In Figure P1.42/43, a rectangular bar having width w = 1.25 in. and thickness t is subjected to a tension load of P = 30 kips. The normal and shear stresses on plane AB must not exceed 12 ksi and 8 ksi, respectively. Determine the minimum bar thickness t required for the bar.

Question

Accepted Answer

The general equations for normal and shear stresses on an inclined plane in terms of the angle θ are

[latex]\sigma_{n}=\frac{P}{2 A}(1+\cos 2 heta)[/latex] (a)

and

[latex] au_{n t}=\frac{P}{2 A} \sin 2 heta[/latex] (b)

The angle θ for inclined plane AB is calculated from

[latex] an heta=\frac{3}{1}=3 \quad herefore heta=71.5651^{\circ}[/latex]

The normal stress on plane AB is limited to 12 ksi; therefore, the minimum cross-sectional area A required to support P = 30 kips can be found from Eq. (a):

[latex]A \geq \frac{P}{2 \sigma_{n}}(1+\cos 2 heta)=\frac{30 kips }{2(12 ksi )}\left(1+\cos 2\left(71.5651^{\circ} ight) ight)=0.2500 in .^{2}[/latex]

The shear stress on plane AB is limited to 8 ksi; therefore, the minimum cross-sectional area A required to support P = 30 kips can be found from Eq. (b):

[latex]A \geq \frac{P}{2 au_{n t}} \sin 2 heta=\frac{30 kips }{2(8 ksi )} \sin 2\left(71.5651^{\circ} ight)=1.1250 in .{ }^{2}[/latex]

To satisfy both the normal and shear stress requirements, the cross-sectional area must be at least [latex]A_{\min }=1.1250 in .^{2}[/latex]. Since the bar width is 1.25 in., the minimum bar thickness t must be

[latex]t_{\min }=\frac{1.1250 in \cdot^{2}}{1.25 in .}=0.900 in .=0.900 in .[/latex]

Question 1.43: In Figure P1.42/43, a rectangular bar having width w = 1.25 ...

Related Answered Questions