**Problem 869**

Find the value of EIδ at the center of the first span of the continuous beam in Figure P-869 if it is known that M_{2} = -980 lb·ft and M_{3} = -1082 lb·ft.

**Solution 869**

**From Span 1-2**

**Moment at the center of the first span**

$M_A = 5(402) - 1250$

$M_A = 760 ~ \text{lb}\cdot\text{ft}$

**Apply three-moment equation to span A-2-3**

$M_A L_{A2} + 2M_2(L_{A2} + L_{23}) + M_3L_{23} + \left( \dfrac{6A\bar{a}}{L} \right)_{A2} + \left( \dfrac{6A\bar{b}}{L} \right)_{32}$

$= 6EI \left( \dfrac{h_{2A}}{L_{A2}} + \dfrac{h_{23}}{L_{23}} \right)$

$760(5) + 2(-980)(5 + 10) + (-1082)(10) + \dfrac{w_o L^3}{4} + \dfrac{5w_o L^3}{32} = 6EI \left( \dfrac{h_A}{5} + 0 \right)$

$3800 - 29,400 - 10,820 + \dfrac{100(5^3)}{4} + \dfrac{5(160)(10^3)}{32} = \frac{6}{5}EI \, h_A$

$3,800 - 29,400 - 10,820 + 3,125 + 25,000 = \frac{6}{5}EI \, h_A$

$\frac{6}{5}EI \, h_A = -8,295$

$EI \, h_A = -6,912.5 ~ \text{lb}\cdot\text{ft}^3$

The negative sign means that point A is below point 2. Thus,

$EI \, \delta = 6,912.5 ~ \text{lb}\cdot\text{ft}^3 ~ \text{down}$ *answer*