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## Q. 12.11

A shaft of diameter d, bent in the form of a semi-circle of radius R is built-in at A and loaded at B by a force P acting perpendicular to the plane of the shaft as shown in Figure 12.21. Find the value of angle $\phi$ for which the principal stress $\sigma_1$ will be maximum.

## Verified Solution

Let us draw the free-body diagram of the segment BC with angle $\phi(0 \leq \phi \leq \pi)$ as shown in Figure 12.21, of the bent shaft in Figure 12.22.

From the foregoing free-body diagram shown in Figure 12.22, we observe at point C defined by the angle $\phi(0 \leq \phi \leq \pi / 2)$ that the force and moment acting are as follows:
Force P vertically upward
Bending moment $\left(M_{ b }\right)_{ C }$ about OC and
Torsional moment $\left(M_{ t }\right)_{ C }$ about tangent EF drawn at C
Observe that EF and OC are perpendicular to each other at point C. Therefore,

$\left(M_{ b }\right)_{ C }=P \cdot BD =P R \sin \phi \quad(\text { as from } \Delta OBD , \quad OB =R \quad \text { and } \quad BD \perp OD )$

and          $\left(M_{ t }\right)_{ C }=P \cdot CD =P( OC – OD )=P R(1-\cos \phi) \quad(\text { as } OD =R \cos \phi)$

So the principal stress $\sigma_1$ at C is defined as:

$\sigma_1=\frac{\sigma_n}{2}+\sqrt{\left(\frac{\sigma_n}{2}\right)^2+\tau^2}$             (1)

where $\sigma_n$ is the normal stress caused due to $\left(M_{ b }\right)_{ C }$ is

$\sigma_n=\frac{32\left(M_{ b }\right)_{ C }}{\pi d^3}$

and τ is the shear stress caused due to $\left(M_{ t }\right)_{ C }$

$\sigma_n=\frac{16\left(M_{ t }\right)_{ C }}{\pi d^3}$

Putting these values in Eq. (1), we get:

$\sigma_1=\left\lgroup \frac{16}{\pi d^3} \right\rgroup\left[\left(M_{ b }\right)_{ C }+\sqrt{\left(M_{ b }\right)_{ C }^2+\left(M_{ t }\right)_{ C }^2}\right]$

Putting the expressions of $\left(M_{ b }\right)_{ C } \text { and }\left(M_{ t }\right)_{ C }$, we get:

$\sigma_1=\left\lgroup \frac{16 P R}{\pi d^3} \right\rgroup \left[\sin \phi+\sqrt{\sin ^2 \phi+(1-\cos \phi)^2}\right]$

$=\left\lgroup \frac{16 P R}{\pi d^3} \right\rgroup [\sin \phi+\sqrt{2-2 \cos \phi}]$

$=\left\lgroup \frac{16 P R}{\pi d^3}\right\rgroup [\sin \phi+2 \sin (\phi / 2)]$

Clearly $\sigma_1$ will be maximum when $\sin \phi+2 \sin (\phi / 2)$ becomes maximum, that is, when

$\frac{ d }{ d \phi}\left\lgroup \sin \phi+2 \sin \frac{\phi}{2} \right\rgroup =0$

or            $\cos \phi+\cos \frac{\phi}{2}=0$

or            $\cos \phi=-\cos \frac{\phi}{2}=\cos \left\lgroup \pi-\frac{\phi}{2} \right\rgroup$

Therefore,

$3 \phi=2 \pi \Rightarrow \phi=\frac{2 \pi}{3}=120^{\circ}$

We check that the following differential is negative.

$\left[\frac{ d ^2}{ d \phi^2}\left\lgroup \sin \phi+2 \sin \frac{\phi}{2} \right\rgroup\right]_{\phi=2 \pi / 3}$

Hence $\sin \phi+2 \sin (\phi / 2)$ is maximum when $\phi=120^{\circ}$. Thus, $\sigma_1$ the principal stress at C becomes maximum when $\phi=120^{\circ}$.