Page 132 - Shigley's Mechanical Engineering Design

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Load and Stress Analysis 107
Thus the maximum bending moment is 8246 lbf · in and the maximum bending stress
at pulley B is
Md/2 32M 32(8246)
σ = = = = 24 890 psi = 24.89 kpsi
4
3
πd /64 πd 3 π(1.5 )
The maximum torsional shear stress occurs between B and C and is
Td/2 16T 16(1600)
τ = = = = 2414 psi = 2.414 kpsi
4
3
πd /32 πd 3 π(1.5 )
The maximum bending and torsional shear stresses occur just to the right of pulley
B at points E and F as shown in Fig. 3–24e. At point E, the maximum tensile stress will
be σ 1 given by

2 2
σ σ 24.89 24.89

Answer σ 1 = + + τ = + + 2.414 = 25.12 kpsi
2
2
2 2 2 2
At point F, the maximum compressive stress will be σ 2 given by

2 2

−σ −σ −24.89 −24.89

Answer σ 2 = − + τ = − + 2.414 =−25.12 kpsi
2
2
2 2 2 2
The extreme shear stress also occurs at E and F and is

2 2
±σ ±24.89
Answer τ 1 = + τ = + 2.414 = 12.68 kpsi
2
2
2 2
Closed Thin-Walled Tubes (t r) 7
In closed thin-walled tubes, it can be shown that the product of shear stress times thickness
of the wall τt is constant, meaning that the shear stress τ is inversely proportional to the
wall thickness t. The total torque T on a tube such as depicted in Fig. 3–25 is given by

T = τtr ds = (τt) rds = τt(2A m ) = 2A m tτ

where A m is the area enclosed by the section median line. Solving for τ gives
T
τ = (3–45)
2A m t
For constant wall thickness t, the angular twist (radians) per unit of length of the tube
θ 1 is given by

TL m
θ 1 = (3–46)
2
4GA t
m
7 See Sec. 3–13, F. P. Beer, E. R. Johnston, and J. T. De Wolf, Mechanics of Materials, 5th ed., McGraw-Hill,
NewYork, 2009.