ch04

STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS

59

Chapter 4 4-1 An aluminum tube with an outside diameter of 1.000 in. will be used to support a 10-kip load. If the axial

stress in the member must be limited to 30 ksi T or C, determine the wall thickness required for the tube. SOLUTION

10,000 30,000 psiP

A A

2 2 210,0001 in.4 30,000iA d

0.75867 in. 1 2id t

0.1207 in.t .......................................................................................................................... Ans.

4-2 Three steel bars with 25 15-mm cross sections are welded to a gusset plate as shown in Fig. P4-2. Determine the normal stresses in the bars when the forces shown are being applied to the plate.

SOLUTION

6 240,000 106.7 10 N/m 106.7 MPa0.025 0.015A

PA

........................................Ans.

6 250,000 133.3 10 N/m 133.3 MPa0.025 0.015B

PA

.........................................Ans.

6 220,000 53.3 10 N/m 53.3 MPa0.025 0.015C

PA

.............................................Ans.

4-3 Two ¼-in. diameter steel cables A and B are used to support a 220-lb traffic light as shown in Fig. P4-3. Determine the normal stress in each of the cables.

SOLUTION From a free-body diagram of the ring, the equations of equilibrium

0 :xF cos 25 cos 20 0B AT T

0 :yF sin 20 sin 25 220 0A BT T

are solved to get

1.03684B AT T

281.977 lbAT 292.365 lbBT

2281.977 5740 psi0.25 4

AA

A

TA

......................................................................................Ans.

2292.365 5960 psi0.25 4

BB

B

TA

......................................................................................Ans.

4-4 A system of steel pipes is loaded and supported as shown in Fig. P4-4. If the normal stress in each pipe must not exceed 150 MPa, determine the cross-sectional areas required for each of the sections.

SOLUTION From free-body diagrams of the pipes, the equations of equilibrium give

0 :yF 650 0AP 650 kNAP

0 :yF 650 850 0BP 1500 kNBP


60

0 :yF 650 850 1500 0CP 3000 kNCP

3

6 2 26

650 10 4333 10 m 4330 mm150 10

AA

A

PA .......................................................Ans.

3

6 2 26

1500 10 10,000 10 m 10,000 mm150 10

BB

B

PA ..............................................Ans.

3

6 2 26

3000 10 20,000 10 m 20,000 mm150 10

CC

C

PA ............................................Ans.

4-5 Two 1-in. diameter steel bars are welded to a gusset plate as shown in Fig. P4-5. Determine the normal stresses in the bars when forces F1 = 550 lb and F2 = 750 lb are applied to the plate.

SOLUTION

12

550 700 psi1 4A

A

FA

...............................................................................................Ans.

22

750 955 psi1 4B

B

FA

...............................................................................................Ans.

4-6 Two strips of a plastic material are bonded together as shown in Fig. P4-6. The average shearing stress in the glue must be limited to 950 kPa. What length L of splice plate is needed if the axial load carried by the joint is 50 kN?

SOLUTION From a free-body diagram of the middle plate, the equations of equilibrium

0 :xF 32 50 10 0V

gives

3

3

25 10 N

950 10 0.3 2

V A

L

0.1754 m 175.4 mmL .....................................................................................................Ans.

4-7 A coupling is used to connect a 2-in. diameter plastic rod to a 1.5-in. diameter rod, as shown in Fig. P4-7. If the average shearing stress in the adhesive must be limited to 500 psi, determine the minimum lengths L1 and L2 required for the joint if the applied axial load P is 8000 lb.

SOLUTION From a free-body diagram of the rod, the equation of equilibrium

0 :xF 8000 0V

gives

1

2

8000 lb500 2

500 1.5

V AL

L

1 2.55 in.L ................................................................Ans.

2 3.40 in.L ................................................................Ans.


61

4-8 Three plates are joined with a 12-mm diameter pin as shown in Fig. P4-8. Determine the maximum load P that can be transmitted by the joint if

(a) The maximum normal stress on a cross section at the pin must be limited to 350 MPa. (b) The maximum bearing stress between a plate and the pin must be limited to 650 MPa. (c) The maximum shearing stress on a cross section of the pin must be limited to 240 MPa. (d) The punching shear resistance of the material in the top and bottom plates is 300 MPa. SOLUTION (a) For the outside plates

2 6 230 12 10 180 mm 180 10 mnA

6 6 32 2 350 10 180 10 126 10 NnP A

For the middle plate

2 6 225 12 25 325 mm 325 10 mnA

6 6 3350 10 325 10 113.7 10 NnP A

Therefore

max 113.7 kNP ......................................................................................................................... Ans.

(b) For the bearing stress on the hole in the middle plate

2 6 212 25 300 mm 300 10 mbA dt

6 6 3650 10 300 10 195 10 NbP A

For the outer plates

2 6 212 10 120 mm 120 10 mbA dt

6 6 32 2 650 10 120 10 156 10 NbP A

Therefore

max 156 kNP ............................................................................................................................. Ans.

(c) For the shearing stress on the cross section of the pin (which is in double shear)

22

2 6 2

4 12 4

113.10 mm 113.10 10 msA d

6 62 2 240 10 113.10 10sP A

354.3 10 N 54.3 kN .....................................................................................................Ans. (d) For the punching shear stress in the outside plates

2 6 2

2 2 18 10

360 mm 360 10 msA Lt

6 62 2 300 10 360 10sP A

3216 10 N 216 kN ..........................................................................................................Ans.


62

4-9 A 100-ton hydraulic punch press is used to punch holes in a 0.50-in.-thick steel plate, as illustrated schematically in Fig. P4-9. If the average punching shear resistance of the steel plate is 40 ksi, determine the maximum diameter hole that can be punched.

SOLUTION

340 10 0.5

100 2000 lb

sP A dt

d

3.18 in.d ............................................................................................. Ans.

4-10 A body with a mass of 250 kg is supported by five 15-mm diameter cables, as shown in Fig. P4-10. Determine the normal stress in each of the cables.

SOLUTION The tension in cable E is equal to the weight of the hanging body. The normal stress in cable E is then

6 22

250 9.8113.88 10 N/m 13.88 MPa

0.015 4E

ETA

...........................................Ans.

From a free-body diagram of the lower ring, the equations of equilibrium

0 :xF cos60 0C DT T

0 :yF sin 60 250 9.81 0CT

give

22831.90 N 0.015 4C CT

6 216.03 10 N/m 16.03 MPaC ....................................................................................Ans.

21415.95 N 0.015 4D DT

6 28.01 10 N/m 8.01 MPaD ........................................................................................Ans.

Then, from a free-body diagram of the upper ring, the equations of equilibrium

0 :xF cos30 cos 40 cos 60 0B A CT T T

0 :yF sin 30 sin 40 sin 60 0B A CT T T

give

21506.83 N 0.015 4A AT

6 28.53 10 N/m 8.53 MPaA ........................................................................................Ans.

22967.85 N 0.015 4B BT

6 216.79 10 N/m 16.79 MPaB ...................................................................................Ans.


63

4-11 The joint shown in Fig. P4-11 is used in a steel tension member which has a 2 1-in. rectangular cross section. If the allowable normal, bearing, and punching-shearing stresses in the joint are 13.5 ksi, 18.0 ksi, and 6.50 ksi, respectively, determine the maximum load P that can be carried by the joint.

SOLUTION For normal stress at the narrowest section

20.5 1 0.5 in.nA

313.5 10 0.5 6750 lbnP A

For bearing stress

22 3 8 1 0.75 in.bA

318.0 10 0.75 13,500 lbbP A

For punching shearing stress

22 3 8 1 0.75 in.sA

36.5 10 0.75 4880 lbsP A

Therefore

max 4880 lbP ............................................................................................................................ Ans.

4-12 A vertical shaft is supported by a thrust collar and bearing plate, as shown in Fig. P4-12. Determine the maximum axial load that can be applied to the shaft if the average punching shear stress in the collar and the average bearing stress between the collar and the plate are limited to 75 and 100 MPa, respectively.

SOLUTION For bearing stress between the collar and the plate

2 2 3 20.15 0.10 4 9.8175 10 mbA

6 3 3100 10 9.8175 10 982 10 NbP A

For punching shearing stress in the collar

3 20.10 0.025 7.854 10 msA dL

6 3 375 10 7.854 10 589 10 NsP A

Therefore

max 589 kNP ............................................................................................................................ Ans.

4-13 A device for determining the shearing strength of wood is shown in Fig. P4-13. The dimensions of the wood specimen are 6-in. wide by 8-in. high by 2-in. thick. If a load of 16,800 lb is required to fail the specimen, determine the shearing strength of the wood.

SOLUTION

16,800 1050 psi

8 2s

VA

...................................................................................................Ans.


64

4-14 The 75-kg traffic light shown in Fig. P4-14 is supported by three cables of equal diameter. Determine the minimum diameter required if the normal stress in any cable must not exceed 160 MPa.

SOLUTION

2 2 2

4 8 5 0.39036 0.78072 0.487954 8 5

A A A A AT T T Ti j kT i j k

2 2 2

6 8 5 0.53666 0.71554 0.447216 8 5

B B B B BT T T Ti j kT i j k

2 2

8 5 0.84800 0.530008 5

C C C CT T Tj kT j k

The equations of equilibrium

0 :xF 0.39036 0.53666 0A BT T

0 :yF 0.78072 0.71554 0.84800 0A B CT T T

0 :zF 0.48795 0.44721 0.53000 75 9.81 0A B CT T T

give

1.37478A BT T 2.10950C BT T

6 2452.353 N 160 10 4A AT d 0.00190 mAd

6 2329.037 N 160 10 4B BT d 0.00162 mBd

6 2694.103 N 160 10 4C CT d 0.00235 mCd

Therefore

min 2.35 mmd .......................................................................................................................... Ans.

4-15 A 1000-lb load is securely fastened to a hoisting rope as shown in Fig. P4-15. The force in the weightless flexible cable does not change as it passes around the small frictionless pulley at support C. The sag distance d is 4 ft. Cables AB and BC have the same diameter. The normal stresses in these cables must not exceed 24 ksi, and the shearing stress in pin A (double shear) must not exceed 12 ksi. Determine

(a) The minimum diameter of cables AB and BC. (b) The minimum diameter of the pin at A. SOLUTION

11

4sin 23.57810

110cos 9.16515 fta

30 20.83485 ftb a

12

4tan 10.868b

Then, from a free-body diagram of the ring, the equations of equilibrium

0 :xF 2 1cos cos 0AP T

0 :yF 2 1sin sin 1000 0AP T

give


65

1.07152AT P

1620.35 lbP 1736.24 lbAT

(a) For the normal stress in the cables

3 21620.35 lb 24 10 4BCP d

0.293 in.BCd ........................................................................................................................... Ans.

3 21736.24 lb 24 10 4A ABT d

0.304 in.ABd ........................................................................................................................... Ans.

(b) For the pin at A, which is in double shear,

3 21736.24 lb 12 10 2 4A pT d

0.304 in.pd ............................................................................................................................. Ans.

4-16 A flat steel bar 100 mm wide by 25 mm thick has axial loads applied with 40 mm-diameter pins in double shear at points A, B, C, and D as shown in Fig. P4-16. Determine

(a) The axial stress in the bar on a cross section at pin B. (b) The average bearing stress on the bar at pin B. (c) The shearing stress on the pin at A. SOLUTION (a) For axial stress in the bar

3

6 2350 10 233 10 N/m 233 MPa0.06 0.025n

n

PA

....................................................Ans.

(b) For bearing stress at the pin

3

6 2250 10 250 10 N/m 250 MPa0.04 0.025b

b

PA

...................................................Ans.

