FRAME ANALYSIS. Dr. Izni Syahrizal bin Ibrahim. Faculty of Civil Engineering Universiti Teknologi Malaysia

Transcription

1 FRAME ANALYSIS Dr. Izni Syahrizal bin Ibrahim Faculty of Civil Engineering Universiti Teknologi Malaysia

2 Introduction 3D Frame: Beam, Column & Slab 2D Frame Analysis Building Construction 3D Frame: Beam & Column ONLY 2D Frame Analysis

3 Introduction: Substitute Frame 2D Frame Analysis Sub-frame: Roof Level Sub-frame: Other Floor Level

4 Types of Frame Braced Frame NOT contribute to the overall structural stability None lateral actions are transmitted to columns & beams Support vertical actions only Shear wall as bracing members Unbraced Frame Contribute to the overall structural stability All lateral actions are transmitted to columns & beams Support vertical & lateral actions

5 Methods of Analysis (i) One-level Sub-frame (ii) Two-points Sub-frame K b1 0.5K b2 0.5K b1 K b2 0.5K b3

6 Methods of Analysis (iii) Continuous Beam and One-point Sub-frame 0.5K b 0.5K b 0.5K b 0.5K b

7 Combination of Actions BS EN : Cl Load Set 1: Alternate or adjacent spans loaded Alternate span carrying the design variable and permanent load (1.35G k + 1.5Q k ), other spans carrying only the design permanent loads (1.35G k ) Any two adjacent spans carrying the design variable and permanent loads (1.35G k + 1.5Q k ), all other spans carrying only the design permanent load (1.35G k )

8 Combination of Actions Load Set 1 Alternate Span Loaded Adjacent Span Loaded 1.5Q k 1.5Q k 1.35G k

9 Combination of Actions MALAYSIAN / UK NATIONAL ANNEX Load set 2: All or alternate spans loaded All span carrying the design variable and permanent loads (1.35G k + 1.5Q k ) Alternate span carrying the design variable and permanent load (1.35G k + 1.5Q k ), other spans carrying only the design permanent loads (1.35G k )

10 Combination of Actions All Spans Loaded Load Set 2 1.5Q k 1.35G k Alternate Span Loaded

11 Example 1 ACTION ON BEAMS

12 Example 1 2@4 m = 8 m 2@4 m = 8 m 7@5 m = 35 m 6@3.5 m = 21 m 4 m 1 m 6 m 6 m 6 m Level 7 Level 6 Level 5 Level 4 Level 3 Level 2 Level 1 6 m 6 m 6 m

13 Example 1: Action on Beams Loading Data Finishes, ceiling & services = 0.75 kn/m 2 Concrete density = 25 kn/m 3 Variable action = 4 kn/m 2 Partition = 0.50 kn/m 2 Structural Dimensions Slab thickness, h Beam size, b h Column size, b h = 150 mm = mm = mm

14 Example 1: Action on Beams Load on Slab Slab self weight = 0.15 m 25 kn/m 3 = 3.75 kn/m 2 Finishes, ceiling, services = 1.25 kn/m 2 Characteristics permanent action, g k = 5.00 kn/m 2 Imposed load = 4.00 kn/m 2 Partition = 0.50 kn/m 2 Characteristics variable action, q k = 4.50 kn/m 2 Other Loads on Beam Beam self weight = 0.25 m ( ) m 25 kn/m 3 = 2.81 kn/m

15 Example 1: Action on Beams L y = 6 = 1.50 (Case 2) L x 4 L y = 6 = 1.50 (Case 1) L x 4 L y = 6 = 1.50 (Case 2) L x m m 6 m 6 m 6 m L y = 6 = 1.20 (Case 2) L x 5 L y = 6 = 1.20 (Case 1) L x 5 L y = 6 = 1.20 (Case 2) L x 5

16 Example 1: Action on Beams G k = Permanent action on slab + Beam self weight = ( ) + ( ) = kn/m Q k = Variable action on slab = ( ) + ( ) = kn/m G k = Permanent action on slab + Beam self weight = ( ) + ( ) = kn/m Q k = Variable action on slab = ( ) + ( ) = kn/m

17 Example 1: Action on Beams w max = 1.35G k + 1.5Q k = kn/m w min = 1.35G k = kn/m w max = 1.35G k + 1.5Q k = kn/m w min = 1.35G k = kn/m w max = 1.35G k + 1.5Q k = kn/m w min = 1.35G k = kn/m 6 m 6 m 6 m

