L  kc   3 14.5 × 106  = (9425) Fi = Fo  6 6  30 ×10 + 3 14.5 × 10  kb + k c  kb =

(

(

)

)

  = 5578 lb 

 kb   Fe ∆Fb =   kb + k c  ∆Fb1 = 0  kb   30 × 106  Fe =  ∆Fb 2 =  6 6  30 ×10 + 3 14.5 × 10  kb + k c 

(

 π  (8)2 (750 ) = 1923 lb   8  4 

)

1 (∆Fb 2 + ∆Fb1 ) = 5578 + 1 (1923 + 0) = 6540 lb 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (1923 − 0 ) = 962 lb 2 2

Fm = Fi +

Fm 6540 = As As F 962 sa = a = As As (a) For AISI 3140 steel, OQT 1000 oF su = 153 ksi sm =

s y = 134 ksi

sn = 0.5su = 0.5(153) = 76.5 ksi = 76,500 psi For axial loading with size factor sn = (0.8)(0.85)(76,500 psi ) = 52,000 psi Table AT 12, K f = 3.3 (hardened) Soderberg Equation 1 sm K f sa = + sn N sy 1 6540 (3.3)(962) = + 2 134,000 As 52,000 As As = 0.2197 sq.in.

Page 26 of 42



Table AT 14, UNC 5 Use D = in. , As = 0.226 sq.in. 8 (b) Goodman Equation 1 sm K f sa = + sn N su 1 6540 (3.3)(962) = + 2 153,000 As 52,000 As As = 0.2076 sq.in. Table AT 14, UNC 5 Use D = in. , As = 0.226 sq.in. 8 (c) T = 0.2 DFi 5 T = 0.2 (5578) = 697 in − lb. 8 Fi 5578 = = 24,681 psi As 0.226 s 24,681 Ratio = i = = 0.184 s y 134,000

(d) si =

i5.8 si = 0.85s y = 0.85(134,000 ) = 113,900 psi

Factor of safety Fi = si As = (113,900 )(0.226 ) = 25,742 lb 1 Fm = 25,742 + (1923) = 26,704 lb 2 1 Fa = (1923) = 962 lb 2 F 26,704 sm = m = = 118,159 psi As 0.226 F 962 sa = a = = 4257 psi As 0.226 Soderberg Equation

Page 27 of 42


SECTION 3 – SCREW FASTENINGS 1 sm K f sa = + sn N sy

1 118,159 (3.3)(4257 ) = + 52,000 N 134,000 N = 0.87 246.

A 30,000-lb. body is to be mounted on a shaker (vibrator). The shaker will exert a harmonic force of F = 30,000 sin 2π t f lb. on the body where f cps is the frequency and t sec. is the time. The frequency can be varied from 5 to 10,000 cps. The harmonic force will exert a tensile load on the bolts that attach the body to the shaker when F is positive. Determine the minimum number of ½-in.-UNF bolts that must be used for N = 2 based on Soderberg line. The material of the bolts is to be AISI 8630, WQT 1100 oF; the material of the body that is to be vibrated is aluminum alloy, 2014-T6 and the joint is not to open for an external force that is 1.25 times the maximum force exerted by the shaker. It may be assumed that the equivalent diameter of the material in compression is twice the bolt diameter.

Solution: Fe min = 0 Fe max = 30,000 lb  kc   Fi = QFe  + k k  b c Q = 1.25 AE kb = b b Lb Ab =

π

D2

4 Eb = 30 ×10 6 psi

Lb = L

Page 28 of 42



kc = Ac =

Ac Ec Lc

π

De2 −

4 De = 2 D

4

D2

3π 2 D = 3 Ab 4 4 4 Ec = 10.6 × 10 6 psi , (Aluminum 2014-T6) Ac =

π

π

(2 D )2 − π

D2 =

Lc = L kb =

Ab Eb Ab (30× 106 ) = Lb L

kc =

Ac Ec 3 Ab (10.6 ×106 ) = Lc L

 kc   Fi = QFe   kb + kc    3(10.6 ×106 ) Fi = (1.25)(30,000 ) = 19,296 lb 6 6   30 ×10 + 3(10.6 × 10 )  kb   Fe ∆Fb =   kb + k c  ∆Fb1 = 0  kb    30 × 106  Fe =  (30,000) = 14,563 lb ∆Fb 2 =  6 6   30 ×10 + 3(10.6 × 10 )  kb + k c  1 (∆Fb 2 + ∆Fb1 ) = 19,296 + 1 (14,563 + 0) = 26,578 lb 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (14,563 − 0 ) = 7282 lb 2 2

Fm = Fi +

Fm nAs F sa = a nAs For ½-in.-UNF (Table AT 14) As = 0.1419 sq.in. F 26,578 187,300 sm = m = = nAs 0.1419n n sm =

Page 29 of 42



sa =

Fa 7282 51,318 = = nAs 0.1419n n

For AISI 8630, WQT 1100 oF K f = 3 .3 su = 137 ksi s y = 125 ksi sn = 0.5su = 0.5(137 ) = 68.5 ksi = 68,500 psi For axial loading with size factor sn = (0.8)(0.85)(68,500 psi ) = 46,580 psi Soderberg Equation, N = 2 1 sm K f sa = + sn N sy 1 187,300 (3.3)(51,318) = + 46,580n 2 125,000n n = 10.3 Minimum number of bolts = 10 bolts 248.

The maximum external load on the cap bolts of an automotive connecting rod end, imposed by inertia forces at top dead center, is taken to be 4000 lb.; the minimum load is zero at bottom dead center. The material is AISI 4140, OQT 1100oF (qualifying for SAE grade 5); assume that sn′ = 0.45su . The grip for through bolts is 1.5 in. For design purposed, let each bolt take half the load, 3 and use an equivalent De = 1 in. for the connected parts. The threads extend 8 a negligible amount into the grip. For the initial computation, use an opening load Fo = 1.75 Fe . Considering the manner in which the bolt is loaded, we decide that a design factor of 1.4 (Soderberg) should be quite adequate. (a) Does a 5/16-24 UNF satisfy this situation? If not, what size do you recommend? (b) Experience suggests that, in situations such as this, an initial stress of the order suggested in i5.8, Text, is good insurance against fatigue failure. Decide upon such an si and recomputed N . How does it change? Would you be concerned about the safety in this case? Consider the variation of si as a consequences of the use of torque wrench and also the stress relaxation with time (due to seating and other factors), and discuss. Compute the required tightening torque for each si .

Page 30 of 42


SECTION 3 – SCREW FASTENINGS Solution: Fo = 1.75Fe = 1.75(4000) = 7000 lb  kc   Fi = Fe   kb + k c  AE kb = b b Lb Ab =

π

D2

4 Eb = 30 ×10 6 psi

Lb = 1.5 in. AE kc = c c Lc Ac =

π 4

De2 −

π  3

π 4 2

D2

π

π

2 2 1  − D = 1.485 − D = 1.485 − Ab 4  8 4 4 6 Ec = 10.6 × 10 psi , (Aluminum 2014-T6)

Ac =

Lc = 1.5 in. As ≈ Ab kb =

Ab Eb Ab (30 × 106 ) = 1.5 Lb

Ac Ec (1.485 − As )(30 × 106 ) = 1.5 Lc 1.485 − As  Fi = 7000   = 7000 − 4714 As  1.485 

kc =

 kb   Fe ∆Fb =   kb + k c  ∆Fb1 = 0  kb  A  Fe =  s (4,000 ) = 2694 As ∆Fb 2 =   1.485   kb + k c  1 (∆Fb 2 + ∆Fb1 ) = 7000 − 4714 As + 1 (2694 As + 0) = 7000 − 3367 As 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (2694 As − 0 ) = 1347 As 2 2

Fm = Fi +

Page 31 of 42


SECTION 3 – SCREW FASTENINGS Table 5.2 su = 120 ksi s y = 88 ksi sn′ = 0.45su = 0.45(120 ) = 54 ksi = 54,000 psi 7000 F sm = m = − 3367 As As F sa = a = 1347 As K f = 3.3 (hardened, Table AT 12) Soderberg Equation, N = 1.4 1 sm K f sa = + sn N sy 1 7000 3367 (3.3)(1347 ) = − + 54,000 1.4 88,000 As 88,000 As = 0.1187 sq.in. 7 Table At14, we D = in , As = 0.1187 sq.in. 16 (a) 5/16-24 UNF will

not

satisfy the situation. Instead use

As = 0.1187 sq.in.

(b) i5.8, Text si = 0.85s y = 0.85(88) = 74.8 ksi = 74,800 psi

Fi = si As = (74,800 )(0.1187 ) = 8879 lb 1 1 Fm = Fi + (∆Fb 2 + ∆Fb1 ) = 8879 + (2694 As + 0) = 8879 + 1347 As 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (2694 As − 0 ) = 1347 As 2 2 Fm 8879 sm = = + 1347 As As F sa = a = 1347 As 1 sm K f sa = + N sy sn

Page 32 of 42

D=

7 in , 16


SECTION 3 – SCREW FASTENINGS  8879  + 1347   1  0.1187  + (3.3)(1347 ) = 54,000 N 88,000 N = 1.06 , it decreases N > 1 , therefore, safe.

Considering variation of si , si tends to exceeds the limiting stress therefore reduces the factor of safety. While stress relaxation tends to reduce the limiting stress approaching the si and causing lower design factor. (c) Fi = 7000 − 4714 As = 7000 − 4714(0.1187 ) = 6440 lb 7 T = 0.2 DFi = 0.2 (6440) = 564 in − lb  16 

at Fi = 8879 lb 7 T = 0.2 DFi = 0.2 (8879 ) = 777 in − lb  16 

CHECK PROBLEMS 249.

A 1-in. steel bolt A (normalized AISI 1137, cold-rolled threads) passes through a yellow brass tube B (B36-8, ½ hard) as shown. The tube length is 30 in., its cross-sectional area is 2 sq. in. and the UNC bolt threads extend a negligible amount below the nut. The steel washers are ¼ in. thick and are assumed not to bend (clearances are exaggerated). The nut is turned ¼ turn. (a) If an external tensile axial load, varying from 0 to 5 kips, is repeatedly applied to the bolt, what is the factor of safety of the bolt by the Soderberg criterion? (b) What is the external load on the bolt at the instant that the load on the tube becomes zero.

Problem 249, 250 Solution: For 1-in. UNC As = 0.606 sq.in. Page 33 of 42


SECTION 3 – SCREW FASTENINGS Dr = 0.8466 in. Th in. = 8 1 1 = in. Th in 8 δ 1 = p 4 1 1 1 δ =   = in. 4  8  32 1 Lb = 30 + 2  = 30.5 in. 4 Eb = 30 ×10 6 psi 1 Fi (30.5) δ= = 32 π (1)2 30 × 106 4 Fi = 24,141 lb p=

(

)

 kb   ∆Fb = Fe  + k k  b c

π AE kb = b b = 4 Lb AE kc = c c Lc Ac = 2 sq.in. Lc = 30 in.

(1)2 (30 ×106 ) 30.5

= 772,525

Ec = 15× 10 6 psi (Yellow Brass)

kc =

2(15 × 106 ) = 1,000,000 30

(a) ∆Fb1 = 0 772,525   ∆Fb 2 = (5000 )  = 2179 lb  772,525 + 1,000,000  1 1 Fm = Fi + (∆Fb 2 + ∆Fb1 ) = 24,141 + (2179 + 0 ) = 25,230 lb 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (2179 − 0 ) = 1090 lb 2 2 Fm 25,230 sm = = = 38,227 psi As 0.606

Page 34 of 42


SECTION 3 – SCREW FASTENINGS Fa 1090 = = 1799 psi As 0.606 For normalized AISI 1137, cold-rolled thread su = 98 ksi sa =

s y = 58 ksi

sn = 0.5su = 0.5(98) = 49 ksi = 49,000 psi K f = 1.4 (Table AT 12) For axial loading, sn = 0.8(49,000 ) = 39,200 psi 1 sm K f sa = + sn N sy 1 38,227 (1.4 )(1799 ) = + 39,200 N 58,000 N = 1.38  kc   (b) Fi = Fo   kb + k c  1,000,000   24,141 = Fo    772,525 + 1,000,000  Fo = 42,790 lb

250.

A ¾-in. fine-thread bolt, made of AISI 1117, cold drawn, with rolled threads, passes through a yellow brass tube and two steel washers, as shown. The tube is 4 in. long, 7/8 in. internal diameter, 1 ¼-in. external diameter. The washers are each ¼-in. thick. The unthreaded part of the bolt is 3 in. long. Assume that there is no stretching of the bolt inside the nut in finding its k . The unlubricated bolt is tightened by a torque of 1800 in-lb. The external load, varying from 0 to 4 kips, is axially applied to the washers an indefinite number of times. (a) Compute the factor of safety of the bolt by the Soderberg criterion. Is there any danger of failure of the bolt? (b) What pull must be exerted by the washers to remove all load from the brass tube?

Solution: T = 0.2 DFi 3 1800 = 0.2  Fi 4 Fi = 12,000 lb

Page 35 of 42


SECTION 3 – SCREW FASTENINGS  kb   ∆Fb = Fe   kb + kc  1 1 1 = + kb kb1 kb 2 AE kb1 = b b Lb1 Lb1 = 3 in.

π 3

2

  = 0.4418 sq.in. 4 4 Eb = 30 ×10 6 psi

Ab =

kb1 =

(0.4418)(30 ×106 ) = 4,418,000

3 As Eb kb 2 = Lb 2 For ¾-in. UNF (Table AT 14) As = 0.373 sq.in.

1 Lb 2 = 4 + 2  − 3 = 1.5 in. 2 (0.373) 30 ×106 = 7,460,000 kb 2 = 1.5 1 1 1 = + kb kb1 kb 2 1 1 1 = + kb 4,418,000 7,460,000 kb = 2,774,733 AE kc = c c Lc

(

)

2 2 π  1   7  

1  −    = 0.6259 sq.in. 4  4   8   Ec = 15× 10 6 psi Lc = 4 in. Ac =

kc =

(0.6259)(15 ×106 ) = 2,347,125 4

∆Fb1 = 0 2,774,733   ∆Fb 2 = (4000 )  = 2167 lb  2,774,733 + 2,347,125  Page 36 of 42


SECTION 3 – SCREW FASTENINGS 1 (∆Fb 2 + ∆Fb1 ) = 12,000 + 1 (2167 + 0) = 13,084 lb 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (2167 − 0 ) = 1084 lb 2 2 F 13,084 sm = m = = 35,078 psi As 0.373 F 1084 sa = a = = 2906 psi As 0.373 For AISI 111, cold drawn, rolled threads sn = 40 ksi Fm = Fi +

s y = 68 ksi K f = 1.4

sn = 0.8(40 ) = 32 ksi = 32,000 psi , axial loading (a)

1 sm K f sa = + sn N sy

1 35,078 (1.4 )(2906 ) = + 32,000 N 68,000 N = 1.56  kc   (b) Fi = Fo   kb + k c  2,347,125   12,000 = Fo    2,774,733 + 2,347,125  Fo = 26,186 lb

251.

A coupling bolt (i5.13, Text) is used to connect two parts made of cast-iron, class 35. The diameter of the coarse-thread bolt is ½-in.; its grip is 2 in., which is also nearly the unthreaded length. The bolt tightened to have an initial tension of 4000 lb. The parts support an external load Fe that tends to separate them and it varies from zero to 5000 lb. What is the factor of safety, (Soderberg)?

Solution: Fi = 4000 lb  kb   ∆Fb = Fe   kb + kc 

Page 37 of 42



kb =

Ab Eb Lb

π 1

2

  = 0.19635 sq.in. (unthreaded length) 4 2 Eb = 30 ×10 6 psi

Ab =

Lb = 2 in. kb =

(0.19635)(30 ×106 ) = 2,945,250

2 Table AT 14, UNC 1 D = in. 2 As = 0.1419 sq.in. 3 A = in. 4 h De = A + 2 h = 2 in. 3 2 3 De = + = 1 in. 4 2 4 AE kc = c c Lc

π

De2 −

π

D2 =

2 2 π  3   1  

1  −    = 2.209 sq.in. 4  4   2   Ec = 14.5 × 10 6 psi , (Cast iron, class 35) Lc = 2 in. Ac =

kc =

4

4

(2.209)(14.5 ×106 ) = 16,015,250 2

∆Fb1 = 0 2,945,250   ∆Fb 2 = (5000 )  = 777 lb  2,945,250 + 16,015,250  1 1 Fm = Fi + (∆Fb 2 + ∆Fb1 ) = 4000 + (777 + 0 ) = 4389 lb 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (777 − 0 ) = 389 lb 2 2 Fm 4389 sm = = = 30,930 psi As 0.1419 F 389 sa = a = = 2741 psi As 0.1419

Page 38 of 42



For ASTM 354 BC (Table 5.2), D =

1 in. 2

su = 125 ksi s y = 109 ksi sn = 0.5su For axial loading sn = (0.8)(0.5)(125) = 50 ksi = 50,000 psi K f = 1.8 Soderberg Line 1 sm K f sa = + sn N sy 1 30,930 (1.8)(2741) = + 50,000 N 109,000 N = 2.6 252.

The cap on the end of a connecting rod (automotive engine) is held on by two 5/16-in. bolts that are forged integrally with the main connecting rod. These bolts have UNF threads with a 5/8-in. on an unthreaded length of virtually 5/8 in. The nuts are to be tightened with a torque of 20 ft-lb. and the maximum external load on one bolt is expected to be 2330 lb. Let the equivalent diameter of the connected parts be ¾ in. (a) Estimate the maximum force on the bolt. (b) Compute the opening load. Is this satisfactory? (c) If the bolt material is AISI 4140, OQT 1000 oF, what is the factor of safety based on the Soderberg criterion?

Solution: T = 20 ft − lb = 240 in − lb T = 0.2 DFi 5 240 = 0.2  Fi  16  Fi = 3840 lb

 kb   ∆Fb = Fe  + k k  b c Ab Eb kb = Lb

π5

2

  = 0.0767 sq.in. (unthreaded length) 4  16  Eb = 30 ×10 6 psi 5 Lb = in. 8

Ab =

Page 39 of 42



kb =

kc =

Ac =

(0.0767 )(30 ×106 ) = 3,681,600 5   8

Ac Ec Lc

π 4

2 e

D −

π 4

2

D =

2 2 π  3   5  

  −    = 0.3651 sq.in. 4  4   16  

Ec = 30 ×10 6 psi , (Cast iron, class 35) 5 Lc = in. 8 (0.3651) 30 ×106 = 17,524,800 kc = 5   8 3,681,600   ∆Fb = (2330 )  = 405 lb  3,681,600 + 17,524,800 

(

)

(a) Fmax = Fi + ∆Fb = 3840 + 405 = 4245 lb  kc   (b) Fi = Fo   kb + k c  17,524,800   3840 = Fo    3,681,600 + 17,524,80  Fo = 4647 lb < Fmax 405 ∆Fb = 3840 + = 4042 lb 2 2 405 ∆F Fa = b = = 202 lb 2 2 For AISI 4140, OQT 1000 oF su = 170 ksi

(c) Fm = Fi +

s y = 155 ksi

Table AT 12, K f = 2.6 sn = 0.5su For axial loading sn = (0.8)(0.5)(170 ) = 68 ksi = 68,000 psi Soderberg Line 1 sm K f sa = + N sy sn Page 40 of 42


SECTION 3 – SCREW FASTENINGS For 5/16-in.-UNF, Table AT 14, As = 0.0580 sq.in. F 4042 sm = m = = 69,690 psi As 0.0580 F 202 sa = a = = 3843 psi As 0.0580 1 sm K f sa = + sn N sy 1 69,690 (2.6 )(3483) = + 68,000 N 155,000 N = 1.72 SET SCREWS 254.

A 6-in. pulley is fastened to a 1 ¼ in. shaft by a set screw. If a net tangential force of 75 lb, is applied to the surface of the pulley, what size screw should be used when the load is steady?

Solution:  6  Tangential force = (75 lb )  = 365 lb  1.25  Assume tangential force = holding force Table 5.3, use Screw size 8, Holding force = 385 lb.

255.

An eccentric is to be connected to a 3-in. shaft by a setscrew. The center of the eccentric is 1 ¼ in. from the center of the shaft when a tensile force of 1000 lb. is applied to the eccentric rod perpendicular to the line of centers. What size set screw should be used for a deign factor of 6?

Solution:  1.25   = 833 lb Tangential force = (1000 lb ) 32 Holding force = (6 )(833) = 5000 lb Table 5.3, use Screw size ¾ in. 256.

A lever 16 in. long is to be fastened to a 2-in. shaft. A load of 40 lb. is to be applied normal to the lever at its end. What size of set screw should be used for a design factor of 5?

Solution: Torque = (16 )(40 ) = 640 in − lb 2(640 ) Tangential force = = 640 lb 2

Page 41 of 42


SECTION 3 – SCREW FASTENINGS Holding force = (5)(640 ) = 3200 lb Table 5.3, use Screw size 9/16 in. 257.

A 12-in. gear is mounted on a 2-in. shaft and is held in place by a 7/16 in. setscrew. For a design factor of 3, what would be the tangential load that could be applied to the teeth and what horsepower could be transmitted by the screw.

Solution: Table 5.3, 7/16 in. Holding force = 2500 lb 2500 = 833 lb Tangential force = 3 2 Tangential load on gear = 833  = 139 lb  12  Assume vm = 4500 fpm (139)(4500) = 19 hp Hp transmitted = 33,000 - end -

Page 42 of 42


SECTION 4 - SPRINGS HELICAL COMPRESSION SPRINGS DESIGN – LIGHT, MEDIUM SERVICE 271.

A solenoid brake (Fig. 18.2, Text) is to be actuated by a helical compression spring. The spring should have a free length of approximately 18 in. and is to exert a maximum force of 2850 lb. when compressed to a length of 15 in. The outside diameter must not exceed 7 in. Using oil-tempered wire, design a spring for this brake, (wire diameter, coil diameter, number of active coils, pitch, pitch angle, “solid stress”). General Electric used a spring made of 1 in. wire, with an outside diameter of 6 in., and 11 ½ free coils for a similar application.

Solution: For oil tempered wire, Table AT 17 146 su = 0.19 ksi , [0.032 < Dw < 0.5] Dw “solid stress” = 0.6 s u design stress, (average service) ssd = 0.324su 0.324(146) 47.304 ssd = ksi = Dw0.19 Dw0.19 Dw + Dm ≤ 7 F = 2850 lb = 2.85 kips

 8FDm  47.304 = ss = K  3  Dw0.19  πDw  say K = 1.3  8(2.85)(7 − Dw )  47.304 ss = 1.3 = πDw3 Dw0.19   Dw = 1.062 in > 0.5 in 47.304 ksi = 54 ksi use ssd = (0.5)0.19  8(2.85)(7 − Dw )  ss = 1.3  = 54 πDw3   Dw = 1.015 in say Dw = 1.0 in  8(2.85)Dm  ss = 1.3  = 54 3  π (1)  Dm = 5.72 in Page 1 of 70


SECTION 4 - SPRINGS say Dm = 5.0 in OD = Dm + Dw = 5.0 + 1.0 = 6 in < 7 in. D 5 .0 C= m = =5 Dw 1.0 δ = Free length – Compressed length = 18 in – 15 in = 3 in.

δ=

8FC 3 N c GDw

G = 10,500 ksi , Dw >

3 in 8

8(2.85)(5) N c (10,500)(1) N c = 11.05 say N c = 11.5 3

δ =3=

8(2.85)(5) (11.5) = 3.12 in (10,500)(1) Free length = 15 + 3.12 = 18.12 in 3

δ=

At 2.85 kips  8FDm   ss = K  3   πDw  C =5 4C − 1 0.615 4(5) − 1 0.615 K= + = + = 1.3105 C 5 4C − 4 4(5) − 4  8(2.85)(5)  ss = 1.3105  = 47.55 ksi 3  π (1)  Permissible solid stress 0.6(146 ) ksi = 99.93 ksi sso = 0.6 su = (0.5)0.19 F k= δ using or let δ T = Free length – Solid height 47.55 99.93 = 3.12 δT δ T = 6.56 in δ T = Free length – Solid height = (P − Dw )N c 6.56 = (P − 1)(11.5) P = 1.570 in

Page 2 of 70


SECTION 4 - SPRINGS 1 use P = 1 in 2 Pitch angle,  1.5  P o o = tan −1  λ = tan −1  = 5.5 < 12 , o.k. πD π ( 5 )  

For actual solid stress δ T = (1.5 − 1)(11.5) = 5.75 in. 47.55 sso = 3.12 5.75 sso = 87.63 ksi < 99.93 ksi , ok Summary of answer: Dw = wire diameter = 1 in. Dm = coil diameter = 5 in. N c = no. of active coils = 11 1/2 P = pitch = 1 ½ in. γ = pitch angle = 5.5o sso = solid stress = 87.63 ksi 272.

A coil spring is to be used for the front spring of a automobile. The spring is to have a rate of 400 lb./in., an inside diameter of 4 3/64 in., and a free length of 14 1/8 in., with squared-and-ground ends. The material is to be oiltempered chrome vanadium steel. Decide upon the diameter of the wire and the number of free coils for a design load of F = 1500 lb . Be sure “solid stress” is all right. How much is the pitch angle?

Solution: Table AT 17 Cr-V steel 168 su = 0.166 ksi , [0.032 < Dw 0.437] Dw average service ssd = 0.324su 0.324(168) 54.432 ssd = = 0.166 ksi Dw0.166 Dw Max “solid stress” = 0.6su

Page 3 of 70


SECTION 4 - SPRINGS 3 in = 4.046875 in 64 Dm = Dw + 4.046875 in

ID = Dm − Dw = 4

 8FDm   = ssd ss = K  3   πDw  Assume K = 1.3 F = 1500 lb = 1.5 kips

ssd =

 8(1.5)(Dw + 4.046875)  54.432 = 1.3  0.166 Dw πDw3  

Dw = 0.747 in > 0.437 in use 54.432 ssd = ksi = 62.45 ksi (0.437 )0.19  8(1.5)(Dw + 4.046875)  ssd = 1.3  = 62.45 πDw3   Dw = 0.724 in 3 use Dw = in 4 3 3 51 Dm = + 4 = 4 in 4 64 64  8FDm   ss = K  3   πDw  4C − 1 0.615 K= + 4C − 4 C  51  4  Dm  64  C= = ≈ 6.4 Dw 3   4 4(6.4) − 1 0.615 K= + = 1.235 4(6.4) − 4 6.4

  51    8(1.5) 4    64   = 53.64 ksi < 62.45 ksi , (o.k.) ss = 1.235 3   3 π     4  

δ=

8FC 3 N c GDw

Page 4 of 70


SECTION 4 - SPRINGS 3 G = 10,500 ksi , D w > in 8 F 1500 = 3.75 in δ= = k 400 3 8(1.5)(6.4 ) N c δ = 3.75 = (10,500) 3  4 N c = 9.4

Table AT 16, Total coils = N c + 2 = 9.4 + 2 = 11.4 for square and grounded end. Summary of answer: Dw = wire diameter = ¾ in. No. of free coils = 11.4 To check for solid stress. Permissible solid stress =

0.6(168) = 115.65 ksi (0.437 )0.166

Free length = PN c + 2 Dw 3 Solid height = Dw ( N c + 2 ) =  (11.4 ) = 8.55 in 4  1   14 − 8.55   = 78.74 ksi < 115.65 ksi (safe) Solid stress = (53.64 ) 8  3.75     

Pitch: 1 PN c + 2 Dw = 14 in 8 1 3 P(9.4) + 2  = 14 8 4 11 P = 1.343 in = 1 in 32 Pitch angle,  11   1  −1 P −1 = tan  32  = 5.1o < 12o , o.k. λ = tan πD  π  4 51     64  

Page 5 of 70


SECTION 4 - SPRINGS 273.

A coiled compression spring is to fit inside a cylinder 5/8 in. in diameter. For one position of the piston, the spring is to exert a pressure on the piston equivalent to 5 psi of piston area, and in this position, the overall length of the spring must not exceed (but may be less than) 2 in. A pressure of 46 psi on the piston is to compress the spring ¾ in. from the position described above. Design a spring for medium service. Specify the cheapest suitable material, number of total and active coils for square-and-ground ends, and investigate the pitch angle, and “solid stress”.

Solution:  8FDm   ss = K  3  D π w   OD = Dm + Dw = Dm + 1.5Dw =

5 D in − w 8 2

5 in 8 2

 π  5  F1 = (5)   = 1.534 lb  4  8  2

 π  5  F2 = (46 + 5)   = 15.647 lb  4  8 

Using hard-drawn spring wire, Cost Index = 1 ssd = 0.324su (0.85) 140 su = 0.19 ksi Dw , [0.028 < Dw < 0.625] 70 ksi Dw0.19 140 38.556 ssd = 0.324(0.85) 0.19 = Dw Dw0.19

Max “solid stress” =

 8FC  38.556 38,556 = ksi = 0.19 psi ss = K  2  0.19 Dw Dw  πDw   8(15.647 )C  K = 38,556 Dw1.81  π   K (39.845)C = 38,556 Dw1.81 Dm + 1.5Dw = 0.625 CDw + 1.5Dw = 0.625 0.625 Dw = C + 1 .5

Page 6 of 70


SECTION 4 - SPRINGS

K=

4C − 1 0.615 + 4C − 4 C 1.81

 4C − 1 0.615   0.625  +  (39.845)C = 38,556  C   4C − 4  C + 1.5   4C − 1 0.615  1.81 +  (C + 1.5) C = 413.3 C   4C − 4 C = 7.035 0.625 0.625 = = 0.0732 in Dw = C + 1.5 7.035 + 1.5 Table AT 15, Dw = 0.0720 in , W & M 15

Dm = 7.035(0.0720) = 0.5065 in For N c

δ 2 − δ1 =

8(F2 − F1 )C 3 N c GDw

G = 11.5 × 106 psi 3 8(15.647 − 1.534)(7.035) N c δ 2 − δ1 = = 11.5 × 106 (0.0720) 4 N c = 15.8 Table AT 16, Total coils = N c + 2 = 15.8 + 2 = 17.8 Solid height = ( N c + 2)Dw = (15.8 + 2)(0.0720) = 1.28 in Free length = PN c + 2 Dw Free length = 2 + δ1 3

(

δ1 =

)

8(F1 )C 3 N c GDw

8(1.534 )(7.035) (15.8) = 0.082 in. 11.5 × 106 (0.0720) Free length = 2 + 0.082 = 2.082 in 3

δ1 =

δ2 =

(

)

8(F2 )C 3 N c GDw

8(15.647 )(7.035) (15.8) = 0.832 in. 11.5 × 106 (0.0720) Solid Height ≤ Free Length - δ 2 Solid Height ≤ 2.082 − 0.832 in Solid Height ≤ 1.25 in But Solid Height > 1.25 in. 3

δ2 =

(

)

Therefore change material to Oil-tempered spring wire, Cost Index = 1.5 Page 7 of 70


SECTION 4 - SPRINGS

Table AT 17 146 su = 0.19 ksi , 0.028 < Dw < 0.5 Dw Max “solid stress” = ssd = 0.324

87.5 ksi D w0.19

146 47.304 = Dw0.19 Dw0.19

 8FC  47.304 47,304 = ksi = psi ss = K  2  0.19 0.19 D D π D w w  w ( )  8 15.647 C  K = 47,304 Dw1.81  π   K (39.845)C = 47,304 Dw1.81 0.625 Dw = C + 1 .5 4C − 1 0.615 K= + 4C − 4 C 1.81

 4C − 1 0.615   0.625  +  (39.845)C = 47,304  C   4C − 4  C + 1.5   4C − 1 0.615  1.81 +  (C + 1.5) C = 507.1 C   4C − 4 C = 7.684 0.625 0.625 Dw = = = 0.0680 in C + 1.5 7.684 + 1.5 Table AT 15, Dw = 0.0625 in , W & M 16

Dm = 7.684(0.0625) = 0.48025 in 15 say Dm = = 0.46875 in 32 D 0.46875 C= m = = 7.5 Dw 0.0625  8FC   ss = K  2   πDw  4C − 1 0.615 4(7.5) − 1 0.615 K= + = + = 1.1974 C 4C − 4 4(7.5) − 4 7.5  8(15.647 )(7.5)  ss = 1.1974 = 91,600 psi = 91.6 ksi 2   π (0.0625)  For N c

Page 8 of 70


SECTION 4 - SPRINGS

δ 2 − δ1 =

8(F2 − F1 )C 3 N c GDw

G = 11.5 × 106 psi 3 8(15.647 − 1.534)(7.5) N c = 11.5 × 106 (0.0625) 4 N c = 11.32 Table AT 16, squared and ground ends Total coils = N c + 2 = 11.32 + 2 = 13.32 Solid height = ( N c + 2 )Dw = (11.32 + 2)(0.0625) = 0.8325 in Free length = PN c + 2 Dw Free length = 2 + δ1 3

δ 2 − δ1 =

δ1 =

(

)

8(F1 )C 3 N c GDw

8(1.534 )(7.5) (11.32) = 0.082 in. 11.5 × 106 (0.0625) Free length = 2 + 0.082 = 2.082 in = P(11.32) + 2(0.0625) 3

δ1 =

(

)

P = 0.1729 in ≈

11 in 64

Pitch angle,  0.1729  P o o = tan −1  λ = tan −1  = 6.7 < 12 , o.k. πD π ( ) 0 . 46875   Solid stress  2 − 0.8325  sso = (91.6 )  = 142.6 ksi  0.75  87.5 Permissible solid stress = = 148.8 ksi > 137.5 ksi , safe. (0.0625)0.19 Summary of answer: Suitable material = Oil-Tempered Spring Wire Total Coils = 13.32 Active Coils, N c = 11.32 274.

A helical spring is to fit about a 11/16-in. rod with a free length of 2 ¾ in. or less. A maximum load of 8 lb. is to produce a deflection of 1 ¾ in. The spring is expected to be compressed less than 5000 times during its life, but is subjected to relatively high temperatures and corrosive atmosphere. Select a material and determine the necessary wire size, mean coil diameter, and number of coils. Meet all conditions advised by Text.

Page 9 of 70


SECTION 4 - SPRINGS Solution: For 5000 cycles < 104 cycles, use light service Use stainless steel, type 302 (Cr-Ni), ASTM A313 – for relative high temperature and corrosive atmosphere, Table AT 17. ssd = 0.32 su (i) 170 ksi , [0.01 < Dw < 0.13] Dw0.14 97 su = 0.41 ksi , [0.13 < Dw < 0.375] Dw Maximum “solid” so = 0.47 su su =

 8FDm   ss = K  3  D π w   F = 8 lb 11 D Dm − Dw = + w 16 2 Dm − 1.5Dw = 0.6875 in CDw − 1.5 Dw = 0.6875 0.6875 Dw = C − 1 .5 4C − 1 0.615 K= + 4C − 4 C 170 assume su = 0.14 ksi Dw 0.32(170 ) 54.4 54,400 psi ssd = = 0.14 ksi = 0.14 Dw Dw0.14 Dw  4C − 1 0.615   8(8)C  + = 54,400 D1w.86    C  π   4C − 4 1.86

 4C − 1 0.615  64C   0.6875  +    = 54,400  C  π   4C − 4  C − 1.5   4C − 1 0.615  1.86 +  (C − 1.5) C = 1330 C   4C − 4 C = 12.919 0.6875 Dw = = 0.0602 in 12.919 − 1.5 Use Table AT 15, Dw = 0.0625 in , 16 W & M

Dm = 12.919(0.0602) = 0.8074 in 25 say Dm = in = 0.78125 in 32

Page 10 of 70


SECTION 4 - SPRINGS 11 16 0.78125 − 0.0625 > 0.6875 0.71875 > 0.6875 Dm − Dw >

0.71875 − 0.6875 = 0.03125 = C=

Dw , o.k. 2

Dm 0.71875 = = 12.5 Dw 0.0625

[0.0625 < 0.13], therefore,

su =

170 ksi is o.k. Dw0.14

 8FC   ss = K  2  D π  w 4(12.5) − 1 0.615 K= + = 1.1144 4(12.5) − 4 12.5  8(8)(12.5)  ss = 1.1144 = 72,648 psi 2  π (0.0625)  8(F )C 3 N c δ= GDw G = 10.6 × 106 psi 3 8(8)(12.5) N c 4 10.6 × 106 (0.0625) N c = 9.3 To check for solid stress and pitch Minimum solid height = Dw N c = (0.0625)(9.3) = 0.58125 in 3

δ =1 =

(

)

(72,648) 2 3 − 0.58125 

 4  = 90,000 psi = 90 ksi 3 1 4 (0.47 )(170) = 117.8 ksi > 90 ksi , o.k. Permissible solid stress = (0.0625)0.14 Free length = PN c , minimum 3 P (9.3) = 2 4 P = 0.2957 in Pitch angle,  0.2957  P = tan −1  = 7.5o < 12o , o.k. λ = tan −1  πD  π (0.71825)  Summary of answer Material, Stainless Steel, Cr-Ni. ASTM A313

Solid stress =

Page 11 of 70


SECTION 4 - SPRINGS Dw = 0.0625 in , 16 W & M 25 Dm = in 32 N c = 9.3 275.

In order to isolate vibrations, helical compression springs are used to support a machine. The static load on each spring is 3500 lb., under which the deflection should be about 0.5 in. The solid deflection should be about 1 in. and the outside coil diameter should not exceed 6 in. Recommend a spring for this application; include scale, wire size, static stress, material, number of coils, solid stress, and pitch of coils.

Solution: Use Music wire (The best material) Table AT 17 190 su = 0.154 ksi , [0.004 < Dw < 0.192] Dw Maximum “solid” sso = 0.5su Light service, ssd = 0.405su 0.405(190 ) 76.95 76,950 ssd = = 0.154 ksi = 0.154 psi 0.154 Dw Dw Dw  8FC   ss = K  2   πDw  F = 3500 lb OD = Dm + Dw = 6 in (C + 1)Dw = 6 6 Dw = C +1    76,950  4C − 1 0.615  8(3500 )C  = ss =  +  C    6    6  0.154  4C − 4 π   C + 1    C + 1     4C − 1 0.615  1.846 + = 235.9   C (C + 1) C   4C − 4 C = 5.635 6 Dw = = 0.9043 in > 0.192 in 5.635 + 1 76.950 use ss = = 99,216 psi (0.192)0.154

[

Page 12 of 70

]


SECTION 4 - SPRINGS     4C − 1 0.615  8(3500 )C   = 99,216 ss =  +  C   6    4C − 4 π   C + 1    4C − 1 0.615  2 +  C (C + 1) = 400.8 C   4C − 4 C = 6.205 6 = 0.8328 in Dw = 6.205 + 1 13 Say Dw = in = 0.8125 in 16 Dm = (6.205)(0.8125) = 5.042 in

Say Dm = 5 in D 5 C= m = = 6.154 Dw 0.8125 4(6.154) − 1 0.615 K= + = 1.2455 4(6.154) − 4 6.154  8FC   ss = K  2   πDw   8(3500)(6.154)  ss = 1.2455  = 103,481 psi > 99,216 psi , not o.k. 2  π (0.8125)  Use Dm = 4.5 in D 4.5 C= m = = 5.5385 Dw 0.8125 4(5.5385) − 1 0.615 K= + = 1.2763 45.5385 − 4 5.5385  8(3500)(5.5385)  ss = 1.2763  = 95,435 psi > 99,216 psi , o.k. 2  π (0.8125)  To check for solid stress (0.5)(190) = 122.488 ksi = 122,488 psi Permissible solid stress = (0.192 )0.154  1  Solid stress =  (95,435) = 190,870 psi > 122,488 psi , not ok  0.5  Use  0.5  ssd = 122,488  = 61,244 psi  1  .

Page 13 of 70


SECTION 4 - SPRINGS     4C − 1 0.615  8(3500 )C   = 61,244 ss =  +  C   6    4C − 4 π   C + 1    4C − 1 0.615  2 +  C (C + 1) = 247.4 C   4C − 4 C = 5.1 6 = 0.9836 in Dw = 5 .1 + 1 Say Dw = 1.0 in

Dm = (5.1)(1.0 ) = 5.1 in Say Dm = 5 in D 5 C = m = =5 Dw 1  8FC   ss = K  2   πDw  4(5) − 1 0.615 K= + = 1.3105 5 4(5) − 4  8(3500)(5)  ss = 1.3105 = 58,400 psi > 61,244 psi o.k. 2  ( ) π 1 . 0   Use Dw = 1.0 in , Dm = 5 in  1  Solid stress =  (58,400 ) = 116,800 psi < 122,488 psi , o.k.  0.5  8(F )C 3 N c δ= GDw (Table AT 17) G = 12× 106 psi

8(3500 )(5) N c δ = 0.5 = 12 ×10 6 (1.0) N c = 1.7143 say N c = 1.75 Free length – Solid length = Solid Deflection PN c − Dw N c = 1 in P(1.75) − (1)(1.75) = 1 9 P = 1.5714 in ≈ 1 in 16 Pitch angle, 3

(

Page 14 of 70

)


SECTION 4 - SPRINGS  9  1  P = tan −1  16  = 5.68o < 12o , o.k. λ = tan −1 πD  π (5)    Summary of answer. F 3500 = 7000 lb in Scale, k = = δ 0 .5 Wire size, Dw = 1.0 in Material = Music Wire Solid sress = 116,800 psi 9 Pitch of stress = P = 1 in 16 CHECK PROBLEMS – LIGHT, MEDIUM SERVICE 276.

The front spring of an automobile has a total of 9 ½ coils, 7 3/8 active coils (square-and-ground ends), an inside diameter of 4 3/64 in., and a free length of 14 ¼ in. It is made of SAE 9255 steel wire, OQT 1000oF, with a diameter of 43/64 in. Compute (a) the rate (scale) of the spring; (b) the “solid stress” and compare with a permissible value (is a stop needed to prevent solid compression?). (c) Can 95 % of the solid stress be repeated 105 times without danger of failure? Would you advise shot peening of the spring?

Solution: 8FC 3 N c GDw 43 3 Dw = in > in 64 8 6 G = 10.5 × 10 psi D C= m Dw Dm − Dw = ID 43 3 Dm − = 4 in 64 64 23 Dm = 4 in 32 23 4 C = 32 = 7.0233 43 64

(a) δ =

Page 15 of 70


SECTION 4 - SPRINGS

Nc = 7

3 8

43  (10.5 ×10 ) 64  6

GD k = rate = = 3 w = δ 8C N c F

  = 345 lb in 3 3  8(7.0233)  7   8

(b) “Solid Stress”  43  1  Solid height = N c = (Dw )(Total Coils ) =   9  = 6.3828 in  64  2  Solid deflection = Free length – Solid height = 14 ¼ - 6.3828 = 7.8672 in.

Solid Force = Fso = 7.8672(345) = 2714 lb  8F C  Solid Stress = K  so2   πDw  4C − 1 0.615 K= + 4C − 4 C 4(7.0233) − 1 0.615 K= + = 1.212 4(7.0233) − 4 7.0233     8(2714)(7.0233)   ss = 1.212 = 130,322 psi 2    43  π     64    Permissible value, ss = s ys = 0.6 sy , [Dw > 0.5 in] SAE 9255, OQT 1000 oF s y = 160 ksi su = 180 ksi , s ys = 0.6(160 ) = 96 ksi = 96,000 psi < 130,322 psi Therefore a stop is needed to prevent solid compression.

(c) ssd = 0.324su (105 cycles) ssd = 0.324(180) = 58.32 ksi 0.95sso = 0.95(130,322) = 123,800 psi = 123.8 ksi > 58.32 ksi There is a danger of failure, shot peening is advisable s ys = 1.25(96,000 ) = 120,000 psi ≈ 0.95sso 277.

An oil-tempered steel helical compression spring has a wire size of No. 3 W & M, a spring index of 4.13, 30 active coils, a pitch of 0.317 in., ground-andsquared ends; medium service. (a) What maximum load is permitted if the recommended stress is not exceeded (static approach)? Compute (b) the

Page 16 of 70


SECTION 4 - SPRINGS corresponding deflection, (c) “solid stress,”. (d) pitch angle, (e) scale, (f) the energy absorbed by the spring from a deflection of 0.25 in. to that of the working load. (g) Is there any danger of this spring buckling? (h) What maximum load could be used if the spring were shot peened? Solution: Table AT 17, oil-tempered 146 su = 0.19 ksi , [0.032 < Dw < 0.5] Dw 87.5 Maximum “solid” sso = 0.19 ksi Dw ssd = 0.324su (medium service) Table AT 15, No. 3 W & M Dw = 0.2437 in C = 4.13 Dm = CDw = 4.13(0.2437 ) = 1.0 in  8F C  (a) ss = K  s 2   πDw  4C − 1 0.615 K= + 4C − 4 C 4(4.13) − 1 0.615 K= + = 1.3885 4(4.13) − 4 4.13 0.324(146 ) ss = ssd = = 61.858 ksi = 61,858 psi (0.2437 )0.19  8(F )(4.13)  ss = 61,858 = 1.3885 2  π (0.2437 )  F = 252 lb 8FC 3 N c δ= GDw G = 11.5 × 106 psi N c = 30 8(252)(4.13) (30) = 1.52 in 11.5 × 106 (0.2437 ) 3

δ=

(

)

(c) For solid stress . Square-and-ground end) Free length = PN c + 2 Dw = (0.317 )(30) + 2(0.2437 ) = 9.9974 in

Page 17 of 70


SECTION 4 - SPRINGS Solid height = Dw N c + 2 Dw = (30 + 2)(0.2437 ) = 7.7984 in Solid deflection = 9.9974 – 7.7984 = 2.199 in.  2.199  Solid stress = (61,858)  = 89,491 psi  1.52  Maximum “solid” sso =

87.5 87.5 ksi = 114.4 ksi > 89.491 ksi , o.k. safe ksi = 0.19 Dw (0.2437 )0.19

 0.317  P = tan −1  = 5.76o < 12o , o.k.  πD  π (1)  F 252 = 166 lb in (e) scale = k = = δ 1.52 1 (f) U s = k (δ 22 − δ12 ) 2 k = 166 lb in δ1 = 0.25 in δ 2 = 1.52 in 1 2 2 U s = (166 ) (1.52 ) − (0.25) = 186.6 in − lb 2 (d) λ = tan −1

[

]

(g) i 6.18 Free length = 9.9974 in Mean Diameter = Dm = 1.0 in Free length 9.9974 = = 9.9974 > 4 Mean Diameter 1.0 There is a danger for spring buckling (h) Shot peened, Table AT 17 ssd = (61,858)(1.25) = 77,322 psi  8(F )(4.13)  ss = 77,322 = 1.3885 2  π (0.2437 )  F = 314 lb 280.

It is desired to isolate a furnace, weighing 47,300 lb., from the surroundings by mounting it on helical springs. Under the weight, the springs should deflect approximately 1 in., and at least 2 in. before becoming solid. It has been decided to use springs having a wire diameter of 1 in., an outside diameter of 5 3/8 in., 4.3 free coils. Determine (a) the number of springs to be used, (b) the stress caused by the weight, (c) the “solid stress”. (d) What steel should be used?

Solution:

Page 18 of 70


SECTION 4 - SPRINGS Dw = 1 in 3 Dm + Dw = 5 in 8 3 Dm = 4 in 8 3 4 D C = m = 8 = 4.375 1 Dw

8FC 3 N c (a) δ = GDw Assume N c = 4.3 G = 10.5 × 106 psi , Dw >

3 in 8

8F (4.375) (4.3) 10.5 × 106 (1) F = 3645 lb W 47,300 No. of springs = = = 13 F 3645 3

δ = 1.0 =

(

)

W 47,300 = = 3638 lb 13 13  8F C  ss = K  s 2   πDw 

(b) F =

4C − 1 0.615 + 4C − 4 C 4(4.375) − 1 0.615 K= + = 1.3628 4(4.375) − 4 4.375 K=

 8(3638)(4.375)  ss = 1.3628  = 55,235 psi π (1.0)2   2 (c) “Solid Stress” = ss = 55,235  = 110,470 psi 1 (d) s ys ≈ 110,470 psi

s ys

110,470 = 184,117 psi = 184.117 ksi 0.6 0.6 From Table AT 7, Use AISI 8760, OQT 800 oF, s y = 200 ksi sy =

=

Page 19 of 70


SECTION 4 - SPRINGS

VARYING STRESS APPROACH DESIGN PROBLEMS 282.

A spring, subjected to a load varying from 100 lb. to 250 lb., is to be made of oiltempered, cold-wound wire. Determine the diameter of the wire and the mean diameter of the coil for a design factor of 1.25 based on Wahl’s line. The spring index is to be at least 5. Conform to good practice, showing checks for all significant parameters. Let the free length be between 6 and 8.

Solution: Fmax = 250 lb Fmin = 100 lb 1 1 Fm = (Fmax + Fmin ) = (250 + 100 ) = 175 lb = 0.175 kip 2 2 1 1 Fa = (Fmax − Fmin ) = (250 − 100 ) = 75 lb = 0.075 kip 2 2 Wahl’s line 1 sms − sas 2sas = + N s ys sno 8 KFa Dm 8 KFa C sas = = πDw3 πDw2 8 KFm Dm 8 KFmC sms = = K cπDw3 K cπDw2 C =5 4C − 1 0.615 K= + 4C − 4 C 4(5) − 1 0.615 K= + = 1.31 5 4(5) − 4 Fig. AF 15, C = 5 K c = 1.19 For oil-tempered wire, 87.5 s ys = 0.19 Dw , [0.032 < Dw < 0.5] 47 Dw0.1 , [0.041 < Dw < 0.15] 30 sno = 0.34 Dw , [0.15 < Dw < 0.625] sno =

N = 1.25

Page 20 of 70


SECTION 4 - SPRINGS 8(1.31)(0.075)(5) 1.251 = πDw2 Dw2 8(1.31)(0.175)(5) 2.453 sms = = (1.19)πDw2 Dw2 30 say sno = 0.34 ksi Dw 1 sms − sas 2sas = + N s ys sno sas =

 1.251   2.453 − 1.251   2 2   2 Dw 1  +  Dw  = 1.25  30   87.5   0.34   0.19   Dw   Dw  1 1 1 = + 1.81 11.99 Dw1.66 1.25 72.8 Dw Dw = 0.2857 in > 0.15 in Table AT 15, use No. 1, W & M Dw = 0.2830 in

Dm = CDw = 5(0.2830 ) = 1.415 in 7 say Dm = 1 in 16 Check for Free length 6 in < Free length < 8 in  7 Free length = 4 Dm = 41  = 5.75 in  16  Increase Dm 1 Dm = 1 in 2  1 Free length = 4 Dm = 41  = 6 in , o.k.  2 Summary of answer Dw = 0.2830 in 1 Dm = 1 in 2 283.

A carbon-steel spring is to be subjected to a load that varies from 500 to 1200 lb. The outside diameter should be between 3.5 and 4 in., the spring index between 5 to 10; approximate scale of 500 lb./in. Choose a steel and for a design factor of 1.4 by the Wahl line, find the wire diameter. Also determine the number of active

Page 21 of 70


SECTION 4 - SPRINGS coils and the free length for squared-and-ground ends. Conform to the general conditions specified in the Text. Solution: For carbon steel, Table AT 17 91 s ys = 0.1 ksi , [0.093 < Dw < 0.25] Dw 49 sno = 0.15 ksi , [0.093 < Dw < 0.25] Dw Fmax = 1200 lb Fmin = 500 lb 1 1 Fm = (Fmax + Fmin ) = (1200 + 500 ) = 850 lb = 0.85 kip 2 2 1 1 Fa = (Fmax − Fmin ) = (1200 − 500 ) = 350 lb = 0.35 kip 2 2 OD = 3.5 ~ 4.0 in C = 5 ~ 10 Wahl’s line 1 sms − sas 2sas = + N s ys sno Figure AF 15, C = 5 ~ 10 Assume K = 1.2 , K c = 1.125 8 KFa Dm sas = πDw3 8 KFm Dm sms = K cπDw3 OD ≈ 3.75 in Dm = 3.75 − Dw 8(1.2 )(0.35)(3.75 − Dw ) 1.0695(3.75 − Dw ) = sas = πDw3 Dw3 8(1.2)(0.85)(3.75 − Dw ) 2.3088(3.75 − Dw ) = sms = (1.125)πDw3 Dw3 

Dw   91   0.1   Dw  1 3.75 − Dw 3.75 − Dw = + 2.9 1.4 73.4285 Dw 22.9079 Dw2.85 Dw = 0.6171 in > 0.25 in Use 1 = 1.4



 3.75 − Dw   2(1.0695) Dw3 +    49   0.15   Dw 

(2.3088 − 1.0695) 3.75 −3 Dw 

Page 22 of 70


SECTION 4 - SPRINGS 91 = 104.53 ksi (0.25)0.1 49 sno = = 60.33 ksi (0.25)0.15

s ys =



 3.75 − Dw   2(1.0695) Dw3   + 60.33 

(2.3088 − 1.0695) 3.75 −3 Dw 

Dw 1  = 1 .4 104.53 1 3.75 − Dw 3.75 − Dw = + 1.4 84.346 Dw3 28.205 Dw3 1 3.75 − Dw = 1.4 21.137 Dw3 Dw = 0.5935 in use 19 Dw = in 32 3 Dm + Dw ≈ 3 in 4 19 3 = 3 in Dm + 32 4 5 Dm = 3 in 32  5  3  Dm  32  C= = = 5.316 Dw  19     32  . o.k. 19 in , Carbon Steel Wire Diameter Dw = 32 Number of coils: 8FC 3 N c δ= GDw 3 G = 10.5 ×10 6 psi = 10,500 ksi , Dw > in 8 GDw F =k = 3 δ 8C N c  (10.5 ×10 ) 19  32 6

  8(5.316 ) N c N c = 10.4 Table AT 16, square-and-ground ends 500 =

Page 23 of 70

3


SECTION 4 - SPRINGS Free length = PN c + 2 Dw Solid height = Dw N c + 2 Dw Total Coils = N c + 2  19  Solid height = Dw N c + 2 Dw = (10.4 + 2 )  = 7.3625 in  32  F 1200 = 2.4 in δ= = k 500 Min. Free length = 2.4 + 7.3625 in = 9.7625 in Use Free length = 10 in

To check for pitch angle. Free length = PN c + 2 Dw  19  P(10.4 ) + 2  = 10  32  P = 0.8474 in    P 0.8474   = 4.885o < 12o , o.k. = tan −1  λ = tan −1 πDm π  3 5     32  

Solid stress: δ T = solid deflection = 10 − 7.3625 = 2.6375 in F = kδ T = (500 )(2.6375) = 1319 lb 4C − 1 0.615 K= + 4C − 4 C 4(5.316 ) − 1 0.615 K= + = 1.29 4(5.316 ) − 4 5.316  5 8(1.29 )(1319 ) 3  8 KFDm  32  = 23,033 psi = 23.033 ksi < s (= 104.53 ksi ) ss = = ys 3 3 πDw  19  π   32  284.

A helical compression spring, made of oil-tempered, cold-wound carbon steel, is to be subjected to a working load varying from 100 to 300 lb. for an indefinite time (severe). A mean coil diameter of 2 in. should be satisfactory. (a) Using the static approach, compute a wire diameter. (b) For this wire size, compute the factor of safety as given by the Wahl line.

Solution:

Page 24 of 70


SECTION 4 - SPRINGS Table AT 16, For carbon steel, 182 su = 0.1 ksi , [0.093 < Dw < 0.25] Dw 91 s ys = 0.1 ksi 91 Dw Max. “solid” s ys = 0.1 ksi Dw 49 sno = 0.15 ksi , [0.093 < Dw < 0.25] Dw Dm = 2 in. Fmax = 300 lb Fmin = 100 lb (a) F = 300 lb = 0.3 kip severe service, ssd = 0.263su =

(0.263)(182) = 47.866 ksi Dw0.1

 8F C  ss = K  s 2   πDw  4C − 1 0.615 K= + 4C − 4 C Dm C= Dw D 2 Dw = m = C C      4C − 1 0.615   8(0.3)(2 ) 47.866 ss =  + = C    2 3   2  0.1  4C − 4  π      C   C   4C − 1 0.615  2.9  4C − 4 + C C = 233.84   C = 6.075 2 Dw = = 0.3292 in > 0.25 in 6.075 47.866 Therefore use ssd = = 54.984 ksi (0.25)0.1

Page 25 of 70

Dw0.1


SECTION 4 - SPRINGS      4C − 1 0.615   8(0.3)(2 ) ss =  + = 54.984 C    2 3   4C − 4  π    C   4C − 1 0.615  3  4C − 4 + C C = 287.9   C = 6.136 2 = 0.3259 in Dw = 6.136 21 say Dw = in 64 91 = 104.53 ksi (0.25)0.1 49 sno = = 60.33 ksi (0.25)0.15 1 1 Fm = (Fmax + Fmin ) = (300 + 100 ) = 200 lb = 0.2 kip 2 2 1 1 Fa = (Fmax − Fmin ) = (300 − 100 ) = 100 lb = 0.1 kip 2 2 Dm 2 C= = = 6.095 Dw  21     64  Figure AF 15 K c = 1.15 K = 1.25 K  8Fm Dm  sms =   K c  πDw3 

(b) s ys =

sms

    1.25  8(0.2 )(2 )  = = 31.34 ksi 1.15   21 3  π     64  

sas =

8 KFa Dm πDw3

Page 26 of 70


SECTION 4 - SPRINGS     8(0.1)(2)   sas = 1.25 = 18.02 ksi   21 3  π     64  

Wahl’s line 1 sms − sas 2sas = + N s ys sno 1 31.34 − 18.02 2(18.02 ) = + N 104.53 60.33 N = 1.38 285.

A helical spring of hard-drawn wire with a mean diameter of 1 ½ in. and squareand-ground ends is to be subjected to a maximum load of 325 lb. (a) Compute the wire diameter for average service. (b) How many total coils are required if the scale is 800 lb./in.? (c) For a minimum load of 100 lb., what is the factor of safety according to Wahl line? Would it be safe for an indefinite life?

Solution: Table AT 17, Hard-drawn wire, 140 su = 0.19 ksi , [0.028 < Dw < 0.625] Dw 70 Maximum “solid” ss = s ys = 0.19 ksi Dw (0.9)(47 ) ksi , [0.041 < D < 0.15] sno = w Dw0.1 (0.9)(30) ksi , [0.15 < D < 0.625] sno = w Dw0.34 Average service (a) ssd = 0.85(0.324 )su = 0.2754 su = F = 325 lb = 0.325 kip 1 Dm = 1 in 2  8 FDm  ss = K  3   πDw 

Page 27 of 70

0.2754(140 ) 38.556 ksi = Dw0.19 Dw0.19


SECTION 4 - SPRINGS 4C − 1 0.615 + 4C − 4 C 1.5 Dw = C K=

     4C − 1 0.615   8(0.325)(1.5)  38.556 ss =  + =  3 0.19 C   4C − 4  1.5    1.5   π     C   C    4C − 1 0.615  2.81 +  C = 97.05 C   4C − 4 C = 4.586 1.5 1.5 Dw = = = 0.3271 in < 0.625 in C 4.586 21 Dw = in 64 Dm 1. 5 = = 4.57 Dw  21     64  4(4.57 ) − 1 0.615 K= + = 1.345 4(4.57 ) − 4 4.57

(b) C =

8FC 3 N c GDw GD F =k = 3 w δ 8C N c k = 800 lb in = 0.8 kip in (11,500) 21   64  0.8 = 3 84.57 N c N c = 6.2

δ=

(c) s ys =

70 0.19

 21     64  (0.9)(30)

= 86.5 ksi

= 39.44 ksi , Dw > 0.15 in 0.19  21     64  1 Fm = (325 + 100 ) = 212.5 lb = 0.2125 kip 2

sno =

Page 28 of 70


SECTION 4 - SPRINGS 1 (325 − 100) = 112.5 lb = 0.1125 kip 2 K c = 1.212 , Fig. AF 15 K = 1.345     K  8 Fm Dm  1.345  8(0.2125)(1.5) sms = = = 25.5 ksi   3  K c  πDw3  1.212   21   π    64         8Fm Dm  8(0.1125)(1.5)   sas = K  = 1.345 = 16.36 ksi 3 3     21   πDw   π    64    1 sms − sas 2 sas 25.5 − 16.36 2(16.36) = + = + 39.44 N s ys sno 86.5 Fa =

N = 1.07 < 1.15[N min ] Not safe for indefinite life. 286.

A helical spring is to be subjected to a maximum load of 200 lb. (a) Determine the wire size suitable for medium service if the material is carbon steel ASTM A230; C = 6 . Determine the factor of safety of this spring according to the Wahl line (b) If the minimum force is 150 lb., (c) if the minimum force is 100 lb., (d) if the minimum force is 25 lb.

Solution: For carbon steel ASTM A230 Table AT 17 182 su = 0.1 ksi , [0.093 < Dw < 0.25] Dw 91 s ys = 0.1 ksi , [0.093 < Dw < 0.25] Dw 49 sno = 0.15 ksi , [0.093 < Dw < 0.25] Dw Medium Service ssd = 0.324 su  182  58.968 58,968 ksi = psi (a) ssd = 0.324 0.1  = 0.1 Dw Dw0.1  Dw 

Page 29 of 70


SECTION 4 - SPRINGS  8 FDm  ss = K  3   πDw  4C − 1 0.615 K= + 4C − 4 C 4(6) − 1 0.615 K= + = 1.2525 6 4(6) − 4 F = 200 lb  8(200 )(6 )  58,968 ss = 1.2525 = 2 Dw0.1  πDw  Dw = 0.2371 in

Table At 15, use Dw = 0.2437 in , No. 3 W & M Dw = 0.2437 in < 0.25 in , o.k. Factor of safety. 91 91 ksi = 104.8 ksi s ys = 0.1 ksi = Dw (0.2437 )0.1 49 49 ksi = 60.56 ksi sno = 0.15 ksi = Dw (0.2437 ) (a) Fm =

1 (200 + 150) = 175 lb = 0.175 kip 2

1 (200 − 150) = 25 lb = 0.025 kip 2 Figure AF 15, K c = 1.156 Fa =

K  8 FmC  1.2525  8(0.175)(6 )    = 48.8 ksi  = K c  πDw2  1.156  π (0.2437 )2   8(0.025)(6 )   8F C  sas = K  a 2  = 1.2525 = 8.1 ksi 2  πDw   π (0.2437 )  1 sms − sas 2sas = + N s ys sno 1 48.8 − 8.1 2(8.1) = + N 104.8 60.56 N = 1.525 sms =

(b) Fm = Fa =

1 (200 + 100) = 150 lb = 0.15 kip 2

1 (200 − 100) = 50 lb = 0.05 kip 2

Page 30 of 70


SECTION 4 - SPRINGS Figure AF 15, K c = 1.156  8 FmC  1.2525  8(0.15)(6 )  = = 41.8 ksi   2 2   πDw  1.156  π (0.2437 )   8(0.05)(6)   8F C  sas = K  a 2  = 1.2525 = 16.11 ksi 2  πDw   π (0.2437 )  1 sms − sas 2sas = + N s ys sno 1 41.8 − 16.11 2(16.11) + = N 104.8 60.56 N = 1.287

sms =

K Kc

(c) Fm =

1 (200 + 25) = 112.5 lb = 0.1125 kip 2

1 (200 − 25) = 87.5 lb = 0.0875 kip 2 Figure AF 15, K c = 1.156 Fa =

K  8 FmC  1.2525  8(0.1125)(6)    = 31.36 ksi  = K c  πDw2  1.156  π (0.2437 )2   8(0.0875)(6)   8F C  sas = K  a 2  = 1.2525 = 28.2 ksi 2  D π ( ) 0 . 2437 π  w    1 sms − sas 2sas = + N s ys sno 1 31.36 − 28.20 2(28.20 ) = + N 104.8 60.56 N = 1.04 sms =

CHECK PROBLEMS A Diesel valve spring is made of 3/8-in. chrome-vanadium steel wire, shot-peened; inside diameter is 3 in., 7 active coils, free length is 7 3/8 in., solid length is 4 1/8 in., length with valve closed, 6 ¼ in., length when open, 5 1/8 in. (a) Compute the spring constant and the factor of safety as defined by the Wahl criterion (see § 6.13, Text). (b) Is there any danger of damage to the spring if it is compressed solid? (c) What is the natural frequency? If this spring is used on a 4-stroke Diesel engine at 450 rpm, is there any danger of surge? Compute the change of stored energy between working lengths.

Solution: For chrome-vanadium steel wire, shot-peened, Table AT 17 Page 31 of 70


SECTION 4 - SPRINGS

(1.25)(168) ksi , [0.032 < D

w < 0.437 ] Dw0.166 (1.25)(100 ) ksi , [0.032 < D < 0.437] s ys = w Dw0.166 (1.25)(49) ksi , [0.028 < D < 0.5] sno = w Dw0.15 3 Dw = in = 0.375 in 8 (1.25)(100) ksi = 147.1 ksi s ys = (0.375)0.166 (1.25)(49) ksi = 70.96 ksi sno = (0.375)0.15

su =

8FC 3 N c GDw GD F =k = 3 w δ 8C N c

(a) δ =

G = 11.5 × 106 psi Nc = 7 Dw = 0.375 in Dm − Dw = ID = 3 in Dm = 3.375 in D 3.375 C= m = =9 Dw 0.375 k = spring constant GDw 11.5 ×10 6 )(0.375) ( k= 3 = = 105.64 lb in 3 8C N c 8(9 ) (7 ) 3 1 δ1 = 7 − 4 = 3.25 in 8 8 F1 = kδ1 = (105.64 )(3.25) = 343.33 lb 3 1 δ 2 = 7 − 6 = 1.125 in 8 4 F2 = kδ 2 = (105.64 )(1.125) = 118.85 lb 1 Fm = (343.33 + 118.85) = 231.09 lb = 0.231 kip 2 1 Fa = (343.33 − 118.85) = 112.24 lb = 0.11224 kip 2 K  8FmC  sms =   K c  πDw2 

Page 32 of 70


SECTION 4 - SPRINGS 4C − 1 0.615 + 4C − 4 C 4(9) − 1 0.615 K= + = 1.162 9 4(9 ) − 4 Figure AF 15, K c = 1.10 K=

 8 FmC  1.162  8(0.231)(9)  = = 39.8 ksi   2  2   πDw  1.10  π (0.375)   8(0.11224 )(9)   8F C  sas = K  a 2  = 1.162   = 21.3 ksi 2  πDw   π (0.375)  1 sms − sas 2sas = + N s ys sno 1 39.8 − 21.3 2(21.3) = + N 147.1 70.96 N = 1.377

sms =

K Kc

(b) max. “solid” ss = s ys = 147.1 ksi Min. Solid Height = Dw N c = (0.375)(7 ) = 2.625 in 3 Solid deflection = 7 − 2.625 = 4.75 in. 8 F = kδ = (105.64 )(4.75) = 501.8 lb = 0.5018 kip  8(0.5018)(9)   8 FC  Solid stress = ss = K  2  = 1.162  = 95 ksi < 147.1 ksi 2   πDw   π (0.375)  There is no danger of damage

(c) Natural frequency For steel 14,050 Dw cps φ= N c Dm2 14,050 cps φ= N c C 2 Dw 14,050 cps = 66 cps φ= (7 )(9)2 (0.375)  2π  For 450 rpm, φ = 450  = 47 cps  60  66 = 1.4 < 12 , there is danger of surging. 47

Page 33 of 70


SECTION 4 - SPRINGS

(d) U s =

[

]

1 1 2 2 k δ12 − δ 22 = (105.64 ) (3.25) − (1.125) = 491 in − lb 2 2

289.

(

)

A helical spring is hot wound from 5/8-in. carbon-steel wire with an outside diameter of 3 ¼ in. A force of 3060 lb. is required to compress the spring 1 ¾ in to the solid heigh. In service the spring is compressed so that its deformation varies form ½ in. to1 1/8 in. (a) What is the factor of safety by the Wahl criterion? (b) Is the “solid stress” safe? Compute (c) the pitch angle, (d) the change of stored energy between the working lengths, (e) the factor of safety if the spring is peened?

Solution: For hot-wound carbon steel wire 5 Dw = in 8 Table AT 17 91 s ys = 0.1 ksi , [0.093 < Dw < 0.25] Dw 91 s ys = ksi = 104.5 ksi , Dw > 0.25 in. (0.25)0.1 49 sno = 0.15 ksi , [0.093 < Dw < 0.25] Dw 49 sno = ksi = 60.33 ksi , Dw > 0.25 in. (0.25)0.15 Permissible solid stress = ss = ss =

117 ksi , [Dw > 0.375 in.] § 6.3 Dw0.31

117 ksi = 35.4 ksi (0.625)0.31

3060 = 1748.6 lb in δ 1.75 1 F1 = kδ1 = (1748.6 )  = 874.3 lb 2  1 F2 = kδ 2 = (1748.6 )1  = 1967.2 lb  8 1 Fm = (1967.2 + 874.3) = 1420.7 lb = 1.4207 kip 2 1 Fa = (1967.2 − 874.3) = 546.4 lb = 0.5464 kip 2

(a) k =

F

=

Page 34 of 70


SECTION 4 - SPRINGS 5 in = 0.625 in 8 1 Dm + Dw = 3 in 4 Dm = 2.625 in D 2.625 C= m = = 4.2 Dw 0.625 4C − 1 0.615 K= + 4C − 4 C 4(4.2) − 1 0.615 K= + = 1.3808 4(4.2) − 4 4.2 K c = 1.234 Dw =

K  8 FmC  1.3808  8(1.4207 )(4.2)    = 43.5 ksi  = K c  πDw2  1.234  π (0.625)2   8(0.5464 )(4.2)   8F C  sas = K  a 2  = 1.3808  = 20.7 ksi 2  πDw   π (0.625)  1 sms − sas 2sas = + N s ys sno 1 43.5 − 20.7 2(20.7 ) + = N 104.5 60.33 N = 1.106 sms =

(b) Permissible solid stress = 135.4 ksi F = 3.060 kip

 8(3.060 )(4.2)   8 FC  Solid stress, ss = K  2  = 1.3808 = 115.7 ksi < 135.4 ksi , safe 2   πDw   π (0.625)  3 (c) Solid deflection = 1 in 4 (P − Dw )N c = 1.75 in

8FC 3 N c δ= GDw G = 10.5 × 106 psi , hot-wound F GD k= = 3w δ 8C N c

Page 35 of 70


SECTION 4 - SPRINGS

(10.5 ×10 )(0.625) 6

1748.6 =

8(4.2 ) N c 3

N c = 6.332 (P − 0.625)(6.332) = 1.75 P = 0.9014 in Pitch angle P P tan λ = = πDm πCDw

λ = tan −1

 0.9014  P = tan −1  = 6.24o  πCDw  π (4.2 )(0.625) 

[

]

1 1 2 2 k δ12 − δ 22 = (1748.6 ) (1.125) − (0.5) = 888 in − lb 2 2 (e) When peened s ys = 12.5(104.5) = 130.6 ksi

(

(d) U s =

)

sno = 1.25(60.33) = 75.4 ksi 1 43.5 − 20.7 2(20.7 ) = + N 130.6 75.4 N = 1.38

ENERGY STORAGE 293.

A 10-lb. body falls 10 in. and then strikes a helical spring. Design a harddrawn carbon steel spring that will absorb this shock occasionally without permanent damage. Determine appropriate values of wire diameter, coil diameter, pitch, free length, closed length, and the maximum stress under the specified conditions, and scale. Let C = 7 .

Solution: For hard-drawn carbon steel, Table AT 17 182 su = 0.1 ksi , [0.093 < Dw < 0.25] Dw 91 Max. “solid” ss = 0.1 ksi , [0.093 < Dw < 0.25] Dw 36.855 ssd = (0.50 )(0.405)su = ksi Dw0.1 Us =

ss2V 4 K 2G

Page 36 of 70


SECTION 4 - SPRINGS 4C − 1 0.615 + 4C − 4 C 4(7 ) − 1 0.615 K= + = 1.213 7 4(7 ) − 4 K=

 πD 2  V ≈  w (πDm )N c  4  π 2 Dw2 Dm N c V= 4 Dm = CDw

V=

π 2CDw3 N c

4 8FC 3 N c δ= GDw U s = W (h + δ )  8 FC  ss = K  2   πDw  s πD 2 F= s w 8KC  s πD 2   8C 3 N c  δ =  s w    8KC   GDw 

s sπDwC 2 N c KG 2 2 3  ssπDwC 2 N c  s s π CD w N c U s = W h + = KG 16 K 2G   Wh Nc = 2 2 2 ss π CDw s sπDwC 2W − 16 K 2G KG 36.855 ksi when ss = Dw0.1 Wh Nc = 2 2 2.8 (36.855) π CDw − 36.855πD 0w.9C 2W 16 K 2G KG 0.010)(10) ( Nc = (36.855)2 π 2 (7)Dw2.8 − 36.855πD w0.9 (7 )2 (0.010) 2 (1.213)(11,500) 16(1.213) (11,500) 0.10 Nc = 2.8 0.3466 Dw − 0.004067 D 0w.9

δ=

Page 37 of 70


SECTION 4 - SPRINGS combination of Dw and N c Gage No. W & M 12 11 10 9 8 7 6 5 4 3

Dw 0.1055 0.1205 0.1350 0.1483 0.1620 0.1770 0.1920 0.2070 0.2253 0.2437

Nc 991.2 312.1 166.1 108.0 75.2 53.7 40.2 31 23.4 18.1

Dw N c 105 37.6 22.4 16.0 12.2 9.5 7.7 6.4 5.3 4.4

Use Dw = 0.2437 in < 0.25 in , N c = 18.1 45 Dm = 7 Dw = 7(0.2437 ) = 1.7059 in = 1 in 64 0.9 2 0.9 2 s sπDwC 2 N c 36.855πD w C N c 36.855π (0.2437 ) (7 ) (18.1) = = = 2.066 in (1.213)(11,500) KG KG 36.855 ss = = 42.44 ksi (0.2437 )0.1 91 91 sso = 0.1 = = 104.8 ksi Dw (0.2437 )0.1 Solid deflection  104.8  = (2.066) = 5.1 in  42.44  (P − Dw )N c = 5.1

δ=

(P − 0.2437)(18.1) = 5.1 P = 0.5255 in 17 = 0.53125 in say P = 32 Minimum Solid Height = Dw N c = (0.2437 )(18.1) = 4.41 in Assume squared and ground end Solid height = Dw N c + 2 Dw = (0.2437 )(18.1) + 2(0.2437 ) = 5.0 in

Solid deflection = (0.53125 − 0.2437 )(18.1) = 5.2 in Free length = 5.0 in + 5.2 in = 10.2in

Summary of answer: Dw = 0.2437 in , No. 3 W & M Page 38 of 70


SECTION 4 - SPRINGS

Dm = 1

45 in 64

17 in 32 Free length = 10.2 in Closed length = 5 in Maximum stress = 42.44 ksi P=

294.

