Page 130 - Shigley's Mechanical Engineering Design

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Load and Stress Analysis 105
The torsional stress is

−T −16T −16(1200)
Answer τ xz = = = =−14 500 psi =−14.5 kpsi
J/c πd 3 π(0.75) 3
where the reader should verify that the negative sign accounts for the direction of τ xz .
(d) Point A is in a state of plane stress where the stresses are in the xz plane. Thus
the principal stresses are given by Eq. (3–13) with subscripts corresponding to the
x, z axes.
Answer The maximum normal stress is then given by

2

σ x + σ z σ x − σ z 2
σ 1 = + + τ xz
2 2

47.1 + 0 47.1 − 0 2
2
= + + (−14.5) = 51.2 kpsi
2 2
Answer The maximum shear stress at A occurs on surfaces different than the surfaces contain-
ing the principal stresses or the surfaces containing the bending and torsional shear
stresses. The maximum shear stress is given by Eq. (3–14), again with modified sub-
scripts, and is given by

2 2

σ x − σ z 47.1 − 0
2
2
τ 1 = + τ xz = + (−14.5) = 27.7 kpsi
2 2

EXAMPLE 3–9 The 1.5-in-diameter solid steel shaft shown in Fig. 3–24a is simply supported at the ends.
Two pulleys are keyed to the shaft where pulley B is of diameter 4.0 in and pulley C is of
diameter 8.0 in. Considering bending and torsional stresses only, determine the locations
and magnitudes of the greatest tensile, compressive, and shear stresses in the shaft.

Solution Figure 3–24b shows the net forces, reactions, and torsional moments on the shaft.
Although this is a three-dimensional problem and vectors might seem appropriate, we
will look at the components of the moment vector by performing a two-plane analysis.
Figure 3–24c shows the loading in the xy plane, as viewed down the z axis, where bend-
ing moments are actually vectors in the z direction. Thus we label the moment diagram
as M z versus x. For the xz plane, we look down the y axis, and the moment diagram is
M y versus x as shown in Fig. 3–24d.
The net moment on a section is the vector sum of the components. That is,

2
M = M + M 2 (1)
y z
At point B,

2
2
M B = 2000 + 8000 = 8246 lbf · in
At point C,

2
2
M C = 4000 + 4000 = 5657 lbf · in