As shown in Fig. 3–16a, beam OC is loaded in the xy plane by a uniform load of 50 lbf/in, and in the xz plane by a concentrated force of 100 lbf at end C. The beam is 8 in long.
(a) For the cross section shown determine the maximum tensile and compressive bending stresses and where they act.
(b) If the cross section was a solid circular rod of diameter, d = 1.25 in, determine the magnitude of the maximum bending stress.
a) The reactions at O and the bending-moment diagrams in the xy and xz planes are shown in Figs. 3–16b and c, respectively. The maximum moments in both planes occur at O where
\left(M_{z}\right)_{o}=-\frac{1}{2}(50) 8^{2}=-1600 lbf – in \quad\left(M_{y}\right)_{o}=100(8)=800 lbf -\text { in }
The second moments of area in both planes are
I_{z}=\frac{1}{12}(0.75) 1.5^{3}=0.2109 in ^{4} \quad I_{y}=\frac{1}{12}(1.5) 0.75^{3}=0.05273 in ^{4}
The maximum tensile stress occurs at point A, shown in Fig. 3–16a, where the maximum tensile stress is due to both moments. At A, y_{A}=0.75 in and z_{A}=0.375 in. Thus, from Eq. (3–27)
\sigma_{x}=-\frac{M_{z} y}{I_{z}}+\frac{M_{y} z}{I_{y}} (3–27)
\left(\sigma_{x}\right)_{A}=-\frac{-1600(0.75)}{0.2109}+\frac{800(0.375)}{0.05273}=11380 psi =11.38 kpsi
The maximum compressive bending stress occurs at point B where, y_{B}=-0.75 \text { in and } z_{B}=-0.375 \text { in. Thus }
\left(\sigma_{x}\right)_{B}=-\frac{-1600(-0.75)}{0.2109}+\frac{800(-0.375)}{0.05273}=-11380 psi =-11.38 kpsi
(b) For a solid circular cross section of diameter, d = 1.25 in, the maximum bending stress at end O is given by Eq. (3–28) as
\sigma_{m}=\frac{M c}{I}=\frac{\left(M_{y}^{2}+M_{z}^{2}\right)^{1 / 2}(d / 2)}{\pi d^{4} / 64}=\frac{32}{\pi d^{3}}\left(M_{y}^{2}+M_{z}^{2}\right)^{1 / 2} (3–28)
\sigma_{m}=\frac{32}{\pi(1.25)^{3}}\left[800^{2}+(-1600)^{2}\right]^{1 / 2}=9329 psi =9.329 kpsi