ch04
Short Description
William F. Riley, Leroy D. Sturges, and Don H. Morris (2002), “Statics and Mechanics of Materials: An Integrated Approac...
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STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
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
P 10, 000 30, 000 psi A A 10, 000 S 2 1 di2 in.2 A 4 30, 000
V
0.75867 in. 1 2t
di
0.1207 in. .......................................................................................................................... Ans.
t 4-2
Three steel bars with 25 u 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
P A P A P A
VA VB VC
40, 000 106.7 u 106 N/m 2 106.7 MPa ........................................ Ans. 0.025 u 0.015 50, 000 133.3 u 106 N/m 2 133.3 MPa ......................................... Ans. 0.025 u 0.015 20, 000 53.3 u 106 N/m 2 53.3 MPa ............................................. Ans. 0.025 u 0.015
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
o 6Fx n 6Fy
0:
TB cos 25q TA cos 20q 0 TA sin 20q TB sin 25q 220 0
0:
are solved to get
TB
1.03684TA
TA
281.977 lb
VA
TA AA
VB
TB AB
TB
281.977 S 0.25 4 2
292.365 S 0.25 4 2
292.365 lb
5740 psi ...................................................................................... Ans. 5960 psi ...................................................................................... 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
n 6Fy
0:
PA 650 0
PA
650 kN
n 6Fy
0:
PB 650 850 0
PB
1500 kN
59
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
n 6Fy
4-5
PC 650 850 1500 0
0:
RILEY, STURGES AND MORRIS
PC
3000 kN
AA
PA VA
650 u 103 150 u106
4333 u 106 m 2 # 4330 mm 2 ....................................................... Ans.
AB
PB VB
1500 u103 150 u 106
10, 000 u 106 m 2 # 10, 000 mm 2 .............................................. Ans.
AC
PC VC
3000 u103 150 u 106
20, 000 u 106 m 2 # 20, 000 mm 2 ............................................ Ans.
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
VA
F1 AA
VB
F2 AB
550
700 psi ............................................................................................... Ans.
S 1 4 2
750
955 psi ............................................................................................... Ans.
S 1 4 2
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
o 6Fx
0:
2V 50 u103
0
gives
V
25 u 103 N W A
950 u10 0.3 L 2 3
L
0.1754 m 175.4 mm ..................................................................................................... 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
o 6Fx
0:
V 8000 0
gives
V
8000 lb W A
500 S 2 L1 500 S 1.5 L2 L1
2.55 in. ................................................................Ans.
L2
3.40 in. ................................................................Ans.
60
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
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
An
30 12 10
180 mm 2
180 u106 m 2
2 V An 2 350 u 106 180 u 106 126 u 103 N
P
For the middle plate
An
25 12 25
P V An
325 u106 m 2
325 mm 2
350 u10 325 u10 6
6
113.7 u 103 N
Therefore
Pmax (b)
113.7 kN ......................................................................................................................... Ans.
For the bearing stress on the hole in the middle plate
Ab
dt
P V Ab
12 25
300 u 106 m 2
300 mm 2
650 u10 300 u10 6
6
195 u 103 N
For the outer plates
Ab
dt 2V Ab
P
12 10
120 mm 2
120 u 106 m 2
2 650 u 106 120 u106 156 u 103 N
Therefore
Pmax (c)
156 kN ............................................................................................................................. Ans.
For the shearing stress on the cross section of the pin (which is in double shear)
As
S d 2 4 S 12 4 2
113.10 mm 2
113.10 u 106 m 2
2 W As 2 240 u 106 113.10 u 106
P
54.3 u 103 N 54.3 kN ..................................................................................................... Ans. (d)
For the punching shear stress in the outside plates
As
2 Lt
2 18 10
360 mm 2
P
360 u 106 m 2
2 W As 2 300 u 106 360 u106 216 u103 N
216 kN .......................................................................................................... Ans.
61
STATICS AND MECHANICS OF MATERIALS, 2nd Edition 4-9
RILEY, STURGES AND MORRIS
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
P W As
W S dt
40 u10 S d 0.5 3
100 2000 lb d
3.18 in. ............................................................................................. 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
TE A
VE
250 9.81 S 0.015 4 2
13.88 u 106 N/m 2
13.88 MPa ........................................... Ans.
From a free-body diagram of the lower ring, the equations of equilibrium
o 6Fx n 6Fy
0:
TC cos 60q TD
0
TC sin 60q 250 9.81 0
0:
give
TC
2 2831.90 N V C ªS 0.015 4º ¬ ¼
VC
16.03 u 106 N/m 2
TD
2 1415.95 N V D ªS 0.015 4 º ¬ ¼
VD
8.01u106 N/m 2
16.03 MPa .................................................................................... Ans.
8.01 MPa ........................................................................................ Ans.
Then, from a free-body diagram of the upper ring, the equations of equilibrium
o 6Fx n 6Fy
0: 0:
TB cos 30q TA cos 40q TC cos 60q 0 TB sin 30q TA sin 40q TC sin 60q 0
give 2 TA 1506.83 N V A ªS 0.015 4 º ¬ ¼
VA
8.53 u106 N/m 2
TB
2 2967.85 N V B ªS 0.015 4º ¬ ¼
VB
16.79 u 106 N/m 2
8.53 MPa ........................................................................................ Ans.
16.79 MPa ................................................................................... Ans.
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STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
4-11 The joint shown in Fig. P4-11 is used in a steel tension member which has a 2 u 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
An
0.5 1
0.5 in.2
P V An d 13.5 u 103 0.5 6750 lb For bearing stress
Ab
2 3 8 1 0.75 in.2
P V Ab d 18.0 u 103 0.75 13,500 lb For punching shearing stress
As
2 3 8 1 0.75 in.2
P W As d 6.5 u 103 0.75 4880 lb Therefore
4880 lb ............................................................................................................................ Ans.
Pmax
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
Ab
S 0.152 0.102 4 9.8175 u 103 m 2
P V Ab d 100 u106 9.8175 u 10 3 982 u103 N For punching shearing stress in the collar
As
S dL S 0.10 0.025 7.854 u 103 m 2
P W As
75 u10 7.854 u10 6
3
589 u 103 N
Therefore
589 kN ............................................................................................................................ Ans.
Pmax
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
W
V As
16,800 1050 psi ................................................................................................... Ans. 8u 2
63
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
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
TA
TA
TB
TB
4i 8 j 5k 42 82 52 6i 8 j 5k
0.39036TAi 0.78072TA j 0.48795TAk
0.53666TB i 0.71554TB j 0.44721TB k 6 2 82 52 8 j 5k 0.84800TC j 0.53000TC k TC 82 52
TC
The equations of equilibrium
6Fx
0:
0.39036TA 0.53666TB
0
6Fy
0:
0.78072TA 0.71554TB 0.84800TC
0
6Fz
0:
0.48795TA 0.44721TB 0.53000TC 75 9.81 0
give
TA 1.37478TB TA
452.353 N
TB
329.037 N
TC
694.103 N
2.10950TB
TC
160 u10 S d 4 160 u10 S d 4 160 u10 S d 4 6
2 A
dA
0.00190 m
6
2 B
dB
0.00162 m
6
2 C
dC
0.00235 m
Therefore
2.35 mm .......................................................................................................................... Ans.
d min
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
4 23.578q 10 a 10 cosT1 9.16515 ft
T1
sin 1
b 30 a
T2
tan 1
20.83485 ft
4 10.868q b
Then, from a free-body diagram of the ring, the equations of equilibrium
o 6Fx n 6Fy
0: 0:
P cosT 2 TA cosT1
0
P sin T 2 TA sin T1 1000 0
give
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STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
TA 1.07152 P TA 1736.24 lb
P 1620.35 lb (a)
For the normal stress in the cables
P 1620.35 lb
3
2 BC
4
0.293 in. ........................................................................................................................... Ans.
d BC
TA 1736.24 lb
24 u10 S d 3
4
2 AB
0.304 in. ........................................................................................................................... Ans.
d AB (b)
24 u10 S d
For the pin at A, which is in double shear,
TA 1736.24 lb
12 u10 ª¬2 S d 3
2 p
4 º¼
0.304 in. ............................................................................................................................. Ans.
dp
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
P An
Vn (b)
233 u 106 N/m 2
233 MPa .................................................... Ans.
250 u 106 N/m 2
250 MPa ................................................... Ans.
For bearing stress at the pin
P Ab
Vb (c)
350 u103 0.06 u 0.025 250 u103 0.04 u 0.025
For shearing stress on the pin (which is in double shear)
W
V 2 As
350 u 103 2 ªS 0.04 4 º ¬ ¼ 2
139.3 u 106 N/m2
139.3 MPa ....................................... 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
o 6Fx n 6Fy
0: 0:
TCD cos 45q TBC cos D
0
(a)
TCD sin 45q TBC sin D 8 0
(b)
From a free-body diagram for the lower ring
o 6Fx n 6Fy
0: 0:
TBC cos D TAB cos 45q 0
(c)
TAB sin 45q TBC sin D 10 0
(d)
Adding equations (a) and (c) gives
TCD cos 45q TAB cos 45q 0
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STATICS AND MECHANICS OF MATERIALS, 2nd Edition
TCD
RILEY, STURGES AND MORRIS
TAB
Then adding Eqs. (b) and (d) gives
TCD sin 45q TAB sin 45q 18 lb TCD
12.7279 lb
TAB
Now Eqs. (b) and (a) can be written
TBC sin D 1.0000
(e)
TBC cos D
(f)
9.0000
Dividing Eq. (e) by Eq. (f) gives
tan D
D
19
6.340q
9.0554 lb
TBC
P A
V max
12.7279 d 18 u 103 psi Sd2 4
d t 0.0300 in. ............................................................................................................................ 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 V as a function of the diameter do, (20 mm d do d 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
di
d o 2 0.1d o 0.8d o
V
P A
40 u 103 S d o2 di2 4
141, 471.06 N/m 2 2 do
For standard steel pipe,
V
P A
40 u103 d 250 u 106 N/m 2 A
A 160 u 10 6 m 2 160 mm 2 d 13 mm .................................................... 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. d do d 4 in.). What diameter would be required for a solid aluminum shaft? SOLUTION
V di t
P A
10, 000 S do2 di2 4
30, 000 psi
d o2 0.42441 d o 2t d o d o2 0.42441 2
For a solid aluminum shaft
66
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
10, 000 Sd2 4
RILEY, STURGES AND MORRIS
V
P A
d
0.65 in. ................................................................................................................................. Ans.