18 Example 2 ANALYSIS OF ONE-LEVEL SUB-FRAME

19 Example 2 2@4 m = 8 m 2@4 m = 8 m 7@5 m = 35 m 6@3.5 m = 21 m 4 m 1 m 6 m 6 m 6 m Level 7 Level 6 Level 5 Level 4 Level 3 Level 2 Level 1 6 m 6 m 6 m

20 Example 2: Analysis of One Level Sub- Frame Analysis using LOAD SET m w max = kn/m w min = kn/m w max = kn/m w min = kn/m w max = kn/m w min = kn/m K c,u K AB K BC K CD 4 m K c,l 6 m 6 m 6 m A B C D

21 Example 2: Analysis of One Level Sub- Frame Structural Dimension Beam: 250 mm 600 mm Column: 300 mm 400 mm Moment of Inertia, I = bh 3 /12 Beam: Column: = mm 4 = mm 4 Stiffness, K = I/L Beam: K AB = K BC = K CD = = mm 3 Column: K c,u = = mm 3 K c,l = = mm 3

22 Example 2: Analysis of One Level Sub- Frame Distribution Factor, F = K/ K Joint A & Joint D: F AB, CD = F c,u = F c,l = Joint B & Joint C: F BA, CB = F BC, CD = F c,u = F c,l = K AB, CD K AB, CD +K c,u +K c,l = 0.47 K c,u K AB, CD +K c,u +K c,l = 0.28 K c,l K AB, CD +K c,u +K c,l = 0.25 K AB, BC K AB, BC +K BC, CD +K c,u +K c,l = 0.32 K BC, CD K AB, BC +K BC, CD +K c,u +K c,l = 0.32 K c,u K AB, BC +K BC, CD +K c,u +K c,l = 0.19 K c,l K AB, BC +K BC, CD +K c,u +K c,l = 0.17

23 Example 2: Analysis of One Level Sub- Frame CASE 1: Span 1 & 2 = w max, Span 3 = w min 3.5 m w max = 57.5 kn/m w max = 54.4 kn/m w min = 30.7 kn/m Span 1 Span 2 Span 3 4 m 6 m 6 m 6 m A B C D Fixed End Moment M AB = M BA = wl2 12 M BC = wl2 12 M CD = wl = = 12 = = knm = knm = 92.0 knm

24 Example 2: Analysis of One Level Sub- Frame Moment Distribution Method A B 0.19 C 0.19 D

25 Example 2: Analysis of One Level Sub- Frame Moment in the COLUMN at each joint A B 0.19 C 0.19 D

26 Example 2: Analysis of One Level Sub- Frame Moment in the BEAM at each joint A B 0.19 C 0.19 D

27 Example 2: Analysis of One Level Sub- Frame Reaction Calculation at Support w max = 57.5 kn/m w max = 54.4 kn/m w min = 30.7 kn/m m 6 m 6 m V AB V BA V BC V CB V CD V DC Solve for each span using equilibrium method of analysis: V AB = kn V BA = kn V BC = kn V CB = kn V CD = kn V DC = 78.5 kn

28 Example 2: Analysis of One Level Sub- Frame Shear Force and Bending Moment Diagram SFD (kn) 2.7 m 3.1 m 3.4 m BMD (knm)

29 Example 2: Analysis of One Level Sub- Frame Do the same for Load CASE 2, CASE 3 & CASE 4 SELF STUDY!!! CASE 2: Span 2 & 3 = w max, Span 1 = w min 3.5 m w min = 30.7 kn/m w max = 54.4 kn/m w max = 57.5 kn/m Span 1 Span 2 Span 3 4 m 6 m 6 m 6 m A B C D

30 Example 2: Analysis of One Level Sub- Frame Do the same for Load CASE 2, CASE 3 & CASE 4 SELF STUDY!!! CASE 3: Span 1 & 3 = w max, Span 2 = w min 3.5 m w max = 57.5 kn/m w min = 29.1 kn/m w max = 57.5 kn/m Span 1 Span 2 Span 3 4 m 6 m 6 m 6 m A B C D

31 Example 2: Analysis of One Level Sub- Frame Do the same for Load CASE 2, CASE 3 & CASE 4 SELF STUDY!!! CASE 4: Span 1 & 3 = w min, Span 2 = w max 3.5 m w min = 30.7 kn/m w max = 54.4 kn/m w min = 30.7 kn/m Span 1 Span 2 Span 3 4 m 6 m 6 m 6 m A B C D

32 Example 2: Analysis of One Level Sub- Frame Shear Force and Bending Moment Envelope SFD (kn) Case 1 Case 2 Case 3 Case 4 BMD (knm) 161.9, , , , , , ,

33 Example 2: Analysis of One Level Sub- Frame Shear Force and Bending Moment Envelope SFD (kn) Case 1 Case 2 Case 3 Case , , , , , , BMD (knm)

