Answer:
w = 10.437 kips
deflection at 1/4 span 20.83\E ft
at mid span = 1.23\E ft
shear stress 7.3629 psi
Explanation:
area of cross section = 18*76
length of span = 32 ft
moment = 334 kips-ft
we know that
moment = load *eccentricity
334 = w * 32
w = 10.437 kips
deflection at 1/4 span
[tex]\delta = \frac{wa^2b^2}{3EI}[/tex]
[tex]= \frac{10.4375*8^2 *24^2}{3E \frac{BD^3}{12}}[/tex]
[tex]=\frac{10.437 *8^2*24^2}{3E \frac{18*16^3}{12}}[/tex]
= 20.83\E ft
at mid span
[tex]\delta = \frac{wl^3}{48EI}[/tex]
[tex]= \frac{10.43 *32^3}{48 *E*\frac{18*16^3}{12}}[/tex]
[tex]\delta = 1.23\E ft[/tex]
shear stress
[tex]\tau = \frac{w}{A} = \frac{10.43 7*10^3}{18*76} =7.3629 psi[/tex]