30, 000 psi
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 Vmax above the column and decreasing linearly to Vmax/5 at the outside edge rp of the bearing plate), calculate and plot Vmax versus the radius rp of the bearing plate (75 mm d rp d 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 Vmax for a 400 mmdiameter 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
V
150 u103 d 3.25 u106 N/m 2 Sd2 4
P A
d t 0.242 m
242 mm ........................................................................................................... Ans.
For the non-rigid bearing plate
V V max
c1r c2
where the constants c1 and c2 are chosen such that
1 c1 0.075 c2 0.2 c1rp c2 Therefore,
c1
0.8 0.075 rp
c2
1
0.8 0.075 0.075 rp
Integrating the stress to get the axial force
F
³ V dA
gives
150 u103
³
0.075
0
V max 2S r dr ³
rp
0.075
V max c1r c2 2S r dr
ª 0.0752 § rp3 0.0753 · § rp2 0.0752 · º 2SV max « c1 ¨ ¸ c2 ¨¨ ¸» ¨3 3 ¸¹ 2 ¸¹ »¼ «¬ 2 © © 2
67
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
150 u 103
V max
ª 0.075 § r 0.075 2S « c1 ¨¨ 3 «¬ 2 ©3 2
3 p
3
· §r 0.075 ¸¸ c2 ¨¨ 2 ¹ ©2 2 p
2
·º ¸¸ » ¹ »¼
N/m 2
When
rp
0.075 m
V max
8.49 u 106 N/m 2
When
rp
0.2 m
V max
1.976 MPa (a 76.7% decrease) ................................ Ans.
When
rp
0.3 m
V max
0.965 MPa (an 88.6% decrease) .............................. Ans.
8.49 MPa ...................................... 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 Vmax under the 2.4-in.-diameter restraining nut and decreasing as 1/r to the outside edge rw of the washer), calculate and plot Vmax versus the radius rw of the washer (1 in. d rw d 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 Vmax 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
V
18 u 103 d 400 psi S d 2 1.52 4
P A
d t 7.72 in. .................................................. Ans. For the non-rigid washer
V
1.2V max r
Integrating the stress to get the axial force
F
³ V dA
gives
18 u 103
1.2V max 2S r dr r SV max ª¬ 1.22 0.752 2.4 rw 1.2 º¼ SV max ª¬0.87750 2.4 rw 1.2 º¼
³
1.2
0.75
V max 2S r dr ³
rw
1.2
18 u 103 psi S ª¬0.87750 2.4 rw 1.2 º¼
V max When
V max
When
d
2rw
2.4 in. V max
When
d
2rw
8 in.
When
d
2rw 12 in.
400 psi
d
2rw 13.61 in. ...................................................................... Ans. 6529 psi
V max
754 psi
an 88.5% decrease.................................. Ans.
V max
462 psi
a 38.7% decrease.................................... Ans.
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STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
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
6Fn
0:
N 400 cos 37q 0
6Ft
0:
V 400sin 37 q 0
give
319.45 kN
N
240.73 kN
V
The areas are related by
A 0.075 u 0.045
An cos 37q
An
4.226 u103 m 2
Therefore
V
N An
319.45 u 103 4.226 u10 3
75.6 u106 N/m 2
75.6 MPa ................................................... Ans.
W
V An
240.73 u 103 4.226 u 103
57.0 u 106 N/m 2
57.0 MPa .................................................... Ans.
4-23 An axial load P is applied to a timber block with a 4 u 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
6Fn
0:
N 5000sin14 q 0
6Ft
0:
V 5000 cos14 q 0
give
N 1209.61 lb V 4851.48 lb The areas are related by
A 4u 4
An sin14q
66.137 in.2
An Therefore
V
N An
1209.61 18.29 psi ............................................................................................... Ans. 66.137
W
V An
4851.48 66.137
73.4 psi ................................................................................................. Ans.
4-24 A structural steel bar with a 20 u 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
V max W
P 2A
P A
55 u 103 110.0 u106 N/m 2 0.02 u 0.025
V max 2
110.0 MPa ........................................... Ans.
55.0 MPa .................................................................................................... Ans.
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STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
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
50 u 103 d 25 u 103 psi 2 Sd 4
V max
P A
W max
P 2A
d min
1.596 in. .......................................................................................................................... Ans.
d t 1.596 in.
50 u103 d 15 u 103 psi 2 2 S d 4
d t 1.457 in.
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
6Fn
0:
P sin 60 q N
0
6Ft
0:
P cos 60q V
0
The areas are related by
A 0.2 u 0.12 Therefore
P
N sin 60q
An sin 60q
An
0.027713 m 2
13.60 u 106 0.027713 V An d sin 60q sin 60q
P d 435 u 103 N
5.25 u 106 0.027713 V W An P P d 291u103 N d cos 60q cos 60q cos 60q Pmax 291 kN ............................................................................................................................. Ans. 4-27 A steel bar with a 4 u 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 T and the applied load P. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are
6Fn
0:
N P cos T
0
6Ft
0:
V P sin T
0
The areas are related by
A 4 u1
An cos T
Therefore
N
P cos T
V
P sin T
V An W An
An
12 u10 4 cosT 9 u10 4 cosT 3
3
Therefore
70
4 cos T in.2
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
T
9 P sin T P cos T 12 36.870q ................................................................................................................................. Ans.
P
75 u 103 lb
tan T
75 kip ........................................................................................................... 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. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are
6Fn
0:
N 400sin 57 q 0
6Ft
0:
V 400 cos 57 q 0
The areas are related by
A 0.1t Therefore
V
N An
W
V An
An sin 57q
An
400 u10 sin 57q d 70 u10
0.11924t m 2
3
6
N/m 2
t t 0.0402 m
6
N/m 2
t t 0.0406 m
0.11924t
400 u10 cos 57q d 45 u10 3
0.11924t
40.6 mm ........................................................................................................................... Ans.
tmin
4-29 The shearing stress on plane AB of the 4 u 8-in. rectangular block shown in Fig. P4-29 is 2 ksi when the axial load P is applied. If the angle I is 35q, 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
6Fn
0:
P sin 35q N
0
6Ft
0:
P cos 35 q V
0
The areas are related by
A 4u8 An (a)
(c)
55.790 in.2
Solving the second equation for P gives
P (b)
An sin 35q
V cos 35q
W An cos 35q
2 u10 55.790 3
136, 215 lb # 136.2 kip ........................ Ans.
cos 35q
Solving the first equation for N gives
N
P sin 35q 78,129.6 lb V An
V
1400 psi ............................................................................................................................... Ans.
Then the maximum normal and shearing stresses are given by
71
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
P 136, 215 4260 psi ............................................................................................ Ans. 4u8 A P V max 2130 psi ................................................................................................. Ans. 2A 2
V max W max
4-30 A wood tension member with a 50 u 100-mm rectangular cross section will be fabricated with an inclined glued joint (45q d I d 90q) 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 I 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
6Fn
0:
N P sin I
0
6Ft
0:
V P cos I
0
The areas are related by
A 0.05 u 0.1 An sin I 0.005 sin I m 2
An Therefore
N
P sin I
V An
V
P cos I
W An
5 u10 0.005 sin I 3 u10 0.005 sin I 6
6
and dividing the first equation by the second gives
P sin I P cos I
tan I
I
5 1.66667 3
59.04q ................................................................................................................................... Ans. 34.0 u 103 N 34.0 kN ............................................................................................... Ans.
Pmax
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. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are
6Fn
0:
2 N P cos 30 q 0
6Ft
0:
2V P sin 30q 0
The cross sectional areas of the bolts are
Ab
S 0.5 4 0.19635 in.2 2
The first equation gives
P
2N cos 30q
3 2 V Ab 2 12 u 10 0.19635 d cos 30q cos 30q
while the second equation gives
72
5441 lb
STATICS AND MECHANICS OF MATERIALS, 2nd Edition 3 2 W Ab 2 8 u10 0.19635 d sin 30q sin 30q
2V sin 30q
P
RILEY, STURGES AND MORRIS
6283 lb
Therefore
5441 lb # 5440 lb ......................................................................................................... Ans.
