The deflection y at the center of a uniform loaded simply supported beam is given by: $$y=\frac{5}{384}*\frac{wl^2}{EI}$$ Where $w$ is the uniform load, $l$ is the beam length, $E$ is the young modulus and $I$ is the moment of inertia of beam cross-section. If $w$ increases by $0.02%$, $l$ increases by $0.03%$, $E$ decreases by $0.02$, and $I$ decreases by $0.01%$, using the total differential to find the percentage increase in $y$?

how can I solve this problem using the total differential?

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    – saulspatz
    May 28 '19 at 13:31

$y = \frac{5}{384}\frac{wl^2}{EI}$

$\ln(y)= \ln(5) - \ln(384) + \ln(w) + 2\ln(l)-\ln(E)-\ln(I)$

Taking differentials on both sides,

$\frac{dy}{y} = 0 - 0 +\frac{dw}{w}+2\frac{dl}{l}-\frac{dE}{E}-\frac{dI}{I}$

Given: $$dw = 0.02 , dl = 0.03 , dE = -0.02 , dI = -0.01$$


$$\frac{dy}{y} = \frac{0.02}{w} + \frac{0.06}{l} +\frac{0.02}{E} +\frac{0.01}{I} = \bigg[\frac{2}{w}+\frac{6}{l}+\frac{2}{E}+\frac{1}{I}\bigg]\frac{1}{100}$$

  • $\begingroup$ thanks, but can you please tell me why we should add the numbers together? why did we turn the minuses into pluses? $\endgroup$ May 28 '19 at 14:05
  • $\begingroup$ There are already minuses in $-\frac{dE}{E}$ and $-\frac{dI}{I}$, so $-(-0.02) = + 0.02$ and $-(-0.01) = +0.01$ :) $\endgroup$
    – 19aksh
    May 28 '19 at 14:07
  • $\begingroup$ you're right thanks $\endgroup$ May 28 '19 at 14:09
  • $\begingroup$ You're welcome! $\endgroup$
    – 19aksh
    May 28 '19 at 14:09
  • 1
    $\begingroup$ Sorry, first I thought them to be relative errors, now I've corrected it. $\endgroup$
    – 19aksh
    May 28 '19 at 16:23

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