Chapter 2 PROBLEM 2.1 2.1 A steel rod is 2.2 m long and must not stretch more than 1.2 mm when a 8.5 kN load is applied to it. Knowing that E = 200 GPa, determine (a) the smallest diameter rod which should be used, (b) the corresponding normal stress caused by the load. SOLUTION (a) d =
A=
(b) s =
PL AE p 2 d 4
\
A=
d
ia f
8.5 ¥ 10 3 2.2 PL = = 77.92 ¥ 10–6 m2 Ed 200 ¥ 10 9 1.2 ¥ 10 -3
\ d=
d
4A = p
id
i
( 4 ) ( 77.92 ¥ 10 -6 ) = 9.96 ¥ 10–3 m p = 9.96 mm
8.5 ¥ 10 3 p = = 109.1 ¥ 106 Pa = 109.1 MPa -6 A 77.92 ¥ 10
PROBLEM 2.2 2.2 A 1.5 m long steel wire of 6 mm diameter steel wire is subjected to a 3.4 kN tensile load. Knowing that E = 200 GPa, determine (a) the elongation of the wire, (b) the corresponding normal stress. SOLUTION (a) L = 1.5 m = 1500 mm d= (b) s =
A=
p 2 p d = (6)2 = 28.27 mm2 4 4
(3.4 ¥ 1000 N ) (1500 mm) PL = = 0.902 mm = 0.902 mm AE ( 28.27 mm 2 ) (200 ¥ 1000 N / mm 2 ) 3400 N P = = 120.26 N/mm2 = 120.26 MPa 28.27 mm 2 A
74 Mechanics of Materials
PROBLEM 2.3 2.3 Two gage marks are placed exactly 250 mm apart on a 12 mm diameter aluminium rod with E = 70 GPa and an ultimate strength of 110 MPa. Knowing that the distance between the gage marks is 250.230 mm after a load is applied, determine (a) the stress in the rod, (b) the factor of safety. SOLUTION (a) d = 250.230 – 250 = 0.230 mm
d s = L E
\
ES ( 70000 N / mm 2 ) (0.230 mm) = = 64.4 N/mm2 L 250 mm
s=
= 64.4 MPa (b) F.S. =
sU 110 = = 1.708 s 64.4
PROBLEM 2.4 2.4 A control rod made of yellow brass must not stretch more than 3 mm when the tension in the wire is 4 kN. Knowing that E = 105 GPa and that the maximum allowable normal stress is 180 MPa, determine (a) the smallest diameter that can be selected for the rod, (b) the corresponding maximum length of the rod. SOLUTION (a) s =
A=
(b) d =
P A p 2 d 4
PL AE
\
\
A=
4 ¥ 10 3 P = = 22.222 ¥ 10–6 m2 180 ¥ 10 6 s
d=
\ L=
d
( 4) 22.222 ¥ 10 -6
4A = p
p
d
id
i = 5.32 ¥ 10
–3
= 5.32 mm
id
22.222 ¥ 10 -6 105 ¥ 10 9 3 ¥ 10 -3 AEd = P 4 ¥ 10 3
i
= 1.750 m PROBLEM 2.5 2.5 A 9 m length of 6 mm diameter steel wire is to be used in a hanger. It is noted that the wire stretches 18 mm when a tensile force P is applied. Knowing that E = 200 GPa, determine (a) the magnitude of the force P, (b) the corresponding normal stress in the wire.
Chapter 2
75
SOLUTION (a) A =
d=
p 2 p d = (0.006)2 = 28.274 ¥ 10–6 m2 4 4 PL AE
\ P=
d
i
d
28.274 ¥ 10 -6 (200 ¥ 10 9 ) 18 ¥ 10 -3 AEd = L 9
i
= 11.31 ¥ 103 N = 11.31 kN (b) s =
11.31 ¥ 10 3 P = = 400 ¥ 106 Pa = 400 MPa -3 A 28.274 ¥ 10
PROBLEM 2.6 2.6 A 1.4 m aluminum pipe should not stretch more than 1.2 mm when it is subjected to a tensile load. Knowing that E = 70 GPa and that the allowable tensile strength is 96 MPa, determine (a) the maximum allowable length of the pipe, (b) the required area of the pipe if the tensile load is 560 kN. SOLUTION PL (a) d = AE
(b) s =
P A
\ \
d
id
i
70 ¥ 10 9 Pa 1.2 ¥ 10 -3 m EAd Ed L= = = = 0.875 m 96 ¥ 10 6 Pa P s A=
560 ¥ 103 N P = = 5833.3 mm2 96 N / mm 2 s
PROBLEM 2.7 2.7 A nylon thread is subjected to a 8.5 N tension force. Knowing that E = 3.3 GPa and that the length of the thread increases by 1.1%, determine (a) the diameter of the thread, (b) the stress in the thread. SOLUTION (a)
d 11 . L = \ = 90.909 L 100 d (8.5) (90.909) PL PL d= \ A= = = 234.16 ¥ 10–9 m2 AE Ed 3.3 ¥ 10 9
