A cantilever of length L has a solid circular cross section of diameter D and carries a vertically downward loadWat its free end. The modulus of rigidity of the cantilever is G. Calculate the shear stress distribution across a section of the cantilever and hence determine the deflection due to shear at its free end.
Ans. $\tau=16 W\left(1-4 y^{2} / D^{2}\right) / 3 \pi D ^{2}, 40 W L / 9 \pi G D^{2}$ (downwards).

Question

A cantilever of length L has a solid circular cross section of diameter D and carries a vertically downward loadWat its free end. The modulus of rigidity of the cantilever is G. Calculate the shear stress distribution across a section of the cantilever and hence determine the deflection due to shear at its free end.
Ans. [latex] au=16 W\left(1-4 y^{2} / D^{2} ight) / 3 \pi D ^{2}, 40 W L / 9 \pi G D^{2}[/latex] (downwards).

Accepted Answer

The shear stress distribution has been calculated in Ex. 10.5 and is given by Eq. (10.19),

[latex] au=-\frac{16 S_{y}}{3 \pi D^{2}}\left(1-\frac{4 y^{2}}{D^{2}} ight)[/latex] (10.19)

i.e.

[latex] au=\left(\frac{16 W}{3 \pi D^{2}} ight)\left(1-\frac{4 y^{2}}{D^{2}} ight)[/latex]

Now substituting in Eq. (13.20)

[latex]\beta=\frac{A}{S_{y}^{2}} \int_{y_{1}}^{y_{2}} au^{2} b_{0} d y[/latex] (13.20)

[latex]\beta=\left(\frac{\pi D^{2}}{4 W^{2}} ight)\left(\frac{16 W}{3 \pi D^{2}} ight)^{2} \int_{-D / 2}^{D / 2}\left(1-\frac{4 y^{2}}{D^{2}} ight)^{2} b_{0} d y[/latex] (i)

Changing to angular coordinates, (see Fig. 10.10)

[latex]y=\left(\frac{D}{2} ight) \sin heta, \quad d y=\left(\frac{D}{2} ight) \cos heta d heta, \quad b_{0}=D \cos heta[/latex]

Substituting in Eq. (i) and simplifying

[latex]\beta=\frac{32}{9 \pi} \int_{-\pi / 2}^{\pi / 2} \cos ^{6} heta d heta[/latex]

The value of the integral may be found using a reduction formula. Then

[latex]\beta=\frac{10}{9}[/latex]

The shear force at any section of the cantilever is W. Substituting in Eq. (13.22)

[latex]v_{s}=\frac{\beta}{G} \int_{L}\left(\frac{S_{y}}{A} ight) d x[/latex] (13.22)

[latex]v_{s}=\left(\frac{10}{9 G} ight) \int_{0}^{L} \frac{W}{\left(\frac{\pi D^{2}}{4} ight)} d x[/latex]

i.e.

[latex]v_{ s }=\frac{40 W L}{9 \pi G D^{2}}[/latex].

Question 13.P.22: A cantilever of length L has a solid circular cross section ...

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