Deflection of Fixed Rotated Beam

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Homework Help Overview

The discussion revolves around calculating the deflection of a fixed beam that has been rotated 90 degrees, specifically a W14 x 30 I-beam supporting a uniformly distributed load from a pipe filled with water. The participants are examining the relevant equations and parameters involved in the calculation, including the moment of inertia and deflection formulas.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the weight of water, the moment of inertia of the beam, and the resulting deflection. Questions arise regarding the interpretation of the moment of inertia terms, particularly the inclusion of the Iinside term and its calculation.

Discussion Status

Some participants have provided clarifications on the calculations and assumptions, particularly regarding the dimensions used for the moment of inertia. There is an acknowledgment of potential mistakes in the initial setup, and discussions are ongoing about the appropriateness of unit conversions and precision in the final answer.

Contextual Notes

There is a mention of the original poster's choice to use SI units despite the context of USA standard beams, which has led to some confusion. Additionally, the precision of the given data is noted as being limited to two significant figures.

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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.
 
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roldy said:
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.
 
roldy said:
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.
 
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.