34 Example 3 ANALYSIS OF TWO FREE- JOINT SUB-FRAME

35 Example 3 2@4 m = 8 m 2@4 m = 8 m 7@5 m = 35 m 6@3.5 m = 21 m 4 m 1 m 6 m 6 m 6 m Level 7 Level 6 Level 5 Level 4 Level 3 Level 2 Level 1 6 m 6 m 6 m

36 Span AB Example 3: Analysis of Two Free-Joint Sub- Frame CASE 1: Span 1 & 2 = w max 3.5 m w max = 57.5 kn/m w max = 54.4 kn/m 4 m Span 1 Span 2 6 m 6 m C A B Stiffness, K = I/L Beam: K AB = = mm 3 K BC = = mm 3 Column: K c,u = = mm 3 K c,l = = mm 3

37 Span AB Example 3: Analysis of Two Free-Joint Sub- Frame Fixed End Moment M AB = wl2 12 M BC = wl2 12 Distribution Factor, F = K/ K Joint A: F AB = F c,u = F c,l = Joint B: F BA = = = 12 F BC = F c,u = F c,l = = = = = = = knm = knm = = 0.20 Moment Distribution Method A B

38 Span AB Example 3: Analysis of Two Free-Joint Sub- Frame 3.5 m CASE 2: Span 1 = w max, Span 2 = w min w max = 57.5 kn/m w min = kn/m Span 1 Span 2 4 m 6 m 6 m A B C Moment Distribution Method A B

39 Span AB Example 3: Analysis of Two Free-Joint Sub- Frame Reaction Calculation at Support CASE 1 w max = 57.5 kn/m CASE 2 w max = 57.5 kn/m m 6 m V AB V BA V AB V BA Solve for each span using equilibrium method of analysis: V AB = kn V BA = kn Solve for each span using equilibrium method of analysis: V AB = kn V BA = kn

40 Span AB Example 3: Analysis of Two Free-Joint Sub- Frame Shear Force and Bending Moment Diagram SFD (kn) BMD (knm)

41 Span BC Example 3: Analysis of Two Free-Joint Sub- Frame CASE 1: Span 1 & 2 = w max, Span 3 = w min 3.5 m w max = 57.5 kn/m w max = 54.4 kn/m w min = 30.7 kn/m Span 1 Span 2 Span 3 4 m A D 6 m 6 m 6 m B C

42 Span BC Example 3: Analysis of Two Free-Joint Sub- Frame Fixed End Moment M AB = M BA = wl2 12 M BC = M CB = wl2 12 M CD = M DC = wl2 12 Stiffness, K = I/L = = = 12 = knm = knm = 92.0 knm Beam: K AB = K BC = = mm 3 K BC = = mm 3 Column: K c,u = = mm 3 K c,l = = mm 3 Distribution Factor, F = K/ K Joint B: F BA = F BC = F c,u = F c,l = Joint C: F CB = F CD = F c,u = F c,l = = = = = = = = = 0.20

43 Span BC Example 3: Analysis of Two Free-Joint Sub- Frame Moment Distribution Method B 0.23 C

44 Span BC Example 3: Analysis of Two Free-Joint Sub- Frame CASE 2: Span 2 = w max, Span 1 & 3 = w min 3.5 m w max = 30.7 kn/m w max = 54.4 kn/m w min = 30.7 kn/m Span 1 Span 2 Span 3 4 m A D 6 m 6 m 6 m B C

45 Span BC Example 3: Analysis of Two Free-Joint Sub- Frame Fixed End Moment M AB = M BA = wl2 12 M BC = M CB = wl2 12 M CD = M DC = wl2 12 Stiffness, K = I/L = = = 12 = 92.0 knm = knm = 92.0 knm Beam: K AB = K BC = = mm 3 K BC = = mm 3 Column: K c,u = = mm 3 K c,l = = mm 3 Distribution Factor, F = K/ K Joint B: F BA = F BC = F c,u = F c,l = Joint C: F CB = F CD = F c,u = F c,l = = = = = = = = = 0.20

46 Span BC Example 3: Analysis of Two Free-Joint Sub- Frame Moment Distribution Method B 0.23 C

47 Span BC Example 3: Analysis of Two Free-Joint Sub- Frame Reaction Calculation at Support CASE 1 w max = 54.4 kn/m CASE 2 w max = 54.4 kn/m m 6 m V BC V CB V BC V CB Solve for each span using equilibrium method of analysis: V BC = kn V CB = kn Solve for each span using equilibrium method of analysis: V CB = kn V CB = kn

