Plugin Chapter16

Chapter 16

Fasteners and Power Screws Section 16.3 16.1 An Acme-thread Acme-threaded ed power screw with a crest diameter diameter of 1.125 in. and single thread thread is used to raise a load of 25,000 25,000 lb. The collar collar mean diameter diameter is 1.5 in. The coefficient coefficient of friction friction is 0.12 for both the thread and the collar. Determine the following: (a) Pitch Pitch diameter diameter of the screw (b) Screw torque torque required required to raise the load Ans. T r = 4637 in.-lb. (c) Maximum thread coefficient of friction allowed to prevent the screw from self-locking if collar friction is eliminated Notes: Recognizing that this is an Acme screw, some data is obtained from Table 16.2. In addition,

Eqs. (16.4), (16.8) and (16.10) are used in the solution of this problem. Solution: This is an Acme screw, so referring to Table 16.2 with a crest diameter of 1 1/8 in., n = 5

per inch. inch. Therefore Therefore,, from Eq. (16.1), (16.1), p = 1/n = 0.2 in. The pitch pitch diameter diameter is calculat calculated ed from Eq. (16.4): d p = d c − 0.5 p − 0.01 in. = 1. 1 .125 in. − 0.5(0. 5(0.2 in.) − 0.01 = 1. 1.015 in. If this is a single thread then the lead, l, l , equals the pitch, or l = l = 0.2 in. From Eq. (16.5), α is α is given by α = tan

1

−

  l πd p

= tan

1

−





0.2 in. = 3. 3 .589 π (1. (1.015 in.)

◦

◦

From Fig. 16.5, it can be seen that β = β = 29 , so from Eq. (16.8), θ n is θn = tan

1

−



β cos α tan 2



= tan

1

−



29 cos3. cos3.589 tan 2

◦

◦



◦

= 14. 14 .47

The torque to raise the load is given by Eq. (16.10): T r





(d p /2)(cos θn tan α + µ + µ)) = W + rc µc cos θn − µ tan α (1. (1.015 in./ in./2)(cos 2)(cos 14 14..47 tan3. tan3.589 + 0. 0.12 = (25 (25 kip) kip) + cos14. cos14.47 − 0.12tan3. 12tan3.589 = 4637 4637 in.in.-lb lb



◦

◦

◦

◦

   1.5 in. 2

(0. (0.12)

If collar friction is eliminated, then the load will lower if the numerator of Eq. (16.12) is zero, or ◦

◦

µ = cos θn tan α = cos(14. cos(14.47 ) tan(3 tan(3..589 ) = 0.0607 383

384

CHAPTER CHAPTER 16.

FASTENERS ASTENERS AND POWER POWER SCREWS SCREWS

16.22 A car jack 16. jack consis consists ts of a screw screw and a nut, nut, so that that the car is lifted lifted by turning turning the screw. screw. Calcul Calculate ate the torque needed to lift a load with a mass of 1000 kg if the lead of the thread l = 9 mm, its pitch diameter is 22 mm, and its thread angle is 30 . The coefficient coefficient of friction friction is 0.09 in the threads threads and zero elsewhere. Ans. T r = 24 24..40 Nm. ◦

Notes: The torque is calculate calculated d from Eq. (16.10), (16.10), which requires requires calculation calculation of α and θ from Eqs.

(16.5) and (16.8), respectively. Solution: From Eq. (16.5), α is given by

α = tan

1

−

  l πd p

= tan



1

−



9 mm = 7. 7 .42 π (22 mm)

◦

From Eq. (16.8), θ (16.8), θ n is θn = tan

1

−



β cos α tan 2



= tan

1

−



30 cos7. cos7.42 tan 2

◦

◦



= 14. 14 .9

◦






(d p /2)(cos θn tan α + µ + µ)) = W + rc µc cos θn − µ tan α (0. (0.022 m/ m/2)(cos 2)(cos 14 14..9 tan7. tan7.42 + 0. 0.09 = (100 (10000 kg)( kg)(99.81 m/s2 ) +0 cos14. cos14.9 − 0.09tan7. 09tan7.42 = 24 24..40 Nm



◦

◦

◦

◦



16.3 A power power screw gives the axial axial tool motions motions in a numerica numerically lly controlle controlled d lathe. lathe. To get high accuracy accuracy in the motions, the heating and power loss in the screw have to be low. Determine the power efficiency of the screw if the coefficient of friction is 0.12, pitch diameter is 30 mm, lead is 6 mm, and thread angle is 25 . Ans. η = 33 33..8%. ◦

Notes: The efficiency is the output work divided by the input work. If this is evaluated over a given

distan distance, ce, such as one screw revolu revolutio tion, n, then then the torque torque equati equation on can be used. used. In this this regard regards, s, the analysis is similar to Problem 16.2. Note however, that we’re worried about the efficiency of the screw, not the collar, so we’ll ignore the collar losses, that is, we’ll take µ c = 0. Solution: The input work work is the product of the torque and the rotation. rotation. Any reference reference rotation rotation can

be used, so use one revolution or 2 π radians as the reference. The torque is given by Eq. (16.10), but first, from Eq. (16.5), α is given by α = tan

1

−

  l πd p

= tan

1

−





6 mm = 3. 3 .65 π (30 mm)

◦

From Eq. (16.8), θ (16.8), θ n is θn = tan

1

−



β cos α tan 2



= tan

1

−



25 cos3. cos3.65 tan 2 ◦

◦



◦

= 12. 12 .47


 



(d p /2)(cos θn tan α + µ + µ)) = W + rc µc cos θn − µ tan α (0. (0.030 m/ m/2)(cos 2)(cos 12 12..47 tan3. tan3.65 + 0. 0.12 = W +0 cos12. cos12.47 − 0.12tan3. 12tan3.65 = (0. (0.00282 Nm ◦

◦

◦

◦



385

If the lead is 6 mm, this is the distance the load will be moved in one revolution. The output work is the product of the load and the distance. distance. Therefore, Therefore, the ratio of the output work to the input work is: e =

Wl W (0 W (0..006 m) = = 0. 0 .338 = 33. 33 .8% Tθ W (0 W (0..00282 m)(2π m)(2π)

16.4 Sketch Sketch a shows shows a stretchi stretching ng device for steel wires used to stabilize stabilize the mast of a sailing boat. Both front and side views are shown and all dimensions are in millimeters. A screw with square threads ( β = 0), a lead and pitch of 4 mm, and an outer diameter of 20 mm is used. The screw can move axially but is prevented from rotating by flat guiding pins (side view in sketch a). Derive Derive an expression expression for the tightening torque as a function of the stretching force P when P when coefficient of friction of all surface contact contactss is 0.20. Also, calculate calculate the torque needed when the tightening tightening force is 1000 N. 40 20

20

5

20

10

P

(a) Front view

(b) Side view

Sketch a Sketch a,, for Problem 16.4 Notes: This problem requires a derivation derivation similar to that on in the text for the particular circumstances

of this problem. Solution: From Eq. (16.5),

tan α =

l 4 mm = = 0. 0 .07074 2πrm 2π (9 mm)

Equilibrium in the vertical direction of the free body diagram gives:



F y = 0 = −P ax + P n cos α − µP n sin α ax + P

→

P ax = P n cos α − µP n sin α ax = P

386

CHAPTER CHAPTER 16.

FASTENERS ASTENERS AND POWER POWER SCREWS SCREWS

and in the horizontal direction,



F x = 0 = P t − P n sin α − µP n cos α

→

P t = P n sin α + µP + µP n cos α

Therefore we can write

P t sin α + µ + µ cos α tan α + µ + µ = = P ax cos α − µ sin α 1 − µ tan α ax Vertical equilibrium of the screw gives

 

F y = 0 = P ax ax − P −

µP t rm L

→

P ax ax = P +

µP t rm L

Torque equilibrium of the nut gives T = 0 = T − µP ax ax rm − P t rm

Eliminating P Eliminating P ax and P t gives ax and P

→

T = µP ax + P t rm ax rm + P

rm (µ + tan α) + µrm 1 − µ tan α T = P µrm (tan α + µ + µ)) 1− L(1 − µ tan α) If P = P = 1000 N, then substituting µ = 0.2, r 2, r m = 9 mm, r m = 15 mm, L = 20 mm and tan α = 0.0707, this expression yields T = 5. 5 .61 Nm.

 

 

16.5 A flywheel of a motorcycl motorcyclee is fastened fastened by a thread thread manufactured manufactured directly directly in the center center of the flywheel as shown in sketch b sketch b.. The flywheel is mounted by applying a torque T . T . The cone angle is γ . γ . Calculate the tensile force W in W in the shaft between the contact line at N and N and the thread as a function of D of D 1 , D 2 , γ , and T and T .. The lead angle is α at the mean diameter D 1 . The shaft is assumed to be rigid. γ N T

γ

T

1

2

D D

Sketch b Sketch b,, for Problem 16.5 Notes: This problem is solved by performing torque equilibrium on the flywheel and then using Eq.

(16.10) (16.10) for the screw torque. torque. Solution: The forces forces acting acting on the shaft shaft are shown shown above above right right.. Note Note that that when when the flywheel flywheel is

mounted, the contact slides in the circumferential circumferential and the axial direction. Torque equilibrium on the flywheel gives D2 − T t T = 0 = T − µN cos cos φ 2



 

387

where T t is the thread torque and φ is the angle as shown defined by tan φ =

D1 tan α D2 cos γ

Horizontal force equilibrium gives



F x = 0 = W − N sin γ − µN sin φ cos γ

→

W = N (sin γ + µ sin φ cos γ )

Substituting this into Eq. (16.10) and then the resulting expression into the torque equation given above yields µD2 W cos φ/2 W Dt (cos θn tan α + µ) /2 − T − =0 sin γ + µ sin φ cos γ cos θn − µ tan α This could be solved for W if desired. 16.6 To change its oil, a 20,000 lb truck is lifted a height of 5 feet by a screw jack. The power screw has Acme threads and a crest diameter of 5 in. with two threads per inch, and the lead equals the pitch. Calculate how much energy has been used to lift and lower the truck if the only friction is in the threads, where the coefficient of friction is 0.10. Ans. E r = 409 kip-ft. Notes: The approach is similar to Problems (16.1) to (16.3), but one must also use Eq. (16.12) to

calculate the torque needed to lower the load. Solution: For a crest diameter of 5 in., Table 16.2 gives the number of threads per inch as 2, so that

the pitch is 0.5 in. From Eq. (16.4), the pitch diameter is d p = d c − 0.5 p − 0.01 in. = 5 in. − 0.5(0.5 in.) − 0.01 = 4.74 in. Since the lead, l, equals the pitch, l = 0.5 in. From Eq. (16.5), α is given by tan α =

l 0.5 in. = = 1.923 2πr m π(4.74 in.)

◦

From Figure 16.5, it can be seen that β = 29 , so from Eq. (16.8), θ n is ◦

θn = tan

1

−



β cos α tan 2



= tan

1

−



29 cos1.923 tan 2 ◦

◦



◦

= 14.49

The torque to raise the load is given by Eq. (16.10) (note that ? c = 0): T r





(d p /2)(cos θn tan α + µ) = W + rc µc cos θn − µ tan α (4.74 in./2)(cos 14.49 tan1.923 + 0.10 = (20 kip) +0 cos14.49 − 0.10tan1.923 = 6510 in.-lb



◦

◦

◦

◦



The energy is the product of the torque and the rotation. To travel 5 feet = 60 in., the screw needs to rotate 60/l = 60/0.5 = 120 rev = 754.0 rad. The energy needed to raise the load is then E r = (754 rad)(6510 in-lb) = 4908 kip-in. or 409 kip-ft The torque to lower the load is obtained from Eq. (16.12): T l





(d p /2)(µ − cos θn tan α) = −W + rc µc cos θn + µ tan α (4.74 in./2)(0.10 − cos 14.49 tan1.923 ) = −(20 kip) cos14.49 + 0.10tan1.923 = −3293 in.-lb



◦

◦

◦

◦



388

CHAPTER 16.