4A p 2 d \ d= = 0.546 ¥ 10–3 m = 0.546 mm p 4 8.5 P (b) s = = = 36.3 ¥ 106 Pa = 36.3 MPa 234.16 ¥ 10 -9 A A=
76 Mechanics of Materials
PROBLEM 2.8 2.8 A cast-iron tube is used to support a compressive load. Knowing that E = 70 GPa and that the maximum allowable change in length is 0.025 percent, determine (a) the maximum normal stress in the tube, (b) the minimum wall thickness for a load of 7 kN if the outside diameter of the tube is 50 mm. SOLUTION
d 0.025 = = 0.00025 L 100 Ed = (70 ¥ 109) (0.00025) = 17.5 ¥ 106 Pa = 17.5 MPa s= L P P 7000 (b) s = \ A= = = 400 mm2 A s 17.5 p A= do2 - di2 4 4A ( 4 ) ( 400 ) = 502 – = 1990.7 mm2 \ di = 44.6 mm di2 = do2 – p p 1 1 t = (do – di) = (50 – 44.6) = 2.7 mm 2 2 (a)
d
i
PROBLEM 2.9 2.9 A block of 250 mm length and 45 ¥ 40 mm cross section is to support a centric compressive load P. The material to be used is a bronze for which E = 95 GPa. Determine the largest load which can be applied, knowing that the normal stress must not exceed 124 MPa and that the decrease in length of the block should be at most 0.12 percent of its original length. SOLUTION Considering allowable stress
s = 124 MPa = 124 N/mm2 A = (45)(40) = 1800 mm2
s=
P A
P = s A = (124)(1800) = 223.2 kN Considering allowable deformation
d 0.12 = = 0.0012 L 100 d=
Smaller value governs
PL AE
\ P = AE
P = 205.2 kN
d = (1800)(95 ¥ 109) (0.0012) = 205.2 kN L
Chapter 2
77
PROBLEM 2.10 2.10 A 9 kN tensile load will be applied to a 50 m length of steel wire with E = 200 GPa. Determine the smallest diameter wire which can be used, knowing that the normal stress must not exceed 150 MPa and that the increase in the length of the wire should be at most 25 mm. SOLUTION Considering allowable stress s = 150 ¥ 106 Pa s=
P A
\
A=
P 9 ¥ 10 3 = = 60 ¥ 10–6 m2 150 ¥ 10 6 s
Considering allowable elongation d = 25 ¥ 10–3 m d=
PL AE
\ A= –6
d
ia f
9 ¥ 10 3 50 PL = = 90 ¥ 10–6 m2 -3 9 Ed 200 ¥ 10 25 ¥ 10
d
id
i
2
A = 90 ¥ 10 m
Larger area governs
p A = d2 4
d=
4A = p
d
( 4) 90 ¥ 10 -6 p
i = 10.70 ¥ 10
–3
m
= 10.70 mm PROBLEM 2.11 2.11 The 4-mm-diameter cable BC is made of a steel with E = 200 GPa. Knowing that the maximum stress in the cable must not exceed 190 MPa and that the elongation of the cable must not exceed 6 mm, find the maximum load P that can be applied as shown.
B 2.5 m P 3.5 m
SOLUTION
A
LBC = 6 2 + 4 2 = 7.2111 m
4.0 m
Use bar AB as a free body S MA = 0
3.5 P – (6)
C
F 4 F I =0 H 7.2111 K BC
P
FBC
P = 0.9509 FBC Considering allowable stress
s = 190 ¥ 106 Pa A=
s=
FBC A
\
p 2 p d = (0.004)2 = 12.566 ¥ 10–6 m2 4 4
Ax Ay
FBC = sA = (190 ¥ 106)(12.566 ¥ 10–6) = 2.388 ¥ 103 N
78 Mechanics of Materials
Considering allowable elongation d = 6 ¥ 10–3 m d= \
FBC =
Fµ LBC AB
d
id
id
12.566 ¥ 10 -6 200 ¥ 10 9 6 ¥ 10 -3 AES = LBC 7.2111
i
= 2.091 ¥ 103 N FBC = 2.091 ¥ 103 N
Smaller value governs
P = 0.9509 FBC = (0.9509)(2.091 ¥ 103) = 1.988 ¥ 103 N = 1.988 kN P= 580 kN
PROBLEM 2.12 2.12 Rod BD is made of steel (200 GPA) and is used to brace the axially compressed member ABC. The maximum force that can be developed in member BD is 0.02 P. If the stress must not exceed 124 MPa and the maximum change in length of BD must not exceed 0.001 times the length of ABC, determine the smallest diameter rod that can be used for member BD.
