Question 14.3.1: Derive an expression for the bending moment in a rectangular...
Derive an expression for the bending moment in a rectangular section when the stress distribution is partly plastic and partly elastic (Fig. 14.3-3).
Force in each plastic zone = \sigma _{y}b\left(1-\alpha \right)h/2
Moment lever arm for plastic zones = \alpha h+\left(h-\alpha h\right)/2,
\qquad \qquad \qquad =\left(1+\alpha \right)h/2\begin{array}{ll}\therefore \ M \ \left(plastic \ zones\right) & = \left[\sigma _{y}b\left(1-\alpha \right)\frac{h}{2} \right] \left[\frac{\left(1+\alpha \right)h}{2} \right] \\[0.5 cm] & = \sigma _{y}\frac{bh^{2}}{4} \left(1-\alpha ^{2} \right) \qquad \qquad \qquad\\[0.5 cm] & = M_{p} \left(1-\alpha ^{2} \right) from \ Eqn \ 14.3-3 \\[0.5 cm]M\left(elastic \ core\right) &= \sigma _{y} \left[elastic \ section \ modulus \ of \ elastic \ core\right] \\[0.5 cm] &=\sigma _{y} \frac{bh^{2}}{6} \alpha ^{2} \\[0.5 cm] &=M_{y} \alpha ^{2} \qquad from \ Eqn \ 14.3-2 \end{array}
\begin{array}{ll}\therefore \ Total \ M &= \sigma_{y}\frac{bh^{2}}{4} \left(1-\alpha ^{2} \right)+\sigma _{y} \frac{bh^{2} }{6} \alpha ^{2} \\[0.5 cm] & = \sigma _{y}\frac{bh^2}{4} \left(1-\alpha ^2 /3\right) \\[0.5 cm] & = M_{p}\left(1-\alpha ^2/3\right) \qquad from \ Eqn \ 14.3-3\end{array} (14.3-6)
M_{p} = \left(\sigma _{y}\frac{bh}{2} \right)\left(\frac{h}{2} \right)=\sigma _{y}\frac{bh^{2} }{4} (14.3-3)
\sigma _{y} =\frac{M_{y}}{I}\left(\frac{h}{2} \right) or M_{y} =\sigma _{y} \frac{I}{\left(h/2\right) } =\sigma _{y} \frac{bh^{2}}{6} (14.3-2)
COMMENTS
Theoretically (see Fig. 14.3-3(c)) the full plastic moment M_P can be reached only when the extreme fibre strain \epsilon_{max} is infinity—an impossible condition. However,M approaches M_{P} for reasonably small values of α , as can be seen from Table 14.3-1, where the M values have been calculated from Eqn 14.3-6.
| Table 14.3-1 Relation between M and α | |
|
α |
M |
|
1 |
0.667M_{P}{}^\dagger |
|
1/10 |
0.997M_{P} |
|
1/15 |
0.999M_{P} |
| ^\dagger M=0.667 M_{p} = M_y (Eqn 14.3-6) | |
Thus, if structural mild steel strain hardens at \varepsilon_{max} =15 \varepsilon_{y} (so that α = 1/15), then the moment M would have reached 0.999 M_{P} by the time strain hardening occurs; even at \varepsilon_{max}= 10 \varepsilon _{y}(so that α = 1/10), M already reaches 0.997 M_P. Hence, for practical purposes, it is wholly acceptable to assume that the full plastic moment M_pcan be developed. When the moment at a section of a beam reaches M_p,a plastic hinge is said to have formed at that section, because (see Examples 14.3-2 and 3) the local curvature approaches infinity and the rotation increases indefinitely under a constant value of the moment M_p.