Question 3.9: The 1.5-in-diameter solid steel shaft shown in Fig. 3–24a is...

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 bending 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,

M =\sqrt {M^{2}_{y} + M^{2}_{z}}        (1)

At point B,

M_{B} =\sqrt {2000^{2} + 8000^{2}} = 8246  lbf · in

At point C,

M_{C} =\sqrt {4000^{2} + 4000^{2}} = 5657  lbf · in

Thus the maximum bending moment is 8246 lbf · in and the maximum bending stress at pulley B is

σ =\frac {M d/2}{πd^{4}/64} =\frac {32M}{πd^{3}} =\frac {32(8246)}{π(1.5^{3})} = 24 890 psi

The maximum torsional shear stress occurs between B and C and is

τ =\frac {T d/2}{πd^{4}/32} =\frac {16T}{πd^{3}} =\frac {16(1600)}{π(1.5^{3})} = 2414 psi

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

σ_{1} =\frac {σ}{2} +\sqrt {(\frac {σ}{2})^{2}+ τ^{2}} =\frac {24 890}{2} +\sqrt{(\frac {24 890}{2})^{2}+ 2414^{2}} = 25 120  psi

At point F, the maximum compressive stress will be σ_{2} given by

σ_{2} =\frac {-σ}{2} -\sqrt {(\frac {-σ}{2})^{2}+ τ^{2}} =\frac {-24 890}{2} -\sqrt{(\frac {-24 890}{2})^{2}+ 2414^{2}} = -25 120  psi

The extreme shear stress also occurs at E and F and is

τ_{1} =\sqrt {(\frac {±σ}{2})^{2}+ τ^{2}} =\sqrt{(\frac {±24 890}{2})^{2}+ 2414^{2}} = 12680  psi