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P2.11. A mechanical support framing system isshown in Figure P2.11. The framing consists ofsteel floor grating over steel beams and entirelysupported by four tension hangers that areconnected to floor framing above it. It supportslight machinery with an operating weight of4000 lbs, centrally located. (a) Determine theimpact factor I from the Live Load ImpactFactor, Table 2.3. (b) Calculate the total liveload acting on one hanger due to the machineryand uniform live load of 40 psf around themachine. (c) Calculate the total dead load actingon one hanger. The floor framing dead load is25 psf. Ignore the weight of the hangers. Lateralbracing is located on all four edges of themechanical floor framing for stability andtransfer of lateral loads.3ʹ10ʹ3ʹhangerhanger2.5ʹvertical lateral bracing, located on4 sides of framing (shown dashed)mechanicalunit5ʹhanger2.5ʹedge of mechanical support framinghangerMechanical Floor Plan(beams not shown)(a)floor framing above supportsvertical lateral bracing beyondhangerhangermechanicalunitfloor gratingmechanicalsupport framingSection(b)P2.11a) Live Load Impact Factor 20%b) Total LLMachinery 1.20 (4 kips) 4.8kUniform LL ((10′ 16′) ‒ (5′ 10′)) (0.04 ksf) 4.4kkTotal LL 9.2 Total′ LL Acting on One Hanger 9.2 /4 Hangers 2.3kklpsc) Total DLFloor Framing 10′ 16′ (0.025 ksf) 4k Total DL Acting on one Hanger 4 /4 Hangers 1 kipkkk Total DL LL on One Hanger 2.3 l 3.3 kips2-12Copyright 2018 McGraw-Hill Education. All rights reserved.No reproduction or distribution without the prior written consent of McGraw-Hill Education.

qz GCpP2.12. The dimensions of a 9-m-high warehouseare shown in Figure P2.12. The windward andleeward wind pressure profiles in the longdirection of the warehouse are also shown.Establish the wind forces based on the followinginformation: basic wind speed 40 m/s, windexposure 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 inTable 2.4. What is the total wind force acting inthe long direction of the warehouse?Use I 1qhGCp20 m9m40 m(not to scale)P2.12Total Windforce, FW, Windward Wallqs 0.613 V (Eq. 2.4b)2 0.613(40) 980.8 N/m2FW 481.8[4.6 20] 510.2[1.5 20] 532.9[1.5 20] 555.6[1.4 20]2qz qs IK z K zt K dFW 91,180 Nqz 980.8(1)(K z )(1)(0.85) 833.7 K zFor Leeward Wall0 - 4.6 m: qz 833.7(0.85) 708.6 N/m 24.6 - 6.1m: qz 833.7(0.90) 750.3 N/m26.1 7.6 m: qz 833.7(0.94) 783.7 N/m7.6 9 m: qz 833.7(0.98) 817.1 N/mFor the Windward Wallp qz GC p (Eq. 2.7)where GC p 0.85(0.8) 0.68p qh GC p qh (0.85)(- 0.2)qh q z at 9 m 817.1 N/m 2 (above)2p 817.1 (0.85)( - 0.2) -138.9 N/m 22Total Windforce, FL, on Leeward WallFL (20 9)( -138.9) -25,003 N*Total Force FW FL 91,180 N 25,003p 0.68 qz 116,183.3 N0 - 4.6 m p 481.8 N/m 24.6 - 6.1 m p 510.2 N/m 2*Both FL and FN Act in Same Direction.6.1- 7.6 m p 532.9 N/m 27.6 - 9 m p 555.6 N/m 22-13Copyright 2018 McGraw-Hill Education. All rights reserved.No reproduction or distribution without the prior written consent of McGraw-Hill Education.

