INSTRUCTOR
SOLUTIONS
MANUAL
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
Chapter 1
1-1*
From a free-body diagram of the forearm,
the equilibrium equations give
↑ ΣFy = 0 :
T − F − W − 20 = 0
4 ΣM F = 0 :
1.5T − 5.5W − 11.5 ( 20 ) = 0
T = 3.667W + 153.33 lb ............................... Ans.
F = 2.667W + 133.33 lb ............................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-2*
From a free-body diagram of the ring, the equations of equilibrium
→ ΣFx = 0 :
↑ ΣFy = 0 :
T2 cos10° − T1 sin10° = 0
T1 cos10° − T2 sin10° − 175 ( 9.81) = 0
are solved to get
T1 = 5.67128T2
T1 = 1799 N .............................................................................. Ans.
T2 = 317 N ................................................................................ Ans.
T3 = 175 ( 9.81) = 1717 N ....................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-3
The equations of equilibrium
→ ΣFx = 0 :
↑ ΣFy = 0 :
N A sin 30° − N B sin 30° = 0
N A cos 30° + N B cos 30° − 800 = 0
are solved to get
N A = NB
N A = 462 lb
60° ................................................................ Ans.
N B = 462 lb
60° ................................................................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-4*
The equations of equilibrium
Ax − N B sin 45o = 0
→ ΣFx = 0 :
↑ ΣFy = 0 :
4 ΣM A = 0 :
Ay + N B cos 45o − 300 = 0
1.5 ( N B cos 45o ) − 1.5 ( 300 ) = 0
are solved to get
N B = 424.264 N ≅ 424 N
Ax = 300 N
A = 300 N →
45............................. Ans.
Ay = 0 N
........................................................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-5
The equations of equilibrium
→ ΣFx = 0 :
↑ ΣFy = 0 :
4 ΣM A = 0 :
Ax = 0
Ay − 250 = 0
M A − 3 ( 250 ) = 0
are solved to get
Ax = 0 lb
A = 250 lb ↑
Ay = 250 lb
.....................................................Ans.
M A = 750 lb ⋅ ft 4 .................................................Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-6
From an overall free-body diagram, the equations of equilibrium
→ ΣFx = 0 :
Ax = 0
Ay − 10 − 15 + N F = 0
↑ ΣFy = 0 :
4 ΣM A = 0 :
9 N F − 3 (10 ) − 6 (15 ) = 0
are solved to get
Ax = 0 kN
Ay = 11.6667 kN ↑
N F = 13.3333 kN ↑
Then, from a free-body diagram of the right hand section of the truss,
the equations of equilibrium
4 ΣM C = 0 :
3 (13.3333) − 3TDE = 0
4 ΣM D = 0 :
3TBC − 3 (15 ) + 6 (13.3333) = 0
↑ ΣFy = 0 :
13.3333 − 15 − TCD cos 45o = 0
are solved to get
TDE = 13.33 kN (T) ............................................................................................................ Ans.
TBC = −11.67 kN = 11.67 kN (C) .................................................................................. Ans.
TCD = −2.36 kN = 2.36 kN (C) ...................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
1-7*
4 ΣM A = 0 :
RILEY, STURGES AND MORRIS
27 N F − 13 ( 33cos15° ) + 4 ( 35sin15° ) = 0
N F = 14.93561 lb ≅ 14.94 lb
Z ΣFx = 0 :
P cos 30° − 35sin15° = 0
P = 10.46005 lb ≅ 10.46 lb
^ ΣFy = 0 :
N R = 24.1 lb
75° ............. Ans.
15° ................ Ans.
N R + N F − P sin 30° − 35cos15° = 0
75° .......................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-8
↑ ΣFy = 0 :
P + 4060 − 5210 = 0
P = 1150 N .......................................................................................Ans.
4 ΣM A = 0 :
C − 16 ( 4060 ) − 37 ( 5210 ) = 0
C = 257, 730 N ⋅ mm ≅ 258 N ⋅ m ...............................................Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-9*
From an overall free-body diagram, the equations of equilibrium
→ ΣFx = 0 :
Ax = 0
Ay − 10 − 20 + N E = 0
↑ ΣFy = 0 :
4 ΣM A = 0 :
45 N E − 15 (10 ) − 30 ( 20 ) = 0
are solved to get
Ax = 0 kip
Ay = 13.3333 kip ↑
N E = 16.6667 kip ↑
Then, from a free-body diagram of the right hand section
of the truss, the equations of equilibrium
4 ΣM F = 0 :
15 N E + 15 (TDE sin 30° ) = 0
4 ΣM E = 0 :
15 ( 20 ) − 15 (TCF sin 60° ) = 0
4 ΣM C = 0 :
22.5 N E − 7.5 ( 20 ) − (15cos 30° ) TFG = 0
are solved to get
TCD = −33.3333 kip ≅ 33.3 kip (C) ............................................................................... Ans.
TCF = +23.094 kip ≅ 23.1 kip (T) .................................................................................. Ans.
TFG = +17.32 kip = 17.32 kip (T) .................................................................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
1-10*
RILEY, STURGES AND MORRIS
W = 2000 ( 9.81) = 19, 620 N
Z ΣFx = 0 :
4 ΣM A = 0 :
^ ΣFy = 0 :
P − W sin 30° = 0
2 (W sin 30° ) − 2 (W cos 30° ) + 3 N F − 1P = 0
N R + N F − W cos 30° = 0
P = 9810 N = 9.81 kN
30° ............................... Ans.
N R = 8933.806 N ≅ 8.93 kN
60° .................... Ans.
N F = 8057.612 N ≅ 8.06 kN
60° ................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-11
From a free-body diagram of the brake pedal, the equilibrium
equations are solved to get the forces
4 ΣM A = 0 :
5.5Q − ( 30 cos 30° )(11) − ( 30sin 30° )( 4 ) = 0
Q = 62.871 lb
→ ΣFx = 0 :
Ax − Q + 30 cos 30° = 0
Ax = 36.890 lb
↑ ΣFy = 0 :
Ay − 30sin 30° = 0
Ay = 15.00 lb
A = 39.8 lb
22.13° ........................................................... Ans.
Q = 62.9 lb ← ........................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-12*
From a free-body diagram of the beam, the equilibrium equations are solved to get the forces and moment
→ ΣFx = 0 :
Ax = 0
Ay − 2 = 0
↑ ΣFy = 0 :
Ay = 2 kN
4 ΣM A = 0 :
M A − 2 ( 4) − 3 = 0
M A = 11 kN ⋅ m
A = 2 kN ↑ ................................................................................................................................ Ans.
M A = 11 kN ⋅ m 4 ..................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-13
From a free-body diagram of the beam, the equilibrium equations are solved to get the forces
→ ΣFx = 0 :
Ax = 0
4 ΣM A = 0 :
15 B − 3 ( 500 ) − 6 ( 800 )
− 9 ( 700 ) − 12 ( 400 ) = 0
B = 1160 lb
↑ ΣFy = 0 :
Ay + B − 500 − 800 − 700 − 400 = 0
Ay = 1240 lb
A = 1240 lb ↑ ........................................................................................................................... Ans.
B = 1160 lb ↑ ............................................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
1-14
RILEY, STURGES AND MORRIS
W = 450 ( 9.81) = 4414.50 N
From a free-body diagram of the lower pulley,
vertical equilibrium equation gives the tension
↑ ΣFy = 0 :
2T1 − 4414.50 = 0
T1 = 2207.25 N
Then, from a free-body diagram of the upper pulley,
moment equilibrium equation gives the force F
4 ΣM axle = 0 :
100T1 − 90T1 − 100 F = 0
F = 221 N ............................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-15*
From a free-body diagram of the bracket, the equilibrium equations
4 ΣM A = 0 :
18 B − (12 )(10 ) 2  (12 3) = 0
→ ΣFx = 0 :
Ax + B = 0
↑ ΣFy = 0 :
Ay − (12 )(10 ) 2  = 0
Are solved to get the forces
B = 13.333 lb
Ax = −13.333 lb
Ay = 60.0 lb
A = 61.464 lb ≅ 61.5 lb
77.47° ..................................... Ans.
B = 13.33 lb → ....................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-16
(a) From a free-body diagram of the plane, the force
equilibrium equations are solved to get the forces
→ ΣFx = 0 :
V − 40 cos 70° − 70 cos16° = 0
↑ ΣFy = 0 :
N − 40sin 70° − 70sin16° = 0
V = 81.0 N .........................................Ans.
N = 56.882 N ≅ 56.9 N .................Ans.
(b)
Then the moment equilibrium equation gives
the required location of the normal force
4 ΣM A = 0 :
( 70 cos16° )( 75) − ( 70sin16° )( 280 ) + Nd
+ ( 40 cos 70° )( 60 ) − ( 40sin 70° )( 60 ) = 0
d = 31.5 mm
31.5 mm (from the left end of plane) ....................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
1-17*
4 ΣM A = 0 :
RILEY, STURGES AND MORRIS
1.25 N D − 9 ( 25 ) = 0
N D = 180.0 lb ....................................Ans.
→ ΣFx = 0 :
↑ ΣFy = 0 :
Ax + N D sin 38° = 0
Ay − 25 − N D cos 38° = 0
Ax = −110.8 lb
A = 200.3 lb
Ay = 166.8 lb
56.4° ................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-18*
From a free-body diagram of the upper handle,
moment equilibrium gives
FA cos θ + Bx = 0
→ ΣFx = 0 :
FA sin θ − 100 + By = 0
↑ ΣFy = 0 :
4 ΣM B = 0 :
93 (100 ) − 28 ( FA sin θ ) + 5 ( FA cos θ ) = 0
30
= 30.964°
50
FA = 919.116 N
θ = tan −1
Bx = −788.136 N
By = −372.881 N
Then a free body diagram of the upper jaw gives
4 ΣM C = 0 :
Fd + 12 By + 35 Bx = 0
d = 352 + 152 = 38.0789 mm
Therefore
F = 842 N
F = 842 N
66.8° on the jaw
66.8° on the block.......................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-19
Use W = 3750 lb (the weight carried by one truss). Then by symmetry (or from equilibrium of a free-body
diagram of the entire truss) each support carries half of the total weight
E y = Ay = W 2 = 1875 lb
Also by symmetry (or equilibrium of a free-body diagram of the truck), the truck’s weight is divided equally
between its front and rear wheels.
N F = N R = W 2 = 1875 lb
Then equilibrium of the floor panel between pins G and H gives
10G1 − 4 (1875 ) = 0
4 ΣM H = 0 :
G1 = 750 lb
6 (1875 ) − 10 H1 = 0
4 ΣM G = 0 :
H1 = 1125 lb
Next, from a free-body diagram of a section of the left side of the truss
θ = tan −1
4 ΣM B = 0 :
5
= 26.565°
10
φ = tan −1
8
= 38.660°
10
8TGH − 5 (1875 ) = 0
TGH = 1171.875 lb
4 ΣM G = 0 :
−15 (1875 ) − 8 (TBC cos θ ) − 10 (TBC sin θ ) + 10 (1125 ) = 0
TBC = −1451.295 lb ≅ 1451 lb (C) ............................................. Ans.
↑ ΣFy = 0 :
1875 + TBC sin θ − TBG sin φ − 1125 = 0
TBG = +161.6 lb = 161.6 lb (T) ................................................... Ans.
Finally, from a free-body diagram of pin C
→ ΣFx = 0 :
↑ ΣFy = 0 :
TCD cos θ − TBC cos θ = 0
−TBC sin θ − TCD sin θ − TCG = 0
TCD = −1451.295 lb
TCG = +1298 lb = 1298 lb (T) ...................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-20*
Cut a section through CD, DG, and FG, and draw a free-body diagram
of the upper-portion of the truss. The equilibrium equations give
4 ΣM G = 0 :
4 ΣM D = 0 :
( 5cos 30° )( 9 ) − ( 5sin 30° )( 4 ) + TCD ( 3) = 0
( 5cos 30° )( 6 ) − TFG ( 3) = 0
TCD = −9.66 kN = 9.66 kN (C) ................... Ans.
TFG = 8.66 kN (T) .......................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-21
A free-body diagram of cylinder A gives
→ ΣFx = 0 :
↑ ΣFy = 0 :
N AB cos 60° − N AC cos 60° = 0
N AB sin 60° + N AC sin 60° − 100 = 0
N AB = N AC = 57.73503 lb
Then, from a free-body diagram of cylinder B
→ ΣFx = 0 :
B sin θ − N AB cos 60° − N BC = 0
↑ ΣFy = 0 :
B cos θ − N AB sin 60° − 200 = 0
The minimum force occurs when N BC = 0 , therefore
N AB cos 60°
B sin θ
= tan θ =
B cos θ
N AB sin 60° + 200
θ = 6.59° ........................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
1-22
RILEY, STURGES AND MORRIS
W = 250 ( 9.81) = 2452.50 N
From a free-body diagram of the pulley
T cos φ − T cos θ = 0
→ ΣFx = 0 :
↑ ΣFy = 0 :
T sin φ + T sin θ − 2452.50 = 0
φ =θ
From the geometry of the cable
a + b = 42 m
( a + b ) cos θ = 40 m
θ = cos −1
40
= 17.7528°
42
Also from the geometry of the cable
h = a sin θ = 6 + b sin θ
Therefore
( a − b ) = 6 sin θ = 19.67789 m
which together with a + b = 42 m gives
a = 30.83895 m
b = 11.16105 m
x = a cos θ = 29.4 m ...................................................................... Ans.
T = 4020 N ...................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
1-23*
RILEY, STURGES AND MORRIS
(
Max pull occurs when either the rear wheels begin to slip B = 0.8 By
(
)
) or when the front wheels start to lift off the
ground Ay = 0 . The force which makes the front wheels lift off of the ground is
4 ΣM B = 0 :
8 (15, 000 ) − 5 ( P cos 30° ) − 10 ( P sin 30° ) = 0
P = 12,862 lb ≅ 12.86 kip ................................................................................................... Ans.
Checking the amount of friction required and the amount of friction available for this pulling force
→ ΣFx = 0 :
P sin 30° − Bx = 0
↑ ΣFy = 0 :
Ay + By − 15, 000 − P cos 30° = 0
Ay = 0
Bx = 6431 lb (friction required)
By = 26,138 lb
0.8 ( 26138 ) = 20,911 lb (friction available)
Since the friction required is much less than the friction available, we made the correct guess.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-24
Divide the weight by 2 since there are two frames
W = 200 ( 9.81) 2 = 981 N
Then from a free-body diagram of the drum
→ ΣFx = 0 :
↑ ΣFy = 0 :
N1 cos 45° − N 2 cos 45° = 0
0.4 N1 sin 45° + N 2 sin 45° − 981 = 0
N1 = N 2 = 693.672 N ≅ 694 N ................................................Ans.
Finally from a free-body diagram of one leg
4 ΣM C = 0 :
1T − 1A − 0.8 N 2 = 0
→ ΣFx = 0 :
T + Cx + N 2 sin 45° = 0
↑ ΣFy = 0 :
A + C y − N 2 cos 45° = 0
where by symmetry (or from overall equilibrium)
A = 981 2 = 490.5 N .............................................................. Ans.
and then
T = 1045.4376 N ≅ 1045 N .................................................. Ans.
Cx = −1535.94 N ≅ 1536 N ← ............................................ Ans.
C y = 0 N .................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-25*
The components of the three tension forces are
TA = TA
20i + 30 j − 50k
202 + 302 + 502
= 0.32444TAi + 0.48666TA j − 0.81111TAk
TB = TB
16i − 25 j − 50k
162 + 252 + 502
= 0.27517TB i − 0.42995TB j − 0.85990TB k
TC = TC
−25i − 15 j − 50k
252 + 152 + 502
= −0.43193TC i − 0.25916TC j − 0.86387TC k
Then the x-, y-, and z-components of the force equilibrium equation give
x:
0.32444TA + 0.27517TB − 0.43193TC = 0
y:
0.48666TA − 0.42995TB − 0.25916TC = 0
z:
−0.81111TA − 0.85990TB − 0.86387TC + 900 = 0
TA = 418.214 lb ≅ 418 lb ..................................................................................................... Ans.
TB = 205.219 lb ≅ 205 lb ..................................................................................................... Ans.
TC = 444.876 lb ≅ 445 lb .................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-26*
The components of the forces are
W = −100 ( 9.81) k = −981k N
TA = TA
4i − 8 j + 5k
4 2 + 82 + 5 2
= 0.39036TAi − 0.78072TA j + 0.48795TAk
TB = TB
−6i − 8 j + 5k
6 2 + 8 2 + 52
= −0.53666TB i − 0.71554TB j + 0.44721TB k
TC = TC
8 j + 5k
82 + 5 2
= 0.84800TC j + 0.53000TC k
Then the x-, y-, and z-components of the force equilibrium equation give
x:
0.39036TA − 0.53666TB = 0
y:
−0.78072TA − 0.71554TB + 0.84800TC = 0
z:
0.48795TA + 0.44721TB + 0.53000TC − 981 = 0
TA = 603.139 N ≅ 603 N ..................................................................................................... Ans.
TB = 438.716 N ≅ 439 N ..................................................................................................... Ans.
TC = 925.473 N ≅ 925 N ..................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-27
From a free-body diagram of pin A
→ ΣFx = 0 :
↑ ΣFy = 0 :
TAF cos 45° − TAB cos 45° = 0
20 − TAB sin 45° − TAF sin 45° = 0
TAB = TAF = 14.14214 lb ≅ 14.14 lb .................................... Ans.
Finally from a free-body diagram of BCD
→ ΣFx = 0 :
TAB cos 45° + Cx + Dx = 0
↑ ΣFy = 0 :
TAB sin 45° + C y − Dy = 0
4 ΣM C = 0 :
2 Dx − 1.5 (10 ) − 2 (TAB cos 45° ) − 1(TAB sin 45° ) = 0
where by symmetry (or from overall equilibrium)
Dy = 20 2 = 10 lb ↓
and then
Cx = −32.500 lb ≅ 32.5 lb ←
C y = 0 lb
Dx = 22.500 lb ≅ 22.5 lb →
C = 32.5 lb ← ................................................................................. Ans.
D = 24.6 lb
24.0° ..................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-28
From a free body diagram of the wheel and arm BC
→ ΣFx = 0 :
TCD + Bx = 0
2700 − By = 0
↑ ΣFy = 0 :
4 ΣM B = 0 :
−150 ( 2700 ) − 325TCD = 0
Bx = 1246 N
By = 2700 N
B = 2970 N
65.2° (on AB) ....................................................... Ans.
TCD = −1246.154 N ≅ 1246 N (C) ............................................. Ans.
Then from a free-body diagram of the arm AB (and assuming that the
spring pushes perpendicularly against the arm)
4 ΣM A = 0 :
−100 (1246.154 ) − 500 ( 2700 ) + bFS = 0
→ ΣFx = 0 :
Ax − 1246.154 + FS sin φ = 0
2700 − FS cos φ + Ay = 0
↑ ΣFy = 0 :
φ = tan −1
50
= 11.310°
250
b = 502 + 2502 = 254.951 mm
FS = 5783.92 N ≅ 5780 N (C) .................................................... Ans.
Ax = 111.827 N
A = 2970 N
Ay = 2971.60 N
87.8° .................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-29*
First equilibrium of an overall free-body diagram gives
→ ΣFx = 0 :
Ax − 100 = 0
↑ ΣFy = 0 :
200 − Ay = 0
12 N F − 24 (100 ) = 0
4 ΣM A = 0 :
Ax = 100 lb
Ay = 200 lb
N F = 200 lb
A = 224 lb
63.4° ...............................Ans.
Then from a free-body diagram of the bar ABCD
→ ΣFx = 0 :
Ax + Cx − FBE cos 45° − 100 = 0
C y + FE sin 45° − Ay = 0
↑ ΣFy = 0 :
4 ΣM C = 0 :
−18 Ax + 6 ( FBE cos 45° ) − 6 (100 ) = 0
Cx = 400 lb
C y = −200 lb
FBE = 565.685 lb ≅ 566 lb
C = 447 lb
45° .................................. Ans.
26.6° ............................................................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-30 The total weight of water carried by each truss is
WT = 1000 ( 9.81)( 0.2 )( 3.6 )( 2 ) = 14,126.40 N
By symmetry (or overall equilibrium) the support reactions at A and F are
Ay = F = WT 2 = 7063.20 N
The weight of water carried by each roof panel is
WP = 1000 ( 9.81)( 0.2 )(1.2 )( 2 ) = 4708.80 N
and by symmetry (or equilibrium of the panel) half of this load is carried by the pins at each end of the panel
B = E = WP 2 = 2354.40 N
sin θ = 0.8000
C = D = 2 (WP 2 ) = 4708.80 N
cos θ = 0.6000
Then, equilibrium of Pin B gives
→ ΣFx = 0 :
↑ ΣFy = 0 :
TBC = 0
−TAB − 2354.40 = 0
TAB = −2354.40 N
Pin E:
→ ΣFx = 0 :
↑ ΣFy = 0 :
−TDE = 0
−TEF − 2354.40 = 0
TEF = −2354.40 N
Pin A:
→ ΣFx = 0 :
↑ ΣFy = 0 :
TAH + TAC cos θ = 0
7063.20 + ( −2354.40 ) + TAC sin θ = 0
TAH = 3531.60 N
TAC = −5886.00 N
Pin H:
→ ΣFx = 0 :
↑ ΣFy = 0 :
TGH − ( 3531.60 ) = 0
TGH = 3531.60 N
TCH = 0
Pin C:
→ ΣFx = 0 :
TCD + TCG cos θ − ( −5886 ) cos θ − ( 0 ) = 0
↑ ΣFy = 0 : −4708.80 − ( −5886 ) sin θ − ( 0 ) − TCG sin θ = 0
TCD = −3531.60 N
TCG = 0 N
Pin G:
→ ΣFx = 0 :
↑ ΣFy = 0 :
TFG − ( 3531.60 ) − ( 0 ) cos θ = 0
( 0 ) sin θ + TDG = 0
TFG = 3531.60 N
TDG = 0 N
Pin D:
↑ ΣFy = 0 :
−4708.80 − ( 0 ) − TDF sin θ = 0
TDF = −5886.00 N
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-30 (cont.)
The member forces are
AB: 2350 N (C) ...............................DE:
0 N ............................................................ Ans.
AC: 5890 N (C) ................................DF: 5890 N (C) .................................................... Ans.
AH: 3530 N (T) ................................DG:
BC:
0 N ............................................................ Ans.
0 N .......................................EF: 2350 N (C) .................................................... Ans.
CD: 3530 N (C) ................................FG: 3530 N (T) .................................................... Ans.
CG:
0 N .......................................GH: 3530 N (T) .................................................... Ans.
CH:
0 N ............................................................................................................................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-31
The equations of equilibrium for the two blocks are
T sin θ − N 2 cos θ = 0
→ ΣFx = 0 :
−T cos θ + N1 sin θ = 0
↑ ΣFy = 0 :
T cos θ + N 2 sin θ − 150 = 0
T sin θ + N1 cos θ − 200 = 0
Adding the second and third equation together gives
N1 sin θ + N 2 sin θ = 150
while subtracting the first equation from the last equation gives
N1 cos θ + N 2 cos θ = 200
Dividing these two equations gives
( N1 + N 2 ) sin θ
( N1 + N 2 ) cos θ
= tan θ =
150
200
θ = 36.87°
N1 + N 2 = 250 lb
Now the first two equations can be rewritten
T sin 2 θ − N 2 sin θ cos θ = 0
−T cos 2 θ + N1 sin θ cos θ = 0
and subtracting the second equation from the first gives
T ( sin 2 θ + cos 2 θ ) = ( N1 + N 2 ) sin θ cos θ
T (1) = ( 250 )( 0.6000 )( 0.8000 )
T = 120 lb
(a)
(b)
(c)
N1 = 160 lb ......................................... N 2 = 90 lb ............................................................... Ans.
T = 120 lb ................................................................................................................................ Ans.
θ = 36.87° ............................................................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-32*
ABC is an equilateral triangle with a height of
h = 950sin 60° = 822.724 mm , and
W = −60 ( 9.81) k = −588.60k N
−475 j + 822.724k
950
= −0.50000 FAB j + 0.86603FAB k
FAB = FAB
Moment equilibrium about C
ΣM C = 0 :
( −0.95i ) × ( By j + Bz k )
( 0.95 j) × ( −0.50000 FAB j + 0.86603FABk )
+ ( −0.475i + 0.23750 j + 0.411362k ) × ( −588.6k ) = 0
has components
x:
0.82272 FAB − 139.7925 = 0
y:
0.95Bz − 279.585 = 0
z:
0 − 0.95By = 0
FAB = 169.915 N ≅ 169.9 N (C) ........................................................................................ Ans.
By = 0 N .................................................................................................................................. Ans.
Bz = 294.300 N ≅ 294 N .................................................................................................... Ans.
Then the x-, y-, and z-components of the force equilibrium equation give
x:
Cx = 0
y:
C y + ( 0 ) − 0.5 (169.915 ) = 0
z:
294.300 + Cz − 588.6 + 0.86603 (169.915 ) = 0
Cx = 0 N .................................................................................................................................. Ans.
C y = 84.958 N ≅ 85.0 N ...................................................................................................... Ans.
Cz = 147.15 N ≅ 147.2 N .................................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
1-33
RILEY, STURGES AND MORRIS
P = − Pj
Moment equilibrium about C
ΣM C = 0 :
( −50k ) × ( Dx i + Dy j)
+ ( −20k + 2.5i ) × ( −40 j)
+ (11k − 17.32051i + 10 j) × ( − Pj) = 0
has components
x:
+50 Dy − 800 + 11P = 0
y:
−50 Dx = 0
z:
−100 + 17.32051P = 0
P = 5.77350 lb ≅ 5.77 lb ..................................................................................................... Ans.
Dx = 0 lb .................................................................................................................................. Ans.
Dy = 14.7298 lb ≅ 14.73 lb ................................................................................................. Ans.
Then the x-, y-, and z-components of the force equilibrium equation give
x:
Cx + 0 = 0
Cx = 0 lb ............................................ Ans.
y:
C y + 14.7298 − 40 − 5.77350 = 0
C y = 31.044 lb ≅ 31.0 lb ............... Ans.
z:
Cz = 0
Cz = 0 lb ............................................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-34*
From a free-body diagram of the platform the equilibrium equations give
FAC cos θ − FBD cos θ = 0
→ ΣFx = 0 :
FAC sin θ + FBD sin θ − P = 0
↑ ΣFy = 0 :
FBD = FAC
P = 2 FAC sin θ
Then from a free-body diagram of the screw-block A,
the equilibrium equations give
→ ΣFx = 0 :
800 − FAC cos θ − FAE cos θ = 0
FAE sin θ − FAC sin θ = 0
↑ ΣFy = 0 :
FAE = FAC =
400
cos θ
Therefore
θ = 15°
FAC = 414.110 N
P = 214 N .................... Ans.
θ = 30°
FAC = 461.880 N
P = 462 N .................... Ans.
θ = 45°
FAC = 565.685 N
P = 800 N .................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-35*
From an overall free-body diagram, the equations of equilibrium give
→ ΣFx = 0 :
Ax − E = 0
↑ ΣFy = 0 :
Ay − 25 = 0
4 ΣM A = 0 :
3 ( 25 ) − 21E = 0
Ax = E = 3.57143 lb
Ay = 25 lb
Assume that the weight of the seat back is very small compared to
the weight of the seat. Then the center of gravity of the seat and the
center of gravity of the entire chair are the same point. A free-body
diagram of the seat gives
TBD cos θ − Cx = 0
→ ΣFx = 0 :
C y + TBD sin θ − 25 = 0
↑ ΣFy = 0 :
4 ΣM C = 0 :
3 ( 25 ) − 12 (TBD sin θ ) + 1(TBD cos θ ) = 0
10
= 39.806°
12
= 10.8476 lb
Cx = 4.76188 lb
θ = tan −1
in which
TBD
A = 25.3 lb
TBD = 10.85 lb
C = 18.67 lb
C y = 18.0555 lb
81.87° .......................................................................................................... Ans.
39.81° .................................................................................................... Ans.
75.23° ........................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-36
From an overall free-body diagram, the equations of equilibrium give
Ax + ( 500 × 3) + 750 = 0
→ ΣFx = 0 :
Ay − 900 + C = 0
↑ ΣFy = 0 :
4 ΣM A = 0 :
4C − ( 3 × 500 )( 3) − 4 ( 900 ) − 3 ( 750 ) = 0
C = 2587.50 N
Ax = −2250 N
Ay = −1687.50 N
Next, draw a free-body diagram of the bar ABC. Note that the forces
Ax and Ay are the forces exerted on bar ABC by the support (the same
forces that are shown on the overall free-body diagram) and that the forces
FAx and FAy are the forces exerted on bar ABC by the bar
ADE. The equations of equilibrium give
→ ΣFx = 0 :
Ax + Bx − FAx = 0
Ay − FAy − By + C = 0
↑ ΣFy = 0 :
4 ( 2587.50 ) − 2 By = 0
4 ΣM A = 0 :
By = 5175.00 N
FAy = −4275.00 N
Finally, from a free-body diagram of the bar ADE
→ ΣFx = 0 :
FAx + Dx + 750 = 0
↑ ΣFy = 0 :
FAy + Dy − 900 = 0
4 ΣM D = 0 :
1.5 FAx − 2 ( −4275 )
−2 ( 900 ) − 1.5 ( 750 ) = 0
FAx = −3750 N
Dx = 3000 N
A = 5690 N
D = 5980 N
E = 1170 N
Dy = 5175 N
48.7° ........................................................................................................... Ans.
59.9° ........................................................................................................... Ans.
50.2° ............................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-37
First draw a free-body diagram of the lower jaw,
and write the equations of equilibrium
→ ΣFx = 0 :
−C x = 0
D − Cy − E = 0
↑ ΣFy = 0 :
4 ΣM D = 0 :
3C y − 2 E = 0
Cx = 0
E = 1.5C y
Next, from a free-body diagram of the lower handle,
the equations of equilibrium give
→ ΣFx = 0 :
0 − Bx = 0
50 + C y − By = 0
↑ ΣFy = 0 :
4 ΣM B = 0 :
1C y − 20 ( 50 ) = 0
C y = 1000 lb
Bx = 0 lb
By = 1050 lb
E = 1500 lb ............................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
1-38*
RILEY, STURGES AND MORRIS
W = 100 ( 9.81) = 981 N
2500 − 1875
1200
300 − 150sin θ
tan φ =
1875 − 150 cos θ
cos θ =
θ = 58.612°
φ = 5.466°
There are two rollers (one on each side of the door) at B and D,
hence the 2 N B and 2 N C on the free-body diagram
→ ΣFx = 0 :
↑ ΣFy = 0 :
4 ΣM D = 0 :
T cos φ − 2 N B = 0
T sin φ + 2 N C − 981 = 0
(1050 cos θ )( 981) − (150 cos θ )( 2 NC ) − (1350sin θ )( 2 N B ) = 0
Substituting the first two equations into the third gives
(1050 cos θ )( 981) − (150 cos θ )( 981 − T sin φ ) − (1350sin θ )(T cos φ ) = 0
T = 403.455 N ≅ 403 N ...................................................................................................... Ans.
N B = 200.812 N ≅ 201 N ................................................................................................... Ans.
N C = 471.283 N ≅ 471 N ................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-39*
(a) First draw an overall free-body diagram. The force of the
ground on the track of the crane is equivalent to a single
concentrated force N acting at some location on the treads.
As the load increases, the distance d gets smaller and
smaller. The maximum load that the crane can lift
corresponds to d = 0 . Then, summing moments about
the point where the normal force acts gives
9 (12, 000 ) − (12 cos 30° − 1)( 600 )
− ( 24 cos 30° − 1 + 1)(W ) = 0
W = 4930 lb ..................................................... Ans.
(b) Next, from a free-body diagram of the pulley at the end of the boom
the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
Bx − 3600 cos10° = 0
By − 3600 − 3600sin10° = 0
Bx = 3545.3079 lb
By = 4225.1334 lb
Finally, from a free-body diagram of the boom the equations of
equilibrium give
Ax − ( 3545.3079 ) − T cos10° = 0
→ ΣFx = 0 :
↑ ΣFy = 0 :
Ay − 600 − T sin10° − ( 4225.1334 ) = 0
4 ΣM A = 0 :
24 (T sin 20° ) + ( 24sin 30° )( 3545.3079 )
− (12 cos 30° )( 600 ) − ( 24 cos 30° )( 4225.1334 ) = 0
T = 6275.1466 lb ≅ 6280 lb ............................................................................................... Ans.
(c)
Ax = 9725.1209 lb
A = 11,380 lb
Ay = 5914.8012 lb
31.3° ........................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-40
From an overall free-body diagram, the equations of equilibrium give
4 ΣM A = 0 :
(1.8cos 30° ) P − ( 0.9sin 30° )( 2 )
− ( 2.7 sin 30° + 0.6sin 30° )(1)
− ( 2.7 sin 30° + 1.8sin 30° + 0.3)(10 ) = 0
P = 17.99 kN .......................................................... Ans.
Next, from a free-body diagram of the bucket,
the equations of equilibrium give
→ ΣFx = 0 :
Gx − TEI = 0
↑ ΣFy = 0 :
G y − 10 = 0
4 ΣM G = 0 :
(1.2 cos 30° ) TEI − 0.3 (10 ) = 0
TEI = 2.88675 kN ≅ 2.89 kN (T) ........................................ Ans.
Gx = 2.88675 kN
G y = 10.00 kN
Finally, from a free-body diagram of the arm DEFG,
the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
4 ΣM D = 0 :
( 2.88675) − Dx − ( 2.88675) + FBF cos φ = 0
Dy − (10 ) − 1 + FBF sin φ = 0
( 0.6 cos 30° )( 2.88675) − ( 0.6sin 30° )(1)
+ (1.2 cos 30° )( FBF cos φ ) + (1.2sin 30° )( FBF sin φ )
− (1.8cos 30° )( 2.88675 ) − (1.8sin 30° )(10 ) = 0
in which
1.8cos 30° − 1.2 cos 30°
φ = 19.107°
1.8sin 30° + 1.2sin 30°
= 10.43807 kN ≅ 10.44 kN (C) .....................................Ans.
tan φ =
FBF
Dx = 9.86303 kN
D = 12.44 kN
G = 10.41 kN
Dy = 7.58627 kN
37.6° ........................................................................................................ Ans.
73.9° ........................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-41
First draw a free-body diagram of the wheel. It is stated that
the pin B is at or near the surface of the wheel. Then, the
equations of equilibrium give
4 ΣM axle = 0 :
2P − 2B = 0
P=B
Next, from a free-body diagram of the platform,
the equations of equilibrium give
FDE cos θ − Cx = 0
→ ΣFx = 0 :
C y + FDE sin θ − 80 = 0
↑ ΣFy = 0 :
3 ( FDE cos θ ) − 2 ( 80 ) = 0
4 ΣM C = 0 :
2
= 30°
4
= 61.5840 lb
θ = sin −1
FDE
Cx = 53.3333 lb
C y = 49.2080 lb
Finally, from a free-body diagram of the arm ABC,
the equations of equilibrium give
→ ΣFx = 0 :
Ax + ( 53.3333) − B sin θ = 0 = 0
Ay + B cos θ − ( 49.2080 ) = 0
↑ ΣFy = 0 :
4 ΣM A = 0 :
2 B − ( 4 cos θ )( 49.2080 ) − ( 4sin θ )( 53.3333) = 0
B = 138.5641 lb
Ax = 15.9487 lb
A = 72.6 lb
B = 138.6 lb
C = 72.6 lb
Ay = 70.7920 lb
77.3° ............................................................................................................ Ans.
60° .............................................................................................................. Ans.
42.7° ............................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-42*
The weight of the bar AB is
W = 25 ( 9.81) = 245.25 N
From a free-body diagram of the bar AB,
the equations of equilibrium give
→ ΣFx = 0 :
A sin 30° − B sin 45° = 0
↑ ΣFy = 0 :
A cos 30° + B cos 45° − W = 0
4 ΣM B = 0 :
( 0.5cos θ )W − (1cos θ )( A cos 30° )
+ (1sin θ )( A sin 30° ) = 0
Since sin 45° = cos 45° adding the first two equations together gives
A ( sin 30° + cos 30° ) = W
Substituting this result into the third equation gives
A ( cos 30° cos θ − sin 30° sin θ ) = 0.5cos θ  A ( sin 30° + cos 30° ) 
Dividing by A cos θ gives
2 ( cos 30° − sin 30° tan θ ) = ( sin 30° + cos 30° )
2 cos 30° − ( sin 30° + cos 30° )
2sin 30°
θ = 20.10° ............................................................................................................................... Ans.
tan θ =
(Actually, this result is independent of both the length and weight of the bar.)
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
1-43
RILEY, STURGES AND MORRIS
W = 25 ( 9.81) = 245.25 N
FCD = FCD
0.65 j + 0.95k
0.652 + 0.952
= 0.56468 FCD j + 0.82531FCD k
Moment equilibrium about B
ΣM B = 0 :
(1.2i ) × ( Ay j + Az k ) + ( 0.6i + 0.5 j) × ( −245.25k )
+ (1.6i + 0.65 j) × ( 0.56468 FCD j + 0.82531FCD k ) = 0
has components
x:
−122.625 + 0.53645 FCD = 0
y:
−1.2 Az + 147.150 − 1.32050 FCD = 0
z:
1.2 Ay + 0.90349 FCD = 0
FCD = 228.586 N ≅ 229 N .................................................................................................. Ans.
Ay = −172.104 N ≅ −172.1 N ............................................................................................ Ans.
Az = −129.915 N ≅ 129.9 N ............................................................................................... Ans.
Then the x-, y-, and z-components of the force equilibrium equation give
x:
y:
z:
Bx = 0
( −172.104 ) + By + 0.56468 ( 228.586 ) = 0
( −129.915) + Bz + 0.82531( 228.586 ) − 245.25 = 0
Bx = 0 N .................................................................................................................................. Ans.
By = 43.026 N ≅ 43.0 N ..................................................................................................... Ans.
Bz = 185.511 N ≅ 185.5 N .................................................................................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-44
From a free-body diagram of pipe A, the equations of equilibrium are
N A − FAB sin θ = 0
→ ΣFx = 0 :
↑ ΣFy = 0 :
FAB cos θ − 5 ( 9.81) = 0
where
θ = sin −1
b − 150
150
Therefore
49.05
N
cos θ
N A = FAB sin θ N
FAB =
(a)
N A < 50 N for b < 260 mm ..................Ans.
(b)
FAB < 100 N for b < 280 mm ...............Ans.
(c)
FAB < 200 N for b < 295 mm ...............Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-45
From a free-body diagram of the ring, the equations of equilibrium are
TAB sin φ − TBD cos θ = 0
→ ΣFx = 0 :
TAB cos φ − TBD sin θ − 250 = 0
↑ ΣFy = 0 :
where
φ = sin −1
b
5
θ = sin −1
5 (1 − cos φ )
3
Therefore
(a)
TAB =
TBD cos θ
sin φ
TBD =
250sin φ
lb
cos θ cos φ − sin θ sin φ
TAB =
250 cos θ
lb
cos θ cos φ − sin θ sin φ
bmax occurs when TBD goes negative
(after it goes to infinity);
bmax ≅ 3.90 ft ................................................... Ans.
(At this point, the rope will straight from D to B to A.)
(c)
To pull further to the side, the worker needs a longer rope to pull on or he needs to attach
his rope lower - closer to the crate. .......................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
1-46
RILEY, STURGES AND MORRIS
W = 50 ( 9.81) = 490.50 N
The cable is continuous, therefore the tension in the cable is continuous
(equal to the force P ); and from a free-body diagram of the pulley,
the equations of equilibrium give
P cos θ − P cos φ = 0
→ ΣFx = 0 :
P sin θ + P sin φ − W = 0
↑ ΣFy = 0 :
φ = θ = tan −1
P=
(a)
(b)
(c)
d
1.5
W
2sin θ
P < 2W
P < 4W
P < 8W
d > 387 mm ............................. Ans.
d > 189 mm .............................. Ans.
d > 94 mm ................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-47
θ AB = tan −1
d
20
θ BC = tan −1
d
10
From a free-body diagram of the stop light,
the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
TBC cos θ BC − TAB cos θ AB = 0
TAB sin θ AB + TBC sin θ BC − 75 = 0
Solving yields
TAB =
75cos θ BC
sin θ BC cos θ AB + cos θ BC sin θ AB
TBC =
75cos θ AB
sin θ BC cos θ AB + cos θ BC sin θ AB
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MECHANICS OF MATERIALS, 6th Edition
1-48
RILEY, STURGES AND MORRIS
W = 6800 ( 9.81) = 66, 708 N
From a free-body diagram of the truck,
the equilibrium equations are
→ ΣFx = 0 :
P sin θ − Bx = 0
↑ ΣFy = 0 :
Ay + By − 66, 708 − P cos θ = 0
4 ΣM B = 0 :
( 66, 708 )( 2.4 ) − 4.4 Ay − ( P cosθ )(1.5)
− ( P sin θ )( 3) = 0
The fourth equation needed to solve for the four unknowns is either Ay = 0 (the front wheels are on the verge of
lifting off the ground) or Bx = 0.8 By (the rear wheels are on the verge of slipping). Guessing that the front wheels
are on the verge of lifting off the ground gives the solution
Ay = 0 N
66, 708 ( 2.4 )
N
1.5cosθ + 3sin θ
Bx = P sin θ N
P=
By = 66, 708 + P cos θ N
The forces Bx and 0.8 By are plotted on the same
graph as the force P . Since Bx is always less than
0.8By , the guess that the front wheels are on the verge
of lifting off the ground was the correct guess, and the
solution is valid for all values of θ .
E = 1500 lb ............................................................................................................................. Ans.
TBD = 10.85 lb
C = 18.67 lb
39.81° .................................................................................................... Ans.
75.23° ........................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-49
From a free-body diagram of the I-beam,
the equations of equilibrium give
4 ΣM D = 0 :
400 ( 3) + 1000 ( 7 − d ) + 6 F = 0
4 ΣM F = 0 :
6 D − 1000 ( d − 1) − 400 ( 3) = 0
D=
1000 ( d − 1) + 1200
6
F=
1000 ( 7 − d ) + 1200
6
Next, from a free-body diagram of the joint D,
the equations of equilibrium give
→ ΣFx = 0 :
−TCD cos φ − TDE = 0
TCD sin φ − D = 0
↑ ΣFy = 0 :
TCD =
D
sin φ
TDE =
−D
tan φ
in which
1
3
φ = tan −1 = 18.435°
By inspection, members CE and CF are both zero-force
members. Therefore the tension force in member BC
will be the same as the tension force in member CD and
the compression force in member EF will be the same as
the compression force in member DE,
TBC =
D
sin φ
TEF =
−D
tan φ
TCF = 0
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-50
From an overall free-body diagram of the light pole,
θ = tan −1
1.75
b
φ = tan −1
2.75 − b
5
the moment equation of equilibrium gives
4 ΣM A = 0 :
TGH =
2 ( 7500 ) − 2.75 (TGH cos θ ) = 0
2 ( 7500 )
2.75cos θ
It will be assumed that the cable DG supports the end of the
arm BG and that the connection of the horizontal arm BG to the
vertical pole ABCD exerts negligible moment on the arm.
(If the connection could provide a sufficient moment, then the
cable between D and G would not be necessary.)
Then, from a free-body diagram of the pin G,
the equations of equilibrium give
FBG + TGH sin φ − TDG cos θ = 0
→ ΣFx = 0 :
TDG sin θ − TGH cos φ = 0
↑ ΣFy = 0 :
TGH cos φ
sin θ
cos φ cos θ − sin φ sin θ
= TGH
sin θ
TDG =
FBG
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-51
From an overall free-body diagram of the crane,
the equations of equilibrium give
N − 12, 000 − 600 − W = 0
↑ ΣFy = 0 :
( 9 )(12, 000 ) − (12 cos θ − 1)( 600 )
− ( 24 cos θ − 1 + 1) W − Nd = 0
N = (12, 600 + W ) lb
108, 600 − ( 7200 + 24W ) cos θ
d=
ft
4 ΣM C = 0 :
12, 600 + W
(a) For W = 3600 lb
d=
108, 600 − 93, 600 cos θ
ft
16, 200
(b) From a free-body diagram of the pulley at B,
tan φ =
24sin θ − 6
24 cos θ + 9
and the equations of equilibrium give
Bx − 3600 cos φ = 0
→ ΣFx = 0 :
By − 3600 − 3600sin φ = 0
↑ ΣFy = 0 :
By = 3600 (1 + sin φ )
Bx = 3600 cos φ
From a free-body diagram of the boom,
the equations of equilibrium give
Ax − Bx − T cos φ = 0
→ ΣFx = 0 :
Ay − By − T sin φ − 600 = 0
↑ ΣFy = 0 :
4 ΣM A = 0 :
 24sin (θ − φ )  T − (12 cos θ )( 600 )
+ ( 24sin θ ) Bx − ( 24 cos θ ) By = 0
T=
( 7200 + 24B ) cosθ − 24B sin θ
y
24sin (θ − φ )
x
Ax = Bx + T cos φ
Ay = By + T sin φ + 600
A = Ax2 + Ay2
Note that the tension force becomes negative for
an angle of about 66° . Since negative forces in
the cable are not possible, the boom would topple
over onto the top of the cab of the crane if the
operator tried to lift higher than 66° .
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-51 (cont.)
(c) For d = 1 ft
d=
108, 600 − ( 7200 + 24W ) cos θ
= 1 ft
12, 600 + W
Wmax =
96, 000 − 7200 cos θ
lb
1 + 24 cos θ
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
1-52
RILEY, STURGES AND MORRIS
W = 250 ( 9.81) = 2452.50 N
(a) From a free-body diagram of the post AB,
moment equilibrium gives
6 (TBC sin 60° ) − ( 3cos θ ) W = 0
4 ΣM B = 0 :
TBC =
( 2452.50 )( 3cos θ )
6sin 60°
Since the pin at A is frictionless and the weight of the brace AC is
neglected, the brace AC is a two-force member and from a free-body
diagram of the pin C, the equations of equilibrium give
→ ΣFx = 0 :
FAC cos ( 60° + θ ) + TBC cos ( 60° − θ ) − TCD cos φ = 0
↑ ΣFy = 0 :
FAC sin ( 60° + θ ) − TBC sin ( 60° − θ ) − TCD sin φ = 0
FAC =
cos ( 60° − θ ) sin φ + sin ( 60° − θ ) cos φ
TBC
sin ( 60° + θ ) cos φ − cos ( 60° + θ ) sin φ
TCD =
FAC cos ( 60° + θ ) + TBC cos ( 60° − θ )
=P
cos φ
in which
tan φ =
6sin ( 60° + θ )
6sin β
=
7 + 6 cos β 7 + 6 cos ( 60° + θ )
For this arrangement, the force P necessary to start
raising the post ( 2300 N ) is almost as large as the
weight of the post ( 2450 N ) .
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-52 (cont.)
(b) From a free-body diagram of the post AB,
moment equilibrium now gives
4 ΣM B = 0 :
TBC =
6 (TBC sin 67.5° ) − ( 3cos θ ) W = 0
( 2452.50 )( 3cos θ )
6sin 67.5°
Again, since the pin at A is frictionless and the weight of the brace
AC is neglected, the brace AC is a two-force member and from a
free-body diagram of the pin C, the equations of equilibrium give
→ ΣFx = 0 :
FAC cos ( 45° + θ ) + TBC cos ( 67.5° − θ ) − TCD cos ( 22.5° − θ ) = 0
↑ ΣFy = 0 :
FAC sin ( 45° + θ ) − TBC sin ( 67.5° − θ ) + TCD sin ( 22.5° − θ ) = 0
FAC =
sin ( 67.5° − θ ) cos ( 22.5° − θ ) − cos ( 67.5° − θ ) sin ( 22.5° − θ )
TBC
sin ( 45° + θ ) cos ( 22.5° − θ ) + cos ( 45° + θ ) sin ( 22.5° − θ )
TCD =
FAC cos ( 45° + θ ) + TBC cos ( 67.5° − θ )
cos ( 22.5° − θ )
Finally, since the weight of the brace AD is also neglected, the brace
AD is also a two-force member and from a free-body diagram of the
pin D, the equations of equilibrium give
→ ΣFx = 0 : − FAD cos ( 90° − θ ) + TCD cos ( 22.5° − θ ) − TDE cos φ = 0
↑ ΣFy = 0 :
FAD sin ( 90° − θ ) − TCD sin ( 22.5° − θ ) − TDE sin φ = 0
FAD =
cos ( 22.5° − θ ) sin φ + sin ( 22.5° − θ ) cos φ
TCD
cos ( 90° − θ ) sin φ + sin ( 90° − θ ) cos φ
TDE =
TCD cos ( 22.5° − θ ) − FAD cos ( 90° − θ )
=P
cos φ
Note that the force in the brace AD goes to zero at
about θ = 75° . For 75° ≤ θ ≤ 90° , the solution
becomes similar to that of part a (with the angle
between the post and the brace AC 45° rather
than 60° ).
For this arrangement, the force P necessary to start
raising the post (1700 N ) is about 25% less than
the force required using a single brace (part a).
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-53
a = 12.5cos θ − 2
b = 12.5sin θ + 0.5
b
φ = tan −1
a
From a free-body diagram of the truck box,
the equations of equilibrium give
Ax − C cos φ = 0
→ ΣFx = 0 :
Ay + C sin φ − 22, 000 = 0
↑ ΣFy = 0 :
4 ΣM A = 0 :
8.5 ( 22, 000 cos θ ) − 2.5 ( 22, 000sin θ )
−12.5 C sin (φ − θ )  = 0
C=
187, 000 cos θ − 55, 000sin θ
12.5sin (φ − θ )
Ax = C cos φ
Ay = 22, 000 − C sin φ
A = Ax2 + Ay2
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MECHANICS OF MATERIALS, 6th Edition
1-54*
→ ΣFx = 0 :
RILEY, STURGES AND MORRIS
−TAB − 75 + 100 − 50 = 0
TAB = −25 kN = 25 kN (C) ............................................... Ans.
→ ΣFx = 0 :
−TBC + 100 − 50 = 0
TBC = +50 kN = 50 kN (T) ............................................... Ans.
→ ΣFx = 0 :
−TCD − 50 = 0
TCD = −50 kN = 50 kN (C) ............................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
1-55*
→ ΣFx = 0 :
↑ ΣFy = 0 :
4 ΣM cut = 0 :
RILEY, STURGES AND MORRIS
−V = 0
90 − 120 − P = 0
− M − 90 ( 5 8 ) − 120 (1.5 ) = 0
V = 0 lb .................................................................................................Ans.
P = −30 lb = 30 lb (C) ......................................................................Ans.
M = −236 lb ⋅ in. .................................................................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-56
Next, from a free-body diagram of the man standing on the beam,
the equations of equilibrium give
→ ΣFx = 0 :
Cx = 0
↑ ΣFy = 0 :
T − 75 ( 9.81) + T − 40 ( 9.81) + C y = 0
4 ΣM C = 0 :
3 75 ( 9.81)  + 1.5  40 ( 9.81)  − 3T − 1.5T = 0
T = 621.300 N
C y = −114.450 N
Next, draw a free-body diagram of the portion of the beam
between section a–a and the right end of the beam. Note that
since one-fourth of the beam has been “cut away,” only threefourths of the total beam weight is included on the free-body
diagram. The equations of equilibrium give
→ ΣFx = 0 :
− Pa + 0 = 0
↑ ΣFy = 0 :
Va + ( 621.3) − 30 ( 9.81) − 114.450 = 0
4 ΣM a = 0 :
− M a + 0.75 ( 621.3) − 1.125 30 ( 9.81)  − 2.25 (114.450 ) = 0
Pa = 0 N ........................ Va = −213 N ................................................................................. Ans.
M a = −122.6 N ⋅ m ............................................................................................................... Ans.
Finally, draw a free-body diagram of the portion of the beam between
section b–b and the right end of the beam. This time three-fourths of the
beam has been “cut away” and only one-fourth of the total beam weight is
included on the free-body diagram. The equations of equilibrium give
→ ΣFx = 0 :
− Pb + 0 = 0
↑ ΣFy = 0 :
Vb − 10 ( 9.81) − 114.450 = 0
4 ΣM b = 0 :
− M b − 0.375 10 ( 9.81)  − 0.75 (114.450 ) = 0
Pb = 0 N ........................ Vb = 213 N .................................................................................... Ans.
M b = −122.6 N ⋅ m ................................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-57*
From an overall free-body diagram of the bracket,
the equations of equilibrium give
→ ΣFx = 0 :
Ax + B cos 45° = 0
↑ ΣFy = 0 :
Ay + B sin 45° − 500 = 0
4 ΣM A = 0 :
5 ( 500 ) − 15 ( B cos 45° ) − 15 ( B sin 45° ) = 0
B = 117.85113 lb
Ax = −83.3333 lb
Ay = 416.6667 lb
Next, draw a free-body diagram of the portion of the bracket between
section a–a and pin A. The equations of equilibrium give
→ ΣFx = 0 :
V − 83.3333 = 0
↑ ΣFy = 0 :
P + 416.6667 = 0
4 ΣM a = 0 :
M − 8 ( 83.3333) = 0
P = −417 lb = 417 lb (C) .................................................. Ans.
V = 83.3 lb ............................................................................ Ans.
M = 667 lb ⋅ in. .................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-58
From a free-body diagram of the lower half of the clamp,
the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
4 ΣM cut = 0 :
V =0
P − 2000 = 0
M + 0.075 ( 2000 ) = 0
P = 2000 N (T) .................................................................... Ans.
V = 0 N .................................................................................. Ans.
M = −150.0 N ⋅ m ............................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-59*
Draw free-body diagrams of the sections of the tower between
the points of interest and the top. For each section, the vertical
component of the equation of equilibrium gives
↑ ΣFy = 0 :
PA − 2000 − 3 (1500 cos 30° ) − W = 0
W = 40 ( 50 ) = 2000 lb
PA = 7900 lb .................................................... Ans.
↑ ΣFy = 0 :
PB − 2000 − 3 (1500 cos 30° ) − 3 (1250 cos 40° ) − W = 0
W = 40 (150 ) = 6000 lb
PB = 14, 770 lb .....................................................................................Ans.
↑ ΣFy = 0 :
PC − 2000 − 3 (1500 cos 30° ) − W
−3 (1250 cos 40° ) − 3 (1000 cos 50° ) = 0
W = 40 ( 250 ) = 10, 000 lb
PC = 20, 700 lb ............................................ Ans.
↑ ΣFy = 0 :
PD − 2000 − 3 (1500 cos 30° ) − W
−3 (1250 cos 40° ) − 3 (1000 cos 50° )
−3 ( 750 cos 60° ) = 0
W = 40 ( 350 ) = 14, 000 lb
PD = 25,800 lb ....................................................................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-60*
The weights of the two cylinders are the same
W = 50 ( 9.81) = 490.5 N
First draw a free-body diagram of the upper cylinder,
and write the equations of equilibrium
Z ΣFx = 0 :
N 2 − W sin 27° = 0
^ ΣFy = 0 :
N1 − W cos 27° = 0
N1 = 437.0387 N
N 2 = 222.6823 N
Next, from a free-body diagram of the lower cylinder, the equations of equilibrium give
Z ΣFx = 0 :
N 4 − N 2 − W sin 27° = 0
^ ΣFy = 0 :
N 3 − W cos 27° = 0
N 3 = 437.0387 N
N 4 = 445.3647 N
Next, from a free-body diagram of the rack,
the equations of equilibrium give
→ ΣFx = 0 :
Ax − N 4 cos 27° + ( N1 + N 3 ) sin 27° − B sin 27° = 0
↑ ΣFy = 0 :
Ay − N 4 sin 27° − ( N1 + N 3 ) cos 27° + B cos 27° = 0
4 ΣM A = 0 :
120 ( 445.3647 ) − 320 ( 437.0387 )
−120 ( 437.0387 ) + 580 B = 0
B = 239.4022 N
Ax = 902.3320 N
Ay = 767.6911 N
Finally, from a free-body diagram of the upper portion of the rack (between section a and the roller support B),
the equations of equilibrium give
Z ΣFx = 0 :
−P = 0
^ ΣFy = 0 :
B − V − N1 = 0
4 ΣM cut = 0 :
0.360 B − M − 0.100 N1 = 0
P = 0 N ...............................................................................Ans.
V = −197.6 N ....................................................................Ans.
M = 42.5 N ⋅ m .................................................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-61
From a free-body diagram of the lower portion of
the crutch (between section a and the bottom B),
the equations of equilibrium give
Z ΣFx = 0 :
V − 35sin 25° = 0
^ ΣFy = 0 :
P + 35cos 25° = 0
4 ΣM cut = 0 :
− M − 2 ( 35sin 25° ) = 0
P = −31.7 lb = 31.7 lb (C) ............................................Ans.
V = 14.79 lb .......................................................................Ans.
M = 29.6 lb ⋅ ft ..................................................................Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-62
From a free-body diagram of the wheel, the equations of equilibrium give
4 ΣM B = 0 :
325 FCD − 150 ( 2700 ) = 0
FCD = 1246 N
Since CD is a two-force member, the axial force on every cross section is
the same
P = 1246 N (C) ........................................................................ Ans.
and the shear force and the bending moment are both zero
V = 0 N ...................................................................................... Ans.
M = 0 N ⋅ m .............................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-63
From a free-body diagram of the handle, the equations of equilibrium give
4 ΣM F = 0 :
30 (1000 ) − 8 FDE = 0
FDE = 3750 lb
Since DE is a two-force member, the axial force on every cross section is
the same
P = 3750 lb (C) ................................................................... Ans.
and the shear force and the bending moment are both zero
V = 0 lb .................................................................................................Ans.
M = 0 lb ⋅ ft ..........................................................................................Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-64*
First draw an overall free-body diagram of the shaft
and sum moments about the bearing B
ΣM C = 0 :
( −800 j + 250i ) × ( −5k ) + ( −800 j − 250i ) × ( −30k )
+ ( −2000 j + 250k ) × ( 30i ) + ( −2000 j − 250k ) × ( 5i )
+ ( −2800 j) × ( Bx i + Bz k ) = 0
The x-, y-, and z-components of this equation give
x:
4000 + 24, 000 − 2800 Bz = 0
y:
1250 − 7500 + 7500 − 1250 = 0
z:
60, 000 + 10, 000 + 2800 Bx = 0
Bx = −25.00 kN
Bz = 10.00 kN
Next draw a free-body diagram of the portion of the shaft between the
bearing B and the section at A and write the equations of equilibrium
ΣFx = 0 :
Vx + 30 + 5 − 25 = 0
ΣFy = 0 :
Py = 0
ΣFz = 0 :
Vz + 10 = 0
Vx = −10 kN .......................................................................... Ans.
Py = 0 kN .............................................................................. Ans.
Vz = −10 kN .......................................................................... Ans.
ΣM cut = 0 :
( M i + T j + M k ) + ( −0.4 j + 0.25k ) × ( 30i )
x
y
z
+ ( −0.4 j − 0.25k ) × ( 5i ) + ( −1.2 j) × ( −25i + 10k ) = 0
The x-, y-, and z-components of this equation give
x:
M x − 12 = 0
M x = 12 kN ⋅ m ................................. Ans.
y:
Ty + 7.5 − 1.25 = 0
Ty = −6.25 kN ⋅ m ............................ Ans.
z:
M z − 30 + 12 + 2 = 0
M z = 16 kN ⋅ m ................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-65*
First draw a free-body diagram of the blocks,
and write the equations of equilibrium
→ ΣFx = 0 :
NC − N A = 0
FC + FA − 30 = 0
↑ ΣFy = 0 :
8 ( 30 ) − 16 ( FC ) = 0
4 ΣM A = 0 :
NC = N A
FC = FA = 15 lb
Next, from a free-body diagram of the left handle,
the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
4 ΣM B = 0 :
Bx − N C = 0
30 − By − (15 ) = 0
4 ( 30 ) + 4 (15 ) − 4 N C = 0
N C = 45 lb
Bx = 45 lb
By = 15 lb
Next, from a free-body diagram of the lower section of the left handle,
the equations of equilibrium give
→ ΣFx = 0 :
P − 45 = 0
↑ ΣFy = 0 :
V − 15 = 0
4 ΣM cut = 0 :
M − 4 ( 45 ) = 0
P = 45 lb ........................................................................................Ans.
V = 15 lb ......................................................................................................Ans.
M = 180 lb ⋅ in. ...........................................................................................Ans.
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MECHANICS OF MATERIALS, 6th Edition
1-66
RILEY, STURGES AND MORRIS
W = 360 ( 9.81) = 3531.60 N
From a free-body diagram of the bar ABC,
the equations of equilibrium give
− Ax + FBD cos φ = 0
→ ΣFx = 0 :
↑ ΣFy = 0 :
− Ay + FBD sin φ − W = 0
4 ΣM A = 0 :
1430 ( FBD sin β ) − ( 2700 cos16° ) W = 0
h 900 + 1430sin16°
=
b 1430 cos16° − 890
φ = 69.471°
β = φ − 16° = 53.471°
tan φ =
FBD = 7976.730 N
Ax = 2797.2911 N
Ay = 3938.5663 N
Then, from a free-body diagram of the left section of the bar,
the equations of equilibrium give
Z ΣFx = 0 :
P − Ax cos16° − Ay sin16° = 0
^ ΣFy = 0 :
V + Ax sin16° − Ay cos16° = 0
4 ΣM cut = 0 :
M − ( 0.530sin16° ) Ax + ( 0.530 cos16° ) Ay = 0
P = 3770 N .................................................................... Ans.
V = 3010 N .................................................................................................Ans.
M = −1598 N ⋅ m .......................................................................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-67
From a free-body diagram of the pipe the moment equilibrium equation
ΣM cut = 0 :
( M i + T j + M k ) + ( −7i + 18j + 10k ) × ( −50k ) = 0
x
y
z
has x-, y-, and z-components
x:
M x − 900 = 0
M x = 900 lb ⋅ in. .............. Ans.
y:
Ty − 350 = 0
Ty = 350 lb ⋅ in. ................ Ans.
z:
Mz = 0
M z = 0 lb ⋅ in. .................. Ans.
and the force equilibrium equation has components
ΣFx = 0 :
Vx = 0
Vx = 0 lb ................................................................................. Ans.
ΣFy = 0 :
Py = 0
Py = 0 lb ................................................................................. Ans.
ΣFz = 0 :
Vz − 50 = 0
Vz = 50 lb .............................................................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-68*
From a free-body diagram of the bar ABC,
the equations of equilibrium give
→ ΣFx = 0 :
Ax = 0
− Ay + FBD − 3 = 0
↑ ΣFy = 0 :
4 ΣM A = 0 :
200 FBD − 400 ( 3) = 0
FBD = 6 N
Ax = 0 N
Ay = 3 N
Then, from a free-body diagram of the left section of the bar,
the equations of equilibrium give
Z ΣFx = 0 :
P − ( 3) sin θ = 0
^ ΣFy = 0 :
V − ( 3) cos θ = 0
4 ΣM cut = 0 :
( 0.1)( 3) − M = 0
in which
θ = tan −1
120
= 30.964°
200
Then
P = 1.543 kN ................................................................. Ans.
V = 2.57 kN ................................................................................................Ans.
M = 0.3 kN ⋅ m ..........................................................................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-69
From a free-body diagram of a part of the hook,
the equations of equilibrium give
Z ΣFx = 0 :
V − 10 cos θ = 0
^ ΣFy = 0 :
P − 10sin θ = 0
4 ΣM cut = 0 :
− M − 10 (10sin θ ) = 0
P = 10sin θ kip .................................... Ans.
V = 10 cos θ kip ................................... Ans.
M = −100sin θ kip ⋅ in. ...................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-70
From a free-body diagram of the top handle
4 ΣM A = 0 :
93 (100 ) − 38 ( C sin θ )
+ ( 35 − d )( C cos θ ) = 0
C=
9300
38sin θ − ( 35 − d ) cos θ
Note that tan θ = d 40 ; therefore d = 40 tan θ and
9300
9300
=
38sin θ − 35cos θ + 40sin θ 78sin θ − 35cos θ
→ ΣFx = 0 :
C cos θ − Ax = 0
Ax = C cos θ
C=
↑ ΣFy = 0 :
C sin θ − Ay − 100 = 0
Ay = C sin θ − 1000
Next, from a free-body diagram of the part of the top handle
to the right of section a–a the equations of equilibrium give
[ ΣFx = 0 :
P + Ax cos φ + Ay sin φ = 0
^ ΣFy = 0 :
V + Ax sin φ − Ay cos φ = 0
4 ΣM cut = 0 :
Ax (18 tan φ ) − M − 18 Ay = 0
P = ( − Ax cos φ − Ay sin φ ) N ..................... Ans.
V = ( Ay cos φ − Ax sin φ ) N ........................ Ans.
M =  Ax (18 tan φ ) − 18 Ay  N ⋅ mm ......... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-71
(a) From a free-body diagram of the complete beam,
the equations of equilibrium give
4 ΣM A = 0 :
↑ ΣFy = 0 :
20 RB − ( x + 2.5 )( 4000 ) = 0
RA + RB − 4000 = 0
RB = 200 ( x + 2.5 ) = ( 200 x + 500 ) lb
RA = ( 3500 − 200 x ) lb
Load, shear force, and bending moment diagrams are as shown.
The moment under the left wheel is
M C = xRA = ( 3500 x − 200 x 2 ) lb ⋅ ft
and the moment under the right wheel is
M D = (15 − x ) RB = (15 − x )( 200 x + 500 ) lb ⋅ ft
These moments are graphed below. Notice that the maximum
moment occurs under the right wheel when 0 ≤ x ≤ 7.5 m
(and the right wheel is closer to the center of the beam) and
under the right wheel when 7.5 ≤ x ≤ 15 m (and the left
wheel is closer to the center of the beam).
(b) The graph of maximum moment is also shown below.
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MECHANICS OF MATERIALS, 6th Edition
1-72
RILEY, STURGES AND MORRIS
W = 250 ( 9.81) = 2452.50 N
(a) From a free-body diagram of the post AB,
the equations of equilibrium give
Ax − TBC cos ( 60° − θ ) = 0
→ ΣFx = 0 :
Ay + TBC sin ( 60° − θ ) − W = 0
↑ ΣFy = 0 :
6 (TBC sin 60° ) − ( 3cos θ ) W = 0
4 ΣM B = 0 :
 ( 2452.50 )( 3cos θ ) 
TBC = 
 N
6sin 60°


Ax = TBC cos ( 60° − θ )  N
 Ay = 2452.50 − TBC sin ( 60° − θ )  N
Next, draw a free-body diagram of the lower portion of ABC.
The weight of this portion is proportional to its length
Wb =
Wb 2452.50b
=
= ( 408.75b ) N
L
6
Then the equations of equilibrium give
Z ΣFx = 0 :
P + Ax cos θ + Ay sin θ − Wb sin θ = 0
^ ΣFy = 0 :
V − Ax sin θ + Ay cos θ − Wb cos θ = 0
4 ΣM cut = 0 :
M + Wb ( b 2 ) cos θ 
+ Ax ( b sin θ ) − Ay ( b cos θ ) = 0
M =  Ay ( b cos θ ) − Ax ( b sin θ ) − 204.375b 2 cos θ  N ⋅ m .............Ans.
P = ( 408.75b − Ay ) sin θ − Ax cos θ  N .....Ans.
V = ( 408.75b − Ay ) cos θ + Ax sin θ  N .....Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-73*
From a free-body diagram of the collar, the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
T cos 20° − N BC sin 30° = 0
N BC cos 30° − T sin 20° − 900 = 0
T = 700 lb
N BC = 1316 lb
20° ..............................................................................Ans.
60° ........................................................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
1-74*
RILEY, STURGES AND MORRIS
W = 500 ( 9.81) = 4905 N
First draw a free-body diagram of joint A, and write the equations of equilibrium
→ ΣFx = 0 :
↑ ΣFy = 0 :
TAC + TAB cos 30° = 0
TAB sin 30° − W = 0
TAB = 9810 N = 9.81 kN (T) ............................................................Ans.
TBC = −8495.709 N ≅ 8.50 kN (C) ................................................Ans.
Next, from a free-body diagram of joint C,
the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
TAC + TBC cos 60° + TCD cos 45° = 0
TBC sin 60° − TCD sin 45° = 0
TBC = −6219.291 N = 6.22 kN (C) ................................................Ans.
TCD = −7617.044 N ≅ 7.62 kN (C) ................................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-75
From a free-body diagram of the motor and support,
the equations of equilibrium give
→ ΣFx = 0 :
Ax − 21 − 1 = 0
↑ ΣFy = 0 :
Ay + B − 25 = 0
4 ΣM D = 0 :
12 B − 8 ( 25 ) + 10.5 ( 21) + 5.5 (1) = 0
B = −2.16667 lb ≅ 2.17 lb ↓ ............................. Ans.
Ax = 22 lb
Ay = 27.16667 lb
A = 35.0 lb
51.0° ............................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
1-76*
RILEY, STURGES AND MORRIS
W = 135 ( 9.81) = 1324.50 N
φ = cos −1
220 − 75
= 48.769°
220
First draw a free-body diagram of the cylinder. When the cylinder just starts
to rotate about the step, the normal force N becomes zero. Then moment
equilibrium gives
4 ΣM C = 0 :
220 W sin ( 20° + φ )  − 220  P cos ( 20° + φ )  = 0
P = 3410 N ...................................................................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-77
(a) From a free-body diagram of the entire beam,
the equations of equilibrium give
↑ ΣFy = 0 :
RA + RB − 3000 − 500 ( 6 ) = 0
4 ΣM B = 0 :
500 ( 6 )  ( 3) + 3000 (12 ) + 1800 − 18RA = 0
RA = 2600 lb ............................................................................................................ Ans.
RB = 3400 lb ............................................................................................................. Ans.
(b) Next, from a free-body diagram of the left end of the beam,
the equations of equilibrium give
↑ ΣFy = 0 :
4 ΣM cut = 0 :
( 2600 ) − 3000 − V = 0
M + 1800 + 3000 ( 4 ) − ( 2600 )(10 ) = 0
V = −400 lb ............................................................ Ans.
M = 12, 200 lb ⋅ ft .................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-78
First draw a free-body diagram of the upper jaw. The resultant of the
distributed load is a 300 N force acting through the centroid of the
distributed load. The equations of equilibrium give
4 ΣM D = 0 :
225 ( 300 ) − 100 FC = 0
4 ΣM C = 0 :
125 ( 300 ) − 100 FD = 0
FC = 675 N ↓ ........................................................ Ans.
FD = 375 N ↑ ........................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-79*
(a) First, from a free-body diagram of the entire frame,
the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
4 ΣM D = 0 :
Dx − 400 = 0
A + Dy − 1000 = 0
8 ( 400 ) + 2 (1000 ) − 10 A = 0
A = 520 lb
Dx = 400 lb
Dy = 480 lb
Next, from a free-body diagram of bar ABCD,
the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
4 ΣM B = 0 :
( 400 ) − Bx + FCE cos θ = 0
( 520 ) + By − FCE sin θ + 480 = 0
8 ( 480 ) − 2 ( 520 ) − 6 ( FCE sin θ ) = 0
sin θ = 4 5
cos θ = 3 5
FCE = 583.333 lb
Bx = 750 lb
By = −533.333 lb
Next, from a free-body diagram of the lower end of bar BEF,
the equations of equilibrium give
Z ΣFx = 0 :
Pa + 750 cos θ + 533.333sin θ = 0
^ ΣFy = 0 :
Va − 750sin θ + 533.333cos θ = 0
4 ΣM B = 0 :
M a + 2.5 ( 750sin θ ) − 2.5 ( 533.333cos θ ) = 0
Pa = −876.667 lb ≅ 877 lb (C) ................................................Ans.
Va = 280 lb ....................................................................................Ans.
M a = −700 lb ⋅ ft ..........................................................................Ans.
(b) Finally, from a free-body diagram of the left end of bar ABCD,
the equations of equilibrium give
Z ΣFx = 0 :
Pb − 750 = 0
^ ΣFy = 0 :
Vb + 520 − 533.333 = 0
4 ΣM B = 0 :
M b + 3 ( 533.333) − 5 ( 520 ) = 0
Pb = 750 lb (T) .............................................................................Ans.
Vb = 13.33 lb .................................................................................Ans.
M b = 1000 lb ⋅ ft ...........................................................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-80
Next, from a free-body diagram of the upper handle,
the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
4 ΣM B = 0 :
Bx = 0
By − 5 − D = 0
70 ( 5 ) − 40 D = 0
D = 8.75 N ↓ ....................................................................... Ans.
Bx = 0 N
By = 13.75 N
B = 13.75 N ↓ (on the pin) ............................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-81*
Draw a free-body diagram of the shaft.
The moment equation of equilibrium
ΣM A = 0 :
(18 j) × ( Bx i + By j + Bz k ) + (14i + 32 j) × ( 200 j − Pk )
+ ( −12 j + 6k ) × ( −500i ) + ( −12 j − 6k ) × ( −150i ) = 0
has components
x:
18 Bz − 32 P = 0
Bz = 266.667 lb ≅ 267 lb .............. Ans.
y:
−14 P − 3000 + 900 = 0
P = 150 lb .......................................... Ans.
z:
−18Bx + 2800 − 6000 − 1800 = 0
Bx = −277.778 lb ≅ 278 lb ............ Ans.
Then the x-, y-, and z-components of the force equilibrium equation give
x:
Ax + ( −277.778 ) − 500 − 150 = 0
Ax = 928 lb ........................................ Ans.
y:
200 + By = 0
By = −200 lb ..................................... Ans.
z:
Az + ( 266.667 ) − (150 ) = 0
Az = −116.7 lb .................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
1-82*
RILEY, STURGES AND MORRIS
W1 = 300 ( 9.81) = 2943 N
W2 = 100 ( 9.81) = 981 N
W3 = 200 ( 9.81) = 1962 N
WP = 500 ( 9.81) = 4905 N
Draw a free-body diagram of the platform.
The moment equation of equilibrium
ΣM O = 0 :
( i + j) × ( −2943k ) + ( 2i + j) × ( −981k )
+ ( 3i + j) × (TB k ) + ( i + 3 j) × ( −1962k )
+ ( i + 4 j) × (TC k ) + (1.5i + 2 j) × ( −4905k ) = 0
has components
x:
−2943 − 981 + TB − 5886 + 4TC − 9810 = 0
y:
2943 + 1962 − 3TB + 1962 − TC + 7357.5 = 0
TB = 3388.909 N ≅ 3390 N .................................................................................. Ans.
TC = 4057.773 N ≅ 4060 N .................................................................................. Ans.
and the z-component of the force equilibrium equation gives
z:
TA + ( 3388.909 ) + ( 4057.773) − ( 2943) − ( 981) − (1962 ) − ( 4905 ) = 0
TA = 3344.318 N ≅ 3340 N .................................................................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
1-83
From a free-body diagram of the upper half of the clamp,
the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
4 ΣM cut = 0 :
V =0
300 − P = 0
M − 3 ( 300 ) = 0
P = 300 lb (T) ...................................................................... Ans.
V = 0 lb .................................................................................. Ans.
M = 900 lb ⋅ ft ...................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
1-84
RILEY, STURGES AND MORRIS
T = W = 100 ( 9.81) = 981 N
WP = 50 ( 9.81) = 490.50 N
First draw a free-body diagram of the arm AB and the pulley.
The moment equation of equilibrium gives
4 ΣM A = 0 :
1500 By − 300 ( 981) − 350 ( 981) − 650 ( 490.5 ) = 0
By = 637.650 N
Next, from a free-body diagram of the bar BC,
the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
4 ΣM C = 0 :
Cx − ( 981) + Bx = 0
C y − ( 637.5 ) = 0
600 Bx − 300 ( 981) − 300 ( 637.650 ) = 0
Bx = 809.325 N
Cx = 171.675 N
C y = 637.650 N
Finally, from a free-body diagram of the right end of bar AB,
the equations of equilibrium give
→ ΣFx = 0 :
− P − ( 809.325 ) = 0
( 637.650 ) − V = 0
4 ΣM B = 0 : 300 ( 637.650 ) − M = 0
↑ ΣFy = 0 :
P = −809 N = 809 N (C) ..................................................................................................... Ans.
V = 638 N ............................................................................................................................... Ans.
M = 191.3 N ⋅ m ..................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-1*
N 10
= ≤ 30 ksi
A A
π
1
A = (12 − di2 ) ≥ in 2
4
3
di ≤ 0.75867 in.
σ=
t=
d o − di
≥ 0.1207 in. .......................................................................................... Ans.
2
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MECHANICS OF MATERIALS, 6th Edition
2-2*
σA =
RILEY, STURGES AND MORRIS
40 (103 )
NA
=
= 106.6 (106 ) N m 2 = 106.6 MPa (T) ....................... Ans.
AA ( 0.025 )( 0.015 )
50 (103 )
NB
σB =
=
= 133.3 (106 ) N m 2 = 133.3 MPa (T) ........................ Ans.
AB ( 0.025 )( 0.015 )
20 (103 )
NC
σC =
=
= 53.3 (106 ) N m 2 = 53.3 MPa (T) ............................ Ans.
AC ( 0.025 )( 0.015 )
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MECHANICS OF MATERIALS, 6th Edition
2-3
RILEY, STURGES AND MORRIS
↑ ΣFy = 0 :
N AB − 90 − 2 ( 75 ) + 2 ( 60 ) − 2 (125 ) = 0
N AB = 370 kip (C)
↑ ΣFy = 0 :
N BC − 90 − 2 ( 75 ) + 2 ( 60 ) = 0
N BC = 120 kip (C)
↑ ΣFy = 0 :
N CD − 90 − 2 ( 75 ) = 0
N CD = 240 kip (C)
σ AB =
N AB 370
=
= 23.1 ksi (C) ................................................................................... Ans.
16
AAB
σ BC =
N BC 120
=
= 30.0 ksi (C) ................................................................................... Ans.
4
ABC
σ CD =
N CD 240
=
= 20.0 ksi (C) ................................................................................... Ans.
12
ACD
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MECHANICS OF MATERIALS, 6th Edition
2-4*
RILEY, STURGES AND MORRIS
W = 130 ( 9.81) = 1275.30 N
From a free-body diagram of joint B, the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
TAB sin 8° − P cos 20° = 0
TAB cos8° − P sin 20° − (1275.3) = 0
TAB = 1357.261 N
σ AB =
TAB
1357.261
=
= 7.68 (106 ) N m 2 = 7.68 MPa (T) ...................... Ans.
AAB π ( 0.015 )2 4
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-5
From a free-body diagram of a portion of the wood block,
the equations of equilibrium give
↑ ΣFy = 0 :
V − 16,800 = 0
V = 16,800 lb
τ=
V 16800
=
= 1050 psi ........................................................Ans.
A (8 × 2)
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-6
From a free-body diagram of the lever, moment
equilibrium about the center of the shaft gives
4 ΣM = 0 :
25V − 625 (1) = 0
V = 25 kN
3
V 25 (10 )
τ= =
≤ 125 (106 ) N m 2
A 0.020a
a ≥ 0.0100 m = 10.00 mm ........................................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-7*
τ=
V
8000
8000
=
=
≤ 500 psi
A π ( 2 ) L1 π (1.5 ) L2
L1 ≥ 2.55 in. .............................................................................................................. Ans.
L2 ≥ 3.40 in. .............................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-8
τ=
σ=
V
P
=
≤ 75 (106 ) N m 2
A π ( 0.100 )( 0.025 )
N
P
=
≤ 100 (106 ) N m 2
2
2
A π ( 0.150 − 0.100 ) 4
P ≤ 589 (103 ) N
P ≤ 982 (103 ) N
Pmax = 589 kN .......................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
2-9*
τ=
RILEY, STURGES AND MORRIS
V 100 ( 2000 )
=
= 40 (103 ) psi
A
π d ( 0.5 )
d = 3.18 in. ............................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-10*
Assuming that the pulleys rotate freely, the tension will be the same throughout the length of the cord, and
T = W = 45 ( 9.81) = 441.45 N
σ=
T
441.45
=
= 5.62 (106 ) N m 2 = 5.62 MPa (T) .............................. Ans.
2
A π ( 0.010 ) 4
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-11
Assume that the girl’s arms are vertical. Each arm will carry half of the girl’s weight and
(a)
σ=
N
125 2
=
= 79.6 psi (T) ...................................................................................... Ans.
A π (1)2 4
(b)
σ=
N
125 2
=
= 124.3 psi (T) ........................................................................ Ans.
2
A π (1 − 0.62 ) 4
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
2-12
RILEY, STURGES AND MORRIS
W = 225 ( 9.81) = 2207.25 N
From a free-body diagram of pulley 4,
the equations of equilibrium give
↑ ΣFy = 0 :
2T + TB − ( 2207.25 ) = 0
Next, from a free-body diagram of the double pulley
(1 and 2), the equations of equilibrium give
4 ΣM axle = 0 :
100TA − 300TB = 0
Finally, from a free-body diagram of the pulley 3,
the equations of equilibrium give
↑ ΣFy = 0 :
TA − 2T = 0
Combining these three equations gives
TA = 1655.4375 N
TB = 551.8125 N
T = 827.7188 N
T
827.7188
σC = =
= 4.68 (106 ) N m 2 = 4.68 MPa (T) ........................... Ans.
2
A π ( 0.015 ) 4
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-13*
From free-body diagrams of the rings B and C,
the equations of equilibrium give
−TAB cos 45° + TBC cos α = 0
→ ΣFx = 0 :
−TBC cos α + TCD cos 45° = 0
↑ ΣFy = 0 :
TAB sin 45° + TBC sin α − 10 = 0
−TBC sin α + TCD sin 45° − 8 = 0
Adding the first pair of equations together gives
TAB = TCD
Then, adding the second pair of equations together gives
TAB = TCD = 12.72792 lb
Now, the third equation can be written
TBC sin α = 10 − TAB sin 45°
and the first equation can be written
TBC cos α = TAB cos 45°
Dividing this pair of equations gives
10 − (12.72792 ) sin 45°
TBC sin α
= tan α =
TBC cos α
(12.72792 ) cos 45°
or
α = 6.340°
TBC = 9.05539 lb
σ max =
TAB 12.72792
=
≤ 18 (103 ) psi
2
πd 4
A
d ≥ 0.0300 in. ................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-14*
θ = tan −1
1.5
= 36.870°
2
sin θ = 3 5
cos θ = 4 5
From symmetry (or overall equilibrium), each support carries half of the total load
RA = RC = 18 2 = 9 kN
Then, from a free-body diagram of the joint A,
the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
TAD cos θ + TAB = 0
TAD sin θ + 9 = 0
TAD = −15.00 kN
TAB = 12.00 kN
Finally, from a free-body diagram of the bottom chord,
the equations of equilibrium give
→ ΣFx = 0 :
(12.00 ) − V = 0
V = 12 kN
V 12 (10 )
τ= =
≤ 2.25 (106 ) N m 2
A 0.100a
a ≥ 0.0533 m = 53.3 mm ............................................................................................. Ans.
3
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
2-15
RILEY, STURGES AND MORRIS
V = τ A = 225 ( 5 × 4 ) = 4500 lb
From a free-body diagram of member AB and the end
of the bottom chord, the equations of equilibrium give
→ ΣFx = 0 :
V − (12 13) P = 0
P = 4870 lb ................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-16*
From a free-body diagram of the block A,
the equations of equilibrium give
→ ΣFx = 0 :
700 − FAC cos θ − FAE cos θ = 0
FAE sin θ − FAC sin θ = 0
↑ ΣFy = 0 :
FAC = FAE =
τ=
700
N
2 cos θ
FAC
V
=
N m2
2
A π ( 0.010 ) 4
(a)
θ = 15°
FAC = 362.347 N
τ = 4.61 MPa ........................................... Ans.
(b)
θ = 30°
FAC = 404.145 N
τ = 5.15 MPa ........................................... Ans.
(c)
θ = 45°
FAC = 494.975 N
τ = 6.30 MPa ........................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-17
From a free-body diagram of the upper handle,
the equations of equilibrium give
→ ΣFx = 0 :
D sin 38° − Ax = 0
Ay − D cos 38° − 25 = 0
↑ ΣFy = 0 :
4 ΣM A = 0 :
1.25 D − 9 ( 25 ) = 0
D = 180 lb
Ax = 110.8191 lb
Ay = 166.8419 lb
A = Ax2 + Ay2 = 200.2925 lb
τ=
V
200.2925
=
= 4080 psi ................................................................................ Ans.
A π ( 0.25 )2 4
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-18
From an overall free-body diagram of the truss,
the equations of equilibrium give
9 F − 3 (10 ) − 6 (15 ) = 0
4 ΣM A = 0 :
F = 13.3333 kN
Then, from a free-body diagram of the right half of the truss,
the equations of equilibrium give
3 (13.3333) − 3TDE = 0
4 ΣM C = 0 :
TDE = 13.3333 kN
(a)
σ DE =
13.3333 (103 )
4 ΣM D = 0 :
750 (10
−6
)
= 17.78 (106 ) N m 2 = 17.78 MPa
Ans.
3TBC − 3 (15 ) + 6 (13.3333) = 0
TBC = −11.6667 kN
σ BC =
(b)
11.6667 (103 )
ABC
≤ 30 (106 ) N m 2
ABC ≥ 388.89 (10−6 ) m 2 = 389 mm 2 ......................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-19*
From an overall free-body diagram of the seat,
the equations of equilibrium give
→ ΣFx = 0 :
Ax − E = 0
↑ ΣFy = 0 :
Ay − 30 = 0
4 ΣM A = 0 :
3 ( 30 ) − 21E = 0
E = 4.28571 lb
Ax = 4.28571 lb
Ay = 30 lb
Next, from a free-body diagram of the seat back,
the equations of equilibrium give
4 ΣM C = 0 :
9 (TBD cos θ ) − 21( 4.28571) = 0
10
= 39.806°
12
= 13.01707 lb
θ = tan −1
TBD
τ=
VB 13.01707
=
= 117.9 psi ..............................................Ans.
A π ( 3 8)2 4
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-20
From a free-body diagram of the pin E,
the equations of equilibrium give
→ ΣFx = 0 :
−TDE − TEF cos θ = 0
↑ ΣFy = 0 :
−10 − TEF sin θ = 0
4
= 53.130°
3
= 7.500 kN
θ = tan −1
TDE
TEF = −12.500 kN
Next, from a free-body diagram of the pin D,
the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
( 7.5) − TCD cos 45° = 0
−12 − TCD sin 45° − TDF = 0
TCD = 10.60660 kN
TDF = −19.500 kN
T 10.6066 (10 )
= =
= 17.00 (106 ) N m 2 = 17.00 MPa (T) ..................... Ans.
−6
A
624 (10 )
3
(a)
σ CD
σ DF
(b)
3
T 19.5 (10 )
= =
≤ 25 (106 ) N m 2
A
A
ADF ≥ 780 (10−6 ) m 2 = 780 mm 2 ............................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-21
From a free-body diagram of the pin A, the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
TAF cos 45° − TAB cos 45° = 0
20 − TAF sin 45° − TAB sin 45° = 0
TAB = TAF = 14.14214 lb
τB =
VB 14.14214
=
= 1152 psi .............................................Ans.
A π (1 8 )2 4
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-22*
From a free-body diagram of the upper handle,
the equations of equilibrium give
→ ΣFx = 0 :
FC cos θ − Ax = 0
↑ ΣFy = 0 :
FC sin θ − 100 − Ay = 0
4 ΣM A = 0 :
93 (100 ) − 28 ( FC sin θ ) + 5 ( FC cos θ ) = 0
θ = tan −1
FC = 919.1059 N
Ax = 788.1248 N
τA =
Ay = 372.8794 N
30
= 30.964°
50
A = Ax2 + Ay2 = 871.8829 N
VA
871.8829
=
= 34.7 (106 ) N m 2 = 34.7 MPa
2
A 2π ( 0.004 ) 4
Ans.
Then, from a free-body diagram of the upper jaw, the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
4 ΣM A = 0 :
( 788.1248) − Bx − N sin φ = 0
( 372.8794 ) − By + N cos φ = 0
Nd − 35 ( 788.1248 ) − 12 ( 372.8794 ) = 0
N = 841.9084 N
Bx = 456.4753 N
τB =
d = 152 + 352 = 38.0789 mm
By = 1146.7130 N
φ = tan −1
15
= 23.199°
35
B = Bx2 + By2 = 1234.229 N
VB
1234.229
=
= 31.4 (106 ) N m 2 = 31.4 MPa
2
A 2π ( 0.005 ) 4
Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-23*
From a free-body diagram of the platform, the equations of equilibrium give
4 ΣM C = 0 :
3 ( FDE cos 30° ) − 2 ( 80 ) = 0
FDE = 61.58403 lb
(a) Member DE is a two-force member and the stress on every cross
section is the same
FDE 61.58403
=
= 49.3 psi ..........................................Ans.
1.25
A
V
61.58403
τE = E =
= 627 psi ....................................Ans.
A 2 π ( 0.25 )2 4 


σ DE =
(b)
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-24
From a free-body diagram of the bucket, the equations of equilibrium give
→ ΣFx = 0 :
Gx − TEI = 0
↑ ΣFy = 0 :
G y − 10 = 0
4 ΣM G = 0 :
(1.2 cos 30° ) TEI − 0.3 (10 ) = 0
TEI = 2.88675 kN
Gx = 2.88675 kN
G y = 10 kN
Then, from a free-body diagram of the arm DEFG,
the equations of equilibrium give
→ ΣFx = 0 :
Dx + ( 2.88675 ) − ( 2.88675 ) + FBF cos φ = 0
↑ ΣFy = 0 :
Dy − (1) + FBF sin φ − (10 ) = 0
4 ΣM D = 0 :
( 0.6 cos 30° )( 2.88675) − ( 0.6sin 30° )(1)
+ (1.2 cos 30° )( FBF cos φ ) + (1.2sin 30° )( FBF sin φ )
− (1.8cos 30° )( 2.88675 ) − (1.8sin 30° )(10 ) = 0
φ = tan −1
1.8cos 30° − 1.2 cos 30°
= 19.107°
1.8sin 30° + 1.2sin 30°
FBF = 10.43809 kN
Dx = −9.86304 kN
Dy = 7.58327 kN
D = Dx2 + Dy2 = 12.44128 kN
3
VD 12.44128 (10 )
τD =
=
≤ 120 (106 ) N m 2
2
A
2 π d 4 
d ≥ 0.00812 m = 8.12 mm ................................................................................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-25*
First, from an overall free-body diagram of the frame,
the equations of equilibrium give
→ ΣFx = 0 :
Bx − 500 + 1500 = 0
By − 1000 = 0
↑ ΣFy = 0 :
4 ΣM B = 0 :
8 ( 500 ) + 5 (1000 ) − 6 A = 0
A = 1500 lb
Bx = −1000 lb
By = 1000 lb
Next, from a free-body diagram of the horizontal member ACDE,
the equations of equilibrium give
4 ΣM E = 0 :
8 (1000 ) − 3Dy = 0
Dy = 2666.667 lb
Finally, from a free-body diagram of the vertical member BDF,
the equations of equilibrium give
4 ΣM F = 0 :
4 Dx − 10 (1000 ) = 0
Dx = 2500 lb
D = Dx2 + Dy2 = 3655.285 lb
τD =
VD 3655.285
=
≤ 7500 psi
A
πd2 4
d ≥ 0.788 in. .................................................................................Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-26
From a free-body diagram of the wheel,
the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
4 ΣM B = 0 :
Bx − FCD = 0
2700 − By = 0
325 FCD − 150 ( 2700 ) = 0
FCD = 1246.15385 N
Bx = 1246.15385 N
By = 2700 N
B = Bx2 + By2 = 2973.701 N
Then from a free-body diagram of the arm AB (and assuming
that the spring pushes perpendicularly against the arm)
4 ΣM A = 0 :
→ ΣFx = 0 :
−100 (1246.154 ) − 500 ( 2700 ) + bFS = 0
Ax − (1246.154 ) + FS sin φ = 0
2700 − FS cos φ + Ay = 0
↑ ΣFy = 0 :
b = 502 + 2502 = 254.951 mm
φ = tan −1
50
= 11.310°
250
FS = 5783.917 N
Ax = 111.8278 N
Ay = 2971.5958 N
A = Ax2 + Ay2 = 2973.699 N
τA =
VA 2973.699
=
≤ 125 (106 ) N m 2
2
A 2 π d A 4 
d A ≥ 0.00389 m = 3.89 mm ............................................................................................... Ans.
τB =
VB
2973.701
=
≤ 125 (106 ) N m 2
2
A 2 π d B 4 
d B ≥ 0.00389 m = 3.89 mm ............................................................................................... Ans.
The forces on pins C and D are equal (both equal to the force
in the member CD) and their diameters will be the same
τC =
TCD 1246.154
=
≤ 125 (106 ) N m 2
2
A 2 π d C 4 
dC = d D ≥ 0.00252 m = 2.52 mm ..................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
2-27
RILEY, STURGES AND MORRIS
di = d o − 2 ( 0.1d o ) = 0.8d o
σ=
P
9000
11, 459.156
psi
=
=
2
2
A π ( d o − di ) 4
d o2
For standard steel pipe,
σ=
P 9000
=
≤ 12, 000 psi
A
A
A = 0.75 in.2
d = 1.5 in. ...........................................Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-28
For a uniformly distributed load
P
P
σ= = 2
A πd 4
d =
2
4P
πσ
=
4 (150 ) (103 )
π ( 3.25 ) (106 )
= 0.05876
d = 0.242 m = 242 mm ................................................................................................ Ans.
For the flexible bearing plate:
σ = σ max
σ=
0 ≤ r ≤ rc
σ max ( rp − 0.2rc − 0.8r )
rc ≤ r ≤ rp
rp − rc
P = ∫ σ dA = ∫ σ max ( 2π rdr ) + ∫
A
=
π
3
rc
rp
0
rc
σ max ( rp − 0.2rc − 0.8r )
rp − rc
( 2π rdr )
σ max ( 0.8rc2 + 0.8rp rc + 1.4rp2 ) = 150 kN
σ max =
For
rc = 75 mm
450 (103 )
π ( 0.8r + 0.8rp rc + 1.4r
2
c
2
p
)
N m2
σ max = 3.25 MPa
d = 2rp ≅ 296 mm ............................... Ans.
(
For d p = 150 mm , rp = 75 mm
)
σ max = 8.49 MPa
(
For d p = 400 mm , rp = 200 mm
)
σ max = 1.98 MPa
(
For d p = 600 mm , rp = 300 mm
σ max = 0.94 MPa
D=
8.49 − 1.98
(100 ) = 76.7% ................................. Ans.
8.49
D=
8.49 − 0.94
(100 ) = 89.0% ................................. Ans.
8.49
)
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copyright owner is unlawful.45
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-29
For a uniformly distributed load
σ=
P
P
=
A π ( d c2 − 22 ) 4
d c2 =
4P
πσ
+4=
4 ( 50 )
π (10 )
+ 4 = 10.366
d c = 3.22 in. ..................................................................................................................... Ans.
σ=
For the flexible collar:
P = ∫ σ dA = ∫
A
=
π
4
4
σ max ( 5 − r )
4
1


