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Chapter 11

Q. 11.7

Refer Figure 11.26 wherein a 100-mm diameter steel rod with \sigma_{y p}=420 MPa is loaded at its free end such that the line of action of the applied force vector makes equal angles with the positive directions of the coordinate axes. Calculate the magnitude of P that will initiate yielding. Assume octahedral shear stress criterion of failure by yielding.

11.26

Step-by-Step

Verified Solution

Let the point of application of \vec{P} be point A whose coordinates with respect to x-y-z reference system are (0, 300, 400) mm. Clearly,

\left.\vec{P}=\frac{P}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k}) N \quad \text { (as } l=m=n=1 / \sqrt{3}\right)

Thus, the reaction forces at O are:

R_{ O x}=-\frac{P}{\sqrt{3}}, \quad R_{ O y}=-\frac{P}{\sqrt{3}} \text { and } R_{ O z}=-\frac{P}{\sqrt{3}}

and \vec{r}_{ A }=(300 \hat{j}+400 \hat{k}) mm. The moment of this force about O is given by

\vec{M}_{ O }=\vec{r}_{ A } \times \vec{P}=(300 \hat{j}+400 \hat{k}) \times \frac{P}{\sqrt{3}}(i+\hat{j}+\hat{k})  N mm

=\left\lgroup -\frac{300 P}{\sqrt{3}} \right\rgroup \hat{k}+\left\lgroup \frac{300 P}{\sqrt{3}} \right\rgroup \hat{i}+\left\lgroup \frac{400 P}{\sqrt{3}} \right\rgroup \hat{j}+\left\lgroup -\frac{400 P}{\sqrt{3}} \right\rgroup \hat{i}

Therefore,

\vec{M}_{ O }=\left\lgroup -\frac{100 P}{\sqrt{3}} \right\rgroup \hat{i}+\left\lgroup \frac{400 P}{\sqrt{3}} \right\rgroup \hat{j}+\left\lgroup -\frac{300 P}{\sqrt{3}} \right\rgroup \hat{k}

=\left\lgroup \frac{100 P}{\sqrt{3}} \right\rgroup (-\hat{i}+4 \hat{j}-3 \hat{k})   Nmm

Thus, the reaction moments at fixed-end O are:

M_x=\frac{100 P}{\sqrt{3}}, \quad M_y=-\frac{400 P}{\sqrt{3}} \text { and } \quad M_z=\frac{300 P}{\sqrt{3}}

The free-body diagram of the rod is now shown in Figure 11.27:

Clearly, net bending moment at O is

M(\text { say })=\sqrt{M_x^2+M_z^2}=\frac{316.23 P}{\sqrt{3}}  N mm

and torsional moment at O is

M_y=-\frac{400 P}{\sqrt{3}} N mm =T \text { (say) }

Ignoring stresses caused by other loads except those due to R_{ O y} , we obtain

\sigma_n=\frac{32 M^{\prime}}{\pi d^3} \text { and } \tau=\frac{16 T}{\pi d^3}

where            M^{\prime}=M+\left\lgroup\frac{d}{8} \right\rgroup R_{ O y}

This is because the net normal stress at O is

\sigma_n=\frac{32 M}{\pi d^3}+\frac{4 R_{ O y}}{\pi d^3}=\frac{32 M}{\pi d^3}\left\lgroup M+\frac{d}{8} R_{ Oy } \right\rgroup =\frac{32 M}{\pi d^3} M^{\prime}

Therefore, according to maximum octahedral shear stress theory

\sigma_{ yp }=\sqrt{\sigma_n^2+3 \tau^2}

=\frac{16}{\pi d^3} \sqrt{4 M^{\prime 2}+3 T^2}

Therefore,

d=\left[\left\lgroup \frac{16}{\pi \sigma_{y p}} \right\rgroup \sqrt{4 M^{\prime 2}+3 T^2}\right]^{1 / 3}

or            (P)^{1 / 3}\left[\frac{16}{\pi(420)} \sqrt{\frac{4(328.73)^2}{3}+3 \frac{(400)^2}{3}}\right]^{1 / 3}=100

Putting various values, we get

P=\frac{100^3}{\left[\frac{16}{\pi(420)} \sqrt{\frac{4}{3}(328.73)^2+400^2}\right]}  N

= 149548.48 N = 149.55 kN

Hence, the applied load is P = 149.55 kN

11.27