Question 6.10: Determine the yield moment, plastic modulus, plastic moment,......

Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
1, 2. Conceptualize, Categorize: As a preliminary matter, note that since the cross section is doubly symmetric, the neutral axis passes through the center of the circle for both linearly elastic and elastoplastic behavior.
3. Analyze: The yield moment M_Y is found from the flexure formula [Eq. (6-87)] as

M_{Y}\,=\,{\frac{\sigma_{Y}I}{c}}\,=\,{\frac{\sigma_{Y}\,(\pi d^{4}/64)}{d/2}}\,=\,\sigma_{Y}\,{\Bigg\lgroup} \frac{\pi d^{3}}{32} {\Bigg\rgroup}\quad\quad\quad (6-107)
The plastic modulus Z is found from Eq. (6-94), in which A is the area of the circle and \overset{—}{y_{1}} ~ and ~\overset{—}{y_{2}} are the distances to the centroids c_1 ~and ~c_2 of the two halves of the circle (Fig. 6-54). Use Cases 9 and 10 of Appendix E to get

Z={\frac{A({\overline{{y}}}_{1}+{\overline{{y}}}_{2})}{2}}\quad\quad (6-94)
\quad\quad A\,=\,{\frac{\pi d^{2}}{4}}\quad{\overline{{y}}_{1}}={\overline{{y}}_{2}}\,=\,{\frac{2d}{3\pi}}
Now substitute into Eq. (6-94) for the plastic modulus to find
\quad\quad Z\,=\,\frac{A(\overset{—}{y}_{1} + \overset{—}{y}_{2})}{2}\,=\,\frac{d^{3}}{6}\quad\quad (6-108)
Therefore, the plastic moment M_P [Eq. (6-93)] is
\quad\quad M_{P}\ =\,\sigma_{Y}Z\,=\,\frac{\sigma_{Y}d^{3}}{6}\quad\quad (6-109)
and the shape factor f[Eq. (6-95)] is

\quad\quad f\,=\,{\frac{M_{P}}{M_{Y}}}=\frac{Z}{S}\quad \quad (6-95)\\
\quad\quad f\,=\,{\frac{M_{P}}{M_{Y}}}\,= \,{\frac{16}{3\pi}}\,\,\approx\,\,1.70 \quad\quad (6-110)
4. Finalize: This result shows that the maximum bending moment for a circular beam of elastoplastic material is about 70% larger than the bending moment when the beam first begins to yield.