P2.13. The dimensions of an enclosed gabledbuilding are shown in Figure P2.13a. Theexternal pressures for the wind loadperpendicular to the ridge of the building areshown in Figure P2.13b. Note that the windpressure can act toward or away from thewindward roof surface. For the particularbuilding dimensions given, the Cp value for theroof based on the ASCE standard can bedetermined from Table P2.13, where plus andminus signs signify pressures acting toward andaway from the surfaces, respectively. Where twovalues of Cp are listed, this indicates that thewindward roof slope is subjected to eitherpositive or negative pressure, and the roofstructure should be designed for both loadingconditions. The ASCE standard permits linearinterpolation for the value of the inclined angleof roof 𝜃. But interpolation should only becarried out between values of the same sign.Establish the wind pressures on the buildingwhen positive pressure acts on the windwardroof. Use the following data: basic wind speed 100 mi/h, wind exposure category B, Kd 0.85, Kzt 1.0, G 0.85, and Cp 0.8 forwindward wall and 0.2 for leeward pSection(b)P2.13TABLE P2.13 Roof Pressure Coefficient Cp*θ defined in Figure P2.13WindwardLeewardAngle θ10152025303545 60Cp 0.9 0.7 0.4 0.3 0.2 0.20.00.00.20.20.30.40.01θ*1015 0.5 0.5 20 0.6Consider Positive Windward Pressure on Roof, i.e. left side.Interpolate in Table P2.10C p 0.2 (33.69 - 30) 0.1(35 - 30)C p 0.2738(Roof only)From Table 2.4 (see p48 of text)K z 0.57, 0 -15 0.62, 15 - 20 Mean Roof Height, h 24 ft 0.66, 20 - 25 0.70, 25 - 30 æ 16 öθ tan-1 çç 33.69 (for Table 2.10)çè 24 ø 0.76, 30 - 32 2-14Copyright 2018 McGraw-Hill Education. All rights reserved.No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Wind Pressure on Leeward SideP2.13. ContinuedK zt 1.0, K d 0.85, I 1P qh G C pFor h 24 ; qh qz 14.36 lb/ft 2For Wallqs 0.00256 V (Eq 2.4a)2qs 0.00256(100)2 25.6 lb/ft 2qz qs IK z K zt K d0 –15 ; qz 25.6 (1)(0.57)(1)(0.85) 12.40 lb/ftCp ‒0.2 for wall 0.6 for roofFor Wall P 14.36 (0.85)(0.2)P 2.44 lb/ft 2215–16 ; qz 13.49 lb/ft 2For Roof P 14.36 (0.85)(- 0.6) -7.32 lb/ft 2 (uplift)h 24; qz 14.36 lb/ft 2Wind Pressure on Windward Wall & RoofP qz GCPWall 0 –15 P 12.40 0.85 0.80P 8.43 psfWall, 15 –16 P 13.49 0.85 0.8 9.17 psfRoof, P 14.36 0.85 0.2738P 3.34 psf 2-15Copyright 2018 McGraw-Hill Education. All rights reserved.No reproduction or distribution without the prior written consent of McGraw-Hill Education.

P2.14. Establish the wind pressures on thebuilding in Problem P2.13 when the windwardroof is subjected to an uplift wind CpSection(b)P2.13TABLE P2.13 Roof Pressure Coefficient Cp*θ defined in Figure P2.11WindwardLeewardAngle θ10152025303545Cp 0.9 0.7 0.4 0.3 0.2 0.20.00.00.20.20.30.4 600.01θ*1015 0.5 0.5See P2.13 SolutionWindward Roof (Negative Pressure) 33.7 Interpolate between 30 and 35 for negative Cp value in Table P2.12C p -0.274p qh GC p 21.76(0.66) 0.85(- 0.274) -3.34 lb/ft 2 (Suction)Note: Wind pressures on other 3 surfaces are the same as in P2.13.2-16Copyright 2018 McGraw-Hill Education. All rights reserved.No reproduction or distribution without the prior written consent of McGraw-Hill Education. 20 0.6

BP2.15. (a) Determine the wind pressuredistribution on the four sides of the 10-storyhospital shown in Figure P2.15. The building islocated near the Georgia coast where the windvelocity contour map in the ASCE Standardspecifies a design wind speed of 140 mph. Thebuilding, located on level flat ground, isclassified as stiff because its natural period isless than 1 s. On the windward side, evaluate themagnitude of the wind pressure every 35 ft inthe vertical direction. (b) Assuming the windpressure on the windward side varies linearlybetween the 35-ft intervals, determine the totalwind force on the building in the direction of thewind. Include the negative pressure on theleeward side.ACDFwind140 mphqs 0.00256VG160ʹH80ʹP2.15Compute Wind Pressure on Leeward Wallp qz GCp; Use Value of qz at 140 ft. i.c. Kz 152Face2140ʹE(a) Compute Variation of Wind Pressure on Windwardqz qs IK z K zt K dleewardC p -0.5qz 49.05(1.52) 74.556p 74.556 GC p 74.556(0.85)(-0.5)Eq 2.8Eq 2.6a 0.00256(140)2p -31.68 psfqs 50.176 psf; Round to 50.18 psfANS.Wind Pressure on Side WallsI 1.15 for hospitalsp qz GC p 49.05(1.52)(0.85)(-0.7)K zt 1; K d 0.85p -44.36 psfKz, Read in Table 2.4Elev. (ft)035 70 105140Kz1.031.191.341.441.52(b) Variation of Wind Pressure on Windward andLeeward Sidesqz 50.18 (1.15)(K z ) 1 (0.85)qz 49.05 K zCompute Wind Pressure “p” on Windward Facep qz GC p 49.05 K z GC pwhere G 0.85 for natural period less than 1 sec.C p 0.8 on windward sidep 49.05 k z (0.85)(0.8) 33.354 K zCompute “p” for Various ElevationsElev. (ft)03570105140p (psf)34.3639.6944.6948.0350.702-17Copyright 2018 McGraw-Hill Education. All rights reserved.No reproduction or distribution without the prior written consent of McGraw-Hill Education.

