# Solution to Problem 670 | Deflections in Simply Supported Beams

**Problem 670**

Determine the value of EIδ at the left end of the overhanging beam shown in Fig. P-670.

**Solution 670**

$3R_1 = 600 + \frac{1}{2}(3)(900)(1)$

$R_1 = 650 \, \text{ N}$

$\Sigma M_{R2} = 0$

$3R_2 + 600 = \frac{1}{2}(3)(900)(2)$

$R_2 = 700 \, \text{ N}$

$EI \, t_{C/B} = (Area_{BC}) \, \bar{X}_C$

$EI \, t_{C/B} = \frac{1}{2}(3)(1950)(1) - 3(600)(\frac{3}{2}) - \frac{1}{4}(3)(1350)(\frac{3}{5})$

$EI \, t_{C/B} = -382.5 \, \text{ N}\cdot\text{m}^3$

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

$EI \, t_{A/B} = - 1(600)(\frac{1}{2})$

$EI \, t_{A/B} = -300 \, \text{ N}\cdot\text{m}^3$

The negative signs above indicates only the location of elastic curve relative to the reference tangent. It does not indicate magnitude. It shows that the elastic curve is below the reference tangent at points A and C.

By ratio and proportion

$\dfrac{\delta_A - t_{A/B}}{1} = \dfrac{t_{C/B}}{3}$

$\delta_A = \frac{1}{3}t_{C/B} + t_{A/B}$

$EI \, \delta_A = \frac{1}{3}EI \, t_{C/B} + EI \, t_{A/B}$

$EI \, \delta_A = \frac{1}{3}(382.5) + 300$

$EI \, \delta_A = 427.5 \, \text{ N}\cdot\text{m}^3$ *answer*