Solution to Problem 677 | Midspan Deflection

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Problem 677
Determine the midspan deflection of the beam loaded as shown in Fig. P-677.

Simple beam loaded with triangular load over half the span

Solution 677

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

$EI \, t_{A/B} = \frac{1}{4}(\frac{1}{2}L)(\frac{1}{24}w_oL^2)\frac{4}{5}(\frac{1}{2}L) + \frac{1}{2}(\frac{1}{2}L)(\frac{1}{8}w_oL^2)\frac{2}{3}(\frac{1}{2}L) - \frac{1}{3}(\frac{1}{2}L)(\frac{1}{8}w_oL^2)\frac{3}{4}(\frac{1}{2}L)$

$EI \, t_{A/B} = \frac{1}{480}w_oL^4 + \frac{1}{96}w_oL^4 - \frac{1}{128}w_oL^4$

$EI \, t_{A/B} = \frac{3}{640}w_oL^4$

$t_{A/B} = \dfrac{3w_oL^4}{640EI}$

$\delta_{midspan} = \frac{1}{2}t_{A/B}$

$\delta_{midspan} = \dfrac{1}{2}\left( \dfrac{3w_oL^4}{640EI} \right)$

$\delta_{midspan} = \dfrac{3w_oL^4}{1280EI}$ answer

Another Solution

$M = \frac{1}{4}w_oL(\frac{1}{2}L) - \frac{1}{2}(\frac{1}{2}w_oL)\frac{2}{3}(\frac{1}{2}L)$

$M = \frac{1}{24}w_oL^2$

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

$EI \, t_{A/B} = \frac{1}{2}L(\frac{1}{24}w_oL^2)\frac{1}{2}(\frac{1}{2}L) - \frac{1}{4}(\frac{1}{2}L)(\frac{1}{24}w_oL^2)\frac{1}{5}(\frac{1}{2}L)$

$EI \, t_{A/B} = \frac{1}{192}w_oL^4 - \frac{1}{1920}w_oL^4$

$EI \, t_{A/B} = \frac{3}{640}w_oL^4$

$t_{A/B} = \dfrac{3w_oL^4}{640EI}$ (okay!)

‹ Solution to Problem 676 | Midspan Deflection up Solution to Problem 678 | Midspan Deflection ›

Submitted by Romel Verterra on September 11, 2010 - 11:58pm

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Strength of Materials

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