# Problem 737 | Fully restrained beam with one support settling

**Problem 737**

In the perfectly restrained beam shown in Fig. P-737, support B has settled a distance Δ below support A. Show that M_{B} = -M_{A} = 6EIΔ/L^{2}.

**Solution 737**

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$\theta_{AB} = \dfrac{1}{EI}(Area_{AB}) = 0$

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

$\frac{1}{2}R_AL^2 + M_AL = 0$

$R_AL + 2M_A = 0$

$R_A = -\dfrac{2M_A}{L}$ → equation (1)

Deviation of B from tangent through A

$t_{B/A} = \dfrac{1}{EI}(Area_{AB}) \cdot \bar{X}_B = -\Delta$

$\frac{1}{2}L(R_AL)(\frac{1}{3}L) + M_AL(\frac{1}{2}L) = -EI\Delta$

$\frac{1}{6}R_AL^3 + \frac{1}{2}M_AL^2 = -EI\Delta$

$R_AL^3 + 3M_AL^2 = -6EI\Delta$

Substitute R_{A} defined in equation (1)

$\left( -\dfrac{2M_A}{L} \right)L^3 + 3M_AL^2 = -6EI\Delta$

$-2M_AL^2 + 3M_AL^2 = -6EI\Delta$

$M_AL^2 = -6EI\Delta$

$M_A = -6EI\Delta / L^2$

Substitute M_{A} to equation (1)

$R_A = -\dfrac{2(-6EI\Delta / L^2)}{L}$

$R_A = \dfrac{12EI\Delta}{L^3}$

Moment at B (See moment diagram by parts)

$M_B = M_A + R_AL$

$M_B = -6EI\Delta / L^2 + (12EI\Delta / L^3)L$

$M_B = -6EI\Delta / L^2 + 12EI\Delta / L^2$

$M_B = 6EI\Delta / L^2$

Thus,

$M_B = -M_A = 6EI\Delta / L^2$ *okay!*

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