(c) For shearing stress on the pin (which is in double shear)

3

6 22

350 10 139.3 10 N/m 139.3 MPa2 2 0.04 4s

VA

.......................................Ans.

4-17 Two flower pots are supported with steel wires of equal diameter. Pot A weighs 10 lb and pot B weighs 8 lb. Determine the minimum required diameter of the wires if the normal stress in the wires must not exceed 18 ksi.

SOLUTION From a free-body diagram of the upper ring

0 :xF cos 45 cos 0CD BCT T (a)

0 :yF sin 45 sin 8 0CD BCT T (b)

From a free-body diagram for the lower ring

0 :xF cos cos 45 0BC ABT T (c)

0 :yF sin 45 sin 10 0AB BCT T (d)

Adding equations (a) and (c) gives

cos 45 cos 45 0CD ABT T


66

CD ABT T

Then adding Eqs. (b) and (d) gives

sin 45 sin 45 18 lbCD ABT T

12.7279 lbCD ABT T

Now Eqs. (b) and (a) can be written

sin 1.0000BCT (e)

cos 9.0000BCT (f)

Dividing Eq. (e) by Eq. (f) gives

tan 1 9 6.340

9.0554 lbBCT

3max 2

12.7279 18 10 psi4

PA d

0.0300 in.d ............................................................................................................................ Ans.

4-18 A steel pipe will be used to support a 40-kN load. If the wall thickness of the pipe is 10 percent of the pipe’s outside diameter do, calculate and plot the normal stress in the pipe as a function of the diameter do, (20 mm

do 75 mm). If the axial stress in the pipe must be limited to 250 MPa, what is the smallest size standard steel pipe (see Appendix A) that could be used.

SOLUTION

2 0.1 0.8i o o od d d d

3

222 2

40 10 141, 471.06 N/m4 oo i

PA dd d

For standard steel pipe,

3

6 240 10 250 10 N/mPA A

6 2 2160 10 m 160 mmA

13 mmd .................................................... Ans.

4-19 An aluminum tube with an outside diameter do will be used to support a 10-kip load. If the axial stress in the tube must be limited to 30 ksi T or C, calculate and plot the required wall thickness t as a function of do, (0.75 in. do 4 in.). What diameter would be required for a solid aluminum shaft?

SOLUTION

2 2

10,000 30,000 psi4o i

PA d d

2 0.42441 2i o od d d t

2 0.424412

o od dt

For a solid aluminum shaft


67

2

10,000 30,000 psi4

PA d

0.65 in.d .................................................................................................................................Ans.

4-20 The steel pipe column shown in Fig. P4-20a has an outside diameter of 150 mm and a wall thickness of 15 mm. The load imposed on the column by the timber beam is 150 kN. If the bearing stress between the circular steel bearing plate and the timber beam is not to exceed 3.25 MPa, determine the minimum diameter bearing plate that must be used between the column and the beam. Assume that the bearing stress is uniformly distributed over the surface of the plate.

If the bearing plate is not rigid, the stress between the bearing plate and the timber beam will not be uniform. If the stress varies as shown in Fig. P4-20b (a uniform value of max above the column and decreasing linearly to max/5 at the outside edge rp of the bearing plate), calculate and plot max versus the radius rp of the bearing plate (75 mm rp 500 mm). Now what minimum diameter bearing plate must be used if the bearing stress must not exceed 3.25 MPa? What is the percent decrease in max for a 400 mm-diameter bearing plate compared with a 150 mm-diameter bearing plate? For a 600 mm-diameter bearing plate compared with a 150 mm-diameter bearing plate?

SOLUTION For uniform stress over a rigid bearing plate

3

6 22

150 10 3.25 10 N/m4

PA d

0.242 m 242 mmd ...........................................................................................................Ans. For the non-rigid bearing plate

1 2max

c r c

where the constants c1 and c2 are chosen such that

1 21 0.075c c

1 20.2 pc r c

Therefore,

10.8

0.075 p

cr

2

0.8 0.0751

0.075 p

cr

Integrating the stress to get the axial force

F dA

gives

0.0753max max 1 20 0.075

3 22 3 2

max 1 2

150 10 2 2

0.075 0.075 0.07522 3 3 2 2

pr

p p

r dr c r c r dr

r rc c


68

3

2max 3 22 3 2

1 2

150 10 N/m0.075 0.075 0.0752

2 3 3 2 2p pr r

c c

When 0.075 mpr 6 2max 8.49 10 N/m 8.49 MPa ......................................Ans.

When 0.2 mpr max 1.976 MPa (a 76.7% decrease) ................................Ans.

When 0.3 mpr max 0.965 MPa (an 88.6% decrease) ..............................Ans.

4-21 The tie rod shown in Fig. P4-21a has a diameter of 1.50 in and is used to resist the lateral pressure against the walls of a grain bin. The force imposed on the wall by the rod is 18,000 lb. If the bearing stress between the washer and the wall must not exceed 400 psi, determine the minimum diameter washer that must be used between the head of the bolt and the grain bin wall. Assume that the bearing stress is uniformly distributed over the surface of the washer.

If the washer is not rigid, the stress between the washer and the wall will not be uniform. If the stress varies as shown in Fig. P4-21b (a uniform value of max under the 2.4-in.-diameter restraining nut and decreasing as 1/r to the outside edge rw of the washer), calculate and plot max versus the radius rw of the washer (1 in. rw 8 in.). Now what minimum diameter washer must be used if the bearing stress must not exceed 400 psi? What is the percent decrease in max for an 8-in.-diameter washer compared with a 4-in.-diameter washer? For a 12-in.-diameter washer compared with an 8-in.-diameter washer?

SOLUTION For uniform stress over a rigid washer

3

2 2

18 10 400 psi1.5 4

PA d

7.72 in.d .................................................. Ans. For the non-rigid washer

max1.2r

Integrating the stress to get the axial force

F dA

gives

1.23 maxmax0.75 1.2

2 2max

max

1.218 10 2 2

1.2 0.75 2.4 1.2

0.87750 2.4 1.2

wr

w

w

r dr r drrr

r

3

max18 10 psi

0.87750 2.4 1.2wr

When max 400 psi 2 13.61 in.wd r ......................................................................Ans.

When 2 2.4 in.wd r max 6529 psi

When 2 8 in.wd r max 754 psi an 88.5% decrease.................................. Ans.

When 2 12 in.wd r max 462 psi a 38.7% decrease.................................... Ans.


69

4-22 The steel bar shown in Fig. P4-22 will be used to carry an axial tensile load of 400 kN. If the thickness of the bar is 45 mm, determine the normal and shearing stresses on plane AB.

SOLUTION From a free-body diagram of the bar, the equations of equilibrium

0 :nF 400cos37 0N

0 :tF 400sin 37 0V

give

319.45 kNN 240.73 kNV The areas are related by

0.075 0.045 cos37nA A 3 24.226 10 mnA

Therefore

3

6 23

319.45 10 75.6 10 N/m 75.6 MPa4.226 10n

NA

...................................................Ans.

3

6 23

240.73 10 57.0 10 N/m 57.0 MPa4.226 10n

VA

....................................................Ans.

4-23 An axial load P is applied to a timber block with a 4 4-in. square cross section, as shown in Fig. P4-23. Determine the normal and shear stresses on the planes of the grain if P = 5000 lb.

SOLUTION From a free-body diagram of the bar, the equations of equilibrium

0 :nF 5000sin14 0N

0 :tF 5000cos14 0V

give

1209.61 lbN

4851.48 lbV The areas are related by

4 4 sin14nA A

266.137 in.nA

Therefore

1209.61 18.29 psi66.137n

NA

...............................................................................................Ans.

4851.48 73.4 psi66.137n

VA

.................................................................................................Ans.

4-24 A structural steel bar with a 20 25-mm rectangular cross section is subjected to an axial tensile load of 55 kN. Determine the maximum normal and shear stresses in the bar.

SOLUTION

3

6 2max

55 10 110.0 10 N/m 110.0 MPa0.02 0.025

PA

...........................................Ans.

max 55.0 MPa2 2PA

....................................................................................................Ans.


70

4-25 A steel rod of circular cross section will be used to carry an axial tensile load of 50 kip. The maximum stresses in the rod must be limited to 25 ksi in tension and 15 ksi in shear. Determine the minimum diameter d for the rod.

SOLUTION

3

3max 2

50 10 25 10 psi4

PA d

1.596 in.d

3

3max 2

50 10 15 10 psi2 2 4PA d

1.457 in.d

min 1.596 in.d ..........................................................................................................................Ans.

4-26 Determine the maximum axial load P that can be applied to the wood compression block shown in Fig. P4-26 if specifications require that the shear stress parallel to the grain not exceed 5.25 MPa, the compressive stress perpendicular to the grain not exceed 13.60 MPa, and the maximum shear stress in the block not exceed 8.75 MPa.

SOLUTION From a free-body diagram of the bar, the equations of equilibrium are

0 :nF sin 60 0P N

0 :tF cos 60 0P V

The areas are related by

0.2 0.12 sin 60nA A 20.027713 mnA

Therefore

613.60 10 0.027713

sin 60 sin 60 sin 60nANP 3435 10 NP

65.25 10 0.027713

cos60 cos 60 cos 60nAVP 3291 10 NP

max 291 kNP .............................................................................................................................Ans.

4-27 A steel bar with a 4 1-in. rectangular cross section is being used to transmit an axial tensile load, as shown in Fig. P4-27. Normal and shear stresses on plane AB of the bar are 12 ksi tension and 9 ksi shear. Determine the angle and the applied load P.


0 :nF cos 0N P

0 :tF sin 0V P


4 1 cosnA A 24 cos in.nA

Therefore

3cos 12 10 4 cosnN P A

3sin 9 10 4 cosnV P A

Therefore


71

sin 9tancos 12PP

36.870 .................................................................................................................................Ans.

375 10 lb 75 kipP ...........................................................................................................Ans.

4-28 A steel bar with a butt-welded joint, as shown in Fig. P4-28 will be used to carry an axial tensile load of 400 kN. If the normal and shear stresses on the plane of the weld must be limited to 70 MPa and 45 MPa, respectively, determine the minimum thickness t required for the bar.


0 :nF 400sin 57 0N

0 :tF 400cos57 0V


0.1 sin 57nA t A 20.11924 mnA t

Therefore

3

6 2400 10 sin 57

70 10 N/m0.11924n

NA t

0.0402 mt

3

6 2400 10 cos57

45 10 N/m0.11924n

VA t

0.0406 mt

min 40.6 mmt ...........................................................................................................................Ans.

4-29 The shearing stress on plane AB of the 4 8-in. rectangular block shown in Fig. P4-29 is 2 ksi when the axial load P is applied. If the angle is 35 , determine

(a) The load P. (b) The normal stress on plane AB. (c) The maximum normal and shearing stresses in the block. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are

0 :nF sin 35 0P N

0 :tF cos35 0P V


4 8 sin 35nA A

255.790 in.nA

(a) Solving the second equation for P gives

32 10 55.790

136, 215 lb 136.2 kipcos35 cos35 cos35

nAVP ........................Ans.

(b) Solving the first equation for N gives

sin 35 78,129.6 lb nN P A

1400 psi ...............................................................................................................................Ans.

(c) Then the maximum normal and shearing stresses are given by


72

max136, 215 4260 psi

4 8PA

............................................................................................Ans.

maxmax 2130 psi

2 2PA

.................................................................................................Ans.