FASTENERS AND POWER SCREWS

and the energy needed to lower the load is then E l = (754 rad)(3293 in.-lb) = 2483 kip-in. = 207 kip-ft 16.7 A single-threaded M32×3.5 power screw is used to raise a 12-kN load at a speed of 25 mm/s. The coefficients of friction are 0.08 for the thread and 0.12 for the collar. The collar mean diameter is 55 mm. Determine the power required. Also determine how much power is needed for lowering the load at 40 mm/s. Notes: The power is calculated from the torque needed to raise the load or lower the load, obtained

from Eqs. (16.10) and (16.12), respectively. Solution: The bolt designation gives dc = 32 mm and p = 3.5 mm. From Eq. (16.4) for metric

threads, d p = d c − 0.5 p − 0.25 = 32 mm − (0.5)(3.5 mm) − 0.25 mm = 30 mm Since there is a single thread, the lead is the same as the pitch, or l = 3.5 mm. The torque is given by Eq. (16.10), but first, from Eq. (16.5), α is given by tan α =

l 3.5 mm = = 2.13 2πrm π(30 mm)

◦

Note from Fig. 16.5 that β = 29 . From Eq. (16.8), θ n is ◦

θn = tan

1

−



β cos α tan 2



= tan

1

−



29 cos2.13 tan 2 ◦

◦



◦

= 14.49






(d p /2)(cos θn tan α + µ) = W + rc µc cos θn − µ tan α (0.030 m/2)(cos 14.49 tan2.13 + 0.08 (0.055 m)(0.12) = (12 kN) + cos14.49 − 0.08tan2.13 2 = 61.23 Nm



◦

◦

◦

◦



Since the lead is l = 3.5 mm, and the load is raised at 25 mm/s, the screw must be rotating at a speed of ωr = 25/3.5 = 7.14 rev/s = 44.88 rad/s. Therefore the power is the product of torque and angular velocity, or h pr = T r ωr = (61.23 Nm)(44.88 rad/s) = 2748 W The torque to lower the load is given by Eq. (16.12) as T l





(d p /2)(µ − cos θn tan α) = −W + rc µc cos θn + µ tan α (0.030 m/2)(0.08 − cos 14.49 tan2.13 ) (0.055 m)(0.12) = −(12 kN) + cos 14.49 + 0.08tan2.13 2 = −47.75 Nm



◦

◦

◦

◦



The lowering speed is ω l = 40/3.5 = 11.43 rev/s = 71.81 rad/s, so the power needed is h pl = T l ωl = (47.75 Nm)(71.81 rad/s) = 3429 W 16.8 A double-threaded Acme power screw is used to raise a 1350-lb load. The outer diameter of the screw is 1.25 in. and the mean collar diameter is 2.0 in. The coefficients of friction are 0.13 for the thread and 0.16 for the collar. Determine the following:

389

(a) Required torque for raising and lowering the load (b) Geometrical dimensions of the screw (c) Efficiency in raising the load (d) Load corresponding to the efficiency if the efficiency in raising the load is 18% Notes: This problem is similar to the previous problems, especially Problem 16.2 and 16.7. The new

concepts introduced are the use of a double thread and more in-depth use of the equation for efficiency [Eq. (16.13)]. Solution: From Table 16.2, for a crest diameter of 1.25 in., there are n = 5 threads per inch, so that

p = 1/n = 0.2 in., Since this is a double thread, the lead is twice the pitch, or l = 0.4 in. The pitch diameter is obtained from Eq. (16.4) as d p = d c − 0.5 p − 0.01 = 1.25 in. − 0.5(0.2 in.) − 0.01 in. = 1.14 in. The torque is given by Eq. (16.10), but first, from Eq. (16.5), α is given by tan α =

l 0.4 in. = = 6.376 2πr m π(1.14 in.)

◦

Note from Fig. 16.5 that β = 29 . From Eq. (16.8), θ n is ◦

θn = tan

1

−



β cos α tan 2



= tan

1

−



29 cos6.376 tan 2 ◦

◦



= 14.4

◦






(d p /2)(cos θn tan α + µ) = W + rc µc cos θn − µ tan α (1.14 in./2)(cos 14.4 tan6.376 + 0.13 = (1350 lb) + (1.0 in.)(0.16) cos14.4 − 0.13tan6.376 = 408.1 in.-lb



◦

◦



◦

◦

The torque to lower the load is given by Eq. (16.12) as





(d p /2)(µ − cos θn tan α) T l = −W + rc µc = −233 in.-lb cos θn + µ tan α From Eq. (16.13), the efficiency is given by e =

100W l (100)(1350 lb)(0.4 in.) = = 21.06% 2πT 2π(408 in.-lb)

If the efficiency is 18% at the same torque, the load is obtained from Eq. (16.13) as e =

100W l 2πT

→

W =

2πeT 2π(408 in.-lb) = = 1154 lb 100l 100(0.4 in.)

16.9 A 25-kN load is raised by two Acme-threaded power screws with a minimum speed of 35 mm/s and a maximum power of 1750 W per screw. Because of space limitations the screw diameter should not be larger than 45 mm. The coefficient of friction for both the thread and the collar is 0.09. The collar mean diameter is 65 mm. Assuming that the loads are distributed evenly on both sides, select the size of the screw to be used and calculate its efficiency.

390

CHAPTER 16.


Notes: This problem requires selection of a screw from Table 16.2, then analysis of this screw. Solution: For a thread diameter of 45 mm=1.77 in., the largest screw which can be used is, from

Table 16.2, a 1.75 in. crest diameter screw with 4 threads per inch. Therefore, the pitch is 0.25 in. = 6.35 mm. The pitch diameter is calculated from Eq. (16.4) as d p = d c − 0.5 p − 0.01 in. = 1.75 in − 0.5(0.25) − 0.01 in. = 1.615 in. = 41.02 mm At first, use a single thread so that l = p = 6.35 mm. The torque is given by Eq. (16.10), but first, from Eq. (16.5), α is given by tan α =

l 6.35 mm = = 2.82 2πr m π(41.02 mm)

◦

Note from Figure 16.5 that β = 29 . From Eq. (16.8), θ is ◦

θn = tan

1

−



β cos α tan 2



= tan

1

−



29 cos2.82 tan 2 ◦

◦



◦

= 14.48

The torque to raise the load is given by Eq. (16.10), using W = 12, 500 N since there are two screws: T r





(d p /2)(cos θn tan α + µ) = W + rc µc cos θn − µ tan α (0.04102 m/2)(cos 14.48 tan2.82 + 0.09 = (12.5 kN) + (0.0325 m)(0.09) cos14.48 − 0.09tan2.82 = 73.2 Nm



◦

◦



◦

◦

To raise the load at 35 mm/s, the angular velocity ω = 35/6.35 = 5.51 rev/s = 34.63 rad/s. Therefore the power is h p = T ω = (73.2 Nm)(34.63 rad/s) = 2540 W This horsepower is too high. Using double threads, the lead is l = 2 p = 12.70 mm. Using the same equations, one obtains α = 5.6 , θ n = 14.4 , T r = 86.24 Nm, ω = 17.31 rad/s, h p = 1490 W. Therefore, a double thread screw satisfies the power requirement. ◦

◦

16.10 The lead screw of a small lathe is made from a 1/2 in. crest diameter Acme threaded shaft. The lead screw has to exert a force on the lathe carriage for a number of operations, and it is powered by a belt drive from the motor. If a force of 500 lb is desired, what is the torque required if the collar is twice the pitch diameter of the screw? Use µ = µ c = 0.25. With what velocity does the lead screw move the crosshead if the lead screw is single threaded and is driven at 500 rpm? Notes: This is a straightforward problem, requiring Eq. (16.10) for its solution. Solution: From the stated crest diameter, the pitch diameter can be obtained from Eq. (16.4), since

from Table 16.2, n=10 threads/in so p=0.1: d p = d c − 0.5 p − 0.01 = 0.5 − 0.5(0.1) − 0.01 = 0.44 in. Also, since the collar diameter is twice the pitch diameter, then the collar radius is r c = d p = 0.44 in. From Eq. (16.5), l (1)(0.1 in.) α = tan 1 = tan 1 = 4.13 πd p π(0.44 in.) −

 



−



◦

And for an Acme thread, β = 29 (see Figure 15.5). Therefore θ n is given by Eq. (16.8) as ◦

θn = tan

1

−



β cos α tan 2



= tan

1

−

◦

◦

◦

(cos4.14 tan14.5 ) = 14.46

391

The torque needed to raise the load (move the crosshead) is given by Eq. 16.10 as



(d p /2)(cos θn tan α + µ T = W + rc µc cos θn − µ tan α



◦



◦



(0.22 in.)(cos 14.46 tan4.13 + 0.25) = (500 lb) + 0.44(0.25) = 92 in-lb cos 14.46 − 0.25tan4.13 ◦

◦

With every revolution of the lead screw, the cross head moves (0.1in.)( p). If the lead screw moves at 500 rpm, the cross head moves at 50 in./min, or 0.833 in./s.

Section 16.4 16.11 A screw with Acme thread can have more than one entrance to the thread per screw revolution. A single thread means that the pitch and the lead are equal, but for double and triple threads the lead is larger than the pitch. Determine the relationship between the number of threads per inch n, the pitch p, and the lead l. Notes: If a student has difficulty visualizing this problem, the concept can be illustrated by wrapping

a single string around a pencil, a double thread by wrapping two strings (preferably of different colors) and a triple thread by wrapping three strings around a pencil. Solution: If there are m threads, then the lead is related to the pitch and the threads per inch by

l = mp =

m n

16.12 A section of a bolt circle on a large coupling is shown in sketch c. Each bolt is loaded by a repeated force P =6000 lb. The members are steel, and all bolts have been carefully preloaded to P i =25,000 lb each. The bolt is to be an SAE Grade 5, 0.75 inch crest diameter with fine threads, (so that d r =0.674 in) and the nut which fits on this bolt has a thickness of 0.50 in. The threads have been manufactured through rolling, and use a survival probability of 90%. (a) If hardened steel washers 0.134 in thick are to be used under the bolt and nut, what length of bolts should be used? (b) Find the stiffness of the bolt, the members and the joint constant. (c) What is the factor of safety guarding against a fatigue failure?

1.5 in.

Sketch c, for Problem 16.12 Solution:

(a) The length of the members, washers and nut are 1.5 in.+2(0.134 in.)+0.5 in.=2.268 in. To use a standard sized bolt, specify a 2.5 in. long bolt.

392

CHAPTER 16.


(b) Note that E = 30 Mpsi. From Eq. (16.23), the length of the threaded section on the bolt is Lt = 2dc + 0.25 in. = 1.75 in. Therefore, the crest is 0.75 in. long. Therefore, the length of the threads inside the connection is 1 in. From Table 16.9, for d c = 0.75 in., n = 16 threads/in. and A t = 0.373 in2 for fine threads. Therefore, p = 1/n = 0.0625 in. Equation (16.2) gives ht = 0.8660 p = 0.05412 in. From Eq. (16.4), d p = d c − 0.5 p − 0.01 = 0.75 in. − 0.5(0.0625 in.) − 0.01 in. = 0.70785 in. From Fig 16.4, dr = d p − 0.625ht = 0.70785 in. − 0.625(0.05412 in.) = 0.674 in. Equation (15.21) gives the bolt stiffness as:







1 4 Ls + 0.4dc L t + 0.4dr 4 0.75 in. + 0.4(0.75 in.) 0.75 in. + 0.4(0.674 in.) = + = + kb πE d2c d2r π(30 Mpsi) ( in.)2 (0.674 in.)2 or kb = 5.798 Mlb/in. The stiffness of the joint is obtained from the Wileman method, although the conical fustrums give the same answer. For steel, Table 16.6 gives A i = 0.78715 and Bi = 0.62873, so that the stiffness of each member is given by Eq. (15.26) as: ki = Edc Ai eB d i

c

/Li

= (30 Mpsi)(0.75 in.)(0.78715)e(0.62873)(0.75/0.75) = 33.2 Mlb/in.

