Question 4.CA.8: A 1600-lb⋅in. couple is applied to a wooden beam, of rectang......
A 1600-lb⋅in. couple is applied to a wooden beam, of rectangular cross section 1.5 by 3.5 in., in a plane forming an angle of 30° with the vertical (Fig. 4.55a). Determine (a) the maximum stress in the beam and (b) the angle that the neutral surface forms with the horizontal plane.
a. Maximum Stress. The components M _z \text { and } M _y of the couple vector are first determined (Fig. 4.55b):
\begin{aligned} & M_z=(1600 \text{lb} \cdot \text { in. }) \cos 30^{\circ}=1386 \text{lb} \cdot \text { in. } \\ & M_y=(1600 \text{lb} \cdot \text { in. }) \sin 30^{\circ}=800 \text{lb} \cdot in . \end{aligned}
Compute the moments of inertia of the cross section with respect to the z and y axes:
\begin{aligned} & I_z=\frac{1}{12}(1.5 \text { in. })(3.5 \text { in. })^3=5.359 \text { in }^4 \\ & I_y=\frac{1}{12}(3.5 \text { in. })(1.5 \text { in. })^3=0.9844 \text { in }^4 \end{aligned}
The largest tensile stress due to M _z occurs along AB and is
\sigma_1=\frac{M_z y}{I_z}=\frac{(1386 lb \cdot in .)(1.75 in .)}{5.359 in ^4}=452.6 psi
The largest tensile stress due to M _y occurs along AD and is
\sigma_2=\frac{M_y z}{I_y}=\frac{(800 lb \cdot in .)(0.75 in .)}{0.9844 in ^4}=609.5 psi
The largest tensile stress due to the combined loading, therefore, occurs at A and is
\sigma_{\max }=\sigma_1+\sigma_2=452.6+609.5=1062 psi
The largest compressive stress has the same magnitude and occurs at E.
b. Angle of Neutral Surface with Horizontal Plane. The angle \phi that the neutral surface forms with the horizontal plane (Fig. 4.55c) is obtained from Eq. (4.57):
\tan \phi=\frac{I_z}{I_y} \tan \theta (4.57)
\begin{aligned} =\frac{5.359 in ^4}{0.9844 in ^4} \tan 30^{\circ}=3.143 \\ \phi =72.4^{\circ} \end{aligned}
The distribution of the stresses across the section is shown in Fig. 4.55d.
