Exam (elaborations) TEST BANK FOR Fundamentals of Structural Analysis 5th Edition By Kenneth M. Leet, Chia-Ming Uang, Joel T. Lanning, Anne M. Gilber (Solution Manual)

Exam (elaborations) TEST BANK FOR Fundamentals of Structural Analysis 5th Edition By Kenneth M. Leet, Chia-Ming Uang, Joel T. Lanning, Anne M. Gilber (Solution Manual) FUNDAMENTALS OF STRUCTURAL ANALYSIS 5th Edition Kenneth M. Leet, Chia-Ming Uang, Joel T. Lanning, and Anne M. Gilbert SOLUTIONS MANUAL CHAPTER 2: DESIGN LOADS AND STRUCTURAL FRAMING 2-2 Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Compute the weight/ft. of cross section @ 120 lb/ft3. Compute cross sectional area: ( ) ( ) ( ) 2 1 Area 0.5 6 2 0.5 2.67 0.67 2.5 1.5 1 2 7.5 ft æç ÷ö = ¢´ ¢ + ç ´ ¢´ ¢÷÷+ ¢´ ¢ + ¢´ ¢ çè ÷ø = Weight of member per foot length: 2 3 wt/ft = 7.5 ft ´120 lb/ft = 900 lb/ft. P2.1. Determine the deadweight of a 1-ft-long segment of the prestressed, reinforced concrete tee-beam whose cross section is shown in Figure P2.1. Beam is constructed with lightweight concrete which weighs 120 lbs/ft3. 18ʺ Section 24ʺ 12ʺ 6ʺ 6ʺ 48ʺ 72ʺ 8ʺ P2.1 2-3 Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. See Table 2.1 for weights ( ) 3 2 3 wt / 20 unit 20 Plywood: 3 psf 1 5 lb 12 20 Insulation: 3 psf 1 5 lb 12 20 9.17 lb Roof’g Tar & G: 5.5 psf 1 12 19.17 lb lb 1.5 15.5 1 5.97 lb Wood Joist 37 ft 14.4 in / ft Total wt of 20 unit 19 ¢¢ ¢¢ ´ ´ ¢ = ¢¢ ´ ´ ¢ = ¢¢ ´ ´ ¢ = ¢¢´ ¢¢´ ¢ = = ¢¢ = .17 + 5.97 = 25.14 lb. Ans. P2.2. Determine the deadweight of a 1-ft-long segment of a typical 20-in-wide unit of a roof supported on a nominal 2 × 16 in. southern pine beam (the actual dimensions are 1 2 in. smaller). The 3 4 -in. plywood weighs 3 lb/ft2. 20ʺ 20ʺ 2ʺ insulation three ply felt tar and gravel 3/4ʺ plywood 1 1/2ʺ 15 1/2ʺ Section P2.2 2-4 Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Uniform Dead Load WDL Acting on the Wide Flange Beam: Wall Load: 9.5 (0.09 ksf) 0.855 klf Floor Slab: 10 (0.05 ksf) 0.50 klf Steel Frmg, Fireproof’g, Arch’l Features, Floor Finishes, & Ceiling: 10 (0.024 ksf ) 0.24 klf Mech’l, Piping & Electrical Systems: 10 (0.006 ks ¢ = ¢ = ¢ = ¢ f ) 0.06 klf Total DL 1.66 klf W = = P2.3. A wide flange steel beam shown in Figure P2.3 supports a permanent concrete masonry wall, floor slab, architectural finishes, mechanical and electrical systems. Determine the uniform dead load in kips per linear foot acting on the beam. The wall is 9.5-ft high, non-load bearing and laterally braced at the top to upper floor framing (not shown). The wall consists of 8-in. lightweight reinforced concrete masonry units with an average weight of 90 psf. The composite concrete floor slab construction spans over simply supported steel beams, with a tributary width of 10 ft, and weighs 50 psf. The estimated uniform dead load for structural steel framing, fireproofing, architectural features, floor finish, and ceiling tiles equals 24 psf, and for mechanical ducting, piping, and electrical systems equals 6 psf. concrete floor slab piping wide flange steel beam with fireproofing ceiling tile and suspension hangers Section mechanical duct 8ʺ concrete masonry partition 9.5ʹ P2.3 2-5 Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.   