Question 10.21: A cantilever beam with circular section of radius r and leng...

From Eq. (10.17), we get strain energy due to shear is

U_i=\int_0^{L_i} \frac{P^2}{2( AE )_i} d x ; \quad 1 \leq i \leq 3              (10.17)

U=\underset{\sout{V}}{\iiint} \frac{\tau^2}{2 G} d \sout{V} ; \quad \text { where } \tau=\frac{V Q}{b I}                  (1)

Referring to Figure 10.39, we can find τ as follows:

Therefore,

\tau=\left\lgroup\frac{V}{I} \right\rgroup \frac{Q}{2 \sqrt{r^2-y^2}}

where        Q=\int_y^r 2 \sqrt{r^2-y^2} y d y \quad \text { and } \quad b=2 \sqrt{r^2-y^2}

that is,          Q=-\int_y^r \sqrt{r^2-y^2}(-2 y) d y=-\frac{2}{3}\left[\left(r^2-y^2\right)\right]_y^{3 r / 2}

or          Q=\frac{2}{3}\left(r^2-y^2\right)^{3 / 2}

Therefore,

\tau=\left\lgroup \frac{V}{I} \right\rgroup \frac{(2 / 3)\left(r^2-y^2\right)^{3 / 2}}{2\left(r^2-y^2\right)^{1 / 2}}=\left\lgroup \frac{V}{3 I} \right\rgroup\left(r^2-y^2\right)

From Eq. (1),

U=\int_0^L\left\lgroup \underset{A}{\iint} \frac{\tau^2}{2 G} d A \right\rgroup d x=\int_0^L\left\lgroup\frac{V^2}{9 I^2} \int_{-r}^{+r} \frac{1}{2 G}\left(r^2-y^2\right)^2(2)\left(r^2-y^2\right)^{1 / 2} d y \right\rgroup d x

=\left\lgroup \frac{V^2 L}{9 G I^2} \right\rgroup_{-r}^{+r}\left(r^2-y^2\right)^{5 / 2} d y

Therefore,

U=\frac{2}{9}\left\lgroup \frac{V^2 L}{G I^2} \right\rgroup_0^r\left(r^2-y^2\right)^{5 / 2} d y              (2)

Let          A=\int_0^r\left(r^2-y^2\right)^{5 / 2} d y

Putting y = r sinθ ⇒ dy = r cosθ

A=\int_0^{\pi / 2} r^6 \cos ^6 d \theta=r^6 \int_0^{\pi / 2} \sin ^6 \theta d \theta

or            2 A=r^6 \int_0^{\pi / 2}\left(\sin ^6 \theta+\cos ^6 \theta\right) d \theta

=r^6 \int_0^{\pi / 2}\left\lgroup 1-\frac{3}{4} \sin ^2 2 \theta \right\rgroup d \theta

=r^6 \int_0^{\pi / 2}\left\lgroup 1-\frac{3}{8}\{1-\cos 4 \theta\} \right\rgroup d \theta

=r^6 \int_0^{\pi / 2}\left\lgroup\frac{5}{8}+\frac{3}{8} \cos 4 \theta  \right\rgroup d \theta=\frac{5 \pi r^6}{16}

Thus,

A=\int_0^r\left(r^2-y^2\right)^{5 / 2} d y=\frac{5 \pi r^6}{32}

From Eq. (2), we get

U=\frac{2}{9} \frac{V^2 L}{G I^2} \times \frac{5 \pi r^6}{32}

=\frac{5}{(9)(16)} \pi r^6 \frac{V^2 L}{G} \frac{16}{\pi^2 r^8}=\frac{5}{9} \frac{V^2 L}{G A}

=\frac{10}{9}\left\lgroup \frac{V^2}{A^2} \right\rgroup\left\lgroup \frac{1}{2 G} \right\rgroup (A L)

\text { If } V / A=\tau_{\text {avg }} is average shear stress, then

U=\left\lgroup \frac{10}{9} \right\rgroup \tau_{\text {avg }}^2 \frac{A L}{2 G}

Hence, energy shape factor λ for circular beam section is 10/9.