Initials

$2.$ Consider the straight $"$T$-$Beam" with dimensions in mm$,$ as shown. If the allowable stress is ${}_a=400$MPa, use a safety factor of $sf=2$ to compute the maximum allowable load $F_{?}$ for each case below: Case $($a$)$: $F_2=0,F_1=F$

$($:$=\frac{M_c}{T}+\frac{B}{A}\}$

$=\frac{40F}{?},F$ OK

$100=$

$q,$

$I_x=\frac{1}{2}$

set $\frac{a}{si}=1$

Rk$!$

A

$20$

$,3$

$q,$

Case $($b$)$: $F_1=0,F_2=F$

Case $($c$)$: $F_1=F_2=F$