Beam Bending Stresses and Shear Stress

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1 Beam Bending Stresses and Shear Stress Notation: A = name or area Aweb = area o the web o a wide lange section b = width o a rectangle = total width o material at a horizontal section c = largest distance rom the neutral axis to the top or bottom edge o a beam d = calculus symbol or dierentiation = depth o a wide lange section dy = dierence in the y direction between an area centroid ( ) and the centroid o the composite shape ( ) DL = shorthand or dead load E = modulus o elasticity or Young s modulus b = bending stress c = compressive stress max = maximum stress t = tensile stress v = shear stress Fb = allowable bending stress Fconnector = shear orce capacity per connector h = height o a rectangle = moment o inertia with respect to neutral axis bending x = moment o inertia with respect to an x-axis L = name or length LL = shorthand or live load M = internal bending moment = name or a moment vector y ŷ n = number o connectors across a joint n.a. = shorthand or neutral axis (N.A.) O = name or reerence origin p = pitch o connector spacing P = name or a orce vector q = shear per length (shear low) Q = irst moment area about a neutral axis Qconnected = irst moment area about a neutral axis or the connected part R = radius o curvature o a deormed beam S = section modulus Sreq d = section modulus required at allowable stress tw = thickness o web o wide lange = internal shear orce longitudinal = longitudinal shear orce T = transverse shear orce w = name or distributed load x = horizontal distance y = vertical distance y = the distance in the y direction rom a reerence axis (n.a) to the centroid o a shape = the distance in the y direction rom a reerence axis to the centroid o a composite shape = calculus symbol or small quantity = elongation or length change = strain = arc angle = summation symbol ŷ 185

y Maximum Moment and Stress Distribution x n a member o constant cross section, the maximum bending moment will govern the design o the section size when we know what kind o normal stress is caused

2 Pure Bending in Beams With bending moments along the axis o the member only, a beam is said to be in pure bending. Normal stresses due to bending can be ound or homogeneous materials having a plane o symmetry in the y axis that ollow Hooke s law. y Maximum Moment and Stress Distribution x n a member o constant cross section, the maximum bending moment will govern the design o the section size when we know what kind o normal stress is caused by it. For internal equilibrium to be maintained, the bending moment will be equal to the M rom the normal stresses the areas the moment arms. Geometric it helps solve this statically indeterminate problem: 1. The normal planes remain normal or pure bending.. There is no net internal axial orce. 3. Stress varies linearly over cross section. 4. Zero stress exists at the centroid and the line o centroids is the neutral axis (n. a) 186

Relations or Beam Geometry and Stress Pure bending results in a circular arc delection. R is the distance to the center o the arc; is the angle o the arc (radians); c is the distance rom the n.a. to the extreme iber; is a length change; max is the maximum normal stress at the extreme iber; y is a distance in y rom the n.

3 Relations or Beam Geometry and Stress Pure bending results in a circular arc delection. R is the distance to the center o the arc; is the angle o the arc (radians); c is the distance rom the n.a. to the extreme iber; is a length change; max is the maximum normal stress at the extreme iber; y is a distance in y rom the n.a.; M is the bending moment; is the moment o inertia; S is the section modulus. L R M i A i My b L R M max c y i A i max Now: or a rectangle o height h and width b: E y A y c c S ½ c R L y ½ Mc max 3 bh bh S 1 h 6 M S RELATONS: 1 * M R E My b S c b max Mc M S S required M F b *Note: y positive goes DOWN. With a positive M and y to the bottom iber as positive, it results in a TENSON stress (we ve called positive) Transverse Loading in Beams We are aware that transverse beam loadings result in internal shear and bending moments. We designed sections based on bending stresses, since this stress dominates beam behavior. There can be shear stresses horizontally within a beam member. t can be shown that horizontal vertical 187

Equilibrium and Derivation n order or equilibrium or

neutral axis o the area above or below the horizontal

Q is a maximum when y = 0 (at the neutral axis).