P2.15. ContinuedCompute Total Wind Force (kips)50.7 48.02 é 35 160 ùêú 276.42 kipsêë 1000 úû248.03 44.69 é 35 160 ùêú 259.62 kF2 êë 1000 úû2F1 44.69 39.69 é 35 160 ùêú 236.26 kêë 1000 úû239.69 34.36 é 35 160 ùêú 207.39 kF4 êë 1000 úû2F3 F5 31.68(140 160)1000 709.63 kTotal Wind Force Σ F1 F2 F3 F4 F5 1689.27 kips2-18Copyright 2018 McGraw-Hill Education. All rights reserved.No reproduction or distribution without the prior written consent of McGraw-Hill Education.

P2.16. Consider the five-story building shownin Figure P2.10. The average weights of the22floor and roof are 90 lb/ft and 70 lb/ft ,respectively. The values of SDS and SD1 are equalto 0.9g and 0.4g, respectively. Since steelmoment frames are used in the north-southdirection to resist the seismic forces, the valueof R equals 8. Compute the seismic base shearV. Then distribute the base shear along theheight of the building.12345A3 @ 30ʹ 90ʹNBCD4 @ 25ʹ 100ʹT Ct hn4 / 4[Ct 0.035 for steel moment frames](b)(c) 0.0441 (1)(0.9)(3870) 153.6 kipsT 0.75 sec.W 4(100 90) 90 lb/ft 2 (100 90) 70 lb/ft 2 3,870,000 lbs 3,870 kipsV V VmaxSD1 WT (R / I )0.4 (3870)I 1 for office bldgs.13wind pressuresin lb/ft2Vmin 0.0441 I SDS WT 0.035(60)3/415Building SectionP2.10Fundamental Period203 @ 20ʹ 60ʹ5 @ 12ʹ 60ʹPlan(a)Therefore, Use V 258 kipsI - 0.5 1.1252Wx hxkFx Vnå i 1Wi hikk 1 258 kips0.75(8/1)SDS W 0.9(3870) 8/1R/ I 435 kips2-19Copyright 2018 McGraw-Hill Education. All rights reserved.No reproduction or distribution without the prior written consent of McGraw-Hill Education.

P2.16. ContinuedForces at Each Floor LevelWx hxkFloorRoofWeight Wi, (kips)Floor Height hi (ft)Wi hikΣWi hikFx 01213,2610.06216.0 3,870 213,9612-20Copyright 2018 McGraw-Hill Education. All rights reserved.No reproduction or distribution without the prior written consent of McGraw-Hill Education. 258

P2.17. When a moment frame does not exceed 12 stories in height and the story height is at least 10 ft,the ASCE standard provides a simpler expression to compute the approximate fundamental period:T 0.1Nwhere N number of stories. Recompute T with the above expression and compare it with thatobtained from Problem P2.16. Which method produces a larger seismic base shear?ASCE Approximate Fundamental Period:T 0.1NN 5\ T 0.5seconds0.3 6750V 810 kips0.5(5/1)The simpler approximate method produces a larger value of base shear.2-21Copyright 2018 McGraw-Hill Education. All rights reserved.No reproduction or distribution without the prior written consent of McGraw-Hill Education.