2 2 20 ft Method 2 8 8 ( ) Method 1: 40 3 2 2 1 : 320 4 4(4) 288 2 ft T T T T a A A A A                       2 2 ft Method 2: 6.67 ( ) Method 1: 20 66.7 2 1 66.7 2 3.33(3.33) 55.6 ft 2 T T T T b A A A A                      2 2 ft Method 2 6.67 ( ) Method 1: 20 10(10) 2 166.7 1 1 166.7 2 3.33(3.33) 5(5) 2 2 180 : .6 ft 2 T T T T A c A A A                            2 2 ft Metho 40 20 ( ) Method 1: 36 2 2 1080 1 d 2: 1080 2 4(4) 2 1096 ft T T T T A d A A A                   2 2 ; 40 20 ( ) 200 2 2 40 20 40 2 ft 0 ( ) ; 900 2 f 2 2 2 t T T T T e A f A A A                         P2.4. Consider the floor plan shown in Figure P2.4. Compute the tributary areas for (a) floor beam B1, (b) floor beam B2, (c) girder G1, (d) girder G2, (e) corner column C1, and (f ) interior column C A 40ʹ 20ʹ B1 G2 G4 G3 C3 C1 C4 C2 B2 G1 B4 6 @ 6.67ʹ = 40ʹ B3 2 @ 10ʹ = 20ʹ 5 @ 8ʹ = 40ʹ 1 2 3 B C P2.4 5 ft G1 10 ft 5 ft 6.67 ft 6.66 ft 6.67 ft G1 10 ft Right Side Left Side 4 ft 4 ft 36 ft G2 36 ft G2 B1 4 ft B1 36 ft 4 ft 6.67 ft 6.66 ft 6.67 ft B4 AT,C2 AT,C1 2-6 Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.    2 2 2 0 ft Method 2: ( ) Method 1: 10 20 20 1 20 f 0 4 5 2 150 t T T T T A A a A A               2 2 2 ft Metho ( ) Method 1: d 2: f 6.67 20 133.4 1 133.4 4 3.33 2 111.2 t T T T T b A A A A                2 2 2 ft Method 2: ( ) Method 1: 36 20 720 1 t 2 f T T T T A A c A A                 2 2 2 2 ft Method 2 ( ) Method 1: 4 40 33.33(10) 493.4 1 1 493.4 2 4 2 3.33 2 2 : 488.5 ft T T T T A A d A A                          2 2 ( ) 30 20 600 ( ) 10 10 10 ; ft ; 0 ft T T T T e A f A A A     A 40ʹ 20ʹ B1 G2 G4 G3 C3 C1 C4 C2 B2 G1 B4 6 @ 6.67ʹ = 40ʹ B3 2 @ 10ʹ = 20ʹ 5 @ 8ʹ = 40ʹ 1 2 3 B C P2.4 P2.5. Refer to Figure P2.4 for the floor plan. Calculate the tributary areas for (a) floor beam B3, (b) floor beam B4, (c) girder G3, (d) girder G4, (e) edge column C3, and (f ) corner column C4. B4 B3 5 ft B4 B3 10 ft 5 ft 6.67 ft 6.66 ft 6.67 ft 3.33 ft G3 33.33 ft G3 33.33 ft Right Side Left Side 4 ft 4 ft 36 ft G4 36 ft G4 3.33 ft 4 ft 36 ft 4 ft B4 AT,C3 AT,C4 2-7 Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2 ( ) 8(40) 320 ft , 2, 640 400 15 60 60 0.25 50.6 psf 640 2 8(50.6) 404.8 lb/ft 0.40 kips/ft , ok T LL T LL a A K A K L w = = = = = + = = = = æç ÷öçç ÷÷ è ø 2 6.67 ( ) (20) 66.7 ft , 2, 133.4 400, No Reduction 2 6.67 (60) 200.1 lb/ft 0.20 kips/ft 2 T LL T LL b A K A K w = = = = = = = 2 trib beam 6.67 ( ) (20) 10(10) 166.7 ft , 2, 333.4 400, No Reduction 2 6.67 (60) 200.1 lb/ft 0.20 kips/ft 2 ( )( ) 60(10)(20) 6000 lbs 6 kips 2 2 T LL T LL c A K A K w q W L P = + = = = = = = = = = = 2 40 20 ( ) 36 1080 ft , 2, 2160 400 2 2 15 60 60 0.25 34.4 2160 2 34.4 psf 40 20 8(34.4) 