Answer:
correct answer is option C)
Step-by-step 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 =
![(5)/(384)(wl^4)/(EI)](https://img.qammunity.org/2020/formulas/engineering/college/fw420efb0ol5e52utkkrdzyoxrldb70b15.png)
=
![(5)/(384)* (192^4)/(6* 29000* 238)](https://img.qammunity.org/2020/formulas/engineering/college/tmkfp6qq0lorsdq5d4mluyvvwvblsrj4bz.png)
= 0.427 in.
deflection due to self weight =
![(wl^4)/(48EI)](https://img.qammunity.org/2020/formulas/engineering/college/kzvsl32e8dw0dvn0j0r120on82bpe8e2ak.png)
=
![(2.5 * 10^(-3)* 192^4)/(48* 29000* 238)](https://img.qammunity.org/2020/formulas/engineering/college/aeuze4qb2wnytg33zarolef62kmrhs4bka.png)
= 0.01025 inch
total deflection = 0.427 in.+ 0.01025 inch
total deflection = 0.4375 in
correct answer is option C)