σ max 5 ( rc2 − 1) −
σ max =
For
rc
σ max ( 5 − r )
( 2π rdr )
2 3

rc − 1)  = 50 kip
(
3

200
2


π 5 ( rc2 − 1) − ( rc3 − 1)
3


ksi
σ max = 10 ksi
d c = 2rc ≅ 3.32 in. ............................... Ans.
For d c = 2.4 in. , ( rc = 1.2 in.)
σ max = 37.13 ksi
For d c = 3.2 in. , ( rc = 1.6 in.)
σ max = 11.10 ksi
D=
37.13 − 11.10
(100 ) = 70.1% ............................ Ans.
37.13
D=
37.13 − 6.16
(100 ) = 83.4% ............................... Ans.
37.13
D=
37.13 − 3.86
(100 ) = 89.6% ............................... Ans.
37.13
For d c = 4.0 in. , ( rc = 2.0 in.)
σ max = 6.16 ksi
For d c = 5.0 in. , ( rc = 2.5 in.)
σ max = 3.86 ksi
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-30
For a uniformly distributed load
P
P
σ= =
2
A π ( d w − 0.0402 ) 4
d =
2
w
4P
πσ
+ 0.040 =
2
4 ( 80 ) (103 )
π ( 2.8 ) (106 )
+ 0.0016
d w = 0.1949 m = 194.9 mm ........................................................................................ Ans.
For the flexible washer:
σ = σ max
σ=
0.020 mm ≤ r ≤ 0.030 mm
0.030σ max
r
0.030 mm ≤ r ≤ rw
0.030σ max
( 2π rdr )
r
= πσ max ( 0.0302 − 0.0202 ) + 0.060σ maxπ ( rw − 0.030 )
P = ∫ σ dA = ∫
0.030
0.020
A
σ max ( 2π rdr ) + ∫
rw
0.030
= πσ max 0.0005 + 0.060 ( rw − 0.030 )  = 80 kN
σ max =
For
80 (103 )
π 0.0005 + 0.060 ( rw − 0.030 ) 
N m2
σ max = 2.8 MPa
d w = 2rw ≅ 346 mm ............................. Ans.
For no washer
σ max =
80 (103 )
π ( 0.0302 − 0.0202 )
= 50.93 (106 ) N m 2 = 50.93 MPa
For d w = 200 mm , ( rw = 100 mm )
σ max = 5.42 MPa
D=
50.93 − 5.42
(100 ) = 89.4% ............................... Ans.
50.93
D=
50.93 − 3.31
(100 ) = 93.5% ............................... Ans.
50.93
For d w = 300 mm , ( rw = 150 mm )
σ max = 3.31 MPa
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MECHANICS OF MATERIALS, 6th Edition
2-31*
RILEY, STURGES AND MORRIS
P = −5000 lb
A = 4 × 4 = 16 in.2
θ = 90° − 14° = 76°
( −5000 ) 1 + cos 2 76°  = −18.29 psi = 18.29 psi (C) ...................Ans.
( )
2 (16 ) 
− ( −5000 )
sin 2 ( 76° ) = +73.4 psi ...........................................................Ans.
τn =
2 (16 )
σn =
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.48
MECHANICS OF MATERIALS, 6th Edition
2-32*
RILEY, STURGES AND MORRIS
P = 400 kN
A = 75 × 45 mm 2
θ = 37°
σn =
( 400 ) (103 )
1 + cos 2 ( 37° ) 
2 ( 0.075 × 0.045 ) 
σ n = +75.6 (106 ) N m 2 = 75.6 MPa (T) ................................................................ Ans.
τn =
− ( 400 × 103 )
2 ( 0.075 × 0.045 )
sin 2 ( 37° )
τ n = −57.0 (106 ) N m 2 = −57.0 MPa
..................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.49
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-33
P 270
=
= 22.5 ksi (T) .................................................................................. Ans.
A 2×6
P
270
=
=
= 11.25 ksi ............................................................................... Ans.
2 A 2 ( 2 × 6)
σ max =
τ max
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MECHANICS OF MATERIALS, 6th Edition
2-34
RILEY, STURGES AND MORRIS
P = −80 kN
A = π ( 75 ) 4 = 4417.865 mm 2
2
θ = 57°
( −80 ) (103 )
σn =
1 + cos 2 ( 57° ) 
2 ( 4417.865 ) (10−6 )
σ n = −5.37 (106 ) N m 2 = 5.37 MPa (C) ....................................... Ans.
τn =
− ( −80 × 103 )
2 ( 4417.865 ) (10−6 )
sin 2 ( 57° )
τ n = +8.27 (106 ) N m 2 = +8.27 MPa
..................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.51
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-35*
σn =
P
[1 + cos 2θ ] = 12 ksi
2 ( 4 ×1)
τn =
−P
sin 2θ = −9 ksi
2 ( 4 × 1)
sin 2θ
9
=
1 + cos 2θ 12
Solve by trial and error to get
θ = 36.870° ...................................................................................................................... Ans.
P = 75.0 kip .................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.52
MECHANICS OF MATERIALS, 6th Edition
2-36*
σn =
( 400 ) (103 )
1 + cos 2 ( −33° )  ≤ 70 (106 )
2 ( 0.1t )
RILEY, STURGES AND MORRIS
N m2
t ≥ 0.0402 m = 40.2 mm
τn =
− ( 400 × 103 )
2 ( 0.1t )
sin 2 ( −33° ) ≤ 45 (106 ) N m 2
t ≥ 0.0406 m = 40.6 mm
tmin = 40.6 mm ................................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.53
MECHANICS OF MATERIALS, 6th Edition
2-37
− ( −P)
sin 2 ( 55° ) = 2 ksi
2 ( 4 × 8)
τa =
(a)
RILEY, STURGES AND MORRIS
P = 136.21475 kip ≅ 136.2 kip (C) ...............................................Ans.
σa =
( −136.21475) 1 + cos 2 55° 
( )
2 ( 4 × 8) 
(b)
σ a = −1.400 ksi = 1.400 ksi (C) .....................................................Ans.
(c)
σ max =
τ max
P 136.21475
=
= 4.26 ksi .............................................................................. Ans.
A
4×8
P 136.21475
=
=
= 2.13 ksi ............................................................................. Ans.
2A
2 ( 4 × 8)
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-38
σn =
P
1 + cos 2 ( −30° )  ≤ 13.60 (106 ) N m 2
2 ( 0.200 )( 0.120 )
P ≤ 435.2 (103 ) N
τn =
−P
sin 2 ( −30° ) ≤ 5.25 (106 ) N m 2
2 ( 0.200 )( 0.120 )
P ≤ 291.0 (103 ) N
τ max =
P
P
=
≤ 8.75 (106 ) N m 2
2 A 2 ( 0.200 )( 0.120 )
P ≤ 420 (103 ) N
Pmax = 291 kN .................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-39*
σn =
P
1 + cos 2 ( −θ )  ≤ 12 ksi
2 ( 4 ×1) 
τn =
−P
sin 2 ( −θ ) ≤ 9 ksi
2 ( 4 ×1)
Optimum angle occurs when
σ n = 12 ksi
and τ n = 9 ksi
sin 2θ
9
=
1 + cos 2θ 12
Solve by trial and error to get
θ = 36.870°
(a)
φ = 90° − θ = 53.13°
(b)
P = 75.0 kip .................................................................................................................... Ans.
...................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.56
MECHANICS OF MATERIALS, 6th Edition
2-40*
TAB = 500 kN (T)
TBC = 300 kN (T)
RILEY, STURGES AND MORRIS
Ttop = 250 kN (T)
(a) On plane a–a
( 500 ) (103 )
σn =
1 + cos 2 ( 30° ) 
2 ( 0.200 )( 0.100 ) 
σ n = 18.75 (106 ) N m 2 = 18.75 MPa (T) ............................Ans.
τn =
− ( 500 ) (103 )
2 ( 0.200 )( 0.100 )
sin 2 ( 30° )
τ n = −10.83 (106 ) N m 2 = −10.83 MPa ..............................Ans.
(b) The maximum stresses in the bar occur in the section with the maximum load
Pmax = TAB = 500 kN (T)
σ max
500 (103 )
P
= =
= 25.0 (106 ) N m 2 = 25.0 MPa (T) .................... Ans.
A ( 0.200 )( 0.100 )
τ max
500 (103 )
P
=
=
= 12.50 (106 ) N m 2 = 12.50 MPa ................... Ans.
2 A 2 ( 0.200 )( 0.100 )
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-41
There are two bolts and they each carry a normal force of N
and a shear force of V . Equilibrium of the eyebar gives
Z ΣFn = 0 :
2 N − P cos 30° = 0
^ ΣFt = 0 :
2V + P sin 30° = 0
 π ( 0.5 )2 
P cos 30°
N=
≤ 12 
 kip
2
 4 
P ≤ 5.441 kip
 π ( 0.5 )2 
P sin 30°
V=
≤ 8
 kip
2
 4 
P ≤ 6.283 kip
Pmax = 5.44 kip ................................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.58
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-42
σn =
P
1 + cos 2 ( −θ )  ≤ 5 (106 ) N m 2
2 ( 0.050 )( 0.100 )
τn =
−P
sin 2 ( −θ ) ≤ 3 (106 ) N m 2
2 ( 0.050 )( 0.100 )
Optimum angle occurs when
σ n = 5 (106 ) N m 2
( )
and τ n = 3 10
6
N m2
sin 2θ
30
=
1 + cos 2θ 50
Solve by trial and error to get
θ = 30.964°
(a)
φ = 90° − θ = 59.036° .................................................................................................... Ans.
(b)
P = 34.0 (103 ) N = 34.0 kN ........................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.59
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-43
σn =
P
1 + cos 2 ( −θ )  psi
2 ( 3 × 3) 
τn =
−P
sin 2 ( −θ ) psi
2 ( 3 × 3)
σn
800
τn
500
When
=
P
[1 + cos 2θ ]
14, 400
=
P
sin 2θ
9000
θ = 37°
the shear stress reaches its maximum value
(τ n
500 = 1) first at which point
Pmax ≅ 9.36 kip ................................................................................................................ Ans.
When
θ = 25°
the normal stress reaches its maximum value
(σ n
800 = 1) first at which point
Pmax ≅ 8.77 kip ................................................................................................................ Ans.
For simultaneous control
σn
800
=
τ
P
P
sin 2θ = n = 1
[1 + cos 2θ ] =
14, 400
9000
500
sin 2θ
2sin θ cos θ
9000
=
= tan θ =
2
1 + cos 2θ
2 cos θ
14, 400
θ = 32.01°
........................................................................................................................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
2-44
σn =
RILEY, STURGES AND MORRIS
250 (103 )
1 + cos 2 ( −θ )  N m 2
2 ( 0.100 )( 0.025 )
θ = 90° − φ
= 50 (1 + cos 2θ ) MPa
τn =
−250 (103 )
2 ( 0.100 )( 0.025 )
sin 2 ( −θ ) N m 2
= 50sin 2θ MPa
For
σ n ≤ 80 MPa
and τ n ≤ 60 MPa
P = 250 kN
30° ≤ φ ≤ 63°
P = 305 kN
30° ≤ φ ≤ 40°
50° ≤ φ ≤ 54°
P = 350 kN
φ < 30°
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-45*
The given values are
σ x = 20 ksi
σ y = −10 ksi
τ xy = 0 ksi
θ ab = −26°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( 20 ) cos 2 ( −26° ) + ( −10 ) sin 2 ( −26° ) + 2 ( 0 ) sin ( −26° ) cos ( −26° )
σ ab = +14.23 ksi = 14.23 ksi (T) ................................................................................ Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − ( 20 ) − ( −10 )  sin ( −26° ) cos ( −26° ) + ( 0 ) cos 2 ( −26° ) − sin 2 ( −26° ) 
τ ab = +11.82 ksi .............................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.62
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-46*
The given values are
σ x = 95 MPa
σ y = 125 MPa
τ xy = 0 MPa
θ ab = 110°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( 95 ) cos 2 (110° ) + (125) sin 2 (110° ) + 2 ( 0 ) sin (110° ) cos (110° )
σ ab = +121.5 MPa = 121.5 MPa (T) ......................................................................... Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − ( 95 ) − (125 )  sin (110° ) cos (110° ) + ( 0 ) cos 2 (110° ) − sin 2 (110° ) 
τ ab = −9.64 MPa
............................................................................................................ Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.63
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-47
The given values are
σ x = 0 ksi
σ y = 0 ksi
τ xy = 15 ksi
θ ab = 30°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( 0 ) cos 2 ( 30° ) + ( 0 ) sin 2 ( 30° ) + 2 (15 ) sin ( 30° ) cos ( 30° )
σ ab = +12.99 ksi = 12.99 ksi (T)
............................................................................... Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − ( 0 ) − ( 0 )  sin ( 30° ) cos ( 30° ) + (15 ) cos 2 ( 30° ) − sin 2 ( 30° ) 
τ ab = +7.50 ksi ................................................................................................................ Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.64
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-48*
The given values are
σ x = −65 MPa
σ y = −125 MPa
τ xy = 75 MPa
θ ab = 145°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( −65 ) cos 2 (145° ) + ( −125 ) sin 2 (145° ) + 2 ( 75 ) sin (145° ) cos (145° )
σ ab = −155.2 MPa = 155.2 MPa (C) ........................................................................ Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − ( −65 ) − ( −125 )  sin (145° ) cos (145° ) + ( 75 ) cos 2 (145° ) − sin 2 (145° ) 
τ ab = +53.8 MPa
............................................................................................................ Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.65
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-49
The given values are
σ x = 18 ksi
σ y = 6 ksi
τ xy = 15 ksi
θ ab = 155°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= (18 ) cos 2 (155° ) + ( 6 ) sin 2 (155° ) + 2 (15 ) sin (155° ) cos (155° )
σ ab = +4.37 ksi = 4.37 ksi (T) .................................................................................... Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − (18 ) − ( 6 )  sin (155° ) cos (155° ) + (15 )  cos 2 (155° ) − sin 2 (155° ) 
τ ab = +14.24 ksi .............................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.66
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-50
The given values are
σ x = −170 MPa
σ y = 0 MPa
τ xy = −70 MPa
θ ab = 145°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( −170 ) cos 2 (145° ) + ( 0 ) sin 2 (145° ) + 2 ( −70 ) sin (145° ) cos (145° )
σ ab = −48.3 MPa = 48.3 MPa (C) ............................................................................. Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − ( −170 ) − ( 0 )  sin (145° ) cos (145° ) + ( −70 )  cos 2 (145° ) − sin 2 (145° ) 
τ ab = −103.8 MPa
.......................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.67
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-51*
The given values are
σ x = 0 psi
σy =
−5000
= −312.5 psi
( 4 )( 4 )
τ xy = 0 psi
θ = 166°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( 0 ) cos 2 (166° ) + ( −312.5 ) sin 2 (166° ) + 2 ( 0 ) sin (166° ) cos (166° )
σ n = −18.29 psi = 18.29 psi (C) ................................................................................. Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − ( 0 ) − ( −312.5 )  sin (166° ) cos (166° ) + ( 0 )  cos 2 (166° ) − sin 2 (166° ) 
τ n = +73.4 psi
................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.68
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-52*
The given values are
σx =
400 (103 )
( 0.100 )( 0.040 )
= 100 (106 ) N m 2 = 100 MPa
σ y = 0 MPa
τ xy = 0 MPa
θ = −33°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= (100 ) cos 2 ( −33° ) + ( 0 ) sin 2 ( −33° ) + 2 ( 0 ) sin ( −33° ) cos ( −33° )
σ = +70.3 MPa = 70.3 MPa (T) ............................................................................... Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − (100 ) − ( 0 )  sin ( −33° ) cos ( −33° ) + ( 0 ) cos 2 ( −33° ) − sin 2 ( −33° ) 
τ ab = +45.7 MPa
............................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.69
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-53
The given values are
σ x = 0 psi
σy =
−P
psi
9
τ xy = 0 ksi
θ aa = tan −1
4
= 53.130°
3
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( 0 ) + ( P 9 ) sin 2 ( 53.13° ) + ( 0 ) ≤ 800 psi
P ≤ 11, 250 lb
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= ( P 9 ) sin ( 53.13° ) cos ( 53.13° ) + ( 0 ) ≤ 500 psi
P ≤ 9375 lb
P ≤ 9.37 kip ..................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.70
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-54
The given values are
σ x = τ xy = 0 MPa
σy =
−P
= −50 P N m 2
( 0.100 )( 0.200 )
θ ab = −35°
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − ( 0 ) − ( −50 P )  sin ( −35° ) cos ( −35° ) + ( 0 ) = 15 (106 ) N m 2
(a)
P = 638.51(103 ) N ≅ 639 kN (C) ............................................................................. Ans.
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( 0 ) + ( −50 )( 638.51) (103 ) sin 2 ( −35° ) + ( 0 )
(b)
σ ab = −10.52 (106 ) N m 2 = 10.52 MPa (C) .......................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.71
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-55*
The given values are
σ x = τ xy = 0 ksi
σ y = −32 b 2 ksi
θ = −20°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( 0 ) + ( −32 b 2 ) sin 2 ( −20° ) + ( 0 ) ≤ 3.5 ksi
b ≥ 1.034 in.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − ( 0 ) − ( −32 b 2 )  sin ( −20° ) cos ( −20° ) + ( 0 ) ≤ 0.8 ksi
b ≥ 3.59 in. ....................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.72
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-56
The given values are
σx =
−P
= −41.667 P N m 2
( 0.2 )( 0.12 )
σ y = τ xy = 0 MPa
θ = −30°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( −41.667 P ) cos 2 ( −30° ) + ( 0 ) + ( 0 ) ≤ 13.60 (106 ) N m 2
P ≤ 435.2 (103 ) N
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − ( −41.667 P ) − ( 0 )  sin ( −30° ) cos ( −30° ) + ( 0 ) ≤ 5.25 (106 ) N m 2
P ≤ 291(103 ) N
Pmax = 291 kN .................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.73
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-57*
The given values are
σ x = 0 ksi
θ = − tan −1
4
= −53.130°
3
sin θ = −0.8000
cos θ = 0.6000
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( 0 ) + σ y ( −0.8 ) + 2τ xy ( −0.8 )( 0.6 ) = 4800 psi
2
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
2
2
= − ( 0 ) − σ y  ( −0.8 )( 0.6 ) + τ xy ( 0.6 ) − ( −0.8 )  = 1500 psi


0.64σ y − 0.96τ xy = 4800 psi
−0.48σ y − 0.28τ xy = 1500 psi
(a)
τ xy = −5100 psi ............................................................................................................... Ans.
(b)
σ y = −150 psi = 150 psi (C) ........................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.74
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-58*
The given values are
σ y = 0 MPa
τ xy = 25 MPa
θ ab = 90° + tan −1
5
= 112.620°
12
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= σ x cos 2 θ ab + ( 0 ) + 2 ( 25 ) sin θ ab cos θ ab = 15 MPa
(a)
σ x = 221.398 MPa ≅ 221 MPa (T) ........................................................................... Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − ( 221.398 ) − ( 0 )  sin θ ab cos θ ab + ( 25 ) ( cos 2 θ ab − sin 2 θ ab )
(b)
τ ab = +61.0 MPa
............................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.75
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-59
(a) The given values are
σ x = 18 ksi
σ y = 13 ksi
τ xy = 6 ksi
θ ab = −19°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= (18 ) cos 2 ( −19° ) + (13) sin 2 ( −19° ) + 2 ( 6 ) sin ( −19° ) cos ( −19° )
σ ab = +13.78 ksi = 13.78 ksi (T) ................................................................................ Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − (18 ) − (13)  sin ( −19° ) cos ( −19° ) + ( 6 ) cos 2 ( −19° ) − sin 2 ( −19° ) 
τ ab = +6.27 ksi ................................................................................................................ Ans.
(b) The given values are
σ x = 18 ksi
σ y = 13 ksi
τ xy = 6 ksi
θ n = 26°
θt = 116°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= (18 ) cos 2 ( 26° ) + (13) sin 2 ( 26° ) + 2 ( 6 ) sin ( 26° ) cos ( 26° )
σ n = +21.77 ksi = 21.77 ksi (T)
................................................................................ Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − (18 ) − (13)  sin ( 26° ) cos ( 26° ) + ( 6 )  cos 2 ( 26° ) − sin 2 ( 26° ) 
τ nt = +1.724 ksi
.............................................................................................................. Ans.
σ t = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= (18 ) cos 2 (116° ) + (13) sin 2 (116° ) + 2 ( 6 ) sin (116° ) cos (116° )
σ t = +9.23 ksi = 9.23 ksi (T) ...................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.76
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-60*
(a) Using x-y-coordinates rotated to align with the n-t-coordinates
σ n = σ x = 200 MPa
σ t = σ y = 50 MPa
τ nt = τ xy = 0 MPa
θ aa = 135°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( 200 ) cos 2 (135° ) + ( 50 ) sin 2 (135° ) + ( 0 )
σ aa = +125.0 MPa = 125.0 MPa (T)
........................................................................ Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − ( 200 ) − ( 50 )  sin (135° ) cos (135° ) + ( 0 )
τ aa = +75.0 MPa
............................................................................................................ Ans.
(b) Using the original coordinate system with the x- and y-axes horizontal and vertical
θ n = 18°
σ n = 200 MPa
σ t = 50 MPa
τ nt = 0 MPa
σ x = 185.7 MPa (T)
θt = 108°
200 = σ x cos 2 (18° ) + σ y sin 2 (18° ) + 2τ xy sin (18° ) cos (18° )
50 = σ x cos 2 (108° ) + σ y sin 2 (108° ) + 2τ xy sin (108° ) cos (108° )
0 = − (σ x − σ y ) sin (18° ) cos (18° ) + τ xy cos 2 (18° ) − sin 2 (18° ) 
.........
σ y = 64.3 MPa (T)
........... τ xy = 44.1 MPa ........... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.77
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-61
The given values are
σ x = 8 ksi
σ n = 8 ksi
(a)
σ y = 0 ksi
θ aa = 90° + tan −1
3
= 126.870°
4
8 = ( 8 ) cos 2 126.870° + ( 0 ) + 2τ xy sin126.870° cos126.870°
τ xy = τ h = τ v = −5.3333 ksi ≅ −5.33 ksi
................................................................... Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − ( 8 ) − ( 0 )  sin θ aa cos θ aa + ( −5.3333) cos 2 θ aa − sin 2 θ aa 
(b)
τ a = +5.33 ksi
................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-62
The given values are
σ y = 0 MPa
σ n = 15 MPa
(a)
(b)
τ xy = 25 MPa
θ ab = 90° + tan −1
5
= 112.620°
12
15 = σ x cos 2 (112.620° ) + ( 0 ) + 2 ( 25 ) sin (112.620° ) cos (112.620° )
σ x = +221.40 MPa ≅ 221 MPa (T) ........................................................................... Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − ( −221.40 ) − ( 0 )  sin θ ab cos θ ab + ( 25 )  cos 2 θ ab − sin 2 θ ab 
τ ab = +61.0 MPa
............................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-63*
The given values are
σ y = 2σ x
τ xy = 0 ksi
θ = 35°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= σ x cos 2 ( 35° ) + ( 2σ x ) sin 2 ( 35° ) + ( 0 ) ≤ 10 ksi
σ x ≤ 7.52451 ksi
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos2 θ − sin 2 θ )
= − σ x − ( 2σ x )  sin ( 35° ) cos ( 35° ) + ( 0 ) ≤ 7 ksi
σ x ≤ 14.8985 ksi
(σ x )max = 7.52 ksi .......................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
2-64
The given values are
θbb = 126.870°
RILEY, STURGES AND MORRIS
θ cc = 36.870°
σ bb = 125 MPa
125 = σ x cos 2 (126.87° ) + σ y sin 2 (126.87° ) + ( 0 )
σ cc = −225 MPa
−225 = σ x cos 2 ( 36.87° ) + σ y sin 2 ( 36.87° ) + ( 0 )
τ bc = 0 MPa
0 = − (σ x − σ y ) sin (126.87° ) cos (126.87° ) + ( 0 )
0.3600σ x + 0.6400σ y − 0.9600τ xy = 125
0.6400σ x + 0.3600σ y + 0.9600τ xy = −225
0.4800σ x − 0.4800σ y − 0.2800τ xy = 0
(a)
σ x = −99.0 MPa = 99.0 MPa (C) .............................................................................. Ans.
τ xy = −168.0 MPa
(b)
.......................................................................................................... Ans.
σ y = −1.000 MPa = 1.000 MPa (C)
τ xy = −168.0 MPa
......................................................................... Ans.
.......................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-65
The given values are
σx =
P
ksi
4
σ y = 0 ksi
σ ab = 12 ksi
12 = σ x cos 2 θ + ( 0 ) + ( 0 )
τ ab = −9 ksi
−9 = − σ x − ( 0 )  sin θ cos θ + ( 0 )
τ xy = 0 ksi
σ x sin θ cos θ
9
= tan θ =
2
σ x cos θ
12
θ = 36.870° ...................................................... Ans.
σ x = 18.750 ksi = P 4
P = 75.0 kip .................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-66
(a)
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
σ n = ( 60 ) cos2 θ + ( 0 ) + 2 ( −40 ) sin θ cos θ  MPa
.............................................. Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
{
(b)
}
τ nt = − ( 60 ) − ( 0 )  sin θ cos θ + ( −40 ) ( cos 2 θ − sin 2 θ ) MPa ........................ Ans.
At θ = 153.435°
σ max = 80 MPa .........................................Ans.
τ = 0 MPa .................................................Ans.
At θ = 63.435°
σ min = −20 MPa .......................................Ans.
(c)
τ = 0 MPa .................................................Ans.
At θ = 108.435°
τ max = 50 MPa ..........................................Ans.
σ = 30 MPa ..............................................Ans.
At θ = 18.435°
τ min = −50 MPa .............................................................................................................. Ans.
σ = 30 MPa ..................................................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-67
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
(a)
σ n = (18 ) cos 2 θ + (13) sin 2 θ + 2 ( 6 ) sin θ cos θ  ksi ..................................... Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
{
}
= − (18 ) − (13)  sin θ cos θ + ( 6 ) ( cos 2 θ − sin 2 θ ) ksi ................................ Ans.
(b) At
θ = 33.69°
σ max = 22 ksi
............................................Ans.
τ = 0 MPa .................................................Ans.
At θ = 123.69°
σ min = 9 ksi ...............................................Ans.
(c)
τ = 0 MPa .................................................Ans.
At θ = 168.69°
τ max = 6.5 ksi ............................................Ans.
σ = 15.5 ksi ..............................................Ans.
At θ = 78.69°
τ max = −6.5 ksi ................................................................................................................. Ans.
σ = 15.5 ksi
..................................................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-68
Note that the θ on the figure is for the surface rather
than for the normal to the surface. Therefore, need to
use φ = 90° − θ in the transformation equations.
σ n = σ x cos 2 θ + σ y sin 2 θ
+2τ xy sin θ cos θ
= ( −10 ) cos 2 φ + ( −70 ) sin 2 φ
+2 ( 40 ) sin φ cos φ
τ nt = − (σ x − σ y ) sin θ cos θ
+τ xy ( cos 2 θ − sin 2 θ )
= − ( −10 ) − ( −70 )  sin φ cos φ
+40 ( cos 2 φ − sin 2 φ )
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-69
The given values are
σ x = −15 ksi
1
2
θ p = tan −1
When
σ y = 10 ksi
2τ xy
σ x −σ y
=
τ xy = 8 ksi
2 (8)
1
= −16.310°, 73.690°
tan −1
2
( −15) − (10 )
θ p = −16.310°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( −15 ) cos 2 θ p + (10 ) sin 2 θ p + 2 ( 8 ) sin θ p cos θ p
= −17.341 ksi = σ p 2
σ p1 = σ x + σ y − σ p 2 = 12.341 ksi
τ max = τ p = (σ p1 − σ p 2 ) 2 = 14.841 ksi
σ n 45 = (σ p1 + σ p 2 ) 2 = −2.500 ksi
(a)
σ p1 = 12.34 ksi (T)
73.69° ..................................................................................... Ans.
σ p 2 = 17.34 ksi (C)
16.31° ..................................................................................... Ans.
τ max = τ p = 14.841 ksi .................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-70
The given values are
σ x = 50 MPa
1
2
θ p = tan −1
When
σ y = 20 MPa
2τ xy
σ x −σ y
=
τ xy = 40 MPa
2 ( 40 )
1
= 34.722°, − 55.278°
tan −1
2
( 50 ) − ( 20 )
θ p = 34.722°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( 50 ) cos 2 θ p + ( 20 ) sin 2 θ p + 2 ( 40 ) sin θ p cos θ p
= 77.720 MPa = σ p1
σ p 2 = σ x + σ y − σ p1 = −7.720 MPa
τ max = τ p = (σ p1 − σ p 2 ) 2 = 42.720 MPa
σ n 45 = (σ p1 + σ p 2 ) 2 = 35.00 MPa
(a)
σ p1 = 77.7 MPa (T)
34.72° ................................................................................... Ans.
σ p 2 = 7.72 MPa (C)
55.28° ................................................................................... Ans.
τ max = τ p = 42.7 MPa .................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-71
The given values are
σ y = 10 ksi
σ x = −15 ksi
1
2
θ p = tan −1
When
2τ xy
σ x −σ y
=
τ xy = 8 ksi
2 (8)
1
= −16.310°, 73.690°
tan −1
2
( −15) − (10 )
θ p = −16.310°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( −15 ) cos 2 θ p + (10 ) sin 2 θ p + 2 ( 8 ) sin θ p cos θ p
= −17.341 ksi = σ p 2
σ p1 = σ x + σ y − σ p 2 = 12.341 ksi
τ max = τ p = (σ p1 − σ p 2 ) 2 = 14.841 ksi
σ n 45 = (σ p1 + σ p 2 ) 2 = −2.500 ksi
(a)
σ p1 = 12.34 ksi (T)
73.69° ..................................................................................... Ans.
σ p 2 = 17.34 ksi (C)
16.31° ..................................................................................... Ans.
τ max = τ p = 14.841 ksi .................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-72
The given values are
σ x = 200 MPa
1
2
θ p = tan −1
When
σ y = 0 MPa
2τ xy
σ x −σ y
=
τ xy = 25 MPa
2 ( 25 )
1
tan −1
= 7.018°, − 82.982°
2
( 200 ) − ( 0 )
θ p = 7.018°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( 200 ) cos 2 θ p + ( 0 ) + 2 ( 25 ) sin θ p cos θ p
= 203.078 MPa = σ p1
σ p 2 = σ x + σ y − σ p1 = −3.078 MPa
τ max = τ p = (σ p1 − σ p 2 ) 2 = 103.078 MPa
σ n 45 = (σ p1 + σ p 2 ) 2 = 100.00 MPa
(a)
σ p1 = 203 MPa (T)
7.02° ....................................................................................... Ans.
σ p 2 = 3.08 MPa (C)
82.98° ................................................................................... Ans.
τ max = τ p = 103.1 MPa
.................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-73
The given values are
σ x = 12 ksi
1
2
θ p = tan −1
When
σ y = −4 ksi
2τ xy
σ x −σ y
=
τ xy = −6 ksi
2 ( −6 )
1
= −18.435°, 71.565°
tan −1
2
(12 ) − ( −4 )
θ p = −18.435°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= (12 ) cos 2 θ p + ( −4 ) sin 2 θ p + 2 ( −6 ) sin θ p cos θ p
= 14.000 ksi = σ p1
σ p 2 = σ x + σ y − σ p1 = −6.000 ksi
τ max = τ p = (σ p1 − σ p 2 ) 2 = 10.000 ksi
σ n 45 = (σ p1 + σ p 2 ) 2 = 4.000 ksi
(a)
σ p1 = 14.00 ksi (T)
18.43° ..................................................................................... Ans.
σ p 2 = 6.00 ksi (C)
71.57° ...................................................................................... Ans.
τ max = τ p = 10.00 ksi ...................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-74
The given values are
σ x = 75 MPa
1
2
θ p = tan −1
When
σ y = −25 MPa
2τ xy
σ x −σ y
=
τ xy = −35 MPa
2 ( −35 )
1
= −17.496°, 72.504°
tan −1
2
( 75 ) − ( −25)
θ p = −17.496°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( 75 ) cos 2 θ p + ( −25 ) sin 2 θ p + 2 ( −35 ) sin θ p cos θ p
= 86.033 MPa = σ p1
σ p 2 = σ x + σ y − σ p1 = −36.033 MPa
τ max = τ p = (σ p1 − σ p 2 ) 2 = 61.033 MPa
σ n 45 = (σ p1 + σ p 2 ) 2 = 25.00 MPa
(a)
σ p1 = 86.0 MPa (T)
17.50° .................................................................................... Ans.
σ p 2 = 36.0 MPa (C)
72.50° ................................................................................... Ans.
τ max = τ p = 61.0 MPa .................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-75
The given values are
σ x = 25 ksi
1
2
θ p = tan −1
When
σ y = 12 ksi
2τ xy
σ x −σ y
=
τ xy = −10 ksi
2 ( −10 )
1
= −28.488°, 61.512°
tan −1
2
( 25) − (12 )
θ p = −28.488°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( 25 ) cos 2 θ p + (12 ) sin 2 θ p + 2 ( −10 ) sin θ p cos θ p
= 30.427 ksi = σ p1
σ p 2 = σ x + σ y − σ p1 = 6.573 ksi
(a)
τ p = (σ p1 − σ p 2 ) 2 = 11.927 ksi
σ n = 18.500 ksi
τ max = (σ max − σ min ) 2 = 15.213 ksi
σ n = 15.213 ksi
σ p1 = 30.43 ksi (T)
28.49° ..................................................Ans.
σ p 2 = 6.57 ksi (T)
61.51° ...................................................Ans.
τ p = 11.93 ksi ............................. τ max = 15.21 ksi ...................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-76
The given values are
σ x = 36 MPa
1
2
θ p = tan −1
When
σ y = 26 MPa
2τ xy
σ x −σ y
=
τ xy = 12 MPa
2 (12 )
1
tan −1
= 33.690°, − 56.310°
2
( 36 ) − ( 26 )
θ p = 33.690°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( 36 ) cos 2 θ p + ( 26 ) sin 2 θ p + 2 (12 ) sin θ p cos θ p
= 44.000 MPa = σ p1
σ p 2 = σ x + σ y − σ p1 = 18.000 MPa
(a)
τ p = (σ p1 − σ p 2 ) 2 = 13.00 MPa
σ n = 31.00 MPa
τ max = (σ max − σ min ) 2 = 22.00 MPa
σ n = 22.00 MPa
σ p1 = 44.0 MPa (T)
σ p 2 = 18.00 MPa (T)
τ p = 13.00 MPa
33.69° ................................................Ans.
56.31° ................................................................................. Ans.
......................... τ max = 22.0 MPa .................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-77
The given values are
σ x = −2 ksi
1
2
θ p = tan −1
When
σ y = −14 ksi
2τ xy
σ x −σ y
=
τ xy = −8 ksi
2 ( −8 )
1
= −26.565°, 63.435°
tan −1
2
( −2 ) − ( −14 )
θ p = −26.565°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( −2 ) cos 2 θ p + ( −14 ) sin 2 θ p + 2 ( −8 ) sin θ p cos θ p
= 2.000 ksi = σ p1
σ p 2 = σ x + σ y − σ p1 = −18.00 ksi
τ max = τ p = (σ p1 − σ p 2 ) 2 = 10.00 ksi
σ n 45 = (σ p1 + σ p 2 ) 2 = −8.00 ksi
(a)
σ p1 = 2.00 ksi (T)
26.57° ....................................................................................... Ans.
σ p 2 = 18.00 ksi (C)
63.43° .................................................................................... Ans.
τ max = τ p = 10.00 ksi ...................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-78
The given values are
σ x = −170 MPa
1
2
θ p = tan −1
When
2τ xy
σ x −σ y
σ y = 0 MPa
=
τ xy = −70 MPa
2 ( −70 )
1
= 19.736°, − 70.264°
tan −1
2
( −170 ) − ( 0 )
θ p = 19.736°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( −170 ) cos 2 θ p + ( 0 ) + 2 ( −70 ) sin θ p cos θ p
= −195.114 MPa = σ p 2
σ p1 = σ x + σ y − σ p 2 = 25.114 MPa
τ max = τ p = (σ p1 − σ p 2 ) 2 = 110.114 MPa
σ n 45 = (σ p1 + σ p 2 ) 2 = −85.00 MPa
σ p1 = 25.1 MPa (T)
70.26° ................................................................................... Ans.
σ p 2 = 195.1 MPa (C)
19.74° ................................................................................. Ans.
τ max = τ p = 110.1 MPa
.................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-79
The given values are
σ x = σ y = 0 ksi
1
2
θ p = tan −1
When
2τ xy
σ x −σ y
τ xy = 15 ksi
=
2 (15 )
1
tan −1
= 45.00°, − 45.00°
2
(0) − (0)
θ p = 45.00°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( 0 ) + ( 0 ) + 2 (15 ) sin θ p cos θ p
= 15.00 ksi = σ p1
σ p 2 = σ x + σ y − σ p1 = −15.00 ksi
τ max = τ p = (σ p1 − σ p 2 ) 2 = 15.00 ksi
σ n 45 = (σ p1 + σ p 2 ) 2 = 0 ksi
σ p1 = 15.00 ksi (T)
45.00° ..................................................................................... Ans.
σ p 2 = 15.00 ksi (C)
45.00° .................................................................................... Ans.
τ max = τ p = 15.00 ksi ...................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-80
(a) The given values are
σ x = −27 MPa
σ y = 45 MPa
τ xy = 0 MPa
θ ab = −57°
σ n = ( −27 ) cos 2 ( −57° ) + ( 45 ) sin 2 ( −57° ) + ( 0 )
σ ab = 23.6 MPa(T) ...................................................... Ans.
τ nt = − ( −27 ) − ( 45 )  sin ( −57° ) cos ( −57° ) + ( 0 )
τ ab = −32.9 MPa .......................................................... Ans.
(b) Since there are no shear stresses on the horizontal and vertical surfaces, they are principal surfaces and the
stresses on them are principal stresses.
σ p1 = σ y = 45 MPa
σ p 2 = σ x = −27 MPa
τ max = τ p = (σ p1 − σ p 2 ) 2 = 36.00 MPa
σ n 45 = (σ p1 + σ p 2 ) 2 = 9.00 MPa
σ p1 = 45.0 MPa (T) ↑ .................................................................................................. Ans.
σ p 2 = 27.0 MPa (C) → ............................................................................................... Ans.
τ max = τ p = 36.0 MPa
.................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-81
The given values are
σ x = (T)
τ xy = 8 ksi
σ p1 = 12 ksi
σ p 2 = −20 ksi
σ x + σ y = σ p1 + σ p 2 = −8 ksi
We know that
σ p1 =
and that
σ y = (C)
σx +σ y
2
 σ −σ y 
 σ −σ y 
−8
2
2
+  x
+  x
 + τ xy =
 + ( 8 ) = 12 ksi
2
 2 
 2 
2
2
σ x − σ y = 27.71281 ksi
which gives
Therefore
σ x = 9.85641 ksi
1
2
θ p = tan −1
When
2τ xy
σ x −σ y
σ y = −17.85641 ksi
=
2 (8)
1
= 15.00°, − 75.00°
tan −1
2
27.71281
θ p = 15.00°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( 9.85641) cos 2 θ p + ( −17.85641) sin 2 θ p + 2 ( 8 ) sin θ p cos θ p
= 12 ksi = σ p1
σ x = 9.86 ksi (T) ............................................................................................................ Ans.
σ y = 17.86 ksi (C) .......................................................................................................... Ans.
θ p1 = 15.00° ...................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-83
The given values are
σ y = −σ C
σ x = −4σ C = 4σ y
τ xy = 0 ksi
θ = −30°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( −4σ C ) cos 2 ( −30° ) + ( −σ C ) sin 2 ( −30° ) + ( 0 ) ≤ −300 psi
σ C ≤ 92.3 psi
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − ( −4σ C ) − ( −σ C )  sin ( −30° ) cos ( −30° ) + ( 0 ) ≤ 125 psi
σ C ≤ 96.2 psi
(σ C )max = 92.3 psi .......................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-84
(a) The given values are
σ x = −10 MPa
σ y = −70 MPa
τ xy = 40 MPa
θ ab = −28°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( −10 ) cos 2 ( −28° ) + ( −70 ) sin 2 ( −28° ) + 2 ( 40 ) sin ( −28° ) cos ( −28° )
σ ab = −56.4 MPa = 56.4 MPa (C) ............................................................................. Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − ( −10 ) − ( −70 )  sin ( −28° ) cos ( −28° ) + ( 40 ) cos 2 ( −28° ) − sin 2 ( −28° ) 
τ ab = 47.2 MPa
1
2
..........................................................................................Ans.
2τ xy
(b)
θ p = tan −1
When
θ p = 26.565°
σ x −σ y
=
2 ( 40 )
1
= 26.565°, − 63.435°
tan −1
2
( −10 ) − ( −70 )
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( −10 ) cos 2 θ p + ( −70 ) sin 2 θ p + 2 ( 40 ) sin θ p cos θ p
= 10.00 MPa = σ p1
σ p 2 = σ x + σ y − σ p1 = −90.00 MPa
τ max = τ p = (σ p1 − σ p 2 ) 2 = 50.00 MPa
σ n 45 = (σ p1 + σ p 2 ) 2 = −40.00 MPa
σ p1 = 10.00 MPa (T)
26.565° ............................................................................... Ans.
σ p 2 = 90.0 MPa (C)
63.565° ................................................................................ Ans.
τ max = τ p = 50.0 MPa
.................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-85
The given values for use in
drawing Mohr’s circle are
σ x = σ p1 = 10 ksi
σ y = σ p 2 = 0 ksi
σ z = σ p 3 = 0 ksi
10 + 0
= 5 ksi
2
10 − 0
R=
= 5 ksi
2
σ aa = 5 + 5cos 90° = 5 ksi (T) ................................................................Ans.
a=
τ aa = 5sin 90° = 5 ksi (CW) = −5 ksi ...................................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-86
The given values for use in
drawing Mohr’s circle are
σ x = σ y = 0 MPa
τ xy = 40 MPa
σ z = σ p 3 = 0 MPa
R = 40 MPa
σ aa = σ p 2 = −40 MPa = 40 MPa (C) ...................................................Ans.
τ aa = 0 MPa
................................................................................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-87
The given values for use in
drawing Mohr’s circle are
σ x = σ p1 = 12 ksi
σ y = σ p 2 = −16 ksi
σ z = σ p 3 = 0 ksi
a=
R=
σ ab
12 + ( −16 )
= −2 ksi
2
12 − ( −16 )
= 14 ksi
2
= −2 + 14 cos 52° = 6.62 ksi (T) .....................................................Ans.
τ ab = 14sin 52° = 11.03 ksi (CCW) = +11.03 ksi
.............................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-88
The given values for use in drawing Mohr’s circle are
σ x = σ p 2 = −27 MPa
σ y = σ p1 = 45 MPa
σ z = σ p 3 = 0 MPa
a=
R=
σ ab
( −27 ) + 45 = 9 MPa
2
45 − ( −27 )
= 36 MPa
2
= 9 + 36 cos 66° = 23.6 MPa (T) ....................................................Ans.
τ ab = 36sin 66° = 32.9 MPa (CW) = −32.9 MPa
.............................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-89
The given values for use in drawing Mohr’s circle are
σ x = 25 ksi
σ y = 7 ksi
τ xy = 12 ksi
σ z = σ p 3 = 0 ksi
a=
25 + 7
= 16.00 ksi
2
R = 92 + 122 = 15.00 ksi
φ
1
12
tan −1 = 26.57° (CCW)
2 2
9
σ p1 = 16.00 + 15.00 = 31.0 ksi (T) 26.57° ......................................................... Ans.
θ p1 =
=
σ p 2 = 16.00 − 15.00 = 1.0 ksi (T)
63.43° ............................................................ Ans.
τ p = R = 15 ksi ................................................................................................................ Ans.
τ max = (σ max − σ min ) 2 = ( 31 − 0 ) 2 = 15.5 ksi (out of plane) ............................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-90
The given values for use in
drawing Mohr’s circle are
σ x = 50 MPa
σ y = 20 MPa
τ xy = 40 MPa
σ z = σ p 3 = 0 MPa
a=
50 + 20
= 35.00 MPa
2
R = 152 + 402 = 42.72 MPa
τ max
φ
1
40
= 34.72° (CCW)
tan −1
2 2
15
= τ p = R = 42.7 MPa .........................................................Ans.
θ p1 =
=
σ p1 = 35.00 + 42.7 = 77.7 MPa (T)
34.72° ....................Ans.
σ p 2 = 35.00 − 42.72 = −7.72 MPa = 7.72 MPa (C)
55.28° .......................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-91
The given values for use in drawing Mohr’s circle are
σ x = −15 ksi
σ y = 10 ksi
τ xy = 8 ksi
σ z = σ p 3 = 0 ksi
a=
( −15) + 10 = −2.50 ksi
2
R = 12.52 + 82 = 14.841 ksi
φ
1
8
= 16.31° (CW)
tan −1
2 2
12.5
σ p1 = −2.50 + 14.84 = 12.34 ksi (T) 73.69° ........... Ans.
θ p2 =
=
σ p 2 = −2.50 − 14.84 = −17.34 ksi = 17.34 ksi (C)
16.31° ............................. Ans.
τ max = τ p = R = 14.84 ksi .............................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-92
The given values for use in drawing Mohr’s circle are
σ x = 50 MPa
σ y = 0 MPa
τ xy = 25 MPa
σ z = σ p 3 = 0 MPa
a = ( 50 + 0 ) 2 = 25.00 MPa
R = 252 + 252 = 35.355 MPa
θ p1 =
φ
2
=
1
25
= 22.50° (CCW)
tan −1
2
25
τ max = τ p = R = 35.4 MPa
.........................................................Ans.
σ p1 = 25.0 + 35.4 = 60.4 MPa (T)
22.50° .......................Ans.
σ p 2 = 25.00 − 35.36 = −10.36 MPa = 10.36 MPa (C)
67.50° ...................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-93
The given values for use in
drawing Mohr’s circle are
σ x = 25 ksi
σ y = 12 ksi
τ xy = −10 ksi
σ z = σ p 3 = 0 ksi
a=
25 + 12
= 18.50 ksi
2
R = 6.52 + 102 = 11.927 ksi
φ
1
10
= 28.488° (CW)
tan −1
2 2
6.5
σ p1 = 18.5 + 11.9 = 30.4 ksi (T) 28.49° ................... Ans.
θ p1 =
=
σ p 2 = 18.50 − 11.93 = 6.57 ksi (T)
61.51° .............. Ans.
τ p = R = 11.93 ksi .....................................................................................Ans.
τ max = (σ max − σ min ) 2 = ( 30.427 − 0 ) 2 ≅ 15.21 ksi (out of plane) .................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-94
The given values for use in
drawing Mohr’s circle are
σ x = 80 MPa
σ y = −100 MPa
τ xy = −60 MPa
σ z = σ p 3 = 0 MPa
a=
80 − 100
= −10.00 MPa
2
R = 902 + 602 = 108.17 MPa
θ p1 =
(a)
φ
2
=
1
60
= 16.845° (CW)
tan −1
2
90
σ p1 = −10.0 + 108.2 = 98.2 MPa (T)
16.85° ...................................................... Ans.
σ p 2 = −10.0 − 108.2 = −118.2 MPa = 118.2 MPa (C)
τ max = τ p = R = 108.2 MPa
(b)
73.15° ...................... Ans.
.....................................................................Ans.
σ ab = −10 − 108.17 cos 62.310° = −60.3 MPa = 60.3 MPa (C)
τ ab = 108.17 sin 62.310° = 95.8 MPa (CW) = −95.8 MPa
....Ans.
.............Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-95
The given values for use in
drawing Mohr’s circle are
σ x = 8 ksi
σ y = 0 ksi
τ xy = −5 ksi
σ z = σ p 3 = 0 ksi
a=
8+0
= 4.00 ksi
2
R = 42 + 52 = 6.403 ksi
(a)
φ
1
5
tan −1 = 25.670° (CW)
2 2
4
σ p1 = 4.00 + 6.40 = 10.40 ksi (T) 25.67° ......................................Ans.
θ p1 =
=
σ p 2 = 4.00 − 6.40 = −2.40 ksi = 2.40 ksi (C)
64.33° .................Ans.
τ max = τ p = R = 6.40 ksi ...........................................................................Ans.
(b)
σ ab = 4.00 + 6.403cos 54.920° = 7.68 ksi (T)
...................................Ans.
τ ab = 6.403sin 54.920° = 5.24 ksi (CCW) = +5.24 ksi ...................Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-96
The given values for use in
drawing Mohr’s circle are
σ x = 20 MPa
σ y = 120 MPa
τ xy = −80 MPa
σ z = σ p 3 = 0 MPa
a=
20 + 120
= 70.00 MPa
2
R = 502 + 802 = 94.340 MPa
θ p1 =
(a)
φ
2
=
1
80
= 28.997° (CCW)
tan −1
2
50
σ p1 = 70.00 + 94.3 = 164.3 MPa (T)
61.00° ...................................................... Ans.
σ p 2 = 70.00 − 94.3 = −24.3 MPa = 24.3 MPa (C)
τ max = τ p = R = 94.3 MPa
(b)
29.00° ............................. Ans.
................................................................... Ans.
σ ab = 70 + 94.340 cos 5.995° = 163.82 MPa (T) ........................... Ans.
τ ab = 94.340sin 5.995° = 9.85 MPa (CCW) = 9.85 MPa ........... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-97
The given values for use in
drawing Mohr’s circle are
σ x = 12 ksi
σ y = −6 ksi
τ xy = 10 ksi
σ z = σ p 3 = 0 ksi
a=
12 + ( −6 )
= 3.00 ksi
2
R = 92 + 102 = 13.454 ksi
(a)
φ
1
10
tan −1 = 24.006° (CCW)
2 2
9
σ p1 = 3.00 + 13.45 = 16.45 ksi (T) 24.01° .......................................................... Ans.
θ p1 =
=
σ p 2 = 3.00 − 13.45 = −10.45 ksi = 10.45 ksi (C)
65.99° ................................ Ans.
τ max = τ p = R = 13.45 ksi .............................................................................................. Ans.
(b)
σ ab = 3.00 − 13.454 cos 71.987° = −1.160 ksi = 1.160 ksi (C) ......Ans.
τ ab = 13.454sin 71.987° = 12.79 ksi (CCW) = 12.79 ksi
...............Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-98*
The given stress values are:
σ x = 40 MPa
σ y = −20 MPa
σ z = 20 MPa
τ xy = 40 MPa
τ yz = 0 MPa
τ zx = 30 MPa
θ x = 40°
θ y = 75°
θ z = 54°
Then
S x = σ x cos θ x + τ yx cos θ y + τ zx cos θ z
= 40 cos 40° + 40 cos 75° + 30 cos 54° = 58.628 MPa
S y = τ xy cos θ x + σ y cos θ y + τ zy cos θ z
= 40 cos 40° − 20 cos 75° + 0 = 25.465 MPa
S z = τ xz cos θ x + τ yz cos θ y + σ z cos θ z
= 30 cos 40° + 0 + 20 cos 54° = 34.737 MPa
S = S x2 + S y2 + S z2 =
( 58.628) + ( 25.465) + ( 34.737 )
2
2
2
= 72.749 MPa
σ n = S x cos θ x + S y cos θ y + S z cos θ z
= 58.628cos 40° + 25.465cos 75° + 34.737 cos 54°
σ n = 71.920 MPa ≅ 71.9 MPa (T) ............................................................................ Ans.
τ n = S 2 − σ n2 =
( 72.749 ) − ( 71.920 )
2
2
= 10.95 MPa ...................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-99*
The given stress values are:
σ x = 14 ksi
σ y = 12 ksi
σ z = 10 ksi
τ xy = 4 ksi
τ yz = −4 ksi
τ zx = 0 ksi
θ x = 40°
θ y = 60°
θ z = 66.2°
Then
S x = σ x cos θ x + τ yx cos θ y + τ zx cos θ z
= 14 cos 40° + 4 cos 60° + 0 = 12.7246 ksi
S y = τ xy cos θ x + σ y cos θ y + τ zy cos θ z
= 4 cos 40° + 12 cos 60° − 4 cos 66.2° = 7.4500 ksi
S z = τ xz cos θ x + τ yz cos θ y + σ z cos θ z
= 0 − 4 cos 60° + 10 cos 66.2° = 2.0355 ksi
S = S x2 + S y2 + S z2 =
(12.7246 ) + ( 7.4500 ) + ( 2.0355)
2
2
2
= 14.8849 ksi
σ n = S x cos θ x + S y cos θ y + S z cos θ z
= 12.7246 cos 40° + 7.4500 cos 60° + 2.0355cos 66.2°
σ n = 14.2939 ksi ≅ 14.29 ksi (T) ............................................................................... Ans.
τ n = S 2 − σ n2 =
(14.8849 ) − (14.2939 )
2
2
= 4.15 ksi ........................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-100
The given stress values are:
σ x = 60 MPa
σ y = 90 MPa
σ z = 60 MPa
τ xy = 120 MPa
τ yz = 75 MPa
τ zx = 90 MPa
θ x = 60°
θ y = 70°
θ z = 37.3°
Then
S x = σ x cos θ x + τ yx cos θ y + τ zx cos θ z
= 60 cos 60° + 120 cos 70° + 90 cos 37.3° = 142.6347 MPa
S y = τ xy cos θ x + σ y cos θ y + τ zy cos θ z
= 120 cos 60° + 90 cos 70° + 75cos 37.3° = 150.4421 MPa
S z = τ xz cos θ x + τ yz cos θ y + σ z cos θ z
= 90 cos 60° + 75cos 70° + 60 cos 37.3° = 118.3797 MPa
S = S x2 + S y2 + S z2 =
(142.6347 ) + (150.4421) + (118.3797 )
2
2
2
= 238.7283 MPa
σ n = S x cos θ x + S y cos θ y + S z cos θ z
= 142.6347 cos 60° + 150.4421cos 70° + 118.3797 cos 37.3°
σ n = 216.9390 MPa ≅ 217 MPa (T) ........................................................................ Ans.
τ n = S 2 − σ n2 =
( 238.7283) − ( 216.9390 )
2
2
= 99.6 MPa ............................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-101*
The given stress values are:
σ x = σ y = σ z = 0 ksi
τ xy = 6 ksi
θx = θ y = θz
τ yz = 10 ksi
τ zx = 8 ksi
But
cos 2 θ x + cos 2 θ y + cos 2 θ z = 1
Therefore
cos θ x = cos θ y = cos θ z = 1
3
S x = σ x cos θ x + τ yx cos θ y + τ zx cos θ z = ( 0 + 6 + 8)
3 = 8.0829 ksi
S y = τ xy cos θ x + σ y cos θ y + τ zy cos θ z = ( 6 + 0 + 10 )
3 = 9.2376 ksi
S z = τ xz cos θ x + τ yz cos θ y + σ z cos θ z = ( 8 + 10 + 0 )
3 = 10.3923 ksi
S = S x2 + S y2 + S z2 =
(8.0839 ) + ( 9.2376 ) + (10.3923)
2
2
2
= 16.0831 ksi
σ n = S x cos θ x + S y cos θ y + S z cos θ z = ( 8.0829 + 9.2376 + 10.3923)
3
σ n = 16.00 ksi = 16.00 ksi (T) .................................................................................... Ans.
τ n = S 2 − σ n2 =
(16.0831) − (16.00 )
2
2
= 1.633 ksi ........................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-102
The given stress values are:
σ x = 72 MPa
σ y = −32 MPa
σ z = 0 MPa
τ xy = 21 MPa
τ yz = 0 MPa
τ zx = 21 MPa
But
cos 2 θ x + cos 2 θ y + cos 2 θ z = 1
Therefore
cos θ x = cos θ y = cos θ z = 1
3
S x = σ x cos θ x + τ yx cos θ y + τ zx cos θ z = ( 72 + 21 + 21)
S y = τ xy cos θ x + σ y cos θ y + τ zy cos θ z = ( 21 − 32 + 0 )
S z = τ xz cos θ x + τ yz cos θ y + σ z cos θ z = ( 21 + 0 + 0 )
S = S x2 + S y2 + S z2 =
θx = θ y = θz
3 = 65.8179 MPa
3 = −6.3509 MPa
3 = 12.1244 MPa
( 65.8179 ) + ( 6.3509 ) + (12.1244 )
2
2
2
= 67.2260 MPa
σ n = S x cos θ x + S y cos θ y + S z cos θ z = ( 65.8179 − 6.3509 + 12.1244 )
3
σ n = 41.3333 MPa ≅ 41.3 MPa (T) .......................................................................... Ans.
τ n = S 2 − σ n2 =
( 67.2260 ) − ( 41.3333)
2
2
= 53.0 MPa ................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-103*
The given stress values are:
σ x = 12 ksi
τ xy = 8 ksi
σ y = −10 ksi
τ yz = −10 ksi
σ z = 8 ksi
τ zx = 12 ksi
σ x + σ y + σ z = 10 ksi
σ xσ y + σ yσ z + σ zσ x − τ xy2 − τ yz2 − τ zx2 = −412 ksi 2
σ xσ yσ z − σ xτ yz2 − σ yτ zx2 − σ zτ xy2 + 2τ xyτ yzτ zx = −3152 ksi3
σ 3p − (10 ) σ p2 + ( −412 ) σ p − ( −3152 ) = 0
σ p1 = σ max = 22.1706 ksi ≅ 22.2 ksi (T) .................................................................. Ans.
σ p 2 = σ int = 7.3013 ksi ≅ 7.30 ksi (T) ...................................................................... Ans.
σ p 3 = σ min = −19.4719 ksi ≅ 19.47 ksi (C)
τ max =
σ max − σ min
2
=
............................................................. Ans.
22.1706 − ( −19.4719 )
= 20.8 ksi ........................................ Ans.
2
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-104*
The given stress values are:
σ x = 40 MPa
τ xy = 40 MPa
σ y = −20 MPa
τ yz = 0 MPa
σ z = 20 MPa
τ zx = 30 MPa
σ x + σ y + σ z = 40 MPa
σ xσ y + σ yσ z + σ zσ x − τ xy2 − τ yz2 − τ zx2 = −2900 MPa 2
σ xσ yσ z − σ xτ yz2 − σ yτ zx2 − σ zτ xy2 + 2τ xyτ yzτ zx = −30, 000 MPa 3
σ 3p − ( 40 ) σ p2 + ( −2900 ) σ p − ( −30, 000 ) = 0
σ p1 = σ max = 73.7908 MPa ≅ 73.8 MPa (T) ........................................................... Ans.
σ p 2 = σ int = 9.4107 MPa ≅ 9.41 MPa (T) ............................................................... Ans.
σ p 3 = σ min = −43.2014 MPa ≅ 43.2 MPa (C) ........................................................ Ans.
τ max =
σ max − σ min
2
=
73.7908 − ( −43.2014 )
= 58.5 MPa .................................... Ans.
2
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-105
The given stress values are:
σ x = 14 ksi
σ y = 12 ksi
σ z = 10 ksi
τ xy = 4 ksi
τ yz = −4 ksi
σ x + σ y + σ z = 36 ksi
τ zx = 0 ksi
σ xσ y + σ yσ z + σ zσ x − τ xy2 − τ yz2 − τ zx2 = 396 ksi 2
σ xσ yσ z − σ xτ yz2 − σ yτ zx2 − σ zτ xy2 + 2τ xyτ yzτ zx = 1296 ksi3
σ 3p − ( 36 ) σ p2 + ( 396 ) σ p − (1296 ) = 0
σ p1 = σ max = 18.00 ksi ≅ 18.00 ksi (T)
..................................................................... Ans.
σ p 2 = σ int = 12.00 ksi ≅ 12.00 ksi (T) ....................................................................... Ans.
σ p 3 = σ min = 6.00 ksi ≅ 6.00 ksi (T)
τ max =
σ max − σ min
2
=
......................................................................... Ans.
18 − ( 6 )
= 6.00 ksi ................................................................... Ans.
2
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-106*
The given stress values are:
σ x = 60 MPa
σ y = 90 MPa
σ z = 60 MPa
τ xy = 120 MPa
τ yz = 75 MPa
τ zx = 90 MPa
σ x + σ y + σ z = 210 MPa
σ xσ y + σ yσ z + σ zσ x − τ xy2 − τ yz2 − τ zx2 = −13, 725 MPa 2
σ xσ yσ z − σ xτ yz2 − σ yτ zx2 − σ zτ xy2 + 2τ xyτ yzτ zx = 13,500 MPa 3
σ 3p − ( 210 ) σ p2 + ( −13, 725 ) σ p − (13,500 ) = 0
σ p1 = σ max = 262.485 MPa ≅ 262 MPa (T) ............................................................ Ans.
σ p 2 = σ int = −1.000 MPa ≅ 1.000 MPa (C)
............................................................ Ans.
σ p 3 = σ min = −51.485 MPa ≅ 51.5 MPa (C) ........................................................... Ans.
τ max =
σ max − σ min
2
=
262.485 − ( −51.485 )
= 157.0 MPa ..................................... Ans.
2
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-107
The given stress values are:
σ x = σ y = σ z = 0 ksi
τ xy = 6 ksi
τ yz = 10 ksi
τ zx = 8 ksi
σ x + σ y + σ z = 0 ksi
σ xσ y + σ yσ z + σ zσ x − τ xy2 − τ yz2 − τ zx2 = −200 ksi 2
σ xσ yσ z − σ xτ yz2 − σ yτ zx2 − σ zτ xy2 + 2τ xyτ yzτ zx = 960 ksi3
σ 3p − ( 0 ) σ p2 + ( −200 ) σ p − ( 960 ) = 0
σ p1 = σ max = 16.1116 ksi ≅ 16.11 ksi (T) ................................................................. Ans.
σ p 2 = σ int = −5.7511 ksi ≅ 5.75 ksi (C)
................................................................... Ans.
σ p 3 = σ min = −10.3605 ksi ≅ 10.36 ksi (C)
τ max =
σ max − σ min
2
=
............................................................. Ans.
16.1116 − ( −10.3605 )
= 13.24 ksi ...................................... Ans.
2
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-108
The given stress values are:
σ x = 72 MPa
σ y = −32 MPa
σ z = 0 MPa
τ xy = 21 MPa
τ yz = 0 MPa
τ zx = 21 MPa
σ x + σ y + σ z = 40 MPa
σ xσ y + σ yσ z + σ zσ x − τ xy2 − τ yz2 − τ zx2 = −3186 MPa 2
σ xσ yσ z − σ xτ yz2 − σ yτ zx2 − σ zτ xy2 + 2τ xyτ yzτ zx = 14,112 MPa 3
σ 3p − ( 40 ) σ p2 + ( −3186 ) σ p − (14,112 ) = 0
σ p1 = σ max = 81.3151 MPa ≅ 81.3 MPa (T) ............................................................ Ans.
σ p 2 = σ int = −4.7457 MPa ≅ 4.75 MPa (C) ............................................................ Ans.
σ p 3 = σ min = −36.5695 MPa ≅ 36.6 MPa (C)
τ max =
σ max − σ min
2
=
........................................................ Ans.
81.3151 − ( −36.5695 )
= 58.9 MPa ..................................... Ans.
2
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-109*
The given stress values are:
(a)
σ x = −18 ksi
σ y = −15 ksi
σ z = −12 ksi
τ xy = −15 ksi
τ yz = 12 ksi
τ zx = −9 ksi
σ x + σ y + σ z = −45 ksi
σ xσ y + σ yσ z + σ zσ x − τ xy2 − τ yz2 − τ zx2 = 216 ksi 2
σ xσ yσ z − σ xτ yz2 − σ yτ zx2 − σ zτ xy2 + 2τ xyτ yzτ zx = 6507 ksi3
σ 3p − ( −45 ) σ p2 + ( 216 ) σ p − ( 6507 ) = 0
σ p1 = σ max = 9.1477 ksi ≅ 9.15 ksi (T) ..................................................................... Ans.
σ p 2 = σ int = −22.4191 ksi ≅ 22.4 ksi (C) ................................................................. Ans.
σ p 3 = σ min = −31.7286 ksi ≅ 31.7 ksi (C)
τ max =
(b)
For
σ max − σ min
2
=
............................................................... Ans.
9.1477 − ( −31.7286 )
= 20.4 ksi .......................................... Ans.
2
σ p 3 = −31.7286 ksi
( −31.7286 ) − ( −18 )  l − ( −15 ) m − ( −9 ) n = 0
(1)
( −31.7286 ) − ( −15 )  m − ( −15 ) l − (12 ) n = 0
(2)
( −31.7286 ) − ( −12 )  n − ( −9 ) l − (12 ) m = 0
(3)
From Eqs. (1) and (2)
From Eqs. (2) and (3)
m = 1.01021l
n = −0.15827l
l 2 + m 2 + n 2 = l 2 + (1.01021l ) + ( −0.15827l ) = 1
2
2
l = 0.69919
θ x 3 = 45.64° .............................................. Ans.
m = 0.70632
θ y 3 = 45.06° .............................................. Ans.
n = −0.11066
θ z 3 = 96.35° .............................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-110*
The given stress values are:
(a)
σ x = 75 MPa
σ y = 35 MPa
σ z = 55 MPa
τ xy = 45 MPa
τ yz = 28 MPa
τ zx = 36 MPa
σ x + σ y + σ z = 165 MPa
σ xσ y + σ yσ z + σ zσ x − τ xy2 − τ yz2 − τ zx2 = 4570 MPa 2
σ xσ yσ z − σ xτ yz2 − σ yτ zx2 − σ zτ xy2 + 2τ xyτ yzτ zx = 19,560 MPa 3
σ 3p − (165 ) σ p2 + ( 4570 ) σ p − (19,560 ) = 0
σ p1 = σ max = 131.3380 MPa ≅ 131.3 MPa (T)
σ p 2 = σ int = 28.4218 MPa ≅ 28.4 MPa (T)
............................................................ Ans.
σ p 3 = σ min = 5.2399 MPa ≅ 5.24 MPa (T)
............................................................. Ans.
τ max =
(b)
....................................................... Ans.
For
σ max − σ min
2
=
131.33880 − ( 5.2399 )
= 63.0 MPa ..................................... Ans.
2
σ p1 = 131.3380 MPa
(131.3380 ) − ( 75 )  l − ( 45 ) m − ( 36 ) n = 0
(1)
(131.3380 ) − ( 35 )  m − ( 45 ) l − ( 28 ) n = 0
(2)
(131.3380 ) − ( 55 )  n − ( 36 ) l − ( 28 ) m = 0
From Eqs. (1) and (2)
m = 0.67626l
From Eqs. (2) and (3)
n = 0.71963l
(3)
l 2 + m 2 + n 2 = l 2 + ( 0.67626l ) + ( 0.71963l ) = 1
2
2
l = 0.71153
θ x1 = 44.64° .............................................. Ans.
m = 0.48118
θ y1 = 61.24° .............................................. Ans.
n = 0.51204
θ z1 = 59.20°
.............................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-111
The given stress values are:
(a)
σ x = 18 ksi
σ y = 12 ksi
σ z = 6 ksi
τ xy = 12 ksi
τ yz = −6 ksi
τ zx = 9 ksi
σ x + σ y + σ z = 36 ksi
σ xσ y + σ yσ z + σ zσ x − τ xy2 − τ yz2 − τ zx2 = 135 ksi 2
σ xσ yσ z − σ xτ yz2 − σ yτ zx2 − σ zτ xy2 + 2τ xyτ yzτ zx = −2484 ksi3
σ 3p − ( 36 ) σ p2 + (135 ) σ p − ( −2484 ) = 0
σ p1 = σ max = 28.0170 ksi ≅ 28.0 ksi (T) .................................................................. Ans.
σ p 2 = σ int = 14.2186 ksi ≅ 14.22 ksi (T) .................................................................. Ans.
σ p 3 = σ min = −6.2355 ksi ≅ 6.24 ksi (C) .................................................................. Ans.
τ max =
(b)
For
σ max − σ min
2
=
28.0170 − ( −6.2355 )
= 17.13 ksi ........................................ Ans.
2
σ p1 = 28.0170 ksi
( 28.0170 ) − (18 )  l − (12 ) m − ( 9 ) n = 0
(1)
( 28.0170 ) − (12 )  m − (12 ) l − ( −6 ) n = 0
(2)
( 28.0170 ) − ( 6 )  n − ( 9 ) l − ( −6 ) m = 0
From Eqs. (1) and (2)
m = 0.66384l
From Eqs. (2) and (3)
n = 0.22787l
(3)
l 2 + m 2 + n 2 = l 2 + ( 0.66384l ) + ( 0.22787l ) = 1
2
2
l = 0.81852
θ x1 = 35.06°
m = 0.54336
θ y1 = 57.09° .............................................. Ans.
n = 0.18652
θ z1 = 79.25°
.............................................. Ans.
.............................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-112
The given stress values are:
(a)
σ x = 100 MPa
σ y = −100 MPa
σ z = 80 MPa
τ xy = 50 MPa
τ yz = −70 MPa
τ zx = −64 MPa
σ x + σ y + σ z = 80 MPa
σ xσ y + σ yσ z + σ zσ x − τ xy2 − τ yz2 − τ zx2 = −21, 496 MPa 2
σ xσ yσ z − σ xτ yz2 − σ yτ zx2 − σ zτ xy2 + 2τ xyτ yzτ zx = −632, 400 MPa 3
σ 3p − ( 80 ) σ p2 + ( −21, 496 ) σ p − ( −632, 400 ) = 0
σ p1 = σ max = 179.9330 MPa ≅ 179.9 MPa (T) ....................................................... Ans.
σ p 2 = σ int = 27.5659 MPa ≅ 27.6 MPa (T)
............................................................ Ans.
σ p 3 = σ min = −127.4990 MPa ≅ 127.5 MPa (C) .................................................... Ans.
τ max =
(b)
For
179.9330 − ( −127.4990 )
= 153.7160 MPa ....................... Ans.
2
2
= −127.4990 MPa
σ max − σ min
σ p3
=
( −127.4990 ) − (100 )  l − ( 50 ) m − ( −64 ) n = 0
(1)
( −127.4990 ) − ( −100 )  m − ( 50 ) l − ( −70 ) n = 0
(2)
( −127.4990 ) − ( 80 )  n − ( −64 ) l − ( −70 ) m = 0
From Eqs. (1) and (2)
m = −7.31291l
From Eqs. (2) and (3)
n = −2.15885l
(3)
l 2 + m 2 + n 2 = l 2 + ( 7.31591l ) + ( 2.15885l ) = 1
2
2
l = 0.13004
θ x 3 = 82.53° .............................................. Ans.
m = −0.95094
θ y 3 = 161.98° ............................................ Ans.
n = −0.28073
θ z 3 = 106.30° ............................................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-113*
From a free-body diagram of the ring C,
the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
TCE sin15° − TBC = 0
TCE cos15° − 2000 = 0
TCE = 2070.552 lb
TBC = 535.898 lb
Then, from a free-body diagram of the ring B
the equations of equilibrium give
→ ΣFx = 0 :
↑ ΣFy = 0 :
TBC − TAB cos15° = 0
TAB sin15° − P = 0
TAB = 554.803 lb
P = 143.594 lb
σ AB =
TAB
554.803
=
= 11,302 psi ≅ 11.30 ksi (T) ............................................... Ans.
A π ( 0.25 )2 4
σ CE =
TCE
2070.552
=
= 42,181 psi ≅ 42.2 ksi (T) ................................................. Ans.
A π ( 0.25 )2 4
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-114*
From a free-body diagram of the pin B, the equations of equilibrium give
→ ΣFx = 0 :
−TAB + TBC cos 60° = 0
TBC sin 60° − 75 = 0
↑ ΣFy = 0 :
TAB = 43.30127 kN
σ AB =
σ BC
TBC = 86.60254 kN
3
TAB 43.30127 (10 )
=
≤ 75 (106 ) N m 2
AAB
AAB
(86.6025410
T
= BC =
ABC
ABC
3
) ≤ 75 10
( )
6
N m2
AAB ≥ 577 (10−6 ) m 2 = 577 mm 2 ...................................................................................... Ans.
ABC ≥ 1155 (10−6 ) m 2 = 1155 mm 2 ................................................................................... Ans.
τA =
V 43,301.27
=
≤ 100 (106 ) N/m 2
2
π dA 4
A
τC =
V 86, 602.54
=
≤ 100 (106 ) N/m 2
π dC2 4
A
d A ≥ 23.5 (10−3 ) m = 23.5 mm .......................................................................................... Ans.
dC ≥ 33.2 (10−3 ) m = 33.2 mm .......................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-115
From a free-body diagram of the pin B,
the equations of equilibrium give
→ ΣFx = 0 :
TAB cos 30° + TBC cos 30° = 0
TAB sin 30° − TBC sin 30° − 7500 = 0
↑ ΣFy = 0 :
TAB = 7500 lb
TBC = −7500 lb
Then, from a free-body diagram of the pin B
the equations of equilibrium give
→ ΣFx = 0 :
TAC cos 60° + TCD cos 30° − TBC cos 30° = 0
TAC sin 60° − TCD sin 30° + TBC sin 30° − 9000 = 0
↑ ΣFy = 0 :
TAC = 7794.229 lb
P = −12, 000 lb
TAC 7794.229
=
= 5277 psi ≅ 5.28 ksi (T) ........................................................ Ans.
A
1.477
T
12, 000
= CD =
≤ 3500 psi
A
ACD
σ AC =
σ CD
ACD ≥ 3.43 in.2 ........................................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-116*
σn =
P
1 + cos 2 ( 65° )  ≤ 12 (106 ) N m 2
2 ( 0.150 )( 0.180 )
P ≤ 1.814 (106 ) N
τn =
P
sin 2 ( 65° ) ≤ 1.40 (106 ) N m 2
2 ( 0.150 )( 0.180 )
P ≤ 98.7 (103 ) N
Pmax = 98.7 kN .............................................................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-117
The given values are
σ x = 12 ksi
σ y = 28 ksi
τ xy = 7 ksi
θ ab = 23°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= (12 ) cos 2 ( 23° ) + ( 28 ) sin 2 ( 23° ) + 2 ( 7 ) sin ( 23° ) cos ( 23° )
σ ab = +19.48 ksi = 19.48 ksi (T) ...........................................................Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − (12 ) − ( 28 )  sin ( 23° ) cos ( 23° ) + ( 7 ) cos 2 ( 23° ) − sin 2 ( 23° ) 
τ ab = +10.62 ksi .........................................................................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-118
The given values are
σ x = −35 MPa
1
2
θ p = tan −1
When
σ y = 45 MPa
2τ xy
σ x −σ y
=
τ xy = −18 MPa
2 ( −18 )
1
= 12.114°, − 77.886°
tan −1
2
( −35) − ( 45)
θ p = 12.114°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( −35 ) cos 2 θ p + ( 45 ) sin 2 θ p + 2 ( −18 ) sin θ p cos θ p
= −38.863 MPa = σ p 2
σ p1 = σ x + σ y − σ p 2 = 48.863 MPa
τ max = τ p = (σ p1 − σ p 2 ) 2 = 43.863 MPa
σ n 45 = (σ p1 + σ p 2 ) 2 = 5 MPa
σ p1 = 48.9 MPa (T)
77.89° ................................. Ans.
σ p 2 = 38.9 MPa (C)
12.11° ................................. Ans.
τ max = τ p = 43.9 MPa .................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-119*
The given values for use in
drawing Mohr’s circle are
σ x = 13 ksi
σ y = 7 ksi
τ xy = −4 ksi
σ z = σ p 3 = 0 ksi
a=
13 + 7
= 10.00 ksi
2
R = 32 + 42 = 5.00 ksi
(a)
φ
1
4
tan −1 = 26.57° (CW)
2 2
3
σ p1 = 10.00 + 5.00 = 15.00 ksi (T)
θ p1 =
=
σ p 2 = 10.00 − 5.00 = 5.0 ksi (T)
26.57° ......................................................... Ans.
63.43° ............................................................. Ans.
τ p = R = 5 ksi .................................................................................................................. Ans.
τ max = (σ max − σ min ) 2 = (15 − 0 ) 2 = 7.50 ksi (out of plane) ............................. Ans.
(b)
σ ab = 10 − 5.00 cos 66.870° = 8.04 ksi = 8.04 ksi (T) ......................Ans.
τ ab = 5.00sin 66.870° = 4.60 ksi (CW) = −4.60 ksi
........................Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-120
The given stress values are:
(a)
σ x = 53 MPa
σ y = −28 MPa
σ z = 36 MPa
τ xy = 24 MPa
τ yz = −18 MPa
τ zx = 46 MPa
θ x = 40°
θ y = 75°
θ z = 54°
S x = σ x cos θ x + τ yx cos θ y + τ zx cos θ z
= 53cos 40° + 24 cos 75° + 46 cos 54° = 73.8501 MPa
S y = τ xy cos θ x + σ y cos θ y + τ zy cos θ z
= 24 cos 40° − 28cos 75° − 18cos 54° = 0.5578 MPa
S z = τ xz cos θ x + τ yz cos θ y + σ z cos θ z
= 46 cos 40° − 18cos 75° + 36 cos 54° = 51.7385 MPa
S = S x2 + S y2 + S z2 =
( 73.8501) + ( 0.5578) + ( 51.7395)
2
2
2
= 90.1728 MPa
σ n = S x cos θ x + S y cos θ y + S z cos θ z
= 73.8501cos 40° + 0.5578cos 75° + 51.7395cos 54°
σ n = 87.1285 MPa ≅ 87.1 MPa (T) .......................................................................... Ans.
τ n = S 2 − σ n2 =
( 90.1728) − ( 87.1285)
2
2
= 23.2 MPa ................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
2-120 (cont.)
(b)
σ x + σ y + σ z = 61 MPa
σ xσ y + σ yσ z + σ zσ x − τ xy2 − τ yz2 − τ zx2 = −3600 MPa 2
σ xσ yσ z − σ xτ yz2 − σ yτ zx2 − σ zτ xy2 + 2τ xyτ yzτ zx = −71,828 MPa 3
σ 3p − ( 61) σ p2 + ( −3600 ) σ p − ( −71,828 ) = 0
σ p1 = σ max = 91.7133 MPa ≅ 91.7 MPa (T)
........................................................... Ans.
σ p 2 = σ int = 16.5662 MPa ≅ 16.57 MPa (T) ........................................................... Ans.
σ p 3 = σ min = −47.2785 MPa ≅ 47.3 MPa (C)
91.7133 − ( −47.2785 )
= 69.5 MPa ..................................... Ans.
2
2
σ p1 = 91.7133 MPa
τ max =
(c)
For
........................................................ Ans.
σ max − σ min
=
( 91.7133) − ( 53)  l − ( 24 ) m − ( 46 ) n = 0
(1)
( 91.7133) − ( −28 )  m − ( 24 ) l − ( −18 ) n = 0
(2)
( 91.7133) − ( 36 )  n − ( 46 ) l − ( −18 ) m = 0
From Eqs. (1) and (2)
m = 0.08023l
From Eqs. (2) and (3)
n = 0.79973l
(3)
l 2 + m 2 + n 2 = l 2 + ( 0.08023l ) + ( 0.79973l ) = 1
2
2
l = 0.77944
θ x1 = 38.79°
m = 0.06253
θ y1 = 86.41° .............................................. Ans.
n = 0.62334
θ z1 = 51.44°
.............................................. Ans.
.............................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
2-121
RILEY, STURGES AND MORRIS
Equation 2-22 is
σ 3p − (σ x + σ y + σ z ) σ p2 + (σ xσ y + σ yσ z + σ zσ x − τ xy2 − τ yz2 − τ zx2 ) σ p
− (σ xσ yσ z − σ xτ yz2 − σ yτ zx2 − σ zτ xy2 + 2τ xyτ yzτ zx ) = 0
For plane stress
σ z = τ zx = τ yz = 0
σ 3p − (σ x + σ y ) σ p2 + (σ xσ y − τ xy2 ) σ p = 0
σ p σ p2 − (σ x + σ y ) σ p + (σ xσ y − τ xy2 )  = 0
σp =
(σ
(σ
=
x
+σ y ) ±
(σ
+ σ y ) − 4 (σ xσ y − τ xy2 )
2
x
2
x
+σ y )
2
±
σ x2 − 2σ xσ y + σ y2
4
+τ
2
xy
(σ
=
(σ
=
x
x
+σ y )
2
+σ y )
2
±
(σ
+ σ y ) − 4 (σ xσ y − τ xy2 )
2
x
4
 σ x −σ y 
2
± 
 + τ xy
 2 
2
which is Eq. 2-15.
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MECHANICS OF MATERIALS, 6th Edition
3-1*
ε=
RILEY, STURGES AND MORRIS
∆L
0.625
=
= 0.00208 in./in. = 2080 µ in./in. ................................................. Ans.
L ( 25 )(12 )
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MECHANICS OF MATERIALS, 6th Edition
3-2*
RILEY, STURGES AND MORRIS
∆L
∆L
1200 (10−6 ) =
400
L
∆L = 0.480 mm ..................................................................................................................... Ans.
ε=
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MECHANICS OF MATERIALS, 6th Edition
3-3
ε avg =
RILEY, STURGES AND MORRIS
∆L 1.5 + 0.450 ( 2 )
=
= 0.300 in./in. ..................................................................... Ans.
L
8
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MECHANICS OF MATERIALS, 6th Edition
3-4
(a)
ε avg =
∆L 5 + 5.5 + 6.5 + 9 + 19.5 + 7 + 6 + 5
=
L
200
(b)
ε avg =
9 + 19.5
50
RILEY, STURGES AND MORRIS
ε avg = 0.317 m/m
(ε )
avg max
.................. Ans.
= 0.570 m/m .................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
3-5*
γ=
RILEY, STURGES AND MORRIS
∆y 0.1
=
= 0.0200 in./in. = 0.0200 rad ................................................................... Ans.
Lx 0.5
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-6*
0.380
0.200
+ tan −1
= 0.08938° = 0.001560 rad
500
250
= 1560 µ rad ..................................................................................................................... Ans.
γ xy = tan −1
γ xy
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MECHANICS OF MATERIALS, 6th Edition
3-7
RILEY, STURGES AND MORRIS
γ xy = 90° − 89.92° = 0.0800° = 0.001396 rad = 1396 µ rad ........................................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-8
(a)
LAB = 22 + 1992 = 199.01005 mm
ε avg =
(b)
γ=
LAB − 200 −0.98995
=
= −0.00495 m/m = −4950 µ m/m .............................. Ans.
200
200
∆x
2
=
= 0.0100 rad ................................................................................................ Ans.
Ly 200
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-9*
(a)
δ BC = δ DE = ε DE LDE = 800 (10−6 ) ( 70 ) = 0.056000 in.
ε BC =
δ BC
LBC
=
0.056000
= 0.001400 in./in.
40
ε BC = 1400 µ in./in. ................................................................................................................ Ans.
(b)
δ BC = δ DE − 0.006 = 0.050000 in.
ε BC =
δ BC
LBC
=
0.050000
= 0.001250 in./in.
40
ε BC = 1250 µ in./in. ................................................................................................................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
3-10*
(a)
RILEY, STURGES AND MORRIS
δ D = ε D LD = 0.0075 (150 ) = 1.1250 mm
e
b
=
100 50
b = δ D + 0.09 mm
e = 2b
e = δ CE
δ CE = 2 (δ D + 0.09 ) = 2 (1.1250 + 0.09 ) = 2.4300 mm
ε CE =
(b)
δ CE
LCE
=
2.4300
= 0.00810 m/m = 8100 µ m/m .................................................... Ans.
300
b = δ D + 0.09 mm
e = δ CE + 0.10 mm
(δ CE + 0.10 ) = 2 (δ D + 0.09 ) = 2 (1.12500 + 0.09 ) = 2.4300 mm
δ CE = 2.3300 mm
ε CE =
δ CE
LCE
=
2.3300
= 0.00777 m/m = 7770 µ m/m .................................................... Ans.
300
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
3-11
(a)
ε avg =
(b)
ε circ =
ε circ
RILEY, STURGES AND MORRIS
∆d 2.15 − 2.00
=
= 0.0750 in./in. .....................................Ans.
2.00
d
∆circ circ′ − circ 2π d ′ − 2π d d ′ − d ∆d
=
=
=
=
circ
circ
2π d
d
d
∆d
=
= ε d = 0.0750 in./in. .......................................................Ans.
d
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
3-12*
1250 (10−6 ) = k (1500 )
ε = kx 2
k = 5.55556 (10−10 ) /mm 2
(a)
RILEY, STURGES AND MORRIS
∆L = ∫
3000
0
kx 3
kx dx =
3
2
ε = 5.55556 (10−10 ) x 2 m/m
3000
2
= 9 (109 ) k = 5.00 mm ................................................. Ans.
0
∆L 5.00
=
= 0.001667 m/m = 1667 µ m/m ............................................... Ans.
L 3000
(b)
ε avg =
(c)
ε max = ε x =3000 = k ( 3000 ) = 0.00500 m/m = 5000 µ m/m .................................... Ans.
2
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MECHANICS OF MATERIALS, 6th Edition
3-13
RILEY, STURGES AND MORRIS
δ B = ε B LB = 0.0014 (15 ) = 0.02100 in.
a b
=
10 6
a = δ AD
a=
5b
3
b = δ B + 0.009 in.
5 (δ B + 0.009 ) 5 ( 0.02100 + 0.009 )
=
= 0.05000 in.
3
3
δ
0.05000
= AD =
= 0.001000 in./in. = 1000 µ in./in. ...............Ans.
LAD
50
δ AD =
ε AD
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
3-14
RILEY, STURGES AND MORRIS
1.5° = 0.02618 rad
a = 55 ( 0.02618 ) = 1.43990 mm
b = 42.5 ( 0.02618 ) = 1.11265 mm
(a)
1.43990
a
= tan −1
= 6.5710° = 0.1147 rad
12.5
12.5
1.11265
b
β = tan −1
= tan −1
= 5.0866° = 0.0888 rad
12.5
12.5
γ rθ = α = 0.1147 rad ............................................................................................................ Ans.
(b)
γ rθ = β = 0.0888 rad
α = tan −1
............................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
3-15*
α = R P′QR = tan −1
RILEY, STURGES AND MORRIS
20sin 45° + 0.4
20 cos 45°
= 45.7989°
θ ′ = R QP′R = 180° − 2α
= 180° − 2 ( 45.7989° ) = 88.4021°
γ = 90° − θ ′ = 90° − ( 88.4021° ) = 1.5979°
γ = 0.0279 rad
..................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-16*
The given values are
ε x = −2000 µ m/m
ε y = −1500 µ m/m
γ xy = 1250 µ rad
θ n = −35°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( −2000 ) cos 2 ( −35° ) + ( −1500 ) sin 2 ( −35° ) + (1250 ) sin ( −35° ) cos ( −35° )
ε n = −2420 µ m/m ................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
3-17*
The given values are
ε x = 880 µ in./in.
(a)
ε y = 960 µ in./in.
RILEY, STURGES AND MORRIS
γ xy = −750 µ rad
θ AC = tan −1 ( 2 4 ) = 26.565°
ε AC = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 880 ) cos 2 ( 26.565° ) + ( 960 ) sin 2 ( 26.565° ) + ( −750 ) sin ( 26.565° ) cos ( 26.565° )
ε AC = 596 µ in./in. .................................................................................................................. Ans.
(b)
θ BD = − tan −1 ( 2 4 ) = −26.565°
ε BD = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 880 ) cos 2 ( −26.565° ) + ( 960 ) sin 2 ( −26.565° ) + ( −750 ) sin ( −26.565° ) cos ( −26.565° )
ε BD = 1196 µ in./in. ................................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
3-18
The given values are
(a)
ε x = 1750 µ m/m
ε y = −2200 µ m/m
θ n = −45°
θt = +45°
RILEY, STURGES AND MORRIS
γ xy = −800 µ rad
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= (1750 ) cos 2 ( −45° ) + ( −2200 ) sin 2 ( −45° ) + ( −800 ) sin ( −45° ) cos ( −45° )
ε n = 175 µ m/m
...................................................................................................................... Ans.
ε t = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= (1750 ) cos 2 ( 45° ) + ( −2200 ) sin 2 ( 45° ) + ( −800 ) sin ( 45° ) cos ( 45° )
ε t = −625 µ m/m .................................................................................................................... Ans.
(b)
γ nt = −2 ( ε x − ε y ) sin θ cos θ + γ xy ( cos 2 θ − sin 2 θ )
= −2 (1750 ) − ( −2200 )  sin ( −45° ) cos ( −45° ) + ( −800 ) cos 2 ( −45° ) − sin 2 ( −45° ) 
γ nt = 3950 µ rad
..................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
3-19*
RILEY, STURGES AND MORRIS
∆L 0.06
=
= 0.00200 in./in. = 2000 µ in./in. ........................................................ Ans.
L
30
∆L −0.03
εy =
=
= −0.00100 in./in. = 1000 µ in./in. ................................................... Ans.
L
30
γ xy = 0 µ rad ............................................................................................................................ Ans.
(a)
εx =
(b)
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 2000 ) cos 2 ( 30° ) + ( −1000 ) sin 2 ( 30° ) + ( 0 )
ε n = 1250 µ in./in. .................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-20*
Lx′ =
(100.5) + ( 0.1)
L′y =
( 0.02 ) + ( 49.99 )
2
2
2
= 100.0500500 mm
2
= 49.9900040 mm
0.02   −1 0.1 

+ tan
= 90.03438°
50  
100 

∆L
0.0500500
εx = x =
= 501(10−6 ) m/m = 500 µ m/m ............................................. Ans.
100
Lx
θ ′ xy = 90° −  tan −1
εy =
∆Ly
Ly
=
49.9900040 − 50
= −199.9 (10−6 ) m/m = −199.9 µ m/m ..................... Ans.
50
γ xy = 90° − θ xy′ = −0.03438° = −600 (10−6 ) rad = −600 µ rad
.................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-21
The given values are
ε x = −800 µ in./in.
θ n = 42°
ε y = 640 µin./in.
γ xy = −960 µ rad
θt = 132°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( −800 ) cos 2 ( 42° ) + ( 640 ) sin 2 ( 42° ) + ( −960 ) sin ( 42° ) cos ( 42° )
ε n = −633 µ in./in. ................................................................................................................. Ans.
ε t = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( −800 ) cos 2 (132° ) + ( 640 ) sin 2 (132° ) + ( −960 ) sin (132° ) cos (132° )
ε t = 473 µ in./in. ..................................................................................................................... Ans.
γ nt = −2 ( ε x − ε y ) sin θ cos θ + γ xy ( cos 2 θ − sin 2 θ )
= −2 ( −800 ) − ( 640 )  sin ( 42° ) cos ( 42° ) + ( −960 ) cos 2 ( 42° ) − sin 2 ( 42° ) 
γ nt = 1332 µ rad ...................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-22
The given values are
ε x = 720 µ m/m
ε y = −480 µ m/m
θ n = −30°
γ xy = 360 µ rad
θt = 60°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 720 ) cos 2 ( −30° ) + ( −480 ) sin 2 ( −30° ) + ( 360 ) sin ( −30° ) cos ( −30° )
ε n = 264 µ m/m ...................................................................................................................... Ans.
ε t = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 720 ) cos 2 ( 60° ) + ( −480 ) sin 2 ( 60° ) + ( 360 ) sin ( 60° ) cos ( 60° )
ε t = −24.1 µ m/m ................................................................................................................... Ans.
γ nt = −2 ( ε x − ε y ) sin θ cos θ + γ xy ( cos 2 θ − sin 2 θ )
= −2 ( 720 ) − ( −480 )  sin ( −30° ) cos ( −30° ) + ( 360 ) cos 2 ( −30° ) − sin 2 ( −30° ) 
γ nt = 1219 µ rad ...................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-23
Equation (b) is
(1 + ε n )
2
dn 2 = (1 + ε x ) dn 2 cos 2 θ + (1 + ε y ) dn 2 sin 2 θ
2
2
+2dn 2 sin θ cos θ (1 + ε x ) (1 + ε y ) sin γ xy
1 + 2ε n + ε n2 = (1 + 2ε x + ε x2 ) cos 2 θ + (1 + 2ε y + ε y2 ) sin 2 θ
+2sin θ cos θ ( sin γ xy + ε x sin γ xy + ε y sin γ xy + ε xε y sin γ xy )
But sin
2
θ + cos 2 θ = 1 , therefore
2ε n + ε n2 = ( 2ε x + ε x2 ) cos 2 θ + ( 2ε y + ε y2 ) sin 2 θ
+2sin θ cos θ ( sin γ xy + ε x sin γ xy + ε y sin γ xy + ε xε y sin γ xy )
ε n2 = ε n , ε x2 = ε x ,and ε y2 = ε y . Also, sin γ xy ≅ γ xy ,
are all = γ xy , and Eq. (b) can be written
and then the small strain approximation
therefore
ε xγ xy , ε yγ xy , and ε xε yγ xy
2ε n = 2ε x cos 2 θ + 2ε y sin 2 θ + 2γ xy sin θ cos θ
which upon dividing through by 2 is Eq. 3-7a.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-24*
The given values are
ε x = ε AD = −600 µ m/m
ε AB = −1200 µ m/m
θ AB = tan −1
ε BD = 750 µ m/m
θ BD = −θ AB
240
= 50.194°
200
= −50.194°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε AB = ( −600 ) cos 2 θ AB + ε y sin 2 θ AB + γ xy sin θ AB cos θ AB = −1200
ε BD = ( −600 ) cos 2 θ BD + ε y sin 2 θ BD + γ xy sin θ BD cos θ BD = 750
0.59016ε y + 0.49180γ xy = −954.098
0.59016ε y − 0.49180γ xy = 995.902
ε y = 35.4 µ m/m ..................................................................................................................... Ans.
γ nt = −1983 µ rad ................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-25*
The given values are
ε x = 1250 µ in./in.
ε n = 1575 µ in./in.
θ n = 45°
ε t = 1350 µ in./in.
θt = 135°
ε n = (1250 ) cos 2 ( 45° ) + ε y sin 2 ( 45° ) + γ xy sin ( 45° ) cos ( 45° ) = 1575
ε t = (1250 ) cos 2 (135° ) + ε y sin 2 (135° ) + γ xy sin (135° ) cos (135° ) = 1350
0.5ε y + 0.5γ xy = 950
0.5ε y − 0.5γ xy = 725
ε y = 1675 µ in./in.
(a)
γ xy = 225 µ rad
γ nt = −2 (1250 ) − (1675 )  sin ( 45° ) cos ( 45° ) + ( 225 ) cos 2 ( 45° ) − sin 2 ( 45° ) 
γ nt = 425 µ rad ........................................................................................................................ Ans.
(b)
ε y = 1675 µ in./in. .................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-26
The given values are
ε x = 1950 µ m/m
ε y = −1625 µ m/m
sin θ n = 3 5
(a)
cos θ n = 4 5
ε n = (1950 ) cos 2 θ n + ( −1625 ) sin 2 θ n + γ xy sin θ n cos θ n = −1275
γ xy = −4037.500 µ rad ≅ −4040 µ rad
(b)
ε n = −1275 µ m/m
.............................................................................. Ans.
ε QR = (1950 ) cos 2 ( −θ n ) + ( −1625 ) sin 2 ( −θ n ) + ( −4037.5 ) sin ( −θ n ) cos ( −θ n )
ε QR = 2600 µ m/m
................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-27*
The given values are
ε x = 600 µ in./in.
1
2
θ p = tan −1
When
ε y = −200 µ in./in.
γ xy = −480 µ rad
γ xy
( −480 ) = −15.482°, 74.518°
1
= tan −1
εx −εy 2
( 600 ) − ( −200 )
θ p = −15.482°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 600 ) cos 2 θ p + ( −200 ) sin 2 θ p + ( −480 ) sin θ p cos θ p
= 666.476 µ in./in. = ε p1
ε p 2 = ε x + ε y − ε p1 = −266.476 µ in./in.
γ max = γ p = ε p1 − ε p 2 = 932.952 µ rad
ε p1 = +666 µ in./in.
15.48° ................................... Ans.
ε p 2 = −266 µ in./in.
74.52° ............................................................................................ Ans.
ε p 3 = 0 µ in./in. ....................................................................................................................... Ans.
γ max = γ p = 933 µ rad ............................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-28*
The given values are
ε x = 960 µ m/m
1
2
θ p = tan −1
When
ε y = −320 µ m/m
γ xy = 500 µ rad
γ xy
( 500 )
1
= tan −1
= 10.668°, − 79.332°
εx −εy 2
( 960 ) − ( −320 )
θ p = 10.668°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 960 ) cos 2 θ p + ( −320 ) sin 2 θ p + ( 500 ) sin θ p cos θ p = 1007.095 µ m/m = ε p1
ε p 2 = ε x + ε y − ε p1 = −367.095 µ m/m
γ max = γ p = ε p1 − ε p 2 = 1374.190 µ rad
ε p1 = +1007 µ m/m
10.67° ..............Ans.
ε p 2 = −367 µ m/m
79.33° ................Ans.
ε p 3 = 0 µ m/m ......................................................................................................................... Ans.
γ max = γ p = 1374 µ rad .......................................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-29
The given values are
ε x = 900 µ in./in.
1
2
θ p = tan −1
When
ε y = −300 µ in./in.
γ xy = 480 µ rad
γ xy
( 480 )
1
= tan −1
= 10.901°, − 79.099°
εx −εy 2
( 900 ) − ( −300 )
θ p = 10.901°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 900 ) cos 2 θ p + ( −300 ) sin 2 θ p + ( 480 ) sin θ p cos θ p = 946.220 µ in./in. = ε p1
ε p 2 = ε x + ε y − ε p1 = −346.220 µ in./in.
γ max = γ p = ε p1 − ε p 2 = 1292.440 µ rad
ε p1 = +946 µ in./in.
10.90° ..................... Ans.
ε p 2 = −346 µ in./in.
79.10° ..................... Ans.
ε p 3 = 0 µ in./in. ....................................................................................................................... Ans.
γ max = γ p = 1292 µ rad .......................................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-30
The given values are
ε x = −900 µ m/m
1
2
θ p = tan −1
When
ε y = 600 µ m/m
γ xy = −420 µ rad
γ xy
( −420 ) = 7.821°, − 82.179°
1
= tan −1
εx −εy 2
( −900 ) − ( 600 )
θ p = 7.821°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( −900 ) cos 2 θ p + ( 600 ) sin 2 θ p + ( −420 ) sin θ p cos θ p
= −928.845 µ m/m = ε p 2
ε p1 = ε x + ε y − ε p 2 = 628.845 µ m/m
γ max = γ p = ε p1 − ε p 2 = 1557.690 µ rad
ε p1 = +629 µ m/m
82.18° ...........................Ans.
ε p 2 = −929 µ m/m
7.82° .............................Ans.
ε p 3 = 0 µ m/m ......................................................................................................................... Ans.
γ max = γ p = 1558 µ rad .......................................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-31*
The given values are
ε x = 750 µ in./in.
1
2
θ p = tan −1
When
ε y = −1000 µ in./in.
γ xy = 360 µ rad
γ xy
( 360 )
1
= tan −1
= 5.812°, − 84.188°
εx −εy 2
( 750 ) − ( −1000 )
θ p = 5.812°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 750 ) cos 2 θ p + ( −1000 ) sin 2 θ p + ( 360 ) sin θ p cos θ p
= 768.322 µ in./in. = ε p1
ε p 2 = ε x + ε y − ε p1 = −1018.322 µ in./in.
γ max = γ p = ε p1 − ε p 2 = 1786.64 µ rad
ε p1 = +768 µ in./in.
ε p 2 = −1018 µ in./in.
5.81° ................... Ans.
84.19° .............. Ans.
ε p 3 = 0 µ in./in. ....................................................................................................................... Ans.
γ max = γ p = 1787 µ rad .......................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-32
The given values are
ε x = −750 µ m/m
1
2
θ p = tan −1
When
ε y = 410 µ m/m
γ xy = −250 µ rad
γ xy
( −250 ) = 6.081°, − 83.919°
1
= tan −1
εx −εy 2
( −750 ) − ( 410 )
θ p = 6.081°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( −750 ) cos 2 θ p + ( 410 ) sin 2 θ p + ( −250 ) sin θ p cos θ p
= −763.317 µ m/m = ε p 2
ε p1 = ε x + ε y − ε p 2 = 423.317 µ m/m
γ max = γ p = ε p1 − ε p 2 = 1186.634 µ rad
ε p1 = +423 µ m/m
83.92° ................... Ans.
ε p 2 = −763 µ m/m
6.08° ..................... Ans.
ε p 3 = 0 µ m/m ......................................................................................................................... Ans.
γ max = γ p = 1187 µ rad .......................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-33*
The given values are
ε x = 720 µ in./in.
1
2
θ p = tan −1
When
ε y = 520 µin./in.
γ xy = 480 µ rad
γ xy
( 480 ) = 33.690°, − 56.310°
1
= tan −1
εx −εy 2
( 720 ) − ( 520 )
θ p = 33.690°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 720 ) cos 2 θ p + ( 520 ) sin 2 θ p + ( 480 ) sin θ p cos θ p
= 880 µ in./in. = ε p1
ε p 2 = ε x + ε y − ε p1 = 360 µ in./in.
γ p = ε p1 − ε p 2 = 520 µ rad
γ max = ε max − ε min = 880 − 0 = 880 µ rad
ε p1 = +880 µ in./in.
33.69° ..................... Ans.
ε p 2 = +360 µ in./in.
56.31° ..................... Ans.
ε p 3 = 0 µ in./in. ....................................................................................................................... Ans.
γ p = 520 µ rad
........................................................................................................................ Ans.
γ max = 880 µ rad (out-of-plane) .......................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-34*
The given values are
ε x = −540 µ m/m
1
2
θ p = tan −1
When
ε y = −980 µ m/m
γ xy = 560 µ rad
γ xy
( 560 )
1
= tan −1
= 25.921°, − 64.079°
εx −εy 2
( −540 ) − ( −980 )
θ p = 25.921°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( −540 ) cos 2 θ p + ( −980 ) sin 2 θ p + ( 560 ) sin θ p cos θ p
= −403.910 µ m/m = ε p1
ε p 2 = ε x + ε y − ε p1 = −1116.090 µ m/m
γ p = ε p1 − ε p 2 = 712.180 µ rad
ε p1 = −404 µ m/m
ε p 2 = −1116 µ m/m
25.92° ...........................Ans.
64.08° ........................Ans.
ε p 3 = 0 µ m/m ......................................................Ans.
γ p = 712 µ rad .....................................................Ans.
γ max = ε max − ε min = 0 − ( −1116 ) = 1116 µ rad (out-of-plane) ..................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-35
The given values are
ε x = 864 µ in./in.
1
2
θ p = tan −1
When
ε y = 432 µ in./in.
γ xy = 288 µ rad
γ xy
( 288) = 16.845°, − 73.155°
1
= tan −1
εx −εy 2
(864 ) − ( 432 )
θ p = 16.845°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 864 ) cos 2 θ p + ( 432 ) sin 2 θ p + ( 288 ) sin θ p cos θ p
= 907.600 µ in./in. = ε p1
ε p 2 = ε x + ε y − ε p1 = 388.400 µ in./in.
γ p = ε p1 − ε p 2 = 519.200 µ rad
ε p1 = +908 µ in./in.
16.85° ..................... Ans.
ε p 2 = +388 µ in./in.
73.15° ..................... Ans.
ε p 3 = 0 µ in./in. ................................................ Ans.
γ p = 519 µ rad
................................................. Ans.
γ max = ε max − ε min = 908 − 0 = 908 µ rad (out-of-plane) ................................................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-36*
The given values are
ε x = 900 µ m/m
1
2
θ p = tan −1
When
ε y = 650 µ m/m
γ xy = 300 µ rad
γ xy
( 300 ) = 25.097°, − 64.903°
1
= tan −1
εx −εy 2
( 900 ) − ( 650 )
θ p = 25.097°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 900 ) cos 2 θ p + ( 650 ) sin 2 θ p + ( 300 ) sin θ p cos θ p
= 970.256 µ m/m = ε p1
ε p 2 = ε x + ε y − ε p1 = 579.744 µ m/m
γ p = ε p1 − ε p 2 = 390.512 µ rad
ε p1 = +970 µ m/m
25.10° ..................... Ans.
ε p 2 = +580 µ m/m
64.90° ..................... Ans.
ε p 3 = 0 µ m/m ................................................ Ans.
γ p = 391 µ rad
............................................... Ans.
γ max = ε max − ε min = 970 − 0 = 970 µ rad (out-of-plane) ................................................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-37
The given values are
ε x = −325 µ in./in.
1
2
θ p = tan −1
When
ε y = −625 µ in./in.
γ xy = 680 µ rad
γ xy
( 680 )
1
= tan −1
= 33.097°, − 56.903°
εx −εy 2
( −325) − ( −625 )
θ p = 33.097°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( −325 ) cos 2 θ p + ( −625 ) sin 2 θ p + ( 680 ) sin θ p cos θ p
= −103.382 µ in./in. = ε p1
ε p 2 = ε x + ε y − ε p1 = −846.618 µ in./in.
γ p = ε p1 − ε p 2 = 743.236 µ rad
ε p1 = −103.4 µ in./in.
ε p 2 = −847 µin./in.
33.10° ........................ Ans.
56.90° ............................ Ans.
ε p 3 = 0 µ in./in. ....................................................... Ans.
γ p = 743 µ rad
........................................................ Ans.
γ max = ε max − ε min = 0 − ( −847 ) = 847 µ rad (out-of-plane)
......................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-38
The given values are
ε x = −900 µ m/m
1
2
θ p = tan −1
When
ε y = −650 µ m/m
γ xy = −600 µ rad
γ xy
( −600 )
1
= tan −1
= 33.690°, − 56.310°
εx −εy 2
( −900 ) − ( −650 )
θ p = 33.690°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( −900 ) cos 2 θ p + ( −650 ) sin 2 θ p + ( −600 ) sin θ p cos θ p
= −1100 µ m/m = ε p 2
ε p1 = ε x + ε y − ε p 2 = −450 µ m/m
γ p = ε p1 − ε p 2 = 650 µ rad
ε p1 = −450 µ m/m
ε p 2 = −1100 µ m/m
56.31° .............................. Ans.
33.69° ........................... Ans.
ε p 3 = 0 µ m/m ......................................................... Ans.
γ p = 650 µ rad ........................................................................................................................ Ans.
γ max = ε max − ε min = 0 − ( −1100 ) = 1100 µ rad (out-of-plane) ..................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-39*
The given values are
ε x = 480 µ in./in.
ε y = −1200 µ in./in.
ε p 2 = −1400 µ in./in.
ε p1 = ε x + ε y − ε p 2 = +680 µ in./in.
480 + ( −1200 )
 480 − ( −1200 )   γ xy 
680 =
+ 

 +
2
2

  2 
2
γ xy = ±1226.377 µ rad ≅ ±1226 µ rad
1
2
θ p = tan −1
2
.............................................................................. Ans.
γ xy
( ±1226.377 ) = ±18.064°, m71.936°
1
= tan −1
εx −εy 2
( 480 ) − ( −1200 )
γ max = γ p = ε p1 − ε p 2 = 2080 µ rad
Using
θ p = +18.064°
ε p1 = +680 µ in./in.
ε p 2 = −1400 µ in./in.
18.06° .........................Ans.
71.94° ......................Ans.
ε p 3 = 0 µ in./in. ....................................................Ans.
γ max = γ p = 2080 µ rad
......................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-40*
The given values are
ε x = 300 µ m/m
ε y = −800 µ m/m
ε p1 = 1500 µ m/m
ε p 2 = −2000 µ m/m ............................................... Ans.
300 + ( −800 )
 300 − ( −800 )   γ xy 
1500 =
+ 

 +
2
2

  2 
2
γ xy = ±3322.650 µ rad ≅ ±3320 µ rad
1
2
θ p = tan −1
2
.............. Ans.
γ xy
( ±3322.650 )
1
= tan −1
εx −εy 2
( 300 ) − ( −800 )
= ±35.841°, m54.159°
γ max = γ p = ε p1 − ε p 2 = 3500 µ rad
ε p 3 = 0 µ m/m ......................................................... Ans.
γ max = γ p = 3500 µ rad
......................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-41
The given values are
ε x = 800 µ in./in.
Assuming that
γ max = γ p
ε p1 = 1280 µ in./in.
γ max = 2400 µ rad
gives
γ max = γ p = (1280 ) − ε p 2 = 2400
ε p 2 = −1120 µ in./in.
Then
ε y = ε p1 + ε p 2 − ε x = (1280 ) + ( −1120 ) − ( 800 )
ε y = −640 µ in./in. .......................................... Ans.
800 + ( −640 )
 800 − ( −640 )   γ xy 
1280 =
+ 

 +
2
2

  2 
2
2
γ xy = ±1920.00 µ rad ≅ ±1920 µ rad .......... Ans.
1
2
θ p = tan −1
Using
γ xy
( ±1920 ) = ±26.565°, m63.435°
1
= tan −1
εx −εy 2
(800 ) − ( −640 )
θ p = +26.565°
ε p 2 = −1120 µ in./in.
71.94° ......................................................................................... Ans.
ε p 3 = 0 µ in./in. ....................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-42
The given values are
ε x = −450 µ m/m
Assuming that
γ max = γ p
ε p 2 = −780 µ m/m
γ max = 960 µ rad
gives
γ max = γ p = ε p1 − ( −780 ) = 960
ε p1 = 180.00 µ m/m
Then
ε y = ε p1 + ε p 2 − ε x = (180 ) + ( −780 ) − ( −450 )
ε y = −150.0 µ m/m
............................................... Ans.
( −450 ) + ( −150 ) +
180 =
2
 ( −450 ) − ( −150 )   γ xy 


 +
2

  2 
2
2
γ xy = ±911.921 µ rad ≅ ±912 µ rad .................... Ans.
1
2
θ p = tan −1
Using
γ xy
( ±911.921) = m35.895°, ± 54.105°
1
= tan −1
εx −εy 2
( −450 ) − ( −150 )
θ p = −35.895°
ε p1 = +180.0 µ m/m
54.10° .......................................................................................... Ans.
ε p 3 = 0 µ m/m ......................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-43*
The given values are
γ xy = −1800 µ rad
tan 2 ( −30° ) =
ε p1 = 225 µ in./in.
θ p = −30°
( −1800 )
εx −ε y
ε x − ε y = 1039.230 µ in./in.
225 =
εx +εy
2
 (1039.230 )   −1800 
+ 
 +

