# Solution to Problem 667 | Deflections in Simply Supported Beams

**Problem 667**

Determine the value of *EI*δ at the right end of the overhanging beam shown in Fig. P-667. Is the deflection up or down?

**Solution 667**

$9R_1 + 4(60) = \frac{1}{2}(6)(100)(2)$

$R_1 = 40 \, \text{ lb}$

$\Sigma M_{R1} = 0$

$9R_2 = \frac{1}{2}(6)(100)(7) + 13(60)$

$R_2 = 320 \, \text{ lb}$

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

$EI \, t_{A/B} = \frac{1}{2}(9)(360)(6) - \frac{1}{4}(6)(600)(\frac{39}{5})$

$EI \, t_{A/B} = 2700 \, \text{ N}\cdot\text{m}^3$

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

$EI \, t_{C/B} = -\frac{1}{2}(4)(240)(\frac{8}{3})$

$EI \, t_{C/B} = -1280 \, \text{ N}\cdot\text{m}^3$

The negative sign indicates that the elastic curve is below the tangent line. It is shown in the figure indicated as *t _{C/B}*. See Rules of Sign for Area-Moment Method.

$\dfrac{y_C}{4} = \dfrac{t_{A/B}}{9}$

$y_C = \frac{4}{9}t_{A/B}$

$EI \, y_C = \frac{4}{9}EI \, t_{A/B}$

$EI \, y_C = \frac{4}{9}(2700)$

$EI \, y_C = 1200 \, \text{ N}\cdot\text{m}^3$

Since the absolute value of *EI t _{C/B}* is greater than the absolute value of

*EI y*, the elastic curve is below the undeformed neutral axis (NA) of the beam.

_{C}Therefore,

$EI \, \delta_C = 1280 - 1200$

$EI \, \delta_C = 80 \, \text{ N}\cdot\text{m}^3$ below *C* (deflection is down) *answer*

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