PCI 6th Edition

1. PCI 6th Edition Building Systems (Snow + Wind)

2. Presentation Outline Building System Loads Snow Uniform loading and drifting Example Wind Main lateral wind resisting system Component example

3. Structural Systems Gravity Load Systems Beams Columns Floor Member – Double Tees, Hollow Core Spandrels

4. Structural Systems Lateral Load Systems Shear Walls Moment Resisting Frames Cantilever Columns Braced Frames – K Frames

5. System Loads Dead Loads Live Loads Snow Loads Roof / Ground Drifting Wind Loads Earthquake Loads This session willplace emphasison these forces.

6. Snow Loads Based on ASCE 7 Different than other live loads due to transient nature

7. Load Combinations U = 1.2D +1.6(Lr or S or R) + (1.0L or 0.8W) U = 1.2D + 1.6W + 1.0L + 0.5(Lr or S or R) U = 1.2D + 1.0E + f1L + 0.2S

8. Snow Loads Roof Snow Load, pf is now based on the Ground Snow Load, pg pf = 0.7·Ce·Ct·?·pg Where pf = flat roof snow load (psf) Ce = exposure factor Ct = thermal factor ? = importance factor pg = ground snow load (psf)

9. Snow Loads Limited by pf = ?·pg where pg = 20 psf pf = 20·? where pg > 20 psf

10. Exposure Factor, Ce – Page 3-106

11. Thermal Factor, Ct – Page 3-106

12. Drifting Loads Consideration for Windward and Leeward

13. Snow Drift Configuration If hc/hb = 0.2, drift loads need not be applied

14. Otherwise (Example) Given: A flat roofed office building 450 ft long has a 50 ft long, 8 ft high penthouse centered along the length. The building is located in downtown Milwaukee, Wisconsin.

15. Snow Load Example Problem: Determine the following Roof Snow Load. pf Drift Load and Location

16. Solution Steps Step 1 – Calculate the Ground Snow Load Step 2 – Calculate Drift Requirements Step 3 – Calculate Balanced Snow Height Step 4 – Determine if drifting is considered Step 5 – Determine Drift Height Step 6 – Drift Force

17. Step 1 – Roof Snow Load, pf Recall - pf = 0.7·Ce ·Ct ·I

18. Step 1 – Terrain Category Assumption – Within the life of the structure, taller buildings may be built around it Exposure B Ce =1.2

19. Step 1 – Thermal Condition Office - Heated Structure Ct = 1.0

20. Step 1 – Importance factor For office buildings ? = 1.0 From Figure 3.10.1 pg 3-103 Building Category II

21. Step 1 – Determine Ground Snow Load Downtown Milwaukee, Wisconsin (pg 3-105 figure 3.10.2) Space bar brings up map zoom and highlight circle for city on map.Space bar brings up map zoom and highlight circle for city on map.

22. Step 1 - Alternate Special Case Study Region

23. Step 1 – Roof Snow Load, pf pf = 0.7·Ce ·Ct ·I ·pg = 0.7(1.2)(1.0)(1.0)(30) = 25.2 psf

24. Step 2 – Calculate Drift Requirements Balanced snow load height - hb

25. Step 3 – Balanced Snow Height, hb hb = Roof Snow Load / Unit Weight = pf/g Unit Weight of Snow, g g = 0.13·pg + 14 = 30 pcf = 0.13(30 pcf ) + 14 pcf = 17.9 pcf 17.9 = 30 pcf hb = pf/ g = 25.2 psf /17.9 pcf = 1.40 ft Balanced snow load Balanced snow load

26. Step 4 – Determine if Drifting is Considered hc – penthouse height in this case = 8.0 ft hc/hb = (8.0 -1.4)/1.4 = 4.7 4.7 > 0.2 Drifting must be considered! Recall the limit was 0.2Recall the limit was 0.2

27. Step 5 – Determine Drift Height Leeward lu = 50ft hd ~ 2.5ft Windward lu = 200ft hd = 75% (Graph value) hd ~ 0.75 ( 4.8) = 3.6ft