Therefore, the total joint stiffness is, from Eq. (15.25), 1 1 1 1 1 = + = + kj k1 k2 33.2 Mlb/in. 33.2 Mlb/in

kj = 16.6 Mlb/in.

→

From Eq. (15.17), the joint constant is C k =

kb 5.798 Mlb/in. = = 0.26 kb + kj 5.798 Mlb/in. + 16.6 Mlb/in.

(c) If the load is a repeated force of 6 kip, then P a = P m = 3 kip. From Table 16.7, S u = 120 ksi. From Table 16.11, K f = 3.0 (rolled threads). From Eq. (7.7), S e = 0.45S u = 54 ksi. kr = 0.9, so S e = 0.9(54 ksi) = 48.6 ksi. from Eq. (16.16). The prestress is 

σi =

P i 25, 000 lb = = 67.0 ksi At 0.373 in.2

The mean and alternating stresses are: σa = σm =

3 kip = 8.04 ksi 0.373 in.2

Note that C k is not used, since it is used in Eq. (16.40), which gives: ns =

S ut − σi

   

C k K f σa

S ut S e

+ σm

=

120 ksi − 67.0 ksi = 3.14 120 0.26 (3.0)(8.04 ksi) 48.6 + 8.04 ksi



 



16.13 An M12, coarse-pitch, class-5.8 bolt with a hexagonal nut assembly is used to keep two machine parts together as shown in sketch d. Determine the following: (a) Bolt stiffness and clamped member stiffness.



393

(b) Maximum external load that the assembly can support for a load safety factor of 2.5 (c) Safety factor guarding against separation of the members (d) Safety factor guarding against fatigue if a repeated external load of 10 kN is applied to the assembly MI2 coarse-pitch, class-5.8 bolt

35 Aluminum

5

25

Aluminum

Sketch d, for Problem 16.13 Notes: This problem is long only because of the many parts; each part is very straightforward. The

equations used are for part (a), (16.21), (16.25) and (16.26). For part (b), (16.17) and (16.31), for part (c), (16.32), and for part (d), (16.40). Solution:

(a) Bolt and Member Stiffness. Note from the inside front cover that for steel, E s = 207 GPa and for aluminum E al = 72 GPa. From the sketch we see that Lt = 20 mm and Ls = 40 mm. From Table 16.10, for a crest diameter of 12 mm and coarse threads, p = 1.75 mm, At = 84.3 mm2 . From Eq. (16.2), h t is ht =

0.5 p 0.5(1.75 mm) = = 1.516 mm tan30 tan30 ◦

◦

From Fig. 16.4, the root diameter is: dr = d c − 2(0.625ht ) = 12 mm − 2(0.624)(1.516 mm) = 10.105 mm Therefore, the bolt stiffness is, from Eq. (16.21), 1 kb

= =





4 Ls + 0.4dc L t + 0.4dr + πE d2c d2r 4 0.04 m + 0.4(0.012 m) 0.02 m + 0.4(0.010105 m) + π(207 GPa) (0.012 m)2 (0.010105 m)2





or k b = 297.5 MN/m. For the members, we can use the Wileman method given by Eq. (16.26): kj = E i dc Ai exp(Bi dc /Li ) = (71.0 GPa)(0.012 m)(0.79670) exp[(0.62873)(0.012)(0.060)] = 769.7×106 N/m

394

CHAPTER 16.


The stiffness parameter C k is

C k =

kb 297.5 MN/m = = 0.279 kb + km 297.5 MN/m + 769.7 MN/m

(b) Maximum Load. From Table 16.8, for a 5.8 grade bolt, the proof strength is S p = 380 MPa, S ut = 520 MPa and S y = 415 MPa. If we assume this is a reused connection, then from Eq. (16.33), P i = 0.75P p = 0.75S pAt = 0.75(380 MPa)(84.3 mm2 ) = 24, 025 N ≈ 24 kN The maximum load is obtained from Eq. (16.31): At S p − P i nsb = P max,b C k

At S p − P i (84.3 mm2 )(380 MPa) − 24 kN P max,b = = = 11.5 kN nsb C k 2.5(0.279)

→

(c) Joint Separation The safety factor against joint separation is given by Eq. (16.32) as

nsj =

P i 24 kN = = 2.89 P max,j (1 − C k ) (11.5 kN)(1 − 0.279)

(d) Fatigue Analysis. Assuming the threads are rolled, then K f = 2.2 from Table 16.11. We don’t know how the load is applied, but if we assume the loading is axial, then from Eq. (7.7) the endurance limit is S e = 0.45S u = 0.45(520 MPa) = 234 MPa. The prestress is σi = P i /At = (24 kN)/(84.3 mm2 ) = 285 MPa. The alternating stress is given by Eq. (16.36) as

σa =

C k P a 0.279(5000 N) = = 16.54 MPa At 84.3 mm2

and the mean stress is calculated from Eq. (16.37) as

σm =

P i + C k P m 24 kN + (0.279)(5000 N) = = 301.2 MPa At 84.3 mm2

Therefore, the safety factor against fatigue failure is given by Eq. (16.40) as

ns =

S ut − σi P a S ut P m C k K f + At S e At

   

=

520 MPa − 285 MPa = 2.20 520 (0.279) 2.2(16.54 MPa) + 301.2 MPa 234



16.14 Repeat Problem 16.13 if the 25-mm-thick member is made of steel.

 



395

MI2 coarse-pitch, class-5.8 bolt

Region 1 35 Aluminum Region 2 5 Region 3

25 Steel

Notes: This is complicated since the two materials are different, and the Wileman method cannot be

used. Solution:

(a) Bolt and Member Stiffness. Note from the inside front cover that for steel, E s = 207 GPa and for aluminum E al = 72 GPa. From the sketch we see that Lt = 20 mm and Ls = 40 mm. From Table 16.10, for a crest diameter of 12 mm and coarse threads, p = 1.75 mm, At = 84.3 mm2 . From Eq. (16.2), h t is ht =

0.5 p 0.5(1.75 mm) = = 1.516 mm tan30 tan30 ◦

◦

From Fig. 16.4, the root diameter is: dr = d c − 2(0.625ht ) = 12 mm − 2(0.624)(1.516 mm) = 10.105 mm Therefore, the bolt stiffness is, from Eq. (16.21), 1 kb

= =





4 Ls + 0.4dc L t + 0.4dr + πE d2c d2r 4 0.04 m + 0.4(0.012 m) 0.02 m + 0.4(0.010105 m) + π(207 GPa) (0.012 m)2 (0.010105 m)2





or k b = 297.5 MN/m. For the member, we need to use the regions defined in the modified sketch. Note that d c = 0.012 m, and also note the following: Region

Material

I II

Aluminum Aluminum

III

Steel

di 1.5(0.012 m) = 0.018 m 0.018 + 2(0.025m)tan30 = 0.04087 m 0.018 m

◦

Li 0.030 m 0.005 m 0.025 m

396

CHAPTER 16.


◦

The stiffness of member 1 is then, from Table 16.8 with α = 30 , kj1

=

1.813E j dc (1.15Li + di − dc )(di + dc ) ln 1.15Li + di + dc )(di − dc )



= 1.350 × 109 N/m

=

1.813(72 GPa)(0.012 m) [1.15(0.030) + 0.018 − 0.012](0.018 + 0.012) ln [1.15(0.030) + 0.018 + 0.012](0.018 − 0.012)

 



Similarly, for the other members, kj2

=




= 10.7053 × 109 N/m

kj3

=




= 1.4245 × 109 N/m

=

1.813(72 GPa)(0.012 m) [1.15(0.005) + 0.04087 − 0.012](0.04087 + 0.012) ln [1.15(0.005) + 0.04087 + 0.012](0.04087 − 0.012)

  =

1.813(207 GPa)(0.012 m) [1.15(0.025) + 0.018 − 0.012](0.018 + 0.012) ln [1.15(0.025) + 0.018 + 0.012](0.018 − 0.012)

 



Therefore, the joint stiffness, from Eq. (16.25) is 1 1 1 1 1 1 1 = + + = + + 9 9 kj kj1 kj2 kj3 1.350 × 10 10.7053 × 10 1.4245 × 109

→

kj = 669.6 MN/m

The stiffness parameter C k is C k =

kb 297.5 MN/m = = 0.3076 kb + km 297.5 MN/m + 669.6 MN/m

(b) Maximum Load. From Table 16.8, for a 5.8 grade bolt, the proof strength is S p = 380 MPa, S ut = 520 MPa and S y = 415 MPa. If we assume this is a reused connection, then from Eq. (16.33), P i = 0.75P p = 0.75S pAt = 0.75(380 MPa)(84.3 mm2 ) = 24, 025 N ≈ 24 kN The maximum load is obtained from Eq. (16.31): nsb =

At S p − P i P max,b C k

→

P max,b =

At S p − P i (84.3 mm2 )(380 MPa) − 24 kN = = 10.4 kN nsb C k 2.5(0.3076)

(c) Joint Separation The safety factor against joint separation is given by Eq. (16.32) as nsj =

P i 24 kN = = 3.41 P max,j (1 − C k ) (10.4hboxkN )(1 − 0.3076)

(d) Fatigue Analysis. Assuming the threads are rolled, then K f = 2.2 from Table 16.11. We don’t know how the load is applied, but if we assume the loading is axial, then from Eq. (7.7) the endurance limit is S e = 0.45S u = 0.45(520 MPa) = 234 MPa. The prestress is σi = P i /At = (24 kN)/(84.3 mm2 ) = 285 MPa. The alternating stress is given by Eq. (16.36) as σa =

C k P a 0.3076(5000 N) = = 18.24 MPa At 84.3 mm2



397

and the mean stress is calculated from Eq. (16.37) as σm =

P i + C k P m 24 kN + (0.3076)(5000 N) = = 302.9 MPa At 84.3 mm2

Therefore, the safety factor against fatigue failure is given by Eq. (16.40) as ns =


   

=

520 MPa − 285 MPa = 1.95 520 (0.3076) 2.2(18.24 MPa) + 302.9 MPa 234



 



16.15 Find the total shear load on each of the three bolts for the connection shown in sketch e. Also, compute the shear stress and the b earing stress. Find the area moment of inertia for the 8-mm-thick plate on a section through the three bolt holes. Holes for M12

× 1.75

8 mm thick 12 kN

36 32 64

36

200 Column

Sketch e, for Problem 16.15 Notes: An assumption must be made that manufacturing tolerances are such that bolts are loaded

evenly in the vertical direction and that the moment reactions are equally shared by the top and bottom bolts. Also, the bolts can be taken as points so that the shear stress is uniform over the cross section. Solution: The vertical component of load is 4 kN for each bolt if it is shared equally. The moment

of (12 kN)(0.2 m)=2400 Nm is shared by the top and bottom bolts, requiring each to generate a horizontal force of 2400 Nm/((2)(0.032 m))=37.5 kN. The direction of the force is opposite for the top and bottom bolts, but this does not matter in terms of the maximum stress. The shear load on the center bolt is simply 4 kN. The top and bottom bolts see a shear of P =



(4 kN)2 + (37.5 kN)2 = 37.7 kN

The bolts should be loaded on the shank, not on the threads for such an application, so that the loaded area is πd 2c π(0.012 m)2 A = = = 1.13 × 10 4 m2 4 4 The central bolt sees a shear stress of −

τ c =

4 kN = 35.4 MPa 1.13 × 10 4 m2 −

while the outer bolts see a stress of τ o =

37.7 kN = 334 MPa 1.13 × 10 4 m2 −

398

CHAPTER 16.


