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Solutions Manual for Fundamentals of Structural Analysis 5th Edition by Leet IBSN 0073398004

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    2-1 Copyright © 2018 McGraw-Hill Education. All rights reserved.  No reproduction or distribution without the prior written consent of McGraw-Hill Education.    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 Solutions Manual for Fundamentals of Structural Analysis 5th Edition by Leet IBSN 0073398004 Full Download: http://downloadlink.org/product/solutions-manual-for-fundamentals-of-structural-analysis-5th-edition-by-leet-i Full all chapters instant download please go to Solutions Manual, Test Bank site: downloadlink.org    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/ft 3 . Compute cross sectional area: ( ) ( ) ( ) 2 1Area0.5620.52.670.672.51.5127.5ft æ ö÷ç¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢= ´ + ´ ´ + ´ + ´÷ç ÷ç ÷è ø=  Weight of member per foot length: 23 wt/ft7.5ft120lb/ft900 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/ft 3 . 18ʺ Section24ʺ 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 ( ) 323 wt / 20 unit20Plywood: 3 psf15 lb1220Insulation: 3 psf15 lb12209.17 lbRoof’g Tar & G: 5.5 psf11219.17 lb1.515.515.97 lblbWood Joist37ft14.4in/ftTotal wt of 20 unit19 ¢¢¢¢¢´ ´ =¢¢¢´ ´ =¢¢¢´ ´ =¢¢¢¢ ¢´ ´ ==¢¢ = .17 + 5.9725.14 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 12 in. smaller). The 34 -in.  plywood weighs 3 lb/ft 2 . 20ʺ20ʺ 2ʺ insulationthree ply felttar and gravel3/4ʺ plywood15 1  /  2 ʺ 1 1  /  2 ʺ 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 W   DL  Acting on the Wide Flange Beam: Wall Load:9.5(0.09ksf)0.855klf Floor Slab:10(0.05ksf)0.50klf Steel Frmg, Fireproof’g, Arch’l Features, Floor Finishes, & Ceiling:10(0.024ksf)0.24klf Mech’l, Piping & Electrical Systems:10(0.006 ks ¢ =¢ =¢ =¢ f)0.06 klf Total1.66klf   DL 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 slabpipingwide flange steelbeam with fireproofingceiling tile and suspension hangersSectionmechanicalduct8ʺ concrete masonrypartition9.5ʹ    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.     22 20 ftMethod 288() Method 1: 40322132044(4)28: 82ft TT TT  aAA A A                   22  ftMethod 2:6.67() Method 1: 2066.72166.723.33(3.33)55.6ft2 TT TT  bA A A A                  22  ftMethod 26.67() Method 1: 2010(10)2166.711166.723.33(3.33)5(5)22180:. ft6  2 T T T T   AcA A A                      22  ftMetho4020() Method 1: 362210801108024(4d 2: )210ft96 T T T T   AdA A A             22 ; 4020() 200224020402 ft0(; ) 9002f 222t TT TT  eA fA 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 A40ʹ20ʹ B1G2G4G3C3C1C4C2G1B2B46 @ 6.67ʹ = 40ʹ B32 @ 10ʹ = 20ʹ 5 @ 8ʹ = 40ʹ 123BC   P2.4 5 ft G1 10 ft5 ft6.66ft6.67 ft6.67 ft G1 10 ftRightSideLeftSide4 ft4 ft36ft G2 36 ft G2 B1 4 ft B1 36 ft4 ft6.66ft6.67 ft6.67 ft B4    A T,C2  A T,C1
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