48 Span BC Example 3: Analysis of Two Free-Joint Sub- Frame Shear Force and Bending Moment Diagram SFD (kn) BMD (knm)

49 Example 4 CONTINUOUS BEAM + ONE FREE-JOINT SUB-FRAME

50 Example 4 2@4 m = 8 m 2@4 m = 8 m 7@5 m = 35 m 6@3.5 m = 21 m 4 m 1 m 6 m 6 m 6 m Level 7 Level 6 Level 5 Level 4 Level 3 Level 2 Level 1 6 m 6 m 6 m

51 Example 4: Continuous Beam + One Free- Joint Sub-Frame Load Set 1 6 m 6 m 6 m A B C D Moment of Inertia I beam = bh3 12 = Stiffness, K = I/L = mm 4 K AB = K CD = 0.75 I beam = 0.75 L = mm 3 K BC = I beam = L = mm 3 Distribution Factor, F = K/ K Joint A: F BA = = 1.00 Joint B: F BA = = F BC = = Joint C: F BC = = 0.57 F CD = = 0.43 Joint D: F DC = = 1.00

52 Example 4: Continuous Beam + One Free- Joint Sub-Frame CASE 1: Span 1 & 2 = w max, Span 3 = w min w max = 57.5 kn/m w max = 54.4 kn/m w min = 30.7 kn/m Span 1 Span 2 Span 3 6 m 6 m 6 m A B C D Fixed End Moment, FEM M AB = M BA = wl2 12 M BC = M CB = wl2 12 M CD = M DC = wl = = = 12 = knm = knm = 92.0 knm

53 Example 4: Continuous Beam + One Free- Joint Sub-Frame Moment Distribution Method A B C D

54 Example 4: Continuous Beam + One Free- Joint Sub-Frame Reaction Calculation at Support w max = 57.5 kn/m w max = 54.4 kn/m w min = 30.7 kn/m 6 m 6 m 6 m V AB V BA V BC V CB V CD V DC Solve for each span using equilibrium method of analysis: V AB = kn V BA = kn V BC = kn V CB = kn V CD = kn V DC = 69.1 kn

55 Example 4: Continuous Beam + One Free- Joint Sub-Frame Do the same for Load CASE 2, CASE 3 & CASE 4 SELF STUDY!!! CASE 2: Span 2 & 3 = w max, Span 1 = w min w min = 30.7 kn/m w max = 54.4 kn/m w max = 57.5 kn/m Span 1 Span 2 Span 3 6 m 6 m 6 m A B C D

56 Example 4: Continuous Beam + One Free- Joint Sub-Frame Do the same for Load CASE 2, CASE 3 & CASE 4 SELF STUDY!!! CASE 3: Span 1 & 3 = w max, Span 2 = w min w max = 57.5 kn/m w min = 29.1 kn/m w max = 57.5 kn/m Span 1 Span 2 Span 3 6 m 6 m 6 m A B C D

57 Example 4: Continuous Beam + One Free- Joint Sub-Frame Do the same for Load CASE 2, CASE 3 & CASE 4 SELF STUDY!!! CASE 4: Span 2 = w max, Span 1 & 3 = w min w min = 30.7 kn/m w max = 54.4 kn/m w min = 30.7 kn/m Span 1 Span 2 Span 3 6 m 6 m 6 m A B C D FINALLY DO THE SFD & BMD ENVELOPE!!!

58 Example 4: Continuous Beam + One Free- Joint Sub-Frame Sub-Frame: Bending Moment in Column Column A 3.5 m K c,u w max = 57.5 kn/m Fixed End Moment, FEM M AB = wl2 12 = Moment of Inertia, I = bh 3 /12 = knm 4 m K c,l 0.5K AB 6 m Beam: Column: = mm 4 = mm 4 Stiffness, K = I/L Beam: K AB = K BC = K CD = = mm 3 Column: K c,u = = mm 3 K c,l = = mm 3

59 Example 4: Continuous Beam + One Free- Joint Sub-Frame Sub-Frame: Bending Moment in Column Moment in upper column, M upper K M upper = FEM c,u = K c,u +K c,l +0.5K AB = knm Moment in lower column, M lower K M upper = FEM c,l = K c,u +K c,l +0.5K AB = knm

60 Example 4: Continuous Beam + One Free- Joint Sub-Frame Sub-Frame: Bending Moment in Column Column B 3.5 m w max = 57.5 kn/m K c,u w min = 29.1 kn/m 0.5K AB 0.5K BC Fixed End Moment, FEM M BA = wl2 12 M BC = wl = = 12 = knm = 87.3 knm 4 m K c,l 6 m 6 m FEM = M AB M BC = knm