The bearing stress (see Eq. (16.47)) for the outer bolts is σ =

P 37.7 kN =− = −393 MPa td (0.008 m)(0.012 m)

where the negative sign indicates a compressive stress. The moment of inertia of the cross section is easily calculated using the parallel-axis theorem: th3 td3 td3 − −2 I = + a2 td 12 12 12 (0.008 m)(0.136 m)3 (0.008 m)(0.012 m)3 − = 12 12 3 (0.008 m)(0.012 m) 2 −2 − 2(0.032 m) (0.008 m)(0.012 m) 12





6

−

= 1.48 × 10



m4



The bending stress of the plate is σ =

M c (2400)(0.068) = = 110.3 MPa I 1.48 × 10 6 −

16.16 A coarse-pitch, SAE grade-5 bolt with a hexagonal nut assembly is used to keep two machine parts together as shown in sketch f . The major diameter of the bolt is 0.5 in. The bolt and the bottom member are made of carbon steel. Assume that the connection is to be reused. Length dimension is in inches. Determine the following: (a) Length of the bolt (b) Stiffnesses of the bolt and the member, assuming a washer ensures d i = 1.5dc under the bolt and the washer. (c) Safety factor guarding against separation of the members when the maximum external load is 5000 lb (d) Safety factor guarding against fatigue if the repeated maximum external load is 2500 lb in a released-tension loading cycle

Region 1 1.2

Aluminum

Region 2

Region 3

1.0

Carbon steel

Sketch f , for Problem 16.16

399

Notes: This problem uses the information in previous problems regarding calculation of bolt and joint

stiffness and introduces safety factor calculations from Eqs. (16.32) and (16.40). Solution: The members require 2.2 in of bolt length. Note that we must also allow for the nut and a

few threads beyond the nut. Allowing an extra half-inch of length would lead to a length of 2.7 inches, so we specify a 2.75 in. length to use a standard bolt size. Therefore, the threaded length is given by Eq. (16.23) as Lttot = 2dc + 0.25 in. = 2(0.5 in.) + 0.25 = 1.25 in. Therefore, Ls = L − Lt = 2.75 in. − 1.25 in. = 1.50 in. Since the members are a total of 2.2 inches in thickness, there is only 0.7 in. of thread length in the members, so referring to Fig. 16.12, we take Lt = 0.7 in. From the inside front cover that for steel, E s = 30 Mpsi and for aluminum E al = 10.5 Mpsi. From Table 16.9, for a crest diameter of 0.50 in. and coarse threads, n = 13 threads/in, so p = 1/13 in. = 0.0769 in, A t = 0.1419 in.2 From Eq. (16.2), h t is 0.5 p 0.5(0.0769 in.) = = 0.0666 in. tan30 tan30

ht =

◦

◦

From Fig. 16.4, the root diameter is: dr = d c − 2(0.625ht ) = 0.5 in. − 2(0.625)(0.0666 in.) = 0.4167 Therefore, the bolt stiffness is, from Eq. (16.21), 1 kb

= =





4 Ls + 0.4dc L t + 0.4dr + πE d2c d2r 4 1.5 in. + 0.4(0.5 in.) 0.7 in. + 0.4(0.4167 in.) + π(30 Mpsi) (0.5 in.)2 (0.4167 in.)2





or k b = 2.00 Mlb/in. For the member, we need to use the regions defined in the modified sketch. Note that d c = 0.50 in., and also note the following: Region

Material

1 2 3

Aluminum Aluminum Carbon Steel

di 1.5(0.50 in.) = 0.75 in. 1.6547 in. 0.75 in.

Li 1.1 in. 0.1 in. 1.0 in.

◦


=




= 8.633 × 106 lb/in.

=

1.813(10.5 Mpsi)(0.50 in.) [1.15(1.1) + 0.75 − 0.5](0.75 + 0.50) ln [1.15(1.1) + 0.75 + 0.50](0.75 − 0.50)

 





=



= 221.6 × 106 lb/in. kj3

=




= 25.41 × 106 lb/in.

=

1.813(10.5 Mpsi)(0.50 in.) [1.15(0.1) + 1.6547 − 0.50](1.6547 + 0.50) ln [1.15(0.1) + 1.6547 + 0.50](1.6547 − 0.50)

  =

1.813(30 Mpsi)(0.50 in.) [1.15(1.0) + 0.75 − 0.50](0.75 + 0.50) ln [1.15(1.0) + 0.75 + 0.50](0.75 − 0.50)

 





400

CHAPTER 16.



→

kj = 6.26 × 106 lb/in.


kb 2.0 MN/m = = 0.2421 kb + km 2.0 + 6.26

For an SAE grade 5 bolt, Table 16.7 gives S u = 120 ksi, S y = 92 ksi and S p = 85 ksi. Since the connection is to be reused, Eq. (16.33) gives P i = 0.75P p = 0.75(At S p ) = 0.75(0.1419 in.2 )(85 ksi) = 9.046 kip Therefore, from Eq. (16.32), the safety factor against separation is nsj =

P i 9.046 kip = = 2.39 P max,j (1 − C k ) (5 kip)(1 − 0.2421)

If the repeated maximum external load is 2500 lb, in a released-tension cycle, then P a = P m = 1250 lb. Referring to Table 16.11, we use K f = 3.0 since we are using a standard size where the threads can be rolled. (A specialty bolt is more likely to be cut.) Therefore, from Eq. (7.7), and using this factor,



1 1 = (0.45)(120) S e = 0.45S u K f 3

= 18.0 ksi

From Eq. (16.40),

ns =


   

=

120 ksi − (9.046 kip/0.1419 in.2 ) 1.250 kip 120 1.250 kip 0.2421 + 2 18 0.1419 in. 0.1419 in.2



 



or n s = 3.44.

16.17 An electric-motor-driven press (sketch g) has the total press force P = 5000 lb. The screws are Acme type with β = 29 , d p = 3 in., p = l = 0.5 in., and µ = 0.05. The thrust bearings for the screws have dc = 5 in. and µc = 0.06. The motor speed is 1720 rpm, the total speed ratio is 75:1, and the mechanical efficiency e = 0.95. Calculate ◦

(a) Press head speed (b) Power rating needed for the motor

401

Motor

Bearings Worm Spur gears

Bronze bushings

B C

Foot

Collar bearing

A

Sketch g , for Problem 16.17 Notes: This problem builds on the approach in Problems 16.1 to 16.8. Solution: The speed of the power screw is

N a = 1720/75 = 22.93 rpm = 144.1 rad/s Therefore the speed of the press head is v = N a l = (22.93 rev/min)(0.5 in.) = 11.47 in./min The load per screw is W = 5000 lb/2 = 2500 lb. The pitch diameter is obtained from Eq. (16.4): d p = d c − 0.5 p − 0.01 = 3 in. − 0.5(0.5 in.) − 0.01 in. = 2.74 in. The torque is given by Eq. (16.10), but first, from Eq. (16.5), α is given by tan α =

l 0.5 in. = = 3.32 2πrm π(2.74 in.)

◦

Note from Figure 16.5 that β = 29 . From Eq. (16.8), θ is ◦

θn = tan

1

−



β cos α tan 2



= tan

1

−



29 cos3.32 tan 2 ◦

◦



◦

= 14.47






(d p /2)(cos θn tan α + µ) = W + rc µc cos θn − µ tan α (2.74 in./2)(cos 14.47 tan3.32 + 0.05 = (2500 lb) + (2.5 in.)(0.06) cos14.47 − 0.05tan3.32 = 752 in.-lb



◦

◦

◦

◦

The motor torque is, including the inefficiency in the power transmission, T motor =

(752 in.-lb/screw)(2 screws) 1 = 21.10 in.-lb 75 0.95



402

CHAPTER 16.

The power required is h p =


T motor N a = 0.577 hp 63, 000

16.18 A valve for high-pressure air is shown in sketch h. The spindle has thread M12 with a pitch diameter of 10.9 mm, lead l = 1.75 mm, and a thread angle of 60 . Derive the relationship between torque and axial thrust force, and calculate the axial force against the seating when the applied torque is 10 N-m during tightening. The coefficient of friction is 0.12. ◦

T r

Thread

Airflow Seating

Sketch g , for Problem 16.18 Notes: This bolt will prevent flow unless an inlet pressure develops a force on the bolt larger than the

axial force during tightening. The analysis requires the use of Eqs. (16.5), (16.8) and (16.10). Solution: The torque is given by Eq. (16.10), but first, from Eq. (16.5), α? is given by

tan α =

l 1.75 mm = = 2.92 2πr m π(10.9 mm)

◦

From Eq. (16.8), θ n is θn = tan

1

−



β cos α tan 2



= tan

1

−



60 cos2.92 tan 2 ◦

◦



◦

= 29.97


 



(d p /2)(cos θn tan α + µ) = W + rc µc cos θn − µ tan α (0.0109 m/2)(cos 29.97 tan2.92 + 0.12 = W cos29.97 − 0.12tan2.92 = 0.00104 mW ◦

◦

◦

◦



And if T = 10 Nm, then W = 9.61 kN 16.19 Derive the expression for the power efficiency of a lead screw with a flat thread (thread angle β = 0 ) and find the lead angle α that gives maximum efficiency in terms of the coefficient of friction. Also give results if µ = 0.1. ◦

403

Notes: This problem uses Eqs. (16.5), (16.10) and (16.13) to obtain an expression of e in terms of

α. The maximum efficiency is found by taking the derivative of this expression, setting the derivative equal to zero and solving for α. Solution: Note from Eq. (16.8) that θ n = 0. Therefore, from Eq. (16.10),



(d p /2)(tan α + µ) T r = W + rc µc 1 − µ tan α



The collar’s contribution to the lead screw efficiency has noting to do with the thread, and we will neglect the second term in the parentheses. Therefore, the efficiency is, from Eq. (16.13),





Wl l 1 − µ tan α l e = = = 2πT r 2π (d p /2)(tan α + µ) πd p tan α



1 − µ tan α 1 + µ cot α



But note from Eq. (16.5) that tan α = l/πd p , so that e =






To find the optimum efficiency, take the derivative with respect to α, set to zero and solve for α: ∂e ∂ = ∂α ∂α






µ −(1 + µ cot α) − (1 − µ tan α) cos2 α = (1 + µ cot α)2

 



µ − sin2 α



=0

Therefore the numerator must be zero, so 1 + µ cot α) sin2 α = (1 − µ tan α) cos2 α

 

 

cos2 α − sin2 α = 2µ sin α cos α cos2α = µ sin2α 1 ∴ tan 2α = µ For µ = 0.1, then α = 42 . ◦

16.20 A car manufacturer has problems with the cylinder head studs in a new high-power motor. After a relatively short time the studs crack just under the nuts, the soft cylinder head gasket blows out, and the motor stops. To be able to analyze the problem, the car manufacturer experimentally measures the stiffnesses of the various components. The stiffness for all bolts together is 400 N/ µm, the stiffness of the gasket is 600 N/µm, and the stiffness of the cylinder head that compresses the gasket is 10,000 N/µm. By comparing the life-stress relationships with those for rolling-element bearings, the car manufacturer estimates that the stress amplitude in the screws needs to be halved to get sufficient life. How can that be done? Notes: This problem can be solved with a good understanding of the concepts of bolt and joint

stiffnesses and the dimensionless joint parameter. Solution: When an extra load ∆P is applied to the cylinder head, the force in the stud increases to

∆P b and the compression of the gasket and head decrease to ∆ P − ∆P b . The extra force in the studs is ∆P b = kb δ and the force decrease in the gasket is ∆P G = ∆P − ∆P b =

δ



1 1 + kg kb



404

CHAPTER 16.