A helical spring, of hard-drawn steel wire, is to absorb 75 in-lb of energy without being stressed beyond the recommended value of average service. Let C = 6 . Decide upon satisfactory dimensions; Dw , Dm , N c , free length, pitch angle, solid stress, volume of metal, possibility of spring buckling.

Solution: For hard-drawn steel wire, shock load, average service 140 su = 0.19 ksi , [0.028 < Dw < 0.625] Dw 70 Max. “solid” ss = 0.19 ksi , [0.028 < Dw < 0.625] Dw  140  19.278 ssd = (0.50)(0.85)(0.324)su = (0.50)(0.85)(0.324) 0.19  = 0.19 ksi  Dw  Dw s 2V s 2π 2 Dw3 CN c Us = s 2 = s 4K G 16 K 2G C=6 4C − 1 0.615 K= + 4C − 4 C 4(6) − 1 0.615 K= + = 1.2525 6 4(6 ) − 4 U s = 75 in − lb = 0.075 in − kip 19.278  π 2 Dw3 (6)N c U s = 0.075 =  0.19  2  Dw  16(1.2525) (11,500) 0.9837 = Dw2.62 N c

Table AT-15 W&M

Page 39 of 70

Dw

Nc

Dw N c


SECTION 4 - SPRINGS 9 8 7 6 5 4 3 2 1 0 2-0 3-0 4-0 5-0

0.1483 0.1620 0.1770 0.1920 0.207 0.2253 0.2437 0.2625 0.2830 0.3065 0.3310 0.3625 0.3938 0.4305

146 116 92 74 61 49 40 32.7 26.9 21.8 17.8 14.0 11.3 8.95

Use Dw = 0.4305 in , 5-0 W & M Nc ≈ 9 9 Dm = 6(0.4305) = 2.583 in ≈ 2 in 16 19.278 ss = = 22.63 ksi (0.4305)0.19 70 Max. Solid Stress = sso = = 82.16 ksi (0.4305)0.19 s sπDwC 2 N c (22.63)(π )(0.4305)(6) (9) = = 0.6885 in KG (1.2525)(11,500)  82.16  Solid deflection =  (0.6885) = 2.5 in  22.63  (P − Dw )N c = 2.5 (P − 0.4305)(9) = 2.5 P = 0.7083 in 45 say P = = 0.703125 in 64 Solid deflection = (0.703125 − 0.4305)(9) = 2.453625 in  2.453625  Solid stress = 22.63  = 80.65 ksi  0.6885  2

δ=

7 Minimum Solid Height = Dw N c = (0.4305)(9 ) = 3.8745 in ≈ 3 in 8 45 21   Minimum Free Length = PN c =  (9 ) = 6.328125 in ≈ 6 in 64  64  Pitch Angle

Page 40 of 70

21.65 18.79 16.28 14.21 12.63 11.04 9.75 8.58 7.61 6.68 5.89 5.075 4.45 3.85


SECTION 4 - SPRINGS  45    P = tan −1  64  = 5o < 12o λ = tan −1 πDm π  2 9     16   Volume  π (0.4305)2    9   πDw2  3 (πDm )N c =  V ≈   π  2 (9 ) = 10.55 in 4  4      16 

Summary of answer: Dw = 0.4305 in , No. 5-0 W & M 9 Dm = 2 in 16 Nc = 9 21 Free length = 6 in 64 Pitch Angle = λ = 5o Solid Stress = 80.65 ksi Volume of metal = 10.55 in3 Possibility of spring buckling 21 6 64 = 2.47 < 4 , no possibility 9 2 16 CONCENTRIC HELICAL SPRINGS 297.

Two concentric helical springs are to be subjected to a load that varies from a maximum of 235 lb. to a maximum of 50 lb. They are to fit inside a 1 5/8 in. cylinder. The maximum deflection is to be ¾ in., and the deflection when compressed solid is to be approximately 1 in. Using the “static approach” for severe service (maximum load), determine the wire diameter, mean coil diameter, number of coils, solid length, and free length of both springs. (Start with oil-tempered wire and assume a diametral clearance between the outer D spring and the cylinder of w , assume a similar clearance between springs. 2 Search for a suitable spring index and wire size.)

Solution: For oil-tempered wire Table AT 17

Page 41 of 70


SECTION 4 - SPRINGS 146 ksi , [0.032 < Dw < 0.5] Dw0.19 87.5 Max. “solid” ss = 0.19 ksi , [0.032 < Dw < 0.5] Dw Severe service 0.263(146) 38.398 ssd = 0.263su = ksi = Dw0.19 Dw0.19 F = 235 lb = 0.235 kip δo = δi su =

8 FoCo3 N co 8 Fi Ci3 N i = GDwo GDwi Assume, Co = Ci 3GDwo Fo = 32C 3 N co 3GDwi Fi = 32C 3 N ci  8F C  sso = K  o2   πDwo   8F C  ssi = K  i 2   πDwi  D D C = mo = mi Dwo Dwi 4C − 1 0.615 K= + 4C − 4 C D Dmo − Dwo − wi = Dmi + Dwi 2 Dmo − Dmi = Dwo + 1.5Dwi D 1.625 − wo = Dmo + Dwo 2 Dmo + 1.5Dwo = 1.625 CDwo + 1.5Dwo = 1.625 1.625 Dwo = C + 1 .5 1.625C Dmo = C + 1 .5 CDwo − CDwi = Dwo + 1.5Dwi

(C − 1)Dwo = (C + 1.5)Dwi Page 42 of 70


SECTION 4 - SPRINGS 1.625(C − 1) (C + 1.5)2 1.625C (C − 1) Dmi = (C + 1.5)2 Dwi =

 8F C  38.398 sso = K  o2  = 0.19 ksi Dwo  πDwo  1.81 15.08Dwo Fo = KC  8F C  38.398 ssi = K  i 2  = 0.19 ksi Dwi  πDwi  1.81 15.08Dwi KC Fo + Fi = F = 0.235 kip

Fi =

1.81 1.81 15.08Dwo 15.08Dwi + = 0.235 KC KC 1.81 1.81 15.08Dwo + 15.08 Dwi = 0.235KC 1.81

1.81

 1.625  1.625(C − 1)  15.08 = 0.235 KC  + 15.08  C + 1.5   C + 1.5  1.81   ( C − 1) 1  4C − 1 0.615  + = 0.235 + 154.52  C 1.81 3.62  C  (C + 1.5)   4C − 4  (C + 1.5) C = 5.328 1.625(5.328 − 1) Dwi = = 0.1509 in (5.328 + 1.5)2 1.625 Dwo = = 0.2380 in 5.328 + 1.5 Table AT 15, use Dwi = 0.1620 in , No. 8 W & M and Dwo = 0.2625 in , No. 2 W & M 13 Dmo = CDwo = (5.328)(0.2625) = 1.3986 in ≈ 1 in 32 7 Dmi = CDwi = (5.328)(0.1620) = 0.8631 in ≈ in 8 7   Dmi 8 Ci = =   = 5.401 Dwi 0.1620

Co =

Dmo Dwo

 13  1  32  = = 5.357 0.2625

Page 43 of 70


SECTION 4 - SPRINGS 1.81 15.08Dwo K o Co 4(5.357 ) − 1 0.615 Ko = + = 1.287 4(5.357 ) − 1 5.357

Fo =

15.08(0.2625) = 0.194 kip (1.287)(5.357 ) 1.81

Fo =

1.81 15.08Dwi K i Ci 4(5.401) − 1 0.615 Ki = + = 1.2843 4(5.401) − 1 5.401 15.08(0.1620) Fi = = 0.081 kip (1.2843)(5.401) Fo + Fi = 0.194 + 0.071 = 0.275 kip > 0.235 kip , ok 3GDwo Fo = 32C 3 N co 3(11,500)(0.2625) 0.194 = 3 32(5.357 ) N co N co = 9.5 3GDwi Fi = 32C 3 N ci 3(11,500)(0.1620) 0.071 = 3 32(5.401) N ci N ci = 15.6 87.5 Max. solid stress, sss = 0.19 ksi , Dw 87.5 ssso = = 112.82 ksi (0.2625)0.19 87.5 sssi = = 123.65 ksi (0.1620)0.19 Stress 38.398 ssi = = 54.26 ksi (0.1620)0.19 38.398 sso = = 49.51 ksi (0.2625)0.19

Fi =

Solid stress  1  sso = 49.51  = 66.01 ksi < 112.82 ksi  0.75 

Page 44 of 70


SECTION 4 - SPRINGS  1  ssi = 54.26  = 72.35 ksi < 123.65 ksi  0.75  Solid length Dwo N co = (0.2625)(9.5) = 2.5 in

Dwi N ci = (0.1620 )(15.6) = 2.53 in assume solid length = 3 in Dwi ( N ci + xi ) = (0.1620)(15.6 + xi ) = 3 in xi = 2.92 Total coils = 15.6 + 2.92 = 18.52 Dwo ( N co + xo ) = (0.2625)(9.5 + xo ) = 3 in xo = 1.93 Total coils = 9.5 + 1.93 = 11.43 Free Length = 3 in + 1 in = 4 in Summary of answer: Outside wire. Dwo = 0.2625 in , No. 2 W & M 13 Dmo = 1 in 32 N to = 11.43 Solid length = 3 in Free length = 4 in Inside wire. Dwi = 0.1620 in , No. 8 W & M 7 Dmi = in 8 N ti = 18.52 Solid length = 3 in Free length = 4 in 298.

Two concentric, helical compression springs are used on a freight car. The larger spring has an outside diameter of 7 in., a free length of 7 1/8 in., and is made of a 1 ¾ in. steel bar. The smaller has an outside diameter of 4 1/8 in., a free length of 6 13/16 in. , and is made of 7/8 in. steel bar. The solid height of each spring is 5 ¼ in. and the forces required to compress them solid are 15,530 lb. and 7,000 lb., respectively. The working load on the two springs is 11,350 lb. Determine (a) the number of free coils in each spring, (b) the stress in each spring when compressed solid, (c) the stresses induced by the working

Page 45 of 70


SECTION 4 - SPRINGS load. Notice that the outer spring deflects 5/16 in. before the inner one takes a load. (d) What energy is absorbed while changing deflection from that at the working load to that when the springs are compressed “solid”? Solution: ODo = 7 in 3 Dwo = 1 in 8 1 FLo = 7 in 8 1 ODi = 4 in 8 7 Dwi = in 8 13 FLi = 6 in 16 1 (a) Solid height = Dw N T = 5 in 4 5.25 = 3.82 N To = 1.375 5.25 =6 N Ti = 0.875

(b) Fo = 15,530 lb Fi = 7000 lb  8 FC  ss = K  2   πDw  3 Dmo = 7 − 1 = 5.625 in 8 Dmo 5.625 Co = = = 4.091 Dwo 1.375 4(4.091) − 1 0.615 Ko = + = 1.393 4(4.091) − 4 4.091 1 7 Dmi = 4 − = 3.25 in 8 8 D 3.25 Ci = mi = = 3.714 Dwi 0.875 4(3.714) − 1 0.615 Ki = + = 1.442 4(3.714) − 4 3.714

Page 46 of 70


SECTION 4 - SPRINGS Solid stress  8(15,530)(4.091)  sso = 1.393  = 119,203 psi 2 ( ) π 1 . 375    8(7000 )(3.714 )  ssi = 1.442  = 124,689 psi 2  π (0.875)  (b) Stresses induced by working load Fi + Fo = 11,350 lb 15,530 ko = = 8283 lb in 1  1 7 − 5  4  8 7000 ko = = 4480 lb in 1  13 6 − 5  4  16 5 δ o − δ i = = 0.3125 in 16 Fi = kiδ i = 4480δ i Fo = koδ o = 8283(0.3125 + δ i ) Fi + Fo = koδ o = 4480δ i + 8283(0.3125 + δ i ) = 11,350 lb δ i = 0.6865 in δ o = 0.3125 + 0.6865 = 0.9990 in Fi = (4480)(0.6865) = 3076 lb Fo = (8283)(0.9990) = 8275 lb Stresses  8(8275)(4.091)  sso = 1.393  = 63,516 psi 2  π (1.375)   8(3076)(3.714)  ssi = 1.442  = 54,792 psi 2  π (0.875) 

(d) Energy 1 U so = ko (δ o22 − δ o21 ) 2 1 1 δ o 2 = 7 − 5 = 1.875 in 8 4 δ o1 = 0.9990 in 1 2 2 U so = (8283)(1.875) − (0.999) = 10,427 in − lb 2

[

Page 47 of 70

]


SECTION 4 - SPRINGS 1 ki δ i22 − δ i21 2 13 1 δ i 2 = 6 − 5 = 1.5625 in 16 4 δ o1 = 0.6865 in 1 2 2 U si = (4480 )(1.5625) − (0.6865) = 4,413 in − lb 2 U si =

(

)

[

]

TORSION-BAR SPRINGS 299.

A torsion-bar similar to that shown is to be used for the front spring of an automobile. Its rate should be 400 lb./in. of deflection of the end of the arm which is e = 10 in. long. It is made of AISI 9261,OQT 900 oF, and the maximum repeated load is 1500 lb. perpendicular to the centerline of the arm. The support is such that bending of the bar is negligible. (a) Determine its diameter and length so that no permanent set occurs due to a 30 % overload (limited by a stop). Use s ys = 0.6 s y , but check with equation (c) § 6.3, Text, if appropriate. (b) Determine the factor of safety according to the Soderberg criterion if the load varies from 1200 lb. to 1500 lb.; minimum r d = 0.1 , D d = 3 . (c) The same as (b) except that the bar is shot-peened all over. What other steps may be taken to improve the fatigue strength?

Problem 299, 300 Solution: e = 10 in For AISI 9261, OQT 900 oF s y = 192 ksi su = 215 ksi s ys = 0.6s y = 115.2 ksi = 115,200 psi

Page 48 of 70


SECTION 4 - SPRINGS 16T πd3 T = Fe = (1.3)(1500)(10) = 19,500 in − lb 16(19,500) 115,200 = π d3 d = 0.95 in use d = 1 in 117 117 § 6.3 ss = 0.3 = 0.3 = 117 ksi ≈ s ys Dw (1)

(a) ss =

(b) Soderberg Criterion 1 sms K f sas = + N s ys sns sns = (0.6)(0.5)(215) = 64.5 ksi Figure AF 12, r d = 0.1 , D d = 3 K t = 1.45 K f ≈ K t = 1.45 1 (1500 + 1200) = 1350 lb 2 Tm = (1350)(10) = 13,500 in − lb = 13.5 in − kips 16(13.5) sms = = 68.8 ksi π (1)3 Fm =

1 (1500 − 1200) = 150 lb 2 Ta = (150)(10) = 1500 in − lb = 1.5 in − kips 16(1.5) sas = = 7.64 ksi π (1)3 1 68.8 (1.45)(7.64 ) = + N 115.2 64.5 N = 1.30 Fa =

(c) Shot-peened s ys = 1.25(115.2 ) = 144 ksi sns = 1.25(64.5) = 80.6 ksi 1 68.8 (1.45)(7.64 ) = + N 144 80.6 N = 1.625

Page 49 of 70


SECTION 4 - SPRINGS

300.

A solid steel torsion bar is loaded through a 10 in. arm as shown. The load F perpendicular to the center-line of the arm varies from 500 to 1000 lb., 7 200,000 cycles. The bar is d = in. in diameter and 30 in. long; let D d = 3 ; 8 r d = 0.1 ; (a) Determine the maximum stress in the bar, the angular deflection, and the scale (lb./in.) where F is applied. The support is such that bending of the bar is negligible. (b) Select a material and heat treatment for this bar for a minimum N = 1.2 , Soderberg criterion.

Problem 299, 300 Solution: Fig. AF 12, K f = 1.45 1 (1000 + 500) = 750 lb 2 1 Fa = (1000 − 500 ) = 250 lb 2 Tm = (750)(10) = 7500 in − lb = 7.5 in − kips Fm =

Ta = (250)(10) = 2500 in − lb = 2.5 in − kips K fl =

n

(log K f ) 3

Kf

=

(200,000)(log1.45) 3 1.33 1.45

16T πd3 16(7.5) sms = = 57 ksi 3 7 π  8 16(2.5) sas = = 19 ksi 3 7 π  8 ss =

(a) smax = sms + K fl sas = 57 + (1.33)(19 ) = 82.27 ksi

Page 50 of 70


SECTION 4 - SPRINGS TL 64TL = JG π d 4G 64(500)(10)(30 ) θ min = = 0.4533 rad 4 7 6 π   11.5 ×10 8 64(1000 )(10)(30 ) θ max = = 0.9066 rad 4 7 6 π   11.5 ×10 8 F 1000 scale = = = 110.3 lb in θ e (0.9066)(10)

θ=

(c)

(

)

(

)

1 sms K f sas = + N s ys sns

 10 6   sns = (0.6 )(0.5)su   200,000  s ys = 0.6 s y

0.085

= 0.344su

1 57 (1.33)(19) = + 1.2 0.6s y 0.344su Use AISI 8760, OQT 800 oF s y = 200 ksi su = 220 ksi N = 1.24 HELICAL SPRINGS – NON CIRCULAR SECTION 301.

A spring is to be designed of square oil-tempered steel wire and subjected to a repeated maximum load of 325 lb.; mean coil diameter, 1 ½ in.; deflection, 13/32 in. Determine (a) the wire size for average service, (b) the required number of active coils, (c) the solid height, free length, and pitch (the ends are squared and ground, the “solid stress” must be satisfactory, and the pitch angle not excessive). (d) What amount of energy is stored when the load is 325 lb.? Express in in-lb. and Btu.

Solution: For oil-tempered wire, 146 su = 0.19 , [0.032 < Dw < 0.5] Dw Max. “solid” ss = 0.6su

Page 51 of 70


SECTION 4 - SPRINGS (a) average service, b = Dw , t = b K FD (3b + 1.8t ) 2.4 FDm ss= q m 2 2 = Kq 2b t b3 s sd = 0.324su , average service 2.4 FDm 0.324(146) Kq = 2b 3 b 0.19 F = 0.325 kip 1 Dm = 1 in 2 K q = 1.25 (assumed)  2.4(0.325)(1.5) 0.324(146) 1.25 = b3 b 0.19   b = 0.2902 in Table AT 15, use b = 0.313 in , # 1 wire size D 1.5 C= m = = 4.8 b 0.313 Figure AF 15, K q = 1.275

(b) δ =

2.45 FDm3 N c 2.45FDm3 N c = 0.44Gb 4 Gt 3 (b − 0.56t )

13 2.45(0.325)(1.5) N c = 32 0.44(11,500)(0.313)4 N c = 7.34 3

(c) Solid height = b( N c + 2) = 0.313(7.34 + 2) = 2.92 in Free length = PN c + 2b F = 0.325 lb  2.4(0.325)(1.5)  2.4 FDm  ss = Kq  = 1.275  = 48.65 ksi 3 3   b   (0.313)  0.6(146) solid stress = = = 109.2 ksi (0.313)0.19 109.2  13  solid deflection = =   = 0.91 in 48.65  32  (P − b )N c = 0.91 (P − 0.313)(7.34) = 0.91 P = 0.437 in

Page 52 of 70


SECTION 4 - SPRINGS

use P =

7 in 16

27 7 Free length = PN c + 2b =  (7.34) + 2(0.313) = 3.837 in ≈ 3 in 32  16 

7   P 16 tan λ = =   πDm π (1.5)

λ = 5.3o < 10o (d) U s = Us =

1 2 1 1  13  kδ = Fδ = (0.325)  = 0.066 in − kip = 66 in − lb 2 2 2  32 

66 = 0.085 Btu 778

302.

A coil spring, of hard-drawn carbon steel, is to deflect 1 in. under a load of 100 lb. The outside coil diameter is to be 1 in. Compute the number of active coils, (a) if the wire is round, 5/32 in. in diameter, (b) if the wire is square, 5/32 in. on the side, (c) if the wire is rectangular 1/8 x 3/16 in., long dimension parallel to the axis, (d) If the wire is rectangular 3/16 x 1/8 in., short dimension parallel to the axis. (e) What is the maximum stress in each of the above springs under the 100-lb load? (f) What is the ratio of the approximate volumes, square- or rectangular-wire to round wire spring?

Solution: Dm + Dw = 1 in 5 in (a) Dw = 32 5 27 Dm = 1 − in = 32 32  27    Dm  32  = 5.4 C= = Dw  5     32  8FC 3 N c δ= GDw 8(100)(5.4 ) N c  5  11.5 × 10 6    32  N c = 14.3 3

1=

(

Page 53 of 70

)


SECTION 4 - SPRINGS 5 in 32 5 27 Dm = 1 − = in 32 32 2.45FD m3 N c = δ 0.44Gb 4

(b) Square, b =

3

 27  2.45(100)  N c  32  1= 4 6  5  0.44 11.5 × 10    32  N c = 20.5

(

)

3 1 in , t = in 16 8 1 7 Dm = 1 − t = 1 − = in 8 8 3 2.45FD m N c δ= 3 Gt (b − 0.56t )

(c) b =

3

7 2.45(100)  N c 8 1= 3 1  3  1  11.5 × 10 6    − 0.56   8  16  8 

(

)

N c = 16.1 1 3 in , t = in 8 16 3 13 Dm = 1 − t = 1 − = in 16 16 3 2.45FD m N c δ= 3 Gt (b − 0.56t )

(d) b =

3

 13  2.45(100)  N c  16  1= 3  3  1  3  11.5 × 10 6    − 0.56   16   8  16 

(

)

N c = 11.5 (e) Maximum Stress

Page 54 of 70


SECTION 4 - SPRINGS 8 FC π Dw3 4(5.4) − 1 0.615 K= + = 1.284 4(5.4) − 4 5.4

For (a) ss = K

    8(100)(5.4)  ss = 1.284 = 72,320 psi   5 3   π     32   FD m (3b + 1.8t ) 2.4 FD m = Kq 2 2 2b t b3  27    Dm  32  = 5.4 C= = b  5     32  K q = 1.25

For (b) ss = K q

  27    2.4(100)    32   = 66,355 psi ss = 1.25 3    5       32    FD m (3b + 1.8t ) For (c) ss = K q 2b 2t 2 7   D 8 C = m =   =7 t 1   8 K q = 1 .1  7   (100)    8   3 3  + 1.8 1  = 68,992 psi ss = 1.1     2   3   1  2    16   8   2       16   8   FD m (3b + 1.8t ) For (d) ss = K q 2b 2t 2  13    Dm  16  C= = = 4.33 t 3    16 

Page 55 of 70


SECTION 4 - SPRINGS K q = 1.2

  13    (100)    16   3 1  + 1.8 3  = 63,232 psi ss = 1.2     2   1   3  2    8   16   2       8   16  

(e) Ratio of the approximate volumes For (a) Round wire π  Va =  Dw2 (π Dm )N c 4   π  5  2    27  Va =     π   (14.3) = 0.727 in 3  4  32     32  For (b) Square wire Vb = b 2 (π Dm )N c 2

 5    27  Vb =   π   (20.5) = 1.327 in 3  32    32  For (c) rectangular wire Vc = bt (π Dm )N c  3  1    7  Vc =    π   (16.1) = 1.037 in3  16  8    8  For (d) rectangular wire Vd = bt (π Dm )N c  1  3    13  Vd =    π   (11.5) = 0.688 in 3  8  16    16 

Ratio of volume Square to round wire V 1.327 = b = = 1.825 Va 0.727 Rectangular to round wire (long dimension parallel to the axis) V 1.037 = c = = 1.426 Va 0.727 Rectangular to round wire (short dimension parallel to the axis) V 0.688 = d = = 0.946 Va 0.727

Page 56 of 70


SECTION 4 - SPRINGS TENSION SPRINGS 305.

Design two tension springs for a spring balance with a capacity of 200 lb. Each spring supports a maximum load of 100 lb. The outside diameter must not exceed 1 ¼ in. and the total length including end loops must not exceed 9 ½ in. Select a material and determine the dimension, including wire diameter, number of coils, and free length.

Solution: Table AT 17, assume oil tempered wire 146 su = 0.19 ksi Dw 87.5 s ys = 0.19 ksi Dw 0.8(87.5) 70 ssd = = 0.19 ksi , [0.032 < Dw < 0.5] 0.19 Dw Dw F = Fi + kδ 8 K cFDm ss = π Dw3 2r D C= m = m Dw Dw 8 FaC 3 N c G Dw GD k = 3w 8C N 3  G D  8 F C N c  kδ =  3 w  a = Fa  8C N  GDw  Fa = 100 lb = 0.10 kip Figure AF 15, assume K c = 1.2 8 K cFi Dm 8 K cFa Dm ss = + π Dw3 π Dw3 8 K cFaC s s = K c si + π Dw2 OD = Dm + Dw = 1.25 in 1.25 Dw = C +1 § 6.21, assume si = 18 ksi ssd = ss

δ=

Page 57 of 70


SECTION 4 - SPRINGS 70 8(1.2 )(0.1)C = (1.2 )(18) + 0.19 Dw π Dw2 70(C + 1) (1.25)0.19

8(1.2 )(0.1)C (C + 1) π (1.25)2

2

0.19

= 21.6 +

67.1(C + 1)

0.19

= 21.6 + 0.1956C (C + 1)

2

67.1(C + 1) − 0.1956C (C + 1) = 21.6 C = 6.7 1.25 1.25 = = 0.1623 in Dw = C + 1 6 .7 + 1 Table AT 15, use Dw = 0.1620 in , 8 W & M 0.19

2

Dm = CDw = (6.7 )(0.1620 ) = 1.085 in say Dm = 1.0 in D 1.0 C= m = = 6.17 Dw 0.1620 si = 17.7 ksi To check, Fig. AF 15, K c = 1.15 8(1.15)(0.10 )(6.17 ) ss = 1.15(17.7 ) + = 89.20 ksi π (0.1620)2 70 ssd = = 98.92 ksi > 89.20 ksi , o.k. (0.1620)0.19 Total length = Dw N c + 2(Dm + Dw ) 9.5 = (0.162)N c + 2(1.0 + 0.162) N c = 44.3 coils Free length = Dw N c = (0.1620)(44.3) = 7.18 in Summary of answer: Material, oil-tempered wire Dw = 0.1620 in , 8 W & M N c = 44.3 coils Free length = 7.18 in. 306.

Two helical tension springs are to be used in scales for weighing milk. The capacity of the scales is 30 lb., each spring carries 15 lb. with a deflection of 3 9/16 in. The springs are made of No. 14, W & M steel wire, outside diameter, 29/32 in. (a) how many coils should each spring have? (b) What is the maximum stress in the wire? What material should be used?

Page 58 of 70


SECTION 4 - SPRINGS Solution: kδ = 15 lb 9 δ = 3 in 16 (a) Table AT 15, No. 14 W &M

Dw = 0.0800 in 29 − 0.0800 = 0.82625 in 32 D 0.82625 C= m = = 10.328 Dw 0.0800 Dm = OD − Dw =

δ=

8(kδ )C 3 N c GDw

9 8(15)(10.328) N c = 16 11.5 ×106 (0.080) N c = 24.8 3

3

(

)

(b) F = Fi + kδ Fi =

π si Dw3

8Dm § 6.21, C = 10.328 si = 11,272 psi Fi =

π (11,272)(0.08)3

= 2.743 lb 8(0.82625) F = 2.743 + 15 = 17.743 lb Figure AF 15, K c = 1.09 8 K c FDm 8(1.09)(17.743)(0.82625) ss = = = 79,476 psi πDw3 π (0.080)3 s 79,476 s ys ≈ s = = 99,345 psi = 99.345 ksi 0 .8 0 .8 Table AT 17, use Hard drawn wire 70 70 s ys = 0.19 = = 113 ksi > 99.345 ksi Dw (0.080)0.19 307.

A tension spring for a gas-control lever is made of Dw = 0.078 in steel wire; inside diameter, 0.609 in.; number of coils, 55; free length including end loops, 5 9/16 in. When the spring is extended to a length of 6 5/16 in., it must exert a force 5 ½ lb.; it must extend to (a) the initial tension, (b) the stress in the spring caused by the initial tension (compare with the recommended

Page 59 of 70


SECTION 4 - SPRINGS maximum values), (c) the stress caused by the 5 ½-lb load, (d) the maximum stress. What material should be used? (e) What energy is absorbed from the point where the load is the initial tension until the spring’s length is 6 5/16 in.? (Data courtesy Worthington Corporation.) Solution: Dw = 0.078 in Dm − Dw = 0.609 in Dm = 0.609 + 0.078 = 0.687 in D 0.687 C= m = = 8.8 Dw 0.078 N c = 55 8FC 3 N c GDw 1 F = 5 lb 2 5 9 δ = 6 − 5 = 0.75 in 16 16 3 8(kδ )(8.8) (5) δ = 0.75 = 11.5 × 106 (0.078) kδ = 2.244 lb

δ=

(

)

(a) Fi = F − kδ = 5.5 − 2.244 = 3.256 lb 8F C 8(3.256)(8.8) (b) si = i 2 = = 12,000 psi π Dw π (0.078)2 § 6.21, C = 8.8 si = 13,300 psi > 12,000 psi , ok (c) F = 5.5 lb 8 K c FC ss = π Dw2 C = 8.8 Figure AF 15 K c = 1.1 8(1.1)(5.5)(8.8) ss = = 22,284 psi π (0.078)2 (d) maximum stress kδ 2.244 k= = = 2.992 lb in δ 0.75

Page 60 of 70


SECTION 4 - SPRINGS F = kδ ′ 5 9 δ ′ = 9 − 5 = 3.75 in 16 16 F = Fi + kδ ′ = 3.256 + (2.992)(3.75) = 14.476 lb 8 K c FC 8(1.1)(14.476)(8.8) ss = = = 58,651 psi π Dw2 π (0.078)2 Table AT 16 s 58,651 s ys ≈ s = = 73,300 psi = 73.3 ksi 0 .8 0 .8 Table AT 17, use Hard drawn wire 70 70 s ys = 0.19 = = 113.658 ksi > 73.3 ksi Dw (0.078)0.19

(e) U s =

1 2 1 2 kδ = (2.992)(0.75) = 0.8415 in − lb 2 2

TORSION SPRINGS 308.

A carbon-steel (ASTM A230) torsion spring is to resist a force of 55 lb. at a radius of 2 in.; the mean diameter is to be 2 ½ in. Compute (a) the diameter of the wire for average service, (b) the number of coils for a deflection of 180o under the given torque, (c) the energy the spring has absorbed when the force is 55 lb.

Solution: T = M = Fa F = 55 lb a = 2 in T = M = (55)(2) = 110 in − lb Dm = 2.5 in 182 Table AT 17, su = 0.1 ksi , [0.093 < Dw < 0.25] Dw Average service  182  117.936 117,936 ksi = psi sd = (1.6 )(0.405)su = 0.648 0.1  = 0.1 Dw0.1 Dw  Dw  KMc I For round wire, assume K c = K ci = 1.08 , Table AT 18 D c= w 2

(a) ss =

Page 61 of 70


SECTION 4 - SPRINGS I π Dw3 = c 32 (1.08)(110)(32) = 117,936 ss = π Dw3 Dw0.1 Dw = 0.2060 in < 0.25 in Table AT 15, use Dw = 0.2070 in , No. 5 W & M r D 2 = 9.66 > 9 , ok To check: = m = c Dw 0.2070 Table AT 18, K = 1.08 (1.08)(110)(32) = 136,430 psi ss = π (0.2070 )3 117,936 ssd = = 138,054 psi > 136,430 psi (0.2070)0.1 Therefore, use No. 5 W & M, Dw = 0.2070 in MπDm N c EI 6 E = 30× 10 psi

(b) θ =

I=

π Dw4

64 θ = 180o = π 64 MDm N c θ= EDw4 64(110 )(2 )N c π= (30 ×106 )(0.2070)4 N c = 12.29 1 1 (c) U s = Tθ = (110)(π ) = 172.8 in − lb 2 2

312.

A pivoted roller follower is held in contact with the cam by a torsion spring. The moment exerted by the spring varies from 20 lb-in to 50 lb-in. as the follower oscillates through 30o. The spring is made of AISI 6152 steel, OQT 1000 oF. What should be the value of Dw , Dm , and N c if the factor of safety is 1.75 based on the Soderberg line? Would this be a conservative or risky approach?

Solution: AISI 6152, OQT 1000 oF su = 184 ksi

Page 62 of 70


SECTION 4 - SPRINGS s y = 173 ksi

sn = 0.5su = 92 ksi 1 M m = (50 + 20) = 35 lb − in 2 1 M a = (50 − 20) = 15 lb − in 2 assume K = 1.08 32 KM m 32(1.08)(35) 385 sm = = = 3 psi Dw πDw3 πDw3 32 KM a 32(1.08)(15) 165 sa = = = 3 psi Dw πDw3 πDw3 1 s m sa = + N s y sn 1 385 165 = + 3 1.75 173,000 Dw 92,000 Dw3 Dw = 0.1916 in Table AT 15, use Dw = 0.1920 in , No. 6 W & M To solve for K 32(35)K sm = = 50,369 K psi π (0.1920)3 32(15)K sa = = 21,587 K psi π (0.1920)3 1 50,369 K 20,587 K = + 92,000 1.75 173,000 K = 1.0868 Table AT 18 K ci = K = 1.0868 r Dm = = 9.32 > 9 , ok c Dw Dm = 9.32(0.1920) = 1.7894 in 7 use Dm = 1 in = 1.875 in 8 ∆MπDm N c 64∆MDm N c ∆θ = = EI EDw4 30π 64(50 − 20)(1.875)N c = 180 (30 ×106 )(0.1920)4 N c = 5.93 Summary of answer: Dw = 0.1920 in , No. 6 W & M

Page 63 of 70


SECTION 4 - SPRINGS 7 Dm = 1 in 8 N c = 5.93 , N > 1.4 , therefore conservative.

FLAT AND LEAF SPRINGS 315.

A cantilever flat spring of uniform strength, Fig. 6.20, Text, is to absorb an energy impact of 500 ft-lb. Let the thickness of the steel, AISI 1095, OQT 900 o F, be ½ in. and let the maximum stress be half of the yield strength. (a) Find the width b of the spring at the widest point in terms of the length L . Determine values of b for lengths of 36 in., 48 in., 60 in., and 72 in. (b) Determine the deflection of the spring for each set of values found in (a).

Solution. Fig. 6/20

6 FL bh 2 6 FL3 δ= Ebh3 AISI 1095, OQT 900 oF, s y = 104 ksi , Table AT 9 sB =

s B = 0.5s y = 0.5(104) = 52 ksi = 52,000 psi 1 Fδ 2 s bh 2 F= B 6L  s bh 2  L3 s B L2  δ = 6 B = 3 Eh  6 L  Ebh

Us =

1  s bh 2  s B L2  1  s B2 bhL   =   U s =  B 2  6 L  Eh  12  E  U s = 500 ft − lb = 6000 in − lb

Page 64 of 70


SECTION 4 - SPRINGS

(52,000)2 b 1  L

1 2 12 30 × 106 bL = 1598 in 2 6000 =

b=

1598 in 2 L

L = 36 in , b =

1598 in 2 = 44.4 in 36 in

L = 48 in , b =

1598 in 2 = 33.3 in 48 in

L = 60 in , b =

1598 in 2 = 26.6 in 60 in

L = 72 in , b =

1598 in 2 = 22.2 in 72 in

(b) δ =

s B L2 Eh

2 52,000 )(36 ) ( L = 36 in , δ =

= 4.4928 in 1 30 × 10   2 2 52,000 )(48) ( = 7.9872 in L = 48 in , δ = 6 1 30 ×10   2 2 52,000)(60 ) ( L = 60 in , δ = = 12.48 in 6 1 30 ×10   2 2 52,000)(72 ) ( L = 72 in , δ = = 17.9712 in 6 1 30 × 10   2

(

317.

6

)

(

)

(

)

(

)

One of the carbon contacts on a circuit breaker is mounted on the free end of a phosphor-bronze beam ( µ = 0.35 ). This beam has the shape of the beam 9 1 1 shown in Fig. 6.24, Text, with b = 1 in. , b′ = in. , L = 4 in. , and h = in. 16 2 16 When the contacts are closed, the beam deflects ¾ in. Compute (a) the force on the contacts, (b) the maximum stress.

Solution:

Page 65 of 70


SECTION 4 - SPRINGS Figure 6.24 6 FL 3WL s= 2 = bh bh 2 K FL3 1 − µ 2 K WL3 1 − µ 2 = 1 δ= 1 3EI 6 EI 9   b′  16  = = 0.5625 b 1 Figure 6.25, K1 = 1.14

(

)

(

)

(a) Force on contacts = F K FL3 1 − µ 2 δ= 1 3EI E = 16× 106 psi (phosphor bronze)

(

)

bh3 12 3 4 K FL 1 − µ 2 δ= 1 Ebh 3 3 2 3 4(1.14 )F (4.5) 1 − (0.35) = 3 4 (16 ×106 )(1) 161    F = 8 lb

as a beam, I =

(

)

[

(b) s =

318.

]

6 FL 6(8)(4.5) = = 55,296 psi 2 bh 2 1 (1)   16 

A cantilever leaf spring 26 in. long is to support a load of 175 lb. The construction is similar to that shown in Fig. 6.22 (a), Text. The leaves are to be 2 in. wide, 3/16 in. thick; SAE 9255 steel, OQT 1000 oF; 107 cycles (§ 6.26). (a) How many leaves should be used if the surfaces are left as rolled? (b) The same as (a) except that the leaves are machined and the surfaces are not decarburized. (c) The same as (b), except that the surface is peened all over. (d) Which of these springs absorbs the most energy? Compute for each: (e) What are the load and deflection of the spring in (b) when the maximum stress is the standard-test yields strength?

Solution: Figure 6.22 (a) 6 FL sA = 2 bh

Page 66 of 70


SECTION 4 - SPRINGS 6 FL3 Ebh3 b = N1b′ F = 175 lb b′ = 2 in 3 h = in 16 L = 26 in § 6.26, SAE 9255, OQT 1000 oF su = 180 ksi s y = 160 ksi

δA =

3 in = 0.1875 in 16 sd = 83.75 ksi

t=

(a) As rolled, Figure AF 5 Surface factor = 0.275 sd = 0.275(83.75) = 23 ksi = 23,000 psi 6 FL sA = 2 bh 6(175)(26 ) 23,000 = 2 3 N 1 (2)   16  N1 = 16.88 say N1 = 17 (b) Machined, Figure AF 5 Surface factor = 0.75 sd = 0.75(83.75) = 62.8 ksi = 62,800 psi 6 FL sA = 2 bh 6(175)(26 ) 62,800 = 2 3 N 1 (2 )   16  N1 = 6.2 say N1 = 7 (c) Peened surface, (b)

Page 67 of 70


SECTION 4 - SPRINGS

sd = 1.25(62.8) = 78.5 ksi = 78,500 psi 6 FL sA = 2 bh 6(175)(26 ) 78,500 = 2 3 N 1 (2 )   16  N1 = 4.95 say N1 = 5 1 Fδ 2 F = 175 lb

(d) U s =

6 FL3 EN1b′h 3 For (a) N1 = 17

δ=

6(175)(26 )

3

δ=

= 2.745 in 3 3 30 × 10 (17 )(2 )   16  1 U s = (175)(2.745) = 240 in − lb 2 For (b) N1 = 7

(

6

)

6(175)(26 )

3

δ=

= 6.666 in 3 3   30 ×10 6 (7 )(2 )   16  1 U s = (175)(6.666 ) = 583 in − lb 2 For (c) N1 = 5

(

)

6(175)(26 )

3

δ=

= 9.332 in 3 3 30 × 10 (5)(2 )   16  1 U s = (175)(9.332 ) = 817 in − lb 2

(

6

)

answer – spring (c) (e) sd = s y = 160 ksi , N1 = 7 (b)

Page 68 of 70


SECTION 4 - SPRINGS 6 FL 6 FL = 2 bh N1b′h 2 6 F (26 ) 160,000 = 2 (7 )(2) 3   16  load F = 505 lb sd =

6(505)(26 )

3

δ=

3 30 × 10 (7 )(2 )   16 

( 319.

6

)

3

= 19.24 in

The rear spring of an automobile has 9 leaves, each with an average thickness of 0.242 in. and a width of 2 in.; material is SAE 9261, OQT 1000 oF. The length of the spring is 56 in. and the total weight on the spring is 1300 lb. Assume the spring to have the form shown in Fig. 6.22 (b), Text. Determine (a) the rate of the spring, (b) the maximum stress caused by the dead weight. (c) What approximate repeated maximum force (0 to Fmax ) would cause impending fatigue in 105 cycles, the number of applications of the maximum load expected during the ordinary life of a car? (If the leaves are cold rolled to induce a residual compressive stress on the surfaces, the endurance limit as su 2 should be conservative.)

Solution: Figure 6.22 (b) 3FL sA = 2bh 2 3FL3 δA = 8Ebh 3 F = 1300 lb h = 0.242 in N1 = 9 b′ = 2 in L = 56 in (a) Rate , k = k=

F

δA

=

(

8Ebh3 3L3

)

8 EN1b′h 3 8 30 × 10 6 (9 )(2 )(0.242 ) = = 116.21lb in 3 3L3 3(56 )

(b) s A =

3

3FL 3(1300 )(56 ) = = 103,590 psi 2 2 2 N1b′h 2(9 )(2 )(0.242 )

Page 69 of 70


SECTION 4 - SPRINGS (c) SAE 9261. OQT 1000 oF su = 192 ksi 192 sn = = 96 ksi 2 3FL sA = 2 N1b′h 2 3F (56 ) 96,000 = 2 2(9 )(2 )(0.242 ) F = 1200 lb 321.

The front spring of an automobile is made similar to Fig. 6.23, Text. The average thickness for each of the 6 leaves, 0.213 in.; material is SAE 9255, OQT 1000 oF. The load caused by the weight of the car is 775 lb. (a) What stress is caused by a force of twice the dead weight? (b) What load would stress the spring to the yield strength?

Solution: Figure 6.23 3WL 6 FL 3WL s= 2 = = 2 bh bh N1b′h 2 W = 775 lb , N1 = 6 , b′ = 2 in , h = 0.213 in 36 in = 18 in L= 2 (a) W = 2(775) = 1550 lb 3(1550 )(18) s= = 153,740 psi (6)(2)(0.213)2 (b) SAE 9255, OQT 1000 oF s y = 160 ksi 3WL 6 FL 3WL = = 2 2 bh bh N1b′h 2 3W (18) 160,000 = (6)(2)(0.213)2 W = 1613 lb s=

- end -

Page 70 of 70


SECTION 5 – COLUMNS DESIGN PROBLEMS 334.

A round steel rod made of structural steel, AISI C1020, as rolled, is to be used as a column, centrally loaded with 10 kips; N = 3 . Determine the diameter for (a) L = 25 in. , (b) L = 50 in. (c) The same as (a) and (b) except that the material is AISI 8640, OQT 1000 F. Is there any advantage in using this material rather than structural steel?

Solution: For AISI C1020,as rolled s y = 48 ksi F = 10 kips N =3 (a) Le = L = 25 in. Consider first J.B. Johnson 2   Le    sy    k Fc = NF = s y A1 −  2    4π E      2 πD A= 4 D k= 4 E = 30× 103 ksi 2         (48) 25 D      2 (3)(10) = (48) πD 1 − 2  4  3  4   4π 30 ×10    4   30 = 12πD 2 1 − 2 2   π D  48 30 = 12πD 2 −

(

π

D = 1.096 in 1 say D = 1 in = 1.0625 in 16

Page 1 of 18

         

)


SECTION 5 – COLUMNS Le 25 = = 94 < 120 ∴ o.k. k  1.0625     4  (b) Le = L = 50 in. Consider Euler’s Equation π 2 EA Fc = NF = 2  Le    k   πD 2   π 2 30 ×103  4   (3)(10) = 2    50     D   4 3 4 30 = 0.1875π D D = 1.507 in 1 say D = 1 in = 1.5 in 2 Le 50 = = 133 > 120 ∴ o.k. k  1.5     4 

(

)

(c) For AISI 8640, OQT 1000 F s y = 150 ksi 1

Le  2π 2 E  2 = k  s y 

(

)

1

Le  2π 2 30 × 103  2 =  = 62.83 150 k   For (a) Le = L = 25 in. Consider first J.B. Johnson 2   Le    sy    k Fc = NF = s y A1 −  2    4π E     

Page 2 of 18


SECTION 5 – COLUMNS 2         (150) 25 D      2 (3)(10) = (150) πD  1 − 2  4  3  4   4π 30 ×10    12.5   30 = 37.5πD 2 1 − 2 2   π D  468.75 30 = 37.5πD 2 −

(

         

)

π

D = 1.23 in say D = 1.25 in Le 25 = = 80 > 62.83 ∴ use Euler’s equation k  1.25     4  π 2 EA Fc = NF = 2  Le    k   πD 2   π 2 30 ×103  4   (3)(10 ) = 2    25     D   4  3 4 30 = 0.75π D D = 1.0657 in 1 say D = 1 in = 1.0625 in 16 Le 25 = = 94 > 62.83 ∴ ok k  1.0625     4  For (b) Le = L = 50 in. Consider Euler’s Equation π 2 EA Fc = NF = 2  Le    k 

(

Page 3 of 18

)


SECTION 5 – COLUMNS  πD 2    4 

π 2 (30 ×103 )

(3)(10 ) =

2

   50     D   4 3 4 30 = 0.1875π D D = 1.507 in 1 say D = 1 in = 1.5 in 2 Le 50 = = 133 > 62.83 ∴ o.k. k  1.5     4  There is no advantage. 335.

A hollow circular column, made of AISI C1020, structural steel, as rolled, is to support a load of 10,000 lb. Let L = 40 in , Di = 0.75Do , and N = 3 . Determine Do by (a) using either Euler’s or the parabolic equation; (b) using the straightline equation. (c) What factor of safety is given by the secant formula for the dimensions found in (a)?

Solution: For AISI C1020, as rolled s y = 48 ksi Le = L = 40 in F = 10,000 lb = 10 kips N =3 Di = 0.75Do

I A π (Do4 − Di4 ) 4 I= = π Do4 − (0.75Do ) = 0.033556 Do4 64 π (Do2 − Di2 ) π Do2 − (0.75Do )2 A= = = 0.343612 Do2 4 4

k=

[ [

k=

0.033556 Do4 = 0.3125Do 0.343612 Do2

Page 4 of 18

] ]


SECTION 5 – COLUMNS (a) Consider parabolic equation 2   Le    sy    k Fc = NF = s y A1 −  2    4π E          25  (48)  0.3125Do  (3)(10) = (48)(0.343612)Do2 1 −  2 4 3 4π 30 × 10     30 = 16.493376 Do2 − 10.9519 Do = 1.576 in 9 say Do = 1 in = 1.5625 in 16 Le 40 = = 82 < 120 ∴ o.k. k 0.3125(1.5625)

(

(b) Straight-line equation F L = 16,000 − 70 A k   10,000 40  = 16,000 − 70 2 0.343612 Do  0.3125 Do  10,000 = 5498 Do2 − 3078 Do

Do = 1.6574 in 5 say Do = 1 in = 1.625 in 8 Le 40 = = 78.8 < 120 ∴ o.k. k 0.3125(1.625) (c) Secant formula NF  ec  Le NF sy = 1 + sec A  k 2  2k EA Do = 1.5625 in k = 0.3125 Do = 0.4883 in A = 0.343612 Do2 = 0.8389 in 2

Page 5 of 18

   

)

    

2

         


SECTION 5 – COLUMNS ec = 0.25 , (i7.8) k2

48 =

 N (10 )  40 1 + 0.25 sec  0.8389   2(0.4883)

[

(

48 = 11.92 N 1 + 0.25 sec 0.81645 N N = 2.289

336.

)]

A column is to be built up of ½-in., AISI C1020, rolled-steel plates, into a square box-section. It is 6 ft long and centrally loaded to 80,000 lb. (a) Determine the size of section for N = 2.74 . (b) Compute N from the secant formula for the size found and compare with 2.74.

Solution: For AISI C1020, rolled-steel plate s y = 48 ksi b 4 (b − 1) b 4 − (b − 1) − = 12 12 12 2 2 A = b − (b − 1) 4

4

I=

I b 4 − (b − 1) = 2 A 12 b 2 − (b − 1) 4

k=

  10 N (30 ×103 )(0.8389)  

[

]

Le = L = 6 ft = 72 in F = 80,000 lb = 80 kips (a) N = 2.74 Consider J.B. Johnson 2   Le    sy    k NF = s y A1 −  2    4π E      2   72   (48)     (2.74)(80) = (48)A 1 − 2  k  3   4π (30 ×10 )     10.085 A 219.2 = 48 A − k2

Page 6 of 18


SECTION 5 – COLUMNS try b = 3.23 in 4 4 ( 3.23) − (3.23 − 1) k= 2 2 = 1.1331 in 12[(3.23) − (3.23 − 1) ] 2 2 2 A = b 2 − (b − 1) = (3.23) − (3.23 − 1) = 5.46 in 2 10.085(5.46 ) 219.2 = 48(5.46 ) − = 219.2 ok (1.1331)2

Therefore use b = 3.23 in Le 72 = = 63.54 < 120 ∴ o.k. k 1.1331 1 b = 3.23 in or b = 3 in 4 (b) s y =

NF  ec  Le 1 + sec A  k 2  2k

NF EA

   

ec = 0.25 , (i7.8) k2

48 =

 72 N (80)  1 + 0.25 sec  5.46   2(1.1331)

[

(

  80 N (30 ×103 )(5.46) 

48 = 14.652 N 1 + 0.25 sec 0.70214 N N = 2.2 < 2.74

337.

)]

A column is to be made of ½-in structural steel plates (AISI 1020, as rolled), welded into an I-section as shown in Table AT 1 with G = H . The column, 15 ft long, is to support a load of 125 kips. (a) Determine the cross-sectional dimensions from the straight-line equation. (Using either Johnson’s or Euler’s equation, compute the equivalent stress and the factor of safety. (c) Compute N from the secant formula.

Solution: For AISI C1020, as rolled s y = 48 ksi Le = L = 15 ft = 180 in F = 125 kips

Page 7 of 18


SECTION 5 – COLUMNS Table AT 1.

G=H A = GH − gh = H 2 − (H − 0.5)(H − 1) = H 2 − H 2 − 1.5H + 0.5 = 1.5H − 0.5 = 0.5(3H − 1)

(

k=

)

3 1  GH 3 − gh3  1  H 4 − (H − 0.5)(H − 1)    =  = 12  GH − gh  12  0.5(3H − 1) 

H 4 − (H − 0.5)(H − 1) 6(3H − 1)

(a) Straight-line equation L  F = 16,000 A1 − 0.0044  k 

(180)  125,000 = 16,000 A1 − 0.0044 k    0.792  7.8125 = A1 −  k   use H = 7.37 in

(7.37 )4 − (7.37 − 0.5)(7.37 − 1)3 6(3(7.37 ) − 1) A = 0.5[3(7.37 ) − 1] = 10.555 in

k=

= 3.04527 in

0.792   7.8125 ≈ 10.5551 −  = 7.81  3.04527  Therefore use H = 7.37 in 3 Or H = 7 in = 7.375 in 8 L 180 (b) Consider J.B. Johnson, e = = 59 < 120 k 3.04527 s se = y N F 125 se = = = 13.8 ksi 2 2    Le    180      sy     (48) k   3.04527       A 1− 10.555 1 −   4π 2 E  4π 2 30 ×103         

(

Page 8 of 18

)

3


SECTION 5 – COLUMNS

N=

sy 48 = = 3.48 se 13.8

(c) From secant formula NF  ec  Le NF  sy = 1 + sec  A  k 2  2k EA  ec = 0.25 , (i7.8) k2 48 =

 180 N (125)  1 + 0.25 sec  10.555   2(3.04527 )

[

(

48 = 11.843 N 1 + 0.25 sec 0.5872 N N = 2.8

338.

  125 N 3 (30 ×10 )(10.555) 

)]

The link shown is to be designed for N = 2.5 to support an axial compressive load that varies from 0 to 15 kips; L = 20 in ; Material AISI 1030, as rolled. (a) Determine the diameter considering buckling only. (b) Determine the diameter considering varying stresses and using the Soderberg line (perhaps too conservative). Estimate an appropriate strength-reduction factor (see Fig. AF 6). (c) Keeping in mind that the stress is always compressive, do you think that the answer from (a) will do? Discuss.

Problem 338. Solution: For AISI C1030, as rolled s y = 51 ksi

su = 80 ksi 1

(

)

1

Le  2π 2 E  2  2π 2 30 ×103  2 = =  = 108 51 k  s y    L = 20 in N = 2.5 (a) F = 15 kips Consider J.B. Johnson

Page 9 of 18


SECTION 5 – COLUMNS 2   Le   s    y k NF = s y A1 −  2    4π E      D k= 4 πD 2 A= 4 Le = L = 20 in 2   20    (51) D     2 (2.5)(15) = (51) πD 1 − 2  4  3   4   4π (30 ×10 )       2.72   37.5 = 12.75π D 2 1 − 2 2   π D  34.68 37.5 = 12.75π D 2 −

π

D = 1.101 in 3 say D = 1 in = 1.1875 in 16 Le 20 = = 68 < 108 ∴ o.k. k 1.1875 4

(b) Variable stresses sn = 0.5su = 0.5(80 ) = 40 ksi Size factor = 0.85 sn = 0.85(40 ) = 34 ksi K f = 2.8 (Figure AF 6) 1 sm K f sa = + sn N sy F = 0 to 15 kips Fm = Fa = 7.5 kips sem = sea (2.8)sem 1 s = em + 2.5 51 34 sem = 3.923 ksi

Page 10 of 18


SECTION 5 – COLUMNS 2   Le   s    em  k Fm = sem A1 −  2    4π E      2   20    (51) D   2   πD    1 − 2  4  3  7.5 = (3.923)  4   4π (30 ×10 )       2.72   7.5 = 0.98π D 2 1 − 2 2   π D  2.67 7.5 = 0.98π D 2 −

π

D = 1.65 in 5 say D = 1 in = 1.625 in 8 Le 20 = = 49 < 108 ∴ o.k. k 1.625 4

(c) The answer in (a) will not do because it is lower than (b) 339.

The connecting link for a machine (see figure) is subjected to a load that varies fro + 450 (tension) to –250 lb. The cross section is to have the proportions G = 0.4 H , t = 0.1H , fillet radius r ≈ 0.05 H ; L = 10 in ; material, AISI C1020, as rolled. (a) Considering buckling only, determine the dimensions for a design factor of 2.5. (b) For the dimension found compute the factor of safety from the Soderberg criterion.

Problems 339, 340 Solution: For AISI C1030, as rolled s y = 48 ksi

su = 65 ksi

Page 11 of 18


SECTION 5 – COLUMNS Table AT 1

G = 0.4 H t = 0.1H r ≈ 0.05 H A = GH − gh g = G − t = 0.4 H − 0.1H = 0.3H h = H − 2(0.1H ) = 0.8 H

A = (0.4 H )(H ) − (0.3H )(0.8H ) = 0.16 H 2 k=

3 3 1  GH 3 − gh3  1  (0.4 H )(H ) − (0.3H )(0.8 H )    =   = 0.35824 H 12  GH − gh  12  0.16 H 2 

(

)

(a) Consider J.B. Johnson 2   Le   s    y k NF = s y A1 −  2    4π E      F = 350 lb = 0.35 kip Le = 10 in 2     10    48 2 ( 0.35824 H )   2   (2.5)(0.35) = (48) 0.16 H 1 −  4π 2 30 × 103     0.875 = 7.68 H 2 − 0.2425 H = 0.3815 in Le 10 = = 73 < 120 ∴ ok k (0.35824)(0.3815) 15 in = 0.46875 in say H = 32 3 G = 0.4 H = 0.1875 in = in 16 3 t = 0.1H = 0.046875 in = in 64

(

Page 12 of 18

)

(

)


SECTION 5 – COLUMNS (b) with H = 0.46875 in A = 0.16(0.46875) = 0.0352 in 2 k = 0.35824(0.46875) = 0.1679 in − 350 Fmin 0.0352 A = smin = s e = = −11,600 psi = −11.6 ksi 2 2     Le   10      s y     48 1 −  k   1 −  0.1679    4π 2 E   4π 2 30 × 103          F + 450 smax = max = = +12,800 psi = +12.8 psi A 0.0352 1 sm = (12.8 − 11.6 ) = 0.6 ksi 2 1 sa = (12.8 + 11.6 ) = 12.1 ksi 2 su = 0.5sn = 0.5(65) = 32.5 ksi Size factor = 0.85 su = 0.85(32.5) ksi = 27.62 ksi 2

(

)

Figure AF 9, r = 0.05H = 0.05(0.46875) = 0.023 h = 1.5H = 15.(0.46875) = 0.7031 in d = H = 0.4688 in r 0.05 H = = 0.05 d H h 1.5 H = = 1.5 d H K t = 2.65 1 1 q= = = 0.70 0.01 0.01 1+ 1+ r 0.023 K f = 0.70(2.65 − 1) + 1 = 2.2 1 sm K f sq = + sn N sy 1 0.6 (2.2 )(12.1) = + N 48 27.62 N = 1.024

CHECK PROBLEMS 341.

The link shown is subjected to an axial compressive load of 15 kips. Made of AISI C1030, as rolled, it has sectional length of 20 in. Assume a loose fit with the

Page 13 of 18


SECTION 5 – COLUMNS pins. What is (a) the critical load for this column, (b) the design factor, (c) the equivalent stress under a load of 15 kips? What material does the secant formula indicate as satisfactory for the foregoing load, when (e) ec k 2 = 0.25 , (f) L e= e . 400

Problem 341, 342 Solution: For AISI C1030, as rolled s y = 51 ksi b = 0.75 in h = 1.75 in

A = bh = (0.75)(1.75) = 1.3125 in 2 For loose fit bh3 I= 12 bh 3 h I 1.75 k= = = = = 0.5052 in A 12bh 12 12 Le 20 = 39.6 < 108 for AISI C1030, as rolled = k 0.5052 use J.B. Johnson equation 2 2    Le    20      sy     51 k   0.5052       (a) Fc = s y A 1 − = (51)(1.3125) 1 − = 62.42 kips   4π 2 (30,000)  4π 2 E          (b) Fc = NF F 62.42 N= c = = 4.16 F 15 s 51 (c) se = y = = 12.26 ksi N 4.16 F 15 = 11.43 ksi (d) Actual s = = A 1.3125

Page 14 of 18


SECTION 5 – COLUMNS

Secant Formula NF  ec  Le NF  sy = 1 + sec  A  k 2  2k EA  ec (e) 2 = 0.25 k    62.42  20 62.42 sy = 1 + 0.25 sec    = 64.4 ksi 3 1.3125   2(0.5052) 30 × 10 (1.3125)   use AISI C1020, cold drawn, s y = 66 ksi

(

)

Le 20 = = 0.05 in 400 400 h 1.75 c= = = 0.875 in 2 2 ec (0.05)(0.875) = = 0.1714 k2 (0.5052)2

(f) e =

sy =

 62.42  20 1 + 0.1714 sec  1.3125   2(0.5052 )

  62.42   = 59.12 ksi 3 30 × 10 (1.3125)  

(

)

use AISI C1045, cold drawn, s y = 59 ksi 343.

A schedule-40, 4-in. pipe is used as a column. Some of its properties are: Do = 4.5 in , Di = 4.026 in , I = 3.174 sq.in. , L = 15 ft ; material equivalent to AISI C1015, as rolled. The total load to be carried is 200 kips. (a) What minimum number of these columns should be used if a design factor of 2.5 is desired and the load evenly distributed among them? For the approximately fixed ends, use Le = 0.65L as recommended by AISC. (b) What is the equivalent stress in the column?

Solution: For AISI C1015, as rolled s y = 45.5 ksi 1

(

)

1

Le  2π 2 E  2  2π 2 30 × 103  2 = =  = 114 k  s y  45.5   L = 15 ft = 180 in Le = 0.65L = 0.65(180) = 117 in Le 117 = = 77.5 < 114 k 1.509

Page 15 of 18


SECTION 5 – COLUMNS Use J.B. Johnson equation 2   Le   sy    2 s y A  k   (45.5)(3.174)  45.5(77.5)   (a) F = 1− = 1 − 2  = 44.4 kips N  4π 2 E  2.5  4π (30,000)      No. of columns 200 = = 4.5 say 5 columns 44.4 F A (b) se = 2   Le   s    y 1 −  k    4π 2 E      200 F= = 40 kips 5 40 3.174 se = = 16.4 ksi 2   77.5      45.5  k   1 −  4π 2 (30,000)     

344.

A generally loaded column is a 10-in. x 49 lb., wide-flange I-beam whose properties are (see figure); k x = 4.35 in , k y = 2.54 in , area A = 14.4 sq.in. , I x = 272.9 in 4 , I y = 93.0 in 4 ; length L = 30 ft , material AISI 1022, as rolled. Let

the ends be a “little” fixed with Le = 0.8L and determine the critical load (a) according to the Johndon or the Euler equation; (b) according to the secant formula if ec k 2 is assumed to be 0.25.

Page 16 of 18


SECTION 5 – COLUMNS Solution: For AISI C1022, as rolled s y = 52 ksi 1

1

Le  2π 2 E  2  2π 2 (30 ×103 ) 2 = =  = 107 52 k  s y    (a) k = 2.54 in I = 93.0 in 4 Le = 0.8(30)(12) = 288 in Le 288 = 113.4 > 107 = k 2.54 Use Euler’s Equation π 2 EA π 2 (30,000)(14.4) Fc = = = 332 kips 2 (113.4)2  Le    k  (b) Secant formula NF  ec  Le NF sy = 1 + sec A  k 2  2k EA 113.4 F  52 = c 1 + 0.25 sec  14.4   2 F 52 = c 1 + 0.25 sec 0.0863 14.4 Fc = 273 kips

{

348.

[

      Fc 3 (30 ×10 )(14.4)   Fc

]}

A 4 x 3 x ½-in. angle is used as a flat-ended column, 5 ft. long, with the resultant load passing through the centroid G (see figure); k x = 1.25 in , k y = 0.86 in , ku = 1.37 in , kv = 0.64 in , A = 3.25 sq.in. Find the safe load if N = 2.8 and the material is (a) structural steel, (b) magnesium alloy AZ 91C (i7.12.\, Text), (c) magnesium alloy AZ 80A, (d) magnesium alloy AZ 80A as before, but use the Johnson formula and compare.

Page 17 of 18


SECTION 5 – COLUMNS Solution: L (5)(12 ) Le = = = 30 in 2 2 k = k min = 0.64 in Le 30 = = 46.875 k 0.64 (a) Structural steel, s y = 48 ksi Le = 46.875 < 120 k use J.B. Johnson 2   Le   sy    2 s y A  k   (48)(3.25)  48(46.875)   F= 1− = 1 − 2  = 50.75 kips N  4π 2 E  2.8  4π (30,000 )     (b) magnesium alloy AZ 91C NF C = 2 A  Le  C  k 1+   6 64.4 × 10 C = 57,000 57,000 (2.8)(F ) = psi 2 3.25 57,000(46.875) 1+ 64.4 ×106 F = 22,467 lb = 22.467 kips (c) magnesium alloy AZ 80A C = 82,900 82,900 (2.8)(F ) = psi 2 3.25 82,900(46.875) 1+ 64.4 × 106 F = 25,134 lb = 25.134 kips

(d) By J.B. Johnson For magnesium alloy AZ 80A, s y = 36 ksi 2   Le   s   y s y A  (36)(3.25) 1 − 36(46.875)2  = 39 kips > 25.134 kips k F= 1−  2   =   2 N  4π E  2.8  4π (30,000)      - end -

Page 18 of 18


SECTION 7 – SHAFT DESIGN 471.

A short stub shaft, made of SAE 1035, as rolled, receivers 30 hp at 300 rpm via a 12-in. spur gear, the power being delivered to another shaft through a flexible coupling. The gear is keyed (profile keyway) midway between the bearings. The pressure angle of the gear teeth φ = 20 o ; N = 1.5 based on the octahedral shear stress theory with varying stresses. (a) Neglecting the radial component R of the tooth load W , determine the shaft diameter. (b) Considering both the tangential and the radial components, compute the shaft diameters. (c) Is the difference in the results of the parts (a) and (b) enough to change your choice of the shaft size?

Problem 471. Solution: For SAE 1035, as rolled s y = 55 ksi su = 85 ksi sn = 0.5su = 0.5(85) = 42.5 ksi A = W cos φ 63,000hp 63,000(30 ) T= = = 6300 in − lb n 300 AD T= 2 A(12 ) 6300 = 2 A = 1050 lb A = W cos φ 1050 = W cos 20 W = 1118 lb Shear stress 16T 16(6300) ss = = π d3 π d3 100,800 ss = sms = π d3 sas = 0

Page 1 of 76


SECTION 7 – SHAFT DESIGN bending stress

From Table AT 2 FL M= 4 (a) Negligible R : AL (1050)(16 ) M= = = 4200 in − lb 4 4 32M 32(4200) 134,400 s= = = π d3 π d3 π d3 sm = 0 134,400 sa = s = π d3 K f sa s se = n s m + sy SF For profile keyway K f = 2.0 K fs = 1.6 SF = 0.85 K s (2.0)(134,400) = 100,661 se = f a = SF (0.85) π d 3 d3 K fs sas s ses = ns sms + s ys SF

(

)

1 sns sn 42.5 = = = 55 1.294 s ys s y sns  1  100,800  24,796  = sms =   3 s ys d3  1.294  π d  Octahedral-shear theory ses =

1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     2 1  100,661   24,796 =   +  3 1.5  42,500d   0.577 42,500d 3 d = 1.569 in

(

Page 2 of 76

1

)

  

2

2  


SECTION 7 – SHAFT DESIGN

use d = 1

11 in 16

(b) Considering both radial and tangential component. WL (1118)(16 ) M= = = 4472 in − lb 4 4 32M 32(4472) 143,104 s= = = π d3 π d3 π d3 sm = 0 143,104 sa = s = π d3 K s (2.0)(143,104) = 107,180 se = f a = SF (0.85)(π d 3 ) d3 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     2 1  107,180   24,796 =   +  3 1.5  42,500d   0.577 42,500d 3 d = 1.597 in 11 use d = 1 in 16

(

1

)

  

2

2  

(c) The difference in the results of the parts (a) and (b) is not enough to change the choice of the shaft size. 472.

A cold-finished shaft, AISI 1141, is to transmit power that varies from 200 to 100 and back to 200 hp in each revolution at a speed of 600 rpm. The power is received by a 20-in. spur gear A and delivered by a 10-in. spur gear C. The tangential forces have each been converted into a force ( A and C shown) and a couple (not shown). The radial component R of the tooth load is to be ignored in the initial design. Let 2 and, considering varying stresses with the maximum shear theory, compute the shaft diameter.

Problems 472 – 474

Page 3 of 76


SECTION 7 – SHAFT DESIGN Solution: For AISI 1141, cold-finished s y = 90 ksi sn = 50 ksi sn 1 = s y 1.8 SF = 0.85 63,000hp T= n 63,000(200 ) Tmax = = 21,000 in − lb 600 63,000(100 ) Tmin = = 10,500 in − lb 600 1 1 Tm = (Tmax + Tmin ) = (21,000 + 10,500) = 15,750 in − lb 2 2 1 1 Ta = (Tmax − Tmin ) = (21,000 − 10,500 ) = 5,250 in − lb 2 2 16T ss = πd3 16(15,750 ) 252,000 sms = = π d3 π d3 16(5250) 24,000 sas = = π d3 π d3 K fs sas s ses = ns sms + s ys SF For profile keyway K f = 2 .0 K fs = 1.6

1 sns sn = = s ys s y 1.8  1  252,000  (1.6 )(84,000 ) 94,894  + ses =   = 3 0.85π d 3 d3  1.8  π d  Bending stress, negligible radial load T = 21,000 in − lb at 200 hp

For A:  20  A  = T  2  A(10 ) = 21,000

Page 4 of 76


SECTION 7 – SHAFT DESIGN A = 2100 lb at 200 hp For C:  10  C  = T 2 C (5) = 21,000 C = 4200 lb at 200 hp

[∑ M

]

= 0 A(10 ) + D(25) = C (15) at 200 hp (2100)(10) + D(25) = (4200)(15) D = 1680 lb

[∑ F

B

]

=0 A+C = B + D at 200 hp 2100 + 4200 = B + 1680 B = 4620 lb At 200 hp: A = 2100 lb , B = 4620 lb , C = 4200 lb , D = 1680 lb Shear Diagram V

Maximum moment at B M = (2100)(10 ) = 21,000 in − lb 32M 32(21,000) 672,000 s= = = π d3 π d3 π d3 sm = 0 672,000 sa = s = π d3 K f sa (2.0)(672,000) = 503,304 s se = n s m + = 0+ sy SF 0.85π d 3 d3

Page 5 of 76


SECTION 7 – SHAFT DESIGN 94,894 d3 Maximum Shear Theory ses =

1

2 2 2 1  se   ses     =   +  N  sn   0.5sn     2 1  503,304   94,894 =   +  3 2  50,000d   0.5 50,000d 3 d = 2.78 in 3 use d = 2 in 4

(

475.

1

)

  

2

2  

A shaft S, of cold-drawn AISI 1137, is to transmit power received from shaft W, which turns at 2000 rpm, through the 5-in. gear E and 15-in. gear A. The power is delivered by the 10-in. gear C to gear G, and it varies from 10 hp to 100 hp and back to 10 hp during each revolution of S. The design is to account for the varying stresses, with calculations based on the octahedral shear stress theory. Let N = 1.8 and compute the shaft diameter, using only the tangential driving loads for the first design.

Problem 475 – 477 Solution. For AISI 1137, cold drawn s y = 93 ksi su = 103 ksi sn = 0.5su = 0.5(103) = 51.5 ksi 1 sn sns 51.5 = = = 93 1.806 s y s ys 63,000hp T= n 5 in. E n= (2000 rpm) = 667 rpm 15 in. A

Page 6 of 76


SECTION 7 – SHAFT DESIGN 63,000(100) = 9450 in − lb 667 63,000(10 ) Tmin = = 945 in − lb 667 1 1 Tm = (Tmax + Tmin ) = (9450 + 945) = 5197.5 in − lb 2 2 1 1 Ta = (Tmax − Tmin ) = (9450 − 945) = 4252.5 in − lb 2 2 16T ss = πd3 16(5197.5) 83,160 sms = = π d3 π d3 16(4252.5) 68,040 sas = = π d3 π d3 K fs sas s ses = ns sms + s ys SF For profile keyway K f = 2.0 Tmax =

K fs = 1.6 SF = 0.85  1  83,160  (1.6)(68,040) 55,425 + ses =  =  3  0.85π d 3 d3  1.806  π d  Bending stress, using only tangential loads

For 100 hp: T = 9450 in − lb  15  A  = T 2 A(7.5) = 9450 A = 1260 lb For C:  10  C  = T 2

Page 7 of 76


SECTION 7 – SHAFT DESIGN C (5) = 9450 C = 1890 lb

[∑ M

B

]

= 0 6 A + 20 D = 14C

6(1260 ) + 20 D = 14(1890) D = 945 lb

[∑ F

]

=0 A+C = B + D 1260 + 1890 = B + 945 B = 2205 lb Shear diagram V

Maximum moment at B M = (1260)(6) = 7560 in − lb 32M 32(7560 ) 241,920 s= = = π d3 π d3 π d3 sm = 0 241,920 sa = s = π d3 K f sa (2.0)(241,920) 181,189 s se = n s m + = = sy SF 0.85π d 3 d3 55,425 d3 Octahedral Shear Theory ses =

1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     2 1  181,189   55,425 =   +  3 2  51,500d   0.577 51,500d 3

(

d = 1.997 in use d = 2 in

Page 8 of 76

1

)

  

2

2  



A shaft made of AISI 1137, cold rolled, for a forage harvester is shown. Power is supplied to the shaft by a vertical flat belt on the pulley A. At B, the roller chain to the cutter exerts a force vertically upwards, and the V-belt to the blower at C exerts a force vertically upwards. At maximum operating conditions, the flat belt supplies 35 hp at 425 rpm, of which 25 hp is delivered to the cutter and 10 hp to the blower. The two sections of the shaft are joined by a flexible coupling at D and the various wheels are keyed (sled-runner keyway) to the shafts. Allowing for the varying stresses on the basis of the von Mises-Hencky theory of failure, decide upon the diameters of the shafts. Choose a design factor that would include an allowance for rough loading.

Problem 478. Solution: For AISI 1137, cold rolled s y = 93 ksi su = 103 ksi sn = 0.5su = 0.5(103) = 51.5 ksi 1 sn sns 51.5 = = = 93 1.806 s y s ys Pulley, 63,000hp 63,000(35) TA = = = 5188 in − lb n 425 For flat-belt  2T  4(5188) FA = F1 + F2 = 2(F2 − F1 ) = 2 A  = = 692 lb 30  DA  Sprocket, 63,000hp 63,000(25) TB = = = 3706 in − lb n 425 For chain, 2T 2(3706) FB = B = = 741 lb DB 10 Sheave, 63,000hp 63,000(10 ) TC = = = 1482 in − lb n 425

Page 9 of 76


SECTION 7 – SHAFT DESIGN For V-belt,  2T FC = F1 + F2 = 1.5(F2 − F1 ) = 1.5 C  DC Consider shaft ABD.