Pmax
4-32 The bar shown in Fig. P4-32 has a 200 u 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
6Fn
0:
N 500 cos 30q 0
6Ft
0:
V 500sin 30q
0
The areas are related by
A 0.2 u 0.1
An cos 30q
23.094 u10 3 m 2
An Therefore
V
500 u10 cos 30q 500 u10 sin 30q
V
18.75 u 106 N/m 2
18.75 MPa ...................................................................................... Ans.
W
10.83 u 106 N/m2
10.83 MPa ....................................................................................... Ans.
433.01 kN V An
3
N
250.00 kN W An
3
P 500 u103 25.0 u106 N/m 2 25.0 MPa ..................................................... Ans. A 0.2 u 0.1 P V max 12.50 MPa ............................................................................................. Ans. 2A 2
V max W max
4-33 A steel eyebar with a 4 u 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 (45q d I d 90q), 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 I 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
6Fn
0:
N P sin I
0
6Ft
0:
V P cos I
0
The areas are related by
A 4 u1
An sin I
Therefore
N
P sin I
V An d 12 u103 4 sin I
V
P cos I
W An d 9 u103 4 sin I
An
73
4 sin I in.2
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
and dividing the first equation by the second gives
P sin I P cos I
tan I
12 1.33333 9
I
53.13q ................................................. Ans.
75, 000 lb ........................................................................................................................ Ans.
Pmax
4-34 A steel eyebar with a 100 u 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 Vn and the shear stress Wn in the weld material for weld angles I (30q d I d 90q). If the stresses in the weld material must be limited to 80 MPa in tension and 60 MPa in shear, what ranges of I would be acceptable for the joint? Repeat for P = 305 kN and for P = 350 kN. Are weld angles I < 30q reasonable? Why or why not? SOLUTION From a free-body diagram of the bar, the equations of equilibrium are
6Fn
0:
N P sin I
0
6Ft
0:
V P cos I
0
The areas are related by
A 0.1u 0.025 An
An sin I
0.0025 sin I m 2
Therefore
V
N An
P sin I 0.0025 sin I
P sin 2 I N/m 2 ............................................................................ Ans. 0.0025
W
V An
P cos I 0.0025 sin I
P sin I cos I N/m 2 ..................................................................... Ans. 0.0025
When
P
250 kN
0q I 64q .................................................................................. Ans.
When
P
305 kN
0q I 40q or 50q I 54q .................................................. Ans.
When
P
350 kN
0q I 30q .................................................................................. Ans.
Angles less than about I 30q are not very practical since the weld gets long quickly as the angle decreases................................................................................................................................. Ans. 4-35 Specifications for the rectangular (3 u 3 u 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 T = 37q with the horizontal, calculate and plot the ratios V/Vmax and W/Wmax as a function of the load P (0 d P d 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 T = 25q. For what angle T will the normal stress and the shear stress both reach their limiting values at the same time? SOLUTION From a free-body diagram of the bar, the equations of equilibrium are
6Fn
0:
N P cos T
0
74
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
6Ft
V P sin T
0:
RILEY, STURGES AND MORRIS
0
The areas are related by
A 3u 3
An cos T
9 cos T in.2
An Therefore
V V max W
P sin T cosT 9 psi ......... Ans. 500
V An W max
W max When T
P cos 2 T 9 psi .................................................. Ans. 800
N An V max
37q , shear stress controls and
Pmax # 9360 lb ............................................. Ans. When T
25q , normal stress controls and
Pmax # 8770 lb ............................................. Ans. The normal stress and the shear stress will reach their maximum values at the same time if
V N
tan T opt
T opt
W max An V max An
500 800
0.62500
32.01q ................................................................................................................................ Ans.
4-36 Compression tests of concrete indicate that concrete fails when the axial compressive strain is 1200 Pm/m. Determine the maximum change in length that a 200-mm diameter u 400-mm long concrete test specimen can tolerate before failure occurs. SOLUTION
G
HL
1200 u10 400 6
0.480 mm .......................................................................... Ans.
4-37 The 0.5 u 2 u 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
J
G L
0.01 0.02 in./in. 0.02 rad ................................................................................. Ans. 0.5
4-38 A structural steel bar was loaded in tension to fracture. A 200 mm-length of the bar was marked off in 25mm 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
H H max
G L
L f Li Li 78.5 50 50
263.5 200 200
0.317 mm/mm ............................................................. Ans.
0.570 mm/mm ........................................................................................ Ans.
75
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
4-39 Mutually perpendicular axes in an unstressed member were found to be oriented at 89.92q when the member was stressed. Determine the shearing strain associated with these axes in the stressed member. SOLUTION
S T 2
J
90q 89.92q
S 1.396 u 103 rad .......................................................... Ans. 180q
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
LPcQ
500 mm
LMQ
LMPc
LMP
LMPc 10 363.5534 mm
I
tan 1
J
S T 2
500 cos 45q 353.5534 mm
LMQ LMP
44.20107q
S 2I 2
90q 88.40213q J
S 180q
27.9 u 103 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)
HD
(b)
HC
2.15 2.00 75.0 u 103 in./in. ................................................................................... Ans. 2.00 S 2.15 S 2.00 75.0 u 103 in./in. ..................................................................... Ans. S 2.00
4-42 A thin rectangular plate is uniformly deformed as shown by PRSQ in Fig. P4-42. Determine the shearing stain Jxy at P. SOLUTION
I1 I2 J J
0.380 0.04354q 500 0.200 tan 1 0.04584q 250 S T I1 I2 2 0.08938q tan 1
1.560 u103 rad ................................................................................................................... Ans.
76
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
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 1.5q with respect to the sleeve. Assume that radial lines in the unstressed state remain straight as the rubber deforms. Determine the shearing strain JrT in the rubber insert (a) At the inside surface. (b) At the outside surface. SOLUTION
b (b)
I1
§ 3.3 · § 1.5S · ¨ ¸¨ ¸ 0.0431969 in. © 2 ¹ © 180 ¹ b tan 1 4.93774q 0.5
rT
86.2 u 103 rad J outside .................................................. Ans. (a)
I2
I1 1.5q 6.43774q 112.4 u 10 3 rad J 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 (H = kx2). If the strain is 1250 Pm/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
H
kx 2
H
dG dx
1250 u10 6
k 1.5
2
k
555.556 u106
kx 2
³ dG ³ H dx
3
3
k ³ x dx
ª kx3 º « 3 » ¬ ¼0
(a)
G
(b)
H avg
0.005 1.667 u103 m/m 1667 P m/m ................................................................. Ans. 3
(c)
H max
H3
k 3
2
0
9k
2
9k
5.00 u 103 m 5.00 mm ................... Ans.
5.00 u10 3 m/m 5000 P m/m ................................................ 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 Jy/3E, where J 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 J, 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
H (a)
G
Jy 3E
³ dG
dG dy
J L y dy 3E ³0
L
ª J y2 º « 3E 2 » ¬ ¼0
J L2 .......................................................................... Ans. 6E
77
STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b)
H avg
G L
(c)
H max
HL
RILEY, STURGES AND MORRIS
JL in./in. ................................................................................................................ Ans. 6E JL in./in. ............................................................................................................... Ans. 3E
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 Pm/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 Pm/m. Determine (a) The total elongation of the cable. (b) The maximum height that the balloon could achieve SOLUTION
(a) (b)
H
500 u 106 cy
H
dG dy
G
³ dG
500 u 106 c 150
350 u 10 6
c 1u 106
500 u10 6 1u106 y 10
6
³
L
0
L
ª y2 º 500 y dy 10 «500 y » 2 ¼0 ¬ 6
500 L 0.5L u10 2
6
300 m G 0.1050 m 105.0 mm ..................................................................................................... Ans. If y ! 500 m , then H 0 . If L ! 1000 m , then G 0 . Neither is possible. Therefore If L
ymax
500 m .............................................................................................................................. 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 Pin./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 Pin./in., determine (a) The strain in the cable at a depth of 500 ft. (b) The total elongation of the cable. SOLUTION
H (a)
H 500
250 u106 cy
700 u 106
250 u 106 c 2000
c
0.22500 u 106
250 u10 6 0.22500 u 106 1500 587 u 106 in./in. 587 P in./in. .................................................................................. Ans.
(b)
H
dG dy
G
³ dG
250 u 106 0.22500 u 106 y 10
6
³
2000
0
2000
ª 0.225 y 2 º 250 0.225 y dy 10 «250 y 2 »¼ 0 ¬ 6
ª 250 2000 0.1125 2000 2 º u 106 ¬ ¼ 0.9500 ft 11.40 in. ......................................................................................................... Ans.