For a given ∆P and δ , ∆P b has to be halved by changing the stiffness. From this equation, we can write ∆P = ∆P G + ∆P b = δ

∆P b = ∆P

 

1 + kb 1 1 + kg kb

 

kb δ 400 = = 0.4141 1 1 + 400 δ + kb 1/600 + 1/10, 000 1/kg + 1/kb





To decrease this to one-half, or 0.2070, the gasket stiffness has to increase. Using the same equation,





1 + 400 (0.2070) = 400 1/x + 1/10, 000

x = 1809 N/m

→

16.21 A pressure vessel of compressed air is used as an accumulator to make it possible to use a small compressor that works continuously. The stiffness parameter for the lid around each of the 10-mm bolt diameters is 900 MN/m. The shank length is 20 mm. Because the air consumption is uneven, the air pressure in the container varies between 0.2 and 0.8 MPa many times during a week. After 5 years of use one of the bolts holding down the top lid of the pressure vessel cracks off. A redesign is then made, decreasing the stress variation amplitude by 25%, to increase the life of the bolts to at least 50 years. The stress variation amplitude is decreased by lengthening the bolts and using circular tubes with the same cross-sectional area as the solid circular cross section of the bolt to transfer the compressive force from the bolt head to the lid. Calculate how long the tubes should be. Notes: The solution of this problem requires the use of Eqs. (16.12), (16.16), and (16.21). Solution: From Eq. (16.12),



1 4 0.02 + 0.4(0.01) 0.4 = + kb π(207) (109 ) 0.012 0.0859



→

kb = 664.5 × 106 N/m

From Eq. (16.16), ∆P b = P b,max − P b,min =

kb (P max − P min ) kb + kj

∆P b is decreased by 25%, so the new bolt stiffness, k bn , is kbn kb (664.5)(106 ) = 0.75 = 0.75 = 0.3186 (664.5 + 900)106 kbn + kj kb + kj Solving for k bn yields k bn = 420.7 × 106 N/m. Equation (16.21) then gives 1 4 = kbn πE



Ls + 2L + 0.4dc 0.4 + d2c dr



where L is the length of the tube and the extra screw length. Therefore, 1 4 = 420.7 × 106 π(207 × 109 )



0.02 + 2L + 0.4(0.01) 0.4 + (0.01)2 0.0859



→

The tube should be 7.089 mm long and have a cross sectional area of 78.54 mm 2 .

3

−

L = 7.089 × 10

m

405

16.22 A loading hook of a crane is fastened to a block hanging in six steel wires. The hook and block are bolted together with four 10-mm-diameter screws prestressed to 20 000 N each. The shank length is 80 mm and the thread length is 5 mm. The stiffness of the material around each screw is 1 GN/m. One of the screws of the crane cracks due to fatigue after a couple of years of use. Will it help to change the screws to 12-mm diameter while other dimensions are unchanged if the stress variation needs to be decreased by at least 20%? Notes: This problem uses Eq. (16.21) to calculate the bolt stiffness and Eq. (16.16) to calculate the

force variation in the bolt. Solution: From Table 16.10, the tensile stress area for a 10 mm crest diameter screw is 58.0 mm

2

and

the pitch is p = 1.5 mm. Therefore, from Eq. (16.2), h t is ht =

0.5 p 0.5(1.5 mm) = = 1.299 mm tan30 tan30 ◦

◦

From Fig. 16.4, the root diameter is: dr = dc − 2(0.625ht ) = 10 mm − 2(0.625)(1.299 mm) = 8.376 mm For a 12-mm crest diameter, A t = 84.3 mm2 , p = 1.75 mm and d r = 10.11 mm. From Eq. (16.21) 1 k10

4 πE





Ls + 0.4dc L t + 0.4dr = + d2c d2r 4 0.08 m + 0.4(0.01 m) 0.005 m + 0.4(0.008376 m) = + π(207 GPa) (0.01 m)2 (0.008376 m)2 = 169.5 MN/m





Similarly for the 12 mm bolt, 1 k12

4 πE










The bolt force variation is given by Eq. (16.16): ∆P b10 =

kb ∆P 169.5 MN/m = ∆P = 0.145∆P kb + kj 169.5 MN/m + 1000 MN/m

For the 12 mm bolt: ∆P b12 =

kb ∆P 240.0 MN/m = ∆P = 0.194∆P kb + kj 240.0 MN/m + 1000 MN/m

The stress variation is the force variation divided by the stress area: ∆P b10 0.145∆P ∆σ10 = = At 58.0 mm2 ∆P b12 0.194∆P ∆σ12 = = At 84.3 mm2

 

1000 mm 1m 1000 mm 1m

2

   

2

−

= 2500 m

2

2

−

= 2300 m

 

∆P

∆P

However, if the stress variation is supposed to be reduced 20%, then the stress variation should be (0.8)(2500)∆P = (2000 m 2 )∆P . The twelve millimeter bolts are still insufficient. −

406

CHAPTER 16.


16.23 Depending on the roughness of the contacting surfaces of a b olted joint, some plastic deformation takes place on the tops of the roughness peaks when the joint is loaded. The rougher the surfaces are, the more pressure in the bolted joint is lost by plastic deformation. For a roughness profile depth of 20 µm on each of the surfaces a plastic deformation of 6.5 µm can be expected for the two surfaces in contact. For a bolt-and-nut assembly as shown in Fig. 16.13 three sets of two surfaces are in contact. The stiffness of the two steel plates together is 700 MN/m when each is 40 mm thick. The bolt diameter is 16 mm with metric thread. The shank length is 70 mm. The bolt is prestressed to 25 kN before plastic deformation sets in. Calculate how much of the prestress is left after the asperities have deformed. Notes: To solve this problem, one must assume that the bolt is prestressed very quickly, and that

plastic deformation in the joint surface asperities takes place much more slowly. Then one can calculate the stress in the bolt after plastic deformation has relaxed the stress from the initial state. Solution: From Table 16.10, for a 16 mm bolt, the tensile stress area is At = 157 mm2 , p = 2 mm.

Therefore, from Eq. (16.2), h t is ht =

0.5 p 0.5(2 mm) = = 1.732 mm tan30 tan30 ◦

◦

From Fig. 16.4, the root diameter is: dr = dc − 2(0.625ht ) = 16 mm − 2(0.625)(1.732 mm) = 13.83 mm Therefore the bolt stiffness is, from Eq. (16.21), 1 kb

4 πE










The deformation of the joint before plastic deformation occurs is: δ e = F e



1 1 + kb kj



= (25 kN)



1 1 + 428.3 MN/m 700 MN/m



= 94.1 µm

The plastic deformation reduces this deformation in the bolt. The remaining deformation is (note that there are three surfaces involved): δ p = δ e − 3(6.5 µm) = 94.1 µm − 3(6.5 µm) = 74.6 µm Therefore the remaining force is δ p = F p



1 1 + kb kj



→

F p =

δ p 74.6 µm = = 19.8 kN (1/kb + 1/kj ) (1/428.3 MN/m + 1/700 MN/m)

16.24 An ISO M12 × 1.75 class = 12.9 bolt is used to fasten three members as shown in sketch i. The first member is made of cast iron, the second is low-carbon steel, and the third is aluminum. The static loading safety factor is 2.5. Dimensions are in millimeters. Determine (a) Total length, threaded length, and threaded length in the joint ◦

(b) Bolt-and-joint stiffness using a 30 cone (c) Preload for permanent connections (d) Maximum static load that the bolt can support

407

25

I

1

II

2

10

III

IV

3

30

Sketch i, for Problem 16.24 Notes: The student has to determine a reasonable bolt length, and there can be some difference in

the bolt length assumed. This solution uses a bolt length of 80 mm. Calculating the bolt stiffness is as done in previous problems. The joint stiffness is calculated using Eq. (16.24) instead of Eq. (16.26) as was done previously. Solution: The cone has been sketched in the figure with the sections labeled. The bolt length must

be 65 mm plus the length of the nut and space for a few threads beyond the nut. Therefore, we take L = 80 mm. From Table 16.10, A t = 84.3 mm2 and p = 1.75 mm. Therefore, from Eq. (16.2), h t is 0.5 p 0.5(1.75 mm) = = 1.516 mm tan30 tan30

ht =

◦

◦

From Fig. 16.4, the root diameter is: dr = d c − 2(0.625ht ) = 12 mm − 2(0.625)(1.516 mm) = 10.105 mm From Table 16.8, S p = 970 MPa. From Eq. (16.22), Lt = 2dc + 6 mm = 2(12 mm) + 6 mm = 30 mm Therefore the shank length is L s = L − Lt = 80 mm − 30 mm = 50 mm. Therefore the bolt stiffness is, from Eq. (16.21), 1 kb

4 πE










For the joint stiffness, we apply Eq. (16.24) for each joint section: kji =

πE i dc tan αf (2Li tan αf + di − dc )(di + dc ) ln (2Li tan αf + di + dc )(di − dc )





For section I: cast iron, use Nodular cast iron, E = 172 GPa (inside front cover), dc = 0.012 m, di = 1.5dc = 0.018 m (see text explaining the member closest to the nut always has d i = 1.5dc ), L i =

408

CHAPTER 16.


0.025 m. Inserting these values and evaluating, kjI = 3.448 GN/m. For section II, low carbon steel, E = 207 GPa, d c = 0.012 m, d i = 0.018m +2(0.025 m) tan 30 = 0.0469 m, L i = 0.0075 m. Therefore, kjI I = 53.3 GN/m. For section IV, aluminum, E = 69 GPa, dc = 0.012 m, di = 1.5dc = 0.018 m, Li = 0.030 m. Therefore, kjI V = 1.311 GN/m. For section III, low carbon steel, E = 207 GPa, dc = 0.012, di = 0.018 m + 2(0.030 m) tan 30 = 0.0526 m, Li = 0.0025 m. Therefore, kjIII = 180.4 GN/m. Therefore, the joint stiffness is, from Eq. (16.25), ◦

◦

1 1 1 1 1 1 1 1 1 = + + + = + + + kj kjI kjI I kjIII kjI V 3.448 GN/m 53.3 GN/m 1.311 GN/m 180.4 GN/m or k j = 0.928 GN/m. Therefore, from Eq. (16.17), C k =

kb 286.8 MN/m = = 0.236 kb + kj 286.8 MN/m + 0.928 GN/m

The proof load is P p = At S p = (84.3 mm2 )(970 MPa) = 81.8 kN. For a permanent connection, Eq. (16.33) gives P i = 0.90P p = 73.6 kN. Therefore, from Eq. (16.31), ns =

At S p − P i P max,b C k

→

P max,b =

At S p − P i (84.3 mm2 )(970 MPa) − 73.6 kN = = 13.8 kN ns C k 2.5(0.236)

16.25 A pressurized cast iron cylinder shown in sketch j is used to hold pressurized gas at a static pressure of 8 MPa. The cylinder is joined to a low-carbon-steel cylinder head by bolted joints. The bolt to be used is metric grade 12.9 with a safety factor of 3. Dimensions are in millimeters. Determine the bolt dimensions and the required number of bolts. Use grade 12.9 bolts, with M36 × 100 mm coarse threads. 850

25 30

Sketch j , for Problem 16.25 Notes: The bolt stiffness is determined from Eq. 16.21 after the bolt geometry is determined. The

joint stiffness is found from Eq. (16.26), although Eq. (16.24) could be used. Solution: Using M36

× 100 mm grade 12.9 bolts, from Table 16.8, S u = 1220 MPa, S y = 1100 MPa, S p = 970 MPa, and from Table 16.10, A t = 817 mm2 and p = 4 mm. Therefore, from Eq. (16.2), ht is

ht =

0.5 p 0.5(4 mm) = = 3.464 mm tan30 tan30 ◦

◦

From Fig. 16.4, the root diameter is: dr = dc − 2(0.625ht ) = 36 mm − 2(0.625)(3.464 mm) = 31.67 mm For the 100 mm bolt, the thread length is, from Eq. 16.22, Lt = 2dc + 6 = 78 mm

409

Therefore the shank length is 22 mm. From the figure, 33 mm of the thread is exposed in the member. The bolt stiffness is obtained from Eq. (16.21): 1 kb

4 πE





Ls + 0.4dc L t + 0.4dr = + d2c d2r 4 0.022 m + 0.4(0.036 m) 0.033 m + 0.4(0.03167 m) = + π(207 GPa) (0.036 m)2 (0.03167 m)2 = 2.208 GN/m





For the member, we need to use the regions defined in the sketch below which shows a detail of a single bolt and the joint.

