A 90 mm wide bar will be used to carry an axial tension load of 280 kN as shown in Figure P1.41. The normal and shear stresses on plane AB must be limited to 150 MPa and 100 MPa, respectively. Determine the minimum 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 plane AB is 50°.

The normal stress on plane AB is limited to 150 MPa; therefore, the minimum cross-sectional area A required to support P = 280 kN can be found from Eq. (a):

[latex]A \geq \frac{P}{2 \sigma_{n}}(1+\cos 2 heta)=\frac{(280 kN )(1,000 N / kN )}{2\left(150 N / mm ^{2} ight)}\left(1+\cos 2\left(50^{\circ} ight) ight)=771.2617 mm ^{2}[/latex]

The shear stress on plane AB is limited to 100 MPa; therefore, the minimum cross-sectional area A required to support P = 280 kN can be found from Eq. (b):

[latex]A \geq \frac{P}{2 au_{n t}} \sin 2 heta=\frac{(280 kN )(1,000 N / kN )}{2\left(100 N / mm ^{2} ight)} \sin 2\left(50^{\circ} ight)=1,378.7309 mm ^{2}[/latex]

To satisfy both the normal and shear stress requirements, the cross-sectional area must be at least [latex]A_{\min }=1,379.7309 mm ^{2}[/latex] . Since the bar width is 90 mm, the minimum bar thickness t must be

[latex]t_{\min }=\frac{1,378.7309 mm ^{2}}{90 mm }=15.3192 mm =15.32 mm[/latex]

Question 1.41: A 90 mm wide bar will be used to carry an axial tension load...

Related Answered Questions