61 Example 4: Continuous Beam + One Free- Joint Sub-Frame Sub-Frame: Bending Moment in Column Moment in upper column, M upper K M upper = FEM c,u = 85.3 K c,u +K c,l +0.5K AB +0.5K BC = knm Moment in lower column, M lower K M upper = FEM c,l = 85.3 K c,u +K c,l +0.5K AB +0.5K BC = knm

62 Example 4: Continuous Beam + One Free- Joint Sub-Frame Sub-Frame: Bending Moment in Column Column C Fixed End Moment, FEM 3.5 m w min = 29.1 kn/m K c,u w max = 57.5 kn/m 0.5K BC 0.5K CD M BC = wl2 12 M CD = wl = = 12 = 87.3 knm = knm 4 m K c,l 6 m 6 m FEM = M CD M BC = knm

63 Example 4: Continuous Beam + One Free- Joint Sub-Frame Sub-Frame: Bending Moment in Column Moment in upper column, M upper K M upper = FEM c,u = 85.3 K c,u +K c,l +0.5K BC +0.5K CD = knm Moment in lower column, M lower K M upper = FEM c,l = 85.3 K c,u +K c,l +0.5K BC +0.5K CD = knm

64 Example 4: Continuous Beam + One Free- Joint Sub-Frame Sub-Frame: Bending Moment in Column Column D Fixed End Moment, FEM w max = 57.5 kn/m K c,u 3.5 m M DC = wl2 12 = = knm 0.5K CD 6 m K c,l 4 m

65 Example 4: Continuous Beam + One Free- Joint Sub-Frame Sub-Frame: Bending Moment in Column Moment in upper column, M upper K M upper = FEM c,u = K c,u +K c,l +0.5K CD = knm Moment in lower column, M lower K M upper = FEM c,l = K c,u +K c,l +0.5K CD = knm

66 Wind Action Intensity of wind action on a structure related to: a) Square of wind velocity b) Member dimensions that resist the wind Wind velocity dependent on: a) Geographical location b) Height of structure c) Topography area d) Roughness of the surrounding terrain Response of the structure from the variable wind action: a) Background component b) Resonant component

67 Wind Action Background Component Static deflection of the structure under wind pressure Resonant Component Dynamic vibration of the structure under wind pressure Relatively small Normally treated using static method of analysis

68 Wind Action Plan View Side View Wind Separation point Separation Zone ( ) 15 ( ) Windward Stagnation point Wake Leeward (+) ( ) Effect of Wind on Buildings

69 Wind Action Calculation of Wind Pressure (MS 1553: 2002) Simplified Method Building of rectangular in plan Height 15 m Analytical Procedure Rectangular buildings Height 200 m Roof span 100 mm Wind Tunnel Procedure Complex buildings

70 Wind Action: Simplified Method Appendix A, MS 1553: Determined basic wind speed, V s (Figure A1, MS 1553: 2002) 2. Determine the terrain/height multiplier, M z,cat (Table A1, MS 1553: 2002) 3. Determine external pressure coefficient, C p,e for surface of enclose building (A2.3 and A2.4, MS 1553: 2002) 4. Determine internal pressure coefficient, C p,i for surface of enclose building. Taken as +0.6 or 0.3 and shall be considered to determine the critical load requirements. 5. Design wind pressure, p (in Pa): p = V s 2 M z,cat 2 Cp,e C p,i

71 Wind Action: Simplified Method Figure A1, MS 1553: 2002

72 Wind Action: Simplified Method

73 Wind Action: Simplified Method

74 Wind Action: Simplified Method

75 Wind Action: Analytical Procedure Section 2, MS 1553: 2002 Design wind pressure (in Pa): p = V des 2 C fig C dyn V des = V sit l C dyn = 1. 0 if > 1 Hertz (unless the structure is wind sensitive) C fig = C p,e K a K c K l K p V sit = V s M d M z,cat M s M h If others, see Next

76 Wind Action: Analytical Procedure V des = Design wind speed l = Importance factor (Table 3.2, MS 1553: 2002) V s = Basic wind speed (33.5 m/s: Zone 1 & 32.5 m/s: Zone 2) M d = Wind directional multiplier = 1.0 M z, cat = Terrain/Height multiplier (Table 4.1, MS 1553: 2002) M s = Shielding multiplier (Table 4.3, MS 1553: 2002) = 1.0 if effects of shielding is ignored or not applicable M h = Hill shape multiplier = 1.0 except for particular cardinal direction in the local topographic zones C fig = Aerodynamic shape factor C p,e = External pressure coefficient (Table 5.2(a) & 5.2(b), MS 1553: 2002) C dyn = Dynamic response factor K a = Area reduction factor = 1.0 (in most cases) K c = Combination factor = 1.0 (in most cases) K l = Local pressure factor = 1.0 (in most cases) = Porous cladding reduction factor = 1.0 (in most cases) K p

