Question

Q3) A simply supported beam is subjected to a uniform service dead load of 2.3 kips/ft (excluding the weight of the beam), a uniform service live load of 3.0 kips/ft. The beam is 30 feet long, and deflection not to exceed L/360. The beam has continuous lateral support, and A992 steel is used. Is a W27 x 84 adequate?

Answer 1

Answer) Given Dead Load (WD)= 2.3k/ft

Live load (LL) = 3.0 k/ft

Given beam is W27x84 (Ix=2850 in4 )

self weight of given beam = 84lb/ft = 0.084k/ft

Factored load Wu= 1.2 x Wd + 1.6 x WL

Wu=1.2 x (2.3 +0.084) + 1.6 x (3)

Wu=7.66K/ft

Permissible deflection $\Delta _{p}$ =L/360

$\Delta _{p}$= (30 x 12)/360

$\Delta _{p}$=1in

The Given beam Adequacy can be determined by 2 ways

1) $\Delta _{c}= \frac{5Wul^{4}}{384EI_{x}}$

Where Wu is a factored Load Wu= 7.66k/ft = 7660/12 =638.33lb/in

L=span of the beam = 12 x 30 = 360 in

E = Modulus of elasticity = 29 x 106 lb/in2

Now  $\Delta _{c}= \frac{5*638.33*360^{4}}{384*29 * 10^{6}*2850}$

$\Delta _{c}=$1.69 > $\Delta _{p}$ Not Adequate

2) $\Delta _{c}= \frac{M *L^{2} }{C_{1}*I_{x} }$

Where M = Factored moment in K-ft= Wul2/8

M= (7.66 x 302)/8

M= 861.75 K-ft

L= Span of the Beam in ft =30ft

C1= constant = 161 (for simply supported uniform loading)

Ix= Moment of inertia w.r.t x axis in in4 = 2850 in4

Now  $\Delta _{c}= \frac{861.75 *30^{2} }{161*2850 }$

$\Delta _{c}=$ 1.69> $\Delta _{p}$ Not Adequate