## Re: Deflection of beam. Efect of % errors in span and load data.

What you are heading for is a general formula, one into which you can plug any small changes for L and w, not just the specific values given. They go in right at the end.
Your second line should look like this,

$\displaystyle E+\delta E=5(w+\delta w)(L^4+4L^3\delta L+\dots)/384ly$.

Now multiply out the brackets on the RHS and you get (ignoring the second order term),

$\displaystyle E+\delta E=5(wL^4+4wL^3\delta L+L^4\delta w + \dots)/384ly.$

The next step is to split the RHS into three separate fractions. You will find that the E on the LHS will cancel with an E on the RHS. Having done that, divide both sides by E, on the RHS replacing it with $\displaystyle 5wL^4/384ly.$
Cancel down the fractions on the RHS and finally multiply both sides by 100 to arrive at the percentage changes.
On the LHS for example you should finish up with $\displaystyle 100\frac{\delta E}{E}$ which is the percentage change in E.
If all goes well you should finish up with a statement that says that the percentage error ( change) in E is approximately 4 times the percentage error (change) in L plus the percentage error (change) in w. (Approximate because you have to own up to the fact that you have ignored second and higher order terms on the RHS).
You can now plug in the values for the errors given in the question.
Notice that the formula is quite useful in the information it gives you. For example, it tells you that if you want to improve your results, its better to try to improve your measurement of L rather than w. It also says that it is better to overestimate one of the values and to underestimate the value of the other rather than to over or underestimate both of them.