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Chapter 11

Q. 11.9

The member shown in Figure 11.28 has 20 mm diameter and is subjected to static loads P and Q as shown in Figure 11.28. Calculate the magnitude of the yield stress, \sigma_{y p}  of the material using maximum shear stress criterion.

11.28

Step-by-Step

Verified Solution

Let us show the free-body diagram of the member in Figure 11.29. For equilibrium

\left(M_x \hat{i}+M_y \hat{j}+M_z \hat{k}\right)+(150 \hat{i}+200 \hat{j}) \times(600 \hat{i}+1500 \hat{k})=\overrightarrow{0}

\left(M_x \hat{i}+M_y \hat{j}+M_z \hat{k}\right)+(-225000 \hat{j}-120000 \hat{k}+300000 \hat{i})=\overrightarrow{0}

Therefore, we get

M_x=-3,00,000  Nmm , \quad M_y=2,25,000  Nmm \quad \text { and } \quad M_z=+1,20,000  Nmm

Ignoring stress caused due to 1500 N load, we get resultant bending moment as

M=\sqrt{M_y^2+M_z^2}

or            M=\sqrt{225^2+120^2} \times 10^3  N mm =255 \times 10^3  N mm

and torsional moment T = M_x = 300 × 10³ N mm. Therefore, the resultant normal stress on the section is sum of stress due to bending and stress due to force Q

\sigma_n=\frac{4 Q}{\pi d^2}+\frac{32 M}{\pi d^3}

=\frac{(4)(600)}{\pi(20)^2}+\frac{(32)(255)\left(10^3\right)}{\pi(20)^3}  MPa

= 326.59 MPa

and the shear stress on the shaft is

\tau=\frac{16 T}{\pi d^3}=\frac{(16)(300)\left(10^3\right)}{\pi(20)^3}  N / mm ^2

= 190.99 MPa

According to the maximum shear stress criterion of failure by yielding, we get

\tau_{\max }=\tau_{ yp }=\frac{\sigma_{ yp }}{2}

Therefore,

\sigma_{ yp }=2 \tau_{\max }=2 \sqrt{\left\lgroup \frac{\sigma_n}{2} \right\rgroup^2}+\tau^2

=\sqrt{\sigma_n^2+4 \tau^2}

=\sqrt{(326.59)^2+(4)(190.99)^2}  MPa

= 502.56 MPa

So, the required yield stress for the material is 502.56 MPa.

11.29