2

  2 
2
2
ε x + ε y = −1628.461 µ in./in.
ε x = −294.615 µ in./in. ≅ −295 µ in./in. ........... Ans.
ε y = −1333.845 µ in./in. ≅ −1334 µ in./in. ...................................................................... Ans.
ε p 2 = ε x + ε y − ε p1 = −1853.461 µ in./in. ≅ −1853 µ in./in.
ε p 2 = −1853 µ in./in.
60° ............................................................................................... Ans.
ε p 3 = 0 µ in./in. ....................................................................................................................... Ans.
γ max = γ p = ε p1 − ε p 2 = 2078 µ rad
.................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-44
The given values are
γ xy = 840 µ rad
tan 2 ( 20° ) =
ε p1 = 1100 µ m/m
θ p = 20°
( 840 )
εx −εy
ε x − ε y = 1001.073 µ m/m
1100 =
εx +εy
2
 (1001.073)   840 
+ 
 +

2

  2 
2
2
ε x + ε y = 893.192 µ m/m
ε x = +947.133 µ m/m ≅ 947 µ m/m .................................................................................. Ans.
ε y = −53.940 µ m/m ≅ −53.9 µ m/m
................................................................................ Ans.
ε p 3 = 0 µ m/m ......................................................................................................................... Ans.
ε p 2 = ε x + ε y − ε p1 = −206.808 µ m/m ≅ −207 µ m/m
ε p 2 = −207 µ m/m
70° .................................................................................................... Ans.
γ max = γ p = ε p1 − ε p 2 = 1306.808 µ rad ≅ 1307 µ rad
.................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-45
The given values are
ε y = −750 µ in./in.
−1500 =
ε x + ( −750 )
2
γ xy = −750 µ rad
 ε − ( −750 )   −750 
+  x
 +

2

  2 
2
ε x = −1312.500 µ in./in. ≅ −1313 µ in./in.
1
2
θ p = tan −1
ε p 2 = −1500 µ in./in.
2
...................................................................... Ans.
γ xy
( −750 )
1
= tan −1
= 26.565°, − 63.435°
εx −εy 2
( −1312.5) − ( −750 )
ε p1 = ε x + ε y − ε p 2 = −562.500 µ in./in.
γ p = ε p1 − ε p 2 = 937.500 µ rad
γ max = ε max − ε min = ( 0 ) − ( −1500 ) = 1500 µ rad
ε p1 = −563 µ in./in.
63.43° .......................... Ans.
ε p 3 = 0 µ in./in. ..................................................... Ans.
γ p = 937 µ rad ........................................................................................................................ Ans.
γ max = 1500 µ rad (out-of-plane) ........................................................................................ Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-46*
The given values are
ε y = 750 µ m/m
1000 =
ε x + ( 750 )
2
γ xy = −750 µ rad
 ε − ( 750 )   −750 
+  x
 +

2

  2 
2
ε p1 = 1000 µ m/m
2
ε x = 437.50 µ m/m ≅ +437 µ m/m .................... Ans.
1
2
θ p = tan −1
γ xy
( −750 )
1
= tan −1
εx −εy 2
( 437.5) − ( 750 )
θ p = 33.690°, − 56.310° ................................Ans.
ε p 2 = ε x + ε y − ε p1 = 187.50 µ m/m
ε p 2 ≅ +187.5 µ m/m ...........................................Ans.
γ p = ε p1 − ε p 2 = 813 µ rad
................................................................................................... Ans.
γ max = ε max − ε min = 1000 − 0 = 1000 µ rad (out-of-plane) ............................................ Ans.
ε p 3 = 0 µ m/m ......................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-47*
The given values for use in
drawing Mohr’s circle are
ε p1 = 400 µ in./in.
ε p 2 = −600 µ in./in.
θp =
φ
2
= 18.43°
a=
400 + ( −600 )
= −100 µ in./in.
2
R=
400 − ( −600 )
= 500 µ in./in.
2
ε x = −100 + 500 cos 36.86° = +300 µ in./in. ......... Ans.
ε y = −100 − 500 cos 36.86° = −500 µ in./in.
........ Ans.
γ xy = 2 ( 500sin 36.86° ) = 600 µ rad (CCW)
γ xy = +600 µ rad .......................................................... Ans.
γ max = γ p = 2 R = 1000 µ rad
............................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-48*
The given values for use in
drawing Mohr’s circle are
ε p1 = 945 µ m/m
ε p 2 = −785 µ m/m
θp =
φ
2
= 16.85°
a=
945 + ( −785 )
= 80 µ m/m
2
R=
945 − ( −785 )
= 865 µ m/m
2
ε x = 80 + 865cos 33.70° = +800 µ m/m ................ Ans.
ε y = 80 − 865cos 33.70° = −640 µ m/m ................ Ans.
γ xy = 2 ( 865sin 33.70° ) = 960 µ rad (CCW)
γ xy = +960 µ rad .......................................................... Ans.
γ max = γ p = 2 R = 1730 µ rad
............................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-49
The given values for use in
drawing Mohr’s circle are
ε p1 = 708 µ in./in.
ε p 2 = −104 µ in./in.
θp =
φ
2
= −34.10°
a=
708 + ( −104 )
= 302 µ in./in.
2
R=
708 − ( −104 )
= 406 µ in./in.
2
ε x = 302 + 406 cos 68.20° = +453 µ in./in. ........... Ans.
ε y = 302 − 406 cos 68.20° = +151.2 µ in./in. ........ Ans.
γ xy = 2 ( 406sin 68.20° ) = 754 µ rad (CW)
γ xy = −754 µ rad .......................................................... Ans.
γ max = γ p = 2 R = 812 µ rad
................................................................................................. Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-50
The given values for use in
drawing Mohr’s circle are
ε p1 = −114 µ m/m
ε p 2 = −903 µ m/m
θp =
φ
2
= 19.26°
a=
( −114 ) + ( −903) = −508.5 µ m/m
R=
( −114 ) − ( −903) = 394.5 µ m/m
2
2
ε x = −508.5 + 394.5cos 38.52° = −199.8 µ m/m
Ans.
ε y = −508.5 − 394.5cos 38.52° = −817 µ m/m .... Ans.
γ xy = 2 ( 394.5sin 38.52° ) = 491 µ rad (CCW)
γ xy = +491µ rad
........................................................... Ans.
γ p = 2 R = 789 µ rad .............................................................................................................. Ans.
γ max = ( ε max − ε min ) = 0 − ( −903) = 903 µ rad (out-of-plane) ...................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-51*
The given values for use in
drawing Mohr’s circle are
ε x = 950 µ in./in.
ε y = −225 µ in./in.
γ xy = −275 µ rad
a=
950 + ( −225 )
= 362.5 µ in./in.
2
R=
( 587.5 ) + (137.5)
2
φ
2
= 603.38 µ in./in.
1
137.5
= 6.59° (CW)
tan −1
2 2
587.5
ε p1 = 362.5 + 603.38 = +965.88 µ in./in.
θ p1 =
=
ε p1 ≅ +966 µ in./in.
6.59° ............................ Ans.
ε p 2 = 362.5 − 603.38 = −240.88 µ in./in.
ε p 2 ≅ −241 µ in./in.
83.41° ............................................................................................ Ans.
γ max = γ p = 2 R = 1207 µ rad
............................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-52*
The given values for use in
drawing Mohr’s circle are
ε x = 900 µ m/m
ε y = −333 µ m/m
γ xy = 982 µ rad
a=
900 + ( −333)
= 283.5 µ m/m
2
R=
( 616.5) + ( 491)
2
φ
2
= 788.133 µ m/m
1
491
= 19.267° (CCW)
tan −1
2 2
616.5
ε p1 = 283.5 + 788.133 = +1071.633 µ m/m
θ p1 =
=
ε p1 ≅ +1072 µ m/m
19.27° .......................... Ans.
ε p 2 = 283.5 − 788.133 = −504.633 µ m/m
ε p 2 ≅ −505 µ m/m
70.73° .............................................................................................. Ans.
γ max = γ p = 2 R = 1576 µ rad
............................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-53
The given values for use in
drawing Mohr’s circle are
ε x = 750 µ in./in.
ε y = 390 µin./in.
γ xy = −900 µ rad
a=
750 + 390
= 570 µ in./in.
2
R=
(180 ) + ( 450 )
2
φ
2
= 484.66 µ in./in.
1
450
= 34.100° (CW)
tan −1
2 2
180
ε p1 = 570 + 484.66 = +1054.66 µ in./in.
θ p1 =
=
ε p1 ≅ +1055 µ in./in.
34.10° ..............Ans.
ε p 2 = 570 − 484.66 = +85.34 µ in./in.
ε p 2 ≅ +85.3 µ in./in.
55.90° ..............Ans.
γ p = 2 R = 969 µ rad .............................................................................................................. Ans.
γ max = ε max − ε min = 1055 − 0 = 1055 µ rad (out-of-plane) ............................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-54
The given values for use in
drawing Mohr’s circle are
ε x = 600 µ m/m
ε y = 480 µ m/m
γ xy = 480 µ rad
a=
600 + 480
= 540 µ m/m
2
R=
( 60 ) + ( 240 )
2
φ
2
= 247.386 µ m/m
1
240
= 37.982° (CCW)
tan −1
2 2
60
ε p1 = 540 + 247.386 = +787.386 µ m/m
θ p1 =
=
ε p1 ≅ +787 µ m/m
37.98° ..................Ans.
ε p 2 = 540 − 247.386 = +292.614 µ m/m
ε p 2 ≅ +293 µ m/m
52.02° ..................Ans.
γ p = 2 R = 495 µ rad .............................................................................................................. Ans.
γ max = ε max − ε min = 787 − 0 = 787 µ rad (out-of-plane)
............................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-55*
The given values for use in
drawing Mohr’s circle are
ε x = −680 µ in./in.
ε y = 320 µin./in.
ε p1 = 414 µ in./in.
a=
( −680 ) + ( 320 ) = −180 µin./in.
2
R = 414 + 180 = 594 µ in./in.
θp =
φ
1
500
= cos −1
= 16.34°
2 2
594
γ xy = 2 ( 594 ) sin 32.68°
γ xy = ±641 µ rad ..................................................Ans.
ε p 2 = ( −180 ) − 594
ε p 2 = −774 µ in./in. ............................................Ans.
ε p 3 = 0 µ in./in. ....................................................Ans.
γ max = γ p = 2 R = 1188 µ rad
............................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-56*
The given values for use in
drawing Mohr’s circle are
ε x = 450 µ m/m
ε y = 150 µ m/m
ε p1 = 780 µ m/m
a=
( 450 ) + (150 )
2
= 300 µ m/m
R = 780 − 300 = 480 µ m/m
θp =
φ
1
150
= cos −1
= 35.895°
2 2
480
γ xy = 2 ( 480 ) sin 71.790°
γ xy = ±912 µ rad ..................................................Ans.
ε p 2 = ( 300 ) − ( 480 ) = −180 µ m/m ................Ans.
ε p 3 = 0 µ m/m ......................................................Ans.
γ max = γ p = 2 R = 960 µ rad
................................................................................................. Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-57
The given values for use in
drawing Mohr’s circle are
ε x = 360 µ in./in.
ε y = 750 µin./in.
ε p 2 = 120 µ in./in.
a=
( 360 ) + ( 750 )
2
= 555 µ in./in.
R = 555 − 120 = 435 µ in./in.
θp =
φ
1
195
= cos −1
= 31.683°
2 2
435
γ xy = 2 ( 435 ) sin 63.367°
γ xy = ±778 µ rad ..................................................Ans.
ε p1 = 555 + 435 = +990 µ in./in. ......................Ans.
ε p 3 = 0 µ in./in. ....................................................Ans.
γ p = 2 R = 870 µ rad ...........................................Ans.
γ max = ε max − ε min = 990 − 0 = 990 µ rad (out-of-plane) ................................................ Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-58
The given values for use in
drawing Mohr’s circle are
ε x = −300 µ m/m
ε y = 600 µ m/m
ε p 2 = −450 µ m/m
a=
( −300 ) + ( 600 )
2
= 150 µ m/m
R = 150 + 450 = 600 µ m/m
θp =
φ
1
450
= cos −1
= 20.705°
2 2
600
γ xy = 2 ( 600 ) sin 41.410°
γ xy = ±794 µ rad ..................................................Ans.
ε p1 = (150 ) + ( 600 ) = +750 µ m/m .................Ans.
ε p 3 = 0 µ m/m ......................................................Ans.
γ max = γ p = 2 R = 1200 µ rad
............................Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
3-59*
(a) The given values are
RILEY, STURGES AND MORRIS
ε a = ε x = 750 µ in./in.
ε b = ε 45° = −125 µin./in.
ε c = ε y = −250 µ in./in.
ν = 0.30
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε b = ( 750 ) cos 2 ( 45° ) + ( −250 ) sin 2 ( 45° ) + γ xy sin ( 45° ) cos ( 45° ) = 125
ε x = +750 µ in./in. .............. ε y = −250 µ in./in.
Therefore:
.................................. Ans.
γ xy = −750 µ rad ........................................................................................ Ans.
(b)
1
2
θ p = tan −1
When
γ xy
( −750 ) = −18.435°, 71.565°
1
= tan −1
εx −εy 2
( 750 ) − ( −250 )
θ p = 71.565°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 750 ) cos 2 θ p + ( −250 ) sin 2 θ p + ( −750 ) sin θ p cos θ p
= −375 µ in./in. = ε p 2
ε p1 = ε x + ε y − ε p 2 = 875 µ in./in.
ε p3 = ε z =
−ν
−0.30
( 750 ) + ( −250 )  = −214 µ in./in.
εx +ε y ) =
(
1 −ν
1 − 0.30 
ε p1 = +875 µ in./in.
18.43° ............................................................................................ Ans.
ε p 2 = −375 µ in./in.
71.57° ............................................................................................ Ans.
ε p 3 = −214 µ in./in. ............................................................................................................... Ans.
γ max = γ p = ε p1 − ε p 2 = 1250 µ rad ..................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
3-60*
(a) The given values are
RILEY, STURGES AND MORRIS
ε a = ε x = −555 µ m/m
ε b = ε120° = 925 µ m/m
ε c = ε 240° = 740 µ m/m
ν = 0.30
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε b = ( −555 ) cos 2 (120° ) + ε y sin 2 (120° ) + γ xy sin (120° ) cos (120° ) = 925
ε c = ( −555 ) cos 2 ( 240° ) + ε y sin 2 ( 240° ) + γ xy sin ( 240° ) cos ( 240° ) = 740
0.75000ε y − 0.43301γ xy = 1063.750
0.75000ε y + 0.43301γ xy = 878.750
ε x = −555 µ m/m ....................................................................................... Ans.
Therefore:
ε y = 1295.00 µ m/m = +1295 µ m/m ................................................................................ Ans.
γ xy = −213.620 µ rad ≅ −214 µ rad ................................................................................... Ans.
(b)
1
2
θ p = tan −1
When
γ xy
( −213.620 ) = 3.293°, − 86.707°
1
= tan −1
εx −εy 2
( −555) − (1295)
θ p = 3.293°
ε n = ( −555 ) cos 2 θ p + (1295 ) sin 2 θ p + ( −213.620 ) sin θ p cos θ p
= −561.146 µ m/m = ε p 2
ε p1 = ε x + ε y − ε p 2 = 1301.146 µ m/m
ε p3 = ε z =
−ν
−0.30
εx +ε y ) =
( −555 ) + (1295 )  = −317 µ m/m
(
1 −ν
1 − 0.30 
ε p1 = +1301 µ m/m
86.71° ............................................................................................ Ans.
ε p 2 = −561 µ m/m
3.29° ................................................................................................ Ans.
ε p 3 = −317 µ m/m
................................................................................................................. Ans.
γ max = γ p = ε p1 − ε p 2 = 1862 µ rad ..................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
3-61
(a) The given values are
RILEY, STURGES AND MORRIS
ε a = ε x = 800 µ in./in.
ε b = ε n = 950 µ in./in.
θb = tan −1
ε c = ε y = 600 µ in./in.
3
= 36.870°
4
ν = 0.33
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε b = ( 800 ) cos 2 θb + ( 600 ) sin 2 θb + γ xy sin θb cos θ b = 950
ε x = +800 µ in./in. .............. ε y = 600 µin./in. ..................................... Ans.
Therefore:
γ xy = 462.500 µ rad ≅ 463 µ rad ............................................................ Ans.
(b)
1
2
θ p = tan −1
When
γ xy
( 462.5 ) = 33.307°, − 56.693°
1
= tan −1
εx −εy 2
( 800 ) − ( 600 )
θ p = 33.307°
ε n = ( 800 ) cos 2 θ p + ( 600 ) sin 2 θ p + ( 462.5 ) sin θ p cos θ p
= 951.946 µ in./in. = ε p1
ε p 2 = ε x + ε y − ε p1 = 448.054 µ in./in.
−ν
−0.33
εx +εy ) =
( 800 ) + ( 600 )  = −689.55 µ in./in.
(
1 −ν
1 − 0.33 
ε p1 = +952 µ in./in. 33.31° ............................................................................................ Ans.
ε p3 = ε z =
ε p 2 = +448 µ in./in.
56.69° ............................................................................................ Ans.
ε p 3 = −690 µ in./in. ............................................................................................................... Ans.
γ p = ε p1 − ε p 2 = 504 µ rad
................................................................................................... Ans.
γ max = ε p1 − ε p 3 = 1641 µ rad (out-of-plane)
................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
3-62*
(a) The given values are
RILEY, STURGES AND MORRIS
ε a = ε x = 780 µ m/m
ε b = ε120° = 345 µ m/m
ε c = ε 60° = −332 µ m/m
ν = 0.33
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε b = ( 780 ) cos 2 (120° ) + ε y sin 2 (120° ) + γ xy sin (120° ) cos (120° ) = 345
ε c = ( 780 ) cos 2 ( 60° ) + ε y sin 2 ( 60° ) + γ xy sin ( 60° ) cos ( 60° ) = −332
0.75000ε y − 0.43301γ xy = 150
0.75000ε y + 0.43301γ xy = −527
ε x = +780 µ m/m ....................................................................................... Ans.
Therefore:
ε y = −251.333 µ m/m ≅ −251 µ m/m ................................................................................ Ans.
γ xy = −781.732 µ rad ≅ −782 µ rad ................................................................................... Ans.
(b)
1
2
θ p = tan −1
When
γ xy
( −781.732 ) = −18.581°, 71.419°
1
= tan −1
εx −εy 2
( 780 ) − ( −251.333)
θ p = −18.581°
ε n = ( 780 ) cos 2 θ p + ( −251.333) sin 2 θ p + ( −781.732 ) sin θ p cos θ p
= 911.395 µ m/m = ε p1
ε p 2 = ε x + ε y − ε p1 = −382.728 µ m/m
−ν
−0.33
εx +ε y ) =
( 780 ) + ( −251.333)  = −260.388 µ m/m
(
1 −ν
1 − 0.33 
ε p1 = +911 µ m/m 18.58° ............................................................................................... Ans.
ε p3 = ε z =
ε p 2 = −383 µ m/m
71.42° .............................................................................................. Ans.
ε p 3 = −260 µ m/m
................................................................................................................. Ans.
γ max = γ p = ε p1 − ε p 2 = 1294 µ rad ..................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
3-63*
The given values are
RILEY, STURGES AND MORRIS
ε a = ε x = 36 µ in./in.
ε b = ε 45° = 310 µ in./in.
ε c = ε y = 150 µ in./in.
ν = 0.30
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε b = ( 36 ) cos 2 ( 45° ) + (150 ) sin 2 ( 45° ) + γ xy sin ( 45° ) cos ( 45° ) = 310
γ xy = 434.00 µ rad
1
2
θ p = tan −1
When
γ xy
( 434 ) = −37.641°, 52.359°
1
= tan −1
εx −εy 2
( 36 ) − (150 )
θ p = −37.641°
ε n = ( 36 ) cos 2 θ p + (150 ) sin 2 θ p + ( 434 ) sin θ p cos θ p
= −131.361 µ in./in. = ε p 2
ε p1 = ε x + ε y − ε p 2 = 317.361 µ in./in.
−ν
−0.30
εx +εy ) =
( 36 ) + (150 ) 
(
1 −ν
1 − 0.30 
= −79.714 µ in./in.
ε p3 = ε z =
ε p1 = +317 µ in./in.
ε p 2 = −131.4 µ in./in.
52.36° ................................Ans.
37.64° ............................Ans.
ε p 3 = −79.7 µin./in. .............................................................................................................. Ans.
γ max = γ p = ε p1 − ε p 2 = 449 µ rad ....................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
3-64
The given values are
RILEY, STURGES AND MORRIS
ε a = ε x = 525 µ m/m
ε b = ε 45° = 450 µ m/m
ε c = ε135° = 1425 µ m/m
ν = 0.30
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε b = ( 525 ) cos 2 ( 45° ) + ε y sin 2 ( 45° ) + γ xy sin ( 45° ) cos ( 45° ) = 450
ε c = ( 525 ) cos 2 (135° ) + ε y sin 2 (135° ) + γ xy sin (135° ) cos (135° ) = 1425
0.5000ε y + 0.5000γ xy = 187.5
0.5000ε y − 0.5000γ xy = 1162.5
ε y = 1350 µ m/m
1
2
θ p = tan −1
γ xy = −975 µ rad
γ xy
( −975)
1
= tan −1
εx −εy 2
( 525) − (1350 )
= 24.882°, − 65.118°
When
θ p = 24.882°
ε n = ( 525 ) cos 2 θ p + (1350 ) sin 2 θ p + ( −975 ) sin θ p cos θ p
= 298.898 µ m/m = ε p 2
ε p1 = ε x + ε y − ε p 2 = 1576.102 µ m/m
−ν
−0.30
εx + εy ) =
( 525 ) + (1350 )  = −803.571 µ m/m
(
1 −ν
1 − 0.30 
ε p1 = +1576 µ m/m 65.12° ............................................................................................ Ans.
ε p3 = ε z =
ε p 2 = +299 µ m/m
24.88° .............................................................................................. Ans.
ε p 3 = −804 µ m/m
................................................................................................................. Ans.
γ p = ε p1 − ε p 2 = 1277 µ rad
................................................................................................. Ans.
γ max = ε p1 − ε p 3 = 2380 µ rad .............................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
3-65*
(a) The given values are
RILEY, STURGES AND MORRIS
ε a = ε x = 875 µ in./in.
ε b = ε n = 700 µ in./in.
θb = tan −1
ε c = ε y = 350 µ in./in.
4
= 53.130°
3
ν = 0.30
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε b = ( 875 ) cos 2 θb + ( 350 ) sin 2 θb + γ xy sin θ b cos θb = 700
γ xy = 335.417 µ rad
1
2
θ p = tan −1
When
γ xy
( 335.417 ) = 16.287°, − 73.713°
1
= tan −1
εx −εy 2
(875 ) − ( 350 )
θ p = 16.287°
ε n = ( 875 ) cos 2 θ p + ( 350 ) sin 2 θ p + ( 335.417 ) sin θ p cos θ p
= 924.00 µ in./in. = ε p1
ε p 2 = ε x + ε y − ε p1 = 301.00 µ in./in.
−ν
(ε x + ε y )
1 −ν
−0.30
=
( 875 ) + ( 350 ) 
1 − 0.30 
= −525.00 µ in./in.
ε p3 = ε z =
ε p1 = +924 µ in./in.
16.29° ..................... Ans.
ε p 2 = +301 µin./in.
73.71° ..................... Ans.
ε p 3 = −525 µ in./in. ......................................... Ans.
γ p = ε p1 − ε p 2 = 623 µ rad
................................................................................................... Ans.
γ max = ε p1 − ε p 3 = 1449 µ rad (out-of-plane) ................................................................... Ans.
(b)
ε n = ( 875 ) cos 2 (120° ) + ( 350 ) sin 2 (120° ) + ( 335.417 ) sin (120° ) cos (120° )
ε n = +336 µ in./in. ................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
3-66
(a) The given values are
RILEY, STURGES AND MORRIS
ε a = ε x = 875 µ m/m
ε b = ε120° = 700 µ m/m
ε c = ε 60° = −650 µ m/m
ν = 0.33
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε b = ( 875 ) cos 2 (120° ) + ε y sin 2 (120° ) + γ xy sin (120° ) cos (120° ) = 700
ε c = ( 875 ) cos2 ( 60° ) + ε y sin 2 ( 60° ) + γ xy sin ( 60° ) cos ( 60° ) = −650
0.75000ε y − 0.43301γ xy = 481.25
0.75000ε y + 0.43301γ xy = −868.75
ε y = −258.333 µ m/m
γ xy = −1558.846 µ rad
1
2
θ p = tan −1
γ xy
( −1558.846 )
1
= tan −1
εx −εy 2
(875 ) − ( −258.333)
= −26.991°, 63.009°
When
θ p = −26.991°
ε n = ( 875 ) cos 2 θ p + ( −258.333) sin 2 θ p + ( −1558.846 ) sin θ p cos θ p
= 1271.978 µ m/m = ε p1
ε p 2 = ε x + ε y − ε p1 = −655.312 µ m/m
−ν
−0.33
εx +ε y ) =
( 875 ) + ( −258.333)  = −303.732 µ m/m
(
1 −ν
1 − 0.33 
ε p1 = +1272 µ m/m 26.99° ............................ Ans.
ε p3 = ε z =
ε p 2 = −655 µ m/m
63.01° .............................. Ans.
ε p 3 = −304 µ m/m
................................................. Ans.
γ max = γ p = ε p1 − ε p 2 = 1927 µ rad ..................... Ans.
(b)
γ nt = −2 ( ε x − ε y ) sin θ cos θ + γ xy ( cos 2 θ + sin 2 θ )
= −2 ( 875 ) − ( 258.333)  sin ( 40° ) cos ( 40° )
+ ( −1558.846 )  cos 2 ( 40° ) + sin 2 ( 40° ) 
γ nt = −1387 µ rad ................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
3-67*
(a) The given values are
RILEY, STURGES AND MORRIS
ε a = ε x = 800 µ in./in.
ε b = ε y = 950 µin./in.
ε c = ε120° = 600 µ in./in.
ν = 0.33
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε c = ( 800 ) cos 2 (120° ) + ( 950 ) sin 2 (120° ) + γ xy sin (120° ) cos (120° ) = 600
γ xy = 721.688 µ rad
1
2
θ p = tan −1
When
γ xy
( 721.688 ) = −39.129°, 50.871°
1
= tan −1
εx −εy 2
( 800 ) − ( 950 )
θ p = −39.129°
ε n = ( 800 ) cos 2 θ p + ( 950 ) sin 2 θ p + ( 721.688) sin θ p cos θ p
= 506.444 µ in./in. = ε p 2
ε p1 = ε x + ε y − ε p 2 = 1243.556 µ in./in.
−ν
−0.33
εx +εy ) =
( 800 ) + ( 950 ) 
(
1 −ν
1 − 0.33 
= −861.94 µ in./in.
ε p3 = ε z =
ε p1 = +1244 µ in./in.
50.87° ..............................Ans.
ε p 2 = +506 µ in./in.
39.13° ................................Ans.
ε p 3 = −862 µ in./in. ...................................................Ans.
γ p = ε p1 − ε p 2 = 737 µ rad
................................................................................................... Ans.
γ max = ε p1 − ε p 3 = 2110 µ rad .............................................................................................. Ans.
(b)
ε n = ( 800 ) cos 2 ( 200° ) + ( 950 ) sin 2 ( 200° ) + ( 721.688 ) sin ( 200° ) cos ( 200° )
ε n = +1049 µ in./in. ............................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
3-68*
RILEY, STURGES AND MORRIS
δ BF = ε BF LBF = 400 (10−6 ) (1000 ) = 0.400 mm
b
c
=
80 240
c = 3b = 3δ BF = 1.200 mm
ε CE =
δ CE
LCE
=
c
= 2000 (10−6 ) m/m
600
ε CE = 2000 µ m/m
................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
3-69*
δ AB = ε AB LAB = 0.0015 (15 ) = 0.02250 in.
b f
=
6 2
(a)
RILEY, STURGES AND MORRIS
f =
b δ AB
=
= 0.00750 in.
3
3
f = δ EF = 0.00750 in.
ε EF =
δ EF
LEF
=
0.00750
= 937 (10−6 ) in./in.
8
ε EF = 937 µ in./in. .................................................................................................................. Ans.
(b)
f = 0.005 + δ EF = 0.00750 in.
ε EF =
δ EF
LEF
=
δ EF = 0.00250 in.
0.00250
= 313 (10−6 ) in./in.
8
ε EF = 313 µ in./in. .................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-70
γ A = θA =
R2φ
.................................................. Ans.
R2 − R1
γ B = θB =
R1φ
.................................................. Ans.
R2 − R1
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-71*
The given values are
ε x = 3200 µ in./in.
(a)
ε y = 1500 µ in./in.
γ xy = 1000 µ rad
θ n = 45°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε n = ( 3200 ) cos 2 ( 45° ) + (1500 ) sin 2 ( 45° ) + (1000 ) sin ( 45° ) cos ( 45° )
ε n = 2850 µ in./in.
(b)
................................................................................................................. Ans.
γ nt = −2 ( ε x − ε y ) sin θ cos θ + γ xy ( cos 2 θ − sin 2 θ )
γ nt = −2 ( 3200 ) − (1500 )  sin ( 45° ) cos ( 45° ) + (1000 ) cos 2 ( 45° ) − sin 2 ( 45° ) 
γ nt = −1700 µ rad ................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-72
The given values are
ε x = 1500 µ m/m
ε y = −1250 µ m/m
θ BD = − tan −1
γ xy = 1000 µ rad
150
= −36.870°
200
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε BD = (1500 ) cos 2 θ BD + ( −1250 ) sin 2 θ BD + (1000 ) sin θ BD cos θ BD
ε BD = 30.0 µ m/m ................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-73
The given values are
ε x = 1000 µ in./in.
1
2
θ p = tan −1
When
ε y = −800 µ in./in.
γ xy = −800 µ rad
γ xy
( −800 )
1
= tan −1
= −11.981°, 78.019°
εx −εy 2
(1000 ) − ( −800 )
θ p = −11.981°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= (1000 ) cos 2 θ p + ( −800 ) sin 2 θ p + ( −800 ) sin θ p cos θ p
= 1084.886 µ in./in. = ε p1
ε p 2 = ε x + ε y − ε p1 = −884.886 µ in./in.
γ max = γ p = ε p1 − ε p 2 = 1969.772 µ rad
ε p1 = +1085 µ in./in.
11.98° ................... Ans.
ε p 2 = −885 µ in./in.
78.02° ..................... Ans.
ε p 3 = 0 µ in./in. ....................................................................................................................... Ans.
γ max = 1970 µ rad
................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-74*
The given values are
ε x = −600 µ m/m
1
2
θ p = tan −1
When
ε y = 1200 µ m/m
γ xy = 2000 µ rad
γ xy
( 2000 )
1
= tan −1
= −24.006°, 65.996°
εx −εy 2
( −600 ) − (1200 )
θ p = −24.006°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( −600 ) cos 2 θ p + (1200 ) sin 2 θ p + ( 2000 ) sin θ p cos θ p
= −1045.362 µ m/m = ε p 2
ε p1 = ε x + ε y − ε p 2 = 1645.362 µ m/m
γ max = γ p = ε p1 − ε p 2 = 2690.724 µ rad
ε p1 = +1645 µ m/m
65.99° .......................... Ans.
ε p 2 = −1045 µ m/m
24.01° ......................... Ans.
ε p 3 = 0 µ m/m ....................................................... Ans.
γ max = γ p = 2690 µ rad
....................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
3-75
The given values are
RILEY, STURGES AND MORRIS
ε a = ε x = 600 µ in./in.
ε b = ε 45° = 500 µ in./in.
ε c = ε y = −200 µ in./in.
ν = 0.30
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε b = ( 600 ) cos 2 ( 45° ) + ( −200 ) sin 2 ( 45° ) + γ xy sin ( 45° ) cos ( 45° ) = 500
γ xy = 600.00 µ rad
1
2
θ p = tan −1
When
γ xy
( 600 )
1
= tan −1
= 18.435°, − 71.565°
εx −εy 2
( 600 ) − ( −200 )
θ p = 18.435°
ε n = ( 600 ) cos 2 θ p + ( −200 ) sin 2 θ p + ( 600 ) sin θ p cos θ p
= 700.00 µ in./in. = ε p1
ε p 2 = ε x + ε y − ε p1 = −300.00 µ in./in.
ε p1 = +700 µ in./in.
18.43° ............................................................................................ Ans.
ε p 2 = −300 µ in./in.
71.57° ............................................................................................ Ans.
−ν
−0.30
εx +ε y ) =
( 600 ) + ( −200 )  = −171.4 µ in./in. ................ Ans.
(
1 −ν
1 − 0.30 
= γ p = ε p1 − ε p 2 = 1000 µ rad ..................................................................................... Ans.
ε p3 = ε z =
γ max
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
3-76*
The given values for use in
drawing Mohr’s circle are
ε x = −800 µ m/m
ε y = 640 µ m/m
γ xy = −960 µ rad
a=
( −800 ) + ( 640 ) = −80 µ m/m
R=
θ p1 =
2
( 720 ) + ( 480 )
2
φ
2
=
2
= 865.332 µ m/m
1
480
= 16.845° (CCW)
tan −1
2
720
ε p1 = ( −80 ) + ( 865 ) = +785 µ m/m
73.15° ................................................................ Ans.
ε p 2 = ( −80 ) − ( 865 ) = −945 µ m/m
16.85° ............................................................... Ans.
ε p 3 = 0 µ m/m
73.15° ..................................................................................................... Ans.
γ max = γ p = 2 R = 1731 µ rad
............................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
3-77
The given values are
RILEY, STURGES AND MORRIS
ε a = ε x = 800 µ in./in.
ε b = ε120° = 960 µ in./in.
ε c = ε 240° = 800 µ in./in.
ν = 0.33
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε b = ( 800 ) cos 2 (120° ) + ε y sin 2 (120° ) + γ xy sin (120° ) cos (120° ) = 960
ε c = ( 800 ) cos 2 ( 240° ) + ε y sin 2 ( 240° ) + γ xy sin ( 240° ) cos ( 240° ) = 800
0.75000ε y − 0.43301γ xy = 760
0.75000ε y + 0.43301γ xy = 600
ε y = 906.667 µin./in.
γ xy = −184.752 µ rad
1
2
θ p = tan −1
When
γ xy
( −184.752 ) = 30.00°, − 60.00°
1
= tan −1
εx −εy 2
( 800 ) − ( 906.667 )
θ p = 30.00°
ε n = ( 800 ) cos 2 θ p + ( 906.667 ) sin 2 θ p + ( −184.752 ) sin θ p cos θ p
= 746.667 µ in./in. = ε p 2
ε p1 = ε x + ε y − ε p 2 = 960.00 µ in./in.
3-77 (cont.)
ε p3 = ε z =
−ν
−0.33
εx +ε y ) =
( 800 ) + ( 906.667 )  = −840.597 µ m/m
(
1 −ν
1 − 0.33 
ε p1 = +960 µ m/m
60.00° ................... Ans.
ε p 2 = +747 µ m/m
30.00° ................... Ans.
ε p 3 = −841 µ m/m ....................................... Ans.
γ p = ε p1 − ε p 2 = 213 µ rad
........................ Ans.
γ max = ε p1 − ε p 3 = 1801 µ rad .................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-1*
A=
πd2
=
π (1.5 )
2
= 1.76715 in.2
4
4
∆L
0.48
=
= 0.002000 in./in.
ε=
L ( 20 × 12 )
E=
ν=
G=
F
53
=
= 29.9919 ksi
A 1.76715
∆d −0.001
εt =
=
= 666.667 (10−6 ) in./in.
d
1.5
σ=
σ 29.9919
=
= 15, 000 ksi ....................................................................................... Ans.
ε 0.002000
−ε t
εl
=
−666.667 (10−6 )
0.002000
= 0.333 .................................................................................. Ans.
E
15, 000
=
= 5630 ksi .......................................................................... Ans.
2 (1 + ν ) 2 (1 + 0.333)
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-2*
A=
πd2
4
=
π ( 0.015 )
2
4
62.6 (103 )
= 176.715 (10−6 ) mm 2
σ=
F
=
= 354.244 (106 ) N/m 2 = 354.244 MPa
−6
A 176.715 (10 )
ε=
∆L 0.90
=
= 0.004500 m/m
L
200
σ 354.244 (10
E= =
0.004500
ε
ν=
−ε t
εl
=
6
εt =
) = 78.721 10 N/m
( )
− ( −1.46667 ) (10−3 )
0.004500
9
2
∆d −0.022
=
= −1.46667 (10−3 ) m/m
d
15
≅ 78.7 GPa ......................................... Ans.
= 0.326 ........................................................................... Ans.
σ = σ PL = 354 MPa .............................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-3
(a)
σ=
F
10
=
= 20.0 ksi ........................................................................................... Ans.
A ( 2 )( 0.25 )
(b)
E=
σ
20
=
= 15, 000 ksi ................................................................................. Ans.
ε 0.08 ( 5 × 12 )
(c)
ν = −ε t ε l
0.25 =
− (δ 0.25 0.25 )
0.08 ( 5 ×12 )
δ 0.25 = −0.0000833 in. ........................................................ Ans.
0.25 =
− (δ 2 2 )
0.08 ( 5 ×12 )
δ 2 = −0.000667 in.
............................................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
4-4*
RILEY, STURGES AND MORRIS
Ao = π ( 5.64 ) 4 = 24.98 mm 2
2
= 24.98 (10−6 ) m 2
∆D = νε frac D = 0.30 ( 0.115 )( 5.64 )
= 0.19458 mm
Af = π ( 5.64 − 0.19458 ) 4
2
= 23.29 mm 2 = 23.29 (10−6 ) m 2
From the
σ −ε
diagram:
(b)
6
∆σ (139 − 0 ) (10 )
E=
≅
= 69.5 (109 ) N/m 2 = 69.5 GPa ........................................ Ans.
∆ε
0.002 − 0
σ PL ≅ 139 MPa ...................................................................................................................... Ans.
(c)
σ ult ≅ 450 MPa
(d)
σ ys ( 0.05% ) ≅ 220 MPa ...................................................................................................... Ans.
(e)
σ ys ( 0.20% ) ≅ 278 MPa ...................................................................................................... Ans.
(f)
σ t = σ ult ≅ 450 MPa ............................................................................................................. Ans.
(g)
σ ft =
(a)
(h)
(i)
Pf
Af
=
...................................................................................................................... Ans.
σ f Ao
Af
≅
450 ( 24.98 )
= 483 MPa ................................................................. Ans.
23.29
∆σ ( 410 − 393) (10
Et =
≅
∆ε
0.06 − 0.04
6
) = 850 10 N/m
( )
6
2
= 850 MPa .................................... Ans.
6
∆σ ( 410 − 0 ) (10 )
Es =
≅
= 6.83 (109 ) N/m 2 = 6.83 GPa ....................................... Ans.
∆ε
0.06
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-5
Ao = π ( 0.25 ) 4 = 0.04909 in.2
2
Af = π ( 0.212 ) 4 = 0.03530 in.2
2
From the
σ −ε
(a)
E=
(b)
σ PL
(c)
σ ult ≅ 73 ksi
(d)
σ ys ( 0.05% ) ≅ 43 ksi ............................................................................................................ Ans.
(e)
σ ys ( 0.20% ) ≅ 43 ksi ............................................................................................................ Ans.
(f)
σ t ≅ 65 ksi
(g)
σ ft =
(h)
Et =
(i)
diagram:
∆σ
34.5 − 0
≅
= 27, 600 ksi ............Ans.
∆ε 0.00125 − 0
≅ 36 ksi .........................................................Ans.
Pf
Af
=
.........................................................Ans.
.............................................................................................................................. Ans.
σ f Ao
Af
≅
65 ( 0.04909 )
= 90 ksi ..................................................................... Ans.
0.03530
∆σ
64 − 50
≅
= 467 ksi ...................................................................................... Ans.
∆ε 0.06 − 0.03
∆σ 56 − 0
≅
= 1400 ksi ............................................................................................. Ans.
Es =
∆ε
0.04
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MECHANICS OF MATERIALS, 6th Edition
4-6
RILEY, STURGES AND MORRIS
Ao = π ( 5.64 ) 4 = 24.98 mm 2
2
= 24.98 (10−6 ) m 2
Af = π ( 4.75 ) 4
2
= 17.72 mm 2 = 17.72 (10−6 ) m 2
From the
(a)
σ −ε
diagram:
∆σ ( 225 − 0 ) (10
≅
E=
∆ε
0.0012 − 0
6
)
E ≅ 187 (109 ) N/m 2 = 187 GPa ........................................................................................ Ans.
(b)
σ PL ≅ 270 MPa ...................................................................................................................... Ans.
(c)
σ ult ≅ 510 MPa
(d)
σ ys ( 0.05% ) ≅ 305 MPa
...................................................................................................... Ans.
(e)
σ ys ( 0.20% ) ≅ 328 MPa
...................................................................................................... Ans.
(f)
σ t ≅ 450 MPa
(g)
σ ft =
(h)
(i)
Pf
Af
=
...................................................................................................................... Ans.
........................................................................................................................ Ans.
σ f Ao
Af
≅
450 ( 24.98 )
= 634 MPa ................................................................. Ans.
17.72
∆σ ( 460 − 410 ) (10
≅
Et =
∆ε
0.04 − 0.02
∆σ ( 440 − 0 ) (10
≅
Es =
∆ε
0.03
6
6
) = 2.50 10 N/m
( )
9
) = 14.67 10 N/m
( )
9
2
2
= 2.50 GPa .................................. Ans.
= 14.67 GPa .................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-7*
Ao = π ( 0.505 ) 4 = 0.200 in.2
2
Af = π ( 0.425 ) 4 = 0.142 in.2
First calculate stresses and strains from the given data and draw the
σ − ε diagram:
σ −ε
diagram (next page). Then, from the
∆σ
32 − 0
≅
= 26, 700 ksi .................................................................................. Ans.
∆ε 0.0012 − 0
≅ 38 ksi ............................................................................................................................ Ans.
(a)
E=
(b)
σ PL
(c)
σ ult ≅ 73 ksi
(d)
σ ys ( 0.05% ) ≅ 43 ksi ................................. Ans.
(e)
σ ys ( 0.20% ) ≅ 47 ksi ................................. Ans.
(f)
σ t ≅ 65 ksi
(g)
σ ft =
(h)
Et =
(i)
2
Pf
Af
=
................................................. Ans.
................................................... Ans.
σ f Ao
Af
≅
13
= 91 ksi ........ Ans.
0.142
∆σ
7.75
≅
= 2400 ksi ................ Ans.
∆ε 0.0032
∆σ
46 − 0
≅
= 14, 400 ksi ....................................................................................... Ans.
Es =
∆ε 0.0032
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-8
Ao = π (11.28 ) 4 = 99.93 mm 2
2
Af = π ( 9.50 ) 4 = 70.88 mm 2
2
First calculate stresses and strains from the given
data and draw the σ − ε diagram (next page). Then,
from the σ − ε diagram:
(a)
∆σ ( 222 − 0 ) (10
≅
E=
∆ε
0.0012 − 0
6
)
E ≅ 185 (109 ) N/m 2 = 185 GPa ........................................................................................ Ans.
(b)
σ PL ≅ 270 MPa ...................................................................................................................... Ans.
(c)
σ ult ≅ 510 MPa
(d)
σ ys ( 0.05% ) ≅ 305 MPa
...................................................................................................... Ans.
(e)
σ ys ( 0.20% ) ≅ 328 MPa
...................................................................................................... Ans.
(f)
σ t ≅ 450 MPa
(g)
σ ft =
(h)
(i)
Pf
Af
=
...................................................................................................................... Ans.
........................................................................................................................ Ans.
σ f Ao
Af
≅
450 ( 99.93)
= 634 MPa ................................................................. Ans.
70.88
6
∆σ 80 (10 )
≅
= 16.7 (109 ) N/m 2 = 16.7 GPa .................................................... Ans.
Et =
∆ε
0.0048
6
∆σ 315 (10 )
≅
= 109 (109 ) N/m 2 = 109 GPa .................................................... Ans.
Es =
∆ε
0.0029
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-9*
The given values are
ε x = 900 µ in./in.
ε y = −300 µ in./in.
E = 10, 000 ksi
γ xy = −400 µ rad
ν = 0.30
E
10, 000
ε +νε y ) =
 900 ) + 0.30 ( −300 )  (10−6 )
2 (
2 ( x
1 −ν
1 − ( 0.30 )
σx =
σ x = +8.90 ksi = 8.90 ksi (T)
σy =
............................................................................................ Ans.
E
10, 000
ε +νε x ) =
 −300 ) + 0.30 ( 900 )  (10−6 )
2 (
2 ( y
1 −ν
1 − ( 0.30 )
σ y = −0.330 ksi = 0.330 ksi (C) ....................................................................................... Ans.
G=
E
10, 000
=
= 3846.15 ksi
2 (1 + ν ) 2 (1 + 0.30 )
τ xy = Gγ xy = 3846.15 ( −400 ×10−6 ) = −1.538 ksi .......................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-10*
The given values are
ε x = 1175 µ m/m
ε y = −1250 µ m/m
E = 190 GPa
γ xy = 850 µ rad
ν = 0.25
190 ×103 )
(
E
σx =
 1175 ) + 0.25 ( −1250 )  (10−6 )
(ε x +νε y ) =
2 (
1 −ν 2
1 − ( 0.25 )
σ x = +174.8 MPa = 174.8 MPa (T) ................................................................................. Ans.
190 × 103 )
(
E
σy =
ε +νε x ) =
 −1250 ) + 0.25 (1175 )  (10−6 )
2 (
2 ( y
1 −ν
1 − ( 0.25 )
σ y = −193.8 MPa = 193.8 MPa (C)
G=
................................................................................ Ans.
E
190
=
= 76.00 GPa
2 (1 + ν ) 2 (1 + 0.25 )
τ xy = Gγ xy = ( 76.0 × 103 )( 850 ×10−6 ) = +64.6 MPa ..................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-11
The given values are
ε x = 500 µ in./in.
ε y = 250 µ in./in.
E = 15, 000 ksi
γ xy = 150 µ rad
ν = 0.34
E
15, 000
ε +νε y ) =
 500 ) + 0.34 ( 250 )  (10−6 )
2 (
2 ( x
1 −ν
1 − ( 0.34 )
σx =
σ x = +9.92 ksi = 9.92 ksi (T)
σy =
............................................................................................ Ans.
E
15, 000
ε +νε x ) =
 250 ) + 0.34 ( 500 )  (10−6 )
2 (
2 ( y
1 −ν
1 − ( 0.34 )
σ y = +7.12 ksi = 7.12 ksi (T) ............................................................................................ Ans.
G=
E
15, 000
=
= 5597.01 ksi
2 (1 + ν ) 2 (1 + 0.34 )
τ xy = Gγ xy = 5597.01(150 × 10−6 ) = +0.840 ksi ............................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-12
The given values are
ε x = 1000 µ m/m
ε y = 400 µ m/m
γ xy = 800 µ rad
ν = 0.25
E = 210 GPa
210 ×103 )
(
E
σx =
 1000 ) + 0.25 ( 400 )  (10−6 )
(ε x +νε y ) =
2 (
1 −ν 2
1 − ( 0.25 )
σ x = +246 MPa = 246 MPa (T) ........................................................................................ Ans.
( 210 ×10 )  400 + 0.25 1000  10−6
E
σy =
ε +νε x ) =
)
(
) ( )
2 (
2 ( y
1 −ν
1 − ( 0.25 )
3
σ y = +145.6 MPa = 145.6 MPa (T) ................................................................................. Ans.
G=
E
210
=
= 84.00 GPa
2 (1 + ν ) 2 (1 + 0.25 )
τ xy = Gγ xy = ( 84.0 ×103 )( 800 × 10−6 ) = +67.2 MPa ..................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-13*
The given values are
σ x = 15, 000 psi
σ y = 5000 psi
σ z = 7500 psi
E = 30, 000 ksi
τ xy = 5500 psi
τ yz = 4750 psi
τ zx = 3200 psi
ν = 0.30
G=
εx =
E
30, 000
=
= 11,538.46 ksi
2 (1 + ν ) 2 (1 + 0.30 )
σ x −ν (σ y + σ z ) 15, 000 − 0.30 ( 5000 + 7500 )
E
=
( 30 ×10 )
6
ε x = +375 (10−6 ) in./in. = +375 µ in./in. .......................................................................... Ans.
σ y −ν (σ x + σ z )
εy =
E
=
5000 − 0.30 (15, 000 + 7500 )
( 30 ×10 )
6
ε y = −58.0 (10−6 ) in./in. = −58.0 µ in./in. ....................................................................... Ans.
εz =
σ z − ν (σ x + σ y )
E
=
7500 − 0.30 (15, 000 + 5000 )
( 30 ×10 )
6
ε z = +50.0 (10−6 ) in./in. = +50.0 µ in./in. ........................................................................ Ans.
γ xy =
γ yz =
γ zx =
τ xy
G
τ yz
G
τ zx
G
=
5500
= +477 (10−6 ) rad = +477 µ rad ................................... Ans.
6
×
11.53846
10
(
)
=
4750
= +412 (10−6 ) rad = +412 µ rad ................................... Ans.
6
(11.53846 ×10 )
=
3200
= +277 (10−6 ) rad = +277 µ rad ................................... Ans.
6
(11.53846 ×10 )
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-14*
The given values are
σ x = 120 MPa
σ y = −85 MPa
σ z = 45 MPa
E = 73 GPa
τ xy = 35 MPa
τ yz = 48 MPa
τ zx = 76 MPa
ν = 0.33
G=
εx =
E
73
=
= 27.444 GPa
2 (1 + ν ) 2 (1 + 0.33)
σ x −ν (σ y + σ z ) 120 − 0.33 ( −85 + 45 )
E
=
( 73 ×10 )
9
ε x = +1825 (10−6 ) m/m = +1825 µ m/m .......................................................................... Ans.
σ y − ν (σ x + σ z )
εy =
E
=
( −85) − 0.33 (120 + 45)
( 73 ×10 )
9
ε y = −1910 (10−6 ) m/m = −1910 µ m/m
εz =
σ z − ν (σ x + σ y )
E
=
45 − 0.33 (120 − 85 )
( 73 ×10 )
9
ε z = +458 (10−6 ) m/m = +458 µ m/m
γ xy =
γ yz =
γ zx =
τ xy
G
τ yz
G
τ zx
G
......................................................................... Ans.
.............................................................................. Ans.
=
35 ×106
= 1275 (10−6 ) rad = 1275 µ rad ......................................... Ans.
9
27.444
×
10
(
)
=
48 ×106
= 1749 (10−6 ) rad = 1749 µ rad ......................................... Ans.
9
( 27.444 ×10 )
=
76
= 2770 (10−6 ) rad = 2770 µ rad .................................... Ans.
6
(11.53846 ×10 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
4-15
γ xy =
∆y
Lx
=
RILEY, STURGES AND MORRIS
0.001
= 0.00200 in./in.
0.5
τ xy = Gγ xy = ( 3000 )( 0.00200 ) = 6 psi
P = 2τ xy A = 2 ( 6 )( 2 × 4 ) = 96 lb ................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
4-16*
The given values are
RILEY, STURGES AND MORRIS
ν = 0.33
E = 73 GPa
ε a = ε x = 875 µ m/m
ε b = ε120° = 700 µ m/m
ε c = ε 60° = −650 µ m/m
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε b = ( 875 ) cos 2 (120° ) + ε y sin 2 (120° ) + γ xy sin (120° ) cos (120° ) = 700
ε c = ( 875 ) cos 2 ( 60° ) + ε y sin 2 ( 60° ) + γ xy sin ( 60° ) cos ( 60° ) = −650
0.75000ε y − 0.43301γ xy = 481.25
0.75000ε y + 0.43301γ xy = −868.75
ε y = −258.33 µ m/m
σx =
γ xy = −1558.85 µ rad
( 73 ×103 )  875 + 0.33 −258.33  10−6
E
ε
νε
+
=
)
(
) ( )
( x y)
2 (
1 −ν 2
1 − ( 0.33)
σ x = +64.7 MPa = 64.7 MPa (T) ..................................................................................... Ans.
73 × 103 )
(
E
σy =
 −258.33) + 0.33 ( 875 )  (10−6 )
(ε y +νε x ) =
2 (
1 −ν 2
1 − ( 0.33)
σ y = +2.49 MPa = 2.49 MPa (T) ..................................................................................... Ans.
G=
E
73
=
= 27.444 GPa
2 (1 + ν ) 2 (1 + 0.33)
τ xy = Gγ xy = ( 27.444 × 103 )( −1558.85 × 10−6 ) = −42.8 MPa ..................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-17
The given values are
E = 30, 000 ksi
ν = 0.30
ε a = ε x = 650 µ in./in.
ε b = ε 45° = 475 µ in./in.
ε c = ε y = −250 µ in./in.
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε b = ( 650 ) cos 2 ( 45° ) + ( −250 ) sin 2 ( 45° ) + γ xy sin ( 45° ) cos ( 45° ) = 475
γ xy = +550 µ rad
E
30, 000
ε + νε y ) =
 650 ) + 0.30 ( −250 )  (10−6 )
2 (
2 ( x
1 −ν
1 − ( 0.30 )
σx =
σ x = +18.96 ksi = 18.96 ksi (T) ........................................................................................ Ans.
σy =
E
30, 000
ε + νε x ) =
 −250 ) + 0.30 ( 650 )  (10−6 )
2 (
2 ( y
1 −ν
1 − ( 0.30 )
σ y = −1.813 ksi = 1.813 ksi (C) ........................................................................................ Ans.
G=
E
30, 000
=
= 11,538.46 ksi
2 (1 + ν ) 2 (1 + 0.30 )
τ xy = Gγ xy = (11,538.46 ) ( 550 × 10−6 ) = +6.35 ksi ........................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
4-18
The given values are
RILEY, STURGES AND MORRIS
ν = 0.30
E = 200 GPa
ε a = ε x = 540 µ m/m
ε b = ε 45° = 930 µ m/m
ε c = ε y = 20 µ m/m
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε b = ( 540 ) cos 2 ( 45° ) + ( 20 ) sin 2 ( 45° ) + γ xy sin ( 45° ) cos ( 45° ) = 930
γ xy = +1300 µ rad
( 200 ×10 )  540 + 0.30 20  10−6
E
σx =
ε + νε y ) =
)
( ) ( )
2 (
2 ( x
1 −ν
1 − ( 0.30 )
3
σ x = +120.0 MPa = 120.0 MPa (T) ................................................................................. Ans.
200 × 103 )
(
E
σy =
ε + νε x ) =
 20 ) + 0.30 ( 540 )  (10−6 )
2 (
2 ( y
1 −ν
1 − ( 0.30 )
σ y = +40.0 MPa = 40.0 MPa (T) ..................................................................................... Ans.
G=
E
200
=
= 76.923 GPa
2 (1 + ν ) 2 (1 + 0.30 )
τ xy = Gγ xy = ( 76.923 × 103 )(1300 × 10−6 ) = +100.0 MPa ............................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
4-19*
The given values are
σ x = 8 ksi
εx =
εy =
G=
σ x −νσ y
E
σ y −νσ x
E
RILEY, STURGES AND MORRIS
ν = 0.30
τ xy = −5 ksi
E = 30, 000 ksi
σ y = 0 ksi
=
( 8) − 0.30 ( 0 ) = 267
(10 ) = 267 µin./in. ........................................ Ans.
=
( 0 ) − 0.30 ( 8) = −80
(10 ) = −80 µin./in. ....................................... Ans.
30, 000
30, 000
−6
−6
E
30, 000
=
= 11,538.46 ksi
2 (1 + ν ) 2 (1 + 0.30 )
γ xy =
τ xy
G
=
−5
= −433 (10−6 ) = −433 µ rad ....................................................... Ans.
11,538.46
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
4-20*
The given values are
RILEY, STURGES AND MORRIS
ν = 0.30
E = 200 GPa
ε a = ε x = −555 µ m/m
ε b = ε120° = 925 µ m/m
ε c = ε −120° = 740 µ m/m
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε b = ( −555 ) cos 2 (120° ) + ε y sin 2 (120° ) + γ xy sin (120° ) cos (120° ) = 925
ε c = ( −555 ) cos 2 ( −120° ) + ε y sin 2 ( −120° ) + γ xy sin ( −120° ) cos ( −120° ) = 740
0.75000ε y − 0.43301γ xy = 1063.75
0.75000ε y + 0.43301γ xy = 878.75
ε y = 1295.0 µ m/m
γ xy = −213.620 µ rad
( 200 ×10 )  −555 + 0.30 1295  10−6
E
ε +νε y ) =
)
(
) ( )
2 (
2 ( x
1 −ν
1 − ( 0.30 )
3
(a)
σx =
σ x = −36.593 MPa ≅ 36.6 MPa (C)
................................................................................ Ans.
200 × 103 )
(
E
σy =
 1295 ) + 0.30 ( −555 )  (10−6 )
(ε y +νε x ) =
2 (
1 −ν 2
1 − ( 0.30 )
σ y = +248.022 MPa ≅ 248 MPa (T) ............................................................................... Ans.
τ xy = Gγ xy =
E
200 × 103
γ xy =
( −213.620 ×10−6 )
2 (1 + ν )
2 (1 + 0.30 )
τ xy = −16.4323 MPa ≅ −16.43 MPa ................................................................................ Ans.
(b)
1
2
θ p = tan −1
When
2τ xy
σ x −σ y
=
2 ( −16.4323)
1
= 3.293°, − 86.707°
tan −1
2
( −36.593) − ( 248.022 )
θ p = 3.293°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( −36.593) cos 2 θ p + ( 248.022 ) sin 2 θ p + 2 ( −16.4323) sin θ p cos θ p
= −37.539 MPa = σ p 2
σ p1 = σ x + σ y − σ p 2 = 248.968 MPa
τ max = τ p = (σ p1 − σ p 2 ) 2 = 143.3 MPa
σ n 45 = (σ p1 + σ p 2 ) 2 = 105.7 MPa
σ p1 = 249 MPa (T)
σ p 2 = 37.5 MPa (C)
τ max = τ p = 143.3 MPa
86.71° ........................Ans.
3.29° ............................................................................................ Ans.
......................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
4-21*
The given values are
RILEY, STURGES AND MORRIS
ν = 0.30
E = 30, 000 ksi
ε a = ε x = 1000 µ in./in.
ε b = ε 60° = 2000 µin./in.
ε c = ε120° = 1200 µ in./in.
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε b = (1000 ) cos 2 ( 60° ) + ε y sin 2 ( 60° ) + γ xy sin ( 60° ) cos ( 60° ) = 2000
ε c = (1000 ) cos 2 (120° ) + ε y sin 2 (120° ) + γ xy sin (120° ) cos (120° ) = 1200
0.75000ε y + 0.43301γ xy = 1750
0.75000ε y − 0.43301γ xy = 950
ε y = 1800 µ in./in.
εp =
εx +εy
2
γ xy = 923.760 µ rad
 ε x − ε y   γ xy 
± 
 +

 2   2 
2
2
(1000 ) + (1800 ) ±
=
 (1000 ) − (1800 )   923.760 

 +

2
2
2


 
= 1400 ± 611.01 µ in./in.
2
ε p1 = 1400 + 611.01 µ in./in. = 2011.01 µ in./in.
2
( < 2200 µin./in. - okay )
ε p 2 = 1400 − 611.01 µ in./in. = 788.99 µ in./in.
ε p3 = ε z =
−ν ( ε x + ε y )
1 −ν
=
−0.30 (1000 ) + (1800 ) 
1 − 0.30
γ max = ( 2011) − ( −1200 ) µ rad = 3211 µ rad
σ p1 =
= −1200 µ in./in.
( > 2500 µ rad - design fails )
( 30, 000 )  2011.01 + 0.30 788.99  10−6
E
ε +νε p 2 ) =
)
(
) ( )
2 (
2 ( p1
1 −ν
1 − ( 0.30 )
( > 74 ksi - design fails )
( 30, 000 ) 3211×10−6
E
τ max = Gγ max =
γ max =
(
)
2 (1 +ν )
2 (1 + 0.30 )
= 37.05 ksi ( < 40 ksi - okay )
= 74.1 ksi
Design fails since both γ max and σ p1 are above the design limits. ........................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-22
The given values are
δ x = 0 mm
Then
εx = 0 =
εz =
δ y = 0 mm
σ x − 0.4 (σ y + σ z )
1400
δ z = −0.4 mm
εy = 0 =
σ z = −P A
σ y − 0.4 (σ x + σ z )
1400
−0.4 σ z − 0.4 (σ x + σ y )
=
25.4
1400
σ x − 0.4σ y − 0.4σ z = 0 MPa
−0.4σ x + σ y − 0.4σ z = 0 MPa
−0.4σ x − 0.4σ y + σ z = −22.04724 MPa
σ x = σ y = −31.496 MPa
σ z = −47.244 MPa
P = −σ z A = − ( 22.04724 × 106 ) ( 0.010 × 0.010 ) = 4.72 (103 ) N
P = 4.72 kN ............................................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
4-23
The given values are
E = 10, 600 ksi
ε a = ε x = 875 µ in./in.
RILEY, STURGES AND MORRIS
ν = 0.33
ε b = ε135° = 700 µ in./in.
ε c = ε −135° = −350 µ in./in.
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
ε b = ( 875 ) cos 2 (135° ) + ε y sin 2 (135° ) + γ xy sin (135° ) cos (135° ) = 700
ε c = ( 875 ) cos 2 ( −135° ) + ε y sin 2 ( −135° ) + γ xy sin ( −135° ) cos ( −135° ) = −350
0.5000ε y − 0.5000γ xy = 262.5
0.5000ε y + 0.5000γ xy = −787.5
ε y = −525.00 µ in./in.
γ xy = −1050.00 µ rad
(a)
1
2
θ p = tan −1
When
γ xy
( −1050 ) = −18.435°, 71.565°
1
= tan −1
εx −εy 2
(875 ) − ( −525)
θ p = −18.435°
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 875 ) cos 2 θ p + ( −525 ) sin 2 θ p + ( −1050 ) sin θ p cos θ p
= 1050.00 µ in./in. = ε p1
ε p 2 = ε x + ε y − ε p1 = −700.00 µ in./in.
ε p1 = 1050 µ in./in.
18.43° ............................................................................................. Ans.
ε p 2 = −700 µ in./in.
71.57° ............................................................................................ Ans.
−ν
−0.33
εx + εy ) =
( 875 ) + ( −525 )  = −172.4 µ in./in. ......................... Ans.
(
1 −ν
1 − 0.33 
= γ p = ε p1 − ε p 2 = 1750 µ rad ..................................................................................... Ans.
ε p3 =
γ max
(b)
σ p1 =
(10, 600 )  1050 + 0.33 −700  10−6
E
ε +νε p 2 ) =
)
(
) ( )
2 (
2 ( p1
1 −ν
1 − ( 0.33)
σ p1 = +9.7423 ksi ≅ 9.74 ksi (T) ...................................................................................... Ans.
σ p2 =
(10, 600 )  −700 + 0.33 1050  10−6
E
ε +νε p1 ) =
)
(
) ( )
2 (
2 ( p2
1 −ν
1 − ( 0.33)
σ p 2 = −4.2050 ksi ≅ 4.21 ksi (C) ...................................................................................... Ans.
σ p 3 = 0 ksi
.............................................................................................................................. Ans.
τ max = τ p = (σ p1 − σ p 2 ) 2 = 6.97 ksi ................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
4-24
The given values are
RILEY, STURGES AND MORRIS
ε ai = 500 µ m/m
ε hi = 750 µ m/m
ν = 0.30
σ ri = − p = −100 MPa
ε ao = 500 µ m/m
ε ho = 100 µ m/m
σ ro = 0 MPa
E = 200 GPa
On the inside:
σ ri =
−100 (10
6
E
(1 −ν ) ε ri + ν ( ε ai + ε hi ) 
(1 +ν )(1 − 2ν ) 
200 (109 )
) = (1 + 0.30 )(1 − 0.60 ) (1 − 0.30 ) ε
ri
+ 0.30 ( 500 + 750 )  (10−6 )
ε ri = −907.1 µ m/m
σ ai =
=
E
(1 −ν ) ε ai +ν ( ε ri + ε hi ) 
(1 +ν )(1 − 2ν ) 
200 (109 )
(1 − 0.30 )( 500 ) + 0.30 ( −907.1 + 750 )  (10−6 )
(1 + 0.30 )(1 − 0.60 )
σ ai = 116.5 (106 ) N/m 2 = 116.5 MPa (T) ....................................................................... Ans.
σ hi =
=
E
(1 −ν ) ε hi +ν ( ε ri + ε ai ) 
(1 +ν )(1 − 2ν ) 
200 (109 )
(1 − 0.30 )( 750 ) + 0.30 ( −907.1 + 500 )  (10−6 )
(1 + 0.30 )(1 − 0.60 )
σ hi = 155.0 (106 ) N/m 2 = 155.0 MPa (T) ....................................................................... Ans.
On the outside:
σ ao
200 (109 )
E
=
ε + νε ho ) =
 500 ) + 0.30 (100 )  (10−6 )
2 (
2 ( ao
1 −ν
1 − ( 0.30 )
σ ao = 116.5 (106 ) N/m 2 = 116.5 MPa (T)
σ ho
...................................................................... Ans.
200 (109 )
E
=
ε + νε ao ) =
 100 ) + 0.30 ( 500 )  (10−6 )
2 (
2 ( ho
1 −ν
1 − ( 0.30 )
σ ho = 54.9 (106 ) N/m 2 = 54.9 MPa (T)
.......................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
4-25*
300 − τ i (π 3.25 )( 8 ) = 0
RILEY, STURGES AND MORRIS
τ i = 3.67281 psi = 2000γ i
γ i = 1836.403 µ rad
300 − τ o (π 4.25 )( 8 ) = 0
τ o = 2.80862 psi = 2000γ o
γ o = 1404.308 µ rad
γ avg =
1836.403 + 1404.308
∆x
= 1620.356 µ rad ≅
2
0.5
∆x ≅ 0.5 (1620.356 × 10−6 ) = 8.10 (10−4 ) in. ................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-26*
(a)
δ L = α∆TL = (12.1×10−6 ) ( 70 )( 2500 ) = 2.12 mm ....................................................... Ans.
δ do = α∆Td o = (12.1× 10−6 ) ( 70 )(105 ) = 0.0889 mm
δ di = α∆Tdi = (12.1×10−6 ) ( 70 )( 70 ) = 0.0593 mm
(b)
.................................................. Ans.
..................................................... Ans.
δ L = α∆TL = (12.1×10−6 ) ( −85 )( 2500 ) = −2.57 mm .................................................. Ans.
δ do = α∆Td o = (12.1×10−6 ) ( −85 )(105 ) = −0.1080 mm ............................................. Ans.
δ di = α∆Tdi = (12.1×10−6 ) ( −85 )( 70 ) = −0.0720 mm ................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-27*
δ =εL =
0.05 =
PL
+ α ∆T L
AE
3000 ( 4 ×12 )
(1× 2 ) ( 30 ×10
6
)
+ ( 6.6 × 10−6 ) ( ∆T )( 4 × 12 )
∆T = 150.3 °F ........................................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-28
PL
+ α∆TL
AE
= 0 + ( 22.5 ×10−6 ) ( −80 ) ( 40 × 103 )
δ =εL =
δ = −72.0 mm ........................................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-29
δ =εL =
PL
+ α ∆T L
AE
δ L = 0 + ( 6.5 ×10−6 ) ( 250 )( 225 × 12 ) = 4.39 in. ............................................................. Ans.
δ d = 0 + ( 6.5 × 10−6 ) ( 250 )(12 × 12 ) = 0.234 in. ............................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-30*
99.8 + δ br = 100 + δ st
δ =εL =
PL
+ α ∆T L
AE
99.8 + 0 + (16.9 ×10−6 ) ( ∆T )( 99.8 )  = 100 + 0 + (11.9 × 10−6 ) ( ∆T )(100 ) 
∆T = 403 °C ........................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-31*
δ =εL =
PL
+ α ∆T L
AE
δ a = 0 + (12.5 × 10−6 ) ( 80 )( 20 ) = 0.020000 in.
δ s = 0 + ( 6.6 × 10−6 ) ( 80 )( 20 ) = 0.0105600 in.
b = 5 (δ a − δ s ) = 5 ( 0.02 − 0.01056 ) = 0.0472 in. ↑ ..................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-32
δ =εL =
PL
+ α ∆T L
AE
δ a = 0 + ( 22.5 × 10−6 ) ( 75 )( 300 ) = 0.50625 mm
δ s = 0 + (11.9 ×10−6 ) ( 75 )( 300 ) = 0.26775 mm = b
250
250
(δ a − δ s ) =
( 0.50625 − 0.26775)
25
25
a = 2.12 mm ← ..........................................................................Ans.
a+b =
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
4-33
ε d = εσ + εT =
=
σ d −νσ a
E
0 − 0.33 ( 4 )
(10, 000 ) π ( 0.25)
RILEY, STURGES AND MORRIS
+ α ∆T
2
4

+ (12.5 ×10−6 ) ( 60 ) =
δd
0.25
δ d = +0.000860 in. ............................................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
4-34*
(a)
(b)
P = W = 2500 ( 9.81) = 24,525 N
P
24,525
=
= 49.962 (106 ) N/m 2 ≅ 50.0 MPa ......................................... Ans.
A π ( 0.025 )2 4
σ=
ε=
σ
E
+ α ∆T =
49.962 (106 )
73 (10
9
)
+ ( 22.5 × 10−6 ) ( −50 )
ε = −441(10−6 ) = −441 µ m/m
(c)
RILEY, STURGES AND MORRIS
ε d = εσ + εT =
εd =
σ d −νσ a
E
........................................................................................... Ans.
+ α ∆T
0 − 0.33 ( 49.962 × 106 )
+ ( 22.5 × 10−6 ) ( −50 ) =
δd
73 ×10
25
δ d = −0.0338 mm ................................................................................................................. Ans.
9
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
4-35
εd =
RILEY, STURGES AND MORRIS
( 25 − 10 )
P
+ α ∆T =
+ ( 6.5 × 10−6 ) (100 − 72 ) = 214 (10−6 )
6
AE
(1 2 )(1 32 ) ( 30 ×10 )
∆L = ε L = ( 214 ×10−6 ) (100 × 12 ) = +0.257 in.
correction = −0.257 in. ........................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
4-36
(a)
RILEY, STURGES AND MORRIS
TAB = TBC = P = 25 kN
δ AB =
( 25, 000 )( 200 )
PL
+ α∆TL =
+ (12 × 10−6 ) ( 20 )( 200 )
AE
π ( 0.050 )2 4  ( 200 ×109 )


δ AB = 0.0607 mm .................................................................................................................. Ans.
(b)
δ BC =
( 25, 000 )(150 )
PL
+ α∆TL =
+ ( 22.5 × 10−6 ) ( 20 )(150 )
2
9
AE
π 0.025 ) 4  ( 70 × 10 )
 (

δ BC = 0.1766 mm .................................................................................................................. Ans.
(c)
∆ C = δ AB + δ BC = 0.237 mm → ........................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-37*
From Table 4-1 for the T300/5208 material
E1 = 26,300 ksi
ν 21 =
E2 = 1494 ksi
G12 = 1040 ksi
ν 12 = 0.28
E2
1494
ν 12 =
( 0.28) = 0.01591
E1
26,300
The given data are
ε1 = 2000 µ in./in.
σ1 =
E1
1 −ν 12ν 21
ε 2 = 4000 µ in./in.
(ε1 +ν 21ε 2 ) =
γ 12 = 1500 µ rad
26,300
( 2000 ) + 0.01591( 4000 )  (10−6 )
1 − ( 0.28 )( 0.01591)
σ 1 = +54.5 ksi = 54.5 ksi (T) ............................................................................................. Ans.
σ2 =
E2
1 −ν 12ν 21
(ε 2 +ν 12ε1 ) =
1494
( 4000 ) + 0.28 ( 2000 )  (10−6 )
1 − ( 0.28 )( 0.01591) 
σ 2 = +6.84 ksi = 6.84 ksi (T)
............................................................................................ Ans.
τ 12 = G12γ 12 = (1040 ) (1500 × 10−6 ) = +1.560 ksi ........................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-38*
From Table 4-1 for the Scotchply 1002 Glass/Epoxy material
E1 = 36.8 GPa
E2 = 8.27 GPa
G12 = 4.14 GPa
ν 12 = 0.26
E2
8.27
ν 12 =
( 0.26 ) = 0.05843
E1
36.8
ν 21 =
The given data are
σ 1 = 30 MPa
ε1 =
σ1
E1
−ν 21
σ 2 = −2 MPa
σ2
E2
ε1 = 829 µ m/m
ε2 =
σ2
E2
−ν 12
σ1
E1
τ 12
G12
=
30
−2
− ( 0.05843)
= 829 (10−6 ) m/m
3
3
36.8 (10 )
8.27 (10 )
=
...................................................................................................................... Ans.
=
ε 2 = −454 µ m/m
γ 12 =
τ 12 = 0.3 MPa
−2
30
− ( 0.26 )
= −454 (10−6 ) m/m
3
3
8.27 (10 )
36.8 (10 )
................................................................................................................... Ans.
0.3
= +72.5 (10−6 ) rad = +72.5 µ rad ............................................ Ans.
3
4.14 (10 )
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-39
From Table 4-1 for the T300/5208 material
E1 = 26,300 ksi
ν 21 =
E2 = 1494 ksi
G12 = 1040 ksi
ν 12 = 0.28
E2
1494
ν 12 =
( 0.28) = 0.01591
E1
26,300
The given data are
σ 1 = 40 ksi
ε1 =
σ1
E1
σ 2 = −10 ksi
−ν 21
σ2
E2
=
τ 12 = 2 ksi
( −10 ) = 1627 10−6 in./in.
40
− ( 0.01591)
( )
26,300
1494
ε1 = 1627 µ in./in. ................................................................................................................... Ans.
ε2 =
σ2
E2
−ν 12
σ1
E1
=
( −10 ) −
1494
( 0.28)
40
= −7120 (10−6 ) in./in.
26,300
ε 2 = −7120 µ in./in. ............................................................................................................... Ans.
γ 12 =
τ 12
G12
=
2
= +1923 (10−6 ) rad = +1923 µ rad ................................................... Ans.
1040
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-40
From Table 4-1 for the Boron/epoxy material
E1 = 200 GPa
E2 = 20 GPa
ν 21 =
G12 = 6 GPa
ν 12 = 0.23
E2
20
ν 12 =
( 0.23) = 0.02300
E1
200
The given data are
ε1 = 1000 µ m/m
σ1 =
E1
1 −ν 12ν 21
ε 2 = 500 µ m/m
(ε1 +ν 21ε 2 ) =
200 (103 )
(1000 ) + 0.023 ( 500 )  (10−6 )
1 − ( 0.23)( 0.023) 
σ 1 = +203 MPa = 203 MPa (T)
σ2 =
E2
1 −ν 12ν 21
(ε 2 +ν 12ε1 ) =
γ 12 = 300 µ rad
........................................................................................ Ans.
20 (103 )
( 500 ) + 0.23 (1000 )  (10−6 )
1 − ( 0.23)( 0.023) 
σ 2 = +14.68 MPa = 14.68 MPa (T) ................................................................................. Ans.
τ 12 = G12γ 12 = ( 6 × 103 )( 300 × 10−6 ) = +1.800 MPa
...................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-41*
From Table 4-1 for the T300/5208 material
E1 = 26,300 ksi
ν 21 =
E2 = 1494 ksi
G12 = 1040 ksi
ν 12 = 0.28
E2
1494
ν 12 =
( 0.28) = 0.01591
E1
26,300
The given data are
σ 1 = 5 ksi
ε1 =
ε2 =
σ1
E1
σ2
E2
−ν 21
−ν 12
σ 2 = 5 ksi
σ2
E2
σ1
E1
τ 12 = 1 ksi
=
5
5
− ( 0.01591)
= 0.00013687 in./in.
26,300
1494
=
5
5
− ( 0.28 )
= 0.003293 in./in.
1494
26,300
δ x = ε 2 L = ( 0.003293)(10 ) = 0.0329 in. ......................................................................... Ans.
δ y = ε1 L = ( 0.00013687 )(10 ) = 0.001369 in. ................................................................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-42
From Table 4-1 for the Scotchply 1002 Glass/Epoxy material
E1 = 36.8 GPa
ν 21 =
E2 = 8.27 GPa
G12 = 4.14 GPa
ν 12 = 0.26
E2
8.27
ν 12 =
( 0.26 ) = 0.05843
E1
36.8
The given data are
σ 1 = 5 MPa
ε1 =
σ1
E1
−ν 21
σ 2 = −2 MPa
σ2
E2
=
( −2 ) = 150.00 10−6 m/m
5
− ( 0.05843)
( )
3
36.8 (10 )
8.27 (103 )
δ x = ε1 L = ( 0.00015000 )(125 ) = 0.01875 mm
ε2 =
σ2
E2
−ν 12
σ1
E1
=
τ 12 = 0 MPa
( −2 )
8.27 (10
3
)
− ( 0.26 )
............................................................. Ans.
5
= −277.16 (10−6 ) m/m
3
36.8 (10 )
δ y = ε 2 L = ( −0.00027716 )(100 ) = −0.0277 mm
......................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
4-43
Since
σy = 0,
ε x = σ x Ex
or
RILEY, STURGES AND MORRIS
σ x = Exε x
The total load carried by the composite is carried partially by the fibers and partially by the matrix. Thus,
P = σ x A = σ f Af + σ m Am = E f ε f Af + Emε m Am
where
ε f = εm = ε x
σx =
since the fibers and the matrix are bonded together. Therefore,
E f Af + Em Am
A
ε x = E xε x
from which
Ex =
E f Af + Em Am
A
=
E f Af
A
+
Em Am
= E f V f + EmVm ................................................... Ans.
A
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-44
The total load carried by the composite is carried partially by the fibers and partially by the matrix. Thus,
P = Pf + Pm = σ f Af + σ m Am
Therefore,
Pf
P
where
=
σ f Af
E f ε f Af
=
σ f Af + σ m Am E f ε f Af + Emε m Am
ε f = εm = ε x
since the fibers and the matrix are
bonded together. Therefore,
Pf
P
=
E f Af
E f Af + Em Am
=
E f Af
E f Af + Em ( A − Af
)
Dividing both the numerator and the denominator by the total area A gives
Pf
P
=
E f ( Af A )
E f ( Af A ) + Em ( Am A )
=
E fVf
E f V f + Em (1 − V f
)
................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
4-45*
(a)
50 − 0
= 25, 000 ksi .....................................................................................................Ans.
0.002 − 0
26 − 0
=
= 13, 000 ksi ...................................................................................................Ans.
0.002 − 0
E75 =
E1600
(b)
RILEY, STURGES AND MORRIS
σ y ( 75° ) ≅ 57 ksi ...........................................................................................................................Ans.
σ y (1600° ) ≅ 21 ksi .......................................................................................................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-46*
The given values are
σ x = 120 MPa
σ y = −80 MPa
E = 70 GPa
εx =
εy =
G=
σ x − ν (σ y + σ z )
E
σ y −ν (σ x + σ z )
E
τ xy = 60 MPa
ν = 0.33
=
(120 ) − 0.33 ( −80 ) + 0 
= +0.00209143 = +2091.43 µ m/m
=
( −80 ) − 0.33 (120 ) + 0
= −0.00170857 = −1708.57 µ m/m
70 (103 )
70 (103 )
E
70
=
= 26.3158 GPa
2 (1 + ν ) 2 (1 + 0.33)
γ xy =
τ xy
G
=
60
= +0.002280 rad = +2280 µ rad
26.3158 (103 )
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 2091.43) cos 2 ( −30° ) + ( −1708.57 ) sin 2 ( −30° ) + ( 2280 ) sin ( −30° ) cos ( −30° )
ε n = 154.2 µ m/m ................................................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-47
The given values are
σ x = 12 ksi
σ y = −4 ksi
E = 30, 000 ksi
εx =
σ x − ν (σ y + σ z )
E
=
τ xy = −6 ksi
ν = 0.30
(12 ) − 0.30 ( −4 ) + 0
30, 000
ε x = ε a = +0.000440 in./in. = +440 µ in./in. ................................................................... Ans.
εy =
σ y −ν (σ x + σ z )
E
=
( −4 ) − 0.30 (12 ) + 0
30, 000
ε y = ε b = −0.00025333 in./in. ≅ −253 µ in./in. .............................................................. Ans.
G=
E
30, 000
=
= 11,538.46 ksi
2 (1 + ν ) 2 (1 + 0.30 )
γ xy =
τ xy
G
=
−6
= −0.000520 rad = −520 µ rad
11,538.46
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 440 ) cos 2 (120° ) + ( −253.33) sin 2 (120° ) + ( −520 ) sin (120° ) cos (120° )
ε n = ε c = 145.2 µ in./in. ........................................................................................................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-48*
The given values are
σ x = −100 MPa
σ z = 0 MPa
E = 2 (1 + ν ) G
210 = 2 (1 +ν )( 80 )
εy =
σ y −ν (σ x + σ z )
E
=
E = 210 GPa
σ y − 0.31250 ( −100 ) + 0 
210 (103 )
G = 80 GPa
ν = 0.31250
=0
σ y = −31.250 MPa
εz =
σ z − ν (σ x + σ y )
E
=
0 − 0.31250 ( −100 ) + ( −31.250 ) 
210 (103 )
= 0.0001953 m/m
δ = ε z L = ( 0.0001953)(10 ) = 0.001953 mm ................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-49
E = 29, 000 ksi
G = 11, 000 ksi
29, 000 = 2 (1 + ν )(11, 000 )
δ xa = ε xa Lxa = ε xa ( 2 ) =
δ xb = ε xb Lxb = ε xb ( 3) =
But
E = 2 (1 + ν ) G
ν = 0.31818
6 − 0.31818σ y
29, 000
( 2)
5
( 3)
29, 000
δ xa = δ xb , therefore
12 − 0.63636σ y = 15
σ y = −4.71 ksi ........................................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
4-50
400 (106 ) − 0
(a)
E1 = E2 =
(b)
σ y1 = 350 MPa
0.002 − 0
RILEY, STURGES AND MORRIS
= 200 (109 ) N/m 2 = 200 GPa ............................................... Ans.
...................................................................................................................... Ans.
σ y 2 = 1000 MPa
.................................................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-51*
Assume series of rails all initially separated by 0.125 in.
When heated, rails expand from center in both directions.
(a)
δ = 0.125 = α ∆T L = ( 6.6 × 10−6 ) ( ∆T )( 55 × 12 )
∆T = 28.7 °F
Rails touch when T = 60 + 28.7 = 88.7 °F .................................................................. Ans.
(b)
δ = ( 6.6 ×10−6 ) ( −50 )( 55 × 12 ) = −0.21780 in.
gap = 0.125 + 0.2178 = 0.3428 in. ≅ 0.343 in. .............................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-52
The given values are
σ x = 72 MPa
σ y = 36 MPa
E = 100 GPa
εx =
σ x −νσ y
E
=
G=
σ y −νσ x
E
=
ν = 0.28
( 72 ) − 0.28 ( 36 ) = +0.00061920 = +619.20 µ m/m
100 (103 )
ε x = ε a = +619 µ m/m
εy =
τ xy = −24 MPa
.......................................................................................................... Ans.
( 36 ) − 0.28 ( 72 ) = 0.00015840 = 158.40 µ m/m
100 (103 )
E
100
=
= 39.0625 GPa
2 (1 + ν ) 2 (1 + 0.28 )
γ xy =
τ xy
G
=
−24
= −0.00061440 rad = −614.40 µ rad
39.0625 (103 )
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 619.20 ) cos 2 ( 45° ) + (158.40 ) sin 2 ( 45° ) + ( −614.40 ) sin ( 45° ) cos ( 45° )
ε n = ε b = +81.6 µ m/m ......................................................................................................... Ans.
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 619.20 ) cos 2 (135° ) + (158.40 ) sin 2 (135° ) + ( −614.40 ) sin (135° ) cos (135° )
ε n = ε c = +696 µ m/m ........................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
4-53
The given values are
σ x = 8.5 ksi
σ y = 4.5 ksi
E = 30, 000 ksi
εx =
εy =
G=
σ x −νσ y
E
σ y −νσ x
E
τ xy = 6 ksi
ν = 0.30
=
(8.5) − 0.30 ( 4.5) = +0.00023833 = +238.33 µin./in.
=
( 4.5) − 0.30 ( 8.5) = 0.00006500 in./in. ≅ +65.00 µin./in.
30, 000
30, 000
E
30, 000
=
= 11,538.46 ksi
2 (1 + ν ) 2 (1 + 0.30 )
γ xy =
τ xy
G
=
6
= 0.00052000 rad = 520.00 µ rad
11,538.46
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 238.33) cos 2 ( 20° ) + ( 65.00 ) sin 2 ( 20° ) + ( 520.00 ) sin ( 20° ) cos ( 20° )
ε n = 385 µ in./in. .................................................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
4-54*
The given values are
ν = 0.33
E = 73 GPa
ε x = 825 µ m/m
(a)
RILEY, STURGES AND MORRIS
ε y = 950 µ m/m
γ xy = 680 µ rad
73 × 103 )
(
E
σx =
ε +νε y ) =
 825 ) + 0.33 ( 950 )  (10−6 )
2 (
2 ( x
1 −ν
1 − ( 0.33)
σ x = +93.2673 MPa ≅ +93.3 MPa (T) ............................................................................ Ans.
73 × 103 )
(
E
σy =
ε +νε x ) =
 950 ) + 0.33 ( 825 )  (10−6 )
2 (
2 ( y
1 −ν
1 − ( 0.33)
σ y = +100.1282 MPa ≅ 100.1 MPa (T) .......................................................................... Ans.
τ xy = Gγ xy
E
73 × 103
=
γ xy =
680 ×10−6 )
(
2 (1 + ν )
2 (1 + 0.33)
τ xy = +18.6617 MPa ≅ +18.66 MPa ................................................................................ Ans.
(b)
1
2
θ p = tan −1
When
2τ xy
σ x −σ y
=
2 (18.6617 )
1
= −39.792°, 50.208°
tan −1
2
( 93.2673) − (100.1282 )
θ p = −39.792
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( 93.2673) cos 2 θ p + (100.1282 ) sin 2 θ p + 2 (18.6617 ) sin θ p cos θ p
= +77.7234 MPa = σ p 2
σ p1 = σ x + σ y − σ p 2 = 115.6721 MPa
σ p1 = 115.7 MPa (T)
50.21° ................. Ans.
σ p 2 = 77.7 MPa (T)
39.79° ................... Ans.
σ p 3 = 0 MPa .................................................... Ans.
τ p = (σ p1 − σ p 2 ) 2 = 18.97 MPa
............... Ans.
σ n 45 = (σ p1 + σ p 2 ) 2 = 96.7 MPa
τ max = (σ p1 − σ p 3 ) 2 = 57.8 MPa (out-of-plane)
.......................................................... Ans.
σ n 45 = (σ p1 + σ p 3 ) 2 = 57.8 MPa
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-1*
AA = π ( 62 − 4.52 ) 4 = 12.37002 in.2
RILEY, STURGES AND MORRIS
PA = PB = 120 kip
AB = π 42 4 = 12.56637 in.2
δ total = ∑
(120 )( 3 ×12 ) + (120 )( 4 ×12 )
PL
=
AE (12.37002 )( 30, 000 ) (12.56637 )(10, 600 )
δ = 0.0549 in.
................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-2*
(a)
RILEY, STURGES AND MORRIS
A = ( 0.100 )( 0.025 ) = 0.002500 m 2
δ AB
350 × 103 ) ( 800 )
(
PL
=
=
= 0.560 mm ........................................................ Ans.
AE ( 0.002500 ) ( 200 × 109 )
(100 ×10 ) (1200 ) = 0.240 mm ........... Ans.
=
( 0.002500 ) ( 200 ×10 )
(180 ×10 ) (1000 ) = 0.360 mm
=
( 0.002500 ) ( 200 ×10 )
3
(b)
δ BC
9
3
(c)
δ CD
9
δ total = 0.560 + 0.240 + 0.360 = 1.160 mm
............. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-3
(a)
RILEY, STURGES AND MORRIS
Apipe = π ( 62 − 4.82 ) 4 = 10.17876 in.2
σ avg =
P
30
=
= 2.95 ksi (C) .................................................................................... Ans.
A 10.17876
( −30 )( 24 )
PL
=
= −0.0024392 in. ≅ 0.00244 in. (shorten) ............. Ans.
AE (10.17876 )( 29, 000 )
(b)
δ=
(c)
ε avg =
δ
L
=
−0.0024392
= −101.6 (10−6 ) = −101.6 µ in./in. ............................................ Ans.
24
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-4*
AT = π ( 25 ) 4 = 490.9 mm 2
2
(a)
δ a−a
AB = π (15 ) 4 = 176.71 mm 2
2
3
3
PL ( 55 ×10 ) (1200 ) + ( 35 × 10 ) (1500 )
=∑
=
AE
( 490.9 ×10−6 )( 73 ×109 )
δ a − a = 3.30693 mm ≅ 3.31 mm ........................................ Ans.
(b)
δ b −b
20 ×103 ) (1500 )
(
PL
=∑
= 3.30693 +
AE
(176.71×10−6 )( 73 ×109 )
δ b −b = 0.563 mm .................................................................. Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-5
(a)
δ =∑
(10 )( 4 ×12 )
PL
=
AE π (1.5 )2 4  ( 30, 000 )


+
δ = 0.0498 in.
(b)
( 20 )( 4 ×12 )
π (1)2 4  ( 30, 000 )


........................................................ Ans.
2
 π (1.5 )2

π (1)
π d 2

W = γV = γ 
48
48
+
=
γ

( )
( )
( 96 )

4
 4

 4

δ =∑
RILEY, STURGES AND MORRIS
d = 1.27475 in.
PL (10 )( 4 × 12 ) + ( 20 )( 4 × 12 )
=
= 0.0376 in. ............................................... Ans.
AE π (1.27475 )2 4  ( 30, 000 )


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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-6
σ=
N
P
=
≤ 100 (106 ) N/m 2
A ( 0.025 × 0.075 )
P ≤ 187.5 (103 ) N
δ=
P ( 2000 )
PL
=
≤ 4 mm
AE ( 0.025 × 0.075 ) ( 70 × 109 )
P ≤ 262.5 (103 ) N
Pmax = 187.5 kN ..................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-7*
(a)
δ=
RILEY, STURGES AND MORRIS
PL ( 230 )(11×12 )
=
AE
( 9 )( 29, 000 )
+
( 20 )( 4 ×12 )
π (1)2 4  ( 30, 000 )


δ = 0.11632 in. ≅ 0.1163 in. .............................. Ans.
(b)
δ =∑
(85)(11×12 )
PL
= 0.11632 +
AE
( 9 )( 29, 000 )
δ = 0.1593 in. ......................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-8
δ =∑
RILEY, STURGES AND MORRIS
P (1000 )
PL
=
AE π ( 0.030 )2 4  ( 200 × 109 )


+
P ( 2000 )
π ( 0.0602 − 0.0402 ) 4  (180 ×109 )


= 3 mm
P = 212 (103 ) N = 212 kN ................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-9*
PL ( 5 )( 8 ) + (10 )( 6 ) + ( 4 )(10 )
=
AE
(1× 2 )(10, 000 )
(a)
δ =∑
(b)
δ = 0.00700 in. ...................................................... Ans.
σ xA = τ xyA = 0 .......................................................... Ans.
σ yA =
RILEY, STURGES AND MORRIS
P
4
=
= 2.00 ksi (T) ........................ Ans.
A (1× 2 )
σ xB = τ xyB = 0 .......................................................... Ans.
σ yB =
P
10
=
= 5.00 ksi (T) ........................ Ans.
A (1× 2 )
σ xC = τ xyC = 0
σ yC =
......................................................................................................................... Ans.
P
5
=
= 2.50 ksi (T) ....................................................................................... Ans.
A (1× 2 )
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MECHANICS OF MATERIALS, 6th Edition
5-10*
RILEY, STURGES AND MORRIS
AA = π ( 602 − 502 ) 4 = 863.938 mm 2
AB = π ( 50 ) 4 = 1963.495 mm 2
2
AC = π ( 25 ) 4 = 490.874 mm 2
2
(a)
135 × 103 ) ( 400 )
(
PL
=
δA =
AE ( 863.938 × 10−6 )( 200 × 109 )
δ A = 0.31252 mm ≅ 0.313 mm ............................... Ans.
(b)
δ total
265 × 103 ) ( 500 )
45 ×103 ) ( 500 )
(
(
PL
=∑
= 0.31252 +
+
AE
(1963.435 ×10−6 )( 73 ×109 ) ( 490.874 ×10−6 )( 73 ×109 )
δ total = 1.865 mm
(c)
(d)
εA =
δ
L
=
................................................................................................................... Ans.
0.31252
= 781(10−6 ) = 781 µ m/m ................................................................. Ans.
400
3
N 135 (10 )
σA = =
= 156.3 (106 ) N/m 2 = 156.3 MPa
A 863.938
3
N 265 (10 )
σB = =
= 135.0 (106 ) N/m 2 = 135.0 MPa
A 1963.495
3
N 45 (10 )
σC = =
= 91.7 (106 ) N/m 2 = 91.7 MPa
A 490.874
σ max = 156.3 MPa ................................................................................................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
5-11
RILEY, STURGES AND MORRIS
AA = π ( 62 − 4.52 ) 4 = 12.3700 in.2
AB = π ( 4 ) 4 = 12.5664 in.2
2
(a)
δA =
( 205 )( 50 )
PL
=
AE (12.3700 )( 29, 000 )
δ A = 0.02857 in. ≅ 0.0286 in. .......................................... Ans.
(b)
δ total = ∑
(120 )( 40 )
PL
= 0.02857 +
AE
(12.5664 )(10, 600 )
δ total = 0.0646 in. ................................................................................................................... Ans.
(c)
σ max B =
τ max B =
(d)
ε longB =
N
120
=
= 9.5493 ksi ≅ 9.55 ksi .............................................................. Ans.
A 12.5664
σ max B
2
σ
E
=
= 4.7747 ksi ≅ 4.77 ksi ............................................................................ Ans.
9.5493
= 900.8774 (10−6 )
10, 600
ε latB = −νε longB = − ( 0.33) ( 900.8774 × 10−6 ) = −297.2895 (10−6 )
δ dB = ε latB d = ( −297.2895 ×10−6 ) ( 4 ) = −1.189 (10−3 ) in. ........................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
5-12*
RILEY, STURGES AND MORRIS
AA = ( 25 )( 25 ) = 625 mm 2
AB = π ( 25 ) 4 = 490.874 mm 2
2
(a)
δ =∑
( 50 ×103 ) (1000 )
PL
=
AE ( 625 ×10−6 )( 73 ×109 )
( 50 ×10 ) ( 2000 )
+
( 490.874 ×10 )( 73 ×10 )
3
−6
δ = 3.89 mm
(b)
σ xA
9
.......................................................................................................................... Ans.
50 (103 )
N
= =
= 80 (106 ) N/m 2 = 80 MPa ....................................................... Ans.
−6
A 625 (10 )
σ yA = τ xyA = 0 MPa
σ xB =
............................................................................................................... Ans.
50 (103 )
N
=
= 101.859 (106 ) N/m 2 ≅ 101.9 MPa .............................. Ans.
−6
A 490.874 (10 )
σ yB = τ xyB = 0 MPa
σ
............................................................................................................... Ans.
101.859 (106 )
= 1395.331(10−6 ) ≅ 1395 µ m/m ......................................... Ans.
(c)
ε xB =
(d)
ε yB = −νε xB = − ( 0.33) (1395.331× 10−6 ) = −460.459 (10−6 )
E
=
73 (10
9
)
δ d = ε yB d = ( −460.459 ×10−6 ) ( 25 ) = −0.01151 mm ................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-13*
A = π ( 0.75 ) 4 = 0.44179 in.2
2
(a)
δA =
(b)
σx =
(c)
εx =
( 5)( 3 ×12 )
PL
=
= 0.01405 in. .............................................................. Ans.
AE ( 0.44179 )( 29, 000 )
N
5
=
= 11.3177 ksi ≅ 11.32 ksi ............................................................... Ans.
A 0.44179
σ y = τ xy = 0 ksi ...................................................................................................................... Ans.
σ
E
=
11.31769
= 390.265 (10−6 ) ≅ 390 µ in./in. .................................................... Ans.
29, 000
ε y = −νε x = − ( 0.30 ) ( 390.265 × 10−6 ) = −117.1(10−6 ) ≅ −117.1 µ in./in.
γ xy =
τ xy
G
............... Ans.
= 0 µ rad ................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-14
RILEY, STURGES AND MORRIS
3
3
PL ( 260 ×10 ) ( 3500 ) + ( 930 × 10 ) ( 3500 )
=
AE
( 9485 ×10−6 )( 200 ×109 )
(a)
δ =∑
(b)
δ = 2.20 mm ........................................................................ Ans.
σ xA = τ xyA = 0 MPa ............................................................. Ans.
σ yA =
260 (103 )
N
=
= 27.4117 (106 ) N/m 2
−6
A 9485 (10 )
σ yA ≅ 27.4 MPa (C)
σ xB = τ xyB = 0 MPa
σ yB
(c)
........................................................... Ans.
............................................................................................................... Ans.
930 (103 )
N
= =
= 98.0496 (106 ) N/m 2 ≅ 98.0 MPa (C) .............................. Ans.
−6
A 9485 (10 )
E = 2 (1 + ν ) G
ε yA =
σ
E
=
200 = 2 (1 +ν )( 76 )
−27.4117 (106 )
200 (10
9
)
ν = 0.31579
= −137.059 (10−6 ) ≅ −137.1 µ m/m ................................ Ans.
ε xA = −νε yA = − ( 0.31579 )( −137.059 ) = +43.3 µ m/m ................................................ Ans.
γ xyA = 0 µ m/m ........................................................................................................................ Ans.
ε yB =
σ
E
=
−98.0496 (106 )
200 (10
9
)
= −490.2478 (10−6 ) ≅ −490 µ m/m .................................. Ans.
ε xB = −νε yB = − ( 0.31579 )( −490.2478 ) = +154.8 µ m/m ........................................... Ans.
γ xyB = 0 µ m/m ........................................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-15
(a)
AA = π ( 42 − 22 ) 4 = 9.42478 in.2
δ a−a =
RILEY, STURGES AND MORRIS
AB = π ( 2 ) 4 = 3.14159 in.2
2
(130 )( 50 )
PL
=
AE ( 9.42478 )(15, 000 )
δ a − a = 0.045978 in. ≅ 0.0460 in. ↑ ......................................... Ans.
(b)
δ b −b = 0.045978 +
(c)
σ yA =
σ yB
σ xB
(d)
(80 )( 60 )
= 0.0969 in.
( 3.14159 )( 30, 000 )
↑ ..... Ans.
N
130
=
= 13.7934 ksi ≅ 13.79 ksi (C) ............. Ans.
A 9.42478
80
=
= 25.4648 ksi ≅ 25.5 ksi (C) ................................................................. Ans.
3.14159
= τ xyB = 0 ksi .............................. σ xA = τ xyA = 0 ksi ................................................... Ans.
E = 2 (1 + ν ) G
15, 000 = 2 (1 + ν )( 5600 )
δ d = ε x d = ( −νε y ) d =
−νσ y
E
ν = 0.33929
d
δ do =
− ( 0.33929 )( −13.7934 )
( 4 ) = 0.000624 in. ........................................................ Ans.
30, 000
δ di =
− ( 0.33929 )( −13.7934 )
( 2 ) = 0.000312 in. ......................................................... Ans.
30, 000
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-16*
δ =∑
RILEY, STURGES AND MORRIS
( 35 ×103 ) ( 900 )
( 35 ×103 ) ( 300 ) ≤ 0.40 mm
PL
=
+
AE π ( 0.025 )2 4  ( 200 × 109 )
A ( 73 × 109 )


A ≥ 1.81740 (10−3 ) m 2 = 1817.40 mm 2
π ( 752 − di2 )
4
≥ 1817.40 mm 2
di ≤ 57.5414 mm
tmin = ( 75 − 57.5414 ) 2 = 8.73 mm ................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-17*
For the uniform section:
W1 = γ V = 0.284 ( 2 × 5 ) y = 0.284 y
P1 = 30, 000 + 0.284 y
δU = ∫
25
0
( 30, 000 + 0.284 y ) dy
( 2 × 5) ( 29 ×106 )
 252 
= (1.03448 ×10 ) ( 25 ) + ( 9.79310 × 10 ) 

 2 
= 0.02587 in.
−3
−9
For the tapered section:
W2 = γ V = 0.284 ( y + 0.0125 y 2 )
P2 = 30, 007 + 0.284 ( y + 0.0125 y 2 )
y 

Ay = by t =  2 +  ( 0.5 ) = 1 + 0.025 y
20 

dδ =
Py dy
Ay E
δT = ∫
60
0
=
30, 007 + 0.284 y + 0.00355 y 2
dy
(1 + 0.025 y ) ( 29 ×106 )
30, 007 + 0.284 y + 0.00355 y 2
dy
(1 + 0.025 y ) ( 29 ×106 )
= 0.04139 ln (1 + 0.025 y ) 0 +
60
0.284 (1600 )
( 29 ×10 )
6
(1 + 0.025 y ) − ln (1 + 0.025 y )  0
2

0.00355 ( 64, 000 )  (1 + 0.025 y )
+
−
+
+
+
2
1
0.025
y
ln
1
0.025
y


(
)
(
)
2
( 29 ×106 ) 

60
60
0
= 0.03793 in.
δ total = δU + δ T = 0.0638 in. ................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-18
γ π r 2
Py =

3  L2

3
( L + y ) − π r 2 L

2
2
r
 πr
Ay = π  ( L + y )  = 2 ( L + y )
L
L

2
dδ =
δ=
Py dy
γ
Ay E
=
γ 
L3 
dy
( L + y ) −
2
3E 
( L + y ) 
L
L
3E ∫0
2

γ (L + y)
γ L2
L3 
L3 
.................. Ans.
dy
=
+
=
( L + y ) −



2
3E 
2
( L + y )  0 3E
( L + y ) 

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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-19*
Py = W = γ Ay
ε = (γ y K )
σ = P A = γ y = Kε 1 2
2
γ y
dδ = ε dy = 
 dy
K 
2
δ =∫
L
0
L
γ 2  y 3  γ 2 L3
γ y
...........................................Ans.
dy
=
=