4-30 A wood tension member with a 50 100-mm rectangular cross section will be fabricated with an inclined glued joint (45 90 ) at its midsection, as shown in Fig. P4-30. If the allowable stresses for the glue are 5 MPa in tension and 3 MPa in shear, determine

(a) The optimum angle for the joint. (b) The maximum safe load P for the member. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are

0 :nF sin 0N P

0 :tF cos 0V P


0.05 0.1 sinnA A

20.005 sin mnA

Therefore

6sin 5 10 0.005 sinnN P A

6cos 3 10 0.005 sinnV P A

and dividing the first equation by the second gives

sin 5tan 1.66667cos 3PP

59.04 ...................................................................................................................................Ans.

3max 34.0 10 N 34.0 kNP ...............................................................................................Ans.

4-31 The two parts of the eyebar shown in Fig. P4-31 are connected with two ½-in. diameter bolts (one on each side). Specifications for the bolts require that the axial tensile stress not exceed 12.0 ksi and that the shearing stress not exceed 8.0 ksi. Determine the maximum load P that can be applied to the eyebar without exceeding either specification.


0 :nF 2 cos30 0N P

0 :tF 2 sin 30 0V P

The cross sectional areas of the bolts are

2 20.5 4 0.19635 in.bA

The first equation gives

32 12 10 0.1963522 5441 lb

cos30 cos30 cos30bANP

while the second equation gives


73

32 8 10 0.1963522 6283 lb

sin 30 sin 30 sin 30bAVP

Therefore

max 5441 lb 5440 lbP .........................................................................................................Ans.

4-32 The bar shown in Fig. P4-32 has a 200 100-mm rectangular cross section. Determine (a) The normal and shearing stresses on plane a-a. (b) The maximum normal and shearing stresses in the bar. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are

0 :nF 500cos30 0N

0 :tF 500sin 30 0V


0.2 0.1 cos30nA A

3 223.094 10 mnA

Therefore

3500 10 cos30 433.01 kN nN A

3500 10 sin 30 250.00 kN nV A

6 218.75 10 N/m 18.75 MPa ......................................................................................Ans.

6 210.83 10 N/m 10.83 MPa .......................................................................................Ans.

3

6 2max

500 10 25.0 10 N/m 25.0 MPa0.2 0.1

PA

.....................................................Ans.

maxmax 12.50 MPa

2 2PA

.............................................................................................Ans.

4-33 A steel eyebar with a 4 1-in. rectangular cross section has been designed to transmit an axial tensile load. The length of the eyebar must be increased by welding a new center section in the bar (45 90 ), as shown in Fig. P4-33. The stresses in the weld material must be limited to 12 ksi in tension and 9 ksi in shear. Determine

(a) The optimum angle for the joint. (b) The maximum safe load P for the redesigned member. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are

0 :nF sin 0N P

0 :tF cos 0V P


4 1 sinnA A 24 sin in.nA

Therefore

3sin 12 10 4 sinnN P A

3cos 9 10 4 sinnV P A


74

and dividing the first equation by the second gives

sin 12tan 1.33333cos 9PP

53.13 .................................................Ans.

max 75,000 lbP ........................................................................................................................Ans.

4-34 A steel eyebar with a 100 25-mm rectangular cross section has been designed to transmit an axial tensile load P. The length of the eyebar must be increased by welding a new center section in the bar, as shown in Fig. P4-33. If P = 250 kN, calculate and plot the normal stress n and the shear stress n in the weld material for weld angles (30 90 ). If the stresses in the weld material must be limited to 80 MPa in tension and 60 MPa in shear, what ranges of would be acceptable for the joint? Repeat for P = 305 kN and for P = 350 kN. Are weld angles < 30 reasonable? Why or why not?


0 :nF sin 0N P

0 :tF cos 0V P


0.1 0.025 sinnA A

20.0025 sin mnA

Therefore

2

2sin sin N/m0.0025 sin 0.0025n

N P PA

............................................................................Ans.

2cos sin cos N/m0.0025 sin 0.0025n

V P PA

.....................................................................Ans.

When 250 kNP 0 64 ..................................................................................Ans.

When 305 kNP 0 40 or 50 54 ..................................................Ans.

When 350 kNP 0 30 ..................................................................................Ans.

Angles less than about 30 are not very practical since the weld gets long quickly as the angle decreases.................................................................................................................................Ans.

4-35 Specifications for the rectangular (3 3 21-in.) block shown in Fig. P4-35 require that the normal and shearing stresses on plane A-A not exceed 800 psi and 500 psi, respectively. If the plane A-A makes an angle

= 37 with the horizontal, calculate and plot the ratios / max and / max as a function of the load P (0 P 13 kip). What is the maximum load Pmax that can be applied to the block? Which condition controls what the maximum load can be? Repeat for = 25 . For what angle will the normal stress and the shear stress both reach their limiting values at the same time?


0 :nF cos 0N P


75

0 :tF sin 0V P


3 3 cosnA A

29 cos in.nA

Therefore

2

max max

cos 9 psi800

nN A P.................................................. Ans.

max max

sin cos 9 psi500

nV A P......... Ans.

When 37 , shear stress controls and

max 9360 lbP ............................................. Ans.

When 25 , normal stress controls and

max 8770 lbP ............................................. Ans.

The normal stress and the shear stress will reach their maximum values at the same time if

max

max

500tan 0.62500800

nopt

n

AVN A

32.01opt ................................................................................................................................Ans.

4-36 Compression tests of concrete indicate that concrete fails when the axial compressive strain is 1200 m/m. Determine the maximum change in length that a 200-mm diameter 400-mm long concrete test specimen can tolerate before failure occurs.

SOLUTION

61200 10 400 0.480 mmL ..........................................................................Ans.

4-37 The 0.5 2 4-in. rubber mounts shown in Fig. P4-37 are used to isolate the vibrational motion of a machine from its supports. Determine the average shearing strain in the rubber mounts if the rigid frame displaces 0.01 in. vertically relative to the support.

SOLUTION

0.01 0.02 in./in. 0.02 rad0.5L

.................................................................................Ans.

4-38 A structural steel bar was loaded in tension to fracture. A 200 mm-length of the bar was marked off in 25-mm lengths before loading. After the rod broke, the 25-mm segments were found to have lengthened to 30.0, 30.5, 31.5, 34.0, 44.5, 32.0, 31.0, and 30.0 mm, consecutively. Determine

(a) The average strain over the 200-mm length. (b) The maximum average strain over any 50-mm length. SOLUTION

263.5 200 0.317 mm/mm

200f i

i

L LL L

.............................................................Ans.

max78.5 50 0.570 mm/mm

50........................................................................................Ans.


76

4-39 Mutually perpendicular axes in an unstressed member were found to be oriented at 89.92 when the member was stressed. Determine the shearing strain associated with these axes in the stressed member.

SOLUTION

390 89.92 1.396 10 rad2 180

..........................................................Ans.

4-40 A thin triangular plate is uniformly deformed as shown in Fig. P4-40. Determine the shearing strain at P associated with the two edges (PQ and PR) that were orthogonal in the undeformed plate.

SOLUTION

500 mmP QL

500cos 45 353.5534 mmMQ MPL L

10 363.5534 mmMP MPL L

1tan 44.20107MQ

MP

LL

2

2 2

90 88.40213180

327.9 10 rad ..................................................................................................................... Ans.

4-41 The sanding-drum mandrel shown in Fig. P4-41 is made for use with a hand drill. The mandrel is made from a rubber-like material which expands when the nut is tightened to secure the sanding drum placed over the outside surface. If the diameter D of the mandrel increases from 2.00 in. to 2.15 in. as the nut is tightened, determine

(a) The average normal strain along a diameter of the mandrel. (b) The circumferential strain at the outside surface of the mandrel. SOLUTION

(a) 32.15 2.00 75.0 10 in./in.2.00D ...................................................................................Ans.

(b) 32.15 2.0075.0 10 in./in.

2.00C .....................................................................Ans.

4-42 A thin rectangular plate is uniformly deformed as shown by PRSQ in Fig. P4-42. Determine the shearing stain xy at P.

SOLUTION

11

0.380tan 0.04354500

12

0.200tan 0.04584250

1 220.08938

31.560 10 rad ................................................................................................................... Ans.


77

4-43 A steel sleeve is connected to a steel shaft with a flexible rubber insert, as shown in Fig. P4-43. The insert has an inside diameter of 3.3 in. and an outside diameter of 4.3 in. When the unit is subjected to a torque T, the shaft rotates 1.5 with respect to the sleeve. Assume that radial lines in the unstressed state remain straight as the rubber deforms. Determine the shearing strain r in the rubber insert

(a) At the inside surface. (b) At the outside surface. SOLUTION

3.3 1.5 0.0431969 in.2 180

b r

(b) 11 tan 4.93774

0.5b

386.2 10 rad outside ..................................................Ans.

(a) 2 1 1.5 6.43774

3112.4 10 rad inside .................................................Ans.

4-44 A steel rod is subjected to a nonuniform heating that produces an extensional (axial) strain that is proportional to the square of the distance from the unheated end ( = kx2). If the strain is 1250 m/m at the midpoint of a 3.00-m rod, determine

(a) The change in length of the rod. (b) The average axial strain over the length L of the rod. (c) The maximum axial strain in the rod. SOLUTION

2kx 261250 10 1.5k 6555.556 10k

2d kxdx

(a)

333 2 3

00

9 5.00 10 m 5.00 mm3kxd dx k x dx k ...................Ans.

(b) 30.005 1.667 10 m/m 1667 m/m3avg .................................................................Ans.

(c) 2 3

max 3 3 9 5.00 10 m/m 5000 m/mk k ................................................Ans.

4-45 The axial strain in a suspended bar of material of varying cross section due to its own weight, as shown in Fig. P4-45, is given by the expression y/3E, where is the specific weight of the material, y is the distance from the free (bottom) end of the bar, and E is a material constant. Determine, in terms of , L, and E,

(a) The change in length of the bar due to its own weight. (b) The average axial strain over the length L of the bar. (c) The maximum axial strain in the bar. SOLUTION

3y dE dy

(a) 2 2

003 3 2 6

LL y Ld y dy

E E E..........................................................................Ans.


78

(b) in./in.6avgL

L E................................................................................................................Ans.

(c) max in./in.3LLE

...............................................................................................................Ans.

4-46 A steel cable is used to tether an observation balloon. The force exerted on the cable by the balloon is sufficient to produce a uniform strain of 500 m/m in the cable. In addition, at each point in the cable, the weight of the cable reduces the axial strain by an amount that is proportional to the length of the cable between the balloon and the point. When the balloon is directly overhead at an elevation of 300 m, the axial strain at the midlength of the cable is 350 m/m. Determine

(a) The total elongation of the cable. (b) The maximum height that the balloon could achieve SOLUTION

6500 10 cy 6 6350 10 500 10 150c 61 10c

6 6500 10 1 10d ydy

2

6 6 2 6

00

10 500 10 500 500 0.5 102

LL yd y dy y L L

(a) If 300 mL

0.1050 m 105.0 mm .....................................................................................................Ans.

(b) If 500 my , then 0 . If 1000 mL , then 0 . Neither is possible. Therefore

max 500 my .............................................................................................................................. Ans.

4-47 A steel cable is used to support an elevator cage at the bottom of a 2000 ft-deep mine shaft. A uniform axial strain of 250 in./in. is produced in the cable by the weight of the cage. At each point the weight of the cable produces an additional axial strain that is proportional to the length of the cable below the point. If the total axial strain in the cable at the cable drum (upper end of the cable) is 700 in./in., determine

(a) The strain in the cable at a depth of 500 ft. (b) The total elongation of the cable. SOLUTION

6250 10 cy 6 6700 10 250 10 2000c 60.22500 10c

(a) 6 6500 250 10 0.22500 10 1500

6587 10 in./in. 587 in./in. ..................................................................................Ans.

