Question

An A992 beam, simply supported at both ends, spans 20 ft and is loaded at mid-span with a dead load of 8.o kips and live load of 24.0 kips, in addition to its self-weight. Assume full lateral support and a compact section. Select the lightest weight wide-flange members with respect to bending capacity. Does the member selected in part (a), satisfy the live load deflection criteria of L/360.? Does the member selected in part (a), satisfy the live load deflection criteria of L/1000? If not, select the lightest wide flange that does meet the live load deflection criteria of L/1000. Design the bearing plate for the concentrated load and the support for the members selected in part (c) considering web crippling and web local yielding. a. b. c. d.

Answer 1

Given:

Span=20 ft

D.L=8 kips, L.L=24 kips, self weight=not given

(a)

Step 1:

P_u=1.2(8)+1.6(24)

P_u=48 kips

M=48x20/4

M=240 ft-kips

Step 2:

Calculate the section modulus required

Zx=M/(0.9Fy)

Zx=240x12/(0.9x50)

Zx=64 in³

As the self weight is neglected take any flange which will have section modulus greater than 64 in³ and for self weight assume a slightly higher W section

Step 3:

Consider a W18X35 section Zx=66.5 in³

Now verify it for self weight

D.L=35/1000=0.035 kip/ft

P_u=1.2(8.035)+1.6(24)=48.042 ft-kips

M=48.042x20/4=240.21 ft-kips

Zx=240.21x12/(0.9x50)

Zx=64.056 in³ is required section modulus W18X35is safe

(b)

L.L deflection permissilbe=L/360=20x12/360=0.66 in

L.L actual=WL³/48EI

=24x(20x12)³/(48x29000x510)

=0.47 in

Actual is less than permssible so W18X35 satisfies it

(c)

Actual deflection=0.47 in

Permissible=24x12/1000=0.29 in, modify this by selecting a W section having greater moment of inertia

Assume permissilbe=actual and find MOI

L.L deflection=WL³/48EI

0.29=24x(240)³/48x29000xI_x

I_x=827 in⁴

Provide W21X44 section having I_x=843 in⁴

(d)

Assumptions: Fy=50 ksifor beam, Fy=36 ksi for plate fc=4ksi, 21X44 section

P_u=1.2(0.044+8)+1.6(24)=48.0576

R_u=P_u/2

R_u=24.0288 kips

For web yielding

R_n=(2.5k+N)Fyt_w

24.0288=(2.5(0.95)+N)x50x0.35

N=negative value, negative value is not possible in dimensions

Provide N=6 in (Assume)

For web crippling

We get same negative value for N

So provide N=4 in

From AISC equation

$\phi$ (0.85fcA)=R_u

0.6x0.85x4xA=24.0288

A=11.77

A=BN

11.7=B(4)

B=3.92 in, provide 4 in, but as the flange width of W21x44 section is 6.5 in, provide 8in, B=8 in

n=B-2k/2

n=8-2(0.95)/2

n=3.05

t=(2.2R_un²/BNfy)^0.5

t=0.65 in, provide 1 in

Use PL1x4x8