K 2  3  0 3K 2
K 
2
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-21
Py = W = γ V = γ
Py dy
dδ =
Ay E
δ =∫
L
0
=
πρ 2 y
Ay = πρ 2
3
γπρ y 3
γy
dy
dy =
2
3E
πρ E
2
2 L
γy
γ y

 dy =
6E
 3E 
0
=
γ L2
6E
................................................Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-22*
RILEY, STURGES AND MORRIS
AA = π ( 502 − 302 ) 4 = 1256.6 mm 2
AC = π ( 202 ) 4 = 314.16 mm 2
75 ×103 ) ( 400 )
(
PL
δA =
=
= 0.3411 mm
AE (1256.6 × 10−6 )( 70 × 109 )
75 × 103 ) ( 500 )
(
PL
δC =
=
= 1.7052 mm
AE ( 314.16 ×10−6 )( 70 × 109 )
For the tapered section:
ρ = 0.01(1 + 2 y )
A = πρ 2 = π ( 0.01) (1 + 2 y )
2
2
3.4105 (10 ) dy
( 75 ×10 ) dy
Pdy
=
=
2
2
2
Ay E π ( 0.01) (1 + 2 y ) ( 70 × 109 )
(1 + 2 y )
3
dδ =
δ = 3.4105 (10
−3
)∫
−3
 dy 
−1

 = 3.4105 (10−3 )
2


2 (1 + 2 y ) 0
 (1 + 2 y ) 
0.75
0.75
0
= 1.0231(10−3 ) m = 1.0231 mm
δ total = δ A + δ B + δ C = 0.3411 + 1.0231 + 1.7052
δ total 3.0694 mm ≅ 3.07 mm
............................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-23
(a)
RILEY, STURGES AND MORRIS
P = τπ dL = π ( 300 )( 3) L = 900π L lb
0 in. ≤ L ≤ 15 in.
(b)
Px = τπ d ( L − x ) = 900π (15 − x ) lb
σx =
Px 900π (15 − x )
=
=400 (15 − x ) psi
2
A
π ( 3) 4
0 in. ≤ x ≤ 15 in.
(c)
εx =
(Both
σ x −νσ y −νσ z
E
σx
and
σy
=
( 400 )(15 − x ) − 2 ( 0.30 )( 600 )
30 (106 )
are compressive, therefore
εx
will be a shrink.)
5640 (15 − x ) − 200 ( 225 − x
1 15
1 15
δ = ∫ ε x ds = ∫ ( 5640 − 400s ) ds =
E x
E x
30 (106 )
2
) in.
0 in. ≤ x ≤ 15 in.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-24
Px = ∫ τπ ( 0.300 ) ds
8
(a)
x
σx =
8
Px
1
=
π ( 400s )(1 − sin 28° )( tan 28° )( 0.300 ) ds
2
A π ( 0.300 ) 4 ∫x
 82 x 2 
−  = 752.23 ( 64 − x 2 ) N/m 2
2 2
σ x = 1504.46 
0 m≤ x≤8 m
(b)
εx =
σ x −νσ y −νσ z
E
=
( 752.23) ( 64 − x 2 ) − 2 ( 0.30 )( 400 )(1 − sin 28° )
13 (109 )
ε x = 3.6935 (10−6 ) − 5.7864 (10−8 ) x 2
δ = ∫ ε x ds = ∫ 3.6935 (10−6 ) − 5.7864 (10−8 ) s 2  ds
8
8
x
x
δ = 3.6935 (10−6 ) ( 8 − x ) − 1.9288 (10−8 )( 83 − x3 ) m
(c)
0 m≤ x≤8 m
P = σ x =0 π ( 0.300 ) 4
2
= 752.23L2 π ( 0.300 ) 4
2
P = 53.1721L2 N
0 m≤ L≤8 m
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MECHANICS OF MATERIALS, 6th Edition
5-25*
→ ΣFx = 0 :
TBC cos 60° − TAB = 0
↑ ΣFy = 0 :
TBC sin 60° − 16 = 0
RILEY, STURGES AND MORRIS
TAB = 9.2376 kip (T)
TBC = 18.4752 kip (T)
(a)
δ AB =
PL ( 9.2376 )( 80 )
=
= 0.11620 in.
AE ( 0.6 )(10, 600 )
δ AB ≅ 0.1162 in. (stretch)
δ BC =
(b)
................................................................................................... Ans.
(18.4752 )(160 ) = 0.08155 in.
(1.25)( 29, 000 )
δ BC ≅ 0.0815 in. (stretch)
................................................................................................... Ans.
sin 60° = a δ BC
a = 0.07062 in.
cos 60° = b δ BC
b = 0.04077 in.
tan 60° =
b + δ AB
c
c = 0.09063 in.
uB = δ AB = 0.1162 in. → ........................................... Ans.
vB = a + c = 0.1613 in. ↓ ........................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-26
→ ΣFx = 0 :
TAB cos 45° − FBC + 50 cos 55° = 0
TAB sin 45° − 50sin 55° = 0
↑ ΣFy = 0 :
TAB = 57.9228 kN (T)
(a)
RILEY, STURGES AND MORRIS
FBC = 69.6364 kN (C)
LAB = 12 + 12 = 1.41421 m
δ AB
3
PL ( 57.9228 × 10 ) (1414.21)
=
=
= 1.89619 mm
AE
( 450 ×10−6 )( 96 ×109 )
δ AB ≅ 1.896 mm (stretch) .................................................. Ans.
( 69.6364 ×10 ) (1000 ) = 0.26681 mm
(1450 ×10 )(180 ×10 )
3
δ BC =
−6
9
δ BC ≅ 0.267 mm (shrink) .................................................. Ans.
(b)
sin 45° = a δ AB
a = 1.34081 mm = b
cos 45° = b δ AB
b = 1.34081 mm
tan 45° =
uB = δ BC
b + δ BC
c = 1.60761 mm
c
= 0.267 mm → ........................................... Ans.
vB = a + c = 2.95 mm ↓ ............................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-27
→ ΣFx = 0 :
TBC cos 25° − TAC cos 20° = 0
TBC sin 25° + TAC sin 20° − 220 = 0
↑ ΣFy = 0 :
TAC = 281.9768 lb (T)
(a)
σ AC =
TBC = 292.3648 lb (T)
P ( 281.9768 )
=
= 18, 798 psi ≅ 18.80 ksi (T) ............. Ans.
A
( 0.015 )
σ BC =
(b)
RILEY, STURGES AND MORRIS
( 292.3648 ) = 19, 491 psi ≅ 19.49 ksi (T) ........................................................... Ans.
( 0.015)
LAC = 20 cos 20° = 21.28356 ft = 255.4027 in.
LBC = 20 cos 25° = 22.06956 ft = 264.8107 in.
δ AC =
PL ( 281.9768 )( 255.4027 )
=
= 0.16556 in.
AE ( 0.015 ) ( 29, 000 × 103 )
δ AC ≅ 0.1656 in. (stretch) ................................................................................................... Ans.
δ BC =
( 292.3648)( 264.8107 ) = 0.17798 in.
( 0.015) ( 29, 000 ×103 )
δ BC ≅ 0.1780 in. (stretch)
(c)
................................................. Ans.
a = δ AC cos 45° = 0.11707 in.
b = δ AC sin 45° = 0.11707 in.
d = δ AC sin 20° = 0.05662 in.
b + δ BC
c = 0.41726 in.
c
e = c cos 20° = 0.39210 in.
cos 45° =
vC = d + e = 0.449 in. ↓ ............................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-28*
θ = tan −1 ( 2.5 6 ) = 22.6199°
LAB = 2.52 + 62 = 6.5000 m
φ = tan −1 ( 4.5 6 ) = 36.8699°
LAC = 4.52 + 62 = 7.5000 m
→ ΣFx = 0 :
↑ ΣFy = 0 :
TAB cos θ − FAC cos φ = 0
TAB sin θ + FAC sin φ − 100 = 0
TAB = 92.8571 kN (T)
FAC = 107.1432 kN (C)
P ( 92.8571× 10 )
= =
= 149.8 (106 ) N/m 2
−6
A
( 620 ×10 )
3
(a)
σ AB
σ AB = 149.8 MPa (T)
........................................................................................................... Ans.
(107.1432 ×10 ) = 107.1 10 N/m = 107.1 MPa (C) .................................. Ans.
=
( )
(1000 ×10 )
PL ( 92.8571× 10 ) ( 6500 )
=
=
= 4.86751 mm
AE ( 620 × 10 )( 200 ×10 )
3
σ AC
6
2
−6
3
(b)
δ AB
−6
9
δ AB ≅ 4.87 mm (stretch) ...................................................................................................... Ans.
(107.1432 ×10 ) ( 7500 ) = 4.01787 mm
=
(1000 ×10 )( 200 ×10 )
3
δ AC
−6
δ AC ≅ 4.02 mm (shrink)
(c)
9
........................................... Ans.
β = 180° − θ − φ − 90° = 30.5102°
a = δ AC cos φ = 3.21430 mm
b = δ AC sin φ = 2.41072 mm
d = δ AC sin β = 2.03984 mm
δ AB + d
= 8.01746 mm
cos β
f = e cos φ = 6.41397 mm
g = e sin φ = 4.81048 mm
e=
u A = a − g = −1.596 mm = 1.596 mm ← ...................................................................... Ans.
v A = b + f = 8.82 mm ↓ ..................................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
5-29
→ ΣFx = 0 :
↑ ΣFy = 0 :
RILEY, STURGES AND MORRIS
TBC sin 30° − TAC sin 45° = 0
TBC cos 30° + TAC cos 45° − 10 = 0
TAC = 5.17638 kip (T)
TBC = 7.32051 kip (T)
(a)
δ AC =
PL ( 5.17638 )(10 ×12 )
=
= 0.17976 in.
AE
( 0.326 )(10, 600 )
δ AC ≅ 0.1798 in. (stretch)
δ BC =
( 7.32051)(15 ×12 ) = 0.08944 in.
( 0.508)( 29, 000 )
δ BC ≅ 0.0894 in. (stretch)
(b)
................................................................................................... Ans.
................................................Ans.
a = δ AC cos 45° = 0.12711 in.
b = δ AC sin 45° = 0.12711 in.
c = δ AC sin15° = 0.04653 in.
d = δ AC cos15° = 0.17363 in.
e = δ BC − c = 0.04291 in.
f = e cos15° = 0.04443 in.
g = f cos 45° = 0.03142 in.
uC = a − h = 0.0957 in. → .............................. Ans.
h = f sin 45° = 0.03142 in.
vC = b + g = 0.1585 in. ↓ ................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-30
(a)
εB =
δ
=
δB
RILEY, STURGES AND MORRIS
= 1500 (10−6 )
L 375
δ B = b = 0.56250 mm
vC = c = ( 250 150 ) b = 0.937 mm ↓ ..................Ans.
(b)
δ A = a = ( 50 150 ) b = 0.18750 mm
δA =
TA ( 200 )
(1250 ×10 )( 200 ×10 )
−6
9
= 0.18750 mm
TA = 234,375 N
δB =
TB ( 375 )
( 940 ×10 )(100 ×10 )
−6
9
= 0.56250 mm
TB = 141, 000 N
4 ΣM D = 0 :
250 P − 150 (141.000 ) − 50 ( 234.375 ) = 0
P = 131.5 kN .......................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-31*
RILEY, STURGES AND MORRIS
δ CD = ε CD LCD = ( 680 ×10−6 ) ( 72 ) = 0.048960 in. ≅ 0.0490 in. .................................. Ans.
δ CD =
T ( 6 × 12 )
PL
= CD
= 0.048960 in.
AE ( 2.5 )( 29, 000 )
TCD = 49.300 kip (T)
sin θCD = 4 5
cos θCD = 3 5
sin θ AD = 12 13
cos θ AD = 5 13
δ CD = δ BD sin θ BD
δ AD = δ BD sin θ AD
δ BD = 0.061200 in. ≅ 0.0612 in. ............................................Ans.
δ AD = 0.056492 in. ≅ 0.0565 in. ............................................Ans.
δ BD =
TBD ( 6 ×12 )
= 0.061200 in.
( 2.5)( 6500 )
TBD = 13.8125 kip (T)
δ AD =
TAD ( 6 × 12 )
= 0.056492 in.
( 2.5)( 26, 000 )
TAD = 51.0000 kip (T)
↑ ΣFy = 0 :
TCD sin θCD + TAD sin θ AD + TBD − P = 0
P = 100.3 kip .......................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-32
RILEY, STURGES AND MORRIS
δ A = ε A LA = ( 625 × 10−6 ) ( 5000 ) = 3.12500 mm
δ A = a sin θ A = ( 3 5 ) a
a = 5.20833 mm
d = ( 5 4 ) a = 6.51 mm ↓ .................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-33
→ ΣFx = 0 :
30 − FBC cos 45° − TAB cos 60° = 0
FBC sin 45° − TAB sin 60° = 0
↑ ΣFy = 0 :
TAB = 21.9615 kip (T)
(a)
σ max AB =
FBC = 26.8973 kip (C)
P ( 21.9615 )
=
= 17.57 ksi (T) .......................... Ans.
A
(1.25)
τ max AB = σ max AB 2 = 8.78 ksi
(b)
RILEY, STURGES AND MORRIS
σ max BC =
.............................................................................................. Ans.
( 26.8973) = 10.76 ksi (C) ................................................................................. Ans.
( 2.50 )
τ max BC = σ max BC 2 = 5.38 ksi .............................................................................................. Ans.
(c)
LAB = 30 sin 60° = 34.6410 in.
δ AB =
PL ( 21.9615 )( 34.6410 )
=
= 0.0574166 in.
AE
(1.25)(10, 600 )
δ AB ≅ 0.0574 in. (stretch)
δ BC =
LBC = 30 sin 45° = 42.4264 in.
................................................................................................... Ans.
( 26.8973)( 42.4264 ) = 0.0157401 in.
( 2.5)( 29, 000 )
δ BC ≅ 0.01574 in. (shrink) .................................................................................................. Ans.
(d)
a = δ AB cos 30° = 0.0497242 in.
b = δ AB sin 30° = 0.0287083 in.
c = δ AB sin15° = 0.0148605 in.
c + δ BC
= 0.0316800 in.
cos15°
e = d cos 30° = 0.0274657 in.
f = d sin 30° = 0.0158400 in.
d=
uB = b + e = 0.0561 in. → ......................................... Ans.
vB = a − f = 0.0339 in. ↑ .......................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-34*
AA =
RILEY, STURGES AND MORRIS
π ( 0.1002 − 0.0502 )
AB =
π ( 0.050 )
2
4
= 1.96350 (10−3 ) m 2
4
= 5.89049 (10−3 ) m 2
FA + TB − 500 = 0
↑ ΣFy = 0 :
FA + TB = 500 kN
(a)
δA = δB
FA (1500 )
TB ( 2000 )
=
( 5.89049 ×10 )(100 ×10 ) (1.96350 ×10 )( 200 ×10 )
−3
9
−3
9
FA = 2.000TB
(b)
FA = 333.333 kN (C)
(a)
TB = 166.6667 kN (T)
3
P ( 333.333 × 10 )
σA = =
= 56.6 (106 ) N/m 2 = 56.6 MPa (C) .............................. Ans.
A ( 5.89049 ×10−3 )
(166.6667 ×10 ) = 84.9 10 N/m = 84.9 MPa (T) ..................................... Ans.
=
( )
(1.96350 ×10 )
(166.6667 ×10 ) ( 2000 ) = 0.849 mm ↓ ............................................... Ans.
=δ =
(1.96350 ×10 )( 200 ×10 )
3
σ BC
6
2
−3
3
(b)
vC
B
−3
9
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-35*
↑ ΣFy = 0 :
2 FS + FW − 700 = 0
2 FS + FW = 700 kip
(a)
δ S = δW
FS ( 20 )
FW ( 20 )
=
( 2 × 7.5 )( 29, 000 ) ( 7.5 × 7.5 )(1800 )
FS = 4.29630 FW
FW = 72.9730 kip (C)
(a)
σW =
(b)
FS = 313.5138 kip (C)
P ( 72.9730 )
=
= 1.297 ksi (C) ..................................... Ans.
A ( 7.5 × 7.5 )
( 313.5138) = 20.9 ksi (C) ........................................................................................ Ans.
( 2 × 7.5)
( 313.5138)( 20 ) = 0.01441 in. .............................................................................. Ans.
δS =
( 2 × 7.5)( 29, 000 )
σS =
(b)
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-36
RILEY, STURGES AND MORRIS
TS + TC = 90 N
(a)
δ S = δC

π ( 2 − 1.5
2
TS L
2
) 4 (14 ×10 )
TS = 1.55556TC
9
=
TC L
π (1.5 ) 4  ( 7 × 109 )


2
(b)
TC = 35.2 N (T) ...................................................................................................................... Ans.
TS = 54.8 N (T) ...................................................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-37*
2TB + TC = 40 kip
RILEY, STURGES AND MORRIS
(a)
δ B = δC
TB ( 36 )
TC ( 36 )
=
( 2 )(10, 600 ) ( 2 )( 28, 000 )
TC = 1.32075TB
(b)
TB = 12.0455 kip (T)
TC = 15.9090 kip (T)
σB =
P (12.0455 )
=
= 6.02 ksi (T) .................................................................................. Ans.
A
( 2)
σC =
(15.9090 ) = 7.95 ksi (T) .......................................................................................... Ans.
( 2)
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-38
4 ΣM D = 0 :
RILEY, STURGES AND MORRIS
250 ( 5 ) − 50TA − 150TB = 0
TA + 3TB = 25 kN
(a)
b = 3a
δ B = 3δ A
TB ( 400 )
(80 ×10 )( 200 ×10 )
−6
TB = 1.5TA
(a)
9
=3
TA ( 200 )
(80 ×10 )( 200 ×10 )
−6
9
(b)
TA = 4.5455 kN ≅ 4.55 kN (T) .....................................Ans.
TB = 6.82 kN (T) ................................................................................................................... Ans.
(b)
 ( 4.5455 ×103 ) ( 200 ) 
 = 1.136 mm ↓ ....................................... Ans.
vC = 5a = 5δ A = 5 
−6
9
 ( 80 ×10 )( 200 × 10 ) 
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-39
RILEY, STURGES AND MORRIS
TAB + FBC = 150 kip
(a)
δ AB = δ BC
TAB ( 24 )
π ( 2.25 ) 4  (10, 600 )


2
=
FBC = 2.02272TAB
TAB = 49.6242 kip (T)
FBC ( 24 )
π ( 3.2 )2 4  (10, 600 )


(b)
FBC = 100.3758 kip (C)
( 49.6242 )( 24 ) = 0.02826 in. ≅ 0.0283 in.
π ( 2.25 )2 4  (10, 600 )


(a)
u B = δ AB =
(b)
ε AB =
(c)
ε d = −νε AB = − ( 0.33) (1177.425 × 10−6 ) = −388.550 (10−6 ) in./in.
δ AB
LAB
=
→ ......................... Ans.
0.02826
= 1177.425 (10−6 ) ≅ 1177 µ in./in. ............................................ Ans.
24
δ d = ε d d = ( −388.550 × 10−6 ) ( 2.25 ) = −0.874 (10−3 ) in. ............................................ Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-40*
RILEY, STURGES AND MORRIS
W = 4500 ( 9.81) = 44,145 N = 44.145 kN
TA + TB − 44,145 = 0
↑ ΣFy = 0 :
3000TB − 1000 ( 44.145 ) = 0
4 ΣM A = 0 :
TA = 29.430 kN
TB = 14.715 kN
δA = δB
( 29.430 ×10 ) (1200 ) = (14.715 ×10 ) (1800 )
π d 4  ( 70 ×10 )
π d 4  ( 200 × 10 )
3
2
A
3
9
2
B
9
d A d B = 1.952 ........................................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-41*
TB = TA + 40 kip
(a)
TC = TA − 10 kip
(b)
δ A + δ B + δC = 0
(c)
TA ( 20 )
π ( 2.5 − 2

2
TA = 0 kip
(a)
RILEY, STURGES AND MORRIS
σA =
2
)
(TA + 40 )( 24 ) + (TA − 10 )( 24 ) = 0
2
2
4  ( 30, 000 ) π ( 2 ) 4  (10, 000 ) π (1) 4  (10, 000 )




+
TB = 40 kip (T)
TC = −10 kip = −10 kip (C)
P
= 0 ksi ....................................................................................................................... Ans.
A
( 40 ) = 12.73 ksi (T) ........................................................................................... Ans.
2
π ( 2) 4
( −10 ) = −12.73 ksi = 12.73 ksi (C) ................................................................... Ans.
σC =
2
π (1) 4
( −10 )( 24 )
ua − a = δ C =
= −0.0306 in. = 0.0306 in. → .............................. Ans.
π (1)2 4  (10, 000 )


σB =
(b)
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-42*
Fs + Fa = 30 kN
RILEY, STURGES AND MORRIS
(a)
δs = δa
Fs L
Fa L
=
9
( 0.025 × 0.100 ) ( 200 ×10 ) ( 0.100 × 0.100 ) ( 73 ×109 )
Fa = 1.4600 Fs
(b)
Fs = 12.1951 kN (C)
Fa = 17.8049 kN (C)
x=
12.1951(12.5 ) + 17.8049 ( 75 )
= 49.6 mm ...................................Ans.
30
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MECHANICS OF MATERIALS, 6th Edition
5-43
TA + FB = 95 kip
RILEY, STURGES AND MORRIS
(a)
δ A = δ B + 0.015 in.
TA ( 72 )
FB ( 24 )
=
+ 0.015
(1.25)( 30, 000 ) ( 3.75)(15, 000 )
TA = 0.22222 FB + 7.8125 kip
(b)
TA = 23.6648 kip (T)
FB = 71.3352 kip (C)
(a)
(b)
N 23.6648
=
= 18.93 ksi (T) .................................................................................. Ans.
A
1.25
N 71.3352
σB = =
= 19.02 ksi (C) .................................................................................. Ans.
A
3.75
( 23.6648)( 72 ) = 0.0454 in. ↓ ..................................................................... Ans.
vC = δ A =
(1.25)( 30, 000 )
σA =
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MECHANICS OF MATERIALS, 6th Edition
5-44*
RILEY, STURGES AND MORRIS
TP = TB
(a)
δ P + δ B = 0.15 mm
(b)
TP ( 200 )
π ( 0.15 ) 4  ( 2.1×109 )


2
+
TB ( 400 )
π ( 0.045 )2 4  (100 × 109 )


= 0.15
TP = TB = 18,976.74 N
(a)
σP =
N
18,976.74
=
= 1.074 (106 ) N/m 2
2
A π ( 0.15 ) 4
σ P = 1.074 MPa (T) ............................................................... Ans.
σB =
(b)
N
18,976.74
=
= 11.93 (106 ) N/m 2 = 11.93 MPa (T) ................................ Ans.
2
A π ( 0.045 ) 4
x = 200 + δ P = 200 +
(18,976.74 )( 200 ) = 200.1023 mm .............................. Ans.
π ( 0.15 )2 4  ( 2.1×109 )


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MECHANICS OF MATERIALS, 6th Edition
5-45*
4 ΣM A = 0 :
RILEY, STURGES AND MORRIS
2TB + 4TC − 6 (1000 ) = 0
TB + 2TC = 3000 = 0
(a)
δ B = b cos θ = ( L∆θ ) cos θ
LB = (1 3)( 48 ) = 16 in.
δ C = c cos θ = ( 2 L∆θ ) cos θ
LC = ( 2 3)( 48 ) = 32 in.
δ C = 2δ B
(b)
TC ( 32 )
TB (16 )
=2
( 0.3)( 29, 000 ) ( 0.3)( 29, 000 )
TB = TC = 1.000 kip (T)
δB =
(1.000 ) 1( 48) 3
= 0.001839 in. ......................... Ans.
( 0.3)( 29, 000 )
δ C = 2δ B = 0.00368 in. ...................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
5-46
→ ΣFx = 0 :
RILEY, STURGES AND MORRIS
Cx = 0
C y + TAD + FB − 225 = 0
↑ ΣFy = 0 :
500TAD + 1350 FB − 1350 ( 225 ) = 0
4 ΣM C = 0 :
10TAD + 27 FB = 6075 kN
(a)
δ B = (1350 500 ) δ AD

TAD ( 450 )
 1350  


=
( 4500 ×10−6 )(12 ×109 )  500   ( 300 ×10−6 )( 200 ×109 ) 
FB ( 375 )
FB = 2.91600TAD
(b)
TAD = 68.4646 kN
Cx = 0 kN
(a)
σ AD
FB = 199.6427 kN
C y = −43.1073 kN
C = Cx2 + C y2 = 43.1073 kN
3
N 68.4646 (10 )
= =
= 228 (106 ) N/m 2
−6
A
300 (10 )
σ AD = 228 MPa (T) .............................................................................................................. Ans.
σB =
199.6427 (103 )
4500 (10
−6
)
= 44.4 (106 ) N/m 2 = 44.4 MPa (C) .......................................... Ans.
(b)
43.1073 (103 )
V
τC = =
= 68.6 (106 ) N/m 2 = 68.6 MPa ..................................... Ans.
A 2 π ( 0.020 )2 4 


(c)
( 68.4646 ×10 ) ( 450 ) = 0.513 mm ↓ ..................................................... Ans.
=
( 300 ×10 )( 200 ×10 )
3
vD = δ AD
−6
9
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MECHANICS OF MATERIALS, 6th Edition
5-47
RILEY, STURGES AND MORRIS
5FS + FC = 200 kip
(a)
δ S = δC
π (1)

2
FS L
FC L
=
4  ( 29, 000 ) AC ( 4500 )

(b)
2
AC = b 2 − 5 π (1) 4 


If FS = FS max = σ S A = (18 ) π (1)

FC = 129.3142 kip
2
(c)
4  = 14.1372 kip , then

AC = 46.2977 in.2
b = 7.09 in.
F 129.3142
=
= 2.79 ksi > σ C max = 1.4 ksi
A 46.2977
If σ C = FC AC = σ C max = 1.4 ksi , then
and
σC =
FS = 7.0860 kip
and
σS =
Therefore
FC = 164.5698 kip
F 7.0860
=
= 9.02 ksi < σ S max = 18 ksi
A π (1)2 4
(not correct guess)
AC = 117.5499 in.2
b = 11.02 in.
(correct guess)
b = 11.02 in. .............................................................................................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
5-48*
RILEY, STURGES AND MORRIS
2Ts = Fa
(a)
δ s + δ a = 1 mm
(b)
Ts ( 330 )
Fa ( 251)
+
(120 ×10 )(190 ×10 ) ( 625 ×10 )( 73 ×10 )
T = 39.2520 (10 ) N = 39.2520 kN
F = 78.5040 (10 ) N = 78.5040 kN
N 39.2520 (10 )
σ = =
= 327 (10 ) N/m
A
120 (10 )
−6
9
−6
9
= 1 mm
3
s
3
a
3
(a)
6
2
−6
s
σ s = 327 MPa (T) ................................................................................. Ans.
σa =
78.5040 (103 )
625 (10
−6
)
= 125.6 (106 ) N/m 2 = 125.6 MPa (C) ........................................ Ans.
( 78.5040 ×10 ) ( 251) = 0.432 mm (shrink) ........................................................ Ans.
=
( 625 ×10 )( 73 ×10 )
3
(b)
δa
−6
9
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MECHANICS OF MATERIALS, 6th Edition
5-49*
Ts = Fb
RILEY, STURGES AND MORRIS
(a)
δ b + δ s = 0.020 in.
Fb ( 6 )
Ts ( 6 )
+
= 0.02 in.
2
( 0.375 )(15, 000 ) π ( 0.5) 4 ( 30, 000 )


Ts = Fb = 9.59114 kip
(b)
σs =
N
9.59114
=
= 48.8 ksi (T) .............................................Ans.
A π ( 0.5 )2 4
σb =
9.59114
= 25.6 ksi (C) ...........................................................Ans.
0.375
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-50
Initially, bar CE is not connected to bar ABCD (TCE = 0 ) and
the load P = 50 kN is applied to the end of bar ABCD . Then
4 ΣM A = 0 :
320 ( 50 ) − 80TBF = 0
TBF = 200 kN
( 200 ×10 ) (1000 ) = 0.79365 mm
=
(1200 ×10 )( 210 ×10 )
3
δ BF
−6
9
At this point, bar CE is connected to bar ABCD (TCE ≠ 0 ) and
the applied load P is removed ( P = 0 ) . Then
4 ΣM A = 0 :
240TCE − 80TBF = 0
TBF = 3TCE
(a)
c = 3b = 3δ BF
c = 3 ( 0.79365 ) − δ CE = 2.38095 − δ CE
3δ BF = 2.38095 − δ CE
(b)


TBF (1000 )
TCE ( 600 )

=
−
3
2.38095
−6
9
 (1200 ×10 )( 210 × 10 ) 
( 900 ×10−6 )( 73 ×109 )
TBF = 159.2727 (103 ) N = 159.2727 kN
TCE = 53.0909 (103 ) N = 53.0909 kN
(a)
σ CE
3
N 53.0909 (10 )
= =
= 59.0 (106 ) N/m 2
−6
A
900 (10 )
σ CE
3
N 53.0909 (10 )
= =
= 59.0 (106 ) N/m 2 = 59.0 MPa (T) ................................. Ans.
−6
A
900 (10 )
δ BF
(159.2727 ×10 ) (1000 ) = 0.63203 mm
=
(1200 ×10 )( 210 ×10 )
3
(b)
−6
9
vD = d = 4b = 4δ BF = 2.53 mm ↑ ..................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
5-51
4 ΣM C = 0 :
RILEY, STURGES AND MORRIS
8TA − 5TB = 0
TB = 1.6TA
(a)
δ A = ( 8 5 ) b = ( 8 5 )( 0.1 − δ B )
δ A = 1.6 ( 0.1 − δ B ) in.
(b)
TA ( 60 )

TB ( 60 ) 
= 1.6  0.1 −

0.5 ( 29, 000 )
(1.2 )(15, 000 ) 

TA = 12.6270 kip
(a)
(b)
TB = 20.2032 kip
N 12.6270
=
= 25.3 ksi (T) .........................Ans.
A
0.5
(12.6270 )( 60 ) = 0.0523 in. ↑ ....................................................................... Ans.
vA = δ A =
( 0.5)( 29, 000 )
σA =
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-52*
RILEY, STURGES AND MORRIS
A = π ( 0.030 ) 4 = 706.858 (10−6 ) m 2
2
TBC = (TAB − 6 ) kN
(a)
TCD = (TAB − 3) kN
(b)
δ AB + δ BC + δ CD = 0
(c)
TAB (1000 ) + TBC (1000 ) + TCD (1000 )
( 706.858 ×10 )( 73 ×10 )
−6
9
=0
TAB + (TAB − 6 ) + (TAB − 3) = 0
TAB = 3 kN
(a)
σ AB =
TBC = −3 kN
TCD = 0 kN
N
3000
=
= 4.24 (106 ) N/m 2 = 4.24 MPa (T) ............................... Ans.
−6
A 706.858 (10 )
σ BC = 4.24 MPa (C) ........................ σ CD = 0 MPa .......................................................... Ans.
(b)
τ max = σ max 2 = 2.12 MPa
(c)
δ BC =
(d)
E = 2 (1 + ν ) G
ε BC =
.................................................................................................. Ans.
( 3000 )(1000 )
( 706.858 ×10 )( 73 ×10 )
−6
δ
L
=
9
= 0.058139 mm ≅ 0.0581 mm (shrink) ............ Ans.
73 = 2 (1 + ν )( 28 )
ν = 0.30357
−0.058139
= −58.139 (10−6 ) m/m
1000
ε dBC = −νε BC = − ( 0.30357 ) ( −58.139 ×10−6 ) = +17.649 (10−6 ) m/m
δ d = ε d d = ( +17.649 × 10−6 ) ( 30 ) = +0.529 (10−3 ) mm (expands)
(e)
........................... Ans.
δ AB = −δ BC = +0.0581 mm (stretch)
ε AB =
δ
L
=
+0.058139
= +58.139 (10−6 ) m/m ≅ +58.1 µ m/m ................................... Ans.
1000
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MECHANICS OF MATERIALS, 6th Edition
5-53*
→ ΣFx = 0 :
RILEY, STURGES AND MORRIS
Cx = 0
FA + FB − 10 + C y = 0
↑ ΣFy = 0 :
4 ΣM C = 0 :
5.5 (10 ) − 8 FA − 3FB = 0
8FA + 3FB = 55 kip
(a)
δ A = ( 8 3) δ B

FA ( 48 )
FB ( 36 )
8
= 

( 2.25 )(10, 600 ) 3  (1.75)( 28, 000 ) 
FB = 1.02725 FA
FA = 4.96311 kip
(b)
FB = 5.09835 kip
C y = −0.06146 kip
(b)
N 4.96311
=
= 2.2058 ksi ≅ 2.21 ksi (C) ............................................................ Ans.
A
2.25
5.09835
σB =
= 2.9133 ksi ≅ 2.91 ksi (C) ..................................................................... Ans.
1.75
τ max A = σ A 2 = 1.103 ksi ....................... τ max B = σ B 2 = 1.457 ksi .............................. Ans.
(c)
C = Cx2 + C y2 = 61.46 lb
(a)
σA =
τC =
(d)
vD =
V
61.46
=
= 156.5 psi ................................................................................
A 2 π ( 0.5 )2 4 


Ans.
10
10δ A 10 ( 4.96311)( 48 )
a=
=
= 0.01249 in. ↓ ............................................. Ans.
8
8
8 ( 2.25 )(10, 600 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-54
4 ΣM F = 0 :
RILEY, STURGES AND MORRIS
300 P − 50 FD − 100TC = 0
FD + 2TC = 6 P
(a)
δ C = 2 (δ D + 0.09 ) mm


FD (150 )


=
+
.09
2
0

( 600 ×10−6 )( 200 ×109 )  ( 2500 ×10−6 )(100 ×109 )
TC ( 300 )
TC = 0.48000 FD + 72, 000 N
(b)
Guess that
TC = TC max = σ A
= ( 215 × 106 )( 600 ×10−6 ) = 129, 000 N
Then
FD = 118, 750 N
and
σD =
Since
P = 62, 792 N
N
118, 750
=
= 47.5 (106 ) N/m 2 = 47.5 MPa
−6
A 2500 (10 )
σ D = 47.5 MPa < σ max = 95 MPa , the guess was correct and
Pmax = 62.8 kN ....................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-55
Ts = Fb
δ s + δ b = ∆ nut =
RILEY, STURGES AND MORRIS
(a)
0.125θ
360
(b)
Ts (14 )
Fb (12 )
0.125θ
+
=
( 0.785 )( 30, 000 ) (1.767 )(15, 000 ) 360
Ts = Fb = 0.33156θ kip
N 0.33156θ
=
= 0.42237θ ksi (T)
A
0.785
0.33156θ
σb =
= 0.18764θ ksi (C)
1.767
(a)
σs =
(b)
δs =
(c)
L = 12 − δ b = 12 − 0.15011θ (10−3 ) in.
( 0.33156θ )(14 ) = 0.19711θ 10−3 in.
( )
( 0.785)( 30, 000 )
( 0.33156θ )(12 ) = 0.15011θ 10−3 in.
δb =
( )
(1.767 )(15, 000 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-56
4 ΣM C = 0 :
RILEY, STURGES AND MORRIS
200TA − 125TB = 0
TB = 1.6TA
(a)
δ A = ( 200 125 ) b = ( 200 125 )( ∆ nut − δ B )
δ A = 1.6 ( ∆ nut − δ B ) mm
 2.5θ

TB (1500 )


=
−
1.6
( 350 ×10−6 )( 200 ×109 )  360 ( 750 ×10−6 )(100 ×109 ) 
TA (1500 )
TA + 1.49333TB = 518.5185θ N
(b)
TA = 152.9854θ N
TB = 244.7766θ N
(a)
σA =
N 152.9854θ
=
= 0.43710θ (106 ) N/m 2
−6
A 350 ×10
N 244.7766θ
=
A 750 × 10−6
= 0.0.32637θ (106 ) N/m 2
σA =
(b)
vA = δ A =
vA
(c)
(152.9854θ )(1500 )
( 350 ×10 )( 200 ×10 )
= 3.27826θ (10 ) mm
tan φ =
−6
9
−3
δA
200
= 16.6913θ (10−6 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-57
4 ΣM C = 0 :
RILEY, STURGES AND MORRIS
5P − 10TA − 6 FB = 0
10TA + 6 FB = 5 P
(a)
δ A = (10 6 )(δ B + 0.009 ) = (10δ B 6 ) + 0.015 in.
If
δ A ≤ 0.015 in. , then
and
If
δB = 0
FB = 0
TA = P 2
σ A = TA 2 = P 4
δA =
σB = 0
( P 2 )( 50 ) = 0.00125P in.
( 2 )(10, 000 )
δ A ≥ 0.015 in. , then
10TA + 6 FB = 5 P
(a)
δ A = (10δ B 6 ) + 0.015 in.
(b)
TA ( 50 )
10  FB (15 ) 
= 
 + 0.015
( 2 )(10, 000 ) 6  (12 )(15, 000 ) 
TA = ( 0.042373P + 5.4915 ) kip
FB = ( 0.76271P − 9.15254 ) kip
0.042373P + 5.4915
= ( 0.02119 P + 2.74575 ) ksi (T)
2
0.76271P − 9.15254
σB =
12
= ( 0.06356 P − 0.76271) ksi (C)
(a)
σA =
(b)
δA =
(c)
( 0.042373P + 5.4915) ( 50 )
( 2 )(10, 000 )
( 0.76271P − 9.15254 ) ( 50 )
δB =
(12 )(15, 000 )
θ = tan −1
in.
in.
δA
10
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MECHANICS OF MATERIALS, 6th Edition
5-58
4 ΣM F = 0 :
RILEY, STURGES AND MORRIS
300 P − 50 FD − 100TC = 0
FD + 2TC = 6 P
(a)
δ C = 2 (δ D + 0.09 ) mm
If
δ C ≤ 0.18 mm , then
δD = 0
σD = 0
FD = 0
TC = 3P
σC =
and
If
δC =
3P
= 0.00500 P (106 ) N/m 2
−6
600 (10 )
( 3P )( 300 )
( 600 ×10 )( 73 ×10 )
−6
9
= 20.5480 P (10−6 ) mm
δ C ≤ 0.18 mm , then
FD + 2TC = 6 P
(a)
δ C = 2 (δ D + 0.09 ) mm


FD (150 )

 mm
=
+
.09
2
0

( 600 ×10−6 )( 73 ×109 )  ( 2500 ×10−6 )(100 ×109 )
TC ( 300 )
TC = ( 0.17520 FD + 26, 280 ) N
(b)
TC = ( 0.77844 P + 19,831) N
FD = ( 4.44313P − 39, 662 ) N
(b)
δC =
δD =
(c)
( 0.77844 P + 19,831)( 300 )
( 600 ×10 )( 73 ×10 )
−6
9
( 4.44313P − 39, 662 )(150 )
( 2500 ×10 )(100 ×10 )
θ = tan −1
−6
9
mm
mm
δC
100
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-59*
 σ

σ

+ α ∆T  L = 
+ (12.5 × 10−6 ) ( −100 )  ( 80 )
E

10, 600

(a)
δ =0=
(b)
σ = 13.25 ksi .......................................................................................................................... Ans.
δ = 0 (rigid supports) ........................................................................................................... Ans.
(c)
σ = 0 (nothing to exert a force) ........................................................................................ Ans.
(d)
σ

δ =  + α∆T  L = 0 + (12.5 ×10−6 ) ( −100 )  ( 80 )
E

δ = −0.1000 in. ...................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-60*
(a)
 σ

σ

−6
 ( 6000 ) = −1 mm
+
×
−
22.5
10
55
δ =  + α ∆T  L = 
(
)
(
)
9
E

 70 (10 )


σ = 74.9583 (10
(b)
6
) N/m

2
≅ 75.0 MPa ............................................................................. Ans.
 − ( 0.346 ) ( 74.9583 ×106 )

 −νσ

−6
 ( 50 )
+ α ∆T  d = 
+
×
−
δd = 
22.5
10
55
(
)
(
)
70 (109 )
 E



δ d = −0.0804 mm
................................................................................................................. Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-61
FA = FS
AA = π 32 4 = 7.06858 in.2
δ total = δ A + δ S = 0
AS = π 42 4 = 12.5664 in.2
 F

+ α ∆T  L
 AE

δ =


( − FA )
+ (12.5 × 10−6 ) ( 80 )  ( 30 )

 ( 7.06858 )(10, 600 )



( − FS )
+
+ ( 6.6 × 10−6 ) ( 80 )  ( 20 ) = 0
 (12.5664 )( 30, 000 )

FA = FS = 89.4493 kip (both C)
F 89.4493
=
= 12.65 ksi (C) .................................................................................. Ans.
A 7.06858
89.4493
σS =
= 7.12 ksi (C) .............................................................................................. Ans.
12.5664
(a)
σA =
(b)
 − ( 0.30 )( −89.4493)

 −νσ

δd = 
+ α ∆T  d = 
+ ( 6.6 × 10−6 ) ( 80 )  ( 4 )
 E

 (12.5664 )( 30, 000 )

δ d = +0.00240 in. .................................................................................................................. Ans.
(c)

( −89.4493) + 12.5 ×10−6 80  30
uB = δ A = 
(
) ( ) ( )
7.06858
10,
600
(
)(
)


uB = −0.00581 in. = 0.00581 in. ← ................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-62
δ summer
RILEY, STURGES AND MORRIS
 15 (106 )

σ

−6
L
=  + α ∆T  L = 
+
×
11.9
10
0
(
)
(
)
9
E

 200 (10 )



−6
 L = δ summer
+
×
−
11.9
10
40
(
)
(
)
9
 200 (10 )

δ winter = 
σ



 15 (106 )
σ
−6
−6



+
×
−
=
+
×
11.9
10
40
L
11.9
10
0
(
)
(
)
(
)
(
) L
9
9

 200 (10 )

 200 (10 )
σ = 110.2 (106 ) N/m 2 = 110.2 MPa
................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-63*
AS = π ( 0.75 ) 4 = 0.44179 in.2
2
AC = (10 ) − 9 ( 0.44179 ) = 96.0239 in.2
2
9 FS + FC = 150 kip
(a)
δ S = δC
( − FS )( 24 ) + 6.6 ×10−6 100 24
(
) ( )( )
( 0.44179 )( 30, 000 )
( − FC )( 24 ) + 6.0 ×10−6 100 24
=
(
) ( )( )
( 96.0239 )( 4500 )
32.60305FS − FC = 25.92646 kip
FC = 111.9428 kip
(a)
(c)
(b)
FS = 4.22869 kip
111.9418
= 1.166 ksi (C) .......................................................................................... Ans.
96.0239
4.22869
σS =
= 9.57 ksi (C) ............................................................................................. Ans.
0.44179
( −4.22869 )( 24 ) + 6.6 ×10−6 100 24
δ = δS =
(
) ( )( )
( 0.44179 )( 30, 000 )
σC =
δ = +0.00818 in. .................................................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-64*
AAB = ABC = π ( 50 ) 4 = 1963.495 mm 2
2
TAB = TBC + 100 kN
(a)
δ AB + δ BC = 0
(T
AB
× 103 ) ( 200 )
+ (11.9 ×10−6 ) ( −20 )( 200 )
(1963.495 ×10 )( 200 ×10 )
(T ×10 ) ( 200 )
+
+ (11.9 × 10 ) ( −20 )( 200 ) = 0
(1963.495 ×10 )( 200 ×10 )
−6
9
3
−6
BC
−6
TAB + TBC = 186.9248 kN
TAB = 143.4624 kN
σ AB =
σ BC =
(b)
TBC = 43.4624 kN
143.4624 (103 )
1963.495 (10
9
−6
43.4624 (103 )
1963.495 (10
−6
)
)
= 73.1(106 ) N/m 2 = 73.1 MPa .............................................. Ans.
= 22.1(106 ) N/m 2 = 22.1 MPa .............................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-65
RILEY, STURGES AND MORRIS
TC + 2TB = W
If W = 0 , then
TC = −2TB
(a)
But since wires cannot sustain compressive
forces, neither of the tensions can be negative
and the only solution of Eq. (a) is
TC = TB = 0
regardless of any temperature change. And if the tensions are all zero, the stresses are also all zero
σ B = σC = 0
............................................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-66*
→ ΣFx = 0 :
RILEY, STURGES AND MORRIS
Cx = 0
C y + TA + TB − 100 = 0
↑ ΣFy = 0 :
150TA + 450TB − 450 (100 ) = 0
4 ΣM C = 0 :
TA + 3TB = 300 kN
(a)
δ B = ( 450 150 ) δ A


TB ( 500 )
−6


+
×
−
22.5
10
25
500
(
)(
)
(
)
−6
9
 ( 300 × 10 )( 70 × 10 )


TA ( 250 )
 450  
−6


=
+
×
−
11.9
10
25
250
(
)(
)
(
)

 150   (1200 ×10−6 )( 210 × 109 )

TA − 8TB = −19,530 N = −19.530 kN
TA = 212.8555 kN
Cx = 0 kN
(a)
σA =
TB = 29.0482 kN
C y = −141.904 kN
212.8555 (103 )
1200 (10
(b)
= 177.4 (106 ) N/m 2 = 177.4 MPa (T) ...................................... Ans.
)
29.0482 (10 )
=
= 96.8 (10 ) N/m
300 (10 )
−6
C = Cx2 + C y2 = 141.904 kN
3
σB
(b)
6
−6
2
= 96.8 MPa (T) ............................................ Ans.
τ max A = σ max A 2 = 88.7 MPa
.............................................................................................. Ans.
τ max B = σ max B 2 = 48.4 MPa
.............................................................................................. Ans.
(c)
141.904 (103 )
V
τC = =
= 226 (106 ) N/m 2 = 226 MPa ....................................... Ans.
A 2 π ( 0.020 )2 4 


(d)
vE = δ B =
( 29.0482 )( 500 )
( 300 ×10 )( 70 ×10 )
−6
9
+ ( 22.5 × 10−6 ) ( −25 )( 500 )
vE = 0.410 mm ↓ ................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-67
RILEY, STURGES AND MORRIS
TC + 2TB = W = 5 kip
Assume that the wires are adjusted such that the weight is evenly
supported by the three wires prior to the temperature change. Then
TC = TB = 5 3 = 1.66667 kip
(1.66667 )( 3 ×12 ) = 0.0082759 in.
( 0.25)( 29, 000 )
(1.66667 )( 5 ×12 ) = 0.0188679 in.
δC =
( 0.5)(10, 600 )
δB =
After the temperature change of +50 °F
TC + 2TB = 5 kip
(a)
and assuming that the wires stay taught (in tension)
the additional stretch of the wires must be equal
δ B − 0.0082759 = δ C − 0.0188679


TB ( 36 )
+ ( 6.6 ×10−6 ) ( 50 )( 36 )  − 0.0082759

 ( 0.25 )( 29, 000 )

 TC ( 60 )

=
+ (12.5 × 10−6 ) ( 50 )( 60 )  − 0.0188679
 ( 0.5 )(10, 600 )

TB − 2.27987TC = 3.02647 kip
TB = 2.59469 kip
(b)
TC = −0.18939 kip
But since wires cannot sustain compressive forces, neither of the tensions can be negative. Therefore, the wire C
must have become slack, all of the weight is being carried by the wires B , and
TB = 5 2 = 2.500 kip
TC = 0 kip
2.500
= 10.00 ksi ....................................................................................................... Ans.
0.25
σ C = 0 ksi ................................................................................................................................ Ans.
σB =
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-68
→ ΣFx = 0 :
RILEY, STURGES AND MORRIS
Ax = 0
Ay + TCE − TBF = 0
↑ ΣFy = 0 :
4 ΣM A = 0 :
240TCE − 80TBF = 0
TBF = 3TCE
(a)
b = δ BF
c = 3b
c = −δ CE
( −δ CE ) = 3δ BF


TCE ( 600 )
−6

−
+
×
−
22.5
10
60
600
(
)(
)
(
)
9
−6
 ( 900 × 10 )( 73 × 10 )



TBF (1000 )
−6

= 3
+
×
−
11.9
10
60
1000
(
)(
)
(
)
9
−6
 (1200 × 10 )( 210 × 10 )

TCE + 1.30357TBF = 323, 244.00 N
TBF = 197, 472.87 N
(a)
(b)
Ay = 131, 648.58 N
σ BF =
197, 472.87
= 164.6 (106 ) N/m 2 = 164.6 MPa (T) .......................................... Ans.
−6
1200 (10 )
σ CE =
65,824.29
= 73.1(106 ) N/m 2 = 73.1 MPa (T) ................................................. Ans.
−6
900 (10 )
A = Ax2 + Ay2 = 131, 648.58 N
τA =
τB =
τC =
(c)
TCE = 65,824.29 N
(b)
131, 641.58
π ( 0.030 ) 4
2
= 186.2 (106 ) N/m 2 = 186.2 MPa ................................................. Ans.
197, 472.87
= 139.7 (106 ) N/m 2 = 139.7 MPa .......................................... Ans.
2
2 π ( 0.030 ) 4 


65,824.29
π ( 0.030 ) 4
δ BF =
2
= 93.1(106 ) N/m 2 = 93.1 MPa ...................................................... Ans.
(197, 472.87 )(1000 )
(1200 ×10 )( 210 ×10 )
−6
9
+ (11.9 ×10−6 ) ( −60 )(1000 ) = 0.069623 mm
vD = d = 4b = 4δ BF = 0.278 mm ↑ .................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-69*
Initially,
δ init =
( 200 )(120 ) + 12.5 ×10−6 0 120
PL
+ α ∆T L =
(
) ( )( )
AE
( 0.15) (10.6 ×106 )
= 0.01509434 in.
First determine the temperature rise required to close the gap so the weight rests on the floor
δ=
( 200 )(120 ) + 12.5 ×10−6 ∆T 120
(
) ( )( )
( 0.15) (10.6 ×106 )
= δ init + 0.08 = 0.02309434 in.
∆T = 53.33 °F
(a) Not touching the floor. Therefore, T = W = 200 lb and
200
= 1333 psi (T) ...................................................................................................... Ans.
0.15
(b) Now the weight is partially resting on the floor so T < W and
σ=
δ=
σ (120 )
(10.6 ×10 )
6
+ (12.5 × 10−6 ) ( 60 )(120 ) = 0.02309434 in.
σ = −5910 psi
But the wire cannot support a compression. Therefore, at this temperature the wire has become slack, and the
weight rests totally on the floor.
σ = 0 psi .................................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-70*
RILEY, STURGES AND MORRIS
dδ σ
P
= + α∆T =
+ α∆T
dx E
AE
A = π r 2 = π (1 + x ) ro 
2
= π (1 + x ) ro2 = (1 + x ) Ao
2
1
2
 P

+ α∆T  dx = 0
δ = ∫ dδ = ∫ 
0 AE



P 1
dx
P  −1 
P
α
+
∆
=
+ α∆T =
+ α∆T = 0
T
1


(
)


2
∫
0
E  (1 + x ) Ao 
EAo  (1 + x )  0
2 EAo
P
= −2 Eα ∆T
Ao
1
9
−6
P
P
−2 Eα ∆T −2 ( 74 ×10 )(12.5 × 10 ) ( 50 )
σ= =
=
=
2
2
A (1 + x )2 Ao
(1 + x )
(1 + x )
σ=
−92.5 (106 )
(1 + x )
2
N/m 2 =
92.5
(1 + x )
2
MPa (C) ............................................................ Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-71*
sin θ = 4 5
P = 200 kip
TA + 2 ( 4 5 ) TB = 200 kip
(a)
δ B = ( 4 5) δ A
TB ( 5 × 12 )
+ ( 6.6 × 10−6 ) ( 30 )( 5 ×12 )
( 2.5 )( 30, 000 )

 4   T ( 4 ×12 )
=   A
+ ( 9.4 ×10−6 ) ( −50 )( 4 × 12 ) 
 5   ( 3)(15, 000 )

TB = 1.06667TA − 37.41000 kip
TA = 96.0055 kip (T)
(b)
TB = 64.9966 kip (T)
96.0055
= 32.0 ksi .........................................................Ans.
3
64.9966
σB =
= 26.0 ksi .........................................................Ans.
2.5
(a)
σA =
(b)
vC = δ A =
( 96.0055)( 4 ×12 ) + 9.4 ×10−6 −50 4 ×12
(
) ( )( )
( 3)(15, 000 )
vC = 0.0799 in. ...........................................................................Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-72
2Ts = Fa
RILEY, STURGES AND MORRIS
(a)
δ s + δ a = 0.5 mm


Ts ( 330 )
−6


+
×
17.3
10
100
330
(
)(
)
(
)
−6
9
 (115 × 10 )(190 ×10 )



Fa ( 250.5 )
−6
 = 0.5 mm
+
−
×
22.5
10
100
250.5
(
)(
)
(
)
−6
9
 ( 625 ×10 )( 73 × 10 )

Fa + 2.75079Ts = 89, 742.83 N
Ts = 18,890.08 N
σs =
(b)
Fa = 37, 780.16 N
18,890.08
= 164.3 (106 ) N/m 2
−6
115 (10 )
σ s = 164.3 MPa (T) .............................................................................. Ans.
σa =
37, 780.16
= 60.4 (106 ) N/m 2 = 60.4 MPa (C) .................................................. Ans.
−6
625 (10 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-73
TS = V
RILEY, STURGES AND MORRIS
FA = 2V = 2TS
(a,b)
δS = δ A
TS ( 7 )
+ ( 6.6 ×10−6 ) ( −40 )( 7 )
( 0.5 ×1)( 29, 000 )

( − FA )( 7 ) + 12.5 ×10−6 −40 7 
=
(
) ( )( )
 ( 0.5 ×1)(10, 600 )

TS + 2.73585FA = −3.42200 kip
TS = V = −0.52876 kip
(c)
FA = 2V = −1.05752 kip
(The negative means that the steel is actually in compression and the aluminum is actually in tension.)
τ=
0.52876
π ( 0.5 ) 4
2
= 2.69 ksi ............................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-74
 100 x 2 
dδ σ
σ
= + α ∆T = + α  2 
dx E
E
 L 
δ =0
L σ
100α x 2 
σ L 100α L3
=
+
δ =∫  +
dx
=0

0
L2 
E
3L2
E
100α E 100 ( 22.5 × 10
σ=
=
3
3
−6
)( 70 ×10 )
9
σ = 52.5 (106 ) N/m 2 = 52.5 MPa .............................................................................. Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-75
RILEY, STURGES AND MORRIS
Ta = Fs
(a)
δ a + δ s = ∆ nut
(b)


Ta (14 )
+ (12.5 × 10−6 ) ( ∆T )(14 ) 

 (1.4 )(10, 000 )



Fs (12 )
+
− ( 6.6 × 10−6 ) ( ∆T )(12 )  = ∆ nut
 ( 0.400 )( 30, 000 )

Initially,
Ts = Fb = 3.500 kip
∆T = 0
∆ nut = 0.00700 in.
As the temperature rises,


Ta (14 )
+ (12.5 × 10−6 ) ( ∆T )(14 ) 

 (1.4 )(10, 000 )



Fs (12 )
+
− ( 6.6 × 10−6 ) ( ∆T )(12 )  = 0.00700 in.
 ( 0.400 )( 30, 000 )

Ta = Fs = ( 3.5 − 0.0479000∆T ) kip (bolt - tension; sleeve - compression)
Note that at a temperature of about 73 °F the force in the bolt and in the sleeve both go to zero. Beyond this point,
the two pieces separate and no longer exert forces on each other – the forces and stresses both become zero
(a)
σa =
σs =
N ( 3.5 − 0.0479000∆T )
=
= ( 2.5 − 0.0342143∆T ) ksi (T)
A
1.4
0 °F ≤ ∆T ≤ 73 °F
( 3.5 − 0.0479000∆T ) =
0 °F ≤ ∆T ≤ 73 °F
0.4
( 8.75 − 0.119750∆T )
ksi (C)
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-75 (cont.)
σ a = σ s = 0 ksi
(b)
δa =
73 °F ≤ ∆T ≤ 100 °F
( 3.5 − 0.0479000∆T )(14 ) + 12.5 ×10−6 ∆T 14
(
) ( )( )
(1.4 )(10, 000 )
δ a = ( 3.500 + 0.12710∆T ) (10−3 ) in. (stretch)
( 3.5 − 0.0479000∆T )(12 ) − 6.6 ×10−6 ∆T 12
(
) ( )( )
( 0.400 )( 30, 000 )
= ( 3.500 − 0.12710∆T ) (10−3 ) in. (shrink)
δs =
0 °F ≤ ∆T ≤ 73 °F
0 °F ≤ ∆T ≤ 73 °F
δ a = (12.5 × 10−6 ) ( ∆T )(14 ) = 0.17500∆T (10−3 ) in. (stretch)
73 °F ≤ ∆T ≤ 100 °F
δ s = − ( 6.6 ×10−6 ) ( ∆T )(12 ) = −0.0.07920∆T (10−3 ) in. (shrink)
73 °F ≤ ∆T ≤ 100 °F
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MECHANICS OF MATERIALS, 6th Edition
5-76
RILEY, STURGES AND MORRIS
P = 35 kN
4 ΣM F = 0 :
300 ( 35 ) − 50 FD − 100TC = 0
FD + 2TC = 210 kN
(a)
δ C = 2 (δ D + 0.09 ) mm


TC ( 300 )
−6

+
×
∆
11.9
10
T
300
(
) ( )( )
9
−6
 ( 600 × 10 )( 200 × 10 )



FD (150 )
−6

+
= 2
−
×
∆
22.5
10
T
150
0.09
(
)(
)
(
)
9
−6
 ( 2500 × 10 )( 73 × 10 )

TC = ( 0.65753FD − 4.12800∆T + 72.000 ) kN
TC = ( 90.7455 − 1.78311∆T ) kN
(a)
σC =
( 90.7455 − 1.78311∆T ) (103 )
600 (10−6 )
(b)
FD = ( 28.5090 + 3.56621∆T ) kN
= (151.2425 − 2.97185∆T ) MPa (T)
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MECHANICS OF MATERIALS, 6th Edition
5-76 (cont.)
σD =
(b)
δC =
RILEY, STURGES AND MORRIS
( 28.5090 + 3.56621∆T ) (103 )
2500 (10−6 )
= (11.4036 + 1.42648∆T ) MPa (C)
( 90.7455 − 1.78311∆T ) (103 ) ( 300 )
( 600 ×10 )( 200 ×10 )
−6
9
+ (11.9 × 10−6 ) ( ∆T )( 300 )
δ C = ( 0.22686 − 0.00088778∆T ) mm
δD =
( 28.5090 + 3.56621∆T ) (103 ) (150 )
( 2500 ×10 )( 73 ×10 )
−6
9
− ( 22.5 × 10−6 ) ( ∆T )(150 )
δ D = ( 0.02343 + 0.00044387 ∆T ) mm
(c)
θ = tan −1
δC
100
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MECHANICS OF MATERIALS, 6th Edition
5-77
4 ΣM C = 0 :
RILEY, STURGES AND MORRIS
8TA + 3TB = 0
8σ A (1.75 ) + 3σ B ( 2.25 ) = 0
σ B = −2.07407σ A
(a)
δ A = ( 8 3) δ B
 σ A ( 48 )

+ ( 6.6 ×10−6 ) ( ∆T )( 48 ) 

 ( 28, 000 )


8  σ ( 36 )
=  B
+ (12.5 × 10−6 ) ( ∆T )( 36 ) 
3  (10, 600 )

σ A − 5.2830σ B = 0.51520∆T ksi
(a)
(b)
σ A = 0.043087∆T ksi (T)
σ B = −0.089365∆T = 0.089365∆T ksi (C)
(b)
δ A = 390.7 (10−6 ) ∆T in. (stretch)
δ B = 146.51(10−6 ) ∆T in. (stretch)
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-78
4 ΣM A = 0 :
RILEY, STURGES AND MORRIS
240TCE − 80TBF = 0
TBF = 3TCE
σ BF (1200 ×10−6 ) = 3σ CE ( 900 × 10−6 )
σ BF = 2.2500σ CE
(a)
c = 3b
(b)
b = δ BF =
σ BF (1000 )
( 210 ×10 )
9
= 4.76190 (10−9 ) σ BF mm
(The temperature of BF never changes.)
 σ ( 600 )

−6

c = −  CE
T
22.5
10
600
+
×
∆
(
)(
)
(
)
9
 ( 73 ×10 )

=  −8.21918 (10−9 ) σ CE − 13.500 (10−3 ) ( ∆T )  mm
When ∆T = 0 (after CE has been heated and the pin inserted),
b = c = σ CE = σ BF = 0
δ CE = ( 22.5 ×10−6 ) ( 80 )( 600 ) = 1.0800 mm
As CE cools down
c =  −8.21918 (10−9 ) σ CE − 13.500 (10−3 ) ( ∆T )  = 3b = 3  4.76190 (10−9 ) σ BF 
14.28570σ BF + 8.21918σ CE = −13.500 (106 ) ( ∆T )
(b)
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-78 (cont.)
(a)
σ BF = −0.75256 (106 ) ∆T N/m 2
........................................................................................ Ans.
σ CE = −0.33447 (106 ) ∆T N/m 2
........................................................................................ Ans.
(Note that since ∆T is negative the stresses will be positive or tension stresses.)
(b)
δ BF
 −0.75256 (106 ) ∆T  (1000 )

=
= −3.58364 (10−3 ) ∆T mm ................................ Ans.
9
( 210 ×10 )
c = −8.21918 (10−9 )  −0.33447 (106 ) ∆T  − 13.500 (10−3 ) ∆T
= −10.75093 (10−3 ) ∆T
δ CE = 1.080 − c = 1.080 + 10.75093 (10−3 ) ∆T  mm ................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
5-79*
For the fillet:
D d = 3 2 = 1.50
From Fig. 5-20(c):
σ = Kt
r d = 0.4 2 = 0.20
K t ≅ 1.73
P
P
= 1.73
≤ 66 ksi
At
( 2 )( 0.25 )
P ≤ 19.08 kip
d w = 0.5 3 = 0.1667
For the hole:
From Fig. 5-20(b):
σ = Kt
RILEY, STURGES AND MORRIS
K t ≅ 2.48
P
P
= 2.48
≤ 66 ksi
At
( 3 − 0.5)( 0.25)
P ≤ 16.63 kip
Pmax = 16.63 kip ..................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-80
For the small hole:
From Fig. 5-20(b):
σ = Kt
d w = 16 160 = 0.10
K t ≅ 2.6
P
P
= 2.6
≤ 760 (106 ) N/m 2
At
0.160
−
0.016
0.010
(
)(
)
P ≤ 421(103 ) N
d w = 64 160 = 0.40
For the large hole:
From Fig. 5-20(b):
σ = Kt
RILEY, STURGES AND MORRIS
K t ≅ 2.2
P
P
= 2.2
≤ 760 (106 ) N/m 2
At
( 0.160 − 0.064 )( 0.010 )
P ≤ 332 (103 ) N
Pmax = 332 kN ........................................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-81
(a) With no hole:
σA =
d w = (1 64 ) 4 = 0.0039
With a hole:
K t ≅ 3.00
From Fig. 5-20(b):
σ A = 3.00
500
= 3012 psi ................................................................. Ans.
( 4 − 1 64 )(1 8)
θ = 0°
(b) At A:
Eq. 5-6:
500
= 1000 psi ....................................................................... Ans.
( 4 )(1 8)
σr = 0
τ rθ = 0
σ θ = σ (1 + 2 cos 2θ ) = 1000 1 + 2 cos 2 ( 0° )  = 3000 psi
(c) With a 1-in. hole:
d w = 1 4 = 0.25
σ A = 2.35
K t ≅ 2.35
500
= 3133 psi ....................................................................... Ans.
( 4 − 1)(1 8 )
σ θ = σ (1 + 2 cos 2θ ) = 1000 1 + 2 cos 2 ( 0° )  = 3000 psi
(d) With a 2-in. hole:
d w = 2 4 = 0.50
σ A = 2.12
............................ Ans.
............................ Ans.
K t ≅ 2.12
500
= 4240 psi ...................................................................... Ans.
( 4 − 2 )(1 8)
σ θ = σ (1 + 2 cos 2θ ) = 1000 1 + 2 cos 2 ( 0° )  = 3000 psi
............................ Ans.
(Since Eq. 5-6 assumes a plate of infinite width, the size of the hole has no effect in the equation.)
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-82*
For the fillet:
RILEY, STURGES AND MORRIS
D d = 80 40 = 2.0
r d = r 40
σ = Kt
100 (103 )
P
= Kt
≤ 205 (106 ) N/m 2
At
( 0.040 )( 0.020 )
K t ≤ 1.64
From Fig. 5-20(c):
r d = r 40 ≥ 0.3
rmin ≅ 12.00 mm ..................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-83*
d
r
0.5
=1
=
r
b B −1
P
10
σ = Kt = Kt
≤ 20 ksi
At
( B − 1)( 0.5)
For the grooves:
(a)
Solve by systematic trial and error.
First try K t ≅ 2 :
then
B −1 = 2
r b = 0.5 2 = 0.25
Fig. 5-20(a):
K t ≅ 1.92
Eq. (a) gives
Next try K t ≅ 1.92 :
Eq. (a) gives
B − 1 = 1.92
then
r b = 0.5 1.92 = 0.26
Fig. 5-20(a):
K t ≅ 1.92
Therefore
Bmin = 1.92 + 1 = 2.92 in. ........................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-84
At A:
RILEY, STURGES AND MORRIS
τ xy = 0
σx = 0
d w = 80 200 = 0.40
K t ≅ 2.2
From Fig. 5-20(b):
180 (103 )
P
σ y = K t = 2.2
= 132.0 (106 ) N/m 2
At
( 0.200 − 0.080 )( 0.025)
ν=
εx =
E
73
−1 =
− 1 = 0.30357
2G
2 ( 28 )
σ x −νσ y
E
=
0 − ( 0.30357 ) (132.0 × 106 )
73 (109 )
ε x = −549 (10−6 ) = −549 µ m/m .................................................................................. Ans.
εy =
σ y −νσ x
E
=
132.0 (106 ) − 0
73 (109 )
ε y = 1808 (10−6 ) = +1808 µ m/m ................................................................................ Ans.
γ xy =
τ xy
G
= 0 µ rad .......................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-85*
4 ΣM B = 0 :
(a)
( 35 ) = 35.0 ksi ................................................................Ans.
1.00
TCD = 1.25 P = 1.25 ( 35 ) = 43.75 kip
σ CD =
(c)
8TCD − 10 P = 0
TCD = 1.25 P = 1.25 ( 28 ) = 35 kip
σ CD =
(b)
RILEY, STURGES AND MORRIS
( 43.75) = 43.75 ksi ≅ 43.7 ksi ...................................Ans.
1.00
ε CD = 0.004 +
43.75 − 42
= 0.0052500 in./in.
1400
δ CD = ε CD LCD = ( 0.0052500 )( 8 ) = 0.0420 in. ................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-86*
A=
(a)
π
4
σ top =
( 20 )
2
= 314.1593 mm 2
50 (103 )
314.1593 (10
−6
)
σ top = 159.2 MPa
σ bottom =
σ top =
30 (103 )
314.1593 (10
95 (103 )
314.1593 (10
σ bottom =
= 159.2 (106 ) N/m 2
.................................................. Ans.
−6
σ bottom = 95.5 MPa
(b)
RILEY, STURGES AND MORRIS
−6
)
......................................................................................................... Ans.
= 302 (106 ) N/m 2 = 302 MPa ................................................ Ans.
30 (103 )
314.1593 (10
)
= 95.5 (106 ) N/m 2
−6
)
= 95.5 (106 ) N/m 2 = 95.5 MPa .......................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-87
(a)
σ AB =
σ BC =
10
2
10
π ( 0.75 ) 4
2
 8.14873
( < 42 ksi )
= 8.14873 ksi
π (1.25 ) 4
= 22.63537 ksi
  22.63537
( < 42 ksi )

δ =
(10 )  + 
(10 ) = 0.0293 in.
 10,500
  10,500

(b)
σ AB =
σ BC =
20
π (1.25 ) 4
2
20
π ( 0.75 ) 4
1629747
2
= 16.29747 ksi
( < 42 ksi )
= 45.27074 ksi
( > 42 ksi )
  42
δ =
+
(10 )  + 
 10,500
 10,500
................................................... Ans.
45.27074 − 42 
 (10 ) = 0.0789 in. ........................ Ans.
1400

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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-88*
→ ΣFx = 0 :
RILEY, STURGES AND MORRIS
Cx = 0
C y + TA + TB − 160 = 0
↑ ΣFy = 0 :
4 ΣM C = 0 :
1.5TB + 4TA − 5 (160 ) = 0
1.5TB + 4TA = 800 kN
(a)
δ A = ( 4 1.5 ) δ B
(b)
Try elastic solution …

TB (1500 )
 4 


=
( 500 ×10−6 )( 73 ×109 )  1.5   ( 750 ×10−6 )( 210 ×109 ) 
TA ( 2000 )
TB = 2.15753TA
TA = 110.55372 kN
σA =
σB =
(a) Therefore,
110.55362 (103 )
500 (10
)
−6
238.52343 (103 )
750 (10
−6
)
TB = 238.52343 kN
= 221(106 ) N/m 2
( < 330 MPa - elastic )
= 318 (106 ) N/m 2
( > 275 MPa - plastic )
σ B = 275 MPa (T) .................................................................................................. Ans.
TB = ( 275 × 106 )( 750 ×10−6 ) = 206, 250 N
σA =
(b)
(c)
(d)
122, 656
= 245 (106 ) N/m 2 = 245 MPa (T) (still elastic) ........................... Ans.
500 (10−6 )
Cx = 0 kN
τC =
TA = 122, 656 N
C y = −168.906 kN
168,906
2 π ( 0.030 ) 4 


2
C = Cx2 + C y2 = 168,906 kN
= 119.5 (106 ) N/m 2 = 119.5 MPa .......................................... Ans.
τ max A = σ max A 2 = 122.5 MPa
............................................................................................ Ans.
τ max B = σ max B 2 = 137.5 MPa
............................................................................................ Ans.
vD =
5δ A 5  (122, 656 )( 2000 ) 
 = 8.40 mm ↓ ................................................. Ans.
= 
4
4  ( 500 × 10−6 )( 73 × 109 ) 


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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-89*
2TS + TA = P
RILEY, STURGES AND MORRIS
2 (1.6σ S ) + ( 3.2σ A ) = P
(a)
σ S = 3.41981σ A
(b)
δS = δ A
σ S (8)
=
σ A (10 )
( 29, 000 ) (10, 600 )
σ S = 100 ksi
TS = 100 (1.6 ) = 160 kip (T)
σ A = 29.2414 ksi
TA = 29.2414 ( 3.2 ) = 93.572 kip (T)
Pmax = 93.572 + 2 (160 ) = 414 kip ..................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-90
RILEY, STURGES AND MORRIS
sin θ = 4 5
TA + 2 ( 4 5 ) TB = 1110 kN
(a)
δ B = ( 4 5) δ A
(b)
Try elastic solution …

TA (1200 )
 4


=
(1500 ×10−6 )( 200 ×109 )  5   (1500 ×10−6 )( 72 ×109 ) 
TB (1500 )
TB = 1.77778TA
TA = 288.7283 kN
σA =
σB =
(a) Therefore,
TB = 513.2948 kN
288.7283 (103 )
1500 (10−6 )
513.2948 (103 )
1500 (10
−6
)
= 192.5 (106 ) N/m 2
= 342 (106 ) N/m 2
( < 380 MPa - elastic )
( > 250 MPa - plastic )
σ B = 250 MPa (T) .................................................................................................. Ans.
TB = ( 250 ×106 )(1500 ×10−6 ) = 375, 000 N
σA =
(b)
TA = 510, 000 N
510, 000
= 340 (106 ) N/m 2 = 340 MPa (T) (still elastic) ......................... Ans.
−6
1500 (10 )
vC = δ A =
( 510, 000 )(1200 )
(1500 ×10 )( 72 ×10 )
−6
9
= 5.67 mm ↓ .......................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-91*
Cx + ( 3 5 ) TA = 0
→ ΣFx = 0 :
C y + TB + ( 4 5 ) TA − 50 = 0
↑ ΣFy = 0 :
4 ΣM C = 0 :
6TB + 15 ( 4 5 ) TA  − 18 ( 50 ) = 0
TB + 2TA = 150 kip
(a)
δ A = ( 4 5 ) a = ( 4 5 ) (15 6 ) δ B 
δ A = 2δ B
(b)
Try elastic solution …
 TB ( 20 )

TA ( 40 )
= 2

(1.5)(10,500 )  (1.5 )( 29, 000 ) 
TB = 2.76190TA
TA = 31.50 kip
TB = 87.00 kip
31.5
= 21.0 ksi ( < 55 ksi - elastic )
1.5
87.00
σB =
= 58.0 ksi ( > 36 ksi - plastic )
1.5
(a) Therefore, σ B = 36 ksi (T) ........................................................................................................ Ans.
σA =
TB = ( 36 )(1.5 ) = 54.00 kip
σA =
(b)
48.00
= 32.0 ksi (T) (still elastic) ........................................................................ Ans.
1.5
Cx = −28.80 kip
τC =
TA = (150 − 54 ) 2 = 48.00 kip
C y = −42.40 kip
C = Cx2 + C y2 = 51.2562 kip
51.2562
= 32.6 ksi ............................................................................................... Ans.
2
2 π (1) 4 


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MECHANICS OF MATERIALS, 6th Edition
5-92
EA =
350 (106 )
RILEY, STURGES AND MORRIS
= 70.0 (109 ) N/m 2
ES =
0.005
TA + TS = 530 kN
(a)
δ A = δS
(b)
1400 (106 )
0.007
= 200.0 (109 ) N/m 2
Try elastic solution …
TA ( 750 )
=
TS ( 750 )
( 315 ×10 )( 70 ×10 ) ( 315 ×10 )( 200 ×10 )
−6
9
−6
9
TA = 137.407 kN
TS = 392.593 kN
σA =
137, 407
= 436 (106 ) N/m 2
315 (10−6 )
σS =
392,593
= 1246 (106 ) N/m 2
315 (10−6 )
Therefore,
TS = 2.85714TA
( > 350 MPa - plastic )
( < 1400 MPa - elastic )
TA = ( 350 ×106 )( 315 × 10−6 ) = 110, 250 N
σ A = 350 MPa (T)
TS = 530, 000 − 110, 250 = 419, 750 N
σS =
419, 750
= 1333 (106 ) N/m 2 = 1333 MPa (T) (still elastic)
−6
315 (10 )
v = δS =
( 419, 750 )( 750 )
( 315 ×10 )( 200 ×10 )
−6
9
= 5.00 mm ↓ ............................................................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
5-93
4 ΣM D = 0 :
10 P − 5TA − 2TB = 0
5TA + 2TB = 10 P
(a)
RILEY, STURGES AND MORRIS
(a)
δ A = ( 5 2 ) δ B = 2.5δ B
(b)
c = (10 5 ) δ A = 2δ A
(c)
P = 40 kip
Try elastic solution …


TA (10 )
TB (10 )
= 2.5 

( 2 )(10, 000 )
 ( 2.5 )( 30, 000 ) 
5TA + 2TB = 10 ( 40 ) = 400 kip
TB = 1.5TA
TA = 50.0 kip
50.0
= 25.0 ksi
2
75.0
σB =
= 30.0 ksi
2.5
P = 60 kip
σA =
(b)
TB = 75.0 kip
( < 34 ksi - elastic ) ...................................................................... Ans.
( < 36 ksi - elastic ) ...................................................................... Ans.
Try elastic solution …
TB = 1.5TA
TA = 75.0 kip
5TA + 2TB = 10 ( 60 ) = 600 kip
TB = 112.5 kip
75
= 37.5 ksi ( > 34 ksi - yields )
2
112.5
σB =
= 45.0 ksi ( > 36 ksi - plastic )
2.5
Therefore, σ B = 36 ksi (T) ........................................................................................................ Ans.
σA =
TB = ( 36 )( 2.5 ) = 90.0 kip
σA =
TA = 84.0 kip
84.00
= 42.0 ksi (T) .................................................................................................. Ans.
2
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MECHANICS OF MATERIALS, 6th Edition
5-93 (cont.)
(c)
RILEY, STURGES AND MORRIS
P = 65 kip
5TA + 2TB = 10 ( 65 ) = 650 kip
σ B = 36 ksi (T)
TB = ( 36 )( 2.5 ) = 90.0 kip
TA = 94.0 kip
σA =
94.0
= 47.0 ksi
2
ε A ≅ 0.006 in./in.
vC = 2δ A = 2ε A LA = 2 ( 0.006 )(10 ) = 0.1200 in. ↓ ....................................................... Ans.
25
= 0.00250 in./in. ........................................................ Ans.
10, 000
(d)
P = 40 kip
εA =
(e)
P = 65 kip
ε A ≅ 0.006 in./in. ............................................................................... Ans.
εB =
δB
LB
=
δ A 2.5
10
=
εA
2.5
= 0.00240 in./in. .................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
5-94
4 ΣM C = 0 :
RILEY, STURGES AND MORRIS
150TA + 450TB − 450 P = 0
TA + 3TB = 3P
σ A ( 500 × 10−6 ) + 3σ B ( 750 × 10−6 ) = 3P
σ A + 4.5σ B = 6000 P
(a)
δ B = 3δ A
If
(b)
ε B ≤ 0.0013095 m/m , then
 σ ( 250 ) 
 A

=
3
9
×
210
10
(
)  ( 73 ×109 ) 
σ B ( 500 )
σ B = 4.3151σ A
(b)
From Eqs. (a) and (b):
σ A = 293.9 P N/m 2 (T)
σ B = 1268.0 P N/m 2 (T)
δA =
δB =
If
σ A ( 250 )
( 73 ×10 )
9
mm
σ B ( 500 )
( 210 ×10 )
9
mm
ε B ≥ 0.0013095 m/m , then
σ B = 275 (106 ) N/m 2 (T)
and from Eq. (a):
σ A = 6000 P − 1237.5 (106 ) N/m 2 (T)
δA =
σ A ( 250 )
( 73 ×10 )
9
mm
δ B = 3δ A
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-95
Ast = π ( 0.75 ) 4 = 0.44179 in 2
2
Fal = Tst
σ al ( 0.40 ) = σ st ( 0.44179 )
σ al = 1.10447σ st
δ st + δ al = ∆ nut =
δ st =
If
ε al ≤ 0.004
δ al =
σ st (12 )
( 30, 000 )
(a)
0.125θ
360
(b)
= 0.400 (10−3 ) σ st
then
σ al (10 )
(10,500 )
= 0.95238 (10−3 ) σ al
0.400 (10−3 ) σ st + 0.95238 (10−3 ) σ al =
0.125θ
360
(b)
From Eqs. (a) and (b)
(a)
σ st = 0.23915θ ksi (T)
σ al = 1.10447σ st = 0.26414θ ksi (C)
(b)
δ st = 0.400 (10−3 ) σ st = 0.09566 (10−3 ) θ in.
δ al = 0.95238 (10−3 ) σ al = 0.25156 (10−3 ) θ in.
If
ε al ≥ 0.004
then

δ al = 0.004 +

(σ al − 42 )  10 = 7.14286 10−3 σ − 0.2600
( ) al
( )
(1400 ) 
0.125θ
0.400 (10−3 ) σ st +  7.14286 (10−3 ) σ al − 0.2600  =
360
(b)
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-95 (cont.)
Now from Eqs. (a) and (b)
(a)
σ st = ( 31.3666 + 0.041889θ ) ksi (T)
σ al = 1.10447σ st = ( 34.6434 + 0.046265θ ) ksi (C)
(b)
δ st = 0.400 (10−3 ) σ st = (12.5466 + 0.01676θ ) (10−3 ) in.
δ al = 7.14286 (10−3 ) σ al − 0.2600 = ( −12.5470 + 0.33046θ ) (10−3 ) in.
(c)
L = 10 − δ al
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MECHANICS OF MATERIALS, 6th Edition
5-96*
σ=
RILEY, STURGES AND MORRIS
pr (100 )(148 )
=
= 3700 kPa = 3.70 MPa ............................................................ Ans.
2t
2 ( 2)
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-97*
RILEY, STURGES AND MORRIS
pr ( 800 )( 5 )
=
≤ 10, 000 psi
t
t
t ≥ 0.400 in. ..................................................................................................................... Ans.
σh =
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-98
RILEY, STURGES AND MORRIS
σa =
pr p (1625 − 22 )
=
≤ 45.0 MPa
2t
2 ( 22 )
p ≤ 1.235 MPa
σh =
pr p (1625 − 22 )
=
≤ 100.0 MPa
t
( 22 )
p ≤ 1.372 MPa
pmax = 1.235 MPa ................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-99
σ=
RILEY, STURGES AND MORRIS
pr (100 ) (17.5 × 12 ) − 7 8
=
= 11,950 psi
2t
2 ( 7 8)
σ = 11.95 ksi ................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-100*
RILEY, STURGES AND MORRIS
σx = σa =
pr ( 950 )( 500 − 50 )
=
= 4275 kPa = 4.275 MPa
2t
2 ( 50 )
σ y = σh =
pr ( 950 )( 500 − 50 )
=
= 8550 kPa = 8.550 MPa
t
( 50 )
τ xy = 0 MPa
θ = −30°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= 4.275cos 2 ( −30° ) + 8.55sin 2 ( −30° ) + 0
σ n = 5.34 MPa (T)
............................................................................................................... Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − ( 4.275 − 8.55 ) sin ( −30° ) cos ( −30° ) + 0
τ nt = −1.851 MPa .................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-101*
(a)
(b)
RILEY, STURGES AND MORRIS
σa =
pr ( 200 )( 3 × 12 )
=
= 7200 psi = 7.20 ksi ............................................................ Ans.
2t
2 ( 0.5 )
σh =
pr ( 200 )( 3 × 12 )
=
= 14, 400 psi = 14.40 ksi ...................................................... Ans.
t
( 0.5)
E = 2 (1 + ν ) G
εa =
εh =
σ a −νσ h
E
σ h −νσ a
E
ν=
29, 000
− 1 = 0.31818
2 (11, 000 )
=
7.2 − 0.31818 (14.4 )
= 90.3 (10−6 ) = 90.3 µ in./in. ......................... Ans.
29, 000
=
14.4 − 0.31818 ( 7.2 )
= 418 (10−6 ) = 418 µ in./in. ........................... Ans.
29, 000
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-102
RILEY, STURGES AND MORRIS
p = γ h = ( 850 × 9.81)( 6 ) = 50, 031.00 N/m 2
pr ( 50, 031.00 )(10 )
=
≤ 80 (106 ) N/m 2
t
t
t ≥ 0.00625 m = 6.25 mm ............................................................................................ Ans.
σh =
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-103
(a)
p = γ h = ( 62.4 )( 50 ) = 3120 psf = 21.6667 psi
σ a = 0 ksi (not including the weight of the tank)
σh =
(b)
RILEY, STURGES AND MORRIS
..................................................... Ans.
pr ( 21.6667 )( 6 ×12 − 0.5 )
=
= 3098 psi ≅ 3.10 ksi ......................................... Ans.
t
( 0.5)
p = γ h = ( 62.4 )( 25 ) = 1560 psf = 10.8333 psi
σ a = 0 ksi (not including the weight of the tank)
σh =
..................................................... Ans.
pr (10.8333)( 6 × 12 − 0.5 )
=
= 1549 psi = 1.549 ksi ........................................ Ans.
t
( 0.5)
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-104*
RILEY, STURGES AND MORRIS
pr ( 2800 × 10 ) ( 600 )
σx = σh =
=
t
( 20 )
3
(a)
= 84.00 (106 ) N/m 2 = 84.00 MPa
pr ( 2800 × 10 ) ( 600 )
σ y = σa =
=
2t
2 ( 20 )
3
= 42.00 (106 ) N/m 2 = 42.00 MPa
τ xy = 0 MPa
φ = tan −1
3
= 36.870°
4
θ = 90° + φ = 126.870°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= 84 cos 2 (126.870° ) + 42sin 2 (126.870° ) + 0
σ n = 57.1 MPa (T) ................................................................................................................ Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
= − ( 84 − 42 ) sin (126.870° ) cos (126.870° ) + 0
τ nt = +20.2 MPa .................................................................................................................... Ans.
(b)
E = 2 (1 + ν ) G
εa =
εh =
(c)
σ a −νσ h
E
σ h −νσ a
E
=
84 − 0.31579 ( 42 )  (106 )
=
= 354 (10−6 ) = 354 µ m/m ................... Ans.
9
200 (10 )
σ max − σ min
2
=
σ max = σ h = 84.00 MPa
τ max =
200
− 1 = 0.31579
2 ( 76 )
 42 − 0.31579 ( 84 )  (106 )
= 77.4 (10−6 ) = 77.4 µ m/m ................. Ans.
9
200 (10 )
σ max = σ h = 84.00 MPa
τ max =
(d)
ν=
σ max − σ min
2
=
σ min = σ z = 0 MPa
84.00 − 0
= 42.0 MPa ...................................................... Ans.
2
σ min = σ z = −2.800 MPa
84.00 − ( −2.800 )
= 43.4 MPa ....................................... Ans.
2
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-105*
E = 2 (1 + ν ) G
(a)
ν=
30, 000
− 1 = 0.29310
2 (11, 600 )
E
30, 000
ε +νε 2 ) =
619 + 0.29310 ( 330 )  (10−6 )
2 ( 1
2 
1 −ν
1 − 0.29310
σ 1 = 23.490 ksi ≅ 23.5 ksi .................................................................................................. Ans.
σ1 =
E
30, 000
ε +νε1 ) =
330 + 0.29310 ( 619 )  (10−6 )
2 ( 2
2 
1 −ν
1 − 0.29310
σ 2 = 16.785 ksi ≅ 16.79 ksi ................................................................................................ Ans.
σ2 =
(b)
σx = σa =
pr p ( 9.875 )
=
= 39.5 p ksi
2t 2 ( 0.125 )
σ y = σh =
pr p ( 9.875 )
=
= 79.0 p ksi
t
( 0.125)
τ xy = 0 ksi
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
σ 2 = ( 39.5 p ) cos 2 ( 30° ) + ( 79.0 p ) sin 2 ( 30° ) + 0 = 16.785 ksi
p = 0.342 ksi = 342 psi ....................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-106
At point A:
pAp − σ h Aσ = 0
↑ ΣFy = 0 :
2
p π R 2 − π ( R − r )  − σ h  2π ( R − r ) t  = 0


σh =
pr ( 2 R − r )
............................................ Ans.
2t ( R − r )
σh
σa
rr
+
σa =
ra
=
p
t
rh = r
ra = − ( R − r )
p ( 2R − r )
σ
p
p
pr
...................... Ans.
 − ( R − r )  =
( ra ) − h ( ra ) =  − ( R − r ) −
2t ( R − r )
2t
t
rh
t
At point B:
p π ( R + r ) − π R 2  − σ h  2π ( R + r ) t  = 0


↑ ΣFy = 0 :
2
σh =
pr ( 2 R + r )
................................................................................................................... Ans.
2t ( R + r )
σh
σa
rr
+
σa =
ra
=
p
t
rh = r
ra = R + r
p ( 2R + r )
σ
p
p
pr
( ra ) − h ( ra ) = ( R + r ) −
( R + r ) = ..................................... Ans.
2t ( R + r )
2t
t
rh
t
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-107
p =γ y
θ = sin −1
r 2
= 30°
r
W = ∫ γ dV = ∫ γπ ( r 2 − y 2 ) dy
r
r 2
r

y3 
5γπ r 3
= γπ  r 2 y −  =
3 r 2
24

x = r 2 − y 2 = r 2 − ( r 2) =
2
r 3
2
σ m Am cos 30° − W − pAp = 0
↑ ΣFy = 0 :
σ m ( 2π xt ) cos 30° − W − p (π x 2 ) = 0
2
r 3
 5γπ r 3 
 γ r  r 3 
σ m ( 2π t ) 
 cos 30° − 
 =0
 − π   
 2   2 
 24 
 2 
7γ r 2
........................................................................................................................ Ans.
18t
σm σt p
+ =
rm = rt = r
rm
rt
t
σm =
7γ r σ t γ r
+ =
18t
r
2t
σt =
γ r2
9t
................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-108
RILEY, STURGES AND MORRIS
pr p ( 600 )
=
= 30 p N/m 2
t
20
pr p ( 600 )
σ y = σa =
=
= 15 p N/m 2
2t
2 ( 20 )
σx = σh =
τ xy = 0 N/m 2
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ = 30 p cos 2 ( −53° ) + 15 p sin 2 ( −53° )
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ ) = − ( 30 p − 15 p ) sin ( −53° ) cos ( −53° )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-109
σx = σh =
pr ( 200 )( 24 )
=
= 6400 psi
t
( 0.75)
σ y = σa =
pr ( 200 )( 24 )
=
= 3200 psi
2t
2 ( 0.75 )
RILEY, STURGES AND MORRIS
τ xy = 0 psi
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
σ n = ( 6400 ) cos 2 (θ − 90° ) + 3200sin 2 (θ − 90° )
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
τ nt = − ( 6400 − 3200 ) sin (θ − 90° ) cos (θ − 90° )
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MECHANICS OF MATERIALS, 6th Edition
5-110*
σx =
RILEY, STURGES AND MORRIS
6
( 40 ×103 )
pr P ( 2 × 10 ) ( 500 )
− =
−
2t A
2 ( 20 )
π (1.0402 − 12 ) 4
σ x = 24.4 (106 ) N/m 2 = 24.4 MPa
........................................................................... Ans.
pr ( 2 ×10 ) ( 500 )
=
t
( 20 )
6
σy =
σ y = 50.0 (106 ) N/m 2 = 50.0 MPa ............................................................................ Ans.
τ xy = 0 MPa
..................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-111
σx =
RILEY, STURGES AND MORRIS
( P)
pr P ( 300 )(1.5 × 12 )
+ =
+
≤ 18 (103 ) psi
2
2
2t A
2 ( 0.375 )
π (18.75 − 18 ) 4
P ≤ 234 (103 ) lb = 234 kip .......................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-112
RILEY, STURGES AND MORRIS
pr ( 2800 × 10 ) ( 600 )
σx = σh =
=
t
( 20 )
3
= 84.00 (106 ) N/m 2 = 84.00 MPa
3
130 (103 )
pr P ( 2800 × 10 ) ( 600 )
σy =
− =
−
2t A
2 ( 20 )
π (1.282 − 1.22 ) 4
= 41.1657 (106 ) N/m 2 = 41.1657 MPa
τ xy = 0 MPa
(a)
φ = tan −1
3
= 36.870°
4
θ = 90° + φ = 126.870°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
σ n = 84 cos 2 (126.870° ) + 41.1657 sin 2 (126.870° ) = 56.6 MPa (T) ....................... Ans.
(b)
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
τ nt = − ( 84 − 41.1657 ) sin (126.870° ) cos (126.870° ) = +20.6 MPa ......................... Ans.
(c)
σ max = σ h = 84.00 MPa
τ max =
(d)
σ max − σ min
2
=
σ max = σ h = 84.00 MPa
τ max =
σ max − σ min
2
=
σ min = σ z = 0 MPa
84.00 − 0
= 42.0 MPa ...................................................... Ans.
2
σ min = σ z = −2.800 MPa
84.00 − ( −2.800 )
= 43.4 MPa ....................................... Ans.
2
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MECHANICS OF MATERIALS, 6th Edition
5-113*
σx =
RILEY, STURGES AND MORRIS
(10, 000 )
pr P p ( 20 )
+ =
−
2t A 2 ( 0.4 ) π ( 20.82 − 202 ) 4
= 25 p + 390.08564 psi
σy =
pr p ( 20 )
=
= 50 p
t
( 0.4 )
τ xy = 0 MPa
φ = tan −1
3
= 36.870°
4
θ = 90° + φ = 126.870°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= ( 25 p + 390.08564 ) cos 2 (126.870° ) + ( 50 p ) sin 2 (126.870° ) + 0
= 41 p + 140.43150 ≤ 11, 000
p = 265 psi ............................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-114*
σ axial
RILEY, STURGES AND MORRIS
50 (103 )
pr P ( 2500 )( 500 )
=
+ =
+
2t A
2 ( 20 )
π (1.042 − 12 ) 4
= 811.421(103 ) N/m 2 = 811.421 kPa
σ hoop =
pr ( 2500 )( 500 )
=
= 62.500 (103 ) N/m 2 = 62.500 kPa
t
20
( )
E = 2 (1 + ν ) G
ε axial =
σ axial −νσ hoop
E
ν=
=
200
− 1 = 0.31579
2 ( 76 )
811, 421 − 0.31579 ( 62,500 )
200 (109 )
ε axial = 3.96 (10−6 ) = 3.96 µ m/m ........................................................................................ Ans.
ε hoop =
σ hoop −νσ axial
E
=
62,500 − 0.31579 ( 811, 421)
200 (109 )
ε hoop = −0.969 (10−6 ) = −0.969 µ m/m .............................................................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
5-115
RILEY, STURGES AND MORRIS
σx =
( 5000 )
pr P (100 )( 2 × 12 )
+ =
+
= 4931.94 psi
2t A
2 ( 0.25 )
π ( 48.52 − 482 ) 4
σy =
pr (100 )( 2 × 12 )
=
= 9600.00 psi
t
( 0.25)
σ max = 9600 psi
Outside:
τ max =
σ max − σ min
2
=
9600 − 0
= 4800 psi ................................................................ Ans.
2
σ max = 9600 psi
τ max =
σ max − σ min
2
=
σ min = 0 psi
σ min = − p = −100 psi
9600 − ( −100 )
= 4850 psi ..................................................... Ans.
2
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-116
σx = σh =
RILEY, STURGES AND MORRIS
pr p ( 600 )
=
= 30 p N/m 2
t
( 20 )
130 (103 )
pr P p ( 600 )
σy =
− =
−
2t A 2 ( 20 ) π (1.282 − 1.22 ) 4
= 15 p − 834, 280 N/m 2
τ xy = 0 N/m 2
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
σ n = ( 30 p ) cos 2 ( −53° ) + (15 p − 834, 280 ) sin 2 ( −53° ) N/m 2
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
τ nt = − ( 30 p ) − (15 p − 834, 280 )  sin ( −53° ) cos ( −53° ) N/m 2
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MECHANICS OF MATERIALS, 6th Edition
5-117
σx = σh =
σy =
RILEY, STURGES AND MORRIS
pr ( 200 )( 24 )
=
= 6400 psi
t
( 0.75)
pr P ( 200 )( 24 )
30, 000
− =
−
= 2934.74 psi
2t A
2 ( 0.75 ) π ( 4 × 12 )( 0.75 )
τ xy = 0 psi
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
σ n = ( 6400 ) cos 2 (θ − 90° ) + 2934.74sin 2 (θ − 90° )
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
τ nt = − ( 6400 − 2934.74 ) sin (θ − 90° ) cos (θ − 90° )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-118*
RILEY, STURGES AND MORRIS
(a)
σt =
2
a 2 pi  b 2  125 ( 75 )  2002 
1
+
=


1 +
 = 171.2 MPa .................................. Ans.
b 2 − a 2  a 2  2002 − 1252  1252 
(b)
σt =
2
a 2 pi  b 2  125 ( 75 )  2002 
1
+
=


1 +
 = 96.2 MPa .................................... Ans.
b 2 − a 2  b 2  2002 − 1252  2002 
(c)
τ max =
σ max − σ min
2
=
σ t − ( − p ) 171.2 − ( −75 )
2
=
2
= 123.1 MPa ........................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-119
t = b − a = 0.1a
RILEY, STURGES AND MORRIS
b = 1.1a
From thick-walled cylinder equations:
 (1.1a )2 
a2 p  b2 
a2 p
σ t = 2 2 1 + 2  =
1 +
 = 10.524 p
b − a  a  (1.1a )2 − a 2 
a 2 
From thin-walled cylinder equations:
pr
pa
=
= 10 p
2t 0.1a
10.524 p − 10 p
Error =
(100 ) = 4.979 ≅ 5% ................................................................. Ans.
10.524 p
σh = σt =
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-120
σ max = σ t
σt =
at
RILEY, STURGES AND MORRIS
ρ =a
a 2 pi  b 2 
1002 pi  1502 
+
=
1


1 +
 ≤ 430 MPa
b 2 − a 2  a 2  1502 − 1002  1002 
pi ≤ 165.4 MPa .............................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-121*
σ max = σ t
σt =
at
RILEY, STURGES AND MORRIS
ρ =a
a 2 pi  b 2 
22 pi  42 
+
=
1


1 +  = 1.6667 pi
b 2 − a 2  a 2  42 − 22  22 
τ max =
σ max − σ min
=
σ t − ( − p ) 1.6667 pi − ( − pi )
=
≤ 24 ksi
2
2
2
pi ≤ 18.00 ksi .................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-122
(a)
σt =
(b)
At
RILEY, STURGES AND MORRIS
2
a 2 pi  b 2  100 ( 75 )  1502 
1
+
=


1 +
 = 195.0 MPa ................................... Ans.
b 2 − a 2  a 2  1502 − 1002  1002 
ρ = 125 mm
σt =
2
a 2 pi  b 2  100 ( 75 )  1502 
1
+
=


1 +
 = 146.4 MPa
b 2 − a 2  ρ 2  1502 − 1002  1252 
σr =
2
a 2 pi  b 2  100 ( 75 )  1502 
1
−
=


1 −
 = −26.4 MPa
b 2 − a 2  ρ 2  1502 − 1002  1252 
τ max =
σ max − σ min
2
=
σt −σr
2
=
146.4 − ( −26.4 )
= 86.4 MPa ..................................... Ans.
2
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-123*
(a)
σt =
2
a 2 pi  b 2  4 ( 25 )  7 2 
1
+
=


1 +  = 49.242 ksi
b 2 − a 2  a 2  7 2 − 42  42 
τ max =
(b)
RILEY, STURGES AND MORRIS
σ max − σ min
2
At ρ = 5.5 in.
=
σt − (− p)
2
=
49.242 − ( −25 )
= 37.1 ksi ..................................... Ans.
2
σt =
2
a 2 pi  b 2  4 ( 25 ) 
72 
1
1
+
=
+



 = 31.8 ksi ................................................. Ans.
b 2 − a 2  ρ 2  7 2 − 42  5.52 
σr =
2
a 2 pi  b 2  4 ( 25 ) 
72 
1
1
−
=
−



 = −7.51 ksi .............................................. Ans.
b 2 − a 2  ρ 2  7 2 − 42  5.52 
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-124*
RILEY, STURGES AND MORRIS
2 2
a 2 pi − b 2 po a b ( pi − po )
σt =
+
b2 − a 2
(b2 − a 2 ) ρ 2
252 ( 85 ) − 1252 ( 30 ) ( 25 ) (125 ) ( 85 − 30 ) 
35,807 
MPa
=
+
=  −27.708 +
2
2
2
2
2
ρ 2 
125 − 25
(125 − 25 ) ρ

2
2
75
F =∫
75
50


35,807 
35,807 
( −27.708 ) + ρ 2  d ρ = ( −27.708 ) ρ − ρ 



 50
F = −454 (103 ) N/m = −454 kN/m .......................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-125
(a)
(b)
RILEY, STURGES AND MORRIS
2
a 2 pi  b 2  8 ( 20 )  b 2 
1
+
=


1 +  ≤ 36 ksi
b 2 − a 2  a 2  b 2 − 8 2  82 
b ≥ 14.967 in. ≅ 14.97 in. ............................................................................................. Ans.
30, 000
E = 2 (1 + ν ) G
ν=
− 1 = 0.29310
2 (11, 600 )
σt =
δa =
=
a 2 pi
(1 −ν ) a 2 + (1 + ν ) b 2 
2
2
( b − a ) Ea
82 ( 20 )
(1 − 0.29310 )( 8 )2 + (1 + 0.29310 )(14.967 )2 

(14.967 − 8 ) ( 30, 000 )(8) 
2
δ a = 0.01162 in.
2
∆Di = 2δ a = 2 ( 0.01162 ) = 0.0223 in. .................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-126*
From Table B-18
RILEY, STURGES AND MORRIS
σ y = 250 MPa
3
6
P 100 (10 ) 250 (10 )
σ= =
≤
A
A
1.6
A ≥ 640 (10−6 ) m 2 = 640 mm 2
From Table B-14
d min = 51 mm ............................................................................................. Ans.
`
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-127*
From Table B-17
σ=
RILEY, STURGES AND MORRIS
σ y = 36 ksi
P 80 36
=
≤
A A 3
A ≥ 6.667 in.2
From Table B2, sections with A ≥ 6.667 in. include W6 × 25,
2
The lightest section is
W8 × 24, W10 × 30, W12 × 30
W8 × 24 ....................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-128
RILEY, STURGES AND MORRIS
W = 2000 ( 9.81) = 19, 620 N
→ ΣFx = 0 :
TB cos 30° − TA cos 50° = 0
TB sin 30° + TA sin 50° − 19, 620 = 0
↑ ΣFy = 0 :
TA = 17, 253.54 N (T)
From Table B-18
TB = 12,806.05 N (T)
σ y = 250 MPa
6
P 17, 253.54 250 (10 )
σA = =
≤
A
A
1.75
A ≥ 120.771(10−6 ) m 2
6
12,806.05 250 (10 )
σB =
≤
A
1.75
A ≥ 89.642 (10−6 ) m 2
A = π d 2 4 ≥ 120.771 mm 2
d ≥ 12.40 mm
d min = 13 mm ................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-129*
From Table B-17
σ AB =
σ y = 50 ksi
P
20
50
=
≤
2
A π d AB 4 1.5
d AB ≥ 0.874 in. = d min AB
σ BC =
RILEY, STURGES AND MORRIS
Ans.
P
30
50
=
≤
2
A π d BC 4 1.5
d BC ≥ 1.070 in. = d min BC
Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-130*
From Table B-18
RILEY, STURGES AND MORRIS
concrete
σ f = 34 MPa
E = 31 GPa
steel
σ y = 250 MPa
E = 200 GPa
AS = bS2 − bC2
AC = bC2
AC = 10 AS
δC = δ S
FC + FS = 1000 kN
FC L
FS L
=
9
AC ( 31× 10 ) AS ( 200 ×109 )
FS = 392,156.86 N
Try
P 607,843.14 34 (10
σC = =
≤
A
AC
1.4
Then
AS = 2.50288 (10−3 ) m 2
σS =
FC = 607,843.14 N
6
)
AC ≥ 25.0288 (10−3 ) m 2
392,156.86
= 156.7 (106 ) N/m 2 ≤ 250 MPa (correct guess)
−3
2.50288 (10 )
bC = 25.0288 (10−3 ) = 0.1582 m = 158.2 mm ..................................................... Ans.
AS = bS2 − bC2 = 2.50288 (10−3 )
bS = 25.0288 (10−3 ) + 2.50288 (10−3 ) = 0.1659 m = 165.9 mm ...................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-131
From Table B-17
σ y = 53 ksi
P 20 53
=
≤
A 1w 1.75
10
53
σ BC =
≤
1w 1.75
50 53
σ CD =
≤
1w 1.75
wmin = 1.651 in.
σ AB =
RILEY, STURGES AND MORRIS
w ≥ 0.66038 in.
w ≥ 0.33019 in.
w ≥ 1.65094 in.
Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
5-132
→ ΣFx = 0 :
↑ ΣFy = 0 :
RILEY, STURGES AND MORRIS
N cos 30° + V cos 60° − 85 = 0
N sin 30° − V sin 60° = 0
N = 73.6122 kN
V = 42.500 kN
6
N 73, 612.2 1035 (10 )
σ= =
≤
A 2 (π d 2 4 )
1.5
(There are two bolts)
d ≥ 8.24 (10−3 ) m
620 (10
V
42,500
τ= =
≤
2
A 2 (π d 4 )
1.5
6
)
d ≥ 8.09 (10−3 ) m
d min = 8.24 mm ............................................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-133*
σ AB =
σ BC =
σ CD =
P
20
=
= 4.07 ksi (C) .......................... Ans.
A π ( 2.50 )2 4
60
π (1.50 ) 4
2
10
π (1) 4
δ total = ∑
2
= 34.0 ksi (T) ................................... Ans.
= 12.73 ksi (T) ........................................ Ans.
( −20 )(15 )
PL
=
AE π ( 2.50 )2 4  ( 30, 000 )