(b) 6 6250 10 0.22500 10d ydy

2000220006 6

00

2 6

0.22510 250 0.225 10 2502

250 2000 0.1125 2000 10

yd y dy y

0.9500 ft 11.40 in. .........................................................................................................Ans.


79

4-48 At the proportional limit, a 200 mm-gage length of a 15 mm-diameter alloy bar has elongated 0.90 mm and the diameter has been reduced 0.022 mm. The total axial load carried was 62.6 kN. Determine the modulus of elasticity, Poisson’s ratio, and the proportional limit for the material.

SOLUTION E

3

262.6 10 0.90

2000.015 4E

9 278.7 10 N/m 78.7 GPaE ......................................................................................Ans.

0.022 15 0.3260.9 200

t

a

.........................................................................................Ans.

3

6 22

62.6 10 354 10 N/m 354 MPa0.015 4p

PA

...............................................Ans.

4-49 A 1.50 in.-diameter rod 20 ft long elongates 0.48 in. under a load of 53 kip. The diameter of the rod decreases 0.001 in. during the loading. Determine the modulus of elasticity, Poisson’s ratio, and the modulus of rigidity for the material.

SOLUTION E

3

253 10 0.48

20 121.5 4E

615.00 10 psi 15,000 ksiE ........................................................................................Ans.

0.001 1.5 0.333

0.48 20 12t

a

..................................................................................Ans.

15,000 5625 ksi

2 1 2 1 0.333EG ..........................................................................Ans.

4-50 A tensile test specimen having a diameter of 5.64 mm and a gage length of 50 mm was tested to fracture. Stress and strain values, which were calculated from load and deformation data obtained during the test, are shown in Fig. P4-50. Determine

(a) The modulus of elasticity. (b) The proportional limit. (c) The ultimate strength. (d) The yield strength (0.05% offset). (e) The yield strength (0.2% offset). (f) The fracture stress. (g) The true fracture stress if Poisson’s ratio = 0.30 remains constant. (h) The tangent modulus at a stress level of 400 MPa. (i) The secant modulus at a stress level of 400 MPa. SOLUTION From the stress-strain diagram:

(a) 3140 70.0 10 MPa 70.0 GPa0.002

E ...........................................................................Ans.

(b) 150 MPapl ........................................................................................................................... Ans.

(c) 450 MPault .......................................................................................................................... Ans.


80

(d) .05 220 MPa ............................. Ans.

(e) .2 275 MPa .............................. Ans.

(f) 450 MPaf .............................. Ans.

(g) 0.3 0.115t add

0.19458 mmd

5.64 0.19458 5.4454 mmfd

2

2

450 5.64 4483 MPa

5.4454 4f f i

tff f

P AA A

......................................................Ans.

(h) 430 350 1000 MPa 1.000 GPa0.08 0tE ..........................................................................Ans.

(i) 400 8890 MPa 8.89 GPa

0.045sE ...................................................................................Ans.

4-51 A tensile test specimen having a diameter of 0.250 in. and a gage length of 2.000 in. was tested to fracture. Stress and strain values, which were calculated from load and deformation data obtained during the test, are shown in Fig. P4-51. Determine

(a) The modulus of elasticity. (b) The proportional limit. (c) The ultimate strength. (d) The yield strength (0.05% offset). (e) The yield strength (0.2% offset). (f) The fracture stress. (g) The true fracture stress if the final diameter of the specimen at the location of the fracture was 0.212 in. (h) The tangent modulus at a stress level of 56 ksi. (i) The secant modulus at a stress level of 56 ksi. SOLUTION From the stress-strain diagram:

(a) 3

628 10 28.0 10 psi 28,000 ksi0.001

E ........................................................................Ans.

(b) 34 ksipl ................................... Ans.

(c) 74 ksiult ................................... Ans.

(d) .05 43 ksi ................................... Ans.

(e) .2 43 ksi .................................... Ans.

(f) 65 ksif .................................... Ans.

(g)

2

2

65 0.25 4

0.212 4f f i

tff f

P AA A

90.4 ksi ............................................................................................................................. Ans.

(h) 78 35 537 ksi0.08 0tE ............................................................................................................Ans.


81

(i) 56 1400 ksi

0.04sE ................................................................................................................Ans.

4-52 A tensile test specimen having a diameter of 11.28 mm and a gage length of 50 mm was tested to fracture. Determine

(a) The modulus of elasticity. (b) The proportional limit. (c) The ultimate strength. (d) The yield strength (0.05% offset). (e) The yield strength (0.2% offset). (f) The fracture stress. (g) The true fracture stress if the final diameter of the specimen at the location of the fracture was 9.50 mm. (h) The tangent modulus at a stress level of 315 MPa. (i) The secant modulus at a stress level of 315 MPa. SOLUTION From the stress-strain diagram:

(a) 3222.1 185 10 MPa0.0012

E

185 GPaE ...................Ans.

(b) 280 MPapl ...............Ans.

(c) 507 MPault ...............Ans.

(d) .05 300 MPa ...............Ans.

(e) .2 330 MPa ................Ans.

(f) 450 MPaf ................Ans.

(g) 3

6 22

45.1 10 636 10 N/m 636 MPa0.0095 4

ftf

f

PA

..............................................Ans.

(h) 3320 306 17.5 10 MPa 17.5 GPa0.0032 0.0024tE ........................................................Ans.

(i) 3315 105 10 MPa 105 GPa0.003sE .............................................................................Ans.

4-53 A tensile test specimen having a diameter of 0.505 in. and a gage length of 2.00 in. was tested to fracture. Determine

(a) The modulus of elasticity. (b) The proportional limit. (c) The ultimate strength. (d) The yield strength (0.05% offset). (e) The yield strength (0.2% offset). (f) The fracture stress. (g) The true fracture stress if the final diameter of the specimen at the location of the fracture was 0.425 in. (h) The tangent modulus at a stress level of 46,000 psi. (i) The secant modulus at a stress level of 46,000 psi. SOLUTION From the stress-strain diagram:

(a) 3

632 10 26.7 10 psi 26,700 ksi0.0012

E ........................................................................Ans.


82

(b) 40 ksipl .....................Ans.

(c) 73 ksiult .....................Ans.

(d) .05 44 ksi .....................Ans.

(e) .2 47 ksi ......................Ans.

(f) 65 ksif ......................Ans.

(g) 3

213 100.425 4

ftf

f

PA

391.6 10 psi 91.6 ksi ..........................................................................Ans.

(h) 47.43 45.93 1875 ksi

0.004 0.0032tE ........................................................................Ans.

(i) 46 14,370 ksi

0.0032sE ..................................................................................Ans.

4-54 A cast iron pipe has an inside diameter of 70 mm and an outside diameter of 105 mm. The length of the pipe is 2.5 m. The coefficient of thermal expansion for cast iron is = 12.1(10 6)/ C. Determine the dimension changes caused by

(a) An increase in temperature of 70 C. (b) A decrease in temperature of 85 C. SOLUTION

(a) 612.1 10 70 2500 2.12 mmL ..........................................................................Ans.

612.1 10 70 105 0.0889 mmo .........................................................................Ans.

612.1 10 70 70 0.0593 mmi ............................................................................Ans.

(b) 612.1 10 85 2500 2.57 mmL .....................................................................Ans.

612.1 10 85 105 0.108 mmo ......................................................................Ans.

612.1 10 85 70 0.072 mmi .........................................................................Ans.

4-55 A large cement kiln has a length of 225 ft and a diameter of 12 ft. Determine the change in length and diameter of the structural steel shell [ = 6.5(10 6)/ F] caused by an increase in temperature of 250 F.

SOLUTION

66.5 10 250 225 12 4.39 in.L .......................................................................Ans.

66.5 10 250 12 12 0.234 in.d ........................................................................Ans.

4-56 An airplane has a wing span of 40 m. Determine the change in length of the aluminum alloy [ = 22.5(10 6)/ C] wing spar if the plane leaves the ground at a temperature of 40 C and climbs to an altitude where the temperature is 40 C.

SOLUTION

6 322.5 10 80 40 72.0 10 m 72.0 mm ...........................................Ans.


83

4-57 Determine the movement of the pointer of Fig. P4-57 with respect to the scale zero when the temperature increases 80 F. The coefficients of thermal expansion are 6.6(10 6)/ F for the steel and 12.5(10 6)/ F for the aluminum.

SOLUTION The steel post and the steel scale each stretch the same amount

66.6 10 80 20 0.01056 in.s o

The aluminum post stretches

612.5 10 80 20 0.02000 in.a

Therefore, the motion of the pointer relative to the zero on the steel scale is

5 1p a s

0.0472 in. p ..................................................................................................................... Ans.

4-58 A bronze [ B = 16.9(10 6)/ C] sleeve with an inside diameter of 99.8 mm is to be placed over a solid steel [ S = 11.9(10 6)/ C] cylinder, which has an outside diameter of 100 mm. If the temperatures of the cylinder and sleeve remain equal, how much must the temperature be increased in order for the bronze sleeve to slip over the steel cylinder?

SOLUTION

6

6

100 11.9 10 100

99.8 16.9 10 99.8

d T

T

403 CT ................................................................................................................................Ans.

4-59 A steel [E = 30,000 ksi and = 6.5(10 6)/ F] surveyor’s tape ½-in. wide 1/32-in. thick is exactly 100 ft long at 72 F and under a pull of 10 lb. What correction should be introduced if the tape is used to make a 100-ft measurement at a temperature of 100 F and under a pull of 25 lb.

SOLUTION

1L L LTE L L L

where L is the length at 72 F and zero force. When 10 lbP , 10 640 psi

1 2 1 32, 100 ftL , and

6

100 6401 030 10L

99.99787 ftL

Then, when 25 lbP , 25 1600 psi

1 2 1 32, and 28 FT

66

16001 6.5 10 2830 10

LL

100.02140 ftL

0.02140 ft 0.257 in.L

correction 0.257 in. ............................................................................................................Ans.


84

4-60 A 25-mm diameter aluminum [ = 22.5(10 6)/ C, E = 73 GPa, and = 0.33] rod hangs vertically while suspended from one end. A 2500-kg mass is attached at the other end. After the load is applied, the temperature decreases 50 C. Determine

(a) The axial stress in the rod. (b) The axial strain in the rod. (c) The change in diameter of the rod. SOLUTION

(a) 6 22

2500 9.8149.962 10 N/m 50.0 MPa

0.025 4......................................................Ans.

(b) 6

6 69

49.962 10 22.5 10 50 441 10 m/m73 10a T

E......................Ans.

(c) 6

6 69

49.962 100.33 22.5 10 50 1350.86 10 m/m73 10t T

E

61350.86 10 25 0.0338 mmd td ...........................................................Ans.

4-61 The rigid yokes B and C of Fig. P4-61 are securely fastened to the 2-in. square steel (E 30,000 ksi) bar AD. Determine

(a) The maximum normal stress in the bar. (b) The change in length of segment AB. (c) The change in length of segment BC. (d) The change in length of the complete bar. SOLUTION

(a) 3

maxmax

82 10 20,500 psi 20.5 ksi2 2

PA

.................................................................Ans.

(b) 3

6

82 10 8 120.0656 in.

30 10 2 2ABPLEA

........... Ans.

(c) 3

6

12 10 5 120.00600 in.

30 10 2 2BC .............. Ans.

3

6

45 10 4 120.01800 in.