78
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
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
V
EH
62.6 u103
E
S 0.015 4 2
0.90 200
78.7 u 109 N/m 2 78.7 GPa ...................................................................................... Ans. H 0.022 15 0.326 ......................................................................................... Ans. t 0.9 200 Ha
E
Q
P A
Vp
62.6 u103
S 0.015 4 2
354 u 106 N/m 2
354 MPa ............................................... 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
V
EH
53 u 103
E
S 1.5 4 2
0.48 20 u12
E 15.00 u 106 psi 15, 000 ksi ........................................................................................ Ans.
Ht Ha
Q
G
E 2 1 Q
0.001 1.5 0.48 20 u 12 15, 000 2 1 0.333
0.333 .................................................................................. Ans. 5625 ksi .......................................................................... 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 Q = 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:
140 70.0 u103 MPa 70.0 GPa ........................................................................... Ans. 0.002 # 150 MPa ........................................................................................................................... Ans.
(a)
E#
(b)
V pl
(c)
V ult # 450 MPa .......................................................................................................................... Ans. 79
STATICS AND MECHANICS OF MATERIALS, 2nd Edition (d)
V .05 # 220 MPa ............................. Ans.
(e)
V .2 # 275 MPa .............................. Ans.
(f)
V f # 450 MPa .............................. Ans.
(g)
Ht
QH a
0.3 0.115
RILEY, STURGES AND MORRIS
'd d
'd # 0.19458 mm d f 5.64 0.19458 5.4454 mm
Pf
V tf (h) (i)
Af
V f Ai 450 ª¬S 5.64 4º¼ # 2 Af S 5.4454 4 2
483 MPa ...................................................... Ans.
430 350 1000 MPa 1.000 GPa .......................................................................... Ans. 0.08 0 400 8890 MPa 8.89 GPa ................................................................................... Ans. Es # 0.045 Et #
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:
28 u 103 28.0 u 106 psi 28, 000 ksi ........................................................................ Ans. 0.001 # 34 ksi ................................... Ans.
(a)
E#
(b)
V pl
(c)
V ult # 74 ksi ................................... Ans.
(d)
V .05 # 43 ksi ................................... Ans.
(e)
V .2 # 43 ksi .................................... Ans.
(f)
V f # 65 ksi .................................... Ans.
(g)
(h)
V tf
Pf Af
2 V f Ai 65 ª¬S 0.25 4º¼ # 2 Af S 0.212 4
# 90.4 ksi ............................................................................................................................. Ans. 78 35 537 ksi ............................................................................................................ Ans. Et # 0.08 0 80
STATICS AND MECHANICS OF MATERIALS, 2nd Edition (i)
Es #
RILEY, STURGES AND MORRIS
56 1400 ksi ................................................................................................................ Ans. 0.04
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:
(b)
222.1 185 u 103 MPa 0.0012 E # 185 GPa ...................Ans. V pl # 280 MPa ...............Ans.
(c)
V ult # 507 MPa ...............Ans.
(d)
V .05 # 300 MPa ...............Ans.
(e)
V .2 # 330 MPa ................Ans.
(f)
V f # 450 MPa ................Ans.
(g)
V tf
(h)
Et #
(a)
(i)
E#
Pf Af
#
45.1u 103
S 0.0095 4 2
636 u106 N/m 2
636 MPa .............................................. Ans.
320 306 17.5 u 103 MPa 17.5 GPa ........................................................ Ans. 0.0032 0.0024 315 105 u 103 MPa 105 GPa ............................................................................. Ans. Es # 0.003
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)
E#
32 u 103 0.0012
26.7 u 106 psi
26, 700 ksi ........................................................................ Ans.
81
STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b)
V pl # 40 ksi .....................Ans.
(c)
V ult # 73 ksi .....................Ans.
(d)
V .05 # 44 ksi .....................Ans.
(e)
V .2 # 47 ksi ......................Ans.
(f)
V f # 65 ksi ......................Ans.
(g)
V tf
Pf Af
#
RILEY, STURGES AND MORRIS
13 u 103
S 0.425 4 2
# 91.6 u 103 psi 91.6 ksi ..........................................................................Ans. (h) (i)
47.43 45.93 1875 ksi ........................................................................Ans. 0.004 0.0032 46 14, 370 ksi ..................................................................................Ans. Es # 0.0032 Et #
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 D = 12.1(106)/qC. Determine the dimension changes caused by (a) An increase in temperature of 70qC. (b) A decrease in temperature of 85qC. SOLUTION (a)
GL Go Gi
(b)
GL Go Gi
12.1u10 12.1u10 12.1u10 12.1u10 12.1u10 12.1u10
70 2500 2.12 mm .......................................................................... Ans. 70 105 0.0889 mm ......................................................................... Ans. 70 70 0.0593 mm ............................................................................ Ans. 85 2500 2.57 mm ..................................................................... Ans. 85 105 0.108 mm ...................................................................... Ans. 85 70 0.072 mm ......................................................................... Ans.
6
6
6
6
6
6
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 [D = 6.5(106)/qF] caused by an increase in temperature of 250qF. SOLUTION
GL Gd
6.5 u10 250 225 u12 4.39 in. ....................................................................... Ans. 6.5 u10 250 12 u12 0.234 in. ........................................................................ Ans. 6 6
4-56 An airplane has a wing span of 40 m. Determine the change in length of the aluminum alloy [D = 22.5(106)/qC] wing spar if the plane leaves the ground at a temperature of 40qC and climbs to an altitude where the temperature is 40qC. SOLUTION
G
22.5 u10 80 40 6
72.0 u 103 m
82
72.0 mm ........................................... Ans.
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
4-57 Determine the movement of the pointer of Fig. P4-57 with respect to the scale zero when the temperature increases 80qF. The coefficients of thermal expansion are 6.6(106)/qF for the steel and 12.5(106)/qF for the aluminum. SOLUTION The steel post and the steel scale each stretch the same amount
Gs
Go
6.6 u10 80 20 6
0.01056 in.
The aluminum post stretches
Ga
12.5 u10 80 20 6
0.02000 in.
Therefore, the motion of the pointer relative to the zero on the steel scale is
Gp 5
Ga Gs 1
Gp
0.0472 in. n ..................................................................................................................... Ans.
4-58 A bronze [DB = 16.9(106)/qC] sleeve with an inside diameter of 99.8 mm is to be placed over a solid steel [DS = 11.9(106)/qC] 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
d
100 11.9 u10 6 'T 100
99.8 16.9 u 106 'T 99.8 'T
403qC ................................................................................................................................ Ans.
4-59 A steel [E = 30,000 ksi and D = 6.5(106)/qF] surveyor’s tape ½-in. wide u 1/32-in. thick is exactly 100 ft long at 72qF 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 100qF and under a pull of 25 lb. SOLUTION
H
V D'T E
G L
Lc L L
Lc 1 L
where L is the length at 72qF and zero force. When P
10 lb , V
10 1 2 1 32
640 psi , Lc 100 ft , and
100 640 1 0 30 u 106 L L 99.99787 ft Then, when P
25 lb , V
25 1600 psi , and 'T 1 2 1 32
28qF
1600 Lc 1 6.5 u106 28 6 30 u 10 L Lc 100.02140 ft 'L 0.02140 ft 0.257 in. correction 0.257 in. ............................................................................................................ Ans.
83
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
4-60 A 25-mm diameter aluminum [D = 22.5(106)/qC, E = 73 GPa, and Q = 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 50qC. 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)
V
2500 9.81 2 S 0.025 4
(b)
Ha
V D'T E
(c)
Ht
Q
Gd
Ht d
49.962 u106 N/m 2 # 50.0 MPa ...................................................... Ans.
49.962 u 106 22.5 u 106 50 441u106 m/m ...................... Ans. 73 u109 § 49.962 u106 · 6 6 0.33 ¨ ¸ 22.5 u 10 50 1350.86 u 10 m/m 9 © 73 u 10 ¹
V D'T E
1350.86 u10 25 6
0.0338 mm ........................................................... 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)
(b)
V max G AB
Pmax A
82 u 103 2u 2
20, 500 psi
20.5 ksi ................................................................. Ans.
82 u10 8 u12 0.0656 in. ........... Ans. 30 u10 2 u 2 12 u10 5 u12 0.00600 in. .............. Ans. 30 u10 2 u 2 45 u10 4 u12 0.01800 in. 30 u10 2 u 2 3
PL EA
6
3
(c)
G BC
6
3
G CD (d)
G total
6
G AB G BC G CD
0.0776 in. ...................................................................................... Ans.
4-62 A structural tension member of aluminum alloy (E = 70 GPa) has a rectangular cross section 25 u 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
G P Pmax
PL EA
P 2000
d 4 mm
70 u10 0.025 u 0.075 V A d 100 u 10 0.025 0.075 9
6
P d 262,500 N
187,500 N
187.5 kN ......................................................................................................................... Ans.
84
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
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
PA
120 kip
PB
120 u10 3 u12 30 u10 ª¬S 6 4.5 3
120 u10 4 u12 4 º¼ 10.6 u 10 ª¬S 4 4 º¼ 3
G
GA GB
G
0.01164 0.04324 0.0549 in. ..................................................................................... Ans.