+
( 60 )(15)
(10 )(15 )
+
2
π (1.50 ) 4  ( 30, 000 ) π (1)2 4  ( 30, 000 )




δ = +0.0213 in. ...................................................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-134*
σB =
P
P
=
≤ 200 (106 )
2
A π ( 0.050 ) 4
P ≤ 392.7 (103 ) N
σS =
P
P
=
≤ 500 (106 )
2
A π ( 0.032 ) 4
P ≤ 402.1(103 ) N
δ total = δ B + δ S
P (1500 )
π ( 0.050 ) 4  (100 × 109 )


2
+
P (1000 )
π ( 0.032 )2 4  (190 × 109 )


≤ 5.60 mm
P ≤ 394.8 (103 ) N
Pmax = 393 kN .......................................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-135
→ ΣFx = 0 :
FBC cos 30° + TAB cos 60° − 3000 = 0
FBC sin 30° − TAB sin 60° = 0
↑ ΣFy = 0 :
TAB = 1500 lb
FBC = 2598.076 lb
→ ΣFx = 0 :
FBC cos 30° − TAC = 0
TAC = 2250 lb
σ AC =
N
2.250
=
= 11.46 ksi .........................................Ans.
A π ( 0.5 )2 4
δ AC =
( 2.250 )( 30 )
PL
=
= +0.01146 in. .................................................... Ans.
AE π ( 0.5 )2 4  ( 30, 000 )


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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-136*
TBD − FCE − 50 = 0
↑ ΣFy = 0 :
900 ( 50 ) − 300TBD = 0
4 ΣM C = 0 :
TBD = 150 kN (T)
FCE = 100 kN (C)
σ BD
150 (103 )
N
= =
= 120.0 (106 ) N/m 2 = 120.0 MPa (T) ................................. Ans.
−6
A 1250 (10 )
σ CE =
100 (103 )
750 (10
= 133.3 (106 ) N/m 2 = 133.3 MPa (C) ............................................. Ans.
)
(150 ×10 ) ( 600 ) = 0.98630 mm
PL
=
=
AE 1250 × 10  ( 73 ×10 )
(100 ×10 ) ( 400 ) = 0.26667 mm
PL
=
=
AE 750 ×10  ( 200 ×10 )
−6
3
δ BD
−6
9
3
δ CE
δ CE + δ BD
300
−6
=
9
δ CE + a
900
v A = a = 3.49 mm ↓ ............................................................................................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
5-137
4 ΣM C = 0 :
RILEY, STURGES AND MORRIS
5P − 10TA − 6 FB = 0
10TA + 6 FB = 5 P
(a)
δ A = (10 6 )(δ B + 0.009 ) = (10δ B 6 ) + 0.015 in.
TA ( 50 )
(1.24 )( 30, 000 )
If
=
10  FB (15 ) 

 + 0.015
6  ( 4 )(15, 000 ) 
(b)
σ A = σ max = 30 ksi , then
TA = 30 (1.24 ) = 37.20 kip
FB = 84.00 kip
P = 175.2 kip
84.00
= 21 ksi > 20 ksi (wrong guess)
4
= 20 ksi , then
σB =
If
σ B = σ max
FB = 20 ( 4 ) = 80.00 kip
TA = 35.9600 kip
P = 167.9 kip
35.9600
= 29.00 ksi < 230ksi (correct guess)
1.24
= 167.9 kip
σA =
Pmax
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MECHANICS OF MATERIALS, 6th Edition
5-138
4 ΣM F = 0 :
RILEY, STURGES AND MORRIS
300 P − 50 FD − 100TC = 0
FD + 2TC = 6 P
(a)
δ C = 2 (δ D + 0.09 ) mm


FD (150 )


.09
=
+
2
0

( 625 ×10−6 )( 73 ×109 )  ( 2500 ×10−6 )(12 ×109 )
TC ( 300 )
TC = (1.52083FD + 2737.50 ) N
(b)
Guess that
TC = TC max = σ A
= (100 × 106 )( 625 × 10−6 ) = 62,500 N
Then
FD = 39, 295.98 N
and
σD =
Since
P = 27,383 N
N 39, 295.98
=
= 15.72 (106 ) N/m 2 = 15.72 MPa
−6
A 2500 (10 )
σ D = 15.72 MPa < σ max = 30 MPa , the guess was correct and
Pmax = 27.4 kN ....................................................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-139*
TA + TB − 8 − 16 = 0
↑ ΣFy = 0 :
8 ( 8 ) − 4TA − 2 (16 ) = 0
4 ΣM B = 0 :
TA = 8.00 kip (T)
TB = 16.00 kip (T)
δ A = δ B + 4 tan θ = δ B + 4 ( 5 10, 000 ) = (δ B + 0.00200 ) in.
σ A (10 )
=
σ B (16 )
(10, 600 ) (15, 000 )
+ 0.00200
σ A = 1.13067σ B + 2.12000
Guess that
σ B = σ B max = 15 ksi
Then
Since
(a)
σ A = 19.0802 ksi
σ A < 20 ksi = σ A max , the guess was correct and
σA =
N 8
= ≤ 19.0802 ksi
A A
AA ≥ 0.419 in.2 = AA min ........................................................................................................ Ans.
σB =
(b)
N 16
= ≤ 15 ksi
A A
AB ≥ 1.067 in.2 = AB min ............................................... Ans.
(19.0802 )(10 ) = 0.01800 in. ........................................................................ Ans.
E
(10, 600 )
(15)(16 ) = 0.01600 in. .............................................................................................. Ans.
δB =
(15, 000 )
δA =
σL
=
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-140
RILEY, STURGES AND MORRIS
TB = TS
(a)
δB + δS = 0
TB ( 800 )
π ( 0.090 ) 4  (100 × 10


2
+
9
)
+ (17.6 × 10−6 ) ( −70 )( 800 )
TS ( 480 )
π ( 0.050 ) 4  ( 200 ×10


2
TB + 0.97200TS = 1.101724 (106 ) N
9
)
+ (11.9 × 10−6 ) ( −70 )( 480 ) = 0
(b)
TB = TS = 558, 683 kN (both T)
σB =
σS =
N
558, 683
=
= 87.8 (106 ) N/m 2 = 87.8 MPa (T) .................................... Ans.
2
A π ( 0.090 ) 4
558, 683
π ( 0.050 ) 4
2
= 285 (106 ) N/m 2 = 285 MPa (T) ............................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-141
RILEY, STURGES AND MORRIS
sin θCD = 4 5
cos θCD = 3 5
sin θ AD = 12 13
cos θ AD = 5 13
AAD = ABD = ACD = ( 50 × 25 ) = 1250 mm 2
(12 13) TAD + TBD + ( 4 5) TCD = 650 kN
(12 13) σ AD + σ BD + ( 4 5 ) σ CD  (1250 × 10−6 ) = 650 kN
(12 13) σ AD + σ BD + ( 4 5) σ CD = 520 (106 ) N/m 2 = 520 MPa
δ AD = δ BD sin θ AD = (12 13) δ BD
(b)
δ CD = δ BD sin θCD = ( 4 5 ) δ BD
(c)
(a)
Assume all bars elastic:
 12   σ L 
=    BD 
(180 )  13   ( 40 ) 
σ AD = 4.15385σ BD
(b)
 4 σ L 
=    BD 
A ( 200 )  5   A ( 40 ) 
σ CD = 4.00000σ BD
(c)
σ AD L
σ CD L
σ AD = 268.8468 MPa < 400 MPa (elastic)
σ BD = 64.7223 MPa < 100 MPa (elastic)
σ CD = 258.8893 MPa > 240 MPa (plastic)
Therefore
(a)
σ CD = σ y = 240 MPa (T)
.................................................................................................... Ans.
and from Eqs. (a) and (b)
σ BD = 67.8482 MPa (T) ≅ 67.8 MPa (T) ....................................................................... Ans.
σ AD = 282 MPa (T) .............................................................................................................. Ans.
(b)
vD = δ BD =
σL
E
( 67.8482 ×10 ) ( 4000 ) = 6.78 mm ↓ ............................................ Ans.
=
( 40 ×10 )
6
9
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MECHANICS OF MATERIALS, 6th Edition
5-142*
4 ΣM F = 0 :
RILEY, STURGES AND MORRIS
300 (100 ) − 50 FD − 100TC = 0
FD + 2TC = 600 kN
σ D ( 2000 ×10−6 ) + 2σ C ( 600 ×10−6 ) = 600 kN
5σ D + 3σ C = 1500 (106 ) N/m 2 = 1500 MPa
(a)
δ C = 2 (δ D + 0.09 ) mm
(b)
Assume both bars elastic. Then
 σ (150 )

D


2
0
.09
=
+

( 200 ×109 )  (100 ×109 )
σ C ( 300 )
σ C = 2σ D + 120 (106 ) N/m 2 = 2σ D + 120 MPa
(b)
σ C = 327.273 MPa > 240 MPa (plastic)
σ D = 103.636 MPa < 410 MPa (elastic)
Therefore
(a)
σ C = σ y = 240.0 MPa .......................................................................................................... Ans.
and from Eq. (a)
(b)
σ D = 156.0 MPa
.............................................................................................. Ans.
 (156.0 ×106 ) (150 )

 = 1.944 mm ↑ .................. Ans.
+
0.09
v A = a = 6 (δ D + 0.09 ) = 6 


(100 ×109 )
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-143*
θ = tan −1
4
= 26.565°
8
p = γ y = ( 62.4 )( 8 ) = 499.20 psf = 3.46667 psi
W = γV =
γπ r 2 h
3
( 62.4 ) π ( 4 ) ( 8) = 8364.176 lb
=
2
3
↑ ΣFy = 0 : pAp − W − σ m Aσ cos 26.565° = 0
( 499.20 ) π ( 4 )
2
− 8364.176 − σ m 2π ( 48)(1 8 ) cos 26.565° = 0
σ m = 496.109 psi ≅ 496 psi (T) .......................................Ans.
σm
rm
+
σt =
σt
rt
=
p
t
rm = ∞
rt = r cos θ = 53.6656 in.
prt ( 3.46667 )( 53.6656 )
=
= 1488 psi (T) ........................................................... Ans.
t
18
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-144
σa =
pr
=σx
2t
σ a νσ h
εa =
E
−
E
σh =
=
pr
=σy
t
pr ν pr
−
2tE tE
( 300 ×10 ) ( 20 ) ( 200 ×10
ε a 2tE
=
r (1 − 2ν )
(1000 − 10 )(1 − 0.6 )
−6
(a)
p=
τ xy = 0
9
) = 3.030303 10 N/m
( )
6
2
p = 3.030303 MPa ≅ 3.03 MPa ........................................................................................ Ans.
(b)
(c)
σa =
pr ( 3.030303)(1000 − 10 )
=
= 150.0 MPa ......................................................... Ans.
2t
2 (10 )
σh =
pr ( 3.030303)(1000 − 10 )
=
= 300.0 MPa ......................................................... Ans.
t
(10 )
σ z = − p = −3.030303 MPa
τ max =
(d)
εh =
σ max − σ min
2
σ h −νσ a
E
=
( 300 ) − ( −3.030303) = 151.5 MPa ............................................ Ans.
2
( 300 ×10 ) − 0.3 (150 ×10 ) = 1275 10
=
( )
200 (10 )
6
6
−6
9
ε h = 1275 µ m/m .................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
5-145
σt =
a 2 pi  b 2 
1.52 pi  3.52 
+
=
1


1 +
 = 50 ksi
b 2 − a 2  a 2  3.52 − 1.52  1.52 
pi = 34.5 ksi .................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
5-146*
RILEY, STURGES AND MORRIS
(a)
σt =
2
a 2 pi  b 2  100 (125 )  2252 
1
+
=


1 +
 = 186.5 MPa .................................. Ans.
b 2 − a 2  a 2  2252 − 1002  1002 
(b)
δa =
a 2 pi
(1 −ν ) a 2 + (1 + ν ) b 2 
2
2
( b − a ) Ea
δa =
1002 (125 )
(1 − 0.30 )(100 )2 + (1 + 0.30 )( 225 )2 

( 225 − 100 ) ( 210, 000 )(100 ) 
2
2
= 0.106685 mm
∆Di = 2δ a = 2 ( 0.106685 ) = 0.213 mm ........................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-1*
(a) Free-body diagrams for parts of the shaft to the left of sections
in the intervals AB, BC, CD, and DE of the shaft are shown.
From the free-body diagrams:
4 ΣM = 0 :
TAB − 80 = 0
TAB = +80 kip ⋅ ft .................................................Ans.
4 ΣM = 0 :
TBC − 80 + 100 = 0
TBC = −20 kip ⋅ ft .................................................Ans.
4 ΣM = 0 :
TCD − 80 + 100 − 40 = 0
TCD = +20 kip ⋅ ft .................................................Ans.
4 ΣM = 0 :
TDE − 80 + 100 − 40 − 25 = 0
TDE = +45 kip ⋅ ft .................................................Ans.
(b)
A torque diagram for the shaft is shown below the
free-body diagrams.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-2*
Free-body diagrams for parts of the shaft to the left of sections
in the intervals AB, BC, CD, and DE of the shaft are shown.
From the free-body diagrams:
4 ΣM = 0 :
TAB − 30 = 0
TAB = +30 kN ⋅ m = Tmax .....................................Ans.
(b)
4 ΣM = 0 :
TBC − 30 + 40 = 0
TBC = −10 kN ⋅ m ................................................Ans.
4 ΣM = 0 :
TCD − 30 + 40 − 15 = 0
TCD = +5 kN ⋅ m ...................................................Ans.
4 ΣM = 0 :
TDE − 30 + 40 − 15 − 5 = 0
TDE = +10 kN ⋅ m ................................................Ans.
(a)
A torque diagram for the shaft is shown below the
free-body diagrams.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-3
(a) Free-body diagrams for parts of the shaft to the left of sections
in the intervals AB, BC, CD, and DE of the shaft are shown.
From the free-body diagrams:
4 ΣM = 0 :
TAB − 10 = 0
TAB = +10 kip ⋅ ft .................................................Ans.
4 ΣM = 0 :
TBC − 10 − 15 = 0
TBC = +25 kip ⋅ ft = Tmax .....................................Ans.
4 ΣM = 0 :
TCD − 10 − 15 + 30 = 0
TCD = −5 kip ⋅ ft ...................................................Ans.
4 ΣM = 0 :
TDE − 10 − 15 + 30 + 15 = 0
TDE = −20 kip ⋅ ft .................................................Ans.
A torque diagram for the shaft is shown below the
free-body diagrams.
(b)
Tmax = TBC = +25 kip ⋅ ft ............................................Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-4*
(a) Free-body diagrams for parts of the shaft to the left of sections
in the intervals AB, BC, and CD of the shaft are shown.
From the free-body diagrams:
4 ΣM = 0 :
TAB − 500 = 0
TAB = +500 N ⋅ m ................................................Ans.
4 ΣM = 0 :
TBC − 500 + 100 = 0
TBC = +400 N ⋅ m ................................................Ans.
4 ΣM = 0 :
TCD − 500 + 100 + 150 = 0
TCD = +250 N ⋅ m ................................................Ans.
(b)
A torque diagram for the shaft is shown below the
free-body diagrams.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-5
J = π d 4 32 = π ( 2 ) 32 = 1.57080 in.4
4
(a)
τ=
Tc (18 )(1)
=
= 11.46 ksi ........................................................................................... Ans.
J (1.5708 )
(b)
θ=
(18)( 6 ×12 ) = 0.0688 rad ..................................................................... Ans.
TL
=
JG (1.5708 )(12, 000 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-6
RILEY, STURGES AND MORRIS
J = π d 4 32 = π (1204 − 804 ) 32 = 16.33628 (106 ) mm 4
Tc ( 28, 000 )( 0.060 )
=
= 102.8 (106 ) N/m 2 = 102.8 MPa ................................ Ans.
J (16.33628 ×10−6 )
(a)
τo =
(b)
τi =
(c)
θ=
(d)
A = π (1202 − 802 ) 4 = π r 2
( 28, 000 )( 0.040 ) = 68.6
(16.33628 ×10 )
−6
(10 ) N/m
6
2
= 68.6 MPa .............................................. Ans.
( 28, 000 )( 2 )
TL
=
= 0.0429 rad ................................................... Ans.
JG (16.33628 × 10−6 )( 80 × 109 )
r = 44.72136 mm
J = π r 4 2 = π ( 44.72136 ) 2 = 6.28319 (106 ) mm 4
4
θ=
( 28, 000 )( 2 )
TL
=
= 0.1114 rad ..................................................... Ans.
JG ( 6.28319 ×10−6 )( 80 ×109 )
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MECHANICS OF MATERIALS, 6th Edition
6-7*
RILEY, STURGES AND MORRIS
τ=
Tc ( 2.2 × 12 )( d 2 )
=
≤ 14.5 ksi
J
(π d 4 32 )
d ≥ 2.10 in.
θ=
TL ( 2.2 × 12 )( 6.5 × 12 ) 5π
=
≤
rad
JG
(π d 4 32 ) ( 4000 ) 180
d ≥ 2.78 in.
d min = 2.78 in. ........................................................................................................................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
6-8
RILEY, STURGES AND MORRIS
J AB = π d 4 32 = π ( 80 ) 32 = 4.02124 (106 ) mm 4
4
J AC = π d 4 32 = π ( 65 ) 32 = 1.75248 (106 ) mm 4
4
(a)
τ AB =
Tc ( 6000 )( 0.040 )
=
= 59.7 (106 ) N/m 2 = 59.7 MPa ................................... Ans.
J ( 4.02124 × 10−6 )
τ AC =
Tc ( 4000 )( 0.0325 )
=
= 74.2 (106 ) N/m 2 = 74.2 MPa ................................... Ans.
−6
J (1.75248 × 10 )
 TL 
( 6000 )( 2.25 )
(b)
θB/ A =   =
= 0.04197 rad ≅ 0.0420 rad
−6
9
 JG  AB ( 4.02124 × 10 )( 80 × 10 )
(c)
θC / A = 
........... Ans.
( 4000 )(1.60 )
 TL 
=
= 0.04565 rad
−6
9
 JG  AC (1.75248 × 10 )( 80 × 10 )
θC / B = θC / A + θ B / A = 0.04565 − 0.04197 = 0.00368 rad ............................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-10*
RILEY, STURGES AND MORRIS
J = π d 4 32 = π (100 ) 32 = 9.81748 (106 ) mm 4
4
(a)
Tmax = 14 kN ⋅ m
τ max =
Tc (14, 000 )( 0.050 )
=
= 71.3 (106 ) N/m 2
−6
J
( 9.81748 ×10 )
τ max = 71.3 MPa
(b)
θD/ B =
(c)
θE / A =
.................................................Ans.
TL ( −7000 )(1.50 ) + ( +8000 )(1.50 )
=
= +0.001910 rad ............................. Ans.
JG
( 9.81748 ×10−6 )(80 ×109 )
(8000 − 7000 + 8000 + 14, 000 )(1.50 ) = +0.0439 rad .................................... Ans.
( 9.81748 ×10 )(80 ×10 )
−6
9
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MECHANICS OF MATERIALS, 6th Edition
6-11
(a)
(b)
RILEY, STURGES AND MORRIS
J = π d 4 32 = π ( 44 − 24 ) 32 = 23.56194 in.4
Tmax = TBC = 11 kip ⋅ ft
τ max =
Tc (11× 12 )( 2 )
=
= 11.20 ksi ......................... Ans.
J ( 23.56194 )
θD/ B =
TL ( −11×12 )( 5 ×12 ) + ( +10 × 12 )( 4 × 12 )
=
JG
( 23.56194 )(12, 000 )
θ D / B = −0.00764 rad ............................................................................................................. Ans.
(c)
θD/ A =
( +9 ×12 )( 3 ×12 ) + −0.00764 = 0.00611 rad .......................................... Ans.
(
)
( 23.56194 )(12, 000 )
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MECHANICS OF MATERIALS, 6th Edition
6-12*
(a)
RILEY, STURGES AND MORRIS
Tmax = TAB = 5.5 kN ⋅ m
τ max =
Tc ( 5500 )( d 2 )
=
≤ 100 (106 ) N/m 2
4
J
(π d 32 )
d ≥ 0.0654 m = 65.4 mm .......................................... Ans.
(b)
θD/ A =
( 5500 )( 2 ) + ( 2500 )( 2 ) + (1000 )( 2 )
π ( 0.075 )4 32  ( 80 ×109 )


θ D / A = 0.0724 rad .................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-13
RILEY, STURGES AND MORRIS
T ( 0.875 )
Tc
T ≤ 8.42 kip ⋅ in.
=
≤ 8 ksi
J π (1.75 )4 32 


T ( 3 × 12 )
T ( 4 × 12 )
TL
θ=
=
+
≤ 0.04 rad
JG π ( 2.5 )4 32  ( 4000 ) π (1.75 )4 32  ( 4000 )




T ≤ 2.60 kip ⋅ in.
Tmax = 2.60 kip ⋅ in. .................................................................................................. Ans.
τ=
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-14
RILEY, STURGES AND MORRIS
J AB = J BC = π (160 ) 32 = 64.33982 (106 ) mm 4
4
J CD = π (100 ) 32 = 9.81749 (106 ) mm 4
4
J DE = π ( 50 ) 32 = 0.61359 (106 ) mm 4
4
(a)
τ3 =
Tc ( 25, 000 )( 0.050 )
=
= 127.3 (106 ) N/m 2
−6
J
( 9.81749 ×10 )
τ 3 = 127.3 MPa ...................................................................................................................... Ans.
TL ( 40, 000 )(1) + ( 75, 000 )(1)
=
= 0.0223 rad ................................................. Ans.
JG ( 64.33982 × 10−6 )( 80 × 109 )
(b)
θ2 =
(c)
θE / A =
( 40, 000 )(1) + ( 75, 000 )(1.5) +
( −25, 000 )( 2 )
( 64.33982 ×10 )(80 ×10 ) ( 9.81748 ×10 )(80 ×10 )
−6
+
θ E / A = −0.0850 rad
−6
9
9
( −5000 )( 0.5)
( 0.61359 ×10 )(80 ×10 )
−6
9
= −0.0850 rad
............................................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
6-15*
(a)
TAB = T
RILEY, STURGES AND MORRIS
TCD = 2T
τ=
T ( 0.75 )
Tc
=
≤ 15 ksi
J π (1.5 )4 32
T ≤ 9.94020 kip ⋅ in.
τ=
( 2T )(1) ≤ 15 ksi
Tc
=
J π ( 2 )4 32
T ≤ 11.78 kip ⋅ in.
Tmax = 9.94 kip ⋅ in. ................................................................................................................. Ans.
θCD =
TL ( 2 × 9.94020 )( 3 × 12 )
=
= 0.03797 rad
JG π ( 2 )4 32  (12, 000 )


θ A = 2 ( 0.03797 ) +
( 9.94020 )( 4 ×12 ) = 0.1559 rad .......................................... Ans.
π (1.5 )4 32  (12, 000 )


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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-16
(a)
RILEY, STURGES AND MORRIS
τ = Tc J = ( 2000 )( 0.040 ) J ≤ 50 (106 ) N/m 2
J ≥ 1.600 (10−6 ) m 4 = 1.600 (106 ) mm 4
J=
(b)
θ=
π ( 804 − di4 )
32
≥ 1.600 (106 ) mm 4
di ≤ 70.5 mm .................................... Ans.
( 2000 )( 3.5)
( 2000 )( 2.5) ≤ 0.25 rad
TL
=
+
4
JG π ( 0.050 ) 32  ( 80 × 109 ) J ( 28 × 109 )


J ≥ 1.66272 (10−6 ) m 4 = 1.66272 (106 ) mm 4
J=
π ( 804 − di4 )
32
≥ 1.66272 (106 ) mm 4
di ≤ 70.0 mm .................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-17*
Tc ( 30 ×12 )( d 2 )
=
≤ 12, 000 psi
J
(π d 4 32 )
(a)
τ=
(b)
TCD = ( 2 5 )( 30 ) = 12 lb ⋅ ft
(c)
τ=
Tc (12 × 12 )( d 2 )
=
≤ 12, 000 psi
J
(π d 4 32 )
θ=
(12 ×12 ) L
≤ 0.5 rad
π ( 0.394 )4 32  ( 3.8 × 106 )


RILEY, STURGES AND MORRIS
d AB ≥ 0.535 in. .................................. Ans.
dCD ≥ 0.394 in. ................................. Ans.
L ≤ 31.2 in. ............................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-18*
Tc ( 45, 000 )( 0.075 )
=
= 67.9 (106 ) N/m 2 = 67.9 MPa ................................. Ans.
4
J
π ( 0.075 ) 2
(a)
τ max =
(b)
TC = (150 450 )( 45 ) = 15 kN ⋅ m
τ max = τ CE =
TD = 15 − 8 = 7 kN ⋅ m
(8000 )( 0.040 ) = 79.6 106 N/m 2
( )
4
π ( 0.040 ) 2
τ max = 79.6 MPa
(c)
RILEY, STURGES AND MORRIS
.................................................................. Ans.
TL ( 8000 )( 2.5 ) + ( −7000 )(1.5 )
=
JG
π ( 0.040 )4 2  ( 80 ×109 )


θ = 0.0295 rad ..................................................................... Ans.
θ=
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-19
(a)
τ AB =
Tc ( 3.6 × 12 )( D1 2 )
=
= 18 ksi
J
π D14 32
D1 = 2.30 in. .................................................................. Ans.
τ BC =
(8.8 ×12 )( D2 2 ) = 18 ksi
π D24 32
D2 = 3.10 in. .................................................................................................................... Ans.
(b)
θ=
TL ( 3.6 ×12 )( 5 ×12 ) + ( 8.8 × 12 )( 4 × 12 )
=
= 0.15 rad
JG
(π D 4 32 ) (11, 600 )
D = 2.59 in. ..................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-20*
With the left end of the shaft at x = 0
and the right end of the shaft at x = L
ρ = r + mx
J = πρ 4 2 = π ( r + mx ) 2
4
dθ =
T dx
T dx
2T dx
=
=
4
JG π ( r + mx ) 2  G π G ( r + mx )4


θ = ∫ dθ = ∫
L
0
θ=
2T
3π Gmr 3
2T dx
π G ( r + mx )
4
 ( r + mL ) − r

3
 ( r + mL )
3

2T L
dx
2T 
−1
=
=


4
3
π G ∫0 ( r + mx ) π G  3m ( r + mx ) 
3
L
0

 ........................................................................................... Ans.

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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-21
With the left end of the shaft at x = 0
and the right end of the shaft at x = L
dθ =
T = qx
θ = ∫ dθ = ∫
L
0
2qx dx
2q
=
4
π Gc
π Gc 4
∫
L
0
( qx ) dx = 2qx dx
T dx
=
JG (π c 4 2 ) G π Gc 4
x dx =
qL2
................................................................ Ans.
π Gc 4
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-22
With the left end of the shaft at x = L
and the right end of the shaft at x = 2 L
ρ = rx L
J=
π
(ρ
2
dθ =
4
π
2 L4
(r
4
x 4 − R 4 L4 )
T dx 2TL4 
dx

=
 4 4
4 4 
JG
πG  r x − R L 
θ = ∫ dθ =
θ=
− R4 ) =
2TL4
πG
∫
2L
L
dx


 4 4
4 4 
r x −R L 
TL   2r − R  r + R 
−1  2r 
−1  r  
ln 

 − 2 tan   + 2 tan    .......................... Ans.
3 
2π GR r   r − R  2r + R 
 R
 R 
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-23*
With the left end of the shaft at x = L
and the right end of the shaft at x = 2 L
ρ = rx L
2π r 3tx3
 rx 
J = ρ A = ρ ( 2πρ t ) = 2πρ t = 2π   t =
L3
L
3
2
dθ =
2
T dx
TL3  dx 
=
 
JG 2π r 3tG  x 3 
3
θ = ∫ dθ =
2 L dx
TL3
3TL
=
....... Ans.
3
3
∫
L
2π r tG
x 16Gtπ r 3
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-24
With the left end of the shaft at x = 0
and the right end of the shaft at x = L
T dx ( qx 2 L ) dx qx 2 dx
=
=
dθ =
JG
(π c4 2 ) G π GLc 4
2
qx 2
T=
2L
q
θ = ∫ dθ =
π GLc 4
RILEY, STURGES AND MORRIS
∫
L
0
qL2
................................................................................ Ans.
x dx =
3π Gc 4
2
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-25
A = π ( ro2 − ri 2 ) = 3 in.2
RILEY, STURGES AND MORRIS
ri 2 = ro2 − ( 3 π )
J = π ( ro4 − ri 4 ) 2 = π ro4 2 − π  ro2 − ( 3 π ) 
2
2 = 3ro2 − ( 4.5 π )
(a)
θ=
 0.246758 
( 3 ×12 )( 3 ×12 )
TL
=
=
 rad
2
JG 3ro − ( 4.5 π )  (11, 000 )  2π ro2 − 3 
(b)
τc =
( 3 ×12 )( ro ) =  75.3982ro  ksi
Tc
=


J 3ro2 − ( 4.5 π )   2π ro − 3 
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-26
RILEY, STURGES AND MORRIS
4
J = π ( ro4 − ri 4 ) 2 = π ro4 1 − ( ri ro )  2


4
4
4
= π ( 50 ) 1 − ( ri ro )  2 = ( 9.81748 × 106 ) 1 − ( ri ro )  mm 4




θ=
( 7500 )( 2 )
0.0381972
TL
rad
=
=
4
6
9
−
JG ( 9.81748 × 10 ) 1 − ( ri ro )  ( 40 ×10 ) 1 − ( ri ro )4 




τc =
( 7500 )( 0.050 )
38.1972 ×106
Tc
N/m 2
=
=
J ( 9.81748 ×10−6 ) 1 − ( ri ro )4  1 − ( ri ro )4 

 

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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-27
RILEY, STURGES AND MORRIS
ri = ( ro − 0.25 ) in.
4
J = π ( ro4 − ri 4 ) 2 = π ro4 2 − π ( ro − 0.25 ) 2  in.4


(a)
θ=
 0.0750054 
( 3 ×12 )( 3 ×12 )
TL
=
=

 rad
JG π ro4 2 − π ( ro − 0.25 )4 2  (11, 000 )  ro4 − ( ro − 0.25 )4 


(b)
τc =
 22.91831r 
( 3 ×12 )( ro )
Tc
o
=
=

 ksi
J π ro4 2 − π ( ro − 0.25 )4 2   ro4 − ( ro − 0.25 )4 


(c)
As the outside radius increases, the shear stress decreases. Between ro = 0.5 in. and ro = 1.5 in. the
decrease in shear stress is very dramatic. Between ro = 1.5 in. and ro = 2.0 in. the decrease in shear stress
is much less dramatic. And beyond ro = 2.0 in. the decrease in shear stress is very slight. Therefore, a
reasonable maximum value for the outside radius would be around ro = 1.5 in.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-28
θD/ A =
TLAB T (1.5 − LAB ) T ( 0.75 )
+
+
J AB G
J BC G
J CD G
RILEY, STURGES AND MORRIS
TAB = TBC = TCD = T = 2500 N ⋅ m
GAB = GBC = GCD = G = 28 (109 ) N/m 2
J BC = π d 4 32 = π (100 ) 32 = 9.81748 (106 ) mm 4
4
J CD = π ( 75 ) 32 = 3.10631(106 ) mm 4
4
(a)
For d = 75 mm
J AB = π (1004 − 754 ) 32
= 6.71117 (106 ) mm 4
θ D / A = ( 4.20948LAB + 28.0135 ) (10−3 ) rad
(b)
For d = 90 mm
J AB = π (1004 − 904 ) 32
= 3.37623 (106 ) mm 4
θ D / A = (17.3508LAB + 28.0135 ) (10−3 ) rad
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-29*
RILEY, STURGES AND MORRIS
(a)
σ max = τ c =
Tc (15 )(1.5 )
=
= 2.83 ksi (T) .................................................................... Ans.
J π (1.5 )4 2
(b)
σ max = τ c =
Tc (15 )(1)
=
= 9.55 ksi (C) ........................................................................ Ans.
J π (1)4 2
(c)
θ=
( −15)( 3 ×12 ) + ( −15)( 4 ×12 ) = −0.0439 rad ................. Ans.
TL
=
JG π (1.5 )4 2  (12, 000 ) π (1)4 2  (12, 000 )




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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-30*
RILEY, STURGES AND MORRIS
σ max = τ max = Tc J ≤ 75 MPa ≤ 80 MPa
Tmax ( 0.045 )
π ( 0.090 − 0.050
4
4
)
32
= 75 (106 ) N/m 2
Tmax = 9710 N ⋅ m = 9.71 kN ⋅ m ......................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-31
σ x = σ y = 0 ksi
τ xy =
(a)
RILEY, STURGES AND MORRIS
θ = −55°
− (1.00 )( 0.75 )
Tc
=
= −4.38797 ksi
J π (1.54 − 1.354 ) 32
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
σ n = 0 + 0 + 2 ( −4.38797 ) sin ( −55° ) cos ( −55° ) = 4.12 ksi (T) ................................ Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ )
τ nt = 0 + ( −4.38797 ) cos 2 ( −55° ) − sin 2 ( −55° )  = +1.501 ksi
(b)
σ max T = τ xy = 4.39 ksi (T)
................................ Ans.
................................................................................................... Ans.
σ max C = τ xy = 4.39 ksi (C) ................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-32
RILEY, STURGES AND MORRIS
J 60 = π (1204 − 604 ) 32 = 19.08517 (106 ) mm 4
J100 = π (1204 − 1004 ) 32 = 10.54004 (106 ) mm 4
Tc ( 7500 )( 0.060 )
=
= 23.6 (106 ) N/m 2 = 23.6 MPa ........................... Ans.
−6
J 19.08517 (10 )
(a)
σ max = τ =
(b)
σ max =
(c)
θ 60 =
( −7500 )( 2 )
TL
=
= −0.00982 rad ........................................... Ans.
JG (19.08517 ×10−6 )( 80 ×109 )
θ100 =
( −7500 )( 2 )
TL
=
= −0.01779 rad .......................................... Ans.
JG (10.54004 ×10−6 )( 80 ×109 )
( 7500 )( 0.060 )
10.54004 (10
−6
)
= 42.7 (106 ) N/m 2 = 42.7 MPa ............................................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
6-33*
RILEY, STURGES AND MORRIS
(a)
TBC = ( 4 12 )(1500 ) = 500 lb ⋅ ft ........................................................................................ Ans.
(b)
TCD = 500 − 250 = 250 lb ⋅ ft ............................................................................................... Ans.
(c)
τ max ( motor ) =
Tc (1500 × 12 )(1)
=
= 11, 459 psi ≅ 11.46 ksi
4
J
π (1) 2
σ max ( motor ) = τ max ( motor ) = 11.46 ksi (T&C) ........................................................... Ans.
( 500 ×12 )( 0.625) = 15, 646 psi ≅ 15.65 ksi
4
π (1.25 ) 32
σ max ( power ) = τ max ( power ) = 15.65 ksi (T&C) ......................................................... Ans.
τ max ( power ) =
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MECHANICS OF MATERIALS, 6th Edition
6-34*
τ AB =
RILEY, STURGES AND MORRIS
Tc ( 4T )( 0.080 )
=
≤ 150 (106 ) N/m 2
4
J π ( 0.080 ) 2
T ≤ 30.2 (103 ) N ⋅ m
σ AB = τ AB = 150 (106 ) N/m 2 ≤ 260 (106 ) N/m 2
σ BC = τ BC =
T ( 0.050 )
π ( 0.050 ) 2
4
≤ 75 (106 ) N/m 2 ≤ 125 (106 ) N/m 2
T ≤ 14.73 (103 ) N ⋅ m
θ=
( 4T )( 2 )
(T )(1.5 )
TL
2.5π
rad
=
+
≤
4
4
9
9
JG π ( 0.080 ) 2  ( 45 ×10 ) π ( 0.050 ) 2  ( 76 × 10 ) 180




T ≤ 9140 N ⋅ m
Tmax = 9.14 kN ⋅ m ................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-35
4 ΣM = 0 :
RILEY, STURGES AND MORRIS
−9 + Q − 21 + 10 = 0
Q = 20 kip ⋅ ft
Tmax = TBC = 11 kip ⋅ ft
(a)
τ BC =
Tc (11× 12 )( 2 )
=
= 11.20 ksi
J π ( 24 − 14 ) 2
σ max = τ BC = 11.20 ksi (C) ......................................... Ans.
(b)
τ BC =
Tc (11×12 )(1)
=
= 5.60 ksi
J π ( 24 − 14 ) 2
σ max = τ BC = 5.60 ksi (C)
(c)
θ=
........................................... Ans.
TL ( 9 × 3 − 11× 5 + 10 × 4 )(12 × 12 )
=
= +0.00611 rad ......................................... Ans.
JG
π ( 24 − 14 ) 2  (12, 000 )


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MECHANICS OF MATERIALS, 6th Edition
6-36
RILEY, STURGES AND MORRIS
J = π d 4 32 = π ( 20 ) 2 = 0.251327 (106 ) mm 4
4
(a)
τ AB =
( 600 )( 0.020 ) = 47.7 106 N/m 2
Tc
=
( )
J ( 0.251327 × 10−6 )
τ AB = 47.7 MPa
τ BC =
(120 )( 0.020 )
( 0.251327 ×10 )
−6
τ BC = 9.55 MPa
τ CD =
τ DE =
..................................................Ans.
= 9.55 (106 ) N/m 2
................................................... Ans.
( 840 )( 0.020 )
( 0.251327 ×10 )
−6
(120 )( 0.020 )
( 0.251327 ×10 )
−6
= 66.8 (106 ) N/m 2 = 66.8 MPa ........................................... Ans.
= 9.55 (106 ) N/m 2 = 9.55 MPa ........................................... Ans.
(b)
σ max T = σ max C = τ max T = 66.8 MPa (T&C) ...................................................................... Ans.
(c)
θ=
TL ( 600 + 120 + 840 + 120 )(1.5 )
=
= 0.1319 rad ................................................. Ans.
JG ( 0.251327 × 10−6 )( 76 ×109 )
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-37
(a)
θ=
TL 0.172
rad
=
6
JG
(T1 ×12 )( 24 ) + (T1 − 9 ) ×12 ( 48) = 0.172
6
π (1)4 2  (12, 000 ) π ( 2 )4 2  (12, 000 )




rad
T1 = 2.66776 kip ⋅ ft ≅ 32.0 kip ⋅ in. ................................Ans.
(b)
τ BC =
Tc ( 2.66776 × 12 )(1)
=
= 20.4 ksi
4
J
π (1) 2
σ BC = τ BC = 20.4 ksi (T)
(c)
..................................................Ans.
( 2.66776 − 9 ) × 12  ( 2 )
= 6.05 ksi (C)
σ CD = τ CD = 
4
π ( 2) 2
................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
6-38*
Power = T ω =
RILEY, STURGES AND MORRIS
2π NT 2π (180 ) T
=
= 240 (103 ) N ⋅ m/s
60
60
T = 12.73240 (103 ) N ⋅ m
τ=
Tc (12, 732.40 )( d 2 )
=
= 80 (106 ) N/m 2
J
π d 4 32
d ≥ 0.0932 m = 93.2 mm ............................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-39
RILEY, STURGES AND MORRIS
T ( 20 × 12 )
TL
=
= 0.06 rad
JG π ( 4 )4 32  (12, 000 )


T = 75.39822 kip ⋅ in. = 6.28319 kip ⋅ ft
θ=
Power = T ω =
2π NT 2π ( 270 )( 6283.19 )
=
= 323 hp .............................................. Ans.
60
( 60 )( 550 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-40*
τ=
T ( 0.050 )
Tc
=
= 80 (106 ) N/m 2
4
4
J π ( 0.050 − 0.030 ) 2
RILEY, STURGES AND MORRIS
T = 13, 672.211 N ⋅ m
2π NT 2π ( 200 )(13, 672.211)
=
= 286 (103 ) N ⋅ m/s
60
60
Power = 286 kW ........................................................................................................... Ans.
(a)
Power = T ω =
(b)
θ=
(13, 672.211)( 3)
TL
=
= 0.0600 rad ...................................... Ans.
JG π ( 0.0504 − 0.0304 ) 2  ( 80 ×109 )


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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-41*
RILEY, STURGES AND MORRIS
Power = T ω = 2π NT 60 = 2π ( 60 ) T 60 = ( 20, 000 × 550 ) lb ⋅ ft/s
T = 1.750704 (106 ) lb ⋅ ft = 21.00845 (106 ) lb ⋅ in.
(a)
(b)
3
Tc ( 21.00845 ×10 ) (15 )
= 3.96 ksi ...................................................................... Ans.
τ= =
4
J
π (15 ) 2
3
TL ( 21.00845 × 10 ) ( 20 × 12 )
=
= 0.00528 rad .................................................... Ans.
θ=
JG
π (15 )4 2  (12, 000 )


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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-42
(a)
RILEY, STURGES AND MORRIS
Power = T ω = 2π ( 400 ) T 60 = 200 (103 ) N ⋅ m/s
T = 4774.648 N ⋅ m
τ=
Tc ( 4774.648 )( d 2 )
=
= 70 (106 ) N/m 2
4
J
π d 32
d = 0.0703 m
θ=
( 4774.648)(1.5) = 0.045 rad
TL
=
JG π d 4 32  ( 80 × 109 )
d = 0.0671 m
d min = 70.3 mm ............................................................................................................... Ans.
(b)
τ=
T ( 0.075 2 )
π ( 0.075 ) 32
Power =
4
= 50 (106 ) N/m 2
2π N ( 4141.748 )
= 200 (103 ) N ⋅ m/s
60
T = 4141.748 N ⋅ m
N = 461 rpm ...................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-43
RILEY, STURGES AND MORRIS
Power = T ω = 2π NT 60 = 2π ( 3800 ) T 60 = (162 × 550 ) lb ⋅ ft/s
T = 223.90588 lb ⋅ ft = 2686.8705 lb ⋅ in.
(a)
τ=
( 2686.8705 )( d 2 ) = 5000 psi
(b)
τ=
Tc ( 2686.8705 )(1.5 )
=
= 5000 psi
J
J
J=
(c)
π d 4 32
π ( 34 − di4 )
32
= 0.80606 in.4
d = 1.39878 in. ≅ 1.399 in. ...................... Ans.
J = 0.80606 in.4
di = 2.92090 in. ≅ 2.92 in. ....................... Ans.
π ( 32 − 2.920902 ) 4  − π (1.39878 )2 4 
Wh − Ws
 
 (100) = −76.1
(100 ) = 
2
Ws
π (1.39878 ) 4 


%reduction = 76.1% .............................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-44
RILEY, STURGES AND MORRIS
2π NT 2π (1800 ) T
=
= 1200 (103 ) N ⋅ m/s
60
60
T = 6366.1977 N ⋅ m
Power = T ω =
τ=
( 6366.1977 )( d 2 ) = 100 106 N/m 2
Tc
=
( )
J π  d 4 − ( 0.75d )4  32


d = 0.0780 m
θ=
( 6366.1977 )( 3)
TL
=
= 0.20 rad
4
4
9
JG


π d − ( 0.75d ) 32 ( 80 × 10 )


d = 0.0844 m
{
}
d min = 84.4 mm ............................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-45*
RILEY, STURGES AND MORRIS
2π NT 2π (1800 ) T
=
= ( 350 × 550 ) lb ⋅ ft/s
60
60
T = 1021.24422 lb ⋅ ft = 12, 254.93 lb ⋅ in.
Motor shaft:
Power = T ω =
τ=
Tc (12.25493)( d 2 )
=
= 15 ksi
J
π d 4 32
d = 1.61 in.
θ=
TL (12.25493)(10 × 12 )
=
= 0.10 rad
JG (π d 4 32 ) (12, 000 )
d = 1.880 in.
d min ( motor ) = 1.880 in. ............................................................................................... Ans.
2π NT 2π ( 200 ) T
=
= ( 350 × 550 ) lb ⋅ ft/s
60
60
T = 9191.19796 lb ⋅ ft = 110, 294.4 lb ⋅ in.
Power shaft:
Power = T ω =
τ=
Tc (110.2994 )( d 2 )
=
= 15 ksi
J
π d 4 32
d = 3.35 in.
θ=
TL (110.2994 )(10 ×12 )
=
= 0.10 rad
JG (π d 4 32 ) (12, 000 )
d = 3.26 in.
d min ( power ) = 3.35 in. ................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-46*
(a)
RILEY, STURGES AND MORRIS
Power = T ω = 2π NT 60
2π ( 250 ) TAB 60 = 200 (103 ) N ⋅ m/s
TAB = 7639.4373 N ⋅ m
2π ( 250 ) TBC 60 = 75 (103 ) N ⋅ m/s
TBC = 2864.7890 N ⋅ m
τ=
Tc ( 7639.4373)( d 2 )
=
= 75 (106 ) N/m 2
4
J
π d 32
d = 0.0804 m
d1 = 80.4 mm ................................................................................................................... Ans.
(b)
τ=
Tc ( 2864.7890 )( d 2 )
=
= 75 (106 ) N/m 2
4
J
π d 32
d = 0.0579 m
d 2 = 57.9 mm .................................................................................................................. Ans.
(c)
θ=
TL ( 7639.4373)(1) + ( 2864.7890 )( 2 )
=
= 0.0538 rad ...................................... Ans.
JG
π ( 0.075 )4 32  ( 80 ×109 )


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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-47
(a)
2π NT 2π ( 360 ) T1
=
= (100 × 550 ) lb ⋅ ft/s
60
60
T1 = 1458.92031 lb ⋅ ft = 17,507.0 lb ⋅ in.
Motor shaft:
τ=
(b)
RILEY, STURGES AND MORRIS
Power = T ω =
Tc (17.5070 )( d1 2 )
=
= 12 ksi
J
π d14 32
Power shaft:
d1 = 1.951 in. ........................................ Ans.
N power = ( 96 16 ) N motor = 6 ( 360 ) = 2160 rpm
Power = T ω =
2π NT 2π ( 2160 ) T
=
= (100 × 550 ) lb ⋅ ft/s
60
60
T2 = 243.15339 lb ⋅ ft = 2917.84 lb ⋅ in. ( = T1 6 )
τ=
( 2.91784 )( d 2 2 ) = 12 ksi
π d 24 32
d 2 = 1.074 in. ........................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-48
(a)
Shaft D:
N D = ( 48 24 ) N E = 2 ( 400 ) = 800 rpm
Power = T ω =
τ=
(b)
Shaft E:
τ=
(c)
Shaft F:
2π NT 2π ( 800 ) TD
=
= 40 (103 ) N ⋅ m/s
60
60
( 477.4648 )( d D 2 ) = 70
π d 32
4
D
(10 ) N/m
6
TD = 477.4648 N ⋅ m
d D = 0.0326 m = 32.6 mm ........... Ans.
2
2π NT 2π ( 400 ) TD
=
= 180 (103 ) N ⋅ m/s
60
60
TE = 4297.1835 N ⋅ m
Power = T ω =
( 4297.1835 )( d E 2 ) = 70
π d 32
4
E
(10 ) N/m
6
2
d E = 0.0679 m = 67.9 mm ............ Ans.
N F = ( 48 24 ) N E = 2 ( 400 ) = 800 rpm
Power = T ω =
τ=
RILEY, STURGES AND MORRIS
2π NT 2π ( 800 ) TE
=
= 80 (103 ) N ⋅ m/s
60
60
( 954.9297 )( d E 2 ) = 70
π d E4 32
(10 ) N/m
6
2
TE = 954.9297 N ⋅ m
d E = 0.0411 m = 41.1 mm ............. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-49*
Equilibrium:
Torque diagram
Deformations:
θ AB + θ BC = 0
RILEY, STURGES AND MORRIS
θ = TL JG
( −TA )( 2 ) + ( 3.5 ×12 ) − TA  ( 3) = 0
JG
JG
TA = 25.200 kip ⋅ in. = −TAB
TBC = ( 3.5 × 12 ) − TA = 16.800 kip ⋅ in.
TAB c ( 25.200 )(1)
=
= 16.04 ksi ................... Ans.
4
J
π ( 2 ) 32
(a)
τ max =
(b)
θ = θ AB =
( −25.200 )( 2 ×12 ) = −0.0321 rad ................................................. Ans.
TL
=
JG π ( 2 )4 32  (12, 000 )


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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-50*
RILEY, STURGES AND MORRIS
J S = π d 4 32 = π (1254 − 1004 ) 32 = 14.15097 (106 ) mm 4
J M = π (1754 − 1254 ) 32 = 68.10875 (106 ) mm 4
Equilibrium:
TS + TM = 10 kN ⋅ m
Deformations:
θS = θM
(a)
θ = TL JG
TS L
TM L
=
−6
9
(14.15097 ×10 )(80 ×10 ) ( 68.10875 ×10−6 )( 65 ×109 )
TM = 3.91057TS
(b)
TS = 2.03642 kN ⋅ m
(a)
τS =
τM =
(b)
TM = 7.96358 kN ⋅ m
Tc ( 2036.42 )( 0.0625 )
=
= 8.99 (106 ) N/m 2 = 8.99 MPa ................................ Ans.
−6
J
(14.15097 ×10 )
( 7963.58 )( 0.0875) = 10.23 106
( 68.10875 ×10 )
θ = θS =
−6
( ) N/m
2
= 10.23 MPa ..................................... Ans.
( 2036.42 )( 2 )
TL
=
= 0.00360 rad ........................................ Ans.
JG (14.15097 × 10−6 )( 80 × 109 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-51
T (1.5 )
Tc
=
≤ 15 ksi
J π ( 3)4 32
(a)
τ=
(b)
θS = θ A
π 3
 ( )
4
T ≤ 79.5 kip ⋅ in. ..................................................... Ans.
θ = TL JG
TS L
TA L
=
4
4
32  (11, 600 ) π ( 3.5 − 3 ) 32  ( 4000 )

TS = 3.40127TA
T (1.5 )
Tc
= S 4
= τ max = 15 ksi
J π ( 3) 32
Guess that
τS =
Then
TS = 79.52156 kip ⋅ in.
and
τA =
Therefore
RILEY, STURGES AND MORRIS
TA = 23.37998 kip ⋅ in.
( 23.37998 )(1.75 ) = 6.03 ksi ≤ 12 ksi = τ
π ( 3.54 − 34 ) 32
max
(correct guess)
Tmax = TA + TS = 102.9 kip ⋅ in. ...................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
6-52*
RILEY, STURGES AND MORRIS
J B = π d 4 32 = π ( 804 − 604 ) 32 = 2.74889 (106 ) mm 4
J A = π ( 60 ) 32 = 1.272345 (106 ) mm 4
4
TB L
TA L
=
−6
9
( 2.74889 ×10 )( 45 ×10 ) (1.272345 ×10−6 )( 28 ×109 )
TB = 3.47222TA
(a)
τB =
TB ( 0.040 )
Tc
=
= 150 (106 ) N/m 2
−6
J ( 2.74889 ×10 )
TB = 10,308.35 N ⋅ m
TA = 2968.807 N ⋅ m
Tmax = TB + TA = 13.28 kN ⋅ m ...................................................................................... Ans.
(b)
τA =
Tc ( 2968.807 )( 0.030 )
=
= 70.0 (106 ) N/m 2 = 70.0 MPa ............................... Ans.
−6
J
(1.272345 ×10 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-53
Equilibrium:
Torque diagram
Deformations:
θ AB + θ BC = 0
RILEY, STURGES AND MORRIS
θ = TL JG
( −TA )( 5 ×12 ) + (T − TA )(8 ×12 ) = 0
J (11, 600 )
J ( 6500 )
T = 1.35022TA
Tc (T − TA )(1.5 )
=
= τ max = 6 ksi
4
J
π ( 3) 32
Guess that
τB =
Then
T − TA = 31.80863 kip ⋅ in.
TA = 90.82586 kip ⋅ in. = −TS
and
Therefore
( 90.82586 )(1.5) = 17.13 ksi ≤ 18 ksi = τ (correct guess)
max
4
π ( 3) 32
Tmax = 1.35022 ( 90.82586 ) = 122.6 kip ⋅ in. .............................................................. Ans.
τS =
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MECHANICS OF MATERIALS, 6th Edition
6-54*
RILEY, STURGES AND MORRIS
J B = π d 4 32 = π ( 40 ) 32 = 0.251327 (106 ) mm 4
4
J S = π ( 804 − 404 ) 32 = 3.76991(106 ) mm 4
J A = π (1004 − 804 ) 32 = 5.79624 (106 ) mm 4
Equilibrium:
T = TA + TS + TB
Deformations:
θ A = θS = θB
(a)
θ = TL JG
TS L
TA L
TB L
=
=
−6
9
−6
9
( 5.79624 ×10 )( 28 ×10 ) ( 3.76991×10 )(80 ×10 ) ( 0.251327 ×10−6 )( 39 ×109 )
(a)
TA = 16.5577TB
(b)
TS = 30.7693TB
(c)
16.5577TB + 30.7693TB + TB = 15 kN ⋅ m
TB = 0.31039 kN ⋅ m
τB =
(b)
TA = 5.13928 kN ⋅ m
TS = 9.55034 kN ⋅ m
Tc ( 310.39 )( 0.020 )
=
= 24.7 (106 ) N/m 2 = 24.7 MPa .................................. Ans.
−6
J ( 0.251327 ×10 )
τA =
( 5139.28)( 0.050 ) = 44.3 106
τS =
( 9550.34 )( 0.040 ) = 101.3 106
( 5.79624 ×10 )
−6
( 3.76991×10 )
−6
( ) N/m
2
( ) N/m
= 44.3 MPa ............................................ Ans.
2
= 101.3 MPa ........................................ Ans.
16.5577TB + 30.7693TB + TB = 10 kN ⋅ m
TB = 0.206924 kN ⋅ m
θ = θB =
TA = 3.42618 kN ⋅ m
TS = 6.36690 kN ⋅ m
( 206.924 )( 3)
TL
=
= 0.0633 rad .......................................... Ans.
JG ( 0.251327 × 10−6 )( 39 × 109 )
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-55*
Equilibrium:
TS + TP = T = 1000 lb ⋅ in.
Deformations:
θ AB , P = θ AB , S
(a)
TP (12 )
π ( 2 )4 32  (150 )


=
θ = TL JG
TS (12 )
π ( 2.254 − 24 ) 32  (12, 000 )


TS = 48.1445TP
TP = 20.3482 lb ⋅ in.
θ=
(a)
(b)
TS = 979.6518 lb ⋅ in.
TL ( 20.3482 )(12 ) + (1000 )( 4 )
=
= 0.01801 rad .......................................... Ans.
JG
π ( 2 )4 32  (150 × 103 )


(b)
θ=
(c)
%error =
(1000 )( 4 )
π ( 2 ) 32  (150 ×103 )


4
= 0.01698 rad .................................................................... Ans.
0.01801 − 0.01698
(100 ) = 5.75% ................................................................. Ans.
0.01801
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MECHANICS OF MATERIALS, 6th Edition
6-56
RILEY, STURGES AND MORRIS
J S = π (1004 − 504 ) 32 = 9.20388 (106 ) mm 4
J M = π (1254 − 1024 ) 32 = 13.34170 (106 ) mm 4
θS = θM
θ = TL JG
TS L
TM L
=
−6
9
( 9.20388 ×10 )(80 ×10 ) (13.34170 ×10−6 )( 65 ×109 )
TS ( 0.05 )
Tc
=
= τ max = 70 (106 ) N/m 2
−6
J ( 9.20388 × 10 )
Guess that
τS =
Then
TS = 12,885.4 N ⋅ m
and
τM =
TS = 0.84906TM
TM = 15,176.2 N ⋅ m
(15,176.2 )( 0.0625 ) = 71.1 106
(13.34170 ×10 )
−6
( ) N/m
2
≤ 85 MPa = τ max (correct guess)
(a)
Tmax = TS + TM = 12.885 + 15.176 = 28.061 kN ⋅ m ≅ 28.1 kN ⋅ m .............................. Ans.
(b)
θ = θS =
(12,885.4 )( 2.5)
TL
=
= 0.0437 rad ............................................ Ans.
JG ( 9.20388 × 10−6 )( 80 × 109 )
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MECHANICS OF MATERIALS, 6th Edition
6-57*
Equilibrium:
T = TS + TA = 1.5TA
Deformations:
θ A = θS
TS = 0.5TA
θ = TL JG
( 0.5TA ) L
TA L
=
π ( 2 ) 32  ( 4000 ) J S (12, 000 )


4
π ( d 4 − 24 )
RILEY, STURGES AND MORRIS
J S = 0.26180 in.4
d = 2.07858 in. = 2 (1 + t )
32
t = 0.0393 in. .......................................................................................................................... Ans.
JS =
= 0.26180 in.4
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-58
RILEY, STURGES AND MORRIS
TB = TS
θB = θS
θ = TL JG
TS L
TB L
=
9
J B ( 45 ×10 ) J S ( 80 ×109 )
JB =
π (1004 − d 4 )
32
= (1.77778 )
J B = 1.77778 J S
πd4
32
= 1.77778 J S
d = 77.5 mm .......................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-59
Equilibrium:
Torque diagram
Deformations:
θ AB + θ BC = 0
RILEY, STURGES AND MORRIS
θ = TL JG
( −TA )( 5 ×12 ) + ( 25 ×12 ) − TA  (10 ×12 ) = 0
π ( 4 )4 32  G
π ( 44 − 24 ) 32  G




TA = 204.255 kip ⋅ in. = −TAB
TBC = ( 25 ×12 ) − TA = 95.7447 kip ⋅ in.
(a)
τ AB =
τ BC =
(b)
Tc ( 204.255 )( 2 )
=
= 16.25 ksi
4
J
π ( 4 ) 32
( 95.7447 )( 2 )
π ( 44 − 24 ) 32
θ = θ AB =
= 8.13 ksi
τ max = τ AB = 16.25 ksi
............................ Ans.
( −204.255)( 5 ×12 ) = −0.0406 rad ................................................ Ans.
TL
=
JG π ( 4 )4 32  (12, 000 )


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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-60*
RILEY, STURGES AND MORRIS
J B = π d 4 32 = π ( 75 ) 32 = 3.10631(106 ) mm 4
4
J M = π ( 60 ) 32 = 1.272345 (106 ) mm 4
4
Equilibrium:
Torque diagram
Deformations:
θ AB + θ BC = 0
θ = TL JG
( −TA )( 0.300 )
+
(15, 000 − TA ) [ 0.450]
( 3.10631×10 )( 39 ×10 ) (1.272345 ×10 )( 65 ×10 )
−6
9
−6
TA = 10,308.49 N ⋅ m = −TAB
(a)
τ AB =
τ BC =
9
=0
TBC = 15, 000 − TA = 4691.51 N ⋅ m
Tc (10,308.49 )( 0.0375 )
=
= 124.4 (106 ) N/m 2 = 124.4 MPa ...................... Ans.
−6
J
( 3.10631×10 )
( 4691.51)( 0.030 ) = 110.6
(1.272345 ×10−6 )
(10 ) N/m
6
2
= 110.6 MPa ...................................... Ans.
( −10,308.49 )( 0.300 ) = −0.0255 rad ........................................ Ans.
TL
=
JG ( 3.10631×10−6 )( 39 ×109 )
(b)
θ = θ AB =
(c)
σ max T = σ max C = τ AB = 124.4 MPa ..................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-61
J = π d 4 32 = π ( 6 ) 32 = 127.2345 in 4
4
(a)
θ=
T ( 6 × 12 )
TL
=
= 0.0018 rad
GJ (127.2345 )(12, 000 )
T = 38.17035 kip ⋅ in. ≅ 3.18 kip ⋅ ft ...............................................Ans.
(b)
θ=
Tb ( 9 ×12 )
TL
=
= 0.0018 rad
GJ (127.2345 )(12, 000 )
Tb = 25.44690 kip ⋅ in.
τ=
Tc ( 25.44690 )( 3)
=
= 0.600 ksi
J
(127.2345)
τ = 600 psi ..................................................................... Ans.
(c)
Equilibrium:
Torque diagram
Deformations:
θ A + θ AB + θ BC = 0
θ = TL JG
( −TA )( 3 ×12 ) + (T − TA )( 6 ×12 ) = 0
(127.2345)(12, 000 ) (127.2345)(12, 000 )
(TA )( 3) = τ = 10 ksi
Tc
τ AB = =
max
J (127.2345 )
−0.0018 +
Guess that
Then
TA = 424.1150 kip ⋅ in. = −TAB
T − TA = 250.2279 kip ⋅ in. = TBC
( 250.2279 )( 3) = 5.90 ksi ≤ 10 ksi = τ
max
(127.2345)
and
τ BC =
Therefore
Tmax = 250.2279 + 424.1150 = 674.3429 kip ⋅ in.
(correct guess)
Tmax = 56.2 kip ⋅ ft ........................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-62*
RILEY, STURGES AND MORRIS
TA − 4 ( 0.075V ) = 0
4 ΣM = 0 :
TA = 0.300V = 0.300 (τ A )
= 0.300 ( 60 ×106 )(150 ×10−6 ) = 2700 N ⋅ m
Equilibrium:
Torque diagram
Deformations:
θ AB + θ BC + θCD = 0
θ = TL JG
( −2700 )( 0.3)
( −2700 )( 0.6 )
(T − 2700 )( 0.3)
=0
+
+
π ( 0.075 )4 32  ( 28 ×109 ) π ( 0.075 )4 32  ( 80 ×109 ) π ( 0.075 )4 32  ( 80 ×109 )






(a)
T = 15,814.3 N ⋅ m ≅ 15.81 kN ⋅ m .................................. Ans.
(b)
TCD = 15,814.3 − 2700 = 13,114.3 N ⋅ m
Tmax ( steel ) = TCD = 13,114.3 N ⋅ m
τ max = τ CD =
Tc (13,114.3)( 0.0375 )
=
J
π ( 0.075 )4 32 


τ max = 158.3 (106 ) N/m 2 = 158.3 MPa ........................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-63*
The torque of the shear force VC is TC = VC (1) = VC and the
torque of the shear force VD is TD = VD (1) = VD .
Equilibrium:
Torque diagrams
Deformations:
θCD ,h = θCD , s
(1000 − VC ) L
π ( 24 − 14 ) 32  (12 ×106 )


Therefore
and
θ = TL JG
=
(VC ) L
π (1)4 32  (12 ×106 )


VC = 62.500 lb
VD = 1000 − VC = 1000 − 62.5 = 937.500 lb
VC = τ A = ( 25, 000 ) (π dC2 4 ) = 62.500 lb
dC = 0.0564 in. ...................................................... Ans.
VD = τ A = ( 25, 000 ) (π d D2 4 ) = 937.500 lb
d D = 0.219 in. ........................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-64
RILEY, STURGES AND MORRIS
J S = π d 4 32 = π ( 80 ) 32 = 4.02124 (106 ) mm 4
4
J B = π (1204 − 804 ) 32 = 16.33628 (106 ) mm 4
Equilibrium:
TS + TB = 27 kN ⋅ m
Deformations:
θ AB , S = θ AB , B
(a)
θ = TL JG
TS (1.5 )
TB (1.5 )
=
( 4.02124 ×10 )(86 ×10 ) (16.33628 ×10 )( 39 ×10 )
−6
9
−6
TB = 1.84230TS
9
(b)
TB = 17.50065 kN ⋅ m
TS = 9.49935 kN ⋅ m
Tc ( 9, 499.35 )( 0.040 )
=
= 94.5 (106 ) N/m 2 = 94.5 MPa ................................ Ans.
−6
J
( 4.02124 ×10 )
(a)
τS =
(b)
τB =
(c)
σ max = τ max = 64.3 MPa (C) ................................................................................................ Ans.
(d)
θ=
(17,500.65)( 0.060 ) = 64.3 106
(16.33628 ×10 )
−6
( ) N/m
2
= 64.3 MPa ......................................... Ans.
TL ( 9, 499.35 )(1.5 ) + ( −9000 )(1)
=
= 0.01518 rad ............................................ Ans.
JG
( 4.02124 ×10−6 )(86 ×109 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-65
Equilibrium:
Torque diagram
Deformations:
θ AB + θ BC + θCD = 0
RILEY, STURGES AND MORRIS
θ = TL JG
TA ( 24 )
T ( 24 )
(T − T )( 24 ) = 0
+ A
+ A
J ( 6000 ) J (12, 000 )
J (12, 000 )
Guess that
Then
τ CD =
Tc (TA − T )( 2 )
=
= τ max = −12 ksi
4
J
π ( 4 ) 32
TA − T = −150.79645 kip ⋅ in. = TCD
TA = 50.26548 kip ⋅ in. = TAB = TBC
( 50.26548)( 2 ) = 4.00 ksi ≤ 5 ksi = τ (okay)
max
4
π ( 4 ) 32
( 50.26548 )( 2 ) = 4.00 ksi ≤ 12 ksi = τ (okay)
τ BC =
max
4
π ( 4 ) 32
τ AB =
and
Therefore
Tmax = 50.26548 + 150.79645 = 201.062 kip ⋅ in. ≅ 201 kip ⋅ in. .......................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-66*
RILEY, STURGES AND MORRIS
J = π d 4 32 = π (100 ) 32 = 9.81748 (106 ) mm 4
4
Equilibrium:
Torque diagram
Deformations:
θ AB + θ BC + θCD = 0
( −TA )( 2 )
J ( 80 × 10
9
)
+
θ = TL JG
( 20 − TA )( 2 ) + ( 40 − TA )(1.5) = 0
J ( 80 × 109 )
J ( 40 ×109 )
TA = 22.85714 kN ⋅ m = −TAB
TBC = 20 − TA = −2.85715 kN ⋅ m
TCD = 40 − TA = 17.14286 kN ⋅ m
(a)
τB =
(b)
τS =
(c)
θ=
Tc (17,142.86 )( 0.050 )
=
= 87.3 (106 ) N/m 2 = 87.3 MPa ............................... Ans.
−6
J
9.81748
10
×
(
)
( 22,857.14 )( 0.050 ) = 116.4
( 9.81748 ×10 )
−6
(10 ) N/m
6
2
= 116.4 MPa .................................... Ans.
( −22,857.14 )( 2 )
TL
=
= −0.0582 rad ................................................... Ans.
JG ( 9.81748 ×10−6 )( 80 × 109 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-67
RILEY, STURGES AND MORRIS
J AB = π d 4 32 = π ( 64 − 44 ) 32 = 102.1018 in 4
J BC = π ( 4 ) 32 = 25.1327 in 4
4
Initially
θi =
( 40 ×12 )( 6 ×12 ) = 0.02821 rad
TL
=
GJ (102.1018 )(12, 000 )
After the torque is removed
Deformations:
TAB = TCD , and
θ B / A + θC / D + θ slip = θi
θ = TL JG
TAB ( 6 × 12 )
TBC ( 4 × 12 )
+
+ 0.010 = 0.02821 rad
(102.1018)(12, 000 ) ( 25.1327 )( 4000 )
TAB = TCD = 33.9540 kip ⋅ in.
(a)
τA =
(b)
τS =
(c)
Tc ( 33.9540 )( 2 )
=
= 2.70 ksi .................................................................................. Ans.
J
( 25.1327 )
( 33.9540 )( 3) = 0.998 ksi .......................................................................................... Ans.
(102.1018)
TL ( −33.9540 )( 6 × 12 )
θ=
=
= −0.001995 rad ......................................................... Ans.
GJ (102.1018 )(12, 000 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-68
RILEY, STURGES AND MORRIS
J S = π d 4 32 = π ( 80 ) 32 = 4.02124 (106 ) mm 4
4
J B = π (1604 − 1404 ) 32 = 26.62500 (106 ) mm 4
θi =
Initially
(10, 000 )( 0.800 )
TL
=
= 0.02487 rad
GJ ( 4.02124 × 10−6 )( 80 × 109 )
After the torque is removed
Deformations:
TS = TB , and
θ S + θ B = θi
θ = TL JG
TS ( 0.800 )
TB ( 0.800 )
+
( 4.02124 ×10 )(80 ×10 ) ( 26.62500 ×10 )( 40 ×10 )
−6
9
−6
9
= 0.02487 rad
TS = TB = 7680.1087 N ⋅ m
(a)
τB =
(b)
τS =
(c)
θ=
Tc ( 7680.1087 )( 0.080 )
=
= 23.1(106 ) N/m 2 = 23.1 MPa ............................. Ans.
−6
J
( 26.62500 ×10 )
( 7680.1087 )( 0.040 ) = 76.4
( 4.02124 ×10 )
−6
(10 ) N/m
6
2
= 76.4 MPa ....................................... Ans.
( −7680.1087 )( 0.800 ) = −0.01910 rad ................................................ Ans.
TL
=
GJ ( 4.02124 × 10−6 )( 80 × 109 )
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MECHANICS OF MATERIALS, 6th Edition
6-69
RILEY, STURGES AND MORRIS
4
4
J = π d 4 32 = π ( ro4 − ri 4 ) 2 = π ro4 1 − ( ri ro )  2 = 8π 1 − ( ri ro )  in.4




Equilibrium:
Deformations:
Torque diagram
θ AC + θCB = 0
θ = TL JG
( −TA )( 3 ×12 ) + ( 7.5 − TA )( 2 ×12 ) = 0
J ( 4000 )
J ( 4000 )
TA = 3.000 kip ⋅ ft = TAC
TCB = 7.5 − TA = 4.500 kip ⋅ ft
(a)
θ=
( −3 ×12 )( 3 ×12 ) rad
TL
=
JG 8π 1 − ( ri ro )4  ( 4000 )


(b)
τ=
( 4.5 ×12 )( 2 ) ksi
Tc
=
J 8π 1 − ( ri ro )4 


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MECHANICS OF MATERIALS, 6th Edition
6-70
Equilibrium:
Deformations:
Torque diagram
θ AC + θCB = 0
( −TA )( 2 )
J s ( 80 ×10
9
)
+
RILEY, STURGES AND MORRIS
θ = TL JG
(T − TA )( 2 )
J b ( 40 × 109 )
=0
TA 1 + ( J b 2 J s )  = T
J s = π d s4 32
J b = π db4 32
db + d s = 200 mm
In terms of the diameter ratio
r = ( db d s )
d s = 200 (1 + r )
db = rd s = 200r (1 + r )
Method of solution:
Guess
τs =
Ts ( d s 2 )
= τ s ( max )
Js
Compute
TA = −Ts , T − TA = Tb , T , τ b , and θ AC
Check
τ b ≤ τ b ( max )
Guess
τb =
and
θ AC ≤ θ max
Tb ( db 2 )
= τ b ( max )
Jb
Compute
T − TA = Tb , Ts = −TA , T , τ s , and θ AC
Check
τ s ≤ τ s ( max )
Guess
and
θ AC ≤ θ max
θ AC = θ max
Compute
TA = −Ts , T − TA = Tb , T , τ b , and τ s
Check
τ b ≤ τ b ( max )
and τ s ≤ τ s ( max )
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MECHANICS OF MATERIALS, 6th Edition
6-71
JS =
Equilibrium:
πd4
=
π ( 44 − d a4 )
J Al =
in.4
32
32
TSt + TAl = 6.00 kip ⋅ ft = 72.00 kip ⋅ in.
θ ST = θ Al
Deformations:
RILEY, STURGES AND MORRIS
π d a4
32
in.4
(a)
θ = TL JG
TSt L
TAl L
=
J St (12, 000 ) J Al ( 4000 )
TSt =
3 J StTAl
J Al
(b)
72 J Al
36d a4
TAl =
=
kip ⋅ in.
J Al + 3J St 384 − d a4
108 ( 256 − d a4 )
216 J St
=
TSt =
kip ⋅ in.
J Al + 3 J St
384 − d a4
τ Al =
TAl ( d a 2 ) 183.3465d a
=
ksi
J Al
384 − d a4
τ St =
TSt ( d a 2 )
72 J Al
550.0395d a
=
=
ksi
J St
J Al + 3 J St
384 − d a4
θ=
TAl 36
3.30024
rad
=
J Al ( 4000 ) 384 − d a4
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-72
J=
πd4
=
32
Ab =
πd2
=
4
= 20.3575 (106 ) mm 4
32
π (18 )
2
= 254.4690 mm 2
Equilibrium:
4
TA + TC = T
Deformations:
θtotal = θ slip + θ AB + θ BC = 0
θ AB =
4
π (120 )
(a)
θ = TL JG
( −TA )( 0.75)
( 20.3575 ×10 )(80 ×10 )
= −0.460518 (10 ) T rad
−6
9
−6
A
θ BC =
(T − TA )(1.50 )
( 20.3575 ×10 )(80 ×10 )
= 0.921036 (10 ) (T − T ) rad
−6
9
−6
A
If
θ BC ≤ 1° = 0.0174533 rad , then
TA = θ AB = 0
TBC = T − TA = T = 1.08573 (106 ) θ BC N
If
θ BC = 1° = 0.0174533 rad , then
TA = θ AB = 0
TBC = T = 18,949.63 N
If
θ BC ≥ 1° = 0.0174533 rad , then
( −0.0174533) +  −0.460518 (10−6 ) TA  + 0.921036 (10−6 ) (T − TA )  = 0
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MECHANICS OF MATERIALS, 6th Edition
6-72 (cont.)
RILEY, STURGES AND MORRIS
TA = ( 0.666667T − 12, 633.09 ) N = −TAB
TBC = T − TA = ( 0.333333T + 12, 633.09 ) N
(a)
τ AB =
(TA )( 0.060 ) = 2947.32T N/m 2
Tc
=
(
A)
J ( 20.3575 × 10−6 )
τ BC =
Tc (T − TA )( 0.060 )
=
= 2947.32 (T − TA ) N/m 2
−6
J ( 20.3575 ×10 )
TA = 8 [ 0.150Vb ] = 8 [ 0.150τ b ] ( 254.4690 ×10−6 ) = 305.3628 (10−6 )τ b
(b)
τb =
τb
(c)
θB
(TA )
( 305.3628 ×10 )
=
( 0.666667T − 12, 633.09 )
( 305.3628 ×10 )
=  2183.197T − 41.37076 (10 )  N/m
= θ = 0.921036 (10 ) (T − T ) rad
−6
−6
6
2
−6
BC
A
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-73
Assume that the steel shaft extends all the way through the aluminum shell and attaches to the wall at D.
Ja =
Equilibrium:
Deformations:
πd4
π ( 34 − 24 )
=
32
32
Ts + Ta = T
θCD , s = θCD ,a
= 6.38136 in.
4
Ja =
π ( 24 )
32
= 1.57080 in.4
(a)
θ = TL JG
Ts (12 )
Ta (12 )
=
(1.57080 )(12, 000 ) ( 6.38136 )( 4000 )
Ta = 1.35416Ts
Ta = 0.575221T
(a)
τa =
τs =
(b)
Ts ,CD = 0.424779T
Ts ,CB = T
Td ( 0.575221T )(1.5 )
=
= 0.135211T ksi
J
( 6.38136 )
(T )(1)
= 0.636618T ksi
(1.57080 )
Ta = 2V = 2 (τ b Ab )
τb =
(c)
(b)
θ=
V
0.575221T
=
= 1.46479T ksi
2 Ap 2 π ( 0.5 )2 4 


( 0.424779T )(12 ) +
(T )(18)
= 0.00122535T
(1.57080 )(12, 000 ) (1.57080 )(12, 000 )
rad
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-74
Js =
Ja =
Jb =
πd4
32
=
π (120 )
32
π (1204 − 604 )
32
π ( 60 )
4
= 20.35752 (106 ) mm 4
= 19.08518 (106 ) mm 4
4
= 1.272345 (106 ) mm 4
32
Ts = −TC
TD − TC = Ta + Tb
Equilibrium:
θtotal = θCD + θ slip + θ EF = 0
θ EF ,a = θ EF ,b
Deformations:
θCD =
θ = TL JG
(a)
(b)
(c)
(d)
( −TC )( 2 )
( 20.35752 ×10 )(80 ×10 )
= −1.22805 (10 ) (T ) rad
−6
9
−6
C
θ EF ,a =
(Ta )(1.40 )
(19.08518 ×10 )( 28 ×10 )
= 2.61983 (10 ) (T ) rad
−6
9
−6
a
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MECHANICS OF MATERIALS, 6th Edition
6-74 (cont.)
θ EF ,b =
RILEY, STURGES AND MORRIS
(Tb )(1.40 )
(1.272345 ×10 )( 45 ×10 )
−6
If
θCD ≤ 3° = 0.0523599 rad , then
If
θCD ≥ 3° = 0.0523599 rad , then
9
= 24.45179 (10−6 ) (Tb ) rad
Ta = Tb = θ EF = 0
TC = TD
2.61983 (10−6 ) (Ta ) = 24.45179 (10−6 ) (Tb )
(d)
Ta = 9.33335Tb
−1.22805 (10−6 ) TC +
3π
+ 2.61983 (10−6 ) (Ta ) = 0
180
(c)
TC = [ 42, 636.60 + 2.13333Ta ] N = [ 42, 636.60 + 19.91107Tb ] N
TD = ( 42, 636.60 + 19.91107Tb ) + ( 9.33335Tb ) + Tb
Tb = ( 0.0330639TD − 1409.734 ) N
(b)
Ta = ( 0.308597TD − 13,157.54 ) N
Ts = − ( 0.658338TD + 14,567.29 ) N
τa =
(Ta )( 0.060 )
τb =
(19.08518 ×10 )
θ D = θCD =
−6
(Tb )( 0.030 )
(1.272345 ×10 )
−6
τs =
(Ts )( 0.030 )
( 20.35752 ×10 )
−6
(Ts )( 2 )
( 20.35752 ×10 )(80 ×10 )
−6
9
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-75*
σx =
N
50
=
= 3.97887 ksi
A π ( 4 )2 4
τ xy =
( 30 )( 2 ) = 2.38732 ksi
Tc
=
J π ( 4 )4 32
1
2
θ p = tan −1
σ p1, p 2 =
2τ xy
σ x −σ y
σx +σ y
2
=
σ y = 0 ksi
2 ( 2.38732 )
1
= 25.10°, − 64.90°
tan −1
2
( 3.97887 ) − ( 0 )
 σ −σ y 
2
±  x
 + τ xy
 2 
2
2
σ p1, p 2
3.97887
2
 3.97887 
=
± 
 + ( 2.38732 ) = 1.98944 ± 3.10760 ksi
2
2


σ p1 = 1.98944 + 3.10760 = 5.10 ksi (T)
25.10° ...................................................... Ans.
σ p 2 = 1.98944 − 3.10760 = −1.118 ksi = 1.118 ksi (C)
64.90° ........................... Ans.
σ p 3 = σ z = 0 ksi ..................................................................................................................... Ans.
τ max = τ p = 3.10760 ksi ≅ 3.11 ksi .................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-76*
RILEY, STURGES AND MORRIS
1500 (103 )
N
=
= 27.2837 (106 ) N/m 2 ≅ 27.3 MPa ................... Ans.
2
2
A π ( 0.400 − 0.300 ) 4
σx =
350 × 103 ) ( 0.200 )
(
Tc
τ xy = =
= 40.7437 (106 ) N/m 2 ≅ 40.7 MPa ............... Ans.
4
4
J π ( 0.400 − 0.300 ) 32
σ y = σ z = τ xz = τ yz = 0 MPa ............................................................................................... Ans.
1
2
θ p = tan −1
σ p1, p 2 =
2τ xy
σ x −σ y
σx +σ y
2
=
2 ( 40.7437 )
1
= 35.74°, − 54.26°
tan −1
2
( 27.2837 ) − ( 0 )
 σ x −σ y 
2
± 
 + τ xy
 2 
2
2
σ p1, p 2
27.2837
2
 27.2837 
=
± 
 + ( 40.7437 ) = 13.6419 ± 42.9668 MPa
2
2


σ p1 = 13.6419 + 42.9668 = 56.6 MPa (T)
35.74° ................................................... Ans.
σ p 2 = 13.6419 − 42.9668 = −29.3 MPa = 29.3 MPa (C)
σ p 3 = σ z = 0 Mpa
54.26° ........................ Ans.
................................................................................................................. Ans.
τ max = τ p = 42.9668 MPa ≅ 43.0 MPa
............................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-77
σx =
N
2800
=
= 891.268 psi
A π ( 2 )2 4
σ y = 0 psi
2π NT 2π (1500 ) T
=
= ( 360 × 550 ) lb ⋅ ft/s
60
60
T = 1260.50715 lb ⋅ ft = 15,126.09 lb ⋅ in.
Power = T ω =
τ xy =
Tc (15,126.09 )(1)
=
= 9629.565 psi
4
J
π ( 2 ) 32
1
2
θ p = tan −1
σ p1, p 2 =
2τ xy
σ x −σ y
σx +σ y
2
=
2 ( 9629.565 )
1
tan −1
= 43.68°, − 46.32°
2
(891.268 ) − ( 0 )
 σ x −σ y 
2
± 
 + τ xy
2


2
2
σ p1, p 2 =
0.891268
2
 0.891268 
± 
 + ( 9.629565 ) = 0.44563 ± 9.63987 ksi
2
2


σ p1 = 0.44563 + 9.63987 = 10.08 ksi (T)
43.68° .................................................... Ans.
σ p 2 = 0.44563 − 9.63987 = −9.19 ksi = 9.19 ksi (C)
46.32° ............................... Ans.
σ p 3 = σ z = 0 ksi ..................................................................................................................... Ans.
τ max = τ p = 9.63987 ksi ≅ 9.64 ksi .................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-78*
RILEY, STURGES AND MORRIS
σx =
125 (10 )
N
=
= 44.20971(106 ) N/m 2
2
A π ( 0.060 ) 4
τ xy =
T ( 0.030 )
Tc
=
= 23,578.51T N/m 2
J π ( 0.060 )4 32
3
σ p1 =
σx +σ y
2
σ y = 0 N/m 2
 σ x −σ y 
2
+ 
 + τ xy
2


2
2
σ p1 =
44.20971
 44.20971 
−3 2
+ 
 + ( 23.57851T × 10 ) ≤ 100 MPa
2
2


T ≤ 3167.83 N ⋅ m
Tmax = 3.17 kN ⋅ m ........................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-79
τ xy =
σ p1 =
Tc (100 )( 2 )
=
= 7.95775 ksi
J π ( 4 )4 32
σx +σ y
2
RILEY, STURGES AND MORRIS
σ y = 0 ksi
 σ x −σ y 
σx
2
σ 
2
+ 
+  x  + ( 7.95775 ) ≤ 18 ksi
 + τ xy =
2
 2 
 2 
2
2
σ x ≤ 14.48190 ksi
Pmax
 π ( 4 )2 
= σ x A = (14.48190 ) 
 = 182.0 kip ............................................................. Ans.
 4 
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-80*
τ xy =
RILEY, STURGES AND MORRIS
3
Tc ( 35 × 10 ) ( 0.075 )
=
= 52.81586 N/m 2
4
J
π ( 0.150 ) 32
σ y = 0 N/m 2
 σ −σ y 
2
σx 
2
τp =  x
 + τ xy =   + ( 52.81586 ) ≤ 60 MPa
 2 
 2 
σ x ≤ 56.93802 MPa
2
σ p2 =
Pmax
Pmax
2
σx +σ y
 σ x −σ y 
σx
2
σ 
2
− 
−  x  + ( 52.81586 ) ≥ −96 MPa
 + τ xy =
2
2
 2 
 2 
σ x ≥ −66.94255 MPa
2
2
 π ( 0.150 )2 
= σ x A = ( 56.93802 × 10 ) 
 = 1.006 (106 ) N
4


= 1006 kN (C) ............................................................................................................... Ans.
6
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-81*
σx =
N
175
=
= 6.18936 ksi
A π ( 6 )2 4
τ xy =
Tc (175 )( 3)
=
= 4.12624 ksi
J π ( 6 )4 32
1
2
θ p = tan −1
σ p1, p 2 =
2τ xy
σ x −σ y
σx +σ y
2
=
σ y = 0 ksi
2 ( 4.12624 )
1
tan −1
= 26.57°, − 63.43°
2
( 6.18936 ) − ( 0 )
 σ x −σ y 
2
± 
 + τ xy
 2 
2
2
σ p1, p 2 =
6.18936
2
 6.18936 
± 
 + ( 4.12624 ) = 3.09468 ± 5.15780 ksi
2
2


σ p1 = 3.09468 + 5.15780 = 8.25 ksi (T)
26.57° ...................................................... Ans.
σ p 2 = 3.09468 − 5.15780 = −2.06 ksi = 2.06 ksi (C)
63.43° ............................... Ans.
σ p 3 = σ z = 0 ksi ..................................................................................................................... Ans.
τ max = τ p = 5.15780 ksi ≅ 5.16 ksi .................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-82
AAB =
J AB =
πd2
4
=
π (100 )
4
π (100 )
σ AB
2
4
= 7853.98 mm 2
= 9.81748 (10
6
) mm
π (1602 − 1002 )
4
π (160 − 1004 )
4
4
32
P
P
=
=
= 127.324 P
AAB 7853.98 (10−6 )
σ BC =
ABC =
J BC =
32
= 12, 252.21 mm 2
= 54.5223 (106 ) mm 4
P
P
=
= 81.6180 P
ABC 12, 252.21(10−6 )
τ AB =
Tc (10, 000 )( 0.050 )
=
= 50.9296 (106 ) N/m 2 = 50.9296 MPa
−6
J
( 9.81748 ×10 )
τ BC =
Tc ( 30, 000 )( 0.080 )
=
= 44.0187 (106 ) N/m 2 = 44.0187 MPa
−6
J
( 54.5223 ×10 )
Since both
σ AB
σ p1 =
and τ AB are larger than
σx +σ y
2
σ BC
and τ BC , the maximum stresses will occur in section AB,
 σ x −σ y 
σx
2
σ 
2
+ 
+  x  + ( 50.9296 ) ≤ 140 MPa
 + τ xy =
2
 2 
 2 
2
2
σ x ≤ 121.473 MPa
 σ −σ y 
2
σx 
2
τp =  x
 + τ xy =   + ( 50.9296 ) ≤ 80 MPa
 2 
 2 
2
2
σ x ≤ 123.389 MPa
Pmax = σ x A = (121.473 ×106 )( 7853.98 × 10−6 ) = 954 (103 ) N
Pmax = 954 kN (C) ................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-83*
πd2
A4 =
4
πd4
J4 =
=
2
4
π ( 4)
= 12.5664 in.2
A6 =
4
π ( 6)
2
4
π (6)
= 28.2743 in.2
4
= 25.1327 in.
J6 =
σ4 =
P
125
=
= 9.94716 ksi
A4 12.5664
τ4 =
T ( 2)
Tc
=
= 0.0795776T
J ( 25.1327 )
σ6 =
P
125
=
= 4.42098 ksi
A6 28.2743
τ6 =
( 3T )( 3) = 0.0707355T
Tc
=
J (127.2345 )
Since both
32
σ4
=
π ( 4)
32
and τ 4 are larger than
4
σ6
32
= 127.2345 in.4
and τ 6 ,
the maximum stresses will occur in the 4-in. section
σ p1 =
σx +σ y
2
 σ x −σ y 
9.94716
 9.94716 
2
2
+ 
+ 
 + τ xy =
 + τ xy ≤ 15 ksi
2
2


 2 
2
2
τ xy ≤ 8.70589 ksi
 σ −σ y 
 9.94716 
2
2
τp =  x
 + τ xy = 
 + τ xy ≤ 10 ksi
2


 2 
2
2
τ xy ≤ 8.67545 ksi
Tc (T )( 2 )
=
= 8.67545 ksi
J 25.1327
= 109.0 kip ⋅ in. ............................................................................................................... Ans.
τ xy =
Tmax
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-84
RILEY, STURGES AND MORRIS
σx =
( −200 ) (10 )
N
−339,530.545
=
=
N/m 2
2
2
2
A π  d − ( 0.5d )  4
d


τ xy =
( 7500 )( 0.5d ) = 40, 743.665 N/m 2
Tc
=
J π  d 4 − ( 0.5d )4  32
d3


3
σ p2 =
σx +σ y
2
σ y = 0 N/m 2
 σ x −σ y 
2
− 
 + τ xy
 2 
2
−169, 765.273
 −169, 765.273   40, 743.665 
6
2
=
− 
 +
 ≤ −100 (10 ) N/m
2
2
3
d
d
d

 

2
2
Simplifying yields
10 (106 ) d 6 − 33.9531(103 ) d 4 − 1.66005 = 0
from which
d = 0.0830 m = 83.0 mm ...................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-85
σx =
N
20
25.46479
=
=
ksi
2
A πd 4
d2
σ y = 0 ksi
2π NT 2π (1800 ) T
=
= ( 240 × 550 ) lb ⋅ ft/s
60
60
T = 700.28175 lb ⋅ ft = 8403.3810 lb ⋅ in.
Power = T ω =
τ xy =
Tc ( 8.403381)( 0.5d ) 42.79807
=
=
ksi
J
d3
π d 4 32
σ p1 =
12.73240
 12.73240   42.79807 
+ 
 +
 ≤ 15 ksi
2
d
d2
d3

 

2
2
Simplifying yields
225d 6 − 381.9720d 4 − 1831.6746 = 0
from which
d = 1.662 in. ........................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-86*
σ 45 =
RILEY, STURGES AND MORRIS
210 (103 )
E
ε
+
νε
=
1084 + ( 0.30 )( −754 )  (10−6 ) = 197.9539 MPa
b]
2 
2 [ a
1 −ν
1 − ( 0.30 )
210 (103 )
E
=
ε + νε a ] =
 −754 ) + ( 0.30 )(1084 )  (10−6 ) = −98.9539 MPa
2 (
2 [ b
1 −ν
1 − ( 0.30 )
σ −45
σ 45 = σ x cos 2 ( 45° ) + σ y sin 2 ( 45° ) + 2τ xy sin ( 45° ) cos ( 45° )
σ −45 = σ x cos 2 ( −45° ) + σ y sin 2 ( −45° ) + 2τ xy sin ( −45° ) cos ( −45° )
For a shaft subjected to an axial load P and a torque T :
σ y = 0 N/m 2
0.50σ x + τ xy = 197.9539 MPa
Therefore
0.50σ x − τ xy = −98.9539 MPa
σ x = 99.0000 MPa
Solving yields
P = σ x A = ( 99 × 10
T=
τ xy J
c
6
)
π ( 0.025 )
2
4
= 48.6 (103 ) N = 48.6 kN ...................................... Ans.
(148.4539 ×10 ) π ( 0.025)
=
6
32 ( 0.0125 )
τ xy = 148.4539 MPa
4
= 455 (103 ) N ⋅ m = 455 kN ⋅ m ............... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-87*
ε n = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cos θ
= ( 36 ) cos 2 ( 45° ) + (150 ) sin 2 ( 45° ) + γ xy sin ( 45° ) cos ( 45° ) = 310
γ xy = 434.00 µ rad
σa = σx =
E
30, 000
ε +νε b ] =
36 + ( 0.30 )(150 )  (10−6 ) = 2.67033 ksi
2 
2 [ a
1 −ν
1 − ( 0.30 )
σb = σ y =
E
30, 000
ε + νε a ] =
 150 ) + ( 0.30 )( 36 )  (10−6 ) = 5.30110 ksi
2 (
2 [ b
1 −ν
1 − ( 0.30 )
p=
2σ a t 2 ( 2.67033)( 0.375 )
=
= 200 psi .............................................................. Ans.
r
10
( 30, 000 ) ( 434.00 ×10−6 )
τ xy =
=
= 5.00769 ksi
2 (1 + ν )
2 (1 + 0.30 )
Eγ xy
J = π d 4 32 = π ( 20.754 − 204 ) 32 = 2492.075 in 4
T=
τ xy J
c
=
( 5007.69 )( 2492.075 )
(10.375 )
T = 1.20285 (106 ) lb ⋅ in. = 100.2 kip ⋅ ft ................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-88
σ 45 =
RILEY, STURGES AND MORRIS
200 (103 )
E
ε
+
νε
=
1414 + ( 0.30 )( −212 )  (10−6 ) = 296.7912 MPa
b]
2 
2 [ a
1 −ν
1 − ( 0.30 )
200 (103 )
E
=
ε + νε a ] =
 −212 ) + ( 0.30 )(1414 )  (10−6 ) = 46.6374 MPa
2 (
2 [ b
1 −ν
1 − ( 0.30 )
σ −45
σ 45 = σ x cos 2 ( 45° ) + σ y sin 2 ( 45° ) + 2τ xy sin ( 45° ) cos ( 45° )
σ −45 = σ x cos 2 ( −45° ) + σ y sin 2 ( −45° ) + 2τ xy sin ( −45° ) cos ( −45° )
For a shaft subjected to an axial load P and a torque T :
σ y = 0 N/m 2
0.50σ x + τ xy = 296.7912 MPa
Therefore
0.50σ x − τ xy = 46.6374 MPa
σ x = 343.4286 MPa
Solving yields
P = σ x A = ( 343.4286 × 10
T=
τ xy J
c
6
)
π ( 0.050 )
τ xy = 125.0769 MPa
2
4
(125.0769 ×10 ) π ( 0.050 )
=
6
32 ( 0.025 )
4
= 674 (103 ) N = 674 kN ........................... Ans.
= 3.07 (103 ) N ⋅ m = 3.07 kN ⋅ m ............. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-89
(a)
RILEY, STURGES AND MORRIS
σx =
pr ( 360 )( 24 )
=
= 14, 400 psi = 14.400 ksi
t
( 0.6 )
σy =
pr P 14.400
28.000
− =
−
= 5.97741 ksi
2t A
2
π ( 24.62 − 242 ) 4
τ xy =
−Tc − ( 550 ×12 )( 24.6 )
=
= −48.01566 ksi
J
π ( 24.64 − 244 ) 32
θ = − tan −1 ( 4 3) = −53.130°
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= (14.4 ) cos 2 ( −53.130° ) + ( 5.97741) sin 2 ( −53.130° )
+ 2 ( −48.01566 ) sin ( −53.130° ) cos ( −53.130° )
σ n = +55.1 ksi = 55.1 ksi (T) ............................................................................................. Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos2 θ − sin 2 θ )
= − (14.4 ) − ( 5.97741)  sin ( −53.130° ) cos ( −53.130° )
+ ( −48.01566 )  cos 2 ( −53.130° ) − sin 2 ( −53.130° ) 
τ nt = 17.49 ksi
(b)
1
2
θ p = tan −1
σ p1, p 2 =
........................................................................................................................ Ans.
2τ xy
σ x −σ y
σx +σ y
2
=
2 ( −48.01566 )
1
= −42.494°, + 47.506°
tan −1
2
(14.4 ) − ( 5.97741)
 σ x −σ y 
2
± 
 + τ xy
 2 
2
(14.4 ) + ( 5.97741) ±
=
2
 (14.4 ) − ( 5.97741) 
2

 + ( −48.01566 )
2


2
σ p1 = (10.18871) + ( 48.19999 ) = 58.3887 ksi
σ p1 ≅ 58.4 ksi (T)
42.494° .................................................................................... Ans.
σ p 2 = (10.18871) − ( 48.19999 ) = −38.0113 ksi
σ p 2 ≅ 38.0 ksi (C)
47.506° ................................................................................... Ans.
σ p 3 = − p = − ( 0.360 ) ksi = 0.360 ksi (C) ............................................................... Ans.
τ max =
σ max − σ min
2
=
( 58.3887 ) − ( −38.0113) = 48.2 ksi .................................... Ans.
2
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-90
A=
π
4
( do2 − di2 ) =
π d s2
4
J = π d 32 = π ( d − d
4
4
o
=
4
i
)
π ( 0.120 )
2
4
di = d o2 − ( 0.120 )
2
32
250 (103 )
P
σx = =
= 22.10485 (106 ) N/m 2
2
A π ( 0.120 ) 4
3
−Tc − ( 20, 000 )( d o 2 ) −101.85916 (10 ) d o
τ xy =
=
=
N/m 2
4
4
4
4
J
π ( d o − di ) 32
( d o − di )
σ p1 =
σx
2
σ 
+  x  + (τ xy )
2
 2 
2
2
σ 
τ max =  x  + (τ xy )
 2 
2
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-91
RILEY, STURGES AND MORRIS
2π NT 2π (1500 ) T
=
= ( 360 × 550 ) lb ⋅ ft/s
60
60
T = 1260.50715 lb ⋅ ft = 15,126.09 lb ⋅ in.
Power = T ω =
σx =
P 2800 3565.071
=
=
psi
A πd2 4
d2
τ xy =
−Tc − (15,126.09 )( d 2 ) −77, 036.54
=
=
psi
π d 4 32
J
d3
σ p1 =
σx
2
σ 
+  x  + (τ xy )
2
 2 
2
2
σ 
τ max =  x  + (τ xy )
 2 
2
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-92*
RILEY, STURGES AND MORRIS
r d = 12 100 = 0.12
D d = 135 100 = 1.35
Therefore, from Fig. 6-25b
K t ≅ 1.6
J = π d 4 32 = π (100 ) 4 32 = 9.81748 (106 ) mm 4
τ max = K t
(10, 000 )( 0.050 )
Tc
= (1.6 )
J
( 9.81748 ×10−6 )
τ max = 81.5 (106 ) N/m 2 = 81.5 MPa .......................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-93*
RILEY, STURGES AND MORRIS
r d = 0.15 3 = 0.05
D d = 4 3 = 1.3333
K t ≅ 2.0
Therefore, from Fig. 6-25b
J = π d 4 32 = π ( 3) 32 = 7.95216 in.4
4
τ max = K t
( 4 ×12 )(1.5) = 18.11 ksi ................................................................... Ans.
Tc
= ( 2)
J
( 7.95216 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-94
τ = Kt
RILEY, STURGES AND MORRIS
( 614 )( 0.025) = 40 106 N/m 2
Tc
= Kt
( )
J
π ( 0.050 )4 32 


K t = 1.60
Therefore, from Fig. 6-25b
r d = 4.5 50 = 0.09
D d ≅ 1.2
D = 1.2d = 1.2 ( 50 ) = 60 mm ............................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-95
r d = 0.125 6 = 0.021
Therefore, from Fig. 6-25b
τ max = Kt
RILEY, STURGES AND MORRIS
D d = 8 6 = 1.3333
K t ≅ 2.6
T ( 3)
Tc
= ( 2.6 )
= 12 ksi
4
J
π ( 6 ) 32
T = 195.7 kip ⋅ in. ............................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-96*
h r = 5 5 =1
Therefore, from Fig. 6-25a
τ max = K t
RILEY, STURGES AND MORRIS
r d = 5 100 = 0.05
K t ≅ 1.85
T ( 0.050 )
Tc
= (1.85 )
= 60 (106 ) N/m 2
4
J
π ( 0.100 ) 32
T = 6.37 (103 ) N ⋅ m = 6.37 kN ⋅ m ............................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-97
K t = τ max τ nom = 12 8 = 1.50
Therefore, from Fig. 6-25b
RILEY, STURGES AND MORRIS
D d = 5 4 = 1.25
r d ≅ 0.125
r = 0.125d = 0.125 ( 4 ) = 0.50 in. ...................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-98*
τ max = K t
RILEY, STURGES AND MORRIS
( 3270 )( d 2 ) = 60 106 N/m 2
Tc
= Kt
( )
J
π d 4 32
d 3 = 277.56622 (10−6 ) K t
Guess
K t ≅ 2.0
Then Eq. (a)
r d = 5 82.19 = 0.061
and from Fig. 6-25b
2nd guess
K t ≅ 1.8
d = 0.08219 m
D d = 100 82.19 = 1.22
K t ≅ 1.8
Then Eq. (a)
r d = 5 79.35 = 0.063
and from Fig. 6-25b
(a)
d = 0.07935 m
D d = 100 79.35 = 1.26
K t ≅ 1.8
Therefore, the 2nd guess was correct, and
d = 79 mm ..................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-99
h r = 0.5 0.25 = 2
Therefore, from Fig. 6-25a
τ max = K t
RILEY, STURGES AND MORRIS
r d = 0.25 1 = 0.25
K t ≅ 1.65
( 500 )( 0.5 ) = 4201.69 psi = σ y
Tc
= (1.65 )
4
J
3
π (1) 32
σ y = 12, 605.1 ksi ≅ 12.61 ksi
..................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-100*
(a)
τ=
RILEY, STURGES AND MORRIS
T ( 0.050 )
Tc
=
= 120 (106 ) N/m 2
4
J π ( 0.100 ) 32
T = 23.5619 (103 ) N ⋅ m ≅ 23.6 kN ⋅ m .................................Ans.
(b)
Plastic zone
0≤ρ ≤c
T = ∫ ρ (τ dA ) = ∫
0.050
0
τ ρ = τ y = 120 MPa
ρ (120 ×106 ) ( 2πρ ) d ρ
 0.0503 
3
= 240π (106 ) 
 = 31.4159 (10 ) N ⋅ m
3


= 31.4159 kN ⋅ m
31.4159 − 23.5619
% Inc =
(100 ) = 33.3% .................................................................... Ans.
23.5619
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-101*
τ yρ
24 ρ
ksi
ry
Elastic zone
0 ≤ ρ ≤ ry
τρ =
Plastic zone
ry ≤ ρ ≤ c
τ ρ = τ y = 24 ksi
(a)
ry
=
ry = 1.5 in.
2
24 ρ
2πρ ) d ρ + ∫ ρ ( 24 )( 2πρ ) d ρ
(
0
1.5
1.5
4
3
3
 2 − 1.5 
48π 1.5 
=
+ ( 48π ) 


