15 Flexure Section Modulus

Fl e x u r e Fo Fo r m u l a   Defi fine nes s be b end ndin ing g stress

Flexure Formula (derived for rectangular beam) 1 Beam with marked area 2 Bent beam deforms area into rhomboid 3 Stress block due to bending Flexure formula derivation: f y /y = f/c  f y= y f/c (similar triangles) The internal resiting moment is the sum of forces F times lever arm y about NA, the Neutral Axis M=Fy  F = a f y F = a y f/c M = y a y f/c  M =  a y2 f/c f = Mc /  a y2  f = Mc / I I =  a y2 I = Moment of Inertia  Calculus defines the area a as differential area da and the sum sign  as integration sign  I = y 2 da I = Moment of Inertia  Flexure Formula  f = Mc / I f = bending stress at any distance c from NA. For maximum stress the flexure formula simplifies to f=M/S S = I / c = Section Modulus 

Section Modulus (for rectangular beam) 1 Stress block of partial beam C = T = b (d/2) (f/2) Stress block centroids are d/3 from NA Lever arm between C and T is 2/3 d Internal resisting moment: M = C 2/3 d = T 2/3 d Substituting b (d/2) (f/2) for C and T, yields M = 2/3 d b (d/2) (f/2) M = 2/3 f bd2/4  M = f bd2/6 Solving for f (maximum stress) f = M / bd2/6  f = M/ S where S = bd 2/6 (Section Modulus)  Comparing a 2”x12” joist upright and flat: 2 S=2 (12)2/6 S = 48 in3 3 S=12 (2)2/6 S = 8 in3 The upright joist is 6 times stronger !

Moment of Inertia 1 Stress block 2 Moment of Inertia as parabolic volume 3 T-beam with asymmetric stress block 4 L-bar stress blocks about X, Y, and Z-axis The Moment of Inertia formula I = y 2da reveals, resistance of areas da increases quadratic with the distance from NA (parabolic distribution). The Moment of Inertia parabolic volume is 1/3 the volume of a cube of equal dimensions: I = 1/3 bd (d/2)2  I = 1/3 bd3 / 4 I = bd 3/12 I = Moment of Inertia  for rectangular beams only From previous derivation, the flexure formula f=Mc/I defines stress at any distance c from NA (needed for  asymmetrical shapes, such as T or L-shapes).

Moment of Inertia Effect of shapes 1 Upright joist: 2”x12”, I = 2 (12)3/12 I = 288 in4 2 Flat joist: 12”x2”,

I = 12 (2)3/12

I = 8 in4

3 Wide flange beam: effective (flanges far from NA) 4 Cross beam: ineffective (cross bar at NA) Note: • Beams at right deform more than beams at left • Material at NA is least effective (short lever arm) • Moment of Inertia defines strength and stiffness

 Area Method review • Shear at any point is: V =  load area left of the point • Bending at any point is: M =  shear area left of the point • Maximum bending occurs where shear goes through zero • Negative bending causes convex deflection • Positive bending causes concave deflection • Inflection point (0 bending) coincides with change of deflection curvature

Design Defines beam size for actual loads and allowable stress of selected material Analysis Checks if a given beam satisfies allowable stress of the actual material  Assume: Wood  Allowable bending stress  Allowable shear stress (parallel to fiber) Steel Yield strength  Allowable bending stress (0.6 Fy)  Allowable shear stress (0.4 Fy) Note: F = allowable stress f = actual stress

Fb = 1200 psi Fv = 95 psi Fy = 50 ksi Fb = 30 ksi Fv = 20 ksi

Beam analysis  Mc=0 =16 Rb-1000(20)-300(4)18-200(16)8 Rb=(20000+21600+25600)/16 Rb = 4200 lb  Mb=0 =-16Rc-1000(4)-300(4)2+200(16)8 Rc=(-4000-2400+25600)/16 Rc = 1200 lb Shear  Var  = 0 - 1000 Var = -1000 lb Vbl = -1000 - 300 (4) Vbl =-2200 lb Vbr  = -2200 + 4200 Vbr = +2000 lb Vcl = +2000 - 200(16) = - Rc Vcl = -1200 lb Find x (V = 0  Mmax) Vbr - w X = 0; X = Vbr / w = 2000 / 200 X = 10 ft Bending moment Mb = 4(-1000-2200)/2 Mb= -6400 lb’ Mx = -6400+10 (2000)/2 Mx= +3600 lb’ Section modulus S=bd2/6 =(3.5)11.252/6 S = 74 in3 Bending stress f b=M/S= 6400(12)/74 fb=1038psi<1200   Shear stress f v=1.5V/(bd)=1.5(2200)/[3.5(11.25)] f v= 84 psi < 95 →

1

2 30 k 

-36k 30 k  360 k’

Steel beam design 1 Actual Beam 2 Beam diagram - ignore load at supports (has no effect on beam but on columns)  Assume: L = 36’, P = 30 k, Fb = 30 ksi, Fv = 20 ksi Reactions R = 2P/2 = 2 (30)/2 R = 30 k Shear  Var = Vbl = R Var  = Vbl = 30 k Vbr  = Vcl = 30 – 30 Vbr = Vbr = 0 Vcr  = Vdl = 0 -30 Vcr  = Vdl = -30 k Vdr  = -30 + 30 Vdr  = 0 Bending moment Mb = Mc = 30 (12) Mb = 360 k’ Section modulus required S = M/Fb = 360 k’(12”)/ 30 ksi S = 144 in3 Use W18x75 S = 146>144 Shear stress f v = V/(d tw) = 30k/(18.21x0.425) f v = 3.88 ksi<<20 Note: steel beam shear stress is rarely critical

1

2

Concrete beam analysis 1 Actual beam 2 Beam diagram - ignore load at supports (has no effect on beam but on columns)  Assume: Span 30’, point load P = 20k DL = 2’x1.33’ x 150 pcf /1000 w = 0.4 klf Reaction R = (2P+w L)/2 = (2x20+0.4x30)/2 R = 26 k Shear  Var  = R Var  = 26 k Vbl = 26 - 0.4 (10) Vbl = 22 k Vbr  = 22 - 20 Vbr = 2 k Vcl = 2 - 0.4 (10) Vcl = -2 k Vcr  = -2 - 20 Vcr  =-22 k Vdl = -22 - 0.4 (10) Vdl =-26 k Bending moment Mb = 10 (26+22)/2 Mb = 240 k’ Mmax = Mb + 2(5)/2 Mmax = 245 k’ Note: concrete stress will be covered in Arch 313

Fl e x u r e Fo r m u l a   Don’t over stress

or else