30 10 2 2CD

(d) 0.0776 in.total AB BC CD ......................................................................................Ans.

4-62 A structural tension member of aluminum alloy (E = 70 GPa) has a rectangular cross section 25 75 mm and is 2 m long. Determine the maximum axial load that may be applied if the axial stress is not to exceed 100 MPa and the total elongation is not to exceed 4 mm.

SOLUTION

9

20004 mm

70 10 0.025 0.075PPL

EA 262,500 NP

6100 10 0.025 0.075 187,500 NP A

max 187.5 kNP ......................................................................................................................... Ans.


85

4-63 The tension member of Fig. P4-63 consists of a steel (E = 30,000 ksi) pipe A, which has an outside diameter of 6 in. and an inside diameter of 4.5 in., and a solid aluminum alloy (E = 10,600 ksi) bar B, which has a diameter of 4 in. Determine the overall elongation of the member.

SOLUTION

120 kipA BP P

3 3

6 2 2 6 2

120 10 3 12 120 10 4 12

30 10 6 4.5 4 10.6 10 4 4A B

0.01164 0.04324 0.0549 in. .....................................................................................Ans.

4-64 A steel (E = 200 GPa) rod, which has a diameter of 30 mm and a length of 1.0 m, is attached to the end of a Monel (E = 180 GPa) tube, which has an internal diameter of 40 mm, a wall thickness of 10 mm, and a length of 2.0 m, as shown in Fig. P4-64. Determine the load required to stretch the assembly 3.00 mm.

SOLUTION

S MP P P

9 2 9 2 2

1000 20003.00 mm

200 10 0.030 4 180 10 0.060 0.040 4P P

3212 10 N 212 kNP ......................................................................................................Ans.

4-65 The floor of a warehouse is supported by an air-dried, red-oak, timber column (see Appendix A for properties), as shown in Fig. P4-65. Contents of the warehouse subject the 12 12-in. column to an axial load of 200 kip. If the column is 30-in. long, determine

(a) The deformation of the column. (b) The normal stress in the column. (c) The bearing stress between the column and the lower bearing plate. (d) The maximum shearing stress in the column. SOLUTION 1800 ksiE

(a) 3

3

200 10 300.0231 in.

1800 10 12 12PLEA

.....................................................................Ans.

(b) 3200 10 1389 psi

12 12PA

................................................................................................Ans.

(c) 3200 10 1389 psi

12 12b .......................................................................................................Ans.

(d) 3

max200 10 694 psi

2 2 12 12PA

..........................................................................................Ans.

4-66 The roof and second floor of a building are supported by the column shown in Fig. P4-66. The column is a structural steel (see Appendix A for properties) section having a cross sectional area of 5700 mm2. The roof and floor subject the column to the axial forces shown. Determine

(a) The amount that the first floor will settle. (b) The amount that the roof will settle. SOLUTION

200 GPaE 1 1030 kN (C)P 2 380 kN (C)P


86

(a) 3

1 9 6

1030 10 35003.1623 3.16 mm

200 10 5700 10PLEA

.............................................Ans.

(b) 3

1 9 6

380 10 35003.1623 1.1667 4.33 mm

200 10 5700 10r ...............................Ans.

4-67 The tension member of Fig. P4-67 consists of a structural steel pipe A, which has an outside diameter of 6 in. and an inside diameter of 4.5 in., and a solid 2014-T4 aluminum alloy bar B, which has a diameter of 4 in. (see Appendix A for properties). Determine

(a) The change in length of the steel pipe. (b) The overall deflection of the member. (c) The maximum normal and shearing stresses in the aluminum bar. SOLUTION

29,000 ksistE 10,600 ksialE

205 kip (T)AP 120 kip (T)BP

(a) 3

6 2 2

205 10 500.0286 in.

29 10 6 4.5 4A

PLEA

......................................................Ans.

(b) 3

6 2

120 10 400.0360 in.

10.6 10 4 4B

0.0646 in.total A B ....................................................................................................Ans.

(c) 3

max 2120 10 9550 psi

4 4PA

...........................................................................................Ans.

maxmax 4770 psi

2 2PA

.................................................................................................Ans.

4-68 An aluminum alloy (E = 73 GPa) tube A with an outside diameter of 75 mm is used to support a 25-mm diameter steel (E = 200 GPa) rod B, as shown in Fig. P4-68. Determine the minimum thickness t required for the tube if the maximum deflection of the loaded end of the rod must be limited to 0.40 mm.

SOLUTION

total A B

3 3

29 9

35 10 300 35 10 9000.40 mm

73 10 200 10 0.025 4A

6 2 2

2 2

1817.40 10 m 1817.40 mm

75 4i

A

d

57.54 mm 75 2id t

8.73 mmt ................................................................................................................................Ans.


87

4-69 A structural steel (see Appendix A for properties) bar of rectangular cross section consists of uniform and tapered sections as shown in Fig. P4-69. The width of the tapered section varies linearly from 2 in. at the bottom to 5 in. at the top. The bar has a constant thickness of ½ in. Determine the elongation of the bar resulting from application of the 30-kip load P. Neglect the weight of the bar.

SOLUTION 29,000 ksiE

32 in.60yb 0 60 in.y

3 3 60

66 0

6030

3

30 10 25 30 10329 1029 10 2 0.5 2 0.560

0.02586 1.03448 10 40ln 40

0.02586 41.37931 10 4.60517 3.68888

dyy

y

0.0638 in. ............................................................................................................................ Ans.

4-70 Determine the change in length of the homogeneous conical bar of Fig. P4-70 due to its own weight. Express the results in terms of L, E, and the specific weight of the material. The taper of the bar is slight enough for the assumption of a uniform axial stress distribution over a cross section to be valid.

SOLUTION

0 :F 0A W

2

2

3R yR W Vol

3y

ryRL

2 2

003 3 2 6

LLP dy y dy y Ldy

EA E E E E........................................................Ans.

4-71 The bar shown in Fig. P4-71 is made of annealed bronze (see Appendix A for properties). In addition to its own weight, the bar is subjected to an axial tensile load P of 5000 lb at its lower end. Determine the elongation of the bar due to the combined effects of its weight and the load P. Let r = 4 in. and L = 60 in.

SOLUTION

15,000 ksiE 30.320 lb in

60 604 in.

60 15y yR

22 4 6060

3 3R y

W Vol

0 :F 5000 0A W

2 25000 60 9603

R R y


88

2

60

20

5000 60 9603

R yP dy dy dyEA E E R

60 60

26 60 0

60

26 0

5000 22560

15 10 3 15 1060

960 225

3 15 10 60

dy y dyy

dyy

6060 26

0 060

6

0

6010.02387 0.00711 1060 2

1 1536.00 1060

yy

y

3 3 30.19892 10 0.03839 10 0.01280 10

30.225 10 in. ................................................................................................................... Ans.

4-72 A hollow brass (E = 100 GPa) tube A with an outside diameter of 100 mm and an inside diameter of 50 mm is fastened to a 50-mm diameter steel (E = 200 GPa) rod B, as shown in Fig. P4-72. The supports at the top and bottom of the assembly and the collar C used to apply the 500-kN load P are rigid. Determine

(a) The normal stresses in each of the members. (b) The deflection of the collar C. SOLUTION (a) From equilibrium

0 :F 500 0B AT P

2 2 2 30.05 0.1 0.05 500 10 N

4 4B A

where B is a tension stress and A is a compressive stress. Then, expressing that the stretch of the rod is equal to the shrink of the tube

B A in terms of the stresses gives

9 9

2000 1500200 10 100 10B A

1.5B A

Substituting back into the equilibrium equation gives

256.588 N/m 56.6 MPaA ..............................................................................Ans.

1.5 84.9 MPaB A ............................................................................................Ans.

(b) 6

9

56.588 10 15000.849 mm

100 10C A ................................................Ans.


89

4-73 The 7.5 7.5 20-in. oak (E = 1800 ksi) block shown in Fig. P4-73 was reinforced by bolting two 2 7.5 20-in. steel (E = 29,000 ksi) plates to opposite sides of the block. If the stresses in the wood and the steel are to be limited to 4.6 ksi and 22 ksi, respectively, determine

(a) The maximum axial compressive load P that can be applied to the reinforced block. (b) The shortening of the block when the load of part (a) is applied. SOLUTION (a) From equilibrium

0 :F 2 0w sP P P

7.5 7.5 2 2 7.5w s P

where w and s are both compressive stresses. Expressing that

the shrink of the steel is equal to the shrink of the wood s w in terms of the stresses gives

3 329,000 10 1800 10s wL L

16.111s w

If 22 ksis , then 1.36552 ksi 4.6 ksiw , and

1.36552 7.5 7.5 2 22 2 7.5 737 kipP .......................................Ans.

(b) 3

3

22 10 200.01517 in.

29,000 10s .............................................................Ans.

4-74 Five 25-mm diameter steel (E = 200 GPa) reinforcing bars will be used in a 1-m long concrete (E = 31 GPa) pier with a square cross section, as shown in Fig. P4-74. The allowable strengths in compression for steel and concrete are 130 MPa and 9.5 MPa, respectively. Determine the minimum size of pier required to support a 900-kN axial load.

SOLUTION From equilibrium

0 :F 5 900 0s cP P

2 35 0.025 4 900 10 Ns c cA

where s and c are both compressive stresses. Expressing that the

shrink of the steel rods is equal to the shrink of the concrete s c in terms of the stresses gives

9 9200 10 31 10s cL L

6.4516s c

If 9.5 MPac , then 61.29 MPa 130 MPas , and

6 2 2

22

78,902 10 m 78,902 mm

5 25 4cA

b

285 mmb ................................................................................................................................Ans.


90

4-75 A hollow steel (E = 30,000 ksi) tube A with an outside diameter of 2.5 in. and an inside diameter of 2 in. is fastened to an aluminum (E = 10,000 ksi) bar B that has a 2-in. diameter over one-half of its length and a 1-in. diameter over the other half. The assembly is attached to unyielding supports at the left and right ends and is loaded as shown in Fig. P4-75. Determine

(a) The normal stresses in all parts of the bar. (b) The deflection of cross-section a-a. SOLUTION (a) From equilibrium

40 kipB AP P 10 kipC AP P

where AP , BP , and CP are all tension forces. Since the total length of the assembly cannot change, the total stretch must be zero 0A B C . Therefore

3 2 2 3 2 3 2

20 40 24 10 240

30 10 2.5 2 4 10 10 2 4 10 10 1 4A A AP P P

0 kipAP 0 ksiA .......................................................................Ans.

2

0 4012.73 ksi (T)

2 4B ....................................................................................Ans.

2

0 1012.73 ksi 12.73 ksi (C)

1 4C ............................................................Ans.

(b) 3

26

10 10 240.03056 in.

10 10 1 4C

Therefore section a-a moves 0.03056 in. ............................................................................Ans.

4-76 A 150-mm diameter 200-mm long polymer (E = 2.10 GPa) cylinder will be attached to a 45-mm diameter 400-mm long brass (E = 100 GPa) rod by using the flange type of connection shown in Fig. P4-76. A 0.15-mm clearance exists between the parts as a result of a machining error. If the bolts are inserted and tightened, determine

(a) The normal stresses produced in each of the members. (b) The final position of the flange-polymer interface after assembly, with respect to the left support. SOLUTION (a) From equilibrium

p bT T

where pT and bT are both tension forces. When the bolts are tightened, the stretch of the two pieces closes the gap,

0.15 mmp b . Therefore

2 29 9

200 4000.15 mm

2.1 10 0.15 4 100 10 0.045 4p bT T

18,976.74 Np bT T


91

6 22

18,976.741.074 10 N/m 1.074 MPa (T)

0.15 4p ......................................Ans.