1.5  4 
3 

T = ∫ ρ (τ dA ) = ∫ ρ
1.5
T = 360 kip ⋅ in. ...................................................................................................................... Ans.
(b)
ry = 1.0 in.
2
24 ρ
2πρ ) d ρ + ∫ ρ ( 24 )( 2πρ ) d ρ
(
0
1
1
 23 − 13 
48π 14 
=
+
π
48
(
)
 


1 4
 3 
T = ∫ ρ (τ dA ) = ∫ ρ
1
T = 390 kip ⋅ in. ...................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-102
τ yρ
140 (106 ) ρ
Elastic zone
0 ≤ ρ ≤ ry
τρ =
Plastic zone
ry ≤ ρ ≤ c
τ ρ = τ y = 140 (106 ) N/m 2
(a)
ry
=
ry
N/m 2
ry = 40 mm
T = ∫ ρ (τ dA ) = ∫
0.040
0.025
0.040
( 2πρ ) d ρ + ∫0.040 ρ 140 (106 ) ( 2πρ ) d ρ
0.050
280 (10 ) π  0.040 − 0.0254 
 0.0503 − 0.0403 
6
π
+
280
10
( ) 


0.040 
4
3



6
=
ρ
140 (106 ) ρ
4
T = 29.8 (103 ) N ⋅ m = 29.8 kN ⋅ m ................................................................................... Ans.
(b)
ry = 25 mm
T = ∫ ρ (τ dA ) = ∫
0.050
0.025
ρ 140 (106 )  ( 2πρ ) d ρ
 0.0503 − 0.0253 
= 280 (106 ) π 

3


T = 32.1(103 ) N ⋅ m = 32.1 kN ⋅ m .................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-103*
(a)
(b)
c
r
c  τ y  1.50  24 
= 0.004 = 4000 µ rad .......................................... Ans.
=
r  G  0.75  12, 000 
γc = γ = 
Elastic zone
0 ≤ ρ ≤ 0.75 in.
τ ρ = τ y ρ ry = 24 ρ 0.75 = 32.00 ρ ksi
Plastic zone
0.75 in. ≤ ρ ≤ 1.5 in.
τ ρ = τ y = 24 ksi
T = ∫ ρ (τ dA ) = ∫
0.75
0
ρ ( 32 ρ )( 2πρ ) d ρ + ∫
1.50
0.75
ρ ( 24 )( 2πρ ) d ρ
 0.754 
1.503 − 0.753 
= ( 64π ) 
+
48
π
(
)



3
 4 


T = 164.3 kip ⋅ in. ................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-104
(a)
(b)
γ = rθ L
θ = 8° = 0.13963 rad
γ max =
γy =
τy
=
G
RILEY, STURGES AND MORRIS
cθ 0.050 ( 0.13963)
=
= 0.002327 ≅ 2330 µ rad ....................................... Ans.
L
3
140
= 0.00175 rad
80, 000
ry =
γ r L 0.00175 ( 3)
=
= 0.03760 m
θ
0.13963
τ yρ
140 ρ
= 3723.40 ρ MPa
0.03760
Elastic zone
0.025 m ≤ ρ ≤ 0.03760 m
τρ =
Plastic zone
0.03760 m ≤ ρ ≤ 0.050 m
τ ρ = τ y = 140 MPa
T = ∫ ρ (τ dA ) = ∫
0.0376
0.025
ry
=
ρ 3723.40 (106 ) ρ  ( 2πρ ) d ρ + ∫
0.050
0.0376
ρ 140 (106 ) ( 2πρ ) d ρ
= 5848.71(106 ) 0.03764 − 0.0254  + 293.215 (106 ) 0.0503 − 0.03763 
T = 30.5 (103 ) N ⋅ m = 30.5 kN ⋅ m ................................................................................... Ans.
(c)
Elastic zone since 25 mm < 37.6 mm
τρ =
ρτ y
ry
=
25 (140 )
= 93.1 MPa ...................................................................................... Ans.
37.6
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MECHANICS OF MATERIALS, 6th Edition
6-105
(a)
θ = 5° = 0.0872665 rad
γ max =
(b)
γy =
τy
G
=
RILEY, STURGES AND MORRIS
γ = rθ L
cθ 2 ( 0.0872665 )
=
= 0.0029089 ≅ 2910 µ rad ........................................ Ans.
L
( 5 ×12 )
18
= 0.00150 rad
12, 000
ry =
γ r L 0.001500 ( 5 × 12 )
=
= 1.03132 in.
θ
0.0872665
τ yρ
18 ρ
= 17.45330 ρ ksi
1.03132
Elastic zone
0 in. ≤ ρ ≤ 1.03132 in.
τρ =
Plastic zone
1.03132 in. ≤ ρ ≤ 2.0 in.
τ ρ = τ y = 18 ksi
T = ∫ ρ (τ dA ) = ∫
1.03132
0
=
ry
ρ (17.45330 ρ )( 2πρ ) d ρ + ∫
2.0
1.03132
ρ (18 )( 2πρ ) d ρ
= ( 27.41558 ) 1.031324  + ( 37.69911)  2.03 − 1.031323 
T = 291 kip ⋅ in. ...................................................................................................................... Ans.
(c)
Elastic zone since 0.50 in. < 1.03132 in.
τρ =
ρτ y
ri
=
( 0.50 )(18) = 8.73 ksi .................................................................................... Ans.
1.03132
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-106*
(a)
RILEY, STURGES AND MORRIS
γ y ro 7.5 ( 25 )
=
= 15 mm
γo
12.5
ry =
γρ =
G1 =
ργ o
ro
=
0.0125ρ
= 0.500 ρ rad
0.025
210 (106 )
0.0075
= 28.00 (109 ) N/m 2
0 ≤ γ ≤ 0.0075 rad
G2 =
( 230 − 210 ) (106 )
0.0125 − 0.0075
= 4.00 (109 ) N/m 2
0.0075 rad ≤ γ ≤ 0.0125 rad
τ ρ 1 = G1γ ρ = 28 (109 ) ( 0.5ρ ) = 14.00 (109 ) ρ  N/m 2
0 ≤ ρ ≤ 15 mm
τ ρ 2 = 180 (106 ) + G2γ ρ = 180 (106 ) + 2.00 (109 ) ρ  N/m 2
15 mm ≤ ρ ≤ 25 mm
T = ∫ ρ (τ dA ) = ∫
0.015
0
ρ 14 (109 ) ρ  ( 2πρ ) d ρ
+∫
0.025
0.015
ρ 180 (106 ) + 2 (109 ) ρ  ( 2πρ ) d ρ
= 21,991.15 (106 ) 0.0154  + 376.9911(106 )  0.0253 − 0.0153 
+ 3141.5927 (106 )  0.0254 − 0.0154 
T = 6.80 (103 ) N ⋅ m = 6.80 kN ⋅ m ................................................................................... Ans.
(b)
θ=
γ yL
ry
=
0.0075 (1)
= 0.500 rad ≅ 28.6° ...................................................................... Ans.
0.015
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MECHANICS OF MATERIALS, 6th Edition
6-107*
(a)
RILEY, STURGES AND MORRIS
roγ i ( 2.00 )( 0.0075 )
=
= 0.0120 rad < 0.0125 rad
ri
(1.25)
ργ 0.0075ρ
γρ = i =
= 0.00600 ρ rad
ri
1.25
γo =
G2 =
33 − 30
= 600 ksi
0.0125 − 0.0075
0.0075 rad ≤ γ ≤ 0.0125 rad
τ ρ = 25.5 + G2γ ρ = [ 25.5 + 3.600 ρ ] ksi
T = ∫ ρ (τ dA ) = + ∫
2.0
1.25
1.25 in. ≤ ρ ≤ 2.0 in.
ρ [ 25.5 + 3.6 ρ ] ( 2πρ ) d ρ
= 53.40708  2.03 − 1.253  + 5.65487  2.04 − 1.254 
T = 400 kip ⋅ in. ...................................................................................................................... Ans.
(b)
θ=
γ yL
ry
=
0.0075 ( 3 × 12 )
= 0.216 rad ≅ 12.4° .............................................................. Ans.
1.25
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-108
γ=
ry =
θρ
L
=
RILEY, STURGES AND MORRIS
0.300 ρ
= 0.150 ρ rad
2
γ y ρ ( 0.0035 ) ρ
=
= 0.023333 m = 23.333 mm
γρ
( 0.150 ρ )
τ = 2910γ 0.74 = 2910 ( 0.150 ρ )
τ = 533γ 0.44 = 533 ( 0.150 ρ )
T = ∫ ρ (τ dA ) = ∫
0.02333
0
0.74
0.44
0 ≤ ρ ≤ 23.333 mm
=  714.826 ρ 0.74  MPa
=  231.317 ρ 0.44  MPa
23.333 mm ≤ ρ ≤ 40 mm
ρ  714.826 (106 ) ρ 0.74  ( 2πρ ) d ρ
+∫
0.040
0.02333
ρ  231.317 ρ 0.44 (106 )  ( 2πρ ) d ρ
= 1200.905 (106 ) 0.023333.74  + 422.502 (106 )  0.0403.44 − 0.023333.44 
T = 6.48 (103 ) N ⋅ m = 6.48 kN ⋅ m ................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-109
T = ∫ ρ (τ dA ) = ∫ ρ  kγ 1 2  ( 2πρ ) d ρ
0
R
(a)
T =∫
R
0
Solving for
(b)
θ
 ρθ 
ρk  
 L 
12
yields
γ max =
γ = ρθ L
4π k 7 2
θ
R
( 2πρ ) d ρ = 2π k θ L ∫0 ρ 5 2 d ρ =
7
θ=
R
L
49 LT 2
............................................................................................ Ans.
16π 2 k 2 R 7
Rθ
49T 2
=
L 16π 2 k 2 R 6
12
τ max = kγ
12
max
 49T 2 
=k
2 2 6
16π k R 
=
7T
................................................................... Ans.
4π R 3
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-110*
τ b ( max ) = 60 MPa ≤ 84 MPa
RILEY, STURGES AND MORRIS
(all elastic)
J b = π d 4 32 = π ( 80 ) 32 = 4.02124 (106 ) mm 4
4
Tb =
τ max J
c
θb = θ B / C
( 60 ×10 )( 4.02124 ×10 ) = 6.03186 10 N ⋅ m = 6.03186 kN ⋅ m
=
( )
( 0.040 )
−6
6
3
60 ×106 ) ( 2.5 )
(
τL 
=
= 0.083333 rad = θ s
 =
9
 cG b ( 0.040 ) ( 45 × 10 )
γ s ( max ) =
cθ ( 0.040 )( 0.083333)
=
= 0.002222 rad > 0.0015 rad
L
1.5
Therefore, part of the steel shaft will be in plastic deformation,
and the yield surface location is located using
γ y L ( 0.0015 )(1.5 )
=
= 0.02700 m = 27.00 mm
θ
0.083333
τ ρ 120 ρ
Elastic zone
0 m ≤ ρ ≤ 0.027 m
= 4444.44 ρ MPa
τρ = y =
ry =
ry
Plastic zone
0.027 m ≤ ρ ≤ 0.040 m
Ts = ∫ ρ (τ dA ) = ∫
0.027
0
0.0270
τ ρ = τ y = 120 MPa
ρ  4444.44 ρ (106 )  ( 2πρ ) d ρ + ∫
0.040
0.027
ρ 120 (106 )  ( 2πρ ) d ρ
= 6981.32 (106 ) 0.02704  + 251.327 (106 ) 0.0403 − 0.02703 
T = 14.8482 (103 ) N ⋅ m = 14.8482 kN ⋅ m
T = Tb + Ts = 6.03186 + 14.8482 = 20.8801 kN ⋅ m ≅ 20.9 kN ⋅ m ............................ Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-111*
γ b ( max ) =
RILEY, STURGES AND MORRIS
cθ (1)( 0.30 )
=
= 0.00500 rad
L
60
τ b ( max ) = Gγ b ( max ) = 5000 ( 0.005 ) = 25.0 ksi ≤ 30 ksi
(all elastic)
J b = π d 4 32 = π ( 2 ) 32 = 1.57080 in 4
4
( 25.0 )(1.57080 ) = 39.2700 kip ⋅ in.
c
(1)
rθ ( 2 )( 0.30 )
= 0.01000 rad
γ s ( min ) = i =
Tb =
τ max J
=
L
60
Therefore, all of the steel shaft will be in plastic deformation, and
τ s = τ y = 18 ksi
Ts = ∫ ρ (τ dA ) = ∫ ρ [18] ( 2πρ ) d ρ = 37.69911 33 − 23  = 716.283 kip ⋅ in.
3
2
T = Tb + Ts = 39.270 + 716.283 = 755.55 kip ⋅ in. ≅ 756 kip ⋅ in. ................................ Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-112
(a)
RILEY, STURGES AND MORRIS
J AB = π d 4 32 = π (160 ) 32 = 64.3398 (106 ) mm 4
4
4 ΣM = 0 :
τ AB =
200 − 125 − TAB = 0
TAB = +75 kN ⋅ m
Tc ( 75, 000 )( 0.080 )
=
= 93.2549 (106 ) N/m 2 < 120 MPa (elastic)
−6
J
( 64.3398 ×10 )
τ AB ≅ 93.3 MPa
.............................................................................................................. Ans.
TBC = 125 kN ⋅ m > TAB = 75 kN ⋅ m (assume steel is fully plastic)
Ts = ∫ ρ (τ dA ) = ∫
0.080
0.050
ρ 120 (106 )  ( 2πρ ) d ρ = 251.3274 (106 ) 0.0803 − 0.0503 
= 97.2637 (103 ) N ⋅ m = 97.2637 kN ⋅ m
Tb = TBC − Ts = 125 − 97.2637 = 27.7363 kN ⋅ m
J b = π d 4 32 = π (100 ) 32 = 9.81748 (106 ) mm 4
4
τb =
Tc ( 27, 736.3)( 0.050 )
=
= 141.2598 (106 ) N/m 2 < 240 MPa (elastic)
−6
J
( 9.81748 ×10 )
τ b ≅ 141.3 MPa ............................................................................................................... Ans.
At r = 50 mm
For the steel:
γs = γb =
γy =
τy
Gb
τb
Gb
=
=
141.2598 (106 )
40 (109 )
120 (106 )
80 (109 )
= 0.0035315 rad
= 0.001500 rad < 0.0035315 rad
Therefore, the steel is fully plastic in AB as assumed and
Ts ≅ 97.3 kN ⋅ m
(b)
τ s = τ y = 120 MPa ............................................................... Ans.
Since the steel in AB is elastic and the bronze in BC is elastic:
 TAB LAB   Tb LBC 
+
+0
 J AB Gs   J b Gb 
( 75, 000 )( 2 )
( 27, 736.3)(1.5)
=
+
9
−6
( 64.3398 ×10 )(80 ×10 ) ( 9.81748 ×10−6 )( 40 ×109 )
θ D / A = θ B / A + θC / B + θ D / C = 
θ D / A = −0.0769 rad ............................................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-113
(a) For section AB (steel)
γy =
τy
Gs
γ AB =
=
18
= 0.001500 rad
12, 000
rABθ B ( 2 )( 0.072 )
=
LAB
( 4 ×12 )
= 0.003 rad > 0.001500 rad (steel yields)
ry =
γ y LAB ( 0.0015 )( 4 ×12 )
=
= 1.00 in.
θB
( 0.072 )
τ yρ
18 ρ
= 18 ρ ksi
1.00
Elastic zone
0 in. ≤ ρ ≤ 1.00 in.
τρ =
Plastic zone
1.00 in. ≤ ρ ≤ 2.0 in.
τ ρ = τ y = 18 ksi
ry
=
TA = −TAB = ∫ ρ (τ dA ) = ∫ ρ (18 ρ )( 2πρ ) d ρ + ∫ ρ (18 )( 2πρ ) d ρ
1.0
2.0
0
1.0
= ( 28.2743) 1.0  + ( 37.69911)  2.0 − 1.0  = 292.168 kip ⋅ in.
4
3
3
In section BD assume elastic action since τ yb = 35 ksi ?
4 ΣM = 0 :
τ ys = 18 ksi
J = π d 4 32 = π ( 4 ) 32 = 25.1327 in.4
TA + TD − 2T = 0
4
TD = 2T − TA = 2T − 292.168 kip ⋅ in.
 ( 2T − TA ) LCD   (T − TA ) LBC 
+
 − 0.072 rad
JGs
JGs

 

θ D / A = 0 = θ D / C + θC / B + θ B / A = 
( 2T − 292.168 )( 3 ×12 ) + (T − 292.168)( 4 ×12 ) − 0.072 = 0
( 25.1327 )( 6000 )
( 25.1327 )(12, 000 )
T = 295.702 kip ⋅ in. ≅ 296 kip ⋅ in. ................................................................................... Ans.
(b)
TBC = T − TA = 295.702 − 292.168 = 3.534 kip ⋅ in.
τs =
Tc ( 3.534 )( 2 )
=
= 0.2812 ksi < τ ys = 18 ksi (elastic)
J
( 25.1327 )
TCD = 2T − TA = 2 ( 295.702 ) − 292.168 = 299.236 kip ⋅ in.
τb =
Therefore
Tc ( 299.236 )( 2 )
=
= 23.812 ksi < τ b = 35 ksi (elastic)
J
( 25.1327 )
τ max ( steel ) = τ y = 18 ksi (in section AB )
........................................................ Ans.
τ max ( bronze ) = 23.8 ksi (in section CD) ......................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
6-114
RILEY, STURGES AND MORRIS
τ AB = τ b = 50 MPa < τ yb = 60 MPa (elastic)
τ BC = τ s = 50 MPa < τ ys = 120 MPa (elastic)
J = π d 4 32 = π (100 ) 32 = 9.81748 (106 ) mm 4
4
τbJ
TA = TAB =
c
( 50 ×10 )( 9.81748 ×10 )
=
−6
6
( 0.050 )
= 9817.48 N ⋅ m = 9.81748 kN ⋅ m
θ D / A = 0 = θ D / C + θC / B + θ B / A
 (T ) L   (T ) L 
= θ D / C +  A BC  +  A AB 
 JGs   JGb 
( 9,817.48)( 0.600 )
θD/C +
+
( 9,817.48)( 0.600 )
( 9.81748 ×10 )(80 ×10 ) ( 9.81748 ×10 )( 40 ×10 )
−6
9
−6
9
=0
θ D / C = −0.02250 rad
For segment CD:
γ ys =
τ ys
γ CD =
ry =
Gs
=
120 (106 )
80 (109 )
= 0.001500 rad
rθCD ( 50 )( 0.0225 )
=
= 0.001875 rad > 0.001500 rad (part plastic)
LCD
( 600 )
γ y LCD ( 0.0015 )( 0.600 )
=
= 0.0400 m = 40.0 mm
θC
( 0.0225)
τ yρ
120 ρ
= 3000 ρ MPa
0.040
Elastic zone
0 in. ≤ ρ ≤ 40 mm
τρ =
Plastic zone
40 mm ≤ ρ ≤ 50 mm
τ ρ = τ y = 120 MPa
TD = −TCD = ∫ ρ (τ dA ) = ∫
0.040
0
ry
=
ρ 3000 ρ (106 )  ( 2πρ ) d ρ + ∫
0.050
0.040
ρ 120 (106 )  ( 2πρ ) d ρ
= 4712.3889 (106 ) 0.0404  + 251.3274 (106 ) 0.0503 − 0.0403 
= 27.3947 (103 ) N ⋅ m = 27.3947 kN ⋅ m
T = TA + TD = 9.81748 + 27.3947 = 37.2 kN ⋅ m ............................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-115
J = π d 4 32 = π ( 4 ) 32 = 25.13274 in.4
τ y = Ty c J
4
Ty =
τ max J
c
=
(18)( 25.13274 ) = 226.1947 kip ⋅ in. = 18.8496 kip ⋅ ft
( 2)
For 0 ≤ T ≤ 18.85 kip ⋅ ft
τ=
Tc (T × 12 )( 2 )
=
= 0.95493T ksi
J ( 25.13274 )
θ=
(T ×12 )(10 ×12 )
TL
=
JG ( 25.13274 )(12, 000 )
= 4.77465 (10−3 ) T rad
For 18.85 kip ⋅ ft ≤ T ≤ 25 kip ⋅ ft
2.0
18 ρ 
 ( 2πρ ) d ρ + ∫r ρ (18 )( 2πρ ) d ρ
 r 
(T ×12 ) = ∫ ρ (τ dA) = ∫0 ρ 
r
=
9π 4
 r  + 12π  2.03 − r 3  = 96π − 3π r 3
r  
ry = 3 32 − ( 4T π )
θ=
γ yL
ry
=
τ yL
Gry
=
τ max = τ y = 18 ksi
(18)(10 ×12 ) = 0.1800
ry
(12, 000 ) ry
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
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MECHANICS OF MATERIALS, 6th Edition
6-116
RILEY, STURGES AND MORRIS
J = π d 4 32 = π (120 ) 32 = 20.35752 (106 ) mm 4
τ y = Ty c J
4
Ty =
τ max J
c
( 44.3 ×10 )( 20.35752 ×10 ) = 15, 030.636 N ⋅ m
=
−6
6
( 0.060 )
For 0 ≤ T ≤ 15.03 kN ⋅ m
τ=
T ( 0.060 )
Tc
=
= 2947.314T N/m 2
J 20.35752 (10−6 )
θ=
T ( 3)
TL
=
JG ( 20.35752 × 10−6 )(16.66 × 109 )
= 8.84548 (10−6 ) T rad
θ = 0.13295 rad ≅ 7.618°
When T = 15, 030.636 N ⋅ m
For 7.618° ≤ θ
( 44.3 ×10 ) ( 3) = 7.97719 (10 )
θ=
=
=
r
Gr
r
(16.66 ×10 ) r
7.97719 (10 )
7.97719
r =
m=
mm
γ yL
τ yL
−3
6
9
y
y
y
y
−3
y
θ
θ
 ρθ 

 L 
0.44
τ max = 602γ 0.44 = 602 
 0.060θ 
= 602 

 3 
0.44
MPa
r
0.060
 44.3 × 106 ρ 
 ρθ 
T = ∫ ρ (τ dA ) = ∫ ρ 
ρ ( 602 ×106 ) 
2πρ ) d ρ + ∫
(


0
r
r
 3 


θ 
= 22.15 (10 ) π r + 350 (10 ) π  
3
6
3
y
6
0.44
( 2πρ ) d ρ
0.44
 0.0603.44 − ry3.44  N ⋅ m
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-117
J s = π ( 0.5 ) 32 = 0.00613592 in 4
J a = π (1) 32 = 0.09817477 in 4
4
Tya =
Tys =
4
( 30, 000 )( 0.09817477 ) = 5890.49 lb ⋅ in. = 490.874 lb ⋅ ft
( 0.5)
(18, 000 )( 0.00613592 ) = 441.786 lb ⋅ in. = 36.8155 lb ⋅ ft
=
( 0.25)
τ max J a
=
ca
τ max J s
cs
θ = TL JG
For 0 ≤ T ≤ Ty
(Ta ×12 )(10 )
= 0.3055775 (10−3 ) Ta rad
( 0.09817477 )( 30, 000 0.0075)
(Ts ×12 )( 30 )
θs =
= 4.88924 (10−3 ) Ts rad
( 0.00613592 )(18, 000 0.0015 )
θa =
θa = θs
T = Ta + Ts
Ta = 16.000Ts
T = 17.000Ts
If Ta = Tya = 5890.49 lb ⋅ in. , then
(1)
(2)
Ts = 368.1556 lb ⋅ in.
Tmax = Ta + Ts = 6258.6456 lb ⋅ in. = 521.5538 lb ⋅ ft
If Ts ≥ Tys (where r is the yield boundary)
0.25
18, 000 ρ 
 ( 2πρ ) d ρ + ∫r ρ (18, 000 )( 2πρ ) d ρ
r


9000π 4
 r  + 12, 000π  0.253 − r 3  = 187.50π − 3000π r 3
=
r
(Ts ×12 ) = ∫0 ρ 
r
r = 3 62.500 − ( 0.004Ts π )
θs =
τ max L
Gr
=
(18 )( 30 )
(18 0.0015) r
(3)
τ max = τ y = 18 ksi
=
0.04500
rad
r
(4)
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-117 (cont.)
If Ta ≥ Tya (where r is the yield boundary)
τ a = 30 ρ r
τ a = 30 +
τ ≤ 30 ksi
3 ( γ − 0.0075 )
4.5ρ
= 25.5 +
r
0.0125 − 0.0075
τ ≥ 30 ksi
ρ≤r
ρ≥r
4500 ρ 
0.50
30, 000 ρ 

 ( 2πρ ) d ρ + ∫r ρ  25,500 +
 ( 2πρ ) d ρ
r
r 



15, 000π 4
2250π
 r  + 17, 000π  0.503 − r 3  +
 0.504 − r 4 
=
r
r 
140.625π
= 2125.0π +
− 4250π r 3
r
τ L
( 30 )(10 ) = 0.07500 rad
θ a = max =
Gr
r
( 30 0.0075) r
(Ta ×12 ) = ∫0 ρ 
r
(5)
(6)
Computer approach:
θ = θa = θs
1.
Increment
2.
Compute Ts and Ta using Eqs. (1) and (2)
If Ts ≥ 36.82 lb ⋅ ft ,
use Eqs. (3) and (4).
If Ta ≥ 490.9 lb ⋅ ft ,
use Eqs. (5) and (6).
3.
Compute T = Ta + Ts
4.
Plot
5.
Repeat until T = 750 lb ⋅ ft
θ
versus T
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
6-118
RILEY, STURGES AND MORRIS
J a = π ( 20 ) 32 = 15.70796 (103 ) mm 4
4
J s = π ( 25 ) 32 = 38.34952 (103 ) mm 4
4
τ max J a
Tya =
ca
τ max J s
Tys =
cs
θs =
−9
( 0.010 )
(120 ×10 )( 38.34952 ×10 ) = 368.155 N ⋅ m
=
−9
6
( 0.0125)
θ = TL JG
For 0 ≤ T ≤ Ty
θa =
( 210 ×10 )(15.70796 ×10 ) = 329.867 N ⋅ m
6
=
(T )( 0.250 )
(15.70796 ×10 )( 210 ×10
−9
6
(T )( 0.300 )
( 38.34952 ×10 )(120 ×10
−9
6
0.0075 )
0.0015 )
= 568.411(10−6 ) T rad
= 97.7848 (10−6 ) T rad
If T ≥ Tys (where r is the yield boundary)
T =∫
r
0
=
120 (106 ) ρ 
c
 ( 2πρ ) d ρ + ∫ ρ (120 ×106 ) ( 2πρ ) d ρ
ρ
r
r


60 (106 ) π
r
r = 3 4c 3 −
θs =
τ max L
Gr
 r 4  + 80 (106 ) π  c3 − r 3  = 20 (106 ) π ( 4c3 − r 3 )
T
20π (106 )
where
c = 0.0125 m
(120 ×10 ) ( 0.300 ) = 450 (10 ) rad
=
r
(120 ×10 0.0015) r
(1)
−6
6
6
224
(2)
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-118 (cont.)
If T ≥ Tya (where r is the yield boundary)
(ρ ≤ r)
(ρ ≥ r)
τ ≤ 210 MPa
τ ≥ 210 MPa
τ a = 210 ρ r
τ a = 210 + 4000 ( γ − 0.0075 ) = 180 + 4000γ
30 ρ
 0.0075ρ 
= 180 + 4000 
 = 180 +
r
r


r
c
30 ρ  6
 210 ρ  6

T = ∫ ρ
 (10 ) ( 2πρ ) d ρ + ∫r ρ 180 +
 (10 ) ( 2πρ ) d ρ
0
r 
 r 

=
105 (106 ) π
r
 r  + 120 (10 ) π c − r  +
4
6
3
3
15 (106 ) π
r
 c 4 − r 4 


7.5c 4
where c = 0.010 m
= 2π (106 ) 60c 3 +
− 15r 3 
r


For a given torque T , solve for r . For example, using the Newton-Raphson iteration method, let
Tr
+ 15r 4 − 7.5c 4 − 60c 3 r = 0
f (r ) =
6
2π (10 )
f ′(r ) =
Then
T
+ 60r 3 − 60c 3
6
2π (10 )
r ( n +1) = r ( n ) − ( f f ′ )
( 0)
Guess that r = c and iterate until the value for
r no longer changes.
Using the r from the Newton-Raphson solution, calculate
θa =
τ max L
Gr
( 210 ×10 ) ( 0.250 ) = 1.875 (10 ) rad
=
r
( 210 ×10 0.0075) r
Use Eqs. (1) and (2) to determine
−3
6
6
θs
and Eq. (3) to determine
θ = θa + θs
225
θa .
Then,
(3)
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-119*
For the square bar:
The value of
α
b a =1 1=1
α ≅ 0.21
from Fig. 6-30 is:
Tmax = τ maxα a 2b = (12 )( 0.21)(1) (1) = 2.52 kip ⋅ in. ..................................................... Ans.
2
226
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-120*
(a) For the rectangular bar:
The value of
α
b a = 75 40 = 1.875
α ≅ 0.245
from Fig. 6-30 is:
Tmax = τ maxα a 2b = ( 25 × 106 ) ( 0.245 )( 0.040 ) ( 0.075 ) = 735 N ⋅ m .......................... Ans.
2
b a = a a =1 1=1
For the square bar:
The value of
α
and
a=
α ≅ 0.21
from Fig. 6-30 is:
( 40 )( 75) = 54.77 mm
Tmax = τ maxα a 2b = ( 25 × 106 ) ( 0.21)( 0.05477 ) ( 0.05477 ) = 863 N ⋅ m ................... Ans.
2
(b)
β ≅ 0.225
For the rectangular bar:
θ=
( 735)( 0.400 )
TL
=
= 0.00972 rad ............................ Ans.
3
β a bG ( 0.225 )( 0.040 )3 ( 0.075 ) ( 28 ×109 )
β ≅ 0.15
For the square bar:
θ=
(863)( 0.400 )
TL
=
= 0.00913 rad
3
β a bG ( 0.15 )( 0.05477 )3 ( 0.05477 ) ( 28 ×109 )
224
Ans.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-121
(a) For the square bar:
The value of
α
b a = 1.5 1.5 = 1
α ≅ 0.21
from Fig. 6-30 is:
Tmax = τ maxα a 2b = (12 )( 0.21)(1.5 ) (1.5 ) = 8.505 kip ⋅ in. ≅ 8.51 kip ⋅ in. ................ Ans.
2
For the circular bar:
Tmax =
(b)
τ max J
c
=
J = π d 4 32 = π (1.5 ) 32 = 0.4970 in.4
4
(12 )( 0.4970 ) = 7.952 kip ⋅ in. ≅ 7.95 kip ⋅ in. .................................... Ans.
0.75
β ≅ 0.15
For the square bar:
θ=
( 8.505)( 3 ×12 )
TL
=
= 0.1008 rad ................................................ Ans.
3
β a bG ( 0.15 )(1.5 )3 (1.5 )( 4000 )
For the circular bar:
θ=
TL ( 7.952 )( 3 × 12 )
=
= 0.1440 rad ........................................................................ Ans.
JG ( 0.4970 )( 4000 )
229
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-122*
τ=
T
2 At
t=
( 2000 )
T
=
= 0.00318 m = 3.18 mm .............................. Ans.
2 Aτ 2 ( 7854.0 × 10−6 )( 40 ×106 )
A = π r 2 4 = π (100 ) 4 = 7854.0 mm 2
2
230
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-123
τ=
T
2 At
t=
(125) = 0.1229 in. ............................................................................. Ans.
T
=
2 Aτ 2 ( 76.274 )( 8 )
A = ( 6 )( 8 ) + π ( 3) = 76.274 in 2
2
231
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-124
(a) For the circle:
τ=
T
2 At
d = 500 π = 159.155 mm
A = π r 2 4 = π (159.155 ) 4 = 19,894.38 mm 2
2
Tmax = 2 Atτ max = 2 (19.89438 ×10−3 ) ( 0.003) ( 75 × 106 )
Tmax = 8.95 (103 ) N ⋅ m = 8.95 kN ⋅ m ............................................................................... Ans.
(b)
a = 500 3 = 166.667 mm
For the equilateral triangle:
T
τ=
2 At
A=
(166.667 )
Tmax = 2 Atτ max = 2 (12.02813 ×10
2
cos 30°
2
−3
= 12, 028.13 mm 2
) ( 0.003) ( 75 ×10 )
6
Tmax = 5.41(103 ) N ⋅ m = 5.41 kN ⋅ m ............................................................................... Ans.
(c)
a = 500 4 = 125 mm
For the square:
τ=
T
2 At
A = (125 ) = 15, 625 mm 2
2
Tmax = 2 Atτ max = 2 (15.625 × 10−3 ) ( 0.003) ( 75 × 106 )
Tmax = 7.03 (103 ) N ⋅ m = 7.03 kN ⋅ m ............................................................................... Ans.
(d)
For the rectangle:
τ=
T
2 At
a = 100 mm
b = 150 mm
A = (100 )(150 ) = 15, 000 mm 2
Tmax = 2 Atτ max = 2 (15 × 10−3 ) ( 0.003) ( 75 × 106 )
Tmax = 6.75 (103 ) N ⋅ m = 6.75 kN ⋅ m ............................................................................... Ans.
232
MECHANICS OF MATERIALS, 6th Edition
6-125*
a = 2 in.
For the rectangle:
(a)
The value of
α
RILEY, STURGES AND MORRIS
b = 3 in.
b a = 3 2 = 1.5
α ≅ 0.23
from Fig. 6-30 is:
Tmax = TAB = T2 − T1 = 30 − 10 = 20 kip ⋅ in.
τ max =
(b)
Tmax
20
=
= 7.25 ksi ............................................................................ Ans.
2
α a b 0.23 ( 2 )2 ( 3)
The value of
β
β ≅ 0.20
from Fig. 6-30 is:

TL


TL

θC / A = θC / B + θ B / A =  3  +  3 
 β a bG C / B  β a bG  B / A
=
( −10 )( 30 )
( 20 )( 30 )
+
3
3
( 0.20 )( 2 ) ( 3)( 4000 ) ( 0.20 )( 2 ) ( 3)( 4000 )
θ = +0.01563 rad
.................................................................................................................. Ans.
233
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-126*
τ=
TAB
T
2 At
= T2 − T1 = 2T − 2T = 0 kN ⋅ m
A = ( 65 )( 95 ) = 6175 mm 2
Tmax = TBC = T1 = 2T = 2 Atminτ max
= 2 ( 6175 ×10−6 ) ( 0.005 ) ( 80 ×106 ) = 4.940 (103 ) N ⋅ m
T = 4.94 2 = 2.47 kN ⋅ m .................................................................................................... Ans.
234
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-127
(a) For the square bar:
The value of
b a = a a =1
α
α ≅ 0.21
from Fig. 6-30 is:
Tmax = τ maxα a 2b = τ max ( 0.21)( a ) ( a ) = 0.21a 3τ max
2
A = a2 = π d 2 4
For the circular bar:
Tmax =
τ max J
c
d = 1.12838a
τ max π (1.12838a ) 32 
4
=

1.12838a 2
 = 0.28209a 3τ
max
Tcircle 0.28209a 3τ max
=
= 1.34329 ≅ 1.343 ..................................................................... Ans.
Tsquare
0.21a 3τ max
(b)
For the square bar:
θ=
β ≅ 0.15
TL
TL
6.667TL
=
=
3
3
a 4G
β a bG (.15 ) a aG
For the circular bar:
θ=
TL
1.34329TL
8.44009TL
=
=
4
JG π (1.12838a ) 32  G
a 4G


θ circle 8.44009TL a 4G
=
= 1.266 ...................................................................................... Ans.
6.667TL a 4G
θ square
235
MECHANICS OF MATERIALS, 6th Edition
6-128
TAB = T
RILEY, STURGES AND MORRIS
TBC = 2T
TCD = −T
b a = 50 50 = 1
For the square bar:
β ≅ 0.15
α ≅ 0.21
From Fig. 6-30:
For the stress specification:
τ max ≤ 80 MPa
Tmax = TBC = 2T ≤ τ maxα a 2b = ( 80 × 106 ) ( 0.21)( 0.050 ) ( 0.050 ) = 2100 N ⋅ m
2
T ≤ 1050 N ⋅ m
For the deformation specification:
θ ≤ 0.035 rad
θ D / A = θ B / A + θC / B + θ D / C
θD/ A =
(TAB + TBC + TCD ) L =
β a bG
3
(T + 2T − T )( 0.400 )
≤ 0.035 rad
3
(.15)( 0.050 ) ( 0.050 ) ( 28 ×109 )
T ≤ 1148 N ⋅ m
Tmax = 1050 N ⋅ m ................................................................................................................... Ans.
236
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-129*
τ=
T
2 At
A = (12.5 )( 40 ) = 500 in 2
T = 2 Atminτ max = 2 ( 500 )( 0.04 )( 8 ) = 320 kip ⋅ in. ......................................................... Ans.
237
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-130
A = π ( 500 )
2
210
2 150
 500 + 320 
+ π ( 320 )
+ 2
 ( 695cos15° )
360
360
2

= 1.142671(106 ) mm 2 = 1.142671 m 2
200 (103 )
T
=
= 0.001750 m = 1.750 mm ................................. Ans.
t=
2 Aτ 2 (1.142671) ( 50 × 106 )
238
MECHANICS OF MATERIALS, 6th Edition
6-131
Equilibrium:
TA − 8 − 0 + TD = 0
b a = 1.5 1.5 = 1
For the square bar:
From Fig. 6-30:
TA + TD = 8 kip ⋅ in.
TBCD = ( 8 − TA ) kip ⋅ in.
TAB = −TA
β ≅ 0.15
α ≅ 0.21
θD / A = θB / A + θD / B = 0
Deformation:
θD/ A =
RILEY, STURGES AND MORRIS
( −TA )(1.5) + ( 8 − TA )( 3.0 ) = 0 rad
β a 3bG
β a 3bG
4.5TA = 24 kip ⋅ in.
TA = 5.3333 kip ⋅ in. ≅ 5.33 kip ⋅ in. .................................................................................... Ans.
TD = 2.6667 kip ⋅ in. ≅ 2.67 kip ⋅ in. .................................................................................. Ans.
239
MECHANICS OF MATERIALS, 6th Edition
6-132*
RILEY, STURGES AND MORRIS
σ max = σ y FS = 250 1.25 = 200 MPa
τ max = σ max 2 = 100 MPa
For 60 rpm:
2π NT 2π ( 60 ) T
=
= 150 (103 ) N ⋅ m/s
60
60
T = 23,873.24 N ⋅ m
Power = T ω =
τ=
Tc ( 23,873.24 )( d 2 )
=
= 100 (106 ) N/m 2
π d 4 32
J
Use shaft with
d = 0.1067 m
d = 110 mm ..................................................................................... Ans.
For 6000 rpm:
2π NT 2π ( 6000 ) T
=
= 150 (103 ) N ⋅ m/s
60
60
T = 238.7324 N ⋅ m
Power = T ω =
τ=
Tc ( 238.7324 )( d 2 )
=
= 100 (106 ) N/m 2
π d 4 32
J
Use shaft with
d = 0.0230 m
d = 25 mm ....................................................................................... Ans.
Use a speed of 6000 rpm if weight is critical. ...................................................................... Ans.
240
MECHANICS OF MATERIALS, 6th Edition
6-133*
RILEY, STURGES AND MORRIS
σ max = σ y FS = 62 1.5 = 41.3333 ksi
τ=
τ max = σ max 2 = 20.6667 ksi
Tc (1200 × 12 )( c )
=
≤ 20, 666.7 psi
J
J
J c ≥ 0.69677 in 3
J I x + I y 2I
=
=
= 2 S ≥ 0.69677 in 3
c
c
c
S ≥ 0.34839 in 3
From Table A-13
Use a 2-in. diameter pipe. ............................................................................... Ans.
241
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-134
σ max = σ y FS = 250 1.5 = 166.6667 MPa
τ max = σ max 2 = 83.3333 MPa
2π NT 2π ( 60 ) T
=
= 150 (103 ) N ⋅ m/s
60
60
T = 23,873.24 N ⋅ m
Power = T ω =
(a)
τ=
Tc ( 23,873.24 ) c
=
≤ 83.3333 (106 ) N/m 2
J
J
J c ≥ 286.4789 (10−6 ) m3 = 286.4789 (103 ) mm3
J I x + I y 2I
S ≥ 143.239 (103 ) mm3
=
=
= 2 S ≥ 286.4789 (103 ) mm3
c
c
c
Use 203 mm diameter .............................................................................................. Ans.
(b)
τ=
Tc ( 23,873.24 )( d 2 )
=
= 83.3333 (106 ) N/m 2
4
π d 32
J
Use shaft with
(c)
d = 0.1134 m
d = 120 mm ..................................................................................... Ans.
W ( pipe ) = 42.46 kg/m
W ( solid ) = ( 7870 )
π ( 0.120 )
4
2
= 89.01 kg/m
W ( solid ) W ( pipe ) = 89.01 42.46 = 2.10 ............................................................. Ans.
242
MECHANICS OF MATERIALS, 6th Edition
6-135
RILEY, STURGES AND MORRIS
σ max = σ y FS = 36 2 = 18 ksi
τ max = σ max 2 = 9 ksi
2π NT 2π ( 200 ) T
=
= (100 × 550 ) lb ⋅ ft/s
60
60
T = 2626.057 lb ⋅ ft = 31,512.38 lb ⋅ in.
Power = T ω =
(a)
Tc ( 31.51238 )( c )
=
≤ 9 ksi
J
J
J I x + I y 2I
=
=
= 2 S ≥ 3.50141 in 3
c
c
c
τ=
J c ≥ 3.50141 in 3
S ≥ 1.75070 in 3
Use a 3-in. diameter pipe...................................................................................................... Ans.
(b)
τ=
Tc ( 31.51238 )( d 2 )
=
≤ 9 ksi
π d 4 32
J
d ≥ 2.6126 in.
Use shaft with d = 2 5 8 in. ....................................................................................... Ans.
(c)
W ( pipe ) = 7.58 lb/ft
W ( solid ) = ( 0.284 )
π ( 2.625 )
4
2
= 1.54 lb/in. = 18.44 lb/ft
Use hollow pipe if weight is critical..................................................................................... Ans.
243
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-136*
σ max = σ y FS = 360 3 = 120 MPa
τ max = σ max 2 = 60 MPa
Tmax = TAB = 1000 N ⋅ m
τ=
Tc (1000 )( d 2 )
=
= 60 (106 ) N/m 2
4
π d 32
J
d ≥ 0.04395 m = 43.95 mm
Use shaft with d = 50 mm ....................................................................................... Ans.
244
MECHANICS OF MATERIALS, 6th Edition
6-137
RILEY, STURGES AND MORRIS
σ max = σ y FS = 36 2.25 = 16 ksi
τ max = σ max 2 = 8 ksi
Tmax = TAB = 30 kip ⋅ in.
τ=
Tc ( 30 )( d 2 )
=
≤ 8 ksi
π d 4 32
J
d ≥ 2.673 in.
Use shaft with d = 2 3 4 in. ....................................................................................... Ans.
245
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-138*
σ max = σ y FS = 360 3 = 120 MPa
τ max = σ max 2 = 60 MPa
50TB − 650 ( 490 ) = 0
4 ΣM A = 0 :
TB = 6370 N
If the drum is tending to rotate clockwise, then
6370 = TAe
0.2( 3π 2 )
TA = 2482.14 N
and the torque on the axle of the drum is
T = 6370 ( 0.25 ) − 2482.14 ( 0.25 ) = 971.965 N ⋅ m
If the drum is tending to rotate counter-clockwise, then
TA = 6370e
0.2( 3π 2 )
TA = 16,347.54 N
and the torque on the axle of the drum is
T = 16,347.54 ( 0.25 ) − 6370 ( 0.25 ) = 2494.38 N ⋅ m
τ=
Tc ( 2494.38 )( d 2 )
=
≤ 60 (106 ) N/m 2
4
π d 32
J
d ≥ 0.05960 m = 59.60 mm
Use shaft with d = 60 mm ....................................................................................... Ans.
246
MECHANICS OF MATERIALS, 6th Edition
6-139
(a)
RILEY, STURGES AND MORRIS
Tmax = TAB = 380 lb ⋅ ft = 4560 lb ⋅ in.
For the shaft:
τ=
σ max = σ y FS = 53 2 = 26.5 ksi
Tc ( 4.560 )( d 2 )
=
≤ 13.25 ksi
π d 4 32
J
τ max = σ max 2 = 13.25 ksi
d ≥ 1.206 in.
Use shaft with d = 1 1 4 in. ........................................................................................ Ans.
(b)
For the bolts:
TAB
σ max = σ y FS = 36 1.5 = 24 ksi
τ max = σ max 2 = 12 ksi
 π d b2 
τπ db2 d1
 d1 
 d1 
= 4V   = 4 (τ A )   = 2τ 
 d1 =
2
2
2
 4 
4.560 =
(12 ) π db2 ( 3.5)
db = 0.2629 in.
2
Use bolts with d = 5 16 in. ......................................................................................... Ans.
247
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-140
For a given value of power, the values of torque are the same in the shaft and in the collar.
τs =
Tc T ( d s 2 ) 16Td s
=
=
π d s4 32 π d s4
J
τc =
T ( dc 2 )
16Td c
Tc
=
=
J π ( d c4 − d s4 ) 32 π ( d c4 − d s4 )
For the same shear stress in the shaft and in the collar
d c4 − d s4 d c
=
d s4
ds
16Td s
16Td c
=
4
π ds
π ( d c4 − d s4 )
4
 dc   dc 
  −   −1 = 0
 ds   ds 
d c d s = 1.221 (independent of the material) ............................................................... Ans.
248
MECHANICS OF MATERIALS, 6th Edition
6-141*
(a)
τ AB =
Tc ( 48 ×12 )( 3)
=
= 13.58 ksi
4
J
π ( 6 ) 32
τ CD =
Tc (13 ×12 )( 2 )
=
= 12.41 ksi
4
J
π ( 4 ) 32
τ DE =
Tc ( 2 × 12 )(1)
=
= 15.28 ksi
4
J
π ( 2 ) 32
τ max = τ DE = 15.28 ksi
(b)
.......................................Ans.
θ E / A = θ B / A + θC / B + θ D / C + θ E / D
θE / A =
( 48 ×12 )( 3 ×12 ) + ( −37 ×12 )( 4 ×12 )
TL
=
JG π ( 6 )4 32  (12, 000 ) π ( 6 )4 32  (12, 000 )




(13 ×12 )( 5 ×12 ) + ( −2 ×12 )( 2 ×12 )
+
π ( 4 )4 32  (12, 000 ) π ( 2 )4 32  (12, 000 )




θ E / A = 0.1002 (10−3 ) rad
(c)
RILEY, STURGES AND MORRIS
..................................................................................................... Ans.
θC / A = θ B / A + θC / B
θC / A =
( 48 ×12 )( 3 ×12 ) + ( −37 ×12 )( 4 ×12 )
π ( 6 )4 32  (12, 000 ) π ( 6 )4 32  (12, 000 )




θC / A = −0.377 (10−3 ) rad ..................................................................................................... Ans.
249
MECHANICS OF MATERIALS, 6th Edition
6-142*
(a)
τ AB =
RILEY, STURGES AND MORRIS
Tc ( 4000 )( d1 2 )
=
≤ 80 (106 ) N/m 2
4
π d1 32
J
d1 ≥ 0.0634 m = 63.4 mm .......................................Ans.
τ BC =
Tc ( 8000 )( d 2 2 )
=
≤ 80 (106 ) N/m 2
4
π d 2 32
J
d 2 ≥ 0.0799 m = 79.9 mm ......................................Ans.
(b)
θ = TL JG
θC / A =
( 4000 )(1) + ( −8000 )( 2 ) = −0.0483 rad ................................................. Ans.
π ( 0.075 )4 32  ( 80 ×109 )


250
MECHANICS OF MATERIALS, 6th Edition
6-143
(a)
τ=
RILEY, STURGES AND MORRIS
T ( 0.625 )
Tc
=
≤ 8 ksi
J π (1.254 − 1.124 ) 32
T ≤ 1.091 kip ⋅ in. ............................................................................................................ Ans.
(b)
θC / A =
(1)( 3 ×12 )
π (1.25 − 1.124 ) 32  ( 3800 )


4
= 0.1112 rad .................................................... Ans.
251
MECHANICS OF MATERIALS, 6th Edition
6-144*
(a)
τ=
RILEY, STURGES AND MORRIS
(18, 000 )( d 2 ) ≤ 100 106 N/m 2
Tc
=
( )
J π  d 4 − ( 0.5d )4  32


d min = 0.0993 m = 99.3 mm ...................................... Ans.
(b)
θ=
TL ( 8000 )( 2 ) + ( −18, 000 )( 4 ) + ( 4000 )( 3)
=
JG
π 0.1204 − 0.0604  32 (80 ×109 )
{
θ = −0.0288 rad
}
........................................................... Ans.
252
MECHANICS OF MATERIALS, 6th Edition
6-145
RILEY, STURGES AND MORRIS
θ=
T ( 3 × 12 )
TL
=
≤ 0.052 rad
JG π ( 2.5 )4 32  ( 4000 )


T ≤ 22.1575 kip ⋅ in.
= 1.84646 kip ⋅ ft
τ=
T (1.25 )
Tc
=
≤ 10 ksi
J π ( 2.5 )4 32
T ≤ 30.6796 kip ⋅ in.
= 2.55663 kip ⋅ ft
2π NT 2π ( 500 )(1846.46 )
=
= ( 96, 680.3) lb ⋅ ft/s
60
60
Power = 96, 680.3 550 = 175.8 hp ........................................................................... Ans.
Power = T ω =
253
MECHANICS OF MATERIALS, 6th Edition
6-146
(a)
Tc ( 55, 000 )( 0.075 )
=
≤ 83.0 (106 ) N/m 2
4
J
π ( 0.150 ) 32
τ AB =
σ AB = τ AB = 83.0 MPa (T)
(b)
RILEY, STURGES AND MORRIS
τ BC =
......................................Ans.
(15, 000 )( 0.050 ) ≤ 76.4 106 N/m 2
( )
4
π ( 0.100 ) 32
σ BC = τ BC = 76.4 MPa (C) ......................................Ans.
(c)
θ = TL JG
θC =
( −55, 000 )( 0.300 ) + ( −15, 000 )( 0.400 )
π ( 0.150 )4 32  ( 80 × 109 ) π ( 0.100 )4 32  ( 80 × 109 )




θC = −0.01179 rad ......................................................................................................... Ans.
254
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-147*
(a) First look at equilibrium of the aluminum shell and the ends of the pins:
 π ( 0.5 )2 
Ta = V ( 2 ) = ( 5 ) 
 ( 2 ) = 1.963495 kip ⋅ in.
 4 
Ja =
πd4
32
=
π ( 34 − 24 )
32
Js =
= 6.38136 in.4
π ( 24 )
32
= 1.57080 in.4
Assume that the steel shaft extends all the way through the aluminum shell and attaches to the wall at D. Then, the
portion of the shafts between the wall and the pin must rotate the same amount
θCD , s = θCD ,a
θ = TL JG
Ts (12 )
Ta (12 )
=
(1.57080 )(12, 000 ) ( 6.38136 )( 4000 )
Ts = 0.73846Ta = 0.73846 (1.963495 )
= 1.44997 kip ⋅ in.
T = Ta + Ts = 1.963495 + 0.73846 = 3.41346 kip ⋅ in. ≅ 3.41 kip ⋅ in. ........................ Ans.
(b)
τa =
(c)
θ=
Td (1.963495 )(1.5 )
=
= 0.462 ksi = 462 psi ....................................................... Ans.
J
( 6.38136 )
( −1.44997 )(12 ) + ( −3.41346 )(18 ) = −0.00418 rad ................................ Ans.
(1.57080 )(12, 000 ) (1.57080 )(12, 000 )
255
MECHANICS OF MATERIALS, 6th Edition
6-148
RILEY, STURGES AND MORRIS
J AB , s = π d 4 32 = π (160 ) 32 = 64.33982 (106 ) mm 4
4
J BC , s = π (1604 − 1004 ) 32 = 54.52234 (106 ) mm 4
J BC ,b = π (100 ) 32 = 9.81748 (106 ) mm 4
4
Equilibrium:
Ts + Tb = 75 kN ⋅ m
Deformations:
θ BC , s = θ BC ,b
(a)
θ = TL JG
Ts (1.5 )
Tb (1.5 )
=
( 54.52234 ×10 )(80 ×10 ) ( 9.81748 ×10 )( 40 ×10 )
−6
9
−6
9
Ts = 11.10720Tb
(b)
Tb = 6.19466 kN ⋅ m
Ts = 68.80534 kN ⋅ m
(85, 000 )( 0.08) = 105.6888 106 N/m 2 = 105.6888 MPa
Tc
=
( )
J ( 64.33982 ×10−6 )
In AB:
τs =
In BC:
τs =
( 68,805.34 )( 0.08) = 101.0
τb =
( 6194.66 )( 0.05) = 31.5 106
( 54.52234 ×10 )
−6
( 9.81748 ×10 )
−6
(a)
τ max, s = 105.7 MPa
(b)
θ D = θ B / A + θC / B + θ D / C
θD =
(10 ) N/m
6
( ) N/m
2
2
= 101.0 MPa
= 31.5 MPa
....................... τ max,b = 31.5 MPa .............................................. Ans.
θ D = TL JG
(85, 000 )( 2 )
+
( −68,805.34 )(1.5)
( 64.33982 ×10 )(80 ×10 ) ( 54.52234 ×10 )(80 ×10 )
−6
θ D = 0.00937 rad
9
−6
9
+0
........................................................................................................... Ans.
256
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
6-149
τ=
T
2 At
t=
(1850 ×12 )
T
=
= 0.1022 in. ................................................................. Ans.
2 Aτ 2 (13.57080 )( 8000 )
A = ( 6 × 2 ) + π (1) 2 = 13.57080 in 2
2
257
MECHANICS OF MATERIALS, 6th Edition
6-150*
τ=
RILEY, STURGES AND MORRIS
(12, 000 )
T
=
2 At 2 ( 53, 000 ×10−6 ) ( 0.0013)
τ = 87.1(106 ) N/m 2 = 87.1 MPa
............................................................................... Ans.
258
MECHANICS OF MATERIALS, 6th Edition
6-151*
RILEY, STURGES AND MORRIS
2π NT 2π ( 300 ) T
=
= ( 200 × 550 ) lb ⋅ ft/s
60
60
T = 3501.4087 lb ⋅ ft = 42, 016.9 lb ⋅ in.
Power = T ω =
τ=
Tc ( 42.0169 )( d 2 )
=
≤ 15.9 ksi
π d 4 32
J
d ≥ 2.3787 in.
θ=
TL ( 42.0169 )( 3 ×12 )
=
≤ 1.5° = 0.026180 rad
JG (π d 4 32 ) (11, 600 )
d ≥ 2.6689 in.
use d = 2 3 4 in. ............................................................................................................ Ans.
259
MECHANICS OF MATERIALS, 6th Edition
6-152
RILEY, STURGES AND MORRIS
J = π d 4 32 = π (100 ) 32 = 9.81748 (106 ) mm 4
4
θo =
(15, 000 )(1.2 )
TL
=
= 0.04584 rad
JG ( 9.81748 × 10−6 )( 40 ×109 )
θtotal = θ ABC + θ slip + θCD = 0
After the torque at B is removed
( −T )( 2.2 )
( 9.81748 ×10 )( 40 ×10 )
−6
9
+ 0.04584 +
( −T )(1.6 )
( 9.81748 ×10 )(80 ×10 )
−6
9
=0
T = 2149.808 N ⋅ m ............................................................................................................... Ans.
τ=
Tc ( 2149.808 )( 0.030 )
=
= 50.7 (106 ) N/m 2 = 50.7 MPa ................................. Ans.
−6
J
( 9.81748 ×10 )
260
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-1*
bh3 ( 4 )( 6 )
=
= 72.00 in.4
12
12
3
I=
Mr =
σI
c
=
(1000 )( 72.00 ) = 24, 000 lb ⋅ in. = 24.0 kip ⋅ in. ....................................... Ans.
3
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
7-2*
d=
M x (100 ) ( 200 )( 50 )  + ( 25 ) ( 200 )( 50 )  + (100 ) ( 200 )( 50 ) 
=
= 75 mm
A
3 ( 200 )( 50 ) 
(100 )(125)
I=
3
( 300 )( 75)
+
3
3
6
4
= 106.25 (10 ) mm
σ=
RILEY, STURGES AND MORRIS
3
( 200 )( 25)
−
3
3
− M r y − ( −10, 000 )( 0.125 )
=
I
(106.25 ×10−6 )
σ = +11.76 (106 ) N/m 2 = 11.76 MPa (T)
...................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
7-3
d=
RILEY, STURGES AND MORRIS
M x (1) ( 6 )( 2 )  + ( 5 ) ( 6 )( 2 ) 
=
= 3 in.
A
2 ( 6 )( 2 ) 
( 2 )( 5)
I=
3
3
( 6 )( 3)
+
3
( 4 )(1)
−
3
= 136.00 in.4
3
3
− M r y − ( 4000 ×12 )( 5 )
σ=
=
= −667 psi = 667 psi (C) ........................................... Ans.
I
(136.00 )
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-4*
I=
( 200 )( 300 )
12
3
−
(150 )( 200 )
12
3
= 350 (106 ) mm 4
6
−6
−σ A I − ( −7.5 ×10 )( 350 × 10 )
Mr =
=
= 17,500 N ⋅ m
yA
0.150
(a)
σB =
(b)
σC =
(c)
σD =
− M r y − (17,500 )( −0.100 )
=
= 5.00 (106 ) N/m 2 = 5.00 MPa (T) ............... Ans.
−6
I
( 350 ×10 )
− (17,500 )( 0.050 )
( 350 ×10 )
−6
= −2.50 (106 ) N/m 2 = 2.50 MPa (C) ................................ Ans.
− (17,500 )( −0.125 )
( 350 ×10 )
−6
= 6.25 (106 ) N/m 2 = 6.25 MPa (T) ................................ Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-5
I=
( 6 )(10 )
Mr =
12
σI
c
=
3
−
( 4 )( 6 )
12
3
= 428 in.4
(1200 )( 428) = 102, 700 lb ⋅ in. = 102.7 kip ⋅ in. ....................................... Ans.
5
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
7-6
d=
RILEY, STURGES AND MORRIS
M x ( 25 ) (100 )( 50 )  + (150 ) ( 200 )( 37.5 ) 
=
= 100 mm
A
(100 )( 50 ) + ( 200 )( 37.5)
(100 )(100 )
I=
3
( 62.5)( 50 )
−
3
= 72.92 (106 ) mm 4
Mr =
σI
c
3
3
( 37.5)(150 )
+
3
3
( 200 ×10 )( 72.92 ×10 ) = 97.2 10 N ⋅ m = 97.2 kN ⋅ m ................ Ans.
=
( )
0.150
6
−6
3
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-7*
bh3 ( 4 )( 2 )
=
= 2.667 in.4
12
12
3
(a)
I=
Mr =
σI
c
=
(8)( 2.667 ) = 21.3 kip ⋅ in. ..................................Ans.
1
bh3 ( 2 )( 4 )
I=
=
= 10.667 in.4
12
12
3
(b)
Mr =
σI
c
=
(8)(10.667 ) = 85.3 kip ⋅ in. ..............................................Ans.
2
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
7-8
d=
M x (12.5 ) (100 )( 25 )  + (125 ) ( 200 )( 25 ) 
=
= 87.5 mm
A
(100 )( 25) + ( 200 )( 25)
(100 )(87.5)
I=
3
( 50 )( 62.5)
−
3
= 39.93 (106 ) mm 4
(a)
RILEY, STURGES AND MORRIS
σ bot =
3
3
( 25)(137.5)
+
3
3
− M r y − ( −10, 000 )( −0.1375 )
=
= −34.4 (106 ) N/m 2
−6
I
( 39.93 ×10 )
σ bot = 34.4 MPa (C) ............................................................................................................. Ans.
(b)
σ top =
− M r y − ( −10, 000 )( 0.0875 )
=
= +21.9 (106 ) N/m 2
−6
I
( 39.93 ×10 )
σ top = 21.9 MPa (T)
............................................................................................................. Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-9*
(a)
From Table B-3 for an S 24 × 80 section
Mr =
(b)
σI
c
=
d = 2c = 24.00 in.
I = 2100 in.4
(18 )( 2100 ) = 3150 kip ⋅ in. .......................................................................... Ans.
12.00
 ( 8 )( 0.75 )3

2
I = 2100 + 2 
+ ( 8 × 0.75 )(12.375 )  = 3938 in.4
12


Mr =
σI
c
=
(18 )( 3938) = 5560 kip ⋅ in. .......................................................................... Ans.
12.75
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-10*
From Table B-8 for an L 102 × 102 × 12.7-mm angle
S = 32.3 (103 ) mm3
M r = σ S = (120 × 106 )( 2 × 32.3 ×10−6 ) = 7752 N ⋅ m ≅ 7.75 kN ⋅ m ........................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-11
I=
σ top
(16 )( 28)
12
3
−
(15)( 24 )
3
= 11,989 in.4
12
− M r y − (1000 ×12 )(14 )
=
=
= −14.01 ksi = 14.01 ksi (C)
I
(11,989 )
σ bottom =
− M r y − (1000 × 12 )( −14 )
=
= +14.01 ksi = 14.01 ksi (T)
I
(11,989 )
σ max = 14.01 ksi (T, on bottom; C, on top)) ................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
7-12*
d=
RILEY, STURGES AND MORRIS
M x ( 237.5 ) (100 )( 25 )  + (125 ) ( 25 )( 200 )  + (12.5 ) ( 200 )( 25 ) 
=
A
(100 )( 25) + ( 25)( 200 ) + ( 200 )( 25 )
= 102.5 mm
(100 )(147.5)
I=
3
( 75)(122.5)
−
3
= 105.65 (106 ) mm 4
σ top =
3
3
( 200 )(102.5)
+
3
3
(175)( 77.5)
−
3
3
− M r y − ( −3000 )( 0.1475 )
=
= +4.19 (106 ) N/m 2
−6
I
(105.65 ×10 )
σ top = 4.19 MPa (T) ............................................................Ans.
σ bottom =
− ( −3000 )( −0.1025 )
(105.65 ×10 )
−6
σ bottom = 2.91 MPa (C)
= −2.91(106 ) N/m 2
........................................................Ans.
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MECHANICS OF MATERIALS, 6th Edition
7-13
d=
I=
σ top
RILEY, STURGES AND MORRIS
4r 4 ( 3 )
=
= 1.2732 in.
3π
3π
8r 4 π ( 3) 8 ( 3)
=
−
= 8.890 in.4
8
9π
8
9π
− M r y − ( −20 )(1.7268 )
=
=
= +3.88 ksi = 3.88 ksi (T) ..................................... Ans.
I
( 8.890 )
π r4
σ bottom =
4
4
−
− ( −20 )( −1.2732 )
= −2.86 ksi = 2.86 ksi (C) .............................................. Ans.
(8.890 )
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-14
I=
( 50 )( 50 )
12
Mr =
σI
3
−
( 20 )( 20 )
12
3
= 507.5 (103 ) mm 4
(110 ×10 )( 507.5 ×10 ) = 2233 N ⋅ m
=
6
−9
c
0.025
M r ≅ 2.23 kN ⋅ m .................................................................Ans.
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MECHANICS OF MATERIALS, 6th Edition
7-15*
2
2
M x ( 0.5 ) ( 8 )(1)  + ( 3) ( 4 )(1)  + ( 7 ) π ( 4 − 3 ) 4 
=
= 3.113 in.
d=
A
(8)(1) + ( 4 )(1) + π ( 42 − 32 ) 4
(8)( 3.113)
I=
3
3
+
σ top
( 7 )( 2.113)
−
π (4 − 3
4
(a)
RILEY, STURGES AND MORRIS
4
3
) + π (4
2
3
(1)(1.887 )
+
−3
2
)
3
3
( 3.887 )
2
= 152.33 in.4
64
4
− M r y − ( 30 × 12 )( 5.887 )
=
=
= +13.91 ksi
I
(152.33)
σ top = 13.91 ksi (T) ...................................................... Ans.
(b)
σ bottom =
− ( 30 × 12 )( −3.113)
= −7.36 ksi = 7.36 ksi (C) ........................................... Ans.
(152.33)
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MECHANICS OF MATERIALS, 6th Edition
7-16*
RILEY, STURGES AND MORRIS
(a)
σ = Eε = ( 210 × 109 )(1200 × 10−6 ) = 252.00 (106 ) N/m 2 = 252 MPa .................... Ans.
(b)
( 50 )( 50 )
I=
12
Mr =
3
= 0.5208 (106 ) mm 4
6
−6
−σ I − ( 252.00 × 10 )( 0.5208 ×10 )
=
= −5250 N ⋅ m
0.025
y
M r = −5.25 kN ⋅ m ................................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-17
RILEY, STURGES AND MORRIS
ρ = R + ( h 2 ) = 12 + ( h 2 ) in.
σ x = Eε x = E
h=
c
ρ
=
E ( h 2)
12 + ( h 2 )
24 ( 36 )
24σ x
=
= 0.0298 in. ........................................................................... Ans.
E − σ x 29, 000 − 36
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MECHANICS OF MATERIALS, 6th Edition
7-18*
RILEY, STURGES AND MORRIS
ρ = R + ( h 2 ) = R + ( 25 2 ) = ( R + 12.5 ) mm
σ x = Eε x = E
R=
E (12.5 )
σx
c
ρ
=
E (12.5 )
R + 12.5
− 12.5 =
( 73, 000 )(12.5) − 12.5 = 9113 mm ≅ 9.11 m ........................ Ans.
100
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MECHANICS OF MATERIALS, 6th Edition
7-19
4 ΣM cut = 0 :
RILEY, STURGES AND MORRIS
−M r − M = 0
M r = − M = −15 kip ⋅ in.
(The internal resisting moment is the same over the entire length of the beam.)
( 2 )( 2 )
I=
12
σ top =
3
= 1.3333 in.4
− M r y − ( −15 )(1)
=
= +11.25 ksi = 11.25 ksi (T) ........................................... Ans.
I
(1.3333)
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-20
2


M x (12.5 ) ( 250 )( 25 )  + (100 ) ( 25 )(150 )  + ( 225 ) π (100 ) 4 
=
= 124.36 mm
d=
2
A
( 250 )( 25 )  + ( 25 )(150 )  + π (100 ) 4 


( 250 )(124.36 )
I=
3
3
+
π (100 )
σ bottom =
64
( 225)( 99.36 )
−
3
4
+
π (100 )
4
2
3
( 25 )( 50.64 )
+
(100.64 )
3
3
2
= 172.24 (106 ) mm 4
3
− M r y − (100 ×10 ) ( −0.15064 )
=
= +87.5 (106 ) N/m 2
−6
I
(172.24 ×10 )
σ bottom = 87.5 MPa (T)
.................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-21*
4 ΣM C = 0 :
RILEY, STURGES AND MORRIS
20 RD − 3 (1000 ) − 18 ( 2000 ) = 0
RD = 1950 lb ↑
− M r + 5 (1950 ) − 3 ( 2000 ) = 0
4 ΣM cut = 0 :
M r = 3750 lb ⋅ ft = 3.750 kip ⋅ ft
I=
π ( 24 − 1.54 )
4
= 8.590 in.4
σA =
− M r y − ( 3.750 ×12 )( 2 )
=
= −10.48 ksi = 10.48 ksi (C) ................................. Ans.
I
( 8.590 )
σB =
− ( 3.750 × 12 )( −1.5 )
= +7.86 ksi = 7.86 ksi (T) ................................................ Ans.
(8.590 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-22
d=
RILEY, STURGES AND MORRIS
M x (140 ) (120 )( 40 )  + ( 60 ) ( 40 )(120 ) 
=
= 100 mm
A
(120 )( 40 )  + ( 40 )(120 ) 
(120 )( 60 )
I=
3
( 80 )( 20 )
−
3
3
3
6
4
= 21.76 (10 ) mm
4 ΣM cut = 0 :
( 40 )(100 )
+
3
3
−M r − M = 0
M r = −M
(The internal resisting moment is the same over the entire length of the beam.)
At the top of the beam
M = −M r =
σI
y
(σ = 90 MPa T )
( 90 ×10 )( 21.76 ×10 ) = +32.6 kN ⋅ m
=
( 0.060 )
At the bottom of the beam
M = −M r =
σI
y
−6
6
(σ = 140 MPa C )
( −140 ×10 )( 21.76 ×10 ) = +30.5 kN ⋅ m
=
6
−6
( −0.100 )
M max = 30.5 kN ⋅ m 3 ................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-23
(a)
di = d o − 0.5
σ max =
σ max =
(100 )( do 2 )
M rc
=
4
I
π  d o4 − ( d o − 0.5 )  64


32 (100 ) d o
4
π  d o4 − ( d o − 0.5 ) 

(b)
RILEY, STURGES AND MORRIS
ksi

d min ≅ 3.9 in. .............................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-24
bh3 b ( 2b ) 8b 4
=
=
12
12
12
3
I=
(a)
σ max =
M r c ( 6000 )( h 2 ) 12 ( 6000 )( b )
=
=
I
8b 4
(8b4 12 )
12 ( 6000 )( b ) ( 9000 )
N/m 2
=
b3
8b 4
≅ 150 mm .............................................Ans.
σ max =
(b)
hmin
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-25
From Table B-3 for an S 4 × 9.5 section
d = 2c = 4.00 in.
A = 2.79 in.2
I = 6.79 in.4
S = 3.39 in.3
M 1 = σ S = ( 20 )( 3.39 ) = 67.8 kip ⋅ in.
M R = 1.75M = (1.75 )( 67.8 ) = 118.65 kip ⋅ in.
 bt 3
2
I R = 6.79 + 2 
+ ( bt )( 2 + 0.5t ) 
 12

M R cR (118.65 )( 2 + t )
=
≤ 20 ksi
IR
IR
σ=
IR ≥
(118.65 )( 2 + t ) = 5.9325
20
(2 + t )
in.4
 bt 3
2
6.79 + 2 
+ ( bt )( 2 + 0.5t )  ≥ 5.9325 ( 2 + t )
 12

b≥
30.45 + 35.595t
t + 12t ( 2 + 0.5t )
3
2
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-26
RILEY, STURGES AND MORRIS
Aa = ( 200 )( 30 ) = 6000 mm 2
Ab = 2 (150 ) t + ( 200 ) w = 6000 mm 2
w = ( 30 − 1.5t ) mm
Ia
( 30 )( 200 )
=
12
Ma =
σI
c
3
= 20.00 (106 ) mm 4
(150 ×10 )( 20 ×10 ) = 30, 000 N ⋅ m
=
−6
6
0.100
 (150 ) t 3
w ( 200 )
2
+ 2
+ (150t )(100 + 0.5t ) 
12
 12

3
Ib =
= 100t 3 + 30, 000 + 2 (106 ) t + 20 (106 ) mm 4
Mb =
σI
(150 ×10 ) I
=
6
0.100 + t
M − Ma
% Inc = b
(100 )
Ma
c
b
N⋅m
 Mb

=
− 1 (100 )
 30, 000 
 5000 I b

% Inc = 
− 1 (100 )
 0.100 + t 
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-27*
↑ ΣFy = 0 :
−1000 − Vr = 0
Vr = −1000 lb ...............................................................................Ans.
4 ΣM cut = 0 :
M r + 1000 x = 0
M r = ( −1000 x ) lb ⋅ ft .................................................................Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
7-28*
↑ ΣFy = 0 :
RILEY, STURGES AND MORRIS
− ( 2 x ) − Vr = 0
Vr = ( −2 x ) kN .............................................................................Ans.
4 ΣM cut = 0 :
M r + ( 2 x )( x 2 ) = 0
M r = ( − x 2 ) kN ⋅ m .....................................................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-29
4 ΣM B = 0 :
( 200 ×10 )( 5) − ( 500 )( 2 ) − 10 RA = 0
RILEY, STURGES AND MORRIS
RA = 900 lb
0 ≤ x ≤ 10 ft
↑ ΣFy = 0 :
900 − ( 200 x ) − Vr = 0
Vr = ( 900 − 200 x ) lb .......................................................... Ans.
4 ΣM cut = 0 :
M r + ( 200 x )( x 2 ) − 900 x = 0
M r = ( 900 x − 100 x 2 ) lb ⋅ ft ............................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-30*
RA = 45 kN
(15 × 4 )( 6 ) − 8RA = 0
↑ ΣFy = 0 :
45 − (15 x ) − Vr = 0
Vr = ( 45 − 15 x ) kN .....................................................................Ans.
4 ΣM cut = 0 :
M r − 45 x + (15 x )( x 2 ) = 0
4 ΣM B = 0 :
M r = ( 45 x − 7.5 x 2 ) kN ⋅ m .......................................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
7-31
4 ΣM B = 0 :
( 5000 )( 4 ) + ( 2000 )( 8) − 12 RA = 0
RILEY, STURGES AND MORRIS
RA = 3000 lb
4 ft ≤ x ≤ 8 ft
↑ ΣFy = 0 :
3000 − 2000 − Vr = 0
Vr = (1000 ) lb ...................................................................... Ans.
4 ΣM cut = 0 :
M r − 3000 x + ( 2000 )( x − 4 ) = 0
M r = (1000 x + 8000 ) lb ⋅ ft .............................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
7-32
↑ ΣFy = 0 :
4 ΣM A = 0 :
RILEY, STURGES AND MORRIS
RA + ( wL ) − ( 2 wL ) = 0
M A + ( wL )( L 2 ) − ( 2 wL )( 3L 2 ) = 0
RA = wL ↑
M A = 5wL2 2 4
( wL ) + ( wx ) − Vr = 0
Vr = wL + wx = w ( L + x ) ...........................................................Ans.
↑ ΣFy = 0 :
4 ΣM cut = 0 :
Mr =
M r + ( 5wL2 2 ) − ( wL ) x − ( wx )( x 2 ) = 0
wx 2
5wL2 w 2
+ wLx −
= ( x + 2 Lx − 5 L2 ) ...................Ans.
2
2
2
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-33*
4 ΣM A = 0 :
12 RB − ( 2000 )( 4 ) − ( 5000 )( 8 ) = 0
RILEY, STURGES AND MORRIS
RB = 4000 lb
( 4000 )( 2 ) − M r = 0
4 ΣM cut = 0 :
M r = ( 8000 ) lb ⋅ ft
bh3 ( 3)( 8 )
I=
=
= 128.00 in.4
12
12
3
On the bottom of the beam ( y = −4 in.)
σ=
− M r y − ( 8 × 12 )( −4 )
=
= +3.00 ksi = 3.00 ksi (T) ............................................. Ans.
I
(128.00 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-34
From Table B-4 for an S 254 × 52 section
4 ΣM A = 0 :
4 ΣM cut = 0 :
8 RB − (15 × 4 )( 2 ) = 0
− M r + (15 )( 3) = 0
d = 2c = 254.00 mm
I = 61.2 (106 ) mm 4
RB = 15 kN
M r = 45 kN ⋅ m
On the bottom of the beam ( y = −127.0 mm )
σ=
− M r y − ( 45, 000 )( −0.127 )
=
= 93.4 (106 ) N/m 2
−6
I
( 61.2 ×10 )
σ = 93.4 MPa (T) .......................................................................................................... Ans.
On the top of the beam ( y = +127.0 mm )
σ=
− ( 45, 000 )( +0.127 )
( 61.2 ×10 )
−6
= −93.4 (106 ) N/m 2 = 93.4 MPa (C) ............................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-35*
RA = 5600 lb
( 2000 )(13) + (1000 × 6 )( 5 ) − 10 RA = 0
4 ΣM A = 0 :
RB = 2400 lb
( 2000 )( 3) − (1000 × 6 )( 5 ) + 10 RB = 0
(a)
Vr = ( −2000 ) lb ..................................................................................................................... Ans.
M r = −2000 ( x + 3) = [ −2000 x − 6000] lb ⋅ ft ................................................................ Ans.
(b)
Vr = −2000 + 5600 = ( 3600 ) lb ......................................................................................... Ans.
M r = −2000 ( x + 3) + 5600 x = [3600 x − 6000 ] lb ⋅ ft ................................................... Ans.
(c)
Vr = −2000 + 5600 − 1000 ( x − 2 ) = ( −1000 x + 5600 ) lb ............................................ Ans.
M r = −2000 ( x + 3) + 5600 x − 1000 ( x − 2 )( x − 2 ) 2
4 ΣM B = 0 :
M r =  −500 x 2 + 5600 x − 8000  lb ⋅ ft ............................................................................. Ans.
(d)
Vr = −2000 + 5600 − 1000 ( 6 ) = ( −2400 ) lb ................................................................... Ans.
M r = −2000 ( x + 3) + 5600 x − 1000 ( 6 )( x − 5 )
M r = [ −2400 x + 24, 000] lb ⋅ ft .......................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
7-36*
4 ΣM B = 0 :
4 ΣM A = 0 :
(a)
(12 × 6 )( 9 ) + ( 24 )( 4 ) − 10 RA = 0
− (12 × 6 )(1) − ( 24 )( 6 ) + 10 RB = 0
RILEY, STURGES AND MORRIS
RA = 74.4 kN
RB = 21.6 kN
Vr = −12 ( x + 2 ) = ( −12 x − 24 ) kN ................................................................................... Ans.
M r = − (12 )( x + 2 )  ( x + 2 ) 2 =  −6 x 2 − 24 x − 24  kN ⋅ m .................................... Ans.
(b)
Vr = −12 ( x + 2 ) + 74.4 = ( −12 x + 50.4 ) kN ................................................................... Ans.
M r = − (12 )( x + 2 )  ( x + 2 ) 2 + 74.4 x =  −6 x 2 + 50.4 x − 24  kN ⋅ m .................. Ans.
(c)
Vr = −12 ( 6 ) + 74.4 = ( 2.4 ) kN .......................................................................................... Ans.
M r = − 12 ( 6 )  ( x − 1) + 74.4 x = [ 2.4 x + 72] kN ⋅ m ................................................... Ans.
(d)
Vr = −12 ( 6 ) + 74.4 − 24 = ( −21.6 ) kN ............................................................................ Ans.
M r = − 12 ( 6 )  ( x − 1) + 74.4 x − 24 ( x − 6 ) = [ −21.6 x + 216] kN ⋅ m ...................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
7-37
RILEY, STURGES AND MORRIS
4 ΣM B = 0 :
(1 2 )(1500 )(12 )  ( 4 ) − 12 RA = 0
RA = 3000 lb
4 ΣM A = 0 :
12 RB − (1 2 )(1500 )(12 )  ( 8 ) = 0
RB = 6000 lb
(a)
 1   1500 x 
2
Vr = 3000 −   
 ( x ) = ( −62.5 x + 3000 ) lb .................................................. Ans.
2
12
 

 1   1500 x    x 
3
M r = 3000 x −   
 ( x )    =  −20.83 x − 3000 x  lb ⋅ ft ...................... Ans.
 2   12    3 
(b)
dVr
= −125 x = 0
dx
Solving yields:
x=0
Therefore, the maximum shear force occurs either at the beginning or end of the region:
Vx =0 = 3000 lb
Vx =12 = −6000 lb
Vmax = Vx =12 = ( −6000 ) lb .................................................................................................... Ans.
dM r
= −62.5 x 2 + 3000 = 0
dx
M x =0 = 0 lb ⋅ ft
Solving yields:
M x =6.928 = 13,858 lb ⋅ ft
x = 6.928 ft
M x =12 = 0 lb ⋅ ft
M max = M x =6.928 = 13,858 lb ⋅ ft ≅ 13.86 kip ⋅ ft ............................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-38*
RA = 27.0 kN
( 9 )( 4 ) + (15 × 2 )(1.5 ) − 3RA = 0
4 ΣM A = 0 :
3RB + ( 9 )(1) − (15 × 2 )(1.5 ) = 0
RB = 12.0 kN
(a)
Vr = −9 + 27 − 15 ( x − 0.5 ) = ( −15 x + 25.5 ) kN ............................................................. Ans.
4 ΣM B = 0 :
M r = −9 ( x + 1) + 27 x − (15 )( x − 0.5 )  ( x − 0.5 ) 2
M r =  −7.5 x 2 + 25.5 x − 10.88 kN ⋅ m ............................................................................ Ans.
(b) From Table B-4 for an S178 × 30 section
d = 2c = 177.8 mm
I = 17.6 (106 ) mm 4
S = 198 (103 ) mm3
M 1.5 = −9 ( 2.5 ) + 27 (1.5 ) − (15 )(1)  ( 0.5 ) = +10.5 kN ⋅ m
y = −c + 15 = −88.9 + 15 = −73.9 mm
σ=
− M r y − (10,500 )( −0.0739 )
=
= +44.1(106 ) N/m 2
−6
I
(17.6 ×10 )
σ = 44.1 MPa (T)
(c)
σ=
................................................................................................................. Ans.
(10,500 ) = 53.0 106 N/m 2
Mr
=
( )
S
(198 ×10−6 )
σ max = 53.0 MPa (C, top; T, bottom) .............................................................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
7-39
RILEY, STURGES AND MORRIS
4 ΣM B = 0 :
( 500 )( 6 )  ( 7 ) + ( 800 )( 5 )  ( 2.5 ) − 10 RA = 0
RA = 1100 lb
4 ΣM A = 0 :
10 RB − ( 500 )( 6 )  ( 3) − ( 800 )( 5 )  (12.5 ) = 0
RB = 5900 lb
(a)
Vr = 1100 − ( 500 )( 6 ) = ( −1900 ) lb .................................................................................. Ans.
M r = 1100 x − ( 500 )( 6 )  ( x − 3) = [ −1900 x + 9000] lb ⋅ ft ....................................... Ans.
(b)
From Table B-3 for an S 8 × 23 section
d = 2c = 8.00 in.
I = 64.9 in.
4
M 3 = 1100 ( 3) − ( 500 )( 3)  (1.5 ) = 1050 lb ⋅ ft
σ=
(c)
S = 16.2 (103 ) in.3
y = c − 1 = 4 − 1 = +3 in.
− M r y − (1050 × 12 )( 3)
=
= −582 psi = 582 psi (C) ........................................... Ans.
I
( 64.9 )
σ max =
M r (1050 × 12 )
=
= 778 psi (C, top; T bottom) .............................................. Ans.
S
(16.2 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-40
4 ΣM B = 0 :
4 ΣM A = 0 :
d=
Mx
A
RILEY, STURGES AND MORRIS
( 3 × 2 )( 3) − 4 RA = 0
4 RB − ( 3 × 2 )(1) = 0
(140 ) (120 )( 40 ) + ( 60 ) ( 40 )(120 )
=
= 100 mm
(120 )( 40 )  + ( 40 )(120 ) 
(120 )( 60 )
I=
3
3
( 80 )( 20 )
−
3
3
( 40 )(100 )
+
3
3
RA = 4.50 kN
RB = 1.50 kN
= 21.76 (106 ) mm 4
M 3 = RB (1) = (1.50 )(1) = 1.500 kN ⋅ m
At the top of the section ( y = +60 mm )
σ=
− M r y − (1500 )( 0.060 )
=
= −4.14 (106 ) N/m 2
−6
I
( 21.76 ×10 )
σ = 4.14 MPa (C) ................................................................................................................. Ans.
At the bottom of the section ( y = −100 mm )
σ=
− (1500 )( −0.100 )
( 21.76 ×10 )
−6
= +6.89 (106 ) N/m 2 = 6.89 MPa (T) .................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-41*
10
1 10
 −5000
πs 
 π s 
 10, 000 
RA = RB = ∫ 1000sin   ds = 
cos    = 
 lb
2 0
 10 
 10   0  π 
 π
(a)
x
 10, 000  x
πs 
− ∫ 1000sin   ds
Vr = RA − ∫ w ds = 

0
 π  0
 10 
x
 10, 000  10, 000
 π s   10, 000
 π x 
Vr = 
cos    = 
cos 
+
  lb ............................... Ans.
 π   π
 10   0  π
 10  
x
 10, 000 x  x
πs 
M r = RA x − ∫ w ( x − s ) ds = 
 − ∫0 1000 ( x − s ) sin   ds
0
 π

 10 
x
x
x
 10, 000 x  10, 000 x
 π s   100, 000  π s   10, 000 s
 π s 
cos    + 
sin    − 
cos   
=
+
2
 π
  π
 10   0  π
 10   0  π
 10   0
100, 000  π x  

 π x 
Mr = 
sin 
  lb ⋅ ft = 10.13sin 
  kip ⋅ ft ..................................... Ans.
2
 10  
 10  
 π

(b)
Vmax = Vx =0 = Vx =10 =
M max = M x =5 =
10, 000
π
100, 000
π2
= 3183 lb ≅ 3.18 kip ........................................................ Ans.
= 10,132 lb ⋅ ft ≅ 10.13 kip ⋅ ft ............................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-42*
4 ΣM B = 0 :
RA =
∫ w ( 4 − s ) ds − 4R
4
0
A
=0
1 4

 π s 
4 − s )  25cos    dx
(
∫
4 0
 8 

4
4
4
 200  π s    400
 π s    50 s
 π s    400 
sin    −  2 cos    − 
sin    =  2  kN
=
 8 0  π
 8  0  π
 8  0  π 
 π
x
(a)
Vr = RA − ∫
x
0
 400  x
πs 
 400   200  π s  
w ds =  2  − ∫ 25cos   ds =  2  − 
sin   
0
π 
 8 
π   π
 8 0
 400  200  π x  
Vr =  2  −
sin 
  kN .................................................................................... Ans.
 8 
 π  π
x

 400 x  x
 π s 
M r = RA x − ∫ w ( x − s ) ds =  2  − ∫ ( x − s )  25cos    ds
0
0
 π 
 8 

x
x
x
 400 x   200  π s   1600
 π s    200s
 π s 
sin    +  2 cos    + 
sin   
= 2 −
 π   π
 8 0  π
 8  0  π
 8  0
 400 ( x − 4 ) 1600
 π x 
Mr = 
+ 2 cos 
  kN ⋅ m ................................................................. Ans.
2
8
π
π



(b)
Vmax = Vx =0 =
400
π2
= 40.5 kN .............................................................................................. Ans.
dM r 400 200  π x 
= 2 −
sin 
=0
dx
π
π
 8 
x = 1.7573 m
M max = M x =1.7573 = 34.1 kN ⋅ m ........................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-43
4 ΣM B = 0 :
∫
10
0
w (10 − s ) ds − 10 RA = 0
10
10 s 3 s 4 
1 10
2
RA = ∫ (10 − s ) (10 s ) ds = 
−  = ( 833.3) lb
10 0
4 0
 3
(a)
Vr = RA − ∫ w ds = 833.3 − ∫ 10s 2 ds = ( 833 − 3.33 x3 ) lb .......................................... Ans.
x
x
0
0
M r = RA x − ∫ w ( x − s ) ds = 833.3 x − ∫ 10 s 2 ( x − s ) ds
x
x
0
M r = ( 833 x − 0.8333 x
(b)
0
4
) lb ⋅ ft ........................................................................................... Ans.
Vmax = Vx =10 = 833.3 − 3.333 (10 ) = −2500 lb ................................................................ Ans.
3
dM r
x = 6.300 ft
= 833.3 − 3.333x 3 = 0
dx
M max = M x =6.300 = 3.94 kip ⋅ ft ............................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-44
4 ΣM B = 0 :
RA =
(a)
∫ w (8 − s ) ds − 8R
8
0
A
=0
1 8
1 8 3
2
−
s
−
s
ds
=
s − 8s 2 − 64s + 512 ) ds = 213.3 kN
64
8
(
)
(
)
(
∫
∫
0
0
8
8
Vr = RA − ∫ w ds = 213.3 − ∫ ( 64 − s 2 ) ds = 213.3 − 64 x +
x
x
0
0
x3
3
Vr =  0.333 x 3 − 64.0 x + 213 kN ..................................................................................... Ans.
M r = RA x − ∫ w ( x − s ) ds = 213.3 x − ∫ ( 64 − s 2 ) ( x − s ) ds
x
x
0
= 213.3x −
0
4
4
x
x
+ + 32 x 2 − 64 x 2
4 3
M r = 0.0833 x 4 − 32.0 x 2 + 213 x  kN ⋅ m ...................................................................... Ans.
(b)
dVr
= x 2 − 64 = 0
dx
x=8 m
Therefore, the maximum shear force occurs
either at the beginning or end of the region:
Vx =0 = +213.3 kN
Vx =8 = −128.0 kN
Vmax = Vx =0 = 213 kN ............................................................................................................ Ans.
dM r
x = 3.570 m
= 0.3333 x 3 − 64.0 x + 213.3 = 0
dx
M max = M x =3.570 = 367 kN ⋅ m ............................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-45
4 ΣM B = 0 :
(1000 )(10 )  ( 5 ) + 5 P − 10 RA = 0
RILEY, STURGES AND MORRIS
RA = RB = ( 5000 + 0.5 P ) lb
M P = ( 5000 + 0.5P )( 5 ) − (1000 )( 5 )  ( 2.5 ) = (12,500 + 2.5P ) lb ⋅ ft
From Table B-5 for a C10 × 15.3 section
S = 13.5 in.3
M P = σ max S = (16 )( 2 ×13.5 ) = 432 kip ⋅ in. = 36 kip ⋅ ft
P=
M P − 12.5 36 − 12.5
=
= 9.40 kip ............................................................................. Ans.
2.5
2.5
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-46
4 ΣM A = 0 :
5RD − 5b = 0
4 ΣM D = 0 :
5 ( 5 − b ) − 5 RA = 0
(a)
RD = ( b ) kN
R A = ( 5 − b ) kN
M r = ( 5 − b ) x  kN ⋅ m
0 ≤ x ≤ ( b − 0.25 ) m
( b − 0.25) m ≤ x ≤ ( b + 0.25)
M r = ( 5 − b ) x − 2.5 ( x − b + 0.25 )  kN ⋅ m
m
M r = ( 5 − x ) b  kN ⋅ m
( b + 0.25) m ≤ x ≤ 5 m
(Note that for any position of the crane,
the maximum bending moment occurs
under the wheel closest to the center of
the beam.)
(b)
M B = M x =(b −0.25) = ( 5 − b )( b − 0.25 )
M B =  −b 2 + 5.25b − 1.25 kN ⋅ m
M C = M x =(b + 0.25) = ( 4.75 − b ) b  kN ⋅ m
(c)
dM B
= −2b + 5.25 = 0
dx
b = 2.625 m
M B max = 5.64 kN ⋅ m
dM C
= 4.75 − 2b = 0
dx
b = 2.375 m
M C max = 5.64 kN ⋅ m
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-47
RILEY, STURGES AND MORRIS
RB = RC = wL 2 = (1200 )(15 ) 2 = 9000 lb
0 ≤ x ≤ d ft
(a)
d ≤ x ≤ (15 − d ) ft
M r = − (1200 )( x )( x 2 ) =  −600 x 2  lb ⋅ ft
M r = − (1200 )( x )( x 2 ) + 900 ( x − d )
=  −600 x 2 + 9000 x − 9000d  lb ⋅ ft
(15 − d ) ft ≤ x ≤ 15 ft
M r = − (1200 )(15 − x )(15 − x ) 2
=  −600 x 2 + 18, 000 x − 135, 000  lb ⋅ ft
(b)
AB :
BC :
M max AB = M x = d =  −600d 2  lb ⋅ ft
dM
= −1200 x + 9000 = 0
dx
x = 7.5 ft
M max BC = M x =7.5 = [33, 750 − 9000d ] lb ⋅ ft
(c)
− M max AB = M max BC
600d 2 = 33, 750 − 9000d
d = 3.107 ft
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-48
4 ΣM B = 0 : ( 5 − d ) RC − (15 × 6 )( 2 ) = 0
 180 
RC = 
 kN
 5−d 
4 ΣM C = 0 : (15 × 6 )( 3 − d ) − ( 5 − d ) RB = 0
 90 ( 3 − d ) 
RA = 
 kN
 5−d 
(a)
M r = − (15 x )( x 2 ) = ( −7.5 x 2 ) kN ⋅ m
0 ≤ x ≤1 m
1 m ≤ x ≤ (6 − d ) m
M r = − (15 x )( x 2 ) +
90 ( 3 − d )
( x − 1)
5−d
90 ( 3 − d )


M r =  −7.5 x 2 +
( x − 1)  kN ⋅ m
5−d


(6 − d ) m ≤ x ≤ 6 m
M r = −15 ( 6 − x )( 6 − x ) 2
M r =  −7.5 x 2 + 90 x − 270  kN ⋅ m
(b)
M AB max = −7.5 (1) = −7.5 kN ⋅ m
2
90 ( 3 − d )
dM r
= −15 x +
=0
dx
5−d
1 m ≤ x ≤ (6 − d ) m
x=
6 (3 − d )
m
5−d
Note that when d ≥ 2.6 m , this gives a location for the maximum bending moment which is outside the
range for which the bending moment equation is valid ( x ≤ 1.0 m ) . Furthermore, when d = 1.369 m , the
bending moment at the right support is greater than the internal maximum. Therefore, for d ≥ 1.369 m the
maximum bending moment in the region BC is the same as the maximum bending moment in the region CD.
180d 2 − 900d + 1080 
M BC max = 
 kN ⋅ m
2
(5 − d )


d ≤ 1.369 m
M BC max = M CD max =  −7.5d 2  kN ⋅ m
d ≥ 1.369 m
M CD max =  −7.5d 2  kN ⋅ m
(c)
M BC max = − M CD max
180d 2 − 900d + 1080
(5 − d )
2
= 7.50d 2
d = 1.369 m
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-49*
4 ΣM D = 0 :
( 2000 )(12 ) + ( 6000 )( 4 ) − 16 RA = 0
RA = 3000 lb ↑
4 ΣM A = 0 :
16 RD − ( 2000 )( 4 ) − ( 6000 )(12 ) = 0
RD = 5000 lb ↑
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-50*
4 ΣM A = 0 :
(15)( 7.5) + ( 20 )( 2 ) − 5RC = 0
RC = 30.5 kN ↑
4 ΣM C = 0 :
5 RA + (15 )( 2.5 ) − ( 20 )( 3) = 0
RA = 4.50 kN ↑
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-51
4 ΣM C = 0 :
( 2000 ×12 )(10 ) − (8000 )(8) − 16 RA = 0
RA = 11, 000 lb ↑
4 ΣM A = 0 :
16 RC − ( 2000 × 12 )( 6 ) − ( 8000 )( 24 ) = 0
RC = 21, 000 lb ↑
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-52*
4 ΣM D = 0 :
(10 )( 7 ) + ( 20 )( 3) − (15)( 5 ) − 10 RA = 0
RA = 5.50 kN ↑
4 ΣM A = 0 :
10 RD − (10 )( 3) − ( 20 )( 7 ) − (15 )(15 ) = 0
RD = 39.5 kN ↑
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-53
4 ΣM D = 0 :
( 300 × 5 )(13.5) − ( 400 × 5)( 2.5 ) + ( 2000 )( 7 ) − 11RB = 0
RB = 2659 lb ↑
4 ΣM B = 0 :
11RD + ( 300 × 5 )( 2.5 ) − ( 400 × 5 )(13.5 ) − ( 2000 )( 4 ) = 0
RD = 2841 lb ↑
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-54
4 ΣM D = 0 :
( 30 × 4 )( 2 ) + ( 20 )( 6 ) − 8RA = 0
RA = 45 kN ↑
4 ΣM A = 0 :
8 RD − ( 20 )( 2 ) − ( 30 × 4 )( 6 ) = 0
RD = 95 kN ↑
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-55*
4 ΣM D = 0 :
( 250 ×10 )( 5) + (1500 )(15 ) + (1000 ) − 20 RA = 0
RA = 1800 lb ↑
4 ΣM A = 0 :
20 RD + (1000 ) − (1500 )( 5 ) − ( 250 × 10 )(15 ) = 0
RD = 2200 lb ↑
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-56
4 ΣM D = 0 :
(10 )( 9 ) + (10 × 6 )( 3) − ( 20 )( 3) − 6 RB = 0
RB = 35 kN ↑
4 ΣM B = 0 :
6 RD + (10 )( 3) − (10 × 6 )( 3) − ( 20 )( 9 ) = 0
RD = 55 kN ↑
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MECHANICS OF MATERIALS, 6th Edition
7-57*
RILEY, STURGES AND MORRIS
( 400 ×10 )( 5) − (1000 )( 2 ) − 10 RB = 0
4 ΣM C = 0 :
RB = 1800 lb ↑
10 RC − ( 400 × 10 )( 5 ) − (1000 )(12 ) = 0
4 ΣM A = 0 :
RC = 3200 lb ↑
d=
( 7 ) ( 6 )( 2 )  + ( 3) ( 2 )( 6 ) 
( 6 )( 2 )  + ( 2 )( 6 ) 
= 5.00 in.
( 6 )( 3)
I=
3
( 4 )(1)
−
3
= 136.0 in.4
From the moment diagram:
3
3
( 2 )( 5)
+
3
3
M max = +4050 lb ⋅ ft , − 2000 lb ⋅ ft
At the section where M = +4050 lb ⋅ ft
σ top =
− M r y − ( 4050 ×12 )( 3)
=
= −1072 psi = 1072 psi (C)
I
(136.00 )
σ bottom =
− ( 4050 ×12 )( −5 )
= +1787 psi = 1787 psi (T)
(136.00 )
At the section where M = −2000 lb ⋅ ft
σ top =
− ( −2000 × 12 )( 3)
= +529 psi = 529 psi (T) = σ max T ....................................... Ans.
(136.00 )
σ bottom =
− ( −2000 ×12 )( −5 )
= −882 psi = 882 psi (C) = σ max C ................................ Ans.
(136.00 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-58*
4 ΣM E = 0 :
( 3 × 2 )( 7 ) + ( 6 ) + ( 5 × 4 )( 2 ) + ( 3)( 2 ) − 6 RB = 0
RB = 15.667 kN ↑
4 ΣM B = 0 :
6 RE + ( 3 × 2 )(1) + ( 6 ) − ( 5 × 4 )( 4 ) − ( 3)( 4 ) = 0
RE = 13.333 kN ↑
From the moment diagram:
M max = +16.357 kN ⋅ m , − 6 kN ⋅ m
From Table B-2 for a W102 × 19 section :
σ=
S = 89.5 (103 ) mm3
(16.357 ) = +182.8 106 N/m 2
Mr
=
( )
S
(89.5 ×10−6 )
σ = 182.8 MPa (T, bottom; C, top)
................................................................................. Ans.
Both stresses would be less at the section where M = −6 kN ⋅ m .
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-59
4 ΣM D = 0 :
( 3)(17.5) + (10 )( 7.5) − ( 5)( 7.5) − 27.5RA = 0
RA = 3.2727 kip ↑
4 ΣM A = 0 :
27.5 RD − ( 3)(10 ) − (10 )( 20 ) − ( 5 )( 35 ) = 0
RD = 14.7272 kip ↑
From the moment diagram:
M max = +35, 460 lb ⋅ ft , − 37,500 lb ⋅ ft
From Table B-3 for an S18 × 70 section :
d = 2c = 18.00 in.
I = 926 in.4
 (10 )(1)3

2
I = I C + I P = 926 + 2 
+ (10 × 1)( 9.5 )  = 2733 in.4
 12

At the top of the section y = 9 + 1 = 10 in. :
σ top =
− M r y − ( −37.5 ×12 )(10 )
=
= +1.647 ksi = 1.647 ksi (T) ............................. Ans.
I
( 2733)
At the bottom of the section y = −10 in. :
σ bottom =
− ( −37.5 ×12 )( −10 )
= −1.647 ksi = 1.647 ksi (C) ...................................... Ans.
( 2733)
Both stresses would be less at the section where M = +35, 460 lb ⋅ ft .
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-60
RILEY, STURGES AND MORRIS
( 30 × 4 )( 6 ) + ( 40 )( 2 ) − 8RA = 0
4 ΣM D = 0 :
RA = 100 kN ↑
8 RD − ( 30 × 4 )( 2 ) − ( 40 )( 6 ) = 0
4 ΣM A = 0 :
RD = 60 kN ↑
From the moment diagram:
M max = +166.67 kN ⋅ m
From Table B-6 for a C 254 × 45 channel :
d = 2c = 254.0 mm
I X − X = 42.9 (106 ) mm 4
I = IC + I P
 ( 250 )( 25 )3

