Question 6.8: Determine the yield moment, plastic modulus, plastic moment,...
Determine the yield moment, plastic modulus, plastic moment, and shape factor for a beam of circular cross section with diameter d (Fig. 6-44).
As a preliminary matter, we 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.
The yield moment M_{Y} is found from the flexure formula (Eq. 6-74) as follows:
M_{Y}=\frac{\sigma_{Y}I}{c}=\sigma_{Y}S (6-74)
M_{Y}=\frac{\sigma_{Y}I}{c}=\frac{\sigma_{Y}(\pi d^{4}/64)}{d/2}=\sigma_{Y}\left(\frac{\pi d^{3}}{32}\right) (6-87)
The plastic modulus Z is found from Eq. (6-78) in which A is the area of the circle and \overline{y} and \overline{y}_{2} are the distances to the centroids c_{1} and c_{2} of the two halves of the circle (Fig. 6-45). Thus, from Cases 9 and 10 of Appendix D, we get
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Circle (Origin of axes at center)
A=\pi r^{2}=\frac{\pi d^{2}}{4} I_{x}=I_{y}=\frac{\pi r^{4}}{4}=\frac{\pi d^{4}}{64}
I_{xy}=0 I_{P}=\frac{\pi r^{4}}{2}=\frac{\pi d^{4}}{32} I_{BB}=\frac{5\pi r^{4}}{4}=\frac{5\pi d^{4}}{64} |
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Semicircle (Origin of axes at centroid)
A=\frac{\pi r^{2}}{2} \overline{y}=\frac{4r}{3\pi} I_{x}=\frac{\left(9\pi^{2}-64\right)r^{4}}{72\pi}\approx 0.1098r^{4} I_{y}=\frac{\pi r^{4}}{8} I_{xy}=0 I_{BB}=\frac{\pi r^{4}}{8} |
A=\frac{\pi d^{2}}{4} \overline{y}_{1}=\overline{y}_{2}=\frac{2d}{3\pi}
Now substituting into Eq. (6-78) for the plastic modulus, we find
Z=\frac{A\left(\overline{y}_{1}+\overline{y}_{2}\right)}{2} (6-78)
=\frac{d^{3}}{6} (6-88)
Therefore, the plastic moment M_{P} (Eq. 6-77) is
M_{P}=\sigma_{Y}Z=\frac{\sigma_{Y}d^{3}}{6} (6-89)
and the shape factor ƒ (Eq. 6-79) is
f=\frac{M_{P}}{M_{Y}}=\frac{Z}{S} (6-79)
f=\frac{M_{P}}{M_{Y}}=\frac{16}{3\pi}\approx1.70 (6-90)
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.