A 1.8 m D B 1.8 m
SOLUTION
C
FBD = 0.02 P = (0.02) (580) = 11.6 kN Considering stress
s = 124 MPa = 124 ¥ 106 N/m2
F d = BD A Considering deformation
\ Larger area governs
\
or 124 N/mm2 1.35 m
F 11.6 ¥ 10 3 A = BD = = 93.54 mm2 s 124 d = (0.001)(3.6 ¥ 103) = 3.6 mm
d=
FBD LBD AE
A =
11.6 ¥ 10 3 1.35 ¥ 10 3 FBD LBD = = 21.75 mm2 9 6 Ed 200 ¥ 10 /10 3.6
d
id
d
ia f
i
2
A = 93.54 mm A=
p 2 d 4
\ d=
4A = p
( 4) ( 93.54) = 10.91 mm p 30 mm
PROBLEM 2.13 2.13 A single axial load of magnitude P = 58 kN is applied at end C of the brass rod ABC. Knowing that E = 105 GPa, determine the diameter d of portion BC for which the deflection of point C will be 3 mm.
d A 1.2 m
B C 0.8 m
P
Chapter 2
SOLUTION
RS L + L UV TA A W d105 ¥ 10 i d3 ¥ 10 i =
dC = Â
Ed C LBc L = - AB ABC P AAB
Pi Li P = Ai E E
AB
BC
AB
BC
-3
9
1.2 = 3.7334 ¥ 103 m–1 p 2 ( 0.030 ) 4
58 ¥ 10 3
LBC 0.8 = = 214.28 ¥ 10–6 m2 3 3.7334 ¥ 10 3.7334 ¥ 10 3
ABC = ABC =
p 2 d BC 4
4ABC = p
\ dBC =
( 4 ) (214.28 ¥ 10 -6 ) = 16.52 ¥ 10–3 m p = 16.52 mm
PROBLEM 2.14
30 mm
2.14 Both portions of the rod ABC are made of an aluminum for which E = 73 GPa. Knowing that the diameter of portion BC is d = 20 mm, determine the largest force P that can be applied if sall = 160 MPa and the corresponding deflection at point C is not to exceed 4 mm.
d A 1.2 m
B C 0.8 m
SOLUTION p (0.030)2 = 706.86 ¥ 10–6 m2 4 p ABC = (0.020)2 = 314.16 ¥ 10–6 m2 4 s = 160 ¥ 106 Pa
AAB =
Considering allowable stress
s=
P A
\
P = As
Portion AB P = (706.86 ¥ 10–6)(160 ¥ 106) = 113.1 ¥ 103 N P = (314.16 ¥ 10–6)(160 ¥ 106) = 50.3 ¥ 103 N
Portion BC
Considering allowable deflection dC = 4 ¥ 10–3 m dC = Â P = EdC
79
FL GH A
AB AB
L + BC ABC
Smallest value for P governs
I JK
PLi P = AE E
-1
9
FL GH A
LBC ABC
I JK
AB
+
AB
F 1.2 )G H 706.86 ¥ 10
= (73 ¥ 10 ) (4 ¥ 10
–3
= 68.8 ¥ 103 N P = 50.3 ¥ 103 N = 50.3 kN
-6
0.8 + 314.16 ¥ 10 -6
IJ K
-1
P
80 Mechanics of Materials
PROBLEM 2.15 2.15 The specimen shown is made from a 25 mm diameter cylindrical steel rod with two 38 mm outer-diameter sleeves bonded to the rod as shown. Knowing that E = 200 GPa, determine (a) the load P so that the total deformation is 0.05 mm, (b) the corresponding deformation of the central portion BC.
P'
38 mm diameter A 25 mm diameter B 38 mm diameter C 50 mm D 75 mm P
SOLUTION
50 mm
F LI P = Ed G Â J H AK
PL P L (a) d = Â i i = Â i Ai Ei E Ai
Ai =
i
L, mm
d, mm
A, mm2
50 75 50
38 25 38
1134 491 1134
AB BC CD
-1
i
p 2 di 4
L /A, mm–1 0.04409 0.1527 0.04409 0.2409
P = (200 ¥ (b) dBC =
10 9 –1 6 )(0.05)(0.2409) = 41.511 kN = 41.5 kN 10
PL BC P LBC 41.5 ¥ 10 3 = = (0.1527) = 0.0317 mm ABC E E ABC 10 9 200 ¥ 6 10 P
PROBLEM 2.16 2.16 Both portions of the rod ABC are made of an aluminum for which E = 70 GPa. Knowing that the magnitude of P is 4 kN, determine (a) the value of Q so that the deflection at A is zero, (b) the corresponding deflection of B.