35 hp Shaft ABD

[∑ M

D'

=0

]

(6 + 8 + 4)FA = (8 + 4)A'+4FB 18(692) = 12 A'+4(741) A' = 791 lb

[∑ F

V

=0

]

FA + D′ = FB + A′ 692 + D′ = 741 + 791 D′ = 840 lb

Shear Diagram

Maximum M at A’. M = (6)(692) = 4152 in − lb. 32M 32(4152) 132,864 s= = = π d3 π d3 π d3 sm = 0 sa = s =

132,864 π d3

Page 10 of 76

 3(1482 )  = = 445 lb 10 


SECTION 7 – SHAFT DESIGN K f sa sn sm + sy SF For sled-runner keyway (Table AT 13) K f = 1.6 se =

K fs = 1.6 SF = 0.85 K f sa (1.60)(132,864) = 79,610 s se = n s m + = 0+ sy SF 0.85π d 3 d3

at A’ T = TA = 5188 in − lb 16T 16(5188) 83,008 ss = = = π d3 π d3 π d3 sms = ss sas = 0 K fs sas s ses = ns sms + s ys SF  1  83,000  14,630 = ses =   3  d3  1.806  π d  Choose a design factor of 2.0 N = 2.0 von Mises-Hencky theory of failure (Octahedral shear theory) 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     2 1  79,610   14,630 =   +  3 2  51,500d   0.577 51,500d 3 d = 1.48 in 1 use d = 1 in 2

(

1

)

  

2

2  

Consider shaft D-C 63,000hp 63,000(10 ) TC = = = 1482 in − lb n 425 For V-belt,  2T  3(1482 ) FC = F1 + F2 = 1.5(F2 − F1 ) = 1.5 C  = = 445 lb 10  DC 

Page 11 of 76



[∑ M

C'

=0

]

8 D′′ = 3FC 8 D′′ = 3(445) D′′ = 167 lb

[∑ F

V

=0

]

C ′ = D′′ + FC C ′ = 167 + 445 C ′ = 612 lb Shear Diagram

M = (167 )(8) = 1336 in − lb 32M 32(1336) 42,752 s= = = π d3 π d3 π d3 sm = 0 , sa = s K f sa (1.60)(42,752) = 25,616 s se = n s m + = 0+ sy SF 0.85π d 3 d3 at C’, TC = 1482 in − lb 16T 16(1482 ) 23,712 ss = = = π d3 π d3 π d3 sms = ss sas = 0 K fs sas s ses = ns sms + s ys SF 4180  1  23,712  +0 = 3 ses =   3  d  1.806  π d  1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     2 1  25,616   4180 =   +  3 2  51,500d   0.577 51,500d 3

(

Page 12 of 76

1

)

  

2

2  


SECTION 7 – SHAFT DESIGN d = 1.011 in use d = 1 in

479.

A shaft for a punch press is supported by bearings D and E (with L = 24 in.) and receives 25 hp while rotating at 250 rpm, from a flat-belt drive on a 44in. pulley at B, the belt being at 45o with the vertical. An 8-in. gear at A delivers the power horizontally to the right for punching operation. A 1500-lb flywheel at C has a radius of gyration of 18 in. During punching, the shaft slows and energy for punching comes from the loss of kinetic energy of the flywheel in addition to the 25 hp constantly received via the belt. A reasonable assumption for design purposes would be that the power to A doubles during punching, 25 hp from the belt, 25 hp from the flywheel. The phase relations are such that a particular point in the section where the maximum moment occurs is subjected to alternating tension and compression. Sled-runner keyways are used for A, B, and C; material is colddrawn AISI 1137, use a design factor of N = 2.5 with the octahedral shear theory and account for the varying stresses. Determine the shaft diameters.

Problems 479-480 Solution: Flat-Belt Drive (B) 63,000hp 63,000(25) TB = = = 6300 in − lb n 250  2T  4(6300) FB = F1 + F2 = 2(F1 − F2 ) = 2 B  = = 573 lb 44  DB  Gear A, Doubled hp 63,000hp 63,000(25 + 25) TA = = = 12,600 in − lb n 250 2T 2(12,600) FA = A = = 3150 lb DA 8 Loading:

Page 13 of 76



Vertical:

BV = FB cos 45 = 573 cos 45 = 405 lb

[∑ M

D

=0

]

6(1500) + 8BV = 24 EV 6(1500) + 8(405) = 24 EV EV = 510 lb

[∑ F

V

=0

]

1500 + EV = DV + BV 1500 + 510 = DV + 405 DV = 1605 lb Shear Diagram

Page 14 of 76



M DV = (6)(1500 ) = 9000 in − lb M BV = (16)(510 ) = 8160 in − lb M AV = (5)(510 ) = 2550 in − lb

Horizontal:

Bh = FB sin 45 = 573 sin 45 = 405 lb

[∑ M

D

=0

]

8 Bh + 24 Eh = 19 FA 8(405) + 24 Eh = 19(3150) Eh = 2359 lb

[∑ F

h

=0

]

Dh + Bh + Eh = FA Dh + 405 + 2359 = 3150 Dh = 386 lb Shear Diagram

Page 15 of 76



M Dh = 0 in − lb M Bh = (8)(386 ) = 3088 in − lb M Ah = (5)(2359 ) = 11,795 in − lb MA = MB =

(M ) + (M ) (M ) + (M ) 2

Ah

2

Bh

2

=

(11,795)2 + (2550)2

2

=

(3088)2 + (8160)2

AV

BV

= 12,068 in − lb

= 8725 in − lb

M D = 9000 in − lb Therefore M max = 12,068 in − lb 32 M 32(12,068) 386,176 = = s= π d3 π d3 π d3 Maximum moment subjected to alternating tension and compression sm = 0 386,176 sa = s = π d3 K f sa sn sm + sy SF For AISI 1137, cold-drawn, s y = 93 ksi se =

su = 103 ksi sn = 0.5su = 0.5(103) = 51.5 ksi For sled-runner keyway (Table AT 13) K f = 1.6 K fs = 1.6 SF = 0.85 (1.60)(386,176 ) = 231,386 se = 0 + 0.85π d 3 d3

Page 16 of 76



At A, 50 hp max. and 25 hp min. 50 hp 63,000hp 63,000(25 + 25) TA = = 12,600 in − lb = n 250 2T 2(12,600 ) FA = A = = 3150 lb DA 8 16T 16(12,600 ) 201,600 = = ss max = π d3 π d3 π d3 25 hp 63,000hp 63,000(25) TA = = = 6,300 in − lb n 250 2T 2(6,300 ) FA = A = = 1575 lb DA 8 16T 16(6,300 ) 100,800 = = ss min = π d3 π d3 π d3 sms =

 1 ,800  151,200  = (ss max + ss min ) = 1  201,600 + 100 3 2 2 πd π d3 

sas =

 1 ,800  50,400  = (ss max − ss min ) = 1  201,600 − 100 3 πd π d3 2 2 

ses =

K fs sas sns sms + s ys SF

 1  151,200  (1.6)(50,400) 56,848  + ses =  =  3 0.85π d 3 d3  1.806  π d  1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     2 1  231,386   56,848 =   +  3 2  51,500d   0.577 51,500d 3 d = 2.14 in 3 say d = 2 in 16

(

THRUST LOADS

Page 17 of 76

1

)

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2

2  



A cold-drawn monel propeller shaft for a launch is to transmit 400 hp at 1500 rpm without being subjected to a significant bending moment; and Le k < 40 . The efficiency of the propeller is 70 % at 30 knots (1.152 mph/knot). Consider that the number of repetitions of the maximum power at the given speed is 2x 105. Let N = 2 based on the maximum shear theory with varying stress. Compute the shaft diameter.

Solution: For cold-drawn monel shaft, Table AT 10 s y = 75 ksi sn = 42 ksi at 108 at 2 x 105 0.085

 108   sn ≈ 42 = 71.23 ksi 5  10 2 ×   1 sn sns 71.23 = = = 75 1.053 s y s ys 63,000hp 63,000(400) = = 16,800 in − lb n 1500 16T 16(16,800) 268,800 ss = = = π D3 π D3 π D3 s ms = ss

T=

s as = 0 K fs sas sns sms + s ys SF SF = 0.85 assume K f = K fs = 1.0 ses =

81,255  1  268,800   + 0 = ses =   3 D3  1.053  π D  Fvm = ηhp 33,000 vm = (30 knots )(1.152 mph knot )(5280 ft mi )(1 hr 60 min ) = 3041 fpm F (3041) = (0.70)(400) 33,000 F = 3040 lb 4F 4(3040) 12,160 s= = = 2 πD π D2 π D2 sm= s sa = 0

Page 18 of 76



se =

K f sa sn sm + sy SF

3676  1  12,160  +0 = 2 se =   2  D  1.053  π D  Maximum Shear Theory 1

2 2 2 1  se   ses     =   +  N  sn   0.5sn     2 1  3676   81,255 =   +  2 2  71,230 D   0.5 71,230 D 3

(

1

)

  

2

2  

1

2 2 2 1  1   2.2815   =   +   2  19.377 D 2   D 3   By trial and error 11 D = 1.66 in = 1 in 16

482.

A shaft receives 300 hp while rotating at 600 rpm, through a pair of bevel gears, and it delivers this power via a flexible coupling at the other end. The shaft is designed with the average forces ( at the midpoint of the bevel-gear face); the tangential driving force is F , G = 580 lb , Q = 926 lb ; which are the rectangular components of the total reaction between the teeth; Dm = 24 in , L = 36 in , a = 10 in . Let the material be AISI C1045, cold drawn; N = 2 . Considering varying stresses and using the octahedral shear theory, determine the shaft diameter.

Problems 482, 485, 486. Solution: For AISI C1045, cold drawn s y = 85 ksi su = 100 ksi

Page 19 of 76


SECTION 7 – SHAFT DESIGN sn = 0.5su = 0.5(100) = 50 ksi SF = 0.85 sn sns 50 1 = = = s y s ys 85 1.7 63,000hp 63,000(300) T= = = 31,500 in − lb 600 600 16T 16(31,500) 504,000 ss = = = π D3 π D3 π D3 s ms = ss s as = 0 K fs sas s ses = ns sms + s ys SF 94,370  1  504,000   + 0 = ses =   3 D3  1.7  π D  D  F m  = T  2   24  F   = 31,500  2  F = 2625 lb Vertical:

D   24  Q m  = 926  = 11,112 in − lb  2   2  G = 580 lb

[∑ M

B

]

=0

− Av (36) +

QDm − G (10) = 0 2

QDm = G (10) + Av (36) 2 11,112 = 580(10) + Av (36)

Av = 148 lb

Page 20 of 76



[∑ F =0] v

Av + Bv = 580 lb 148 + Bv = 580 lb Bv = 432 lb Shear Diagram

Moment Diagram

M Cv = 11,112 in − lb M Bv = 5328 in − lb

Horizontal:

[∑ M

B

=0

]

Ah (36) = (2625)(10) Ah = 729 lb

[∑ F

h

=0

]

Bh = Ah + F Bh = 725 + 2625

Page 21 of 76


SECTION 7 – SHAFT DESIGN Bh = 3354 lb Shear Diagram

M Ch = 0 M Bh = (36)(729) = 26,244 in − lb

Maximum M M = MB =

(M ) + (M ) 2

Bh

(26,244)2 + (5328)2 = 26,780 in − lb 32(26,780) 4(926) 856,960 3,704 = + = + 2

BV

=

32M 4Q + 3 π D π D2 π D3 π D2 π D3 4Q 32 M 3704 856,960 − = − smin = π D 2 π D3 π D 2 π D3 1 sm = (smax + smin ) 2 1  856,960 3704 3704 856,960  3704 = sm =  + + − π D2 π D2 π D 3  π D 2 2  π D3 1 sa = (smax − smin ) 2 856,960 sa = π D3 K f sa s se = n s m + sy SF smax =

assume K f = 1.0 at B  1  3704   1.0  856,960  964 320,916 +  = 2 + se =    2   3 D3  1.7  π D   0.85  π D  D Octahedral Shear Theory 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn    

Page 22 of 76

π D2


SECTION 7 – SHAFT DESIGN 1 2

 694 320,916   1 2 2 2 2 +       2 3   6.42   3.27  1 94,370  1 D  +    =  =  D + 3  + 3   3  2 2  50,000 D   D     0.577(50,000)D    72 D      By trial and error, use 1 D = 2 in 2 2

483.

The worm shown is to deliver 65.5 hp steadily at 1750 rpm. It will be integral with the shaft if the shaft size needed permits, and its pitch diameter 3 in. The 12in. pulley receives the power from a horizontal belt in which the tight tension F1 = 2.5F2 . The forces (in kips) on the worm are as shown, with the axial force taken by bearing B. The strength reduction factor for the thread roots may be taken as K f = 1.5 , shear or bending. The shaft is machined from AISI 1045, as rolled. (a) For N = 2.2 (Soderberg criterion) by the octahedral-shear theory, compute the required minimum diameter at the root of the worm thread (a first approximation). (b) What should be the diameter of the shaft 2.5 in. to the left of the centerline of the worm? (c) Select a shaft size D and check it at the pulley A.

Problem 483. Solution: For AISI 1045, as rolled s y = 59 ksi su = 96 ksi sn = 0.5su = 48 ksi 1 sn sns 48 = = = s y s ys 59 1.229 63,000hp 63,000(65.5) T= = = 2358 in − lb 1750 1750 (F1 − F2 ) 12  = T 2 (2.5F2 − F2 )(6) = 2358 F2 = 262 lb F1 = 2.5F2 = 655 lb

Page 23 of 76


SECTION 7 – SHAFT DESIGN FA = F1 + F2 = 655 + 262 = 917 lb Horizontal

[∑ M

B

=0

]

(917 )(6) + (1570)(6.5) = 13Eh Eh = 1208 lb

[∑ F

h

=0

]

917 + Eh = Bh + 1570 917 + 1208 = Bh + 1570 Bh = 555 lb Shear Diagram

M Ah = 0 M Bh = (917 )(6) = 5502 in − lb M Ch = (1208)(6.5) = 7852 in − lb

Vertical:

3 M ′ = (2540)  = 3810 in − lb 2 ∑ M E =0

[

]

M ′ + (1170)(6.5) = 13Bv 3810 + (1170)(6.5) = 13Bv Bv = 878 lb

Page 24 of 76



[∑ F =0] v

Ev + Bv = 1170 Ev + 878 = 1170 Ev = 292 lb

Shear Diagram

Moment Diagram

M Av = 0 M Bv = 0 M Cv = 5707 in − lb

M= MA = MB = MC =

(M h )2 + (M v )2

(0)2 + (0)2 = 0 (5502)2 + (0)2 = 5502 in − lb (7852)2 + (5707)2 = 9707 in − lb

(a) Minimum diameter at the root of the warm thread. K f = K fs = 1.5 M = M C = 9707 in − lb F = 2540 lb smax =

32M 4F 32(9707 ) 4(2540) 310,624 10,160 + = + = + 3 2 π Dr π Dr π Dr3 π Dr2 π Dr3 π Dr2

Page 25 of 76



smin = −

310,624 10,160 + π Dr3 π Dr2

1 (smax + smin ) 2 10,160 sm = π Dr2 1 sa = (smax − smin ) 2 310,624 sa = π Dr3 K f sa s se = n s m + sy SF sm =

 1  10,160   1.5  310,624  2632 174,485 se =  +   2   3  = D2 + D3  1.229  π Dr   0.85  π Dr  r r 16T 16(2358) 12,000 ss = = = π Dr3 π Dr3 Dr3 sms = ss sas = 0 K fs sas s ses = ns sms + s ys SF

9764  1  12,000  ses =  +0= 3  3  Dr  1.229  Dr  N = 2.2 , Octahedral shear theory 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

 2632 174,485  2 2 1 + 2 2 2 2   2  3    1  Dr Dr   9764 1 3.635   1      +    = +  = + 2    0.577(48,000)Dr3   2.2  48,000 Dr3   2.84 Dr3    18.24 Dr       

By trial and error Dr = 2.023 in 1 say Dr = 2 in 16 (b) D – shaft diameter 2.5 in. to the left of the center line of worm

Page 26 of 76


SECTION 7 – SHAFT DESIGN 3 in 16 Figure AF 12 3 r 16 = ≈ 0.1 d  3 2.023 − 2 2   16  2.023 D = = 1.2 d  3 2.023 − 2 2   16  K f = K t = 1.65 r=

K fs = K ts = 1.34 at 2.5 in to the shaft M h = (917 )(6) + (362)(6.5 − 2.5) = 6950 in − lb

M v = (878)(6.5 − 2.5) = 3512 in − lb M=

(6950)2 + (3512)2

= 7787 in − lb

10,160 π D2 32 M 32(7787 ) 249,184 sa = = = π D3 π D3 π D3 K f sa s se = n s m + sy SF sm =

 1  10,160   1.65  249,184  2632 153,970 +  = se =  +   2   3 D2 D3  1.229  π D   0.85  π D  9764 ses = D3 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

 2632 153,970  2 2 1 2 2 2 2 +       2 3   1 9764 1 3.21   1   D  +    =  =  D + 3  +   3  2 2.2  48,000 18.24 D D   2.84 D 3     0.577(48,000)D         

By trial and error D = 1.9432 in 15 say D = 1 in 16

Page 27 of 76



15 in = 1.9375 in 16 At the pulley A, or 3 in. right of centerline M h = (917 )(3) = 2751 in − lb

(c) Selecting D = 1

Mv = 0 M = 2751 in − lb For sled runner keyway K f = 1.6 K fs = 1.6

sm = 0 32 M 32(2751) sa = = = 3853 psi 3 πD π (1.9375)3 K f sa s se = n s m + sy SF  1.6  se = 0 +  (3853) = 7253 psi  0.85  9764 ses = = 1343 psi (1.9375)3 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

2 2 2   1  7253   1343   =   + N  48,000   0.577(48,000)   N = 6.30 > 2.2 , therefore o.k.

484.

A propeller shaft as shown is to receive 300 hp at 315 rpm from the right through a flexible coupling. A 16-in. pulley is used to drive an auxiliary, taking 25 hp. The belt pull FB is vertically upward. The remainder of the power is delivered to a propeller that is expected to convert 60% of it into work driving the boat, at which time the boat speed is 1500 fpm. The thrust is to be taken by the right-hand bearing. Let N = 2 ; material cold-worked stainless 410. Use the octahedral shear theory with varying stresses. (a) Determine the shaft size needed assuming no buckling. (b) Compute the equivalent column stress. Is this different enough to call for another shaft size? Compute N by the maximum shear stress theory, from both equations (8.4) and (8.11).

Page 28 of 76



Problem 484. Solution: For stainless 410, cold-worked s y = 85 ksi sn = 53 ksi SF = 0.85 Belt drive 63,000hp 63,000(25) TB = = = 5000 in − lb n 315  2T  4(5000) FB = F1 + F2 = 2(F1 − F2 ) = 2 B  = = 1250 lb 16  DB  Propeller 63,000hp 63,000(300 − 25) TP = = = 55,000 in − lb n 315 Thrust Fvm = ηhp(33,000) F (1500 ) = (0.60)(300 − 25)(33,000) F = 3630 lb Vertical loading

[∑ M

E

=0

]

(20)(1250) = 60C C = 417 lb

[∑ F =0] v

A + C = FB A + 417 = 1250

Page 29 of 76


SECTION 7 – SHAFT DESIGN A = 833 lb

Shear Diagram

M B = (20 )(833) = 16,660 in − lb Maximum T at B T = TB + TP = 60,000 in − lb (a) Shaft size assuming no buckling M = 16,660 in − lb F = 3630 lb 4F 4(3630 ) 14,520 = = sm = 2 πD π D2 π D2 32 M 32(16,660 ) 533,120 sa = = = π D3 π D3 π D3 For sled-runner keyway K f = 1.6 K fs = 1.6

1 sn sns 53 = = = s y s ys 85 1.604 se =


 1  14,520   1.6  533,120  2882 319,430 + = se =  +   2   3  D2 D3  1.604  π D   0.85  π D  16 16(60,000 ) 960,000 ss = sms = = = 3 πD π D3 π D3 sas = 0 K fs sas s ses = ns sms + SF s ys

190,510  1  960,000   + 0 = ses =   2 D3  1.604  π D 

Page 30 of 76


SECTION 7 – SHAFT DESIGN 1 2

1  se   ses   =   +    N  sn   sns     N = 2 , Octahedral Shear Theory, sns = 0.577 sn 2

2

1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1 2

 2882 319,430   1 2 2 2 2 +       2 3   6.027   6.230  1 190,510 1  D  +    =  =  D +  +   3  2 2  53,000 18.39 D D 3   D 3     0.577(53,000 )D          By trial and error D = 2.6 in 5 say D = 2 in = 2.625 in 8 2

(b) Equivalent Column Stress 4F s= α π D2 Le = 12 + 60 + 10 = 82 in 1 1 k = D = (2.625) = 0.65625 in 4 4 Le 82 = 125 > 120 = k 0.65625 Use Euler’s equation 2

L  sy  e  2 85(125) k   = 2 = 4.486 α= π 2E π 30 ×103 4F 4(3630 ) s= (4.486) = 3000 psi α= 2 πD π (2.625)2 Since α > 1 , it is different enough to call for another shaft size.

(

)

Solving for N by maximum shear theory. 2882 319,430 2882 319,430 se = + = + = 18,078 psi 2 2 3 D D (2.625) (2.625)3 190,510 ses = = 10,533 psi (2.625)3 Equation (8.4)

Page 31 of 76


SECTION 7 – SHAFT DESIGN 1

1

2  2  s 2  2   18,078   2 2 τ =  ss +    = (10,533) +    = 13,880 psi  2    2     0.5sn 0.5(53,000) N= = = 1.91 τ 13,880 Equation (8.11) sns = 0.5sn 1

1

2 2 2  18,078  2  10,533  2  2 1  s   ss     =   +    =   +  ( ) , 000 N  sn   sns   53 , 000 0 . 5 53          N = 1.91

CHECK PROBLEMS 485.

A 3-in. rotating shaft somewhat as shown (482) carries a bevel gear whose mean diameter is Dm = 10 in and which is keyed (profile) to the left end. Acting on the gear are a radial force G = 1570.8 lb , a driving force Q = 3141.6 lb . The thrust force is taken by the right-hand bearing. Let a = 5 in and L = 15 in ; material, AISI C1040, annealed. Base calculations on the maximum shearing stress theory with variable stress. Compute the indicated design factor N . With the use of a sketch, indicate the exact point of which maximum normal stress occurs.

Solution: For AISI C1040, annealed, Figure AF 1 s y = 48 ksi su = 80 ksi sn = 0.5su = 40 ksi sn sns 40 1 = = = s y s ys 48 1.2 FDm (6283.2 )(10 ) T= = = 31,416 in − lb 2 2 16T 16(31,416 ) ss = = = 5926 psi π D3 π (3)3 sms = ss sas = 0 K fs sas s ses = ns sms + s ys SF  1  ses =  (5926 ) + 0 = 4940 psi  1.2  Vertical

Page 32 of 76



QDm (3141.6 )(10 ) = = 15,708 in − lb 2 2 ∑ M E =0

[

]

QDm = 5G + 15 AV 2 15,708 = 5(1570.8) + 15 AV

AV = 523.6 lb

[∑ F =0] v

AV + BV = G 523.6 + BV = 1570.8 BV = 1047.2 lb Shear Diagram

Moment Diagram

M CV = 15,708 in − lb M BV = 7854 in − lb

Page 33 of 76


SECTION 7 – SHAFT DESIGN Horizontal

[∑ M

B

=0

]

15 Ah = 5(6283.2) Ah = 2094.4 lb

[∑ F

h

=0

]

Bh = Ah + F Bh = 2094.4 + 6283.2 Bh = 8377.6 lb Shear Diagram

M Ch = 0 M Bh = (15)(2094.4 ) = 31,416 in − lb

Maximum Moment 2 M = M B2h + M Bv =

(31,416)2 + (7854)2

= 32,383 in − lb

Since thrust force is taken by the right-hand bearing sms = 0 32 M 32(32,383) sas = = = 12,217 psi π D3 π (3)3 K f sa s se = n s m + sy SF Assume K f = 1.0 at the bearing B  1.0  se = 0 +  (12,217 ) = 14,373 psi  0.85 

Page 34 of 76


SECTION 7 – SHAFT DESIGN Maximum shear theory sns = 0.5sn 1

2 2 2 1  se   ses     =   +  N  sn   0.5sn     1

2 2 2 1  14,373   4940     =   + N  40,000   0.5(40,000)   N = 2.3

Location of maximum normal stress

487.

A 2 7/16-in. countershaft in a machine shop transmits 52 hp at 315 rpm. It is made of AISI 1117, as rolled, and supported upon bearing A and B, 59-in. apart. Pulley C receives the power via a horizontal belt, and pulley D delivers it vertically downward, as shown. Calculate N based on the octahedral-shearstress theory considering varying stresses.

Problem 487, 488 Solution: For AISI 1117, as rolled s y = 44.3 ksi

Page 35 of 76


SECTION 7 – SHAFT DESIGN su = 70.6 ksi sn = 0.5su = 35.3 ksi 1 sn sns 35.3 = = = s y s ys 44.3 1.255 SF = 0.85 63,000(52 ) T= = 10,400 in − lb 315 Pulley C  2T  4(10,400)  = FC = F1 + F2 = 2(F2 − F1 ) = 2 = 2311 lb D 18  C Pulley D  2T  4(10,400)  = FD = F1 + F2 = 2(F2 − F1 ) = 2 = 1664 lb D 25  D Horizontal

[∑ M

A

=0

]

15(2311) = 59 Bh Bh = 588 lb

[∑ F

h

=0

]

Ah + Bh = 2311 Ah + 588 = 2311 Ah = 1723 lb Shear Diagram

M Ch = (1723)(15) = 25,845 in − lb

Page 36 of 76


SECTION 7 – SHAFT DESIGN M Dh = (1723)(15) − (588)(26 ) = 10,557 in − lb

Vertical

[∑ M

B

=0

]

18(1664) = 59 Av Av = 508 lb

[∑ F =0] v

Av + Bv = 1664 508 + Bv = 1664 Bv = 1156 lb Shear Diagram

M Cv = (508)(15) = 7620 in − lb M Dv = (1156 )(18) = 20,808 in − lb M C = M C2h + M C2v =

(25,845)2 + (7620 )2

M D = M D2 h + M D2 v =

(10,557 )2 + (20,808)2

Maximum M at C M = M C = 26,945 in − lb sm = 0 32 M sa = π D3 7 D = 2 in = 2.4375 in 16 Page 37 of 76

= 26,945 in − lb = 23,333 in − lb


SECTION 7 – SHAFT DESIGN 32(26,945) = 18,952 psi π (2.4375)3 assume K f = K fs = 1.0 sa =

se =


(1.0)(18,952) = 22,300 psi  1  se =  (0 ) + 0.85  1.255  16T 16(10,400 ) ss = = = 3658 psi π D 3 π (2.4375)3 sms = s s = 3658 psi sas = 0 ses =


 1  ses =  (3658) + 0 = 2915 psi  1.255  Octahedral shear theory sns = 0.577 sn 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

2 2 2   1  22,300   2915   =   + N  35,300   0.577(35,300)   N = 1.544

489.

A shaft for a general-purpose gear-reduction unit supports two gears as shown. The 5.75-in. gear B receives 7 hp at 250 rpm. The 2.25-in. gear A delivers the power, with the forces on the shaft acting as shown; the gear teeth have a o 1 A B pressure angle of φ = 14 ( tan φ = h = h ). Both gears are keyed (profile) to Av Bv 2 the shaft of AISI 1141, cold rolled. (a) If the fillet radius is 1/8 in. at bearing D, where the diameter is 1 3/8 in., compute N based on the octahedral-shear-stress theory (Soderberg line). The shaft diameter at A is 1 11/16 in. What is N here?

Page 38 of 76



Problem 489, 490 Solution: For AISI 1141, cold rolled s y = 90 ksi sn = 50 ksi sn sns 50 1 = = = s y s ys 90 1.8 SF = 0.85 63,000(7 ) T= = 1764 in − lb 250 16T sms = πD 3 sas = 0 Gear B:  5.75  Bv   = T = 1764 in − lb  2  Bv = 614 lb Bh = Bv tan φ = 614 tan 14.5 = 159 lb Gear A:  2.25  Av   = T = 1764 in − lb  2  Av = 1568 lb Ah = Av tan φ = 1568 tan 14.5 = 406 lb Vertical

Page 39 of 76



[∑ M

D

=0

]

8Cv = 4(1568) − 3(614) Cv = 554 lb

[∑ F =0] v

Cv + Dv = Av + Bv 554 + Dv = 1568 + 614 Dv = 1628 lb Shear Diagram

M Av = (554 )(4 ) = 2216 in − lb M Dv = (614 )(3) = 1842 in − lb

Horizontal

[∑ M

C

=0

]

4(406) + 8Dh = 11(159) Dh = 16 lb

[∑ F

h

=0

]

Ch + Bh = Ah + Dh Ch + 159 = 406 + 16 Ch = 263 lb Shear Diagram

Page 40 of 76


SECTION 7 – SHAFT DESIGN M Ah = (263)(4 ) = 1052 in − lb M Dh = (159 )(3) = 477 in − lb M A = M A2h + M A2v =

(1052)2 + (2216)2

= 2453 in − lb

M D = M D2 h + M D2 v =

(477 )2 + (1842)2

= 1903 in − lb

(a) At bearing D 1 r = in 8 3 d = 1 in 8 r 0.125 = ≈ 0.10 d 1.375 D 1.375 + 0.25 = ≈ 1.2 d 1.375 K t ≈ K f = 1.6 K ts ≈ K fs = 1.34

M = MD sm = 0 32 M 32(1903) sa = = = 7456 psi π d 3 π (1.375)3 K f sa s se = n s m + sy SF se = 0 + sms

(1.6)(7456) = 14,035 psi

0.85 16T 16(1764 ) = = = 3456 psi 3 π D π (1.375)3

sas = 0 ses =

K fs sas sns sms + SF s ys

 1  ses =  (3456 ) + 0 = 1920 psi  1.8  Octahedral shear theory 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn    

Page 41 of 76


SECTION 7 – SHAFT DESIGN 1

2 2 2   1  14,035   3456   =   +  N  50,000   0.577(50,000)   N = 3.28

(b) At A For profile keyway K f = 2.0 , K fs = 1.6 11 in = 1.6875 in 16 M = M A = 2453 in − lb sm = 0 32 M 32(2453) sa = = = 5200 psi 3 πd π (1.6875)3 K f sa s se = n s m + sy SF

d =1

se = 0 + sms

(2.0)(5200) = 12,235 psi

0.85 16T 16(1764 ) = = = 1870 psi 3 π D π (1.6875)3

sas = 0 ses =


 1  ses =  (1870 ) + 0 = 1040 psi  1.8  Octahedral shear theory

1  se   ses   =   +  N  sn   0.577 sn   2

2

  

1 2

1

2 2 2   1  12,235   1040   =   + N  50,000   0.577(50,000)   N = 4.043

THRUST LOADS 491.

The high-speed shaft of a worm-gear speed reducer, made of carburized AISI 8620, SOQT 450 F, is subjected to a torque of 21,400 in-lb. Applied to the right

Page 42 of 76


SECTION 7 – SHAFT DESIGN end with no bending. The force on the worm has three components: a horizontal force opposing rotation of W = 6180 lb , a vertical radial force S = 1940 lb , and a rightward thrust of F = 6580 lb . The shaft has the following dimensions: a = 6 , 7 9 3 9 b = 4 , c = 10 , d = 4 , e = 2 , f = 13 , g = 11.646 , h = 10.370 , 8 16 4 16 13 D2 = 4 , D3 = 4 , D4 = 3.3469 , D5 = 3.253 , r1 = 0.098 , D1 = 3.740 , 16 3 1 r2 = r3 = , r4 = 0.098 , r5 = , all in inches. The pitch diameter of the worm, 4 16 6.923 in., is the effective diameter for the point of application of the forces. The root diameter, 5.701 in. is used for stress calculations. The left-hand bearing takes the thrust load. Calculate N based on the octahedral-shear-stress theory with varying stresses. (Data courtesy of Cleveland Worm and Gear Company.)

Problem 491 Solution: Table AT 11n For AISI 8620, SOQT 450 F s y = 120 ksi su = 167 ksi sn = 0.5su = 83.5 ksi 1 sn sns 83.5 = = = s y s ys 120 1.437 SF = 0.85 T = 21,400 in − lb Vertical

Page 43 of 76


SECTION 7 – SHAFT DESIGN  6.923   6.923  M ′ = F  = 22,777 in − lb  = 6580  2   2  ∑ M A =0

[

]

22,777 + (11.646)(1940) = (11.646 + 10.370)Gv Gv = 2061 lb

[∑ F =0] v

S + Av = Gv 1940 + Av = 2061 Av = 121 lb Shear Diagram

Moment Diagram

M Av = 0 M Bv = −(121)(1.2035) = −146 in − lb M Cv = −(121)(1.2035 + 4.875) = −736 in − lb M Dv = −(121)(1.2035 + 4.875 + 5.5675) = −1409 in − lb at left side M Dv = −1409 + M ′ = −1409 + 22,777 = 21,368 in − lb at right side M Ev = 21,368 − (2061)(4.4325) = 12,233 in − lb M Fv = 12,233 − (2061)(4.5625) = 2830 in − lb M Gv = 2830 − (2061)(1.375) = 0 Horizontal

Page 44 of 76



[∑ M

A

=0

]

(11.646)(6180) = (11.646 + 10.370)Gh Gh = 3269 lb

[∑ F

h

=0

]

Ah + Gv = W Ah + 3269 = 6180 Ah = 2911 lb Shear Diagram

Moment Diagram

M Ah = 0 M Bh = (2911)(1.2035) = 3500 in − lb M Ch = (2911)(1.2035 + 4.875) = 17,695 in − lb M Dh = 33,900 in − lb M Eh = 33,900 − (3269 )(4.4325) = 19,410 in − lb M Fh = 19,410 − (3269 )(4.5625) = 4495 in − lb M Fh = 4495 − (3269 )(1.375) = 0

Combined M = M h2 + M v2 MA = MB = MC = MD = MD = ME =

(0)2 + (0)2 = 0 in − lb (3500)2 + (146)2 = 3503 in − lb (17,695)2 + (736)2 = 17,710 in − lb (33,900)2 + (1409)2 = 33,930 in − lb (left) (33,900)2 + (21,368)2 = 40,073 in − lb (right) (19,410)2 + (12,233)2 = 22,944 in − lb

Page 45 of 76



MF = MG =

(2830)2 + (4495)2 = 5312 in − lb (0)2 + (0)2 = 0 in − lb

Bending stresses (Maximum) At A, s A = 0 32 M B 32(3503) = = 682 psi At B, s B = π D13 π (3.740 )3 32 M C 32(17,710 ) = = 1618 psi At C, sC = π D23 π (4.8125)3 32 M D 32(40,073) At D, s D = = = 2203 psi π Dr3 π (5.701)3 32 M E 32(22,944 ) At E, s E = = = 3652 psi π D33 π (4)3 32 M F 32(5312 ) At F, s F = = = 1443 psi 3 π D4 π (3.3469 )3 At G, sG = 0 Shear Stresses: 16T 16(21,400 ) ssA = ssB = = = 2083 psi π D13 π (3.740)3 16T 16(21,400 ) ssC = = = 978 psi π D23 π (4.8125)3 16T 16(21,400 ) ssD = = = 588 psi π Dr3 π (5.701)3 16T 16(21,400 ) ssE = = = 1703 psi π D33 π (4)3 16T 16(21,400 ) ssF = ssG = = = 2907 psi π D43 π (3.3469)3 Tensile stresses: F = 6580 lb 4F 4(6580 ) s′A = s′B = = = 599 psi 2 π D1 π (3.740)2 4F 4(6580 ) sC′ = = = 362 psi 2 π D2 π (4.8125)2 4F 4(6580 ) s′D = = = 258 psi 2 π Dr π (5.701)2 4F 4(6580 ) s′E = = = 524 psi 2 π D3 π (4 )2

Page 46 of 76



s′E = s′F =

4F 4(6580 ) = = 748 psi 2 π D4 π (3.3469 )2

r1 0.098 = = 0.03 D1 3.740 D2 4.8125 = = 1.3 D1 3.740 Figure AF 12 K f ≈ K t = 2.3

At B:

K fs ≈ K ts = 1.7 K f sa sn sm + sy SF sm = s′B = 599 psi se =

sa = s B = 682 psi

(2.3)(682 ) = 2262 psi  1  se =  (599 ) + 0.85  1.437  K fs sas s ses = ns sms + s ys SF sms = ssB = 2083 psi sas = 0  1  ses =  (2083) + 0 = 1450 psi  1.437  Octahedral shear theory 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

2 2 2   1  2262   1450   =   + N  83,500   0.577(83,500)   N = 24.7

r2 0.75 = = 0.16 D2 4.8125 Dr 5.701 = = 1.2 D2 4.8125 Figure AF 12 K f ≈ K t = 1.5

At C:

K fs ≈ K ts = 1.2

Page 47 of 76



se =


sm = 362 psi sa = 1618 psi

(1.5)(1618) = 3107 psi  1  se =  (362 ) + 0.85  1.437  K fs sas s ses = ns sms + s ys SF sms = ssC = 978 psi sas = 0  1  ses =  (978) + 0 = 681 psi  1.437  Octahedral shear theory 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

2 2 2   1  3107   681   =   + N  83,500   0.577(83,500)   N = 25.1

At D: Assume K f = 1.5 as in Prob. 483 se =


sm = 258 psi sa = 2203 psi

(1.5)(2203) = 4067 psi  1  se =  (258) + 0.85  1.437  K fs sas s ses = ns sms + s ys SF sms = s sD = 588 psi sas = 0  1  ses =  (588) + 0 = 409 psi  1.437  Octahedral shear theory

Page 48 of 76



1  se   ses   =   +  N  sn   0.577 sn   2

2

  

1 2

1

2 2 2   1  4067   409   =   + N  83,500   0.577(83,500)   N = 20.2

r3 0.75 = = 0.19 4 D3 Dr 5.701 = = 1.43 4 D3 Figure AF 12 K f ≈ K t = 1.45

At E:

K fs ≈ K ts = 1.25 K f sa sn sm + sy SF sm = s′E = 524 psi se =

sa = s E = 3652 psi

(1.45)(3652) = 6595 psi  1  se =  (524 ) + 0.85  1.437  K fs sas s ses = ns sms + s ys SF sms = s sE = 1703 psi sas = 0  1  ses =  (1703) + 0 = 1185 psi  1.437  Octahedral shear theory 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

2 2 2   1  6595   1185   =   + N  83,500   0.577(83,500)  

N = 12

At F:

r4 0.098 = = 0.03 D4 3.3469

Page 49 of 76


SECTION 7 – SHAFT DESIGN D3 4 = = 1.2 D4 3.3469 Figure AF 12 K f ≈ K t = 2.3 K fs ≈ K ts = 1.7 K f sa sn sm + sy SF sm = s′F = 748 psi se =

sa = s F = 1443 psi

(2.3)(1443) = 4425 psi  1  se =  (748) + 0.85  1.437  K fs sas s ses = ns sms + SF s ys sms = s sF = 2907 psi sas = 0  1  ses =  (2907 ) + 0 = 2023 psi  1.437  Octahedral shear theory 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

2 2 2   1  4425   2023   =   + N  83,500   0.577(83,500)  

N = 14.8

Then N = 12 at r3 = 492.

3 in (E) 4

The slow-speed shaft of a speed reducer shown, made of AISI 4140, OQT 1200 F, transmits 100 hp at a speed of 388 rpm. It receives power through a 13.6 in. gear B. The force on this gear has three components: a horizontal tangential driving force Ft = 2390 lb , a vertical radial force S = 870 lb , and a thrust force Q = 598 lb taken by the right-hand bearing. The power is delivered to a belt at F that exerts a downward vertical force of 1620 lb.; sled runner keyways. Use the octahedral shear theory with the Soderberg line and compute N at sections C and D. (Data courtesy of Twin Disc Clutch Company.)

Page 50 of 76



Problem 492, 493 Solution: For AISI 4140, OQT 1200 F s y = 83 ksi su = 112 ksi sn = 0.5su = 56 ksi 1 sn sns 56 = = = s y s ys 83 1.482 SF = 0.85 63,000(100 ) T= = 16,237 in − lb 388 Vertical

 13.6   13.6  M ′ = Q  = (598)  = 4066.4 in − lb  2   2  ∑ M A =0

[

]

Page 51 of 76


SECTION 7 – SHAFT DESIGN 5 5 3 7 11 13 3  3  3 1 + 1 (870 ) + 1 + 1 + 3 + 1 + 1 + + 2 (1620 ) + 4066.4 8 32 32 16 4  16 8   16 8 5 3 7   3 = 1 + 1 + 3 + 1 Gv 8 32   16 8 Gv = 3573 lb

[∑ F =0] v

Av + S + F = Gv Av + 870 + 1620 = 3573 Av = 1083 lb Shear Diagram

Moment Diagram

M Av = 0  3 M Pv = −(1083)1  = −1286 in − lb  16   5 M Bv = −1286 + (− 1083)1  = −3046 in − lb at the left  8 M Bv = −3046 + 4066.4 = 1021 in − lb at the right  3 M Cv = 1021 − (1953) 3  = −5570 in − lb  8  7  M Gv = −5570 − (1953)1  = −7950 in − lb  32 

Page 52 of 76


SECTION 7 – SHAFT DESIGN  11  M Dv = −7950 + (1620 )1  = −5773 in − lb  32   13  M Ev = −5773 + (1620 )  = −4457 in − lb  16   3 M Fv = −4457 + (1620 ) 2  = 0 in − lb  4

Horizontal

[∑ M

A

=0

]

19   13   13  2 (2390 ) +  2 + 4 Gh 32   16   16 Gh = 908 lb

[∑ F

h

=0

]

Ah + Gh = Ft Ah + 908 = 2390 Ah = 1482 lb Shear Diagram

M Ah = 0  3 M Ph = (1482 )1  = 1760 in − lb  16   5 M Bh = 1760 + (1482 )1  = 4168 in − lb  8

Page 53 of 76


SECTION 7 – SHAFT DESIGN  3 M Ch = 4168 − (908) 3  = 1104 in − lb  8  7  M Ch = 1104 − (908)1  = 0 in − lb  32  M Dh = 0 in − lb M Eh = 0 in − lb M Fh = 0 in − lb

Combined M = M h2 + M v2 M A = 0 in − lb MP = MB = MC = MD = ME = MF =

(1760)2 + (1286)2 = 2180 in − lb (4168)2 + (3046)2 = 5163 in − lb (1104)2 + (5570)2 = 5678 in − lb (0)2 + (5773)2 = 5773 in − lb (0)2 + (4457 )2 = 4457 in − lb (0)2 + (0)2 = 0 in − lb

1 in = 0.125 in 8 d = 2.750 in D = 2.953 in r 0.125 = = 0.05 d 2.750 D 2.953 = = 1.10 d 2.750 Figure AF 12 K f 1 ≈ K t = 1.9

at C: r =

K fs1 ≈ K ts = 1.3 For sled runner keyway K f 2 = 1.6 K fs 2 = 1.6 K f = 0.8 K f 1 K f 2 = 0.8(1.9)(1.6) = 2.4 K fs = 0.8 K fs1 K fs 2 = 0.8(1.3)(1.6) = 1.7

Page 54 of 76



se =


4Q 4(598) = = 101 psi 2 πd π (2.750)2 32 M C 32(5678) sa = = = 2781 psi π d3 π (2.750)3 (2.4)(2781) = 7920 psi  1  se =  (101) + 0.85  1.482  K fs sas s ses = ns sms + s ys SF sm =

16T 16(16,237 ) = = 3976 psi π d 3 π (2.750)3 sas = 0 sms =

 1  ses =  (3976 ) + 0 = 2683 psi  1.482  Octahedral shear theory 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

2 2 2   1  7920   2683   =   + N  56,000   0.577(56,000)   N =6

1 in = 0.0625 in 16 d = 2.953 in 3 D = 3 in = 3.375 in 8 r 0.0625 = = 0.02 d 2.953 D 3.375 = = 1.14 d 2.953 Figure AF 12 K f ≈ K t = 2.4

at D: r =

K fs ≈ K ts = 1.6 se =


Page 55 of 76


SECTION 7 – SHAFT DESIGN 4Q 4(598) = = 87.3 psi 2 πd π (2.953)2 32 M C 32(5773) sa = = = 2284 psi π d3 π (2.953)3 (2.4)(2284) = 6508 psi  1  se =  (87.3) + 0.85  1.482  K fs sas s ses = ns sms + s ys SF sm =

16T 16(16,237 ) = = 3211 psi π d 3 π (2.953)3 sas = 0 sms =

 1  ses =  (3211) + 0 = 2167 psi  1.482  Octahedral shear theory 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

2 2 2   1  6508   2167   =   + N  56,000   0.577(56,000)   N = 7.5

TRANSVERSE DEFLECTIONS 494.

The forces on the 2-in. steel shaft shown are A = 2 kips , C = 4 kips . Determine the maximum deflection and the shaft’s slope at D.

Problems 494-496 Solution:

Page 56 of 76



[M B = 0] 2(10) + 25D = 4(15) D = 1.6 kips [Fv = 0] A+C = B + D 2 + 4 = B + 1.6 B = 4.4 kips

Shear Diagram

Moment Diagram

64 M M = EI Eπ D 4

M (in − kip ) M  4 4   D 10  EI 

Page 57 of 76

A 0

B -20

C 16

D 0

0

-135.8

108.6

0


SECTION 7 – SHAFT DESIGN Scale ss = 10 in in

M 200 × 10−4 , Scale sM = per in EI EI D4

Slope θ , Scale sθ = 0.2 D 4 rad in

y deflection, Scale s y = 2.0 D 4 in in

Deflection: At A: y A =

0.625 in D4

Page 58 of 76



At C: yC =

0.375 in D4

Slope: 0.075 rad D4 0.0125 rad At B: θ = D4 0.05625 rad At D: θ = D4

At A: θ =

Maximum deflection: 0.625 y = yA = = 0.04 in (2 )4 Shaft’s slope at D 0.05625 θ= = 0.0035 rad (2)4 495.

The forces on the steel shaft shown are A = 2 kips , C = 4 kips . Determine the constant shaft diameter that corresponds to a maximum deflection of 0.006 in. at section C.

Solution: (see Problem 494) 0.375 yC = = 0.006 D4 D = 2.812 in 7 say D = 2 in 8 496. The forces on the steel shaft shown are A = 2 kips , C = 4 kips . Determine a constant shaft diameter that would limit the maximum deflection at section A to 0.003 in. Solution: (see Problem 494) 0.625 yA = = 0.003 D4 D = 3.80 in 7 say D = 3 in 8 497.

A steel shaft is loaded as shown and supported in bearings at R1 and R2 . Determine (a) the slopes at the bearings and (b) the maximum deflection.

Page 59 of 76



Problem 497 Solution: ∑ M R1 = 0

[

]

(3000 ) 7 + 1 1  − (2100) 7 + 2 1 + 1 =  7 + 2 1 + 2 + 7  R2 8 8 R2 = −444 lb

[∑ F = 0]

R1 + R2 + 2100 = 3000 R1 − 444 + 2100 = 3000 R1 = 1344 lb Loading

Shear Diagram

Moment Diagram

Page 60 of 76

8

4

 8

4

8



M A = 0 in − lb 7 M B = (1344 )  = 1176 in − lb 8 7 1 M C = (1134 ) + 1  = 2688 in − lb 8 8  1 M D = 2688 − (1656 )1  = 825 in − lb  8 M E = 825 − (1656)(1) = −831 in − lb M F = −831 + (444)(1) = −387 in − lb 7 M G = −387 + (444 )  = 0 in − lb 8

M (in − kips ) D(in ) M 10 4 EI

(

)( )

A 0 1½

B1 1.18 1½

B2 1.18 2

C 2.69 2

D1 0.83 2

D2 0.83 1¾

E -0.83 1¾

F1 -0.39 1¾

F2 -0.39 1½

G 0 1½

0

1.58

0.50

1.14

0.35

0.60

-0.60

-0.28

-0.52

0

Scale ss = 2 in in

Page 61 of 76


SECTION 7 – SHAFT DESIGN M 2 × 10−4 , Scale sM = per in EI EI D4

Slope θ , Scale sθ = 4 × 10 −4 D 4 rad in

y deflection, Scale s y = 8 × 10 −4 D 4 in in

(a) Slopes at the bearings

(

)

at R1 , θ A = 0.375 4 × 10 −4 = 1.5 × 10 −4 rad at R2 , θ G = 0 rad (b) Maximum deflection at C, yC = 0.1875(8 × 10 −4 ) = 1.5 × 10 −4 in

Page 62 of 76



(a) Determine the diameter of the steel shaft shown if the maximum deflection is to be 0.01 in.; C = 1.5 kips , A = 1.58 kips , L = 24 in . (b) What is the slope of the shaft at bearing D? See 479.

Problems 498, 505, 506. Solution: Vertical

[∑ M

D

=0

]

6(1.5) + 8(0.424) = 24 Ev Ev = 0.516 kip

[∑ F

v

=0

]

Dv + 0.424 = 1.5 + Ev Dv + 0.424 = 1.5 + 0.516 Dv = 1.592 kip Shear Diagram

M C = 0 ; M D = −6(1.5) = −9 in − kips M B = −9 + 8(0.092) = −8.264 in − kips

Page 63 of 76


SECTION 7 – SHAFT DESIGN M A = −8.264 + 11(0.516) = −2.588 in − kips M E = −2.588 + 5(0.516) = 0 M (in − kips ) M D 4 ×10 4 EI

(

)

C 0 0

D -9 -61.1

Scale ss = 8 in in

M 120 × 10 −4 per in , Scale sM = EI EI D4


Page 64 of 76

B -8.264 -56.1

A -2.588 -17.6

E 0 0


SECTION 7 – SHAFT DESIGN y deflection, Scale s y = 0.768 D 4 in in

Deflections. 0.384 yCv = in D4 0.288 y Bv = in D4 0.168 y Av = in D4 Slope 0.057 rad θ Dv = D4 Horizontal

[∑ M

D

=0

]

8(0.424 ) + 24 Eh = 19(1.58) Eh = 1.1095 kip

[∑ F

h

=0

]

Dh + Eh + 0.424 = 1.58 Dh + 1.1095 + 0.424 = 1.58

Page 65 of 76


SECTION 7 – SHAFT DESIGN Dh = 0.0465 kip Shear Diagram

Moments MC = 0 MD = 0 M B = 8(− 0.0465) = −0.372 in − kip M A = −0.372 + 11(− 0.4705) = −5.5475 in − kips M E = −5.5475 + 5(1.1095) = 0 M (in − kips ) M D 4 ×10 4 EI

(

)

C 0 0

D 0 0

Scale ss = 8 in in

M 4 × 10−4 , Scale sM = per in EI EI D4


Page 66 of 76

B -0.372 -2.53

A -5.5475 -37.7

E 0 0



y deflection, Scale s y = 0.256 D 4 in in

Deflections. 0.064 yC h = in D4 0.072 y Bh = in D4 0.096 y Ah = in D4 Slope 0.012 rad θ Dh = D4 Resultant deflection:

(

y = yh2 + yv2

1 2

)

1 2 2

[(0.064) + (0.384) ] 2

yC =

D

4

[(0.072) + (0.288) ] D

4

[(0.096) + (0.168) ]

Slope: Page 67 of 76

=

0.297 D4

=

0.194 D4

1 2 2

2

yA =

0.390 D4

1 2 2

2

yB =

=

D

4



θ = (θ + θ 2 h

1 2 2 v

)

1 2 2

[(0.012) + (0.057) ] = 2

θD

D

4

=

0.05823 rad D4

(a) Diameter D. Maximum deflection = yC =

0.390 = 0.01 in D4

D = 2.50 in

(b) slope of the shaft at bearing D 0.05823 0.05823 θD = = = 0.0015 rad D4 (2.5)4

CRITICAL SPEED 499.

A small, high-speed turbine has a single disk, weighing 0.85 lb., mounted at the midpoint of a 0.178-in. shaft, whose length between bearings is 6 ½ in. What is the critical speed if the shaft is considered as simply supported? Solution:

Table AT 2 3 WL3 0.85)(6.5) ( y= = = 0.052634 in 4 3EI 6  π (0.178 )  3 30 × 10   64  

(

)

1

1

1

30  g o (∑ Wy ) 2 30  g o  2 30  386  2 nc = = = 818 rpm   = π  ∑ Wy 2  π  y  π  0.052634 

500.

The bearings on a 1 ½-in. shaft are 30 in. apart. On the shaft are three 300-lb disks, symmetrically placed 7.5 in. apart. What is the critical speed of the shaft?

Solution:

Page 68 of 76



Table AT 2 Deflection of B. y B = y B1 + y B2 + y B3 y B1 =

(300)(22.5)(7.5)([ 30)2 − (22.5)2 − (7.5)2 ] = 0.01273 in 4 6  π (1.5 )  6(30 × 10 ) (30)

   64  (300)(15)(7.5)(30)2 − (15)2 − (7.5)2 = 0.01556 in y B2 =  π (1.5)4  6 30 × 106   (30)  64  2 2 2 300 )(7.5)(7.5)(30 ) − (7.5) − (7.5) ( y B3 = = 0.00990 in 4 6  π (1.5 )  6 30 × 10   (30) 64   y B = 0.01273 + 0.00990 + 0.01556 = 0.03819 in

[

(

]

)

[

(

]

)

Deflection of C. yC = yC1 + yC2 + yC3 yC1

2 2 2 300)(7.5)(30 − 15)( 30) − (7.5) − (30 − 15) ] ( [ = = 0.01556 in 4 6  π (1.5 )  6(30 × 10 ) (30)

   64  (300)(15)(30 − 15)(30)2 − (15)2 − (30 − 15)2 = 0.02264 in yC2 = 4 6  π (1.5 )  6 30 ×10   (30)  64  (300)(7.5)(15)(30)2 − (7.5)2 − (15)2 = 0.01556 in yC3 =  π (1.5)4  6 30 × 106   (30)  64  yC = 0.01556 + 0.02264 + 0.01556 = 0.05376 in Deflection of D. y D = y D1 + y D2 + y D3

[

(

)

[

(

Page 69 of 76

]

]

)



y D1 =

(300)(7.5)(30 − 22.5)([ 30)2 − (7.5)2 − (30 − 22.5)2 ] = 0.00990 in 4 6  π (1.5 )  6(30 × 10 ) (30)

   64  (300)(15)(30 − 22.5)(30)2 − (15)2 − (30 − 22.5)2 = 0.01556 in y D2 = 4 6  π (1.5 )  6 30 ×10   (30)  64  (300 )(7.5)(22.5)(30)2 − (7.5)2 − (22.5)2 = 0.01273 in y D3 =  π (1.5)4  6 30 × 106   (30 )  64  y D = 0.00990 + 0.01556 + 0.01273 = 0.03819 in

[

(

]

)

[

(

]

)

1 2

1

30  g o (∑ Wy ) 30  g o ( y B + yC + y D )  2 nc = =   π  ∑ Wy 2  π  y B2 + yC2 + y D2  1

30  386(0.03819 + 0.05376 + 0.03819)  2 nc =   = 888 rpm π  (0.03819)2 + (0.05376)2 + (0.03819)2 

501.

A fan for an air-conditioning unit has two 50-lb. rotors mounted on a 3-in. steel shaft, each being 22 in. from an end of the shaft which is 80 in. long and simply supported at the ends. Determine (a) the deflection curve of the shaft considering its weight as well as the weight of the rotors, (b) its critical speed.

Solution:

W1 = 50 lb W3 = 50 lb π  2 W2 = (0.284 ) (3) (80 ) = 160 lb weight of shaft 4 160 w2 = = 2 lb in 80 Deflection of B. y B = y B1 + y B2 + y B3

Page 70 of 76



y B1 =

(50)(50)(22)([ 80)2 − (58)2 − (22)2 ] = 0.002844 in 4 6  π (35 )  6(30 × 10 ) (80)

   64  (50)(22)(22)(80)2 − (22)2 − (22)2 = 0.002296 in y B3 = 4 6  π (3)  6 30 × 10   (80)  64  (2)(22)(80)3 − 2(80)(22)2 − (22)3 = 0.006843 in y B2 =  π (3)4  6 30 × 106    64  y B = 0.002844 + 0.006843 + 0.002296 = 0.011983 in

[

(

]

)

[

]

(

)

Deflection of C. yC = yC1 + yC2 + yC3 yC1

2 2 2 50)(22)(80 − 40)( 80) − (22) − (80 − 40) ] ( [ = = 0.003317 in 4 6  π (35)  (80) 6(30 ×10 )

   64  2 2 2 50)(22 )(40)(80) − (22) − (40) ( yC3 = = 0.003317 in 4 6  π (3)  6 30 × 10   (80)  64  (2)(40)(80)3 − 2(80)(40)2 − (40)3 = 0.008942 in yC2 = 4 6  π (3)  6 30 × 10    64  yC = 0.003317 + 0.008942 + 0.003317 = 0.015576 in

[

(

]

)

[

]

(

By symmetry y D = y B = 0.011983 in (a) Deflection curve

Page 71 of 76

)


SECTION 7 – SHAFT DESIGN (b) Critical speed 1

30  g o (∑ Wy ) 2 nc =   π  ∑ Wy 2 

∑Wy = (50)(0.011983) + (160)(0.015576) + (50)(0.011983) = 3.69046 ∑Wy = (50)(0.011983) + (160)(0.015576) + (50)(0.011983) = 0.053177 2

2

2

2

1

30  386(3.69046)  2 nc = = 1563 rpm π  0.053177  ASME CODE 502.

A cold-rolled transmission shaft, made of annealed AISI C1050, is to transmit a torque of 27 in-kips with a maximum bending moment of 43 in-kips. What should be the diameter according to the Code for a mild shock load?

Solution: For AISI C1050, annealed s y = 53 ksi su = 92 ksi 0.3s y = 15.9 ksi 0.18su = 16.56 ksi use τ d = 0.3s y = 15.9 ksi M = 43 in − kips T = 27 in − kips 1

2 2   16 α FD 1 + B 2   2 3   (K sT ) +  K m M + D = 8 πτ d 1 − B 4      Reduce to 1 16 2 2 2 3 D = (K sT ) + ( K m M ) πτ d 1 − B 4 For mild shock load, rotating shafts K m = 1.75 K s = 1.25 B=0 1 16 2 2 2 3 D = [(1.25)(27,000)] + [(1.75)(43,000)] π (15,900) D = 2.98 in say D = 3 in

(

)

(

)[

{

Page 72 of 76

(

)

]

}



503.

A machinery shaft is to transmit 82 hp at a speed of 1150 rpm with mild shock. The shaft is subjected to a maximum bending moment of 7500 in-lb. and an axial thrust load of 15,000 lb. The material is AISI 3150, OQT 1000 F. (a) What should be the diameter when designed according to the Code? (b) Determine the corresponding conventional factor of safety (static-approach and maximum-shear theory).

Solution: For AISI 3150, OQT 1000 F s y = 130 ksi su = 151 ksi 0.3s y = 39 ksi 0.18su = 27.18 ksi use τ d = 0.18su = 27.18 ksi 63000(82 ) T= = 4492 in − lb 1150 M = 7500 in − lb F = 15,000 lb   16 α FD 1 + B 2 2 3 (K sT ) +  K m M + (a) D = 8 πτ d 1 − B 4   

(

(

)

1

)   

2

2  

For mild shock load K m = 1.75 K s = 1.25 B=0 α =1 1

2  ( 16 1)(15,000)D   2  2 3 D = [(1.25)(4492)] + (1.75)(7500) +   8 π (27180)   

{

1 2 2

D 3 = 0.1874 31.53 + [13.125 + 1.875D ] D = 1.4668 in say D = 1.5 in

(b) s =

}

32 M 4F 32(7500 ) 4(15,000 ) + = + = 31,124 psi = 31.124 ksi 3 2 πD πD π (1.5)3 π (1.5)2

Page 73 of 76



ss =

16T 16(4492 ) = = 6778.5 psi = 6.7785 ksi π D 3 π (1.5)3

2 2 1  s   ss   + = N  s y   s ys     Maximum shear theory s ys = 0.5s y 2 2 1  31.124   6.7785     =   + N  130   0.5(130)   N = 3.83

504.

short stub shaft, made of SAE 1035, as rolled, receives 30 hp at 300 rpm via a 12-in. spur gear, the power being delivered to another shaft through a flexible coupling. The gear is keyed midway between the bearings and its pressure angle φ = 20o . See the figure for 471. (a) Neglecting the radial component of the tooth load, determine the shaft diameter for a mild shock load. (b) Considering both tangential and radial components, compute the shaft diameter. (c) Is the difference in the foregoing results enough to change your choice of the shaft size?

Solution:

Figure for 471.

For SAE 1035, as rolled s y = 55 ksi su = 85 ksi 0.3s y = 16.5 ksi 0.18su = 15.3 ksi use τ d = 0.18su = 15.3 ksi

Page 74 of 76



Data are the same as 471. From Problem 471. (a) M = 4200 in − lb = 4.2 in − kips T = 6300 in − lb = 6.3 in − kips   16 α FD 1 + B 2 2   ( ) + + D3 = K T K M s  m 8 πτ d 1 − B 4    Reduce to 1 16 2 2 2 K M D3 = K T + ( ) ( ) s m πτ d 1 − B 4 For mild shock load, rotating shafts K m = 1.75 K s = 1.25 B=0 1 16 D3 = [(1.25)(6.3)]2 + [(1.75)(4.2)]2 2 π (15.3) D = 1.5306 in 9 say D = 1 in 16

(

)

(

)[

(

1

)   

2

2  

]

{

}

(b) M = 4472 in − lb = 4.472 in − kips T = 6300 in − lb = 6.3 in − kips 16 D = [(1.25)(6.3)]2 + [(1.75)(4.472)]2 π (15.3) D = 1.5461 in 9 say D = 1 in 16

{

3

1 2

}

(c) Not enough to change the shaft size. 505.

Two bearings D and E, a distance D = 24 in . Apart, support a shaft for a punch press on which are an 8-in. gear A, a 44-in. pulley B, and a flywheel C, as indicated (498). Weight of flywheel is 1500 lb.; pulley B receives the power at an angle of 45o to the right of the vertical; gear A delivers it horizontally to the right. The maximum power is 25 hp at 250 rpm is delivered, with heavy shock. For cold-finish AISI 1137, find the diameter by the ASME Code.

Solution:

Page 75 of 76



Data and figure is the same as in Problem 479. Also figure is the same as in Problem 498. For AISI 1137, cold-finished s y = 93 ksi su = 103 ksi 0.3s y = 27.9 ksi 0.18su = 18.54 ksi use τ d = 0.18s u = 18.54 ksi From Problem 479 M = M B = 14,343 in − lb = 14.343 in − kips T = TA = 12,600 in − lb = 12.6 in − kips For heavy shock load K m = 2.5 K s = 1.75 B=0   16 α FD 1 + B 2 (K sT ) +  K m M + 4 8 πτ d 1 − B    1 16 2 2 2 K M ( ) ( ) D3 = K T + s m πτ d 1 − B 4 D3 =

(

)

(

)[

(

]

16 [(1.75)(12.6)]2 + [(2.5)(14.343)]2 π (18.54) D = 2.2613 in 5 say D = 2 in 16 D3 =

{

2

1 2

}

- end -

Page 76 of 76

)   2

  

1 2


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS LIGHTLY LOADED BEARINGS (a) A 3 x 3 – in. full bearing supports a load of 900 lb., c d D = 0.0015 , n = 400 rpm . The temperature of the SAE 40 oil is maintained at 140 oF. Considering the bearing lightly loaded (Petroff), compute the frictional torque, fhp, and the coefficient of friction. (b) The same as (a) except that the oil is SAE 10W.

551.

Solution. (a) T f =

µπ DLvips  D    2

(cd 2)

L = 3 in D = 3 in πDn π (3)(400) vips = = = 20π ips 60 60 c d D = 0.0015 SAE 40 oil, 140 oF, Figure A16. µ = 7.25 µreyns µπ DLvips 7.25 ×10 −6 (π )(3)(3)(20π ) F= = = 17.173 lb (cd 2) (0.0015 2)

(

)

D 3 T f = F   = (17.173)  = 25.76 in − lb 2 2 Fvm fhp = 33,000 π Dn π (3)(400 ) vm = = = 314.16 fpm 12 12 Fvm (17.173)(314.16) = 0.1635 hp fhp = = 33,000 33,000 F 17.173 f = = = 0.0191 W 900

(b) SAE 10W oil, 140 oF, Figure A16. µ = 2.2 µreyns = 2.2 ×10 −6 reyn µπ DLvips 2.2 ×10 −6 (π )(3)(3)(20π ) F= = = 5.211 lb (cd 2) (0.0015 2)

(

)

D 3 T f = F   = (5.211)  = 7.817 in − lb 2 2 Fvm fhp = 33,000

Page 1 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS

vm =

π Dn

π (3)(400 )

= 314.16 fpm 12 12 Fvm (5.211)(314.16) = 0.0496 hp fhp = = 33,000 33,000 F 5.211 f = = = 0.00579 W 900

553.

=

The average pressure on a 6-in. full bearing is 50 psi, cd = 0.003 in. , L D = 1 . While the average oil temperature is maintained at 160 oF with n = 300 rpm , the frictional force is found to be 13 lb. Compute the coefficient of friction and the average viscosity of the oil. To what grade of oil does this correspond?

Solution: W p= LD D = 6 in. L D =1 L = 6 in. W = pLD = (50)(6)(6 ) = 1800 lb F = 13 lb Coefficient of Friction F 13 f = = = 0.0072 W 1800 µπ DLvips F= (cd 2) πDn π (6)(300) vips = = = 30π ips 60 60 µπ DLvips µ (π )(6)(6)(30π ) F= = = 13 lb (cd 2) (0.003 2)

µ = 1.8 ×10 −6 reyn = 1.8 µreyns Figure AF 16, 160 oF use SAE 10W or SAE 20W FULL BEARINGS 554.

The load on a 4-in. full bearing is 2000 lb., n = 320 rpm ; L D = 1 ; cd D = 0.0011 ; operating temperature = 150 oF; ho = 0.00088 in . (a) Select an oil that will closely accord with the started conditions. For the selected oil determine (b) the frictional loss (ft-lb/min), (c) the hydrodynamic oil flow through the bearing, (d) the amount of end leakage,

Page 2 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS (e) the temperature rise as the oil passes through the bearing, (f) the maximum pressure. Solution: (a) D = 4 in L D =1 L = 4 in cd = 0.0011D = 0.0011(4) = 0.0044 in ho = 0.00088 in 2h 2(0.00088) ε = 1− o = 1− = 0.6 cd 0.0044 Table AT 20 ε = 0.6 , L D = 1 Sommerfield Number

µns  D 

2

  p  cd  320 = 5.333 rps ns = 60 W 2000 p= = = 125 psi LD (4)(4) cd D = 0.0011 S=

µ (5.333) 

2

1  0.121 =   125  0.0011  µ = 3.4 ×10 −6 reyn = 3.4 µreyns Figure AF-16, 150 oF, use SAE 30 or SAE 20 W Select SAE 30, the nearest µ = 3.9 ×10−6 reyn

(b) Table AT 20, L D = 1 , ε = 0.6 r f = 3.22 cr 1 r D = = cr cd 0.0011  1    f = 3.22  0.0011  f = 0.003542 F = f W = (0.003542)(2000 ) = 7.084 lb

Page 3 of 63



vm =

π Dn

=

π (4)(320)

= 335.1 fpm 12 12 Frictional loss = Fvm = (7.084 )(335.1)2374 ft − lb min

(c) Table AT 20, L D = 1 , ε = 0.6 q = 4.33 rcr ns L D r = = 2.0 in 2 c 0.0044 cr = d = = 0.0022 in 2 2 ns = 5.333 rps L = 4 in q = 4.33rcr ns L = 4.33(2.0 )(0.0022 )(5.333)(4 ) = 0.4064 in 3 sec (d) Table AT 20, L D = 1 , ε = 0.6 qs = 0.680 q qs = 0.680q = 0.680(0.4064 ) = 0.2764 in 3 sec (e) Table AT 20, L D = 1 , ε = 0.6 ρ c∆to = 14.2 p ρ c = 112 , p = 125 psi 14.2 p 14.2(125) ∆to = = = 15.85 o F 112 ρc (f) Table AT 20, L D = 1 , ε = 0.6 p = 0.415 pmax 125 pmax = = 301.2 psi 0.415 555.

A 4-in., 360o bearing, with L D = 1.1 (use table and chart values for 1), is to support 5 kips with a minimum film thickness 0.0008 in.; cd = 0.004 in. , n = 600 rpm . Determine (a) the needed absolute viscosity of the oil .(b) Suitable oil if the average film temperature is 160 F, (c) the frictional loss in hp. (d) Adjusting only ho to the optimum value for minimum friction, determine the fhp and compare. (e) This load varies. What could be the magnitude of the maximum impulsive load if the eccentricity ration ε becomes 0.8? Ignore “squeeze” effect.

Page 4 of 63



Solution: D = 4 in L = 1.1D = 1.1(4) = 4.4 in W 5000 p= = = 284 psi LD (4.4)(4) ho = 0.0008 in cd = 0.004 in. 2h 2(0.0008) ε = 1− o = 1− = 0.6 cd 0.004 600 = 10 rps η= 60 (a)

Table AT20, L D = 1 , ε = 0.6

S = 0.121

r S =   cr

2

 µ ns µ n s  D     = p  cd   p

µ (10) 

4  0.121 =   284  0.004  µ = 3.4 ×10−6 reyn

2

2

(b) Figure AF16, 160 F Use SAE 30, µ = 3.2 ×10−6 reyn (c) Table AT 20, L D = 1 , ε = 0.6 r f = 3.22 cr D f = 3.22 cd

 4    f = 3.22  0.004  f = 0.00322 F = f W = (0.00322)(5000 lb ) = 16.1 lb π Dn π (4)(600) vm = = = 628.3 fpm 12 12

Page 5 of 63



fhp =

Fvm (16.1)(628.3) = 0.3065 hp = 33,000 33,000

(d) adjusting ho , cd = 0.004 in. Table AT 20, L D = 1 ho cr = 0.30 optimum value for minimum friction r f = 2.46 cr D f = 2.46 cd  4    f = 2.46  0.004  f = 0.00246 F = f W = (0.00246)(5000 lb ) = 12.3 lb π Dn π (4)(600) vm = = = 628.3 fpm 12 12 Fvm (12.3)(628.3) = 0.234 hp < fhp (c ) fhp = = 33,000 33,000

(e) ε = 0.8 , Table AT 20, L D = 1 S = 0.0446

r S =   cr

2

 µ ns µ n s  = p  p

D    cd 

(3.2 ×10 )(10)  0.0446 = −6

p

2

4    0.004 

2

p = 717.5 psi W = pDL = (717.5)(4)(4.4) = 12,628 lb

556.

For an 8 x 4 – in. full bearing, cr = 0.0075 in. , n = 2700 rpm , average µ = 4 ×10 −6 reyn . (a) What load may this bearing safely carry if the minimum film thickness is not to be less than that given by Norton, i11.14, Text? (b) Compute the corresponding frictional loss (fhp). (c) Complete calculations for the other quantities in Table AT 20, φ , q , qs , ∆to , pmax . Compute the maximum load for an optimum (load) bearing (d) if cr remains the same, (e) if ho remains the same.

Page 6 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS Solution: D × L = 8× 4 L D =1 2 cr = 0.0075 in r = D 2 = 4 in µ = 4 ×10 −6 reyn (a) by Norton, ho = 0.00025D = 0.00025(8) = 0.002 in ho 0.002 = = 0.27 cr 0.0075 h Table AT 20, L D = 1 2 , o = 0.27 cr S = 0.172 2

 r  µns S =    cr  p 2700 ns = = 45 rps 60 2

(

)

−6  4  4 × 10 (45) S = 0.172 =   p  0.0075  p = 298 psi W = pDL = (298)(8)(4) = 9536 lb

(b) Table AT 20, L D = 1 2 ,

ho = 0.27 cr

φ = 38.5o r f = 4.954 cr D f = 4.954 cd

 4    f = 4.954  0.004  f = 0.0093 F = f W = (0.0093)(9536 lb ) = 88.7 lb π Dn π (8)(2700) vm = = = 5655 fpm 12 12 Fvm (88.7 )(5655) = 15.2 hp fhp = = 33,000 33,000

Page 7 of 63



(c) Table AT 20, L D = 1 2 ,

ho = 0.27 cr

φ = 38.5o q = 5.214 rcr ns L q = 5.214rcr ns L = 5.214(4 )(0.0075)(45)(4 ) = 28.2 in 3 sec qs = 0.824 q qs = 0.824(28.2 ) = 23.2 in 3 sec ρ c∆t = 20.26 p 20.26(298) ∆t = = 54 o F 112 p = 0.3013 pmax 298 pmax = = 989 psi 0.3013 h To solve for maximum load, Table AT 20, L D = 1 2 , o = 0.43 cr

r S =   cr

2

 µns  = 0.388  p

(d) cr = 0.0075 in 2

(

)

−6  4  4 × 10 (45) S = 0.388 =   p  0.0075  p = 132 psi W = pDL = (132)(8)(4) = 4224 lb

(e) ho = 0.002 in ho = 0.43 cr 0.002 cr = = 0.00465 in 0.43 2

(

)

−6 4   4 × 10 (45) S = 0.388 =   p  0.00465  p = 343.3 psi W = pDL = (343.3)(8)(4) = 10,986 lb

Page 8 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS A 6 x 6 – in full bearing has a frictional loss of fhp = 11 when the load is 68,500 lb. and n = 1600 rpm ; cr r = 0.001 . (a) Compute the minimum film thickness. Is this in the vicinity of that for an optimum bearing? (b) What is the viscosity of the oil and a proper grade for an operating temperature of 160 F? (c) For the same ho , but for the maximum-load optimum, determine the permissible load and the fhp.

557.

