Question 14.8.3: Using the safe theorem, given an alternative solution to Exa...
Using the safe theorem, given an alternative solution to Example 14.8-2.
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_cwhichever 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.