28. Step 6 – Drift Force Drift Width = w = 4·hd 4(3.6ft)=14.4ft Force = ½·g·hd·w ½ (17.9pcf)(3.6)(14.4) = 464 plf Location - acts 1/3·w from penthouse wall 1/3(14.4) = 4.8ft Space bar bring up for arrow and calculation and then brings up dimension line and calculationSpace bar bring up for arrow and calculation and then brings up dimension line and calculation

29. Wind Load Method Presented ASCE 7 – 02 “Method 1 – Simplified Procedure”

30. Wind Load Limitations of Simplified Procedure Height = 60 ft or least lateral dimension Enclosed building (includes parking structures) Regular shaped No expansion joints Fundamental frequency = 1 Hz. Flat or shallow pitched roof No unusual topography around the building Fundamental frequency = 1 Hz. (Nearly all concrete buildings under 60 ft will qualify) Fundamental frequency = 1 Hz. (Nearly all concrete buildings under 60 ft will qualify)

31. Wind Load Procedure Determine Basic wind speed Directionality Factor Exposure Pressure Zone Load per unit area Importance Factor

32. Determine the Basic Wind Speed Basic Wind Speed Chart (pg 3-108, 3-109)

33. Determine the Directionality Factor The directionality factor is 0.85 for buildings For additional Factors Minimum Design Loads for Buildings and Other Structures, Revision of ASCE 7-98 (SEI/ASCE 7-02), American Society of Civil Engineers, Reston, VA, 2003 (Co-sponsored by the Structural Engineering Institute). includes more detailed descriptions)

34. Determine Exposure Category Applies to upwind direction Exposure B: Urban and suburban areas, wooded areas Exposure D: Flat, unobstructed areas outside hurricane-prone regions Exposure C: All others

35. Determine the Pressure Zones

36. Pressure on Lateral System where: ps = combined windward and leeward net pressures 30 – represents average 30 ft building height

37. Pressure on Lateral System where: l = building height and exposure coefficient from Figure 3.10.6(c) pg 3-110

38. Pressure on Lateral System where: ? = importance factor for wind Figure 3.10.1

39. Pressure on Lateral System where: ps30 = simplified design wind pressure from Figure 3.10.6(a) pg 3-110

40. Wind Load Force The force on the MWFRS is then determined by multiplying the values of ps30 by their respective zone areas Zone A can be at both ends of the structure

41. Cladding Pressure Determined from: pnet = ?·?·pnet 30

42. Cladding Wind Load Example Given: A 114 ft wide by 226 ft long by 54 ft tall hospital building in Memphis, TN. Cladding panels are 7 ft tall by 28 ft long. A 6 ft high window is attached to the top of the panel, and an 8 ft high window is attached to the bottom.

43. Cladding Wind Load Example Problem: Part A Determine the design wind load on the MWFRS Part B Determine the design wind load on the cladding panels. Solution Method: As this is an enclosed building under 60 ft high, Method 1 may be used. Suburban Area - Exposure Category B

44. Part A – Solution Steps (MWFRS) Step 1MWFRS – Determine Wind Speed Step 2MWFRS – Determine Zone Coefficients Step 3MWFRS – Calculate Zone Pressure Step 4MWFRS – Calculate Zone Area Step 5MWFRS – Calculate Zone Force Step 6MWFRS – Calculate Force Location

45. Step 1MWFRS – Determine Wind Speed Figure 3.10.5 (page 3-109) Memphis, TN

46. Step 2MWFRS – Zone Coefficient, l Height / Exposure Coefficient Building Height – 54ft, Exposure - B Figure 3.10.6(c) (pg 3-110) Height Exposure coeficient. Linearly interpolate between 50 and 55 ft for Exposure B. Figure 3.10.6(c) (pg 3-110) Space Bar highlights interpotating values and equation.Height Exposure coeficient. Linearly interpolate between 50 and 55 ft for Exposure B. Figure 3.10.6(c) (pg 3-110) Space Bar highlights interpotating values and equation.