Solution: L = 6 in D = 6 in L D =1 r = D 2 = 3 in cr r = 0.001 n = 1600 rpm π Dn π (3)(1600) vm = = = 2513 fpm 12 12 Fvm fhp = 33,000 33,000(11) F= = 144.45 lb 2513 F 144.45 f = = = 0.00211 W 68,500 r  1  f = (a) (0.00211) = 2.11 cr  0.001  r f = 2.11 Table AT 20, L D = 1 , cr Near the vicinity of optimum bearing cr = 0.001r = 0.001(3) = 0.003 in ho = 0.254cr = 0.254(0.003) = 0.0008 in (b) Table AT 20, L D = 1 ,

r f = 2.11 cr

S = 0.0652 2

 r  µns = 0.388 S =    cr  p 1600 ns = = 26.67 rps 60 W 68,500 p= = = 1902.8 psi LD (6)(6)

Page 9 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS 2

 1  (µ )(26.67 ) S = 0.0652 =    0.001  1902.8 µ = 4.7 ×10 −6 reyn Figure AF 16, 160 F, use SAE 40.

(c) Table AT 20, L D = 1 optimum bearing, maximum load,

ho = 0.53 cr

ho 0.0008 = = 0.0015 in 0.53 0.53 ho r = 0.53 , S = 0.214 , f = 4.89 cr cr

ho the same, cr =

r S =   cr

2

 µns   p 2

(

)

−6  3  4.7 × 10 (26.67 ) S = 0.214 =   p  0.0015  p = 2343 psi W = pDL = (2343)(6 )(6 ) = 84,348 lb r f = 4.89 cr

 3    f = 4.89  0.0015  f = 0.00245 F = f W = 0.00245(84,348) = 206.65 lb vm = 2513 fpm Fvm (206.65)(2513) = 15.74 hp fhp = = 33,000 33,000

558.

The maximum load on a 2.25 x 1.6875 in. main bearing of an automobile is 3140 lb. with wide-open throttle at 1000 rpm. If the oil is SAE 20W at 210 F, compute the minimum film thickness for a bearing clearance of (a) 0.0008 in. and (b) 0.0005 in. Which clearance results in the safer operating conditions? Note: Since a load of this order exists for only 20-25o of rotation, the actual ho does not reach this computed minimum (squeeze effect).

Solution: D × L = 2.25 × 1.6875 in L 1.6875 = = 0.75 D 2.25

Page 10 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS SAE 20 W at 210 oF µ = 0.96 ×10 −6 reyn W = 3140 lb n = 1000 rpm W 3140 p= = = 827 psi DL (2.25)(1.6875) 1000 ns = = 16.67 rps 60 D r = = 1.125 in 2

µ ns  r 

2

  p  cr  (a) cr = 0.0008 in S=

(0.96 ×10 )(16.67 )  1.125  S= −6

2

= 0.038 827  0.0008  Table AT 20, L D = 3 4 , S = 0.038

L D 1 ½

ho cr 0.2 0.2

S 0.0446 0.0923

¾

0.2

0.0685

L D 1 ½

ho cr 0.1 0.1

S 0.0188 0.0313

¾

0.1

0.0251

At L D = 3 4 ho  0.038 − 0.0251  = (0.2 − 0.1) + 0.1 = 0.13 cr  0.0685 − 0.0251  ho = 0.13cr = 0.13(0.0008) = 0.0001 in (b) cr = 0.0005 in

(0.96 ×10 )(16.67 )  1.125  S= −6

827

 0.0005 

2

= 0.098

Table AT 20, L D = 3 4 , S = 0.098

Page 11 of 63



L D 1 ½

ho cr 0.2 0.2

S 0.0446 0.0923

¾

0.2

0.0685

L D 1 ½

ho cr 0.4 0.4

S 0.121 0.319

¾

0.4

0.220

At L D = 3 4 ho  0.098 − 0.0685  = (0.4 − 0.2 ) + 0.2 = 0.239 cr  0.220 − 0.0685  ho = 0.239cr = 0.239(0.0005) = 0.00012 in use cr = 0.0005 in , ho = 0.00012 in A 360o bearing supports a load of 2500 lb.; D = 5 in. , L = 2.5 in. , cr = 0.003 in. , n = 1800 rpm ; SAE 20 W oil entering at 100 F. (a) Compute the average temperature t av of the oil through the bearing. (An iteration procedure. Assume µ ; compute S and the corresponding ∆to ; then the average oil temperature t av = ti + ∆to 2 . If this t av and the assumed µ do not locate a point in Fig. AF 16 on the line for SAE 20 W oil, try again.) Calculate (b) the minimum film thickness, (c) the fhp, (d) the amount of oil to be supplied and the end leakage.

561.

Solution: D = 5 in L = 2.5 in L 2.5 = = 0.5 D 5 cr = 0.003 in (a) Table AT 20 ρ c∆to , ρ c = 112 Parameter, p r S =   cr

2

 µns   p

Page 12 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS W 2500 = = 200 psi DL (5)(2.5) 1800 = 30 rps ns = 60 D r = = 2.5 in 2 cr = 0.003 in Fig. AF 16, SAE 20 W, Table AT 20, L D = 0.5 , ti = 100 o F ρ c∆to ∆to o F Trial µ ( t o F ), reyns S p 3.5 x 10-6 (130 F) 0.365 36.56 65 -6 3.2 x 10 (134 F) 0.333 34.08 61 -6 3.4 x 10 (132 F) 0.354 35.71 64 p=

Therefore, use t av = 132 o F , S = 0.354 (b) Table AT 20, L D = 0.5 , S = 0.354 ho = 0.415 cr ho = 0.415(0.003) = 0.00125 in (c) Table AT 20, L D = 0.5 , S = 0.354 r f = 8.777 cr  2.5    f = 8.777  0.003  f = 0.0105 F = f W = 0.0105(2500 ) = 26.25 lb π Dn π (5)(1800) vm = = = 2356 fpm 12 12 Fvm (26.25)(2356) = 1.874 hp fhp = = 33,000 33,000

(d) Table AT 20, L D = 0.5 , S = 0.354 q = 4.807 rcr ns L q = 4.807rcr ns L = 4.807(2.5)(0.003)(30 )(2.5) = 2.704 in 3 sec qs = 0.7165 q

Page 13 of 63

t av = ti + ∆to 2 o F

132.5 130.5 132.0


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS qs = 0.7165(2.704) = 1.937 in 3 sec

PARTIAL BEARINGS 562.

A 2 x 2-in. bearing has a clearance cr = 0.001 in , and ho = 0.0004 in. , n = 2400 rpm , and for the oil, µ = 3 × 10−6 reyn . Determine the load, frictional horsepower, the amount of oil to enter, the end leakage of oil, and the temperature rise of the oil as it passes through for : (a) a full bearing, partial bearings of (b) 180o, (c) 120o, (d) 90o, (e) 60o.

Solution: D = L = 2 in L D =1 cr = 0.001 in r = D 2 = 1 in n = 2400 rpm ns = 40 rps

µ = 3 ×10−6 reyn ho = 0.004 in. ho 0.0004 = = 0. 4 cr 0.001 πDn π (2)(2400) vm = = = 1257 fpm 12 12 (a) Full bearing Table AT 20, L D = 1 , ho cr = 0.4 S = 0.121 rf = 3.22 cr q = 4.33 rcr ns L qs = 0.680 q ρ c∆to = 14.2 p p = 0.415 pmax Load W

Page 14 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS r S =   cr

2

 µ ns   p 2

(

)

−6  1  3 ×10 (40 ) 0.121 =   p  0.001  p = 992 psi W = pDL = (992 )(2 )(2 ) = 3968 lb fhp: F = fW rf = 3.22 cr

 1    f = 3.22  0.001  f = 0.00322 F = f W = (0.00322 )(3968) = 12.78 lb Fvm (12.78)(1257 ) = 0.4868 hp fhp = = 33,000 33,000 Oil flow, q q = 4.33 rcr ns L q = 4.33 (0.1)(0.001)(40)(2) q = 0.3464 in3 sec End leakage qs = 0.680 q qs = 0.68(0.3464 ) = 0.2356 in3 sec

Temperature rise, ∆to ρ c∆to = 14.2 p (112)∆to = 14.2 992 ∆to = 126 o F (b) 180o Bearing Table AT 21, L D = 1 , ho cr = 0.4 S = 0.128

Page 15 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS rf = 2.28 cr q = 3.25 rcr ns L qs = 0.572 q ρ c∆to = 12.4 p

Load W r S =   cr

2

 µ ns   p 2

(

)

−6  1  3 × 10 (40 ) 0.128 =   p  0.001  p = 937.5 psi W = pDL = (937.5)(2 )(2 ) = 3750 lb fhp: F = fW rf = 2.28 cr

 1    f = 2.28  0.001  f = 0.00228 F = f W = (0.00228)(3750 ) = 8.55 lb Fvm (8.55)(1257 ) = 0.3257 hp fhp = = 33,000 33,000 Oil flow, q q = 3.25 rcr ns L q = 3.25 (0.1)(0.001)(40)(2) q = 0.26 in3 sec End leakage qs = 0.572 q qs = 0.572(0.26 ) = 0.1487 in3 sec

Temperature rise, ∆to

Page 16 of 63



ρ c∆to

= 12.4 p (112)∆to = 12.4 937.5 ∆to = 104 o F (c) 12o Bearing Table AT 22, L D = 1 , ho cr = 0.4 S = 0.162 rf = 2.16 cr q = 2.24 rcr ns L qs = 0.384 q ρ c∆to = 15 p Load W r S =   cr

2

 µ ns   p 2

(

)

−6  1  3 × 10 (40 ) 0.162 =   p  0.001  p = 741 psi W = pDL = (741)(2 )(2 ) = 2964 lb fhp: F = fW rf = 2.16 cr

 1    f = 2.16  0.001  f = 0.00216 F = f W = (0.00216 )(2964 ) = 6.4 lb Fvm (6.4)(1257 ) = 0.2438 hp fhp = = 33,000 33,000 Oil flow, q

Page 17 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS q = 2.24 rcr ns L q = 2.24 (0.1)(0.001)(40)(2) q = 0.1792 in3 sec End leakage qs = 0.384 q q s = 0.384(0.1792) = 0.0688 in 3 sec

Temperature rise, ∆to ρ c∆to = 15 p (112)∆to = 15 741 ∆to = 99 o F (d) 60o Bearing L D = 1 , ho cr = 0.4 S = 0.450 rf = 3.29 cr q = 1.56 rcr ns L qs = 0.127 q ρ c∆to = 28.2 p Load W r S =   cr

2

 µ ns   p 2

(

)

−6  1  3 × 10 (40 ) 0.450 =   p  0.001  p = 267 psi W = pDL = (267 )(2 )(2 ) = 1068 lb fhp: F = fW

Page 18 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS rf = 3.29 cr

 1    f = 3.29  0.001  f = 0.00329 F = f W = (0.00329 )(1068) = 3.514 lb Fvm (3.514 )(1257 ) = 0.1339 hp fhp = = 33,000 33,000 Oil flow, q q = 1.56 rcr ns L q = 1.56 (0.1)(0.001)(40)(2) q = 0.1248 in3 sec End leakage qs = 0.127 q qs = 0.127(0.1248) = 0.0158 in3 sec

Temperature rise, ∆to ρ c∆to = 28.2 p (112)∆to = 28.2 267 ∆to = 67 o F 563.

A 2 x 2 in. bearing sustains a load of W = 5000 lb. ; cr = 0.001 in. ; n = 2400 rpm ; µ = 3 × 10−6 reyn . Using Figs. AF 17 and AF 18, determine the minimum film thickness and the frictional loss (ft-lb/min.) for (a) a full bearing, and for partial bearings of (b) 180o, (c) 120o, (d) 90o, (e) 60o.

Solution: L = 2 in D = 2 in W = 5000 lb cr = 0.001 in. n = 2400 rpm ns = 40 rps

µ = 3 ×10−6 reyn r = D 2 = 1 in

Page 19 of 63



p=

W 5000 = = 1250 psi LD (2)(2) 2

2

(

)

 r  µns  1  3 × 10−6 (40 ) S =   = = 0.10  1250  cr  p  0.001  πDn π (2)(2400) vm = = = 1257 fpm 12 12 Using Fig. AF 17 and AF 18

(a) Full Bearing ho = 0.346 cr r f = 2.8 cr ho = 0.346(0.001) = 0.000346 in

 1    f = 2.8  0.001  f = 0.0028 F = f W = (0.0028)(5000 ) = 14 lb Fvm = (14 )(1257 ) = 17,600 ft − lb min

(b) 180o Bearing ho = 0.344 cr r f = 2.0 cr ho = 0.344(0.001) = 0.000344 in

 1    f = 2.0  0.001  f = 0.0020 F = f W = (0.0020 )(5000 ) = 10 lb Fvm = (10 )(1257 ) = 12,570 ft − lb min

(c) 120o Bearing ho = 0.302 cr

Page 20 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS r f = 1.7 cr ho = 0.302(0.001) = 0.000302 in

 1    f = 1.7  0.001  f = 0.0017 F = f W = (0.0017 )(5000 ) = 8.5 lb Fvm = (8.5)(1257 ) = 10,685 ft − lb min

(d) 60o Bearing ho = 0.20 cr r f = 1.4 cr ho = 0.20(0.001) = 0.0002 in

 1    f = 1.4  0.001  f = 0.0014 F = f W = (0.0014 )(5000 ) = 7 lb Fvm = (7 )(1257 ) = 8,800 ft − lb min

564.

A 120o partial bearing is to support 4500 lb. with ho = 0.002 in. ; L D = 1 ; D = 4 in. ; cd = 0.010 in. ; n = 3600 rpm . Determine (a) the oil’s viscosity,(b) the frictional loss (ft-lb/min), (c) the eccentricity angle, (d) the needed oil flow, (e) the end leakage, (f) the temperature rise of the oil as it passes through, (g) the maximum pressure. (h) If the clearance given is the average, what approximate class of fit (Table 3.1) is it? (i) What maximum impulsive load would be on the bearing if the eccentricity ratio suddenly went to 0.8? Ignore “squeeze” effect.

Solution: W = 4500 lb ho = 0.002 in L D =1 D = 4 in L = 4 in r = D 2 = 2 in cd = 0.010 in. n = 3600 rpm

Page 21 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS 3600 = 60 rps 60 πDn π (2)(3600) vm = = = 3770 fpm 12 12 W 4500 p= = = 281.25 psi LD (4 )(4 ) ho 2ho 2(0.002) = = = 0.4 cr cr 0.010 Table AT 22, L D = 1 , ho cr = 0.4 S = 0.162 φ = 35.65o r f = 2.16 cr q = 2.24 rcr ns L qs = 0.384 q ρ c∆to = 15.0 p p = 0.356 pmax ns =

r (a) S =   cr

2

 µns   p

2

 D  µns S =    cd  p 2

 4  µ (60 ) 0.162 =    0.010  281.25 µ = 4.75 ×10−6 reyn

(b)

r f = 2.16 cr

D f = 2.16 cd

 4    f = 2.16  0.010  f = 0.0054 F = f W = 0.0054(4500 ) = 24.30 lb

Page 22 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS Fvm = (24.30 )(3770 ) = 91,611 ft − lb min (c) φ = 35.65o 4q q = = 2.24 rcr ns L Dcd ns L 4q = 2.24 (4)(0.010)(60)(4) q = 5.4 in3 sec

(d)

qs = 0.384 q qs = 0.384(5.4 ) = 2.07 in3 sec

(e)

ρ c∆to

= 15.0 p (112)∆to = 15.0 281.25 ∆to = 38 o F

(f)

p

= 0.356 pmax 281.25 pmax = = 790 psi 0.356

(g)

(h) cd = 0.010 in , D = 4 in Table 3.1 RC 8, Hole, average = + 0.0025 Shaft, average = - 0.00875 cd = 0.0025 + 0.00875 = 0.01125 ≈ 0.010 in Class of fit = RC 9 (i) ε = 0.80 Table AT 22, , L D = 1 S = 0.162 2

 D  µ ns S =    cd  p 2

(

)

−6  4  3 × 10 (60 ) 0.0531 =   p  0.010  p = 542 psi

Page 23 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS W = pDL = (542 )(4 )(4 ) = 8672 lb A 120o partial bearing is to support 4500 lb., D = 3 in. , cd = 0.003 in. ; n = 3600 rpm ; SAE 20W entering at 110 F. Calculate (a) the average temperature of the oil as it passes through,(b) the minimum film thickness, (c) the fhp, (d) the quantity of oil to be supplied. HINT: In (a) assume µ and determine the corresponding values of S and ∆to ; then tav = ti + ∆to 2 . If assumed µ and tav do not locate a point in Fig. AF 16 that falls on line for SAE 20W, iterate.

565.

Solution: W = 4500 lb D = 3 in L = 3 in L D =1 cd = 0.003 in. 2

 D  µns S =    cd  p 3600 ns = = 60 rps 60 W 4500 p= = = 500 psi DL (3)(3) ρ c∆to , (SAE 20W) p

(a) Using Table AT22, L D = 1 , ρ c = 112 , ti = 110o F Trial µ

t , oF

S

3.5 x 10-6 2.0 x 10-6 2.6 x 10-6 2.35 x 10-6 2.4 x 10-6

130 160 145 150 149

0.42 0.24 0.312 0.282 0.288

∴ Use tav = 149 o F

(b) Table AT 22, L D = 1 , S = 0.288

Page 24 of 63

ρ c∆to p 19.8 15.4 17.7 17.2 17.3

∆to

tav = ti + ∆to 2

88 68 79 76 78

154 144 149.5 148 149


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS ho = 0.513 cr 2ho = 0.513 cd 2ho = 0.513(0.003)

ho = 0.00077 in (c) Table At 22, L D = 1 , S = 0.288 r f = 2.974 cr D 3 f = f = 2.974 cr 0.003 f = 0.002974 F = f W = (0.002974 )(4500 ) = 13.383 lb Fvm fhp = 33,000 πDn π (3)(3600) vm = = = 2827 fpm 12 12 Fvm (13.383)(2827 ) = 1.15 hp fhp = = 33,000 33,000

(d) Table At 22, L D = 1 , S = 0.288 q = 2.528 rcr ns L 4q = 2.528 Dcd ns L 4q = 2.528 (3)(0.003)(60)(3) q = 1.024 in3 sec

566.

The 6000-lb. reaction on an 8 x 4 –in., 180o partial bearing is centrally applied; n = 1000 rpm ; ho = 0.002 in . For an optimum bearing with minimum friction determine (a) the clearance, (b) the oil’s viscosity, (c) the frictional horsepower. (d) Choose a cd D ratio either smaller or larger than that obtained in (a) and show that the friction loss is greater than that in the optimum bearing. Other data remain the same.

Solution: W = 6000 lb

Page 25 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS D = 8 in L = 4 in n = 1000 rpm 1000 ns = = 16.67 rps 60 L D =1 2 ho = 0.002 in

(a) Table AT 21, L D = 1 2 Optimum value (minimum friction) ho cr = 0.23 0.002 cr = = 0.0087 in 0.23 (b) Table AT 21, L D = 1 2 , ho cr = 0.23 S = 0.126 2

 r  µns S =    cr  p W 6000 p= = = 187.5 psi DL (4)(8) D r = = 4 in 2 2

 4  µ (16.67 ) S = 0.126 =    0.0087  187.5 µ = 6.70 ×10−6 reyn

(c) Table AT 21, L D = 1 2 , ho cr = 0.23 r f = 2.97 cr

 4    f = 2.97  0.0087  f = 0.00646 F = f W = (0.00646 )(6000 ) = 38.76 lb πDn π (8)(1000) vm = = = 2094 fpm 12 12

Page 26 of 63



fhp =

Fvm (38.76)(2094) = 2.46 hp = 33,000 33,000

For (a)

cd 2cr 2(0.0087 ) = = = 0.0022 D D 8

cd > 0.0022 D cd = 0.0030 D cd = 0.0030(8) = 0.0240 in

cr = 0.0120 in ho 0.002 = = 0.1667 cr 0.012 Table AT 21, L D = 1 2 r f = 1.67 cr  4    f = 1.67  0.0016  f = 0.00668 F = f W = (0.00668)(6000 ) = 40.08 lb πDn π (8)(1000) vm = = = 2094 fpm 12 12 Fvm (40.08)(2094) = 2.54 hp > 2.46 hp fhp = = 33,000 33,000 cd < 0.0022 D cd = 0.0020 D cd = 0.0020(8) = 0.0160 in

cr = 0.0080 in ho 0.002 = = 0.25 cr 0.008 Table AT 21, L D = 1 2 r f = 3.26 cr  4    f = 3.26  0.0016  f = 0.00652

Page 27 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS F = f W = (0.00652 )(6000 ) = 39.12 lb πDn π (8)(1000) vm = = = 2094 fpm 12 12 Fvm (39.12)(2094) = 2.48 hp > 2.46 hp fhp = = 33,000 33,000 A 120o partial bearing supports 3500 lb. when n = 250 rpm ; D = 5 in. , L = 5 in. ; µ = 3 × 10−6 reyn . What are the clearance and minimum film thickness for an optimum bearing (a) for maximum load, (b) for minimum friction? (c) On the basis of the average clearance in Table 3.1, about what class fit is involved? Would this fit be on the expensive or inexpensive side? (d) Find the fhp for each optimum bearing.

567.

Solution: D = 5 in. L = 5 in. L =1 D n = 250 rpm 250 ns = = 4.17 rps 60 µ = 3 ×10−6 reyn W = 3500 lb W 3500 p= = = 140 psi DL (5)(5)

(a)

Table AT 22,

h L = 1 , max. load o = 0.46 cr D

S = 0.229 2

 r  µns S =    cr  p D r = = 2.5 in 2 2

(

)

 2.5  3.0 × 10−6 (4.17 )  S = 0.229 =  140  cr  cr = 0.00156 in ho = 0.46cr = 0.46(0.00156 ) = 0.00072 in

Page 28 of 63



(b)

Table AT 22,

L h = 1 , min. friction o = 0.40 cr D

S = 0.162 2

 r  µns S =    cr  p D r = = 2.5 in 2 2

(

)

 2.5  3.0 × 10−6 (4.17 )  S = 0.162 =  140  cr  cr = 0.00186 in ho = 0.46cr = 0.40(0.00186 ) = 0.00074 in

(c) cd 1 = 2(0.00156 ) = 0.00312 in cd 2 = 2(0.00186 ) = 0.00372 in Use Class RC4, ave. cd = 0.00320 in , expensive side (d) Table AT 22,

L h = 1 , max. load o = 0.46 cr D

r f = 2.592 cr

 2.5    f = 2.592  0.00156  f = 0.00162 F = f W = (0.00162 )(3500 ) = 5.67 lb πDn π (5)(250) vm = = = 327.25 fpm 12 12 Fvm (5.67 )(327.25) = 0.0562 hp fhp = = 33,000 33,000

For minimum friction,

ho = 0.40 cr

r f = 2.16 cr

 2.5    f = 2.16  0.00186  f = 0.00161 F = f W = (0.00161)(3500 ) = 5.635 lb

Page 29 of 63



vm =

πDn

π (5)(250)

= 327.25 fpm 12 12 Fvm (5.635)(327.25) = 0.0559 hp fhp = = 33,000 33,000

570.

=

A 180o partial bearing is to support 17,000 lb. with p = 200 psi , n = 1500 rpm , ho = 0.003 in , L D = 1 . (a) Determine the clearance for an optimum bearing with minimum friction. (b) Taking this clearance as the average, choose a fit (Table 3.1) that is approximately suitable. (c) Select an oil for an average temperature of 150 F. (d) Compute fhp.

Solution: W = 17,000 lb p = 200 psi n = 1500 rpm 1500 ns = = 25 rps 60 L D =1 L=D W p= DL 17,000 200 = D2 D = L = 9.22 in D 9.22 r= = = 4.61 in 2 2 (a) For optimum bearing with minimum friction

Table AT 21, L D = 1 , ho cr = 0.44 ho cr = 0.44 0.003 = 0.44 cr cr = 0.00682 in (b) Table 3.1, D = 9.22 in cd = 2cr = 2(0.00682 ) = 0.01364 in Use Class RC7, average cd = 0.01065 in Or use Class RC8, average cd = 0.01575 in (c) Table AT 21, L D = 1 , ho cr = 0.44 S = 0.158 Page 30 of 63



2

 µns   p 2

 4.61  µ (25) 0.158 =    0.00682  200 µ = 2.8 ×10−6 reyn Fig. AF 16, at 150 F Use Either SAE 20W or SAE 30.

(d) Table AT 21, L D = 1 , ho cr = 0.44 r f = 2.546 cr  4.61    f = 2.546  0.00682  f = 0.00377 πDn π (9.22)(1500) vm = = = 3621 fpm 12 12 F = f W = (0.00377 )(17,000 ) = 64.09 lb Fvm (64.09)(3621) = 7.0 hp fhp = = 33,000 33,000

571.

The reaction on a 120o partial bearing is 2000 lb. The 3-in journal turns at 1140 rpm; cd = 0.003 in. ; the oil is SAE 20W at an average operating temperature of 150 F. Plot curves for the minimum film thickness and the frictional loss in the bearing against the ratio L D , using L D = 0.25, 0.5, 1, and 2. (Note: This problem may be worked as a class problem with each student being responsible for a particular L D ratio.)

Solution: W = 2000 lb D = 3 in. n = 1140 rpm 1140 ns = = 19 rps 60 cd = 0.003 in cr = 0.0015 in For SAE 20W, 150 F µ = 2.75 ×10−6 reyn (a)

L = 0.25 D

Page 31 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS L = 0.25 D = 0.25(3) = 0.75 in W 2000 p= = = 889 psi DL (3)(0.75) L = 0.25 Table AT 22, D D r = = 1.5 in 2 2

2

(

)

 r  µns  1.5  2.75 ×10−6 (19 ) S =   = = 0.0588  889 c p 0 . 0015    r ho = 0.083 cr ho = 0.083(0.0015) = 0.000125 in r f = 2.193 cr  1.5    f = 2.193  0.0015  f = 0.002193 F = f W = (0.002193)(2000 ) = 4.386 lb πDn π (3)(1140) vm = = = 895 fpm 12 12 Fvm (4.386)(895) = 0.119 hp fhp = = 33,000 33,000 L = 0.5 (b) D L = 0.5 D = 0.5(3) = 1.5 in W 2000 p= = = 444 psi DL (3)(1.5) L = 0.5 Table AT 22, D D r = = 1.5 in 2 2

2

(

)

 r  µns  1.5  2.75 × 10−6 (19 ) S =   = = 0.1177  444  cr  p  0.0015  ho = 0.2159 cr ho = 0.2159(0.0015) = 0.000324 in r f = 2.35 cr

Page 32 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS  1.5    f = 2.35  0.0015  f = 0.00235 F = f W = (0.00235)(2000 ) = 4.7 lb πDn π (3)(1140) vm = = = 895 fpm 12 12 Fvm (4.7 )(895) = 0.1275 hp fhp = = 33,000 33,000 L =1 (c) D L = D = 3 in W 2000 p= = = 222 psi DL (3)(3) L =1 Table AT 22, D D r = = 1.5 in 2 2

2

(

)

 r  µns  1.5  2.75 × 10−6 (19 ) S =   = = 0.2354  222  cr  p  0.0015  ho = 0.4658 cr ho = 0.4658(0.0015) = 0.000699 in r f = 2.634 cr  1.5    f = 2.634  0.0015  f = 0.002634 F = f W = (0.002634 )(2000 ) = 5.268 lb πDn π (3)(1140) vm = = = 895 fpm 12 12 Fvm (5.268)(895) = 0.1429 hp fhp = = 33,000 33,000 L =2 (d) D L = 2 D = 2(3) = 6 in W 2000 p= = = 111 psi DL (3)(6) L Table AT 22, =2 D

Page 33 of 63



r=

D = 1.5 in 2 2

2

(

)

 r  µns  1.5  2.75 × 10−6 (19 ) S =   = = 0.47  111  cr  p  0.0015  ho = 0.718 cr ho = 0.718(0.0015) = 0.00108 in r f = 3.8118 cr  1.5    f = 3.8118  0.0015  f = 0.003812 F = f W = (0.003812 )(2000 ) = 7.624 lb πDn π (3)(1140) vm = = = 895 fpm 12 12 Fvm (7.624)(895) = 0.2068 hp fhp = = 33,000 33,000 L D 0.25 0.5 1.0 2.0

Page 34 of 63

ho , in

fhp

0.000125 0.000324 0.000699 0.001080

0.119 0.128 0.143 0.207


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS STEADY-STATE TEMPERATURE A 180o partial bearing is subjected to a load of 12,000 lb.; D × L = 8 × 8 in. , cr r = 0.0015 , ho ≈ 0.0024 in. , n = 500 rpm . The air speed about the bearing is expected to be in excess of 1000 fpm (on moving vehicle) and the effective radiating area is 20 DL . Determine: (a) the eccentricity factor, (b) µreyns, (c) the frictional loss (ft-lb/min), (d) the estimated temperature of oil and bearing ( a self-contained oil-bath unit) for steady-state operation, and a suitable oil.(e) Compute ∆to of the oil passing through the load-carrying area, remark on its reasonableness, and decide upon whether some redesign is desirable.

572.

Solution: D = 8 in. L = 8 in. L D =1 W = 12,000 lb D r = = 4 in 2 cr = 0.0015r = 0.0015(4) = 0.0060 in ho 0.0024 = = 0. 4 cr 0.0060 n = 500 rpm 500 ns = = 8.33 rps 60 Table AT 21, ho cr = 0.4 , L D = 1 S = 0.128 r f = 2.28 cr ρ c∆to = 12.4 p W 12,000 p= = = 187.5 psi DL (8)(8) (a) ε = 1 −

ho = 1 − 0.4 = 0.6 cr

r (b) S =   cr

2

 µns   p 2

 4  µ (8.33) S = 0.128 =    0.0060  187.5 µ = 6.5 ×10−6 reyn

Page 35 of 63



(c)

r f = 2.28 cr

 4    f = 2.28  0.0060  f = 0.00342 F = f W = (0.00342 )(12,000 ) = 41.04 lb πDn π (8)(500) vm = = = 1047 fpm 12 12 Fvm (41.04)(1047 ) = 1.302 hp fhp = = 33,000 33,000 Frictional loss = 43,000 ft-lb/min

(d) Q = hcr Ab ∆tb ft-lb/min Q = 43,000 ft − lb min hcr = hc + hr hr = 0.108 ft − lb min − sq.in. − F va0.6 , va ≥ 1000 fpm D 0.4 0.6 ( 1000) hc = 0.017 = 0.467 ft − lb min − sq.in. − F (8)0.4 hcr = 0.467 + 0.108 = 0.575 ft − lb min − sq.in. − F Ab = 20 DL = 20(8)(8) = 1280 sq.in. Q = hcr Ab ∆tb 43,000 = (0.575)(1280 )(∆tb ) ∆tb = 58.42 F Oil-bath, 1000 fpm ∆toa ≈ (1.2)(1.3)(∆tb ) ∆toa = (1.2 )(1.3)(58.42 ) = 91.1 F assume 100 F ambient temperature tb = 100 + 58.42 F = 158.42 F tb = 100 + 91.1 F = 191.1 F hc = 0.017

ρ c∆to

= 12.4 p (112)∆to = 12.4 187.5 ∆to = 20.8 F Solve for to 2 (c)

Page 36 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS to1 + to 2 = 2(191.1) = 382.2 F to1 = 382.2 − to 2 to 2 − to1 = 20.8 F to 2 − 362.2 + to 2 = 20.8 to 2 = 201.5 F ≈ 200 F ∴ not reasonable since the oil oxidizes more rapidly above 200 F, a redesign is desireable. 573.

A 2 x 2-in. full bearing (ring-oiled) has a clearance ratio cd D = 0.001 . The journal speed is 500 rpm, µ = 3.4 × 10−6 reyn , and ho = 0.0005 in. The ambient temperature is 100 F; Ab = 25DL , and the transmittance is taken as hcr = 2 Btu hr − sq. ft. − F . Calculate (a) the total load for this condition; (b) the frictional loss, (c) the average temperature of the oil for steady-state operation. Is this temperature satisfactory? (d) For the temperature found, what oil do you recommend? For this oil will ho be less or greater than the specified value? (e) Compute the temperature rise of the oil as it passes through the bearing. Is this compatible with other temperatures found? (f) What minimum quantity of oil should the ring deliver to the bearing?

Solution: L = 2 in. D = 2 in. cd D = 0.001 cd = (0.001)(2) = 0.0020 in

µ = 3.4 ×10−6 reyn ho = 0.0005 in. cr = 0.0010 in ho cr = 0.0005 0.0010 = 0.5 Table AT 20, L D = 1 , ho cr = 0.5 , Full Bearing S = 0.1925 r f = 4.505 cr q = 4.16 rcr ns L ρ c∆to = 19.25 p (a) S = 0.1925 Page 37 of 63



 r  µns S =    cr  p D r = = 1 in 2 500 ns = = 8.33 rps 60 2

(

)

−6  1  3.4 × 10 (8.330) S = 0.1925 =   p  0.0010  p = 147 psi W = pDL = (147 )(2 )(2 ) = 588 lb

(b)

r f = 4.505 cr

 1    f = 4.505  0.001  f = 0.004505 F = f W = (0.004505)(588) = 2.649 lb πDn π (2)(500) vm = = = 261.8 fpm 12 12 U f = Fvm = (2.649 )(261.8) = 693.5 ft − lb min

(c) Q = hcr Ab ∆tb hcr = 2 Btu hr − sq. ft. − F = 0.18 ft − lb min − sq.in. − F Ab = 25 DL = 25(2 )(2 ) = 100 sq.in. Q =Uf

(0.18)(100)(∆tb ) = 693.5 ∆tb = 38.53 F ∆toa = 2∆tb = 2(38.53) = 77 F to = 77 + 100 = 177 F , near 160 F ∴ satisfactory. (d) to = 177 F , µ = 3.4 × 10−6 reyn Figure AF 16 Use SAE 40 oil, µ = 3.3 × 10−6 reyn r S =   cr

2

 µns   p

Page 38 of 63



(

)

−6  1  3.3 × 10 (8.33) S = = 0.187  147  0.0010  Table AT 20, L D = 1 , S = 0.187 ho cr = 0.4923

ho = 0.4923(0.0010 ) = 0.00049 in < ho (= 0.0005 in ) ρ c∆to = 19.25 (e) p (112)∆to = 19.25 147 ∆to = 25.3 F ∆to1 + ∆to 2 = 2(177 ) = 354 F ∆t o 2 − ∆t o1 = 25.3 F 2∆to 2 = 354 + 25.3 ∆to 2 = 190 F < 200 F compatible. ∴ q = 4.16 rcr ns L q = 4.16 (1)(0.001)(8.33)(2) q = 0.0693 in3 sec

(f)

574.

An 8 x 9-in. full bearing (consider L D = 1 for table and chart use only) supports 15 kips with n = 1200 rpm ; cr r = 0.0012 ; construction is medium heavy with a radiating-and-convecting area of about 18DL ; air flow about the bearing of 80 fpm may be counted on (nearby) pulley; ambient temperature is 90 F. Decide upon a suitable minimum film thickness. (a) Compute the frictional loss and the steady state temperature. Is additional cooling needed for a reasonable temperature? Determine (b) the temperature rise of the oil as it passes through the load-carrying area and the grade of oil to be used if it enters the bearing at 130 F, (c) the quantity of oil needed.

Solution: D = 8 in. L = 9 in. W = 15,000 lb. n = 1200 rpm. 1200 ns = = 20 rps 60 cr r = 0.0012

Page 39 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS r = D 2 = 4 in cr = 0.0012(4) = 0.0048 in By Norton: ho = 0.00025 D = 0.00025(8) = 0.002 in ho 0.002 = = 0.4 cr 0.0048 Table AT 20, L D = 1 , ho cr = 0.4 S = 0.121 r f = 3.22 cr q = 4.33 rcr ns L ρ c∆to = 14.2 p (a)

r f = 3.22 cr

 4    f = 3.22  0.0048  f = 0.003864 F = f W = (0.003864 )(15,000 ) = 57.96 lb πDn π (8)(1200) vm = = = 2513 fpm 12 12 U f = Fvm = (57.96 )(2513) = 145,654 ft − lb min

Q = hcr Ab ∆tb hr = 0.108 ft − lb min − sq.in. − F va0.6 ft − lb min − sq.in. − F D 0.4 (80)0.6 = 0.103 ft − lb min − sq.in. − F hc = 0.017 (8)0.4 hcr = hc + hr = 0.103 + 0.108 = 0.211 ft − lb min − sq.in. − F Ab = 18 DL = 18(8)(9 ) = 1296 sq.in. Uf =Q hc = 0.017

145,654 = (0.211)(1296 )∆tb ∆tb = 533 F , very high, additional cooling is necessary. (b)

ρ c∆to p

= 14.2

Page 40 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS W 15,000 = = 208 psi DL (8)(9) (112)∆to = 14.2 208 ∆to = 26 F ti = 130 F to = 156 F tave = 12 (130 + 156 ) = 143 F p=

r S =   cr

2

 µns   p 2

 4  µ (20 ) S = 0.121 =    0.0048  208 µ = 1.8 ×10−6 reyn Figure AF 16, µ = 1.8 µreyns , 143 F Use SAE 10W q = 4.33 rcr ns L q = 4.33 (4)(0.0048)(20)(9) q = 14.96 in3 sec

(c)

575.

A 3.5 x 3.5-in., 360o bearing has cr r = 0.0012 ; n = 300 rpm ; desired minimum ho ≈ 0.0007 in . It is desired that the bearing be self-contained (oilring); air-circulation of 80 fpm is expected; heavy construction, so that Ab ≈ 25DL . For the first look at the bearing, assume µ = 2.8 × 10−6 reyn and compute (a) the frictional loss (ft-lb/min), (b) the average temperature of the bearing and oil as obtained for steady-state operation, (c) ∆to as the oil passes through the load-carrying area (noting whether comparative values are reasonable). (d) Select an oil for the steady-state temperature and decide whether there will be any overheating troubles.

Solution: D = 3.5 in. L = 3.5 in. cr r = 0.0012 r = D 2 = 1.75 in. cr = (0.0012 )(1.75) = 0.0021 in ho ≈ 0.0007 in

Page 41 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS ho cr = 0.0007 0.0021 = 0.333 Table AT 20, 360o Bearing, L D = 1 , ho cr = 0.333 S = 0.0954 r f = 2.71 cr ρ c∆to = 12.12 p 2

 r  µns (a) S =    cr  p 300 ns = = 5 rps 60 µ = 2.8 ×10−6 reyn 2

(

)

−6  1.75  2.8 × 10 (5) S = 0.0954 =   p  0.0021  p = 102 psi W = pDL = (102 )(3.5)(3.5) = 1250 lb r f = 2.71 cr

 1.75    f = 2.71  0.0021  f = 0.00325 F = f W = (0.00325)(1250 ) = 4.0625 lb πDn π (3.5)(300 ) vm = = = 275 fpm 12 12 U f = Fvm = (4.0625)(275) = 1117 ft − lb min

(b) Q = hcr Ab ∆tb hr = 0.108 ft − lb min − sq.in. − F va0.6 ft − lb min − sq.in. − F D 0.4 0.6 ( 80 ) hc = 0.017 = 0.143 ft − lb min − sq.in. − F (3.5)0.4 hcr = hc + hr = 0.143 + 0.108 = 0.251 ft − lb min − sq.in. − F Ab = 25 DL = 25(3.5)(3.5) = 306.25 sq.in. Uf =Q hc = 0.017

1117 = (0.251)(306.25)∆tb

Page 42 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS ∆tb = 14.5 F ∆toa = 2∆tb = 2(14.5) = 29 F assume ambient temperature of 100 F tb = 114.5 F to = 129 F

ρ c∆to

= 12.12 p (112)∆to = 12.12 102 ∆to = 11 F to1 + to 2 = 2(129 ) = 258 F to 2 − to1 = 11 F 2to 2 = 269 F to 2 = 135 F < 140 F ∴ reasonable

(c)

(d) to = 129 F , µ = 2.8 × 10−6 reyn use SAE 10W Figure AF 16, to = 126 F ∆toa = 126 − 100 = 26 F ∆toa = 2∆tb 26 = 13 F ∆t b = 2 Q = hcr Ab ∆tb = (0.251)(306.25)(13) = 999 ft − lb min < U f ∴ there is an overheating problem. 576.

A 10-in. full journal for a steam-turbine rotor that turns 3600 rpm supports a 20-kip load with p = 200 psi ; cr r = 0.00133 . The oil is to have µ = 2.06 ×10−6 reyn at an average oil temperature of 130 F. Compute (a) the minimum film thickness (comment on its adequacy), (b) the fhp, (c) the altitude angle, the maximum pressure, and the quantity of oil that passes through the load-carrying area (gpm).(d) At what temperature must the oil be introduced in order to have 130 F average? (e) Estimate the amount of heat lost by natural means from the bearing (considered oil bath) with air speed of 300 fpm. If the amount of oil flow computed above is cooled back to the entering temperature, how much heat is removed? Is this total amount of heat enough to care for frictional loss? If not, what can be done (i11.21)?

Page 43 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS Solution: D = 10 in. n = 3600 rpm 3600 ns = = 60 rps 60 W = 20,000 lb p = 200 psi W p= DL 20,000 200 = 10 L L = 10 in L D =1 D r = = 5 in 2 cr r = 0.00133 cr = 0.00133(5) = 0.00665 in µ = 2.06 ×10−6 reyn tave = 130 F r S =   cr

2

 µns   p 2

(

)

−6 5   2.06 × 10 (60 ) S = = 0.35  200  0.00665  Table AT 20, L D = 1 , S = 0.35 ho cr = 0.647

φ = 65.66o r f = 7.433 cr q = 3.90 rcr ns L p = 0.495 pmax ρ c∆to = 30.8 p qs = 0.446 q

(a) ho = 0.647cr = 0.647(0.00665) = 0.00430 in

Page 44 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS Norton’s recommendation = 0.00025 D = 0.00025(10 ) = 0.00250 in < 0.00430 in ∴ adequate (b)

r f = 7.433 cr

5     f = 7.433 00665 0 .   f = 0.0099 F = f W = (0.0099 )(20,000 ) = 198 lb πDn π (10)(3600) vm = = = 9425 fpm 12 12 Fvm (198)(9425) = 56.55 hp fhp = = 33,000 33,000

(c) φ = 65.66o p 200 pmax = = = 404 psi 0.495 0.495 q = 3.90rcr ns L q = 3.90(5)(0.00665)(60 )(10 ) = 77.805 in3 sec q = (77.805 in3 sec )(1 gpm 231 in3 )(60 sec min ) = 0.21 gpm

ρ c∆to

= 30.8 p (112)∆to = 30.8 200 ∆to = 55 F ∆t tave = ti + o 2 55 130 = ti + 2 ti = 102.5 F (e) Q = hcr Ab ∆tb hr = 0.108 ft − lb min − sq.in. − F (d)

va0.6 ft − lb min − sq.in. − F D 0.4 0.6 ( 300) hc = 0.017 = 0.207 ft − lb min − sq.in. − F (3.5)0.4 hcr = hc + hr = 0.207 + 0.108 = 0.315 ft − lb min − sq.in. − F Assume hc = 0.017

Page 45 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS Ab = 25 DL = 25(10 )(10 ) = 2500 sq.in. ∆toa = 130 − 100 = 30 F ∆toa = 1.3∆tb 30 = 23 F ∆t b = 1 .3 Q = (0.315)(2500 )(23) = 18,113 ft − lb min Qr = ρ c(q − qs )∆to in − lb sec Qr = (112 )(1 − 0.446 )(77.805)(55)(1 12 )(60 ) = 1,327,602 ft − lb min QT = Q + Qr = 18,113 + 1,327,602 = 1,345,735 ft − lb min U f = Fvm = (198)(9425) = 1,866,150 ft − lb min > QT not enough to care for frictional loss, use pressure feed (i11.21). DESIGN PROBLEMS 578.

A 3.5-in. full bearing on an air compressor is to be designed for a load of 1500 lb.; n = 300 rpm ; let L D = 1 . Probably a medium running for would be satisfactory. Design for an average clearance that is decided by considering both Table 3.1 and 11.1. Choose a reasonable ho , say one that gives ho cr ≈ 0.5 . Compute all parameters that are available via the Text after you have decided on details. It is desired that the bearing operate at a reasonable steady-state temperature (perhaps ring-oiled medium construction), without special cooling. Specify the oil to be used and show all calculations to support your conclusions. What could be the magnitude of the maximum impulsive load if the eccentricity ration ε becomes 0.8, “squeeze” effect ignored?

Solution: L D =1 D = 3.5 in L = 3.5 in W = 1500 lb n = 300 rpm 300 ns = = 5 rps 60 W 1500 p= = = 122.45 psi DL (3.5)(3.5) Table 3.1, medium running fit, D = 3.5 in RC 5 or RC 6 Use RC 6 Average cd = 0.0052 in

Page 46 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS Table 11.1, air-compressor General Machine Practice Average cd = 0.0055 in Using cd = 0.0055 in cr = 0.00275 in ho = 0.5cr = 0.5(0.00275) = 0.001375 in Table AT 20, L D = 1 , ho cr = 0.5 ε = 0.5 S = 0.1925 φ = 56.84o r f = 4.505 cr q = 4.16 rcr ns L ρ c∆to = 19.25 p p = 0.4995 pmax Specifying oil: Q = hcr Ab ∆tb U f = Fvm r f = 4.505 cr

 1.75    f = 4.505  0.00275  f = 0.00708 F = f W = (0.00708)(1500 ) = 10.62 lb πDn π (3.5)(300 ) vm = = = 275 fpm 12 12 U f = Fvm = (10.62)(275) = 2921 ft − lb min

Q = hcr Ab ∆tb Assume hcr = 0.516 ft − lb min − sq.in. − F Medium construction Ab = 15.5 DL = 15.5(3.5)(3.5) = 189.875 sq.in. Oil-ring bearing ∆toa = 2∆tb Q =Uf

(0.516)(189.875)(∆tb ) = 2921 Page 47 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS ∆tb = 30 F ∆toa = 2∆tb = 2(30 ) = 60 F assume ambient temperature = 90 F t o = 150 F r S =   cr

2

 µns   p 2

 1.75  µ (5) S = 0.1925 =    0.00275  122.45 µ = 11.6 ×10−6 reyn

Figure AF 16, 150 F, µ ≈ 11.6 × 10−6 reyn Use SAE 70 oil Maximum load, W with ε = 0.8 Table AT 20, L D = 1 S = 0.0446 r S =   cr

2

 µns   p 2

(

)

−6  1.75  11.6 × 10 (5) S = 0.0446 =   p  0.00275  p = 527 psi W = pDL = (527 )(3.5)(3.5) = 6456

580.

A 2500-kva generator, driven by a water wheel, operates at 900 rpm. The weight of the rotor and shaft is 15,100 lb. The left-hand, 5 –in, full bearing supports the larger load, R = 8920 lb . The bearing should be above medium-heavy construction (for estimating Ab ). (a) Decide upon an average clearance considering both Table 3.1 and 11.1, and upon a minimum film thickness ( ho cr ≈ 0.5 is on the safer side). (b) Investigate first the possibility of the bearing being a self-contained unit without need of special cooling. Not much air movement about the bearing is expected. Then make final decisions concerning oil-clearance, and film thickness and compute all the parameters given in the text, being sure that everything is reasonable.

Solution: n = 900 rpm 900 ns = = 15 rps 60 D = 5 in W = R = 8920 lb

Page 48 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS (a) Table 3.1, D = 5 in RC 5, average cd = 0.0051 in cr = 0.00255 in ho = 0.5cr = 0.5(0.00255) = 0.00128 in (b) Use L D = 1 L = 5 in D r = = 2.5 in 2 W 8920 p= = = 356.8 psi DL (5)(5) Table AT 20, L D = 1 , ho cr = 0.5 S = 0.1925 r f = 4.505 cr q = 4.16 rcr ns L ρ c∆to = 19.25 p r S =   cr

2

 µns   p 2

 2.5  µ (15) S = 0.1925 =    0.00255  356.8 µ = 4.8 ×10−6 reyn r f = 4.505 cr  2.5    f = 4.505  0.00255  f = 0.00460 F = f W = (0.00460 )(8920 ) = 41.032 lb πDn π (5)(900) vm = = = 1178 fpm 12 12 U f = Fvm = (41.032)(1178) = 48,336 ft − lb min

Q = hcr Ab ∆tb Medium-Heavy Ab = 20.25 DL = 20.25(5)(5) = 506.25 sq.in. Assume hcr = 0.516 ft − lb min − sq.in. − F Q =Uf Page 49 of 63



(0.516)(506.25)(∆tb ) = 48,336 ∆tb = 185 F , very high Therefore, special cooling is needed. ρ c∆to = 19.25 p (112)∆to = 19.25 356.8 ∆to = 61 F Assume ti = 100 F 61 tave = 100 + ≈ 130 F 2 Figure AF 16, µ = 4.8 µreyns , 130 F Select SAE 30 oil. µ = 6.0 µreyns r S =   cr

2

 µns   p 2

(

)

−6  2.5  6.0 × 10 (15) S = = 0.242  356.8  0.00255  Table AT 20, L D = 1 , S = 0.242 SAE 30 oil at 130 F ho = 0.569 cr

φ = 61.17o r f = 5.395 cr q = 4.04 rcr ns L ρ c∆to = 22.75 p p = 0.4734 pmax Oil, SAE 30 cr = 0.00255 in ho = 0.569(0.00255) = 0.00145 in

PRESSURE FEED 581.

An 8 x 8-in. full bearing supports 5 kips at 600 rpm of the journal; cr = 0.006 in. ; let the average µ = 2.5 × 10−6 reyn . (a) Compute the frictional loss U f . (b) The

Page 50 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS oil is supplied under a 40-psi gage pressure with a longitudinal groove at the point of entry. Assuming that other factors, including U f , remain the same and that the heat loss to the surrounding is negligible, determine the average temperature rise of the circulating oil. Solution: L = 5 in D = 5 in W = 5000 lb n = 600 rpm 600 ns = = 10 rps 60 cr = 0.006 in µ = 2.5 ×10−6 reyn L D =1 W 5000 p= = = 78.125 psi DL (8)(8) r S =   cr

2

 µns   p 2

(

)

−6  4  2.5 ×10 (10 ) S = = 0.1422  78.125  0.006 

(a) Table AT 20, L D = 1 , S = 0.1422 r f = 3.6 , ε = 0.57 cr  4    f = 3.6  0.006  f = 0.0054 F = f W = (0.0054 )(5000 ) = 27 lb πDn π (8)(600) vm = = = 1257 fpm 12 12 U f = Fvm = (27 )(1257 ) = 33,940 ft − lb min

(b) Longitudinal Groove. 2π r  c3 p  2 3 q = 2.5 r i  tan −1  1 + 1.5ε in sec 3µ  L  pi = 40 psi

(

)

3 0.006 ) (40)  −1 2π (4)  ( q = 2.5 tan [1 + 1.5(0.57)2 ]in3

3(2.5 × 10−6 ) 

Page 51 of 63

8



sec


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS q = 5.41 in3 sec U f = ρ cq∆to

(33,940 ft − lb

min )(12 in ft )(1min 60sec ) = (12)(5.41)∆to

∆to = 11.2 F 583.

A 4-in. 360o bearing, with L D = 1 , supports 2.5 kips with a minimum film of ho = 0.0008 in. , cd = 0.01 in. , n = 600 rpm. The average temperature rise of the oil is to be about 25 F. Compute the pressure at which oil should be pumped into the bearing if (a) all bearing surfaces are smooth, (b) there is a longitudinal groove at the oil-hole inlet. (c) same as (a) except that there is a 360o circumferential groove dividing the bearing into 2-in. lengths.

Solution: D = 4 in L = 4 in r = 2 in W = 2500 lb cd = 0.010 in cr = 0.005 in n = 600 rpm 600 ns = = 10 rps 60 ∆to = 25 F W 2500 p= = = 156.25 psi DL (4)(4) ho = 0.00080 in ho 0.0008 = = 0.16 cr 0.005 Table AT 20, L D = 1 , ho cr = 0.16 r f = 1.44 , ε = 0.84 cr  2    f = 1.44  0.005  f = 0.0036 F = f W = (0.0036 )(2500 ) = 9 lb πDn π (4)(600) vm = = = 628 fpm 12 12 U f = Fvm = (9 )(628) = 5652 ft − lb min = 1130 in − lb sec S = 0.0343

Page 52 of 63



2

 µns   p 2

 2  µ (10 ) S = 0.0343 =    0.005  156.25 µ = 3.35 ×10−6 reyn U f = ρ cq∆to

1130 = (112 )(q )(25) q = 0.404 in3 sec (a) Smooth q=

cr3 pi  −1 2π r  2 3  tan  1 + 1.5ε in sec 3µ  L 

(

0.404 =

)

(0.005)3 ( pi ) tan −1 2π (2) [1 + 1.5(0.84)2 ] in3

3(3.35 × 10−6 )  pi = 12.5 psi (b) Longitudinal groove q=



4

sec

2.5cr3 pi  −1 2π r  2 3  tan  1 + 1.5ε in sec 3µ  L 

(

)

2.5(0.005) ( pi )  −1 2π (2)  2 tan 1 + 1.5(0.84) in3 sec −6   4  3 3.35 × 10  pi = 5 psi (c) Circumferential groove 3

0.404 =

(

)

[

]

2π rcr3 pi 1 + 1.5ε 2 in3 sec q= 3µL

(

)

2π (2 )(0.005) ( pi ) 2 1 + 1.5(0.84) in3 sec 0.404 = −6 3 3.35 × 10 (4 ) pi = 5 psi 3

(

) [

]

BEARING CAPS 584.

An 8-in. journal, supported on a 150o partial bearing, is turning at 500 rpm; bearing length = 10.5 in., c d = 0.0035 in ., ho = 0.00106 in . The average temperature of the SAE 20 oil is 170 F. Estimate the frictional loss in a 160o cap for this bearing.

Solution:

Page 53 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS ho = 0.00106 in c d = 0.0035 in c r = 0.00175 in 2   ho   hav = cr 1 + 0.741 −   in   cr   2   0.00106   hav = (0.00175)1 + 0.741 −   = 0.00195 in  0.00175    For SAE 20, 170 F µ = 1.7 ×10−6 reyn µAvips F= hav 1 A = θDL 2 D = 8 in L = 10.5 in 160 8π θ = 160o = π= 180 9 1  8π  A =  (8)(10.5) = 117.3 sq.in. 2 9   500  vips = π Dns = π (8)  = 209.5 ips  60  1.7 × 10−6 (117.3)(209.5) = 21.424 lb F= 0.00195 πDn π (8)(500) vm = = = 1047 fpm 12 12 U f = Fvm = (21.424)(1047 ) = 22,430 ft − lb min = 1130 in − lb sec

(

585.

)

A partial 160o bearing has a 160o L = 2 in ., cd = 0.002 in ., ho = 0.0007 in ., n = 500 rpm , and For the cap only, what is the frictional loss?

Solution: c d = 0.002 in c r = 0.001 in ho = 0.0007 in ho 0.0007 = = 0.7 cr 0.001

Page 54 of 63

cap; D = 2 in ., µ = 2.5 ×10−6 reyn .



ε = 1−

ho = 1 − 0.7 = 0.3 cr

(

)

[

]

hav = cr 1 + 0.74ε 2 = (0.001) 1 + 0.74(0.3) = 0.001067 in µAvips F= hav πDn π (2)(500) vm = = = 261.8 fpm 12 12  12  vips = (261.8)  = 52.36 ips  60  1  160  1  160  A=  πDL =  (π )(2 )(2 ) = 5.585 sq.in. 2  180  2  180  (2.5 ×10−6 )(5.585)(52.36) = 0.685 lb F= 0.001067 U f = Fvm = (0.685)(261.8) = 179.3 ft − lb min 586.

2

The central reaction on a 120o partial bearing is 10 kips; D = 8 in ., L D = 1 ., cr r = 0.001 . Let n = 400 rpm and µ = 3.4 ×10−6 reyn . The bearing has a 150o cap. (a) For the bearing and the cap, compute the total frictional loss by adding the loss in the cap to that in the bearing. (b) If the bearing were 360o, instead of partial, calculate the frictional loss and compare.

Solution: 2

 r  µns S =    cr  p 400 ns = = 6.67 rps 60 W 10,000 p= = = 156.25 psi DL (8)(8) 2

(

)

−6  1  3.4 × 10 (6.67 ) S = = 0.145  156.25  0.001  (a) Table AT 22, L D = 1 , S = 0.145 r f = 2.021 cr ε = 0.6367 r f = 2.021 cr

 1    f = 2.021  0.001  f = 0.002021

Page 55 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS F = f W = (0.002021)(10,000) = 20.21 lb πDn π (8)(400) vm = = = 838 fpm 12 12 U f 1 = Fvm = (20.21)(838) = 16,936 ft − lb min CAP: hav = cr (1 + 0.74ε 2 ) cr = 0.001r D r = = 4 in 2 cr = 0.001(4) = 0.004 in

(

)

[

]

hav = cr 1 + 0.74ε 2 = (0.004) 1 + 0.74(0.6367 ) = 0.0052 in µAvips F= hav 2

 12  vips = (838)  = 167.6 ips  60  1  150  1  150  A=  πDL =  (π )(8)(8) = 83.78 sq.in. 2  180  2  180  3.4 ×10−6 (83.78)(167.6) F= = 9.18 lb 0.0052 U f 2 = Fvm = (9.18)(838) = 7693 ft − lb min Total Frictional Loss = U f 1 + U f 2 = 16,936 + 7693 = 24,629 ft − lb min

(

)

(b) 360o Bearing, L D = 1 , S = 0.145 r f = 3.65 cr ε = 0.5664 BEARING:  1    f = 3.65  0.001  f = 0.00365 F = f W = (0.00365)(10,000) = 36.5 lb πDn π (8)(400) vm = = = 838 fpm 12 12 U f 1 = Fvm = (36.5)(838) = 30,587 ft − lb min CAP: hav = cr (1 + 0.74ε 2 )

[

]

hav = cr (1 + 0.74ε 2 ) = (0.004) 1 + 0.74(0.5664) = 0.00495 in

Page 56 of 63

2



F=

µAvips hav

(3.4 ×10 )(83.78)(167.6) = 9.645 lb F= −6

0.00495 U f 2 = Fvm = (9.645)(838) = 8083 ft − lb min Total Frictional Loss = U f 1 + U f 2 = 30,587 + 8083 = 38,670 ft − lb min

587.

The central reaction on a 120o partial bearing is a 10 kips; D = 8 in. , L D = 1 , cr r = 0.001 ; n = 1200 rpm . Let µ = 2.5 ×10−6 reyn . The bearing has a 160o cap. (a) Compute ho and fhp for the bearing and for the cap to get the total fhp. (b) Calculate the fhp for a full bearing of the same dimensions and compare. Determine (c) the needed rate of flow into the bearing, (d) the side leakage qs . (e) the temperature rise of the oil in the bearing both by equation (o), i11.13, Text, and by Table AT 22. (f) What is the heat loss from the bearing if the oil temperature is 180 F? Is the natural heat loss enough to cool the bearing? (g) It is desired to pump oil through the bearing with a temperature rise of 12 F. How much oil is required? (h) For the oil temperature in (f), what is a suitable oil to use?

Solution: 2

 r  µns S =    cr  p 1200 ns = = 20 rps 60 W 10,000 p= = = 156.25 psi DL (8)(8) −6  1  (2.5 ×10 )(20 ) S = = 0.32  156.25  0.001  (a) Table AT 22, L D = 1 , S = 0.32 ε = 0.5417 ho = 0.4583 cr r f = 3.18 cr q = 2.60 rcr ns L

2

Page 57 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS qs = 0.305 q ρ c∆to = 17.834 p p = 0.38434 pmax ho = 0.4583cr = 0.4583(0.001)(4) = 0.00183 in BEARING: r f = 3.18 cr  1    f = 3.18  0.001  f = 0.00318 F = f W = (0.00318)(10,000) = 31.8 lb πDn π (8)(1200) vm = = = 2513 fpm 12 12 U f 1 = Fvm = (31.8)(2513) = 79,913 ft − lb min , 2.42 hp CAP: hav = cr (1 + 0.74ε 2 )

cr = 0.001r D r = = 4 in 2 cr = 0.001(4) = 0.004 in

[

]

hav = cr (1 + 0.74ε 2 ) = (0.004) 1 + 0.74(0.5417 ) = 0.00487 in µAvips F= hav 2

 12  vips = (2513)  = 503 ips  60  1  160  1  160  A=  πDL =  (π )(8)(8) = 89.36 sq.in. 2  180  2  180  2.5 ×10−6 (89.36)(5036) F= = 23.1 lb 0.00487 U f 2 = Fvm = (23.1)(2513) = 58,050 ft − lb min , 1.76 hp Total Frictional Loss = U f 1 + U f 2 = 79,913 + 58,050 = 137,963 ft − lb min

(

)

Uf

137,963 = 4.18 hp 33,000 33,000 (b) Full Bearing, L D = 1 , S = 0.32 fhp =

Page 58 of 63

=


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS Table AT 20 ho = 0.6305 cr r f = 6.86 cr ε = 0.3695 ho = 0.6305(0.004 ) = 0.002522 in BEARING: r f = 6.86 cr  1    f = 6.86  0.001  f = 0.00686 F = f W = (0.00686)(10,000) = 68.6 lb U f 1 = Fvm = (68.6)(2513) = 172,392 ft − lb min , 5.224 hp CAP: hav = cr (1 + 0.74ε 2 )

[

]

hav = cr (1 + 0.74ε 2 ) = (0.004) 1 + 0.74(0.3695) = 0.00440 in µAvips F= hav

(2.5 ×10 )(89.36)(503) = 25.54 lb F=

2

−6

0.00440 U f 2 = Fvm = (25.54)(2513) = 64,182 ft − lb min , 1.946 hp Total Frictional Loss = U f 1 + U f 2 = 172,392 + 64,182 = 236,574 ft − lb min fhp =

Uf

33,000

=

236,574 = 7.17 hp 33,000

(c) 120o Bearing q = 2.60 rcr ns L q = 2.60 (4)(0.004)(20)(8) q = 6.656 in3 sec q (d) s = 0.305 q qs = 0.305 6.656

Page 59 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS qs = 2.03 in 3 sec

(e) Equation(o) U f 1 = ρ cq∆to  12  U f 1 = 79,913 ft − lb min = 79,913  in − lb sec = 15,983 in − lb sec  60  U f 1 = 15,983 = (112)(6.656 )∆to

∆to = 21.4 F Table 22. ρ c∆to = 17.834 p 112∆to = 17.834 156.25 ∆to = 24.9 F (f) Q = hcr Ab ∆tb assume hcr = 0.516 ft − lb min − sq.in. − F Ab = 25DL = 25(8)(8) = 1600 sq.in. ∆t ∆tb = oa 2 assume ambient = 100 F 180 − 100 ∆t b = = 40 F 2 Q = (0.516)(1600)(40 ) = 33,024 ft − lb min < U f 1 Therefore not enough to cool the bearing. (g) Qr + Q = U f 1 + U f 2 Qr + 33,024 = 137,963 Qr = 104,939 ft − lb min Qr = 20,988 in − lb sec Qr = ρ cq∆to 20,988 = (112)q(12) q = 15.62 in3 sec (h) Fig. AF 16, 180 F, µ = 2.5 ×10−6 reyn use SAE 30 oil

Page 60 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS IMPERFECT LUBRICATION: 588.

A 0.5 x 0.75-in. journal turns at 1140 rpm. What maximum load may be supported and what is the frictional loss if the bearing is (a) SAE Type I, bronze base, sintered bearing, (b) nylon (Zytel) water lubricated, (c) Teflon, with intermittent use, (d) one with carbon graphite inserts.

Solution: (a) f = 0.12 πDn π (0.5)(1140) vm = = = 149.23 fpm 12 12 pvm = 50,000 p(149.23) = 50,000 p = 335 psi W = pDL = (335)(0.5)(0.75) = 126 lb F = f W = (0.12)(126) = 15.12 lb U f = Fvm = (15.12)(149.23) = 2256 ft − lb min (b) f = 0.14 ~ 0.18 , use f = 0.16 pvm = 2500 , water p(149.23) = 2500 p = 16.75 psi W = pDL = (16.75)(0.5)(0.75) = 6.28 lb F = f W = (0.16 )(6.28) = 1.005 lb U f = Fvm = (1.005)(149.23) = 150 ft − lb min (c) vm > 100 fpm f = 0.25 pvm = 20,000 , intermittent p(149.23) = 20,000 p = 134 psi W = pDL = (134 )(0.5)(0.75) = 50.25 lb F = f W = (0.25)(50.25) = 12.5625 lb U f = Fvm = (12.5625)(149.23) = 1875 ft − lb min (d) pvm = 15,000 p(149.23) = 15,000 p = 100.5 psi W = pDL = (100.5)(0.5)(0.75) = 37.69 lb assume f = 0.20 Page 61 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS F = f W = (0.20)(37.69) = 7.54 lb U f = Fvm = (7.54)(149.23) = 1125 ft − lb min 590.

A bearing to support a load of 150 lb at 800 rpm is needed; D = 1 in. ; semilubricated. Decide upon a material and length of bearing, considering sintered metals, Zytel, Teflon, and graphite inserts.

Solution: πDn π (1)(800) vm = = = 209.44 fpm 12 12 assume, L = D = 1 in W 150 p= = = 150 psi DL (1)(1) pvm = (150)(209.44) = 31,416 Use sintered metal, limit pvm = 50,000 THRUST BEARINGS 592.

A 4-in. shaft has on it an axial load of 8000 lb., taken by a collar thrust bearing made up of five collars, each with an outside diameter of 6 in. The shaft turns 150 rpm. Compute (a) the average bearing pressure, (b) the approximate work of friction.

Solution: (a) p =

4W 4(8000) = = 283 psi πDo2 π (6)2

(b) assume f = 0.065 , average F = f W = (0.065)(8000) = 520 lb πDn π (3)(150) vm = = = 117.81 fpm 12 12 U f = nFvm = (5)(520)(117.81) = 306,306 ft − lb min 593.

A 4-in. shaft, turning at 175 rpm, is supported on a step bearing. The bearing area is annular, with a 4-in. outside diameter and a 3/4 –in. inside diameter. Take the allowable average bearing pressure as 180 psi. (a) What axial load may be supported? (b) What is the approximate work of friction?

Solution: πDn vm = 12

Page 62 of 63


SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS 1 (4 + 0.75) = 2.375 in 2 πDn π (2.375)(175) vm = = = 108.81 fpm 12 12 assume f = 0.065 , average D=

(a) p =

4W π Do2 − Di2

(

)

π  2 3  W = (4) −    (180) = 2182 lb 4   4   (b) U f = f Wvm = (0.065)(2182)(108.81) = 15,433 ft − lb min 2

- end -

Page 63 of 63


SECTION 10 - BALL AND ROLLER BEARINGS 601.

The radial reaction on a bearing is 1500 lb.; it also carries a thrust of 1000 lb.; shaft rotates 1500 rpm; outer ring stationary; smooth load, 8-hr./day service, say 15,000 hr. (a) Select a deep-groove ball bearing. (b) What is the rated 90 % life of the selected bearing? (c) For b = 1.34 , compute the probability of the selected bearing surviving 15,000 hr.

Solution: Fx = 1500 lb Fy = 1000 lb

( )

B10 = (15,000 )(60 )(1500 ) 10−6 = 1350 mr

F e= 0.56Cr Fx + Ct Fz Cr = 1 , outer ring stationary assume Ct = 1.8 F e= 0.56(1)(1500) + (1.8)(1000) = 2640 lb 1

1  B 3 Fr =  10  Fe = (1350)3 (2640) = 29,178 lb  Br  (a) Table 12.3 use 320, Fr = 29,900 lb Fs = 29,900 lb To check: Fz 1000 = = 0.0340 Fs 29,400 Table 12.2, Ct = 1.93 , Q = 0.2286 Fz 1000 = = 0.667 > Q Cr Fx (1.0 )(1500 ) F e= 0.56Cr Fx + Ct Fz F e= 0.56(1)(1500) + (1.93)(1000) = 2770 lb 1

1  B 3 Fr =  10  Fe = (1350)3 (2770) = 30,614 lb  Br  2.4 % higher than 29,900 lb. Safe. Therefore use Bearing 320, Deep-Groove Ball Bearing.

(b) Fr = 29,900 lb F e= 2770 lb 1

 B 3 29,900 =  10  (2770 ) = 30,614 lb  1 mr  B10 = 1258 mr

Page 1 of 17


SECTION 10 - BALL AND ROLLER BEARINGS

( )

B10 = (HR )(60 )(1500 ) 10−6 = 1258 HR ≈ 14,000 hr 1

  1  b  ln   B  P  (c) = B10   1    ln     P10  

 1   1  ln  = ln   0.9   P10  B10 = 1258 mr B = 1350 mr 1

  1   1.34 ln  1350   P    = 1258   1   ln   0.9   P = 0.891 602.

A certain bearing is to carry a radial load of 500 lb. and a thrust of 300 lb. The load imposes light shock; the desired 90 % life is 10 hr./day for 5 years at n = 3000 rpm . (a) Select a deep-groove ball bearing. What is its bore? Consider all bearings that may serve. (b) What is the computed rated 90 % life of the selected bearing? (c) What is the computed probability of the bearing surviving the specified life? (d) If the loads were changed to 400 and 240 lb., respectively, determine the probability of the bearing surviving the specified life, and the 90 % life under the new load.

Solution: Fx = 500 lb Fz = 300 lb Assume Cr = 1 Fz 300 = = 0.6 Cr Fx (1.0 )(500 ) Fz Table 12.2, >Q Cr Fx (a) F e= 0.56Cr Fx + Ct Fz Cr = 1 Assume Ct = 1.8 F e= 0.56(1)(500 ) + (1.93)(300 ) = 820 lb

Page 2 of 17


SECTION 10 - BALL AND ROLLER BEARINGS For light shock, service factor ~ 1.1 F e= (1.1)(820) = 902 lb 1

1  B 3 Fr =  10  Fe = (1350 )3 (2770 ) = 30,614 lb  Br  B10 = (5)(365)(10 )(60 )(3000 )(10−6 ) = 3285 mr 1

1  B 3 Fr =  10  Fe = (3285)3 (902) = 13,409 lb  Br  Table 12.3, Bearing No. Fr , lb Fs , lb 217 14,400 12,000 312 14,100 10,900

Select, Bearing No. 312 Fr = 14,100 lb Fs = 10,900 lb (b) Table 12.2 Fz 300 = = 0.0285 Fs 10,900 Ct = 1.99 Q = 0.22 F e= 0.56Cr Fx + Ct Fz F e= 0.56(1)(500 ) + (1.99)(300) = 877 lb F e= (1.1)(877 ) = 965 lb 1

 B 3 Fr =  10  Fe  Br  1

 B 3 14,100 =  10  (965)  1  B10 = 3119 mr

( )

B10 = (YR)(365)(10 )(60 )(3000 ) 10 −6 = 3119 YR = 4.75 years 1

  1  b  ln   B  P  (c) = B10   1    ln     P10  

Page 3 of 17

Bore 85 mm 60 mm


SECTION 10 - BALL AND ROLLER BEARINGS use b = 1.125 B10 = 3119 mr B = 3285 mr 1

  1   1.125 ln  3285   P     = 3119   1   ln   0.9   P = 0.8943 (d) Fx = 400 lb Fz = 240 lb Cr = 1 Fz 240 = = 0.6 Cr Fx (1.0 )(400 ) Table 12.2 Ct = 2.15 Q = 0.21 < 0.6 F e= 0.56Cr Fx + Ct Fz F e= 0.56(1)(400) + (2.15)(240) = 740 lb F e= (1.1)(740) = 814 lb 1

 B 3 Fr =  10  Fe  Br  1

 B 3 14,100 =  10  (814)  1  B10 = 5197 mr 1

  1  b  ln   B P =    B10   1      ln   P10   1

  1   1.125 ln  3285   P    = 5197   1   ln   0.9   P = 0.939 Life: B10 = (YR)(365)(10 )(60 )(3000 )(10 −6 ) = 5197

Page 4 of 17


SECTION 10 - BALL AND ROLLER BEARINGS YR = 8 years

603.

The smooth loading on a bearing is 500-lb radial, 100 lb. thrust; n = 300 rpm . An electric motor drives through gears; 8 hr./day, fully utilized. (a) Considering deep-groove ball bearings that may serve, choose one end specify its bore. For the bearing chosen, determine (b) the rated 90 % life and (c) the probability of survival for the design lufe.

Solution: Fx = 500 lb Fz = 100 lb Table 12.1, 8 hr./day fully utilized, assume 25,000 hr B10 = (25,000 )(60 )(300 )(10−6 ) = 450 mr (a) assume Cr = 1 Fz 100 = = 0.2 Cr Fx (1.0 )(500 ) F consider Q > z Cr Fx Fe = Cr Fx = (1.0)(500) = 500 lb 1 3

1 B  Fr =  10  Fe = (450)3 (500) = 3832 lb  Br  Table 12.3 Bearing No. Fr , lb Fs , lb 207 4440 3070 306 4850 3340 305 3660 2390 Select 305, Fr = 3660 lb , Fs = 2390 lb Bore (Table 12.4) = 25 mm

Fz 100 = = 0.0418 Fs 2390 Table 12.2, 0.22 < Q0.26 F Q> z Cr Fx Fe = Cr Fx = (1.0)(500) = 500 lb

(a)

1

 B 3 3660 =  10  (500)  1  B10 = 392 mr Rated Life:

Page 5 of 17



( )

B10 = (HR )(60 )(300 ) 10−6 = 392 HR ≈ 22,000 hr 1

  1  b  ln   B P (c) =     B10  1     ln   P10   b = 1.125 1

  1   1.125 ln  450   P    = 392   1   ln   0.9   P = 0.884 605.

A No. 311, single-row, deep-groove ball bearing is used to carry a radial load of 1500 lb. at a speed of 500 rpm. (a) What is the 90 % life of the bearing in hours? What is the approximate median life? What is the probability of survival if the actual life is desired to be (b) 105 hr., (c) 104 hr.?