Region 1

25 mm Carbon steel

Region 2

30 mm

Region 3

Cast iron

Note that d c = 0.036 m, and also note the following: Region

Material

1 2 3

Carbon steel Cast iron Cast iron

di 1.5(0.036 m) = 0.054 m 0.08575 0.054 m

Li 0.025 m 0.0025 m 0.0275 m

The stiffness of member 1 is then, from Table 16.8 with α = 30 , ◦

kj1

=




= 19.95 × 109 N/m

=

1.813(206.8 GPa)(0.036 m) [1.15(0.025) + 0.054 − 0.036](0.054 + 0.036) ln [1.15(0.025) + 0.054 + 0.036](0.054 − 0.036)

 





=



= 198.7 × 109 N/m kj3

=




= 9.154 × 109 N/m

=

1.813(100 GPa)(0.036 m) [1.15(0.0025) + 0.08575 − 0.036](0.08575 + 0.036) ln [1.15(0.005) + 0.08575 + 0.036](0.08575 − 0.036)

  =

1.813(100 GPa)(0.036 m) [1.15(0.0275) + 0.054 − 0.036](0.054 + 0.036) ln [1.15(0.0275) + 0.054 + 0.036](0.054 − 0.036)

 





410

CHAPTER 16.



→

kj = 6.083 × 109 N/m

The stiffness parameter C k is kb 2.28 GN/m = = 0.2726 kb + km 2.28 GN/m + 6.083 GN/m

C k =

Assuming the connection is permanent, which is usually the case for pressure vessels caps, Eq. (16.33) gives P i = 0.9P p = 0.9S p At = 713 kN With the specified loaded diameter, the applied load is P = pA = 8 MPa



π (0.85 m)2 = 4.54 MN 4



The maximum load that can be taken by a bolt is given by Eq. (16.31): nsb =

At S p − P i P max C k

→

P max =

At S p − P i (817 mm2 )(970 MPa) − 713 kN = = 97.23 nsb C k 3(0.2726)

Note that P /P max = 4.54 MN/97.23 kN = 46.7; use 47 bolts. Equation 16.32 then gives nsj = 3 =

P i P max,j (1 − C k ) n

=

713 kN 4.54 MN (1 − 0.2726) n

→

n = 13.9

Therefore, use 47 bolts. 16.26 In the bolted joint shown in sketch k the first member is made of low-carbon steel, the second member is aluminum, and the third member is cast iron. Assuming that the members can be rearranged and the frustum cone angle is 45 , find the arrangement that can support the maximum load. Dimensions are in millimeters. ◦

1

35

2

60

3

40

Sketch k, for Problem 16.26

411

Notes: The maximum load can be achieved by minimizing C k (see Eq.

(16.31)), and therefore maximizing kj . One can solve this problem by considering all alternatives and calculating the joint stiffness, or one can recognize that symmetry reduces the number of alternatives by one-half. Solution: The maximum load occurs from the smallest value of C k [see Eq. (16.31)]. From Eq.

(16.17), C k =

kb kb + kj

So to minimize C k we want to maximize k j . The stiffness of a joint member is kji =

πE i dc tan αf (2Li tan αf + di − dc )(di + dc ) ln (2Li tan αf + di + dc )(di − dc )





We can take any value of d c if our goal is to investigate trends; we take d c = 0.020 m. (a) Steel-Al-Iron. For this case, we have four frustums: 1. low carbon steel, E = 207 GPa, d c = 0.020 m, di = 1.5dc = 0.030 m, L i = 0.035 m, so that k1 = 15.8 GN/m. 2. aluminum, E = 69 GPa, dc = 0.020 m, di = 0.03 + 0.035sin45 = 0.0547, Li = 0.0325, so that k 2 = 13.6 GN/m. 3. aluminum, E = 69 GPa, dc = 0.020, d i = 0.03 + 0.04sin45 = 0.0583 m, L i = 0.0275 m, so that k 3 = 15.88 GN/m 4. cast iron, E = 172 GPa, dc = 0.020 m, di = 1.5dc = 0.030 m, Li = 0.040 m, so that k4 = 12.89 GN/m ◦

◦

Therefore, the stiffness of the member is 1 1 1 1 1 1 1 1 1 = + + + = + + + kj k1 k2 k3 k4 15.8 GN/m 13.6 GN/m 15.88 GN/m 12.89 GN/m or k j = 3.61 GN/m (b) Steel-Iron-Al. 1. same as above, k 1 = 15.8 GN/m. 2. cast iron, E = 172 GPa, d c = 0.020 m, d i = 0.0547 m, L i = 0.0325 m, k 2 = 34.0 GN/m 3. cast iron, E = 172 GPa, d c = 0.020 m, di = 0.03 + 0.06sin45 = 0.0724 m, Li = 0.0075 m, k3 = 118.8 GN/m 4. aluminum, E = 69 GPa, d c = 0.020, d i = 0.03, L i = 0.06, k 4 = 4.9 GN/m ◦

so that k = 3.28 GN/m. (c) Al-Steel-Iron. 1. aluminum, E = 69 GPa, d c = 0.020, d i = 0.03, L i = 0.06, k 1 = 4.9 GN/m. 2. steel, E = 207 GPa, dc = 0.02, d i = 0.03 + 0.06sin45 = 0.0724 m, L = 0.0075 m, k 2 = 143 GN/m 3. steel, E = 207 GPa, d c = 0.02, d i = 0.0 3 + 0.04sin45 = 0.0583 m, L = 0.00275 m, k 3 = 208 GN/m 4. cast iron, E = 172 GPa, dc = 0.02, di = 0.03 m, Li = 0.04 m, k4 = 12.9 GN/m so that kj = 3.41 GN/m. ◦

◦

Note that all other combinations are the same as one of these. The stiffest member occurs with aluminum in the center.

412

CHAPTER 16.


16.27 The cylinder shown in sketch l is pressurized up to 2 MPa and is connected to the cylinder head by sixteen M24 × 3 metric-grade-8.8 bolts. The bolts are evenly spaced around the perimeters of the two circles with diameters of 1.2 and 1.5 m, respectively. The cylinder is made of cast iron and its head is made of high-carbon steel. Assume that the force in each bolt is inversely related to its radial distance from the center of the cylinder head. Calculate the safety factor guarding against failure due to static loading.

1.5 m 1.2 m 25 mm 1m

30 mm

Sketch l, for Problem 16.27 Notes: This problem is similar to Problem 16.22 in the approach of determining the bolt and joint

stiffness. The maximum load can be calculated using the assumption given in the problem statement. This problem is open ended in that the student must choose a bolt length, so answers may vary slightly. This solution uses a bolt length of 80 mm. Solution:

(a) Bolt and Joint Analysis. From Table 16.8, for a grade 8.8 bolt, S p = 600 MPa. From Table 16.10, for d c = 24 mm, A t = 353 mm 2 and p = 3 mm. Therefore from Eq. (16.2), h t is ht =

0.5 p 0.5(3 mm) = = 2.598 mm tan30 tan30 ◦

◦

From Fig. 16.4, the root diameter is: dr = d c − 2(0.625ht ) = 24 mm − 2(0.625)(2.598 mm) = 20.75 mm A reasonable bolt length for the joint and with clearance for the nut is 80 mm. From Eq. (16.22), Lt = 2dc + 6 mm = 2(24 mm) + 6 mm = 54 mm

413

Therefore Ls = 80 − 54 = 26 mm. Since the combined joint thickness is 55 mm, the threaded length in the members is 55 − 26 = 29 mm. The bolt stiffness is calculated from Eq. (16.21): 1 kb

4 πE





Ls + 0.4dc L t + 0.4dr = + d2c d2r 4 0.026 m + 0.4(0.024 m) 0.029 m + 0.4(0.02075 m) = + π(207 GPa) (0.024 m)2 (0.02075 m)2 = 1.095 GN/m





For the member, see the sketch below to identify the regimes.

Region 1

25 mm Carbon steel

Region 2

30 mm

Region 3

Cast iron

Note that d c = 0.024 m, and also note the following: Region

Material

1 2 3

Carbon steel Cast iron Cast iron

di 1.5(0.024 m) = 0.036 m 0.06775 0.036 m

Li 0.025 m 0.0025 m 0.0275 m

◦


=




= 10.84 × 109 N/m

=

1.813(206.8 GPa)(0.024 m) [1.15(0.025) + 0.036 − 0.024](0.036 + 0.024) ln [1.15(0.025) + 0.036 + 0.024](0.036 − 0.024)

 




=




= 132.7 × 109 N/m

kj3

=




= 5.236 × 109 N/m

=

1.813(100 GPa)(0.024 m) [1.15(0.0025) + 0.06775 − 0.024](0.06775 + 0.024) ln [1.15(0.005) + 0.06775 + 0.024](0.06775 − 0.024)

  =

1.813(100 GPa)(0.024 m) [1.15(0.0275) + 0.036 − 0.024](0.036 + 0.024) ln [1.15(0.0275) + 0.036 + 0.024](0.036 − 0.024)

 





414

CHAPTER 16.



→

kj = 3.439 × 109 N/m


kb 1.095 GN/m = = 0.2415 kb + km 1.095 GN/m + 3.439 GN/m

(b) Forces and Safety Factor. The preload for reused connections is obtained from Eq. (16.33), P i = 0.75S p At = (0.75)(600 MPa)(353 mm 2 ) = 158.9 kN The bolts are placed over two diameters, with eight bolts per circle. The load due to the pressure is P = π(1 m)2 (7 MPa)/4 = 5495 kN Refer to the inner bolt force as P 1 , the outer as P 2 . From equilibrium, 8P 1 + 8P 2 = 5495 kN Also, from the assumption given, P 1 D2 1.5 = = = 1.25 P 2 D1 1.2 Substituting this into the equilibrium equation, P 1 = 381.6 kN, P 2 = 305.3 kN. The inner bolts are therefore critical. Using Eq. (16.31), At S p − P i (353 mm2 )(600 MPa) − 158.9 kN ns = = = 0.574 P max,b C k (381.6 kN)(0.2415)

Section 16.5 16.28 A steel plate (sketch m) is riveted to a vertical pillar. The three rivets have a 5/8-in. diameter and carry the load and moment resulting from the external load of 1950 lb. All length dimensions are in inches. The yield strengths of the materials are S y,rivet = 85, 000 psi and S y,plate = 50, 000 psi. y

1950 lb 5/8-in.-diam bolts

2.5

5/8

1 1 _4

B

1 _

22

D 0

1

1 _4

x

5

C 5

16

Sketch m, for Problem 16.28 Calculate the safety factors for (a) Shear of rivet when S sy = 0.57S y (b) Bearing of rivet (c) Bearing of plate

415

(d) Bending of plate State how failure should first occur. Notes: The most time consuming portion of this problem is the statics in order to get the shear

stresses on each rivet. Once the statics is finished, the safety factors are calculated from Eqs. (16.46), (16.47), (16.48) and (16.49). Solution:

(a) Statics. The area of each rivet is A = π(5/8 in.)2 /4 = 0.3068 in.2 . The figure shows a coordinate system relative to which lengths are measured. This coordinate system is arbitrary and a different location can be used without affecting results. The x coordinate of the centroid is: x ¯ =



xi Ai A(2.5 in. − 2.5 in. − 2.5 in.) = = −0.833 in. Ai 3A

The y coordinate is found to be 0. The distance from the force to the centroid is 16 in. + 2.5 in. + 0.833 in. = 19.33 in. The applied torque is T = (1950 lb)(19.33 in.) = 37.7 kip-in. The distances from the centroid to the bolts are as follows: rb = r c =



(2.5 in. − 0.833 in.)2 + (1.25 in.)2 = 2.084 in.

ro = 2.5 in. + 0.833 in. = 3.333 in.