6
2
2
6
2
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
PS
PM
P P 1000
G
200 u10 ª¬S 0.030
P
212 u 103 N
9
2
4 º¼
P 2000
180 u10 ª¬S 0.060 9
2
0.0402 4 º¼
3.00 mm
212 kN ...................................................................................................... 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 u 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
E 1800 ksi (a)
G
PL EA
(b)
V
P A
(c)
Vb
(d)
W max
200 u10 30 1800 u10 12 u12 3
3
200 u 103 12 u12
200 u 103 12 u12 P 2A
0.0231 in. ..................................................................... Ans.
1389 psi ................................................................................................ Ans.
1389 psi ....................................................................................................... Ans.
200 u103 2 12 u12
694 psi .......................................................................................... 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
E
200 GPa
P1 1030 kN (C)
85
P2
380 kN (C)
STATICS AND MECHANICS OF MATERIALS, 2nd Edition (a)
1030 u10 3500 3.1623 # 3.16 mm ............................................. Ans. 200 u10 5700 u10 380 u10 3500 3.1623 1.1667 # 4.33 mm ............................... Ans. 200 u10 5700 u10 3
PL EA
G1
RILEY, STURGES AND MORRIS
6
9
3
(b)
Gr
G1
6
9
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
(a)
Est
29, 000 ksi
Eal
10, 600 ksi
PA
205 kip (T)
PB
120 kip (T)
GA
PL EA
205 u10 50 0.0286 in. ...................................................... Ans. 29 u10 ª¬S 6 4.5 4º¼ 120 u10 40 0.0360 in. 10.6 u10 ª¬S 4 4º¼ 3
6
2
2
3
(b)
(c)
GB
6
G total
GA GB
V max
P A
W max
P 2A
2
0.0646 in. .................................................................................................... Ans.
120 u103
9550 psi ........................................................................................... Ans.
S 4 4 2
V max 2
4770 psi ................................................................................................. 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
G total
GA GB
35 u10 300 35 u10 900 73 u10 A 200 u10 ª¬S 0.025 3
0.40 mm
3
9
A 1817.40 u 10 6 m 2
9
2
4º ¼
1817.40 mm 2
S 752 di2 4 di t
57.54 mm
75 2t
8.73 mm ................................................................................................................................ Ans.
86
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
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
E
29, 000 ksi
b
2
G
30 u10 25 30 u10 29 u10 2 u 0.5 29 u10 ³
3y in. 60
0 y 60 in. 3
3
6
6
60
0
dy 3y · § ¨ 2 60 ¸ 0.5 © ¹
0.02586 1.03448 u10 3 ª¬ 40 ln 40 y º¼ 0
60
0.02586 41.37931u10 3 > 4.60517 3.68888@
G
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 J 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
n 6F
V A W
0:
0
V S R 2 W V R
ry L
G
P dy ³ EA
J Vol J
S R2 y 3
J y dy 3E
ª J y2 º « 3E 2 » ¬ ¼0
Jy 3
V ³ E dy
³
L
0
L
J L2 ........................................................ Ans. 6E
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
J
E 15, 000 ksi R
y 60 · ¸ © 60 ¹
4 ¨§
J Vol
W n 6F
0:
0.320 lb in 3
y 60 in. 15
ª S R 2 y 60 S 42 60 º J« » 3 3 «¬ »¼
V A W 5000 0 JS V S R 2 5000 ª¬ R 2 y 60 960º¼ 3 87
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
G
³
P dy EA
³
V dy E
5000 225
15 u10
6
³
§ JS 5000 ¨ 60 © 3
0
dy
60
S ³ y 60 0
960 225J 3 15 u 10
6
60
2
RILEY, STURGES AND MORRIS
·ª 2 ¸ ¬ R y 60 960 º¼ ¹ dy ES R 2
60 J y 60 dy 6 ³0 3 15 u 10
dy
³ y 60 0
2 60
60 2 ª º ª 1 º 6 y 60 0.02387 « 0.00711 10 u « » » 2 ¬ y 60 ¼ 0 «¬ »¼ 0 60
ª 1 º 1536.00 u 10 « » ¬ y 60 ¼ 0 6
0.19892 u 103 0.03839 u 103 0.01280 u 103
G
0.225 u 103 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
n 6F
0:
TB PA 500 0
S S 2 0.05 V A 0.12 0.052 500 u103 N 4 4 where V B is a tension stress and V A is a compressive stress. Then, VB
expressing that the stretch of the rod is equal to the shrink of the tube G B G A in terms of the stresses gives
V B 2000 V A 1500 200 u 109 100 u 109 V B 1.5V A Substituting back into the equilibrium equation gives
VA
56.588 N/m 2 # 56.6 MPa .............................................................................. Ans.
V B 1.5V A
84.9 MPa ............................................................................................ Ans.
56.588 u10 1500 6
(b)
GC
GA
100 u 109
0.849 mm p ................................................ Ans.
88
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
4-73 The 7.5 u 7.5 u 20-in. oak (E = 1800 ksi) block shown in Fig. P4-73 was reinforced by bolting two 2 u 7.5 u 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
n 6F
Pw 2 Ps P
0:
0
V w 7.5 u 7.5 2 ª¬V s 2 u 7.5 º¼
P
V w and V s are both compressive stresses. Expressing that the shrink of the steel is equal to the shrink of the wood G s G w
where
in terms of the stresses gives
VsL 29, 000 u 103 Vs If
Vs
V wL 1800 u103
16.111V w
22 ksi , then V w 1.36552 ksi 4.6 ksi , and P 1.36552 7.5 u 7.5 2 ª¬ 22 2 u 7.5 º¼
22 u10 20
737 kip ....................................... Ans.
3
G
(b)
Gs
29, 000 u103
0.01517 in. p ............................................................. 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
n 6F
0:
5Ps Pc 900 0
2 5V s ªS 0.025 4 º V c Ac ¬ ¼
900 u 103 N
V s and V c are both compressive stresses. Expressing that the shrink of the steel rods is equal to the shrink of the concrete G s G c where
in terms of the stresses gives
VsL VcL 9 200 u 10 31u 109 V s 6.4516V c If
Vc
9.5 MPa , then V s Ac
b
61.29 MPa 130 MPa , and
78,902 u 106 m 2
78,902 mm 2
2 b 2 5 ªS 25 4 º ¬ ¼ 285 mm ................................................................................................................................ Ans.
89
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
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
PB where
PA 40 kip
PC
PA 10 kip
PA , PB , and PC are all tension forces. Since the total
length of the assembly cannot change, the total stretch must be zero G A G B G C 0 . Therefore
PA 20
30 u10 ª¬S 2.5 3
PA
VB VC
2
22 4 º¼
VA
0 kip
0 40
S 22 4
0 10
S 12 4
PA 40 24
10 u10 ª¬S 2 3
2
4 º¼
PA 10 24
10 u10 ª¬S 1 3
2
4º¼
0
0 ksi ....................................................................... Ans.
12.73 ksi (T) .................................................................................... Ans. 12.73 ksi 12.73 ksi (C) ............................................................ Ans.
10 u10 24 10 u10 ª¬S 1 4º¼ 3
GC
(b)
2
6
0.03056 in.
Therefore section a-a moves 0.03056 in. o ............................................................................ Ans. 4-76 A 150-mm diameter u 200-mm long polymer (E = 2.10 GPa) cylinder will be attached to a 45-mm diameter u 400-mm long brass (E = 100 GPa) rod by using the flange type of connection shown in Fig. P4-76. A 0.15mm 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
Tp where
Tb
Tp and Tb are both tension forces. When the bolts
are tightened, the stretch of the two pieces closes the gap, G p G b 0.15 mm . Therefore
Tp 200
2.1u10 ª¬S 0.15 9
Tp
Tb
2
4º ¼
Tb 400 100 u109 ª¬S 0.045 2
18,976.74 N
90
4º ¼
0.15 mm
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
Vp Vb Gp
(b)
RILEY, STURGES AND MORRIS
18,976.74 1.074 u106 N/m 2 1.074 MPa (T) ...................................... Ans. 2 S 0.15 4 18,976.74 11.93 u106 N/m 2 11.93 MPa (T) .................................... Ans. 2 S 0.045 4 18,976.74 200 0.10227 mm 2 9 ª º 2.1 10 0.15 4 S u ¬ ¼
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,
n 6F
0:
TA PB 95 0
or in terms of stresses
1.25V A 3.75V B
95 u 103 lb
in which V A is a tension stress and V 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, G A G B 0.015 in. Therefore
V A 6 u12 30 u106 VA
V B 2 u12 0.015 in. 15 u106
2 3 V B 6250 psi
Combining the equilibrium equation and the deformation equation gives
V A 18,932 psi # 18.93 ksi (T) ................................................................................. Ans. V B 19, 023 psi # 19.02 ksi (C) ................................................................................. Ans. (b)
GC
GA
18,932 6 u12 30 u106
0.0454 in. p ....................................................................... 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 u 500 mm square cross section, will be made of concrete (E = 20 GPa) and will be reinforced with 50mm 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
91
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
n 6F
RILEY, STURGES AND MORRIS
Pc nPs 4450 0
0: Pc nPs
4450 kN
Expressing that the steel rods and the concrete must shrink the same amount, G c G s , in terms of stresses gives
If
Vc
VcL V sL 9 20 u 10 200 u 109 V max 8 MPa , then V s
8 u10 ª¬0.5 6
2
Vs
10V c
80 MPa 120 MPa , and
2 2 n S 0.05 4 º n 80 u10 6 ªS 0.05 4 º ¼ ¬ ¼
4450 u103
n 17.33 18 .......................................................................... Ans.