Solution: Table 12.3, No. 311 Fs = 9400 lb Fr = 12400 lb Fx = 1500 lb assume Cr = 1 Fe = Cr Fx = (1)(1500 ) = 1500 lb 1

 B 3 (a) Fr =  10  Fe  Br  1

 B 3 12400 =  10  (1500)  1  B10 = 565 mr

( )

B10 = (HR )(60 )(500 ) 10−6 = 565 HR ≈ 18,800 hr

For median life = 5( 90 % life) = 5(18,800) = 94,000 hr

Page 6 of 17



( )

( )

(b) B = 105 (60)(500) 10−6 = 3000 mr  1   ln   B P =    B10   1      ln   P10   b = 1.125

1 b

1

  1   1.125 ln  3000   P    = 565   1   ln   0.9   P = 0.502 (c) 104 hr

( )

( )

B = 104 (60)(500) 10−6 = 300 mr  1   ln   B P =     B10  1     ln   P10   b = 1.125

1 b

1

  1   1.125 ln  300   P     = 565   1   ln   0.9   P = 0.950 606.

The load on an electric-motor bearing is 350 lb., radial; 24 hr. service, n = 1200 rpm ; compressor drive; outer race stationary. (a) Decide upon a deepgroove ball bearing, giving its significant dimensions. Then compute the selected bearing’s 90 % life, and the probable percentage of failures that would occur during the design life. What is the approximate median life of this bearing? (b) The same as (a), except that a 200 series roller bearing is to be selected.

Solution: Fx = 350 lb Fe = Cr Fx outer race stationary, Cr = 1 Fe = (1)(350) = 350 lb Page 7 of 17


SECTION 10 - BALL AND ROLLER BEARINGS Table 12.1 90 % Life, hrs = 50,000 hrs B = (50,000)(60)(1200) 10−6 = 3600 mr

( )

1 3

1 B  (a) Fr =  10  Fe = (3600)3 (350 ) = 5364 lb  Br  Table AT 12.3 earing No. Fr , lb Fs , lb 208 5040 3520 209 5660 4010 306 4850 3340 307 5750 4020 Use No. 209 Fr = 5660 lb Table 12.4, Dimension Bore = 45 mm O.D. = 85 mm Width of Races = 19 mm Max. Fillet r = 0.039 mm

90 % Life: 1 3

B  Fr =  10  Fe  Br  1

 B 3 5660 =  10  (350)  1  B10 = 4229 mr

( )

B10 = (HR )(60 )(1200 ) 10−6 = 4229 HR ≈ 58,740 hr Probability.

 1   ln   B P =     B10  1     ln   P10   b = 1.125

1 b

1

  1   1.125 ln  3600   P    = 4229   1   ln   0.9   P = 0.916 % failures = 1 – 0.916 = 0.084 = 8.4 % Page 8 of 17


SECTION 10 - BALL AND ROLLER BEARINGS Median Life = 5(58,740) = 293,700 hrs (b) Table 12.3, Fr = 5364 lb use No. 207, Fr = 5900 lb Bore = 35 mm O.D. = 72 mm Width of Races = 17 mm 90 % life: 1

 B 3 Fr =  10  Fe  Br  1

 B 3 5900 =  10  (350)  1  B10 = 4790 mr

( )

B10 = (HR )(60 )(1200 ) 10−6 = 4790 HR ≈ 66,530 hr Probability. 1

  1  b  ln   B P =     B10  1     ln   P10   b = 1.125 1

  1   1.125 ln  3600   P    = 4790   1   ln   0.9   P = 0.926 % failures = 1 – 0.926 = 0.074 = 7.4 % Median Life = 5(66,530) = 332,650 hrs 608.

A deep-groove ball bearing on a missile, supporting a radial load of 200 lb., is to have a design life of 20 hr.; with only a 0.5 % probability of failure while n = 4000 rpm . Using a service factor of 1.2 , choose a bearing. ( A 5- or 6- place log table is desirable.)

Solution: No need to use log table. Fx = 200 lb assume Cr = 1

Page 9 of 17


SECTION 10 - BALL AND ROLLER BEARINGS Fe = Cr Fx = (1.0)(200) = 200 lb Fe = (1.2)(200) = 240 lb

( )

B10 = (20 )(60 )(4000 ) 10−6 = 4.8 mr P = 1 − 0.005 = 0.995 1

  1  b  ln   B P =     B10  1     ln   P10   b = 1.125 1

  1   1.125 ln  4.8   0.995    = B10   1   ln   0.9   B10 = 72 mr 1

1  B 3 Fr =  10  Fe = (72)3 (240) = 998.4 lb  Br  Table 12.3 Select No. 201, Fr = 1180 lb

VARIABLE LOADS 610. A certain bearing is to carry a radial load of 10 kip at a speed of 10 rpm for 20 % of the time, a load of 8 kips at a speed of 50 rpm for 50 % of the time, and a load of 5 kips at 100 rpm during 30 % of the time, with a desired life of 3000 hr.; no thrust. (a) What is the cubic mean load? (b) What ball bearings may be used? What roller bearings? Solution: 1

 F 3n + F23n2 + F33n3 + L  3 (a) Fm =  1 1    ∑n ∑ n = n1 + n2 + n3 For 1 min. n1= (0.2)(10) = 2 rev n 2 = (0.5)(50) = 25 rev n3 = (0.3)(100) = 30 rev

∑ n = 2 + 25 + 30 = 57 rev F1 = 10 kips

Page 10 of 17


SECTION 10 - BALL AND ROLLER BEARINGS F2 = 8 kips F3 = 5 kips 1

 (10 )3 (2 ) + (8)3 (25) + (5)3 (30 )  3 Fm =   = 6.88 kips 57  

(b) Fx = 6.88 kips = 6880 lb assume Cr = 1 Fe = (1.0)(6880) = 6880 lb 1 min = 57 rev B10 = (3000 )(60 )(57 )(10−6 ) = 10.26 mr 1

1  B 3 Fr =  10  Fe = (10.26)3 (6880) = 14,950 lb  Br  Table 12.3, Ball Bearing Use Bearing No. 217, Fr = 14,400 lb

(c) Table 12.3 (Roller Bearing) Use Bearing No. 213, Fr = 14,900 lb 612.

A deep-groove ball bearing No. 215 is to operate 30 % of the time at 500 rpm with Fx = 1200 lb and Fz = 600 lb , 55 % of the time at 800 rpm with Fx = 1000 lb and Fz = 500 lb , and 15 % of the time at 1200 rpm with Fx = 800 lb and Fz = 400 lb . Determine (a) the cubic mean load; (b) the 90 % life of this bearing in hours, (c) the average life in hours.

Solution: Bearing No. 215, Fr = 11,400 lb , Fs = 9,250 lb Table 12.2, Fz Fs At 30 % of the time, 500 rpm Fz 600 = = 0.065 Fs 9250 Ct = 1.66 Q = 0.266 Fz 600 = = 0.5 > Q Cr Fx (1)(1200 ) Fe1 = 0.56Cr Fc + Ct Fz = 0.56(1)(1200) + (1.66)(600) = 1668 lb At 55 % of the time, 800 rpm Fz 500 = = 0.054 Fs 9250

Page 11 of 17


SECTION 10 - BALL AND ROLLER BEARINGS Ct = 1.73 Q = 0.257 Fz 500 = = 0.5 > Q Cr Fx (1)(1000 ) Fe 2 = 0.56Cr Fc + Ct Fz = 0.56(1)(1000) + (1.73)(500) = 1425 lb At 15 % of the time, 1200 rpm 400 Fz = = 0.043 Fs 9250 Ct = 1.84 Q = 0.242 Fz 400 = = 0.5 > Q Cr Fx (1)(800 ) Fe1 = 0.56Cr Fc + Ct Fz = 0.56(1)(800) + (1.84)(400) = 1184 lb 1

 F 3n + F23n2 + F33n3 + L  3 (a) Fm =  1 1    ∑n ∑ n = n1 + n2 + n3

F1 = 1668 lb F2 = 1425 lb F3 = 1184 lb For 1 min. n1= (0.3)(500 ) = 150 rev n 2 = (0.55)(800) = 440 rev n3 = (0.15)(1200 ) = 180 rev

∑ n = 150 + 440 + 180 = 770 rev 1 3

 (1668) (150 ) + (1425) (440 ) + (1184 ) (180 ) Fm =   = 1434 kips 770   3

3

(b) Fe = Fm = 1434 lb 1

 B 3 Fr =  10  Fe  Br  1

 B 3 11,400 =  10  (1434 )  1  B10 = 503 mr For 1 min = 770 rev

Page 12 of 17

3


SECTION 10 - BALL AND ROLLER BEARINGS B10 = (HR )(60 )(770 )(10−6 ) = 503 HR ≈ 11,000 hr

(c) Average life = 5(11,000) = 55,000 hrs MANUFACTURER’S CATALOG NEEDED 614.

A shaft for the general-purpose gear-reduction unit described in 489 has radial bearing reactions of RC = 613 lb and RD = 1629 lb ; n = 250 rpm . Assume that the unit will be fully utilized for at least 8 hr./day, with the likelihood of the same uses involving minor shock. (a) Select ball bearings for this shaft. (b) Select roller bearings. (c) What is the probability of both bearings C and D surviving for the design life?

Solution: 3 Problem 489, D = 1 in = 1.375 in 8 Ref: Design of Machine Members, Doughtie and Vallance Fc = (K a K l )K o K p K s K t Fr

at C. Fr = RC = 613 lb K t = 1.0 K p = 1.0 K o = 1.0 Ks = 3

Kr Na Nc

N a = 250 rpm N c = 500 rpm K r = 1.5 Ks = 3

(1.5)(250) = 0.90856 500

K a = 1.0 Ha H c K rel Table 12.1, 8 hr/day, fully utilized, Text H a = 25,000 hr H c = 10,000 hr assume K rel = 1.0 for 90 % reliability Kl = 3

Kl = 3

25,000 = 1.3572 10,000

Page 13 of 17


SECTION 10 - BALL AND ROLLER BEARINGS Fc = (K a K l )K o K p K s K t Fr

Fc = (1.0 )(1.3572 )(1.0 )(1.0 )(0.90856 )(1.0 )(613) = 756 lb Table 9-7, Doughtie and Vallance, Two-row spherical Type, No. 207 Bore = 1.3780 in, Fc = 880 lb At D. Fr = RD = 1629 lb Fc = (K a K l )K o K p K s K t Fr Fc = (1.0 )(1.3572 )(1.0 )(1.0 )(0.90856 )(1.0 )(1629 ) = 2009 lb Table 9-7, Doughtie and Vallance, Two-row spherical Type, No. 407 Bore = 1.3780 in, Fc = 2290 lb (b) at C, Fc = 756 lb Table 9.8, Doughtie and Vallance Use No. 207, Bore = 1.3780 in, Fc = 1540 lb at C, Fc = 2009 lb Table 9.8, Doughtie and Vallance Use No. 307, Bore = 1.3780 in, Fc = 2660 lb (c) For probability: (c.1) at C, Bearing No. 207, Two-row spherical bearing, Fc = 880 lb Fc = 880 lb = (1)K l (1)(1)(0.90856 )(1)(613) K l = 1.58 Kl = 3

Ha H c K rel

1.58 = 3

25,000 10,000 K rel

K rel = 0.634 Table 9-3, Reference Probability = 95.8 % at D, Bearing No. 407, Deep-groove bearing, Fc = 2290 lb Fc = 2290 lb = (1)K l (1)(1)(0.90856 )(1)(1627 ) K l = 1.547 Kl = 3

Ha H c K rel

Page 14 of 17



1.547 = 3

25,000 10,000 K rel

K rel = 0.675 Table 9-3, Reference Probability = 93.3 % (c.2) at C, Roller Bearing No. 207, Fc = 1540 lb Fc = 1540 lb = (1)K l (1)(1)(0.90856 )(1)(613) K l = 2.765 Kl = 3

Ha H c K rel

2.765 = 3

25,000 10,000 K rel

K rel = 0.118 Table 9-3, Reference Probability = 98.8 % at D, Roller Bearing No. 407, Fc = 2660 lb Fc = 2660 lb = (1)K l (1)(1)(0.90856 )(1)(1627 ) K l = 1.80 Kl = 3

Ha H c K rel

1.80 = 3

25,000 10,000 K rel

K rel = 0.43 Table 9-3, Reference Probability = 95.7 % 615.

A shaft similar to that in 478 has the following radial loads on the bearings, left to right: 803 lb, 988 lb, 84 lb, and 307 lb; no thrust. The minimum shaft diameter at the bearings are 1.250 in, 1.125 in, 1.000 in, and 1.0625 in. Assume that the service will not be particularly gentle; intermittently used, with n = 425 rpm . (a) Select ball bearing for this shaft. (b) Select roller bearings.

Solution: Ref: Design of Machine Members by Doughtie and Vallance Fc = (K a K l )K o K p K s K t Fr K a = 1.0

Page 15 of 17



Kl = 3

Ha H c K rel

H c = 10,000 hr Table 12.1, Text, H a = 10,000 hr (intermittent) 90 % reliability, K rel = 1.0 10,000 = 1.0 10,000 K o = 1.0

Kl = 3

K p = 1.0

K r = 1.5 assumed Ks = 3

Kr Na Nc

N a = 425 rpm N c = 500 rpm Ks = 3

(1.5)(425) = 1.0844 500

K t = 1.0 (a) Ball Bearing (a.1) 803 lb, D = 1.250 in Fc = (1.0 )(1.0 )(1.0 )(1.0 )(1.0844 )(1.0 )(803) = 870 lb Table 9-7, Ref. Two-row spherical type, 207 Fc = 880 lb Bore = 1.3780 in (a.2) 988 lb, D = 1.125 in Fc = (1.0 )(1.0 )(1.0 )(1.0 )(1.0844 )(1.0 )(988) = 1071 lb Table 9-7, Ref. Two-row spherical type, 306 Fc = 1050 lb Bore = 1.1811 in (a.3) 84 lb, D = 1.000 in Fc = (1.0 )(1.0 )(1.0 )(1.0 )(1.0844 )(1.0 )(84 ) = 91 lb

Page 16 of 17


SECTION 10 - BALL AND ROLLER BEARINGS Table 9-7, Ref. Deep-groove type, 106 Fc = 544 lb Bore = 1.1811 in (a.4) 307 lb, D = 1.0625 in Fc = (1.0 )(1.0 )(1.0 )(1.0 )(1.0844 )(1.0 )(307 ) = 333 lb Table 9-7, Ref. Deep-groove type, 106 Fc = 544 lb Bore = 1.1811 in (b) Roller Bearing (b.1) 803 lb, D = 1.250 in Fc = 870 lb , Bore = 1.3780 in use No. 207, Fc = 1540 lb (b.2) 988 lb, D = 1.125 in Fc = 1071 lb , Bore = 1.1811 in use No. 206, Fc = 1320 lb (b.3) 84 lb, D = 1.000 in Fc = 91 lb , Bore = 1.1811 in use No. 206, Fc = 1320 lb (b.4) 307 lb, D = 1.0625 in Fc = 333 lb , Bore = 1.1811 in use No. 206, Fc = 1320 lb - end -

Page 17 of 17


SECTION 12 – HELICAL GEARS DESIGN PROBLEMS 701.

For continuous duty in a speed reducer, two helical gears are to be rated at 7.4 hp at a pinion speed of 1750 rpm; m w ≈ 2.75 ; the helix angle 15o ; 20o F.D. teeth in the normal plane; let N p = 21 teeth, and keep b < 2 D p . Determine the pitch, face, N g , and the material and heat treatment. Use through-hardened teeth with a maximum of 250 BHM (teeth may be cut after heat treatment).

Solution: ψ = 15o φn = 20o πD p n p vm = 12 N p 21 Dp = = Pd Pd n p = 1750 rpm  21  (1750 ) Pd  9621  vm = = Pd 12 33,000hp (33,000 )(7.4 ) Ft = = = 25.38 Pd vm  9621    P  d 

π 

b ≤ 2 Dp

 21  42 b = 2  =  Pd  Pd 0.05vm (Ft + Cb cos 2 ψ )cosψ lb Fd = Ft + 1 2 2 0.05vm + (Ft + Cb cos ψ ) Table AT 25 Assume C = 1660 ψ = 15o

Fd = 25.38Pd +

  9621    42   25.38Pd + 1660  cos2 15 cos15 0.05  Pd    Pd     9621    42   + 25.38Pd + 1660  cos 2 15 0.05  Pd    Pd  

Page 1 of 14

1 2

lb


SECTION 12 – HELICAL GEARS

Fd = 25.38 Pd +

465  65050   25.38 Pd +  Pd  Pd  65050  481   +  25.38 Pd + Pd  Pd 

For continuous service: Fw ≥ Fd bD pQK g Fw = cos 2 ψ 2mg 2(2.75) Q= = = 1.467 mg + 1 2.75 + 1 Table At 26, Bhn = 250 Sum of BHN = 500, φn = 20o K g = 131

 42  21  (1.467 )(131) 181,670 Fw =    = 2 Pd2  Pd  Pd  cos 15 Fw ≥ Fd By trial and error method Pd 7 6

Fd 3967 4758

Fw 3708 5046

use Pd = 6 21 21 Dp = = = 3.5 in Pd 6 42 42 b= = = 7 in Pd 6 9621 9621 vm = = = 1604 fpm Pd 6 Fig. AF 19, permissible error = 0.0018 in Fig. AF 20 Use carefully cut gears, Pd = 6 Error = 0.001 in is o.k. For material Strength sbY cosψ Fs = K f Pd

Page 2 of 14

1 2

lb


SECTION 12 – HELICAL GEARS Np

21 = 23 cos ψ cos3 15 Table AT 24, Load near middle N ep = 23 , φn = 20o FD N ep =

3

=

Y = 0.565 assume K f = 2.0 Fs = N sf Fd

assume N sf = 2.0 s (7 )(0.565) cos15 = (4758)(2 ) (2)(6) s = 29,892 psi s use sn = u 3 su = 3(29,892) = 89,676 psi Use C1050, OQT 1100 F, su = 122 ksi , BHN = 248 < 250 Ans. Pd = 6 b = 7 in N g = mw N p = (2.75)(21) = 58 Material. C1050, OQT 1100 F 703.

A pair of helical gears, subjected to heavy shock loading, is to transmit 50 hp at 3 1750 rpm of the pinion.; mg = 4.25 ; ψ = 15o ; minimum D p = 4 in. ; continuous 4 o service, 24 hr/day; 20 F.D. teeth in the normal plane, carefully cut; throughhardened to a maximum BHN = 350. Decide upon the pitch, face width, material and its treatment.

Solution: π (4.75)(1750) vm = = 2176 fpm 12 33,000hp (33,000)(50 ) Ft = = = 758 lb vm (2176) Dynamic load: 0.05vm Ft + Cb cos 2 ψ cosψ lb Fd = Ft + 1 2 2 0.05vm + Ft + Cb cos ψ Fig. AF 19, vm = 2176 fpm Permissible error = 0.0014 in

(

)

(

Page 3 of 14

)


SECTION 12 – HELICAL GEARS Use carefully cut gears, e = 0.001 in , Pd = 5 as standard Table AT 25, Steel and steel, 20o FD C = 1660 Fd = 758 +

(

)

0.05(2176) 758 + 1660b cos 2 15 cos15

(

0.05(2176) + 758 + 1660b cos 15 105.1(758 + 1548.8b ) Fd = 758 + 1 lb 108.8 + (758 + 1548.8b )2 Wear load: bD pQK g Fw = cos 2 ψ 2mg 2(4.25) Q= = = 1.619 mg + 1 4.25 + 1 Table At 26, 20o FD, Sum of BHN =2(350)=700 K g = 270 2

b(4.75)(1.619)(270 ) = 2225b cos 2 15 2π Fw ≥ Fd , bmin = 2 Pa = = 4.69 in. Pd tanψ By trial and error method Fw =

b 5 6

Fd 5203 5811

Fw 11125 13350

use b = 5 in Material: Strength: sbY sbY cosψ = Fs = K f Pdn K f Pd Np N ep = cos3ψ N p = Pd D p = (5)(4.375) = 22 22 = 25 cos3 15 Table AT 24, Load near middle N ep =

Page 4 of 14

1 2

)

lb


SECTION 12 – HELICAL GEARS N ep = 25 , φn = 20o FD Y = 0.580 assume K f = 1.7

s (5)(0.580) cos15 = 0.32955s (1.7 )(5) Fs = N sf Fd for 24 hr/day service, heavy shock loading N sf = 1.75 Fs =

0.32955s = (1.75)(5203) s = 27,629 psi s use sn = u 3 su = 3(27,629) = 82,887 psi Table AT 9 Use 4150, OQT 1200 F, su = 159ksi , BHN = 331 < 350 Ans. Pd = 5 b = 5 in Material. 4150, OQT 1200 F 705.

Design the teeth for two herringbone gears for a single-reduction speed reducer with mw = 3.80 . The capacity is 36 hp at 3000 rpm of the pinion; ψ = 30o ; F.D. teeth with φn = 20o . Since space is at a premium, the initial design is for N p = 15 teeth and carburized teeth of AISI 8620; preferably b < 2 D p .

Solution: N 15 Dp = p = Pd Pd b ≈ 2Dp b = 2 Dp = vm =

30 Pd

πD p n p

12  15  π  (3000) P 11,781 vm =  d  = 12 Pd

Page 5 of 14


SECTION 12 – HELICAL GEARS 33,000hp (33,000)(36 ) = = 101Pd vm  11,781     Pd  Dynamic load 0.05vm Ft + Cb cos 2 ψ cosψ lb Fd = Ft + 1 2 0.05vm + Ft + Cb cos ψ 2 Ft =

(

)

(

φn = 20

)

o

ψ = 30 o Assume C = 1660 , Table AT 25, 20o FD   11,781    30   101Pd + 1660  cos 2 30 cos 30 0.05  Pd    Pd   Fd = 101Pd + lb 1 2   11,781    30   + 101Pd + 1660  cos 2 30 0.05  Pd    Pd   510  37,350  101Pd +  Pd  Pd  Fd = 101Pd + lb 1 2 589  37,350   + 101Pd + Pd  Pd  Wear load bD pQK g Fw = cos 2 ψ 2 mg 2(3.80) Q= = = 1.583 mg + 1 3.80 + 1 For AISI 8620, carburized, 20o FD K g = 750 for 1010 cycles  30  15  (1.583)(750) 712,350 Fw =    = 2 Pd2  Pd  Pd  cos 30 By trial and error, Fw ≥ Fd Pd Fd Fw 5 4433 28,494 4 5454 44,522 6 3817 19,788 8 3173 11,130 9 3008 8794

For carefully cut gears, e = 0.001 vmax = 1400 fpm (Fig. AF 9)

Page 6 of 14



Pd 5 4 6 8 9

11,781 Pd 2356.2 1963.5 1683 1473 1309 fpm vm =

use Pd = 9 Fd = 3008 lb Fw = 5794 lb > Fd 30 30 b= = = 3.3 in Pd 9 use b = 3.0 in To check for strength sbY sbY cosψ = Fs = K f Pdn K f Pd Np N ep = cos3ψ N p = 15 15 = 23 cos3 30 Table AT 24, Load near middle N ep = 23 , φn = 20o FD N ep =

Y = 0.565 assume K f = 1.7

8620, SOQT 450, su = 167 ksi s sn = u 3 su 167 sn = = = 83.5 2 2 (83,500)(3.0)(0.565)cos 30 = 8011 lb > F (= 3008 lb ) Fs = d (1.7)(9) Designed Data: Pd = 9 b = 3.0 in N p = 15 N g = mw N p = (3.8)(15) = 57

Page 7 of 14


SECTION 12 – HELICAL GEARS N p 15 = = 1.67 in Pd 9 N 57 Dg = g = = 6.33 in Pd 9 Dp =

CHECK PROBLEMS 707.

The data for a pair of carefully cut gears are: Pdn = 5 , φn = 20o ,ψ = 12o , b = 3.5 in. , N p = 18 , N g = 108 teeth; pinion turns 1750 rpm. Materials: pinion, SAE 4150, OQT to BHN = 350; gear, SAE 3150, OQT to BHN = 300. Operation is with moderate shock for 8 to 10 hr./day. What horsepower may be transmitted continuously?

Solution: N Dp = p Pd Pd = Pdn cosψ = (5)cos15 = 4.89 18 Dp = = 3.681 in 4.89 Wear load bD pQK g Fw = cos 2 ψ b = 3.5 in. 2Ng 2(108) Q= = = 1.7143 N p + N g 18 + 108 Table AT 26, φn = 20o Sum of BHN = 350 + 300 = 650 K g = 233 Fw =

(3.5)(3.681)(1.7143)(233) = 5379 lb

cos 2 12 Strength of gear sbY lb Fs = K f Pdn For gear: SAE 3150, OQT to BHN = 300 su = 151 ksi

sn = 0.5su = 0.5(151) = 75.5 ksi Ng 108 N eg = = = 116 3 cos ψ cos3 12

Page 8 of 14


SECTION 12 – HELICAL GEARS Table AT 24, Load near middle, φn = 20o Y = 0.763 snY = (75.5)(0.763) = 57.6 For pinion: SAE 4150, OQT to BHN = 350 su = 0.5BHN = 0.5(350) = 175 ksi sn = 0.5su = 0.5(175) = 87.5 ksi Np 18 N ep = = = 19 3 cos ψ cos3 12 Table AT 24, Load near middle, φn = 20o Y = 0.534 snY = (87.5)(0.534) = 46.7 Therefore use pinion as weak Assume K f = 1.7 Fs =

(87,500)(3.5)(0.534) = 19,240 lb (1.7 )(5)

For moderate shock, 8 to 10 hr./day Use N sf = 1.5 Fs ≥ N sf Fd

19,240 = 1.5Fd Fd ≤ 12,827 lb Therefore use Fd = Fw = 5379 lb Fd = Ft +

0.05vm (Ft + Cb cos 2 ψ )cosψ

lb 1 0.05vm + (Ft + Cb cos 2 ψ )2 Fig. AF 20, carefully cut gears, Pdn = 5 , e = 0.001 in Table AT 25, steel and steel, 20o FD C = 1660 πD p n p π (3.681)(1750) vm = = = 1686 fpm 12 12 0.05(1686) Ft + 1660(3.5) cos 2 12 cos12 Fd = Ft + lb 1 2 0.05(1686) + Ft + 1660(3.5) cos 12 2 82.46[Ft + 5559] Fd = Ft + = 5379 lb 1 84.3 + [Ft + 5559]2 Trial and error Ft = 1800 lb Fv (1800)(1686) = 92 hp hp = t m = 33,000 33,000

[

]

[

Page 9 of 14

]



708.

Two helical gears are used in a single reduction speed reducer rated at 27.4 hp at a motor speed of 1750 rpm; continuous duty. The rating allows an occasional 100 % momentary overload. The pinion has 33 teeth. Pdn = 10 , b = 2 in. , φn = 20o ,

ψ = 20o , mw = 2.82 . For both gears, the teeth are carefully cut from SAE 1045 with BHN = 180. Compute (a) the dynamic load, (b) the endurance strength; estimate K f = 1.7 . Also decide whether or not the 100 % overload is damaging. (c) Are these teeth suitable for continuous service? If they are not suitable suggest a cure. (The gears are already cut.) Solution: N Dp = p Pd Pd = Pdn cosψ = (10) cos15 = 9.66 33 Dp = = 3.42 in 9.66 πD p n p π (3.42)(1750) vm = = = 1567 fpm 12 12 33,000hp 33,000(27.4 ) Ft = = = 577 lb vm 1567 (a) Dynamic load 0.05vm (Ft + Cb cos 2 ψ )cosψ lb Fd = Ft + 1 0.05vm + (Ft + Cb cos 2 ψ )2 Fig. AF 20, carefully cut gears, Pdn = 10 , e = 0.001 in Table AT 25, steel and steel, 20o FD C = 1660 b = 2 in Fd = 577 +

[

]

0.05(1567 ) 577 + 1660(2)cos 2 15 cos15

[

(b) Endurance strength sbY lb Fs = K f Pdn For SAE 1045, BHN = 180 su = 0.5BHN = 0.5(180) = 90 ksi sn = 0.5su = 0.5(90) = 45 ksi Np 33 N ep = = = 37 3 cos ψ cos3 15 Table AT 24, Load near middle, φn = 20o

Page 10 of 14

1 2

]

0.05(1567 ) + 577 + 1660(2)cos 15 2

= 2578 lb


SECTION 12 – HELICAL GEARS Y = 0.645 K f = 1.7

sbY (45,000)(2)(0.645) = 3415 lb = (1.7 )(10) K f Pdn For 100 % overload Ft = 2(577 ) = 1154 lb Fs =

Fd = Ft +

0.05vm (Ft + Cb cos 2 ψ )cosψ 1 2

0.05vm + (Ft + Cb cos ψ ) 2

Fd = 1154 +

lb

[

]

0.05(1567 ) 1154 + 1660(2)cos 2 15 cos15

[

(c) Fw =

1 2

]

0.05(1567 ) + 1154 + 1660(2) cos 15 Since Fs ≈ Fd , 100 % overload is not damaging 2

= 3475 lb

bD pQK g cos 2 ψ

b = 2 in. 2mw 2(2.82 ) Q= = = 1.476 mw + 1 2.82 + 1

Table AT 26, φn = 20o Sum of BHN = 2(180) = 360 K g = 62.5 Fw =

(2)(3.42)(1.476)(62.5) = 676 lb < F (= 2578 lb ) d

cos 2 15 Therefore not suitable for continuous service. Cure: Through hardened teeth For Bhn 2578 Kg = (62.5) = 238 676 min Bhn = 0.5(650) = 325

709.

Two helical gears are used in a speed reducer whose input is 100 hp at 1200 rpm, from an internal combustion engine. Both gears are made of SAE 4140, with the pinion heat treated to a BHN 363 – 415, and the gear to 321 – 363; let the teeth be F.D.; 20o pressure angle in the normal plane; carefully cut; helix angle ψ = 15o ; N p = 22 , N g = 68 teeth; Pd = 5 , b = 4 in . Calculate the dynamic load, the endurance strength load, and the limiting wear load for the teeth. Should these gears have long life if they operate continuously? (Data courtesy of the Twin Disc Clutch Co.)

Solution:

Page 11 of 14


SECTION 12 – HELICAL GEARS N p 22 = = 4.4 in Pd 5 πD p n p π (4.4)(1200) vm = = = 1382 fpm 12 12 33,000hp 33,000(100 ) Ft = = = 2388 lb vm 1382 Dynamic load 0.05vm (Ft + Cb cos 2 ψ )cosψ lb Fd = Ft + 1 2 0.05vm + (Ft + Cb cos ψ )2 Fig. AF 20, carefully cut gears, Pdn = 5 , e = 0.001 in Table AT 25, steel and steel, 20o FD C = 1660 b = 4 in Dp =

Fd = 2388 +

[

]

0.05(1382) 2388 + 1660(4) cos 2 15 cos15

[

1 2

]

0.05(1382 ) + 2388 + 1660(4)cos 15 Endurance strength load sbY cosψ lb Fs = K f Pd 2

= 5930 lb

Assume K f = 1.7 Pinion sn = 0.25BHN = 0.25(363) = 90.75 ksi Np 22 N ep = = = 25 3 cos ψ cos3 15 Table AT 24, Load near middle, φn = 20o Y = 0.580 sbY cosψ (90,750)(4 )(0.580) cos15 Fs = = = 23,925 lb (1.7 )(5) K f Pd Gear sn = 0.25BHN = 0.25(321) = 80.25 ksi Np 68 N ep = = = 75 3 cos ψ cos3 15 Table AT 24, Load near middle, φn = 20o Y = 0.735 sbY cosψ (80,250 )(4)(0.735)cos15 Fs = = = 26,811 lb (1.7 )(5) K f Pd use Fs = 23,925 lb

Page 12 of 14


SECTION 12 – HELICAL GEARS Limiting Wear Load bD pQK g Fw = cos 2 ψ Table AT 26, φn = 20o Sum of BHN = 684 to 778 use 700 K g = 270 Q= Fw =

2Ng 2(68) = = 1.511 N p + N g 22 + 68

(4)(4.4)(1.511)(270) = 7696 lb

cos 2 15 Since Fw (= 7696 lb ) > Fd (= 5930 lb ) these gears have long life if they operate continuously.

CROSSED HELICAL 710.

Helical gears are to connect two shafts that are at right angles ( N1 = 20 , N 2 = 40 , Pdn = 10 ,ψ 1 = ψ 2 = 45o ). Determine the center distance.

Solution: πD cosψ 1 N1 = 1 = Pdn D1 cosψ 1 Pcn 20 = (10)(D1 )cos 45 D1 = 2.83 in N 2 = Pdn D2 cosψ 2 40 = (10)(D2 )cos 45 D2 = 5.66 in C = 12 (D1 + D2 ) = 12 (2.83 + 5.66) = 4.25 in 712.

Two shafts that are at right angles are to be connected by helical gears. A tentative design is to use N1 = 20 , N 2 = 60 , Pdn = 10 , and a center distance of 6 in. What must be the helix angles?

Solution: Σ = ψ 1 +ψ 2 = 90o N1 D1 = Pdn cosψ 1 N2 D2 = Pdn cosψ 2 1 C = 2 (D1 + D2 )

Page 13 of 14


SECTION 12 – HELICAL GEARS N1 N2 + Pdn cosψ 1 Pdn cosψ 2 20 60 + 2(6 ) = 10 cosψ 1 10 cosψ 2 2 6 12 = + cosψ 1 cosψ 2 1 3 6= + cosψ 1 cosψ 2 By trial and error method 1 3 + 6= cosψ 1 sinψ 1 2C =

ψ 1 = 39.5o ψ 2 = 50.5o - end -

Page 14 of 14


SECTION 13 – BEVEL GEARS DESIGN PROBLEMS 751.

Decide upon the pitch, face, N g , material, and heat treatment of a pair of straight bevel gears to transmit continuously and indefinitely a uniform loading of 5 hp at 900 rpm of the pinion, reasonable operating temperature, high reliability; mg ≈ 1.75 ; D p ≈ 3.333 in . Pinion overhangs, gear is straddle mounted.

Solution:

(

L = rp2 + rg2

tan γ p =

1 2

)

1 1 = mg 1.75

γ p = 29.75o L sin γ p = rp L sin 29.75 =

3.333 2

L = 3.358 in 33,000hp Ft = lb vm πD p n p π (3.333)(900) vm = = = 785.4 fpm 12 12 33,000(5) Ft = = 210 lb 785.4 Fd = (VF )N sf K m Ft

50 + vm2 50 + (785.4 ) 2 VF = = = 1.56 50 50 One gear straddle, one not K m = 1.2 Table 15.2, uniform N sf = 1.0 1

1

Fd = (1.56)(1.0)(1.2 )(210) = 393 lb Wear load s2  C  Fw = D pbI cd2  l  C e  K t Cr 

2

D p = 3.333 in

b = 0.3L = 0.3(3.358) = 1.0 in Temperature factor K t = 1.0 , reasonable operating temperature Life factor for wear

Page 1 of 17


SECTION 13 – BEVEL GEARS Cl = 1.0 for indefinite life Reliability factor for wear Cr = 1.25 high reliability Geometry factor for wear, Fig. 15.7 Assume I = 0.080 Elastic coefficient (Table 15.4) Steel on steel , Ce = 2800 Fw = Fd 2 (3.333)(1.0)(0.08) (scd ) 2  1.0  (2800)  (1.0)(1.25)

2

= 393

scd = 134,370 psi Table 15.3, use Steel, (300) scd = 135 ksi Strength of bevel gears s bJ K l Fs = d Pd K s K t K r Size factor, assume K s = 0.71 Life factor for strength K l = 1 for indefinite life Temperature factor, K t = 1 good operating condition Reliability factor K r = 1.5 high reliability Geometry factor for strength (Fig. 15.5) Assume J = 0.240 b = 1.0 in sd = design flexural stress Min. BHN = 300 sd = 19 ksi

Fs = F d

(19,000)(1.0)(0.240)  Pd

 1  (0.71)(1)(1.5)  = 393  

Pd = 11 say Pd = 10 10 10 so that b = = = 1.0 in Pd 10 Dg = D p mg = (3.333)(1.75) = 5.833 in

Page 2 of 17


SECTION 13 – BEVEL GEARS N g = Pd Dg = (10 )(5.833) = 58.33

say N g = 58 Use Pd = 10 , b = 1.0 in , N g = 58 Material = steel, min. Bhn = 300 752.

A pair of steel Zerol bevel gears to transmit 25 hp at 600 rpm of the pinion; mg = 3 ; let N p ≈ 20 teeth; highest reliability; the pinion is overhung, the gear straddle mounted. An electric motor drives a multi-cylinder pump. (a) Decide upon the pitch, face width, diameters, and steel (with treatment) for intermittent service. (b) The same as (a) except that indefinite life is desired.

Solution: N 20 Dp = p = Pd Pd  20  (600) πD p n p Pd  1000π  vm = fpm = = Pd 12 12 10 Let b = Pd Dynamic load Fd = (VF )N sf K m Ft

π 

33,000hp lb vm 33,000(25) Ft = = 262.6 Pd  1000π     Pd  Ft =

1

 1000π   50 +  1 2 Pd  50 + vm 1.121 1.121  VF = = = 1 + 12 = 1 + 50 50 Pd Pd Table 15.2, electric motor drives a multi-cylinder pump Service factor, N sf = 1.25 2

One gear straddle, one not, K m = 1.2  1.121    (1.25)(1.2 )(262.6 )Pd = 394 Pd 1 + 1.121  Fd = 1 +   Pd  Pd    (a) Strength of Bevel Gears s bJ K l Fs = d Pd K s K t K r Size factor, assume K s = 0.71

Page 3 of 17


SECTION 13 – BEVEL GEARS Life factor for strength Intermittent service, use K l = 4.6 Temperature factor, say K t = 1.0 Reliability factor, highest reliability K r = 3.0 Geometry factor for strength N mg = g Np N p = 20 N p = 3(20) = 60

Fig. 15.5, J = 0.205 10 b= Pd Design flexural stress, steel Assume sd = 15 ksi Fs = Fd 



(15,000) 10 (0.205)(4.6)

 1.121   = 394 Pd 1 +   P d    1.121  66,408   = P 394 d 1 +  Pd2 P d   Pd = 4.814 say Pd = 5 10 10 b= = = 2.0 in Pd 5  Pd  Pd (0.71)(1.0 )(3)

Wear load for bevel gears s2  C  Fw = D pbI cd2  l  C e  K t Cr 

2

N p 20 = = 4 in Pd 5 K t = 1.0 Dp =

Life factor for wear, intermittent service Cl = 1.5 Reliability factor for wear, highest reliability Cr = 1.25 Geometry factor for wear, Fig. 15.7

Page 4 of 17


SECTION 13 – BEVEL GEARS N p = 20 , N g = 60 I = 0.083 Elastic coefficient, steel on steel (Table 15.4) Ce = 2800

Pd = 5 Fw = Fd 2

2 (4)(2)(0.083) scd 2  1.5  = 394(5)1 + 1.121  (2800)  (1.0)(1.25) 5   scd = 155,730 psi Table 15.3 Use steel, min. BHN = 360, scd = 160 ksi Pd = 5 b = 2 in D p = 4 in

Dg = mg D p = (3)(4) = 12 in steel, min. BHN = 360

(b) For indefinite life, K l = 1.0 , life factor for strength Cl = 1.0 , life factor for wear Strength: s bJ K l Fs = d Pd K s K t K r Fs = Fd 



(15,000) 10 (0.205)(1.0)

 1.121   = 394 Pd 1 +   P d    1.121  14,437  = 394 Pd 1 + 2   Pd P d   Pd = 2.799 say Pd = 3 10 10 b= = = 3.33 in Pd 3  Pd  Pd (0.71)(1.0)(3)

Wear load

Page 5 of 17


SECTION 13 – BEVEL GEARS scd2  Cl   Fw = D pbI 2  C e  K t Cr 

2

N p 20 = 6.67 in = Pd 3 Fw = Fd

Dp =

2

2 (6.67 )(3.33)(0.083) scd 2  1.0  = 394(3)1 + 1.121  (2800)  (1.0)(1.25) 3   scd = 113,744 psi Table 15.3 Use steel, min. BHN = 240, scd = 115 ksi Pd = 3 b = 3.33 in D p = 6.67 in

Dg = mg D p = (3)(6.67 ) = 20 in steel, min. BHN = 240

753.

Decide upon the pitch, face, and number of teeth for two spiral-bevel gears for a speed reducer. The input to the pinion is 20 hp at 1750 rpm; mg ≈ 1.9 ; pinion overhung, gear-straddle mounted. It is hoped not to exceed a maximum D p of 4 3/8-in.; steel gears with minimum 245 BHN on pinion and 210 BHN on gear. The gear is motor-driven, subject to miscellaneous drives involving moderate shock; indefinite life against breakage and wear with high reliability. If the gears designed for the foregoing data are to be subjected to intermittent service only, how much power could they be expected to transmit?

Solution: (a) vm =

πD p n p

π (4.375)(1750)

= 2000 fpm 12 12 33,000hp 33,000(20 ) Ft = = = 330 lb vm 2000 Dynamic load Fd = (VF )N sf K m Ft One gear straddle, one not K m = 1.2 Table 15.2 Motor-driven, moderate shock

Page 6 of 17

=


SECTION 13 – BEVEL GEARS N sf = 1.25 1

 70 + vm2  2 VF =   , spiral  70  1

1

 70 + (2000) 2  VF =   = 1.254 70   Fd = (1.254)(1.25)(1.2)(330) = 621 lb 1

2

Wear load s2  C  Fw = D pbI cd2  l  C e  K t Cr 

2

D p = 4.375 in

Temperature Factor, K t = 1.0 Design contact stresses, BHN = 245 , pinion scd = 116 ksi Life factor for wear Cl = 1.0 , indefinite life Reliability factor for wear Cr = 1.25 , high reliability Geometry factor for wear, Fig. 15.8 Assume I = 0.12 Elastic coefficient, steel on steel (Table 15.4) Ce = 2800

(116,000)2  1.0  Fw = (4.375)(b )(0.12) (2800)2  (1.0)(1.25) Fw = Fd 721b = 621 b = 0.8613 in 7 say b = in = 0.875 in 8

Strength of gear s bJ K l Fs = d Pd K s K t K r sd = design flexural stress min. BHN = 210 sd = 15.4 ksi Size factor, assume K s = 0.71

Page 7 of 17

2

= 721b


SECTION 13 – BEVEL GEARS Life factor for strength K l = 1 for indefinite life Temperature factor, Kt = 1 Reliability factor K r = 1.5 high reliability Geometry factor Fig. 15.6 Assume J = 0.28  3543 (15,400)(0.875)(0.28)  1 Fs =  = P Pd  (0.71)(1)(1.5)  d Fs = Fd 3543 = 621 Pd Pd = 5.7 say Pd = 6 7 Then, Pd = 6 , b = in , N p = D p Pd = (4.375)(6) = 26 8 N g = mw N p = (1.9 )(26 ) = 50 (b) Intermittent service only Strength s bJ K l Fs = d Pd K s K t K r sd = 15,400 psi (Gear) For Pd = 6 , K s = 0.64 For indefinite service, K l = 4.6 K t = 1.0 , K r = 1.5 Geometry factor, Fig. 15.6, N p = 26 , N g = 50 J = 0.292 (15,400)(0.875)(0.292)  4.6  = 3142 lb Fs =  (0.71)(1)(1.5)  6   Wear load s2  C  Fw = D pbI cd2  l  C e  K t Cr  D p = 4.375 in

K t = 1.0 scd = 116 ksi Ce = 2800

Page 8 of 17

2


SECTION 13 – BEVEL GEARS Cl = 1.5 intermittent service Cr = 1.25 Geometry factor for wear, Fig. 15.8 N p = 26 , N g = 50 I = 0.116

(116,000)2  1.5  Fw = (4.375)(0.875)(0.116 ) (2800)2  (1.0)(1.25)

2

= 1098 lb

use Fw = Fd Fd = (VF )N sf K m Ft 1098 = (1.254)(125)(1.2)Ft Ft = 584 lb Fv (584)(2000) = 35 hp hp = t m = 33,000 33,000 CHECK PROBLEMS 755.

A pair of straight-bevel gears are to transmit a smooth load of 45 hp at 500 rpm 3 of the pinion; mg = 3 . A proposed design is Dg = 15 in. , b = 2 in. , Pd = 4 . 8 Teeth are carburized AISI 8620, SOQT 450 F. The pinion overhangs, the gear is straddle-mounted. Would these gears be expected to perform with high reliability in continuous service? If not would you expect more than 1 failure in 100?

Solution: D 15 D p = g = = 5 in mg 3 πD p n p π (5)(500) vm = = = 655 fpm 12 12 33,000hp 33,000(45) Ft = = = 2267 lb vm 655 Dynamic load Fd = (VF )N sf K m Ft 50 + vm2 50 + (655) 2 VF = = = 1.512 50 50 One gear straddle, one not K m = 1.2 1

1

Smooth load, N sf = 1.0 Fd = (1.512)(1.0)(1.2 )(2267 ) = 4113 lb

Page 9 of 17


SECTION 13 – BEVEL GEARS Strength of bevel gears s bJ K l Fs = d Pd K s K t K r Size factor, for Pd = 4 , K s = 0.71 Life factor for strength Kl = 1 Temperature factor, Kt = 1 Geometry factor for strength (Fig. 15.5) N p = Pd D p = (4)(5) = 20 N g = Pd Dg = (4 )(15) = 60 J = 0.205

sd = 30 ksi (55 – 63 Rc) for carburized teeth

Fs =

(30,000)(2.375)(0.205) 

 5143 1  (0.71)(1)(K )  = K r  r 

4

Fs = Fd 5143 = 4113 Kr K r = 1.25 < 1.5 will not perform high reliability. Wear load s2  C  Fw = D pbI cd2  l  C e  K t Cr 

2

D p = 5 in b = 2.375 in Table 15.3, scd = 225 ksi

Table 15.4, Ce = 2800 Geometry factor for wear, Figure 15.7 N p = 20 , N g = 60 I = 0.083 Kt = 1 life factor for wear Cl = 1

(225,000)2  1  Fw = (5)(2.375)(0.083) (2800)2  (1)(Cr ) Fw = Fd

Page 10 of 17

2

=

6364 Cr2


SECTION 13 – BEVEL GEARS 6364 = 4113 Cr2 Cr = 1.244 ≈ 1.25 , high reliability

Since K r < 1.5 ,this will not perform high reliability but 1 in 100, K r ≈ 1.12 < 1.25 756.

A gear catalog rates a pair of cast-iron, straight-bevel gears at 15.26 hp at 800 rpm of the 16-tooth pinion; mg = 3.5 , b = 3 in. , Pd = 3 ; pinion overhangs, straddle-mounted gear. Assume the cast iron to be class 30. If the load is smooth is this rating satisfactory, judging by the design approach of the Text for good reliability (a) when strength alone is considered, (b) when long continuous service is desired?

Solution: N 16 Dp = p = = 5.333 in Pd 3 πD p n p π (5.333)(800) vm = = = 1117 fpm 12 12 33,000hp 33,000(15.26 ) Ft = = = 451 lb vm 1117 Dynamic load Fd = (VF )N sf K m Ft 50 + vm2 50 + (1117 ) 2 VF = = = 1.668 50 50 One gear straddle, one not K m = 1.2 1

1

Smooth load, N sf = 1.0 Fd = (1.668)(1.0)(1.2)(451) = 903 lb (a) Strength s bJ K l Fs = d Pd K s K t K r Pd = 3 , K s = 0.76 Kl = 1 Kt = 1 K r = 1.5 sd = 4.6 ksi , cast-iron class 30 N p = 16

Page 11 of 17


SECTION 13 – BEVEL GEARS N g = mw N p = (3.5)(16) = 56 J = 0.184  (4,600)(3)(0.184)  1 Fs =   = 742 lb < 903 lb(= Fd ) 3  (0.76)(1)(1.5)  with K l = 1.4 for 106 cycles

Fs = (1.4 )(742 ) = 1040 lb > 903 lb(= Fd ) Therefore satisfactory for 106 cycles. (b) Continuous service Wear load s2  C  Fw = D pbI cd2  l  C e  K t Cr 

2

D p = 5.333 in b = 3 in Table 15.3, scd = 50 ksi , cast-iron class 30

Table 15.4, cast-iron and cast-iron Ce = 2250 Cl = 1 Kt = 1 Cr = 1.25 Geometry factor for wear, Figure 15.7 N p = 16 , N g = 56 I = 0.077

(50,000)2  1  Fw = (5.333)(3)(0.077 ) (2250 )2  (1)(1.25)

2

= 389 lb < Fd (= 903 lb )

Therefore, not satisfactory for long continuous service. 757.

An 870-rpm motor drives a belt conveyor through bevel gears having 18 and 72 3 teeth; Pd = 6 , b = 1 in . Both gears are straddle-mounted. What horsepower may 4 these gears transmit for an indefinite life with high reliability if both gears are (a) cast-iron, class 40; (b) AISI 5140, OQT 1000 F; (c) AISI 5140, OQT 1000 F, flame hardened (d) AISI 8620, SOQT 450 F?

Solution: N 18 D p = p = = 3 in Pd 6 πD p n p π (3)(870) vm = = = 683 fpm 12 12 Dynamic load Page 12 of 17


SECTION 13 – BEVEL GEARS Fd = (VF )N sf K m Ft Both gears straddle mounted K m = 1.0

Table 15.2, N sf = 1.0 50 + vm2 50 + (683) 2 = = 1.523 50 50 Fd = (1.523)(1.0)(1.0)Ft = 1.523Ft 1

1

VF =

(a) Cast-iron, class 40 Strength s bJ K l Fs = d Pd K s K t K r sd = 7 ksi , cast-iron class 40 3 b = 1 in 4 K l = 1 , indefinite life Pd = 6 K s = 0.64 Kt = 1 K r = 1.5 , high reliability Figure 15.5, N p = 18 , N g = 72 J = 0.204  (7000)(1.75)(0.204)  1 Fs =   = 434 lb 6  (0.64)(1)(1.5)  Wear: s2  C  Fw = D pbI cd2  l  C e  K t Cr 

2

D p = 3 in 3 b = 1 in 4 Table 15.3, scd = 65 ksi , cast-iron class 40

Table 15.4, cast-iron and cast-iron Ce = 2250 Cl = 1 , indefinite life Kt = 1 Cr = 1.25 , high reliability Geometry factor for wear, Figure 15.7 N p = 18 , N g = 72 I = 0.082 Page 13 of 17


SECTION 13 – BEVEL GEARS 2 ( 65,000 )  1  Fw = (3)(1.75)(0.082 ) (2250)2  (1)(1.25)

2

= 230 lb

Fd = Fw 1.523Ft = 230 Ft = 151 lb Fv (151)(683) = 3 hp hp = t m = 33,000 33,000 (b) AISI 5140, OQT 1000 F, BHN = 300 Strength sd = 19,000 psi Fs =

(19,000)(1.75)(0.204) 

 1  (0.64)(1)(1.5)  = 1178 lb  

6

Wear: s2  C  Fw = D pbI cd2  l  C e  K t Cr 

2

Table 15.3, scd = 135 ksi Table 15.4, steel and steel, Ce = 2800 2 ( 135,000 )  1  Fw = (3)(1.75)(0.082 ) (2800)2  (1)(1.25)

2

= 640 lb

Fd = Fw 1.523Ft = 640 Ft = 420 lb Fv (420)(683) = 8.7 hp hp = t m = 33,000 33,000 (c) AISI 5140, OQT 1000 F, Flame Hardened Strength sd = 13.5 ksi Fs =

(13,500)(1.75)(0.204) 

 1  (0.64 )(1)(1.5)  = 837 lb  

6


2

Table 15.3, scd = 190 ksi Table 15.4, steel and steel, Ce = 2800

Page 14 of 17


SECTION 13 – BEVEL GEARS 2 ( 190,000)  1  Fw = (3)(1.75)(0.082 ) (2800)2  (1)(1.25)

2

= 1269 lb

Fd = Fs 1.523Ft = 837 Ft = 550 lb Fv (550)(683) = 11.4 hp hp = t m = 33,000 33,000 (d) AISI 86200, SOQT 450 F, carburized Strength sd = 30 ksi (55 – 63 Rc) Fs =

(30,000)(1.75)(0.204) 

 1  (0.64)(1)(1.5)  = 1859 lb  

6


2

Table 15.3, scd = 225 ksi Table 15.4, steel and steel, Ce = 2800 2 ( 225,000)  1  Fw = (3)(1.75)(0.082 ) (2800)2  (1)(1.25)

2

= 1779 lb

Fd = Fw 1.523Ft = 1779 Ft = 1168 lb Fv (1168)(683) = 24.2 hp hp = t m = 33,000 33,000 758.

A pair of straight-bevel gears transmits 15 hp at a pinion speed of 800 rpm; Pd = 5 , N p = 20 , N p = 60 , b = 2 in . Both gears are made of AISI 4140 steel, OQT 800 F. What reliability factor is indicated for these gears for strength and for wear (a) for smooth loads, (b) for light shock load from the power source and heavy shock on the driven machine?

Solution: N 20 Dp = p = = 4 in Pd 5 πD p n p π (4)(800) vm = = = 838 fpm 12 12

Page 15 of 17


SECTION 13 – BEVEL GEARS 33,000hp 33,000(15) = = 591 lb vm 838 Fd = (VF )N sf K m Ft

Ft =

50 + vm2 50 + (838) 2 VF = = = 1.579 50 50 assume K m = 1.0 1

1

Fd = (1.579)(N sf )(1.0 )(591) = 933N sf Strength of bevel gear s bJ K l Fs = d Pd K s K t K r For AISI 4140, OQT 800 F, BHN = 429 sd = 24 ksi

assume K l = 1 Kt = 1 Pd = 5 K s = 0.675 Figure 15.5, N p = 20 , N g = 60 J = 0.205  2916 1 (24,000)(2)(0.205)  Fs =  = K 5 r  (0.675)(1)(K r )  Fs = Fd 2916 = 933N sf Kr 3.1254 Kr = N sf Wear load: s2  C  Fw = D pbI cd2  l  C e  K t Cr 

2

BHN = 429 Table 15.3, scd = 190 ksi Table 15.4, steel and steel, Ce = 2800 D p = 4 in b = 2 in Assume Cl = 1.0 , K t = 1.0 Fig. 15.7, N p = 20 , N g = 60 I = 0.083

Page 16 of 17


SECTION 13 – BEVEL GEARS 2 ( 190,000 )  1  ( )( )( ) Fw = 4 2 0.083 (2800)2  (1)(Cr )

2

=

3058 Cr2

Fd = Fw 3058 Cr2 1.810 Cr = N sf

933N sf =

(a) Table 15.2, smooth load N sf = 1.0

For strength, K r = For wear, Cr =

3.1254 3.1254 = = 3.1254 1 N sf

1.810 1.810 = = 1.810 1 N sf

(b) Table 15.2, light shock source, heavy shock driven N sf = 2.0

3.1254 3.1254 = = 1.5627 2 N sf 1.810 1.810 = = 1.2799 For wear, Cr = 2 N sf For strength, K r =

- end -

Page 17 of 17


SECTION 14 – WORM GEARS DESIGN PROBLEMS 791.

(a) Determine a standard circular pitch and face width for a worm gear drive with an input of 2 hp at 1200 rpm of the triple-threaded worm; the 1.58-in. ( Dw ) is steel with a minimum BHN = 250; gear is manganese bronze (Table AT 3); mw = 12 . Consider wear and strength only. Use a φn to match the lead angle λ . (See i16.13, Text.) (b) compute the efficiency.

Solution:  1200 + vmg   Ft lb a) Fd =   1200  33,000hp Ft = vmg πDg ng vmg = 12 n 1200 ng = w = = 100 rpm mw 12 Dg = mw Dw tan λ tan λ =

N t Pc πDw

 N P  N m P (3)(12)Pc Dg = mw Dw  t c  = t w c = = 11.46 Pc π π  πDw  π (11.45Pc )(100) vmg = = 300 Pc 12 33,000(2 ) 220 Ft = = Pc 300 Pc  1200 + 300 Pc  220  55(4 + Pc ) Fd =  = lb  Pc 1200   Pc  Wear load Fw = Dg bK w say b = 2 Pc , Dg = 11.46 Pc Fw = Fd

(11.46 Pc )(2 Pc )(K w ) = 55(4 + Pc ) Pc

22.92 Pc2 K w =

Page 1 of 19

55(4 + Pc ) Pc


SECTION 14 – WORM GEARS Dg 11.46 Pc = = 0.60443Pc mw Dw (12)(1.58)

tan λ =

By trial and error and using Table AT 27 ( φn ≈ λ ) Kw

Pc

Pc (std)

λ

36 50

0.678 0.605

¾ 5/8

24.4 20.7

Use φn = 20o , λ = 20.7o , Pc =

5 in 8

Fw = Fd

(11.46 Pc )(b )(K w ) = 55(4 + Pc ) Pc

5  55 4 +  (11.46) 5 (b )(50) =  5 8  8 8 b = 1.1365 in 5 say b = 1 in 32 To check for strength. sYbPcn sYbPc cos λ Fs = =

π

π

For manganese bronze, s = 30,000 psi

φn = 20o Y = 0.392 λ = 20.7o 5 Pc = in 8 5 b = 1 in 32

(30,000)(0.392)1 5

 5    cos 20.7  32  8  = 2530 lb > Fd

Fs =

π

use Pc = b =1

5 in 8

5 in 32

Page 2 of 19

λmax

i16.11

16 25

φn 14 ½ 20


SECTION 14 – WORM GEARS  cos φn − f tan λ   (b) e = tan λ   cos φn tan λ + f  φn = 20o

λ = 20.7o πDwnw π (1.58)(1200 ) = = 531 fpm > 70 fpm vr = 12 cos λ 12 cos 20.7 0.32 0.32 = = 0.0334 0.36 vr (531)0.36  cos 20 − 0.0334 tan 20.7  e = tan 20.7  = 0.902 = 90.2%  cos 20 tan 20.7 + 0.0334  f =

792.

A high-efficiency worm-gear speed reducer is desired, to accept 20 hp from a 1750-rpm motor. The diameter Dw of the integral worm has been estimated to be 7 1 in. ; the next computations are to be for a steel worm with a minimum BHN = 8 250; phosphor-bronze gear (Table AT 3); mw = 11 . Probably, the worm should not have less than 4 threads. (a) Considering wear and strength only (i16.13), decide upon a pitch and face width that satisfies these requirements (i16.11, Text); specifying the pressure angle, diameters, and center distance. How does Dw used compare with that from equation (m), i16.11, Text? What addendum and dedendum are recommended by Dudley? Compute a face length for the worm. (b) Compute the efficiency. What do you recommend as the next trial for a “better” reducer?

Solution: 33,000hp Ft = vmg πDg ng vmg = 12 n 1750 ng = w = = 159.1 rpm mw 11 PN Pm N P (11)(4 ) Dg = c g = c w t = c = 14 Pc

π

vmg =

π

π (14 Pc )(159.1)

π

= 583Pc 12 33,000(20 ) 1132 Ft = = 583Pc Pc

Page 3 of 19


SECTION 14 – WORM GEARS  1200 + vmg   Ft lb Fd =   1200   1200 + 583Pc  1132  1132(1 + 0.4858 Pc ) Fd =  lb =  Pc 1200   Pc  (a) Wear Fw = Dg bK w b = 2 Pc , Dg = 14 Pc Fw = Fd

(14 Pc )(2 Pc )(K w ) = 1132(1 + 0.4858Pc )

Pc 1132(1 + 0.4858 Pc ) 28 Pc2 K w = Pc Table AT 27, steel, min. BHN = 250, and bronze And by trial and error ethod NP 4(Pc ) tan λ = t c = = 0.6791Pc πDw (π )(1.875)

By trial and error and using Table AT 27 Kw

Pc

Pc (std)

λ

36 50 60

1.213 1.071 1.000

1¼ 1¼ 1.0

40.33 40.33 34.18

Use φn = 25o , λ = 34.18o , Pc = 1 in Fw = Fd (14 Pc )(b )(K w ) = 1132(1 + 0.4858Pc ) Pc (14)(1)(b )(60) = 1132(1 + 0.4858) 1 b = 2 in To check for strength sYbPcn sYbPc cos λ Fs = =

π

π

For phosphor-bronze, s = sn = 31,000 psi

Page 4 of 19

λmax

i16.11

16 25 35

φn 14 ½ 20 25


SECTION 14 – WORM GEARS

For φn = 25o , Y = 0.470 (31,000)(0.470 )(2.0)(1.0)cos 34.18 = 7674 lb > F , ok Fs = d

π

use Pc = 1.0 in b = 2.0 in φn = 25o 7 Dw = 1 in 8  7 Dg = mw Dw tan λ = (11)1  tan 34.18 = 14.0 in  8 1 1 7  C = (Dw + Dg ) = 1 + 14  = 7.9375 in 2 2 8  Equation (m) (7.9375)0.875 = 2.785 in > 1.875 in , ok C 0.875 Dw = in = 2 .2 2 .2 Addendum and dedendum (by Dudley)

Addendum = a = 0.3183Pcn = 0.3183Pc cos λ = 0.3183(1.0) cos 34.18 = 0.2633 in Whole depth = 0.7 Pcn = 0.7 Pc cos λ = 0.7(1.0)cos 34.18 = 0.5791 in Dedendum = whole depth – addendum = 0.5791 in – 0.2633 in = 0.3158 in N   Face length = Pc  4.5 + g  50   N g = mw N p = (11)(4 ) = 44 44   Face length = 1.0 4.5 +  = 5.38 in 50   Or

[

]

Face length = 2 2a(Dg − 2a )

1

2

Dg = 14 in a = 0.2633 in 1

Face length = 2{2(0.2633)[14 − 2(0.2633)]} 2 = 5.33 in Use Face length = 5.38 in  cos φn − f tan λ   (b) e = tan λ   cos φn tan λ + f  πDwnw π (1.875)(1750 ) vr = = = 1038 fpm > 70 fpm 12 cos λ 12 cos 34.18

Page 5 of 19



f =

0.32 0.32 = = 0.0263 ( 70 < vr < 3000 fpm ) 0.36 vr (1038)0.36

φn = 25o , λ = 34.18o ,  cos 25 − 0.0263 tan 34.18  e = tan 34.18  = 0.94 = 94%  cos 25 tan 34.18 + 0.0263  recommendation for next trial φn = 30o

λmax = 45o 793.

The input to a worm-gear set is to be 25 hp at 600 rpm of the worm with m w = 20 . The hardened-steel worm is to be the shell type with a diameter approximately as given in i16.11, Text, and a minimum of 4 threads; the gear is to be chilled phosphor bronze (Table AT 3). (a) Considering wear and strength only determine suitable values of the pitch and face width. Let φn be appropriate to the value of λ . (b) Compute the efficiency. (c) Estimate the radiating area of the case and compute the temperature rise of lubricant. Is special cooling needed?

Solution: 33,000hp Ft = vmg πDg ng vmg = 12 n 600 ng = w = = 30 rpm mw 20 PN Pm N P (20 )(4 ) 80 Pc Dg = c g = c w t = c =

π

π

π

π

 80 Pc  (30) π   vmg = = 200 Pc 12 33,000(25) 4125 Ft = = Pc 200 Pc

π

 1200 + vmg   Ft lb Fd =   1200   1200 + 200 Pc  4125  687.5(6 + Pc ) Fd =  = lb  Pc 1200   Pc 

shell type: Dw = 2.4 Pc + 1.1 in (4)Pc NP 4 Pc tan λ = t c = = πDw (π )(2.4 Pc + 1.1) π (2.4 Pc + 1.1)

Page 6 of 19


SECTION 14 – WORM GEARS (a) Wear load Fw = Dg bK w b = 2 Pc , 80 Pc Dg =

π

Fw = Fd 687.5(6 + Pc )  80 Pc   (2 Pc )(K w ) = Pc  π  687.5(6 + Pc ) 50.93Pc2 K w = Pc Table AT 27, Hardened steel and chilled bronze By trial and error method

By trial and error and using Table AT 27 ( φn ≈ λ ) Kw

Pc

Pc (std)

λ

90 125

1.017 0.907

1.0 1.0

20 20

Use φn = 20o , λ = 20o , Pc = 1 in Fw = Fd 1687.5(6 + 1)  80   (b )(125) = 1 π  b = 1.512 in 5 say b = 1 in 8  cos φn − f tan λ   (b) e = tan λ   cos φn tan λ + f 

φn = 20o λ = 20o Dw = 2.4 Pc + 1.1 = 2.4 + 1.1 = 3.5 in πDwnw π (3.5)(600 ) vr = = = 585 fpm 12 cos λ 12 cos 20 0.32 0.32 f = 0.36 = = 0.0323 ( 70 < vr < 3000 fpm ) vr (585)0.36  cos 20 − 0.0323 tan 20  e = tan 20  = 0.9023 = 90.23%  cos 20 tan 20 + 0.0323 

Page 7 of 19

λmax

i16.11

16 25

φn 14 ½ 20


SECTION 14 – WORM GEARS (c) Radiating area ≈ Amin = 43.2C 1.7 sq. in. 1 (Dw + Dg ) 2 Dw = 3.5 in 80 Pc 80(1) Dg = = = 25.5 in

C=

π

π

1 C = (3.5 + 25.5) = 14.5 in 2 1.7 Amin = 43.2(14.5) = 4072 sq.in.

Temperature rise = ∆t Qc = hcr A∆t ft − lb min Q = (1 − e )(hpi ) = (1 − 0.9023)(25) = 2.4425 hp(33,000 ft − lb min − hp ) = 80,600 ft − lb min Figure AF 21, A = 4072 sq.in. = 28.3 sq. ft. hcr = 0.42 ft − lb min − sq.in. − F Q = Qc 80,600 = (0.42)(4072)(∆t ) ∆t = 47 F with t1 = 100 F t2 = 147 F < 150 F Therefore, no special cooling needed. 794.

A 50-hp motor turning at 1750 rpm is to deliver its power to a worm-gear reducer, whose velocity ratio is to be 20. The shell-type worm is to be made of high-test cast iron; since a reasonably good efficiency is desired, use at least 4 threads; manganese –bronze gear (Table AT 3). (a) Decide upon Dw and φn , and determine suitable values of the pitch and face width. Compute (b) the efficiency, (c) the temperature rise of the lubricant. Estimate the radiating area of the case. Is special cooling needed?

Solution: 33,000hp Ft = vmg πDg ng vmg = 12 n 1750 ng = w = = 87.5 rpm mw 20 PN Pm N P (20 )(4 ) 80 Pc Dg = c g = c w t = c =

π

Page 8 of 19

π

π

π


SECTION 14 – WORM GEARS  80 Pc  (87.5) π   = 583Pc vmg = 12 33,000(50 ) 2830 Ft = = Pc 583Pc

π

 1200 + vmg   Ft lb (a) Fd =   1200   1200 + 583Pc  2830  1375(2.06 + Pc ) Fd =  = lb  Pc 1200   Pc  Wear load Fw = Dg bK w

b = 2 Pc , 80 Pc Dg =

π

Fw = Fd 1375(2.06 + Pc )  80 Pc  (2 Pc )(K w ) =  Pc  π  1375(2.06 + Pc ) 50.93Pc2 K w = Pc NP tan λ = t c πDw Shell-type Dw = 2.4 Pc + 1.1 in 4 Pc tan λ = π (2.4 Pc + 1.1) Table AT 27, high-test cast-iron and manganese bronze By trial and error and using Table AT 27 ( φn ≈ λ ) Kw

Pc

Pc (std)

λ

80 115

1.012 0.885

1.0 7/8

20 19.2

Use λ = 19.2o , φn = 20o , Pc =

7 in 8

7 Dw = 2.4 Pc + 1.1 = 2.4  + 1.1 = 3.2 in 8 Fw = Fd

Page 9 of 19

λmax

i16.11

16 25

φn 14 ½ 20


SECTION 14 – WORM GEARS 7  1375 2.06 +  8  80  7     (b )(115) = 7  π  8  8 b = 1.80 in 7 say b = 1 in 8  cos φn − f tan λ   (b) e = tan λ   cos φn tan λ + f  λ = 19.2o φn = 20o πDwnw vr = 12 cos λ nw = 1750 rpm Dw = 3.2 in πDwnw π (3.2 )(1750 ) vr = = = 1552 fpm 12 cos λ 12 cos19.2 0.32 0.32 f = 0.36 = = 0.0227 ( 70 < vr < 3000 fpm ) vr (1552)0.36  cos 20 − 0.0227 tan 19.2  e = tan 19.2  = 0.9273 = 92.73%  cos 20 tan 19.2 + 0.0227 

(c) Q = (1 − e )(hpi ) = (1 − 0.9273)(50 ) = 3.635 hp = 119,955 ft − lb min Qc = hcr A∆t ft − lb min A = Amin = 43.2C1.7 sq.in. 1 C = (Dw + Dg ) 2 Dw = 3.2 in 7 80  80 Pc 8 Dg = =   = 22.3 in

π

π

1 C = (3.2 + 22.35) = 12.75 in 2 1.7 A = 43.2(12.75) = 3272 sq.in. Figure AF 1 3272 A= = 22.7 ft 2 144 hcr = 0.43 ft − lb min − sq.in. − F

Page 10 of 19


SECTION 14 – WORM GEARS Q = Qc 119,955 = (0.43)(3272)(∆t ) ∆t = 85 F with t1 = 100 F t2 = 185 F > 150 F Therefore, special cooling is needed. CHECK PROBLEMS 795.

A worm-gear speed reducer has a hardened-steel worm and a manganese-bronze gear (Table AT 3); triple-threaded worm with Pc = 1.15278 in. , Dw = 3.136 in. , 1 φn = 25o , b = 2 in. , mw = 12 , nw = 580 rpm . The output is 16 hp. Compute (a) 4 the dynamic load, (b) the endurance strength of the teeth and the indicated service factor on strength, (c) the limiting wear load (is it good for indefinitely continuous service?), (d) the efficiency and input hp, (e) the temperature rise of the oil (estimate case area as Amin , i16.6). (f) Determine the tangential and radial components of the tooth load. (g) Is this drive self-locking?

Solution: 33,000hp Ft = vmg πDg ng vmg = 12 nw 580 ng = = = 48.3 rpm mw 20 PN P m N (1.15278)(12 )(3) Dg = c g = c w t = = 13.21 in

π

vmg =

π

π (13.21)(48.3) 12

π

= 167 fpm

 1200 + vmg   Ft (a) Fd =   1200   1200 + 167  Fd =   Ft  1200  33,000(16 ) Ft = = 3162 lb 167  1200 + 167  Fd =  (3162 ) = 3602 lb  1200 

Page 11 of 19



(b) Fs =

sYbPcn

π

=

sYbPc cos λ

π

N P (3)(1.15278) tan λ = t c = πDw π (3.136 )

λ = 19.34o For manganese-bronze, s = sn = 30,000 psi For φn = 25o , Y = 0.470

(30,000)(0.470) 2 1 (1.15278)cos19.34 Fs =

 4

π

= 10,984 lb

Service factor F 10,984 N sf = s = = 3.05 3602 Fd (c) Fw = Dg bK w Dg = 13.21 in b = 2.25 in Table AT 27, hardened-steel worn and manganese bronze gear φn = 25o

K w = 100 Fw = (13.21)(2.25)(100) = 2972 lb < Fd (= 3602 lb ) Therefore, not good for indefinitely continuous service  cos φn − f tan λ   (d) e = tan λ   cos φn tan λ + f  πDwnw π (1.15278)(580 ) vr = = = 185.5 fpm 12 cos λ 12 cos19.34 0.32 0.32 f = 0.36 = = 0.0488 ( 70 < vr < 3000 fpm ) vr (185.5)0.36  cos 25 − 0.0488 tan 19.34  e = tan 19.34  = 0.85 = 85%  cos 25 tan 19.34 + 0.0488  hp 16 hp hpi = o = = 18.82 hp e 0.85

(e) Temperature rise, ∆t Q = (1 − e )(hpi ) = (1 − 0.85)(18.82)(33,000 ) = 93,159 ft − lb min Qc = hcr A∆t ft − lb min A = Amin = 43.2C1.7 sq.in.

Page 12 of 19


SECTION 14 – WORM GEARS 1 (Dw + Dg ) 2 1 C = (1.15278 + 13.21) = 7.18 in 2 1.7 A = 43.2(7.18) = 1233 sq.in. Figure AF 1 1233 A= = 8.6 ft 2 144 hcr = 0.47 ft − lb min − sq.in. − F C=

Q = Qc 93,159 = (0.47 )(1233)(∆t ) ∆t = 161 F (f) Tangential components on the worm  cos φn sin λ + f cos λ  cos 25 sin 19.34 + 0.0488 cos19.34   = 3162 Wt = Ft   = 1305 lb  cos 25 cos19.34 − 0.0488 sin 19.34   cos φn cos λ − f sin λ  on the gear Ft = 3162 lb radial components Ft sin φn 3162 sin 25 S= = = 1593 lb cos φn cos λ − f sin λ cos 25 cos19.34 − 0.0488 sin 19.34 (g) λ = 19.34o > 5o , not self-locking 797.

A worm-gear speed reducer has a hardened-steel worm and a phosphor-bronze gear. The lead angle of the 5-threaded worm λ = 28o57' , Pc = 1.2812 in. , 1 φn = 25o , b = 2 in. , mw = 8 ; worm speed = 1750 rpm. The gear case is 35 3/8 2 in. high, 22 in. wide, 14 in. deep. Compute (a) the efficiency, (b) the limiting wear load, the strength load, and the corresponding safe input and output horsepowers. (c) The manufacturer rates this reducer at 53-hp input. Is this rating conservative or risky? (d) What is the calculated temperature rise of the oil with no special cooling? (e) The manufacturer specifies that for continuous service power should not exceed 36.5 hp if there is to be no artificial cooling and if ∆t is to be less than 90 F. Make calculations and decide whether the vendor is on the safe side. (Data courtesy of the Cleveland Worm Gear Co.)

