**Summary of key points**

- The three-moment equation can be applied at any three points in any beam. It will determine the relation among the moments at these points.
- The terms $6A\bar{a}/L$ and $6A\bar{b}/L$ refer to the moment diagram by parts resulting from the simply supported loads between any two adjacent points described in (1).
- The heights h
_{1}and h_{3}from the equation

$M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 6EI \left( \dfrac{h_1}{L_1} + \dfrac{h_3}{L_2} \right)$

refer to the respective vertical distance of points 1 and 3 from point 2. The height h is positive if above point 2 and negative if below it.

**Problem 859**

Determine the value of EIδ under P in Fig. P-859. What is the result if P is replaced by a clockwise couple M?

**Solution 859**

$M_2 = -Pb$

$M_3 = 0$

$h_1 = 0$

$| \, h_3 \, | = \delta_3$

$M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 6EI \left( \dfrac{h_1}{L_1} + \dfrac{h_3}{L_2} \right)$

$0 + 2(-Pb)(a + b) + 0 + 0 + 0 = 6EI\left( 0 + \dfrac{h_3}{b} \right)$

$-2PbL = \dfrac{6EI \, h_3}{b}$

$EI \, h_3 = -2PLb^2 / 3$ ← the negative sign indicates that point 3 is below point 2

Thus,

$EI \, \delta_3 = PLb^2 / 3$ *answer*

**If P is replaced by a clockwise couple M**

$M_1 = 0$

$M_2 = -M$

$M_3 = -M$

$h_1 = 0$

$| \, h_3 \, | = \delta$

$M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 6EI \left( \dfrac{h_1}{L_1} + \dfrac{h_3}{L_2} \right)$

$0 + 2(-M)(a + b) + (-M)(b) + 0 + 0 = 6EI\left( 0 + \dfrac{h_3}{b} \right)$

$-2ML - Mb = \dfrac{6EI \, h_3}{b}$

$-M(b + 2L) = \dfrac{6EI \, h_3}{b}$

$-Mb(b + 2L) = 6EI \, h_3$

$EI \, h_3 = -Mb(b + 2L) / 6$ ← point 3 is below point 2

Thus,

$EI \, \delta_3 = Mb(b + 2L) / 6$ *answer*