6 22

18,976.7411.93 10 N/m 11.93 MPa (T)

0.045 4b ....................................Ans.

(b) 29

18,976.74 2000.10227 mm

2.1 10 0.15 4p

Therefore the interface will be located at 200.1023 mm ...........................................................Ans.

4-77 The assembly shown in Fig. P4-77 consists of a steel bar A (Es = 30,000 ksi and As = 1.25 in.2), a rigid bearing plate C that is securely fastened to bar A, and a bronze bar B (EB = 15,000 ksi and AB = 3.75 in.2). A clearance of 0.015 in. exists between the bearing plate C and bar B before the assembly is loaded. After a load P of 95 kip is applied to the bearing plate, determine

(a) The normal stresses in bars A and B. (b) The vertical displacement of the bearing plate C. SOLUTION (a) From equilibrium (assuming that the gap is closed and the bearing plate

pushes on the brass bar,

0 :F 95 0A BT P

or in terms of stresses

31.25 3.75 95 10 lbA B

in which A is a tension stress and B is a compression stress. If the bearing plate presses against the brass bar, the stretch of the steel rod will exceed the shrink of the brass bar by the initial gap, 0.015 in.A B Therefore

6 6

6 12 2 120.015 in.

30 10 15 10A B

2 3 6250 psiA B

Combining the equilibrium equation and the deformation equation gives

18,932 psi 18.93 ksi (T)A .................................................................................Ans.

19,023 psi 19.02 ksi (C)B .................................................................................Ans.

(b) 6

18,932 6 120.0454 in.

30 10C A .......................................................................Ans.

4-78 A column similar to Fig. P4-74 is being designed to carry a load of 4450 kN. The column, which will have a 500 500 mm square cross section, will be made of concrete (E = 20 GPa) and will be reinforced with 50-mm diameter steel (E = 200 GPa) bars. If the allowable stresses are 120 MPa in the steel and 8 MPa in the concrete, determine

(a) The number of steel bars required. (b) The stresses in the steel and concrete when the bars of part (a) are used. (c) The change in length of a 3-m long column when the bars of part (a) are used. SOLUTION (a) From equilibrium


92

0 :F 4450 0c sP nP

4450 kNc sP nP

Expressing that the steel rods and the concrete must shrink the same amount, c s , in terms of stresses gives

9 920 10 200 10c sL L

10s c

If max 8 MPac , then 80 MPa 120 MPas , and

2 26 2 6 38 10 0.5 0.05 4 80 10 0.05 4 4450 10n n

17.33n

Therefore the required number of rods is 18n ..........................................................................Ans.

(b) If 18n , then

2 22 30.5 18 0.05 4 18 10 0.05 4 4450 10c c

6 27.83 10 N/m 7.83 MPac .............................................................................Ans.

10 78.3 MPas c ..................................................................................................Ans.

(c) 6

9 9

78.3 10 300030001.175 mm

200 10 200 10s

s ........................................Ans.

4-79 A ½-in. diameter alloy-steel bolt (E = 30,000 ksi) passes through a cold-rolled brass sleeve (E = 15,000 ksi) as shown in Fig. P4-79. The cross-sectional area of the sleeve is 0.375 in.2 Determine the normal stresses produced in the bolt and sleeves by tightening the nut ¼ turn (0.020 in.).


st brT F


20.5 4 0.375st br

1.90986st br

in which st is a tension stress and br is a compression stress. The stretch of the steel bolt, the shrink of the

brass sleeve, and the movement of the nut are related by 0.02st br . Therefore

6 6

6 60.02

30 10 15 10st br 100,000 2st br


25,576 psi 25.6 ksi (C)br ..................................................................................Ans.

48,847 psi 48.8 ksi (T)st ..................................................................................Ans.


93

4-80 The two faces of the clamp shown in Fig. P4-80 are 250 mm apart when the two stainless-steel (Es = 190 GPa) bolts connecting them are unstretched. A force P is applied to separate the faces of the clamp so that an aluminum alloy (Ea = 73 GPa) bar with a length of 251 mm can be inserted as shown. Each of the bolts has a cross-sectional area of 120 mm2 and the bar has a cross-sectional area of 625 mm2. After the load P is removed, determine

(a) The axial stresses in the bolts and in the bar. (b) The change in length of the aluminum alloy bar. SOLUTION (a) Expressing the equilibrium equation

2a sF T

in terms of stresses gives

6 6625 10 2 120 10a s

2.60417s a

in which s is a tension stress and a is a compression stress. The stretch of the steel bolts and the shrink

of the aluminum bar are related by 1s a Therefore

9 9

330 2511

190 10 73 10s a 6575.76 10 1.97966s a


6 2327 10 N/m 327 MPas ................................................................................Ans.

6 2125.6 10 N/m 125.6 MPaa .........................................................................Ans.

(b) 6

9

125.6 10 2510.432 mm

73 10a .......................................................................Ans.

4-81 A high-strength steel bolt (Es = 30,000 ksi and As = 0.785 in.2) passes through a brass sleeve (Eb = 15,000 ksi and Ab = 1.767 in.2), as shown in Fig. P4-81. As the nut is tightened, it advances a distance of 0.125 in. along the bolt for each complete turn of the nut. Compute and plot

(a) The axial stresses s (in the steel bolt) and b (in the brass sleeve) as functions of the angle of twist of the nut (0 180 ).

(b) The elongations s (of the steel bolt) and b (of the brass sleeve) as function of (0 180 ). (c) The distance L between the two washers as a function of (0 180 ). SOLUTION (a) From equilibrium

st brT F


0.785 1.767st br

2.25096st br

in which st is a tension stress and br is a compression stress. The stretch of the steel bolt, the shrink of

the brass sleeve, and the movement of the nut are related by 0.125 360st br in which is in degrees. Therefore

6 6

14 120.12530 10 360 15 10st br 744.0476 1.71429st br



94

422.374 psi (T)st ................... Ans.

187.642 psi (C)br ................... Ans.

(b) 36

140.19711 10 in.

30 10st

st .......................................................................Ans.

36

120.15011 10 in.

15 10br

br ......................................................................Ans.

(c) 312 12 0.15011 10 in.brL ..................................................................Ans.

4-82 The short pier shown in Fig. P4-82 is reinforced with nine steel (E = 210 GPa) reinforcing bars. An axial compressive load P is applied to the pier through the rigid capping plate. The axial load carried by the matrix material is a function of R, the percentage of the cross section taken up by the steel reinforcement bars. The load is also a function of the modulus ratio ER/EM where ER and EM are the modulus of elasticity for the reinforcement material and the matrix material, respectively. For the three matrix-reinforcement combinations listed, compute and plot the percentage of the load carried by the matrix as a function of R (0 R 100 %).


0 :F 9 0M sP P P

9M sP P P

Since the steel rods and the pier must shrink the same amount, M s ,

sM

M M s s

P LP LE A E A

s ss M

M M

E AP PE A


95


1 9 s sM

M M

E AP PE A

Then, since

9 9 100100 100

9 9 1s s

M s M s

A ARA A A A A

100 1

9M

s

AA R

9 1

100 1 100s

M

A RA R R

and

1

1100

M

s

M

PE RPE R

..............Ans.

4-83 A 3-in. diameter 80-in. long aluminum alloy bar is stress free after being attached to rigid supports, as shown in Fig P4-83. Determine the normal stress in the bar after the temperature drops 100 F. Use E = 10,600 ksi and = 12.5(10 6)/ F.

SOLUTION

66 12.5 10 100 0

10.6 10T

E

313.25 10 psi 13.25 ksi (T) ............................................................................Ans.

4-84 A 6-m long 50-mm diameter rod of aluminum alloy [E = 70 GPa, = 0.346, and = 22.5(10 6)/ C] is attached at the ends to supports that yield to permit a change in length of 1.00 mm in the rod when stressed. When the temperature is 35 C, there is no stress in the rod. After the temperature of the rod drops to 20 C, determine

(a) The normal stress in the rod. (b) The change in diameter of the rod. SOLUTION

(a) 69

600022.5 10 55 6000 1 mm

70 10L T LE

6 274.958 10 N/m 75.0 MPa (T) ...................................................................Ans.

(b) dL T LE

6

69

74.958 10 500.346 22.5 10 55 50

70 10

0.0804 mm 0.0804 mm (shrink) ...............................................................Ans.


96

4-85 A bar consists of 3-in. diameter aluminum alloy [E = 10,600 ksi, = 0.33, and = 12.5(10 6)/ F] and 4-in. diameter steel [E = 30,000 ksi, = 0.30, and = 6.6(10 6)/ F] parts, as shown in Fig. P4-85. If end supports are rigid and the bar is stress free at 0 F, determine

(a) The normal stress in both parts of the bar at 80 F. (b) The change in diameter of the steel part of the bar. SOLUTION (a) From equilibrium

s aP P

in which both sP and aP are compressive forces. Since the end supports are rigid, the total stretch of the rod must be zero 0s a

626

626

206.6 10 80 20

30 10 4 4

30 12.5 10 80 30 0

10.6 10 3 4

s

a

P

P

89, 449 lb (C)s aP P

289, 449 7118 psi 7.12 ksi

4 4s ........................................................................Ans.

289, 449 12,654 psi 12.65 ksi

3 4a ...................................................................Ans.

(b) dsL T LE

66

7118 40.3 6.6 10 80 4 0.00240 in.

30 10..........................Ans.

4-86 A steel tie rod containing a rigid turnbuckle (see Fig. P4-86) has its ends attached to rigid walls. During the summer when the temperature is 30 C, the turnbuckle is tightened to produce a stress in the rod of 15 MPa. Determine the normal stress in the rod in the winter when the temperature is 10 C. Use E = 200 GPa and = 11.9(10 6)/ C.

SOLUTION

30 10 L T LE

6

69 9

15 100 11.9 10 40

200 10 200 10L L L

6 2110.2 10 N/m 110.2 MPa ...........................................................................Ans.

4-87 Nine ¾-in. diameter steel (E = 30,000 ksi) reinforcing bars were used when the short concrete (E = 4500 ksi) pier shown in Fig. P4-87 was constructed. After a load P of 150 kip was applied to the pier, the temperature increased 100 F. The coefficients of thermal expansion for steel and concrete are 6.6(10 6)/ F and 6.0(10 6)/ F, respectively. Determine

(a) The normal stresses in the concrete and in the steel bars after the temperature increases.


97

(b) The change in length of the pier resulting from the combined effects of the temperature change and the load.

SOLUTION (a) The equilibrium equation

0 :F 9 150 0c sP P

is written in terms of stresses

2 2 3100 9 0.75 4 9 0.75 4 150 10 lbc s

24.1504 37,725.62 psis c

in which c and s are both compressive stresses. Expressing that

the steel rods and the concrete must stretch the same amount, c s , in terms of stresses gives

6 66 66.0 10 100 6.6 10 100

4.5 10 30 10c sL LL L

6.6667 1800.00 psis c


1165.770 psi 1.166 ksi (C)c .............................................................................Ans.

9571.805 psi 9.57 ksi (C)s ...............................................................................Ans.

(b) 66

9571.805 246.6 10 100 24 0.00818 in.

30 10............................Ans.