8256 lbs 8.26 kips 2 2 , ok T LL T LL d A K A K L L P = + = = = = + = = = + = = æç ÷öççè ÷÷ø æç ö÷çç ÷÷ è ø æç ÷öççè ÷÷ø A 40ʹ 20ʹ B1 G2 G4 G3 C3 C1 C4 C2 B2 G1 B4 6 @ 6.67ʹ = 40ʹ B3 2 @ 10ʹ = 20ʹ 5 @ 8ʹ = 40ʹ 1 2 3 B C P2.4 P2.6. The uniformly distributed live load on the floor plan in Figure P2.4 is 60 lb/ft2. Establish the loading for members (a) floor beam B1, (b) floor beam B2, (c) girder G1, and (d) girder G2. Consider the live load reduction if permitted by the ASCE standard. w B1 and B2 w P G1 P P P P 5 spaces @ 8’ each G2 2-8 Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2 ( ) 10(20) 200 ft , 2, 400 400 15 60 0.25 60 psf 400 10(60) 600 lb/ft 0.60 kips/ft T LL T LL a A K A K L w = = = = = + = = = = çæ ÷öçç ÷÷ è ø 2 ( ) 6.67(20) 133.4 ft , 2, 266.8 400, No Reduction 6.67(60) 400.2 lb/ft 0.40 kips/ft T LL T LL b A K A K w = = = = = = = 2 trib beam ( ) 36(20) 720 ft , 2, 1440 400 15 60 60 0.25 38.7 psf , ok 1440 2 ( )( ) 38.7(8)(40) 6192 lbs 6.19 kips 2 2 T LL T LL c A K A K L q W L P = = = = = + = = = = = æç ÷öçç ÷÷ è ø 2 8 ( ) 40 33.33(10) 493.3 ft , 2, 986.6 400 2 15 60 60 0.25 43.7 , ok 986.6 2 43.7(4) 174.8 lb/ft 0.17 kips/ft 43.7(6.67(20) 2914.8 lbs 2.91 kips 2 T LL T LL d A K A K L w P = + = = = = + = = = = = = = æç ÷öççè ÷÷ø æç ÷öçç ÷÷ è ø A 40ʹ 20ʹ B1 G2 G4 G3 C3 C1 C4 C2 B2 G1 B4 6 @ 6.67ʹ = 40ʹ B3 2 @ 10ʹ = 20ʹ 5 @ 8ʹ = 40ʹ 1 2 3 B C P2.4 P2.7. The uniformly distributed live load on the floor plan in Figure P2.4 is 60 lb/ft2. Establish the loading for members (a) floor beam B3, (b) floor beam B4, (c) girder G3, and girder G4. Consider the live load reduction if permitted by the ASCE standard. w B3 and B4 P P P P 5 spaces @ 8’ each G3 P P P P 6 spaces @ 6.67’ each P w G4 2-9 Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2 3rd 1st ( ) 900 ft , 4, 3600 400 2 2 2 2 15 60 60 0.25 30 psf ok (minimum permitted) 3600 2 900(30) 27000 lbs 27 kips (3)900(30) 27000 lbs 81 kips , T LL T LL a A K A K L P P = + + = = = = + = = = = = = = = æç ö÷çæ ÷ö ççè ÷÷øççè ÷÷ø çæ ÷öçç ÷÷÷ è ø P2.8. The building section associated with the floor plan in Figure P2.4 is shown in Figure P2.8. Assume a live load of 60 lb/ft2 on all three floors. Calculate the axial forces produced by the live load in column C1 in the third and first stories. Consider any live load reduction if permitted by the ASCE standard. 40ʹ 20ʹ 3 @ 12ʹ = 36ʹ C3 C1 Building Section P2.8 P3rd P1st C2 PLAN ELEVATION B4 AT,C2 AT,C1 2-10 Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2 3rd 1st 40 20 ( ) 20 600 ft , 4, 2400 400 2 2 15 60 60 0.25 33.4 psf , ok 2400 2 600(33.4) 20040 lbs 20.0 kips (3)600(33.4) 60120 lbs 60.1 kips T LL T LL a A K A K L P P = + = = = = + = = = = = = = = æç ö÷ççè ÷÷ø çæ ÷öçç ÷÷ è ø 40ʹ 20ʹ 3 @ 12ʹ = 36ʹ C3 C1 Building Section P2.8 P2.9. The building section associated with the floor plan in Figure P2.4 is shown in Figure P2.8. Assume a live load of 60 lb/ft2 on all three floors. Calculate the axial forces produced by