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Unformatted text preview: “W iii—l “ SOLUTLONS 9—75. The 2in.diameter drive shaft AB on the helicopter
is subjected to an axial tension of 10 000 lb and a torque of SOOlbft. Determine the principal stress and the
maximum in—plane shear stress that act at a point on the surface of the shaft.
P 10 000 .
a — A — ”(1)2 — 3.183 k51
_ Tc _ 300(12)(1) _ _
r— J  %(1)4 —2.29ka:
0', + 0', a"  y
01,2 = 2 i ( 2 )2 + 73y = 3.183 + 0 :I: 3.183 — 0 2 2
2 ( 2 ) + (2.292)
0'1 = 4.38 ksi Ans.
0'; = —1.20 ksi ‘ Ans. '
2%: = ( x 2 ,)2 + 7%,
3.183  o
= (————2 )2 + (2.292)2
= 2.79 ksi Ans. r‘.
[ﬁtr
L, T}? ~9—81. Determine the principal stress at point A on the
cross section of the hanger at section a—a. Specify the orientation of this state of stress and indicate the'result on
an element at the point, Internal Loadings: Considering the equilibrium of the free  body diagram of the 5 mm
hanger’s left out segment, Fig. a, $2F,,=0; 900—N=0 N=900N
+T2F,=0; V—900=0 V=900N
C+2MO = 0; 900(1) — 900(025) — M = o M = 675Nrn Section Properties: The cross  sectional area and the moment of inertia about the
centroidal axis of the hanger’s cross section are Sections a  a and b — b
4 = o.05(o.1_)  o.04(0.09) = 1.4(10'3)m2
_ 1 1 _ _‘
1 — E(0'05)(0'13) — E(0.04)(0.093) — 1.7367(10 6)m4 ‘ Referring to Fig. b, A = 54A; + $.45 = 2[o.0375(o.025)(o.005)] + 0.0475(0.005)(o.04)
= 18.875(10'6) m3 Normal and Shear Stress: The normal stress is a combination of axial and bending 900”
stressesThus, (a)
M I 675 0.025 ~
aA=1V—+ yA=————90—0——+——(——)=9.074MPa
A 1 1.4(103) 1.7367(10‘6) The shear stress is caused by the transverse shear stress. 6
«r. 2 L94 = W = 09782....
. It 1.7367(10—6)(0.01) The state of stress at point A is represented by the element shown in Fig. c. Construction of the Circle: 0,, = 9.074 MPa, 0”,, = 0, and 7x, = 0.9782 MPa.Thus, 0:: + 0y _ 9.074 + 0 ”avg = 2 " ——'2 = 4.537 MPa The coordinates of reference points A and the center C of the circle are A(9.074, 0.9782) C(4.537, 0) Thus, the radius of the circle is  R = CA = V(9.o74 — 4.537)2 + 0.97822 = 4.641 MPa Using these results, the circle is shown in Fig. d. 9—.81. Continued ' In  Plane Principal Stress: The coordinates of point B and D represent 01 and 02,
respectively. 01 = 4.537 + 4.641 = 9.18 MPa Ans.
0'; = 4.537  4.641 = 0.104 MPa Ans. Orientaion of Principal Plane: Referring to the geometry of the circle,F'1g.d, 0.9782 tan 2(9p)1 = m = 0.2156
(9p)1 = 608" (counterclockwise) Ans. The state of. principal stresses is represented on the element shown in Fig. e. 09—85. Draw the three Mohr’s circles that describe the 300 psi
following state of stress. Here, 0min = —300 psi, aim = 0 and on,“ = 400 psi. The three Mohr’s circle for this
state of stress is shown in Fig. a. 99—93. The propane gas tank has an inner diameter of
1500 mm and wall thickness of 15 m. If the tank is
pressurized to 2 MPa, determine the absolute maximum
shear stress in the wall of the tank. Normal Stress: Since : = g = 50 > 10, thin  wall analysis can be used. We have
__ pr _ 2(750) _
01— t — 15 ~100MPa
_ pr _ 2(750) _
2 — 2t — 2(15) —50MPa The state of stress of any point on the wall of the tank can be represented on the
element shown in Fig. a Construction of Three Mohr’s Circles: Referring to the element, am = 100 MPa aim = 50 MPa am = 0 Using these results, the three Mohr’s circles are shown in Fig. b. Absolute Maximum Shear Stress: From the geometry of three circles, ...
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 Fall '10
 Dr.Pollock

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