A
20-mm diameter
0.4 m
SOLUTION B
p 2 p d AB = ( 0.020 ) 2 = 314.16 ¥ 10–6 m2 4 4 p 2 p ABC = d BC = ( 0.060 ) 2 = 2.8274 ¥ 10–3 m2 4 4 Force in member AB is P tension
(a) AAB =
Elongation dAB =
d
ia f
4 ¥ 10 3 0.4 PLAB = EAAB 70 ¥ 10 9 314.16 ¥ 10 -6
d
id
= 72.756 ¥ 10–6 m
Q 0.5 m
i
60-mm diameter
C
Chapter 2
81
Force in member BC is Q – P compression Shortening dBC =
(Q - P) LBC (Q - P)(0.5) = EABC ( 70 ¥ 10 9 )(2.8274 ¥ 10 -3 ) = 2.5263 ¥ 10–9 (Q – P)
For zero deflection at A
dBC = dAB
2.5263 ¥ 10–9 (Q – P) = 72.756 ¥ 10–6
\ Q – P = 28.8 ¥ 103 N
Q = 28.3 ¥ 103 + 4 ¥ 103 = 32.8 ¥ 103 N = 32.8 kN (b) dAB = dBC = dB = 72.756 ¥ 10–6 m = 0.0728 mm PROBLEM 2.17 P
2.17 The rod ABC is made of an aluminum for which E = 70 GPa. Knowing that P = 6 kN and Q = 42 kN, determine the deflection of (a) point A, (b) point B.
A
SOLUTION
20-mm diameter
0.4 m
AAB =
(a) P
p 2 p d AB = (0.020)2 = 314.16 ¥ 10–6 m2 4 4
B
p 2 p ABC = d AC = (0.060)2 = 2.8274 ¥ 10–3 m2 4 4 PAB = P = 6 ¥ 103 N
Q 0.5 m
60-mm diameter
PBC = P – Q = 6 ¥ 103 – 42 ¥ 103 = – 36 ¥ 103 N LAB = 0.4 m dAB
LBC = 0.5 m
d
ia f id
6 ¥ 10 3 0.4 PAB L AB = = AAB E A 314.16 ¥ 10 -6 70 ¥ 10 9
d
C
i
= 109.135 ¥ 10–6 m Q
d
ia f id i
-36 ¥ 10 3 0.5 PBC LBC d BC = = ABC E 2.8274 ¥ 10 -3 70 ¥ 10 9
d
PBC
= – 90.947 ¥ 10–6 m
d A = dAB + dBC = 109.135 ¥ 10–6 – 90.947 ¥ 10–6 m = 18.19 ¥ 10–6 m = 0.01819 mm
(b)
d B = dBC = – 90.9 ¥ 10–6 m = – 0.0909 mm
82 Mechanics of Materials
PROBLEM 2.18 2.18 The 36-mm-diameter steel rod ABC and a brass rod CD of the same diameter are joined at point C to form the 7.5-m rod ABCD. For the loading shown, and neglecting the weight of the rod, determine the deflection of (a) point C, (b) point D.
A 2m Steel: E = 200 GPa
B 50 kN
SOLUTION
3m
p p A = d2 = (0.036)2 = 1.01787 ¥ 10–3 m2 4 4 Portion
Pi
AB BC CD
150 kN 100 kN 100 kN
Li 2m 3m 2.5 m
Ei
C
Pi Li /AEi 1.474 ¥ 10–3 m 1.474 ¥ 10–3 m 2.339 ¥ 10–3 m
200 GPa 200 GPa 105 GPa
D
(a) d C = dAB + dBC = 1.474 ¥ 10–3 + 1.474 ¥ 10–3 = 2.948 ¥ 10–3 m = 2.95 mm –3
2.5 m
Brass: E = 105 GPa
100 kN
–3
(b) dD = dC + dCD = 2.948 ¥ 10 + 2.339 ¥ 10
= 5.287 ¥ 10–3 m = 5.29 mm PROBLEM 2.19 2.19 The brass tube AB (E = 103 GPa) has a cross-sectional area of 142 mm2 and is fitted with a plug at A. The tube is attached at B to a rigid plate which it itself attached at C to the bottom of an aluminum cylinder (E = 72 GPa) with a cross-sectional area of 258 mm2. The cylinder is then hung from a support at D. In order to close the cylinder, the plug must move down through 1.2 mm. Determine the force P that must be applied to the cylinder.
P
D
A
375 mm
SOLUTION Shortening of brass tube AB LAB = 375 + 1.2 = 376.2 mm, AAB = 142 mm2 EAB = 72 ¥ 109 Pa = 72000 MPa dAB =
1.2 mm
PLAB P(376.2) = = 2.572 ¥ 10–5 P EAB AAB (103 ¥ 10 3 )(142)
B C
Lengthening of aluminum cylinder CD LCD = 375 mm,
ACD = 258 mm2, ECD = 72000 ¥ 106 Pa
Chapter 2
dCD =
83
PLCD P(375) = = 2.02 ¥ 10–5 P ECD ACD ( 72000) 258
a f
dA = dAB + dCD
Total deflection
1.2 = (2.57 ¥ 10–5 + 2.02 ¥ 10–5)P
\
P = 26.14 kN = 26.14 kN
PROBLEM 2.20 2.20
B
A 1.2-m section of aluminum pipe of cross-sectional area 1100 mm2 rests on a fixed support at A. The 15-mm-diameter steel rod BC hangs from a rigid bar that rests on the top of the pipe at B. Knowing that the modulus of elasticity is 200 GPa for steel and 72 GPa for aluminum, determine the deflection of point C when a 60 kN force is applied at C.