Solution:

Page 13 of 19


SECTION 14 – WORM GEARS  cos φn − f tan λ   (a) e = tan λ   cos φn tan λ + f  λ = 28o57' = 28.95o φn = 25o πDwnw vr = 12 cos λ nw = 1750 rpm N P Dg = g c

π

N g = mw N t = (8)(5) = 40 Dg =

(40)(1.2812) = 16.31 in π

NP tan λ = t c πDw tan 28.95 =

(5)(1.2812) πDw

Dw = 3.686 in π (3.686 )(150 ) vr = = 1923 fpm 12 cos 28o57' 0.32 0.32 f = 0.36 = = 0.0210 ( 70 < vr < 3000 fpm ) vr (1923)0.36  cos 25 − 0.0210 tan 28.95  e = tan 28.95  = 0.9475 = 94.75%  cos 25 tan 28.95 + 0.0210  (b) Fw = Dg bK w Dg = 16.31 in b = 2.5 in Table AT 27, hardened-steel worn and phosphor bronze gear φn = 25o

K w = 100 Fw = (16.31)(2.5)(100) = 4078 lb Fs =

sYbPcn

π

=

sYbPc cos λ

π

For phosphor-bronze, s = sn = 31,000 psi For φn = 25o , Y = 0.470 (31,000)(0.470 )(2.5)(1.2812 )cos 28.95 = 13,000 lb Fs =

π

Page 14 of 19


SECTION 14 – WORM GEARS For safe input and output  1200 + vmg   Ft Fd =   1200  πDg ng vmg = 12 n 1750 ng = w = = 218.75 rpm mw 8 π (16.31)(218.75) vmg = = 934 fpm 12 Fw = Fd  1200 + 934  4078 =   Ft  1200  Ft = 2293 lb Fv (2293)(934) = 64.9 hp safe output = hpo = t mg = 33,000 33,000 hp 64.9 = 68.5 hp safe input = hpi = o = e 0.9475

(c) 53-hp input < 68.5 hp. ∴ conservative. (d) Q = (1 − e )(hpi ) = (1 − 0.9475)(68.5)(33,000) = 118,676 ft − lb min Qc = hcr A∆t ft − lb min A = 2[(22)(14) + (22)(35.375)] = 2172.5 sq.in. Figure AF 1 2172.5 A= = 15 ft 2 144 hcr = 0.45 ft − lb min − sq.in. − F Q = Qc 118,676 = (0.45)(2172.5)(∆t ) ∆t = 121.4 F (e) ∆t ′ = 90 F hpi′ ∆t ′ = hpi ∆t hpi′ 90 = 68.5 124 hpi′ = 50.8 hp Since 36.5 hp < 50.8 hp, therefore on the safe side.

Page 15 of 19


SECTION 14 – WORM GEARS HEATING 799.

The input to a worm-gear reducer is 50.5 hp at 580 rpm of the 4-threaded worm. The gear case is 22 x 31 x 45 in. in size; φn = 25o , Pc = 1.5 in , Dw = 4.432 in , f = 0.035 , room temperature = 80 F. Compute the steady-state temperature for average cooling.

Solution: NP 4(1.5) tan λ = t c = πDw π (4.432 )

λ = 23.3o  cos φn − f tan λ   e = tan λ   cos φn tan λ + f   cos 25 − 0.035 tan 23.3  e = tan 23.3  = 0.9025  cos 25 tan 23.3 + 0.035  Q = (1 − e )(hpi ) = (1 − 0.9025)(50.5)(33,000) = 162,484 ft − lb min Qc = hcr A∆t ft − lb min A = 2[(22)(31) + (31)(45)] = 4154 sq.in. Figure AF 1 4154 A= = 28.85 ft 2 144 hcr = 0.42 ft − lb min − sq.in. − F Q = Qc 162,484 = (0.42)(4154 )(∆t ) ∆t = 93 F t1 = 80 F t2 = 173 F

801.

A hardened-steel, 4-threaded worm drives a bronze gear; Dw = 1.875 in , Dg ≈ 14 in , Pc = 1.0 in , φn = 25o , area of case ≈ 1500 sq.in. , vr ≈ 1037 fpm ; input = 20 hp at 1750 rpm of the worm; room temperature = 80 F. Compute the steady-state temperature of the lubricant for average ventilation.

Solution: NP 4(1.0 ) tan λ = t c = πDw π (1.875)

λ = 34.2o f =

0.32 0.32 = = 0.0263 ( 70 < vr < 3000 fpm ) 0.36 vr (1037 )0.36

Page 16 of 19


SECTION 14 – WORM GEARS  cos φn − f tan λ   e = tan λ   cos φn tan λ + f   cos 25 − 0.0263 tan 34.2  e = tan 34.2  = 0.94  cos 25 tan 34.2 + 0.0263  Q = (1 − e )(hpi ) = (1 − 0.94)(20)(33,000) = 39,600 ft − lb min Qc = hcr A∆t ft − lb min A = 1500 sq.in. Figure AF 1 1500 A= = 10.4 ft 2 144 hcr = 0.46 ft − lb min − sq.in. − F Q = Qc 39,600 = (0.46 )(1500)(∆t ) ∆t = 57 F t1 = 80 F t2 = 137 F

802.

The input to a 4-threaded worm is measured to be 20.8 hp; Pc = 1.0 in , Dw = 2 in ,

φn = 25o . The area of the case is closely 1800 sq. in.; ambient temperature = 100 F; oil temperature = 180 F. Operation is at a steady thermal state. Compute the indicated coefficient of friction. Solution: Qc = hcr A∆t ft − lb min Figure AF 1 1800 A= = 12.5 ft 2 144 hcr = 0.46 ft − lb min − sq.in. − F A = 1800 sq.in. ∆t = 180 − 100 = 80 F Qc = hcr A∆t = (0.46)(1800)(80) = 66,240 ft − lb min Q = (1 − e )(hpi )(33,000) ft − lb min Q = Qc (1 − e)(hpi )(33,000) = 66,240 e = 0.9035  cos φn − f tan λ   e = tan λ   cos φn tan λ + f 

Page 17 of 19



tan λ =

N t Pc 4(1.0 ) = πDw π (2 )

λ = 32.5o  cos 25 − f tan 32.5   0.9035 = tan 32.5  cos 25 tan 32.5 + f  0.5217 + 0.9035 f = 0.5774 − 0.4059 f f = 0.0425

FORCE ANALYSIS 804.

1 The input to a 4-threaded worm is 21 hp at 1750 rpm; e = 90% , Dw = 2 in , 4 o Dg = 14 in , N g = 44 , φn = 25 . (a) From the horsepowers in and out, compute

the tangential forces on the worm Wt and the gear Ft . (b) Using this value of Ft , compute Wt from equation (k), i16.8, Text. (Check?) (c) Compute the separating force. (d) What is the end thrust on the worm shaft? On the gear shaft? Solution: hpi = 21 hp hpo = (hpi )(e ) = (21)(0.90) = 18.9 hp N Dg mw = g = N t Dw tan λ 44 14 = 4  1  2  tan λ  4 λ = 29.5o  cos φn − f tan λ   e = tan λ   cos φn tan λ + f   cos 25 − f tan 29.5   0.90 = tan 29.5  cos 25 tan 29.5 + f  0.4615 + 0.90 f = 0.5128 − 0.32 f f = 0.0420 πDg ng vmg = 12 ng N t = nw N g

Page 18 of 19


SECTION 14 – WORM GEARS ng 4 = 1750 44 ng = 159 rpm vmg =

π (14)(159)

(a) Ft =

12

= 583 fpm

33,000hpo 33,000(18.9) = = 1070 lb vmg 583

33,000hpi vw πD n π (2.25)(1750) vw = w w = = 1031 fpm 12 12 33,000(21) Wt = = 672 lb 1031 Wt =

 cos φn sin λ + f cos λ  cos 25 sin 29.5 + 0.0420 cos 29.5   = 1070 (b) Wt = Ft   = 672 lb  cos 25 cos 29.5 − 0.0420 sin 29.5   cos φn cos λ − f sin λ  Ft sin φn 1070 sin 25 (c) S = = = 589 lb cos φn cos λ − f sin λ cos 25 cos 29.5 − 0.0420 sin 29.5 (d) End thrust Worm shaft = Ft = 1070 lb Gear shaft = Wt = 672 lb

- end -

Page 19 of 19


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS

LEATHER BELTS DESIGN PROBLEMS 841.

A belt drive is to be designed for F1 F2 = 3 , while transmitting 60 hp at 2700 rpm of the driver D1 ; mw ≈ 1.85 ; use a medium double belt, cemented joint, a squirrel-cage, compensator-motor drive with mildly jerking loads; center distance is expected to be about twice the diameter of larger pulley. (a) Choose suitable iron-pulley sizes and determine the belt width for a maximum permissible s = 300 psi . (b) How does this width compare with that obtained by the ALBA procedure? (c) Compute the maximum stress in the straight port of the ALBA belt. (d) If the belt in (a) stretches until the tight tension F1 = 525 lb ., what is F1 F2 ?

Solution: (a) Table 17.1, Medium Double Ply, Select D1 = 7 in . min. 20 t= in 64 πD n π (7 )(2700) vm = 1 1 = = 4948 fpm 12 12 4000 fpm < 4948 fpm < 6000 fpm (F − F2 )vm hp = 1 33,000 (F − F2 )(4948) 60 = 1 33,000 F1 − F2 = 400 lb F1 = 3F2 3F2 − F2 = 400 lb F2 = 200 lb F1 = 3F2 = 3(200) = 600 lb F1 = sbt sd = 300η For cemented joint, η = 1.0 sd = 300 psi  20  F1 = 600 = (300)(b )   64  b = 6.4 in say b = 6.5 in

Page 1 of 56



(b) ALBA Procedure hp = (hp in., Table 17.1)(bCmC p )(C f1 C f 2 L) Table 17.1, vm = 4948 fpm Medium Double Ply hp in = 12.448 Table 17.2 Squirrel cage, compensator, starting Cm = 0.67 Pulley Size, D1 = 7 in C p = 0.6 Jerky loads, C f = 0.83 hp = 60 = (12.448)(b )(0.67 )(0.6)(0.83) b = 14.5 in say b = 15 in

(c) s =

F1 = ηbt

1

1

600 = 128 psi  20  (1)(15)   64  1

1

1

(d) 2 Fo2 = F12 + F22 = (600)2 + (200)2 Fo = 373.2 lb F1 = 525 lb 1 2

1 2

1 2 2

2(373.2 ) = (525) + F F2 = 247 lb F1 525 = = 2.1255 F2 247 842.

A 20-hp, 1750 rpm, slip-ring motor is to drive a ventilating fan at 330 rpm. The horizontal center distance must be about 8 to 9 ft. for clearance, and operation is continuous, 24 hr./day. (a) What driving-pulley size is needed for a speed recommended as about optimum in the Text? (b) Decide upon a pulley size (iron or steel) and belt thickness, and determine the belt width by the ALBA tables. (c) Compute the stress from the general belt equation assuming that the applicable coefficient of friction is that suggested by the Text. (d) Suppose the belt is installed with an initial tension Fo = 70 lb in . (§17.10), compute F1 F2 and the stress on the tight side if the approximate relationship of the operating tensions 1 2

1 2 2

1 2 o

and the initial tensions is F1 + F = 2 F .

Page 2 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Solution: vm = 4000 to 4500 fpm assume vm = 4250 fpm πD n vm = 1 1 12 πD (1750 ) 4250 = 1 12 D1 = 9.26 in say D1 = 10 in 23 in 64 Minimum pulley diameter for vm ≈ 4250 fpm , D1 = 10 in

(b) Using Heavy Double Ply Belt, t =

Use D1 = 10 in πD n π (10 )(1750) vm = 1 1 = = 4581 fpm 12 12 ALBA Tables hp = (hp in., Table 17.1)(bCm C p ) C f1 C f 2 L

(

hp in = 13.8 Slip ring motor, Cm = 0.4 Pulley Size, D1 = 10 in C p = 0.7 Table 17.7, 24 hr/day, continuous N sf = 1.8 Assume C f = 0.74 hp = (1.8)(20) = (13.8)(b )(0.4)(0.7 )(0.74 ) b = 12.59 in use b = 13 in (c) General belt equation  12 ρvs2  e fθ − 1    F1 − F2 = bt  s − 32.2  e fθ   4581 vs = = 76.35 fps 60 ρ = 0.035 lb cu. in. for leather 23 t= in 64 b = 13 in

Page 3 of 56

)


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 33,000(1.8)(20 ) = 260 lb 4581 f = 0.3 on iron or steel D − D1 θ ≈π ± 2 C C = 8 ~ 9 ft use 8.5 ft  1750  D2 =  (10 ) = 53 in  330  53 − 10 θ =π − = 2.72 rad 8.5(12 ) fθ = (0.3)(2.72 ) = 0.816 F1 − F2 =

e fθ − 1 e 0.816 − 1 = 0.816 = 0.5578 e fθ e 2  23   12(0.035)(76.35)  F1 − F2 = 260 = (13)   s −  (0.5578) 32.2  64    s = 176 psi 1

1

1

(d) F1 2 + F22 = 2 Fo2 Fo = (70 lb in )(13 in ) = 910 lb F1 − F2 = 260 lb F2 = F1 − 260 lb 1 2

1

1

F1 + (F1 − 260 )2 = 2(910 )2 = 60.33 F1 = 1045 lb F2 = 1045 − 260 = 785 lb 1045 = 224 psi  23  (13)   64  F1 1045 = = 1.331 F2 785

s=

F1 = bt

843.

A 100-hp squirrel-cage, line-starting electric motor is used to drive a Freon reciprocating compressor and turns at 1140 rpm; for the cast-iron motor pulley, D1 = 16 in ; D2 = 53 in , a flywheel; cemented joints;l C = 8 ft . (a) Choose an appropriate belt thickness and determine the belt width by the ALBA tables. (b) Using the design stress of §17.6, compute the coefficient of friction that would be needed. Is this value satisfactory? (c) Suppose that in the beginning, the initial tension was set so that the operating F1 F2 = 2 . Compute the maximum stress in a straight part. (d) The approximate relation of the operating tensions and the

Page 4 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 1 2

1 2 2

1 2 o

initial tension Fo is F1 + F = 2 F . For the condition in (c), compute Fo . Is it reasonable compared to Taylor’s recommendation? Solution: (a) Table 17.1 πD n π (16)(1140 ) = 4775 fpm vm = 1 1 = 12 12 Use heavy double-ply belt 23 t= in 64 hp in = 14.1 hp = (hp in., Table 17.1)(bCmC p ) C f1 C f 2 L

(

)

line starting electric motor , Cm = 0.5 Table 17.7, squirrel-cage, electric motor, line starting, reciprocating compressor N sf = 1.4 D1 = 16 in , C p = 0.8 assume, C f = 0.74 hp = (1.4 )(100 ) = 140 hp hp = 140 = (14.1)(b )(0.5)(0.8)(0.74) b = 33.5 in use b = 34 in (b) §17.6, sd = 400η η = 1.00 for cemented joint. sd = 400 psi  12 ρvs2  e fθ − 1    F1 − F2 = bt  s − 32.2  e fθ   4775 vs = = 79.6 fps 60 ρ = 0.035 lb cu. in. for leather 23 t= in 64 b = 34 in 33,000(1.4 )(100) F1 − F2 = = 968 lb 4775 2 12(0.035)(79.6 )  e fθ − 1   23   F1 − F2 = 968 = (34)  400 −  fθ  32.2  64     e

Page 5 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS e fθ − 1 = 0.2496 e fθ fθ = 0.28715 D − D1 θ ≈π ± 2 C C = 8 ft 53 − 16 θ =π − = 2.7562 rad 8(12) f (2.7562 ) = 0.28715 f = 0.1042 < 0.3 Therefore satisfactory. (c) F1 − F2 = 968 lb F1 = 2F2 2 F2 − F2 = 968 lb F1 = 2 F2 = 2(968) = 1936 lb F 1936 s= 1 = = 159 psi bt  23  (34)   64  (d) F1 = 1936 lb , F2 = 968 lb 1

1

1

2 Fo2 = F12 + F22 1 2 o

1

1

2 F = (1936 )2 + (968)2 Fo = 1411 lb Fo =

844.

1411 = 41.5 lb in of width is less than Taylor’s recommendation and is reasonable. 34

A 50-hp compensator-started motor running at 585 rpm drives a reciprocating compressor for a 40-ton refrigerating plant, flat leather belt, cemented joints. The diameter of the fiber driving pulley is 13 in., D2 = 70 in ., a cast-iron flywheel; C = 6 ft.11 in. Because of space limitations, the belt is nearly vertical; the surroundings are quite moist. (a) Choose a belt thickness and determine the width by the ALBA tables. (b) Using recommendations in the Text, compute s from the general belt equation. (c) With this value of s , compute F1 and F1 F2 . (d) 1

1

1

Approximately, F1 2 + F22 = 2 Fo2 , where Fo is the initial tension. For the condition in (c), what should be the initial tension? Compare with Taylor, §17.10. (e) Compute the belt length. (f) The data are from an actual drive. Do you have any recommendations for redesign on a more economical basis?

Page 6 of 56



Solution: (a) vm =

πD1n1

=

π (13)(865)

= 2944 fpm 12 12 Table 17.1, use Heavy Double Ply, Dmin = 9 in for vm = 2944 fpm belts less than 8 in wide 23 t= in 64

(

)

hp = (hp in., Table 17.1)(bCmC p ) C f1 C f 2 L

hp in = 9.86 Table 17.2 Cm = 0.67 C p = 0.8 C f = (0.74)(0.80 ) = 0.592 Table 17.7, electric motor, compensator-started (squirrel cage) and reciprocating compressor N sf = 1.4

hp = (1.4)(50) = 70 hp hp = 70 = (9.86)(b )(0.67 )(0.8)(0.592 ) b = 22.4 in use b = 25 in (b) General Belt Equation  12 ρvs2  e fθ − 1    F1 − F2 = bt  s − 32.2  e fθ   b = 25 in 23 t= in 64 ρ = 0.035 lb cu. in. for leather 2944 vs = = 49.1 fps 60 Leather on iron, f = 0.3 D − D1 θ =π − 2 C 70 − 13 θ =π − = 2.35 rad 6(12) fθ = (0.3)(2.35) = 0.705

Page 7 of 56



e fθ − 1 e 0.705 − 1 = 0.705 = 0.5059 e fθ e 33,000(1.4 )(50 ) = 785 lb F1 − F2 = 2944 2  23   12(0.035)(49.1)  F1 − F2 = 785 = (25)   s −  (0.5059) 32.2  64    s = 204 psi Cemented joint, η = 1.0 s = 204 psi  23  (c) F1 = sbt = (204)(25)  = 1833 lb  64  F2 = 1833 − 785 = 1048 lb F1 1833 = = 1.749 F2 1048 1

1

1

(d) 2 Fo2 = F12 + F22 1 2 o

1

1

2 F = (1833)2 + (1048)2 Fo = 1413 lb 1413 = 56.5 lb in 25 Approximately less than Taylor’s recommendation ( = 70 lb/in.) Fo =

(e) L ≈ 2C + 1.57(D2 + D1 ) +

(D2 − D1 )2

4C 2 70 − 13) ( L = 2(6 )(12 ) + 1.57(70 + 13) + = 286 in 4(6)(12) (f) More economical basis πD n vm = 1 1 12 πD (865) 4500 = 1 12 D1 = 19.87 in use D1 = 20 in

Page 8 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS CHECK PROBLEMS 846.

An exhaust fan in a wood shop is driven by a belt from a squirrel-cage motor that runs at 880 rpm, compensator started. A medium double leather belt, 10 in. wide is used; C = 54 in .; D1 = 14 in . (motor), D2 = 54 in ., both iron. (a) What horsepower, by ALBA tables, may this belt transmit? (b) For this power, compute the stress from the general belt equation. (c) For this stress, what is F1 F2 ? (d) If the belt has stretched until s = 200 psi on the tight side, what is F1 F2 ? (e) Compute the belt length.

Solution: (a) For medium double leather belt 20 t= in 64 hp = (hp in )(b )CmC pC f Table 17.1 and 17.2 Cm = 0.67 C p = 0 .8 C f = 0.74 b = 10 in πD n π (14)(880) vm = 1 1 = = 3225 fpm 12 12 hp in = 6.6625 hp = (6.6625)(10)(0.67 )(0.8)(0.74) = 26.43 hp

 12 ρvs2  e fθ − 1    (b) F1 − F2 = bt  s − 32.2  e fθ   b = 10 in 20 t= in 64 ρ = 0.035 lb cu. in. 3225 vs = = 53.75 fps 60 D − D1 θ =π − 2 C 54 − 14 = 2.4 rad θ =π − 54 Leather on iron f = 0.3 fθ = (0.3)(2.4) = 0.72

Page 9 of 56



e fθ − 1 e 0.72 − 1 = 0.72 = 0.51325 e fθ e 33,000(26.43) = 270 lb F1 − F2 = 3225 2  20   12(0.035)(53.75)  F1 − F2 = 270 = (10)   s −  (0.51325) 32.2  64    s = 206 psi  20  (c) F1 = sbt = (206)(10 )  = 644 lb  64  F2 = 644 − 270 = 374 lb F1 644 = = 1.72 F2 374

(d) s = 200 psi  20  F1 = sbt = (200 )(10 )  = 625 lb  64  F2 = 625 − 270 = 355 lb F1 625 = = 1.76 F2 355

(e)

2 D2 − D1 ) ( L ≈ 2C + 1.57(D2 + D1 ) +

4C (54 − 14 )2 = 222 in L = 2(54 ) + 1.57(54 + 14) + 4(54)

847.

A motor is driving a centrifugal compressor through a 6-in. heavy, single-ply leather belt in a dusty location. The 8-in motor pulley turns 1750 rpm; D2 = 12 in . (compressor shaft); C = 5 ft . The belt has been designed for a net belt pull of F1 − F2 = 40 lb in of width and F1 F2 = 3 . Compute (a) the horsepower, (b) the stress in tight side. (c) For this stress, what needed value of f is indicated by the general belt equation? (d) Considering the original data,what horsepower is obtained from the ALBA tables? Any remarks?

Solution: (a) vm =

πD1n1 12

=

π (8)(1750) 12

b = 6 in F1 − F2 = (40)(6) = 240 lb

Page 10 of 56

= 3665 fpm



hp =

(F1 − F2 )vm = (240)(3665) = 26.65 hp 33,000

33,000

(b) F1 = 3F2 3F2 − F2 = 240 lb F2 = 120 lb F1 = 360 lb F s= 1 bt For heavy single-ply leather belt 13 t= in 64 360 s= = 295 psi  13  (6)   64   12 ρvs2  e fθ − 1    (c) F1 − F2 = bt  s − 32.2  e fθ   ρ = 0.035 lb cu. in. 3665 vs = = 61.1 fps 60 F1 − F2 = 240 lb 2 12(0.035)(61.1)  e fθ − 1   13   F1 − F2 = 240 = (6 )  295 −  fθ  32.2  64     e

e fθ − 1 = 0.7995 e fθ D − D1 θ =π − 2 C 12 − 8 θ =π − = 3.075 rad 5(12) e fθ = 4.9875 fθ = 1.607 f (3.075) = 1.607 f = 0.5226 (d) ALBA Tables (Table 17.1 and 17.2) hp = (hp in )(b )CmC pC f

vm = 3665 fpm hp in = 6.965

Page 11 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS b = 10 in Cm = 1.0 (assumed) C p = 0 .6 C f = 0.74

hp = (6.965)(6)(1.0 )(0.6)(0.74 ) = 18.6 hp < 26.65 hp 848.

A 10-in. medium double leather belt, cemented joints, transmits 60 hp from a 9in. paper pulley to a 15-in. pulley on a mine fab; dusty conditions. The compensator-started motor turns 1750 rpm; C = 42 in . This is an actual installation. (a) Determine the horsepower from the ALBA tables. (b) Using the general equation, determine the horsepower for this belt. (c) Estimate the service factor from Table 17.7 and apply it to the answer in (b). Does this result in better or worse agreement of (a) and (b)? What is your opinion as to the life of the belt?

Solution: πD n π (9 )(1750) vm = 1 1 = = 4123 fpm 12 12 (a) hp = (hp in )(b )CmC pC f Table 17.1 and 17.2 Medium double leather belt 20 t= in 64 vm = 4123 fpm hp in = 11.15 Cm = 0.67 C p = 0 .7 C f = 0.74 b = 10 in hp = (11.15)(10)(0.67 )(0.7 )(0.74) = 38.7 hp

 12 ρvs2  e fθ − 1   fθ  (b) F1 − F2 = bt  s − 32 . 2    e b = 10 in ρ = 0.035 lb cu. in. s = 400η η = 1.0 cemented joint s = 400 psi D − D1 θ =π − 2 C

Page 12 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 15 − 9 = 2.9987 rad 42 Leather on paper pulleys, f = 0.5 fθ = (0.5)(2.9987 ) = 1.5

θ =π −

e fθ − 1 = 0.77687 e fθ 4123 = 68.72 fps vs = 60 2 12(0.035)(68.72 )   20   F1 − F2 = (10 )  400 −  (0.77687 ) = 822 lb 32.2  64    F1 − F2 )vm (822 )(4123) ( hp = = = 102.7 hp 33,000 33,000 (c) Table 17.7 N sf = 1.6 102.7 = 64.2 hp < 102.7 hp 1 .6 Therefore, better agreement hp =

Life of belt, not continuous, 60 hp > 38.7 hp . MISCELLANEOUS 849.

Let the coefficient of friction be constant. Find the speed at which a leather belt may transmit maximum power if the stress in the belt is (a) 400 psi, (b) 320 psi. (c) How do these speeds compare with those mentioned in §17.9, Text? (d) Would the corresponding speeds for a rubber belt be larger or smaller? (HINT: Try the first derivative of the power with respect to velocity.)

Solution:  12 ρvs2  e fθ − 1    F1 − F2 = bt  s − 32.2  e fθ   (F − F2 )vm hp = 1 33,000 60(F1 − F2 )vs hp = 33,000

hp =

60vs bt  12 ρvs2  e fθ − 1  s −   33,000  32.2  e fθ 

60bt  e fθ − 1  12 ρvs2    s − v s hp = 33,000  e fθ  32.2 

Page 13 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS d (hp ) 60bt  e fθ − 1   12 ρvs2  24 ρvs2     s − − = =0 d (vs ) 33,000  e fθ   32.2  32.2 

36 ρvs2 32.2 ρ = 0.035 lb cu. in.

s=

(a) s = 400 psi 36(0.035)vs2 32.2 vs = 101.105 fps vm = 6066 fpm 400 =

(b) s = 320 psi 36(0.035)vs2 320 = 32.2 vs = 90.431 fps vm = 5426 fpm (c) Larger than those mentioned in §17.9 (4000 – 4500 fpm) (d) Rubber belt, ρ = 0.045 lb cu. in. (a) s = 400 psi 36(0.045)vs2 32.2 vs = 89.166 fps vm = 5350 fpm < 6066 fpm 400 =

Therefore, speeds for a rubber belt is smaller. 850.

A 40-in. pulley transmits power to a 20-in. pulley by means of a medium double leather belt, 20 in. wide; C = 14 ft , let f = 0.3 . (a) What is the speed of the 40-in pulley in order to stress the belt to 300 psi at zero power? (b) What maximum horsepower can be transmitted if the indicated stress in the belt is 300 psi? What is the speed of the belt when this power is transmitted? (See HINT in 849).

Solution:  12 ρvs2  e fθ − 1   fθ  F1 − F2 = bt  s − 32 . 2   e 

Page 14 of 56



hp =

60(F1 − F2 )vs 33,000

hp =

60bt  e fθ − 1  12 ρvs2    s − v s 33,000  e fθ  32.2 

d (hp ) 60bt  e fθ − 1   12 ρvs2  24 ρvs2     s − − = =0 d (vs ) 33,000  e fθ   32.2  32.2 

36 ρvs2 s= for maximum power 32.2 (a) At zero power: 12 ρvs2 s= 32.2 s = 300 psi ρ = 0.035 lb cu. in. 12(0.035)vs2 32.2 vs = 151.6575 fps vm = 9100 fpm

300 =

Speed, 40 in pulley, n2 =

12vm 12(9100) = = 869 rpm πD2 π (40)

(b) Maximum power 36 ρvs2 s= 32.2 36(0.035)vs2 300 = 32.2 vs = 87.5595 fps vm = 5254 fpm 60bt  e fθ − 1  12 ρvs2    s − v s hp = 33,000  e fθ  32.2  20 t= in 64 b = 20 in D − D1 θ =π − 2 C 40 − 20 θ =π − = 3.0225 rad 14(12) f = 0.3

Page 15 of 56



fθ = (0.3)(3.0225) = 0.90675 e fθ − 1 = 0.5962 e fθ  20  60(20)  2  64  (0.5962)300 − 12(0.035)(87.5595)  (87.5595) = 118.64 hp =   33,000 32.2   vm = 5254 fpm AUTOMATIC TENSION DEVICES 851.

An ammonia compressor is driven by a 100-hp synchronous motor that turns 1200 rpm; 12-in. paper motor pulley; 78-in. compressor pulley, cast-iron; C = 84 in . A tension pulley is placed so that the angle of contact on the motor pulley is 193o and on the compressor pulley, 240o. A 12-in. medium double leather belt with a cemented joint is used. (a) What will be the tension in the tight side of the belt if the stress is 375 psi? (b) What will be the tension in the slack side? (c) What coefficient of friction is required on each pulley as indicated by the general equation? (d) What force must be exerted on the tension pulley to hold the belt tight, and what size do you recommend?

Solution: (a) F1 = sbt b = 12 in 20 t= in 64  20  F1 = (375)(12)   64  33,000hp vm πD n π (12)(1200 ) vm = 1 1 = = 3770 fpm 12 12 Table 17.7, N sf = 1.2

(b) F1 − F2 =

33,000(1.2)(100) = 1050 lb 3770 F2 = F1 − 1050 = 1406 − 1050 = 356 lb F1 − F2 =

 12 ρvs2  e fθ − 1   fθ  (c) F1 − F2 = bt  s − 32 . 2   e  3770 vs = = 62.83 fps 60

Page 16 of 56



ρ = 0.035 lb cu. in. 12(0.035)(62.83)  e fθ − 1   20   1050 = (12)  375 −  e fθ  32.2  64     e fθ − 1 = 0.8655 e fθ fθ = 2.006 Motor pulley  π  θ = 193o = 193  = 3.3685 rad  180  f (3.3685) = 2.006 f = 0.5955 Compressor Pulley  π  θ = 2403o = 240  = 4.1888 rad  180  f (4.1888) = 2.006 f = 0.4789

(d) Force:

Without tension pulley D − D1 78 − 12 =π − = 2.356 rad θ1 = π − 2 C 84 D − D1 78 − 12 =π + = 3.9273 rad θ2 = π + 2 C 84

Page 17 of 56



α1 = θ1′ − θ1 −

2

= 3.3685 − 2.356 −

π − 2.356 2

= 0.6197 rad = 35.5o

3.9273 − π + 4.1888 − 3.9273 = 0.6544 rad = 37.5o 2 2 Q = F1 (sin α1 + sin α 2 ) = 1406(sin 35.5 + sin 37.5) = 1672 lb of force exerted Size of pulley; For medium double leather belt, vm = 3770 fpm , width = 12 in > 8 in D = 6 + 2 = 8 in

α2 =

θ2 − π

π − θ1

+ θ 2′ − θ 2 =

Page 18 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 852.

A 40-hp motor, weighing 1915 lb., runs at 685 rpm and is mounted on a pivoted 3 base. In Fig. 17.11, Text, e = 10 in ., h = 19 in . The center of the 11 ½-in. 16 motor pulley is 11 ½ in. lower than the center of the 60-in. driven pulley; C = 48 in . (a) With the aid of a graphical layout, find the tensions in the belt for maximum output of the motor if it is compensator started. What should be the width of the medium double leather belt if s = 300 psi ? (c) What coefficient of friction is indicated by the general belt equation? (Data courtesy of Rockwood Mfg. Co.)

Solution: (a)

R = 1915 lb Graphically b ≈ 26 in a ≈ 9 in

[∑ M

B

=0

]

eR = F1a + F2b (10)(1915) = (F1 )(9) + (F2 )(26) 9 F1 + 26 F2 = 19,150 For compensator started hp = 1.4(rated hp ) = 1.4(40 ) = 56 hp 33,000hp F1 − F2 = vm

Page 19 of 56



vm =

πD1n1

π (11.5)(685)

= 2062 fpm 12 33,000(56 ) F1 − F2 = = 896 lb 2062 F2 = F1 − 896 Substituting 9 F1 + 26(F1 − 896) = 19,150 F1 = 1213 lb F2 = 1213 − 896 = 317 lb 12

=

For medium leather belt, t =

20 in 64

F1 = sbt  20  1213 = (300)(b )   64  b = 13 in

 12 ρvs2  e fθ − 1    (c) F1 − F2 = bt  s − 32.2  e fθ   2062 vs = = 34.37 fps 60 ρ = 0.035 lb cu. in.

12(0.035)(34.37 )  e fθ − 1   20   896 = (13)  300 −  e fθ  32.2  64     e fθ − 1 = 0.775 e fθ fθ = 1.492 D − D1 60 − 11.5 =π − = 2.1312 rad θ =π − 2 C 48 f (2.1312 ) = 1.492 f = 0.70 853.

A 50-hp motor, weighing 1900 lb., is mounted on a pivoted base, turns 1140 rpm, 3 and drives a reciprocating compressor; in Fig. 17.11, Text, e = 8 in ., 4 5 h = 17 in . The center of the 12-in. motor pulley is on the same level as the 16 center of the 54-in. compressor pulley; C = 40 in . (a) With the aid of a graphical layout, find the tensions in the belt for maximum output of the motor if it is compensator started. (b) What will be the stress in the belt if it is a heavy double

Page 20 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS leather belt, 11 in. wide? (c) What coefficient of friction is indicated by the general belt equation? (Data courtesy of Rockwood Mfg. Co.) Solution: (a) For compensator-started hp = 1.4(50) = 70 hp 33,000hp F1 − F2 = vm πD n π (12 )(1140) = 3581 fpm vm = 1 1 = 12 12 33,000(70 ) F1 − F2 = = 645 lb 2062

b ≈ 25 in a ≈ 5 in R = 1900 lb eR = F 1a + F2b (8.75)(1900) = F 1 (5) + F2 (25)

F 1+5F2 = 3325 lb 645 + F 2+5F2 = 3325 lb F2 = 447 lb F1 = 645 + F2 = 645 + 447 = 1092 lb (b) For heavy double leather belt 23 t= in 64 b = 11 in

Page 21 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 1092 = 276 psi  20  (11)   64   12 ρvs2  e fθ − 1    (c) F1 − F2 = bt  s − 32.2  e fθ   3581 = 59.68 fps vs = 60 ρ = 0.035 lb cu. in. s=

F1 = bt

12(0.035)(59.68)  e fθ − 1   23   645 = (11)  276 −  e fθ  32.2  64     fθ = 1.241 D − D1 54 − 12 = 2.092 rad =π − θ =π − 2 C 40 f (2.092 ) = 1.492 f = 0.60

RUBBER BELTS 854.

A 5-ply rubber belt transmits 20 horsepower to drive a mine fan. An 8-in., motor pulley turns 1150 rpm; D2 = 36 in ., fan pulley; C = 23 ft . (a) Design a rubber belt to suit these conditions, using a net belt pull as recommended in §17.15, Text. (b) Actually, a 9-in., 5-ply Goodrich high-flex rubber belt was used. What are the indications for a good life?

Solution: (a) θ = π −

D2 − D1 36 − 8 =π − = 3.040 rad = 174o C 23(12 )

Kθ = 0.976 bv N K hp = m p θ 2400 Kθ = 0.976 πD n π (8)(1150) vm = 1 1 = = 2409 fpm 12 12 Np = 5 hp = 20 =

b(2409)(5)(0.976) 2400

b = 4.1 in min. b = 5 in

Page 22 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS (b) With b = 9 in is safe for good life. 855.

A 20-in., 10-ply rubber belt transmits power from a 300-hp motor, running at 650 rpm, to an ore crusher. The center distance between the 33-in. motor pulley and the 108-in. driven pulley is 18 ft. The motor and crusher are so located that the belt must operate at an angle 75o with the horizontal. What is the overload capacity of this belt if the rated capacity is as defined in §17.15, Text?

Solution: bv N hp = m p 2400 b = 20 in πD n π (33)(650) vm = 1 1 = = 5616 fpm 12 12 N p = 10 hp =

(20)(5616)(10 ) = 468 hp 2400

Overlaod Capacity =

468 − 300 (100% ) = 56% 300

V-BELTS NOTE: If manufacturer’s catalogs are available, solve these problems from catalogs as well as from data in the Text.

856.

A centrifugal pump, running at 340 rpm, consuming 105 hp in 24-hr service, is to be driven by a 125-hp, 1180-rpm, compensator-started motor; C = 43 to 49 in . Determine the details of a multiple V-belt drive for this installation. The B.F. Goodrich Company recommended six C195 V-belts with 14.4-in. and 50-in. sheaves; C ≈ 45.2 in .

Solution: Table 17.7 N sf = 1.2 + 0.2 = 1.4 (24 hr/day) Design hp = N sf (transmitted hp) = (1.4)(125) = 175 hp Fig. 17.4, 175 hp, 1180 rpm Dmin = 13 in , D-section D2 1180 50 = = D1 340 14.4 use D1 = 14.4 in > 13 in D2 = 50 in

Page 23 of 56



vm =

πD1n1 12

=

π (14.4)(1180 ) 12

= 4449 fpm

  103  0.09 c vm2  vm    Rated hp = a  − K D − e 106  103   vm   d 1 Table 17.3, D-section a = 18.788 , c = 137.7 , e = 0.0848 D Table 17.4, 2 = 3.47 D1 K d = 1.14 0.09 2   103  137.7 ( 4449)  4449   Rated hp = 18.788  − (1.14)(14.4) − (0.0848) 106  103 = 28.294 hp    4449 

Back to Fig. 17.14, C-section must be used. a = 8.792 , c = 38.819 , e = 0.0416   103  0.09 c v2  v  − Rated hp = a − e m6  m3 K d D1 10  10   vm   0 . 09 2   103  38.819 4449)  4449 (  −  Rated hp = 8.792 − (0.0416) = 20.0 hp (1.14)(14.4) 106  103   4449   Adjusted rated hp = Kθ K L (rated hp ) Table 17.5, D2 − D1 50 − 14.4 = = 0.77 C 46 Kθ = 0.88 Table 17.6 (D − D1 )2 L ≈ 2C + 1.57(D2 + D1 ) + 2 4C 2 50 − 14.4) ( L = 2(46) + 1.57(50 + 14.4) + = 200 in 4(46) use C195, L = 197.9 in K L = 1.07 Adjusted rated hp = (0.88)(1.07 )(20) = 18.83 hp Design hp 175 No. of belts = = = 9.3 belts use 9 belts Adjusted rated hp 18.83 Use 9 , C195 V-belts with 14.4 in and 50 in sheaves

Page 24 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS B + B 2 − 32(D2 − D1 ) C= 16 B = 4 L − 6.28(D2 + D1 ) = 4(197.9) − 6.28(50 + 14.4) = 387.2 in 2

C=

857.

387.2 +

(387.2)2 − 32(50 − 14.4)2 16

= 44.9 in

A 50-hp, 1160-rpm, AC split-phase motor is to be used to drive a reciprocating pump at a speed of 330 rpm. The pump is for 12-hr. service and normally requires 44 hp, but it is subjected to peak loads of 175 % of full load; C ≈ 50 in . Determine the details of a multiple V-belt drive for this application. The Dodge Manufacturing Corporation recommended a Dyna-V Drive consisting of six 5V1800 belts with 10.9-in. and 37.5-in. sheaves; C ≈ 50.2 in .

Solution: Table 17.7, (12 hr/day) N sf = 1.4 − 0.2 = 1.2 Design hp = (1.2)(1.75)(50) = 105 hp Fig. 17.4, 105 hp, 1160 rpm Dmin = 13 in , D-section D2 1160 46.4 = ≈ D1 330 13.2 use D1 = 13.2 in > 13 in D2 = 46.4 in πD n π (13.2)(1160 ) vm = 1 1 = = 4009 fpm 12 12   103  0.09 c vm2  vm   Rated hp = a  − K D − e 106  103   vm   d 1 Table 17.3, D-section a = 18.788 , c = 137.7 , e = 0.0848 D 46.4 Table 17.4, 2 = = 3.5 D1 13.2 K d = 1.14 0.09 2   103  137.7 ( 4009)  4009  −  Rated hp = 18.788 − (0.0848) = 24.32 hp (1.14)(13.2) 106  103   4009  

Back to Fig. 17.14, C-section must be used. a = 8.792 , c = 38.819 , e = 0.0416 Dmin = 9 in

Page 25 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS D2 1160 32 = ≈ D1 330 9.1 use D1 = 9.1 in πD n π (9.1)(1160 ) = 2764 fpm vm = 1 1 = 12 12 0.09   103  (2764)2  2764 = 10.96 hp 38.819  − Rated hp = 8.792 − (0.0416) (1.14)(9.1) 106  103   2764  

Adjusted rated hp = Kθ K L (rated hp ) Table 17.5, D2 − D1 32 − 9.1 = 0.458 = C 50 Kθ = 0.935 Table 17.6 2 D2 − D1 ) ( L ≈ 2C + 1.57(D2 + D1 ) + 4C (32 − 9.1)2 = 167 in L = 2(50) + 1.57(32 + 9.1) + 4(50) use C158, L = 160.9 in K L = 1.02 Adjusted rated hp = (0.935)(1.02)(10.96) = 10.45 hp Design hp 105 No. of belts = = = 10 belts Adjusted rated hp 10.43 B + B 2 − 32(D2 − D1 ) 16 B = 4 L − 6.28(D2 + D1 ) = 4(160.9) − 6.28(32 + 9.1) = 385.5 in 2

C=

C=

385.5 +

(385.5)2 − 32(32 − 9.1)2

16 Use 10-C158 belts, D1 = 9.1 in D2 = 32 in , C = 46.8 in

858.

= 46.8 in

A 200-hp, 600-rpm induction motor is to drive a jaw crusher at 125 rpm; starting load is heavy; operating with shock; intermittent service; C = 113 to 123 in . Recommend a multiple V-flat drive for this installation. The B.F. Goodrich Company recommended eight D480 V-belts with a 26-in. sheave and a 120.175in. pulley; C ≈ 116.3 in .

Solution:

Page 26 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Table 17.7 N sf = 1.6 − 0.2 = 1.4 hp = (1.4)(200) = 280 hp Fig. 17.14, 280 hp, 600 rpm Use Section E But in Table 17.3, section E is not available, use section D Dmin = 13 D2 600 = = 4.8 D1 125 For D1max : D + D2 min C = 1 + D1 2 D + 4.8 D1 113 = 1 + D1 2 D1 = 28 in min C = D2 D2 = 113 in 113 D1 = = 23.5 in 4 .8 1 use D1 ≈ (13 + 23.5) = 18 in 2 D2 = (4.8)(18) = 86.4 in L ≈ 2C + 1.57(D2 + D1 ) +

(D2 − D1 )2

4C 2 86.4 − 18) ( L = 2(118) + 1.57(86.4 + 18) + = 410 in 4(118) using D1 = 19 in , D2 = 91.2 in , C = 118 in 2 91.2 − 19) ( L = 2(118) + 1.57(91.2 + 19) + 4(118)

= 420 in

Therefore use D420 sections D1 = 19 in , D2 = 91.2 in πD n π (19)(600) vm = 1 1 = = 2985 fpm 12 12   103  0.09 c v2  v  − Rated hp = a − e m6  m3 K d D1 10  10   vm   Table 17.3, D-section a = 18.788 , c = 137.7 , e = 0.0848 D Table 17.4, 2 = 4.8 D1

Page 27 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS K d = 1.14 0.09   103  137.7 (2985)2  2985 = 29.6 hp  − Rated hp = 18.788 − (0.0848) (1.14)(19) 106  103   2985   Therefore, Fig. 17.14, section D is used. Adjusted rated hp = Kθ K L (rated hp ) Table 17.5, D2 − D1 91.2 − 19 = = 0.612 C 118 Kθ = 0.83 (V-flat) Table 17.6, D420 L = 420.8 in K L = 1.12 Adjusted rated hp = (0.83)(1.12)(29.6) = 27.52 hp Design hp 280 No. of belts = = = 10 belts Adjusted rated hp 27.52

Use10 , D420, D1 = 19 in , D2 = 91.2 in , C = 118 in 859.

A 150-hp, 700-rpm, slip-ring induction motor is to drive a ball mill at 195 rpm; heavy starting load; intermittent seasonal service; outdoors. Determine all details for a V-flat drive. The B.F. Goodrich Company recommended eight D270 Vbelts, 17.24-in sheave, 61-in. pully, C ≈ 69.7 in .

Solution: Table 17.7, N sf = 1.6 − 0.2 = 1.4 Design hp = (1.4)(150) = 210 hp Fig. 17.4, 210 hp, 700 rpm Dmin = 13 in , D-section   103  0.09 c v2  v  − Rated hp = a − e m6  m3 K d D1 10  10   vm   d (hp ) =0 For Max. Rated hp,  vm  d 3   10  v  Rated hp = a m3   10  v Let X = m3 10

Page 28 of 56

0.91

c  vm   vm  −   − e  K d D1  103   103 

3


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS c X − eX 3 K d D1 v πD1n1 πD1 (700) = X = m3 = 10 12 × 103 12 × 103 12 ×103 X D1 = 700π 700πc hp = aX 0.91 − − eX 3 3 12 ×10 K d d (hp ) = 0.91aX −0.09 − 3eX 2 = 0 d (X ) 0.91a X 2.09 = 3e Table 17.3, D-section a = 18.788 , c = 137.7 , e = 0.0848 hp = aX 0.91 −

2.09

0.91(18.788) v  X 2.09 =  m3  = 3(0.0848)  10  vm = 7488 fpm πD n vm = 1 1 = 7488 12 πD (700) vm = 1 = 7488 12 D1 = 40.86 in max D1 = 40.86 in 1 ave. D1 = (13 + 40.86) = 26.93 in 2 use D1 = 22 in D2 700 79 = ≈ D1 195 22 D1 = 22 in , D2 = 79 in D + D2 22 + 79 + D1 = Min. C = 1 + 22 = 72.5 in 2 2 Or Min. C = D2 = 79 in L ≈ 2C + 1.57(D2 + D1 ) +

(D2 − D1 )2

4C (79 − 22)2 = 327 in L = 2(79) + 1.57(79 + 22) + 4(79) use D330, L = 330.8 in

B + B 2 − 32(D2 − D1 ) C= 16

2

Page 29 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS B = 4 L − 6.28(D2 + D1 ) = 4(330.8) − 6.28(79 + 22) = 689 in C=

689 +

(689)2 − 32(79 − 22)2

= 81.12 in 16 πD n π (22)(700) = 4032 fpm vm = 1 1 = 12 12 K d = 1.14 0.09 2   103  ( 137.7 4032)  4032   ( )  Rated hp = 18.788  − (1.14)(22 ) − 0.0848 106  103 = 39.124 hp 4032    

Adjusted rated hp = Kθ K L (rated hp ) Table 17.5, D2 − D1 79 − 22 = = 0.70 C 81.12 Kθ = 0.84 (V-flat) Table 17.6 D330 K L = 1.07 Adjusted rated hp = (0.84)(1.07 )(39.124 ) = 35.165 hp Design hp 210 No. of belts = = = 5.97 belts use 6 belts Adjusted rated hp 35.165 Use 6 , D330 V-belts , D1 = 22 in , D2 = 79 in , C ≈ 81.1 in 860.

A 30-hp, 1160-rpm, squirrel-cage motor is to be used to drive a fan. During the summer, the load is 29.3 hp at a fan speed of 280 rpm; during the winter, it is 24 hp at 238 rpm; 44 < C < 50 in .; 20 hr./day operation with no overload. Decide upon the size and number of V-belts, sheave sizes, and belt length. (Data courtesy of The Worthington Corporation.)

Solution: Table 17.7 N sf = 1.6 + 0.2 = 1.8 Design hp = (1.8)(30) = 54 hp Speed of fan at 30 hp 30 − 24 n2 = (280 − 238) + 238 = 286 rpm 29.3 − 24 at 54 hp, 1160 rpm. Fig. 17.4 use either section C or section D Minimum center distance: C = D2

Page 30 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS D1 + D2 + D1 2 D2 1160 = = 4.056 D1 286 use C = 4.056 D1 44 in < C < 50 in , use C = 47 in 47 D1max = = 11.6 in 4.056 use C-section, Dmin = 9 in Let D1 = 10 in , D1 = 41 in

or C =

2 D2 − D1 ) ( L ≈ 2C + 1.57(D2 + D1 ) +

4C (41 − 10.1)2 = 179.3 in L = 2(47 ) + 1.57(41 + 10.1) + 4(47 ) use C137, L = 175.9 in

B + B 2 − 32(D2 − D1 ) C= 16 B = 4 L − 6.28(D2 + D1 ) = 4(175.9) − 6.28(41 + 10.1) = 328.7 in 2

C=

382.7 +

(382.7 )2 − 32(41 − 10.1)2

16 C173, satisfies 44 in < C < 50 in 0.91

= 45.2 in ≈ 44 in 3

c  vm   vm  v  Rated hp = a m3  −   − e  K d D1  103   103   10  πD n π (10.1)(1160) = 3067 fpm vm = 1 1 = 12 12 Table 17.4 D2 = 4.056 , K d = 1.14 D1 Table 17.3, D-section a = 8.792 , c = 38.819 , e = 0.0416 0.09   103  38.819 (3067 )2  3067 = 12.838 hp  − Rated hp = 8.792 − (0.0416) (1.14)(10.1) 106  103   3067   Adjusted rated hp = Kθ K L (rated hp ) Table 17.5, D2 − D1 41 − 10.1 = = 0.68 C 45.2 Kθ = 0.90

Page 31 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Table 17.6 L = 175.9 , C173 K L = 1.04 Adjusted rated hp = (0.90)(1.04)(12.838) = 12.02 hp Design hp 54 No. of belts = = = 4.5 belts use 5 belts Adjusted rated hp 12.02 Use 5 , C173 V-belts , D1 = 10.1 in , D2 = 41 in POWER CHAINS NOTE: If manufacturer’s catalogs are available, solve these problems from catalogs as well as from data in the Text. 861.

A roller chain is to be used on a paving machine to transmit 30 hp from the 4cylinder Diesel engine to a counter-shaft; engine speed 1000 rpm, counter-shaft speed 500 rpm. The center distance is fixed at 24 in. The cain will be subjected to intermittent overloads of 100 %. (a) Determine the pitch and the number of chains required to transmit this power. (b) What is the length of the chain required? How much slack must be allowed in order to have a whole number of pitches? A chain drive with significant slack and subjected to impulsive loading should have an idler sprocket against the slack strand. If it were possible to change the speed ratio slightly, it might be possible to have a chain with no appreciable slack. (c) How much is the bearing pressure between the roller and pin?

Solution: (a) design hp = 2(30) = 60 hp intermittent D2 n1 1000 ≈ = =2 D1 n2 500 D2 = 2D1 D C = D2 + 1 = 24 in 2 D 2 D1 + 1 = 24 2 D1 = 9.6 in D2 = 2 D1 = 2(9.6 ) = 19.2 in vm =

πD1n1

=

π (9.6 )(1000)

= 2513 fpm 12 12 Table 17.8, use Chain No. 35, Limiting Speed = 2800 fpm

Page 32 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Minimum number of teeth Assume N1 = 21 N 2 = 2 N1 = 42 [Roller-Bushing Impact] 1.5

 100 N ts  0.8 hp = K r   P  n  Chain No. 35 3 P = in 8 N ts = 21 n = 1000 rpm K r = 29 1.5

100(21)   3  hp = 29     1000   8 

0.8

= 40.3 hp

[Link Plate Fatigue] hp = 0.004 N ts1.08 n 0.9 P 3−0.07 P  3 3− 0.07  

8  3 hp = 0.004(21) (1000)   = 2.91 hp 8 design hp 60 No. of strands = = = 21 rated hp 2.91 3 Use Chain No. 35, P = in , 21 strands 8 1.08

0.9

N + N 2 ( N 2 − N1 ) (b) L ≈ 2C + 1 + pitches 2 40C 24 C= = 64  3   8 N1 = 21 N 2 = 42 2

21 + 42 (42 − 21) L = 2(64) + + = 159.67 pitches ≈ 160 pitches 2 40(64 ) Amount of slack 2

(

h = 0.433 S 2 − L2 L = C = 24 in

Page 33 of 56

1 2

)



(160−159.67) 3 in

[

 = 24.062 in

8

S = 24 in +

2 1 2 2

h = 0.433 (24.062) − (24) 2

]

= 0.75 in =

3 in 4

(c) pb = bearing pressure Table 17.8, Chain No. 25 C = 0.141 in 3 E = in 16 J = 0.05 in 3  A = C (E + 2 J ) = 0.141 + 2(0.05) = 0.04054 in 2 16  FV = 60 hp 33,000 F (2513) = 60 hp 33,000 F = 787.9 lb 787.9 F= = 37.5 lb strand 21 37.5 pb = = 925 psi 0.04054

862.

A conveyor is driven by a 2-hp high-starting-torque electric motor through a flexible coupling to a worm-gear speed reducer, whose mw ≈ 35 , and then via a roller chain to the conveyor shaft that is to turn about 12 rpm; motor rpm is 1750. Operation is smooth, 8 hr./day. (a) Decide upon suitable sprocket sizes, center distance, and chain pitch. Compute (b) the length of chain, (c) the bearing pressure between the roller and pin. The Morse Chain Company recommended 15- and 60-tooth sprockets, 1-in. pitch, C = 24 in ., L = 88 pitches .

Solution: Table 17.7 N sf = 1.2 − 0.2 = 1.0 (8 hr/day) design hp = 1.0(2) = 2.0 hp 1750 n1 = = 50 rpm 35 n2 = 12 rpm Minimum number of teeth = 12

Page 34 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Use N1 = 12 [Link Plate Fatigue] hp = 0.004 N ts1.08 n 0.9 P 3−0.07 P hp 2.0 P 3− 0.07 P ≈ P 3 = = = 1.0 1.08 0.9 1.08 0.9 0.004 N ts n 0.004(12 ) (50 ) Use Chain No. 80, P = 1.0 in To check for roller-bushing fatigue 1.5

 100 N ts  0.8 hp = K r   P  n  K r = 29 100(12 )  hp = 17   1000 

1.5

(1)0.8 = 2747 hp > 2 hp

(a) N1 = 12 n   50  N 2 =  1  N1 =  (12) = 50 teeth  12   n2  D C = D2 + 1 2 PN1 (1.0 )(12 ) D1 ≈ = = 3.82 in

π

D2 ≈

PN1

π

π

(1.0)(50) = 15.92 in = π

3.82 = 17.83 in 2 use C = 18 in C = 18 pitches chain pitch = 1.0 in, Chain No. 80 C ≈ 15.92 +

N1 + N 2 ( N 2 − N1 ) + 2 40C 2 12 + 59 (50 − 12) L ≈ 2(18) + + = 69 pitches 40(18) 2 use L = 70 pitches 2

(b) L ≈ 2C +

(c) pb = bearing pressure Table 17.8, Chain No. 80 C = 0.312 in 5 E = in 8

Page 35 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS J = 0.125 in PN ts n1 (1)(12 )(50 ) vm = = = 50 fpm 12 12 3  A = C (E + 2 J ) = 0.141 + 2(0.05) = 0.04054 in 2 16  FV = 60 hp 33,000 33,000(2 ) = 1320 lb F= 50 F 1320 pb = = = 4835 psi C (E + 2 J ) 5  0.312  + 2(0.125) 8 

863.

A roller chain is to transmit 5 hp from a gearmotor to a wood-working machine, with moderate shock. The 1-in output shaft of the gearmotor turns n = 500 rpm . The 1 ¼-in. driven shaft turns 250 rpm; C ≈ 16 in . (a) Determine the size of sprockets and pitch of chain that may be used. If a catalog is available, be sure maximum bore of sprocket is sufficient to fit the shafts. (b) Compute the center distance and length of chain. (c) What method should be used to supply oil to the chain? (d) If a catalog is available, design also for an inverted tooth chain.

Solution: Table 17.7 N sf = 1.2 design hp = 1.2(5) = 6 hp D2 500 = =2 D1 250 D C = D2 + 1 2 D 16 = 2 D1 + 1 2 D1 = 6.4 in D2 = 2 D1 = 2(6.4) = 12.8 in πD n π (6.4)(500) vm = 1 1 = = 838 fpm 12 12 (a) Link Plate Fatigue hp = 0.004 N ts1.08 n 0.9 P 3−0.07 P πD π (6.4 ) 20.11 N ts = N1 ≈ 1 = = P P P

Page 36 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 1.08

 20.11  0.9 3− 0.07 P hp = 0.004  (500 ) P  P  6 = 27.47 P1.92−0.07 P P = 0.45 in 1 use P = in , Chain No. 40 2 πD1 π (6.4) N1 ≈ = = 40 P 1   2 N 2 = 2 N1 = 80

Size of sprocket, N1 = 40 , N 2 = 80 , P =

1 in . 2

(b) C = 16 in 16 in = 32 pitches C= 1 in 2 2 N1 + N 2 ( N 2 − N1 ) L ≈ 2C + + 2 40C 2 40 + 80 (80 − 40) L ≈ 2(32) + + = 125.25 pitches 40(32) 2 use L = 126 pitches (c) Method: vm = 838 fpm . Use Type II Lubrication ( vmax = 1300 fpm ) – oil is supplied from a drip lubricator to link plate edges. 864.

A roller chain is to transmit 20 hp from a split-phase motor, turning 570 rpm, to a reciprocating pump, turning at 200 rpm; 24 hr./day service. (a) Decide upon the tooth numbers for the sprockets, the pitch and width of chain, and center distance. Consider both single and multiple strands. Compute (b) the chain length, (c) the bearing pressure between the roller and pin, (d) the factor of safety against fatigue failure (Table 17.8), with the chain pull as the force on the chain. (e) If a catalog is available, design also an inverted-tooth chain drive.

Solution: Table 17.7 N sf = 1.4 + 0.2 (24 hr/day) design hp = 1.6(20) = 32 hp n 570 = 2.85 (a) 1 = n2 200

Page 37 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS D2 n1 ≈ = 2.85 D1 n2 Considering single strand hp = 0.004 N ts1.08 n 0.9 P 3−0.07 P

min N ts = 17 hp = 32 = 0.004(17 )

1.08

(570)0.9 P 3−0.07 P

P 3− 0.07 P = 1.24 P = 1.07 in use P = 1.0 in 3− 0.07 (1)

hp = 32 = 0.004( N1 ) (570 ) (1) N1 = 21  570  N2 =  (21) = 60  200  5 Roller width = in 8 D C = D2 + 1 2 PN1 (1)(21) D1 ≈ = = 6.685 in 1.08

π

D2 ≈

PN 2

π

0.9

π

(1)(60) = 19.10 in = π

6.685 = 22.44 in 2 Use C = 23 in 23 C= pitches 1 Considering multiple strands 1 Assume, P = in 2 1.08 0.9 3− 0.07 P hp = 0.004 N ts n P C = 19.10 +

hp = 0.004(21)

(570)0.9 (0.5)3−0.07 (0.5) = 4.148 hp

No. of strands =

32 hp = 7.7 4.148 hp

1.08

Use 8 strands (b) Chain Length 2 N1 + N 2 ( N 2 − N1 ) L ≈ 2C + + 2 40C

Page 38 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 21 + 60 (60 − 21) + = 88.15 pitches 40(23) 2 use L = 88 pitches 2

L ≈ 2(23) +

(c) pb = bearing pressure Table 17.8, P = 1 in 5 E = in 8 J = 0.125 in C = 0.312 in 33,000hp F= vm πD n π (6.685)(570) vm = 1 1 = = 998 fpm 12 12 33,000(32 ) F= = 1058 lb 998 F 1058 pb = = = 3876 psi C (E + 2 J ) 5  0.312  + 2(0.125) 8  (d) Factor of Safety =

Fu , based on fatigue 4F

Fu = 14,500 lb , Table 17.8 F 14,500 = 3.43 Factor of Safety = u = 4 F 4(1058) 865.

A 5/8-in. roller chain is used on a hoist to lift a 500-lb. load through 14 ft. in 24 sec. at constant velocity. If the load on the chain is doubled during the speed-up period, compute the factor of safety (a) based on the chain’s ultimate strength, (b) based on its fatigue strength. (c) At the given speed, what is the chain’s rated capacity ( N s = 20 teeth ) in hp? Compare with the power needed at the constant speed. Does it look as though the drive will have a “long” life?

Solution: Table 17.8 5 P = in 8 Fu = 6100 lb

Page 39 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Fu F F = (500)(2 ) = 1000 lb 6100 = 6.1 Factor of Safety = 1000 F (b) Factor of Safety = u (fatigue) 4F 6100 Factor of Safety = = 1.5 4(1000)

(a) Factor of Safety =

(c) vm =

14 ft  60 sec    = 35 fpm 24 sec  1 min 

N s = 20 5 P = in 8 Rated hp = 0.004 N ts1.08 n 0.9 P 3−0.07 P [Link Plate Fatigue] 5  (20)n PN s n  8  = = 35 fpm vm = 12 12 n = 33.6 rpm 5 3− 0.07  

8 1.08 0.9  5  Rated hp = 0.004(20) (33.6 )   = 0.6 hp 8 Hp needed at constant speed Fvm (500)(35) = 0.53 hp < 0.6 hp hp = = 33,000 33,000 Therefore safe for “long” life.

WIRE ROPES 866.

In a coal-mine hoist, the weight of the cage and load is 20 kips; the shaft is 400 ft. deep. The cage is accelerated from rest to 1600 fpm in 6 sec. A single 6 x 19, IPS, 1 ¾-in. rope is used, wound on an 8-ft. drum. (a) Include the inertia force but take the static view and compute the factor of safety with and without allowances for the bending load. (b) If N = 1.35 , based on fatigue, what is the expected life? (c) Let the cage be at the bottom of the shaft and ignore the effect of the rope’s weight. A load of 14 kips is gradually applied on the 6-kip cage. How much is the deflection of the cable due to the load and the additional energy absorbed? (d) For educational purposes and for a load of 0.2 Fu , compute the energy that this 400-ft rope can absorb and compare it with that for a 400-ft., 1 ¾-in., as-rolled-1045 steel rod. Omit the weights of the rope and rod. What is the energy per pound of material in each case?

Page 40 of 56



Solution: (a)

(1600

 1 min   fpm ) 60 sec   = 4.445 fps 2 6 sec

v2 − v1 = t Wh = 20 kips For 6 x 19 IPS, w ≈ 1.6 Dr2 lb ft  400  2 wL = 1.6 Dr2   kips = 0.64 Dr kips 1000   Ft − wL − Wh = ma a=

m=

20 + 0.64 Dr2 32.2

 20 + 0.64 Dr2  (4.445) Ft − 0.64 Dr2 − 20 =  32.2   2 Ft = 22.76 + 0.73Dr 3 Dr = 1 in 4 2

 3 Ft = 22.76 + 0.731  = 25 kips  4 F − Fb N= u Ft Table AT 28, IPS Fu ≈ 42 Dr2 tons

Fu = 42(1.75) = 129 tons = 258 kips with bending load Fb = sb Am EDw sb = Ds 2

Page 41 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS EAm Dw Ds Table At 28, 6 x 19 Wire Rope Dw = 0.067 Dr = 0.067(1.75) = 0.11725 in Fb =

Ds = 8 ft = 96 in E = 30,000 ksi Am ≈ 0.4 Dr2

Am = 0.4(1.75) = 1.225 sq in (30,000)(1.225)(0.11725) = 45 kips Fb = (96) F − Fb 258 − 45 = = 8.52 N= u Ft 25 without bending load F 258 N= u = = 10.32 Ft 25 2

(b) N = 1.35 on fatigue IPS, su ≈ 260 ksi 2 NFt Dr Ds = ( p su )su (1.75)(96) = 2(1.35)(25) ( p su )(260) p su = 0.0015 Fig. 17.30, 6 x 19 IPS Number of bends to failure = 7 x 105 FL Am Er Am = 1.225 sq in

(c) δ =

Er ≈ 12,000 ksi (6 x 19 IPS) F = 14 kips L = 400 ft = 4800 in (14)(4800) = 4.57 in δ= (1.225)(12,000) 1 1 U = Fδ = (14 )(4.57 ) = 32 in − kips 2 2 (d) F = 0.2 Fu = 0.2(258) = 51.6 kips

Page 42 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS FL Am Er (51.6)(4800) = 16.85 in δ= (1.225)(12,000) 1 1 U = Fδ = (51.6 )(16.85) = 434 in − kips 2 2 For 1 ¾ in, as-rolled 1045 steel rod su = 96 ksi

δ=

π  2 Fu = su A = (96 ) (1.75) = 230.9 kips 4 F = 0.2 Fu = 0.2(230.9) = 46.2 kips FL δ= AE (46.2)(4800 ) = 3.073 in δ= π  2  (1.75) (30,000 ) 4 1 1 U = Fδ = (46.2 )(3.073) = 71 in − kips < U of wire rope. 2 2

868.

A hoist in a copper mine lifts ore a maximum of 2000 ft. The weight of car, cage, and ore per trip is 10 kips, accelerated in 6 sec. to 2000 fpm; drum diameter is 6 ft. Use a 6 x 19 plow-steel rope. Determine the size (a) for a life of 200,000 cycles and N = 1.3 on the basis of fatigue, (b) for N = 5 by equation (v), §17.25, Text. (c) What is the expected life of the rope found in (b) for N = 1.3 on the basis of fatigue? (d) If a loaded car weighing 7 kips can be moved gradually onto the freely hanging cage, how much would the rope stretch? (e) What total energy is stored in the rope with full load at the bottom of te shaft? Neglect the rope’s weight for this calculation. (f) Compute the pressure of the rope on the cast-iron drum. Is it reasonable?

Solution:

a=

v2 − v1 = t

(2000

Page 43 of 56

 1 min   fpm )  60 sec  = 5.56 fps 2 6 sec


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS For 6 x 19 IPS, w ≈ 1.6 Dr2 lb ft  2000  2 wL = 1.6 Dr2   kips = 3.2 Dr kips  1000  Wh = 10 kips  wL + Wh  Ft − wL − Wh =  a  32.2   a   5.56  Ft =  + 1(wL + Wh ) =  + 1 3.2 Dr2 + 10 = 1.17267 3.2 Dr2 + 10 32 . 2 32 . 2     2 NFt (a) Dr Ds = ( p su )su Fig. 17.30, 200,000 cycles, 6 x 19 p su = 0.0028

(

PS: su ≈ 225 ksi Ds = 6 ft = 72 in N = 1.3 2(1.3)(1.17267 ) 3.2 Dr2 + 10 Dr (72) = (0.0028)(225) 45.36 Dr = 9.7566 Dr2 + 30.49 Dr2 − 4.64916 Dr + 3.1251 = 0 Dr = 0.815 in 7 say Dr = in 8

(

(b) by N = 5 , Equation (v) F − Fb N= u Ft EDw sb = Ds Dw = 0.067 Dr (30,000)(0.067 Dr ) = 27.92 D sb = r 72 Fb = sb Am Am = 0.4 Dr2 Fb = (27.92 Dr )(0.4 Dr2 ) = 11.17 Dr3 Fu = 36 Dr2 tons for PS Fu = 72 Dr2 kips

Fu − Fb = NFt

Page 44 of 56

)

)

(

)


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 72 Dr2 − 11.17 Dr3 = (5)(1.17267 )(3.2 Dr2 + 10) 72 Dr2 − 11.17 Dr3 = (5.8634 )(3.2 Dr2 + 10) Dr = 1.216 in 1 use Dr = 1 in 4 2 NFt (c) Dr Ds = ( p su )su

(1.25)(72) = 2(1.3)(1.17267 )[3.2(1.25) ( p su )(225)

2

+ 10

]

p su = 0.00226 Fig. 17.20 Expected Life = 3 x 105 cycles (d) F = 7 kips Er = 12,000 ksi L = 2000 ft = 24,000 in 7 For (a) Dr = in 8 FL δ= Am Er 2

7 Am ≈ 0.4 Dr3 = 0.4  = 0.30625 sq in 8 (7 )(24,000) = 45.7 in δ= (0.30625)(12,000) 1 For (b) Dr = 1 in 4 FL δ= Am Er 2

 1 Am ≈ 0.4 D = 0.41  = 0.625 sq in  4 (7 )(24,000) = 22.4 in δ= (0.625)(12,000) 3 r

1 1 Fδ = (7 )(45.7 ) = 160 in − kips 2 2 1 1 For (b) U = Fδ = (7 )(22.4 ) = 78.4 in − kips 2 2 (f) Limiting pressure, cast-iron sheaves, 6 x19, p = 500 psi .

(e) For (a) U =

Page 45 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS For (a) p su = 0.0028 p = 0.0028(225) = 0.630 kips = 630 psi > 500 psi , not reasonable. For (b) p su = 0.00226 p = 0.00226(225) = 0.5085 kips = 508.5 psi ≈ 500 psi , reasonable. 869.

For a mine hoist, the cage weighs 5900 lb., the cars 2100 lb., and the load of coal in the car 2800 lb.; one car loaded loaded at a time on the hoist. The drum diameter is 5 ft., the maximum depth is 1500 ft. It takes 6 sec. to accelerate the loaded cage to 3285 fpm. Decide on a grade of wire and the kind and size of rope on the basis of (a) a life of 2× 105 cycles and N = 1.3 against fatigue failure, (b) static consideration (but not omitting inertia effect) and N = 5 . (c) Make a final recommendation. (d) If the loaded car can be moved gradually onto the freely hanging cage, how much would the rope stretch? (e) What total energy has the rope absorbed, fully loaded at the bottom of the shaft? Neglect the rope’s weight for this calculation. (f) Compute the pressure of the rope on the cast-iron drum. Is it all right?

Solution:

Wh = 5900 + 2100 + 2800 = 10,800 lb = 10.8 kips

(3285 fpm) 1 min 

 60 sec  = 9.125 fps 2 6 sec  wL + Wh  Ft − wL − Wh =  a  32.2  Assume 6 x 19 IPS, w ≈ 1.6 Dr2 lb ft  1500  2 wL = 1.6 Dr2   kips = 2.4 Dr kips 1000    a   9.125  Ft =  + 1(wL + Wh ) =  + 1(2.4 Dr2 + 10 ) = 3.08 Dr2 + 13.86  32.2   32.2  5 (a) Fig. 17.30, 2 x 10 cycles p su = 0.0028

a=

v2 − v1 = t

Page 46 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 2 NFt ( p su )su Ds = 5 ft = 60 in

Dr Ds =

Ds ≈ 45 Dr 60 Dr max = = 1.33 in 45 1 use Dr = 1 in 4 2

 1 Ft = 3.081  + 13.86 = 18.67 kips  4 2(1.3)(18.67 ) su = = 231 ksi  1 (0.0028)1 (60 )  4

1 Use Plow Steel, 6 x 19 Wire Rope, Dr = 1 in . 4

(b) N = sb =

Fu − Fb Ft

EDw Ds

 1 Dw = 0.067 Dr = 0.0671  = 0.08375 in  4 Ds = 60 in E = 30,000 ksi (30,000)(0.08375) = 41.875 ksi sb = 60 2

 1 Am = 0.4 D = 0.41  = 0.625 in 2  4 Fb = sb Am = (41.875)(0.625) = 26.17 kips N =5 Fu = NFt + Fb = (5)(18.67 ) + 26.17 = 119.52 kips = 59.76 tons Fu 59.76 = = 38.25 2 Dr  1  2 1   4 Table AT 28, 2 r

Use IPS, 6 x 19,

Page 47 of 56

Fu = 42 > 38.25 Dr2


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS (c) Recommendation: 1 6 x 19, improved plow steel, Dr = 1 in 4 FL (d) δ = Am Er F = 2100 + 2800 = 4900 lb Er ≈ 12× 106 psi L = 1500 ft = 18,000 in (4900)(18,000) = 11.76 in δ= (0.625)(12 ×106 ) 1 1 Fδ = (4900 )(11.76 ) = 28,800 in − lb 2 2 (f) p su = 0.0028

(e) U =

su = 231 ksi p = 0.0028(231,000 ) = 646.8 psi For cast-iron sheave, limiting pressure is 500 psi p = 646.8 psi > 500 psi , not al right. 870.

The wire rope of a hoist with a short lift handles a total maximum load of 14 kips each trip. It is estimated that the maximum number of trips per week will be 1000. The rope is 6 x 37, IPS, 1 3/8 in. in diameter, with steel core. (a) On the basis of N = 1 for fatigue, what size drum should be used for a 6-yr. life? (n) Because of space limitations, the actual size used was a 2.5-ft. drum. What is the factor of safety on a static basis? What life can be expected ( N = 1 )?

Solution: (a)  365 days  1 wk  1000 trips     = 312,857 cycles ≈ 3 × 105 cycles No. of cycles = (6 yr )  1 yr  7 days  1 wk  Figure 17.30, 6 x 37, IPS p su = 0.00225 2 NFt Dr Ds = ( p su )su For IPS, su ≈ 260 ksi Ft = 14 kips N = 1.0 Dr = 1.375 in 2 NFt Dr Ds = ( p su )su

Page 48 of 56



(1.375)Ds =

2(1.0)(14) (0.00225)(260)

Ds = 34.8 in (b) Ds = 2 ft = 30 in Static Basis F − Fb N= u Ft Table AT 28, 6 x 37 Dw ≈ 0.048 Dr = 0.048(1.375) = 0.066 in Am ≈ 0.4 Dr2 = 0.4(1.375) = 0.75625 in 2 Fu = su Am = (260 )(0.75625) = 196.6 kips EDw Am (30,000 )(0.066 )(0.75625) Fb = sb Am = = = 49.9 kips Ds 30 F − Fb 196.6 − 49.9 = N= u = 10.5 Ft 10.5 Life: N = 1.0 (fatigue) 2 NFt Dr Ds = ( p su )su (1.375)(30) = 2(1.0)(14) ( p su )(260) p su = 0.0026 2

Figure 17.30, Life ≈ 2.5 × 105 cycles , 6 x 37.

871.

A wire rope passes about a driving sheave making an angle of contact of 540o, as shown. A counterweight of 3220 lb. is suspended from one side and the acceleration is 4 fps2. (a) If f = 0.1 , what load may be noised without slipping on the rope? (b) If the sheave is rubber lined and the rope is dry, what load may be raised without slipping? (c) Neglecting the stress caused by bending about the sheave, find the size of 6 x 19 MPS rope required for N = 6 and for the load found in (a). (d) Compute the diameter of the sheave for indefinite life with say N = 1.1 on fatigue. What changes could be made in the solution to allow the use of a smaller sheave?