2
= 2 ( 42.9 × 10 ) + 2 
+ ( 250 × 25 )(139.5 ) 
12


6
= 329.4 (106 ) mm 4
At the top of the section y = c + 25 = 127 + 25 = 152 mm :
σ top
3
− M r y − (166.67 ×10 ) ( 0.152 )
=
=
= −76.9 (106 ) N/m 2
−6
I
( 329.4 ×10 )
σ top = 76.9 MPa (C) ............................................................................................................. Ans.
At the bottom of the section y = −152 mm :
σ top
3
− M r y − (166.67 × 10 ) ( −0.152 )
=
=
= +76.9 (106 ) N/m 2
−6
I
( 329.4 ×10 )
σ bottom = 76.9 MPa (T)
......................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-61
RILEY, STURGES AND MORRIS
(1000 ×12 )( 6 ) + ( 6000 )( 6 ) − 12 RA = 0
4 ΣM B = 0 :
RA = RB = 9000 lb ↑
From the moment diagram:
M max = +36 kip ⋅ ft
From Table B-11 for a WT 8 × 25 section :
ytop = yC = 1.89 in.
d = 2c = 8.130 in.
I = 42.3 in.4
σ top =
ybottom = − ( d − yC ) = −6.24 in.
− M r y − ( 36.0 × 12 )(1.89 )
=
= −19.30 ksi
I
( 42.3)
σ top = 19.30 ksi (C)
σ bottom =
.............................................. Ans.
− M r y − ( 36.0 ×12 )( −6.24 )
=
= +63.7 ksi
I
( 42.3)
σ bottom = 63.7 ksi (T)
............................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-62*
RA − 3 = 0
↑ ΣFy = 0 :
RA = 3 kN ↑
M A + ( 4 ) − ( 3)( 2 ) = 0
4 ΣM A = 0 :
M A = 2 kN ⋅ m 4
From the moment diagram:
M max = +1 kN ⋅ m , − 3 kN ⋅ m
From Table B-6 for a C 254 × 30 channel :
w f = 69.6 mm
I = 1.17 (106 ) mm 4
ytop = xC = 15.4 mm
ybottom = − ( w f − xC ) = −54.2 mm
At the section where M = −3.00 kN ⋅ m :
σ top =
− ( −3000 )( 0.0154 )
(1.17 ×10 )
−6
σ bottom =
= +39.5 (106 ) N/m 2 = 39.5 MPa (T)
− ( −3000 )( −0.0542 )
(1.17 ×10−6 )
= −139.0 (106 ) N/m 2 = 139.0 MPa (C)
At the section where M = +1.00 kN ⋅ m :
σ top =
− (1000 )( 0.0154 )
(1.17 ×10 )
−6
σ bottom =
Therefore,
= −13.16 (106 ) N/m 2 = 13.16 MPa (C)
− (1000 )( −0.0542 )
(1.17 ×10 )
−6
= +46.3 (106 ) N/m 2 = 46.3 MPa (T)
σ max T = 46.3 MPa ................................................................................................... Ans.
σ max C = 139.0 MPa ................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-63
RILEY, STURGES AND MORRIS
(1.8)( 20 ) + ( 2.2 )(10 ) − 20 RA = 0
4 ΣM C = 0 :
RA = RB = 2.90 kip ↑
From the moment diagram:
M max = +11.00 kip ⋅ ft
From Table B-15 for an 8 × 8-in. timber :
S = 70.3 in.3
At the top and bottom of the timber:
σ max =
M r (11, 000 ×12 )
=
= 1878 psi
S
( 70.3)
σ max = 1878 psi (T, bottom; C, top) ........................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-64*
4 ΣM D = 0 :
( 5)(1) + ( 5) + ( 5)( 3) − ( 5) + (8 ×1.5 )( 3.25) − 4 RA = 0
RA = 14.75 kN ↑
4 ΣM A = 0 :
4 RD − ( 8 ×1.5 )( 0.75 ) − ( 5 )(1) − ( 5 ) + ( 5 ) − ( 5 )( 3) = 0
RD = 7.25 kN ↑
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-65*
RILEY, STURGES AND MORRIS
( w )( 3L ) 2  ( L ) − 2 LRB = 0
4 ΣM C = 0 :
RB = ( 3wL 4 ) ↑
( 2 L ) RC − ( w)( 3L )
4 ΣM B = 0 :
2  ( L ) = 0
RC = ( 3wL 4 ) ↑
The maximum moment occurs
where the shear force goes to zero,
Vr =
3wL ( wx 3L )( x )
−
=0
4
2
x = 3L
M 2.121L
2 ≅ 2.121L
( w 3L )( 2.121L )
 3wL 
=
 (1.121L ) −
6
 4 
2
= 0.3107 wL
From the moment diagram:
3
M max = 0.3107 wL2
From Table B-3 for an S15 × 50 section :
S = 64.8 in.3
Therefore
M 0.3107 w ( 5 × 12 )
σ=
=
≤ 15 ksi
S
64.8
(15)( 64.8 ) = 0.869 kip/in. = 10.43 kip/ft ..................................................... Ans.
w≤
2
0.3107 ( 5 × 12 )
2
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-66
For the complete structure:
4 ΣM D = 0 :
3RC − ( 3)(1.5 ) − (1.5 × 3)(1.5 ) = 0
RC = 3.75 kN = 3.75 kN ↑
RD − ( 3) − (1.5 × 3) + ( 3.75 ) = 0
↑ ΣFy = 0 :
RD = 3.75 kN = 3.75 kN ↑
For the member AB:
− ( 3) − VB = 0
↑ ΣFy = 0 :
VB = −3 kN = 3 kN ↑
4 ΣM B = 0 :
M B + ( 3)(1.5 ) = 0
M B = −4.5 kN ⋅ m = 4.5 kN ⋅ m 3
For the member CD:
( 3.75 ) − (1.5 × 3) − VC = 0
↑ ΣFy = 0 :
VC = −0.75 kN = 0.75 kN ↑
4 ΣM C = 0 :
M C + (1.5 × 3)(1.5 ) − ( 3.75 )( 3) = 0
M C = +4.5 kN ⋅ m = 4.5 kN ⋅ m 4
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-67*
(a) For pipe AB:
64 By − (16 )( 27.5 ) − ( 48 )( 27.5 ) = 0
4 ΣM A = 0 :
By = 27.5 lb ↑
Ay − 27.5 − 27.5 + 27.5 = 0
↑ ΣFy = 0 :
Ay = 27.5 lb ↑
From the moment diagram:
From Table B-13 for a
1
2
M max = 440 lb ⋅ in.
-in. diameter pipe:
S = 0.041 in.3
Therefore
M
440
=
= 10, 732 psi
S 0.041
σ ≅ 10.73 ksi (T, bottom; C top) ...................................................................................... Ans.
σ=
(b)
θ = tan −1 ( 35 64 ) = 28.673°
By = FBC sin θ
FBC = By sin θ = 27.5 sin 28.673° = 57.314 lb
ABC = π ( 3 16 ) 4 = 0.02761 in.2
2
σ BC =
(c)
FBC
57.314
=
= 2076 psi ≅ 2.08 ksi (T) .......................................................... Ans.
ABC 0.02761
→ ΣFx = 0 :
Ax − 57.314 cos 28.673° = 0
Ax = +50.29 lb = 50.29 lb →
FA = Ax2 + Ay2 =
( 50.29 ) + ( 27.5)
2
2
= 57.32 lb
AA = π (1 4 ) 4 = 0.04909 in.2
2
τA =
FA
57.32
=
= 584 psi ............................................................................................ Ans.
AA 0.04909
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-68
4 ΣM D = 0 :
RILEY, STURGES AND MORRIS
1.5 RE − ( 40 )(1.5 ) 2  ( 2.5 ) = 0
RE = 50 kN ↑
RD + 50 − ( 40 )(1.5 ) 2  = 0
↑ ΣFy = 0 :
RD = −20 kN = 20 kN ↓
4 ΣM C = 0 :
2 RB − ( 80 )(1.5 )  ( 2.75 )
− ( 40 )( 2 )  (1) − ( 20 )(1) = 0
RB = 215 kN ↑
4 ΣM B = 0 :
2 RC + ( 80 )(1.5 )  ( 0.75 )
− ( 40 )( 2 )  (1) + ( 20 )( 3) = 0
RC = −35 kN = 35 kN ↓
(a)
Shear force and bending moment graphs are shown
to the right.
(b)
From the moment diagram:
M max = −90 kN ⋅ m
From Table B-4 for an S 457 × 81 section :
σ max =
S = 1465 (103 ) mm3
( 90, 000 ) = 61.4 106 N/m 2
M
=
( )
S (1465 × 10−6 )
σ max = 61.4 MPa (T, top; C, bottom) ....................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-69
4 ΣM B = 0 :
( 4050 )( 40 − x ) + (1010 )( 30.5 − x ) − 40 RA = 0
RA = ( 4820 − 126.5 x ) lb ↑
4 ΣM A = 0 :
40 RB − ( 4050 )( x ) − (1010 )( x + 9.5 ) = 0
RB = ( 239.9 + 126.5 x ) lb ↑
Note that the maximum moment occurs under one of the wheels –
probably under the rear wheels ( C ) but possibly under the front
wheels ( D ) . Finding the position x which gives the maximum
moment under the rear wheels,
M C = RA x = ( −126.5 x 2 + 4820 x ) lb ⋅ ft
dM C
= ( −253x + 4820 ) = 0
dx
x = 19.05 ft
M C max = M C19.05 = −126.5 (19.05 ) + 4820 (19.05 )
2
M C max = 45,914 lb ⋅ ft ≅ 45.9 kip ⋅ ft
Finding the position x which gives the maximum moment under the front wheels,
M D = RB ( 30.5 − x ) = ( 239.9 + 126.5 x )( 30.5 − x ) = ( −126.5 x 2 + 3618 x + 7317 ) lb ⋅ ft
dM D
= ( −253x + 3618 ) = 0
dx
x = 14.30 ft
M D max = M D14.30 = −126.5 (14.30 ) + 3618 (14.30 ) + 7317 = 33,186 lb ⋅ ft
2
M max = 45,914 lb ⋅ ft ≅ 45.9 kip ⋅ ft ............................................................................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
7-70*
(a)
RILEY, STURGES AND MORRIS
P = mg = 1500 ( 9.81) = 14, 715 N = 14.715 kN
4 ΣM B = 0 :
P (10 − x ) − RA (10 ) = 0
RA = ( P − 0.1Px ) kN
4 ΣM A = 0 :
RB (10 ) − Px = 0
RB = ( 0.1Px ) kN
(b)
M = RA x = ( Px − 0.1Px 2 ) kN ⋅ m
dM
= P − 0.2 Px = 0
dx
x = 5.00 m ............................................................... Ans.
M max = M 5 = (14.715 )( 5 ) − 0.1(14.715 )( 5 ) = 36.8 kN ⋅ m
2
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-71
For the combined system of beams:
4 ΣM H = 0 :
22.5 (10 ) − 37.5 RA = 0
RA = 6.00 kip = 6.00 kip ↑
37.5 RH − 15 (10 ) = 0
4 ΣM A = 0 :
RH = 4.00 kip = 4.00 kip ↑
For the beam ABD:
5TBC − 15 ( 6 ) = 0
4 ΣM D = 0 :
TBC = 18.00 kip = 18.00 kip (T)
6.00 − 18.00 − 10 + RD = 0
↑ ΣFy = 0 :
RD = 22.00 kip = 22.00 kip ↑
For the beam EGH:
4 ΣM E = 0 :
15 ( 4 ) − 5TFG = 0
TFG = 12.00 kip = 12.00 kip (T)
↑ ΣFy = 0 :
4.00 − 12.00 + RE = 0
RE = 8.00 kip = 8.00 kip ↑
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MECHANICS OF MATERIALS, 6th Edition
7-72*
(10 )( 2 ) − 4 RB = 0
4 ΣM C = 0 :
5 − VA = 0
↑ ΣFy = 0 :
(150 )( 200 )
I=
3
12
Qa = 0
RILEY, STURGES AND MORRIS
RB = 5 kN = 5 kN ↑
VA = 5.00 kN = 5.00 kN ↓
= 100.0 (106 ) mm 4
τ a = 0 MPa ........................................ Ans.
Qb = yC A = 75 (150 × 50 ) = 562.5 (103 ) mm3
−6
VQ ( 5000 ) ( 562.5 × 10 )
=
= 187.5 (103 ) N/m 2
τb =
−6
It (100.0 × 10 ) ( 0.150 )
τ b = 187.5 kPa ............................................................... Ans.
Qc = yC A = 50 (150 × 100 ) = 750.0 (103 ) mm3
−6
VQ ( 5000 ) ( 750.0 × 10 )
=
= 250 (103 ) N/m 2
τb =
It (100.0 × 10−6 ) ( 0.150 )
τ b = 250 kPa
................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-73*
I=
(a)
−
( 4 )(8 )
12
3
= 981.3 in.4
VQ ( 7000 )( 80.0 )
=
= 142.7 psi .............................. Ans.
It
( 981.3)( 4 )
Q3 = yC A = 5 ( 8 × 2 ) + 3.5 ( 4 × 1) = 94.0 in 3
τ=
(c)
12
3
Q2 = yC A = 5 ( 8 × 2 ) = 80.0 in 3
τ=
(b)
( 8)(12 )
VQ ( 7000 )( 94.0 )
=
= 167.6 psi ................................................................................ Ans.
It
( 981.3)( 4 )
QNA = 5 ( 8 × 2 ) + 2 ( 4 × 4 ) = 112.0 in 3
τ max =
VQ ( 7000 )(112.0 )
=
= 199.7 psi (at neutral axis) .......................................... Ans.
It
( 981.3)( 4 )
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MECHANICS OF MATERIALS, 6th Edition
7-74
4 ΣM C = 0 :
RILEY, STURGES AND MORRIS
( 25)( 3) + ( 25 )(1) − 5RB = 0
RB = 20.0 kN = 20.0 kN ↑
20 − 25 − VA = 0
↑ ΣFy = 0 :
VA = −5.00 kN = 5.00 kN ↑
From Table B-2 for a W 254 × 89 section :
d = 2c = 260 mm
w f = 256 mm
(a)
tw = 10.7 mm
t f = 17.3 mm
QNA = yC A = 121.35 ( 256 ×17.3) + 56.35 (112.7 × 10.7 ) = 605.4 (103 ) mm3
τ max
−6
VQ ( 5000 ) ( 605.4 × 10 )
=
=
= 1.992 (106 ) N/m 2
−6
It
(142 ×10 ) ( 0.0107 )
τ max = 1.992 MPa
(b)
I = 142 (106 ) mm 4
........................................................................Ans.
Aw =  260 − 2 (17.3)  (10.7 ) = 2412 mm 2
τ avg =
V
5000
=
= 2.07 (106 ) N/m 2
−6
Aw 2412 (10 )
τ max = 2.07 MPa ..........................................................................Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-75*
RILEY, STURGES AND MORRIS
RB = RC = (1000 × 8 ) 2 = 4000 lb ↑
↑ ΣFy = 0 :
4000 − (1000 × 2 ) − Vr = 0
Vr = 2000 lb
From Table B-11 for a WT 7 × 34 section :
ts = 0.415 in.
d = 7.020 in.
I = 32.6 in.4
yC = 1.29 in.
Q3 = yC A = 2.865 ( 5.73 × 0.415 ) = 6.813 in 3
τ=
VQ ( 2000 )( 6.813)
=
= 1007 psi (at neutral axis) ................................................ Ans.
It
( 32.6 )( 0.415)
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-76
I=
(180 )( 200 )
3
12
−
(140 )(120 )
12
3
= 99.84 (106 ) mm 4
QJ = yC A = 80 ( 40 ×180 ) = 576 (103 ) mm3
QNA = yC A = 80 ( 40 × 180 ) + 30 ( 40 × 60 ) = 648 (103 ) mm3
Vmax = RA = RB = ( 6 )( 3.5 ) 2 = 10.5 kN
(a)
τJ =
−6
VQ (10,500 ) ( 576 × 10 )
=
= 1.514 (106 ) N/m 2
−6
It ( 99.84 ×10 ) ( 0.040 )
τ J = 1.514 MPa
(b)
..................................................................................................................... Ans.
(10,500 ) ( 648 ×10−6 )
= 1.721(106 ) N/m 2 = 1.721 MPa ................................ Ans.
τ NA =
−6
( 99.84 ×10 ) ( 0.040 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-77
4 ΣM C = 0 :
RILEY, STURGES AND MORRIS
−8 RB − 3 ( 4000 ) = 0
RB = −1500 lb = 1500 lb ↓
−1500 − VA = 0
↑ ΣFy = 0 :
VA = −1500 lb = 1500 lb ↑
From Table B-1 for a W10 × 30 section :
d = 2c = 10.47 in.
w f = 5.810 in.
(a)
QNA = yC A = 4.980 ( 5.810 × 0.510 ) + 2.3625 ( 4.725 × 0.300 ) = 18.105 in 3
τ max =
(b)
I = 170 in.4
tw = 0.300 in.
t f = 0.510 in.
VQ (1500 )(18.105 )
=
= 533 psi (at neutral axis) ............................................ Ans.
It
(170 )( 0.300 )
With the weight of the beam included:
w = 30 lb/ft
4 ΣM C = 0 :
2.5 ( 30 × 11) − 8 RB − 3 ( 4000 ) = 0
RB = −1397 lb = 1397 lb ↓
−1397 − ( 30 × 4 ) − VA = 0
↑ ΣFy = 0 :
VA = −1517 lb = 1517 lb ↑
τ max =
VQ (1517 )(18.105 )
=
= 539 psi (at neutral axis) ........................................... Ans.
It
(170 )( 0.300 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-78*
4 ΣM A = 0 :
RILEY, STURGES AND MORRIS
M A − 2 ( 20 ) = 0
M A = 40.0 kN ⋅ m 4
RA − 20 = 0
↑ ΣFy = 0 :
RA = 20 kN = 20 kN ↑
From Table B-2 for a W 203 × 60 section :
d = 2c = 210 mm
w f = 205 mm
I = 60.8 (106 ) mm 4
tw = 9.1 mm
t f = 14.2 mm
S = 582 (103 ) mm3
From the shear-force and bending-moment diagrams
Vmax = 20 kN (full length of the beam)
M max = 40.0 kN ⋅ m (at the wall)
σ max =
( 40, 000 ) = 68.7 106 N/m 2
Mr
=
( )
S
( 582 ×10−6 )
σ max = 68.7 MPa (T, top; C bottom) .....................................Ans.
QNA = yC A = 97.9 ( 205 × 14.2 ) + 45.4 ( 90.8 × 9.1) = 322.5 (103 ) mm3
VQ ( 20, 000 ) ( 322.5 × 10 )
=
= 11.66 (106 ) N/m 2
−6
It
( 60.8 ×10 ) ( 0.0091)
−6
τ max =
τ max = 11.66 MPa
........................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-79*
From the shear-force and bending-moment diagrams:
Vmax = 1160 lb
M max = 1346 lb ⋅ ft
bh3 ( 2 )( 6 ) ( 2 )( 3)
I=
=
+
= 40.5 in.4
12
12
12
(a) At the bottom of the beam 2.32 ft from the left support:
3
σ max =
3
− M r y − (1346 × 12 )( −3)
=
= +1196 psi
I
( 40.5)
σ max = 1196 psi (T) ...................................................... Ans.
(b)
At the left support:
τ NA =
VQ (1160 )  0.75 (1.5 × 2 ) + 1.5 ( 3 × 2 ) 
=
= 80.6 psi
It
( 40.5)( 4 )
τ 1.5 =
VQ (1160 )  2.25 (1.5 × 2 ) 
=
= 96.7 psi
It
( 40.5)( 4 )
τ max = τ 1.5 = 96.7 psi (1.5 in. above and below NA) .................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-80
RILEY, STURGES AND MORRIS
9 + 2.5 ( 6 ) + 0.5 ( 6 ) − 3RA = 0
4 ΣM B = 0 :
RA = 9.0 kN ↑
9 − 6 − 6 + RB = 0
↑ ΣFy = 0 :
RB = 3.0 kN ↑
From the shear-force and bending-moment diagrams
Vmax = 9.0 kN
(100 )( 240 )
I=
12
(a)
M max = 10.5 kN ⋅ m
3
= 115.2 (106 ) mm 4
QJ = yC A = 95 (100 × 50 ) = 475 (103 ) mm3
( 9000 ) ( 475 ×10−6 )
VQ
=
= 371(103 ) N/m 2 = 371 kPa ............................... Ans.
τJ =
It (115.2 × 10−6 ) ( 0.100 )
(b)
QNA = yC A = 60 (100 ×120 ) = 720 (103 ) mm3
τ NA =
(c)
( 9000 ) ( 720 ×10−6 )
VQ
=
= 563 (103 ) N/m 2 = 563 kPa ............................. Ans.
−6
It (115.2 × 10 ) ( 0.100 )
σ max =
M max c (10,500 )( 0.120 )
=
= 10.94 (106 ) N/m 2
−6
I
115.2
10
×
(
)
σ max = 10.94 MPa (T, bottom; C, top) ............................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-81
4 ΣM B = 0 :
RILEY, STURGES AND MORRIS
( 250 )( 4 ) 2  ( 2 ) − 4 RA = 0
RA = RB = 250 lb ↑
From the shear-force and bending-moment diagrams
Vmax = 250 lb
M max = 333.3 lb ⋅ ft
bh3 ( 8 )( 0.5 )
I=
=
= 0.08333 in.4
12
12
M c ( 333.3 × 12 )( 0.25 )
σ max = r =
= 11,999 psi
I
( 0.08333)
3
σ max ≅ 12.00 ksi (T, bottom; C, top)
........................................................................ Ans.
QNA = yC A = ( 0.125 )( 8 × 0.25 ) = 0.25 in.3
τ max =
VQ ( 250 )( 0.25 )
=
= 93.8 psi (at neutral axis) ............................................. Ans.
It ( 0.08333)( 8 )
The results for τ max are worthless since w d = 16 and Eq. 7-12 gives useful results only when w d < 1 .
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-82*
RILEY, STURGES AND MORRIS
RA = RB = ( 4 × 6 ) 2 = 12.0 kN ↑
From the shear-force and bending-moment diagrams
Vmax = 12.0 kN
V0.5 = 12.0 − 4 ( 0.5 ) = 10.0 kN
M max = 18.0 kN ⋅ m
V1.0 = 12.0 − 4 (1) = 8.0 kN
d=
( 225 ) ( 250 )( 50 )  + (125) ( 50 )(150 ) + ( 25 ) (150 )( 25)
( 250 )( 50 )  + ( 50 )(150 )  + (150 )( 25 ) 
= 143.18 mm
I NA
(150 )(143.18)
=
3
3
(100 )( 93.18)
−
( 250 )(106.82 )
+
3
3
3
(a)
3
( 200 )( 56.82 )
−
3
3
= 209.1(106 ) mm 4
QTJ = yC A = 81.82 ( 250 × 50 ) = 1022.8 (103 ) mm3
τ TJ
−6
VQ ( 8000 ) (1022.8 × 10 )
=
=
It
( 209.1×10−6 ) ( 0.050 )
τ TJ = 783 (103 ) N/m 2 = 783 kPa ......................................Ans.
(b)
QBJ = 118.18 (150 × 50 ) = 886.4 (103 ) mm3
(10, 000 ) (886.4 ×10−6 )
τ BJ =
= 848 (103 ) N/m 2 = 848 kPa ...................................... Ans.
−6
209.1
×
10
0.050
)
(
)(
(c)
QNA = 81.82 ( 250 × 50 ) + 28.41( 56.82 × 50 ) = 1103.5 (103 ) mm3
(12, 000 ) (1103.5 ×10−6 )
= 1267 (103 ) N/m 2 = 1267 kPa .............................. Ans.
−6
( 209.1×10 ) ( 0.050 )
− M max c − (18, 000 )( −0.14318 )
=
=
= +12.33 106 N/m 2
τ max =
(d)
σ max
I
σ max = 12.33 MPa (T)
( 209.1×10 )
−6
( )
.......................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-83
RILEY, STURGES AND MORRIS
( 200 × 6 )( 6 ) − ( 600 )( 3) − 9 RA = 0
4 ΣM B = 0 :
RA = 600 lb = 600 lb ↑
( 600 ) − ( 200 × 6 ) + RB − 600 = 0
↑ ΣFy = 0 :
RB = 1200 = 1200 lb ↑
From the shear-force and bending-moment diagrams:
Vmax = ±600 lb
( 5)( 6 )
I=
12
(a)
( 3)( 4 )
−
12
3
= 74.0 in.4
QJ = yC A = 2.5 ( 5 × 1) = 12.5 in 3
τJ =
(b)
3
M max = ±900 lb ⋅ ft
VQ ( 600 )(12.5 )
=
= 50.7 psi ................................. Ans.
It
( 74.0 )( 2 )
QNA = 2.5 ( 5 ×1) + 2 1( 2 × 1)  = 16.5 in 3
( 600 )(16.5 ) = 66.9 psi (at NA) .......................... Ans.
( 74.0 )( 2 )
M c ( 900 × 12 )( 3)
σ max = r =
= 438 psi
I
( 74.0 )
τ max =
(c)
σ max = 438 psi (T & C)
....................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-84
RILEY, STURGES AND MORRIS
RA = RB = ( 4.5w ) 2 = ( 2.25w ) N ↑
From the shear-force and bending-moment diagrams
Vmax = ( 2.25w ) N
(150 )( 300 )
I=
12
M max = ( 2.53w ) N ⋅ m
3
= 337.5 (106 ) mm 4
QNA = yC A = 75 (150 × 150 ) = 1687.5 (103 ) mm3
(a)
( 2.25w) (1687.5 ×10−6 )
τ max =
= 75.00w
( 337.5 ×10−6 ) ( 0.150 )
τ max = 75.00w = 825 (103 ) N/m 2
w = 11, 000 N/m = 11.00 kN/m .................................................................................. Ans.
(b)
V2 = RA − wx = 2.25 (11, 000 ) − 11, 000 ( 0.6 ) = 18,150 N
Q2 = 125 (150 × 50 ) = 937.5 (103 ) mm3
(18,150 ) ( 937.5 ×10−6 )
τH =
= 336 (103 )
−6
( 337.5 ×10 ) ( 0.150 )
N/m 2
τ H = 336 kPa .......................................................................................................................... Ans.
(c)
M max = 2.53 (11, 000 ) = 27,830 N ⋅ m
σ max =
− M r y − ( 27,830 )( −0.150 )
=
= 12.37 (106 ) N/m 2
−6
I
( 337.5 ×10 )
σ max = 12.37 MPa
................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-85*
RA = RB = 1000 2 = 500 lb = 500 lb ↑
V = constant = 500 lb
( 6 )(12 )
I=
12
(a)
3
( 4 )(8)
−
12
3
= 693.3 in.4
QJ = yC A = 5 ( 6 × 2 ) = 60.0 in 3
τJ =
VQ ( 500 )( 60.0 )
=
= 21.636 psi
It
( 693.3)( 2 )
FJ = τ J AJ = ( 21.636 )(12 × 2 ) = 519.3 lb
FJ ≅ 519 lb ........................................................................... Ans.
(b)
N = FJ FN = 519.3 100 = 5.19 nails
s = 12 5.19 = 2.13 in.
Use 6 nails spaced 2 in. apart in a 12-in. length. ......................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-86*
From Table B-6 for a C 457 × 86 channel :
I = 7.41(106 ) mm 4
A = 11, 030 mm 2
xC = 21.9 mm
tw = 17.8 mm
For the beam:
2
I = 2  7.41(106 ) + 11, 030 ( 21.9 )  = 25.40 (106 ) mm 4


QNA = yC A = 21.9 (11, 030 ) = 241.6 (103 ) mm3
V = constant = 20 kN
(a)
τ NA
−6
VQ ( 20, 000 ) ( 241.6 × 10 )
=
=
It
( 25.40 ×10−6 ) ( 0.4572 )
τ NA = 416 (103 ) N/m 2 = 416 kPa
FB = τ NA AS 2 = ( 416.1× 103 ) ( 0.4572 × 0.300 ) 2 = 28.5 (103 ) N
FB = 28.5 kN ................................................................................................................... Ans.
(b)
τ=
FB 28,540
=
= 60 (106 ) N/m 2
2
AB π d 4
d = 0.02461 m = 24.61 mm
F
28,540
σb = B =
= 125 (106 ) N/m 2
dtw d ( 0.0178 )
d = 0.01283 m = 12.83 mm
Therefore
d min = 24.6 mm ...................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-87
From Table B-1 for a W18 × 97 section :
d = 2c = 18.59 in.
I = 1750 in.4
w f = 11.145 in.
For the beam:
V=
∆M 4600 − 2300
=
= 115 kip
∆x
20
 (10 )( 0.5 )3

2
I = 1750 + 2 
+ (10 × 0.50 )( 9.545 )  = 2661 in.4
12


QJ = yC A = 9.545 (10 × 0.50 ) = 47.73 in 3
τJ =
VQ (115 )( 47.73)
=
= 0.2063 ksi = 206.3 psi
It
( 2661)(10 )
FJ = τ J AJ = ( 206.3)( 20 ×10 ) = 41, 260 lb
N=
FJ
41, 260
=
= 8.596
FW 2 ( 2400 )
Use 5 welds on each side. .................................................................................................. Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-88
RA = RB = 125 2 = 62.5 kN ↑
V = constant = 62.5 kN
From Table B-6 for a C 305 × 45 channel :
d = 2c = 304.8 mm
I = 67.4 (106 ) mm 4
w f = 80.5 mm
For the beam:
 ( 260 )(15 )3

2
I = 2
+ ( 260 × 15 )(159.9 ) 
12


+ 2 ( 67.4 × 106 ) = 334.4 (106 ) mm 4
QJ = yC A = 159.9 ( 260 ×15 ) = 623.6 (103 ) mm3
( 62,500 ) ( 623.6 ×10−6 )
= 723.9 (103 ) N/m 2
τJ =
−6
( 334.4 ×10 ) ( 2 × 0.0805)
FJ = τ J AJ = ( 723.9 × 103 ) ( 2 × 0.0805s ) = 116.55s (103 ) N
2
FB = τ B AB = 2 (150 ×106 ) π ( 0.020 ) 4  = 94.248 (103 ) N


Since FB = FJ
s=
94.248
= 0.809 m = 809 mm ...................................................................................... Ans.
116.55
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-89*
RA = RB = 125 2 = 62.5 kip = 62.5 kip ↑
From Table B-1 for a W 21× 101 section :
d = 2c = 21.36 in.
V = constant = 62.5 kip
I = 2420 in.4
w f = 12.290 in.
For the beam:
 (16 )( 0.75 )3

2
I = 2420 + 2 
+ (16 × 0.75 )(11.055 )  = 5354 in.4
12


QJ = yC A = 11.055 (16 × 0.75 ) = 132.66 in 3
τJ =
VQ ( 62,500 )(132.66 )
=
= 126.01 psi
It
( 5354 )(12.290 )
FJ = τ J AJ = (126.01)(12.290 s ) = (1548.6 s ) lb
2
FB = τ B AB = 2 (17,500 ) π ( 0.75 ) 4  = 15, 463 lb


15, 463
Since FB = FJ
s=
= 9.985 in. ≅ 10.00 in. ...................................Ans.
1548.6
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-90
RA = RB = 96 2 = 48 kN ↑
V = constant = 48 kN
I = 4.58 (106 ) mm 4
From Table B-6 for a C 381× 74 channel :
A = 9485 mm 2
xC = 20.3 mm
tw = 18.2 mm
I = 367 (106 ) mm 4
From Table B-2 for a W 356 × 122 section :
A = 15,550 mm 2
For the beam:
d=
d = 2c = 363 mm
w f = 257 mm
M x ( 360.9 )( 9485 ) + (181.5 )(15,550 )
=
= 249.5 mm
A
( 9485) + (15,550 )
2
2
I = 367 (106 ) + 15,550 ( 68 )  +  4.58 (106 ) + 9485 (111.4 )  = 561.2 (106 ) mm 4

 

QJ = 111.4 ( 9485 ) = 1056.6 (103 ) mm3
VQ ( 48, 000 ) (1056.6 × 10 )
=
= 351.6 (103 ) N/m 2
τJ =
−6
It
( 561.2 ×10 ) ( 0.257 )
−6
(a)
FB = τ J AS 2 = ( 351.64 × 103 ) ( 0.257 × 0.500 ) 2 = 22,593 N
FB = 22.6 kN ................................................................................................................... Ans.
(b)
τ=
FB 22,593
=
= 60 (106 ) N/m 2
2
AB π d 4
d = 0.02190 m
d min = 21.9 mm ...................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-91
I=
(8)(12 )
3
12
−4 in. ≤ y ≤ +4 in. :
−
( 4 )(8)
3
= 981.33 in.4
12
t = 4 in.
(4 + y)

Q = yC A = 5 ( 8 × 2 ) + 2 
( 2 )( 4 − y )  = (112 − 2 y 2 ) in.3
 2

VQ ( 7500 ) (116 − 2 y
τ=
=
It
( 981.33)( 4 )
2
)
= 3.8213 ( 56 − y 2 ) psi
−6 in. ≤ y ≤ −4 in. and 4 in. ≤ y ≤ 6 in. :
t = 8 in.
(6 + y)

Q = yC A = 
( 8)( 6 − y )
 2

= (144 − 4 y 2 ) in.3
VQ ( 7500 ) (144 − 4 y
τ=
=
It
( 981.33)(8 )
2
) = 3.8213 36 − y psi
(
)
2
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-92
4 ΣM B = 0 :
RILEY, STURGES AND MORRIS
( 3 × 2 )( 3) − 4 RA = 0
RA = 4.5 kN
4.5 − ( 3 × .5 ) − Vr = 0
↑ ΣFy = 0 :
Vr = 3.0 kN
d=
M x (100 ) ( 200 )( 25 )  + ( 212.5 ) (100 )( 25 ) 
=
= 137.5 mm
A
( 200 )( 25) + (100 )( 25)
( 25 )(137.5)
I=
3
(100 )(87.5 )
+
3
= 37.89 (106 ) mm 4
3
−137.5 mm ≤ y ≤ +62.5 mm :
Q = yC A =
3
( 75)( 62.5)
−
3
3
t = 25 mm
137.5 + y
( 25)(137.5 − y ) = 12.5 (137.52 − y 2 ) mm3
2
2
2
−9
VQ ( 3000 ) (12.5 ) (137.5 − y )(10 ) 
=
τ=
It
( 37.89 ×10−6 ) ( 0.025)
= 39.59 (137.52 − y 2 ) N/m 2
+62.5 mm ≤ y ≤ +87.5 mm :
t = 100 mm
87.5 + y
(100 )( 87.5 − y )
2
= 50 ( 87.52 − y 2 ) mm3
Q = yC A =
( 3000 ) ( 50 ) (87.52 − y 2 )(10−9 )
τ=
( 37.89 ×10−6 ) ( 0.100 )
= 39.59 ( 87.52 − y 2 ) N/m 2
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-93
d=
RILEY, STURGES AND MORRIS
M x (1) ( 6 )( 2 )  + ( 2 ) ( 4 )(1)  + ( 2 ) ( 4 )(1) 
=
= 1.400 in.
A
( 6 )( 2 ) + ( 4 )(1) + ( 4 )(1)
( 8)(1.4 )
I=
3
3
( 6 )( 0.6 )
+
3
3
(1)( 2.6 )
+
3
3
(1)( 2.6 )
+
3
3
= 19.467 in.4
By symmetry, each support carries half the total load
RA = RB = ( 360 × 18 ) 2 = 3240 lb ↑
↑ ΣFy = 0 :
3240 − ( 360 × 2 ) − Vr = 0
Vr = 2520 lb
−1.4 in. ≤ y ≤ +0.6 in. :
t = 8 in.
 (1.4 + y )

Q = yC A = 
(8)(1.4 − y ) 
2


= 4 (1.42 − y 2 ) in.3
2
2
VQ ( 2520 ) ( 4 ) (1.4 − y ) 
τ=
=
It
(19.467 )(8 )
= 64.725 (1.42 − y 2 ) psi
+0.6 in. ≤ y ≤ +2.6 in.
t = 2 in.
 ( 2.6 + y )

Q = yC A = 2 
(1)( 2.6 − y )
2


= ( 2.62 − y 2 ) in.3
τ=
2
2
VQ ( 2520 ) ( 2.6 − y )
=
It
(19.467 )( 2 )
= 64.725 ( 2.62 − y 2 ) psi
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-94*
RILEY, STURGES AND MORRIS
2 RC − ( 20 )(1.4 ) = 0
4 ΣM B = 0 :
RC = 14.0 kN
VA + (14 ) = 0
↑ ΣFy = 0 :
VA = −14 kN = 14 kN ↓
− M A + (14 )( 0.6 ) = 0
4 ΣM cut = 0 :
M A = +8.40 kN ⋅ m
( 80 )(120 )
I=
12
σA =
3
( 75)(100 )
−
12
3
= 5.270 (106 ) mm 4
− M r y − ( 8400 )( −0.050 )
=
= +79.70 (106 ) N/m 2 = 79.70 MPa (T)
−6
I
( 5.270 ×10 )
QA = yC A = 55 ( 80 × 10 ) = 44.0 (103 ) mm3
−6
VQ (14, 000 ) ( 44.0 × 10 )
τA =
=
= 23.38 (106 ) N/m 2 = 23.38 MPa
−6
It ( 5.270 × 10 ) ( 0.005 )
σ p1, p 2 =
σx +σ y
2
 σ x −σ y 
2
± 
 + τ xy
 2 
2
79.70 + 0
2
 79.70 − 0 
=
± 
 + ( 23.38 )
2
2


= 39.85 ± 46.20 MPa
σ p1 = 39.85 + 46.20 = 86.05 MPa ≅ 86.1 MPa (T) ....................................................... Ans.
2
σ p 2 = 39.85 − 46.20 = −6.350 MPa ≅ 6.35 MPa (C)
.................................................. Ans.
σ p 3 = 0 MPa ........................................................................................................................... Ans.
Since
σ p1
and
σ p2
have opposite signs:
τ max = τ p = (σ p1 − σ p 2 ) 2 = 46.2 MPa
1
2
θ p = tan −1
2τ xy
σ x −σ y
=
............................................................................ Ans.
2 ( 23.38 )
1
= +15.20°, − 74.80° ........................ Ans.
tan −1
2
( 79.70 ) − ( 0 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-95*
VA − 50 = 0
↑ ΣFy = 0 :
VA = +50 kip = +50 kip ↑
− M A − ( 50 )( 40 ) = 0
4 ΣM cut = 0 :
(10 )(15)
I=
3
( 9.5)(13)
−
M A = −2000 kip ⋅ in.
3
= 1073.2 in.4
12
− M r y − ( 2000 )( 6.5 )
σA =
=
= +12.113 ksi = 12.113 ksi (T)
I
(1073.2 )
12
QA = yC A = 7 (10 ×1) = 70.0 in 3
τ=
VQ ( 50.0 )( 70.0 )
=
= 6.523 ksi
It (1073.2 )( 0.5 )
σ p1, p 2 =
σx +σ y
2
 σ x −σ y 
2
± 
 + τ xy
2


2
12.113 + 0
2
 12.113 − 0 
=
± 
 + ( −6.523) = 6.057 ± 8.901 ksi
2
2


2
σ p1 = 6.057 + 8.901 = 14.958 ksi ≅ 14.96 ksi (T)
......................................................... Ans.
σ p 2 = 6.057 − 8.901 = −2.844 ksi ≅ 2.84 ksi (C) .......................................................... Ans.
σ p 3 = 0 ksi
Since
σ p1
and
σ p2
.............................................................................................................................. Ans.
have opposite signs:
τ max = τ p = (σ p1 − σ p 2 ) 2 = 8.90 ksi ................................................................................ Ans.
1
2
θ p = tan −1
2τ xy
σ x −σ y
=
2 ( −6.523)
1
= −23.56°, + 66.44° ...................... Ans.
tan −1
2
(12.113) − ( 0 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
7-96
RILEY, STURGES AND MORRIS
RB = RC = 0.5 P
VA = ( 0.5 P ) N ↓
On a section through A:
M A = ( 0.5 P ) N ⋅ m
( 250 )( 350 )
I=
12
σA =
3
( 235)( 300 )
−
12
3
= 364.5 (106 ) mm 4
− M r y − ( 0.5P )( −0.150 )
=
= ( 205.8 P ) N/m 2
−6
I
( 364.5 ×10 )
QA = yC A = 162.5 ( 250 × 25 ) = 1015.6 (103 ) mm3
−6
VQ ( 0.5P ) (1015.6 ×10 )
τA =
=
= ( 92.88 P ) N/m 2
−6
It
( 364.5 ×10 ) ( 0.015)
205.8 P + 0
2
 205.8 P − 0 
+ 
 + ( 92.88P )
2
2


2
σ max = σ p1 =
= 241.52 P ≤ 120 (106 ) N/m 2
P ≤ 497 (103 ) N
2
 205.8P − 0 
6
2
 + ( 92.88P ) = 138.62 P ≤ 75 (10 ) N/m
2


2
τ max = τ p = 
P ≤ 541(103 ) N
Therefore
Pmax = 497 kN .......................................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-97
4 ΣM C = 0 :
↑ ΣFy = 0 :
3P − 8 RB = 0
RB = 0.375 P lb
0.375P − VA = 0
VA = 0.375 P lb
M A − ( 0.375P )( 2 ) = 0
4 ΣM cut = 0 :
( 6 )( 8)
I=
12
σA =
RILEY, STURGES AND MORRIS
3
( 4 )( 6 )
−
−M r y
=
I
M A = 0.75 P lb ⋅ ft
3
= 184.0 in.4
12
− ( 0.75P × 12 )( +2 )
(184.0 )
= −0.09783 psi
QA = yC A = 3.5 ( 6 × 1) + 2  2.5 (1× 1)  = 26.0 in 3
τ=
VQ ( 0.375 P )( 26.0 )
=
= 0.02649 P psi
It
(184.0 )( 2 )
σ p1, p 2 =
σx +σ y
2
 σ x −σ y 
2
± 
 + τ xy
2


2
−0.09783 + 0
2
 −0.09783 − 0 
=
± 
 + ( 0.02649 P )
2
2


= −0.04892 P ± 0.05563P psi
2
σ p1 = −0.04892 P + 0.05563P psi = +0.00671P psi (T)
σ p 2 = −0.04892 P − 0.05563P psi = −0.10455P psi ≤ −400 psi
P ≤ 3826 lb
from which:
Since
σ p1
and
σ p2
have opposite signs:
τ max = τ p = (σ p1 − σ p 2 ) 2 = 0.05563P psi ≤ 200 psi
from which:
Therefore
P ≤ 3595 lb
Pmax = 3595 lb ≅ 3.60 kip ......................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-98*
From Table B-2 for a W 610 × 155 section :
I = 1290 (106 ) mm 4
S = 4230 (103 ) mm3
d = 2c = 611 mm
tw = 12.7 mm
w f = 324 mm
t f = 19.1 mm
RA − (160 × 3) = 0
↑ ΣFy = 0 :
RA = +480 kN = 480 kN ↑
4 ΣM A = 0 :
M A − (160 × 3)(1.5 ) = 0
M A = +720 kN ⋅ m = 720 kN ⋅ m 4
From the shear-force and bending-moment diagrams:
Vmax = 480 kN
M max = 720 kN ⋅ m
At the top of the beam (at the support):
σ max
720 × 103 )
(
M
=
=
= 170.21(106 ) N/m 2 = 170.21 MPa (T)
S ( 4230 ×10−6 )
τ max = σ max 2 = 170.21 2 = 85.11 MPa
At the junction of the flange and the web (at the support):
QJ = yC A = 296 ( 324 × 19.1) = 1832 (103 ) mm3
3
My ( 720 × 10 ) ( 0.2864 )
σ=
=
= 159.85 (106 ) N/m 2 = 159.85 MPa (T)
−6
I
(1290 ×10 )
−6
VQ ( 480, 000 ) (1832 ×10 )
τ=
=
= 53.67 (106 ) N/m 2 = 53.67 MPa
−6
It
(1290 ×10 ) ( 0.0127 )
159.85 + 0
2
 159.85 − 0 
± 
 + ( 53.67 ) = 79.93 + 96.27 = 176.2 MPa (T)
2
2


2
σ max = σ p1 =
τ max = τ p = 96.27 MPa
At the neutral axis (at the support):
QNA = 296 ( 324 × 19.1) + 143.2 ( 286.4 × 12.7 ) = 2353 (103 ) mm3
( 480, 000 ) ( 2353 ×10−6 )
σ max = τ max =
= 68.94 (106 ) N/m 2 = 68.94 MPa
−6
(1290 ×10 ) ( 0.0127 )
Therefore, the maximum stresses occur at the junction of the flange and the web (at the support):
σ max = 176.2 MPa (T) .......................................................................................................... Ans.
τ max = 96.3 MPa ..................................................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-99*
From Table B-1 for a W18 × 60 section :
I = 984 in.4
d = 2c = 18.24 in.
tw = 0.415 in.
S = 108 in.
w f = 7.555 in.
t f = 0.695 in.
3
From symmetry:
RA = RB = 36 2 = 16 kip = 16 kip ↑
From the shear-force and bending-moment diagrams:
Vmax = 18 kip
M max = 180 kip ⋅ ft
At the bottom of the beam (at midspan):
σ max =
M (180 × 12 )
=
= +20.0 ksi = 20.0 ksi (T)
S
(108 )
τ max = σ max 2 = 20.0 2 = 10.00 ksi
At the junction of the flange and the web (at midspan):
QJ = yC A = 8.773 ( 7.555 × 0.695 ) = 46.06 in.3
σ=
− My − (180 ×12 )( −8.425 )
=
= +18.494 ksi
I
( 984 )
τ=
VQ (18 )( 46.06 )
=
= 2.030 ksi
It ( 984 )( 0.415 )
18.494 + 0
2
 18.494 − 0 
± 
σ max = σ p1 =
 + ( 2.030 )
2
2


= 9.247 + 9.467 = 18.71 ksi
τ max = τ p = 9.467 ksi
2
At the neutral axis (at midspan):
QNA = 8.773 ( 7.555 × 0.695 ) + 4.213 ( 8.425 × 0.415 ) = 60.79 in.3
σ max = τ max =
(18)( 60.79 ) = 2.680 ksi
( 984 )( 0.415)
Therefore, the maximum stresses occur at the top and bottom surfaces (at midspan):
σ max = 20.0 ksi (T, bottom; C, top)
................................................................................. Ans.
τ max = 10.00 ksi ...................................................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-100
RA − P = 0
↑ ΣFy = 0 :
RA = + ( P ) N = ( P ) N ↑
4 ΣM A = 0 :
M A − 3P = 0
M A = + ( 3P ) N ⋅ m = ( 3P ) N ⋅ m 4
From the shear-force and bending-moment diagrams:
Vmax = ( P ) N
M max = ( 3P ) N ⋅ m
From Table B-2 for a W 305 × 97 section :
I = 222 (106 ) mm 4
d = 2c = 308 mm
tw = 9.9 mm
S = 1440 (103 ) mm3
w f = 305 mm
t f = 15.4 mm
At the top of the beam (at the support):
σ max =
( 3P ) = 2083P N/m 2
M
=
(
)
S (1440 ×10−6 )
τ max =
σ max
2
=
2083P
= (1042 P ) N/m 2
2
At the junction of the flange and the web (at the support):
QJ = yC A = 146.3 ( 305 ×15.4 ) = 687.1(103 ) mm3
σ=
My ( 3P )( 0.1386 )
=
= (1873P ) N/m 2
−6
I
222
10
×
(
)
( P ) ( 687.1×10−6 )
VQ
=
= ( 312.6 P ) N/m 2
τ=
It ( 222 × 10−6 ) ( 0.0099 )
1873P + 0
2
 1873P − 0 
2
= σ p1 =
± 
 + ( 312.6 P ) = (1924 P ) N/m
2
2


2
σ max
2
 1873P − 0 
2
 + ( 312.6 P ) = ( 987 P ) N/m
2


2
τ max = τ p = 
At the neutral axis (at the support):
QNA = 146.3 ( 305 ×15.4 ) + 69.3 (138.6 × 9.9 ) = 782.3 (103 ) mm3
( P ) ( 782.3 ×10−6 )
σ max = τ max =
= ( 356 P ) N/m 2
−6
( 222 ×10 ) ( 0.0099 )
Therefore, at the top (at the support):
σ max = ( 2083P ) N/m 2 ≤ 125 (106 ) N/m 2
P ≤ 60.0 (103 ) = 60.0 kN
τ max = 1042 P N/m 2 ≤ 75 (106 ) N/m 2
P ≤ 72.0 (103 ) = 72.0 kN
Pmax = 60.0 kN ....................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-101
From symmetry:
RA = RB = P 2 = ( 0.5 P ) kip ↑
From the shear-force and bending-moment diagrams:
Vmax = ( 0.5 P ) kip
M max = ( 4.5P ) kip ⋅ ft = ( 54 P ) kip ⋅ in.
From Table B-1 for a W 24 × 62 section :
I = 1550 in.4
d = 2c = 23.74 in.
w f = 7.040 in.
S = 131 in.3
tw = 0.430 in.
t f = 0.590 in.
At the bottom of the beam (at midspan):
σ max =
M ( 54 P )
=
= ( 0.4122 P ) ksi
S
(131)
τ max =
σ max
2
=
0.4122 P
= ( 0.2061P ) ksi
2
At the junction of the bottom flange and the web (at midspan):
QJ = yC A = 11.575 ( 7.040 × 0.590 ) = 48.08 in.3
σ=
− My − ( 54 P )( −11.28 )
=
= ( 0.3930 P ) ksi
I
(1550 )
τ=
VQ ( 0.5 P )( 48.08 )
=
= ( 0.03607 P ) ksi
It (1550 )( 0.430 )
0.3930 P + 0
2
 0.3930 P − 0 
= σ p1 =
± 
 + ( 0.03607 P ) = ( 0.3963P ) ksi
2
2


2
σ max
2
 0.3930 P − 0 
=τ p = 
 + ( 0.03607 P ) = ( 0.1998 P ) ksi
2


2
τ max
At the neutral axis (at midspan):
QNA = 11.575 ( 7.040 × 0.590 ) + 5.64 (11.28 × 0.430 ) = 75.43 in.3
σ max = τ max =
( 0.5P )( 75.43) = 0.05659 P ksi
(
)
(1550 )( 0.430 )
Therefore:
σ max = ( 0.4122 P ) ksi ≤ 18 ksi
P ≤ 43.7 kip
τ max = ( 0.2061P ) ksi ≤ 10 ksi
P ≤ 48.5 kip
Pmax = 43.7 kip ....................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-102*
(a)
RA = RB = 700 2 = 350 kN ↑
M max = ( 350 ×1.5 ) = 525 kN ⋅ m
Vmax = 350 kN
From Table B-2 for a W 356 × 179 section :
I = 574 (10
6
) mm
d = 2c = 368 mm
S = 3115 (103 ) mm3
4
tw = 15.0 mm
w f = 373 mm
t f = 23.9 mm
At the bottom of the beam (at midspan):
σ max
525 ×103 )
(
M
=
=
= 168.54 (106 ) N/m 2 = 168.54 MPa (T)
S ( 3115 × 10−6 )
τ max = σ max 2 = 168.54 2 = 84.27 MPa
At the junction of the bottom flange and the web (at midspan):
QJ = yC A = 172.05 ( 373 × 23.9 ) = 1533.8 (103 ) mm3
3
− My − ( 525 ×10 ) ( −0.1601)
σ=
=
= 146.43 (106 ) N/m 2 = 146.43 MPa (T)
−6
I
( 574 ×10 )
VQ ( 350, 000 ) (1533.8 ×10
=
It
( 574 ×10−6 ) ( 0.0150 )
−6
τ=
) = 62.35 10 N/m
( )
6
2
= 62.35 MPa
146.43 + 0
2
 146.43 − 0 
= σ p1 =
± 
 + ( 62.34 ) = 169.38 MPa (T)
2
2


2
σ max
2
 146.43 − 0 
=τ p = 
 + ( 62.34 ) = 96.17 MPa
2


2
τ max
At the neutral axis (at midspan):
QNA = 172.05 ( 373 × 23.9 ) + 80.05 (160.1×15 ) = 1726.0 (103 ) mm3
( 350, 000 ) (1726.0 ×10−6 )
σ max = τ max =
= 70.16 (106 ) N/m 2 = 70.16 MPa
−6
( 574 ×10 ) ( 0.0150 )
Therefore, the maximum stresses occur at the junction of the flange and the web (at midspan):
σ max = 169.4 MPa (T & C) ................................................................................................. Ans.
τ max = 96.2 MPa
.................................................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-102
(cont.)
(b)
RA = RB = 350 2 = 175 kN ↑
RILEY, STURGES AND MORRIS
M max = (175 × 3) = 525 kN ⋅ m
Vmax = 175 kN
At the bottom of the beam (at midspan):
σ max =
( 525 ×103 ) = 168.54 106 N/m 2 = 168.54 MPa (T)
M
=
( )
S ( 3115 × 10−6 )
τ max = σ max 2 = 168.54 2 = 84.27 MPa
At the junction of the bottom flange and the web (at midspan):
σ=
− ( 525 × 103 ) ( −0.1601)
= 146.43 (106 ) N/m 2 = 146.43 MPa (T)
( 574 ×10 )
(175, 000 ) (1533.8 ×10 )
τ=
= 31.17 (10 ) N/m
( 574 ×10 ) ( 0.0150 )
−6
−6
6
−6
2
= 31.17 MPa
146.43 + 0
2
 146.43 − 0 
± 
 + ( 31.17 ) = 152.8 MPa (T)
2
2


2
σ max = σ p1 =
2
 146.43 − 0 
=τ p = 
 + ( 31.17 ) = 79.57 MPa
2


2
τ max
At the neutral axis (at midspan):
(175, 000 ) (1726.0 ×10−6 )
σ max = τ max =
= 35.08 (106 ) N/m 2 = 35.08 MPa
−6
( 574 ×10 ) ( 0.0150 )
Therefore, the maximum stresses occur at the top and bottom surfaces at midspan:
σ max = 168.5 MPa (T, bottom; C, top) ............................................................................ Ans.
τ max = 84.3 MPa ..................................................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-103
(a)
RA = RB = 160 2 = 80 kip ↑
M max = ( 80 × 4 ) = 320 kip ⋅ ft
Vmax = 80 kip
From Table B-1 for a W14 × 120 section :
I = 1380 in.4
d = 2c = 14.48 in.
S = 190 in.3
t f = 0.940 in.
tw = 0.590 in.
w f = 14.67 in.
At the bottom of the beam (at midspan):
σ max =
M ( 320 ×12 )
=
= 20.21 ksi (T)
S
(190 )
τ max =
σ max
2
=
20.21
= 10.11 ksi
2
At the junction of the bottom flange and the web (at midspan):
QJ = yC A = 6.77 (14.67 × 0.940 ) = 93.36 in.3
σ=
− My − ( 320 ×112 )( −6.30 )
=
= 17.53 ksi (T)
I
(1380 )
τ=
( 80 )( 93.36 ) = 9.173 ksi
VQ
=
It (1380 )( 0.590 )
17.530 + 0
2
 17.530 − 0 
= σ p1 =
± 
 + ( 9.173) = 21.45 ksi (T)
2
2


2
σ max
2
 17.530 − 0 
 + ( 9.173) = 12.687 ksi
2


2
τ max = τ p = 
At the neutral axis (at midspan):
QNA = 6.77 (14.67 × 0.940 ) + 3.15 ( 6.30 × 0.590 ) = 105.07 in.3
σ max = τ max =
(80 )(105.07 ) = 10.324 ksi
(1380 )( 0.590 )
Therefore, the maximum stresses occur at the junction of the flange and the web (at midspan):
σ max = 21.5 ksi (T & C)
...................................................................................................... Ans.
τ max = 12.69 ksi ...................................................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
7-103
(b)
RILEY, STURGES AND MORRIS
(cont.)
RA = RA = 80 2 = 40 kip ↑
Vmax = 40 kip
M max = ( 40 × 8 ) = 320 kip ⋅ ft
At the bottom of the beam (at midspan):
σ max =
M ( 320 ×12 )
=
= 20.21 ksi (T)
S
(190 )
τ max =
σ max
2
=
20.21
= 10.11 ksi
2
At the junction of the bottom flange and the web (at midspan):
σ=
− ( 320 × 12 )( −6.30 )
= 17.53 ksi (T)
(1380 )
τ=
( 40 )( 93.36 ) = 4.587 ksi
(1380 )( 0.590 )
17.530 + 0
2
 17.530 − 0 
= σ p1 =
± 
 + ( 4.587 ) = 18.66 ksi (T)
2
2


2
σ max
2
 17.530 − 0 
 + ( 4.587 ) = 9.893 ksi
2


2
τ max = τ p = 
At the neutral axis (at midspan):
σ max = τ max =
( 40 )(105.07 ) = 5.162 ksi
(1380 )( 0.590 )
Therefore, the maximum stresses occur at the top and bottom surfaces at midspan:
σ max = 20.2 ksi (T & C) ...................................................................................................... Ans.
τ max = 10.11 ksi
...................................................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-104*
IZ = Iz =
IY = I y =
( 90 )( 90 )
3
−
3
( 60 )( 90 )
3
+
( 60 )( 30 )
3
3
(180 )( 30 )
+
3
12
12
M ry = −20sin10° = −3.473 kN ⋅ m
( 30 )(150 )
3
3
= 55.08 (106 ) mm 4
= 4.05 (106 ) mm 4
M rz = −20 cos10° = −19.696 kN ⋅ m
(a)
tan β =
M rY I Z ( −3.473)( 55.08 )
=
= +2.398
M rZ IY ( −19.696 )( 4.05 )
β = +67.36° = 67.36° 3 .........................................Ans.
(b)
At point A ( y = −150 mm and z = +15 mm ):
σA =
− M rZ YA M rY Z A − ( −19, 696 )( −0.150 ) ( −3473)( +0.015 )
+
=
+
IZ
IY
( 55.08 ×10−6 )
( 4.05 ×10−6 )
σ A = −66.5 (106 ) N/m 2 = 66.5 MPa (C)
................................................................. Ans.
At point B ( y = +90 mm and z = −45 mm ):
σB =
− M rZ YB M rY Z B − ( −19, 696 )( +0.090 ) ( −3473)( −0.045 )
+
=
+
IZ
IY
( 55.08 ×10−6 )
( 4.05 ×10−6 )
σ B = +70.8 (106 ) N/m 2 = 70.8 MPa (T)
................................................................. Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-105*
IZ = Iz =
IY = I y
( 4 )( 6 )
3
−
12
( 2 )( 4 )
=
3
( 3)( 4 )
3
12
( 4 )(1)
+
= 56.0 in 4
3
= 11.0 in.4
12
12
M ry = 10sin 36.87° = 6.00 kip ⋅ in.
M rz = −10 cos 36.87° = −8.00 kip ⋅ in.
(a)
tan β =
( 6.00 )( 56.0 ) = −3.818
M rY I Z
=
M rZ IY ( −8.00 )(11.0 )
β = −75.32° = 75.32° 4 .............................................................................................. Ans.
(b)
At point A ( y = +3 in. and z = +2 in. ):
σA =
− M rZ YA M rY Z A − ( −8.00 )( 3) ( 6.00 )( +2 )
+
=
+
IZ
IY
( 56.0 )
(11.0 )
σ A = +1.519 ksi = 1.519 ksi (T)
................................................................................ Ans.
At point B ( y = −3 in. and z = −2 in. ):
σB =
− M rZ YB M rY Z B − ( −8.00 )( −3) ( 6.00 )( −2 )
+
=
+
IZ
IY
( 56.0 )
(11.0 )
σ B = −1.519 ksi = 1.519 ksi (C) ................................................................................ Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-106
(a) At point A ( y = −90 mm and z = +85 mm ):
 ( 4.02 )( −90 ) − ( −6.05 )( 85 )   10−9 
 I y y − I yz z 
M
=
−
−8000 )

rz
2 
−12  (
2 
 I y I z − I yz 
 ( 4.02 )(16.64 ) − ( −6.05 )  10 
σA = −
σ A = +40.3 (106 ) N/m 2 = 40.3 MPa (T) ...............................Ans.
(b)
I yz
tan β =
Iy
=
( −6.05) = −1.5050
( 4.02 )
β = −56.40° = 56.40° 4 ...........................................................Ans.
(c)
At point B ( y = −90 mm and z = −5 mm ):
 ( 4.02 )( −90 ) − ( −6.05 )( −5 )   10−9 
 I y y − I yz z 
M
=
−
−8000 )


rz
2 
−12  (
2
 I y I z − I yz 
 ( 4.02 )(16.64 ) − ( −6.05 )  10 
σB = −
σ B = −103.5 (106 ) N/m 2 = 103.5 MPa (C) ............................................................. Ans.
At point D ( y = +90 mm and z = +5 mm ):
 ( 4.02 )( +90 ) − ( −6.05 )( +5 )   10−9 
 I y y − I yz z 
M
=
−
−8000 )


rz
2 
−12  (
2
 I y I z − I yz 
 ( 4.02 )(16.64 ) − ( −6.05 )  10 
σD = −
σ D = +103.5 (106 ) N/m 2 = 103.5 MPa (T) ............................................................. Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-107
(a) At point A ( y = + + 5.63 in. and z = +2.37 in. ):
 I y y − I yz z 
M rz
2 
 I y I z − I yz 
σA = −
 ( 89.0 )( +5.63) − ( 52.5 )( +2.37 ) 
= −
 ( −7.50 ×12 )
2
(89.0 )(89.0 ) − ( 52.5)


σ A = +6.56 ksi = 6.56 ksi (T) ................................... Ans.
(b)
tan β =
I yz
Iy
=
( 52.5) = +0.5899
(89.0 )
β = +30.54° = 30.54° 3 .............................................................................................. Ans.
(c)
At point B ( y = −2.37 in. and z = +2.37 in. ):
 ( 89.0 )( −2.37 ) − ( 52.5 )( +2.37 ) 
 I y y − I yz z 
M
=
−

 ( −7.50 × 12 )

rz
2
2
(89.0 )(89.0 ) − ( 52.5)


 I y I z − I yz 
σ B = −5.84 ksi = 5.84 ksi (C) .................................................................................... Ans.
σB = −
At point D ( y = +5.63 in. and z = +1.37 in. ):
 ( 89.0 )( +5.63) − ( 52.5 )( +1.37 ) 
 I y y − I yz z 
M
=
−

 ( −7.50 × 12 )

rz
2
2
( 89.0 )(89.0 ) − ( 52.5)
 I y I z − I yz 


σ D = +7.48 ksi = 7.48 ksi (T) .................................................................................... Ans.
σD = −
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-108*
Iy =
(180 )( 90 )
36
3
= 3.645 (106 ) mm 4
Iz =
( 90 )(180 )
3
36
= 14.580 (106 ) mm 4
− (180 ) ( 90 )
= −3.645 (106 ) mm 4
72
(a) At point A ( y = +60 mm and z = −60 mm ):
2
2
I yz =
 I y y − I yz z 
M rz
2 
 I y I z − I yz 
 ( 3.645 )( +60 ) − ( −3.645 )( −60 )   10−9 
= −
−2000 )
2 
−12  (
 ( 3.645 )(14.580 ) − ( −3.645 )  10 
σA = −
σ A = 0 N/m 2 = 0 MPa ........................................................Ans.
(b)
tan β =
I yz
Iy
=
( −3.645) = −1.000
( 3.645)
β = −45.00° = 45.00° 4 .............................................................................................. Ans.
(c)
At point B ( y = −120 mm and z = +30 mm ):
 ( 3.645 )( −120 ) − ( −3.645 )( +30 )   10−9 
 I y y − I yz z 
σB = −
M rz = − 
  −12  ( −2000 )
2
2 
 I y I z − I yz 
 ( 3.645 )(14.580 ) − ( −3.645 )  10 
σ B = −16.46 (106 ) N/m 2 = 16.46 MPa (C) ............................................................. Ans.
At point D ( y = +60 mm and z = +30 mm ):
 ( 3.645 )( +60 ) − ( −3.645 )( +30 )   10−9 
 I y y − I yz z 
M
=
−
−2000 )


rz
2 
2
−12  (
 ( 3.645 )(14.580 ) − ( −3.645 )  10 
 I y I z − I yz 
σD = −
σ D = +16.46 (106 ) N/m 2 = 16.46 MPa (T) ............................................................. Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-109*
Iy =
Iz
( 6 )(1.5)
3
+
3
( 2 )( 3.5 )
=
3
3
( 4 )( 0.5)
3
+
3
( 4 )( 2.5 )
+
3
3
( 2 )( 2.5 )
3
= 17.33 in.4
3
( 2 )( 0.5 )
−
3
3
= 49.33 in.4
I yz = ( 2 × 4 )(1.5 )( 0.5 ) + ( 2 × 4 )( −1.5 )( −0.5 ) = 12.00 in.4
(a)
At point A ( y = +3.50 in. and z = +1.50 in. ):
 (17.33)( +3.50 ) − (12.00 )( +1.50 ) 
 I y y − I yz z 
M
=
−

 ( −50 )
rz
2
2 
(17.33)( 49.33) − (12.00 ) 
 I y I z − I yz 

σ A = +3.00 ksi = 3.00 ksi (T) ..................................................................................... Ans.
σA = −
(b)
tan β =
I yz
Iy
=
(12.00 ) = +0.6924
(17.33)
β = +34.70° = 34.70° 3 ..................................................................................................... Ans.
(c)
At point B ( y = +3.50 in. and z = −0.50 in. ):
 (17.33)( +3.50 ) − (12.00 )( −0.50 ) 
 I y y − I yz z 
M
=
−

 ( −50 )

rz
2
2
(17.33)( 49.33) − (12.00 ) 

 I y I z − I yz 
σ B = +4.69 ksi = 4.69 ksi (T) .................................................................................... Ans.
σB = −
At point D ( y = −2.50 in. and z = +1.50 in. ):
 (17.33)( −2.50 ) − (12.00 )( +1.50 ) 
 I y y − I yz z 
M
=
−

 ( −50 )

rz
2
2
(17.33)( 49.33) − (12.00 ) 

 I y I z − I yz 
σ D = −4.31 ksi = 4.31 ksi (C) ..................................................................................... Ans.
σD = −
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-110
IZ =
( 50 )( 300 )
3
12
For symmetric bending:
= 112.50 (106 ) mm 4
σ=
IY =
( 300 )( 50 )
12
3
= 3.125 (106 ) mm 4
− M r y − M ( −0.150 )
=
= (1333.3M ) N/m 2
−6
I
(112.50 ×10 )
For non-symmetric bending:
σ=
− M rZ Y M rY Z − ( M cos 3° )( −0.150 ) ( M sin 3° )( +0.025 )
+
=
+
= (1750.2 M ) N/m 2
−6
−6
IZ
IY
(112.50 ×10 )
( 3.125 ×10 )
∆σ =
1750.2 − 1333.3
(100 ) = 31.3% .............................................................................. Ans.
1333.3
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-111
σ=
− M rZ Y M rY Z
+
IZ
IY
IY =
hb3
12
IZ =
M rY = M sin α
σ max =
− ( M cos α )( −h 2 )
( bh
3
12 )
+
( M sin α )( b 2 ) =
( hb
3
12 )
bh3
12
M rZ = − M cos α
6M
( b cos α + h sin α )
b2 h2
dσ max 6 M
= 2 2 ( −b sin α + h cos α ) = 0
dα
bh
b sin α = h cos α
h
12
α = tan −1 = tan −1 = 63.43° .................................................................................. Ans.
b
6
For σ = 2000 psi :
( 6 ) (12 ) ( 2000 )
b 2 h 2σ
=
M=
6 ( b cos α + h sin α ) 6 ( 6 cos 63.43° + 12sin 63.43° )
2
2
M = 128.8 (103 ) lb ⋅ in. = 128.8 kip ⋅ in. .................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
7-112*
RILEY, STURGES AND MORRIS
A = (140 )( 20 ) + (120 )( 20 ) + ( 80 )( 20 ) = 6800 mm 2
Iy
( 20 )(140 )
=
12
+
Iz
(120 )( 20 )
+
3
3
3
3
3
3
− 6800 ( 7.06 ) = 6.61(106 ) mm 4
(120 )( 60 )
−
( 60 )( 60 )
( 20 )( 70 )
+
2
3
3
−
12
( 20 )(10 )
(140 )(80 )
=
3
3
3
3
( 80 )( 80 )
+
3
3
− 6800 (12.36 ) = 23.5 (106 ) mm 4
2
I yz = 0 − ( 60 × 20 )( 70 )( 40 ) − ( 6800 )( 7.06 )(12.36 ) = −3.95 (106 ) mm 4
 ( 6.61) y − ( −3.95 ) z   10−9 
 I y y − I yz z 
=
−
M
20, 000 )

rz
2
2 
−12  (
 I y I z − I yz 
 ( 6.61)( 23.5 ) − ( −3.95 )  10 
=  −946.1(106 ) y − 565.4 (106 ) z  N/m 2 = [ −946.1y − 565.4 z ] MPa
(a) At point A ( y = +72.36 mm and z = −62.94 mm ):
σ = −
σ A =  −946.1( 72.36 ) − 565.4 ( −62.94 )  = −32.9 MPa = 32.9 MPa (C)
(b)
tan β =
I yz
Iy
=
............... Ans.
( −3.95) = −0.5976
( 6.61)
β = −30.86° = 30.86° 4 ..................................................................................................... Ans.
(b)
At point B ( y = −67.64 mm and z = −62.94 mm ):
σ B =  −946.1( −67.64 ) − 565.4 ( −62.94 )  = +99.6 MPa = 99.6 MPa (T) ............. Ans.
At point D ( y = +92.36 mm and z = +17.06 mm ):
σ B =  −946.1( 92.36 ) − 565.4 ( +17.06 )  = −97.0 MPa = 97.0 MPa (C)
............... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-113*
Iy =
( 6 )( 3)
36
3
= 4.50 in.4
Iz =
( 3)( 6 )
3
36
= 18.00 in.4
− ( 6 ) ( 3)
= −4.50 in.4
72
M ry = 10sin 26.57° = 4.473 kip ⋅ in.
2
2
I yz =
M rz = 10 cos 26.57° = 8.944 kip ⋅ in.
 M rz I y + M ry I yz 
 M ry I z + M rz I yz 
 y+
z
2
2
 I y I z − I yz 
 I y I z − I yz 
σ = −
 ( 8.944 )( 4.50 ) + ( 4.473)( −4.50 ) 
 ( 4.473)(18.00 ) + ( 8.944 )( −4.50 ) 
 y+
z
2
2
( 4.50 )(18.00 ) − ( −4.50 )
 ( 4.50 )(18.00 ) − ( −4.50 )



= ( −0.3312 y + 0.6628 z ) ksi
σ = −
(a)
At point A ( y = −4 in. and z = +1 in. ):
σ A = ( −0.3312 )( −4 ) + ( 0.6628 )( +1) = +1.988 ksi = 1.988 ksi (T)
........................ Ans.
At point B ( y = +2 in. and z = −2 in. ):
σ B = ( −0.3312 )( +2 ) + ( 0.6628 )( −2 ) = −1.988 ksi = 1.988 ksi (C) ........................ Ans.
At point D ( y = +2 in. and z = +1 in. ):
σ D = ( −0.3312 )( +2 ) + ( 0.6628 )( +1) = +0.0004 ksi ≅ 0 ksi ..................................... Ans.
(b)
tan β =
M ry I z + M rz I yz
M rz I y + M ry I yz
=
( 4.473)(18.00 ) + ( 8.944 )( −4.50 ) = +2.001
( 8.944 )( 4.50 ) + ( 4.473)( −4.50 )
β = +63.45° = 63.45° 3 ..................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-114
(a)
σ=
− M rZ Y M rY Z
+
IZ
IY
IY =
hb3
12
IZ =
M rY = M sin α
σ max =
(b)
− ( M cos α )( −h 2 )
12 )
+
( M sin α )( b 2 ) =
( hb
3
12 )
dσ max 6 M
= 2 2 ( −b sin α + h cos α ) = 0
dα
bh
Therefore
(c)
( bh
3
M rZ = − M cos α
6M
( b cos α + h sin α ) ............... Ans.
b2 h2
b sin α = h cos α
α = tan −1 ( h b ) ......................................................................................................... Ans.
α = tan −1 ( 2 ) = +63.43° = +63.43° 3
When h = 2b :
hb3 ( 2b ) b
b4
IY =
=
=
12
12
6
3
tan β =
bh3
12
bh3 b ( 2b )
2b 4
IZ =
=
=
12
12
3
3
IZ
2b 4 3
tan α = 4 tan 63.43° = 8.00
IY
b 6
β = tan −1 8.00 = +82.87° = 82.87° 3
.............................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-115
RILEY, STURGES AND MORRIS
M ry = 20sin 36.87° = 12.00 kip ⋅ in.
M rz = 20 cos 36.87° = 16.00 kip ⋅ in.
 M rz I y + M ry I yz 
 M ry I z + M rz I yz 
 y+
z
2
2
 I y I z − I yz 
 I y I z − I yz 
σ = −
 (16.00 )( 8.83) + (12.00 )(11.3) 
y
2
(8.83)( 25.4 ) − (11.3)


 (12.00 )( 25.4 ) + (16.00 )(11.3) 
+
z
2
( 8.83)( 25.4 ) − (11.3)


= ( −2.8665 y + 5.0273 z ) ksi
σ = −
(a)
At point A ( y = +3 in. and z = +3.3125 in. ):
σ A = ( −2.8665 )( +3) + ( 5.0273)( +3.3125 ) = +8.05 ksi = 8.05 ksi (T)
(b)
tan β =
M ry I z + M rz I yz
M rz I y + M ry I yz
=
.................. Ans.
(12.00 )( 25.4 ) + (16.00 )(11.3) = +1.7538
(16.00 )(8.83) + (12.00 )(11.3)
β = +60.31° = 60.31° 3 ..................................................................................................... Ans.
(c)
At point B ( y = −3 in. and z = +0.1875 in. ):
σ B = ( −2.8665 )( −3) + ( 5.0273)( +0.1875 ) = +9.54 ksi = 9.54 ksi (T)
.................. Ans.
At point D ( y = +3 in. and z = −0.1875 in. ):
σ D = ( −2.8665 )( +3) + ( 5.0273)( −0.1875 ) = −9.54 ksi = 9.54 ksi (C) .................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-116*
σ=
K t Mc
I
w 75
=
= 1.25
h 60
K t = 1.70
From Fig. 7-34:
( 20 )( 60 )
I=
12
M=
σI
Kt c
3
r 6
=
= 0.10
h 60
= 360 (103 ) mm 4
(80 ×10 )( 360 ×10 ) = 565 N ⋅ m ............................................................. Ans.
=
6
−9
(1.70 )( 0.030 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-117*
w 38
=
= 1.50
h 14
For K t = 1.40 Fig. 7-34 gives
r h = 0.25
r = 0.25 (1 4 ) = 0.0625 in. .................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-118
σ=
Mc M ( D 2 ) 32 M
=
=
I
π D 4 64 π D 3
σ = Kt
M R ( d 2 ) 32M R K t
=
π d 4 64
πd3
D 100
=
= 1.25
80
d
From Fig. 7-34:
M=
σπ D 3
MR =
32
σπ d 3
32 K t
r
8
=
= 0.10
d 80
K t = 1.90
 (σπ D 3 32 ) − (σπ d 3 32 K t ) 
M − MR 
 (100 )
%R = 
(100 ) = 
 M 


(σπ D3 32 )
 D3 − ( d 3 Kt ) 
 ( d D )3 
 (100 ) = 1 −
=
 (100 )
D3
K t 



 (1 1.25 )3 
% R = 1 −
 (100 ) = 73.0% ................................................................................ Ans.
1.90 

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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-119
σ=
Mc M ( h 2 ) 6M
=
=
I
Bh3 12 Bh 2
σ = Kt
M R ( h 2 ) 6M R Kt
=
bh3 12
bh 2
d 1 16
=
= 1.00
r 1 16
From Fig. 7-34:
M=
σ Bh 2
MR =
6
σ bh 2
6 Kt
r 1 16
=
= 0.10
b 58
K t = 2.30
 (σ Bh 2 6 ) − (σ bh 2 6 K t ) 
M − MR 
 (100 )
%R = 
(100 ) = 
 M 


(σ Bh2 6 )
 B − ( b Kt ) 
 (b B ) 
=
 (100 ) = 1 −
 (100 )
B
K
t




 ( 0.625 0.75 ) 
% R = 1 −
 (100 ) = 63.8% ........................................................................ Ans.
2.30


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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-120*
σ=
Mc M ( h 2 ) 6 M
=
= 2
I
bh3 12
bh
M=
M R ( h 2)
6M R Kt
σ = Kt
=
3
( b − d ) h 12 ( b − d ) h2
h 200
=
= 8.00
25
d
From Fig. 7-34:
σ bh 2
MR =
6
σ ( b − d ) h2
6 Kt
d 25
=
= 0.17
b 150
K t = 2.55
 b − ( b − d ) Kt 
 b−d
M − MR 
%R = 
(100 ) = 
 (100 ) = 1 −
 (100 )

b
bK t 
 M 




150 − 25 
% R = 1 −
 (100 ) = 67.3% .............................................................................. Ans.
150
2.55
(
)


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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-121
For 04%C hot-rolled steel:
I=
At the wall:
σ=
σ y = 53 ksi
π (1.5 )
4
4
= 3.976 in.4
Mc ( 22 P )(1.5 )
=
≤ 17.667 ksi
3.976
I
At the reduced section:
I=
π (1.365 )
4
4
= 2.727 in.4
w
3
=
= 1.10
h 2.73
From Fig. 7-34:
σ = Kt
Therefore
RILEY, STURGES AND MORRIS
σ all = σ y FS = 53 3 = 17.667 ksi
M = ( 22 P ) kip ⋅ in.
P ≤ 2.13 kip
M = (12 P ) kip ⋅ in.
r 0.25
=
= 0.09
h 2.73
K t = 1.50
(12 P )(1.365) ≤ 17.667 ksi
Mc
= (1.50 )
P ≤ 1.961 kip
2.727
I
Pmax = 1.961 kip ....................................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
7-122*
For 04%C hot-rolled steel:
σ = Kt
σ y = 360 MPa
RILEY, STURGES AND MORRIS
σ all = σ y FS = 360 4 = 90 MPa
M ( h 2)
6 Kt M
Mc
= Kt
=
3
I
( b − d )( h ) 12 ( b − d ) h 2
6 ( 50, 000 )
b − d 6M
=
=
= 0.08333 m
2
σ h ( 90 × 106 ) ( 0.200 )2
Kt
Rearranging gives
h 200
=
= 8.00
25
d
(a)
d 25
=
b b
Solve by trial and error.
Guess that K t ≅ 2.50 .
Then Eq. (a) gives
b − d = 0.208 m
b = 208 + 25 = 233 mm
Then from Fig. 7-34:
d b = 0.107
K t = 2.70
Then Eq. (a) gives
b − d = 0.225 m
b = 212 + 25 = 250 mm
Then from Fig. 7-34:
d b = 0.100
K t = 2.70
Guess that K t ≅ 2.70 .
Therefore
bmin ≅ 250 mm ......................................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-123
I=
( 2 − 0.625)( 4 )
12
σ = Kt
3
( PL 6 ) c
Mc
= Kt
I
I
h
4
=
= 6.40
d 0.625
From Fig. 7-34:
K t ≅ 2.35
L=
 P   L   PL 
M =    = 
 kip ⋅ in.
 2  3   6 
= 7.333 in.4
gives
L=
6σ I
K t cP
d 0.625
=
= 0.3125
2
b
6 ( 20 )( 7.333)
= 37.445 in. = 3.12 ft ........................................................................ Ans.
( 2.35)( 2 )( 5)
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-124
σ = Kt
h ≥
2
d/b
0.05
0.10
0.20
0.25
0.30
0.35
0.40
RILEY, STURGES AND MORRIS
M ( h 2)
6Kt M
Mc
= Kt
=
≤ 200 MPa
3
I
( b − d ) h 12 ( b − d ) h 2
6 K t (10 × 103 )
( b − d ) ( 200 ×10
b (mm)
200
100
50
40
33
29
25
Kt
2.80
2.70
2.50
2.45
2.37
2.30
2.25
6
)
=
K t ( 300 ×10−6 )
( b − 0.010 )
m2
h (mm)
66
95
137
157
176
191
212
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-125
M = ( P 2 )( L 3) = ( 2.25 )(14 ) = 31.5 kip ⋅ in.
σ = Kt
h2 ≥
d/b
0.15
0.20
0.30
0.40
0.50
0.60
0.70
RILEY, STURGES AND MORRIS
M ( h 2)
6 Kt M
Mc
= Kt
=
≤ 15 ksi
3
I
( b − d ) h 12 ( b − d ) h 2
6 K t ( 31.5 )
12.60 K t
=
in.2
( b − d )(15) ( b − 0.75)
b (in.)
5.00
3.75
2.50
1.88
1.50
1.25
1.07
Kt
2.60
2.50
2.35
2.25
2.17
2.10
2.03
h (in.)
2.78
3.24
4.11
5.00
6.04
7.27
8.94
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-126*
For a W 203 × 50 section
w f = 205 mm
RILEY, STURGES AND MORRIS
d = 2c = 210 mm
tw = 9.1 mm
t f = 14.2 mm
S = 582 (103 ) mm3
M e = σ y S = ( 250 ×106 )( 582 × 10−6 )
M e = 145.5 (103 ) N ⋅ m = 145.5 kN ⋅ m ..................................................................... Ans.
M p = 2 ( 250 ×106 ) ( 0.0979 )( 0.205 × 0.0142 ) 
+ 2 ( 250 × 106 ) ( 0.0454 )( 0.0908 × 0.0091) 
M p = 142.49 (103 ) + 18.76 (103 ) = 161.3 (103 ) N ⋅ m = 161.3 kN ⋅ m ....................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
7-127*
For a W 33 × 201 section
w f = 15.745 in.
RILEY, STURGES AND MORRIS
d = 2c = 33.68 in.
tw = 0.715 in.
t f = 1.150 in.
S = 684 in.3
M e = σ y S = ( 36 )( 684 ) = 24, 624 kip ⋅ in. ≅ 24, 600 kip ⋅ in. ....................................... Ans.
M p = 2 ( 36 ) (16.265 )(15.745 ×1.150 )  + 2 ( 36 ) ( 7.845 )(15.69 × 0.715 ) 
M p = 21, 204 + 6337 = 27,541 kip ⋅ in. ≅ 27,500 kip ⋅ in. .................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
7-128
For a W 762 × 196 section
w f = 268 mm
RILEY, STURGES AND MORRIS
d = 2c = 770 mm
tw = 15.6 mm
t f = 25.4 mm
S = 6225 (103 ) mm3
M e = σ y S = ( 250 ×106 )( 6225 × 10−6 )
M e = 1556 (103 ) N ⋅ m = 1556 kN ⋅ m ....................................................................... Ans.
M p = 2 ( 250 ×106 ) ( 0.3723)( 0.268 × 0.0254 ) 
+ 2 ( 250 × 106 ) ( 0.1798 )( 0.3596 × 0.0156 ) 
M p = 1267.2 (103 ) + 504.3 (103 ) = 1772 (103 ) N ⋅ m = 1772 kN ⋅ m ......................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-129*
Me =
σ yI
c
=
σ y ( bh3 12 )
( h 2)
=
RILEY, STURGES AND MORRIS
bh 2σ y
6
 bh   h  bh σ y
M p = 2σ y     =
4
 2  4 
2
Mp
Me
=
14
= 1.5 ............................................... Ans.
16
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-130
Me =
σ yI
c
=
σ y (π R 4 4 )
( R)
=
π R 3σ y
4
3
 π R2   4R  4R σ y
M p = 2σ y 
=


3
 2   3π 
Mp
Me
=
43
= 1.698 ................................................................................................................ Ans.
π 4
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-131*
yC =
RILEY, STURGES AND MORRIS
M x ( 5 ) ( 2 )(10 )  + (1) ( 8 )( 2 )  + ( 5 ) ( 2 )(10 ) 
=
= 3.857 in.
A
( 2 )(10 ) + ( 8)( 2 ) + ( 2 )(10 )
(12 )( 3.857 )
I=
3
3
( 8)(1.857 )
−
3
3
( 4 )( 6.143)
+
3
3
= 521.5 in.4
Elastic action:
Me =
σ yI
c
=
σ y ( 521.5 )
6.143
= 84.89σ y
For fully plastic action:
AT = AC = 56 2 = 28 in.2
4 (10 − c p ) = 28 in.2
c p = 3.00 in.
M p = (1.5 )(12 × 3) − ( 0.5 )( 8 × 1) + ( 3.5 )( 4 × 7 )  σ y
= 148.0σ y
Mp
Me
=
148.0
= 1.743 ............................................................................................................. Ans.
84.89
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-132*
(a)
Me =
σ yI
c
=
σ y ( a 4 12 )
( a 2)
=
RILEY, STURGES AND MORRIS
a 3σ y
6
3
 a2   a  a σ y
M p = 2σ y     =
4
 2  4 
Mp
Me
(b)
=
Me =
14
= 1.500 ........................................................ Ans.
16
σ yI
c
=
σ y ( a 4 12 )
( a sin 45° )
= 0.11785a 3σ y
 a 2   a sin 45° 
3
M p = 2σ y   
 = 0.23570a σ y
3

 2 
Mp
Me
=
0.23570
= 2.00 ......................................................................................................... Ans.
0.11785
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-133
(a) Assume that the lower flange will be fully plastic.
σ a −1 = 36 ( a − 1) a  ksi
 ( 36 )(10 × a )   ( 36 )( a − 1)   ( 9 )( a − 1) 
FC = 
−


a
2
2

 


( 36 )(1× a ) +
FT =
2
( 36 )(1)( 9 − 2a ) + ( 36 )(1)( 6 )
72a 2 − 216a − 162 = 0
Equating FC to FT gives:
a = 3.621 in. ≅ 3.62 in. ............................................................................ Ans.
From which
10 − 2a = 10 − 2 ( 3.621) = 2.758 in. > 1 in.
Therefore, the lower flange will be fully plastic as was assumed.
(b)
σ a −1 = 36 ( 3.621 − 1) 3.621 = 26.06 ksi
 ( 36 )(10 × 3.621)   2 ( 3.621)   ( 26.06 )( 9 × 2.621)   2 ( 2.621) 
M =

−


2
3
2
3


 


 ( 36 )(1× 3.621)   2 ( 3.621) 
+

 + ( 36 )(1× 1.758 )( 4.50 ) + ( 36 )(1× 6 )( 5.879 )
2
3



M = 2748 kip ⋅ in. ≅ 2750 kip ⋅ in. ..................................................................................... Ans.
(c)
yC =
I=
M x ( 9.5 )(10 ×1) + ( 5 )( 8 ×1) + ( 0.5 )( 6 ×1)
=
= 5.75 in.
A
(10 ×1) + (8 ×1) + ( 6 ×1)
(10 )( 4.25)
3
3
−
( 9 )( 3.25)
3
+
( 6 )( 5.75)
3
3
σ I ( 36 )( 354.5 )
=
= 2219 kip ⋅ in.
Me =
c
( 5.75)
For fully plastic action:
AT = ( 6 × 1) + (1) ( c p − 1) = 12 in.2
3
−
( 5)( 4.75)
3
3
= 354.5 in.4
AT = AC = 24 2 = 12 in.2
c p = 7.00 in.
M p = ( 36 )( 6 ×1)( 6.5 ) + ( 36 )( 6 × 1)( 3) + ( 36 )( 2 ×1)(1) + ( 36 )(10 × 1)( 2.5 ) = 3024 kip ⋅ in.
Mp
Me
=
3024
= 1.363 .............................................................................................................. Ans.
2219
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-134
(a) Assume that the lower flange will be only partially plastic.
σ a − 40 =  240 ( a − 40 ) a  MPa
 ( 240 )(120 × a )   ( 240 )( a − 40 )   ( 80 )( a − 40 ) 
FC = 
−


a
2
2

 


FT =
( 240 )(80 × a ) −  ( 240 )(160 − a )   ( 40 )(160 − a ) 


a


+ ( 240 )( 80 )( 200 − 2a )
2
2


a 2 − 120a + 2800 = 0
Equating FC to FT gives:
a = 88.28 mm ≅ 88.3 mm ...................................................................... Ans.
From which
200 − 2a = 200 − 2 ( 88.28 ) = 23.44 mm < 40 mm
Therefore, the lower flange will be only partially plastic as was assumed.
(b)
σ a − 40 =  240 ( 88.28 − 40 ) 88.28 = 131.26 MPa
σ 160− a =  240 (160 − 88.28 ) 88.28 = 194.98 MPa
 ( 240 )(120 × 88.28 )   2 ( 0.08828 )   (131.26 )( 80 × 48.28 )   2 ( 0.04828 ) 
M =

−


2
3
2
3


 


 ( 240 )( 80 × 88.28 )   2 ( 0.08828 )   (194.98 )( 40 × 71.72 )   2 ( 0.07172 ) 
+


−

2
3
2
3



 

+ ( 240 )( 80 × 23.44 )( 0.100 )
M = 148.2 (103 ) N ⋅ m ≅ 148.2 kN ⋅ m ............................................................................. Ans.
(c)
yC =
M x ( 20 )( 80 × 40 ) + (100 )( 40 × 120 ) + (180 )(120 × 40 )
=
= 110 mm
A
( 80 × 40 ) + ( 40 ×120 ) + (120 × 40 )
(120 )( 90 )
I=
3
3
(80 )( 50 )
−
3
3
−
( 40 )( 70 )
( 80 )(110 )
+
3
3
3
= 56.75 (106 ) mm 4
3
σ I ( 240 )( 56.75 )
=
= 123.82 (103 ) N ⋅ m
Me =
c
( 0.110 )
= 123.82 kN ⋅ m
For fully plastic action:
AT = AC = 12,800 2 = 6400 mm 2
AT = ( 40 × 80 ) + ( 40 ) ( c p − 40 ) = 6400 mm 2
c p = 120 mm
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-134
RILEY, STURGES AND MORRIS
(cont.)
M p = ( 240 )( 40 × 80 )( 0.100 ) + ( 240 )( 40 × 80 )( 0.040 )
+ ( 240 )( 40 ×120 )( 0.060 ) + ( 240 )( 40 × 40 )( 0.020 )
= 184.32 (103 ) N ⋅ m = 184.32 kN ⋅ m
Mp
Me
=
184.32
= 1.489 ........................................................................................................... Ans.
123.82
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-135*
(a)
RILEY, STURGES AND MORRIS
σ a − 2 = 36 ( a − 2 ) a  ksi
 ( 36 )( 4 × a )   ( 36 )( a − 2 )   ( 2.5 )( a − 2 ) 
FC = 

−

a
2
2


 

( 36 )(1.5 × a ) +
FT =
2
Equating FC to FT gives:
108a 2 − 360a − 180 = 0
a = 3.775 in. ≅ 3.78 in. ............................................................................ Ans.
From which
(b)
( 36 )(1.5)(10 − 2a ) = 540 − 81a
σ a − 2 = 36 ( 3.775 − 2 ) 3.775 = 16.927 ksi
 ( 36 )( 4 × 3.775 )   2 ( 3.775 )   (16.927 )( 2.5 × 1.775 )   2 (1.775 ) 
M =

−


2
3
2
3


 


 ( 36 )(1.5 × 3.775 )   2 ( 3.775 ) 
+

 + ( 36 )(1.5 × 2.45 )( 5 )
2
3



M = 1558 kip ⋅ in. .................................................................................................................. Ans.
(c)
yC =
M x ( 4 )( 8 ×1.5 ) + ( 9 )( 4 × 2 )
=
= 6.00 in.
A
( 8 ×1.5) + ( 4 × 2 )
(1.5 )( 6 )
I=
3
( 4 )( 4 )
+
3
( 2.5 )( 2 )
−
3
= 186.67 in.4
3
3
σ I ( 36 )(186.67 )
=
= 1120.0 kip ⋅ in.
Me =
c
( 6)
3
AT = AC = 20 2 = 10 in.2
For fully plastic action:
AT = (1.5 ) ( c p ) = 10 in.2
c p = 6.667 in.
M p = ( 36 )(1.5 × 6.667 )( 3.333) + ( 36 )(1.5 × 1.333)( 0.667 )
+ ( 36 )( 4 × 2 )( 2.333) = 1919.9 kip ⋅ in.
Mp
Me
=
1919.9
= 1.714 ........................................................................................................... Ans.
1120.0
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-136
For a WT 305 × 70 section
(a)
d = 2c = 308.7 mm
tw = 13.1 mm
w f = 230.3 mm
t f = 22.2 mm
A = 8905 mm 2
S = 333 (103 ) mm3
σ a − 22.2 =  250 ( a − 22.2 ) a  MPa
 ( 250 )( 230.3 × a )   ( 250 )( a − 22.2 )   ( 217.2 )( a − 22.2 ) 
FC = 

−

a
2
2


 

FT =
( 250 )(13.1× a ) +
2
Equating FC to FT gives:
( 250 )(13.1)( 308.7 − 2a )
a 2 + 29.69a − 2043 = 0
a = 32.73 mm ≅ 32.7 mm ...................................................................... Ans.
From which
308.7 − 2a = 308.7 − 2 ( 32.73) = 243.24 mm
(b)
σ a − 22.2 =  250 ( 32.73 − 22.2 ) 32.73 = 80.44 MPa
 ( 250 )( 230 × 32.73)   2 ( 0.03273)   ( 80.44 )( 217.2 × 10.53)   2 ( 0.01053) 
M =

−


2
3
2
3


 


 ( 250 )(13.1× 32.73)   2 ( 0.03273) 
+

 + ( 250 )(13.1× 243.24 )( 0.15435 )
2
3



M = 144.04 (103 ) N ⋅ m ≅ 144.0 kN ⋅ m ........................................................................... Ans.
(c)
M e = σ S = ( 250 )( 333) = 83.25 (103 ) N ⋅ m = 83.25 kN ⋅ m
For fully plastic action:
AT = AC = 8905 2 = 4453 mm 2
AC = ( 230.3) ( c p ) = 4453 mm 2
c p = 19.34 mm < 22.2 mm
Therefore, the neutral axis for fully plastic action is in the flange.
M p = ( 250 )( 230.3 ×19.34 )( 0.00967 )
+ ( 250 )( 230.3 × 2.86 )( 0.00143)
+ ( 250 )(13.1× 286.5 )( 0.14611)
= 148.10 (103 ) N ⋅ m = 148.10 kN ⋅ m
Mp
Me
=
148.10
= 1.779 ........................................................................................................... Ans.
83.25
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-137*
ε=
εc y
c
=
εc y
( h 2)
=
M = ∫ yσ dA = 2∫
4 h
= b 
5 2
Therefore
52
 2ε c y 

 h 
2ε c y
h
h2
0
12
σ = Kε 1 2 = K 
 2ε 
K c 
 h 
12
h2
4
 2ε 
y bdy = Kb  c   y 5 2 
0
5
 h 
12
32
 σ 
 12 
y 
5 2M
σ=
bh 2
12
 y
 
h
.................................................................................................. Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-138
ε max = 3.5 (10−3 )
σ max = 99.3 MPa
Since
σ = 2792ε 0.59 MPa
(a)
ε=
εc y
c
=
0.0035 y
(c)
 0.0035 y 

c


0.59
σ = 2792 
FC = ∫
0.050
0
 y
99.29 (106 )  
c
c
 y
FT = ∫ 99.29 (106 )  
0
c
Since FC = FT :
 y
= 99.29  
c
∫
0.050
0
0.59
 y
4 
c
0.59
0.59
 y
MPa = 99.29 (106 )  
c
0.59
N/m 2
( 0.100 ) dy
( 0.025) dy
0.59
c y 
dy = ∫  
0 c
 
0.59
dy
c1.59
1.59
c = 0.11957 m ≅ 119.6 mm .................................................................................. Ans.
0.02148 =
Therefore
(b)
M = ∫ yσ dA = ∫
0.05
0
+∫
0.11957
0
= 34.76 (106 ) ∫
Therefore
 y
99.29 (10 )  
c
0.05
0
0.59
6
 y
99.29 (10 )  
c
y ( 0.100 ) dy
0.59
6
y1.59 dy + 8.69 (106 ) ∫
y ( 0.025 ) dy
0.11957
0
y1.59 dy
M = 19.43 (103 ) N ⋅ m = 19.43 kN ⋅ m ............................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-139*
I=
( 4 )( 7 )
12
3
−
( 3.75)( 7 − 2t )
12
3
3

3.75 ( 7 − 2t )  4
= 114.33 −
 in.
12


For a channel section:
( 3.875) ( 7 − t ) t
b 2 h 2t
e=
=
= 1.50 in.
3
4I

3.75 ( 7 − 2t ) 
4 114.33 −

12


2
Solving by trial and error yields:
2
t = 0.250 in. .................................................................................. Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-140*
( 2.5)(100 )
I=
3
+ 2 ( 2.5 × 50 )( 25 )
RILEY, STURGES AND MORRIS
2
12
= 364.6 (103 ) mm 4
(a)
QA = ( 25 )( 50 × 2.5 ) = 3125 mm3
τA =
( 2500 ) ( 3125 ×10−9 )
VQA
=
= 8.571(106 ) N/m 2
−9
It f
( 364.6 ×10 ) ( 0.0025)
F1 = (τ f Af 2 ) = ( 8.571× 106 ) ( 0.0025 × 0.050 ) 2 = 535.7 N
4 ΣM O = 0 :
e=
(b)
Pe − F1d = 0
( 535.7 )( 50 ) = 10.71 mm ............................................................................................. Ans.
2500
QO = ( 25 )( 50 × 2.5 ) + ( 25 )( 50 × 2.5 ) = 6250 mm3
( 2500 ) ( 6250 ×10−9 )
VQO
τO =
=
= 17.14 (106 ) N/m 2
−9
Itw
( 364.6 ×10 ) ( 0.0025)
τ O = 17.14 MPa .............................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-141
RILEY, STURGES AND MORRIS
QA1 = ( 6.5 )( 7 ×1) = 45.5 in.3
QA 2 = ( 6.5 )( 3 ×1) = 19.5 in.3
QAw = 45.5 + 19.5 = 65.0 in.3
QB = ( 65.0 ) + ( 3)( 6 × 1) = 83.0 in.3
bh3 (10 )(14 ) ( 9 )(12 )
I=
=
−
= 990.7 in.4
12
12
12
3
(a)
τ A1 =
3
VQA1 (100 )( 45.5 )
=
= 4.593 ksi
It f
( 990.7 )(1)
F1 = (τ A1 A1 2 ) = ( 4.593)( 7 × 1) 2
= 16.076 kip
τ A2 =
VQA 2 (100 )(19.5 )
=
= 1.968 ksi
It f
( 990.7 )(1)
F2 = (τ A 2 A2 2 ) = (1.968 )( 3 × 1) 2 = 2.952 kip
Pe − ( F1 − F2 ) d = 0
4 ΣM B = 0 :
e=
(b)
(16.076 − 2.952 )(13) = 1.706 in. ..........................Ans.
τ Aw =
τB =
VQAw
Itw
100
(100 )( 65.0 ) = 6.56 ksi
=
( 990.7 )(1)
VQB (100 )( 83.0 )
=
= 8.38 ksi
Itw
( 990.7 )(1)
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-142
F = τ avg A =
(a)
VQA
2 It
F1 =
V  ( 253)( 93 ×14 ) 
V
6

 ( 93 ×14 ) = (15.317 ×10 )
I
2I 
14

F2 =
V  ( 222 )( 93 × 16 ) 
V
6

 ( 93 ×16 ) = (15.360 ×10 )
I
2I 
16

F3 =
V  ( 253)( b − 93)(14 ) 
V
2
3

 ( b − 93)(14 ) = ( b − 93) (1.7710 × 10 )
I
2I 
14

4 ΣM O = 0 :
F3 h3 − F1h1 − F2 h2 = 0
2
V
V
V
3 
6 
6 
 I ( b − 93) (1.7710 ×10 )  ( 474 ) =  I (15.317 × 10 )  ( 474 ) +  I (15.360 × 10 )  ( 444 )
( b − 93)
2
= 16.773 (103 )
b 2 − 186b − 8124.0 = 0
b = 222.5 mm ≅ 223 mm .................................................................................................... Ans.
(b)
QNA = ( 237 )( 223 × 14 ) + ( 222 )( 93 × 16 ) + (111)( 222 ×14 ) = 1.4152 (106 ) mm3
I = 2 ( 223 × 14 )( 237 ) + 2 ( 93 ×16 )( 222 )
2
= 614.1(10
6
) mm
2
(16 )( 444 )
+
3
12
4
VQNA ( 40, 000 ) (1.4152 ×10 )
=
=
= 6.58 (106 ) N/m 2
−6
Itw
( 614.1×10 ) ( 0.014 )
−3
τ max
τ max = 6.58 MPa
............................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-143*
RILEY, STURGES AND MORRIS
QA = ( 6 )( 4 × 0.25 ) = 6.0 in.3
QB = ( 3)( 4 × 0.25 ) = 3.0 in.3
( 0.25 )(12 )
I≅
12
3
+ 2 ( 4 × 0.25 )( 6 )
2
+ 2 ( 4 × 0.25 )( 3) = 126.0 in.4
2
(a)
F1 =
τ A AF
F2 =
4 ΣM O = 0 :
Since V = P :
(b)
2
τ B AF
2

 ( 4 × 0.25 )
V ( 6)
=
= 0.09524V

2
 (126.0 )( 0.25 ) 

 ( 4 × 0.25 )
V ( 3)
=
= 0.04762V

2
 (126.0 )( 0.25 ) 
Pe − F1 (12 ) − F2 ( 6 ) = 0
e = 0.09524 (12 ) + 0.04762 ( 6 ) = 1.429 in. ......................................... Ans.
QNA = ( 6 )( 4 × 0.25 ) + ( 3)( 4 × 0.25 ) + ( 3)( 6 × 0.25 ) = 13.5 in.3
τO =
VQNA (1500 )(13.5 )
=
= 643 psi ............................................................................. Ans.
Itw
(126.0 )( 0.25 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-144*
RILEY, STURGES AND MORRIS
QA = ( 90 )( 90 × 6 ) = 48.6 (103 ) mm3
QB = ( 90 )( 90 × 6 ) + ( 45 )( 90 × 6 ) = 72.9 (103 ) mm3
QC = ( 30 )( 60 × 6 ) = 10.8 (103 ) mm3
( 6 )(180 )
I≅
12
τA =
3
( 6 )(120 )
+
12
3
+ 2 ( 90 × 6 )( 90 ) = 12.528 (106 ) mm 4
2
( 6000 ) ( 48.6 ×10−6 )
VQA
=
= 3.879 (106 ) N/m 2 ≅ 3.88 MPa
−6
It
(12.528 ×10 ) ( 0.006 )
( 6000 ) ( 72.9 ×10−6 )
VQB
τB =
=
= 5.819 (106 ) N/m 2 ≅ 5.82 MPa
−6
It
(12.528 ×10 ) ( 0.006 )
( 6000 ) (10.8 ×10−6 )
VQC
τC =
=
= 0.8621(106 ) N/m 2 ≅ 0.862 MPa
−6
It
(12.528 ×10 ) ( 0.006 )
τ max = τ B = 5.82 MPa ...................................................................................... Ans.
Therefore:
F1 = (τ A A1 2 ) = ( 3.879 ×106 ) ( 0.090 × 0.006 ) 2 = 1047.3 N
F3 = (τ C A3 2 ) = ( 0.8621× 106 ) ( 0.120 × 0.006 ) 2 = 413.8 N
4 ΣM B = 0 : − Pe − F1 (180 ) + F3 (180 ) = 0
Since V = P :
e=
( 413.8)(180 ) − (1047.3)(180 ) = −19.00 mm = 19.00 mm ← ........................... Ans.
6000
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-145
2
2

 th3 
 h   th
I ≅ 2   + 2 ( bt )    =
( h + 3b )
 2   6

 12 
V s
Vs 2
τ 1 = ∫ yt dy =
It 0
2I
F1 = ∫ τ 1 dA = ∫
h2
0
τ 2 = τ 1 s =h 2 +
Vs 2
Vh3t
( t ds ) =
2I
48 I
V s h 
Vh 2 Vhs
t
dx
=
+
)
 (
2I
It ∫0  2 
8I
2
b  Vh
Vhs 
Vhbt
+
F2 = ∫ τ 2 dA = ∫ 
t ds ) =
(
( h + 2b )

0
2I 
8I
 8I
4 ΣM O = 0 :
Since V = P :
Pe − 2 F1b − F2 h = 0
e=
b ( 2h + 3b )
h 2tb
............................................................ Ans.
( 2h + 3b ) =
12 I
2 ( h + 3b )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-146*
(a)
π
( R cos θ ) ( tR dθ ) =
0
I = ∫ y 2 dA = ∫
τ=
2
V s
V
yt ds =
∫
0
It
I
θ
π R 3t
2
∫0 ( R cos θ ) ( R dθ ) =
2
VR 2
sin θ
I
π
 VR 2