4-88 The assembly shown in Fig. P4-88 consists of a steel (E = 210 GPa) cylinder A, a rigid bearing plate C, and an aluminum alloy (E = 71 GPa) bar B. Cylinder A has a cross-sectional area of 1850 mm2, and bar B has a cross-sectional area of 2500 mm2. After an axial load of 600 kN is applied, the temperature of cylinder A decreases 50 C and the temperature of bar B increases 25 C. The coefficients of thermal expansion are 11.9(10 6)/ C for the steel and 22.5(10 6)/ C for the aluminum. Determine

(a) The normal stresses in the cylinder and in the bar after the load is applied and the temperatures change. (b) The displacement of plate C after the load is applied and the temperatures change. SOLUTION (a) The equilibrium equation

0 :F 600 0B AT T

can be written in terms of stresses

6 6 31850 10 2500 10 600 10 NA B

6 21.35135 324.324 10 N/mA B

in which A and B are both tension stresses. The total stretch of the assembly must be zero,

0A B , therefore

6 69 9

750 60011.9 10 50 750 22.5 10 25 600 0

210 10 71 10A B

6 22.36620 30.450 10 N/mA B


6 2217.5 10 N/m 217 MPa (T)A ....................................................................Ans.


98

6 279.051 10 N/m 79.1 MPa (C)B ...............................................................Ans.

(b) 6

69

217.5 10 75011.9 10 50 750

210 10C A

0.331 mm C .........................................................................................................Ans.

4-89 A high-strength steel [Es = 30,000 ksi, As = 0.785 in.2, and s = 6.6(10 6)/ F] bolt passes through a brass [Eb = 15,000 ksi, Ab = 1.767 in.2, and b = 9.8(10 6)/ F] sleeve, as shown in Fig. P4-89. After the unit is assembled at 40 F, the temperature is increased to 100 F. If the unit is free of stress at 40 F, determine the normal stresses in the bolt and in the sleeve at 100 F.

SOLUTION Writing the equilibrium equation

st brT F


0.785 1.767st br

in which st is a tension stress and br is a compressive stress. Expressing that the steel bolt and the brass sleeve must stretch the same amount, st br , in terms of stresses gives

6 66 6

14 126.6 10 60 14 9.8 10 60 12

30 10 15 10st br

14 24 45,360 psist br


1839.29 psi 1839 psi (T)st ................................................................................Ans.

817.10 psi 817 psi (C)br ...................................................................................Ans.

4-90 The two faces of the clamp of Fig. P4-90 are 250 mm apart when the two stainless-steel [Es = 190 GPa, As = 115 mm2 (each), and s = 17.3(10 6)/ C] bolts connecting them are unstretched. A force P is applied to separate the faces of the clamp so that an aluminum alloy [Ea = 73 GPa, Aa = 625 mm2, and a = 22.5(10 6)/ C] bar with a length of 250.50 mm can be inserted as shown. After the load P is removed, the temperature is raised 100 C. Determine the normal stresses in the bolts and in the bar, and the distance between the faces of the clamps.


2a sF T


625 2 115a s

in which s is a tension stress and a is a compressive stress. Expressing that the steel bolts must stretch 0.5 mm more than the aluminum bar must stretch, 0.5s a , in terms of stresses gives


99

69

69

33017.3 10 100 330

190 10

250.5 0.5 22.5 10 100 250.5

73 10

s

a

6 21.73684 3.43151 492.725 10 N/ms a


6 260.448 10 N/m 60.4 MPa (C)a .................................................................Ans.

6 2164.262 10 N/m 164.3 MPa (T)s .............................................................Ans.

Then, the stretch of the aluminum bar is

6

69

60.448 10 250.522.5 10 100 250.5 0.35620 mm

73 10a

and the distance between the faces of the clamp is

250.5 250.856 mmaL ....................................................................................Ans.

4-91 A prismatic bar [E = 10,000 ksi and = 12.5(10 6)/ F], free of stress at room temperature, is fastened to rigid walls at its ends. One end of the bar is heated 200 F above room temperature while the other end is maintained at room temperature. The change in temperature T along the bar is proportional to the square of the distance from the unheated end. Determine the normal stress in the bar after the change in temperature.

SOLUTION

dxd T dxE

2

2

200xTL

2

20 0

200 0L Ldx xd dxE L

3

2

200 03

L LE L

6 6200 12.5 10 10 10200

3 3E

8333 psi 8.33 ksi (C) ......................................................................................Ans.

4-92 The two faces of the clamp shown in Fig. P4-90 are 250 mm apart when the two steel bolts [Es = 190 GPa, As = 115 mm2 (each), and s = 17.3(10 6)/ C] bolts connecting them are unstretched. A force P is applied to separate the faces of the clamp so that an brass bar [Eb = 100 GPa, Ab = 625 mm2, and b = 17.6(10 6)/ C] bar with a length of 250.50 mm can be inserted as shown. After the load P is removed, the temperature of the system is slowly raised. Compute and plot

(a) The axial stress s in the steel bolts and the axial stress b in the brass bar as functions of the temperature rise T (0 T 100 C).

(b) The elongation s of the steel bolts and the elongation b of the brass bar as functions of the temperature rise T (0 T 100 C).

(c) The distance L between the faces of the clamp as a function of the temperature rise T (0 T 100 C).

SOLUTION (a) Writing the equilibrium equation


100

2b sP T


625 2 115b s

in which s is a tension stress and b is a compressive stress. Expressing that the steel bolts must stretch 0.5 mm more than the brass bar must stretch, 0.5s b , in terms of stresses gives

69

69

33017.3 10 330

190 10

250.5 0.5 17.6 10 250.5

100 10

s

b

T

T

6 6 21.73684 2.5050 500 10 1.3002 10 T N/ms b


6 269.207 0.17997 10 N/m (C)b T .........................................................Ans.

6 2188.063 0.48904 10 N/m (T)s T ........................................................Ans.

(b) Then, the elongation of the steel bolts and the brass bar are

6

69

188.063 0.48904 10 33017.3 10 330

190 10sT

T

30.32664 4.860 10 mmT .............................................................................Ans.

6

69

69.207 0.17997 10 250.517.6 10 250.5

100 10bT

T

30.17336 4.860 10 mmT Ans.

(both stretches). (c) Finally, the distance between the faces of

the clamp is

250.5 bL

6250.32664 4.860 10 mmT ..........................................................................Ans.


101

4-93 An aluminum (Eal = 10,000 ksi, al = 12.5 10-6/ F, Aal = 1.4 in.2) bolt passes through a steel (Est = 30,000 ksi, st = 6.6 10-6/ F, Ast = 0.4 in.2) sleeve as shown in Fig. P4-89. Initially, the nut is tightened against the

washer at room temperature until the bolt has a tensile force of 3500 lb. Then the temperature of the assembly is slowly raised. Calculate and plot:

(a) The stress al in the aluminum bolt and the stress st in the steel sleeve as a function of the temperature increase T (0 F < T < 100 F).

(b) The change in length of the aluminum bolt al and the change in length of the steel sleeve st as a function of the temperature increase T (0 F < T < 100 F).

SOLUTION (a) Writing the equilibrium equation

al stT P


1.4 0.4al st

in which al is a tension stress and st is a compressive stress. The aluminum bolt must stretch more than

the steel sleeve stretches by the amount that the nut moves, al st nut . In terms of stresses this gives

6 66 6

14 1212.5 10 14 6.6 10 12

10 10 30 10al st

nutT T

When 0T the tension in the bolt and the compression in the sleeve are each equal to 3500 lb and

3500 2500 psi1.4al

3500 8750 psi0.4st

and the initial movement of the nut is

0.00700 in.nut

Therefore, the deformation equation can be written

14 4 70,000 958.00 psial st T


2500 34.214 psi (T)al T ..............................................................................Ans.

8750 119.750 psi (C)st T ............................................................................Ans.

If the assembly is not welded together, neither stress can be negative. Therefore, the above equations are valid only for T less than about 73 . For T greater than about 73 , the bolt and the sleeve will separate, and both stresses will be zero

0al st for 73T ..................................................................................Ans.


102

(b) Then, the change in length of the aluminum bolt and the steel sleeve are

66

1412.5 10 14 in. (stretch)

10 10al

al T .............................................Ans.

66

126.6 10 12 in. (stretch)

30 10st

st T .............................................Ans.

4-94 A short standard-weight steel pipe (see Appendix A) is used to support an axial compressive load of 100 kN. If yielding ( y = 250 MPa) should not occur, and the factor of safety is to be 1.6, determine the smallest nominal diameter pipe that may be used to support the load.

SOLUTION

3 6100 10 250 10

1.6A 6 2640 10 mA

Use 51 mmd (for which 2693.5 mmA ) ................................................................Ans.

4-95 A short column made of structural steel is used to support the floor beams of a building, as shown in Fig. P4-95. Each floor beam (A and B) transmits a force of 40 kip to the column. The column has the shape of a wide-flange (W) section (see Appendix A). The factor of safety based on failure by yielding is 3.0. Select the lightest wide-flange section that will support the given loads.

SOLUTION

36 ksiy 3.0FS

3 380 10 36 10

3.0A 26.667 in.A

W6 25 , W8 24 , W10 30 , W12 30 , etc. all have sufficiently large area. The lightest section is

W8 24 ....................................................................................................................................Ans.

4-96 The two structural steel (see Appendix A) rods A and B shown in Fig. P4-96 are used to support a mass m = 2000 kg. If failure is by yielding and a factor of safety of 1.75 is specified, determine the diameters of the rods (to the nearest 1 mm) that must be used to support the mass. Both rods are to have the same diameter.

SOLUTION

250 MPay

1.75FS From a free body diagram of the connection, the equilibrium equations

0 :xF cos30 cos50 0B AT T

0 :yF sin 50 sin 30 2000 9.81 0A BT T

give

6250 1017,254 N

1.75AT A 6 2120.8 10 mA

6250 1012,806 N

1.75BT A 6 289.6 10 mA

If the rods are made from the same material and diameter, then

2

2120.8 mm4d A 12.4 mmd

Use 13 mmd ........................................................................................................................ Ans.


103

4-97 The machine component shown in Fig. P4-97 is made of hot-rolled Monel. The forces at B are applied to the component with a rigid collar that is firmly attached to the component. If the mode of failure is yielding and the factor of safety is 1.5, determine the minimum permissible diameter of each segment of the machine component.

SOLUTION

50 ksiy 1.5FS

20 kip (T)ABF 3 3

2

20 10 50 104 1.5d

0.874 in.ABd ................................................................................................................Ans.

30 kip (C)BCF 3 3

2

30 10 50 104 1.5d

1.070 in.BCd .................................................................................................................Ans.

4-98 An axial load P = 1000 kN is applied to the rigid steel bearing plate on the top of the short column shown in Fig. P4-98. The outside segment of the column is made of structural steel. The inside core is made of fairly high strength concrete. Both segments are square. The failure modes are yielding for the steel and fracture for the concrete. The factor of safety is to be 1.4. If the area of the concrete is to be 10 times the area of the steel, determine the required dimensions.

SOLUTION

34 MPafc 250 MPays

1.4FS 10c sA A

Writing the equilibrium equation

1000 kNs cF F


31000 10 Ns s c cA A

in which c and s are both compressive stresses. Expressing that the steel and the concrete must stretch the

same amount, c s , in terms of stresses gives

9 9200 10 31 10s cL L

6.4516s c

If 34 MPac fc , then

6.4516 219.355 MPa 250 MPas c

Therefore

6 6 3219.355 10 34 10 10 1000 10 Ns sA A

6 21787.77 10 msA 6 210 17,877.7 10 mc sA A

and the required sizes are

concrete 133.7 mm 133.7 mm .............................................................................Ans.

2 17,877.7 1787.77b 140.2 mmb

steel 140.2 mm 140.2 mm .............................................................................Ans.