P2.18. (a) A two-story hospital facility shown inFigure P2.18 is being designed in New Yorkwith a basic wind speed of 90 mi/h and windexposure D. The importance factor I is 1.15 andKz 1.0. Use the simplified procedure todetermine the design wind load, base shear, andbuilding overturning moment. (b) Use theequivalent lateral force procedure to determinethe seismic base shear and overturning moment.2The facility, with an average weight of 90 lb/ftfor both the floor and roof, is to be designed forthe following seismic factors: SDS 0.27g andSD1 0.06g; reinforced concrete frames with anR value of 8 are to be used. The importancefactor I is 1.5. (c) Do wind forces or seismicforces govern the strength design of thebuilding?15ʹ100ʹ15ʹ100ʹP2.18(a) Wind Loads Using Simplified Procedure:Design Wind Pressure Ps Kzt IPS30 1.66 Table 2.8, Mean Roof Height 30ʹPS 1.66(1)1.15PS 30 1.909 PS 30Zones PS 30AC12.8 psf8.5 psf24.44 psf16.22 psfResultant Force at Each Level; Where Distance a 0.1(100 ) 10 ; 0.4(30 ) 12 ; 3 a 10 Controls & 2a 20 Region (A)15 24.44 psf215 Zone (C):16.3psf2Froof : Zone (A):(( k k20 3.67) 100080 9.78) 1000Froof Resultant 13.45k(F2nd: Zone (A): 15 24.44 psf(Zone (C): 15 16.3 psf 20 7.33) 1000 k80 19.56)1000kF2nd Resultant 26.89kBase Sheak Vbase Froof F2nd 40.34 kOverturning MomentMO.T. Σ Fi hiMO.T. 13.45 (30 ) 26.89k (15 ) 806.9 ft . kk2-22Copyright 2018 McGraw-Hill Education. All rights reserved.No reproduction or distribution without the prior written consent of McGraw-Hill Education.

P2.18. Continued(b) Seismic Loads by Equivalent Lateral Force Procedure Given: W 90 psf Floor & Roof;SDS 0.27g, SD1 0.06g; R 8, I 1.5Base Shear Vbase SD1 WT (R/I)W Total Building Dead Load WhereWroof 90 psf (100 )2 900 kW2nd 90 psf (100 )2 900 kWtotal 1800 kT CT hnx 0.342 sec.AndCT 0.016 Reinf. Concrete FrameX 0.9 Reinf. Concrete Frameh 30 Building HeightVbase Vmax. Vmin.0.06 (1800 k )(0.342 sec)(8 /1.5 )SDS W 0.033W 59.2 kControls0.27 (1800 k ) 0.051W 91.1kR/I(8/1.5) 0.044 SDS IW 0.044 (0.27)(1.5)(1800 k ) 0.0178W 32.1kForce @ Each Level FX WX hxkΣWi hikVbase, Where Vbase 59.2 kT 0.5 Sec. Thus K 1.0HiWi hikWx hxk Wi hikForce @ Ea. Level:Roofk90030 270000.667Froof 39.5k2nd900k15 135000.333F2nd 19.76kLevelWiΣWi hik 40500ΣFx Vbase 59.2 kOverturning Moment MO.T. ΣFx hiM O.T. 39.5k (30 ) 19.76 k (15 ) 1,481.4 ft k(c) Seismic Forces Govern the Lateral Strength Design.2-23Copyright 2018 McGraw-Hill Education. All rights reserved.No reproduction or distribution without the prior written consent of McGraw-Hill Education.

P2.19. In the gabled roof structure shown inFigure P2.13, determine the sloped roof snowload Ps. The building is heated and is located ina windy area in Boston. Its roof consists ofasphalt shingles. The building is used for amanufacturing facility, placing it in a type IIoccupancy category. Determine the roof slopefactor, Cs using the ASCE graph shown inFigure P2.19. If roof trusses are spaced at 16 fton center, what is the uniform snow load along atruss?1.05 0.8roofs withobstructions ces withthermal resistance,R 30 F·h·ft2/Btu(5.3 C·m2/W) forunventilated roofsor R 20 F·h·ft2/Btu(3.5 C·m2/W) forventilated roofsCs0.40.20030 60 Roof Slope90 Roof slope factor Cswith warm roofs and Ct ed Roof Snow Load PS CS pfFrom Fig. P2.17 Cs is Approximately 0.9 (Non-SlipperyWhere pf Flat Roof Snow Loadpf 0.7 CeCt I pgSurface)Ce 0.7 Windy AreaPs Cs Ps 0.9 (19.6 psf ) 17.64 psfCt 1.0 Heated BuildingI 1.0 Type II OccupancyPg 40 psf for Bostonæ 16 öCs Based on Roof Slope θ Tan-1 çç 33.7 çè 24 øPf 0.7 (0.7)(1.0)(1.0)(40 psf ) 19.6 psfUniform Load Acting on Trusses Spaced @ 16 o.c.Wsnow 17.64 psf (16 ) 282.2 plf2-24Copyright 2018 McGraw-Hill Education. All rights reserved.No reproduction or distribution without the prior written consent of McGraw-Hill Education.

