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Introductory Physics Homework Help
Deflection of Fixed Rotated Beam
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[QUOTE="roldy, post: 6197899, member: 88105"] [B]Homework Statement:[/B] 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 [B]Relevant Equations:[/B] V=A*L A=pi*D^2/4 m=rho*V W=m*g I=1/12 * b*h^3 deflection = wL^3/(24EI) [U]W14 x 30:[/U] d = 13.84" = 0.352 m w = 6.730" = 0.171 m t[SUB]w[/SUB] = 0.270 = 0.00686 m t[SUB]f[/SUB] = 0.385 = 0.00978 m 40' = 12.192 m 18" = 0.4572 m ρ = 997 kg/m[SUP]3[/SUP] E = 1.4*10[SUP]9[/SUP] [U]Calculate Weight of water:[/U] V = A*L = π*(0.4572)[SUP]2[/SUP]/4 *12.192 = 2.0016m[SUP]3[/SUP] W = m*g = ρ*V*g = 997*2.0016*9.81 = 19.576kN uniformally distributed load [U]Calculate moment of inertia of beam:[/U] I = 1/12*b*h[SUP]3[/SUP] = I[SUB]flanges[/SUB] + I[SUB]inside[/SUB] = 2*[1/12*0.00978*0.352[SUP]3[/SUP]] + [1/12*(0.352-0.00978)*0.00686[SUP]3[/SUP]] = 7.1099*10[SUP]-5[/SUP] m[SUP]4[/SUP] [U]Calculate deflection of beam:[/U] δ = wL[SUP]3[/SUP]/(24EI) = 19.576*12.192[SUP]3[/SUP]/(24*1.4*10[SUP]9[/SUP]*7.1099*10[SUP]-5[/SUP]) = 0.149 m Is my process correct? I've never done a rotated I-beam before. [/QUOTE]
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Deflection of Fixed Rotated Beam
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