Assume the fixed end to be equivalent to an imaginary span with imaginary loading. In three-moment equation, all the terms that refer to the imaginary span have zero values.

In the following problems, the ends of the beams are assumed to be perfectly fixed by the walls against rotation. All supports are assumed to remain at the same level.

**Problem 839**

Determine the prop reaction for the beam in Fig. P-839.

**Solution 839**

$L_1 = L$

$L_2 = 0$

$M_1 = -\frac{1}{4}PL$

$\dfrac{6A_1\bar{a}_1}{L_1} = \frac{3}{8}PL^2$

$\dfrac{6A_2\bar{b}_2}{L_2} = 0$

Thus,

$-\frac{1}{4}PL(L) + 2M_2(L + 0) + 0 + \frac{3}{8}PL^2 + 0 = 0$

$2L\,M_2 = -\frac{1}{8}PL^2$

$M_2 = -\frac{1}{16}PL$

$M_2 = \Sigma M_{\text{to the left of }M_2}$

$-\frac{1}{16}PL = RL - P(\frac{5}{4}L) - P(\frac{1}{2}L)$

$-\frac{1}{16}PL = RL - \frac{5}{4}P) - \frac{1}{2}PL$

$RL = \frac{27}{16}PL$

$R = \frac{27}{16}P$ *answer*

**Check using area-moment method**

$EI \, t_{B/C} = 0$

$(Area_{BC}) \cdot \bar{X}_B = 0$

$\frac{1}{2}L(RL)(\frac{2}{3}L) - \frac{1}{2}(\frac{1}{2}L)(\frac{1}{2}PL)[ \, \frac{1}{2}L + \frac{2}{3}(\frac{1}{2}L) \, ]$

$\frac{1}{2}L(\frac{1}{4}PL)(\frac{1}{3}L) - \frac{1}{2}L(\frac{5}{4}PL)(\frac{2}{3}L) = 0$

$\frac{1}{3}RL^3 - \frac{5}{48}PL^3 - \frac{1}{24}PL^3 - \frac{5}{12}PL^3 = 0$

$\frac{1}{3}RL^3 = \frac{9}{16}PL^3$

$R = \frac{27}{16}P$ *okay*

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