An axial load P is applied to the rectangular bar shown in Figure P1.36. The cross-sectional area of the bar is 400 [latex]mm ^{2}[/latex]. 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 = 70 kN.

The angle θ for the inclined plane is 35°. The normal force N perpendicular to plane AB is found from [latex]N=P \cos \theta=(40 kN ) \cos 35^{\circ}=57.3406 kN[/latex] and the shear force V parallel to plane AB is [latex]V=P \sin \theta=(70 kN ) \sin 35^{\circ}=40.1504 kN[/latex] The cross-sectional area of the bar is 400 [latex]mm ^{2}[/latex] , but the area along inclined plane AB is [latex]A_{n}=\frac{A}{\cos \theta}=\frac{400 mm ^{2}}{\cos 35^{\circ}}=488.3098 mm ^{2}[/latex] The normal stress [latex]\sigma_{n}[/latex] perpendicular to plane AB is [latex]\sigma_{n}=\frac{N}{A_{n}}=\frac{(57.3406 kN )(1,000 N / kN )}{488.3098 mm ^{2}}=117.4268 MPa =117.4 MPa[/latex] The shear stress [latex]\tau_{n t}[/latex] parallel to plane AB is [latex]\tau_{n t}=\frac{V}{A_{n}}=\frac{(40.1504 kN )(1,000 N / kN )}{488.3098 mm ^{2}}=82.2231 MPa =82.2 MPa[/latex]

Question 1.36: An axial load P is applied to the rectangular bar shown in F...

Mechanics of Materials: An Integrated Learning System – Solution Manual [EXP-8770]

An axial load P is applied to the rectangular bar shown in Figure P1.36. The cross-sectional area of the bar is 400 $mm ^{2}$ . 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 = 70 kN.

Question Data is a breakdown of the data given in the question above.

Axial load (P) = 70 kN
Cross-sectional area of the bar = 400 mm^2

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Step 1:

Determine the angle θ for the inclined plane, which is given as 35°.

Step 2:

Calculate the normal force N perpendicular to the plane AB. This can be done using the formula N = P cos θ, where P is the given force (40 kN) and θ is the angle (35°). Substituting the values, we get N = (40 kN) cos 35° = 57.3406 kN.

Step 3:

Calculate the shear force V parallel to the plane AB. This can be done using the formula V = P sin θ, where P is the given force (70 kN) and θ is the angle (35°). Substituting the values, we get V = (70 kN) sin 35° = 40.1504 kN.

Step 4:

Determine the cross-sectional area of the bar, which is given as 400 mm^2.

Step 5:

Calculate the area along the inclined plane AB, denoted as An. This can be done using the formula An = A / cos θ, where A is the cross-sectional area (400 mm^2) and θ is the angle (35°). Substituting the values, we get An = (400 mm^2) / cos 35° = 488.3098 mm^2.

Step 6:

Calculate the normal stress σn perpendicular to the plane AB. This can be done using the formula σn = N / An, where N is the normal force (57.3406 kN) and An is the area along the inclined plane. Substituting the values, we get σn = (57.3406 ×1000) / (488.3098 mm^2) = 117.4268 MPa = 117.4 MPa (rounded to one decimal place).

Step 7:

Calculate the shear stress τnt parallel to the plane AB. This can be done using the formula τnt = V / An, where V is the shear force (40.1504 kN) and An is the area along the inclined plane. Substituting the values, we get τnt = (40.1504 × 1000) / (488.3098 mm^2) = 82.2231 MPa = 82.2 MPa (rounded to one decimal place).

Final Answer

The angle θ for the inclined plane is 35°. The normal force N perpendicular to plane AB is found from

N=P \cos \theta=(40  kN ) \cos 35^{\circ}=57.3406  kN

and the shear force V parallel to plane AB is

V=P \sin \theta=(70  kN ) \sin 35^{\circ}=40.1504  kN

The cross-sectional area of the bar is 400 $mm ^{2}$ , but the area along inclined plane AB is

A_{n}=\frac{A}{\cos \theta}=\frac{400  mm ^{2}}{\cos 35^{\circ}}=488.3098  mm ^{2}

The normal stress $\sigma_{n}$ perpendicular to plane AB is

\sigma_{n}=\frac{N}{A_{n}}=\frac{(57.3406  kN )(1,000  N / kN )}{488.3098  mm ^{2}}=117.4268  MPa =117.4  MPa

The shear stress $\tau_{n t}$ parallel to plane AB is

\tau_{n t}=\frac{V}{A_{n}}=\frac{(40.1504  kN )(1,000  N / kN )}{488.3098  mm ^{2}}=82.2231  MPa =82.2  MPa

Question 1.36: An axial load P is applied to the rectangular bar shown in F...

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