Therefore the required number of rods is n (b)
If n
18 , then 2 2 V c ª0.52 18S 0.05 4 º 18 10V c ªS 0.05 4 º ¬ ¼ ¬ ¼
(c)
Vc
7.83 u106 N/m 2
Vs
10V c
G
Gs
4450 u103
7.83 MPa ............................................................................. Ans.
78.3 MPa .................................................................................................. Ans.
V s 3000 200 u109
78.3 u10 3000 6
200 u109
1.175 mm ........................................ 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.). SOLUTION From equilibrium
Tst
Fbr
or in terms of stresses
ªS 0.5 2 4 º V st ¬ ¼
V st
0.375V br
1.90986V br
V 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 G st 0.02 G br . Therefore in which V st is a tension stress and
V st 6 30 u 106
0.02
V br 6 15 u 106
V st
100, 000 2V br
Combining the equilibrium equation and the deformation equation gives
V br
25,576 psi # 25.6 ksi (C) .................................................................................. Ans.
V st
48,847 psi # 48.8 ksi (T) .................................................................................. Ans.
92
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
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
Fa
2Ts
in terms of stresses gives
625 u10 V
2 120 u 106 V s
6
Vs
a
2.60417V a
V a is a compression stress. The stretch of the steel bolts and the shrink of the aluminum bar are related by G s 1 G a Therefore
in which V s is a tension stress and
V s 330 190 u 109
1
V a 251 73 u 109
Vs
575.76 u 106 1.97966V a
Combining the equilibrium equation and the deformation equation gives
Vs
327 u 106 N/m 2
327 MPa ................................................................................ Ans.
V a 125.6 u106 N/m 2 125.6 MPa ......................................................................... Ans.
125.6 u10 251 6
(b)
Ga
73 u 109
0.432 mm ....................................................................... 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 Vs (in the steel bolt) and Vb (in the brass sleeve) as functions of the angle of twist T of the nut (0q d T d 180q). (b) The elongations Gs (of the steel bolt) and Gb (of the brass sleeve) as function of T (0q d T d 180q). (c) The distance L between the two washers as a function of T (0q d T d 180q). SOLUTION (a) From equilibrium
Tst
Fbr
or in terms of stresses
0.785V st
V st
1.767V br
2.25096V br
in which V st is a tension stress and
V 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
G st
0.125 T 360 G br in which T is in
degrees. Therefore
V st 14 30 u 106
0.125T V br 12 360 15 u 106
V st
744.0476T 1.71429V br
Combining the equilibrium equation and the deformation equation gives
93
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
V st
422.374T psi (T) ................... Ans.
V br
187.642T psi (C) ................... Ans.
G st
(b)
G br
V st 14 30 u 106 V br 12 15 u 106
0.19711u10 T in. ....................................................................... Ans. 3
0.15011u10 T in. ...................................................................... Ans. 3
12 0.15011u 103 T in. .................................................................. Ans.
L 12 G br
(c)
RILEY, STURGES AND MORRIS
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 d R d 100 %). SOLUTION From equilibrium
n 6F
PM 9 Ps P
0: PM 9 Ps
0
P
Since the steel rods and the pier must shrink the same amount,
PM L EM AM
Ps L Es As
Ps
94
GM
Gs ,
Es As PM EM AM
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
Combining the equilibrium equation and the deformation equation gives
§ EA · PM ¨1 9 s s ¸ EM AM ¹ ©
P
Then, since
R
9 As 100 A AM 9 As
9 As AM
9 As 100 AM 9 As
100 AM 9 As 1
100 1 R
1 R 100 R 1 100 R
and
PM P
1 E 1 s EM
§ R · ¨ 100 R ¸ ¹ ©
..............Ans.
4-83 A 3-in. diameter u 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 100qF. Use E = 10,600 ksi and D = 12.5(106)/qF. SOLUTION
H
V D 'T E
V 12.5 u 106 100 0 6 10.6 u 10
V
13.25 u 103 psi 13.25 ksi (T) ............................................................................ Ans.
4-84 A 6-m long u 50-mm diameter rod of aluminum alloy [E = 70 GPa, Q = 0.346, and D = 22.5(106)/qC] 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 35qC, there is no stress in the rod. After the temperature of the rod drops to 20qC, determine (a) The normal stress in the rod. (b) The change in diameter of the rod. SOLUTION (a)
(b)
V 6000 22.5 u 106 55 6000 1 mm 9 70 u 10
G
VL D 'T L E
V
74.958 u 106 N/m 2 # 75.0 MPa (T) ................................................................... Ans.
Gd
Q
VL D 'T L E
0.346
74.958 u10 50 6
22.5 u10 55 50 6
70 u 10 0.0804 mm 0.0804 mm (shrink) ............................................................... Ans. 9
95
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
4-85 A bar consists of 3-in. diameter aluminum alloy [E = 10,600 ksi, Q = 0.33, and D = 12.5(106)/qF] and 4-in. diameter steel [E = 30,000 ksi, Q = 0.30, and D = 6.6(106)/qF] parts, as shown in Fig. P4-85. If end supports are rigid and the bar is stress free at 0qF, determine (a) The normal stress in both parts of the bar at 80qF. (b) The change in diameter of the steel part of the bar. SOLUTION (a) From equilibrium
Ps
Pa
in which both
Ps and Pa are compressive forces.
Since the end supports are rigid, the total stretch of the rod must be zero G s G a 0
ª º Ps 20 6 « 6.6 10 80 20 u »» « 30 u 106 ªS 4 2 4º ¬ ¼ ¬« ¼» ª º Pa 30 6 « 12.5 u 10 80 30 » « 10.6 u 106 ªS 3 2 4º » ¬ ¼ ¬« ¼» Ps
Vs Va (b)
G ds
Pa
0
89, 449 lb (C)
89, 449
S 4 4 2
89, 449
S 3 4 2
7118 psi # 7.12 ksi ........................................................................ Ans. 12, 654 psi # 12.65 ksi ................................................................... Ans.
VL D 'T L E 7118 4 6.6 u106 80 4 0.00240 in. .......................... Ans. 0.3 30 u 106
Q
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 30qC, 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 10qC. Use E = 200 GPa and D = 11.9(106)/qC. SOLUTION
G 30
G 10
G
15 u10 L 0
VL D 'T L E
6
G V
VL 11.9 u 106 40 L 200 u 10 200 u 109 110.2 u 106 N/m 2 110.2 MPa ........................................................................... Ans. 9
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 100qF. The coefficients of thermal expansion for steel and concrete are 6.6(106)/qF and 6.0(106)/qF, respectively. Determine (a) The normal stresses in the concrete and in the steel bars after the temperature increases.
96
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
(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
n 6F
Pc 9 Ps 150 0
0:
is written in terms of stresses
ª100 9 S 0.75 2 4 º V c 9 ªS 0.75 2 4 º V s ¬ ¼ ¬ ¼
V s 24.1504V c
150 u103 lb
37, 725.62 psi
V c and V s are both compressive stresses. Expressing that the steel rods and the concrete must stretch the same amount, G c G s ,
in which
in terms of stresses gives
V c L 6.0 u 106 100 L 6 4.5 u 10 V s 6.6667V c 1800.00 psi
V s L 6.6 u 106 100 L 6 30 u10
Combining the equilibrium equation and the deformation equation gives
V c 1165.770 psi # 1.166 ksi (C) ............................................................................. Ans. Vs (b)
G
9571.805 psi # 9.57 ksi (C) ............................................................................... Ans.
9571.805 24 30 u 10
6
6.6 u10 100 24 6
0.00818 in. ............................ 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 50qC and the temperature of bar B increases 25qC. The coefficients of thermal expansion are 11.9(106)/qC for the steel and 22.5(106)/qC 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
o 6F
0:
TB TA 600 0
can be written in terms of stresses
1850 u10 V 2500 u10 V 6
6
A
V A 1.35135V B
B
600 u 103 N
324.324 u 106 N/m 2
V A and V B are both tension stresses. The total stretch of the assembly must be zero, G A G B 0 , therefore
in which
ª V A 750 º ª V 600 º 11.9 u 10 6 50 750 » « B 22.5 u 106 25 600 » « 9 9 ¬ 210 u10 ¼ ¬ 71u10 ¼
V A 2.36620V B
30.450 u 106 N/m 2
Combining the equilibrium equation and the deformation equation gives
VA
217.5 u 106 N/m 2 # 217 MPa (T) .................................................................... Ans. 97
0
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
VB
RILEY, STURGES AND MORRIS
79.051u 106 N/m 2 # 79.1 MPa (C) ............................................................... Ans.
217.5 u10 750 11.9 u10 50 750 210 u10 6
(b)
GC
GA
GC
0.331 mm o ......................................................................................................... Ans.