P2.20. A beam that is part of a rigid frame has end moments and mid-span moments for dead, live,and earth-quake loads shown below. Determine the governing load combination for both negative andpositive moments at the ends and mid-span of the beam. Earthquake load can act in either direction,generating both negative and positive moments in the beam.End Moments (ft-kip)Mid-Span Moments (ft-kip)Dead Load 180 90 150Live LoadEarthquake 80 1500Load Combinations-Factored StrengthEnd Moments(1.4 DL 1.4 -180 ft k)( -252 ft k1.2 DL 1.6 LL 0.5 LR or S)O 1.2 (-180) 1.6 (-150) -456 ft k *1.2 DL 1.0 E LL 0.2 (S ) 1.2 (-180) (-80) (-150) -446 ft kOMid-Span Moments(1.4 DL 1.4 90 ft k()1.2 DL 1.6 LL 0.5 LR or S 126 ft k)O 1.2 ( 90) 1.6 ( 150) 348ft k *O1.2 DL 1.0 E LL 0.2 (S ) 1.2 (90) 0 (150) 258ft kBeam Needs to be Designed for Max. End Moment 456 ft kMax. Mid-Span Moment 348 ft k2-25Copyright 2018 McGraw-Hill Education. All rights reserved.No reproduction or distribution without the prior written consent of McGraw-Hill Education.

P2.21. Calculate the vertical hydrostatic load onthe 5100-lb empty shipping container in FigureP2.19 subjected to a tsunami inundation heightof 3ʹ. Assuming the container is water-tight, willthe tsunami wave be capable of carrying awaythe container as debris?8.6ʹ20ʹ8ʹP2.21FV γsVW 70.4(3)(8)(20)FV 33792 lbsFV 33.8 kips33.8 kips Wcontainer 5.1 kipsYes, the container will be carried away.2-26Copyright 2018 McGraw-Hill Education. All rights reserved.No reproduction or distribution without the prior written consent of McGraw-Hill Education.

30ʹP2.22. Consider the building in Figure P2.22,which has a width into the page of 35 ft.Maximum inundation height, hmax, and flowvelocity, umax, have been determined as 33 ftand 20 ft/sec, respectively. Calculate thehydrodynamic and hydrostatic resultant load andlocation on the walls ABC and IJKL for LoadCases 2 and 3, due to both inflow and outflowdirections. If windows are inundated, calculatethe expected hydrostatic loading on the adjacentoutside walls due to water retained by the floor,or floors. Finally, calculate the debris impactload to be applied to the free-standing columnCD. Assume Itsu 1.0 and Cd 1.25.Load Case 2:23IN Flow30ʹAEIBFJGKHL16ʹOUT Flow 16ʹ3ʹC16ʹOPENDP2.22Load Case 3:hmax hdes 33 fthmax hdes 22 ft1umax udes 20 ft/sec3Jumax udes 6.67 ft/sec3 ftK6 ftHydrodynamic, Load Case 2Hydrodynamic, Load Case 38 fthdes , K Trib height 8 6 14 fthdes , J Trib height 1 8 9 ft8 ftFdK FdK 1212FdJ γ s ( I tsu )(C d )(C cx )( B )(hdes , K )(20 )21270.4(1.0)(1.25)(1.0)(35)(9)(6.67 )2FdJ 616 kips70.4(1.0)(1.25)(1.0)(35)(14)(20 )2hdes , K Trib height 8 8 16 ftFdK 8624 kips12FdK 70.4(1.0)(1.25)(1.0)(35)(16)(6.67 )2FdK 1096.2 kipsHydrostatic on interior wallsFh 1γ s bhdes 2170.4(35)3Load Case 2J1 ft8 ft3 ftKFdK8 ft8 ft22Fh 11.1 kipsLoad Case 3J3 ft33 ftFhFdJ8 ft2Debris Impact on CDFdKKFi 330C o I tsu 330(0.65)(1.0)Fh 214.5 kipsHydrostatic on inside walls2-27Copyright 2018 McGraw-Hill Education. All rights reserved.No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Design Loads andStructural FramingChapter 2Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.1

Chapter OpenerCopyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.2

Figure 2.1Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.3

Figure 2.2Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the pr

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 Fundamentals of Structural Analysis 5th Edition Leet Solutions Manual