CE 331, Spring 2004 Example 1 / 7 Building Analysis: Continuous Joists, Moment Frames The building indicated in the sketches below has two-span continuous joists and a moment frame along the 30-foot side. Calculate the maximum axial force and/or bending moment for each member specified below: 1. joist, girder and column due to gravity loads (DL + LL) 2. girder and column due to wind load (WL) DL = 18 psf, LL = 30 psf, WL = 15 psf 4 4 Girders: W12x50, Izz = 394 in , Columns: W8x24, Izz = 82.8 in t f 22 t f 2 @ 25 Side Elevation 3 @ 10ft Plan View 30ft 22ft Front Elevation Gravity Loads wDL+LL Joist: wDL+LL = (18psf + 30psf)(10ft) = 0.48klf Max MDL+LL = 0.125wL2 = (0.125)(0.48klf)(25ft)2 Max MDL+LL = 37.5k-ft MDL+LL 2 0.125wL CE 331, Spring 2004 Example 2 / 7 Building Analysis: Continuous Joists, Moment Frames Frame: Pext P P Pext ft ft ft k int int Pext = (48psf)(10 /2)(25 /2 + 25 /2) = 6.0 ft ft ft k Pint = (48psf)(10 )(25 /2 + 25 /2) = 12.0 Member-end Moments: Fixed - end Moments : 2 2 FEM = PL = (12 k )(30 ft ) = +80.0 k − ft (+'ve because it' s counter − clockwise (CCW )) girder left end 9 9 Rotational Stiffnesses : 4 4I col 4(82.8in ) Fixed - Base: kcol = = =15.1 (inconsistent units are OK because they will cancel) Lcol 22 ft 4 2I gird 2(394in ) k gird = = = 26.3 Lgird 30 ft Member - end Moments : ⎡ k gird ⎤ k − ft ⎡ 26.3 ⎤ k− ft mgirder left end = FEM girder left end ⎢1− ⎥ = (+80.0 )⎢1− ⎥ = 29.2 (CCW ) ⎣⎢ k gird + kcol ⎦⎥ ⎣ 26.3 +15.1⎦ kcol 15.1 k− ft k− ft mtop col = ()− FEM 23 = ( )(−80.0 ) = −29.2 (CW ) k gird + kcol 26.3 +15.1 k − ft mtop col − 29.2 m = = = −14.6k− ft (CW ) bot col 2 2 CE 331, Spring 2004 Example 3 / 7 Building Analysis: Continuous Joists, Moment Frames Exploded FBD of Frame for DL+LL 12k 12k k-ft 29.2k-ft 29.2 2k 10ft ft 10ft 2k k 10 12 k 18k 12 18k 29.2k-ft 29.2k-ft 2k 2k 22ft 22ft 2k 2k 14.6k-ft 14.6k-ft 18k 18k 12 0 Vgirder, k -12 MDL+LL, k-ft 90.8 (Drawn on compression side) 29.2 29.2 29.2 29.2 14.6 14.6 CE 331, Spring 2004 Example 4 / 7 Building Analysis: Continuous Joists, Moment Frames RISA Results for DL+LL: Axial Forces, k Shear, k Bending Moment, k-ft CE 331, Spring 2004 Example 5 / 7 Building Analysis: Continuous Joists, Moment Frames PWL Sidesway ext Wind Loads: ft ft k Pext = (15psf)(25 /2)(22 /2) = 2.06 WL ft ft k Pint Pint = (15psf)(25 )(22 /2) = 4.13 WL Pext Moment Frame: 4.13k WL k Wind Load (P ) to middle moment frame = Pint = 4.13 Member-end Moments (Fixed-base Columns:) Rotational Stiffnesses : 4 4 I col 82.8in 6I gird 6(394in ) kcol = = ft = 3.76 k gird = = = 78.8 Lcol 22 Lgird 30 ft Member - end Moments : WL ⎡ ⎤ k P Lcol k gird (4.13 )(22 ft) ⎡ 78.8 ⎤ k − ft k − ft mend gird = − ⎢ ⎥ = − ⎢ ⎥ = −(22.7 )(0.954) = −21.7 4 ⎣⎢k gird + kcol ⎦⎥ 4 ⎣78.8 + 3.76⎦ WL ⎡ ⎤ P Lcol kcol k − ft ⎡ 3.76 ⎤ k − ft k − ft mtop col = ⎢1− ⎥ = (22.7 )⎢1− ⎥ = (22.7 )(0.954) = 21.7 4 ⎣⎢ k gird + kcol ⎦⎥ ⎣ 78.8 + 3.76⎦ WL ⎡ ⎤ P Lcol kcol k − ft ⎡ 3.76 ⎤ k − ft k − ft mbot col = ⎢1+ ⎥ = (22.7 )⎢1+ ⎥ = (22.7 )(1.0455) = 23.7 4 ⎣⎢ k gird + kcol ⎦⎥ ⎣ 78.8 + 3.76⎦ CE 331, Spring 2004 Example 6 / 7 Building Analysis: Continuous Joists, Moment Frames Exploded FBD of Frame for WL k-ft 21.7 k 2.07k 1.45 k 2.07k 2.07k 4.13 ft k 30 k-ft k 1.45 21.7 2.07 k k 1.45 FBD of joint shows 1.45 k k-ft horizontal forces only 2.07 k-ft 21.7 21.7 2.07k 22ft 22ft k-ft 2.07k 21.7 2.07k 21.7k-ft 1.45k 1.45k MWL, k-ft (Drawn on compression side) 21.7 21.7 21.7 21.7 23.7 23.7 CE 331, Spring 2004 Example 7 / 7 Building Analysis: Continuous Joists, Moment Frames RISA Results for WL: Axial Force, k Shear, k Bending Moment, k-ft .