# Deflection of Fixed Rotated Beam

## Homework Statement:

What is the deflection of a fixed 40' beam that is rotated 90 degrees so that the flanges are vertical with a loading of a 40' long 18" diameter pipe filled with water? The beam is W14 x 30

## Relevant Equations:

V=A*L
A=pi*D^2/4
m=rho*V
W=m*g
I=1/12 * b*h^3
deflection = wL^3/(24EI)
W14 x 30:
d = 13.84" = 0.352 m
w = 6.730" = 0.171 m
tw = 0.270 = 0.00686 m
tf = 0.385 = 0.00978 m

40' = 12.192 m
18" = 0.4572 m
ρ = 997 kg/m3
E = 1.4*109

Calculate Weight of water:
V = A*L = π*(0.4572)2/4 *12.192 = 2.0016m3
W = m*g = ρ*V*g = 997*2.0016*9.81 = 19.576kN uniformally distributed load

Calculate moment of inertia of beam:
I = 1/12*b*h3 = Iflanges + Iinside = 2*[1/12*0.00978*0.3523] + [1/12*(0.352-0.00978)*0.006863] = 7.1099*10-5 m4

Calculate deflection of beam:
δ = wL3/(24EI) = 19.576*12.1923/(24*1.4*109*7.1099*10-5) = 0.149 m

Is my process correct? I've never done a rotated I-beam before.

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haruspex
Homework Helper
Gold Member
Problem Statement: What is the deflection of a fixed 40' beam that is rotated 90 degrees so that the flanges are vertical with a loading of a 40' long 18" diameter pipe filled with water? The beam is W14 x 30
Relevant Equations: V=A*L
A=pi*D^2/4
m=rho*V
W=m*g
I=1/12 * b*h^3
deflection = wL^3/(24EI)

W14 x 30:
d = 13.84" = 0.352 m
w = 6.730" = 0.171 m
tw = 0.270 = 0.00686 m
tf = 0.385 = 0.00978 m

40' = 12.192 m
18" = 0.4572 m
ρ = 997 kg/m3
E = 1.4*109

Calculate Weight of water:
V = A*L = π*(0.4572)2/4 *12.192 = 2.0016m3
W = m*g = ρ*V*g = 997*2.0016*9.81 = 19.576kN uniformally distributed load

Calculate moment of inertia of beam:
I = 1/12*b*h3 = Iflanges + Iinside = 2*[1/12*0.00978*0.3523] + [1/12*(0.352-0.00978)*0.006863] = 7.1099*10-5 m4

Calculate deflection of beam:
δ = wL3/(24EI) = 19.576*12.1923/(24*1.4*109*7.1099*10-5) = 0.149 m

Is my process correct? I've never done a rotated I-beam before.
I am not going to check all the numbers, but the method looks ok except for one thing I do not understand. What is the Iinside term? If I understand the set-up, it is an H cross-section, with the uprights covered by the Iflanges term. If the Iinside is for the horizontal part then where does (0.352-0.00978) come from?

That value is the "length" of the horizontal part. Depth of the beam subtract the flange widths. I made a mistake, the value should actually be 0.352 - 2×0.00978.

haruspex
Homework Helper
Gold Member
That value is the "length" of the horizontal part. Depth of the beam subtract the flange widths. I made a mistake, the value should actually be 0.352 - 2×0.00978.
Ok. I think I would just have assumed the contribution from that was negligible (which it is).
The given data are to only two significant figures, so I would give the answer to the same precision.

PhanthomJay