Answer:
correct answer is option C)
Explanation:
W 12 x 30 section of the beam dimension
Value of E = 29000 ks i/in²
I = 238 in⁴
weight of beams per length = 30 lbs/ft = 30 × 10⁻³ ks i/ft = 2.5 × 10⁻³ ks i/in.
distribution load = 2 k/ft = 1/6 k/in
L = 16 × 12 = 192 inch.
deflection due to distributed load = [tex]\dfrac{5}{384}\dfrac{wl^4}{EI}[/tex]
= [tex]\dfrac{5}{384}\times \dfrac{192^4}{6\times 29000\times 238}[/tex]
= 0.427 in.
deflection due to self weight = [tex]\dfrac{wl^4}{48EI}[/tex]
= [tex]\dfrac{2.5 \times 10^{-3}\times 192^4}{48\times 29000\times 238}[/tex]
= 0.01025 inch
total deflection = 0.427 in.+ 0.01025 inch
total deflection = 0.4375 in
correct answer is option C)
