Contents AISC 360 10 Example 001
User Manual: AISC-360-10 Example 001
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Software Verification PROGRAM NAME: REVISION NO.: ETABS 3 AISC-360-10 Example 001 COMPOSITE GIRDER DESIGN EXAMPLE DESCRIPTION A typical bay of a composite floor system is illustrated below. Select an appropriate ASTM A992 W-shaped beam and determine the required number of ¾ in.-diameter steel headed stud anchors. The beam will not be shored during construction. To achieve a two-hour fire rating without the application of spray applied fire protection material to the composite deck, 4 ½ in. of normal weight (145 lb/ft3) concrete will be placed above the top of the deck. The concrete has a specified compressive strength, fc′ = 4 ksi. GEOMETRY, PROPERTIES AND LOADING Member Properties W21x50 E = 29000 ksi Fy = 50 ksi Loading w = 800 plf (Dead Load) w = 250 plf (Construction) w = 100 plf (SDL) w = 1000 plf (Live Load) Geometry Span, L = 45 ft AISC-360-10 Example 001 - 1 Software Verification PROGRAM NAME: REVISION NO.: ETABS 3 TECHNICAL FEATURES OF ETABS TESTED Composite beam design, including: Selection of steel section, camber and shear stud distribution Member bending capacities, at construction and in service Member deflections, at construction and in service RESULTS COMPARISON Independent results are referenced from Example I.1 from the AISC Design Examples, Version 14.0. ETABS Independent Percent Difference Pre-composite Mu (k-ft) 344.2 344.2 0.00% Pre-composite ΦbMn (k-ft) 412.5 412.5 0.00% 2.6 2.6 0.00% Required Strength Mu (k-ft) 678.3 678.4 0.01% Full Composite ΦbMn (k-ft) 937.1 937.1 0.00% Partial Composite ΦbMn (k-ft) 763.2 763.2 0.00% Shear Stud Capacity Qn 17.2; 14.6 17.2; 14.6 0.00% Shear Stud Distribution 46 46 0.00% Live Load Deflection (in.) 1.34 1.26 6.0% Required Strength Vu (kip) 60.3 60.3 0.00% ΦVn (k) 237.1 237.1 0.00% Output Parameter Pre-composite Deflection (in.) AISC-360-10 Example 001 - 2 Software Verification PROGRAM NAME: REVISION NO.: ETABS 3 COMPUTER FILE: AISC-360-10 EXAMPLE 001.EDB CONCLUSION The ETABS results show an acceptable comparison with the independent results. The live load deflection differs due to a difference in methodology. In the AISC example, the live load deflection is computed based on a lower bound value of the beam moment of inertia, whereas in ETABS, it is computed based on the approximate value of the beam moment of inertia derived from Equation (C-I3-6) from the Commentary on the AISC Load and Resistance Factor Design Specification – Second Edition. AISC-360-10 Example 001 - 3 Software Verification PROGRAM NAME: REVISION NO.: ETABS 3 HAND CALCULATION Properties: Materials: ASTM A572 Grade 50 Steel E = 29,000 ksi, Fy = 50 ksi, wsteel = 490 pcf 4000 psi normal weight concrete Ec = 3,644 ksi, f c′ = 4 ksi, wconcrete = 145 pcf Section: W21x50 d = 20.8 in, bf = 6.53 in, tf = 0.535 in, tw = 0.38 in, k = 1.04 in Asteel = 14.7 in2, Ssteel = 94.6 in3, Zsteel = 110 in3, Isteel = 984 in4 Deck: tc =4 ½ in., hr = 3 in., sr =12 in., wr = 6 in. Shear Connectors: d = ¾ in, h =4 ½ in, Fu = 65 ksi Design for Pre-Composite Condition: Construction Required Flexural Strength: wD = (10 • 75 + 50) • 10−3 = 0.800 kip/ft wL = 10 • 25 • 10−3 = 0.250 kip/ft wu = 1.2 • 0.800 + 1.6 • 0.250 = 1.36 kip/ft = Mu wu • L2 1.36 • 452 = = 344.25 kip-ft 8 8 Moment Capacity: Φ b M n =Φ b • Z s • Fy =(0.9 • 110 • 50) 12 =412.5 kip-ft AISC-360-10 Example 001 - 4 Software Verification PROGRAM NAME: REVISION NO.: ETABS 3 Pre-Composite Deflection: 5wD L4 = ∆ nc = 384 EI 0.800 4 • ( 45 • 12 ) 12 = 2.59 in. 384 • 29, 000 • 984 5• Camber = 0.8 • ∆ nc = 0.8 • 2.59 = 2.07 in., which is rounded down to 2 in. Design for Composite Flexural Strength: Required Flexural Strength: wu = 1.2 • 0.800 + 1.2 • 0.100 + 1.6 • 1 = 2.68 kip/ft wu • L2 2.68 • 452 = Mu = = 678.38 kip-ft 8 8 Full Composite Action Available Flexural Strength: Effective width of slab: 10.0 45.0 ft beff = • 2 sides =10.0 ft ≤ =11.25 ft 2 8 Resistance of steel in tension: C = Py = As • Fy = 14.7 • 50 = 735 kips controls Resistance of slab in compression: Ac = beff • tc = (10 • 12 ) • 4.5 = 540 in 2 C= 0.85 • f 'c A= 0.85 • 4 • 540 = 1836 kips c Depth of compression block within slab: a = C 735 = = 1.80 in. 0.85 • beff • f 'c 0.85 • (10 • 12 ) • 4 Moment resistance of composite beam for full composite action: a 1.80 d1 = ( tc + hr ) − = ( 4.5 + 3) − = 6.60 in. 2 2 d 20.8 /12 ΦM n = Φ Py • d1 + Py • = 0.9 735 • 6.60 /12 + 735 • 937.1 kip-ft = 2 2 AISC-360-10 Example 001 - 5 Software Verification PROGRAM NAME: REVISION NO.: ETABS 3 Partial Composite Action Available Flexural Strength: Assume 50.9% composite action: C= 0.509 • P= 373.9 kips y Depth of compression block within concrete slab: 373.9 C = = 0.92 in. 0.85 • beff • f 'c 0.85 • (10 • 12 ) • 4 a = d = 1 (tc + hr ) − a2 = ( 4.5 + 3) − 0.92 = 7.04 in. 2 Compressive force in steel section: Py − C 735 − 373.9 = = 180.6 kips 2 2 Steel section flange ultimate compressive force: C flange = b f • t f • Fy = 6.53 • 0.535 • 50 = 174.7 kips Steel section web (excluding fillet areas) ultimate compressive force: Cweb = ( d − 2 • k ) • tw • Fy = (20.8 − 2 • 1.04) • 0.38 • 50 = 355.7 kips Steel section fillet ultimate compressive force: = C fillet Py − (2 • C flange + Cweb ) 735 − (2 • 174.7 + 355.7) = = 14.5 kips 2 2 Assuming a rectangular fillet area, the distance from the bottom of the top flange to the neutral axis of the composite section is: ( P − C ) / 2 − C flange x =(k − t f ) • y C fillet 180.6 − 174.7 =(1.04 − 0.535) • =0.20 in. 14.98 AISC-360-10 Example 001 - 6 Software Verification PROGRAM NAME: REVISION NO.: ETABS 3 Distance from the centroid of the compressive force in the steel section to the top of the steel section: d2 = C flange • t f / 2 + ((Py − C ) / 2 − C flange ) • (t f + x / 2) ( Py − C ) / 2 174.7 • 0.535 / 2 + (180.6 − 174.7) • (0.535 + 0.2 / 2) = 0.279 in. 180.6 Moment resistance of composite beam for partial composite action: ΦM n = Φ C • ( d1 + d 2 ) + Py • ( d 3 − d 2 ) 20.8 12 763.2 kip-ft = 0.9 373.9 • ( 7.04 + 0.279 ) + 735 • − 0.279 = 2 Shear Stud Strength: From AISC Manual Table 3.21, assuming the shear studs are placed in the weak position, the strength of ¾ in.-diameter shear studs in normal weight concrete with f c′ = 4 ksi and deck oriented perpendicular to the beam is: Qn = 17.2 kips for one shear stud per deck flute Qn = 14.6 kips for two shear studs per deck flute Shear Stud Distribution: There are at most 22 deck flutes along each half of the clear span of the beam. ETABS only counts the studs in the first 21 deck flutes as the 22nd flute is potentially too close to the point of zero moment for any stud located in it to be effective. With two shear studs in the first flute, 20 in the next in the next twenty flutes, and one shear stud in the 22nd flute, in each half of the beam, there is a total of 46 shear studs on the beam, and the total force provided by the shear studs in each half span is: ΣQn =2 • 14.6 + 20 • 17.2 =373.9 kip AISC-360-10 Example 001 - 7 Software Verification PROGRAM NAME: REVISION NO.: ETABS 3 Live Load Deflection: Modulus of elasticity ratio: 644 8.0 n E= Ec 29, 000 3,= = Transformed elastic moment of inertia assuming full composite action: Transformed Area A (in2) Moment Arm from Centroid y (in.) Ay (in.3) Ay2 (in,4) I0 (in.4) Slab 67.9 15.65 1,062 16,620 115 W21x50 14.7 0 0 0 984 1,062 16,620 1,099 Element 82.6 Ix = I 0 + Ay 2 = 1, 099 + 16, 620 = 17, 719 in.4 = y 1, 062 = 12.9 in. 82.6 2 I tr = I x − A • y = 17, 719 − 82.6 • 12.92 = 4, 058 in 4 Effective moment inertia assuming partial composite action: I equiv = I s + ΣQn / Py ( I tr − I s ) = 984 + 0.51(4,058 − 984) = 3,176 in 4 I eff = 0.75 • I equiv = 0.75 • 3,176 = 2,382 in 4 = ∆ LL 5wL L4 5 • (1 / 12) • (30 • 12) 4 = = 1.34 in. 384 EI eff 384 • 29, 000 • 2, 382 Design for Shear Strength: Required Shear Strength: wu = 1.2 • 0.800 + 1.2 • 0.100 + 1.6 • 1 = 2.68 kip/ft = Vu wu • L 2.68 • 45 = = 60.3 kip-ft 2 2 AISC-360-10 Example 001 - 8 Software Verification PROGRAM NAME: REVISION NO.: ETABS 3 Available Shear Strength: ΦVn = Φ • 0.6 • d • tw • Fy = 1.0 • 0.6 • 20.8 • 0.38 • 50 = 237.1 kips AISC-360-10 Example 001 - 9
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