1.2 m
A 0.9 m
SOLUTION Rod BC:
LBC = 2.1 m, ABC =
C
EBC = 200 ¥ 109 Pa
p 2 p d = (0.015)2 = 176.715 ¥ 10–6 m2 4 4 dC/B =
P
d
ia f
60 ¥ 10 3 2.1 PLBC = = 3.565 ¥ 10–3 m E BC ABC 200 ¥ 10 9 176.715 ¥ 10 -6
d
9
id
2
i
–6
Pipe AB: LAB = 1.2 m, EAB = 72 ¥ 10 Pa, AAB = 1100 mm = 1100 ¥ 10 m dB/A =
d
ia f
2
60 ¥ 10 3 1.2 PLAB = = 909.1 ¥ 10–6 m2 -6 9 EAB AAB 72 ¥ 10 1100 ¥ 10
d
id
i
–6
dC = dB/A + dC/B = 909.1 ¥ 10 + 3.565 ¥ 10–3 = 4.47 ¥ 10–3 = 4.47 mm P
PROBLEM 2.21 2.21 The steel frame (E = 200 GPa) shown has a diagonal brace BD with an area of 1920 mm2. Determine the largest allowable load P if the change in length of member BD is not to exceed 1.6 mm.
B
C
A
D
6m
SOLUTION dBC = 1.6 ¥ 10–3 m, LBC =
2
5 +6
2
ABD = 1920 mm2 = 1920 ¥ 10–6 m2
= 7.810 m,
9
EBC = 200 ¥ 10 Pa 5m
84 Mechanics of Materials
dBC =
FBC L BC EBC ABC
FBC =
E BC ABCd BC ( 200 ¥ 109 ) (1920 ¥ 10 -6 ) (1.6 ¥ 10 -3 ) = 7.81 LBC
= 78.67 ¥ 103 N Use joint B as a free body:
P
Æ S Fx = 0 +
5 FBC – P = 0 7.810
P=
FBC FAB
( 5)( 78.67 ¥ 10 3 ) 5 FBC = 7.810 7.810
= 50.4 ¥ 103 N = 50.4 kN PROBLEM 2.22 2.22
228 kN
For the steel truss (E = 200 GPa) and loading shown, determine the deformations of members AB and AD, knowing that their cross-sectional areas are 2400 mm2 and 1800 mm2, respectively.
B 2.5 m C
D
A
SOLUTION Statics: Reactions are 114 kN upward at A and C. Member BD is a zero force member
4.0 m
4.0 m
4.0 2 + 2.52 = 4.717 m
LAB = Use joint A as a free body: +≠SFy = 0
114 –
2.5 FAB = 0 4.717
FAB
FAB = 215.10 kN + Æ
Member AB:
S Fx = 0
FAD –
4 FAB = 0 4.717
FAD =
( 4) ( 215.10 ) = 182.4 kN 4.717
dAB =
FAB L AB ( 21510 . ¥ 103 ) ( 4.717) = E AAB ( 200 ¥ 10 9 ) (2400 ¥ 10 -6 )
A
FAD
114 kN
= 2.11 ¥ 10–3 m = 2.11 mm dAD =
FAB L AB (182.4 ¥ 10 3 ) ( 4.0) = = 2.03 ¥ 10–3 m = 2.03 mm E AAD ( 200 ¥ 10 9 ) (1800 ¥ 10 -6 )
85
Chapter 2
PROBLEM 2.23 2.23
1.8 m
1.8 m
Members AB and BC are made of steel (E = 200 GPa) with cross-sectional areas of 516 mm2 and 412 mm2, respectively. For the loading shown, determine the elongation of (a) member AB, (b) member BC.
C
B
1.5 m A D
SOLUTION
E
125 kN
240 kN
(a) LAB = 1.82 + 1.5 2 = 2.34 m = 2340 mm FAB
Use joint A as a free body +≠ SFy = 0
1.5 FAB – 125 = 0 2.34
FAD
A
FAB = 195 kN 125 kN
F L (125 ¥ 10 3 ) (2340) dAB = AB AB = = 2.83 mm E AAB (200 000) (516) (b) Use joint B as a free body + SF = 0 Æ x
FBC –
1.8 FAB = 0 2.34
B
FBC =
(1.8) (195) = 150 kN 2.34
dBC =
FBC LBC (150 ¥ 10 3 )(1800) = = 3.27 mm E ABC (200 000)( 412)
FAB
FBC
FBD
PROBLEM 2.24 2.24
Members AB and CD are 28-mm-diameter steel rods, and members BC and AD are 22-mm-diameter steel rods. When the turnbuckle is tightened, the diagonal member AC is put in tension. Knowing that E = 200 GPa and h = 1.2 m, determine the largest allowable tension in AC so that the deformations in members AB and CD do not exceed 1.0 mm.