Page 49 of 56



Problems 871 – 874. Solution:  4 fps 2   = 2820 lb F2 = (3220 lb )1 − 2   32.2 fps  (a) F1 = F2 e fθ θ = 540o = 3π f = 0.10 F1 = (2820 )e (0.10 )(3π ) = 7237 lb

(b) For rubber lined, dry rope f = 0.495 F1 = (2820 )e (0.495 )(3π ) = 249,466 lb (c) Ft = F1 = 7237 lb F − (Fb ≈ 0 ) Fu N= u = Ft Ft Fu ≈ 32 Dr2 tons for MPS Fu ≈ 64 Dr2 kips Fu = 64,000 Dr2 lb

Fu = NFt 64,000 Dr2 = (6 )(7237 ) Dr = 0.824 in use Dr = 0.875 in

Page 50 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 2 NFt ( p su )su Indefinite life, p su = 0.0015

(d) Dr Ds =

MPS: su ≈ 195 ksi = 195,000 psi (0.875)Ds = 2(1.1)(7237 ) (0.0015)(195,000) Ds = 62.2 in To reduce the size of sheave, increase the size of rope. 872.

A traction elevator with a total weight of 8 kips has an acceleration of 3 fps2; the 6 cables pass over the upper sheave twice, the lower one once, as shown.. Compute the minimum weight of counterweight to prevent slipping on the driving sheave if it is (a) iron with a greasy rope, (b) iron with a dry rope, (c) rubber lined with a greasy rope. (d) Using MPS and the combination in (a), decide upon a rope and sheave size that will have indefinite life ( N = 1 will do). (e) Compute the factor of safety defined in the Text. (f) If it were decided that 5× 105 bending cycles would be enough life, would there be a significant difference in the results?

Solution:  3 fps 2   = 8.745 kips F1 = (8 kips )1 + 2   32.2 fps  θ = 3(180o ) = 3π F F2 = f1θ e Wc = weight of counterweight F2 Wc = = 1.10274 F2 3 1− 32.2 1.10274 F1 Wc = e fθ (a) Iron sheave, greasy rope, f = 0.07 1.10274(8.745) Wc = = 4.986 kips e (0.07 )(3π ) (b) Iron sheave, dry rope, f = 0.12 1.10274(8.745) Wc = = 3.112 kips e (0.12 )(3π ) (c) Rubber lined with a greasy rope, f = 0.205 1.10274(8.745) Wc = = 1.397 kips e (0.205 )(3π )

Page 51 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 2 NFt ( p su )su Indefinite life, p su = 0.0015

(d) Dr Ds =

Ft = F1 = 8.745 kips total 8.745 Ft = = 1.458 kips each rope 6 Ft = 1458 lbs N =1 Table AT 28, 6 x 19 Ds ≈ 45 Dr 2(1)(1458) Dr (45Dr ) = (0.0015)(195,000) Dr = 0.47 in 1 Use Dr = in = 0.5 in 2 Fu − Fb Ft Table AT 28, MPS 2 Fu = 32 Dr2 tons = 64,000 Dr2 lb = 64,000(0.5) lb = 16,000 lb

(e) N =

Fb =

EDw Am Ds

E = 30× 106 psi 6 x 19, Dw = 0.067 Dr Ds ≈ 45 Dr Am = 0.4 Dr2 = 0.4(0.5) = 0.1 sq. in. 2

(30 ×10 )(0.067 )(0.1) = 4467 lb 6

Fb =

45 16,000 − 4467 N= = 7.91 1458

(f) 5 x 105 cycles Fig. 17.30, 6 x 19. p su = 0.0017 2 NFt Dr Ds = ( p su )su

Page 52 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Dr (45Dr ) =

2(1)(1458) (0.0017 )(195,000)

Dr = 0.44 in since Dr = 0.44 in ≈ 0.47 in as in (d), therefore, no significant difference will result. 873.

A 5000-lb. elevator with a traction drive is supported by a 6 wire ropes, each passing over the driving sheave twice, the idler once, as shown. Maximum values are 4500-lb load, 4 fps2 acceleration during stopping. The brake is applied to a drum on the motor shaft, so that the entire decelerating force comes on the cables, whose maximum length will be 120 ft. (a) Using the desirable Ds in terms of Dr , decide on the diameter and type of wire rope. (b) For this rope and N = 1.05 , compute the sheave diameter that would be needed for indefinite life. (c) Compute the factor of safety defined in the Text for the result in (b). (d) Determine the minimum counterweight to prevent slipping with a dry rope on an iron sheave. (e) Compute the probable life of the rope on the sheave found in (a) and recommend a final choice.

Solution: (a)

Ft = 4500 lb Wh = 5000 lb  W + wL  Wh + wL − Ft =  h a  32.2  assume 6 x 19 w = 1.6 Dr2 lb ft

wL = (1.6 Dr2 )(120 ) = 192 Dr2 per rope wL = 6(192 Dr2 ) = 1152 Dr2

 5000 + 1152 Dr2  (4) 5000 + 1152 D − 4500 =  32 . 2   2 2 1152 Dr + 500 = 621.12 + 143.11Dr Dr = 0.3465 in 3 say Dr = 0.375 in = in 8 2 r

Page 53 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 7  3 Ds ≈ 45 Dr = 45  = 16 in 8 8 3 Six – 6 x 19 rope, Dr = in 8

(a) Dr = 0.375 in =

3 in 8

4500 = 750 lb 6 N = 1.05 2 NFt Dr Ds = ( p su )su assume IPS, su = 260 ksi = 260,000 psi Ft =

Indefinite life, p su = 0.0015 (0.375)Ds = 2(1.05)(750) (0.0015)(260,000) Ds = 10.77 in Fu − Fb Ft Ft = 750 lb IPS

(c) N =

2

3 Fu ≈ 42 D tons = 84,000 D lb = 84,000  lb = 11,813 lb 8 2 r

Fb =

2 r

EDw Am Ds

6 x 19, Ds = 10.77 in as in (b) 3 Dw = 0.067 Dr = 0.067  = 0.025 in 8 2

3 Am = 0.4 D = 0.4  = 0.05625 sq. in. 8 6 E = 30× 10 psi 2 r

(30 ×10 )(0.025)(0.05625) = 3917 lb 6

Fb =

Page 54 of 56

10.77



N=

11,813 − 3917 = 10.53 750

(c) F1 = Ft = 4500 lb F1 = F2 e fθ For iron sheave, dry rope, f = 0.12 θ = 540o = 3π F2 =

F1 4500 = (0.12 )(3π ) = 1452 lb fθ e e

a   CW 1 +  = F2  32.2  4   CW 1 +  = 1452  32.2  CW = 1291 lb

874.

A traction elevator has a maximum deceleration of 5 fps2 when being braked on the downward motion with a total load of 10 kips. There are 5 cables that pass twice over the driving sheave. The counterweight weighs 8 kips. (a) Compute the minimum coefficient of friction needed between ropes and sheaves for no slipping. Is a special sheave surface needed? (b) What size 6 x 19 mild-plowsteel rope should be used for N = 4 , including the bending effect? (Static approach.) (c) What is the estimated life of these ropes ( N = 1 )?

Solution:

a = 8.05 fps 2 (a) F1 = 10 kips  8.05  F2 = (8 kips )1 −  = 6 kips  32.2  θ = 3π F1 = e fθ F2

Page 55 of 56


SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 10 = e f (3π ) 6 f = 0.0542

Special sheave surface is needed for this coefficient of friction, §17.21. F − Fb (b) N = u Ft 10 Ft = = 2 kips 5 EDw Am Fb = Ds Table AT 28, 6 x 19, MPS Dw = 0.067 Dr Ds ≈ 45 Dr Am ≈ 0.4 Dr2

E = 30× 106 psi

(30 ×10 )(0.067 D )(0.4 D ) = 17.87 D 6

Fb =

r

45Dr

2 r

2 r

kips

Fu ≈ 32 Dr2 tons = 64 Dr2 kips

64 Dr2 − 17.87 Dr2 2 Dr = 0.4164 in 7 use Dr = in 16 7 (c) Ds ≈ 45 Dr = 45  = 20 in  16  2 NFt Dr Ds = ( p su )su Ft = 2 kips each rope MPS, su = 195 ksi N = 1.0 2(1.0)(2) 7  (20) = ( p su )(195)  16  p su = 0.0023 Expected life, Figure 17.30, 3 x 105 bending cycles. - end N =4=

Page 56 of 56


SECTION 16 – BRAKES AND CLUTCHES

ENERGY TO BRAKES 881.

A motor operates a hoist through a pair of spur gears, with a velocity ratio of 4. The drum on which the cable wraps is on the same shaft as the gear, and the torque cause by the weight of the load and hoist is 12,000 ft-lb. The pinion is on the motor shaft. Consider first on which shaft to mount the brake drum; in the process make trial calculations, and try to think of pros and cons. Make a decision and determine the size of a drum that will not have a temperature rise greater than ∆t = 150o F when a 4000-lb. load moves down 200 ft. at a constant speed. Include a calculation for the frp/sq. in. of the drum’s surface.

Solution: Consider that brake drum is mounted on motor shaft that has lesser torque. 12,000 ft − lb T f= = 3000 ft − lb = 36,000 in − lb 4 From Table AT 29, Assume f = 0.35 , p = 75 psi , max. vm = 5000 fpm Tf =

FD 2

F = fN = N=

2T f D

2T f

fD N p= A A = π Db p=

2T f N 2(36,000) = = = 75 2 π Db π D bf π D 2b(0.35)

D 2b = 873 use D 2b = 873 873 b= 2 D Then, U ft − lb ∆t o F = f Wm c Assume a cast-iron, ρ = 0.253 lb in3 c = 101 Wm = ρ V

Page 1 of 97



 D2   D 2t = π t  Db + 4 4   U f = (4000 )(200 ) = 800,000 ft − lb

V = π Dbt +

π

∆t = 150o F

Wm = ρ V =

Uf

c∆t 800,000 0.253V = (150)(101) V = 208.7 in 3 But  D2   V = π t  Db + 4   873 b= 2 D  873 D 2   + V = π t  4   D For minimum V : dV  − 873 D  = π t 2 +  = 0 2 dD  D 3 D = 2(873) D = 12 in For t :  873 (12 )2  + V = 208.7 = π t   4   12 t = 0.611 in 5 say t = in 8 1 873 = 6.0625 in = 6 in 2 16 (12 ) 5 1 Therefore use D = 12 in , t = in , b = 6 in 8 16 b=

For fhp sq. in. = fhp =

Fvm 33,000

Page 2 of 97

fhp A


SECTION 16 – BRAKES AND CLUTCHES 2(36,000 ) = 6000 lb D 12 vm = 5000 fpm (max.) (6000)(5000) = 909 hp fhp = 33,000  1 A = π Db = π (12 ) 6  in 2  16  fhp 909 = 3.98 (peak value) fhp sq. in. = = A 228.55 F=

882.

2T f

=

A 3500-lb. automobile moving on level ground at 60 mph, is to be stopped in a distance of 260 ft. Tire diameter is 30 in.; all frictional energy except for the brake is to be neglected. (a) What total averaging braking torque must be applied? (b) What must be the minimum coefficient of friction between the tires and the road in order for the wheels not to skid if it is assumed that weight is equally distributed among the four wheels (not true)? (c) If the frictional energy is momentarily stored in 50 lb. of cast iron brake drums, what is the average temperature rise of the drums?

Solution: (a) Solving for the total braking torque. W 2 vs − vs22 U f = −∆KE = 2g 1 W = 3500 lb vs1 = 60 mph = 88 fps

(

)

vs2 = 0 mph = 0 fps

g = 32.2 fps 2 3500 Uf = (882 − 02 ) = 421,000 ft − lb 2(32.2) (T ft − lb )ωm = (T f in − lb )n fhp = f 63,000 33,000 2 2 2 vs − vs1 0 − (88) a= 2 = = −14.892 fps 2 2s 2(260 ) vs − vs1 0 − 88 t= 2 = 5.91 sec = − 14.892 a U − ∆KE 421,000 fhp = = f = = 130 hp (t )(550) 550t 550(5.91)

Page 3 of 97


SECTION 16 – BRAKES AND CLUTCHES 1 (88 fps )(60 sec min ) vm 2 n= = = 336 rpm πD  30  π  ft   12  T n fhp = f 63,000 63,000(130) Tf = = 24,375 in − lb 336 (b) f =

F N

for each wheel, N =

3500 = 875 lb 4

24,375 = 6094 in − lb 4 2T 2(6094 ) F= f = = 406 in − lb D 30 F 406 f = = = 0.464 N 875

Tf =

(c) ∆t =

Uf

Wm c U f = 421,000 ft − lb

Wm = 50 lb c = 101 ft − lb lb − F for cast-iron 421,000 ∆t = = 83.4o F (50)(101)

884.

An overhead traveling crane weighs 160,000 lb. with its load and runs 253 fpm. It is driven by a 25-hp motor operating at 1750 rpm.The speed reduction from the motor to the 18-in. wheels is 32 to 1. Frictional energy other than at the brake is negligible. (a) How much energy must be absorbed by the brake to stop this crane in a distance of 18 ft.? (b) Determine the constant average braking torque that must be exerted on the motor shaft. (c) If all the energy is absorbed by the rim of the cast-iron brake drum, which is 8 in. in diameter, 1 ½ in. thick, with a 3 ¼-in. face, what will be its temperature rise? (d) Compute the average rate at which the energy is absorbed during the first second (fhp). Is it reasonable?

Solution:

Page 4 of 97


SECTION 16 – BRAKES AND CLUTCHES W 2 vs1 − vs22 ) ( 2g W = 160,000 lb g = 32.2 fps 2 vs1 = 253 fpm = 4.22 fps

U f = − ∆KE =

vs2 = 0 fps

Uf =

[

]

160,000 (4.22)2 − 02 = 44,245 ft − lb 2(32.2)

fhp (63,000) n 2 2 vs − vs1 0 − (4.22)2 a= 2 = = −0.495 fps 2 2s 2(18) vs − vs1 0 − 4.22 t= 2 = = 8.53 sec − 0.495 a U 44,245 fhp = f = = 9.43 hp 550t 550(8.53) fhp (63,000 ) (9.43)(63,000 ) Tf = = = 68 in − lb on the motor shaft. 1 n (1750) 2

(b) T f =

(c) ∆t =

Uf

Wm c V = π Dbt (rim only) on the motor shaft D = 8 in b = 3.25 in t = 0.5 in V = π (8)(3.25)(0.5) = 40.84 in 3 Wm = ρ V

ρ = 0.253 lb in3 for cast iron c = 101 ft − lb lb − F for cast-iron Wm = (0.253)(40.84) = 10.33 lb 44,245 ∆t = = 42.4o F (10.33)(101)

(d) First second:

Page 5 of 97


SECTION 16 – BRAKES AND CLUTCHES vs1 = 4.22 fps

a = −0.495 fps 2 vs2 = vs1 + at = 4.22 − 0.495(1) = 3.73 fps

U f = − ∆KE = fhp = 885.

Uf 550t

=

[

]

160,000 (4.22)2 − (3.73)2 = 9680 ft − lb 2(32.2)

9680 = 17.6 hp < 25 hp , therefore reasonable. 550(1)

The diagrammatic hoist shown with its load weighs 6000 lb. The drum weighs 8000 lb., has a radius of gyration k = 1.8 ft ; D = 4 ft . A brake on the drum shaft brings the hoist to rest in 10 ft. from vs = 8 fps (down). Only the brake frictional energy is significant, and it can be reasonably assumed that the acceleration is constant. (a) From the frictional energy, compute the average braking torque. (b) If the average fhp/sq. in. is limited to 0.15 during the first second, what brake contact area is needed?

Problems 885, 886 Solution: 63,000 fhp Tf = n U f = − ∆KE1 − ∆KE2 =

I1 2 W ω1 − ω22 ) + 2 vs21 − vs21 ( 2 2g

(

vs1 = 8 fps , vs2 = 0 fps

ω1 =

2vs1

=

2(8) = 4 rad s , ω2 = 0 rad s 4

D W k2 I1 = 1 g W1 = 8000 lb k = 1.8 ft W2 = 6000 lb

Page 6 of 97

)


SECTION 16 – BRAKES AND CLUTCHES g = 32.2 fps 2 I1 2 W 80000(1.8) ( ω1 − ω22 ) + 2 vs21 − vs21 = (4)2 + 60000 (8)2 = 12,400 ft − lb 2 2g 2(32.2) 2(32.2) 2 2 vs2 − vs1

(

Uf = a=

2

)

2s s = 10 ft

0 2 − 82 = −3.2 fps 2 2(10 ) vs − vs1 0 − 8 t= 2 = = 2.5 sec − 3.2 a U 12,400 fhp = f = = 9 hp 550t 550(2.5) 60ω n= rpm 2π 1 ω = (4 rad s − 0) = 2 rad s − 0 2 60(2 ) n= = 19.1 rpm 2π 63,000 fhp 63,000(9 ) Tf = = = 29,700 in − lb n 19.1 a=

(b) fhp sq. in. = 0.15 (first second) vs2 = vs1 + at = 8 − 3.2(1) = 4.8 fps 2(4.8) = 2.4 rad sec D 4 2 80000(1.8) Uf = (4)2 − (2.4)2 + 60000 (8)2 − (4.8)2 = 6106 ft − lb 2(32.2 ) 2(32.2) U 6106 fhp = f = = 11.10 hp 550t 550(1) fhp 11.10 A= = = 74 in 2 fhp sq.in. 0.15

ω2 =

2vs2

=

[

887.

]

[

]

The same as 885, except that a traction drive, arranged as shown, is used; the counterweight weighs 4000 lb. The ropes pass twice about the driving sheave; the brake drum is on this same shaft.

Page 7 of 97



Problem 887. Solution: WT 2 vs − vs22 2g 1 WT = 4000 lb + 6000 lb = 10,000 lb − ∆KE of pulley is negligible vs1 = 8 fps , vs2 = 0 fps

(a) U f = −∆KE =

(

)

10,000 2 (8) = 9,940 ft − lb 2(32.2) vs22 − vs21 0 2 − 82 a= = = −3.2 fps 2 2s 2(10) vs − vs1 0 − 8 t= 2 = = 2.5 sec − 3.2 a U 9940 fhp = f = = 7.23 hp 550t 550(2.5) D = 4 ft 2v 2(8) = 4 rad sec ω1 = s1 = D 4 2v 2(0) = 0 rad sec ω 2 = s2 = D 4 1 1 ω = (ω1 + ω2 ) = (4 + 0) = 2 rad sec 2 2 60ω 60(2 ) n= = 19.1 rpm 2π 2π

Uf =

Page 8 of 97



Braking torque, T f =

63,000 fhp 63,000(7.23) = = 23,850 in − lb n 19.1

(b) fhp sq. in. = 0.15 (first second) vs1 = 8 fps vs2 − vs1 = at vs2 − 8 = −3.2(1) vs2 = 4.8 fps

[

]

10,000 (8)2 − (4.8)2 = 6360 ft − lb 2(32.2) U 6360 fhp = f = = 11.56 hp 550t 550(1) fhp 11.56 Contact area = A = = = 77.1 in 2 fhp sq.in. 0.15 Uf =

SINGLE-SHOE BRAKES 888.

For the single-shoe, short-block brake shown (solid lines) derive the expressions for brake torque for (a) clockwise rotation, (b) counterclockwise rotation. (c) In which direction of rotation does the brake have self-actuating properties? If f = 0.25 , for what proportions of e and c would the brake be self-actuating?

Problems 888 – 891, 893. Solution: (a) Clockwise rotation (as shown)

Page 9 of 97



FD 2 F = fN

Tf =

[∑ M

]

=0 fN e + Wa = N c N c − fN e = Wa Wa N= c − fe fWa F= c − fe fWaD Tf = 2(c − fe ) H

(b) Counter Clockwise Rotation

FD 2 F = fN

Tf =

[∑ M

H

=0

]

Page 10 of 97


SECTION 16 – BRAKES AND CLUTCHES Wa = fN e + N c Wa N= c + fe fWa F= c + fe fWaD Tf = 2(c + fe )

(c) Clockwise rotation is self-actuating c > fe with f = 0.25 c > 0.25e The same as 888, except that the wheel and brake shoe are grooved, 2θ degrees between the sides of the grooves (as in a sheave, Fig. 17.38, Text).

889.

Solution:

[∑ F

V

=0

]

2 N1 sin θ = N F = 2 f N1

Page 11 of 97


SECTION 16 – BRAKES AND CLUTCHES  N  fN F =2f =  2 sin θ  sin θ

(a) Clockwise rotation Wa N= c − fe fWa F= (c − fe )sin θ fWaD Tf = 2(c − fe )sin θ (b) Counter clockwise rotation Wa c + fe fWa F= (c + fe)sin θ fWaD Tf = 2(c + fe ) sin θ N=

(c) Clockwise rotation is self-actuating c > fe with f = 0.25 c > 0.25e

890.

Consider the single-shoe, short-block brake shown (solid lines) with the drum rotating clockwise; let e be positive measured downward and D = 1.6c . (a) Plot the mechanical advantage MA (ordinate) against f values of 0.1, 0.2, 0.3, 0.4, 0.5 (abscissa) when e c has values 2, 0.5, 0, -0.5, -1. (b) If f may vary from 0.3 to 0.4, which proportions give the more nearly constant brake response? Are proportions good? (c) What proportions are best if braking is needed for both directions of rotation?

Solution:

Page 12 of 97



(a) MA =

Tf

, Clockwise rotation Wa fD MA = 2(c − fe ) D = 1.6c 1.6 fc MA = 2(c − fe ) 0.8 f MA = fe   1 −  c  

Tabulation:

f 0.1 0.2 0.3 0.4 0.5

Page 13 of 97

2 0.100 0.267 0.600 1.600 ∞

Values of MA ec 0.5 0 0.084 0.08 0.178 0.16 0.284 0.24 0.400 0.32 0.533 0.40

-0.5 0.076 0.145 0.209 0.267 0.320

-1 0.073 0.133 0.185 0.229 0.267


SECTION 16 – BRAKES AND CLUTCHES Plot:

(b) f = 0.3 to 0.4 , e c = −1 , with MA ≈ constant . They are good because 1 >

fe except e c = 2 . c

(c) e c = 0 is the best if braking is needed for both directions of rotation with MA the same. 891.

A single-block brake has the dimensions: cast-iron wheel of D = 15 in ., 1 3 11 a = 32 in ., c = 9 in ., e = 4 in ., width of contact surface = 2 in. The brake 2 8 16 block lined with molded asbestos, subtends 80o, symmetrical about the center line; it is permitted to absorb energy at the rate of 0.4 hp/in.2; n = 200 rpm . Assume that p is constant, that F and N act at K , and compute (a) pvm and the approximate braking torque, (b) the force W to produce this torque, (c) the mechanical advantage, (d) the temperature rise of the 3/8-in.-thick rim, if it absorbs all the energy with operation as specified, in 1 min. (e) How long could this brake be so applied for ∆t = 400 o F ? See 893.

Solution:

Page 14 of 97



D = 15 in a = 32.5 in c = 9.375 in e = 4.6875 in b = 2 in

(a) Solving for pvm Fvm = fpAvm ft − lb min Fvm = 0.4 hp in 2 A Fvm (0.4 hp )(33,000 ft − lb hp − min ) 13,200 ft − lb min = = A in 2 in 2 Fvm = fpvm A f = 0.35 from Table AT 29, molded asbestos on cast iron Fvm = 13,200 = 0.35 pvm A pvm = 37,700 ft − lb sq in − min Solving for braking torque Fvm = 13,200 ft − lb sq. in. − min A  15  vm = π Dn = π  (200 ) = 785 fpm  12  θDb A= 2  π  θ = (80)  = 1.3963 rad  180 

Page 15 of 97



A=

θDb (1.3963)(15)(2) = = 21 sq. in. 2 2

F (785) = 13,200 21 F = 353 lb FD (353)(15) Tf = = = 2650 in − lb 2 2

(b) Solving for W f Wa c − fe F = 353 lb f = 0.35 a = 32.5 in e = 4.6875 in c = 9.375 in F (c − fe ) (353)[9.375 − (0.35)(4.6875)] = W= = 240 lb fa (0.35)(32.5) F=

(c) Solving for MA MA =

fD (0.35)(15) = = 0.34 2(c − fe ) 2[9.375 − (0.35)(4.6875)]

(d) Solving for ∆t ∆t o F =

U f , ft − lb

Wm c Wm = ρπ Dbt D = 15 in b = 2 in 3 t = in = 0.375 in 8 ρ = 0.253 lb in3 for cast iron Wm = (0.253)(π )(15)(2 )(0.375) = 8.942 lb c = 101 ft − lb lb − F for cast iron

Page 16 of 97


SECTION 16 – BRAKES AND CLUTCHES U f = 550t ′( fhp ) t ′ = 1 min = 60 sec U f = 550(60 )( fhp ) = 33,000 fhp

fhp =

Tf n

=

(2650)(200) = 8.4127 hp

63,000 63,000 U f = 33,000(8.4127 ) = 277,619 ft − lb ∆t =

Uf Wm c

=

277,619 = 310o F (8.942)(101)

(e) Solving for t ′ , (time) with ∆t = 400o F U f = Wm c∆t U f = (8.942 )(101)(400 ) = 361,260 ft − lb 550( fhp )(t ′) = U f

550(8.4127 )(t ′) = 361,260 t ′ = 78 sec = 1.3 min 892.

1 For a single-block brake, as shown, a = 26 in ., c = 7 in ., e = 3.75 in ., 2 1 D = 15 in ., drum contact width b = 3 in . The molded asbestos lining subtends 2 o θ = 60 , symmetrical about the vertical axis; force W = 300 lb .; n = 600 rpm . Assume that p is constant, that F and N act at K , and compute (a) pvm and the braking torque, (b) the energy rate in fhp/in.2 of contact surface. (c) the mechanical advantage, (d) the temperature of the 3/8-in.-thick rim, if it absorbs all the energy with the operation as specified in 1 min. (e) How long could this brake be so applied for ∆t rim = 400o F ? See 894.

Page 17 of 97


SECTION 16 – BRAKES AND CLUTCHES Problems 892, 894. Solution: For greater braking torque, T f , use counterclockwise rotation

[∑ M

]

=0 aW + efN = cN Wa N= c − ef f Wa F= c − ef From Table AT 29, f = 0.35 for molded asbestos W = 300 lb a = 26 in c = 7.5 in e = 3.75 (0.35)(300)(26) = 442 lb F= 7.5 − (3.75)(0.35) A

(a) Solving for pvm Fvm = fpAvm π Dn π (15)(600) vm = = = 2536 fpm 12 12 θDb A= 2  π  θ = 60  = 1.047 rad  180  (1.047 )(15)(3.5) = 27.5 in 2 A= 2 Fvm = (442 )(2536 ) = (0.35)(27.5) pvm pvm = 116,500 ft − lb sq. in. − min

Page 18 of 97


SECTION 16 – BRAKES AND CLUTCHES Solving for the braking torque, FD (442 )(15) Tf = = = 3315 in − lb 2 2 (b) Energy rate, fhp.in2. fhp =

Tf n

=

(3315)(600) = 31.6 hp

63,000 63,000 2 A = 27.5 in 31.6 hp fhp in 2 = = 1.15 hp in 2 27.5 in 2 T 3315 (c) MA = f = = 0.425 Wa (300)(26) U , ft − lb (d) ∆t o F = f Wm c Wm = ρπ Dbt 3 t = in = 0.375 in 8 D = 15 in b = 3.5 in ρ = 0.253 lb in3 for cast iron c = 101 ft − lb lb − F for cast iron

Wm = (0.253)(π )(15)(3.5)(0.375) = 15.648 lb For 1 min U f = 33,000(1)( fhp ) = 33,000(1)(31.6 ) = 1,042,800 ft − lb

∆t =

1,042,800 = 660o F (15.648)(101)

(e) ∆t rim = 400o F U f = (400 )(15.648)(101) = 632,179 ft − lb ′ = t min

Uf 33,000 fhp

Page 19 of 97

=

632,179 = 0.61 min 33,000(31.6)



LONG-SHOE BRAKES FIXED SHOES 893.

The brake is as described in 891 and is to absorb energy at the same rate but the pressure varies as p = P sin θ . Derive the equations needed and compute (a) the maximum pressure, (b) the moment M F H of F about H , (c) the moment M N H of N about H , (d) the force W , (e) the braking torque, (f) the x and y components of the force at H .

Solution:

p = P sin θ = P sin φ D r= 2 dN = pbrdφ dF = fpbrdφ

Page 20 of 97


SECTION 16 – BRAKES AND CLUTCHES T f = ∫ rdF T f = ∫ fpbr 2 dφ T f = fbr 2 P ∫ sin φdφ

T f = fbr 2 P(cos φ1 − cos φ2 ) (a) Solving for P Tf P= 2 fbr (cos φ1 − cos φ2 ) D r= 2 c tan α = r −e c = 9.375 in 15 r = = 7.5 in 2 e = 4.6875 in 9.375 tan α = 7.5 − 4.6875 α = 73.3o θ = 80o 80 θ φ1 = α − = 73.3 − = 33.3o 2 2 80 θ φ2 = α + = 73.3 + = 113.3o 2 2 f = 0.35 b = 2 in r = 7.5 in Tf P= fbr 2 (cos φ1 − cos φ2 ) Tf Tf P= psi = (0.35)(2)(7.5)2 (cos 33.3 − cos113.3) 48.5 63,000 fhp Tf = n fhp = fhp in 2 ( A) θDb A= 2  π  θ = 80  = 1.396 rad  180 

(

Page 21 of 97

)



A=

(1.396)(15)(2) = 21 in 2

2 fhp in 2 = 0.4 fhp = (0.4)(21) = 8.4 hp n = 200 rpm 63,000(8.4 ) Tf = = 2646 in − lb 200 T 2646 = 55 psi = max .P (φ2 > 90o ) P= f = 48.5 48.5

(b) M F H = ∫ (r − R cos φ )dF φ2

MF H = ∫

φ1

(r − R cos φ ) fbrP sin φdφ φ2

M F H = fbrP ∫

φ1

(r sin φ − R sin φ cos φ )dφ φ2

MF H

R   = fbrP − r cos φ − sin 2 φ  2   φ1

R   M F H = fbrP r (cos φ1 − cos φ2 ) − sin 2 φ2 − sin 2 φ1  2  

(

R = c 2 + (r − e ) = 2

(9.375)2 + (7.5 − 4.6875)2

)

= 9.788 in

9.788   M F H = (0.35)(2 )(7.5)(55)7.5(cos 33.3 − cos113.3) − sin 2 113.3 − sin 2 33.3  2   M F H = 1900 in − lb

(

(c) M N

H

= ∫ R sin φdN φ2

MN

H

= ∫ RP sin 2 φbrdφ

MN

H

= brRP ∫ sin 2 φdφ

MN

H

=

φ1

φ2

φ1

brRP φ2 1 − cos 2 φ dφ 2 ∫φ1

(

)

φ

MN

H

2 brRP  1  sin 2 φ φ = −  2  2 φ1

brRP [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] 4 φ2 − φ1 = θ = 1.396 rad 2φ2 = 2(113.3) = 226.6o MN

H

=

2φ1 = 2(33.3) = 66.6o

Page 22 of 97

)



MN

H

MN

H

(d)

(2)(7.5)(9.788)(55) [2(1.396) − (sin 226.6 − sin 66.6)]

=

4 = 8956 in − lb

∑M

H

=0

Wa + M F H − M N

H

=0

a = 32.5 in W (32.5) + 1900 − 8956 = 0 W = 217 lb

(e) T f = 2646 in − lb (f)

∑F

x

=0

− H x − W cos α + ∫ dN sin φ + ∫ dF cos φ = 0 φ2

φ2

φ1

φ1

− H x = W cos α − Pbr ∫ sin 2 φdφ − fPbr ∫ sin φ cos φdφ brP [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] − fbrP sin 2 φ2 − sin 2 φ1 4 2 2 )(7.5)(55) ( − H x = 217 cos 73.3 − [2(1.396) − (sin 226.6 − sin 66.6)] 4 (0.35)(2)(7.5)(55) sin 2 113.3 − sin 2 113.3 − 2 − H x = −931 lb

(

− H x = W cos α −

(

)

H x = 931 lb

∑F

y

=0

− H y + W sin α − ∫ dN cos φ + ∫ dF sin φ = 0 φ2

φ2

φ1

φ1

− H y = ∫ brP sin φ cos φdφ − ∫ fbrP sin 2 φdφ − W sin α fbrP brP sin 2 φ2 − sin 2 φ1 − [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] − W sin α 2 4 (2)(7.5)(55) sin 2 113.3 − sin 2 33.3 − Hy = 2 (0.35)(2)(7.5)(55) [2(1.396) − (sin 226.6 − sin 66.6)] − 217 sin 73.3 − 4 − H y = −305 lb − Hy =

(

)

(

H y = 305 lb

Page 23 of 97

)

)


SECTION 16 – BRAKES AND CLUTCHES 894.

The brake is as described in 892, but the pressure varies as p = P sin φ . Assume the direction of rotation for which a given W produces the greater T f , derive the equations needed, and compute (a) the maximum pressure, (b) the moment of F about A , (c) The moment of N about A , (d) the braking torque, (e) the x and y components of the force at A .

Solution:

p = P sin φ dN = pbrdφ dN = Pbr sin φdφ dF = fdN = fPbr sin φdφ

Solving for φ1 and φ2

Page 24 of 97



tan α =

c r+e

D = 7.5 in 2 7.5 tan α = 7.5 + 3.75 α = 33.69o r=

θ

60 = 3.69o 2 2 60 θ φ1 = α + = 33.69 + = 63.69o 2 2 M F A = ∫ (R cos φ − r )dF

φ1 = α −

φ2

MF A = ∫

φ1

= 33.69 −

(R cos φ − r ) fPbr sin φdφ φ2

M F A = fPbr ∫

φ1

(R sin φ cos φ − r sin φ )dφ

R  M F A = fPbr  sin 2 φ2 − sin 2 φ1 + r (cos φ2 − cos φ1 ) 2 

(

)

R = c 2 + (e + r ) = 2

(7.5)2 + (3.75 + 7.5)2

= 13.52 in

 13.52   2 2 M F A = (0.35)P(3.5)(7.5) (sin 63.69 − sin 3.69 ) + 7.5(cos 63.69 − cos 3.69)  2   M F A = 11.43P MN

A

= ∫ R sin φdN

MN

A

= ∫ RPbr sin 2 φdφ

φ2

φ1

brPR φ2 1 − cos 2 φ dφ 2 ∫φ1 brPR MN A = [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] 4 φ2 − φ1 = θ = 1.047 rad 2φ2 = 2(63.69 ) = 127.38o MN

A

(

=

)

2φ1 = 2(3.69) = 7.38o (3.5)(7.5)P(13.52) [2(1.047 ) − (sin 127.38 − sin 7.38)] MN A = 4 M N A = 126.68 P (a)

∑M

A

=0

Wa + M F A − M N W = 300 lb

Page 25 of 97

A

=0


SECTION 16 – BRAKES AND CLUTCHES a = 26 in (300 )(26) + 11.43P − 126.68P = 0 P = 67.68 psi

max. p = P sin φ2 = 67.68 sin 63.69 = 60.67 psi (b) M F A = 11.43(67.68) = 774 in − lb (c) M N A = 126.68(67.68) = 8575 in − lb (d) T f = ∫ rdF φ2

T f = ∫ fPbr 2 sin φdφ φ1

T f = fPbr 2 (cos φ1 − cos φ2 ) T f = (0.35)(60.68)(3.5)(7.5) (cos 3.69 − cos 63.69) 2

T f = 2587 in − lb

(e)

[∑ F

x

=0

]

− H x − W cos α + ∫ dN sin φ − ∫ dF cos φ = 0 φ2

φ2

φ1

φ1

− H x = W cos α − Pbr ∫ sin 2 φdφ + fPbr ∫ sin φ cos φdφ Pbr [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] + fPbr sin 2 φ2 − sin 2 φ1 4 2 (67.68)(3.5)(7.5) [2(1.047 ) − (sin 127.38 − sin 7.38)] − H x = 300 cos 33.69 − 4 (0.35)(67.68)(3.5)(7.5) sin 2 63.69 − sin 2 3.69 + 2 − H x = −136 lb

(

− H x = W cos α −

(

)

H x = 136 lb

[∑ F

y

=0

]

H y + W sin α − ∫ dN cos φ − ∫ dF sin φ = 0 φ2

φ2

φ1

φ1

H y = Pbr ∫ sin φ cos φdφ + fPbr ∫ sin 2 φdφ − W sin α fPbr Pbr sin 2 φ2 − sin 2 φ1 + [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] − W sin α 2 4 (67.68)(3.5)(7.5) sin 2 63.69 − sin 2 3.69 Hy = 2 ( 0.35)(67.68)(3.5)(7.5) + [2(1.047 ) − (sin 127.38 − sin 7.38)] − 300 sin 33.69 4 Hy =

(

)

(

Page 26 of 97

)

)


SECTION 16 – BRAKES AND CLUTCHES H y = 766 lb

895.

(a) For the brake shown, assume p = P cos α and the direction of rotation for which a given force W results in the greater braking torque, and derive equations for T f in terms of W , f , and the dimensions of the brake. (b) Under what circumstances will the brake be self-acting? (c) Determine the magnitude and location of the resultant forces N and F .

Solution:

(a) Clockwise rotation has greatest braking torque. p = P cos α dN = pbrdα = Pbr cos αdα dF = fdN = fpbrdα = fPbr cos αdα θ2

MF H = ∫

−θ1

θ2

MF H = ∫

−θ1

(r + c sin α )dF (r + c sin α ) fPbr cos αdα

Page 27 of 97


SECTION 16 – BRAKES AND CLUTCHES θ2

MF H = ∫

−θ1

fPbr (r cos α + c sin α cos α )dα θ

2 1   M F H = fPbr r sin α + c sin 2 α  2   −θ1

1   M F H = fPbr r [sin (θ 2 ) − sin (− θ1 )] + c sin 2 (θ 2 ) − sin 2 (− θ1 )  2   1   M F H = fPbr r (sin θ 2 + sin θ1 ) + c sin 2 θ 2 − sin 2 θ1  2  

[

]

(

)

θ2

MN

H

= ∫ cos αdN

MN

H

= ∫ cPbr cos 2 αdα

−θ1

θ2

−θ1

cPbr θ 2 (1 + cos 2α )dα 2 ∫−θ1 cPbr MN H = [2 + sin 2α ]θ−2θ1 4 cPbr MN H = [2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] 4 ∑MH = 0 MN

=

H

[

]

Wa + M F H − M N

H

=0

1   cPbr [2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] Wa + fPbr r (sin θ 2 + sin θ1 ) + c sin 2 θ 2 − sin 2 θ1  = 2 4   Wa P= cbr [2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] − fbr 2r (sin θ 2 + sin θ1 ) + c sin 2 θ 2 − sin 2 θ1 4 2 4Wa P= br c[2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] − 2 f 2r (sin θ 2 + sin θ1 ) + c sin 2 θ 2 − sin 2 θ1

(

)

[

[

{

)]

(

)]}

(

T f = ∫ rdF θ2

Tf = ∫

−θ1

fPbr 2 cos αdα

T f = fPbr 2 [sin α ]−2θ1 θ

T f = fPbr 2 (sin θ 2 + sin θ1 ) 4 fWabr 2 (sin θ 2 + sin θ1 ) br {c[2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] − 2 f 2r (sin θ 2 + sin θ1 ) + c(sin 2 θ 2 − sin 2 θ1 ) } 4 fWar (sin θ 2 + sin θ1 ) Tf = c[2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] − 2 f 2r (sin θ 2 + sin θ1 ) + c (sin 2 θ 2 − sin 2 θ1 ) D where r = e = 2 Tf =

[

[

Page 28 of 97

]

]



[

]

(b) c[2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] > 2 f 2r (sin θ 2 + sin θ1 ) + c (sin 2 θ 2 − sin 2 θ1 ) 4 fr (sin θ 2 + sin θ1 ) c> 2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 ) − 2 f (sin 2 θ 2 − sin 2 θ1 ) (c) N = ∫ dN θ2

N = ∫ Pbr cos αdα −θ1

N = Pbr [sin α ]−2θ1 θ

N = Pbr (sin θ 2 + sin θ1 ) F = fN F = fPbr (sin θ 2 + sin θ1 ) Solving for the location of F and N . Let A = vertical distance from O . θ2

∑M

F Loc .

=∫

∑M

F Loc .

= ∫ P A cos α − r cos 2 α fbrdα

∑M

F Loc .

= Pfbr ∫

∑M

F Loc .

∑M

−θ1

θ2

−θ1

( A − r cos α )dF

(

θ2

−θ1

)

(A cos α − r cos α )dα 2

θ2  1  = Pfbr ∫  A cos α − r (1 + cos 2α ) dα −θ1 2   θ2

F Loc .

 1  1  = Pfbr  A sin α − r  α + sin 2α  2  2  −θ1 

 1  1  = Pfbr [ A(sin θ 2 + sin θ1 )] − r (θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )  2  2   Then ∑ M F Loc . = 0

∑M

F Loc .

[A(sin θ 2 + sin θ1 )] − 1 r (θ 2 + θ1 ) + 1 (sin 2θ 2 + sin 2θ1 ) = 0 2  2  1  1  A(sin θ 2 + sin θ1 ) = r (θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 ) 2  2  1  1  r (θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 ) 2  2  A= (sin θ 2 + sin θ1 ) r [2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] A= 4(sin θ 2 + sin θ1 )

Page 29 of 97



For the brake shown with θ1 ≠ θ 2 , assume that the direction of rotation is such that a given W results in the greater braking torque and that p = P sin φ . (a) Derive equations in terms of θ1 and θ 2 for the braking torque, for the moment M F H and for M N H . (b) Reduce the foregoing equations for the condition

θ1 = θ 2 . (c) Now suppose that θ , taken as θ = θ1 + θ 2 , is small enough that θ sin θ ≈ θ , cos θ ≈ 1 , θ1 = θ 2 = . What are the resulting equations? 2

Solution:

(a) Use clockwise rotation

p = P sin φ dN = Pbr sin φdφ dF = fdN = fPbr sin φdφ φ1 = 90 − θ1 φ2 = 90 + θ 2 T f = ∫ rdF

Page 30 of 97


SECTION 16 – BRAKES AND CLUTCHES φ2

T f = fPbr 2 ∫ sin φdφ φ1

T f = fPbr (cos φ1 − cos φ2 ) 2

T f = fPbr 2 [cos(90 − θ1 ) − cos(90 + θ 2 )] T f = fPbr 2 (sin θ1 + sin θ 2 ) M F H = ∫ (r − c cos φ )dF φ2

M F H = ∫ fbPr (r − c cos φ )sin φdφ φ1

φ2

M F H = fPbr ∫

φ1

(r sin φ − c sin φ cos φ )dφ φ

MF H

2 1   = fPbr − r cos φ − c sin 2 φ  2   φ1

1   M F H = fPbr r (cos φ1 − cos φ2 ) − c sin 2 φ2 − sin 2 φ1  2   1   M F H = fPbr r [cos(90 − θ1 ) − cos(90 + θ 2 )] − c sin 2 (90 + θ 2 ) − sin 2 (90 − θ1 )  2   1   M F H = fPbr r (sin θ1 + sin θ 2 ) − c cos 2 θ 2 − cos 2 θ1  2   1   M F H = fPbr r (sin θ1 + sin θ 2 ) − c 1 − sin 2 θ 2 − 1 − sin 2 θ1  2   1   M F H = fPbr r (sin θ1 + sin θ 2 ) + c sin 2 θ 2 − sin 2 θ1  2  

(

)

[

(

[(

(

MN

H

= ∫ r sinφdN

MN

H

= bPr 2 ∫ sin 2 φdφ

]

)

)]

) (

)

φ2

φ1

2

MN

H

=

MN

H

=

MN

H

=

MN

H

=

MN

H

=

Pbr φ2 (1 − cos 2φ )dφ 2 ∫φ1 Pbr 2 [2 − sin 2φ ]φφ12 4 Pbr 2 [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] 4 Pbr 2 {2[(90 + θ 2 ) − (90 − θ1 )] − [sin 2(90 + θ 2 ) − sin 2(90 − θ1 )]} 4 Pbr 2 [2(θ 2 + θ1 ) − (− sin 2θ 2 − sin 2θ1 )] 4

Page 31 of 97



MN

H

=

Pbr 2 [2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] 4

(b) θ1 = θ 2 T f = fbPr 2 (sin θ1 + sin θ 2 ) T f = 2 fbPr 2 sin θ1 1   M F H = fPbr r (sin θ1 + sin θ 2 ) + c sin 2 θ 2 − sin 2 θ1  2   2 M F H = 2 fPbr sin θ1

(

MN

H

MN

H

MN

H

MN

H

Pbr 2 = [2(θ 2 + θ1 ) − (− sin 2θ 2 − sin 2θ1 )] 4 bPr 2 = (4θ1 + 2 sin 2θ1 ) 4 bPr 2 = (4θ1 + 4 sin θ1 cos θ1 ) 4 = bPr 2 (θ1 + sin θ1 cos θ1 )

(c) θ = θ1 + θ 2 sin θ ≈ θ cos θ ≈ 1

θ1 = θ 2 =

θ 2

T f = 2 fbPr 2 sin θ1 θ  θ  T f = 2 fbPr 2 sin   = 2 fbPr 2   = fbPr 2θ 2 2

MF

H

= 2 fPbr 2 sin θ1

θ  θ  M F H = 2 fbPr 2 sin   = 2 fbPr 2   = fbPr 2θ 2 2

MN

H

MN

H

= bPr 2 (θ1 + sin θ1 cos θ1 ) θ θ  = bPr 2  + (1) = bPr 2θ 2 2 

Page 32 of 97

)



The brake shown is lined with woven asbestos; the cast-iron wheel is turning at 60 rpm CC; width of contact surface is 4 in. A force W = 1300 lb . is applied via linkage systemnot shown; θ = 90o . Let p = P sin φ . (a) With the brake lever as a free body, take moments about the pivot J and determine the maximum pressure and compare with permissible values. Compute (b) the braking torque, (c) the frictional energy in fhp. (d) Compute the normal force N , the average pressure on the projected area, and decide if the brake application can safely be continuous.

Solution: (a)

dF = fdN p = P sin φ dN = pbrdφ = Pbr sin φdφ dF = fPbr sin φdφ M F J = ∫ (R cos φ − r )dF φ2

M F J = fPbr ∫

φ1

Page 33 of 97

(R cos φ − r )sin φdφ


SECTION 16 – BRAKES AND CLUTCHES φ2

M F J = fPbr ∫

φ1

(R sin φ cos φ − r sin φ )dφ φ

2 1  M F J = fPbr  R sin 2 φ + r cos φ  2 φ1

1  M F J = fPbr  R sin 2 φ2 − sin 2 φ1 + r (cos φ2 − cos φ1 ) 2  12.5 tan β = 10 β = 51.34o

(

φ1 = β − θ = 90

)

θ 2

o

90 = 6.34o 2 90 θ φ1 = β + = 51.34 + = 96.34o 2 2 b = 4 in r = 10 in for woven asbestos f = 0.4 (Table At 29)

φ1 = 51.34 −

(12.5)2 + (10)2

R=

= 16 in

1  M F J = fPbr  R sin 2 φ2 − sin 2 φ1 + r (cos φ2 − cos φ1 ) 2  16  M F J = (0.4 )P(4 )(10 ) sin 2 96.34 − sin 2 6.34 + 10(cos 96.34 − cos 6.34 ) 2  M F J = −51.81P

(

)

(

)

M N J = ∫ R sin φdN φ2

M N J = PbrR ∫ sin 2 φdφ φ1

PbrR [1 − cos 2φ ]φφ12 2 PbrR = [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] 4  P(4)(10)(16)   π  = 2(96.34 − 6.34)  − (sin 2(96.34 ) − sin 2(6.34 ))  4  180    = 572.9 P

MN J = MN

J

MN

J

MN

J

∑M

J

= Wa +M F J − M N J = 0

Page 34 of 97



(1300)(25) + (− 51.81P ) − 572.9P = 0 P = 52 psi max. p = P = 52 psi , φ2 > 90 From Table AT 29, permissible p = 50 psi Therefore pmax ≈ p permissible

(b) T f = ∫ rdF φ2

T f = fPbr ∫ sin φdφ φ1

T f = fPbr (cos φ1 − cos φ2 ) T f = (0.4 )(52 )(4 )(10 )(cos 6.34 − cos 96.34 ) = 9188 in − lb

Tf n

(c) fhp =

, n = 60 rpm 63,000 (9188)(60) = 8.75 hp fhp = 63,000

(d) N = ∫ dN φ2

N = Pbr ∫ sin φdφ φ1

N = Pbr (cos φ1 − cos φ2 ) N = (52 )(4 )(10 )(cos 6.34 − cos 96.34 ) = 2297 lb ave. p =

N 2br sin

θ = 90

θ 2

o

ave. p =

2297 2(4 )(10 )sin

90 2

= 40.6 psi

 π Dn  π  pvm = p  = (40.6 ) (20 )(60 ) = 12,755 ft − lb sq. in. − min  12   12 

since pvm < 28,000 ft − lb sq. in. − min (§18.4) Application is continuous.

Page 35 of 97


SECTION 16 – BRAKES AND CLUTCHES PIVOTED-SHOE BRAKES 898.

In the brake shown, the shoe is lined with flexible woven asbestos, and pivoted at point K in the lever; face width is 4 in.; θ = 90o . The cast-iron wheel turns 60 rpm CL; let the maximum pressure be the value recommended in Table At 29. On the assumption that K will be closely at the center of pressure, as planned, compute (a) the brake torque, (b) the magnitude of force W , (c) the rate at which frictional energy grows, (d) the time of an application if it is assumed that all this energy is stored in the 1-in. thick rim with ∆t rim = 350 F , (e) the average pressure on projected area. May this brake be applied for a “long time” without damage? (f) What would change for CC rotation?

Problem 898. Solution: a = 27 in , b = 4 in , n = 60 rpm CL 2 D sin c=

θ 2

θ + sin θ

D = 20 in , r = 10 in

θ = 90o = 1.571 rad 90 2 = 11.0 in c= 1.571 + sin 90 2(20)sin

(a) T f = 2 fPbr 2 sin

θ

2 For woven asbestos, Table AT 29, f = 0.4 P = 50 psi 90 2 T f = 2(0.4 )(50 )(4 )(10 ) sin = 11,314 in − lb 2

Page 36 of 97


SECTION 16 – BRAKES AND CLUTCHES (b)

 θ + sin θ N = Pbr  2  ∑MJ = 0 Wa = 12 N W (15) = 12(2571) W = 2057 lb

[

  1.571 + sin 90   = (50 )(4 )(10 )  = 2571 lb 2   

]

(c) fhp =

Tf n

=

(11,314)(60) = 10.78 hp

63,000 63,000 rate of frictional energy = 33,000 fhp = 33,000(10.78) = 355,740 ft − lb min (d) Time (min) = ∆t o F =

Uf 33,000 fhp

U f ft − lb

Wm c Wm = ρπ Dbt For cast iron ρ = 0.253 lb in3 c = 101 ft − lb lb − F t = 1 in Wm = (0.253)π (20)(4)(1) = 63.6 lb U f ft − lb ∆t o F = 350 = (63.6 lb )(101 ft − lb lb − F ) U f = 2,248,260 ft − lb Time (min) =

2,248,260 = 6.32 min 33,000(10.78) N

(e) Ave. p =

2br sin

Page 37 of 97

θ 2

=

2571 90 2(4 )(10 ) sin 2

= 45.45 psi


SECTION 16 – BRAKES AND CLUTCHES pπ Dn (45.45)(π )(20 )(60 ) = = 14,280 ft − lb sq. in. − F 12 12 since pvm < 28,000 , this brake may be applied for a long time. pvm =

(f) Since the moment arn of F is zero, no change or CC rotation. The pivoted-shoe brake shown is rated at 450 ft-lb. of torque; θ = 90o ; contact width is 6.25 in.; cast-iron wheel turns at 600 rpm; assume a symmetric sinusoidal distribution of pressure. (a) Locate the center of pressure and compute with the location of K. Compute (b) the maximum pressure and compare with allowable value, (c) the value of force W , (d) the reaction at the pin H , (e) the average pressure and pvm , and decide whether or not the application could be continuous at the rated torque. (f) Compute the frictional work from Tω and estimate the time it will take for the rim temperature to reach 450 F (ambient, 100 F).

899.

Problem 899. Solution: 2 D sin (a) c =

θ 2

θ + sin θ

D = 18 in

θ = 90o = 1.571 rad 90 2 = 9.9011 in c= 1.571 + sin 90 but location of K = 9.8125 in then, c ≈ location K 2(18)sin

(b) T f = 2 fPbr 2 sin

θ

2 T f = 450 ft − lb = 5400 in − lb

Page 38 of 97


SECTION 16 – BRAKES AND CLUTCHES b = 6.25 in r = 9 in use f = 0.4 (on cast-iron) T f = 2 fPbr 2 sin

θ 2

90 2 P = 18.86 psi < allowable (Table AT 9)

5400 = 2(0.4 )P(6.25)(9) sin 2

(c) W (20.375) = N (10.375)  θ + sin θ   1.571 + sin 90  N = Pbr   = (18.86 )(6.25)(9 )  = 1364 lb 2 2     1364 )(10.375) ( W= = 695 lb 20.375 (d) H = N − W = 1364 − 695 = 669 lb ↓ N

(e) Ave. p =

2br sin

θ 2

=

1364 90 2(6.25)(9 ) sin 2

= 17.15 psi

n = 600 rpm pπ Dn (17.15)(π )(18)(600 ) pvm = = = 48,490 ft − lb sq. in. − F 12 12

since pvm > 28,000 , not continuous  2π (600 rpm )  (f) Frictional work = Tω = (450 ft − lb )  = 28,275 ft − lb per sec  60 sec min  U ft − lb ∆t o F = f Wm c Wm = ρπ Dbt For cast iron ρ = 0.253 lb in3 c = 101 ft − lb lb − F 

π (18) Wm = (0.253)π (18)(6.25)t +

2

 t  = 154 t 

4  ∆t = 450 − 100 = 350 F U f = ∆tWm c = (350 )(154 t )(101) = 5,443,900 t ft − lb

Page 39 of 97


SECTION 16 – BRAKES AND CLUTCHES U f = 2,248,260 ft − lb

5,443,900 t = 192.5 t sec 28,275 1 Assume t = in 2 Time = 96 sec Time =

TWO-SHOE BRAKES PIVOTED SHOES 900.

The double-block brake shown is to be used on a crane; the force W is applied by a spring, and the brake is released by a magnet (not shown); θ = 90o ; contact width = 2.5 in. Assume that the shoes are pivoted at the center of pressure. The maximum pressure is the permissible value of Table AT 29. Compute (a) the braking torque, (b) the force W , (c) the rate of growth of frictional energy at 870 rpm, (d) the time it would take to raise the temperature of the 0.5-in.-thick rim by ∆t = 300 F (usual assumption of energy storage), (e) pvm . (f) Where should the pivot center be for the calculations to apply strictly?

Problem 900. Solution:

θ

90 2 = 5.5 in 2 = c= π θ + sin θ + sin 90 2 2 D sin

Page 40 of 97

2(10)sin



[∑ M

R1

=0

]

F1 (5.5 − 0.875) + 12.75W = 6.75 N1 fN1 (4.625) + 12.75W = 6.75 N1 12.75W N1 = 6.25 − 4.625 f

[∑ M

R2

=0

]

12.75W = F2 (5.5 − 0.875) + 6.75 N 2 12.75W = 4.625 fN 2 + 6.75 N 2 12.75W N2 = 6.25 + 4.625 f Assume flexible woven asbestos, f = 0.40 , p = 50 psi 12.75W = 2.898W 6.25 − 4.625(0.40) F1 = fN1 = (0.4)(2.898W ) = 1.16W N1 =

Page 41 of 97


SECTION 16 – BRAKES AND CLUTCHES 12.75W = 1.574W 6.25 + 4.625(0.40) F2 = fN 2 = (0.4)(1.574W ) = 0.63W N2 =

max . T f = T f1 2 fPbr 2 sin

θ 2

= F1c 2

90  10  2(0.40 )(50 )(2.5)  sin = (1.16W )(5.5) 2 2 W = 277 lb

(a) Braking torque = T f1 + T f 2 = (F1 + F2 )c = (1.16 + 0.63)(277 )(5.5) = 2727 in − lb (b) W = 277 lb Tf n (2727 )(870) = 37.66 hp (c) fhp = = 63,000 63,000 (d) Solving for tine: U ft − lb ∆t o F = f Wm c ∆t o F = 300o F c = 101 , ρ = 0.253 for cast iron

Wm = ρV V = π Dbt +

π D 2t

= π (10 )(2.5)(0.5) +

π (10)2 (0.5)

4 Wm = (0.253)(78.54) = 19.87 lb

4

= 78.54 in 3

U f = (300)(19.87 )(101) = 602,061 ft − lb

Time =

Uf 33,000 fhp

=

602,061 = 0.4844 min = 29 sec 33,000(37.66)

(e) pvm : π Dn π (10)(870) vm = = = 2278 fpm 12 12 pvm = (50 )(2278) = 113,900 (f) c = 5.5 in 901.

A pivoted-shoe brake, rated at 900 ft-lb. torque, is shown. There are 180 sq. in. of braking surface; woven asbestos lining; 600 rpm of the wheel; 90o arc of brake contact on each shoe. The effect of spring A is negligible. (a) Is the pin for the

Page 42 of 97


SECTION 16 – BRAKES AND CLUTCHES shoe located at the center of pressure? (b) How does the maximum pressure compare with that in Table AT 29? (c) What load W produces the rated torque? (d) At what rate is energy absorbed? Express in horsepower. Is it likely that this brake can operate continuously without overheating? (e) Does the direction of rotation affect the effectiveness of this brake?


θ

90 2 = 9.9 in 2 = (a) c = π θ + sin θ + sin 90 2 13 19 and 16 ≈ 9.9 in , therefore the pin located at the center of pressure 2 2 D sin

(b)

4 13 19 16 o α = 11.4 tan α =

[∑ M

Q

=0

]

4 FA cos α = 8.5W

Page 43 of 97

2(18) sin


SECTION 16 – BRAKES AND CLUTCHES 4 FA cos 11.4 = 8.5W FA = 2.168W ∑ FV = 0 and ∑ FH = 0

[

]

[

]

Qv = FA sin α + W = (2.168W )sin 11.4 + W = 1.429W Qh = FA cos α = (2.168W ) cos 11.4 = 2.125W

[∑ M

R1

=0

]

N1 (10.375) = 20.375Qh N1 (10.375) = 20.375(2.125W ) N1 = 4.173W F1 = f N1 For woven asbestos lining, f = 0.40 , p = 50 psi F1 = (0.40)(4.173W ) = 1.67W (either direction)

Page 44 of 97



[∑ M

=0

R2

]

10.375 N 2 = 20.375 FA cos α 20.375 N2 = (2.168W ) cos11.4 = 4.174W 10.375 F2 = (0.40)(4.174W ) = 1.67W (either direction) T f = (F1 + F2 )c

(900 )(12) = (1.67 + 1.67 )(W )(9.9) W = 326.6 lb T f1 = T f 2 = Fc = 2 fPbr 2 sin

θ 2

but A = θ br Ar br 2 =

θ

(1.67 )(326.6)(9.9) =

2(0.4)(P )(180)(9)sin

90 2

π 2

P = 9.26 psi < 50 psi

(c) W = 326.6 lb (d) fhp =

Tf n

=

(900)(12)(600) = 103 hp

63,000 63,000 π Dn π (18)(600) vm = = = 2827 fpm 12 12 pvm = (9.26)(2827 ) = 26,178 ft − lb sq.in. − F

since pvm < 28,000 , it is likely to operate continuously. (e) Since the value of F is independent of rotation, the direction doesn’t affect the effectiveness of this brake.

902.

Refer to the diagrammatic representation of the brake of Fig. 18.2, Text, and let 9 9 the dimensions be: a = b = m = t = 4 , c = 14 , D = 15 , h = 9 in ., and the 16 16 o contact width is 4 in.; arc of contact = 90 ; lining is asbestos in resin binder, wheel rotation of 100 rpm CC; applied load W = 2000 lb . (a) Locate the center of pressure for a symmetrical sinusoidal pressure distribution and compare with the actual pin centers. Assume that this relationship is close enough for approximate

Page 45 of 97


SECTION 16 – BRAKES AND CLUTCHES results and compute (b) the dimensions k and e if the braking force on each shoe is to be the same, (c) the normal force and the maximum pressure, (d) the braking torque, (e) pvm . Would more-or-less continuous application be reasonable?

Figure 18.2 Solution:

θ

90 2 = 8.25 in 2 = (a) c = π θ + sin θ + sin 90 2 2 D sin

2(15) sin

On Centers: 9 9 + 4 = 9.125 in > c 16 16 9 9 B : a + b = 4 + 4 = 9.125 in > c 16 16 K :t + m = 4

[∑ M

RC

=0

]

Page 46 of 97


SECTION 16 – BRAKES AND CLUTCHES eRF = (e + c )W e+c RF =  W  e  RC = RF − W cW e+c RC =  W − W = e  e 

[∑ M

RH

=0

]

N1h − F1b = RF a N1h − fN1b = RF a R a N1 = F h − fb fRF a F1 = h − fb fa(e + c )W F1 = e(h − fb )

[∑ M

RE

=0

]

N 2 h + F2t = RC k

Page 47 of 97


SECTION 16 – BRAKES AND CLUTCHES N 2 h + fN 2t = RC k R k N2 = C h + ft fR k F2 = C h − ft fkcW F2 = e(h + ft ) (b) T f 1 = T f 2 F1c = F2c F1 = F2 fkcW fa(e + c )W = e(h + ft ) e(h − fb ) kc a(e + c ) = h − fb h + ft For asbestos in resin binder, f = 0.35 , Table AT 29 9 a = 4 in = 4.5625 in 16 9 b = 4 in = 4.5625 in 16 9 m = 4 in = 4.5625 in 16 9 t = 4 in = 4.5625 in 16 c = 14 in 9 h = 9 in = 9.5625 in 16 k (14) 4.5625(e + 14) = 9.5625 − 0.35(4.5625) 9.5625 + 0.35(4.5625) e + 14 = 2.1903k but k + m = e or e = k + 4.5625 then k + 4.5625 + 14 = 2.1903k k = 15.6 in e = 15.6 + 4.5625 = 20.1625 in (c) N = N1 = N 2 =

Page 48 of 97

kcW (15.6)(14)(2000) = = 2720 lb e(h + ft ) (20.1625)[9.5625 − 0.35(4.5625)]


SECTION 16 – BRAKES AND CLUTCHES (d) T f = T f 1 + T f 2 = f ( N1 + N 2 )c = 0.35(2 )(2720 )(8.25) = 15,708 in − lb (e) vm =

π Dn

=

π (15)(100)

= 393 fpm 12 12 pvm = (64.11)(393) = 25,195 ft − lb sq.in. − F

since pvm < 28,000 , continuous application is reasonable. FIXED SHOES 903.

A double-block brake has certain dimensions as shown. Shoes are lined with woven asbestos; cast-iron wheel turns 60 rpm; applied force W = 70 lb . For each direction of rotation, compute (a) the braking torque, (b) the rate of generating frictional energy (fhp). (c) If the maximum pressure is to be P = 50 psi (Table AT 29), what contact width should be used? (d) With this width, compute pvm and decide whether or not the applications must be intermittent.

Problems 903, 904. Solution:

[∑ M

=0 4Q = 26W B

]

Page 49 of 97


SECTION 16 – BRAKES AND CLUTCHES Q = 6.5W

[∑ M

R

=0

]

2.25S = 6Q = 6(6.5W ) S = 17.33W RH = S = 17.33W RV = Q = 6.5W

e = 10 in R = 12.5 in a = 2.25 + 9 + 12.5 = 23.75 in

∑M ∑M M F1 M N1

H

= Sa − M F1

H

−M N1

H

= 0 (CC)

H

= Sa + M F1

H

−M N1

H

= 0 (CL)

R   = fbrP r (cos φ1 − cos φ2 ) − sin 2 φ2 − sin 2 φ1  2   brRP [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] H = 4

(

H

T f1 = 2 fPbr 2 sin

T f1

Pbr =

2 fr sin r = 10 in

Page 50 of 97

θ 2

θ 2

)



2r sin

θ 2

2(10 ) sin

= 11 in

θ

= 11 in 2 θ = 66.43o = 1.165 rad f = 0.4 for woven asbestos

M F1

H

M F1

H

R   fT f1 r (cos φ1 − cos φ2 ) − (sin 2 φ2 − sin 2 φ1 ) 2   = θ 2 fr sin 2 R   T f1 r (cos φ1 − cos φ2 ) − (sin 2 φ2 − sin 2 φ1 ) 2  =  2r sin

θ

2 66.73 = 56.64o = 0.9886 rad φ1 = 90 − = 90 − 2 2 2φ1 = 113.28o 66.73 θ = 123.36o = 2.1530 rad φ2 = 90 + = 90 + 2 2 o 2φ2 = 246.72 φ2 − φ1 = θ 12.5   T f1 10(cos 56.64 − cos 123.36) − sin 2 123.36 − sin 2 56.64  2  =T M F1 H =  f1 66.73 2(10)sin 2 RT f1 [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] M N1 H = θ  4 2 fr sin  2  12.5T f1 [2(1.165) − (sin 246.72 − sin 113.28)] M N1 H = = 2.96T f1 66.73 8(0.4)(10) sin 2

θ

(

CC: ∑ M H = Sa − M F1

H

−M N1

(17.33)(70)(23.75) − T f

1

H

=0

− 2.960T f1 = 0

T f1 = 7276 in − lb

CL: ∑ M H = Sa + M F1 Page 51 of 97

H

−M N1

H

=0

)



(17.33)(70)(23.75) + T f

1

− 2.960T f1 = 0

T f1 = 14,700 in − lb

e = 10 in d = 12.5 in

CC:

[∑ M

H

=0

]

RH a′ − RV d + M F2

CL:

[∑ M

H

=0

H

− M N2

H

=0

H

− M N2

H

=0

]

RH a′ − RV d − M F2

 T f2    = Tf = M H F1 H 2  Tf   1  Tf  M N 2 H = M N1 H  2  = 2.960T f 2  Tf   1 CC: RH a′ − RV d + M F2 H − M N 2 H = 0

M F2

[(17.33)(21.5) − (6.5)(12.5)](70) + T f T f 2 = 10,405 in − lb CL: RH a′ − RV d − M F2 H − M N 2

H

− 2.960T f 2 = 0

2

− 2.960T f 2 = 0

=0

[(17.33)(21.5) − (6.5)(12.5)](70) − T f T f 2 = 5150 in − lb

(a) Braking Torque = T f1 + T f 2

Page 52 of 97

2


SECTION 16 – BRAKES AND CLUTCHES CC: T f = T f1 + T f 2 = 7276 + 10,405 = 17,681 in − lb CL: T f = T f1 + T f 2 = 14,700 + 5150 = 19,850 in − lb (b) Rate of generating frictional energy T n fhp = f 63,000 (17,681)(60) = 16.84 hp CC: fhp = 63,000 (19,850 )(60) = 18.90 hp CL: fhp = 63,000 (c) p = 50 psi T f1 or T f 2 = 2 fPbr 2 sin

CC: b =

T f1

θ 2

10,405

= 4.73 in 66.73 2 fPr sin 2(0.4 )(50 )(10) sin 2 2 T f2 14,700 = = 6.68 in CL: b = 66.73 θ 2 2 2 fPr sin 2(0.4 )(50 )(10) sin 2 2 2

θ

=

2

(d) pvm πDn π (20)(60) vm = = = 314 fpm 12 12 pvm = (50 )(314 ) = 15,700 < 55,000 pvm = (50 )(314 ) = 15,700 < 28,000 application can be continuous or intermittent. 904.

If the brake shown has a torque rating of 7000 lb-in. for counter-clockwise rotation, what braking torque would it exert for clockwise rotation, force W the same?

Solution: CC: Sa − M F1 H − M N1 M F1

H

= T f1

M N1

H

= 2.960T f1

Page 53 of 97

H

=0


SECTION 16 – BRAKES AND CLUTCHES S = 17.33W a = 23.75 in

(17.33W )(23.75) − T f

1

− 2.96T f1 = 0

T f1 = 103.9W RH a′ − RV d + M F2

H

− M N2

H

=0

RH = 17.33W RV = 6.53W a′ = 21.5 in d = 12.5 in M F2

H

= T f2

M N2

H

= 2.960T f 2

(17.33W )(21.5) − (6.5W )(12.5) + T f

2

− 2.960T f 2 = 0

T f 2 = 148.65W T f = T f1 + T f 2 7000 = 103.9W + 148.65W W = 27.7 lb

CL: Sa + M F1

H

− M N1

H

=0

(17.33)(27.7 )(23.75) − T f

1

− 2.96T f1 = 0

T f1 = 5817 in − lb RH a′ − RV d − M F2

H

− M N2

H

=0

[(17.33)(21.5) − (6.5)(12.5)](27.7 ) + T f

2

− 2.960T f 2 = 0

T f 2 = 2038 in − lb T f = T f1 + T f 2 = 5817 + 2038 = 7855 in − lb (CL)

905.

A double-block brake is shown for which θ = 90o , b = 5 in ., n = 300 rpm , rim thickness = ¾ in., and W = 400 lb . The shoes are lined with asbestos in resin binder. Determine the frictional torque for (a) clockwise rotation, (b) counterclockwise rotation. (c) How much energy is absorbed by the brake? Express in horsepower. (d) Will the brake operate continuously without danger of overheating? How long for a ∆t rim = 300 F ? How does pvm compare with Text values?

Page 54 of 97



Problem 905 Solution:

4 4+4 α = 26.565o tan α =

[∑ M

R

=0

]

(Q cos α )(4) = 16W (Q cos 26.565)(4) = 16(400) Q = 1789 lb

RH = Q cos α = 1789 cos 26.565 = 1600 lb RV = Q sin α + W = 1789 sin 26.565 + 400 = 1200 lb

Page 55 of 97



R   M F H = fbrP r (cos φ1 − cos φ2 ) − sin 2 φ2 − sin 2 φ1  2   brRP MN H = [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] 4

(

T f = 2 fPbr 2 sin

MF H

θ 2

R   T f r (cos φ1 − cos φ2 ) − (sin 2 φ2 − sin 2 φ1 ) 2  =  2r sin

MN

H

)

=

θ

2 RT f [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )]

8 fr sin

θ 2

20 = 10 in 2 4 tan β = 12 β = 18.435o

r=

θ = 90o = 1.571 rad 90 θ φ1 = 90 − − β = 90 − − 18.435 = 26.565o = 0.464 rad

2 2 2φ1 = 2(26.565) = 53.13o 90 θ φ2 = 90 + − β = 90 + − 18.435 = 116.565o = 2.034 rad 2 2 2φ2 = 2(116.565) = 233.13o

R = 4 2 + 12 2 = 12.65 in Asbestos in resin binder f = 0.35

Page 56 of 97



MF H

MN

H

12.65   T f 10(cos 26.565 − cos 116.5656) − ( sin 2 116.565 − sin 2 26.565) 2  = 0.6803T =  f 90 2(10) sin 2 (12.65)T f [2(2.034 − 0.464) − (sin 233.13 − sin 53.13)] = = 3.03T f 90 8(0.35)(10) sin 2

(a) Clockwise

[∑ M

H1

=0

]

(Q sin α )(2.5) + (Q cos α )(24) + M F H − M N H = 0 (1789 sin 26.565)(2.5) + (1789 cos 26.565)(24) + 0.6803T f 1

1

1

1

1

T f1 = 17,195 in − lb

[∑ M

H2

=0

]

2.5 RV − 24 RH + M N 2

H2

+ M F2

H2

=0

2.5(1200 ) − 24(1600 ) + 3.03T f 2 + 0.6803T f 2 = 0 T f 2 = 9541 in − lb

Page 57 of 97

− 3.03T f1 = 0



T f = T f1 + T f 2 = 17,195 + 9541 = 26,736 in − lb

(b) Counterclockwise

[∑ M

H1

=0

]

24Q cos α + 2.5Q sin α − M F1

H1

− M N1

H1

=0

(24)(1789 cos 26.565) + (2.5)(1789 sin 26.565) − 0.6803T f

1

T f1 = 10,890 in − lb

[∑ M

H2

=0

]

2.5 RV − 24 RH − M F2

H2

+ M N2

H2

=0

2.5(1200 ) − 24(1600 ) − 0.6803T f 2 + 3.03T f 2 = 0 T f 2 = 15,066 in − lb T f = T f1 + T f 2 = 10,890 + 15,066 = 25,956 in − lb

Page 58 of 97

− 3.03T f1 = 0


SECTION 16 – BRAKES AND CLUTCHES (c) CL: fhp =

Tf n 63,000

=

(26,736)(300) = 127.3 hp

=

(25,956 )(300) = 123.6 hp

63,000

CC: fhp =

Tf n 63,000

(d) vm =

πDn 12

=

63,000

π (20)(300) 12

= 1571 fpm

For p : T f = 2 fPbr 2 sin

θ 2

= T f1 (CL)

17,195 = 2(0.35)(P )(5)(10 ) sin 2

90 2

P = 69.48 psi pvm = (69.48)(1571) = 109,153 > 28,000 the brake operate continuously with danger of overheating.

For time: U ft − lb ∆t o F = f Wm c c = 101 , ρ = 0.253 Wm = ρV V = πDbt +

πD 2t

4 2  3  π (20)  3  3 V = π (20 )(5)  +   = 471.24 in 4 4 4 Wm = ρV = (0.253)(471.24 ) = 119.22 lb U f = Wm c∆t = (119.22 )(101)(300 ) = 3,612,366 ft − lb

Time =

Uf

33,000 fhp Uf 3,612,366 CL: Time = = = 0.886 min = 53 sec 33,000 fhp 33,000(123.6) Uf 3,612,366 CC: Time = = = 0.860 min = 52 sec 33,000 fhp 33,000(127.3) pvm > 28,000 , not good for continuous application.