Since the radius to O is the greatest, this will have the largest shear stress and is the rivet that will be analyzed. If we assume that the area of the rivets is insignificant, the polar moment of inertia of the rivets is due to the parallel axis theorem, and is J =



Ari2 = 0.3068 in.2



(3.333 in.)2 + 2(2.084 in.)2 = 6.073 in.4





(b) Stress Analysis. The shear stress due to the torque at O is given by Eq. (4.33) as τ t =

(3.333 in.)(37.7 kip-in.) rT = = 20.69 ksi J 6.073 in.4 )

The shear stress due to the force is assumed to be shared equally by the three rivets so that τ p =

P 1950 lb = = 2.119 ksi 3A 3(0.3068 in.2 )

The shear stress at a due to torsion and due to the force are in the same direction, so that the total shear stress at O is τ o = 20.69 ksi + 2.119 ksi = 22.810 ksi. The shear force at O is then P o = τ o A = (22.81 ksi)(0.3068 in.2 ) = 7000 lb (c) Failure Analysis. For shear of rivet, the safety factor is given by Eq. (16.47) as: ns =

S sy 0.4S y,rivet 0.4(85, 000 psi) = = = 1.49 τ o τ o 22, 810 psi

For bearing of the rivet, the normal stress is σ =

P o 7000 lb = = 17.92 ksi td (5/8 in.)(5/8 in.)

416

CHAPTER 16.


From Eq. (16.49), the safety factor against rivet bearing is ns =

0.9S y 0.9(85, 000 psi) = = 4.269 σ 17.92 ksi

For bearing of the plate, note that the stress is the same (same force, same area) as bearing on the rivet, but the strength is different, so that the safety factor is ns =

0.9S y 0.9(50, 000 psi) = = 2.511 σ 17.92 ksi

For bending of the plate at O, we first calculate the moment of inertia as I =

(5/8 in.)(5 in.) 3 − (5/8 in.)(5/8 in.)3 = 6.498 in.4 12

The moment at O is (1950 lb)(16 in.) = 31 , 200 in.-lb. Therefore, the maximum bending stress at O is Mc (31, 200 in.-lb)(2.5 in.) σ = = = 12 ksi I 6.498 in.4 From Eq. (16.44), the safety factor against bending is ns =

0.6S y 0.6(50, 000 psi) = = 2.5 σ 12, 000 psi

The lowest safety factor is shear of rivet, so this is the failure mode which will occur first. 16.29 The flange of a ship’s propeller shaft is riveted in the radial direction against the hollow shaft. The outside diameter is 1 m and there are 180 rivets around the circumference, each with a diameter of 25 mm. The rivets are made of AISI 1020 steel and placed in three rows. Calculate the maximum allowable propeller torque transmitted through the rivets for a safety factor of 3. Notes: The only failure mode which can be analyzed is shear of rivets, since the thickness of the shaft

hasn’t been specified. Equation (16.47) is used for this case. Solution: From the inside front cover for 1020 steel, S y = 295 MPa. From Eq. (16.47),

τ =

4P 0.4S y = 2 πd c ns

→

P =

0.4πS y d2c 0.4π(295 MPa)(0.025 m)2 = = 19.3 kN 4ns 4(3)

Since there are 180 bolts, the maximum shear force that can be supported is P tot = 180(19.3 kN) = 3474 kN The maximum torque is therefore T max = P tot (r) = (3474 kN)(0.5 m) = 1.738 MNm 16.30 A rectangular steel plate is connected with rivets to a steel beam as shown in sketch n. Assume the steel to be low-carbon steel. The rivets have a yield strength of 600 MPa. A load of 24 kN is applied. For a safety factor of 3 calculate the diameter of the rivets. Length dimensions are in millimeters.

417

P

A τ pa τ p

120

B

120

C

320 60

120

120

60

360

15 20

Sketch n, for Problem 16.30 Notes: The problem involves the following steps:

(a) Determine the centroid and polar moment of inertia for the rivet pattern as a function of rivet diameter. (b) Find the rivet with maximum shear and solve for the shear stress. (c) Calculate the shear stress for the critical bolt. Solution: Note the labels added to the figure. The centroid is clearly at the center of the central

rivet. The rivets on the corners have a radius of 120(2) 1/2 = 169.7 mm. The rivets on the center of a side have a radius of 120 mm. Assuming the rivets are small, the polar moment of inertia is J =



ri2 Ai =

4(0.1697 m)

2

πd2 4

 

+ 4(0.12 m)

2

Assuming the shear due to the shear force is distributed evenly, τ p =

πd 2 4

  

= 0.1357 m2 d2



P P 24 kN 3395 N = = = 9A p (πd2 /4) 9 (πd 2 /4) d2

The torque applied to the bolt group is T = P e = (24 kN)(0.32 m + 0.06 m + 0.12 m) = 12 kNm From Eq. (4.33), the shear stress is directly proportional to the distance from the centroid. Therefore, the corner rivets are critical. The shear stress is given by Eq. (4.33) as τ t =

rT (0.1697 m)(12, 000) 15, 000 N = = 2 2 J 0.1358 m d d2

418

CHAPTER 16.


The components in the x- and y- directions is 10, 604 N/d2 . The shear in the rivet in a corner is then τ =



τ p2 + τ t2

=

 

3395 N 10, 604 N + d2 d2

2

  +

10, 604 N d2



2

=

17, 560 N d2

If the yield stress is 600 MPa, then the allowable shear stress is 240 MPa according to Eq. (3.14). Therefore, τ all 240 MPa ns = = = 3 → d = 14.8 mm τ 17, 560 N/d2 This is the minimum rivet dimension; a rivet of 15 mm would be the proper designation to ensure the safety factor while utilizing a standard sized rivet. 16.31 Repeat Problem 16.30 but with the plate and b eam shown in sketch o.

250

20 60

60 20

P

60 60

Sketch o, for Problem 16.31 Notes: The problem is actually very different from Problem 16.27, since the rivets have an axial load

and a shear. This can be solved only if the heads are not the critical part of the rivet, that is, the head must be able to develop the full rivet strength. Solution: The shear stress in each rivet is

τ =

P/4 P 24, 000 N = = = (7639 N)/d2 2 2 πd /4 πd πd 2

If the plate pivots, then the strain, and therefore the stress, is proportional to the distance from the bottom edge. Therefore, if the A rivets are the top two rivets and the B rivets are the bottom, P A = 7P B . From moment equilibrium, 2P A (0.14 m) + 2P B (0.02 m) = 24, 000 N(0.25 m) Therefore, P A = 21, 000 N, P B = 3000 N. Therefore the axial stress in rivet A is σA =

P A 21 kN = = 26.7 kN/d2 A πd2 /4

From Eq. (2.16), σ1,2 =

σA 2

±

   σA 2

2

+ τ A2 =

1 d2

 

26.7 kN 2

±

 

26.7 kN 2



2

+ (7.64 kN)2

 

419

or σ 1 = 28.8 kN/d2 and σ 2 =

kN/d2 . The effective stress is, from Eq. (6.10),

−2.0

σe = Since the safety factor is 3, S y S y d2 ns = = σe 29.8 kN

→



σ12 − σ1 σ2 + σ22 =

d =



29.8 kN d2

(29.8 kN)ns = S y



(29.8 kN)(3) = 0.0122 m 600 MPa

To use a standard size, use rivets with a diameter of thirteen millimeters.

Section 16.6 16.32 The steel plate shown in sketch p is welded against a wall. Length dimensions are in inches. The vertical load W = 4000 lb acts 6.8 in. from the left weld. Both welds are made by AWS electrode number E8000. The allowable shear stress in the weld should be calculated with safety factor ns = 3.0.

6.8 4 B

A y

W x

G ek

6

3/8-in. plate C

Column

Sketch p, for Problem 16.32 Notes: The approach in this problem is to first find the centroid location and distance to the force

from the centroid, then calculate the stresses at the extreme points on the weld. Equation (16.51) then gives an equation for the weld throat depth. This problem is also very similar to Example 16.11. Solution: First of all, from Table 16.13, S y = 67 ksi. Therefore, from Eq. (3.14), S sy = 0.4S y = 26.8

ksi. (a) Statics. The centroid of the weld is: x ¯ =



xi Ai te (x1 L1 + x2 L − 2) (2 in.)(4 in.) + 0 = = = 0.8 in. Ai te (La + L2 ) 4 in. + 6 in.

420

CHAPTER 16.




yi Ai te (y1 L1 + y2 L2 ) 0 + (3 in.)(6 in.) = = = 1.8 in. Ai te (L1 + L2 ) 4 in. + 6 in.

y¯ =

e = 6.8 in. − 0.8 in. = 6.0 in. From Table 16.12, the unit polar moment is J u =

(b + d)4 − 6b2 d2 (4 in. + 6 in.) 4 − 6(4 in.)2 (6 in.)2 = = 54.53 in.3 12(b + d) 12(4 in. + 6 in.)

Therefore from Eq. (16.55), J = 0.707he J u = 0.707he (54.53 in.3 ) = (38.56 in.3 )he The total area is



Ai = 0.707he (6 in. + 4 in.) = (7.07 in.)he

The applied torque is T = (4000 lb)(6 in.) = 24 kip-in. (b) Stress Analysis. The shear stress due to W is τ W =

V 4000 lb 565.8 lb/in. = = A (7.07 in.)he he

The shear stress components due to the torque at point B are: τ BT x =

T r (24 kip-in.)(1.8 in.) = = (1120 lb/in.)/he J (38.56 in.3 )he

τ BT y =


Therefore, the shear stress at B is: τ B =

 

1120 lb/in. he

2

  +

1992 lb/in. 565.8 lb/in. + he he



2

=

2790 lb/in. he

The stress components due to the torque at point C are: τ CT x =


τ CT y =

T r (24 kip-in.)(0.8 in.) = = (497.9 lb/in.)/he J (38.56 in.3 )he

Therefore, the stress at C is: τ C =

 

2614 lb/in. he

2

 

497.9 lb/in. 565.8 lb/in. + − + he he



2

=

2615 lb/in. he

The higher stress is at point B . Therefore, ns =

S sy S sy he = =3 τ B 2790 lb/in.

→

he =

3(2790 lb/in.) = 0.3125 S sy

Which is exactly a 5/16 inch weld. 16.33 Two medium-carbon steel (AISI 1040) plates are attached by parallel-loaded fillet welds as shown in sketch q . E60 series welding rods are used. Each of the welds is 3 in. long. What minimum leg length must be used if a load of 16.0 kN is to be applied with a safety factor of 3.5?

421

P

A B

D

C

P

Sketch q , for Problem 16.33 Notes: The key equation is Eq. (16.49). Solution: The safety factor for a weld is given by Eq. (16.49) as:

ns =

S sy S sy (he Lw ) = = 3.5 τ 1.414P

From Eq. (3.14), S sy = 0.40S y = 0.40(50 ksi) = 20 ksi. Using L w = 6 in. and P = 16 kN = 3.59 kip, he =

3.5(1.414)(3.59 kip) = 0.148 in. (20 ksi)(6 in.)