104

4-99 Four axial forces are applied to the 1-in. thick, 0.4% C hot-rolled steel bar, as shown in Fig. P4-99. The factor of safety for failure by yielding is 1.75. Determine the minimum width w of the constant cross-sectional area bar.

SOLUTION

53 ksiy 1.75FS

20 kip (T)ABF 3 320 10 53 10

1 1.75w 0.660 in.w

10 kip (C)BCF 3 310 10 53 10

1 1.75w 0.330 in.w

50 kip (T)BCF 3 350 10 53 10

1 1.75w 1.651 in.w

1.651 in.w .....................................................................................................................Ans.

4-100 The two parts of the eyebar shown in Fig. P4-100 are connected by two bolts (one on each side of the eyebar). The bolts are made of a grade of steel with a tensile yield strength of 1035 MPa and a shear yield strength of 620 MPa. The eyebar is subjected to the forces P = 85 kN. Determine the minimum bolt diameter required to safely support the forces if the mode of failure is yielding and the factor of safety is 1.5.

SOLUTION Since there are two bolts, there are two normal forces and two shear forces shown on the free-body diagram. The equilibrium equations gives

0 :nF 2 85cos 30 0N

0 :tF 2 85sin 30 0V

give

3 6 2

385 10 1035 10cos30 36.806 102 1.5 4

dN 0.00824 md

3 6 2

385 10 620 10sin 30 21.25 102 1.5 4

dV 0.00809 md

min 8.24 mmd ...............................................................................................................Ans.

4-101 The two solid rods shown in Fig. P4-101 are pin-connected at the ends and support a weight of 10 kip. The rods are made of SAE 4340 heat-treated steel. The factor of safety for failure by yielding is to be 1.5. For a minimum weight of rod design, determine

(a) The optimum angle . (b) The required diameter for the rods. (c) The weight of each rod. Is it reasonable to neglect the weight of the rods in the design? SOLUTION

132 ksiy 1.5FS

30.283 lb/in.

(a) From a free body diagram of the pin connection, the equilibrium equations

0 :xF 2 1cos cos 0T T

0 :yF 1 2sin sin 10,000 0T T


105

give

3

1 2

132 1010,0002sin 1.5

AT T

3256.818 10 in.

sinA

The optimum angle is the angle that makes the volume (and therefore the weight and cost) of the rods a minimum. The volume of each rod is

356.818 10 25 12 17.04545 34.0909

sin cos sin cos sin 2V AL

The minimum volume occurs when sin 2 1

45 .............................................................................................................................. Ans.

(b) When 45

2 3

256.818 10 in.4 sin 45dA

0.31986 in. 0.320 in.d ..........................................................................................Ans.

(c) When 45

34.09090.283 9.65 lbsin 90

W V ....................................................................Ans.

Yes, it is safe to neglect the weight of the rods.....................................................................Ans.

4-102 A tension member consists of a 50-mm diameter brass (E = 100 GPa) bar connected to a 32-mm diameter stainless steel (E = 190 GPa) bar, as shown in Fig. P4-102. For an applied load P = 50 kN, determine

(a) The normal stresses in each segment of the member. (b) The elongation of the member. SOLUTION

50 kNAB BCF F

(a) 3

6 22

50 10 25.5 10 N/m 25.5 MPa0.05 4AB ...............................................Ans.

3

6 22

50 10 62.2 10 N/m 62.2 MPa0.032 4BC .............................................Ans.

(b) 3 3

2 29 9

50 10 1500 50 10 10000.709 mm

100 10 0.05 4 190 10 0.032 4......Ans.

4-103 An alloy steel (E = 30,000 ksi) bar is loaded and supported as shown in Fig. P4-103. The loading collar at B is free to slide on section BC. The diameters of sections AB, BC, and CD are 2.50 in., 1.50 in., and 1.00 in., respectively. The lengths of all three segments are 15 in. Determine the normal stresses in each section and the overall change in length of the bar.

SOLUTION

20 kipABT

60 kipBCT

10 kipCDT


106

3

220 10 4074 psi 4.07 ksi (C)2.5 4AB ........................................................Ans.

3

260 10 33,950 psi 33.9 ksi (T)1.5 4BC .....................................................Ans.

3

210 10 12,730 psi 12.73 ksi (T)

1 4CD .....................................................Ans.

3 3 3

2 2 26 6 6

20 10 15 60 10 15 10 10 15

30 10 2.5 4 30 10 1.5 4 30 10 1 4

0.0213 in. .................................................................................................................. Ans.

4-104 The floor beams of a storage shed are supported as shown in Fig. P4-104. Each of the floor beams B and C transmits a 50 kN load to post A. Post A, the baseplate, and the footing have cross-sectional areas of 15,000 mm2, 30,000 mm2, and 260,000 mm2, respectively. Determine

(a) The normal stress in post A. (b) The bearing stress between the post and the baseplate. (c) The bearing stress between the baseplate and the footing. (d) The bearing stress between the footing and the ground. SOLUTION

(a) 3

6 26

100 10 6.67 10 N/m 6.67 MPa15,000 10n ...............................................Ans.

(b) 3

6 26

100 10 6.67 10 N/m 6.67 MPa15,000 10b ...............................................Ans.

(c) 3

6 26

100 10 3.33 10 N/m 3.33 MPa30,000 10b ...............................................Ans.

(d) 3

3 26

100 10 385 10 N/m 385 kPa260,000 10b .................................................Ans.

4-105 A 1-in.-diameter steel [ = 6.5(10 6)/ F, E = 30,000 ksi, and = 0.30] bar is subjected to a temperature decrease of 150 F. The ends of the bar are supported by two walls that displace a small amount during the temperature change. If the measured strain in the bar is 600 in./in. after the temperature change, determine the load being transmitted to the walls.

SOLUTION

6600 10TE

6 66 6.5 10 150 600 10

30 10

11,250 psi 11.25 ksi (T)

211.25 1 4 8.84 kipP A ...................................................................Ans.


107

4-106 A 90-mm diameter brass (E = 100 GPa) bar is securely fastened to a 50-mm diameter steel (E = 200 GPa) bar. The ends of the composite bar are then attached to rigid supports, as shown in Fig. P4-106. Determine the stresses in the brass and the steel after a temperature drop of 70 C occurs. The thermal coefficients of expansion for the brass and the steel are 17.6(10 6)/ C and 11.9(10 6)/ C, respectively.


b sF F


2 290 4 50 4b s 3.24s b

in which s and b are both tensile stresses. Since the supports are rigid, the total stretch of the bar must be zero,

0b s . In terms of stresses

6 69 9

800 48017.6 10 70 800 11.9 10 70 480 0

100 10 200 10b s

9 28 2.4 1.38544 10 N/mb s


6 287.82 10 N/m 87.8 MPa (T)b ....................................................................Ans.

6 2284.5 10 N/m 285 MPa (T)s .....................................................................Ans.

4-107 A steel (E = 30,000 ksi) pipe column with an outside diameter of 3 in. and an inside diameter of 2.5 in. is attached to unyielding supports at the top and bottom as shown in Fig. P4-107. A rigid collar C is used to apply a 50-kip load P. Determine

(a) The normal stresses in the top and bottom portions of the pipe. (b) The deflection of the collar C. SOLUTION

2 2 23 2.5 4 2.15984 in.A

(a) Writing the equilibrium equation

50 0B AP T


32.15984 2.15984 50 10 lbB A

in which A is a tension stress and B is a compressive stress. Since the supports are rigid, the stretch of the top section must be the same as the shrink of the bottom section, A B . In terms of stresses

6 6

7 12 4 1230 10 30 10A B 1.75B A


8418 psi 8.42 ksi (T)A .......................................................................................Ans.

14,732 psi 14.73 ksi (C)B .................................................................................Ans.

(b) 6

8418 7 120.0236 in.

30 10C A ................................................................Ans.


108

4-108 A 3-mm diameter cord (E = 7 GPa) that is covered with a 0.5-mm thick plastic sheath (E = 14 GPa) is subjected to an axial tensile load P, as shown in Fig. P4-108. The load is transferred to the cord and sheath by rigid blocks attached to the ends of the assembly. The yield strengths for the cord and sheath are 15 MPa and 56 MPa, respectively. Determine the maximum allowable load if a factor of safety of 3 with respect to failure by yielding is specified.

SOLUTION Equilibrium gives that the total force is the sum of the force carried by the cord and the force carried by the sheath

C SP P P

The cord and the sheath must both stretch the same amount, C S . In terms of stresses

9 97 10 14 10C SL L

2S C

If max 15 3 5 MPaC C , then

562 10 MPa 18.667 MPa3S C

Therefore

26 6 2 2

max 5 10 0.003 4 10 10 0.004 0.003 4P

max 90.3 NP ................................................................................................................... Ans.

4-109 The assembly shown in Fig. P4-109 consists of a steel (Es = 30,000 ksi, As = 1.25 in.2) bar A, a rigid bearing plate C that is securely fastened to bar A, and a bronze (Eb = 15,000 ksi, Ab = 3.75 in.2) bar B. A clearance of 0.025 in. exists before the assembly is loaded by a force P = 15 kip. Determine, for each segment of the assembly

(a) The normal stress. (b) The change in length. SOLUTION (a) Writing the equilibrium equation

15 0A BT P


31.25 3.75 15 10 lbA B

in which A is a tension stress and B is a compressive stress. Assuming that the force is sufficient to close the gap, the stretch of bar A must be greater than the shrink of bar B by the amount of the clearance,

0.025 in.A B In terms of stresses

6 6

6 12 2 120.025 in.

30 10 15 10A B

372 48 750 10 psiA B


310.7045 10 psi 10.70 ksi (T)A .....................................................................Ans.

431.8 psi 0.432 ksi (C)B ...................................................................................Ans.

(b) 3

6

10.7045 10 6 120.0257 in. (stretch)

30 10A ..............................................Ans.


109

6

431.8 2 120.000691 in. (shrink)

15 10B .........................................................Ans.

4-110 An 80-kN force P is applied to the 150 180-mm wood block shown in Fig. P4-110. Determine the normal stress perpendicular to the grain of the wood and the shearing stress parallel to the grain of the wood.

SOLUTION From a free-body diagram, the equilibrium equations

0 :nF 80sin 25 0N

0 :tF 80cos 25 0V

give

80sin 25 n nN A

80cos 25 n nV A

where the areas are related by

20.150 0.180 mm sin 25nA A

Therefore

3 2

3 280 10 sin 25

529 10 N/m 0.529 MPa0.150 0.180n .....................................Ans.

3

6 280 10 sin 25 cos 25

1.135 10 N/m 1.135 MPa0.150 0.180n .......................Ans.

4-111 A 4000-lb force P is applied to the square structural steel block shown in Fig. P4-111. Determine (a) The change in length of the block. (b) The maximum normal stress in the block. (c) The maximum shearing stress in the block. (d) The normal stress perpendicular to the plane A-A. (e) The shearing stress parallel to the plane A-A. SOLUTION

sin 3 5 cos 4 5

3 3 cosnA A 211.25 in.nA

(a) 36

4000 210.322 10 in.

29 10 3 3PLEA

.........................................................Ans.

(b) max4000 444 psi3 3

PA

..........................................................................................Ans.

(c) maxmax 222 psi

2 2PA

.......................................Ans.

From a free-body diagram, the equilibrium equations

0 :nF 4000cos 0N

0 :tF 4000sin 0V

give

4000 4 5 11.25nN


110

(d) 284 psin ..................................................................................................................... Ans.

4000 3 5 11.25nV

(e) 213 psin ...................................................................................................................... Ans.

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