A cantilever 50 mm diameter is to carry an axial load 50 kN at an eccentricity of 5 mm below horizontal diameter in the vertical plane of symmetry along with a torque of 1.5 x 10^6 N mm as shown in Fig. 2.24. Calculate the stresses on a plane, normal of which is inclined at 30° clockwise to axis

Question

A cantilever 50 mm diameter is to carry an axial load 50 kN at an eccentricity of 5 mm below horizontal diameter in the vertical plane of symmetry along with a torque of 1.5 x [latex]10^{6}[/latex] N mm as shown in Fig. 2.24. Calculate the stresses on a plane, normal of which is inclined at 30° clockwise to axis of the cantilever at a point on the lower edge of the vertical plane of symmetry.

Accepted Answer

Consider an element at a lowermost point P.

Direct axial stress

[latex]\sigma_x =\frac{-50000}{\pi imes 50^2 / 4} =-25.46 MPa ( ext{Compression})[/latex]

Bending compressive stress at P due to eccentric load,

[latex]\sigma_x =\frac{-M imes y}{I}=\frac{-(50000 imes 5) imes 25}{\pi imes 50^4 / 64} =-20.37 MPa ( ext{Compression}) [/latex]

Total axial load,

[latex]\sigma_x =-25.46-20.37 =-45.83 MPa ( ext{Compression}) [/latex]

Shear stress due to torsion,

[latex] au _{x y}=\frac{T imes y}{J}=\frac{1.5 imes 10^6 imes 25}{\pi imes 50^4 / 30}[/latex] = 61.11 MPa (Clockwise)

(Complementary shear [latex] au _{y x}[/latex], is not shown in Fig. 2.24 for simplicity)

Plane AB shown in Fig. 2.25 has its normal inclined at 30° clockwise to the axial direction [latex]\sigma _x [/latex].

Stresses on plane AB,

[latex]\sigma_{x^{\prime}}= \frac{-45.83}{2}+\frac{-45.83}{2} \cos -60^{\circ}-61.11\sin -60^{\circ}=18.55[/latex] MPa (Tension)

[latex] au_{x^{\prime}y}=\frac{-45.83}{2} \sin -60^{\circ}+61.11 \cos -60^{\circ} [/latex] = 50.4 MPa (Clockwise)

Question 2.15: A cantilever 50 mm diameter is to carry an axial load 50 kN ...

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