However, it is very difficult to lay a bead this small. While this is the minimum bead size for the required conditions, it is better design practice to specify a 3/16 or 1/4 inch weld, as these are easier to produce and would give a larger safety factor. 16.34 Determine the weld size required if only the top (AB) portion is welded in Problem 16.33. Notes: Normally, this would suggest that the weld is loaded in shear due to the force P as well as a

torque T = P e, where e is half of the width of the plate. It is reasonable for a student to take this into account, but no dimensions are given for such an analysis. This solution considers only a shear stress due to the force P , but would have to consider a torque for critical applications. Therefore, it uses the same approach as Problem (16.33). Solution: The safety factor for a weld is given by Eq. (16.51) as:

ns =

S sy S sy (he Lw ) = = 3.5 τ 1.414P

From Eq. (3.14), S sy = 0.40S y = 0.40(50 ksi) = 20 ksi. Using L w = 3 in. and P = 16 kN = 3.59 kip, he =

3.5(1.414)(3.59 kip) = 0.296 in. (20 ksi)(3 in.)

A 3/8 inch weld would be sufficient 16.35 The universal joint on a car axle is welded to the 60-mm-outside-diameter tube and should be able to transfer 1500 N-m of torque from the gearbox to the rear axle. Calculate how large the weld leg should be to give a safety factor of 10 if the weld metal is of class E70. Notes: The weld leg is calculated from the transmitted torque and resulting shear stress from Equation

(16.52), and the polar moment of the fillet weld using Eq. (16.53) and Table 16.12. Solution: From Table 16.13, for a class E70 weld electrode, the yield strength is 57 ksi = 393 MPa.

Therefore, from Eq. (3.14), τ all = 0.40S y = 0.40(393 MPa) = 157 MPa

422

CHAPTER 16.


From Table 16.12, the unit polar moment is J u = π(d3 /4) = π(0.06 m)3 /4 = 1.697 × 10

4

−

m3

From Eq. (16.53), the polar moment of the fillet weld is J = 0.707he J u = 0.707he (1.697 × 10

4

−

m3 ) = 1.20 × 10

4

−

m3 he

Therefore, from Eq. (16.52), τ t =

T r (1500 Nm)(0.06 m/2) = = 157 MPa J (1.20 × 10 4 m3 )he −

Therefore, h e = 0.002389 m or 2.4 mm. A 3 or 5 mm weld leg would be a good design designation. 16.36 The steel bar shown in sketch r is welded by an E60XX electrode to the wall. A 3.5-kN load is applied in the y direction at the end of the bar. Calculate the safety factor against yielding. Also, would the safety factor change if the direction of load P is changed to the z direction? Dimensions are in millimeters. 7

z y

x

7

50

P

20

250

Sketch r, for Problem 16.36 Notes: One must use Table 16.12 for the unit moment of inertia. When the load changes direction, d

and b must be switched in the expression for moment of inertia. Note also that the drawing designates he = 7 mm. Solution: The leg length is given as 7 mm, with weld length of 50 + 20 = 70 mm. The geometry

gives (from Table 16.12): I u =

4bd + d2 4(20)(50) + 20 2 = = 733 mm2 6 6

b2 202 y¯ = = = 2.86 mm 2(b + d) 2(20 + 50) z¯ =

d2 502 = = 17.86 mm 2(b + d) 2(20 + 50)

The moment of inertia is obtained from Eq. 16.55: I = 0.707he I u Lw = (0.707)(0.007 m)(733 mm2 )(0.070 m) = 2.54 × 10 The normal stress is given by Eq. (16.57) as σ =

Mc (3.5 kN)(0.25 m)(0.01714 m) = = 59.0 MPa I 2.54 × 10 7 m4 −

7

−

m4

423

At the same time, the weld sees a shear due to transverse loading, given by Eq. (16.49) as: τ =

1.414P 1.414(3.5 kN) = = 10.1 MPa he Lw (0.007 m)(0.070 m)

Performing a Mohr’s circle analysis, the principal stresses are 60.7 MPa and -1.68 MPa. For an E60XX electrode, Table 16.13 gives the electrode yield strength as 50 ksi which equals 345 MPa. The safety factor is then S y 345 MPa ns = = = 5.53 (σ1 − σ3 ) (60.7 MPa + 1.68 MPa) If the force direction was changed to the z -direction (i.e., acting upwards), then I u can be calculated as 601.9 mm 2 , I = 2.085 × 10 7 m4 and σ = 135 MPa. The shear stress is unchanged. The resultant safety factor is 2.56, which is a large reduction in the safety factor. −

16.37 The bar shown in sketch s is welded to the wall by AWS electrodes. A 40-kN load is applied at the top of the bar. Dimensions are in millimeters. For a safety factor of 2.5 against yielding, determine the electrode number that must be used and the weld throat length, which should not exceed 10 mm.

350

P

35

60

35

Sketch s, for Problem 16.37 Notes: Note that there is a normal stress due to a bending moment and a shear stress due to the

shear force. Therefore, a combined state of stress exists, and the effective stress must be calculated. The unit moment of inertia is obtained from Table 16.12. Solution: Note that the length of the weld is Lw = 130 mm = 0.13 m. The shear stress is then

obtained from Eq. (16.49): τ =

1.414P 1.414(5, 000 N) 54.4 kN/m = = he Lw he (0.13 m) he

From Table 16.12, the unit moment of inertia is I u = bd + d2 /6 = (0.035 m)(0.060 m) + (0.060 m)2 /6 = 0.0027 m2 From Eq. (16.55), the moment of inertia is I = 0.707he I u Lw = 0.707he (0.0027 m2 )(0.13 m) = 0.000248 m3 he The normal stress is then σ =

Mc (5, 000 N)(0.350 m)(0.030 m) 212 kN/m = = I 0.000248 m3 he he

424

CHAPTER 16.


From Eq. (2.16), σ σ1,2 = 2

±

   σ 2

2

+ τ 2

212 kN/m = 2he

±

 

212 kN/m 2he

2

  +



54.4 kN/m he

2

or σ 1 = (225 kN/m)/he and σ 2 = (−13.1 kN/m)/he . The effective stress is , From Eq. (2.31) σe

= = =

σ12 + σ22 − σ1 σ2

 

225 kN/m he



1/2

2

  +

−104.8

kN/m

he

2

  −

225 kN/m he



−13.1

kN/m

he



1/2

232 kN/m he

A relationship can be obtained for the required yield stress from the equation for safety factor as S all ns = σe

→

S y = ns σe = (2.5)



232 kN/m he



=

580 kN/m he

Note that if he = 0.010 m, then S y must be 58 MPa = 8.3 ksi. From Table 16.13, this is met by all electrodes higher than the E80xx series. Therefore, the following can be stated: Electrode Number

E60XX E70XX E80XX E90XX E100XX E120XX

Yield stress, ksi

Yield stress, MPa

Required he , mm

50 57 67 77 87 107

345 393 462 531 600 738

1.68 1.48 1.26 1.09 0.97 0.79

16.38 AWS electrode number E100XX is used to weld a bar, shown in sketch t, to the wall. For a safety factor of 3 against yielding, find the maximum load that can be supported. What is the maximum stress in the weld and where does it occur? Dimensions are in millimeters. 10

P

50

40 6 200

Sketch t, for Problem 16.38 Notes: This problem has two different weld thicknesses. For that reason, one must obtain the moment

of inertia by analyzing each weld thickness and its geometry separately. Solution: From Table 16.13, an E100XX electrode has a yield strength of S y = 87 ksi = 600 MPa.

With a safety factor of 3, the allowable stress is 200 MPa. Refer to the 10 mm leg as weld 1, the 6 mm leg as weld 2. The shear stress from the shear force is, from Eq. (16.49), τ =

1.414P 1.414P = = 955.4 m he1 Lw1 + he2 Lw2 (0.01 m)(0.1 m) + (0.006 m)(0.08 m)



2

−



P

425

The moment of inertia is from Table 16.12 and Eq. (16.55), I = 0.707(he1 I u1 Lw1 + he2 I u2 Lw2 ) (0.05 m)2 = 0.707 (0.01 m) (0.1 m) + (0.006 m)(0.04 m)(0.05 m)(0.08 m) 3





6

m4

−

= 1.268 × 10





The normal stress, from Eq. (16.57) is Mc P (0.2 m)(0.025 m) = = 3943 m I 1.268 × 10 6 m4

σ =



−

From Eq. (2.16), σ σ1,2 = 2 or σ 1 = (4162 m

2

−

±

   σ 2

or σ e = (4275 m

2

−

(3943 m + τ 2 = 2

)P and σ 2 = (−219 m

σe = σ12 + σ22 − σ1 σ2



2

1/2



= P



2

−

2

−

)P

±

 

(3943 m 2

2

−



P

2 )P 2



−

+ [(955.4 m

2 )P ]2

−

)P . The effective stress, from Eq. (2.31), 2 2

−

4162 m

 

2 2

−

+ 219 m

  −

4162 m

2

−



−219

2

−

m

)P . If the maximum allowable stress is 200 MPa, then P = 46.8 kN.



1/2

16.39 A cam is to be attached to an extruded channel using arc welds. Three designs are proposed as shown below. If E60 electrodes are used, find the factor of safety of the welded joint if P =2000 pounds for each case. Use a=15 inches. You can assume that the extrusion fits tightly with the mating hole in the cam so that the loading is pure torsion. Design based on yielding of the weld. P

3 in. diameter

P

3 in.

a

× 3

a

P

in. a

5/16 in. 1/4 in. 1/8 in.

3 in.

× 3

in.

butt welded

Notes: This is straightforward, however, the middle case is deceptively simple - the weld sees no stress

since the torque is transmitted by the geometry of the channel. Solution: Note from Table 16.13 that for this electrode, S y = 50 ksi.

(a) For the case on the left with the solid round shaft, we have from Table 16.12 that the unit polar moment of inertia is (3 in.)3 3 J u = π(d /4) = π = 21.20 in.3 4



Since t e = 0.25, the polar moment of inertia is



J = t e J u = (0.25 in.)(21.20 in.3 ) = 5.301 in.4

426

CHAPTER 16.


Therefore the shear stress is τ =

T ρ (2000 lb)(15 in.)(1.5 in.) = = 8.49 ksi J 5.301 in.4

The factor of safety is, using Eq. (3.14) for shear stress, ns =

τ all 0.4S y (0.4)(50 ksi) = = = 2.36 τ τ 8.49 ksi

(b) For the middle case, the weld sees no load. Notice that a torque is transmitted even if there is not a weld present. In this case, n s = ∞. (c) For the case on the right, we have from Table 16.12, J u =

(b + d)3 3 in. + 3 in.) 3 = = 36 in.3 6 6

Therefore, J = t e J u = (0.3125 in.)(36 in. 3 ) = 11.25 in.4 The extreme location from the centroid on this geometry is 4.24 in. The shear stress is τ =

(2000 lb)(4.24 in.)(15 in.) = 11.31 ksi 11.25 in.4

So that the safety factor is ns =

0.4(50 ksi = 1.77 11.31 ksi

Section 16.7 16.40 When manufacturing the fuselage of a commuter airplane, aluminum plates are glued together with lap joints. Because the elastic deformation for a single plate differs from the deformation for two plates glued together in a lap joint, the maximum shear stress in the glue is twice as high as the average shear stress. The shear strength of the glue is 20 MPa, the tensile strength of the aluminum plates is 95 MPa, and their thickness is 4.0 mm. Calculate the overlapping length needed to make the glue joint twice as strong as the aluminum plate. Notes: This problem requires the use of Eq. (16.50). Solution:The tensile force per width is obtained from:

σ =

P P = A tw

→

P = σt = (95 MPa)(0.004 m) = 380 kN/m w

The glue joint should be twice as strong as the aluminum, or for the glue: P = 2(380 kN/m) = 760 kN/m b From Eq. (16.59) and recalling the maximum stress is twice the average stress, 2P τ max = 2τ ave = bL

→

2P 2 L = = bτ max τ max

or the overlap should be at least 76 mm.

 P b

=

2 (760 kN/m) = 0.076 m 20 MPa

Plugin Chapter16

Recommend Documents