77 Wind Action: Analytical Procedure Section 2, MS 1553: 2002 B s = h s b sh 2 L h S = n ah 1 + g v l h V des n ab 0h 1 + g v l h V des C dyn = 1 + 2l h g v 2 B s + g R 2 SE t ζ 1 + 2g v l h E t = 0. 47N 2 + N 5/6 N = n al h 1 + g v l h V des

78 Wind Action: Analytical Procedure

79 Wind Action: Analytical Procedure

80 Wind Action: Analytical Procedure

81 Wind Action: Analytical Procedure

82 Wind Action: Analytical Procedure

83 Example 5 WIND LOAD CALCULATION

84 Example 5: Wind Load Calculation Wind b = 18 m b = 18 m Wind h = 25 m d = 51 m Side Elevation Plan View Location: Kuala Lumpur (Zone 1) Terrain: Suburban Terrain for All Directions Topography: Ground Slope Les than 1 in 20 for Greater than 5 km in All Directions Construction: Reinforced Concrete Sway Frequencies, n a = n c : 0.2 Hertz Mode Shape: Linear, k: 1.0 Average Building Density: 160 kg/m 3

85 Example 5: Wind Load Calculation Using Analytical Procedure in Section 2, MS 1553: 2002 Design wind pressure, p = V des 2 C fig C dyn (1) Design wind speed, V des = V sit l = = m/s where V sit = V s M d M z,cat M s M h = = 32.5 m/s V s = Basic wind speed (Figure A.1) = 33.5 m/s M d = Wind directional multiplier = 1.00 M z,cat = Terrain/Height multiplier (Table 4.1) = 0.97 by interpolation M s = Shielding multiplier = 1.00 M h = Hill shape multiplier = 1.00 l = Importance factor (Table 3.2) = 1.15

86 25 m Example 5: Wind Load Calculation

87 Example 5: Wind Load Calculation

88 Example 5: Wind Load Calculation (2) Aerodynamic shape factor, C fig = C p,e K a K c K l K p C fig W C fig L = C p,e K a K c K l K p = = C p,e K a K c K l K p = where C p,e = External pressure coefficient Windward wall = 0.80 (for varying z) Leeward wall = by interpolation (for d/b = 51 m/18 m = 2.8) K a = Area reduction factor = 1.00 K c = Combination factor = 1.00 K l = Local pressure factor = 1.00 K p = Porous cladding reduction factor = 1.00

89 Example 5: Wind Load Calculation 2.8

90 Example 5: Wind Load Calculation (3) Dynamic response factor,c dyn = 1+2l h g v 2 B s + g R 2 SEt ξ 1+2g v l h = where l h = Turbulence intensity at z = h (Table 6.1) = by interpolation g v = Peak factor = 3.7 given B s = Background factor = S = Size reduction factor = E t = Spectrum of turbulence = h s 2 +64b sh 2 L h = n ah 1+gvl h V des 0.47N (2+N) 1 5/6 = h = 25 m s = 0 m L h = 1000 h 10 b sh = 51 m 1+ 4n ab 0h 1+gvl h V des = = 1257 m n a = 0.2 Hz given b 0h = b sh = 51 m N = n al h 1 + g v l h V des = = Ratio of structural damping to critical damping (Table 6.2) = 0.05 g R = Peak factor for resonance response = 2 log e 600n a = 3.09

91 Example 5: Wind Load Calculation 25 m

92 Example 5: Wind Load Calculation

93 Example 5: Wind Load Calculation p = V des 2 C fig C dyn (N/m 2 ) Level Height (m) M z,cat (Table 4.1) Windward Leeward Total, p total V des,z (m/s) p W V des,h (m/s) p L p W - p L 7 (Roof)

94 Example 5: Wind Load Calculation Level 7 (Roof) Level Level Level 2 Ground p W (kn/m 2 ) p L (kn/m 2 ) Total (kn/m 2 )

95 Frame Analysis for Lateral Actions Portal Method Frame divided into independent portals Shear in each storey is assumed to be divided between the bay in proportion to their spans Shear in each bay is divided equally between columns Cantilever Method Axial loads in column are assumed to be proportional to the distance from the frame s centre of gravity Assumed column in a storey has equal cross-sectional area & point of contraflexure are located at midspan for all column & beams

96 Lateral load Frame Stability Vertical load Pinned beam-to-column connections Frame is not stable when subjected to lateral forces

