Question 4.CA.11: For the bar of Concept Application 4.10, determine the large......

Use Eq. (4.71) with the given data

\sigma_x=\frac{M(r-R)}{A e r}             (4.71)

M = 8 kip·in.             A = bh = (2.5 in.)(1.5 in.) = 3.75 in²

and the values obtained in Concept Application 4.10 for R and e:

R = 5.969              e = 0.0314 in.

First using r=r_2=6.75 in. in Eq. (4.71), write

\begin{aligned} \sigma_{\max } & =\frac{M\left(r_2-R\right)}{A{e r_2}} \\ & =\frac{(8  kip \cdot in .)(6.75  in . -5.969  in.}{\left(3.75  in ^2\right)(0.0314  in .)(6.75  in .)} \\ & =7.86  ksi \end{aligned}

Now using r=r_1=5.25 in. in Eq. (4.71),

\begin{aligned} \sigma_{\min } & =\frac{M\left(r_1-R\right)}{\operatorname{Aer}_1} \\ & =\frac{(8  kip \cdot in .)(5.25 \text {  in. }-5.969 \text {  in. })}{\left(3.75 \text {  in }^2\right)(0.0314 \text {  in. })(5.25 \text {  in. })} \\ & =-9.30  ksi \end{aligned}

Remark. Compare the values obtained for \sigma_{\max } \text { and } \sigma_{\min } with the result for a straight bar. Using Eq. (4.15) of Sec. 4.2,

\sigma_m=\frac{M c}{I}              (4.15)

\begin{aligned} \sigma_{\max , \min } & =\pm \frac{M c}{I} \\ & =\pm \frac{(8  kip \cdot in .)(0.75  in .)}{\frac{1}{12}(2.5  in .)(1.5  in .)^3}=\pm 8.53  ksi \end{aligned}