AISC LRFD Beam Design in the RAM Structural System

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Chester Baldwin
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1 Model: Verification11_3 Typical Floor Beam #10 W21x44 (10,3,10) AISC LRFD Beam Design in the RAM Structural System Floor Loads: Slab Self-weight: Concrete above flute + concrete in flute + metal deck [(3.25in)(1ft/12in) + (18.0in 2 /ft)(1ft 2 /12in 2 )](115 lb/ft 3 ) +3.0 psf = 48.52psf Imposed Dead Load: 25psf Imposed Live Load: 80psf Office Reducible Point Loads on Girder: Secondary beam size = W14x22 Self-wt = 22 lb/ft Secondary beam spacing = 9 o.c. Secondary beam length = 30 P Dead = [(48.52 psf + 25 psf)(9 )(30 /2) + (22 lb/ft)(30 /2)] (1 k / 1000 lb) = k P Live = (80.0 psf)(9 )(30 /2) (1 k / 1000 lb) = k Live Load Reduction (IBC 2006 Eq ): A T = 2(15 )(18 ) = 540 ft2 K LL = 2 L = L 0 ( /sqrt(2(540))) = % Reduction = 100% ( ) = 29.35% P PrecompDead = [( psf)(9 )(30 /2) + (22 lb/ft)(30 /2)] (1 k/1000 lb) = k P PrecompLive = (20.0 psf)(9 )(30 /2) (1 k / 1000 lb) = 2.70 k Line Load on Girder: w Dead = k/ft (W21x44) Shear: V Dead = (2)( k) + (0.044 k/ft)(27 )/2 = k V Live = (2)(10.80 k)(0.7064) = k Vu = 1.4( k) = k, or = 1.2( k) + 1.6( k) = k Controls Vn = 0.6FyA w C v Eq. G2-1 A w = (20.7 )(0.350 ) = in 2 C v = 1.0 φ v = 1.00 φ v Vn = 1.00(0.6)(50 ksi)( in 2 )(1.0) = k Effective Flange Width: (Section I3.1) Span = 27 Distance to adjacent beams = 9.0 BBeffective = (27 )(12 /1 )/8 = 40.5 each side - Controls BBeffective = (9.0 )(12 /1 )/2 = 54.0 each side BBeffective = 2(40.5 ) = 81.0 RAM Structural System 1 Bentley Systems, Inc.

2 Stud Capacity: (Section I3.2d(3)) Q = 0.5A f ' E R R A F (Eq I3-3) n sc c c g p sc u A sc = π (0.75) 2 /4 = in 2 f c = 3.0 ksi E c = sqrt(3.0) = ksi R g = 1.0 (deck is parallel, wr/hr = 6 /3 = 2.0 > 1.5) R p = 0.75 F u = 65.0 ksi Qn = 0.5(0.4418)sqrt((3.0)( )) = k (1.0)(0.75)(0.4418)(65.0) = k Horizontal Shear: (I3.2d(1)) V = 0.85f c Ac (Eq. I3-1a) Ac = (t c )( B effective ) = (3.25 )(81.0 ) = in 2 V = 0.85(3.0)( ) = k, or V = AsFy (Eq. I3-1b) V = (13.0 in 2 )(50 ksi) = k - Controls Maximum Composite Moment: M Dead = (2)(10.255k)(9 ) + (0.044 k/ft)(27 ) 2 /8 = k-ft M Live = (2)(10.8k)(9 )(0.7064) = k-ft Mu = 1.4( k-ft) = k-ft, or Mu = 1.2( k-ft) + 1.6( k-ft) = midspan = 13.5 Between the point of zero moment (the beam end for this beam) and the point of maximum moment (the midspan for this beam) there are 11 studs available (the stud at the midspan must be ignored since there is no horizontal shear at that point that stud is only there due to the maximum stud spacing requirements). Therefore, the shear carried by the studs is: V = (11 studs)(17.68 k/stud) = kip (Eq. I3-1c) This also represents the compression in the concrete, C. Composite φmn at Maximum Moment: For this beam the plastic neutral axis is in the web: PNA (This form of the equation for calculating PNA considers the contribution of the area of steel within the fillets. It is calculated by setting the compression in the steel, which is onehalf of the total force in the steel and concrete, equal to Fy times one-half the steel area plus the additional area of web between the mid-depth and the PNA, then solving for PNA): 0.5(AsFy +C) = Fy[As/2 + tw(pna - Depth/2)] 0.5(650.0 k k) = (50 ksi)[13.0 in 2 /2 + (0.350 )(PNA 20.7 /2)] PNA = ΣMoments about PNA: Mn = Fy [(tf)(bf)(pna-tf/2) bottom flange + (tw)(pna-tf)(pna-tf)/2 web below PNA + (tw)(d-tf-pna)(d-tf-pna)/2 web above PNA + (tf)(bf)(d-pna-tf/2) top flange + (A fillet )(arm fillet )(2pair)] fillets + C (d-pna+td+tc-a/2) concrete RAM Structural System 2 Bentley Systems, Inc.