length shear low Shearing Stresses q longitudinal x T

Even i Q is a maximum at y = 0, we don t know that

4 Equilibrium and Derivation n order or equilibrium or any element CDD C, there needs to be a horizontal orce H. D da C da Q is a moment area with respect to the neutral axis o the area above or below the horizontal where the H occurs. Q is a maximum when y = 0 (at the neutral axis). longitudinal T Q x q is a horizontal shear per unit length shear low Shearing Stresses q longitudinal x T Q v ave = 0 on the beam s surace. Even i Q is a maximum at y = 0, we don t know that the thickness is a minimum there. v A b x vave Q b Rectangular Sections occurs at the neutral axis: vmax then: 3 bh 1 Q Q 1 8 bh 3 v 3 b 1 bh b bh 1 Ay b h 1 h bh 3 v A 8 188

tension shear across a panel with stieners or the web can buckle.

5 Webs o Beams n steel W or S sections the thickness varies rom the lange to the web. d We neglect the shear stress in the langes and consider the shear stress in the web to be constant: tw vmax 3 A A web vmax t d web Webs o beams can ail in tension shear across a panel with stieners or the web can buckle. Shear Flow Even i the cut we make to ind Q is not horizontal, but arbitrary, we can still ind the shear low, q, as long as the loads on thin-walled sections are applied in a plane o symmetry, and the cut is made perpendicular to the surace o the member. Q q The shear low magnitudes can be sketched by knowing Q. 189

Connectors to Resist Horizontal Shear in Composite Members Typical connections needing to resist shear are plates with nails or rivets or bolts in composite sections or splices.

43 ya 4 1 8 low p p p where p = pitch length longitudinal Q n = number o connectors connecting the connected area to the rest o the cross section F = orce capacity in one connector p nf connector Q

Horizontal Shear in Composite Members Even vertical connectors have shear low across them.

6 Connectors to Resist Horizontal Shear in Composite Members Typical connections needing to resist shear are plates with nails or rivets or bolts in composite sections or splices. The pitch (spacing) can be determined by the capacity in shear o the connector(s) to the shear over the spacing interval, p. x y 4.43 ya low p p p where p = pitch length longitudinal Q n = number o connectors connecting the connected area to the rest o the cross section F = orce capacity in one connector p nf connector Q connected area longitudinal p Q p Qconnected area = Aconnected area yconnected area yconnected area = distance rom the centroid o the connected area to the neutral axis Connectors to Resist Horizontal Shear in Composite Members Even vertical connectors have shear low across them. p p p The spacing can be determined by the capacity in shear o the connector(s) to the shear low over the spacing interval, p. nf p Q connector connected area Unsymmetrical Sections or Shear the section is not symmetric, or has a shear not in that plane, the member can bend and twist. the load is applied at the shear center there will not be twisting. This is the location where the moment caused by shear low = the moment o the shear orce about the shear center. 190

7 Example 1 (pg 37) 191

8 Example * (pg 377), and evaluate the shear stress i F v = 95 psi. Roo: Snow +DL = 00 lb/t Walls: 400 lb on nd loor beams Railing: 100 lb on beam overhang Second Floor: DL + LL = 300 lb/t (including overhang) Roo: *ALSO select the most economical steel section or the second-loor when S req d is 165 in 3 and evaluate the shear stress when = 60 k. Second Floor: 19

9 Example 3 (pg 386) ALSO: Determine the minimum nail spacing required (pitch) i the shear capacity o a nail (F connector) is 50 lb

10 y= 4.5" ARCH 331 Note Set 10.1 Su016abn Example 4 (pg394 Q = Ay = (9")(½")(4.5")+(9")(½")(4.5")+(1.5")(3.5")(8.5") = 83.8 in 3 3 (, 600 #)( 83. 8in. ) 181. psi vmax ( 1, 0. 6in. )( " ") (n) (n) (n) (n)f p (n)f p 194

Beam Stresses Bending and Shear

Beam Stresses Bending and Shear Notation: A = name or area A web = area o the web o a wide lange setion b = width o a retangle = total width o material at a horizontal setion = largest distane rom the