97 Frame Stability Unbraced Rigid beam-tocolumn connection Unstable Frame Pinned base Rigid base

98 Frame Stability Braced Pinned connections Rigid core Unstable Frame Stable Frame

99 Example 6 LATERAL LOAD ANALYSIS CANTILEVER METHOD

100 Example 6: Lateral Load Analysis (Cantilever Method) Design wind load for each floor level of sub-frame 3 = 1.2w k : where w k = Total wind pressure, p total Contact area Roof: kn/m m 1.75 m = 11.3 kn Level 6: kn/m m 3.50 m = 21.8 kn Level 5: kn/m m 3.50 m = 20.8 kn Level 4: kn/m m 3.50 m = 19.8 kn Level 3: kn/m m 3.50 m = 18.3 kn Level 2: kn/m m 3.50 m = 16.7 kn Level 1: kn/m m 3.75 m = 16.7 kn 4 m 5 m

101 4 m 3.5 m 3.5 m 3.5 m 3.5 m 3.5 m 3.5 m 11.3 kn 21.8 kn 20.8 kn 19.8 kn 18.3 kn 16.7 kn 16.7 kn K L M N P Q Example 6: Lateral Load Analysis (Cantilever Method) Roof Level 6 Level 5 Level 4 Level 3 Level 2 Level 1 R x = 9 m A B C D 6 m 6 m 6 m

102 4 m 3.5 m 3.5 m 3.5 m 3.5 m 3.5 m 3.5 m 11.3 kn 21.8 kn 20.8 kn 19.8 kn 18.3 kn 16.7 kn K L M N Example 6: Lateral Load Analysis (Cantilever Method) Roof Level 6 Level 5 Level 4 Level 3 P Column A B C D Level 2 Distance from Centroid Q 16.7 N Column Axial Load 3.0P 1.0P Level 1 1.0P 3.0P R A B C D 6 m 6 m 6 m

103 Example 6: Lateral Load Analysis (Cantilever Method) LEVEL K AND ABOVE

104 Example 6: Lateral Load Analysis (Cantilever Method) Level K and Above 11.3 kn K H 1 H 2 H m H 4 3.0P 6 m 6 m 6 m 1.0P 1.0P 3.0P Axial Force in Columns = P 6 1.0P P 18 = 0 P = 0.33 kn

105 Example 6: Lateral Load Analysis (Cantilever Method) 11.3 kn K F 1 F 2 F m H 1 H 2 H 3 H 4 6 m 6 m 6 m 0.99 kn 0.33 kn 0.33 kn Shear Force in Beams & Columns 0.99 kn Consider sub-frame to the left of F 1 : F v = 0: F = 0 F 1 = 0.99 kn M@F 1 = 0: H = 0 H 1 = 1.70 kn Consider sub-frame to the left of F 2 : F v = 0: F = 0 F 2 = 1.32 kn M@F 2 = 0: H 1 + H = 0 H 2 = 3.96 kn Consider sub-frame to the left of F 3 : F v = 0: F = 0 F 3 = 0.99 kn M@F 3 = 0: H 1 + H 2 + H = 0 H 3 = 3.96 kn F H = 0: 11.3 H 1 H 2 H 3 H 4 = 0 H 4 = 1.68 kn

106 Example 6: Lateral Load Analysis (Cantilever Method) LEVEL L AND ABOVE

107 Example 6: Lateral Load Analysis (Cantilever Method) 3.5 m Level L and Above 11.3 kn 21.8 kn K H 1 H 2 L H 4 H m 6 m 6 m 6 m 3.0P 1.0P 1.0P 3.0P Axial Force in Columns = P 6 1.0P P 18 = 0 P = 1.62 kn

108 Example 6: Lateral Load Analysis (Cantilever Method) 0.99 kn 1.70 K = 33.1 kn 0.33 kn 0.33 kn 0.99 kn F 1 F 2 F m L H 1 H 2 H 3 H m 4.86 kn Shear Force in Beams & Columns 6 m 6 m 6 m 1.62 kn 1.62 kn 4.86 kn Consider sub-frame to the left of F 1 : F v = 0: F = 0 F 1 = 3.87 kn M@F 1 = 0: H = 0 H 1 = 4.93 kn Consider sub-frame to the left of F 2 : F v = 0: F = 0 F 2 = 5.16 kn M@F 2 = 0: H 1 + H = 0 H 2 = kn

