$(20$ pts$)$ A $10-$meters long simply$-$supported beam is loaded axially of $P_{\text{a}\text{x}\text{i}\text{a}\text{l}}=60kN.$ The beam is made of $100250$ Douglas fir. The relevant appendix tables are given as below. Calculate

$(4$ pts$)$ the tensile stress ${}_{\text{a}\text{x}\text{i}\text{a}\text{l}}$ due to the axial load MPa

$(4$ pts$)$ the beam's weight per unit length $w[\frac{N}{m}]$

$(4$ pts$)$ the bending stress ${}_{\text{b}\text{e}\text{n}\text{d}\text{i}\text{n}\text{g}}$ due to beam weight MPa

$(4$ pts$)$ the combined maximum stress ${}_{\text{c}\text{o}\text{m}\text{i}\text{n}\text{e}\text{d}}[MPa]$

$(4$ pts$)$ the maximum deflection ${}_{\text{m}\text{a}\text{x}}[mm]$

Table E$4$: SI Timber Sizes

Table E$2$: SI Materials Properties of Timber

$\backslash$table$[[$Tree species,Allowable Stresses,.$],[$Bending$,$Horizontal shear,Tension parallel to grain,Compression parallel to grain,Density,Young's modulus$],[($MPa$),($MPa$),($MPa$),($MPa$),p_{(k\frac{g}{m^3})},$E $($GPa$)],[$Southern yellow pine,$9\mathrm{.}7,0\mathrm{.}931\mathrm{.}2,1\mathrm{.}9,5\mathrm{.}0,350,7\mathrm{.}6],[$Hem$-$fir $,,6\mathrm{.}9,1\mathrm{.}2,5\mathrm{.}7,8\mathrm{.}6,550,11\mathrm{.}7],[$Douglas fir,$6\mathrm{.}2,1\mathrm{.}2,4\mathrm{.}3,9\mathrm{.}0,400,9\mathrm{.}7],[$Redwood$,6\mathrm{.}9,1\mathrm{.}1,4\mathrm{.}3,7\mathrm{.}27\mathrm{.}2,480,11\mathrm{.}7],[,,1.,4\mathrm{.}1,7\mathrm{.}2,400,9\mathrm{.}0]]$

Uniform distributed load along the entire length of a simply$-$supported beam

$R_A=R_B=V_{\text{m}\text{a}\text{r}}=\frac{wL}{2},V_x=w(\frac{L}{2}-x)$

$M_{\text{m}\text{a}\text{x}}=\frac{w{L}^2}{8}at$ the midspan $M_x=\frac{wx(L-x)}{2}$

${}_m==\frac{5w{L}^4}{384EI}at$ the midspan ${}_x=\frac{wx(L^3-2L{x}^2+x^3)}{24EI}$

${}_A={}_B=\frac{w{L}^3}{24EI},$