3 where: A 4fillets = 13.0in 2 (0.350 )(20.7 2(0.450 )) 2(0.450 )(6.50 ) = in 2 A fillet = (0.220 in 2 ) / 4 = in 2 / fillet r fillet = fillet radius = k tf = = 0.5 arm fillet = moment arm between centroids of fillets = d 2(tf) 2(0.2)(r fillet ) = (0.450 ) 2(0.2)(0.5 ) = a = C / (0.85f cb effective ) = (194.5k) / [(0.85)(3ksi)(81.0 )] = Mn = 50 [ (0.450 )(6.50 )( /2) bottom flange + (0.350 )( )( )/2 web below PNA + (0.350 )( )( )/2 web above PNA + (0.450 )(6.50 )( /2) top flange + (0.055 in 2 )(19.60 )(2) ] fillets + (194.5 k )( /2) concrete = k-in = k-ft (the report shows k-ft, the difference is due to roundoff here) φmn = 0.9( k-ft ) = k-ft Composite Moment at Location of Point Loads: The capacity and moments at the location of Point loads must also be investigated. Note that although the program checked this condition it is not listed on the report because it is not the controlling design condition. If it had been the controlling design condition it would have been shown on the reports. M Dead = (2)(10.255k)(9 ) + (0.044 k/ft)(9 )(27-9 )/2 = k-ft M Live = (2)(10.8k)(9 )(0.7064) = k-ft Mu = 1.4( k-ft) = k-ft, or Mu = 1.2( k-ft) + 1.6( k-ft) = Between the point of zero moment (the beam end for this beam) and the location of the point load there are 10 studs available. Therefore, the shear carried by the studs is: V = (10 studs)(17.68 k/stud) = kip (Eq. I3-1c) This also represents the compression in the concrete, C, at that point. Composite φmn at Location of Point Loads: Use the same procedure as for the Maximum moment. For this beam the plastic neutral axis is in the web: PNA: 0.5(650.0 k k) = (50ksi)[13.0 in 2 / (PNA 20.7 /2)] PNA = a = (176.8 k) / [(0.85)(3ksi)(81.0 )] = ΣMoments about PNA: Mn = 50 [ (0.450 )(6.50 )( /2) bottom flange + (0.350 )( )( )/2 web below PNA + (0.350 )( )( )/2 web above PNA + (0.450 )(6.50 )( /2) top flange + (0.055 in 2 )(19.60 )(2) ] fillets + (176.8 k) ( /2) concrete = k-in = k-ft RAM Structural System 3 Bentley Systems, Inc.