6
9
4-89 A high-strength steel [Es = 30,000 ksi, As = 0.785 in.2, and Ds = 6.6(106)/qF] bolt passes through a brass [Eb = 15,000 ksi, Ab = 1.767 in.2, and Db = 9.8(106)/qF] sleeve, as shown in Fig. P4-89. After the unit is assembled at 40qF, the temperature is increased to 100qF. If the unit is free of stress at 40qF, determine the normal stresses in the bolt and in the sleeve at 100qF. SOLUTION Writing the equilibrium equation
Tst
Fbr
in terms of stresses gives
0.785V st in which
1.767V br
V st is a tension stress and V br is a compressive stress.
Expressing that the steel bolt and the brass sleeve must stretch the same amount, G st G br , in terms of stresses gives
V st 14 6.6 u 10 6 60 14 6 30 u 10 14V st 24V br 45,360 psi
V br 12 9.8 u 106 60 12 6 15 u 10
Combining the equilibrium equation and the deformation equation gives
V st
1839.29 psi # 1839 psi (T) ................................................................................ Ans.
V br
817.10 psi # 817 psi (C) ................................................................................... 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 Ds = 17.3(106)/qC] 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 Da = 22.5(106)/qC] bar with a length of 250.50 mm can be inserted as shown. After the load P is removed, the temperature is raised 100qC. Determine the normal stresses in the bolts and in the bar, and the distance between the faces of the clamps. SOLUTION Writing the equilibrium equation
Fa
2Ts
in terms of stresses gives
625V a in which
2 115V s
V s is a tension stress and V a is a compressive stress.
Expressing that the steel bolts must stretch 0.5 mm more than the aluminum bar must stretch, G s 0.5 G a , in terms of stresses gives
98
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
ª V s 330 º 6 17.3 10 100 330 u « » 9 ¬190 u10 ¼ ª V 250.5 º 0.5 « a 22.5 u 106 100 250.5 » 9 ¬ 73 u 10 ¼ 1.73684V s 3.43151V a
492.725 u 106 N/m 2
Combining the equilibrium equation and the deformation equation gives
Va
60.448 u 106 N/m 2 # 60.4 MPa (C) ................................................................. Ans.
Vs
164.262 u106 N/m 2 # 164.3 MPa (T) ............................................................. Ans.
Then, the stretch of the aluminum bar is
60.448 u10 250.5 6
Ga
73 u109
22.5 u10 100 250.5 6
0.35620 mm
and the distance between the faces of the clamp is
L
250.5 G a
250.856 mm .................................................................................... Ans.
4-91 A prismatic bar [E = 10,000 ksi and D = 12.5(106)/qF], free of stress at room temperature, is fastened to rigid walls at its ends. One end of the bar is heated 200qF 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
dG
G
V dx D 'T dx E
³ dG ³
L
0
'T
200x 2 L2
L 200 x 2 V dx dx 0 ³ D 0 E L2
V L § 200D · § L3 · ¨ ¸ 0 E ¨© L2 ¸¹ © 3 ¹ V V
200 12.5 u 106 10 u 106 200D E 3 3 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 Ds = 17.3(106)/qC] 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 Db = 17.6(106)/qC] 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 Vs in the steel bolts and the axial stress Vb in the brass bar as functions of the temperature rise 'T (0q d 'T d 100qC). (b) The elongation Gs of the steel bolts and the elongation Gb of the brass bar as functions of the temperature rise 'T (0q d 'T d 100qC). (c) The distance L between the faces of the clamp as a function of the temperature rise 'T (0q d 'T d 100qC). SOLUTION (a) Writing the equilibrium equation
99
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
Pb
RILEY, STURGES AND MORRIS
2Ts
in terms of stresses gives
625V b in which
2 115V s
V s is a tension stress and V b is a compressive stress.
Expressing that the steel bolts must stretch 0.5 mm more than the brass bar must stretch, G s 0.5 G b , in terms of stresses gives
ª V s 330 º 17.3 u 106 'T 330 » « 9 ¬190 u10 ¼ ª V 250.5 º 0.5 « b 17.6 u 106 'T 250.5 » 9 ¬ 100 u 10 ¼ 1.73684V s 2.5050V b
500 u 106 1.3002 u 106 'T N/m 2
Combining the equilibrium equation and the deformation equation gives
Vb Vs
(b)
69.207 0.17997 'T u106 N/m 2 (C) ......................................................... Ans. 188.063 0.48904 'T u106 N/m 2 (T) ........................................................ Ans.
Then, the elongation of the steel bolts and the brass bar are
Gs
188.063 0.48904 'T u106 330 190 u10
9
0.32664 4.860 u10 Gb
(c)
3
17.3 u10 'T 330 6
'T mm ............................................................................. Ans.
69.207 0.17997 'T u 106 250.5
17.6 u10 'T 250.5
0.17336 4.860 u10
Ans.
100 u 109
3
'T mm
6
(both stretches). Finally, the distance between the faces of the clamp is
L
250.5 G b
250.32664 4.860 u10
6
'T mm .......................................................................... Ans.
100
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
4-93 An aluminum (Eal = 10,000 ksi, Dal = 12.5u10-6/qF, Aal = 1.4 in.2) bolt passes through a steel (Est = 30,000 ksi, Dst = 6.6u10-6/qF, 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 Val in the aluminum bolt and the stress Vst in the steel sleeve as a function of the temperature increase 'T (0qF < 'T < 100qF). (b) The change in length of the aluminum bolt Gal and the change in length of the steel sleeve Gst as a function of the temperature increase 'T (0qF < 'T < 100qF). SOLUTION (a) Writing the equilibrium equation
Tal
Pst
in terms of stresses gives
1.4V al
0.4V st
V al is a tension stress and V st is a compressive stress. The aluminum bolt must stretch more than the steel sleeve stretches by the amount that the nut moves, G al G st ' nut . In terms of stresses this gives
in which
ª V al 14 º ª V 12 º 12.5 u 106 'T 14 » « st 6 6.6 u106 'T 12 » ' nut « 6 ¬10 u10 ¼ ¬ 30 u10 ¼ When 'T 0 the tension in the bolt and the compression in the sleeve are each equal to 3500 lb and 3500 3500 2500 psi 8750 psi V al V st 1.4 0.4 and the initial movement of the nut is
' nut
0.00700 in.
Therefore, the deformation equation can be written
14V al 4V st
70, 000 958.00 'T
psi
Combining the equilibrium equation and the deformation equation gives
V al V st
2500 34.214 'T psi (T) .............................................................................. Ans. 8750 119.750 'T psi (C) ............................................................................ 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 73q . For 'T greater than about 73q , the bolt and the sleeve will separate, and both stresses will be zero
V al
V st
0
for 'T ! 73q .................................................................................. Ans.
101
STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b)
RILEY, STURGES AND MORRIS
Then, the change in length of the aluminum bolt and the steel sleeve are
G al G st
V al 14 12.5 u 106 'T 14 in. (stretch) ............................................. Ans. 6 10 u 10 V st 12 6.6 u 106 'T 12 in. (stretch) ............................................. Ans. 6 30 u 10
4-94 A short standard-weight steel pipe (see Appendix A) is used to support an axial compressive load of 100 kN. If yielding (Vy = 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
100 u 103 250 u 106 d 1.6 A Use
51 mm
d
A t 640 u106 m 2
(for which
A 693.5 mm 2 ) ................................................................ Ans.
4-95 A short column made of structural steel is used to support the floor beams of a building, as shown in Fig. P495. 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
Vy
36 ksi
FS
3.0
80 u103 36 u103 A t 6.667 in.2 d 3.0 A W6 u 25 , W8 u 24 , W10 u 30 , W12 u 30 , etc. all have sufficiently large area. The lightest section is W8 u 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
Vy
250 MPa
FS 1.75 From a free body diagram of the connection, the equilibrium equations
o 6Fx n 6Fy
TB cos 30q TA cos 50q 0
0:
TA sin 50q TB sin 30q 2000 9.81 0
0:
give
TA 17, 254 N d
250 u106 A 1.75
A t 120.8 u 106 m 2
12,806 N d
250 u106 A 1.75
A t 89.6 u106 m 2
TB
If the rods are made from the same material and diameter, then
Use
Sd2 A t 120.8 mm 2 d t 12.4 mm 4 d 13 mm ........................................................................................................................ Ans.
102
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
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
Vy FAB
50 ksi
1.5
FS
20 u 103 50 u 103 d 1.5 Sd2 4
20 kip (T)
d AB t 0.874 in. ................................................................................................................ Ans. FBC
30 u 103 50 u 103 d 1.5 Sd2 4
30 kip (C)
d BC t 1.070 in. ................................................................................................................. 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
V fc
34 MPa
V ys
FS
1.4
Ac
250 MPa 10 As
Writing the equilibrium equation
Fs Fc
1000 kN
in terms of stresses gives
V s As V c Ac 1000 u103 N V c and V s are both compressive stresses. Expressing that the steel and the concrete must stretch the same amount, G c G s , in terms of stresses gives in which
If
Vc
V fc
VsL VcL 9 200 u 10 31u 109 34 MPa , then Vs
Therefore
6.4516V c
Vs
219.355 MPa 250 MPa
219.355 u10 A 34 u10 10 A 6
1000 u 103 N
6
s
As
6.4516V c
1787.77 u 106 m 2
s
Ac
10 As
17,877.7 u 106 m 2
and the required sizes are concrete
133.7 mm u 133.7 mm ............................................................................. Ans.
b 17,877.7 1787.77 2
steel
b 140.2 mm
140.2 mm u140.2 mm ............................................................................. Ans.
