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## Q. 14.8.3

Using the safe theorem, given an alternative solution to Example 14.8-2.

## Verified Solution

Fig. 14.8-4 shows a bending moment diagram obtained by superposition: ac2b is the simple-beam or free moment diagram for the uniformly distributed load λw per unit length, and aa1b is the moment diagram for a hogging moment of arbitrarily assigned magnitude $M_A$ at the built-in end of the beam. Thus, ac2b is in equilibrium with a transverse load of λw per unit length and aa1b is in equilibrium with zero transverse load; therefore the composite diagram, as shaded in Fig. 14.8-4, must also be in equilibrium with the load λw per unit length, whatever the value assigned to $M_A$.

Let $M_c$ denote the magnitude of the largest sagging moment in the span. It can be concluded from the safe theorem that the beam will not collapse under the load, provided

$\hspace{5em}M_p > M_A \ or \ M_c$

whichever is (numerically) the larger. The value of $M_P$ that satisfies this condition depends on the assigned magnitude of $M_A$. Referring to Fig. 14.8-4. the lowest value of $M_P$ (= 0.0858 λwL2) is obtained if $M_A$ is so chosen that $M_A = M_c$.