Solution to Problem 681 | Midspan Deflection | Strength of Materials Review at MATHalino

Problem 681
Show that the midspan value of EIδ is (w_ob/48)(L³ - 2Lb² + b³) for the beam in part (a) of Fig. P-681. Then use this result to find the midspan EIδ of the loading in part (b) by assuming the loading to exceed over two separate intervals that start from midspan and adding the results.

Solution 681

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Part (a)

Elastic curve and moment diagram by parts of simple beam with uniform load

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

$EI \, t_{A/B} = \frac{1}{2}(\frac{1}{2}L)(\frac{1}{2}Lbw_o)(\frac{1}{3}L) - \frac{1}{3}(b)(\frac{1}{2}b^2w_o)\frac{1}{4}(2L - b)$

$EI \, t_{A/B} = \frac{1}{24}L^3bw_o - \frac{1}{12}Lb^3w_o + \frac{1}{24}b^4w_o$

$EI \, t_{A/B} = \frac{1}{24}w_ob (L^3 - 2Lb^2 + b^3)$

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

$EI \, \delta = \frac{1}{2} [ \, \frac{1}{24} w_ob (L^3 - 2Lb^2 + b^3) \, ]$

$EI \, \delta = \dfrac{w_ob}{48}(L^3 - 2Lb^2 + b^3)$ answer

Part (b)
$EI \, \delta = EI \, \delta_1 + EI \, \delta_2$

single uniform load broken into two loads in analysis

$EI \, \delta = \frac{1}{48} w_o b_1 (L^3 - 2L{b_1}^2 + {b_1}^3) + \frac{1}{48} w_o b_2 (L^3 - 2L{b_2}^2 + {b_2}^3)$

$EI \, \delta = \frac{1}{48} (800)(1) [ \, 6^3 - 2(6)(1^2) + 1^3 \, ] + \frac{1}{48} (800)(2) [ \, 6^3 - 2(6)(2^2) + 2^3 \, ]$

$EI \, \delta = 3416.67 + 5866.67$

$EI \, \delta = 9283.34 \, \text{ N}\cdot\text{m}^3$ answer

Solution to Problem 681 | Midspan Deflection

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