T = ∫ R dF = ∫ Rτ dA = ∫ R 
sin θ  ( tR dθ )
0
 I

4
4
VR t π
2VR t 2VR 4t 4VR
d
=
=
=
=
sin
θ
θ
I ∫0
I
π R 3t 2
π
4 ΣM O = 0 :
Since V = P :
(b)
Pe − T = 0
Pe − 4VR π = 0
e = 4 R π ..................................................................................................... Ans.
I = π R 3t 2 = π ( 25 ) ( 2.5 ) 2 = 61.359 (103 ) mm 4
3
( 440 )( 0.025) sin 90° = 4.48 106 N/m 2
VR 2
τA =
sin θ A =
( )
I
( 61.359 ×10−9 )
2
τ A = 4.48 MPa
....................................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-147*
(a)
I = ∫ y 2 dA = ∫
2π
0
τ=
( R sin θ ) ( tR dθ ) = π R3t
2
V s
V
yt ds =
∫
It 0
I
θ
∫0 ( R sin θ ) ( R dθ ) =
2
VR 2
(1 − cos θ )
I
2π
VR 2

1 − cos θ )  ( tR dθ )
T = ∫ R dF = ∫ Rτ dA = ∫ R 
(
0
 I

4
4
2π VR t 2π VR 4t
VR t 2π
=
=
= 2VR
(1 − cos θ ) dθ =
π R 3t
I ∫0
I
4 ΣM O = 0 :
Since V = P :
(b)
Pe − T = 0
Pe − 2VR = 0
e = 2 R .......................................................................................................... Ans.
I = π R 3t = π ( 2 ) ( 0.1) = 2.51327 in.4
3
(110 )( 2 ) 1 − cos180° = 350 psi ........................................ Ans.
VR 2
τA =
(1 − cos θ A ) =
(
)
I
( 2.51327 )
2
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-148
Let a = 20 mm
QA = (1.5a )( at ) = 1.5a 2t
QB 2 = 1.5a 2t + ( 2a )( 2at ) = 5.5a 2t
QB 3 = ( 2a )( 2at ) = 4a 2t
QBw = 5.5a 2t + 4a 2t = 9.5a 2t
QO = 9.5a 2t + ( a )( 2at ) = 11.5a 2t
( 2t )( 4a )
I=
12
(a)
τ1 =
3
( t )( 2a )
−
12
3
+ 2 ( 4a × t )( 2a ) = 42a 3t
2
V s
V s
V
=
y
t
ds
(
)
( a + s ) t ds =
∫
∫
0
0
It
It
I
F1 = ∫ τ 1 dA = ∫
a
0
V
s2 
2Va 3t
2Va 3t
V
=
+
 as +  ( t ds ) =
3
2
3I
I
3 ( 42a t ) 63
F2 = τ avg A2 =
VQavg
F3 = τ avg A3 =
VQavg
4 ΣM O = 0 :

s2 
as
+


2

It
It
( 2at ) =
V ( 3.5a 2t )
( 2at ) =
V
6
( 42a t ) ( t )
V ( 2a t )
V
( 2at ) =
( 2at ) =
10.5
( 42a t ) ( t )
3
2
3
Pe − 2 F1 ( 2a ) − ( F2 − F3 )( 4a ) = 0
Pe − 2 (V 63)( 2a ) − (V 6 )( 4a ) + (V 10.5 )( 4a ) = 0
Since V = P :
e=
4a 4a 4a 22a 22 ( 20 )
+
−
=
=
= 6.98 mm .............................. Ans.
63 6 10.5 63
63
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-149
RILEY, STURGES AND MORRIS
QA = (1)( 2 × 0.125 ) = 0.25 in.3
QB = 0.25 + (1)( 4 × 0.125 ) = 0.75 in.3
( 0.375)( 4 )
I=
12
3
= 2.00 in.4
τA =
( 300 )( 0.25) = 300 psi
VQA
=
It
( 2.00 )( 0.125)
τB =
( 300 )( 0.75) = 900 psi
VQB
=
It
( 2.00 )( 0.125)
Therefore:
τ max = τ B == 900 psi ...................................................................................................... Ans.
F1 =
τ A A1
3
4 ΣM B = 0 :
=
( 300 )( 2 × 0.125) = 25.00 lb
3
Pe − 2 F1 ( 4 cos 30° ) = 0
Therefore:
e=
2 ( 25.00 )( 4 cos 30° )
= 0.577 in. ......................................................................... Ans.
( 300 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-150*
n=
Es 200
=
= 20
Ew 10
yC =
(175) (150 )( 350 ) + ( 357.5) ( 3000 )(15)
= 259.2 mm
(150 )( 350 ) + ( 3000 )(15)
σw =
−cw  σ s

cs  n
 −259.2  75 
  = −9.19 MPa
=
 105.8  20 
σ w = 9.19 MPa (C) ........................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-151*
n=
Es 30, 000
=
= 20
1500
Ew
yC =
M x ( 6.5 ) ( 6 )(12 )  + ( 0.25 ) (120 )( 0.5 ) 
=
= 3.659 in.
A
( 6 )(12 ) + (120 )( 0.5)
σs =
− cs
−3.659
( nσ w ) =
( 20 )( −1250 ) = +10,347 psi
8.841
cw
σ s ≅ 10.35 ksi (T) ................................................................................................................. Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-152
n=
Ea 73
=
= 9.125
Ew 8
 P   L  PL ( 30 )( 4 )
M =    =
=
= 30 kN ⋅ m
4
 2  2  4
IT
( 200 + 228.1)( 300 )
=
σw =
12
3
= 963.2 (106 ) mm 4
Mc ( 30, 000 )( 0.150 )
=
= 4.672 (106 ) N/m 2 ≅ 4.67 MPa (T) ....................... Ans.
−6
IT
( 963.2 ×10 )
σ a = nσ w = ( 9.125 )( 4.672 ) = 42.6 MPa (T) ................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-153
n=
IT
Es 29, 000
=
= 18.125
1600
Ew
( 4 + 18.125)( 6 )
=
σw =
12
3
= 398.25 in.4
Mc (10, 000 × 12 )( 3)
=
= 903.95 psi ≅ 904 psi (T) ........................................... Ans.
IT
( 398.25)
σ s = nσ w = (18.125 )( 903.95 ) = 16,384 psi ≅ 16,380 psi (T) ................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-154*
n=
Ea 70
=
= 7.00
Ew 10
σw =
Myw ( 3000 )( 0.040 )
=
≤ 15 (106 ) N/m 2
IT
IT
IT ≥ 8.000 (10−6 ) m 4
σa =
IT
nMya 7 ( 3000 )( 0.043)
=
≤ 135 (106 ) N/m 2
IT
IT
( 7 w )( 86 )
=
12
3
( 7 w − 50 )( 80 )
−
72,366w ≥ 5.8667 (10
6
)
12
3
IT ≥ 6.689 (10−6 ) m 4
≥ 8.000 (106 ) mm 4
w ≥ 81.1 mm .................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-155
n=
Es 30, 000
=
= 18.75
1600
Ew
M = PL = ( 4000 )( 6 ) = 24, 000 lb ⋅ ft
IT =
( 75 )(11.5 )
12
3
−
( 71)(10 )
12
3
= 3589 in.4
σw =
Myw ( 24, 000 × 12 )( 5 )
=
= 401.2 psi ≅ 401 psi (T & C) ................................. Ans.
IT
( 3598)
σs =
nMys 18.75 ( 24, 000 ×12 )( 5.75 )
=
= 8651 psi ≅ 8650 psi (T & C) ............... Ans.
IT
( 3598)
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-156
n=
For
Eb 100
=
= 71.43
E p 1.4
M (131 − yC )
IT
σp =
My p
σb =
nMyb 71.43MyC
=
IT
IT
IT
σ b = 10σ p :
=
71.43 yC = 10 (131 − yC )
yC = 16.088 mm
yC =
( 3) ( 6 )( 71.43w ) + ( 68.5) ( 50 )(125 )
= 16.088 mm
( 6 )( 71.43w) + ( 50 )(125)
5609w = 327,575
w = 58.4 mm ........................................................................ Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-157*
n=
Es 30, 000
=
= 18.75
1600
Ew
M=
wL2 w (16 )
=
= 32w
8
8
yC =
M x ( 8 ) ( 8 )(15 )  + ( 0.25 ) (150 )( 0.5 ) 
=
= 5.019 in.
A
( 8)(15) + (150 )( 0.5)
IT
(8)(10.481)
=
3
σw =
Myw
=
IT
σs =
nMys
IT
Therefore:
3
(150 )( 5.019 )
+
3
( 32w ×12 )(10.481)
3
(142 )( 4.519 )
−
≤ 1600 psi
3
3
( 5024 )
18.75 ( 32w × 12 )( 5.019 )
=
≤ 18, 000 psi
( 5024 )
= 5024 in.4
w ≤ 1997.3 lb/ft
w ≤ 2502.5 lb/ft
wmax = 1997 lb/ft ..................................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-158*
n=
Es 200
=
= 15.38
Ew 13
wL2 ( 20 )( 5 )
M=
=
= 62.50 kN ⋅ m
8
8
2
IT
( 2307 )( 330 )
=
12
3
( 2157 )( 300 )
−
12
3
= 2056 (106 ) mm 4
σw =
Myw ( 6250 ×10 ) ( 0.150 )
=
= 4.560 (106 ) N/m 2 ≅ 4.56 MPa (T) ............... Ans.
−6
IT
( 2056 ×10 )
σa =
ys
165
( nσ w ) = (15.38 )( 4.560 ) = 77.1 MPa (T) ................................................ Ans.
150
yw
3
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-159
n=
Es 30, 000
=
= 20
1500
Ew
M = ( 2.50 )(10 ) = 25.0 kip ⋅ ft
IT =
σw =
( 6 )(12 )
12
3
+
( 20w )( t )
12
3
+ ( 20 wt )( 6 + 0.5t )
2
Myw ( 25.0 × 12 )( 6 )
=
≤ 1 ksi
IT
IT
IT ≥ 1800 in.4
IT = 864 + 1.66667 wt 3 + 20wt ( 6 + 0.5t ) ≥ 1800 in.4
2
σs =
nMys 20 ( 25.0 × 12 )( 6 + t )
=
≤ 10 ksi
IT
IT
(a)
IT ≥ 600 ( 6 + t ) in.4
IT = 864 + 1.66667 wt 3 + 20wt ( 6 + 0.5t ) ≥ 600 ( 6 + t ) in.4
2
(b)
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-160
n=
Es 200
=
= 16.667
Ew 12
M = 2 ( P 2) = ( P ) N ⋅ m
yC =
M x ( 8 ) ( 3333)(16 )  + (191) ( 200 )( 350 ) 
=
A
( 3333)(16 ) + ( 200 )( 350 )
= 111.87 mm
IT
( 3333)(111.87 )
=
3
3
( 3133)( 95.87 )
−
3
3
( 200 )( 254.13)
+
3
3
= 1729.4 (106 ) mm 4
σw =
Myw ( P )( 0.25413)
=
≤ 10 (106 ) N/m 2
−6
IT
(1729.4 ×10 )
P ≤ 68.05 (103 ) N
σs =
nMys 16.667 ( P )( 0.11187 )
=
≤ 75 (106 ) N/m 2
−6
IT
(1729.4 ×10 )
P ≤ 69.56 (103 ) N
Therefore:
Pmax = 68.1 kN ......................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-161
bh3 ( 8 )(15 )
=
= 2250 in.4
12
12
3
(a)
Iw =
M=
σ wIw
cw
=
( 2.4 )( 2250 )
7.5
M max = 720 kip ⋅ in. ............................................... Ans.
(b)
n=
IT
Es 30, 000
=
= 18.75
1600
Ew
(8)(15 )
=
12
3
 (150 )( t )3

2
+ 2
+ (150t )( 7.5 + 0.5t ) 
 12

2
=  2250 + 25t 3 + 300t ( 7.5 + 0.5t )  in.4


M=
Either:
σ w IT
cw
=
( 2.4 )( IT )
Choose smaller M and compute
σw =
Mcw M ( 7.5 )
=
IT
IT
7.5
or
∆M =
M=
σ s IT
ncs
=
(18 )( IT )
18.75 ( 7.5 + t )
M − 720
(100 )
720
σs =
nMcs 18.75M ( 7.5 + t )
=
IT
IT
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-162
n=
(a)
IT
Ea 73
=
= 5.615
Ew 13
(150 )( 300 )
=
3
12
 ( nw )( 50 )3

2
+2
+ ( 50nw )(150 + 25 ) 
12


= 337.5 (106 ) + 3.083 (106 ) nw mm 4
Mcw ( 75 × 10 ) ( 0.150 )
σw =
=
IT
IT
3
nMca 5.615 ( 75 × 10
σa =
=
IT
IT
(b)
3
) ( 0.200 )
For
σ w = 15 MPa
w = 23.8 mm
For
σ a = 135 MPa
w = 16.54 mm
Therefore
wmin = 23.8 mm ................................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-163*
2
As = 3 π (1) 4  = 2.356 in.2


AT = nAs = 12 ( 2.356 ) = 28.27 in.2
10h ( h 2 ) = 28.27 (18 − h )
5h 2 + 28.27 h − 508.86 = 0
From which:
h = 7.650 in.
IT
(10 )( 7.650 )
=
σc =
3
3
18 − h = 10.350 in.
+ ( 28.27 )(10.350 ) = 4521 in.4
2
Myc ( M × 12 )( 7.650 )
=
≤ 1000 psi
IT
( 4521)
M ≤ 49, 248 lb ⋅ ft
σs =
nMys 12 ( M × 12 )(10.350 )
=
≤ 18, 000 psi
IT
( 4521)
M ≤ 54, 601 lb ⋅ ft
Therefore:
M max = 49.2 kip ⋅ ft ................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-164*
n=
Es 200
=
= 13.333
Ec 15
2
As = 4 π (15 ) 4  = 706.9 mm 2


AT = nAs = 13.333 ( 706.9 ) = 9425 mm 2
200h ( h 2 ) = 9425 ( 275 − h )
h 2 + 94.25h − 25,919 = 0
From which:
IT
h = 120.62 mm
( 200 )(120.62 )
=
σc =
3
+ ( 9425 )(154.38 ) = 341.6 (106 ) mm 4
2
Myc (15, 000 )( 0.12062 )
=
= 5.30 (106 ) N/m 2
6
IT
( 341.6 ×10 )
σ c = 5.30 MPa (C)
σs =
3
275 − h = 154.38 mm
............................................................................................................... Ans.
ys
(154.38)(13.333 × 5.297 ) = 90.4 MPa (T) ...................................... Ans.
( nσ c ) =
yc
(120.62 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-165
n=
Es 30, 000
=
= 13.636
Ec
2200
wL2 ( 820 )(13)
M max =
=
= 17,322 lb ⋅ ft
8
8
2
As = 3 π ( 0.75 ) 4  = 1.3254 in.2


2
AT = nAs = 13.636 (1.3254 ) = 18.073 in.2
10h ( h 2 ) = 18.073 (15.5 − h )
h 2 + 3.615h − 56.02 = 0
From which:
h = 5.892 in.
IT
(10 )( 5.892 )
=
3
3
15.5 − h = 9.608 in.
+ (18.073)( 9.608 ) = 2350 in.4
2
σc =
Myc (17.322 × 12 )( 5.892 )
=
= 0.5212 ksi ≅ 0.521 ksi (C) .............................. Ans.
IT
( 2350 )
σs =
( 9.608 )(13.636 × 0.5212 ) = 11.59 ksi (T) ....................................... Ans.
ys
( nσ c ) =
yc
( 5.892 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-166*
n=
Es
= 12
Ec
3wL2 3w ( 4 )
=
= (1.5w ) N ⋅ m
M max =
32
32
2
As = 3 π (16 ) 4  = 603.2 mm 2


2
AT = nAs = 12 ( 603.2 ) = 7238 mm 2
200h ( h 2 ) = 7238 ( 300 − h )
h 2 + 72.38h − 21, 714 = 0
From which:
IT
h = 115.55 mm
( 200 )(115.55)
=
3
3
300 − h = 184.45 mm
+ ( 7238 )(184.45 ) = 349.1(106 ) mm 4
2
σc =
Myc (1.5w )( 0.11555 )
=
≤ 6.5 (106 ) N/m 2
−6
IT
( 349.1×10 )
w ≤ 13.092 (103 ) N/m
σs =
nMys 12 (1.5w )( 0.18445 )
=
≤ 120 (106 ) N/m 2
−6
IT
( 349.1×10 )
w ≤ 12.618 (103 ) N/m
Therefore:
wmax = 12.62 kN/m ................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-167
n=
wL2 w (12 )
=
= (18w ) lb ⋅ ft
8
8
2
Es 30, 000
=
= 12.5
Ec
2400
M max =
2
As = 3 π ( 0.875 ) 4  = 1.8040 in.2


AT = nAs = 12.5 (1.8040 ) = 22.55 in.2
8h ( h 2 ) = 22.55 (16 − h )
4h 2 + 22.55h − 360.8 = 0
From which:
h = 7.088 in.
IT =
(8 )( 7.088)
3
3
16 − h = 8.912 in.
+ ( 22.55 )( 8.912 ) = 2741 in.4
2
σc =
Myc (18w ×12 )( 7.088 )
=
≤ 1000 psi
IT
( 2741)
w ≤ 1790.3 lb/ft
σs =
nMys 12.5 (18w ×12 )( 8.912 )
=
≤ 16, 000 psi
IT
( 2741)
w ≤ 1822.6 lb/ft
Therefore:
wmax = 1790 lb/ft ..................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-168*
n=
Es 198
=
= 12
Ec 16.5
300h ( h 2 ) = 12 As ( 500 − h )
12.5h 2
500 − h
Myc Mh
σc =
=
≤ 7 (106 ) N/m 2
IT
IT
As =
σs =
(a)
nMys 12M ( 500 − h )
=
≤ 125 (106 ) N/m 2
IT
IT
(b)
84 ( 500 − h ) = 125h
From Eqs. (a) and (b):
h = 200.95 mm
12.5h 2 12.5 ( 200.95 )
As =
=
= 1688 mm 2 ................................................................... Ans.
500 − h 500 − 200.95
2
(a)
(b)
IT
( 300 )( 200.95)
=
M max =
3
σ c IT
h
3
+ 12 (1688 )( 299.05 ) = 2623 (106 ) mm 4
2
( 7 ×10 )( 2623 ×10 ) = 91,370 N ⋅ m
=
6
−6
( 0.20095 )
M max = 91.4 kN ⋅ m ............................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-169
n=
Es 30, 000
=
= 12.5
Ec
2400
wL2 (1000 )(16 )
M max =
=
= 32, 000 lb ⋅ ft
8
8
Myc ( 32, 000 × 12 ) h
σc =
=
≤ 800 psi
IT
IT
2
σs =
nMys 12.5 ( 32, 000 × 12 ) b
=
≤ 16, 000 psi
IT
IT
IT = 480h = 300b
From Eqs. (a) and (b):
Therefore
(10 )( h )
=
3
3
+ nAs ( b )
2
(b)
h = 0.625b
5 ( 0.625b )
5h 2
=
= 0.15625b
As =
12.5b
12.5b
2
10h ( h 2 ) = nAs b = 12.5 As b
IT
(a)
(10 )( 0.625b )
=
3
IT = 2.767b = 300b
3
3
+ 12.5 ( 0.15625b )( b ) = 2.767b3
2
b = 10.413 in.
h = 0.625b = 0.625 (10.413) = 6.508 in.
(a)
As = 0.15625b = 0.15625 (10.413) = 1.627 in.2 ............................................................. Ans.
(b)
d = h + b = 6.508 + 10.413 = 16.92 in. .............................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-170
From the curved beam formula:
R=
ro − ri
11d − 10d
=
= 10.49206d
ln ( ro ri )
ln (1.1)
yC = R − RC = 10.49206d − 10.5d = −0.00794d
σ max =
M ( 0.49206d )
Myi
−6.197 M
=
=
ri AyC (10d )( td )( −0.00794d )
td 2
From the flexure formula:
Error =
σ max =
− Mc − M ( d 2 ) −6 M
=
=
I
( td 3 12 ) td 2
6.197 − 6
(100 ) = 3.18% ..................................................................................... Ans.
6.197
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
7-171*
σ o = −σ i =
RILEY, STURGES AND MORRIS
M ( R − ro ) − M ( R − ri )
=
ro AyC
ri AyC
ro ( R − ri ) = ri ( ro − R )
14 ( R − 6 ) = 6 (14 − R )
Which gives
R = 8.400 in.
A = 2 ( 8 × 1) + ( b × 2 ) = ( 2b + 16 ) in.2
A = R∫
dA
ρ
Using Table B-20
∫
8
14 


= ( b ) ln  + 2 (1) ln  = 0.28768b + 1.69750
ρ 
6
6

dA
2b + 16 = ( 8.400 )( 0.28768b + 2.26996 )
b = 4.24 in. .............................................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
7-172
RILEY, STURGES AND MORRIS
RC = 450 + 100 = 550 mm
A = 2 ( 50 ×150 ) + ( 75 × 100 ) = 22,500 mm 2
A = R∫
dA
ρ
Using Table B-20
∫
dA
ρ
= (150 ) ln
500
600
650
+ ( 75 ) ln
+ (150 ) ln
= 41.484 mm
450
500
600
22,500 = R ( 41.484 )
R = 542.4 mm
yC = R − RC = 542.4 − 550 = −7.600 mm
σi =
M ( R − ri )
( 20, 000 )( 0.5424 − 0.450 )
=
ri AyC
( 0.450 )( 0.0225)( −0.00760 )
σ i = −24.0 (106 ) N/m 2 = 24.0 MPa (C) .................................................................. Ans.
σo =
M ( R − ro )
( 20, 000 )( 0.5424 − 0.650 )
=
ro AyC
( 0.650 )( 0.0225)( −0.00760 )
σ o = +19.36 (106 ) N/m 2 = 19.36 MPa (T)
............................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-173*
RILEY, STURGES AND MORRIS
A = ( 8 × 2 ) + ( 4 × 2 ) + ( 4 × 2 ) = 32 in.2
RC =
(11)(8 × 2 ) + (14 )( 4 × 2 ) + (17 )( 4 × 2 ) = 13.25 in.
32
A = R∫
dA
ρ
Using Table B-20
∫
dA
ρ
= ( 8 ) ln
12
16
18
+ ( 2 ) ln + ( 4 ) ln = 2.5051 in.
10
12
16
32 = R ( 2.5051)
R = 12.774 in.
yC = R − RC = 12.774 − 13.25 = −0.4760 in.
σi =
M ( R − ri ) ( −30 ×12 )(12.774 − 10 )
=
= +6.56 ksi = 6.56 ksi (T) .................. Ans.
ri AyC
(10 )( 32 )( −0.4760 )
σo =
M ( R − ro ) ( −30 ×12 )(12.774 − 18 )
=
= −6.86 ksi = 6.86 ksi (C) ................. Ans.
ro AyC
(18)( 32 )( −0.4760 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-174*
RC = 75 + 150 = 225 mm
RILEY, STURGES AND MORRIS
A = ( 250 × 300 ) − (150 × 200 ) = 45, 000 mm 2
Using Table B-20
∫
dA
ρ
= ( 250 ) ln
A = R∫
dA
ρ
125
325
375
+ 2 ( 50 ) ln
+ ( 250 ) ln
= 259.0 mm
75
125
325
45, 000 = R ( 259.0 )
R = 173.75 mm
yC = R − RC = 173.75 − 225 = −51.25 mm
σi =
M ( R − ri )
M ( 0.17375 − 0.075 )
=
≤ −140 (106 ) N/m 2
ri AyC
( 0.075)( 0.0450 )( −0.05125)
M ≤ 245 (103 ) N ⋅ m
σo =
M ( R − ro )
M ( 0.17375 − 0.375 )
=
≤ 35 (106 ) N/m 2
ro AyC
( 0.375)( 0.0450 )( −0.05125)
M ≤ 150.4 (103 ) N ⋅ m
Therefore:
M max = 150.4 kN ⋅ m ............................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
7-175
RILEY, STURGES AND MORRIS
A = (1.5 × 0.75 ) + (1.5 × 3) = 5.625 in.2
A = R∫
6
 3 
4ρ  d ρ
ρ  dρ 
= R  ∫  −2 +
+ ∫ 3− 

2.25
3
ρ
3  ρ
3  ρ 



dA


 3  4
6 1
= R  −2 ln 
 + ( 3 − 2.25 ) + 3ln   − ( 6 − 3) 
 2.25  3
3 3


2
= 1.50408R = 5.625 in.
5.625
R=
= 3.7398 in.
1.50408
yC = R − RC = 3.7398 − 4.00 = −0.2602 in.
( −70 )( 3.7398 − 2.25 ) = +31.7 ksi = 31.7 ksi (T) ................ Ans.
ri AyC
( 2.25)( 5.625)( −0.2602 )
M ( R − ro )
( −70 )( 3.7398 − 6.00 ) = −18.02 ksi = 18.02 ksi (C) .......... Ans.
σo =
=
ro AyC
( 6.00 )( 5.625)( −0.2602 )
σi =
M ( R − ri )
=
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MECHANICS OF MATERIALS, 6th Edition
7-176*
F = 1500 kN (T)
RILEY, STURGES AND MORRIS
Vz = 500 kN
M y = − ( 500 × 1) = −500 kN ⋅ m
σ xA =
F Mc
+
A
I
1500 (103 )
( 500 ×10 ) ( 0.200 )
π ( 0.400 − 0.300 ) 4 π ( 0.200 − 0.150 ) 4
= 27.28 (10 ) + 116.41(10 ) N/m = 143.69 MPa (T)
=
2
6
3
2
+
6
4
4
2
σ yA = τ xyA = 0 MPa
σ p1 = 143.7 MPa (T) .......................................................... Ans.
σ p 2 = σ p 3 = 0 MPa
............................................................. Ans.
τ max = τ p = (σ p1 − σ p 2 ) 2 = (143.69 − 0 ) 2 = 71.8 MPa ............................................ Ans.
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MECHANICS OF MATERIALS, 6th Edition
7-177*
RILEY, STURGES AND MORRIS
M = (150 )( 2 ) = 300 kip ⋅ in.
P = 150 kip (C)
A = 2 ( 2 × 6 ) = 24 in 2
xC =
Iy
My
A
=
( 2 )( 5)
=
3
( 3) ( 2 )( 6 ) + ( 7 ) ( 6 )( 2 )
= 5.00 in.
2 ( 6 )( 2 ) 
3
( 6 )( 3)
+
3
3
( 4 )(1)
−
3
3
= 136.00 in.4
σ yC =
− P Mc −150 ( 300 )( 5 )
+
=
+
= 4.78 ksi (T) ..................... Ans.
A
I
24
136
σ yD =
− P Mc −150 ( 300 )( 3)
−
=
−
= −12.87 ksi = 12.87 ksi (C) .......................... Ans.
A
I
24
136
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-178
↑ ΣFy = 0 :
P + 4060 − 5210 = 0
4 ΣM cut = 0 :
M − ( 4060 )( 0.016 ) − ( 5210 )( 0.037 ) = 0
P = 1150 N (C)
(a)
M y = 257.73 N ⋅ m
A = π ( 27 ) 4 = 572.6 mm 2
2
Solid:
I = π ( 27 ) 64 = 26.09 (103 ) mm 4
4
σ xA =
( 257.3)( 0.0135)
− F Mc
−1150
±
=
±
A
I
572.6 (10−6 )
26.09 (10−9 )
= ( −2.0084 ± 133.36 ) (106 ) N/m 2
σ T max = +131.4 MPa = 131.4 MPa (T)
........................................................................... Ans.
σ C max = −135.4 MPa = 135.4 MPa (C) ........................................................................... Ans.
(b)
Hollow:
A = π ( 27 2 − 162 ) 4 = 371.5 mm 2
I = π ( 27 4 − 164 ) 64 = 22.87 (103 ) mm 4
σ xA =
( 257.3)( 0.0135) = −3.096 ± 152.14 106 N/m 2
−1150
±
(
)( )
−6
371.5 (10 )
22.87 (10−9 )
σ T max = +149.0 MPa = 149.0 MPa (T)
........................................................................... Ans.
σ C max = −155.2 MPa = 155.2 MPa (C) ........................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
7-179*
P = 50 kip (T)
RILEY, STURGES AND MORRIS
Vy = 5 kip
M z = ( 5 )( 24 ) = 120 kip ⋅ in.
T = 30 kip ⋅ in.
I z = π ( 4 ) 64 = 12.5664 in.4
4
J = 2 I z = 25.1327 in.4
 4r
Q = yC A = 
 3π
2
3
2 ( 2)
  π r  2r
=
=
= 5.3333 in.3


3
3
 2 
3
σ xA =
( 50 ) = 3.979 ksi (T)
P
=
A π ( 4 )2 4
τ xyA =
( 30 )( 2 ) − ( 5)( 5.3333) = 1.8575 ksi
Tc Vy Q
−
=
J
I z t ( 25.1327 ) (12.5664 )( 4 )
σ p1, p 2 =
σx +σ y
2
σ yA = 0 ksi
 σ x −σ y 
2
± 
 + τ xy
2


2
3.979 + 0
2
 3.979 − 0 
=
± 
 + (1.8575 )
2
2


2
σ p1 = 1.9895 + 2.7218 = +4.7113 ksi ≅ 4.71 ksi (T) .................................................... Ans.
σ p 2 = 1.9895 − 2.7218 = −0.7323 ksi ≅ 0.732 ksi (C) ................................................. Ans.
σ p 3 = 0 ksi
.............................................................................................................................. Ans.
τ max = τ p = (σ p1 − σ p 2 ) 2 = ( 4.7113 + 0.7323) 2 = 2.72 ksi
1
2
θ p = tan −1
2τ xy
σ x −σ y
=
.................................... Ans.
2 (1.8575 )
1
= 21.52° .................................................... Ans.
tan −1
2
( 3.979 ) − 0
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-180
RILEY, STURGES AND MORRIS
M = ( 40 )( 0.65919 ) = 26.37 kN ⋅ m
P = 40 kN (T)
A = (100 × 50 ) + ( 30 × 80 ) + ( 60 × 20 ) = 8600 mm 2
xC =
( 25 )(100 × 50 ) + ( 90 )( 30 × 80 ) + (140 )( 60 × 20 ) = 59.19 mm
8600
100 ( 50 )

2
Iy = 
+ (100 × 50 )( 34.19 ) 
 12

 30 ( 80 )3
  60 ( 20 )3

2
2
+
+ ( 30 × 80 )( 30.81)  + 
+ ( 60 × 20 )( 80.81) 
 12
  12

3
= 18.321(106 ) mm 4
σ CD =
( 26.37 )( 0.09081)
P Mc
40, 000
+
=
+
A I
8600 (10−6 )
18.321(10−6 )
σ CD = 89.845 (106 ) N/m 2 = 89.8 MPa (T) ..................................................................... Ans.
σ EF =
( 26.37 )( 0.05919 )
P Mc
40, 000
−
=
−
−6
A I
8600 (10 )
18.321(10−6 )
σ EF = −126.1(106 ) N/m 2 = 126.1 MPa (C) ................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
7-181
RILEY, STURGES AND MORRIS
P = 15 kip (C)
M = (15 )(12 ) = 180 kip ⋅ in.
A = ( 5 ×12 ) = 60 in 2
I = ( 5 )(12 ) 12 = 720 in.4
3
− P Mc −15 (180 )( 6 )
+
=
+
= 1.250 ksi (T)
A
I
60
720
σ A = 1.250 ksi (T) ........................................................... Ans.
σA =
− P Mc −15 (180 )( 6 )
−
=
−
A
I
60
720
σ B = −1.750 ksi = 1.750 ksi (C) .................................. Ans.
σB =
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-182*
→ ΣFx = 0 :
− P − 25 = 0
↑ ΣFy = 0 :
V − 30 = 0
RILEY, STURGES AND MORRIS
P = −25 kN = 25 kN (C)
V = 30 kN
− M − ( 25 )( 0.350 ) − ( 30 )(1.25 ) = 0
4 ΣM cut = 0 :
M = −46.25 kN ⋅ m = 46.25 kN ⋅ m 4
A = (100 × 150 ) = 15, 000 mm 2
I = (100 )(150 ) 12 = 28.13 (106 ) mm 4
3
σ top =
P Mc ( −25, 000 ) ( 46, 250 )( 0.075 )
+
=
+
= +121.64 (106 ) N/m 2
−3
−6
A I
15.00 (10 )
28.13 (10 )
σ bottom =
Therefore:
P Mc ( −25, 000 ) ( 46, 250 )( 0.075 )
−
=
−
= −124.98 (106 ) N/m 2
−3
−6
A I
15.00 (10 )
28.13 (10 )
σ max = 125.0 MPa (C) ............................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-183*
M = ( 450 )( 3) = 1350 lb ⋅ in.
P = 450 lb (T)
A = ( 0.5 × h ) = ( 0.5h ) in
RILEY, STURGES AND MORRIS
2
I = ( 0.5 )( h ) 12 = ( h3 24 ) in.4
3
σ=
P Mc 450 (1350 )( h 2 )
+
=
+
≤ 16, 000 psi
A I
0.5h
h3 24
16, 000h 2 − 900h − 16, 200 ≥ 0
h ≥ 1.035 in. ....................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-184
RILEY, STURGES AND MORRIS
→ ΣFx = 0 :
P − 30 = 0
P = 30 kN (C)
↑ ΣFy = 0 :
40 − V = 0
V = 40 kN
4 ΣM cut = 0 :
M + ( 30 )( 0.750 ) − ( 40 )(1.20 ) = 0
M = 25.5 kN ⋅ m 4
A = 2 ( 50 ×150 ) + (100 × 75 ) = 22,500 mm 2
(150 )( 200 )
I=
12
σ BB =
Therefore:
3
( 75 )(100 )
−
12
3
= 93.75 (106 ) mm 4
− P Mc − ( 30, 000 ) ( 25,500 )( 0.100 )
±
=
±
A
I
22.5 (10−3 )
93.75 (10−6 )
σ L = σ max T = +25.9 (106 ) N/m 2 = 25.9 MPa (T) ........................................... Ans.
σ R = σ max C = −28.5 (106 ) N/m 2 = 28.5 MPa (C) ........................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-185
RILEY, STURGES AND MORRIS
P = 80 lb (T)
M = ( 80 )( 3) = 240 lb ⋅ in.
A = ( 0.5 × 0.1875 ) = 0.093750 in 2
I = ( 0.1875 )( 0.5 ) 12 = 1.9531(10−3 ) in.4
3
σ=
Therefore:
( 240 )( 0.25) = 853.33 ± 30, 720 psi
P Mc
80
±
=
+
(
)
A I
0.09375 1.9531(10−3 )
σ L = σ max T = +31.6 ksi = 31.6 ksi (T) ............................................................... Ans.
σ R = σ max C = −29.9 ksi = 29.9 ksi (C) .............................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-186*
P = 780 kN (T)
RILEY, STURGES AND MORRIS
Vy = 10 kN
M z = (10 × 0.700 ) = 7.00 kN ⋅ m
T = 14.0 + 5.6 = 19.6 kN ⋅ m
A = π (150 ) 4 = 17, 671 mm 2
2
I = π (150 ) 64 = 24.85 (106 ) mm 4
4
J = 2 I = 49.70 (106 ) mm 4
σ xA
780 (103 )
( 7000 )( 0.075)
P Mc
= −
=
−
−3
A I
17.671(10 ) 24.850 (10−6 )
= 44.14 (106 ) − 21.13 (106 ) N/m 2 = 23.01 MPa (T)
σ yA = 0 MPa
τ xyA =
Tc Vy Q (19, 600 )( 0.075 )
+
=
+0
J
I zt
49.70 (10−6 )
= 29.58 (106 ) N/m 2 = 29.58 MPa
σ p1, p 2 =
σx +σ y
2
 σ −σ y 
23.01 + 0
2
 23.01 − 0 
2
±  x
± 
 + τ xy =
 + ( 29.58 )
2
2


 2 
2
2
σ p1 = 11.505 + 31.74 = 43.245 MPa ≅ 43.2 MPa (T)
................................................. Ans.
σ p 2 = 11.505 − 31.74 = −20.235 MPa ≅ 20.2 MPa (C) .............................................. Ans.
σ p 3 = 0 MPa ........................................................................................................................... Ans.
τ max = τ p =
σ p1 − σ p 2
2
=
( 43.245) − ( −20.235) = 31.7 MPa ...................................... Ans.
2
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-187*
→ ΣFx = 0 :
RILEY, STURGES AND MORRIS
Ax − Q + 30 cos 30° = 0
Ay − 30sin 30° = 0
↑ ΣFy = 0 :
5.5Q − (11)( 30 cos 30° ) − ( 4 )( 30sin 30° ) = 0
4 ΣM A = 0 :
Q = 62.87 lb
Ax = 36.89 lb
Ay = 15.00 lb
θ = tan −1 ( 5.5 4 ) = 53.97°
Z ΣFn = 0 :
36.89 cos θ + 15.00sin θ − N = 0
^ ΣFt = 0 :
V + 15.00 cos θ − 36.89sin θ = 0
M + ( 36.89 )( 2.75 ) − (15.00 )( 2 ) = 0
4 ΣM cut = 0 :
N = 33.83 lb
V = 21.01 lb
M = −71.45 lb ⋅ in. = 71.45 lb ⋅ in. 3
A = ( 0.18750 ×1) = 0.18750 in.2
I = ( 0.18750 )(1) 12 = 0.015625 in.4
3
σ top =
− N Mc − ( 33.83) ( 71.45 )( 0.500 )
+
=
+
= +2106 psi
A
I
0.18750
0.015625
− N Mc − ( 33.83) ( 71.45 )( 0.500 )
−
=
−
= −2467 psi
A
I
0.18750
0.015625
σ top = σ max T = +2.11 ksi = 2.11 ksi (T) ............................................................. Ans.
σ bottom =
Therefore:
σ bottom = σ max C = −2.47 ksi = 2.47 ksi (C) ........................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-188
RILEY, STURGES AND MORRIS
W = mg = ( 360 )( 9.807 ) = 3531 N
a = 1430 cos16° = 1374.6 mm
b = 1430sin16° = 394.2 mm
1374.6 − 890
φ = tan −1
= 20.53°
394.2 + 900
→ ΣFx = 0 :
Ax + B sin 20.53° = 0
Ay + B cos 20.53° − 3531 = 0
↑ ΣFy = 0 :
4 ΣM A = 0 :
( B cos 20.53° )(1374.6 ) − ( B sin 20.53° )( 394.2 ) − ( 3531)( 2700 cos16° ) = 0
B = 7976 N
Ax = −2797 N
Ay = −3938 N
Z ΣFn = 0 :
P − 2797 cos16° − 3938sin16° = 0
^ ΣFt = 0 :
V + 2797 sin16° − 3938cos16° = 0
P = 3774 N
V = 3014 N
M + ( 3938 )( 0.530 cos16° ) − ( 2797 )( 0.530sin16° ) = 0
4 ΣM cut = 0 :
M = −1597.7 N ⋅ m = 1597.7 N ⋅ m 3
A = (100 × 100 ) − ( 60 × 60 ) = 6400 mm 2
(100 )(100 )
I=
12
σ top =
3
( 60 )( 60 )
−
12
3
= 7.253 (106 ) mm 4
( 3774 ) + (1597.7 )( 0.050 ) = +11.60 106 N/m 2
P Mc
+
=
( )
A I
6400 (10−6 )
7.253 (10−6 )
σ top = σ max T = 11.60 MPa (T) ..................................................................................... Ans.
σ bottom =
( 3774 ) − (1597.7 )( 0.050 ) = −10.42 106 N/m 2
P Mc
−
=
( )
A I
6400 (10−6 )
7.253 (10−6 )
σ bottom = σ max C = 10.42 MPa (C) ................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-189*
RILEY, STURGES AND MORRIS
A = ( 6 × 4 ) = 24 in.2
Ix
( 6 )( 4 )
=
Iy
( 4 )( 6 )
=
3
= 32 in.4
12
12
3
= 72 in.4
P = 9600 lb (C)
Vx = 800 lb
M x = ( 9600 )(1) = 9600 lb ⋅ in.
M y = ( 800 )( 24 ) = 19, 200 lb ⋅ in.
σP =
P 9600
=
= 400 psi (C)
A
24
σM =
x
σM =
y
M x c ( 9600 )( 2 )
=
= 600 psi (T & C)
Ix
32
M yc
Iy
=
(19, 200 )( 3) = 800 psi (T & C)
72
σ A = −400 + 600 + 800 = +1000 psi = 1000 psi (T) .............................................. Ans.
σ B = −400 − 600 + 800 = −200 psi = 200 psi (C) .................................................. Ans.
σ C = −400 − 600 − 800 = −1800 psi = 1800 psi (C)
σ D = −400 + 600 − 800 = −600 psi = 600 psi (C)
............................................. Ans.
................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-190
By symmetry, each support carries half of the total weight. Then,
V = 10 kN
M = ( 30 )( 2 ) − (10 × 2 )(1) = 40.00 kN ⋅ m
I=
π ( 604 )
4
−
4
π ( 600 )
 4r
Q = yC A = 
 3π
4
4
= 2741(106 ) mm 4
2
3
2 ( 604 ) 2 ( 600 )
  π r  2r
=
−
=

3
3
3
 2 
3
3
= 2.899 (106 ) mm3
Stresses due to the internal pressure:
σ x = σ axial
3
pr ( 200 × 10 ) ( 0.600 )
=
=
= 15.00 (106 ) N/m 2 (T)
2t
2 ( 0.004 )
σ y = σ hoop = 2σ axial = 30.00 (106 ) N/m 2 (T)
At A:
σ y = 30.00 MPa
σ x = 15.00 MPa
−3
−VQ − (10, 000 ) ( 2.899 × 10 )
=
τ xy =
It
( 2741×10−6 ) ( 0.008)
= −1.3221(106 ) N/m 2 = −1.3221 MPa
σ p1, p 2 =
σx +σ y
2
 σ −σ y 
15 + 30
2
 15 − 30 
2
±  x
± 
 + τ xy =
 + ( −1.3221)
2
 2 
 2 
2
2
σ p1 = 22.500 + 7.616 = +30.116 MPa ≅ 30.1 MPa (T) ............................................... Ans.
σ p 2 = 22.500 − 7.616 = +14.884 MPa ≅ 14.88 MPa (T) ............................................ Ans.
σ p 3 = 0 MPa ........................................................................................................................... Ans.
τ max =
σ max − σ min
2
=
( 30.116 ) − 0 = 15.06 MPa ............................................................ Ans.
2
At B:
σx =
( 40, 000 )( 0.604 )
pr Mc
−
= 15 (106 ) −
2t
I
( 2741×10−6 )
= +6.186 (106 ) N/m 2 = 6.186 MPa (T)
σ y = 30.00 MPa
τ xy = 0 MPa
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-190 (cont.)
Since τ xy = 0 MPa these are principal stresses and
σ p1 = σ y = +30.0 MPa = 30.0 MPa (T) .......................................................................... Ans.
σ p 2 = σ x = +6.186 MPa ≅ 6.19 MPa (T) ....................................................................... Ans.
σ p 3 = 0 MPa ........................................................................................................................... Ans.
τ max =
σ max − σ min
2
=
( 30.0 ) − 0 = 15.00 MPa ................................................................. Ans.
2
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-191
bh3 (1)( t )
=
= ( 0.08333t 3 ) in.4
12
12
3
A = (1)( t ) = ( t ) in 2
I=
P Mc
+
A I
σ = Eε =
(110h )( t 2 )
+
(10.6 ×10 )(1000 ×10 ) = 110
0.08333t
t
−6
6
3
which gives
h (in.)
0
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
1767t 2 − 18.33t = 110h
t (in.)
0.0000
0.1246
0.1713
0.2110
0.2444
0.2738
0.3004
0.3249
0.3477
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-192*
Ix =
( 200 )(150 )
3
12
(150 )( 200 )
=
Iy
12
3
= 56.25 (106 ) mm 4
A = ( 200 ×150 ) = 30, 000 mm 2
= 100.0 (106 ) mm 4
P = 75 kN (C)
M x = ( 75 )( 0.075 ) = 5.625 kN ⋅ m
M y = ( 75 )( 0.050 ) = 3.750 kN ⋅ m
σP =
P 75, 000
=
= 2.50 (106 ) N/m 2 (C)
−3
A 30 (10 )
σM =
x
M x c ( 5625 )( 0.075 )
=
= 7.50 (106 ) N/m 2 (T & C)
−6
Ix
56.25 (10 )
σM =
y
M yc
Iy
=
( 3750 )( 0.100 ) = 3.75 106
100.0 (10
−6
)
( ) N/m
2
(T & C)
σ A = −2.50 + 7.50 + 3.75 = +8.75 MPa = 8.75 MPa (T)
..................................... Ans.
σ B = −2.50 + 7.50 − 3.75 = +1.25 MPa = 1.25 MPa (T) ...................................... Ans.
σ C = −2.50 − 7.50 − 3.75 = −13.75 MPa = 13.75 MPa (C) ................................. Ans.
σ D = −2.50 − 7.50 + 3.75 = −6.25 MPa = 6.25 MPa (C) ..................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-193
I x = ( 6 )( 4 ) 12 = 32 in.4
3
A = ( 6 × 4 ) = 24 in.2
I y = ( 4 )( 6 ) 12 = 72 in.4
3
P = 4000 lb (C)
M x = ( 400 )( 24 ) + ( 4000 )(1) = 13, 600 lb ⋅ in.
M z = ( 500 )( 24 ) = 12, 000 lb ⋅ in.
Neither Vx nor Vz contributes to the stresses at the corners.
σP =
P 4000
=
= 166.67 psi (C)
24
A
σM =
M x c (13, 600 )( 2 )
=
= 850 psi (T & C)
32
Ix
σM =
M z c (12, 000 )( 3)
=
= 500 psi (T & C)
72
Iz
x
z
σ A = −166.67 + 850 + 500 = +1183 psi = 1183 psi (T) ........................................ Ans.
σ B = −166.67 − 850 + 500 = −517 psi = 517 psi (C) ............................................ Ans.
σ C = −166.67 − 850 − 500 = −1517 psi = 1517 psi (C) ........................................ Ans.
σ D = −166.67 + 850 − 500 = +183.3 psi = 183.3 psi (T) ...................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
7-194*
RILEY, STURGES AND MORRIS
A = ( 200 ×150 ) = 30, 000 mm 2
Ix
( 200 )(150 )
=
Iz
(150 )( 200 )
=
3
= 56.25 (106 ) mm 4
3
= 100.0 (106 ) mm 4
12
12
P = 75 kN (C)
M x = ( 75 )( 0.075 ) = 5.625 kN ⋅ m
M z = (15 )( 0.5 ) − ( 75 )( 0.050 ) = 3.750 kN ⋅ m
σP =
P 75, 000
=
= 2.50 (106 ) N/m 2 (C)
A 30 (10−3 )
σM =
M x c ( 5625 )( 0.075 )
=
= 7.50 (106 ) N/m 2 (T & C)
−6
Ix
56.25 (10 )
σM =
M z c ( 3750 )( 0.100 )
=
= 3.75 (106 ) N/m 2 (T & C)
−6
Iz
100.0 (10 )
x
z
σ A = −2.50 + 7.50 − 3.75 = +1.25 MPa = 1.25 MPa (T) ...................................... Ans.
σ B = −2.50 + 7.50 + 3.75 = +8.75 MPa = 8.75 MPa (T)
..................................... Ans.
σ C = −2.50 − 7.50 + 3.75 = −6.25 MPa = 6.25 MPa (C) ..................................... Ans.
σ D = −2.50 − 7.50 − 3.75 = −13.75 MPa = 13.75 MPa (C)
................................ Ans.
1.250 13.25
=
150 − z ′
z′
z ′ = 12.5 mm
8.75
6.25
=
z ′′ 150 − z ′′
z ′′ = 87.5 mm
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MECHANICS OF MATERIALS, 6th Edition
7-195
RILEY, STURGES AND MORRIS
θ = tan1 ( 9 16 ) = 29.357°
4 ΣM A = 0 :
(T sin θ )( 64 ) − ( 27.5)(16 ) − ( 27.5)( 48) = 0
T = 56.10 lb
P − 56.10 cos θ = 0
→ ΣFx = 0 :
V − 27.5 + 56.10sin θ = 0
↑ ΣFy = 0 :
4 ΣM cut = 0 :
− M − ( 27.5 )(16 ) + ( 56.10sin θ )( 32 ) = 0
P = 48.89 lb
(a)
σx =
V = 0 lb
M = 440 lb ⋅ in.
P Mc −48.89 ( 440 )( 0.5 )
−
=
−
= −24.445 − 1320.0
A I
(1× 2 ) ( 2 )(1)3 12
σ x = −1344.4 psi ≅ 1344 psi (C)
σ y = τ xy = 0 lb
Since τ xy = 0 these are principal stresses and
σ p 2 = 1344 psi (C) ........................... σ p1 = σ p 3 = 0 lb ..................................................... Ans.
τ max = τ p =
(b)
σ p1 − σ p 2
2
=
0 − ( −1344.4 )
= 672 psi ......................................................... Ans.
2
P Mc
+
= −24.445 + 1320.0 = +1295.6 psi ≅ 1296 psi (T)
A I
Since τ xy = 0 these are principal stresses and
σx =
σ p1 = 1296 psi (T) ............................ σ p 2 = σ p 3 = 0 lb
τ max = τ p =
(c)
σ p1 − σ p 2
2
=
.................................................... Ans.
1295.6 − 0
= 648 psi ................................................................ Ans.
2
P Mc
+
= −24.445 + 0 = −24.445 psi ≅ 24.4 psi (C)
A I
Because V = 0 , τ xy = 0 and these are principal stresses
σx =
σ y = τ xy = 0 lb
σ y = 0 lb
σ p 2 = 24.4 psi (C) ............................ σ p1 = σ p 3 = 0 lb ..................................................... Ans.
τ max = τ p =
σ p1 − σ p 2
2
=
0 − ( 24.445 )
= 12.22 psi ......................................................... Ans.
2
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-196*
A = π (100 ) 4 = 7854 mm 2
2
I = π (100 ) 64 = 4.909 (106 ) mm 4
4
σ 1 = Eε1 = ( 210 ×109 )( −200 ×10−6 ) = −42.0 (106 ) N/m 2 = 42.0 MPa (C)
σ 2 = Eε 2 = ( 210 × 109 )( 820 ×10−6 ) = 172.2 (106 ) N/m 2 = 172.2 MPa (T)
σ 3 = Eε 3 = ( 210 ×109 )( 600 × 10−6 ) = 126.0 (106 ) N/m 2 = 126.0 MPa (T)
σ 4 = Eε 4 = ( 210 × 109 )( −420 × 10−6 ) = −88.2 (106 ) N/m 2 = 88.2 MPa (C)
σ1 =
P M zc
−
A
I
σ3 =
P M zc
+
A
I
Therefore:
P=
P M yc
+
A
I
P M c
σ4 = − y
A
I
σ2 =
(σ 1 + σ 3 ) A = ( −42.0 + 126.0 ) (106 )( 7854 ×10−6 ) = 329.9
2
2
(10 ) N
3
P ≅ 330 kN ............................................................................................................................. Ans.
Similarly:
My =
(σ 2 − σ 4 ) I = (172.2 − 88.2 ) (106 )( 4.909 ×10−6 ) = 12.783 103 N ⋅ m
( )
2c
2 ( 0.050 )
M y ≅ 12.78 kN ⋅ m ................................................................................................................ Ans.
6
−6
σ 3 − σ 1 ) I (126.0 + 42.0 ) (10 )( 4.909 ×10 )
(
Mz =
=
= 8.247 (103 ) N ⋅ m
2c
2 ( 0.050 )
M z ≅ 8.25 kN ⋅ m ................................................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-197*
I=
π (1)
64
4
−
π ( 0.75 )
64
4
= 0.03356 in.4
J = 2 I = 0.06711 in.4
M x = ( 50 )(18 ) = 900 lb ⋅ in.
T = ( 50 )( 7 ) = 350 lb ⋅ in.
Vz = 50 lb
σ=
Mc ( 900 )( 0.5 )
=
= 13, 409 psi (T, top; C, bottom)
0.003356
I
τ=
Tc ( 350 )( 0.50 )
=
= 2608 psi
0.06711
J
On the top of the pipe:
σ p1, p 2 =
σx +σ y
2
 σ −σ y 
13, 409 + 0
2
 13, 409 − 0 
2
±  x
± 
 + τ xy =
 + ( 2608 )
2
2


 2 
2
2
σ p1 = 6705 + 7194 = +13,899 psi ≅ 13.90 ksi (T) ........................................................ Ans.
σ p 2 = 6705 − 7194 = −489 psi ≅ 0.489 ksi (C) ............................................................. Ans.
σ p 3 = 0 ksi
τ max = τ p =
.............................................................................................................................. Ans.
σ p1 − σ p 2
2
=
(13,899 ) − ( −489 ) = 7194 psi ≅ 7.19 ksi ........................... Ans.
2
On the bottom of the pipe:
σ p1, p 2 =
σx +σ y
2
 σ −σ y 
−13, 409 + 0
2
 −13, 409 − 0 
2
±  x
± 
 + τ xy =
 + ( 2608 )
2
2


 2 
2
2
σ p1 = −6705 + 7194 = +489 psi ≅ 0.489 ksi (T) ........................................................... Ans.
σ p 2 = −6705 − 7194 = −13,899 psi ≅ 13.90 ksi (C) ..................................................... Ans.
σ p 3 = 0 ksi
τ max = τ p =
.............................................................................................................................. Ans.
σ p1 − σ p 2
2
=
( 489 ) − ( −13,899 ) = 7194 psi ≅ 7.19 ksi .......................... Ans.
2
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-198
I=
Qb
π (120 )
4
64
(120 )
=
3
J = 2 I = 20.358 (106 ) mm 4
= 10.179 (106 ) mm 4
= 144.0 (103 ) mm3
12
TA = ( 30 )( 0.250 ) − ( 5 )( 0.250 ) = 6.25 kN ⋅ m
For the horizontal plane:
ΣM B = 0 :
RC ( 2800 ) − ( 35 )( 800 ) = 0
RC = 10.00 kN Z
ΣM C = 0 :
( 35)( 2000 ) − RB ( 2800 ) = 0
RB = 25.00 kN Z
For the horizontal plane:
ΣM B = 0 :
RC ( 2800 ) − ( 35 )( 2000 ) = 0
RC = 25.00 kN ↑
ΣM C = 0 :
( 35)( 2000 ) − RB ( 800 ) = 0
RB = 10.00 kN ↑
From the shear-force and bendingmoment diagrams:
VAx = −10 kN
M Ax = 12 kN ⋅ m
VAz = +10 kN
M Az = 16 kN ⋅ m
(Note that neither VAz nor M Az
affect the stresses at A.)
Therefore:
σx =
− Mc − (12, 000 )( 0.060 )
=
= −70.73 (106 ) N/m 2 = 70.73 MPa (C)
−6
I
10.179 (10 )
−6
Tc VQ ( 6250 )( 0, 060 ) (10, 000 ) (144 ×10 )
τ xy = +
=
+
= 19.60 (106 ) N/m 2 = 19.60 MPa
−6
−6
J
It
20.358 (10 ) (10.179 ×10 ) ( 0.120 )
σ p1, p 2 =
σx +σ y
2
 σ −σ y 
−70.73 + 00
2
 −70.73 − 0 
2
±  x
± 
 + τ xy =
 + (19.60 )
2
2


 2 
2
2
σ p1 = −35.365 + 40.433 = +5.068 MPa ≅ 5.07 MPa (T) ............................................ Ans.
σ p 2 = −35.365 − 40.433 = −75.798 MPa ≅ 76.0 MPa (C) ......................................... Ans.
σ p 3 = 0 MPa ........................................................................................................................... Ans.
τ max = τ p =
σ p1 − σ p 2
2
=
( 5.068) − ( −75.798) = 40.4 MPa ......................................... Ans.
2
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-199*
A = π (1) 4 = 0.7854 in.2
2
I = π (1) 64 = 0.04909 in.4
4
σ A = Eε A = (10 × 106 )( 550 ×10−6 ) = 5500 psi
σ B = Eε B = (10 × 106 )( 400 × 10−6 ) = 4000 psi
σ C = Eε C = (10 × 106 )( −300 ×10−6 ) = −3000 psi
σ A = Eε A =
P ( x + 3)( 0.50 )
Q  P ( x + 3)  c
Q
+
=
+
= 5500 psi
A
I
0.7854
0.04909
(a)
σ B = Eε B =
Px ( 0.50 )
Q ( Px ) c
Q
+
=
+
= 4000 psi
0.7854 0.04909
A
I
(b)
Px ( 0.50 )
Q ( Px ) c
Q
−
=
−
= −3000 psi
(c)
0.7854 0.04909
A
I
From adding Eqs. (b) and (c):
Q = 392.7 lb ≅ 393 lb ................................................. Ans.
σ C = Eε C =
Px = 343.63 lb
and from Eq. (a):
Px + 3P = 490.90 lb
Combining these last two equations gives: P = 49.09 lb ≅ 49.1 lb ................................................ Ans.
and
x = 7.00 in. .................................................................... Ans.
Then from Eq. (b):
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MECHANICS OF MATERIALS, 6th Edition
7-200
P = 125 kN (C)
RILEY, STURGES AND MORRIS
Vx = 50 kN
M z = ( 50 × 0.900 ) = 45.0 kN ⋅ m
A=
I=
π ( 265 )
2
4
π ( 265 )
4
−
4
−
π ( 250 )
2
4
π ( 250 )
4
4
= 6067.2 mm 2
= 805.27 (106 ) mm 4
2r 3 2 ( 265 ) 2 ( 250 )
Q=
=
−
= 1989.75 (103 ) mm3
3
3
3
3
3
125 (103 )
P
σP = =
= 20.60 (106 ) N/m 2 (C)
−6
A 6067.2 (10 )
σ axial
3
pr ( 2500 × 10 ) ( 0.250 )
=
=
= 20.833 (106 ) N/m 2 (T)
2t
2 ( 0.015 )
σ hoop =
pr
= 41.667 (106 ) N/m 2 (T)
t
VQ ( 50, 000 ) (1989.75 × 10 )
τV =
=
= 4.118 (106 ) N/m 2
−6
It
(805.27 ×10 ) ( 0.030 )
−6
Note that both A and B are on the neutral axis for bending and
the bending moment does not affect the stress at either A or B.
The affect of the other stresses is the same at both A and B.
σ x = 41.667 MPa (T)
σ y = 20.833 − 20.60 = 0.233 MPa (T)
τ xy = −4.118 MPa
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MECHANICS OF MATERIALS, 6th Edition
7-200
(a)
RILEY, STURGES AND MORRIS
(cont.)
θ n = −53.13°
σ n = σ x cos 2 θ n + σ y sin 2 θ n + 2τ xy sin θ n cos θ n
= ( 41.667 ) cos 2 ( −53.13° ) + ( 0.233) sin 2 ( −53.13° )
+ 2 ( −4.118 ) sin ( −53.13° ) cos ( −53.13° )
σ n = +19.10 MPa = 19.10 MPa (T) .......................................................................... Ans.
τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos2 θ − sin 2 θ )
= − ( 41.667 ) − ( 0.233)  sin ( −53.13° ) cos ( −53.13° )
+ ( −4.118 )  cos 2 ( −53.13° ) − sin 2 ( −53.13° ) 
τ nt = +21.0 MPa ............................................................................................................. Ans.
41.667 + 0.233
2
 41.667 − 0.233 
± 
 + ( −4.118 )
2
2


2
(b)
σ p1, p 2 =
σ p1 = 20.950 + 21.122 = 42.072 MPa ≅ 42.1 MPa (T) ............................................... Ans.
σ p 2 = 20.950 − 21.122 = −0.172 MPa ≅ 0.172 MPa (C) ........................................... Ans.
σ p 3 = 0 MPa ........................................................................................................................... Ans.
τ max = τ p =
σ p1 − σ p 2
2
=
( 42.072 ) − ( −0.172 ) = 21.1 MPa ......................................... Ans.
2
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-201
A = π ( 4 ) 4 = 12.566 in.2
I z = π ( 4 ) 64 = 12.566 in.4
2
4
Q = 2r 3 3 = 2 ( 2 ) 3 = 5.333 in.3
3
P = 18 kip (C)
J = 2 I = 25.133 in.4
Vx = 2.25 kip
M z = ( 2.25 × 36 ) = 81.0 kip ⋅ in.
T = ( 2.25 × 24 ) = 54.0 kip ⋅ in.
σP =
P
18
=
= 1.4324 ksi (C)
A 12.566
σM =
z
M z c ( 81)( 2 )
=
= 12.892 ksi (T, at A)
12.566
Iz
Tc ( 54 )( 2 )
=
= 4.297 ksi
25.133
J
V Q ( 2.25 )( 5.333)
τV = x =
= 0.2387 ksi
I zt
(12.566 )( 4 )
τT =
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-201 (cont.)
Therefore, at A:
σ x = 12.892 − 1.4324 = 11.459 ksi
σ y = 0 ksi
τ xy = 4.297 ksi
σ p1, p 2 =
σx +σ y
2
 σ −σ y 
11.459 + 0
2
 11.459 − 0 
2
±  x
± 
 + τ xy =
 + ( 4.297 )
2
2


 2 
2
2
σ p1 = 5.730 + 7.162 = +12.892 ksi ≅ 12.89 ksi (T) ...................................................... Ans.
σ p 2 = 5.730 − 7.162 = −1.432 ksi ≅ 1.432 ksi (C)
σ p 3 = 0 ksi
τ max = τ p =
....................................................... Ans.
.............................................................................................................................. Ans.
σ p1 − σ p 2
2
=
(12.892 ) − ( −1.432 ) = 7.16 ksi ............................................. Ans.
2
And at B:
σ x = −1.4324 ksi = 1.4324 ksi (C)
σ y = 0 ksi
τ xy = 4.297 + 0.2387 = 4.536 ksi
−1.4324 + 0
2
 −1.4324 − 0 
=
± 
 + ( 4.536 )
2
2


2
σ p1, p 2
σ p1 = −0.7162 + 4.592 = +3.876 ksi ≅ 3.88 ksi (T) ..................................................... Ans.
σ p 2 = −0.7162 − 4.592 = −5.308 ksi ≅ 5.31 ksi (C) ..................................................... Ans.
σ p 3 = 0 ksi
τ max = τ p =
.............................................................................................................................. Ans.
σ p1 − σ p 2
2
=
( 3.876 ) − ( −5.308) = 4.59 ksi ............................................... Ans.
2
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MECHANICS OF MATERIALS, 6th Edition
7-202*
RILEY, STURGES AND MORRIS
θ = tan −1 ( 3 5 ) = 30.96°
For the entire chair:
↑ ΣFy = 0 :
4 ΣM B = 0 :
A + B − 84 − 28sin θ = 0
( 0.2 )(84 ) − ( 0.5 )( 24 ) − ( 0.4 ) A +  0.3 +
A = 73.82 N
B = 24.58 N

0.5 
 ( 28 ) = 0
cos θ 
For member DF:
→ ΣFx = 0 :
Dx − Fx + 24 = 0
↑ ΣFy = 0 :
Fy + Dy − 84 = 0
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-202* (cont.)
( 0.4 )(84 ) − ( 0.5) Fy = 0
Dx = ( Fx − 24 ) N
4 ΣM D = 0 :
and
Fy = 67.2 N
Dy = 16.80 N
For member BCF:
Fx + Cx − 28cos θ = 0
→ ΣFx = 0 :
24.58 + C y − 67.2 − 28sin θ = 0
↑ ΣFy = 0 :
4 ΣM D = 0 :
0.1667 

 0.3 +
 ( 28 ) + ( 0.1333)( 24.58 )
sin θ 

+ ( 0.1667 )( 67.2 ) − ( 0.2777 ) Fx = 0
Fx = 115.1 N
Cx = −91.0 N
C y = 57.0 N
On a section midway between pins C and F:
a = 222.2 mm
400 500
=
a
b
b = 277.8 mm
b 2 = 138.9 mm
c = 133.3 mm
d = 166.7 mm
d 2 = 83.4 mm
a + b = 500 mm
Therefore
^ ΣFn = 0 :
24.6 cos 30.96° + 57.0 cos 30.96° + 91.0sin 30.96° − P = 0
M + ( 91.0 )(138.9 ) − ( 57.0 )( 83.4 ) − ( 24.6 )( 216.7 ) = 0
4 ΣM cut = 0 :
M = −2555 N ⋅ mm = 2.555 N ⋅ m 4
A = (10 × 30 ) = 300 mm
σ top =
2
(10 )( 30 )
I=
3
12
= 22.5 (103 ) mm 4
− P Mc − (116.79 ) ( 2.555 )( 0.015 )
+
=
+
A
I
( 300 ×10−6 ) ( 22.5 ×10−9 )
σ top = +1.3140 (106 ) N/m 2 ≅ 1.314 MPa (T) ......................................................... Ans.
σ bottom =
− P Mc − (116.79 ) ( 2.555 )( 0.015 )
−
=
−
A
I
( 300 ×10−6 ) ( 22.5 ×10−9 )
σ top = −2.093 (106 ) N/m 2 ≅ 2.09 MPa (C) ............................................................. Ans.
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-203*
A=
I=
π ( 4)
2
G=
= 12.566 in.2
J = 2 I = 25.133 in.2
4
π ( 4)
E
29, 000
=
= 11,154 ksi
2 (1 + ν ) 2 (1 + 0.30 )
= 12.566 in.2
4
64
2r 3 2 ( 2 )
Q=
=
= 5.333 in.3
3
3
Tc VQ
At gage A:
τ xy = −
= σ 45° = −σ −45°
J
It
σ −νσ −45° 1 +ν  Tc VQ 
ε 45° = 45°
= εA
=
−
E
E  J
It 
3
At gage C:
ε 45°
Therefore:
which gives
Tc VQ
+
= σ 45° = −σ −45°
J
It
σ −νσ −45° 1 +ν  Tc VQ 
= εC
= 45°
=
+
E
E  J
It 
τ xy =
 1 + ν   Tc 
ε A + εC = 2 
 
 E  J 
T=
EJ ( ε A + ε C ) ( 29, 000 )( 25.133)( 450 + 550 ) (10
=
2 (1 +ν ) c
2 (1 + 0.30 )( 2 )
−6
)
T = 140.16 kip ⋅ in. ≅ 140.2 kip ⋅ in. .............................................................. Ans.
 1 + ν  VQ 
εC − ε A = 2 


 E  It 
Also
which gives
EIt ( ε C − ε A ) ( 29, 000 )(12.566 )( 4 )( 550 − 450 ) (10
V=
=
2 (1 + ν ) Q
2 (1 + 0.30 )( 5.333)
−6
)
V = 10.513 kip ≅ 10.51 kip ............................................................................ Ans.
At gage B:
τ xy =
σ=
Tc (140.16 )( 2 )
=
= 11.153 ksi
J
( 25.133)
− Mc − M ( 2 )
=
= ( −0.159160 M ) ksi
I
12.566
ε B = ε x cos 2 45° + ε y sin 2 45° + γ xy sin 45° cos 45°
= (σ x E )( 0.5 ) + ( −vσ x E )( 0.5 ) + (τ xy G ) ( 0.5 )
= σ x (1 −ν ) + τ xy 2 (1 +ν )  2 E
2 ( 29, 000 ) ( 325 × 10−6 ) = ( −0.159160M )(1 − 0.30 ) + 2 (11.154 )(1 + 0.30 )
M = 91.1 kip ⋅ in. .............................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-204
A = π ( 4502 − 4402 ) = 27,960 mm 2
π ( 450 − 440
4
I=
4
4
Q=
) = 2768.8 10 mm
( )
6
4
2 ( 4503 − 4403 )
3
= 3960.7 (103 ) mm3
J = 2 I = 5537.5 (106 ) mm 4
F = W = mg = ( 250 )( 9.81) = 2453 N (C)
Vx = P = pA = (1500 )( 8 × 3) = 36, 000 N
M x = W ( 3) = ( 2453)( 3) = 7359 N ⋅ m
M z = P ( 9 ) = ( 36, 000 )( 9 ) = 324 (103 ) N ⋅ m
T = P ( 3) = ( 36, 000 )( 3) = 108, 000 N ⋅ m
σF =
F
2453
=
= 0.08773 (106 ) N/m 2
−6
A 27,960 (10 )
= 0.08773 MPa (C)
σM =
x
σM
M x c ( 7359 )( 0.450 )
=
= 1.1959 (106 ) N/m 2 = 1.1959 MPa
−6
I
2768.8 (10 )
3
M z c ( 324 × 10 ) ( 0.450 )
=
=
= 52.65 (106 ) N/m 2 = 52.65 MPa
−6
I
2768.8 (10 )
z
τT =
Tc (108, 000 )( 0.450 )
=
= 8.776 (106 ) N/m 2 = 8.776 MPa
−6
J
( 5537.5 ×10 )
−6
Vx Q ( 36, 000 ) ( 3960.7 × 10 )
=
= 2.575 (106 ) N/m 2 = 2.575 MPa
τ Vx =
−6
It
( 2768.8 ×10 ) ( 0.020 )
At A:
σ x = 0 MPa
σy =
τ xy =
Tc Vx Q
+
= 8.776 + 2.575 = 11.351 MPa
J
It
F M xc
+
= 0.08773 + 1.1959 = 1.2838 MPa (C)
A
I
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-204 (cont.)
σ p1, p 2 =
σx +σ y
2
 σ −σ y 
0 − 1.2838
2
 0 + 1.2838 
2
±  x
± 
 + τ xy =
 + (11.351)
2
2


 2 
2
2
σ p1 = −0.6418 + 11.369 = +10.7272 MPa ≅ 10.73 MPa (T) ...................................... Ans.
σ p 2 = −0.6418 − 11.369 = −12.0108 MPa ≅ 12.01 MPa (C) ..................................... Ans.
σ p 3 = 0 MPa ........................................................................................................................... Ans.
τ max = τ p =
At B:
σ p1 − σ p 2
2
=
σ x = 0 MPa
σy =
(10.7272 ) − ( −12.0108) = 11.37 MPa ................................ Ans.
2
τ xy =
Tc
= 8.776 MPa
J
F M zc
+
= −0.08773 + 52.65 = 52.56 MPa (T)
A
I
0 + 52.56
2
 0 − 52.56 
=
± 
 + ( 8.776 )
2
2


2
σ p1, p 2
σ p1 = 26.28 + 27.71 = +53.99 MPa ≅ 54.0 MPa (T)
................................................... Ans.
σ p 2 = 26.28 − 27.71 = −1.430 MPa ≅ 1.430 MPa (C) ................................................. Ans.
σ p 3 = 0 MPa ........................................................................................................................... Ans.
τ max = τ p =
σ p1 − σ p 2
2
=
( 53.99 ) − ( −1.430 ) = 27.7 MPa ........................................... Ans.
2
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-205
4 ΣM C = 0 :
RILEY, STURGES AND MORRIS
(1350 )( 5) + ( 670 )( 5 ) − 10 RA = 0
RA = 1010 lb
θ = tan −1 (13 10 ) = 52.431°
At the section containing G and H:
Z ΣFn = 0 :
1010sin θ − P = 0
^ ΣFt = 0 :
1010 cos θ − V = 0
M − (1010 cos θ )( 3.5 ) = 0
4 ΣM cut = 0 :
P = 800.6 lb
V = 615.8 lb
M = 2155.3 lb ⋅ ft
A = ( 4 × 4 ) = 16 in.2
( 4 )( 4 )
I=
QH = yC A = (1.5 )( 4 × 1) = 6.00 in.3
3
12
= 21.333 in.4
At point G:
P Mc −800.6 ( 2155.3 ×12 )( 2 )
τ =0
+
=
−
= −2474.8 psi
A I
16
21.333
σ p1 = σ p 3 = 0 psi ........................ σ p 2 = −2474.8 psi ≅ 2470 psi C ............................ Ans.
σ=
τ max = τ p =
σ p1 − σ p 2
2
=
( 0 ) − ( −2474.8) = 1237 psi .................................................... Ans.
2
At point H:
P Mc −800.6 ( 2155.3 ×12 )(1)
+
=
+
= +1162.3 psi
A I
16
21.333
VQ ( 615.8 )( 6.00 )
τ=
=
= 43.30 psi
It
( 21.333)( 4 )
σ=
1162.3 + 0
2
 1162.3 − 0 
=
± 
 + ( 43.30 )
2
2


2
σ p1, p 2
σ p1 = 581.15 + 582.76 = +1163.91 psi ≅ 1164 psi (T)
σ p 2 = 581.15 − 582.76 = −1.61 psi ≅ 1.61 psi (C)
σ p 3 = 0 psi
τ max = τ p =
................................................ Ans.
........................................................ Ans.
.............................................................................................................................. Ans.
σ p1 − σ p 2
2
=
(1163.91) − ( −1.61) = 583 psi ............................................... Ans.
2
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-206
(a)
P = 360 kN
RILEY, STURGES AND MORRIS
V = 270 kN
M = ( 360 )( 0.150 ) + ( 270 )( 0.450 ) = 175.5 kN ⋅ m
A = 2 (120 × 30 ) + ( 30 × 180 ) = 12, 600 mm 2
(120 )( 240 )
I=
3
12
( 90 )(180 )
−
3
12
= 94.50 (106 ) mm 4
360 (103 )
175.5 ×103 ) y
(
P My
=
+
σx = +
A I
12, 600 (10−6 ) ( 94.50 × 10−6 )
= ( 28.571 + 1857.14 y ) (106 ) N/m 2
σ C = 28.571 + (1857.14 )( 0.120 ) = 251.43 MPa (T)
σ B = 28.571 − (1857.14 )( 0.120 ) = 194.29 MPa (C)
Q = yC A =
(120 + y )
2
(120 )(120 − y ) = 60 (1202 − y 2 ) mm3
Q = (105 )(120 × 30 ) +
( 90 + y )
2
( 30 )( 90 − y )
= 378, 000 + 15 ( 90 − y )  mm
2
3
VQ ( 270 × 10 ) Q
τ xy =
=
It ( 94.50 × 10−6 ) t
2
90 ≤ y ≤ 120 mm
−90 ≤ y ≤ 90 mm
3
t = 120 mm
90 ≤ y ≤ 120 mm
t = 30 mm
−90 ≤ y ≤ 90 mm
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-206
(b)
(cont.)
y
σB
y = 240
(c)
RILEY, STURGES AND MORRIS
σ p1, p 2 =
=
240
σ B +σC
194.29
= 104.62 mm
194.29 + 251.43
σx + 0
τ max = τ p =
2
σ −0
±  x
+ τ xy2

 2 
2
σ p1 − σ p 2
2
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
7-207
RILEY, STURGES AND MORRIS
A = (1× 0.25 ) = 0.25 in 2
bh3 ( 0.25 )(1)
I=
=
= 0.02083 in.4
12
12
3
M = P ( 3) = ( 3P ) lb
(a)
σB =
σC =
(b)
s
σB
s=
=
( 3P )( 0.5) = 76.01P psi
P Mc
P
+
=
+
(
)
A I
0.25 0.02083
( 3P )( 0.5) = −68.01P psi
P Mc
P
−
=
−
(
)
A I
0.25 0.02083
1− s
σC
σB
76.01
=
= 0.5278 in.
σ B + σ C 76.01 + 68.01
The distance s is independent of P .
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-208*
By symmetry each support carries half of the total load:
A = B = 6700 N ↑
From the shear-force and bending-moment diagrams:
Vmax = 6700 N
Then
M max = 13, 400 N ⋅ m
M max 13, 400
=
≤ 9 (106 ) N/m 2
S
S
13, 400
= 1488.9 (10−6 ) m3
S≥
9 (106 )
σ max =
= 1488.9 (103 ) mm3
Try a 203 × 254-mm timber with
S = 1850 (103 ) mm3
A = 46, 000 mm 2
I = 223 (106 ) mm 4
m = 29.4 kg/m
When the weight of the beam is included,
the maximum moment is
M = M load + M weight
( 29.4 × 9.81)( 5)
= 13, 400 +
8
S≥
2
= 14,301 N ⋅ m
14,301
= 1589 (10−6 ) m3 = 1589 (103 ) mm3
6
9 (10 )
which is still okay. Next, check the shear stress,
Vmax = Vload + Vweght = 6700 +
τ max = 1.5
( 29.4 × 9.81)( 5) = 7421 N
2
Vmax 1.5 ( 7421)
=
= 242 (103 ) N/m 2 = 242 kPa
A
46, 000
which is much less than the allowable shear stress of 600 kPa . Therefore, this design is okay.
Use a 203 × 254-mm timber ............................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-209*
A = B = 1800 lb ↑
By symmetry each support carries half of the total load:
From the shear-force and bending-moment diagrams:
Vmax = 1800 lb
M max = 10,800 lb ⋅ ft
M max 10,800 × 12
=
≤ 1900 psi
S
S
10,800 ×12
S≥
= 68.21 in.3
1900
Try an 8 × 8-in. timber with
Then
σ max =
S = 70.3 in.3
A = 56.3 in.2
I = 264 in.4
w = 15.6 lb/ft
When the weight of the beam is included,
the maximum moment is
M = M load + M weight
S≥
(15.6 )( 20 )
= 10,800 +
8
2
= 11,580 lb ⋅ ft
11,580 × 12
= 73.14 in.3
1900
which is bigger than the section modulus of the chosen timber. Next, try a 4 × 12-in. timber with
S = 79.9 in.3
I = 459 in.4
A = 41.7 in.2
w = 11.6 lb/ft
When the weight of the beam is included, the maximum moment is now
(11.6 )( 20 )
M = 10,800 +
8
2
= 11,380 lb ⋅ ft
S≥
11,380 × 12
= 71.87 in.3
1900
which is still okay. Next, check the shear stress,
Vmax = Vload + Vweght = 1800 +
(11.6 )( 20 ) = 1916 lb
2
Vmax 1.5 (1916 )
=
= 68.92 psi
A
41.7
which is less than the allowable shear stress of 90 psi . Therefore, this design is okay.
τ max = 1.5
Use a 4 × 12-in. timber ......................................................................................................... Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-210
RILEY, STURGES AND MORRIS
4 ΣM B = 0 :
1.5 P − ( 0.25 )( 275 × 9.807 ) = 0
P = 449.5 N
4 ΣM C = 0 :
1.75 ( 449.5 ) − 0.25 B = 0
B = 3147 N
M max = ( 449.5 )(1.5 ) = 674.25 N ⋅ m
M max 674.25
=
≤ 135 (106 ) N/m 2
S
S
674.25
= 4.994 (10−6 ) m3 = 4.994 (103 ) mm3
S≥
6
135 (10 )
σ max =
Therefore
Use a 38-mm diameter standard weight pipe ................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-211*
By symmetry each support carries half of the total load:
A = B = ( 4 × 16 ) 2 = 32 kip ↑
From the shear-force and bending-moment diagrams:
Vmax = 32 kip
M max = 128 kip ⋅ ft
M max 128 × 12
=
≤ 22 ksi
S
S
128 × 12
S≥
= 69.82 in.3
22
Try a W 18 × 60 section with
Then
σ max =
S = 108 in.3
tw = 0.415 in.
d = 2c = 18.24 in.
w = 60 lb/ft
When the weight of the beam is included, the maximum moment is
M = M load + M weight
S≥
( 0.060 )(16 )
= 128 +
8
2
= 129.92 lb ⋅ ft
129.92 × 12
= 70.87 in.3
22
which is still okay. Next, check the shear stress,
Vmax = Vload + Vweght = 32 +
τ max =
( 0.060 )(16 ) = 32.48 kip
2
Vmax
32.48
=
= 4.29 ksi
Aweb 0.415 × 18.24
which is less than the allowable shear stress of 14.5 ksi . Therefore, this design is okay.
Use a W 18 × 60 section ...................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-212*
By symmetry each support carries half of the total load:
A = B = 42 kN ↑
From the shear-force and bending-moment diagrams:
Vmax = 42 kN
Then
M max = 82.5 kN ⋅ m
M max 82,500
=
≤ 152 (106 ) N/m 2
S
S
82,500
= 542.8 (10−6 ) m3 = 542.8 (103 ) mm3
S≥
6
152 (10 )
σ max =
Try an S 305 × 47 section with
S = 596 (103 ) mm3
d = 2c = 304.8 mm
tw = 8.9 mm
m = 47 kg/m
When the weight of the beam is included, the maximum moment is
M = M load + M weight
S≥
( 47 × 9.81)( 5.5)
= 82,500 +
8
2
= 84, 243 N ⋅ m
84, 243
= 554.2 (10−6 ) m3 = 554.2 (103 ) mm3
6
152 (10 )
which is still okay. Next, check the shear stress,
Vmax = Vload + Vweght = 42, 000 +
τ max =
( 47 × 9.81)( 5.5) = 43, 268 N
2
Vmax
43, 268
=
= 15.95 (106 ) N/m 2 = 15.95 MPa
Aweb 0.0089 × 0.3048
which is much less than the allowable shear stress of 100 MPa . Therefore, this design is okay.
Use an S 305 × 47 section .................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-213
By symmetry each support carries half of the total load:
A = B = 3000 lb ↑
From the shear-force and bending-moment diagrams:
Vmax = 3 kip
M max = 24 kip ⋅ ft
M max 24 × 12
=
≤ 22 ksi
S
S
24 × 12
S≥
= 13.091 in.3
22
Try a W 6 × 25 section with
Then
σ max =
S = 16.7 in.3
d = 2c = 6.38 in.
tw = 0.320 in.
w = 25 lb/ft
When the weight of the beam is included, the maximum moment is
( 25 )( 24 )
M = 24, 000 +
8
S≥
2
= 25,800 lb ⋅ ft
25.80 × 12
= 14.07 in.3
22
which is still okay. Next, check the shear stress,
Vmax = Vload + Vweght = 3000 +
τ max =
( 25)( 24 ) = 3300 lb
2
Vmax
3300
=
= 1616.4 psi
Aweb 0.320 × 6.35
which is much less than the allowable shear stress of 14.5 ksi . Therefore, this design is okay.
Use a W 6 × 25 section ....................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-214*
4 ΣM B = 0 :
(15)(8 − x ) − 8 A = 0
4 ΣM A = 0 :
8 B − 15 x = 0
RILEY, STURGES AND MORRIS
A = (15 − 1.875 x ) kN ↑
B = (1.875 x ) kN ↑
For any location x , the maximum moment in the beam occurs
under the load and is equal to
M C = Ax = (15 x − 1.875 x 2 ) kN ⋅ m
The location x that makes M C a maximum is found from
dM C
= (15 − 3.75 x ) = 0
dx
which gives
x = 4.00 m
Then
M max = 30 kN ⋅ m
M max 30, 000
=
≤ 152 (106 ) N/m 2
S
S
30, 000
= 197.4 (10−6 ) m3 = 197.4 (103 ) mm3
S≥
6
152 (10 )
σ max =
Try an S 203 × 27 section with
S = 236 (103 ) mm3
d = 2c = 203.2 mm
tw = 6.9 mm
m = 27 kg/m
When the weight of the beam is included, the maximum moment is
M = M load + M weight = 34.386 kN ⋅ m
S≥
34,386
= 226.2 (10−6 ) m3 = 226.2 (103 ) mm3
6
152 (10 )
which is still okay. Next, check the shear stress, The maximum shear stress occurs
when the load is near one of the supports ( x = 0 or x = 8 m ) and is equal to
Vmax = Vload + Vweght = 15, 000 +
τ max =
( 27 × 9.81)(8 ) = 16, 059 N
2
Vmax
16, 059
=
= 11.45 (106 ) N/m 2 = 11.45 MPa
Aweb 0.0069 × 0.2032
which is much less than the allowable shear stress of 100 MPa . Therefore, this design is okay.
Use an S 203 × 27 section .................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-215
Since the joists are spaced 16 in. (1.3333 ft) apart,
the uniformly distributed load is
w = ( 60 )(1.3333) = 80 lb/ft
By symmetry each support carries half of the total load:
A = B = (12 × 80 ) 2 = 480 lb ↑
From the shear-force and bending-moment diagrams:
Vmax = 480 lb
M max = 1440 lb ⋅ ft
M max 1440 ×12
=
≤ 1200 psi
S
S
1440 ×12
S≥
= 14.40 in.3
1200
Try a 2 × 8-in.timber with
Then
σ max =
S = 15.3 in.3
I = 57.1 in.4
A = 12.2 in.2
w = 3.39 lb/ft
When the weight of the beam is included, the maximum moment is
( 3.39 )(12 )
M = 1440 +
8
2
= 1501 lb ⋅ ft
S≥
1501×12
= 15.01 in.3
1200
which is still okay. Next, check the shear stress,
Vmax = 480 +
( 3.39 )(12 ) = 500.3 lb
2
τ max = 1.5
Vmax 1.5 ( 500.3)
=
= 61.51 psi
A
12.2
which is much less than the allowable shear stress of 120 psi . Therefore, this design is okay.
Use a 2 × 8-in. timber section ............................................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-216
4 ΣM A = 0 :
RILEY, STURGES AND MORRIS
6 B − 8829b − ( 8829 )( b + 1.5 ) = 0
B = ( 2943b + 2207 ) N
A + B − 2 ( 8829 ) = 0
↑ ΣFy = 0 :
A = (15, 451 − 2943b ) N
Then
VAC = A = (15, 451 − 2943b ) N
VCD = VAC − 8829 = ( 6622 − 2943b ) N
M C = VAC b = (15, 451b − 2943b 2 ) N ⋅ m
M D = M C + VCD (1.5 )
= ( 9933 + 11, 036b − 2943b 2 ) N ⋅ m
Clearly, the maximum moment occurs under one of the wheels. The
location b which gives the maximum for these moments is found from
dM C
= (15, 451 − 5886b ) = 0
db
b = 2.625 m
M C = 20, 280 N ⋅ m
M D = 18, 624 N ⋅ m
dM D
= (11, 036 − 5886b ) = 0
db
b = 1.875 m
M C = 18, 624 N ⋅ m
M D = 20, 280 N ⋅ m
Therefore, the maximum bending moment occurs
under the wheel closest to the center of the beam
M max = 20, 280 N ⋅ m
The minimum section modulus required is:
S min =
M max
σ all
=
20, 280
= 122.9 (10−6 ) m3 = 122.9 (103 ) mm3
6
165 (10 )
Try a W 127 × 24 section with
S = 139 (103 ) mm3
d = 2c = 127 mm
tw = 6.1 mm
m = 24 kg/m
Next, check the shear stress, The maximum shear stress occurs
when the load is near one of the supports ( x = 0 or x = 6 m )
τ max =
Vmax
15, 451
=
= 19.94 (106 ) N/m 2 = 19.94 MPa
Aweb 0.0061× 0.127
which is much less than the allowable shear stress of 100 MPa .
Therefore, this design is okay.
Use a W 127 × 24 section ................................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-217*
yC =
RILEY, STURGES AND MORRIS
M x 2 ( 4 )( 2 × 8 )  + (1)( 8 × 2 )
=
= 3.00 in.
A
2 ( 2 × 8) + (8 × 2 )
( 2 )( 5)
I =2
3
3
(12 )( 3)
+
3
3
(8)(1)
−
3
3
= 272.00 in.4
(a)
σ max = ( 5 −1) σ A = ( −5 )( 2000 ) = −10, 000 psi = 10 ksi (C)
(b)
Mr =
..................................... Ans.
−σ A I − ( 2000 )( 272.00 )
=
= +544 (103 ) lb ⋅ in. = 544 kip ⋅ in. ...................... Ans.
yA
( −1)
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-218*
yC =
I=
RILEY, STURGES AND MORRIS
M x (100 )( 37.5 × 200 ) + ( 225 )(100 × 50 )
=
= 150.0 mm
A
( 37.5 × 200 ) + (100 × 50 )
( 37.5)(150 )
3
3
+
(100 )(100 )
3
3
−
( 62.5 )( 50 )
− M r y − (100 ×10 ) ( −0.150 )
=
I
( 72.92 ×10−6 )
3
3
= 72.92 (106 ) mm 4
3
σ bottom =
σ bottom = +206 (106 ) N/m 2 = 206 MPa (T) ..................................................................... Ans.
σ top
3
− M r y − (100 ×10 ) ( +0.100 )
=
=
I
( 72.92 ×10−6 )
σ top = −137.1(106 ) N/m 2 = 137.1 MPa (C) ................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-219
For 0 ft ≤ x ≤ 10 ft
V = ( 200 × 6 ) + 760 − 300 x
V = ( −300 x + 1960 ) lb ............................................ Ans.
M = ( 200 × 6 )( x + 3) + 760 x − ( 300 x )( x 2 )
M = ( −150 x 2 + 1960 x + 3600 ) lb ⋅ ft ................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-220*
4 ΣM E = 0 :
(10 )(12 ) + ( 8 × 6 )( 7 ) − 8RB = 0
RB = 57 kN
↑ ΣFy = 0 :
RB + RE − 10 − ( 8 × 6 ) − 20 = 0
RE = 21 kN
For 0 m ≤ x ≤ 4 m
V = − (10 ) − ( 8 )( x + 2 ) + ( 57 )
V = ( 31 − 8 x ) kN ...................................................... Ans.
M = − (10 )( x + 4 ) − ( 8 )( x + 2 ) 
( x + 2) +
2
( 57 x )
M = ( −4 x 2 + 31x − 56 ) kN ⋅ m .............................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-221
yC =
( 4 )(1× 8) + (8.5 )( 4 ×1)
(1× 8 ) + ( 4 ×1)
= 5.50 in.
( 4 )( 3.5 )
I=
3
3
= 97.00 in.4
( 3)( 2.5)
−
3
3
(1)( 5.5)
+
3
3
At M = +14.742 kip ⋅ ft
σ top =
− M r y − (14.742 ×12 )( +3.5 )
=
= −6.38 ksi
I
( 97.00 )
σ bottom =
− M r y − (14.742 × 12 )( −5.5 )
=
= +10.03 ksi
I
( 97.00 )
At M = −3.980 kip ⋅ ft
σ top =
− M r y − ( −3.980 × 12 )( +3.5 )
=
= +1.723 ksi
I
( 97.00 )
σ bottom =
− M r y − ( −3.980 × 12 )( −5.5 )
=
= −2.71 ksi
I
( 97.00 )
σ T ( max ) = 10.03 ksi (T)
........................................................................................................... Ans.
σ C ( max ) = 6.38 ksi (C) ............................................................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-222
RILEY, STURGES AND MORRIS
(10 )( 5) − 4 B + (15 × 2 )( 2 ) + (10 )(1) = 0
4 ΣM E = 0 :
B = 30.0 kN
4 E + (10 )(1) − (15 × 2 )( 2 ) − (10 )( 3) = 0
4 ΣM B = 0 :
E = 20.0 kN
For a WT 305 × 70 section :
d = 2c = 308.7 mm
yC = 75.9 mm
tw = 13.1 mm
w f = 230.3 mm
t f = 22.2 mm
I = 77.4 (106 ) mm 4
(a)
At the bottom of the beam 2.33 m from the left support:
σ bottom
3
− Mc − ( 23.33 × 10 ) ( −0.2328 )
=
=
I
( 77.4 ×10−6 )
σ max T = +70.2 (106 ) N/m 2 = 70.2 MPa (T)
(b)
................................................................... Ans.
At the bottom of the beam at the left support:
σ bottom =
3
− Mc − ( −10.00 × 10 ) ( −0.2328 )
=
I
( 77.4 ×10−6 )
σ max C = −30.1(106 ) N/m 2 = 30.1 MPa (C) .................................................................... Ans.
(c)
QNA = yC A = 116.4 ( 232.8 ×13.1) = 355.0 (103 ) mm3
(d)
( 20 ×10 )( 355 ×10 ) = 7.00 10 N/m = 7.00 MPa .................................. Ans.
τ =
( )
( 77.4 ×10 ) ( 0.0131)
Q = y A = 64.8 ( 230.3 × 22.2 ) = 331.3 (10 ) mm
( 20 ×10 )( 331.3 ×10 ) = 6.53 10 N/m = 6.53 MPa .................................. Ans.
τ =
( )
( 77.4 ×10 ) ( 0.0131)
−6
3
6
2
−6
max
3
J
−6
3
6
J
3
C
2
−6
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
7-223*
RILEY, STURGES AND MORRIS
M ry = −6.00 kip ⋅ ft = −72.00 kip ⋅ in.
tan β =
M ry I z + M rz I yz
M rz I y + M ry I yz
=
M rz = 0
I z 89.0
=
= 1.69524
I yz 52.5
β = +59.46° = 59.46° 3
At point A ( y = −2.37 in. and z = +2.37 in. ):
M ry ( − I yz y + I z z )
 M rz I y + M ry I yz 
 M ry I z + M rz I yz 
y
+
z
=



2
2
I y I z − I yz2
 I y I z − I yz 
 I y I z − I yz 
σA = −
σA =
( −72.00 )  − ( 52.5)( −2.37 ) + (89.0 )( +2.37 )
2
2
(89.0 ) − ( 52.5)
σ A = −4.68 ksi = 4.68 ksi (C)
.................................................................................... Ans.
At point B ( y = −1.37 in. and z = −5.63 in. ):
σB =
( −72.00 )  − ( 52.5)( −1.37 ) + (89.0 )( −5.63)
2
2
( 89.0 ) − ( 52.5)
σ B = +5.98 ksi = 5.98 ksi (T) ..................................................................................... Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-224
I=
( 30 )( 75)
3
= 1.05469 (106 ) mm 4
12
h 75
=
= 3.75
d 20
From Fig. 7-34:
K t ≅ 2.25
σ = Kt
d 20
=
= 0.4
b 50
(1400 )( 0.0375) = 112.0 106 N/m2 = 112.0 MPa ............... Ans.
Mc
= ( 2.25 )
( )
I
1.05469 (10−6 )
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
7-225*
For a W 14 × 120 section
w f = 14.670 in.
RILEY, STURGES AND MORRIS
d = 2c = 14.48 in.
tw = 0.590 in.
t f = 0.940 in.
S = 190 in.3
M e = σ y S = ( 36 )(190 ) = 6840 kip ⋅ in. ............................................................................ Ans.
M p = 2 ( 36 ) (14.670 × 0.940 )( 6.77 )  + 2 ( 36 ) ( 6.30 × 0.590 )( 3.15 ) 
M p = 7564.7 kip ⋅ in. ≅ 7560 kip ⋅ in. ........................................................................ Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-226*
QA = ( 25 )( 50 × 3) = 3.75 (103 ) mm3
RILEY, STURGES AND MORRIS
QB = 0 mm3
QC = ( 50 )(100 × 3) = 15.00 (103 ) mm3
QD = ( 50 )(100 × 3) + ( 25 )( 50 × 3) = 18.75 (103 ) mm3
( 6 )(100 )
I≅
12
τA =
3
+ 2 (100 × 3)( 50 ) = 2.00 (106 ) mm 4
2
−6
VQA ( 2500 ) ( 3.75 × 10 )
=
= 1.5625 (106 ) N/m 2
−6
It
( 2.00 ×10 ) ( 0.003)
−6
VQC ( 2500 ) (15.00 × 10 )
τC =
=
= 6.25 (106 ) N/m 2
−6
It
( 2.00 ×10 ) ( 0.003)
−6
VQD ( 2500 ) (18.75 × 10 )
τD =
=
= 7.81(106 ) N/m 2
−6
It
( 2.00 ×10 ) ( 0.003)
τB = 0
τ max = τ D = 7.81 MPa ...................................................................................... Ans.
Therefore:
F1 = (τ C A1 2 ) = ( 6.25 × 106 ) ( 0.100 × 0.003) 2 = 937.5 N
F3 = 2 (τ A A3 ) 3 = 2 (1.5625 × 106 ) ( 0.100 × 0.003) 3 = 312.5 N
4 ΣM O = 0 :
e=
Pe − F1 (100 ) + F3 (100 ) = 0
( 937.5)(100 ) − ( 312.5)(100 ) = 25.00 mm = 25.00 mm ← ............................... Ans.
2500
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-227
n=
Ea 10, 000
=
= 33.33
Ep
300
σp =
σa =
IT
My p
IT
=
(10 )(1.5) ≤ 1 ksi
IT
IT ≥ 15.00 in.4
nMya 33.33 (10 )(1.625 )
=
≤ 20 ksi
IT
IT
( 33.33w )( 3.25 )
=
12
20.36 w ≥ 22.58
3
( 33.33w − 2 )( 3)
−
IT ≥ 27.08 in.4
3
≥ 27.08 in.4
12
w ≥ 1.109 in. .................................................................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-228
n=
Es 200
=
= 11.765
Ec 17
2
As = 4 π ( 20 ) 4  = 1256.6 mm 2


AT = nAs = 11.765 (1256.6 ) = 14, 784 mm 2
250h ( h 2 ) = 14, 784 ( 400 − h )
h 2 + 118.27 h − 47,310 = 0
From which:
IT =
σc =
h = 166.27 mm
( 250 )(166.27 )
3
3
400 − h = 233.7 mm
+ (14, 784 )( 233.7 ) = 386.5 (106 ) mm 4
2
Myc ( 62,500 )( 0.16627 )
=
= 26.89 (106 ) N/m 2 ≅ 26.9 MPa (C) ................. Ans.
6
IT
( 386.5 ×10 )
σ s = ( ys yc )( nσ c ) = ( 233.7 166.27 )(11.765 × 26.89 ) = 445 MPa (T) ................. Ans.
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purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
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MECHANICS OF MATERIALS, 6th Edition
7-229*
RILEY, STURGES AND MORRIS
A = ( 2.25 × 3) + ( 2.5 ×1) = 9.25 in.2
RC =
( 3.5)( 2 ×1) + (11 3) (1×1)
2  + ( 5.5 )(1.5 × 3) + ( 5 ) (1.5 × 3) 2 
= 4.8468 in.
9.25
7
 4
dρ
dρ 
= R  ∫ ( −1 + ρ )
+ ∫ ( 5 − 0.5 ρ )
4
ρ
ρ
ρ 
 3
= R ( −1) ln ( 4 3) + ( 4 − 3) + ( 5 ) ln ( 7 4 ) 4 − ( 0.5 )( 7 − 4 ) 
A = R∫
dA
= 2.01040 R = 9.25 in.2
9.25
R=
= 4.6011 in.
2.01040
yC = R − RC = 4.6011 − 4.8468 = −0.2457 in.
M = RC P = ( −4.8468 P ) kip ⋅ in.
σi =
( −4.8468P )( 4.6011 − 3) ≤ 12 ksi
P M ( R − ri )
P
+
=
+
A
ri AyC
9.25
( 3)( 9.25)( −0.2457 )
P ≤ 9.63 kip
σo =
( −4.8468P )( 4.6011 − 7 ) ≤ −16 ksi
P M ( R − ro )
P
+
=
+
A
ro AyC
9.25
( 7 )( 9.25)( −0.2457 )
P ≤ 25.7 kip
Pmax = 9.63 kip ....................................................................................................................... Ans.
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MECHANICS OF MATERIALS, 6th Edition
7-230
RILEY, STURGES AND MORRIS
P = ( 320 )( 3 5 ) = 192 kN
V = ( 320 )( 4 5 ) = 256 kN
M = ( 256 )( 0.250 + 0.320 ) + (192 )( 0.240 )
= 192 kN ⋅ m
A = ( 75 × 200 ) = 15, 000 mm 2
bh3 ( 75 )( 200 )
I=
=
= 50.0 (106 ) mm 4
12
12
3
Q = yC A = ( 75 )( 50 × 75 ) = 281.3 (103 ) mm3
At point A:
σ yA
σ xA = 0 MPa
3
3
P Mc 192 (10 ) (192 ×10 ) ( 0.050 )
= +
=
+
50.0 ×10−6
A I
15 (10−3 )
= 204.8 (106 ) N/m 2 = 204.8 MPa
VQ ( 256 × 10 )( 281.3 × 10
=
=
It
( 50.0 ×10−6 ) ( 0.075)
−6
3
τ xyA
σ p1, p 2 =
σx +σ y
2
) = 19.203 10 N/m
( )
6
2
= 19.203 MPa
 σ −σ y 
0 + 204.8
2
 0 − 204.8 
2
±  x
± 
 + τ xy =
 + (19.203)
2
2


 2 
2
2
σ p1 = 102.4 + 104.19 = 206.6 MPa ≅ 207 MPa (T) ..................................................... Ans.
σ p 2 = 102.4 − 104.19 = −1.790 MPa ≅ 1.790 MPa (C) ............................................... Ans.
σ p 3 = 0 MPa ........................................................................................................................... Ans.
τ max = τ p =
σ p1 − σ p 2
2
=
( 206.6 ) − ( −1.790 ) = 104.2 MPa ......................................... Ans.
2
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-231
From symmetry, each of the supports carries half of the total load
RB = RC = 2900 lb
At point A:
M A = 5800 lb ⋅ ft = 69.6 kip ⋅ in.
T = ( 2500 )(12 ) − ( 400 )(12 )
= 25, 200 lb ⋅ in. = 25.20 kip ⋅ in.
I = π d 4 64 = π ( 4 ) 64 = 12.566 in.4
4
J = 2 I = 25.133 in.4
− Mc − ( 69.6 )( 2 )
=
= −11.078 ksi
I
12.566
σ y = 0 ksi
σx =
τ xy =
Tc ( 25.2 )( 2 )
=
= 2.005 ksi
J
25.133
σ p1, p 2 =
σx +σ y
2
 σ −σ y 
−11.078 + 0
2
 −11.078 − 0 
2
±  x
± 
 + τ xy =
 + ( 2.005 )
2
2


 2 
2
2
σ p1 = −5.539 + 5.891 = 0.352 ksi = 0.352 ksi (T) ........................................................ Ans.
σ p 2 = −5.539 − 5.891 = −11.430 ksi = 11.43 ksi (C) ................................................... Ans.
σ p 3 = 0 ksi
τ max = τ p =
.............................................................................................................................. Ans.
σ p1 − σ p 2
2
=
( 0.352 ) − ( −11.430 ) = 5.89 ksi ............................................ Ans.
2
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MECHANICS OF MATERIALS, 6th Edition
RILEY, STURGES AND MORRIS
7-232*
From the bending-moment diagram:
M max = 7.5 kN ⋅ m
S=
M
σ
=
7500
= 125.0 (10−6 ) m3
6
60 (10 )
= 125.0 (103 ) mm3
For each angle:
S=
125 (103 )
= 62.5 (103 ) mm3
2
Use two L 178 ×102 × 9.5-mm angles .............. Ans.
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