SOLUTION dAB = dCD = 1.0 mm
B
C
A
D
h
h = 1.2 m = LCD
p p ACD = d 2 = (28)2 = 615.8 mm2 4 4
0.9 m
86 Mechanics of Materials
dCD =
FCD LCD EACD
FCD =
E ACDd CD ( 200 000 )(615.8)(1.0 ) = LCD 1200
= 102.63 kN
C
FBC
Use joint C as a free body +≠S Fy = 0: FAC
FCD –
4 FAC = 0 \ 5
FAC =
5 FCD 4
5
4 3
FCD
FAC
5 = (102.63) = 128.29 kN 4
PROBLEM 2.25 2.24 Members AB and CD are 28-mm-diameter steel rods, and members BC and AD are 22-mm-diameter steel rods. When the turnbuckle is tightened, the diagonal member AC is put in tension. Knowing that E = 200 GPa and h = 1.2 m, determine the largest allowable tension in AC so that the deformations in members AB and CD do not exceed 1.0 mm. 2.25 For the structure in Prob. of 2.24, determine (a) the distance h so that the deformations in members AB, BC, CD and AD are all equal to 1.0 mm, (b) the corresponding tension in member AC.
B
C
A
D
h
SOLUTION
0.9 m
(a) Statics: Use joint B as a free body FBC
B
FAB
FBD
FBD
FAB
FBC
b
Force triangle
Geometry
From similar triangles
F F FAB = BC = BD b LBC h FAB =
LBD
h
h FBC b
Chapter 2
87
For equal deformations dAB = dBC \
FAB h F b = BC E AAB E ABC
FAB =
b AAB FBC h ABC
Equating expressions for FAB h b AAB FBC = FBC b h ABC
p 2 d AB AAB d2 h2 = 4 = AB 2 = 2 p 2 ABC b d BC d BC 4
d h 28 = AB = = 1.273 b d BC 22
b = 0.9 m
h = 1.273 b = 1.273 (0.9) = 1.146 m (b) Setting dAB = dBC = 1.0 mm
F b dBC = BC E ABC
\
p ( 200 000) ( 22)(1.0) E ABC d BC 4 FBC = = b 900
= 84.473 kN FAB =
h FBC = 1.273 (84.473) = 107.535 kN b
From the force triangle FBD = FAC =
2 2 = 136.746 kN + FAB FBC
PROBLEM 2.26 2.26 Members ABC and DEF are joined with steel links (E = 200 GPa). Each of the links is made of a pair of 25 ¥ 35-mm plates. Determine the change in length of (a) member BE, (b) member CF.
C
F
B
E
A
D
180 mm
260 mm
SOLUTION
C FCF
Use member ABC as a free body S MB = 0
B FBE
A 18 kN
(0.260) (18 ¥ 103) – (0.180) FCF = 0 FCF =
( 0.260)(18 ¥ 10 3 ) = 26 ¥ 103 N 0.180
18 kN
240 mm
18 kN
88 Mechanics of Materials
S MC = 0 FBE = –
(0.440) (18 ¥ 103) + (0.180) FBE = 0
( 0.440)(18 ¥ 103 ) = – 44 ¥ 103 N 0.180
Area for link made of two plates A = (2)(0.025) (0.035) = 1.75 ¥ 10–3 m2 (a) dBE =
FBE LBE ( - 44 ¥ 10 3 )(0.240) = = – 30.2 ¥ 10–6 m = – 0.0302 mm EA ( 200 ¥ 10 9 )(1.75 ¥ 10 –3 )
(b) dCF =
FCF LCF (26 ¥ 10 3 )(0.240) = = 17.83 ¥ 10–6 m = 0.01783 mm EA ( 200 ¥ 10 9 )(1.75 ¥ 10 –3 )
PROBLEM 2.27 2.27 Each of the links AB and CD is made of aluminum (E = 75 GPa) and has a cross-sectional area of 125 mm2. Knowing that they support the rigid member BC, determine the deflection of point E.
A
D P = 5 kN
0.36 m E
SOLUTION B
Use member BC as a free body
0.20 m FAB
0.44 m
C
FCD
B
C 5 ¥ 103 N
S MC = 0
– (0.64)FAB + (0.44)(5 ¥ 103) = 0
FAB = 3.4375 ¥ 103 N
S MB = 0
(0.64)FCD – (0.20)(5 ¥ 103) = 0
FCD = 1.5625 ¥ 103 N
For links AB and CD
A = 125 mm2 = 125 ¥ 10–6 m2 dAB =
(3.4375 ¥ 10 3 )(0.36) FAB L AB = = 132.00 ¥ 10–6 m = dB -6 9 EA ( 75 ¥ 10 )(125 ¥ 10 )
dCD =
FCD LCD (1.5625 ¥ 10 3 )(0.36) = = 60.00 ¥ 10–6 m = dC EA ( 75 ¥ 10 9 )(125 ¥ 10 -6 )
Chapter 2
Slope q =
dB - dC 72.00 ¥ 10 -6 = = 112.5 ¥ 10–6 rad 0.64 l BC
B
E
dE = dC + lEC q
C dC
dE
dB
C¢
q –6
= 60.00 ¥ 10
89
–6
+ (0.44) (112.5 ¥ 10 )
E¢
= 109.5 ¥ 10–6 m = 0.1095 mm
B¢
Deformation diagram
PROBLEM 2.28 2.28 Link BD is made of brass (E = 103 GPa) and has a crosssectional area of 258 mm2. Link CE is made of aluminum (E = 72 GPa) and has a cross-sectional area of 322 mm2. Determine the maximum force P that can be applied vertically at point A if the deflection of A is not to exceed 0.35 mm.
