Specifications for the 50 mm × 50 mm square bar shown in Figure P1.38/39 require that the normal and shear stresses on plane AB not exceed 120 MPa and 90 MPa, respectively. Determine the maximum load P that can be applied without exceeding the specifications.

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 cross-sectional area of the square bar is [latex]A=(50 mm )^{2}=2,500 mm ^{2}[/latex] , and the angle θ for plane AB is 55°.

The normal stress on plane AB is limited to 120 MPa; therefore, the maximum load P that can be supported by the square bar is found from Eq. (a):

[latex]P \leq \frac{2 A \sigma_{n}}{1+\cos 2 heta}=\frac{2\left(2,500 mm ^{2} ight)\left(120 N / mm ^{2} ight)}{1+\cos 2\left(55^{\circ} ight)}=911,882 N[/latex]

The shear stress on plane AB is limited to 90 MPa. From Eq. (b), the maximum load P based the shear stress limit is

[latex]P \leq \frac{2 A au_{n t}}{\sin 2 heta}=\frac{2\left(2,500 mm ^{2} ight)\left(90 N / mm ^{2} ight)}{\sin 2\left(55^{\circ} ight)}=478,880 N[/latex]

Thus, the maximum load that can be supported by the bar is

[latex]P_{\max }=479 kN[/latex]

Question 1.39: Specifications for the 50 mm × 50 mm square bar shown in Fig...

Related Answered Questions