Question 4.11: For the bar of Example 4.10, determine the largest tensile a...
For the bar of Example 4.10, determine the largest tensile and compressive stresses, knowing that the bending moment in the bar is M = 8 kip • in.
We use Eq. (4.71) with the given data
\sigma_{x}=\frac{M(r-R)}{A e r} Eq. (4.71)
M = 8 kip • in A = bh = (2.5 in.)(1.5 in.) = 3.75 in²
and the values obtained in Example 4.10 for R and e,
R = 5.969 e = 0.0314 in
Making first r = r_{2} = 6.75 in. in Eq. (4.71), we write
\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 .)}
\sigma_{\max }=7.86 ksi
Making now r = r_{1} = 5.25 in. in Eq. (4.71), we have
\sigma_{\min }=\frac{M\left(r_{1}-R\right)}{ Aer _{1}}
=\frac{(8 kip \cdot in. )(5.25 in .-5.969 in .)}{\left(3.75 in ^{2}\right)(0.0314 in .)(5.25 in .)}
\sigma_{\min }=-9.30 ksi
Remark. Let us compare the values obtained for \sigma_{max} and \sigma_{min} with the result we would get for a straight bar. Using Eq. (4.15) of Sec. 4.4, we write
\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