Make sure all the math is showen and how you got the answer

Background: The deflection $(\backslash$delta $)$ of a simply$-$supported beam $($of length, L $)$ under a point load $($ P in Newtons$)$ at its

midpoint is:

$\backslash$delta $=($PL$^(3))/(48$EI$)$

where E is known as Young's modulus $($or the modulus of elasticity$)$ and I is the area moment of inertia. The

following image shows two possible cross$-$sections for the beam: a plane rectangle and an I$-$beam.

Rectangular beam

I$-$beam

The formula for I of the rectangular beam is:

I$=(1)/(12)$BH$^(3)$

and the formula for I of the I$-$beam is:

I$=(1)/(12)[$ah$^(3)+$B$($H$^(3)-$h$^(3))]$

QUESTION $1(10$ points$)$

Perform a calculation of $\backslash$delta by hand, showing your work and the cancellation of units, for one value $($each$)$ of P$,$E$,$

and I $.$ Use the rectangular beam model and let m $,$ the mass supported at midspan be $75$ kg $.$ Use L$=3$m$,$B$=4\mathrm{.}5$

cm$,$H$=8\mathrm{.}9$cm$,$g$=9\mathrm{.}81($m$)/($s$^(2))$ and a Young's modulus for wood $($fir$)$ of E$\_($fir$)=10$GPa.

Observation: How does the orientation of the beam affect the deflection? In other words, how does the

deflection change when values for and are switched?

QUESTION $2(10$ points$)$

Repeat the calculations for the preceding problem using the I$-$beam cross$-$section instead of the rectangular

cross$-$section. How does the deflection compare for the rectangular and I$-$beam cross$-$sections? What is the

advantage of an I$-$beam cross$-$section? Use L$=3$m$,$B$=4\mathrm{.}5$cm$,$H$=8\mathrm{.}9$cm$,$h$=3\mathrm{.}5$cm$,$a$=1$cm$,$g$=9\mathrm{.}81($m$)/($s$^(2))$

and a Young's modulus for wood $($fir$)$ of E$\_($fir$)=10$GPa.