109 Example 6: Lateral Load Analysis (Cantilever Method) 0.99 kn 1.70 K = 33.1 kn 0.33 kn 0.33 kn 0.99 kn F 1 F 2 F m L H 1 H 2 H 3 H m 4.86 kn Shear Force in Beams & Columns 6 m 6 m 6 m 1.62 kn 1.62 kn 4.86 kn Consider sub-frame to the left of F 3 : F v = 0: F = 0 F 3 = 3.87 kn M@F 3 = 0: H 1 + H 2 + H = 0 H 3 = kn F H = 0: 33.1 H 1 H 2 H 3 H 4 = 0 H 4 = 5.13 kn

110 Example 6: Lateral Load Analysis (Cantilever Method) LEVEL M AND ABOVE

111 Example 6: Lateral Load Analysis (Cantilever Method) Level M and Above 11.3 kn 21.8 kn 20.8 kn K L M 3.5 m 3.5 m 1.75 m H 1 H 2 H 3 H 4 6 m 6 m 6 m 3.0P 1.0P 1.0P 3.0P Axial Force in Columns = P 6 1.0P P 18 = 0 P = 4.16 kn

112 Example 6: Lateral Load Analysis (Cantilever Method) 4.86 kn 4.93 L = 53.9 kn 1.62 kn 1.62 kn 4.86 kn F 1 F 2 F m M H 1 H 2 H 3 H m kn Shear Force in Beams & Columns 6 m 6 m 6 m 4.16 kn 4.16 kn kn Consider sub-frame to the left of F 1 : F v = 0: F = 0 F 1 = 7.62 kn M@F 1 = 0: H = 0 H 1 = 8.13 kn Consider sub-frame to the left of F 2 : F v = 0: F = 0 F 2 = kn M@F 2 = 0: H 1 + H = 0 H 2 = kn

113 Example 6: Lateral Load Analysis (Cantilever Method) 4.86 kn 4.93 L = 53.9 kn 1.62 kn 1.62 kn 4.86 kn F 1 F 2 F m M H 1 H 2 H 3 H m Shear Force in Beams & Columns kn 6 m 6 m 6 m 4.16 kn 4.16 kn kn Consider sub-frame to the left of F 3 : F v = 0: F = 0 F 3 = 7.62 kn M@F 3 = 0: H 1 + H 2 + H = 0 H 3 = kn F H = 0: 53.9 H 1 H 2 H 3 H 4 = 0 H 4 = 7.85 kn

114 Example 6: Lateral Load Analysis (Cantilever Method) Do the Following Level: LEVEL N AND ABOVE LEVEL P AND ABOVE LEVEL Q AND ABOVE LEVEL R AND ABOVE

115 Example 6: Lateral Load Analysis (Cantilever Method) SHEAR FORCE IN BEAMS & COLUMNS

116 Example 6: Lateral Load Analysis (Cantilever Method) L M K m m m 6 m 6 m 6 m

117 Example 6: Lateral Load Analysis (Cantilever Method) BENDING MOMENTS IN BEAMS & COLUMNS

118 Example 6: Lateral Load Analysis (Cantilever Method) Continues for Level N, P, Q, R

119 Example 7 ANALYSIS OF ONE LEVEL SUB- FRAME VERTICAL LOAD ONLY

120 Example 7: Analysis of One Level Sub- Frame Vertical Load Only 4 m 3.5 m 3.5 m 3.5 m 3.5 m 3.5 m 3.5 m Roof Level 6 Level 5 Level 4 Level 3 Level 2 Level 1 A B C D 6 m 6 m 6 m

121 Example 7: Analysis of One Level Sub- Frame Vertical Load Only CASE 1: 1.2G k + 1.2Q k 3.5 m 48.7 kn/m 46.1 kn/m 48.7 kn/m Span 1 Span 2 Span 3 4 m 6 m 6 m 6 m A B C D Fixed End Moment M AB = M BA = wl2 12 M BC = wl2 12 M CD = wl = = 12 = = knm = knm = knm

122 Example 7: Analysis of One Level Sub- Frame Vertical Load Only Moment Distribution Method A B 0.19 C 0.19 D

123 Example 7: Analysis of One Level Sub- Frame Vertical Load Only Reaction Calculation at Support 48.7 kn/m kn/m kn/m m 6 m 6 m V AB V BA V BC V CB V CD V DC Solve for each span using equilibrium method of analysis: V AB = kn V BA = kn V BC = kn V CB = kn V CD = kn V DC = kn

124 Example 7: Analysis of One Level Sub- Frame Vertical Load Only Shear Force and Bending Moment Diagram SFD (kn) 2.7 m 3.0 m 3.3 m BMD (knm)

125 Example 7: Analysis of One Level Sub- Frame Vertical Load Only Bending Moment due to Vertical Load (1.2G k + 1.2Q k ) Bending Moment due to Wind Load, 1.2W k (from your exercise) = Combination of Vertical & Wind Load