# Solution to Problem 653 | Deflections in Simply Supported Beams

**Problem 653**

Compute the midspan value of EIδ for the beam shown in Fig. P-653. (Hint: Draw the M diagram by parts, starting from midspan toward the ends. Also take advantage of symmetry to note that the tangent drawn to the elastic curve at midspan is horizontal.)

**Solution 653**

By symmetry:

$R_1 = R_2 = 600(2) = 1200 \, \text{N}$

$R_1 = R_2 = 600(2) = 1200 \, \text{N}$

$t_{A/B} = \dfrac{1}{EI}(Area_{AB}) \, \bar{X}_A$

$t_{A/B} = \dfrac{1}{EI} [ \, \frac{1}{2}(2.5)(3000)(\frac{5}{3}) + \frac{1}{3}(0.5)(75)(\frac{19}{8}) - \frac{1}{3}(2.5)(1875)(\frac{15}{8}) \, ]$

$t_{A/B} = \dfrac{3350}{EI}$

From the figure

$\delta_{midspan} = t_{A/B}$

Thus

$EI \, \delta_{midspan} = 3350 \, \text{ N}\cdot\text{m}^3$ *answer*

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