Question 3.15: Plot the distribution of stresses across section A-A of the ...

Since A = bh, we have d A = b dr and, from Eq. (3–63),

r_{n} =\frac {A}{\int {\frac {d A}{r}}}               (3–63)

r_{n} =\frac {A}{\int {\frac {d A}{r}}}=\frac {bh}{\int_{r_{i}}^{r_{o}}{\frac {b}{r} dr }}=\frac {h}{ln\frac {r_{o}}{r_{i}}}                (1)

From Fig. 3–35b, we see that r_{i} = 2  in, r_{o} = 6  in, r_{c} = 4  in, and A = 3  in^{2}. Thus, from Eq. (1),

r_{n} =\frac {h}{ln(r_{o}/r_{i} )} =\frac {4}{ln \frac {6}{2}}= 3.641  in

and so the eccentricity is e = r_{c} − r_{n} = 4 − 3.641 = 0.359 in. The moment M is positive and is M = Fr_{c} = 5000(4) = 20 000 lbf · in. Adding the axial component of stress to Eq. (3–64) gives

σ =\frac{M y}{Ae(r_{n} − y)}              (3–64)

σ =\frac {F}{A} +\frac {My}{Ae(r_{n} − y)} =\frac {5000}{3} +\frac {(20 000)(3.641 − r )}{3(0.359)r}              (2)

Substituting values of r from 2 to 6 in results in the stress distribution shown in Fig. 3–35c. The stresses at the inner and outer radii are found to be 16.9 and −5.63 kpsi, respectively, as shown.