## Chapter 14

## Q. 14.8.3

## Q. 14.8.3

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

## Step-by-Step

## Verified Solution

Fig. 14.8-4 shows a bending moment diagram obtained by superposition: *ac _{2}b* is the simple-beam or free moment diagram for the uniformly distributed load

*λ*

*w*per unit length, and

*aa*is the moment diagram for a hogging moment of arbitrarily assigned magnitude M_A at the built-in end of the beam. Thus,

_{1}b*ac*is in equilibrium with a transverse load of

_{2}b*λ*

*w*per unit length and

*aa*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

_{1}b*λ*

*w*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 λwL ^{2})* is obtained if M_A is so chosen that M_A = M_c.