deflection

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17 Deflection Copyright © G G Schierle, 2001-05 press Esc to end, ↓ for next, ↑ for previous slide 1

De l e c t i onf


DeflectionBeam deflection must be limited to:• Δ = L / 360 for LL • Δ = L / 240 for combined LL+DL.

Beam deflection is caused by both bending and shear• Bending deflection governs long beams• Shear deflection governs short beams,

but is very small and usually ignored

1 Simple beam under uniform load2 Large bending deflection of simple beam3 Small sear deflection of simple beam 4 Small bending deflection of short cantilever beam5 Large shear deflection of short cantilever beam


1 Deformed partial beamAssuming GB is parallel to FOThen FG = AB, the unstressed length and GH/AB = ε, the unit strain. Since E=f/ ε → f=E ε → f=E GH/AB.Due to similar triangles GH/AB = c/r → f = Ec/rSince f=Mc/I → Ec/r=Mc/I → E/r=M/I → 1/r = M / EI

Moment-area methodDeveloped 1868 by Otto Mohr in Dresden

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1 Deformed partial beamAssuming GB is parallel to FOThen FG = AB, the unstressed length and GH/AB = ε, the unit strain. Since E=f/ ε → f=E ε → f=E GH/AB.Due to similar triangles GH/AB = c/r → f = Ec/rSince f=Mc/I → Ec/r=Mc/I → E/r=M/I → 1/r = M / EI2 Elastic curve AB of partial beamAssuming angles in radians: Radial angle dφ between m and n and Tnagent angle θ between A and B, yields: dx= r dφ → dφ/dx = 1/r = M / EI → dφ= Mdx / EIθ = Σdφ → θ = ΣMdx/ EI. Hence

θ = Am/ EIAm=ΣMdx = bending diagram area between A and B. Slope theoremThe angle θ between tangents at A and B on theelastic curve of a beam is the moment diagramarea between A & B, divided by EI


Am= ΣMdx = bending diagram area between A and Bx = lever arm from A to centroid of bending diagramxdφ = x dφThusΔ = Σxdφ = ΣMdx/ EI, orΔ = x Am/ EI

Deflection theoremThe displacement Δ of the tangent at B on theelastic curve equals the moment of the bendingdiagram area between A and B times the lever-arm x from its centroid to A, divided by EI.


Deflection formulas(θ = slope of the tangent of the elastic curve and Δ the maximum deflection.1 Cantilever beam with point load

θ = (PL)(L/2)/(EI) Δ = θ 2/3 Lθ = 1/2 PL2/(EI)Δ = 1/3 PL3/(EI)

2 Cantilever beam with uniform loadθ = (WL/2)(L/3)/(EI) Δ = θ 3/4 Lθ = 1/6 WL2/(EI)Δ = 1/8 WL3/(EI)

3 Simple beam with point loadθ = (PL/4)(L/4)/(EI) Δ = θ 1/3 Lθ = 1/16 PL2/(EI)Δ = 1/48 PL3/(EI)

4 Simple beam with uniform loadθ = (WL/8)(2/3 L/2)/(EI) Δ = θ 5/16 Lθ = 1/24 WL2/(EI)Δ = 5/384 WL3/(EI)

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Beam formulasThe formulas include for common beams:• M = bending moment• V = shear• Δ = deflection


Steel beam with point loadBeam supports joist point load (P=8k). Design for stress(Fb=22 ksi, check deflection and redesign if required.1 Beam diagram2 Load diagram (ignore load at supports which has no effect

on shear, bending, or deflection)Shear

Va = Vbl = 2(8)/2 Va=Vbl = 8 kVbr = Vcl = 8-8 Vbr=Vcl = 0 kVcr = Vd = R = 0-8 Vcr=Vd = 8 k

Bending momentMmax = 8(10) Mmax = 80 k’

Required section modulusS = M/Fb = 80 k’(12”)/22 ksi S = 44 in3

Try W10x45, S =49.1 in3` 49.1 > 44, okFrom AISC table I = 248 in4

Deflection (L = 30’(12”) = 360”) Δ = (23/648)PL3/EIΔ = (23/648)8(3603)/[(30000)248] Δ = 1.78”Δall = 360/240 = 1.5” 1.5<1,78, not okTry W18x35, S = 57.6 in3, I = 510 in4

DeflectionΔ = [(23/648)8(3603)]/[(30000)510] Δ = 0.87”

1.5>0.87, ok Note: the W10x45 of 36 span/depth ratio is too shallow;while the W18x35 has the recommended ratio of 20, islighter and more economical.


Steel beam (with mixed load)Steel beam supports 8k joist load + 4 kfl uniform DL (beam, fir proofing & finish), L = 30’(12”) = 360”, Fb = 22 ksi1 Beam diagram2 Load diagram (ignore load at supports)Shear

Va= R= [2(8)+0.4(30)]/2 Va = 14 kVbl= 14-0.4(10) Vbl = 10 kVbr= 10-8 Vbr = 2 kVcl= 2-0.4(10) Vcl= -2 kVcr= -2-8 Vcr= -10 kVd= -10-0.4(10) Vd= -14 k

Bending momentM = 10(14+10)/2 + 5(2)/2 M =125 k’

Required Section modulus SS = M/Fb= 125 k’(12”)/22 ksi S = 68 in3

Try W18x40, S= 68.4 in3 68.4>68, okFrom AISC table I = 612 in4

DeflectionW = wL = 0.4 (30) W = 12 kΔmax = (23/684) PL3/(EI) + (5/384) WL3/(EI) *

* Combined point load and uniform loadΔmax = [(23/684)8(360)3+(5/384)12(360) 3]/[(30000)612]

Δmax=1.1”Δall=360/240 = 1.5 in 1.5>1.1, ok

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Deflection vs. spanThe formulas• V = w L• M = w L2/8• Δ = (5/384) wL4/EI reveal:V increases linearly with LM increases quadratic with LΔ increases to the 4th power with L

If L doubles Δ increase 16 times !1 Beam with Δ= 1 2 Double span with Δ= 163 Short span: shear V governs4 Medium span: bending M governs5 Long span: deflection Δ governs


Deflection Δ vs. I (moment of inertia)

641/644 Four boards glued

81/83 Twin board glued

21/22 Twin board

111 Single board

IΔType of beam

deflection

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