A rectangular bar having width w = 6.00 in. and thickness t = 1.50 in. is subjected to a tension load P as shown in Figure P1.42/43. The normal and shear stresses on plane AB must not exceed 16 ksi and 8 ksi, respectively. Determine the maximum load P that can be applied without exceeding either stress limit.
The general equations for normal and shear stresses on an inclined plane in terms of the angle θ are
\sigma_{n}=\frac{P}{2 A}(1+\cos 2 \theta) (a)
and
\tau_{n t}=\frac{P}{2 A} \sin 2 \theta (b)
The angle θ for inclined plane AB is calculated from
\tan \theta=\frac{3}{1}=3 \quad \therefore \theta=71.5651^{\circ}The cross-sectional area of the bar is A=w \times t=(6.00 in .)(1.50 in .)=9.0 in .^{2} .
The normal stress on plane AB is limited to 16 ksi; therefore, the maximum load P can be found from Eq. (a):
P \leq \frac{2 A \sigma_{n}}{1+\cos 2 \theta}=\frac{2\left(9.0 in .{ }^{2}\right)(16 ksi )}{1+\cos 2\left(71.5651^{\circ}\right)}=1,440 ksiThe shear stress on plane AB is limited to 8 ksi. From Eq. (b), the maximum load P based the shear stress limit is
P \leq \frac{2 A \tau_{n t}}{\sin 2 \theta}=\frac{2\left(9.0 in .^{2}\right)(8 ksi )}{\sin 2\left(71.5651^{\circ}\right)}=240 kipsThus, the maximum load that can be supported by the bar is
P_{\max }=240 kips