(a) The two free-body diagrams are shown in Fig. 3–23. The results are
At end C of arm BC: F = −300j lbf, T_{C} = −450k lbf · in
At end B of arm BC: F = 300j lbf, M_{1} = 1200i lbf · in, T_{1} = 450k lbf · in
At end B of shaft AB: F = −300j lbf, T_{2} = −1200i lbf · in, M_{2} = −450k lbf · in
At end A of shaft AB: F = 300j lbf, M_{A} = 1950k lbf · in, T_{A} = 1200i lbf · in
(b) For arm BC, the bending moment will reach a maximum near the shaft at B. If we assume this is 1200 lbf · in, then the bending stress for a rectangular section will be
σ =\frac {M}{I/c} =\frac {6M}{bh^{2}} =\frac {6(1200)}{0.25(1.25)^{2}} = 18 400 psi
Of course, this is not exactly correct, because at B the moment is actually being transferred into the shaft, probably through a weldment.
For the torsional stress, use Eq. (3–43). Thus
τ_{max} =\frac {T}{αbc^{2}}\dot{=}\frac {T}{bc^{2}}(3 +\frac {1.8}{b/c}) (3–43)
τ_{max }=\frac {T}{bc^{2}}(3 +\frac {1.8}{b/c})=\frac {450}{1.25(0.25^{2})} (3 +\frac {1.8}{1.25/0.25})= 19 400 psi
This stress occurs at the middle of the 1\frac { 1}{ 4}-in side.
(c) For a stress element at A, the bending stress is tensile and is
σ_{x} =\frac {M}{I/c}=\frac {32M}{πd^{3}} =\frac{32(1950)}{π(0.75)^{3}} = 47 100 psi
The torsional stress is
τ_{xz} = \frac {−T}{J/c} = \frac {−16T}{πd^{3}} = \frac {−16(1200)}{π(0.75)^{3}} = −14 500 psi
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 ubscripts corresponding to the x, z axes.
The maximum normal stress is then given by
σ_{1}, σ_{2}=\frac {σ_{x} + σ_{y}}{2}±\sqrt{(\frac {σ_{x }− σ_{y}}{2})^{2}+ τ^{2}_{xy}} (3-13)
σ_{1} =\frac {σ_{x} + σ_{z}}{2} +\sqrt{(\frac {σ_{x }− σ_{z}}{2})^{2}+ τ^{2}_{xz}}
=\frac {47.1 + 0}{2} +\sqrt{(\frac {47.1− 0}{2})^{2}+ (-14.5)^{2}}=51.2 kpsi
The maximum shear stress at A occurs on surfaces different than the surfaces containing 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 subscripts, and is given by
τ_{1}, τ_{2}=±\sqrt{(\frac {σ_{x }− σ_{y}}{2})^{2}+ τ^{2}_{xy}} (3.14)
τ_{1} =\sqrt{(\frac {σ_{x }− σ_{z}}{2})^{2}+ τ^{2}_{xz}}=\sqrt{(\frac {47.1− 0}{2})^{2}+ (-14.5)^{2}}=27.7 kpsi