4 φmn = 0.9(599.7 k-ft) = k-ft % Composite: There is no longer a Specification requirement that composite beams be at least 25% composite. However, the Commentary indicates that this is good practice. See Commentary I Also, the Commentary indicates that Eq C-I3-3 which is used to calculate I eff is invalid if the percent composite is less than 25%. Therefore, the program enforces that lower limit. Based on the degree of composite at the point of maximum moment: % Composite = V Actual / V Full = k / k = 29.92% Although not explicitly indicated in the Specification, a similar check should be performed at the location of point loads. This is good practice, to ensure that there is at least a minimum degree of composite action available at those locations. The program performs this calculation: % Composite = (V Actual / V Full )( M Max / M AtPointLoad ) = (176.8k / 650.0k) (446.04k-ft / 445.5k-ft) = 27.24% Stud Spacing: Minimum Stud Spacing = 4 x Stud Diameter = (4)(0.75 ) = 3 o.c. Maximum Stud Spacing = 8 x Slab Thickness = (8)( ) = 50 o.c., or = 36 o.c. Controls Actual Stud Spacing: In 1 st and 3 rd Segment: Spacing = 9 / 10 Studs = 0.9 / stud = 10.8 per stud. OK In 2 nd Segment: Spacing = 9 / 3 Studs = 3 / stud = 36 per stud. OK Note that the number of studs in the 2 nd Segment was controlled by maximum spacing req ts. Also note that for this beam, since the deck is parallel to the beam, the spacing of 10.8 can be achieved; if the deck was perpendicular (or at an angle) to the beam, the spacing of the deck ribs would have controlled the spacing of the studs and would have impacted the number of rows of studs. The program checks this condition and designs accordingly. Precomposite Design: For this beam there are three unbraced segments, each 9 long. The program checks each segment per Chapter F. For this beam the middle segment controls; only the calculations for that segment are shown. M Dead = (2)(7.555 k)(9 ) + (0.044 k/ft)(27 ) 2 /8 = k-ft M Live = (2)(2.70 k)(9 ) = 48.6 k-ft Initial Dead Load Mu = 1.4( k-ft) = k-ft Precomposite Pos Mu = 1.2( k-ft) + 1.6(48.6 k-ft) = k-ft midspan = 13.5 Web and Flange are compact. L b = 9 = 108 (Deck parallel to beam doesn t brace top flange). Cb = φmn = 281 k-ft See Table 3-10 (p 3-123) of the AISC manual. RAM Structural System 4 Bentley Systems, Inc.

5 Note that the program rigorously checks the requirements given in Chapter F. Rather than reproducing all of those calculations here, the capacity given in Table 3-10 is listed here, corroborating the value given by the program. Ieff: I eff = I + ΣQ / C )( I I ) (C-I3-3) x ( n f tr x n = E/Ec = /[33.0(115) 1.5 sqrt(3000)] = A c = (3.25 )(81.0 ) / = in 2 y c = 3.25 / = y s = 20.7 /2 = Ybar = [(20.23 in 2 )( ) + (13.0 in 2 )( )] / (20.23 in in 2 ) = Itr = 843 in 4 + (13.0 in 2 )( ) 2 + (20.23 in 2 )( ) 2 + [(81 )(3.25 ) 3 /12] / = in 4 Ieff = 843 in 4 + sqrt(194.5 k / k)( in in 4 ) = in 3 Deflections: Initial Load: Δ = (2)(7.555 k)(108 )[(3)(324 ) 2 (4)(108) 2 ] / [(24)(29000)(843 in 4 )] + (5)(.044 k/ft)(1 ft /12 in)(324 ) 4 / [(384)(29000)(843 in 4 )] = = L / Camber: 0.8 (0.768 ) = < 0.75 Do not camber Live Load: Δ = (2)(10.80 k)(0.7064)(108 )[(3)(324 ) 2 (4)(108) 2 ] / [(24)(29000)( in 4 )] = = L / Post Composite Dead Load: Δ = (2)( k 7.555) (108 )[(3)(324 ) 2 (4)(108) 2 ] / [(24)(29000)( in 4 )] = Post Composite Load: Δ = = = L / Net Total Load: Δ = camber = = L / RAM Structural System 5 Bentley Systems, Inc.

6 RAM Structural System 6 Bentley Systems, Inc.

7 RAM Structural System 7 Bentley Systems, Inc. A. Adams 2/22/08

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