D 225 mm C A
SOLUTION
B 150 mm
P
Use member ABC as a free body.
E
FBD C
A
225 mm
125 mm
B FCE
P
S MC = 0,
350 P – 225 FBD = 0,
FBD = 1.5556 P
S MB = 0,
125 P – 225 FCB = 0,
FCB = 0.5556 P
dB = dBD =
(1.5556 P)(225) FBD LBD = = 13.17 ¥ 10–6 P Ø (103000)(258) EBD ABD
dC = dCE =
FCE LCE ( 0.5556 P)(150) = = 3.5 ¥ 10–6 P ≠ ECE ACE ( 72000)(322)
From the deformation diagram Slope
SA
q=
SB
q
dB + dC 16.67 ¥ 10 -6 P = = 0.074 ¥ 10–6 P 225 lBC
dA = dB + lAB q = 13.17 ¥ 10–6 P + (125)(0.074 ¥ 10–6) P = 22.42 ¥ 10–6 P Apply displacement limit
dA = 0.35 mm = 22.42 ¥ 10–6 P P=
0.35 = 15.611 kN 22.42 ¥ 10 -6
SC
90 Mechanics of Materials
PROBLEM 2.29 2.29 A homogenous cable of length L and uniform cross section is suspended from one end. (a) Denoting by r the density (mass per unit volume) of the cable and by E its modulus of elasticity, determine the elongation of the cable due to its own weight. (b) Assuming now the cable to be horizontal, determine the force that should be applied to each end of the cable to obtain the same elongation as in part a.
y P L L–y
SOLUTION (a) For element at point identified by coordinate y
W
P = weight of portion below the point = rg A (L – y) dd =
Pdy rgA( L - y)dy r ¢g( L - y) = = dy EA EA E
z L
d=
0
=
(b) For d =
rL EA
P=
rg ( L - y) rg dy = E E
rg E
FG L H
2
-
L2 2
IJ = 1 K 2
F Ly - 1 y I H 2 K
L
2
0
rgL2 E
EAd EA rgL2 1 1 = = rgL = W L L 2E 2 2
PROBLEM 2.30 2.30 Determine the deflection of the apex A of a homogenous circular cone of height h, density r, and modulus of elasticity E, due to its own weight. SOLUTION A
Let b = radius of the base and y
h
Element
h r
r= b
Volume of portion above element
V=
r = radius at section with coordinate y.
1 1 b2 p r 2y = p 2 y3 3 3 h
b y h
Chapter 2
prgb 2 y 3 3h 2
P = rgV =
z h
P Dy d=Â = EA
rg y 2 = 3E 2
0
h
= 0
A = pr2 =
P dy = EA
h
Ú
p b2 2 y h2
p r gb 2 y 3
0
3 h2
91
◊
h2 E p b2 y 2
z h
dy =
0
rgy dy 3E
rgh 2 6E
PROBLEM 2.31 2.31 The volume of a tensile specimen is essentially constant while plastic deformation occurs. If the initial diameter of the specimen is d1, show that when the diameter is d, the true strain is Œt = 2 ln (d1/d). SOLUTION If the volume is constant
p 2 p 2 d1 L0 dL= 4 4
F I H K
d d2 L = 12 = 1 d L0 d et = ln
2
F I H K
d L = ln 1 d L0
2
= 2 ln
d1 d
PROBLEM 2.32 2.32 Denoting by e the “engineering strain” in a tensile specimen, show that the true strain is et = ln (l + e). SOLUTION et = ln
L L +d = ln 0 L0 L0
FG H
= ln 1 + Thus
d L0
IJ = ln (1 + e) K
et = ln (1 + e)
PROBLEM 2.33 2.33 An axial force of 60 kN is applied to the assembly shown by means of rigid end plates. Determine
92 Mechanics of Materials
(a) the normal stress in the brass shell, (b) the corresponding deformation of the assembly.
5 mm 20 mm 5 mm
5 mm
20 mm 5 mm
SOLUTION Let
Pb = portion of axial force carried by brass shell
Steel core E = 200 GPa
Ps = portion of axial force carried by steel core
PL d= b Ab Eb d=
Ps L As Es
Brass shell E = 105 GPa
E Ad Pb = b b L Ps =
250 mm
Es Asd L
P = Pb + Ps = (Eb Ab + Es As)
d L
d P =e= L Eb Ab + Es As As = (0.020) (0.020) = 400 ¥ 10–6 m2 Ab = (0.030) (0.030) – (0.020) (0.020) = 500 ¥ 10–6 m2
d 60 ¥ 10 3 =e= = 452.83 ¥ 10–6 L (105 ¥ 10 9 )(500 ¥ 10 –6 ) + (200 ¥ 10 9 )(400 ¥ 10 -6 ) (a) sb = Ebe = (105 ¥ 109) (452.83 ¥ 10–6) = 47.5 ¥ 106 Pa = 47.5 MPa (b) d = Le = (250 ¥ 10–3) (452.83 ¥ 10–6) = 113.2 ¥ 10–6 m = 0.1132 ¥10–3 m = 0.1132 mm PROBLEM 2.34 2.34 The length of the assembly decreases by 0.15 mm when an axial force is applied by means of rigid end plates. Determine (a) the magnitude of the applied force, (b) the corresponding stress in the steel core.