47. Step 2MWFRS – Zone Coefficient, ps 30 Pressure Coefficient From Table 3.10.6(a) (pg 3-110) Zone A ps 30 = 12.8 psf Zone C ps 30 = 8.5 psf Pressure Coeficient From Table 3.10.6(a) (pg 3-110) Space Bar highlights interpotating values and equation. Pressure Coeficient From Table 3.10.6(a) (pg 3-110) Space Bar highlights interpotating values and equation.

48. Step 2MWFRS – Zone Coefficient, I Importance Factor From Table 3.10.1 (page3-103) ? = 1.15 Importance Factor Table 3.10.1 (page3-103) Space highlights value.Importance Factor Table 3.10.1 (page3-103) Space highlights value.

49. Step 3MWFRS – Calculate Zone Pressures ps zone A = ?·?·ps 30 zone A 1.18(1.15)(12.8) = 17.4 psf ps zone C = ?·?·ps 30 zone C 1.18(1.15)(8.5) = 11.5 psf

50. Step 4MWFRS – Calculate Zone A Dimensions Length of building – 226 ft Lesser of 0.2(114) = 22.8 Or 0.8(54) = 43.2 A226 = 22.8 ft

51. Step 4MWFRS – Calculate Zone C Dimensions Length of building – 226 ft C226 = 226 – 22.8 = 203.2 ft

52. Step 5MWFRS – MWFRS Zone Forces F1 = A226 ·h · ps Zone A = 22.8(54)(17.4)/1000 = 21.4 kips F2 = C226 ·h · ps Zone C = 203.2(54)(11.5)/1000 = 126.2 kips Total force = 21.4 + 126.2 = 147.6 kips

53. Step 6MWFRS – MWFRS Forces Location F1 = 21.4 kips, F2 = 126.2 kips Resultant Location, eleft

54. Part B – Solution Steps (Cladding) Step 1Clad – Determine Wind Speed Step 2Clad – Determine Zone Coefficients Step 3Clad – Calculate Tributary Area Step 4Clad – Calculate Zone Pressure Step 5Clad – Calculate Panel Force Step 6Clad – Calculate Window Force

55. Step 3Clad – Calculate the Cladding Tributary Area Tributary area per panel = one-half of upper window + panel + one-half of lower window times the width (6/2 + 7 + 8/2)(28) = 392 ft2

56. Step 4Clad – Cladding pnet 30 Zone Pressure Table 3.10.6(b) (page 3-110) Interpolating panel area between 100 and 500 ft2:

57. Step 4Clad – Cladding pnet 30 Zone Pressure Inward pressure – Zone A 12.4 – 0.73(12.4 – 10.9) = 11.3 psf ?·?·pnet30 = 1.18(1.15)(11.3) = 15.3 psf The largest inward pressure is in Zone AThe largest inward pressure is in Zone A

58. Step 4Clad – Cladding pnet 30 Zone Pressure Outward pressure – Zone A 15.1 – 0.73(15.1 – 12.1) = 12.9 psf ?·?·pnet30 = 1.18(1.15)(12.9) = 17.5 psf Largest outward or suction is in Zone ALargest outward or suction is in Zone A

59. Step 4Clad – Cladding pnet 30 Wind Force Panel Size Height – 7 ft Length – 28 ft Force on panel: Inward: 7.0(28)(15.3) = 2999 lb Outward: 7.0(28)(17.5) = 3430 lb

60. Step 5Clad – Window Forces Force on panel from upper window: Inward: (6.0/2)(28)(15.3) = 1285.2 lb Outward: (6.0/2)(28)(17.5) = 1470 lb

61. Step 5Clad – Window Forces Force on panel from lower window: Inward: (8.0/2)(28)(15.3) = 1713.6 lb Outward: (8.0/2)(28)(17.5)= 1960 lb

62. Step 6Clad – Resultant Cladding FDB

63. Questions?