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Deflection of Fixed Rotated Beam

  • Thread starter roldy
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  • #1
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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.
 

Answers and Replies

  • #2
haruspex
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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?
 
  • #3
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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.
 
  • #4
haruspex
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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.
 
  • #5
PhanthomJay
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I have no idea why you changed to SI units. No need to when you are dealing with the USA standard beams. Otherwise, the depth of a beam is measured between the far edges of each flange. Your beam is rotated along the weak axis, so check your values for appropriate dimensions when calculating the moment of inertia.
 

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