An axial load P is applied to the 1.75 in. by 0.75 in. rectangular bar shown in Figure P1.37. Determine the normal stress perpendicular to plane AB and the shear stress parallel to plane AB if the bar is subjected to an axial load of P = 18 kips.

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The angle θ for the inclined plane is 60°. The normal force N perpendicular to plane AB is found from

[latex]N=P \cos heta=(18 ext { kips }) \cos 60^{\circ}=9.0 ext { kips }[/latex]

and the shear force V parallel to plane AB is

[latex]V=P \sin heta=(18 ext { kips }) \sin 60^{\circ}=15.5885 ext { kips }[/latex]

The cross-sectional area of the bar is (1.75 in.)(0.75 in.) = 1.3125 [latex] ext { in. }^{2}[/latex], but the area along inclined plane AB is

[latex]A_{n}=A / \cos heta=\frac{1.3125 in .{ }^{2}}{\cos 60^{\circ}}=2.6250 in .^{2}[/latex]

The normal stress [latex]\sigma_{n}[/latex] perpendicular to plane AB is

[latex]\sigma_{n}=\frac{N}{A_{n}}=\frac{9.0 kips }{2.6250 in .^{2}}=3.4286 ksi =3.43 ksi[/latex]

The shear stress [latex] au_{n t}[/latex] parallel to plane AB is

[latex] au_{n t}=\frac{V}{A_{n}}=\frac{15.5885 kips }{2.6250 in .{ }^{2}}=5.9385 ksi =5.94 ksi[/latex]

Question 1.37: An axial load P is applied to the 1.75 in. by 0.75 in. recta...

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