the live load in column C3 in the third and first stories. Consider any live load reduction if permitted by the ASCE standard. B4 AT,C3 AT,C4 PLAN ELEVATION P3rd P1st C3 2-11 Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. a) Resulant Wind Forces Roof 20 psf (6 × 90) = 10,800 lb 5th floor 20 psf (12 × 90) = 21,600 lb 4th floor 20 psf (2 × 90) + 15 (10 × 90) = 17,100 lb 3rd floor 15 psf (10 × 90) + 13 (2 × 96) = 15,800 lb 2nd floor 13 psf (12 × 90) = 14,040 lb b) Horizontal Base Shear VBASE = Σ Forces at Each Level = 10.8k + 21.6k + 17.1k + 15.8k + 14.04k = (a) VBASE = 79.34k Overturning Moment of the Building = Σ (Force @ Ea. Level × Height above Base) 10.8k (60′)+ 21.6 (48′) + 17.1 (36′) + 15.8k (24′)+ 14.04k(12′) = M overturning = 2, 848ft.k (a) Plan Building Section N (b) (c) A 1 3 @ 20ʹ = 60ʹ 3 @ 30ʹ = 90ʹ 4 @ 25ʹ = 100ʹ 13 wind pressures in lb/ft2 15 20 5 @ 12ʹ = 60ʹ 2 3 4 5 B C D P2.10 P2.10. A five-story building is shown in Figure P2.10. Following the ASCE standard, the wind pressure along the height on the windward side has been established as shown in Figure P2.10(c). (a) Considering the windward pressure in the east-west direction, use the tributary area concept to compute the resultant wind force at each floor level. (b) Compute the horizontal base shear and the overturning moment of the building. 2-12 Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. a) Live Load Impact Factor = 20% b) Total LL Machinery = 1.20 (4 kips) = 4.8k Uniform LL = ((10′ × 16′) ‒ (5′ × 10′)) (0.04 ksf) = 4.4k Total LL = 9.2k ∴ Total′ LL Acting on One Hanger = 9.2k/4 Hangers = 2.3klps c) Total DL Floor Framing = 10′ × 16′ (0.025 ksf) = 4k ∴ Total DL Acting on one Hanger = 4k/4 Hangers = 1 kip ∴ Total DL + LL on One Hanger = 2.3k + lk = 3.3 kips hanger hanger hanger hanger hanger floor grating mechanical support framing vertical lateral bracing beyond floor framing above supports Mechanical Floor Plan (beams not shown) (a) Section (b) edge of mechanical support framing vertical lateral bracing, located on 4 sides of framing (shown dashed) hanger 2.5ʹ 3ʹ 10ʹ 3ʹ 2.5ʹ 5ʹ mechanical unit mechanical unit P2.11 P2.11. A mechanical support framing system is shown in Figure P2.11. The framing consists of steel floor grating over steel beams and entirely supported by four tension hangers that are connected to floor framing above it. It supports light machinery with an operating weight of 4000 lbs, centrally located. (a) Determine the impact factor I from the Live Load Impact Factor, Table 2.3. (b) Calculate the total live load acting on one hanger due to the machinery and uniform live load of 40 psf around the machine. (c) Calculate the total dead load acting on one hanger. The floor framing dead load is 25 psf. Ignore the weight of the hangers. Lateral bracing is located on all four edges of the mechanical floor framing for stability and transfer of lateral loads. 