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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
t_{w} = 0.270 = 0.00686 m
t_{f} = 0.385 = 0.00978 m
40' = 12.192 m
18" = 0.4572 m
ρ = 997 kg/m^{3}
E = 1.4*10^{9}
Calculate Weight of water:
V = A*L = π*(0.4572)^{2}/4 *12.192 = 2.0016m^{3}
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*h^{3} = I_{flanges} + I_{inside} = 2*[1/12*0.00978*0.352^{3}] + [1/12*(0.3520.00978)*0.00686^{3}] = 7.1099*10^{5} m^{4}
Calculate deflection of beam:
δ = wL^{3}/(24EI) = 19.576*12.192^{3}/(24*1.4*10^{9}*7.1099*10^{5}) = 0.149 m
Is my process correct? I've never done a rotated Ibeam before.
d = 13.84" = 0.352 m
w = 6.730" = 0.171 m
t_{w} = 0.270 = 0.00686 m
t_{f} = 0.385 = 0.00978 m
40' = 12.192 m
18" = 0.4572 m
ρ = 997 kg/m^{3}
E = 1.4*10^{9}
Calculate Weight of water:
V = A*L = π*(0.4572)^{2}/4 *12.192 = 2.0016m^{3}
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*h^{3} = I_{flanges} + I_{inside} = 2*[1/12*0.00978*0.352^{3}] + [1/12*(0.3520.00978)*0.00686^{3}] = 7.1099*10^{5} m^{4}
Calculate deflection of beam:
δ = wL^{3}/(24EI) = 19.576*12.192^{3}/(24*1.4*10^{9}*7.1099*10^{5}) = 0.149 m
Is my process correct? I've never done a rotated Ibeam before.