Page 59 of 97



The double-block brake for a crane has the dimensions: a = 14.3 , b = 2.37 , D = 10 , e = 11.05 , g = 7.1 , h = 12 , j = 6.6 , k = 10.55 , m = 3.5 in ., the width of shoes is 4 in., and the subtended angle is θ = 90o ; wocen asbestos lining. Its rated braking torque is 200 ft-lb. The shoes contact the arms in such a manner that they are virtually fixed to the arms. What force W must be exerted by a hydraulic cylinder to develop the rated torque for (a) counterclockwise rotation, (b) clockwise rotation? Is the torque materially affected by the direction of rotation? (c) Compute the maximum pressure and compare with that in Table AT 29. (Data courtesy of Wagner Electric Corporation.)


b 2.37 = e − c 11.05 − 0.83 α = 13.056o tan α =

Page 60 of 97



[∑ M

]

=0 bQ cos α + cQ sin α = eW (2.37 )(Q cos13.056) + (0.83)(Q sin 13.056) = 14.3W Q = 5.7286W R

RH = Q cos α = 5.7286W cos 13.056 = 5.58W RV = Q sin α − W = 5.7286W sin 13.056 − W = 0.294W

k 5.275 = 6.6 2j o β = 38.63 tan β =

90 θ φ1 = 90 − − β = 90 − − 38.63 = 6.37 o = 0.1112 rad 2 2φ1 = 12.74o

φ2 = 90 +

θ

− β = 90 +

2 2φ2 = 192.74o 2

2

90 − 38.63 = 96.37 o = 1.6820 rad 2 2

k   10.55  2 R =   + j2 =   + (6.6 ) = 8.449 in 2  2  D 10 r= = = 5 in 2 2

Page 61 of 97



MF H

R   T f r (cos φ1 − cos φ2 ) − (sin 2 φ2 − sin 2 φ1 ) 2  =  2r sin

MN

H

=

θ

2 RT f [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )]

8 fr sin

θ

2 For woven asbestos lining, f = 0.40

8.449   T f 5(cos 6.37 − cos 96.37 ) − ( sin 2 96.37 − sin 2 6.37 ) 2  = 0.1985T MF H =  f 90 2(5) sin 2 8.449T f [2(1.682 − 0.1112 ) − (sin 192.74 − sin 12.74 )] MN H = = 2.6755T f 90 8(0.4 )(5) sin 2 (a) CC:

[∑ M

H1

=0

]

RH (12 ) − RV (0.25) − M F1

H1

− M N1

H1

=0

(5.58W )(12) − (0.294W )(0.25) − 0.1985T f

1

T f1 = 23.3W

Page 62 of 97

− 2.6755T f1 = 0



[∑ M

H2

=0

]

12Q cos α + 0.25Q sin α + M F2

H2

− M N2

H2

− 3.5W = 0

12(14.3W ) cos 13.056 + 0.25(14.3W )sin 13.056 − 3.5W = 2.6755T f 2 − 0.1985T f 2 T f 2 = 66.4W T f = T f1 + T f 2 T f = 200 ft − lb = 2400 in − lb 2400 = 23.3W + 66.4W W = 26.8 lb

(b) CL:

Page 63 of 97



[∑ M

H1

=0

]

RH (12 ) − RV (0.25) + M F1

H1

− M N1

H1

=0

(5.58W )(12) − (0.294W )(0.25) + 0.1985T f

1

− 2.6755T f1 = 0

T f1 = 27.0W

[∑ M

H2

=0

]

12Q cos α + 0.25Q sin α − M F2

H2

− M N2

H2

− 3.5W = 0

12(14.3W ) cos 13.056 + 0.25(14.3W )sin 13.056 − 3.5W = 2.6755T f 2 + 0.1985T f 2 T f 2 = 57.2W T f = T f1 + T f 2 2400 = 27.0W + 57.2W W = 28.5 lb

Since W has different values, torque is materially affected by the direction of rotation. (c) T f = 2 fPbr 2 sin

θ

2 For woven asbestos lining, f = 0.40 Use T f = 66.4W = 66.4(26.8) = 1780 in − lb b = 4 in r = 5 in

θ = 90o T f = 1780 = 2(0.4 )P(4 )(5) sin 2

P = 31.47 psi

Page 64 of 97

90 2


SECTION 16 – BRAKES AND CLUTCHES From Table AT 29, pmax = 50 psi 31.47 psi < 50 psi

INTERNAL-SHOE BRAKES 908.

Assuming that the distribution of pressure on the internal shoe shown is given by p = P sin φ , show that the moments M N B , M F B , and TF O of N with respect to B and of F with respect to B and to O are ( b = face width)

MN

B

= (Pbar 2 )[θ − (sin 2φ2 − sin 2φ1 ) 2] ,

[

) ]

(

M F B = fPbr r (cos φ1 − cos φ2 ) − a sin 2 φ2 − sin 2 φ1 2 , TF O = fPbr 2 (cos φ1 − cos φ2 ) .

Problems 908 – 910. Solution: p = P sin φ d (M N B ) = kdN

dN = P sin φ (brdφ ) = Pbr sin φdφ k = a cos(φ − 90) = a sin φ d (M N MN

B

) = (a sin φ )(Pbr sin φ )dφ = Pabr sin φ2

B

= Pabr ∫ sin 2 φdφ = φ1

B

φdφ

Pabr φ2 (1 − cos 2φ )dφ 2 ∫φ1 φ

MN

2

2 Pabr  1 Pabr   (φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 ) = − sin 2 = φ φ    2  2 2  2  φ1 

but φ2 − φ1 = θ Page 65 of 97



MN

B

=

(sin 2φ2 − sin 2φ1 ) Pabr  θ−   2 2  

d (M F B ) = edF dF = fdN = fPbr sin φdφ e = r + a sin (φ − 90) = r − a cos φ d (M F B ) = (r − a cos φ )( fPbr sin φdφ ) = fPbr (r sin φ − a sin φ cos φ )dφ

[

]

φ2

M F B = fPbr − r cos φ − a sin 2 φ φ1

(

)

 a sin 2 φ2 − sin 2 φ1  M F B = fPbr r (cos φ1 − cos φ2 ) −  2  

d (TF O ) = rdF = fPbr 2 sin φdφ TF O = fPbr 2 [− cos φ ]φ12 φ

TF O = fPbr 2 (cos φ1 − cos φ2 ) The same as 908, except that a pressure distribution of p = P cos α is assumed.

909.

[

) ]

(

M N B = Pbr h(2θ + sin 2α 2 + sin 2α1 ) 4 + c sin 2 α 2 − sin 2 α1 2 ,

[

(

)

M F B = fPbr r (sin α 2 + sin α1 ) + h sin 2 α 2 − sin 2 α1 2 − c(2θ + sin 2α 2 + sin 2α1 ) 4 M F O = fPbr 2 (sin α 2 + sin α1 ) . Solution: k = h cos α + c sin α e = r + h sin α − c cos α dN = pbrdα = Pbr cos αdα dF = fdN = fPbr cos αdα dM N

B

= kdN = (h cos α + c sin α )(Pbr cos αdα )

dM N

B

= Pbr h cos 2 α + c sin α cos α dα

MN

(

α2

B

= Pbr ∫

−α1

)

(h cos

2

α + c sin α cos α )dα α2

MN

B

 h(2α + sin 2α ) c sin 2 α  = Pbr  +  4 2  −α 

1

but θ = α1 + α 2

[

]

 h[2(α 2 + α1 ) + sin 2α 2 − sin (− 2α1 )] c sin 2 α 2 − sin 2 (− α1 )  M N B = Pbr  +  4 2  

Page 66 of 97

]



(

)

 h(2θ + sin 2α 2 + sin 2α1 ) c sin 2 α 2 − sin 2 α1  + M N B = Pbr   2 4   dM F B = edF = (r + h sin α − c cos α )( fPbr cos αdα )

(

)

dM F B = fPbr r cos α + h sin α cos α − c cos 2 α dα α2

MF B

 h sin 2 α c(2α + sin 2α )  = fPbr r sin α + −  4 2   −α1

[

]

 h sin 2 α 2 − sin 2 (− α1 ) c[2(α 2 + α1 ) + sin 2α 2 − sin (− 2α1 )] M F B = fPbr r [sin α 2 − sin (− α1 )] + −  2 4  

(

)

 h sin 2 α 2 − sin 2 α1 c(2θ + sin 2α 2 + sin 2α1 )  − M F B = fPbr r (sin α 2 + sin α1 ) +  2 4  

dM F O = rdF = r ( fPbr cos αdα ) = fPbr 2 cos αdα M F O = fPbr 2 [− sin α ]−α2 1 = fPbr 2 [sin α 2 − sin (− α1 )] α

M F O = fPbr 2 (sin α 2 + sin α1 ) The same as 909, except that the α is to be measured from OG , a perpendicular to OB ; limits from − α1 to + α 2 .

910.

Solution: k = a cos α e = r + a sin α dM N

B

= kdN = a cos α (Pbr cos αdα ) = Pbar cos 2 αdα =

Pbar (1 + cos 2α )dα 2

α

MN

B

MN

B

Pbar  2α + sin 2α  2 Pbar [2(α 2 + α1 ) + sin 2α 2 − sin 2(− α1 )] = =   2  2 4  −α1 =

Pbar (2θ + sin 2α 2 − sin 2α1 ) 4

dM F B = edF = (r + a sin α )( fPbr cos αdα ) = fPbr (r cos α + a sin α cos α )dα

[

α2

MF B

]

  a sin 2 α  a sin 2 α 2 − sin 2 (− α1 )  + = fPbr r sin α + = fPbr r [ sin − sin ( − ) ] α α   2 1  2  −α 2    1

(

)

 a sin 2 α 2 − sin 2 α1  M F B = fPbr r (sin α 2 + sin α1 ) +  2   2 dM F O = rdF = r ( fPbr cos αdα ) = fPbr cos αdα

Page 67 of 97


SECTION 16 – BRAKES AND CLUTCHES M F O = fPbr 2 [− sin α ]−α2 1 = fPbr 2 [sin α 2 − sin (− α1 )] α

M F O = fPbr 2 (sin α 2 + sin α1 ) 911.

The following dimensions apply to a two-shoe truck brake somewhat as shown: face b = 5 , r = 8 , h = 5.1 , c = 2.6 , w = u = 6.4 in ., θ = 110 o , φ1 = 15o . Lining is asbestos in rubber compound. For a maximum pressure on each shoe of 100 psi, determine the force Q , and the braking torque for (a) clockwise rotation, (b) counterclockwise rotation. See 908. (Data courtesy of Wagner Electric Corporation.)

Problems 911, 912. Solution: See 908. (sin 2φ2 − sin 2φ1 ) Pbar  MN B = θ−   2 2  

(

)

 a sin 2 φ2 − sin 2 φ1  M F B = fPbr r (cos φ1 − cos φ2 ) − , 2   2 TF O = fPbr (cos φ1 − cos φ2 )

φ1 = 15o , 2φ1 = 30o , φ2 = φ1 + θ = 15 + 110 = 125o , 2φ1 = 250o p = 100 psi b = 5 in r = 8 in

a = h2 + c 2 =

(5.1)2 + (2.6)2

= 5.7245 in

θ = 110o = 1.92 rad For asbestos in rubber compound f = 0.35 Page 68 of 97



(a) Both sides (clockwise rotation)

Q(h + w) + M F B − M N MN

=

B

=0

(100)(5)(5.7245)(8) 1.92 − (sin 250 − sin 30) = 30,224 in − lb

   5.7245 sin 2 125 − sin 2 30 M F B = (0.35)(100 )(5)(8)8(cos 30 − cos 125) − 2  h = 5.1 in , w = 6.4 in B

 

2

2

(

Q(5.1 + 6.4) + 14,436 − 30,224 = 0 Q = 1373 lb TF O = (0.35)(100 )(5)(8) (cos 15 − cos 125) = 17,242 in − lb 2

T f = 2TF O = 2(17,242 ) = 34,484 in − lb

(b) Counterclockwise rotation

Q(h + w) − M N B − M F B = 0

Page 69 of 97

) = 14,436 in − lb  


SECTION 16 – BRAKES AND CLUTCHES Q(5.1 + 6.4) − 30,224 − 14,436 = 0 Q = 3883 lb T f = 2TF O = 2(17,242 ) = 34,484 in − lb 913.

The data are the same as 911, but the shoe arrangement is as shown for this problem. For a maximum pressure on the shoes of 100 psim determine the force Q and TF O for (a) Cl rotation, (b) CC rotation, See 908.

Problem 913. Solution: TF O = T f = fPbr 2 (cos φ1 − cos φ2 ) Pbr =

Tf fr (cos φ1 − cos φ2 )

MN B =

aT f (sin 2φ2 − sin 2φ1 ) = (sin 2φ2 − sin 2φ1 ) Pbar   − − θ θ   2 fr (cos φ − cos φ )  2 2  2 1 2

(

)

 a sin 2 φ2 − sin 2 φ1  M F B = fPbr r (cos φ1 − cos φ2 ) −  2   Tf  a sin 2 φ2 − sin 2 φ1  ( ) − − MF B = r cos cos φ φ 1 2   2 2r (cos φ1 − cos φ2 )   From 911: φ1 = 15o , 2φ1 = 30o , φ2 = φ1 + θ = 15 + 110 = 125o , 2φ1 = 250o θ = 110o = 1.92 rad

(

a = h2 + c 2 = f = 0.35

Page 70 of 97

(5.1)2 + (2.6)2

= 5.7245 in

)



(sin 250 − sin 30) = 1.753T  1.92 − f   2(0.35)(8)(cos 15 − cos 125)  2 Tf  5.7245 sin 2 125 − sin 2 15 ( ) = 8 cos 15 − cos 125 −  28(cos 15 − cos 125)  2 5.7245T f

MN B = MF B

(

(a) CL rotation: Left Side

[∑ M

B

=0

]

Q(h + w) − M N1 B − M F1

B

=0

Q(5.1 + 6.4 ) − 1.753T f1 − 0.43T f1 = 0 T f1 = 5.268Q

Right Side:

[∑ M

B

=0

]

Page 71 of 97

) = 0.43T  

f


SECTION 16 – BRAKES AND CLUTCHES Q(h + w) + M F2 B − M N 2

B

=0

Q(5.1 + 6.4 ) + 0.43T f 2 − 1.753T f 2 = 0 T f 2 = 8.6924Q T f max = T f 2 = 8.6924Q

T f max = fPbr 2 (cos φ1 − cos φ2 ) 8.6924Q = (0.35)(100 )(5)(8) (cos 15 − cos 125) Q = 1984 lb 2

T f1 = 5.268Q = 5.268(1984 ) = 10,452 in − lb T f 2 = 8.6924Q = 8.6924(1984 ) = 17,246 in − lb

Total TF O = T f1 + T f 2 = 10,452 + 17,246 = 27,698 in − lb (b) CC rotation Left Side

[∑ M

B

=0

]

Q(h + w) + M F1 B − M N1

B

=0

Q(5.1 + 6.4 ) + 0.43T f1 − 1.753T f1 = 0 T f1 = 8.6924Q

Right Side:

Page 72 of 97



[∑ M

B

=0

]

Q(h + w) − M F2

B

− M N2

B

=0

Q(5.1 + 6.4 ) − 0.43T f 2 − 1.753T f 2 = 0 T f 2 = 5.268Q Since values are just interchanged Q = 1984 lb Total TF O = 27,698 in − lb as in (a)

914.

A double-shoe internal brake is actuated by an involute cam as shown, where QR is the force on the right shoe at a radius wR and QL is the force on the left shoe at a radius wL . The pressure of each shoe is proportional to the rotation of the shoe about B which is inversely proportional to w ; therefore, the ratio of the maximum pressures is PL PR = wR wL . The dimensions are: face width b = 4 , 9 1 5 5 r = 6 , h = 4 , c = 1 , wL = 9 , wR = 8 in .: for each shoe, θ = 120 o , 16 8 16 16 o φ1 = 30 . The lining is asbestos in rubber compound, Determine the braking torque and forces QR and QL for the maximum permissible pressure for (a) clockwise rotation, (b) counterclockwise rotation.

Page 73 of 97



Problem 914. Solution: (sin 2φ2 − sin 2φ1 ) Pbar  MN B = θ−   2 2  

(

)

 a sin 2 φ2 − sin 2 φ1  M F B = fPbr r (cos φ1 − cos φ2 ) − , 2   2 TF O = fPbr (cos φ1 − cos φ2 ) 2

2

 9   1 a = h + c =  4  + 1  = 4.70 in  16   8  5 8 pL wR = = 16 = 0.8926 pR wL 9 5 16 For asbestos in rubber compound, f = 0.35 , p = 75 psi pR = 75 psi pL = 0.8926(75) = 67 psi 2

2

(a) Clockwise rotation Left Side:

Page 74 of 97



[∑ M

BL

=0

QL wL − M FL

M FL

] BL

− M NL

BL

=0

(

BL

)

 a sin 2 φ2 − sin 2 φ1  = fPLbr r (cos φ1 − cos φ2 ) −  2  

φ1 = 30o 2φ1 = 60o φ2 = θ + φ1 = 120 + 30 = 150o 2φ2 = 300o θ = 120o = 2.094 rad

(

 4.70 sin 2 150 − sin 2 30 ( )( )( )( ) ( ) 0 . 35 67 4 6 6 cos 30 cos 150 = − − BL  2  (sin 2φ2 − sin 2φ1 ) P bar  M NL BL = L θ−   2 2   (67 )(4)(4.7 )(6) 2.094 − (sin 300 − sin 60) = 11,185 in − lb M NL BL =   2 2  

M FL

 5 QL  9  − 5849 − 11,185 = 0  16  QL = 1829 lb

T( F O )L = fPL br 2 (cos φ1 − cos φ2 ) T( F O )L = (0.35)(67 )(4 )(6 ) (cos 30 − cos 150 ) = 5849 in − lb 2

Right side:

Page 75 of 97

) = 5849 in − lb  



[∑ M

BR

=0

]

QR wR + M FR

M FR

BR

M NR

BR

= =

− M NR

BR

M FL

BL

PR

PL M NL

BL

PL

PR

=

BR

=0

(5849)(75) = 6547 in − lb

=

67

(11,185)(75) = 12,520 in − lb 67

 5 QR  8  + 6547 − 12,520 = 0  16  QR = 719 lb T( F O )L PR (5849)(75) T( F O )R = = = 6547 in − lb PL 67

T( F O ) = T( F O )R + T( F O )L = 6547 + 5849 = 12,396 in − lb (b) Counterclockwise rotation Left side:

Page 76 of 97



[∑ M

BL

=0

]

QL wL + M FL

BL

− M NL

BL

=0

 5 QL  9  + 5849 − 11,185 = 0  16  QL = 573 lb T( F O )L = 5849 in − lb

Right Side:

[∑ M

BR

=0

]

QR wR − M FR

BR

− M NR

BR

=0

 5 QR  8  − 6547 − 12,520 = 0  16  QR = 2294 lb T( F O )R = 6547 in − lb

Page 77 of 97



T( F O ) = T( F O )R + T( F O )L = 6547 + 5849 = 12,396 in − lb

BAND BRAKES 915.

The steel band for the brake shown is lined with flexible asbestos and it is expected tha the permissible pressure of Table AT 29 is satisfactory; θ = 245o , 1 a = 20 in ., m = 3 in ., D = 18 in ., and face width b = 4 in .; rotation CL. The 2 cast-iron wheel turns 200 rpm. Set up suitable equations, use the average f given and compute (a) the force in each end of the band, (b) the brake torque and fhp. (c) Determine the mechanical advantage for the limit values of f in Table AT 29 and its percentage variation fron that for the average f . (d) Investigate the overheating problem using relevant information given in the Text.


(1)

F1 = e fθ F2

[∑ M

Fixed point

Wa = F2 m Wa (2) F2 = m Page 78 of 97

=0

]


SECTION 16 – BRAKES AND CLUTCHES (3) F = F1 − F2 = fpA FD (4) T f = 2 From Table AT 29, flexible asbestos Ave. f = 0.40 , p = 50 psi (a) For F1 and F2 : θDb A= 2 θ = 245o = 4.276 rad (4.276)(18)(4) = 154 in 2 A= 2 F = F1 − F2 = fpA = (0.40 )(50 )(154 ) = 3080 lb F1 = e fθ = e (0.40 )(4.276 ) = 5.5312 F2 F1 = 5.5312 F2 F = 5.5312 F2 − F2 = 3080 F2 = 680 lb F1 = 5.5312(680 ) = 3760 lb (b) T f and fhp FD (3080 )(18) = = 27,720 in − lb 2 2 Tf n (27,720 )(200) = 88 hp fhp = = 63,000 63,000 (c) For MA T T MA = f = f Wa F2 m FD Tf = 2 F F2 = fθ e −1  FD    D e fθ − 1 2   MA = = 2m  Fm   fθ   e −1  D = 18 in m = 3.5 in θ = 4.276 rad Tf =

(

Page 79 of 97

)


SECTION 16 – BRAKES AND CLUTCHES Limit values (Table AT 29) f = 0.35 to 0.45 . f = 0.35

[

]

[

]

18 e (0.35 )(4.276 ) − 1 = 8.914 2(3.5) f = 0.45

MA =

18 e (0.45 )(4.276 ) − 1 MA = = 15.042 2(3.5) with f = 0.40 (average)

[

]

18 e (0.40 )(4.276 ) − 1 = 11.652 2(3.5) Percentage variation from f = 0.40 . f = 0.35 11.652 − 8.914 % var = (100% ) = 23.5% 11.652 f = 0.45 15.042 − 11.652 % var = (100% ) = 29.1% 11.652 MA =

(d) Overheating problem fhp 88 = = 0.57 fhp in 2 A 154 Therefore, a problem of overheating is expected as Rasmussen recommends 0.2 to 0.3 fhp per square inch of brake contact area.

916.

(a) For the band brake shown, derive the expressions for the braking torque in terms of W , etc., for CL rotation and for CC rotation, and specify the ratio c b for equal effectiveness in both directions of rotation. Are there any proportions of b and c as shown that would result in the brae being self locking? (b) When θ = 270o , a = 16 in ., b = c = 3 in ., and D = 12 in ., it was found that a force W = 50 lb . Produced a frictional torque of 1000 in-lb. Compute the coefficient of friction.

Page 80 of 97



Problem 916. Solution: (a) CL:

[∑ M

O

=0

]

aW = F1b + F2c F1 = F2 e fθ aW = F2e fθ b + F2 c aW F2 = fθ be + c aWe fθ F1 = F2 e fθ = fθ be + c aWe fθ − aW aW (e fθ − 1) F = F1 − F2 = = be fθ + c be fθ + c

Page 81 of 97



Tf =

FD WaD  e fθ − 1    = 2 2  be fθ + c 

CC:

[∑ M

O

=0

]

aW = F2b + F1c WaD  e fθ − 1    Tf = 2  ce fθ + b  No proportions of b and c as shown that would result in the brake being self-locking. (b) W = 50 lb T f = 1000 in − lb D = 12 in a = 16 in b = c = 3 in θ = 270o = 4.7124 rad T f = 1000 =

(50 )(16)(12)  2

e fθ − 1   3e fθ + 3   

e fθ − 1 = 0.625 e fθ + 1 e fθ = e 4.7124 f = 4.333 f = 0.311 917.

(a) For the brake shown, assume the proper direction of rotation of the cast-iron wheel for differential acion and derive expressions for the braking torque. (b) Let 3 D = 14 in ., n = 1 in ., m = 4 in ., θ = 235o , and assume the band to be lined with 4 woven asbestos. Is there a chance that this brake will be self-acting? If true, will

Page 82 of 97


SECTION 16 – BRAKES AND CLUTCHES it always be for the range of values of f given in Table AT 29? (c) The ratio n m should exceed what value in order for the brake to be self-locking? (d) If the direction of rotation of the wheel is opposite to that taken in (a), what is the braking torque with a force W = 10 lb . at a = 8 in .? (e) Suppose the brake is used as a stop to prevent reverse motion on a hoist. What is the frictional horsepower for the forward motion if the wheel turns 63 rpm?

Problems 917, 918. Solution: (a) Assume CL

[∑ M

O

=0

]

Wa + F1n = F2 m F1 = F2 e fθ Wa = F2 m − F2 ne fθ = F2 (m − ne fθ ) Wa F2 = (m − ne fθ )

Page 83 of 97



F1 =

Wae fθ m − ne fθ

(

)

Wa (e fθ − 1) F = F1 − F2 = , Braking force. m − ne fθ FD WaD  e fθ − 1   , Braking torque.  = Tf = 2 2  m − ne fθ  (b) D = 14 in 3 n = 1 in 4 m = 4 in θ = 235o = 4.10 rad Table AT 29, woven asbestos f = 0.35 to 0.45 There is a chance of self-acting if ne fθ > m m = 4 in use f = 0.40 ne fθ = 1.75e(0.40 )(4.10 ) = 9.0 > m use f = 0.35 ne fθ = 1.75e(0.35 )(4.10 ) = 7.35 > m use f = 0.45 ne fθ = 1.75e (0.45 )(4.10 ) = 11.07 > m Therefore true for the range of values of f . (c) ne fθ > m , f = 0.40 (average) n 1 > fθ m e 1 n > (0.4 )(4.10 ) m e n > 0.2 m

(d) For CC:

Page 84 of 97



[∑ M

O

=0

]

Wa + F2 n = F1m Wa = F1m − F2 n F1 = F2 e fθ Wa = F2 e fθ m − F2 n = F2 (me fθ − n ) Wa F2 = me fθ − n Wae fθ = F1 me fθ − n Wa (e fθ − 1) F = F1 − F2 = me fθ − n FD WaD  e fθ − 1    = Tf = 2 2  me fθ − n  Tf =

2

(e) fhp = 918.

e (0.40 )(4.10 ) − 1   4e (0.40 )(4.10 ) − 1.75  = 123.3 in − lb  

(10)(8)(14)  Tf n 63,000

=

(123.3)(63) = 0.1233 hp 63,000

A differential band brake similar to that shown and lined with woven asbestos, has the dimensions: D = 18 in ., n = 2 in ., m = 12 in ., θ = 195o . (a) Is there a chance that this brake will be self-acting? (b) If W = 30 lb . and a = 26 in . , compute the maximum braking torque and the corresponding mechanical advantage. (c) What is the ratio of the braking torque for CL rotation to the braking torque for CC rotation? (d) A 1/16-in.-thick steel band, SAE 1020 as rolled, carries the asbestos lining. What should be its width for a factor of safety of 8, based on the ultimate stress? What should be the face width if the average pressure is 50 psi?

Solution:

Page 85 of 97


SECTION 16 – BRAKES AND CLUTCHES (a) For CL: ne fθ > m θ = 195o = 3.4 rad m = 12 in n = 2 in f = 0.4 2e 0.4(3.4 ) = 7.8 < m , not self-acting

For CC: n > me fθ me fθ < n 12e 0.4(3.4 ) = 46.8 > n , not self-acting Therefore, there is no change that this brake will be self-acting. (b) T f max = T f (CL) 0.4 (3.4 ) fθ −1   Wad  e − 1   (30)(26)(18)   e   Tf =  =   fθ  0 .  12 − 2e 4(3.4 )  = 4832 in − lb 2  2  m − ne     Tf 4832 MA = = = 6.2 Wa (30)(26)

(c) T f (CL ) = 4832 in − lb 0.4 (3.4 ) fθ −1   Wad  e − 1   (30)(26)(18)   e  T f (CC ) =  =   0.4(3.4 )  = 454 in − lb fθ    2 − 2  2  me − n    12e T f (CL ) 4832 = = 10.64 Ratio = T f (CC ) 454

(d) For SAE 1020, as rolled. su = 65 ksi s 65 s= u = = 8.125 ksi = 8125 psi N 8 F s= 1 bt 1 t = in = 0.0625 in 16 Wae fθ max. F1 = (CL) m − ne fθ (30)(26)e 0.4(3.4 ) = 722.3 lb F1 = 12 − 2e 0.4(3.4 ) Page 86 of 97


SECTION 16 – BRAKES AND CLUTCHES 722.3 b(0.0625) b = 1.422 in With p = 50 psi F = fpA θbD A= 2 Wa (e fθ − 1) (30)(26)(e 0.4(3.4 ) − 1) max. F = = = 536.9 lb m − ne fθ 12 − 2e 0.4(3.4 ) fpθbD F= 2 (0.4)(50)(3.4)(b )(18) 536.9 = 2 b = 0.88 in s = 8125 =

919.

A differential band brake is to be design to absorb 10 fhp at 250 rpm. (a) Compute the maximum and minimum diameters from both equations (z) and (a), p. 495, Text. Decide on a size. (b) The band is to be lined with woven asbestos. The Rasmussen recommendation (§18.4) will help in deciding on the face width. Also check the permissible pressure in Table AT 29. Choose dimensions of the lever, its location and shape and the corresponding θ . Be sure the brake is not self locking. What is the percentage variation of the mechanical advantage from the minimum value ( f min ) for the f limits in Table AT 29?

Solution: 63,000hp 63,000(10 ) Tf = = = 2520 in − lb n 250 (a) Eq. (z) 1

1

1

1

 Tf =   5

 3  2520  3  =   = 7.96 in   5 

 Tf Dmax =   4 Eq. (a)

 3  2520  3  =   = 8.57 in   4 

Dmin

1

1

1

1

Dmin = (60 fhp )3 = [60(10)]3 = 8.44 in Dmin = (80 fhp )3 = [80(10)]3 = 9.28 in use D = 8.5 in (b) By Rasmussen Energy absorption capacity = 0.2 to 0.3 fhp per sq. in. of brake contact area.

Page 87 of 97


SECTION 16 – BRAKES AND CLUTCHES Say 0.25 fhp per sq. in. θ bD A= 2 fhp = 10 hp D = 8.5 in assume θ = 180o = π rad fhp fhp in 2 = A 10 0.25 =  π b(8.5)   2  b = 3 in From Table AT 29, f = 0.40 , p per . = 50 psi F fA π (3)(8.5) A= = 40 in 2 2 2T 2(2520) F= f = = 593 lb D 8.5 593 p= = 37.1 psi < 50 psi (OK) 0.4(40) p=

For MA : Tf D e fθ − 1 MA = = WA 2 c − be fθ Not self-locking c > be fθ c > e fθ b c > e 0.4π b c > 3.5 b c say = 4 or c = 4b b

(

(

) )

For f = 0.40 T D e fθ − 1 8.5 e 0.4π − 1 21.96 MA = f = = = fθ 0.4π WA 2 c − be 2 4b − be b For f = 0.35 = f min

(

Page 88 of 97

(

) ( ) (

) )



(

) ( ) (

) )

D e fθ − 1 8.5 e 0.35π − 1 8.54 = = fθ 0.35π b WA 2 c − be 2 4b − be 21.96 − 8.54 % variation = (100%) = 157% 8.54 MA =

Tf

=

(

DISK CLUTCHES 920.

An automobile engine develops its maximum brake torque at 2800 rpm when the bhp = 200. A design value of f = 0.25 is expected to be reasonable for the asbestos facing and it is desired that the mean diameter not exceed 8.5 in.; permissible pressure is 35 psi. Designing for a single plate clutch, Fig. 18.10, Text, determine the outer and inner diameters of the disk.

Solution: 1 Dm = (Do + Di ) = 8.5 in 2 ro + ri = 8.5 in 63,000hp 63,000(200 ) Tf = = = 4500 in − lb n 2800 p = 35 psi f N (ro + ri ) Tf = 2 (0.25)(N )(8.5) 4500 = 2 N = 4235 lb N ave. p = 2 π (ro − ri 2 ) 4235 35 = π (ro2 − ri 2 ) ro2 − ri 2 = 38.5

ro = 8.5 − ri

(8.5 − ri )2 − ri2 = 38.5 72.25 − 17 ri + ri 2 − ri 2 = 38.5

ri = 1.985 in say ri = 2.0 in ro = 8.5 − 2.0 = 6.5 in Do = 2ro = 2(6.5) = 13 in Di = 2ri = 2(2.0) = 4 in

Page 89 of 97



An automobile engine can develop a maximum brake torque of 2448 in-lb. Which of the following plate clutches, which make up a manufacturer’s standard 7 1 “line,” should be chosen for this car? Facing sizes: (a) Do = 8 , Di = 6 in ., (b) 8 8 1 1 1 Do = 10 , Di = 6 in ., (c) Do = 11 , Di = 6 in . In each case, assume f = 0.3 . 8 16 8 The unit pressures are (a) 34 psi, (b) 30 psi, and (c) 26.2 psi.

Solution: f N (ro + ri ) 2 f N (Do + Di ) Tf = 4 π   N = pave   Do2 − Di2 4 π pave Do2 − Di2 (Do + Di ) Tf = 16

Tf =

(

)

(

)

(a) Do = 8.875 in ; Di = 6.125 in , p = 34 psi , f = 0.3 Tf =

π pave (Do2 − Di2 )(Do + Di )

16 π (0.3)(34 )(8.875)2 − (6.125)2 (8.875 + 6.125) Tf = = 1239 in − lb 16

[

]

(b) Do = 10 in ; Di = 6.125 in , p = 30 psi , f = 0.3 Tf =


16 π (0.3)(30 )(10)2 − (6.125)2 (10 + 6.125) Tf = = 1780 in − lb 16

[

]

(c) Do = 11.0625 in ; Di = 6.125 in , p = 26.2 psi , f = 0.3 Tf =


16 π (0.3)(26.2 )(11.0625)2 − (6.125)2 (11.0625 + 6.125) Tf = = 2251 in − lb 16

[

use (c)

Page 90 of 97

]



A single-disk clutch for an industrial application, similar to that in Fig. 18.11, Text, except that there are two disks attached to one shaft and one attached to the other. The clutch is rated at 50 hp at 500 rpm. The asbestos-in-resin-binder facing 1 3 has a Do = 8 in . and Di = 4 in . What must be the axial force and average 2 4 pressure? How does this pressure compare with that recommended by Table AT 29?

Solution: n = 2 pairs in contact f = 0.35 (Table AT 29) p = 75 psi 63,000hp 63,000(50 ) Tf = = = 6300 in − lb nm 500 Do = 8.5 in Di = 4.75 in nf N (ro + ri ) Tf = 2 nf N (Do + Di ) Tf = 4 (2)(0.35)(N )(8.5 + 4.75) 6300 = 4 N = 2717 lb 4N 4(2717 ) ave. p = = = 69.6 psi < 75 psi 2 2 π (Do − Di ) π (8.5)2 − (4.75)2

[

923.

]

A multiple-disk clutch similar to Fig. 18.11, Text, is rated at 22 hp at 100 rpm. The outside and inside diameters of the disks are 14 and 7 ½ in., respectively. If f = 0.25 , find (a) the axial force required to transmit the rated load, and (b) the unit pressure between the disks.

Solution: (a) Fig. 18-11, n = 4 pairs in contact 63,000hp 63,000(22 ) Tf = = = 13,860 in − lb nm 100 nf N (Do + Di ) Tf = 4 Do = 14 in Di = 7.5 in 4(0.25)( N )(14 + 7.5) 13,860 = 4

Page 91 of 97


SECTION 16 – BRAKES AND CLUTCHES N = 2579 lb

(b) p = 924.

4N 4(2579 ) = = 23.5 psi 2 2 π Do − Di π (14)2 − (7.5)2

(

)

[

]

A multiple-disk clutch for a machine tool operation has 4 phosphor-bronze driving disks and 5 hardened-steel driven disks. This clutch is rated at 5.8 hp at 100 rpm when operated dry. The outside and inside diameters of the disks are 5 ½ and 4 3/16 in., respectively. (a) If the pressure between the disks is that recommended for metal on metal in Table AT 29, what coefficient of friction is required to transmit the rated power? (b) What power may be transmitted for f and p as recommended in Table AT 29?

Solution: Do = 5.5 in Di = 4.1875 in 63,000hp 63,000(5.8) Tf = = = 3654 in − lb nm 100 n = 8 pairs in contact (a) Table AT 29, p = 150 psi , metal to metal nf N (Do + Di ) Tf = 4 π  π  2 2 N = p (Do2 − Di2 ) = 150  (5.5) − (4.1875) = 1498 lb 4 4 (8)( f )(1498)(5.5 + 4.1875) T f = 3654 = 4 f = 0.126

[

]

(b) from Table AT 29, p = 150 psi , f = 0.2 (8)(0.2)(1498)(5.5 + 4.1875) = 5805 in − lb Tf = 4 T n (5805)(100) = 9.2 hp hp = f m = 63,000 63,000 925.

A multiple-disk clutch with three disks on one shaft and two on the other, similar to that in Fig. 18.11, Text, is rated at 53 hp at 500 rpm. (a) What is the largest value of Di if f and p are given by Table AT 29 for asbestos in resin binder and Do = 10.5 in . (b) For the diameter used of Di = 7 in .,what is the required axial force and the average pressure?

Solution: Page 92 of 97


SECTION 16 – BRAKES AND CLUTCHES Table AT 29, asbestos in resin binder, f = 0.3 , p = 75 psi nf N (ro + ri ) Tf = 2 nf N (Do + Di ) Tf = 4 4T f N= nf (Do + Di ) but π  N = p (Do2 − Di2 ) 4 π  4T f = nfp  Do2 − Di2 (Do + Di ) 4 63,000hp 63,000(53) Tf = = = 6678 in − lb nm 500 n = 4 pairs in contact π  4T f = nfp  Do2 − Di2 (Do + Di ) 4 π  2 4(6678) = (4 )(0.3)(75)  (10.5) − Di2 (10.5 + Di ) 4 Di = 9.5607 in

(

)

(

)

[

]

(b) Di = 7 in nf N (Do + Di ) Tf = 4 4 )(0.3)( N )(10.5 + 7 ) ( 6678 = 4 N = 1272 lb 4N 4(1272 ) ave. p = = = 26.44 psi 2 2 π (Do − Di ) π (10.5)2 − (7 )2

[

]

MISCELLANEOUS CLUTCHES AND BRAKES 926.

For the cone brake shown, find an expression for the braking torque for a given applied force W on the bell crank. Consider the force F ′ , Fig. 18.12, Text, in obtaining the expression.

Page 93 of 97



Problems 926-928. Solution: 927.

For the cone brake similar to that shown, certain dimensions are: Dm = 15 in ., 1 c = 2 in ., α = 12o , b = 9 in ., and a = 20 in . The contact surfaces are metal and 2 asbestos. (a) For an applied force W = 80 lb ., what braking torque may be expected of this brake? Consider the resistance F ′ , Fig. 18.12, Text. (b) If the rotating shaft comes to rest from 300 rpm during 100 revolutions, what frictional work has been done? (c) What must be the diameter of the steel pin P , SAE 1020 as rolled, for a factor fo safety of 6 against being sheard off? The diameter 1 of the hub d = 4 in . (d) What is the unit pressure on the face of the brake? 2

Solution: f Dm R f Dm aW = 2(sin α + f cos α ) 2b(sin α + f cos α ) Table AT 29, asbestos on metal, f = 0.40 (0.40)(15)(20)(80) = 890 in − lb Tf = 2(9)(sin 12 + 0.4 cos12)

(a)

Tf =

2π (300 ) = 31.42 rad sec 60 ω2 = 0 rad sec θ = 100(2π ) = 628.3 rad 1 θ = (ω1 + ω2 )t 2 1 628.3 = (31.42 + 0 )t 2

(b) ω1 =

Page 94 of 97


SECTION 16 – BRAKES AND CLUTCHES t = 40 sec 1 nm = (300 + 0 ) = 150 rpm 2 T n (890)(150) = 2.119 hp fhp = f m = 63,000 63,000 U f = 550( fhp )(t ) = 550(2.119 )(40 ) = 46,618 ft − lb

(c) For SAE 1020, as rolled, ssu = 49 ksi s 49 ss = su = = 8.17 ksi N 6 4R ss = πd2 aW (20 )(80 ) R= = = 177.8 lb b 9 4(177.8) ss = 8170 = πd2 d = 0.1665 in 3 say d = in 16 N π Dm c R 177.8 N= = = 297 lb sin α + f cos α sin 12 + 0.4 cos12 297 p= = 2.52 psi π (15)(2.5)

(d) p =

928.

A cone clutch for industrial use is to transmit 15 hp at 400 rpm. The mean diameter of the clutch is 10 in. and the face angle α = 10o ; let f = 0.3 for the cast-iron cup and the asbestos lined cone; permissible p = 35 psi . Compute (a) the needed axial force, (b) the face width, (c) the minimum axial force to achieve engagement under load.

Solution: 63,000hp 63,000(15) Tf = = = 2362.5 in − lb n 400 f Dm R (a) T f = 2(sin α + f cos α )

Page 95 of 97



2362.5 =

(0.3)(10)R

2(sin 10 + 0.3 cos10) R = 739 lb

N π Dm c R 739 N= = = 1575 lb sin α + f cos α sin 10 + 0.3 cos10 1575 35 = π (15)c c = 1.44 in

(b) p =

(c) max. f = 0.4 (Table AT 29) f Dm R Tf = 2(sin α + f cos α ) (0.4)(10)R 2362.5 = 2(sin 10 + 0.4 cos10) R = 670 lb , minimum. 929.

An “Airflex” clutch, Fig. 18.15, Text, has a 16-in drum with a 5-in. face. This clutch is rated at 110 hp at 100 rpm with an air pressure of 75 psi. What must be the coefficient of friction if the effect of centrifugal force is neglected? (Data courtesy of Federal Fawick Corporation.)

Solution: D = 16 in b = 5 in hp = 110 hp rpm = 100 rpm p = 75 psi 63,000hp 63,000(110 ) Tf = = = 69,300 in − lb n 100 FD Tf = 2 F (16 ) 69,300 = 2 F = 8662.5 lb N = p(π Db ) = (75)(π )(16)(5) = 18,850 lb F 8662.5 f = = = 0.46 N 18,850

Page 96 of 97



930.

The same as 929 except that the diameter is 6 in., the face width is 2 in., and the rated horsepower is 3.

Solution: 63,000hp 63,000(3) Tf = = = 1890 in − lb n 100 FD Tf = 2 F (6 ) 69,300 = 2 F = 630 lb N = p(π Db ) = (75)(π )(6)(2 ) = 2827 lb F 630 f = = = 0.22 N 2827 - end -

Page 97 of 97


SECTION 18 – MISCELLANEOUS PROBLEMS THIN SHELLS, EXTERNAL PRESSURE 981.

A closed cylindrical tank is used for a steam heater. The inner shell, 200 in. outside diameter and 50 ft. long, is subjected to an external pressure of 40 psi. The material is equivalent to SA 30 (ASME Pressure-Vessel Code: min. su = 55 ksi ); assume an elastic limit of s y = su 2 ; let N = 5 . (a) What thickness of shell is needed from a stress standpoint? (b) For this thickness, what must be the maximum length of unsupported section to insure against collapse? (c) Choose a spacing L to give a symmetric arrangement and determine the moment of inertia of the steel stiffening rings. (d) For a similar problem, the Code recommends that t ≥ 0.76 in , L = 50 in , and I = 96 in 4 . How do these values check with those obtained above? (e) Without stiffening rings, what thickness would be needed?

Solution: (a) Solving for the thickness of shell,

t=

pc D 2s y

pc = 5 p = 5(40 ) = 200 psi s 55,000 sy = u = = 27,500 psi 2 2 D = 200 in p D (200 )(200 ) t= c = = 0.73 in 2s y 2(27,500 ) say t = 0.75 in (b) Solving for the maximum length of unsupported section, use Eq. (20-1) 5

 t 2 2.60 E   D pc = psi 1 2 L  t  − 0.45  D D 5

 t 2 1 2.60 E   L D  t 2  = + 0.45  D pc D

E = 30× 106 psi pc = 200 psi t = 0.75 in

Page 1 of 25


SECTION 18 – MISCELLANEOUS PROBLEMS D = 200 in 5

 0.75  2 1 2.60 30 × 106   . 75 L 0 2  200  + 0.45 =   200 200  200  L = 72.68 in

(

)

(c) Solving for the moment of inertia of the steel stiffening rings. Choosing L = 60 in for 50 ft long shell 0.035 D 3 Lpc 0.035(200) (60)(200) = = 112 in 4 E 30 ×10 6 3

I=

(d) For t ≥ 0.76 in - above minimum L = 50 in - below maximum I = 96 in 4 - lighter than above. (e) L = 50 ft = 600 in Solving for thickness without stiffening rings By Saunders and Windenburg, Eq. 20-1 5

pc =

 t 2 2.60 E   D

L  t  − 0.45  D D

1 2

psi

1   2 L t    pc − 0.45   5  D D   t 2   =   2.60 E D 1   2 600 t    200 − 0.45   5  200 200     t 2     = 6 30 10 200 2 . 60 ×  

(

5 2

137.886t = 600 − 6.364t 5 2

1 2

)

1 2

137.886t + 6.364t = 600 t = 1.791 in

Page 2 of 25


SECTION 18 – MISCELLANEOUS PROBLEMS

say t = 1 982.

13 in 16

The same as 981, except that p = 175 psi , D = 4 ft , and the length of the tank is 18 ft.

(a) Solving for the thickness of shell, t=

pc D 2s y

pc = 5 p = 5(175) = 875 psi 55,000 s sy = u = = 27,500 psi 2 2 D = 4 ft = 48 in p D (875)(48) t= c = = 1.53 in 2s y 2(27,500) 9 say t = 1 = 1.5625 in 16 (b) Solving for the maximum length of unsupported section, use Eq. (20-1) 5

pc =

 t 2 2.60 E   D

L  t  − 0.45  D D

1 2

psi

5

 t 2 1 2.60 E   2 L t D     + 0.45 =   D pc D

E = 30× 106 psi pc = 875 psi t = 1.5625 in D = 48 in 5

 1.5625  2 1 2.60 30 × 10   L 48   1.5625  2  = + 0.45  48 875  48  L = 822 in

(

6

)

(c) Since length of shell = 18 ft = 216 in < 822 in, there is no need for stiffeners.

Page 3 of 25


SECTION 18 – MISCELLANEOUS PROBLEMS (d) L = 216 in Solving for thickness without stiffening rings By Saunders and Windenburg, Eq. 20-1 5

 t 2 2.60 E   D psi pc = 1 L  t 2 − 0.45  D D 1   2 L t    p 45 − 0 .    c 5  D D   t 2   =   2.60 E D 1   2 216 t    875 − 0.45   5  48  48    t 2     = 6 30 10 2 . 60 48 ×  

(

)

5 2

4886.428t = 3937.5 − 56.833t 5 2

1 2

1 2

4886.428t + 56.833t = 3937.5 t = 0.9122 in 15 say t = in 16 19 but minimum t = 1 in 16 19 use t = 1 in 16 Approximate ratio of weight of this shell to the weight of the shell found in (a) = thickness of shell without stiffening rings / thickness of shell with stiffening rings = 0.34375 / 0.09375 = 3.6667 STEEL TUBES, EXTERNAL PRESSURE 983.

A closed cylindrical tank, 6 ft in diameter, 10 ft long, is subjected to an internal pressure of 1 psi absolute. The atmospheric pressure on the outside is 14.7 psi. The material is equivalent to SA 30 (ASME Pressure Vessel Code: min su = 55 ksi ); assume an elastic limit of s y = su 2 ; let pc p = 5 . (a) What thickness of shell is needed for the specified design stress? (b) For this thickness, what must be the maximum length of unsupported section to insure against collapse? (c) Choose a symmetric spacing L of stiffening rings, and compute

Page 4 of 25


SECTION 18 – MISCELLANEOUS PROBLEMS their moment of inertia and the cross-sectional dimensions h and b if they are rectangular with h = 2b . (d) Suppose that the tank had no stiffening rings. What thickness of shell would be needed? What is the approximate ratio of the weight of the shell found in (a)? Material costs are roughly proportional to the weight. Solution: (a) Solving for the thickness of shell pD t= c 2s y pc = 5 p = 5(14.7 − 1) = 68.5 psi s 55,000 sy = u = = 27,500 psi 2 2 D = 6 ft = 72 in p D (68.5)(72) t= c = = 0.08967 in 2s y 2(27,500) 3 = 0.09375 in say t = 32 (b) Solving for the maximum length of unsupported section, use Eq. (20-1) 5

 t 2 2.60 E   D pc = psi 1 2 L  t  − 0.45  D D 5

 t 2 1 2.60 E   2 L t D     + 0.45 =   D pc D

E = 30× 106 psi pc = 68.5 psi t = 0.09375 in D = 72 in 5

 0.09375  2 1 2.60 30 × 10   L 72   0.09375  2  = + 0.45  72 68.5  72  L = 6.2 in

(

6

)

(c) For the length of shell = 10 ft = 120 in. use L = 6 .0 in

Page 5 of 25


SECTION 18 – MISCELLANEOUS PROBLEMS Moment of inertia of stiffening rings. 0.035 D 3 Lpc 0.035(72) (6)(68.5) I= = = 0.1790 in 4 E 30 ×10 6 Solving for cross-sectional dimension h = 2b bh 3 8b 4 2b 4 I= = = = 0.1790 in 4 12 12 3 b = 0.72 in 3 say b = in 4 1 3 h = 2  = 1 in 2 4 3

(d) L = 10 ft = 120 in Solving for thickness without stiffening rings By Saunders and Windenburg, Eq. 20-1 5

 t 2 2.60 E   D pc = psi 1 L  t 2 − 0.45  D D 1   2 L t    pc − 0.45   5  D D   t 2     = 2.60 E D 1   120 t  2   68 . 5 − 0 . 45   5  72  72    t 2     = 6 30 10 72 2 . 60 ×  

(

5 2

25.8865t = 1.667 − 0.053t 5 2

1 2

)

1 2

25.8865t + 0.053t = 1.667 t = 0.3314 in 11 say t = in = 0.34375 in 32

Page 6 of 25


SECTION 18 – MISCELLANEOUS PROBLEMS 984.

A long lap-welded steel tube, 8-in. OD, is to withstand an external pressure of 120 psi. with N = 5 . (a) What should be the thickness of the wall of the tube? (b) What is the ratio D t ? Is it within the range of the Stewart equation? (c) Assuming the internal pressure to be negligible relative to the external pressure, calculate the maximum principal stress from equation (8.13), p. 255, text. What design factor is given by this stress compared with s y for AISI C1015 annealed? (d) Compute the stress from the thin shell formula.

Solution: (a) Solving for the thickness of the wall Stewart’s formula 3

 t  pc = 50,200,000  D pc = Np = (5)(120) = 600 psi 3

t 600 = 50,200,000  8 t = 0.1829 in 3 say t = in = 0.1875 in 16 D 8 = = 42.67 t 0.1875 or D = 42.67t outside the range of the Steward equation ( D < 40t )

(b) ratio

(c) Using eq. (8.13) (Lame’s formula)

pi ri 2 − po ro2 + ri 2 ro2 ( pi − po ) r 2 ro2 − ri 2 po = 120 psi pi ≈ 0 OD 8 ro = = = 4 in 2 2 ri = 4 − 0.1875 = 3.8125 in r = ri

σt =

− po ro2 + ro2 (0 − po ) − 2 po ro2 − 2(120)(4 ) = 2 2 = 2 = −2621 psi 2 2 ro − ri ro − ri (4) − (3.8125)2 s y of AISI C1015, annealed = 42 ksi 2

σt =

Page 7 of 25



Design factor, N y =

sy

σt

=

42,000 = 16 2621

(d) Solving for the stress from the thin shell formula

s=

pD (120)(8) = 2560 psi = 2t 2(0.1875)

985.

A long lap-welded steel tube, 3 –in. OD, is to withstand an external pressure of 150 psi with N = 5 . Parts (a) – (c) are the same as in 984.

Solution: (a) Solving for the thickness of the wall Stewart’s equation 3

 t  pc = 50,200,000  D pc = Np = (5)(150) = 750 psi 3

t 750 = 50,200,000  3 t = 0.074 in 5 say t = in = 0.078125 in 64 D 3 = = 38.4 t 0.078125 or D = 38.4t within the range of the Steward equation ( D < 40t )

(b) ratio

(c) Using eq. (8.13) (Lame’s formula)

po = 150 psi OD 3 ro = = = 1.5 in 2 2 ri = 1.5 − 0.078125 = 1.421875 in − 2 po ro2 − 2(150)(1.5) = = −2957 psi 2 2 ro − ri (1.5)2 − (1.421875)2 s y of AISI C1015, annealed = 42 ksi 2

σt =


Page 8 of 25

sy

σt

=

42,000 = 14.2 2927



A long lap-welded tube, 3-in. OD, is made of SAE 1015, annealed. Let the shell thickness t = D 40 and N = 5 . (a) What is the corresponding safe external pressure? (b) Compute the maximum principal stress (p. 255, Text), assuming a negligible internal pressure. What design factor is given by this stress compared with s y ? (c) Compare with stress computed from the thin-shell formula.

Solution: (a) Solving for safe external pressure,  t  pc = 50,200,000  D t 1 = D 40

3

3

 1  pc = 50,200,000  = 784 psi  40  p 784 p= c = = 157 psi N 5

(b) Solving for maximum principal stress, neglecting internal pressure po = 157 psi OD 3 ro = = = 1.5 in 2 2 D 3 t= = = 0.075 in 40 40 ri = 1.5 − 0.075 = 1.425 in − 2 po ro2 − 2(157 )(1.5) = = −3221 psi 2 2 ro − ri (1.5)2 − (1.425)2 s y of SAE 1015 annealed = 42 ksi 2

σt =


sy

σt

=

42,000 = 13.0 3221

(c) Solving for stress from the thin-shell formula s=

pD (157 )(3) = = 3140 psi 2t 2(0.075)

FLAT PLATES 987.

A circular plate 24 in. in diameter and supported but not fixed at the edges, is subjected to a uniformly distributed load of 125 psi. The material is SAE 1020, as

Page 9 of 25


SECTION 18 – MISCELLANEOUS PROBLEMS rolled, and N = 2.5 based on the yield strength. Determine the thickness of the plate. Solution: Solving for the thickness of the plate 2

r s = p  psi t for SAE 1020,a s rolled, s y = 48 ksi

sy

48,000 = 19,200 psi N 2.5 p = 125 psi 24 r= = 12 in 2 s=

=

 12  s = 19,200 = 125   t  t = 0.968 in say t = 1 in

988.

2

The cylinder head of a compressor is a circular cast-iron plate (ASTM class 20), mounted on a 12-in. cylinder in which the pressure is 250 psi. Assuming the head to be supported but not fixed at the edges, compute its thickness for N = 6 based on ultimate strength.

Solution: Solving for the thickness of the head 2

r s = p  psi t for cast-iron (ASTM class 20), su = 20 ksi s 20,000 s= u = = 3333 psi N 6 p = 250 psi 12 r= = 6 in 2 2

6 s = 3333 = 250  t t = 1.6432 in 21 say t = 1 in = 1.65625 in 32

Page 10 of 25



A 10x15-in. rectangular opening in the head of a pressure vessel, whose internal pressure is 175 psi, is covered with a flat plate of SAE 1015, annealed. Assuming the plate to be supported at the edges, compute its thickness for N = 6 based on ultimate strength.

Solution: Solving for the thickness of the head a 2b 2 p psi s= 2 2 2t (a + b 2 ) for SAE 1015, annealed, su = 56 ksi s 56,000 s= u = = 9333 psi N 6 a = 10 in b = 15 in p = 175 psi s = 9333 =

(10)2 (15)2 (175) psi 2 2 2t 2 [(10) + (15) ]

t = 0.8056 in 13 say t = in = 0.8125 in 16

CAMS 990.

The force between a 5/8-in. hardened steel roller and a cast-iron (140 BHN) cam is 100 lb.; radius of cam curvature at this point is 1 ¼ in. Compute the contact width.

Solution: Solving for the contact width K cb P= 1 1 N  +   r1 r2  From Table 20-2, hardened steel and a cast-iron, BHN = 400 Use K c2 = 900 P = 100 lb 1 5 r1 =   = 0.3125 in 28 1 r2 = 1 = 1.25 in 4 N = 1.15

Page 11 of 25


SECTION 18 – MISCELLANEOUS PROBLEMS 900b 1   1 + 1.15   0.3125 1.25  b = 0.511 in

100 =

A radial cam is to lift a roller follower 3 in. with harmonic motion during a 150o turn of the cam; 1 1.2-in. roller of hardened steel. The reciprocating parts weigh 10 lb., the spring force is 175 lb., the external force during the lift is 250 lb. The cast-iron (225 BHN) cam turns 175 rpm. The cam curvature at the point of maximum acceleration is 1 ½-in. radius. Compute the contact width.

991.

Solution: Neglecting frictional forces Q + Fg + Fs + Fre = P cos φ Q = external force during lift = 250 lb Fg = weight of reciprocating parts = 10 lb

Fs = spring force = 175 lb Fre = reversed effective force = − qma Fre = −2ma for harmonic motion 2

π  L  πω   cos θ  a = &x& =  2 β  β  L = 3 in 2π (175) = 18.326 rad sec ω= 60 150 β = 150o = π = 2.618 rad 180 at maximum acceleration 2

3  π (18.326 )  &x& =  = 726 in sec 2  2  2.618  P cos φ = Q + Fg + Fs + Fre = 250 + 10 + 175 − 2m&x&

10 lb 10 lb − sec 2 = in g (32.2 ft sec 2 )(12 in ft ) 386.4 assume cos φ ≈ 1  10  P = 435 − 2 (726 ) = 397 lb  386.4  m=

Fg

=

Page 12 of 25


SECTION 18 – MISCELLANEOUS PROBLEMS Solving for the contact width K cb P= 1 1 N  +   r1 r2  From Table 20-2, hardened steel and a cast-iron, BHN = 225 Use K c2 = 2100 1 r1 = 1 in = 1.5 in 2 1 r2 = 1 in = 1.5 in 2 N = 1.15 2100b 397 = 1   1 + 1.15   1.5 1.5  b = 0.29 in

992.

The same as 991, except that the motion of the follower is cycloidal.

Solution: Fre = −1.1ma for cycloidal motion As a continuation of 991, but 2π Lω 2  2π  a = &x& = sin  θ  for cycloidal β2  β  L = 3 in 2π (175) = 18.326 rad sec ω= 60 150 β = 150o = π = 2.618 rad 180 at maximum acceleration 2 2π Lω 2 2π (3)(18.326 ) &x& = = = 924 in sec 2 2 β2 (2.618)  10  P = 435 − 1.1 (924 ) = 409 lb  386.4  Solving for the contact width K cb P= 1 1 N  +   r1 r2  From Table 20-2, hardened steel and a cast-iron, BHN = 225 Use K c2 = 2100

Page 13 of 25


SECTION 18 – MISCELLANEOUS PROBLEMS 1 r1 = 1 in = 1.5 in 2 1 r2 = 1 in = 1.5 in 2 N = 1.15 2100b 409 = 1   1 + 1.15   1.5 1.5  b = 0.30 in

993.

The same as 991, except that the motion of the follower is parabolic.

Solution: Fre = −3ma for parabolic motion As a continuation of 991, but 2

 2ω   for parabolic a = &x& = L  β  L = 3 in 2π (175) = 18.326 rad sec ω= 60 150 β = 150o = π = 2.618 rad 180 at maximum acceleration 2

 2(18.326 )  &x& = (3) = 588 in sec 2   2.618   10  P = 435 − 3 (588) = 389 lb  386.4 

Solving for the contact width K cb P= 1 1 N  +   r1 r2  From Table 20-2, hardened steel and a cast-iron, BHN = 225 Use K c2 = 2100 1 r1 = 1 in = 1.5 in 2 1 r2 = 1 in = 1.5 in 2 N = 1.15

Page 14 of 25


SECTION 18 – MISCELLANEOUS PROBLEMS 2100b 1   1 + 1.15   1.5 1.5  b = 0.284 in

389 =

FLYWHEELS AND DISK 994.

A cast-iron flywheel with a mean diameter of 36 in. changes speed from 400 rpm to 380 rpm while it gives up 8000 ft-lb of energy. What is the coefficient of fluctuation, the weight, and the approximate sectional area of the rim?

Solution: Solving for coefficient of fluctuation n −n Cf = 1 2 n n1 + n2 n= 2 2(n1 − n2 ) Cf = n1 + n2 n1 = 400 rpm n2 = 380 rpm 2(400 − 380 ) Cf = = 0.0513 400 + 380 Solving for the weight 32.2∆KE w= C f vs2 ∆KE = 8000 ft − lb C f = 0.0513 vs =

π Dn

12(60) D = 36 in 400 + 380 n= = 390 rpm 2 π (36)(390) vs = = 61.26 fps 12(60) 32.2(8000 ) w= = 1338 lb 2 0.0513(61.26 )

Page 15 of 25


SECTION 18 – MISCELLANEOUS PROBLEMS Solving for the approximate sectional area of the rim w = ρV assume ρ = 0.254 lb in3 for cast iron V = π DA D = 36 in w = ρπ DA 1338 = 0.254π (36)A A = 36 in 2 995.

The energy required to shear a 1-in. round bar is approximately 1000 ft-lb. In use, the shearing machine is expected to make a maximum of 40 cutting strokes a minute. The frictional losses should not exceed 15 % of the motor output. The shaft carrying the flywheel is to average 150 rpm. (a) What motor horsepower is required? (b) Assuming a size of flywheel and choosing appropriate C f , find the mass and sectional dimensions of the rim of a cast-iron flywheel. The width of the rim is to equal the depth and is not to exceed 3 ½ in. It would be safe to assume that all the work of shearing is supplied by the kinetic energy given up by the flywheel.

Solution: (a) Solving for the horsepower required

(Energy required )(Strokes per min ) (1 − Frictional losses )(33,000 ft − lb hp − min ) (1000)40 hp = = 1.426 hp (1 − 0.15)(33,000 ft − lb hp − min )

hp =

(b) Solving for the mass of the rim and size of section 32.2∆KE C f vs2 w = ρV w=

assume ρ = 0.254 lb in3 for cast iron V = π DA

w = ρπ DA =

32.2∆KE C f vs2

assume C f = 0.06 (Table 20-3) ∆KE = 1000 ft − lb

Page 16 of 25



vs =

π Dn

12(60) n = 150 rpm w = ρπ DA =

D3 =

32.2∆KE

 5π D  Cf    24  32.2∆KE

2

2

 5π  Cf   ρπ A  24  using width = depth = 3 ½ in A = (3.5)(3.5) = 12.25 in 2 32.2(1000 ) D3 = 2  5π  0.06  (0.254 )(π )(12.25)  24  D = 50.42 in assume D = 51 in 5π ( 51) vs = = 33.38 fps 24 32.2(1000 ) w= = 482 lb 2 0.06(33.38) w 482 m= = = 15 slugs g 32.2 482 w A = (depth )(width ) = = = 11.84 in 2 ρπ D 0.254π (51) depth = width = 11.84 in 2 = 3.44 in 1 say depth = width = 3 in 2

996.

The same as 995, except that the capacity of the machine is such as to cut 1 ½-in. round brass rod, for which the energy required is about 400 ft-lb./sq. in. of section.

Solution: (a) Solving for the horsepower required hp =

(Energy required )(Strokes per min ) (1 − Frictional losses )(33,000 ft − lb hp − min )

Page 17 of 25


SECTION 18 – MISCELLANEOUS PROBLEMS π  2 Energy required = 400 ft − lb in 2  (1.5 in ) = 707 ft − lb 4 (707 )(40) hp = = 1.01 hp (1 − 0.15)(33,000 ft − lb hp − min )

(

)

(b) Solving for the mass of the rim and size of section 32.2∆KE D3 = 2  5π  Cf   ρπ A  24  ∆KE = 707 ft − lb C f = 0.06

ρ = 0.254 lb in3 A = (3.5)(3.5) = 12.25 in 2 32.2(707 ) D3 = 2  5π  0.06  (0.254 )(π )(12.25)  24  D = 45 in use D = 45 in 32.2∆KE w= C f vs2 5π D 5π (45) vs = = = 29.45 fps 24 24 32.2(707 ) w= = 438 lb 2 0.06(29.45) w 438 m= = = 13.6 slugs g 32.2 w 438 A = (depth )(width ) = = = 12.20 in 2 ρπ D 0.254π (45) depth = width = 12.20 in 2 = 3.49 in 1 say depth = width = 3 in 2

997.

A 75-hp Diesel engine, running at 517 rpm, has a maximum variation of output of energy of 3730 ft-lb. The engine has three 8 x 10 ½ in. cylinders and is directly connected to an a-c generator. (a) What should be the weight and sectional area of the flywheel rim if it has an outside diameter of 48-in.? (b) The actual flywheel and generator have Wk 2 = 6787 lb − ft 2 . Compute the corresponding coefficient of fluctuation and compare.

Page 18 of 25


SECTION 18 – MISCELLANEOUS PROBLEMS Solution: (a) Solving for the weight ad sectional areas w=

32.2∆KE lb C f vs2

assume C f = 0.0035 , Table 20-3 ∆KE = 3730 ft − lb π Dn vs = 12(60) n = 570 rpm assume D = 48 in π (48)(517 ) vs = = 108.28 rps 12(60) 32.2(3730 ) w= = 2927 lb (0.0035)(108.28)2 w = ρπ DA w A= ρπ D

assume ρ = 0.254 lb in3 (cast iron) 2927 A= = 76.42 in 2 (0.254)(π )( 48) (b) Solving for coefficient of fluctuation Ig = Wk 2 = 6787 lb − ft 2 I (ω12 − ω22 ) ft − lb 2 Wk 2 (ω12 − ω22 ) ft − lb ∆KE = g 2

∆KE =

∆KE =

Wk 2 (ω1 − ω2 ) (ω1 + ω2 ) ft − lb g 2

(ω1 + ω2 ) = ω

2 2π (517 ) ω= = 54.14 rad sec 60 ∆KE = 3730 ft − lb 6787 ∆KE = 3730 = (ω1 − ω2 )54.14 32.2 ω1 − ω2 = 0.327

Page 19 of 25



Cf =

998.

ω1 − ω2 0.327 = = 0.006 > 0.0035 ω 54.14 A 4-ft flywheel, with an rim 4 in. thick and 3 in. wide, rotates at 400 rpm. If there are 6 arms, what is the approximate stress in the rim? Is this a safe stress? At what maximum speed should this flywheel rotate if it is made of cast iron, class 30?

Solution: Solving for the approximate stress, vs2 ρ s= psi 144 g o π Dn π (4)(400) = = 83.78 fps vs = 60 60 ρ = 0.254 lb in3 (class 30, cast iron) ρ = 0.254 lb in 3 = (0.254 )(1728)lb ft 3 = 439 lb ft 3 g o = 32.2 ft s 2

(83.78)2 (439) = 665 psi 144(32.2) since v = (83.78)(60) fpm = 5027 s=

fpm < 6000 fpm (cast iron)

this is a safe stress Solving for maximum speed, max. v = 6000 fpm v = π Dn = π (4)n = 6000 fpm maximum, n = 477 rpm 999.

A hollow steel shaft with Do = 6 in and Di = 3 in rotates at 10,000 rpm. (a) What is the maximum stress in the shaft due to rotation? Will this stress materially affect the strength of the shaft? (b) The same as (a), except that the shaft is solid.

Solution: (a) Solving for maximum stress st =

ρω 2

[(3 + µ )r

o

2

4go where D 6 ro = o = = 3 in 2 2 Di 3 ri = = = 1.5 in 2 2 Page 20 of 25

+ (1 − µ )ri

2

] psi


SECTION 18 – MISCELLANEOUS PROBLEMS for steel, ρ = 0.284 lb in3 , µ = 0.30 g o = 386 in sec  10,000   = 1047 rad sec  60 

ω = 2π  st =

ρω 2 4go

[(3 + µ )r

2

o

+ (1 − µ )ri

2

] psi

2 0.284)(1047 ) ( st = (3 + 0.3)(3)2 + (1 − 0.3)(1.5)2 ] = 6306 psi [ 4(386)

(does not affect the strength of the shaft) (b) Solving for the maximum stress for solid st =

ρω 2 (3 + µ )ro 2

2 2 0.284 )(1047 ) (3 + 0.3)(3) ( = 4(386 )

4go

= 2994 psi

1000. A circular steel disk has an outside diameter Do = 10 in and an inside diameter Di = 2 in . Compute the maximum stress for a speed of (a) 10,000 rpm, (b) 20,000 rpm. (c) What will be the maximum speed without danger of permanent deformation if the material is AISI 3150, OQT at 1000 F? Solution: st =

ρω 2

[(3 + µ )r

o

2

4go where ρ = 0.284 lb in3 µ = 0.30 D 10 ro = o = = 5 in 2 2 D 2 ri = i = = 1 in 2 2 g o = 386 in sec

+ (1 − µ )ri

2

] psi

(a) Solving for maximum stress for a speed of 10,000 rpm  10,000  ω = 2π   = 1047 rad sec  60  2 0.284)(1047 ) ( st = (3 + 0.3)(5)2 + (1 − 0.3)(1)2 = 16,776 psi 4(386)

[

Page 21 of 25

]


SECTION 18 – MISCELLANEOUS PROBLEMS (b) Solving for maximum stress for a speed of 10,000 rpm  20,000  ω = 2π   = 2094 rad sec  60  2 0.284)(2094) ( st = (3 + 0.3)(5)2 + (1 − 0.3)(1)2 = 67,104 psi 4(386 )

[

]

(c) Solving for maximum speed, ω . For AISI 3150, OQT at 1000 F, s y = 130 ksi st = s y =

ρω 2 4go

130,000 =

[(3 + µ )r

o

2

+ (1 − µ )ri

2

] psi

(0.284)ω 2 [(3 + 0.3)(5)2 + (1 − 0.3)(1)2 ] 4(386)

ω = 2914.57 rad sec 60ω 60(2914.57 ) = = 27,832 rpm RPM = 2π 2π 1001. The same as 1000, except that Di = 1 in . Solution: Di = 1 in D 1 ri = i = in = 0.5 in 2 2 use other data as in 1000. (a) Solving for maximum stress for a speed of 10,000 rpm  10,000  ω = 2π   = 1047 rad sec  60  2 0.284)(1047 ) ( st = (3 + 0.3)(5)2 + (1 − 0.3)(0.5)2 = 16,670 psi 4(386)

[

]

(b) Solving for maximum stress for a speed of 10,000 rpm  20,000  ω = 2π   = 2094 rad sec  60  2 0.284)(2094) ( st = (3 + 0.3)(5)2 + (1 − 0.3)(0.5)2 = 66,680 psi 4(386 )

[

(c) Solving for maximum speed, ω .

Page 22 of 25

]


SECTION 18 – MISCELLANEOUS PROBLEMS For AISI 3150, OQT at 1000 F, s y = 130 ksi st = s y =

ρω 2 4go

130,000 =

[(3 + µ )r

2

o

+ (1 − µ )ri

2

] psi

(0.284)ω 2 [(3 + 0.3)(5)2 + (1 − 0.3)(0.5)2 ] 4(386)

ω = 2923.8 rad sec 60ω 60(2923.8) RPM = = = 27,920 rpm 2π 2π 1002. A circular steel disk, with Do = 8 in and Di = 2 in , is shrunk onto a solid steel shaft with an interference of metal i = 0.002 in . (a) At what speed will the pressure in the fit become zero as a result of the rotation? Assume that the shaft is unaffected by centrifugal action. (This effect is relatively small.) (b) Compute the maximum stress in the disk and the pressure at the interface when the speed is 10,000 rpm. Note: The maximum stress in the disk is obtained by adding equations (8.15) of

i8.26,

Text, and (n of i20.9. The resulting equation together

with equation (s) of i8.27 can then be used to obtain pi and σ th ; where

σ ts = − pi for a solid shaft. Solution: (a) Solving for speed, pi = 0 From equation 8.15, i8.26, Text. σ + µ h pi σ ts + µ s pi  − i = Di  th  Eh Es   Di = 2 in i = 0.002 in Eh = E s = 30 × 106 psi µ h = µ s = 0.30 σ ts = − pi = 0 σ + (0.3)(0 ) 0 + (0.3)(0 )  − i = 0.002 = 2  th 6 30 × 10 6   30 × 10 σ th = 30,000 psi

σ th = σ ti + st From Equation 8-15 pi (ro2 + ri 2 ) − 2 po ro2 σ ti = ro2 − ri 2 Page 23 of 25


SECTION 18 – MISCELLANEOUS PROBLEMS but po = 0 pi (ro2 + ri 2 ) ro2 − ri 2 8 D ro = o = = 4 in 2 2 D 2 ri = i = = 1 in 2 2 pi = 0 σ ti = 0

σ ti =

From Equation (n) i20.9 st =

ρω 2 4go

[(3 + µ )r

2

o

+ (1 − µ )ri

2

] psi

ρ = 0.284 lb in3 µ = 0.30 ro = 4 in ri = 1 in g o = 386 in sec σ th = σ ti + st

30,000 = 0 +

(0.284)ω 2 [(3 + 0.3)(4)2 + (1 − 0.3)(1)2 ] 4(386)

ω = 1746 rad sec 60ω 60(1746 ) RPM = = = 16,673 rpm 2π 2π (b) Solving for the maximum stress in the disk and the pressure within the interface. σ + µ h pi σ ts + µ s pi  − i = Di  th  Eh Es   σ + 0.3 pi (− pi + 0.3 pi )  − 0.002 = 2  th 6 30 × 10   30 × 10 30,000 = σ th + pi

σ th = 30,000 − pi σ th = σ ti + st

(

)

 4 2 + 12  17 pi ro2 + ri 2 σ ti = 2 2 = pi  2 2  = pi ro − ri  4 − 1  15 st =

ρω 2 4go

[(3 + µ )r

Page 24 of 25

o

2

+ (1 − µ )ri

2

] psi


SECTION 18 – MISCELLANEOUS PROBLEMS 2π (10,000 ) = 1047 rad sec 60 2 0.284 )(1047 ) ( st = (3 + 0.3)(4)2 + (1 − 0.3)(1)2 = 10,788 psi 4(386 ) σ th = σ ti + st 17 σ th = 30,000 − pi = pi + 10,788 15 pi = 9000 psi (interface pressure)

ω=

[

]

σ th = 30,000 − pi = 30,000 − 9000 = 21,000 psi (maximum stress) - end -

Page 25 of 25

Design of Machine Elements - V.M. Faires.pdf

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