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Chapter 12

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.

12.21

Step-by-Step

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} .

12.22