Ftan mpa mpa mpa mpatmax mpa save mpaproblem solve prob
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¥ | 10 | -3 | )( 80 | ¥ | 10 | -3 | ) | |
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w | cos | b | |||||||
= 800¥106 |
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tw = F A w s | 6 |
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¥ | 10 | 6 | = 0.240 | ||||||||||
¥ | 10 | 6 | ![]() |
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Fn – 100 cos b = 0 |
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A = 800 | ¥ | 10 | -6 |
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s = F n | ¥ | ||||||||||||
A w | ¥ | ![]() |
PROBLEM 7.23
7.23 The steel pipe AB has a 102 mm outer diameter and a 6 mm wall thickness. Knowing that arm CD is rigidly attached to the pipe, determine the principal stresses and the maximum shearing stress at point H.
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Fx = 10 ¥ 103 N | H | ||||||||||||||
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T = My = 2000 N◊m | z | H | K | ||||||||||||
c = ro = 51 ¥ 10–3 m | |||||||||||||||
txy = Tc J = ( | 2000 51 | ¥ | 10 | -3 | ) | ![]() |
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¥ | 10 | H | |||||||||||||
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= 27.684 ¥ 10–6 m3 |
V = Fx = 10 ¥ 103 N
t = (2)(6 mm) = 12 mm = 12 ¥ 10–3 m
save = 1 2(sx + sy) = 0 | R = | ��� | s | x | - | s | y | ��� | 2 | + | t | ![]() |
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T = My = 2000 N◊m | ||||||||||||
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txy = Tc J = ( | 2000 51 | ¥ | 10 | - | 3 | ) | ||||||
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¥ | 10 | 6 | |||||||||
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H | ||||
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D | ||||
50 mm | 1.8 kN |
Equivalent force-couple system at center of shaft in section at point H.
V = 1.8 kN M = (1.8 ¥ 103)(0.150) = 270 N◊m
Chapter 7 | 829 | |
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PROBLEM 7.26
7.26 A mechanical uses a crowfoot wrench to loosen at bolt at E. Knowing that the mechanic applies a vertical 100 N force at A, determine the principal stresses and the maximum shearing stress at point Hlocated as shown on top of the 18 mm diameter shaft.
T = (100)(250) = 25000 N◊mm
sx = 82 MPa, |
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U = s | x | - | s |
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R = U | 2 | + t | 2 | |||||||||||
xy | ||||||||||||||
2 | = ± 100 | 2 | - | 28 | 2 | |||||||||
U = ± R | - t | |||||||||||||
xy |
sy = sx – 2U = 82 ± (2)(96) = – 110 MPa, 274 MPa
U = ± R | 2 | - t | 2 | = ± 50 | 2 | - | 40 | 2 | Chapter 7 | 831 | |
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xy |
sx = sy + 2U = 75 ± (2)(30) = 135 MPa, 15 MPa
60 MPa |
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7.5 through 7.8 | ||
SOLUTION
832 | B | b | G | E | C | A |
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Y: | O | s | ||||||||
C: |
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(MPa) |
qA = 1 2a = 52.97º
R = CG 2 + GX 2 = 10 2 + 35 2 = 36.4 MPa
tmax = R = 36.4 MPa
s¢ = save = – 50 MPa
48 MPa | 16 MPa | ||
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7.5 through 7.8 | |||
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Points: |
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sx = 120 MPa |
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t | ![]() |
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s= s | x | + | s | y | ||||||
ave | 2 | D |
834 | 70 MPa | ![]() |
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R = CF | 2 | + | FX | 2 | = 105 | 2 | + | 60 | 2 | ![]() |
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10 MPa | |||||||||||||
7.34 | |||||||||||||
stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
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R = FC | 2 | + | FX | 2 | = 30 | 2 | + | 20 | 2 | |||
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50 MPa
SOLUTION
sx = 0 | x | + | ||||
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s= s | s | y | ||||
ave | 2 |
been rotated through (a) 25º clockwise, (b) 10º
counterclockwise.
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y | sy = – 80 MPa | 40 MPa | ||||
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sx = 55 MPa | |||||||
s= s | x | + | s | ||||
ave | 2 |
j = 50º – 30.65º = 19.35º
Y | t | ![]() |
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sy¢ = save – R cos j = – 86.6 MPa | C | X¢ |
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20° |
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j = 30.65 + 20º = 50.65º |
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sx¢ = save + R cos j = 37.2 MPa | 110 MPa | ||||||||
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sy¢ = save – R cos j = – 62.2 MPa | |||||||||
Solve Prob. 7.16, using Mohr’s circle. |
wise.
SOLUTION
838 | X� | t | Y |
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R = FC | 2 | + | FX | 2 | = 55 | 2 | + | 70 | 51.84° | ||||||||||
50° | |||||||||||||||||||
= 89 MPa | |||||||||||||||||||
O | C | j | |||||||||||||||||
(a) q = 25º | |||||||||||||||||||
MPa | |||||||||||||||||||
j = 51.84º – 50º = 1.84º | X | ![]() |
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j = 51.84º + 20º = 71.84º
sx¢ =save – R cos j = 27.2 MPa tx¢y¢ = R sin j = 84.5 MPa
sy¢ = save + R cos j = 82.7 MPa.
7.17 and 7.18 The grain of a wooden member forms an angle of
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15° | 3 MPa |
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s (MPa) |
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Y: | X | C |
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C: |
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q = – 15º | |||||||
CX = 0.6 MPa | |||||||