**Problem 738**

A perfectly restrained beam is loaded by a couple M applied where shown in Fig. P-738. Determine the end moments.

**Solution 738**

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$EI \theta = (Area_{AB}) = 0$

$Mb + M_AL - \frac{1}{2}L(R_AL) = 0$

$Mb + M_AL = \frac{1}{2}R_AL^2$

$R_A = \dfrac{2(Mb + M_AL)}{L^2}$ → equation (1)

Deviation of B from a tangent line through A is zero

$EI \, t_{B/A} = (Area_{AB}) \bar{X}_B = 0$

$Mb(\frac{1}{2}b) + M_AL(\frac{1}{2}L) - \frac{1}{2}L(R_AL)(\frac{1}{3}L) = 0$

$\frac{1}{2}Mb^2 + \frac{1}{2}M_AL^2 - \frac{1}{6}R_AL^3 = 0$

$3Mb^2 + 3M_AL^2 - R_AL^3 = 0$

Substitute R_{A} of equation (1) to the equation above

$3Mb^2 + 3M_AL^2 - \left[ \dfrac{2(Mb + M_AL)}{L^2} \right]L^3 = 0$

$3Mb^2 + 3M_AL^2 - 2MbL - 2M_AL^2 = 0$

$M_AL^2 = 2MbL - 3Mb^2$

$M_AL^2 = Mb(2L - 3b)$

$M_AL^2 = Mb \, [ \, 2L - 3(L - a) \, ]$

$M_AL^2 = Mb(3a - L)$

$M_A = \dfrac{Mb}{L^2}(3a - L)$

$M_A = \dfrac{Mb}{L}\left( \dfrac{3a}{L} - 1 \right)$ *answer*

From equation (1)

$R_A = \dfrac{2Mb}{L^2} + \dfrac{2M_A}{L}$

$R_A = \dfrac{2Mb}{L^2} + \dfrac{2}{L} \left[ \dfrac{Mb}{L}\left( \dfrac{3a}{L} - 1 \right) \right]$

$R_A = \dfrac{2Mb}{L^2} + \dfrac{6Mab}{L^3} - \dfrac{2Mb}{L^2}$

$R_A = \dfrac{6Mab}{L^3}$

From moment diagram by parts

$M_B = M + M_A - R_AL$

$M_B = M + \dfrac{Mb}{L}\left( \dfrac{3a}{L} - 1 \right) - \dfrac{6Mab}{L^3}(L)$

$M_B = M + \dfrac{3Mab}{L^2} - \dfrac{Mb}{L} - \dfrac{6Mab}{L^2}$

$M_B = M - \dfrac{Mb}{L} - \dfrac{3Mab}{L^2}$

$M_B = M - \dfrac{M(L - a)}{L} - \dfrac{3Mab}{L^2}$

$M_B = M - M + \dfrac{Ma}{L} - \dfrac{3Mab}{L^2}$

$M_B = \dfrac{Ma}{L} - \dfrac{3Mab}{L^2}$

$M_B = -\dfrac{Ma}{L}\left(\dfrac{3b}{L} - 1 \right)$ *answer*