103
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
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 crosssectional area bar. SOLUTION
Vy
53 ksi
FS 1.75
FAB
20 kip (T)
20 u 103 53 u 103 d 1w 1.75
w t 0.660 in.
FBC
10 kip (C)
10 u 103 53 u 103 d 1w 1.75
w t 0.330 in.
FBC
50 kip (T)
50 u 103 53 u 103 d 1w 1.75
w t 1.651 in.
w t 1.651 in. ..................................................................................................................... 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
6Fn
0:
2 N 85 cos 30 q 0
6Ft
0:
2V 85sin 30 q 0
give
N
85 u 103 1035 u 106 § S d 2 · 3 cos 30q 36.806 u 10 d ¨ ¸ 2 1.5 © 4 ¹
d t 0.00824 m
V
85 u 103 620 u 106 § S d 2 · sin 30q 21.25 u 103 d ¨ ¸ 2 1.5 © 4 ¹
d t 0.00809 m
8.24 mm ............................................................................................................... Ans.
d min
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 T. (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
Vy J (a)
132 ksi
FS
1.5
0.283 lb/in.3
From a free body diagram of the pin connection, the equilibrium equations
o 6Fx n 6Fy
0: 0:
T2 cosT T1 cos T
0
T1 sin T T2 sin T 10, 000 0
104
STATICS AND MECHANICS OF MATERIALS, 2nd Edition give
T1 T2
RILEY, STURGES AND MORRIS
3 10, 000 132 u 10 A d 2sin T 1.5
56.818 u 103 in.2 At sin T
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
§ 56.818 u 103 · § 25 u 12 · 17.04545 34.0909 AL t ¨ ¸¨ ¸ sin T sin 2T © ¹ © cos T ¹ sin T cosT The minimum volume occurs when sin 2T 1 T 45q .............................................................................................................................. Ans. When T 45q V
(b)
S d 2 56.818 u 103 in.2 4 sin 45q d 0.31986 in. # 0.320 in. .......................................................................................... Ans. When T 45q § 34.0909 · W J V 0.283 ¨ ¸ 9.65 lb .................................................................... Ans. © sin 90q ¹ A
(c)
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
FAB (a)
V AB V BC
FBC
50 kN
50 u 103
S 0.05 4 2
50 u 103
S 0.032 4 2
25.5 u 106 N/m 2 62.2 u 106 N/m 2
50 u10 1500 100 u10 ª¬S 0.05 3
(b)
G
9
2
25.5 MPa ............................................... Ans. 62.2 MPa ............................................. Ans.
50 u10 1000 4º 190 u 10 ªS 0.032 ¼ ¬ 3
9
2
4º ¼
0.709 mm ...... 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
TAB
20 kip
TBC
60 kip
TCD
10 kip
105
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
V AB V BC V CD
20 u 103
4074 psi # 4.07 ksi (C) ........................................................ Ans.
S 2.5 4 2
60 u 103
RILEY, STURGES AND MORRIS
33,950 psi # 33.9 ksi (T) ..................................................... Ans.
S 1.5 4 2
10 u 103
12, 730 psi # 12.73 ksi (T) ..................................................... Ans.
S 1 4 2
G
20 u10 15 60 u10 15 30 u10 ª¬S 2.5 4º¼ 30 u10 ª¬S 1.5
G
0.0213 in. .................................................................................................................. Ans.
3
6
3
2
2
6
10 u10 15 4º 30 u 10 ªS 1 4 º ¼ ¬ ¼ 3
6
2
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)
Vn
100 u 103 15, 000 u10 6
6.67 u 106 N/m 2
6.67 MPa ............................................... Ans.
(b)
Vb
100 u 103 15, 000 u 106
6.67 u 106 N/m 2
6.67 MPa ............................................... Ans.
(c)
Vb
100 u103 30, 000 u 106
3.33 u 106 N/m 2
3.33 MPa ............................................... Ans.
(d)
Vb
100 u 103 260, 000 u 106
385 u 103 N/m 2
385 kPa ................................................. Ans.
4-105 A 1-in.-diameter steel [D = 6.5(106)/qF, E = 30,000 ksi, and Q = 0.30] bar is subjected to a temperature decrease of 150qF. 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 Pin./in. after the temperature change, determine the load being transmitted to the walls. SOLUTION
V D 'T 600 u 106 E V 6.5 u 106 150 600 u 106 6 30 u 10 V 11, 250 psi # 11.25 ksi (T) H
P VA
11.25 ª¬S 1
2
4 º 8.84 kip ................................................................... Ans. ¼
106
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
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 70qC occurs. The thermal coefficients of expansion for the brass and the steel are 17.6(106)/qC and 11.9(106)/qC, respectively. SOLUTION Writing the equilibrium equation
Fb
Fs
in terms of stresses gives
ªS 90 2 4 º V b ¬ ¼
ªS 50 2 4 º V s ¬ ¼
Vs
3.24V b
V s and V b are both tensile stresses. Since the supports are rigid, the total stretch of the bar must be zero, G b G s 0 . In terms of stresses
in which
ª V b 800 º ª V 480 º 17.6 u 106 70 800 » « s 11.9 u 106 70 480 » « 9 9 ¬100 u10 ¼ ¬ 200 u10 ¼ 8V b 2.4V s
0
1.38544 u109 N/m 2
Combining the equilibrium equation and the deformation equation gives
Vb
87.82 u106 N/m 2 # 87.8 MPa (T) .................................................................... Ans.
Vs
284.5 u 106 N/m 2 # 285 MPa (T) ..................................................................... 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
A S 32 2.52 4
(a)
2.15984 in.2
Writing the equilibrium equation
PB TA 50 0 in terms of stresses gives
2.15984V B 2.15984V A
50 u 103 lb
in which V A is a tension stress and V 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, G A G B . In terms of stresses
V A 7 u 12 30 u 106
V B 4 u 12 30 u 106
V B 1.75V A
Combining the equilibrium equation and the deformation equation gives
VA
8418 psi # 8.42 ksi (T) ....................................................................................... Ans.
V B 14, 732 psi # 14.73 ksi (C) ................................................................................. Ans. (b)
GC
GA
8418 7 u12 30 u 106
0.0236 in. p ................................................................ Ans.
107
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
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
P
PC PS
The cord and the sheath must both stretch the same amount,
If
VC
GC
VCL VSL 9 7 u 10 14 u 109 V C max 15 3 5 MPa , then VS
2V C
10 MPa d
VS
G S . In terms of stresses 2V C
56 18.667 MPa 3
Therefore
Pmax
5 u10 ª¬S 0.003
Pmax
90.3 N ................................................................................................................... Ans.
6
2
4º 10 u106 ª¬S 0.0042 0.0032 4º¼ ¼
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
TA PB 15 0 in terms of stresses gives
1.25V A 3.75V B
15 u103 lb
in which V A is a tension stress and V 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, G A G B 0.025 in. In terms of stresses
V A 6 u12 30 u106 72V A
V B 2 u12 0.025 in. 15 u106
48V B 750 u 103 psi
Combining the equilibrium equation and the deformation equation gives
V A 10.7045 u 103 psi # 10.70 ksi (T) ..................................................................... Ans. VB
431.8 psi # 0.432 ksi (C) ................................................................................... Ans.
10.7045 u10 6 u12 3
(b)
GA
30 u 106
0.0257 in. (stretch) .............................................. Ans.
108
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
GB
431.8 2 u12
RILEY, STURGES AND MORRIS
0.000691 in. (shrink) ......................................................... Ans.
15 u 106
4-110 An 80-kN force P is applied to the 150 u 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
6Fn
0:
N 80sin 25q 0
6Ft
0:
V 80 cos 25q 0
give
N
80sin 25q V n An
V
80 cos 25q W n An
where the areas are related by
A Therefore
0.150 u 0.180
80 u10 sin 3
Vn
2
An sin 25q
mm 2 25q
529 u 103 N/m 2
0.150 u 0.180
80 u10 sin 25q cos 25q
0.529 MPa ..................................... Ans.
3
Wn
1.135 u 106 N/m 2
0.150 u 0.180
1.135 MPa ....................... 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 T
A 3u 3 (a)
G
(b)
V max
(c)
W max
PL EA
cos T
35 An cos T
4000 21 29 u106 3 u 3
P 4000 A 3u 3 P V max 2A 2
An
11.25 in.2
0.322 u 103 in. ......................................................... Ans.
444 psi .......................................................................................... Ans. 222 psi ....................................... Ans.
From a free-body diagram, the equilibrium equations
6Fn
0:
N 4000 cos T
6Ft
0:
V 4000sin T
0 0
give
N
45
4000 4 5 V n 11.25
109
STATICS AND MECHANICS OF MATERIALS, 2nd Edition (d)
(e)
Vn
RILEY, STURGES AND MORRIS
284 psi ..................................................................................................................... Ans.
V
4000 3 5 W n 11.25
Wn
213 psi ...................................................................................................................... Ans.
110
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