# how can I solve this problem using total differential

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?

• Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? Can you calculate the total differential? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. May 28 '19 at 13:31

## 1 Answer

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

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

Taking differentials on both sides,

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

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

So,

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

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