Chapter 2
5 mm 20 mm 5 mm
5 mm
20 mm 5 mm
Steel core E = 200 GPa
Brass shell E = 105 GPa
250 mm
SOLUTION Pb = portion of axial force carried by brass shell.
Let
Ps = portion of axial force carried by steel core. d=
Pb L Ab Eb
Pb =
Eb Ab d L
d=
Ps L As Es
Ps =
Es Asd L
P = Pb + Ps = (Eb Ab + Es As)
d L
As = (0.020) (0.020) = 400 ¥ 10–6 m2 Ab = (0.030) (0.030) – (0.020) (0.020) = 500 ¥ 10–6 m2 (a) P = [(105 ¥ 109) (500 ¥ 10–6) + (200 ¥ 109) (400 ¥ 10–6)]
0.15 ¥ 10 -3 250 ¥ 10 -3
= 79.5 ¥ 103 N = 75.9 kN (b) ss = Ese =
E sd ( 200 ¥ 10 9 )( 0.15 ¥ 10 -3 ) = = 120 ¥ 106 Pa -3 L 250 ¥ 10 = 120 MPa
93
94 Mechanics of Materials
PROBLEM 2.35
P
2.35 The 1.35 m concrete post is reinforced with six steel bars, each with a 28 mm diameter. Knowing that Es = 200 GPa and Ec = 29 GPa, determine the normal stresses in the steel and in the concrete when a 1560 kN axial centric force P is applied to the post. 1.35 m
SOLUTION Let
0.45 m
Pc = portion of axial force carried by concrete Ps = portion carried by the six steel rods d=
Pc L Ec Ac
Pc =
Ec Acd L
d=
Ps L Es As
Ps =
Es Asd L
P = Pc + Ps = (Ec Ac + Es As)
e=
d P = L Ec Ac + Es As
As = 6
Ac =
d L
p 2 6p ds = (28)2 = 3694.5 mm2 4 4
p 2 p dc – As = (450)2 – 3694.5 = 155348.6 mm2 4 4
L = 1.35 m = 1350 mm e=
-1560 ¥ 10 3
FG 29 ¥ 10 IJ (155348.6)FG 200 ¥ 10 IJ (3694.5) = –297.48 ¥ 10 H 10 K H 10 K 9
6
6
FG 200 ¥ 10 IJ (– 297.48 ¥ 10 ) = – 59.5 N/mm = – 59.5 MPa H 10 K F 10 IJ (– 297.48 ¥ 10 ) = – 8.627 N/mm = – 8.627 MPa = E e = G 29 ¥ H 10 K 9
s s = E se =
–6
2
6
9
sc
–6
9
c
6
–6
2
Chapter 2
PROBLEM 2.36 2.36 An axial centric force of magnitude P = 450 kN is applied to the composite block shown by means of a right end plate. Knowing that h = 10 mm, determine the normal stress in (a) the brass core, (b) the aluminum plates.
95
Brass core (E = 105 GPa) P
Aluminum plates (E = 70 GPa)
Rigid end plate
SOLUTION Let
Pb = portion of axial force carried by brass core
300 mm
Pa = portion carried by two aluminum plates d=
Pb L Eb Ab
Pb =
Eb Abd L
60 mm h
d=
Pa L Ea Aa
Pa =
Ea Aad L
P = Pb + Pa = (Eb Ab + Ea Aa)
e=
40 mm h
d L
d P = L Eb Ab + Ea Aa
Ab = (60) (40) = 2400 mm2 = 2400 ¥ 10–6 m2 Aa = (2) (60) (10) = 1200 mm2 = 1200 ¥ 10–6 m2 e=
450 ¥ 10 3 = 1.3393 ¥ 10–3 -6 9 –6 (105 ¥ 10 )(2400 ¥ 10 ) + ( 70 ¥ 10 )(1200 ¥ 10 ) 9
(a) sb = Ebe = (105 ¥ 109) (1.3393 ¥ 10–3) = 140.6 ¥ 106 Pa = 140.6 MPa (b) sa = Ea e = (70 ¥ 109) (1.393 ¥ 10–3) = 93.75 ¥ 106 Pa = 93.75 MPa PROBLEM 2.37 2.37 For the composite block shown in Prob. 2.36, determine (a) the value of h if the portion of the load carried by the aluminum plates is half the portion of the load carried by the brass core, (b) the total load if the stress in the brass is 80 MPa.