2-13 Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Use I = 1 2 2 2 0.613 (Eq. 2.4b) 0.613(40) = 980.8 N/m 980.8(1)( )(1)(0.85) = 833.7 s z s z zt d z z z q V q q IKK K q K K = = = = 2 2 2 2 0 4.6m: 833.7(0.85) 708.6 N/m 4.6 6.1m: 833.7(0.90) 750.3 N/m 6.1 = 7.6 m: 833.7(0.94) = 783.7 N/m 7.6 = 9 m: 833.7(0.98) = 817.1 N/m z z z z q q q q - = = - = = = = For the Windward Wall = (Eq. 2.7) where = 0.85(0.8) = 0.68 = 0.68 z p p z p q GC GC p q 2 2 2 2 0 4.6 m 481.8 N/m 4.6 6.1 m 510.2 N/m 6.1 7.6 m 532.9 N/m 7.6 9 m 555.6 N/m p p p p - = - = - = - = Total Windforce, FW, Windward Wall 481.8[4.6 × 20] + 510.2[1.5 × 20] + 532.9[1.5 × 20] + 555.6[1.4 × 20] 91,180 N W W F F = = For Leeward Wall 2 2 (0.85)( 0.2) at 9m 817.1 N/m (above) 817.1 (0.85)( 0.2) 138.9 N/m h p h h z p q GC q q q p = = - = = = - = - Total Windforce, FL, on Leeward Wall * (20 9)( 138.9) 25,003 N Total Force 91,180N 25,003 116,183.3 N L W L F F F = ´ - = - = + = + = *Both FL and FN Act in Same Direction. P2.12. The dimensions of a 9-m-high warehouse are shown in Figure P2.12. The windward and leeward wind pressure profiles in the long direction of the warehouse are also shown. Establish the wind forces based on the following information: basic wind speed = 40 m/s, wind exposure category = C, Kd = 0.85, Kzt = 1.0, G = 0.85, and Cp = 0.8 for windward wall and ‒0.2 for leeward wall. Use the Kz values listed in Table 2.4. What is the total wind force acting in the long direction of the warehouse? qzGCp qhGCp 40 m 9 m 20 m (not to scale) P2.12 2-14 Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. TABLE P2.13 Roof Pressure Coefficient Cp *θ defined in Figure P2.13 Windward Leeward Angle θ 35 45 ≥60 10 15 ≥20 Cp −0.9 −0.7 −0.4 −0.3 −0.2 −0.2 0.0 0.01θ* −0.5 −0.5 −0.6 0.0 0.2 0.2 0.3 0.4 Mean Roof Height, h  24 ft 1 16 tan 33.69 (for Table 2.10) 24 θ = - æçç ¢ ö÷÷=  çè ¢÷÷ø Consider Positive Windward Pressure on Roof, i.e. left side. Interpolate in Table P2.10 ( ) ( ) 33.69 30 0.2 0.1 35 30 0.2738(Roof only) p p C C - = + ´ - = From Table 2.4 (see p48 of text) 0.57, 0 15 0.62, 15 20 0.66, 20 25 0.70, 25 30 0.76, 30 32 z K = - ¢ = ¢- ¢ = ¢- ¢ = ¢- ¢ = ¢- ¢ (b) (a) 48ʹ Section 80ʹ wind 16ʹ 16ʹ h qhGCp qzGCp qhGCp qhGCp θ P2.13 P2.13. The dimensions of an enclosed gabled building are shown in Figure P2.13a. The external pressures for the wind load perpendicular to the ridge of the building are shown in Figure P2.13b. Note that the wind pressure can act toward or away from the windward roof surface. For the particular building dimensions given, the Cp value for the roof based on the ASCE standard can be determined from Table P2.13, where plus and minus signs signify pressures acting toward and away from the surfaces, respectively. Where two values of Cp are listed, this indicates that the windward roof slope is subjected to either positive or negative pressure, and the roof structure should be designed for both loading conditions. The ASCE standard permits linear interpolation for the value of the inclined angle of roof