Refer to page 81 of the FE Handbook: "Simply Supported Beam Deflections and Slopes:"

Middle column for deflection:

Vmax = 5wL^4/ 384EI

Givens:

L = 15 ft, so L^4 = (15ft)^4 = 50,625 ft^4

w = 15,000 lb/ft

I = 254 ft4

E = The problem statement says a steel beam, therefore, Refer to page 80 of the FE Handbook to find the value of E for steel:

"Typical Material Properties:"

For steel, E = 29 Mpsi, or 29,000,000 psi

Therefore:

Vmax = 5wL^4/ 384EI

Vmax = [5(15,000 lb/ft) (50,625 ft4)] / [384(29 Mpsi) (254 ft^4)]

Vmax = [3,796,875,000 lb ft3)] / [2.8 x 1012 (lb/in2) ft4)] • (144 in2/ft2) (note: watch your units! we need to convert psi into psf)

Vmax = [3,796,875,000 lb ft3)] / [4 x 1014 (lb ft2)]

Vmax = 9.3 x 10-6 ft (note: the problem statement wants the answer to be in inches, not feet)

Vmax = 9.3 x 10-6 ft (12 in / 1 ft)

Vmax = .00001 in

The answer is .00001 in.