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TRANSCRIPT
FRAME ANALYSIS
Dr. Izni Syahrizal bin Ibrahim Faculty of Civil Engineering
Universiti Teknologi Malaysia
Email: [email protected]
Introduction
Building Construction
3D Frame: Beam, Column & Slab
3D Frame: Beam & Column ONLY
2D Frame Analysis
2D Frame Analysis
Introduction: Substitute Frame
Sub-frame: Roof Level
Sub-frame: Other Floor Level
2D Frame Analysis
Types of Frame
Braced Frame • NOT contribute to the overall structural
stability • None lateral actions are transmitted to
columns & beams • Support vertical actions only
Unbraced Frame • Contribute to the overall structural
stability • All lateral actions are transmitted to
columns & beams • Support vertical & lateral actions
Shear wall as bracing members
Methods of Analysis
(i) One-level Sub-frame
(ii) Two-points Sub-frame
Kb1 0.5Kb2 0.5Kb1 0.5Kb3 Kb2
Methods of Analysis
(iii) Continuous Beam and One-point Sub-frame
0.5Kb 0.5Kb
0.5Kb 0.5Kb
Combination of Actions
BS EN 1992-1-1: Cl. 5.1.3
Load Set 1: Alternate or adjacent spans loaded Alternate span carrying the design variable and permanent load (1.35Gk + 1.5Qk), other spans carrying only the design permanent loads (1.35Gk) Any two adjacent spans carrying the design variable and permanent loads (1.35Gk + 1.5Qk), all other spans carrying only the design permanent load (1.35Gk)
Combination of Actions
Alternate Span Loaded Adjacent Span Loaded
1.35Gk
1.5Qk 1.5Qk
Load Set 1
Combination of Actions
Load set 2: All or alternate spans loaded All span carrying the design variable and permanent loads (1.35Gk + 1.5Qk)
Alternate span carrying the design variable and permanent load (1.35Gk + 1.5Qk), other spans carrying only the design permanent loads (1.35Gk)
MALAYSIAN / UK NATIONAL ANNEX
Combination of Actions
All Spans Loaded
Alternate Span Loaded
1.35Gk
1.5Qk
Load Set 2
Example 1
ACTION ON BEAMS
2@
4 m
= 8
m
2@
4 m
= 8
m
7@
5 m
= 3
5 m
6 m 6 m 6 m
6 m 6 m 6 m
Level 1
Level 2
Level 3
Level 4
Level 5
Level 6
Level 7
6@
3.5
m =
21
m
4 m
1 m
Example 1
Example 1: Action on Beams
Loading Data Finishes, ceiling & services = 0.75 kN/m2
Concrete density = 25 kN/m3
Variable action = 4 kN/m2
Partition = 0.50 kN/m2
Structural Dimensions Slab thickness, h = 150 mm Beam size, b h = 250 600 mm Column size, b h = 300 400 mm
Example 1: Action on Beams
Load on Slab Slab self weight = 0.15 m 25 kN/m3 = 3.75 kN/m2
Finishes, ceiling, services = 1.25 kN/m2
Characteristics permanent action, gk = 5.00 kN/m2
Imposed load = 4.00 kN/m2
Partition = 0.50 kN/m2 Characteristics variable action, qk = 4.50 kN/m2
Other Loads on Beam Beam self weight = 0.25 m (0.60 – 0.15) m 25 kN/m3 = 2.81 kN/m
Example 1: Action on Beams
6 m 6 m 6 m
5 m
4 m 0.47
0.42
0.45
0.39
0.47
0.42
𝐿𝑦𝐿𝑥=6
4= 1.50 (𝐶𝑎𝑠𝑒 2)
𝐿𝑦𝐿𝑥=6
5= 1.20 (𝐶𝑎𝑠𝑒 2)
𝐿𝑦𝐿𝑥=6
4= 1.50 (𝐶𝑎𝑠𝑒 1)
𝐿𝑦𝐿𝑥=6
5= 1.20 (𝐶𝑎𝑠𝑒 1)
𝐿𝑦𝐿𝑥=6
4= 1.50 (𝐶𝑎𝑠𝑒 2)
𝐿𝑦𝐿𝑥=6
5= 1.20 (𝐶𝑎𝑠𝑒 2)
Example 1: Action on Beams
0.47
0.42
0.45
0.39
0.47
0.42
Gk = Permanent action on slab + Beam self weight = (0.47 5.00 4) + (0.42 5.00 5) + 2.81 = 22.71 kN/m
Qk = Variable action on slab = (0.47 4.50 4) + (0.42 4.50 5) = 17.91 kN/m
Gk = Permanent action on slab + Beam self weight = (0.45 5.00 4) + (0.39 5.00 5) + 2.81 = 21.56 kN/m
Qk = Variable action on slab = (0.45 4.50 4) + (0.39 4.50 5) = 16.88 kN/m
Example 1: Action on Beams
wmax = 1.35Gk + 1.5Qk = 57.53 kN/m wmin = 1.35Gk = 30.66 kN/m
wmax = 1.35Gk + 1.5Qk = 54.42 kN/m wmin = 1.35Gk = 29.11 kN/m
6 m 6 m 6 m
wmax = 1.35Gk + 1.5Qk = 57.53 kN/m wmin = 1.35Gk = 30.66 kN/m
Example 2
ANALYSIS OF ONE-LEVEL SUB-FRAME
2@
4 m
= 8
m
2@
4 m
= 8
m
7@
5 m
= 3
5 m
6 m 6 m 6 m
6 m 6 m 6 m
Level 1
Level 2
Level 3
Level 4
Level 5
Level 6
Level 7
6@
3.5
m =
21
m
4 m
1 m
Example 2
Example 2: Analysis of One Level Sub-Frame
6 m 6 m 6 m
3.5 m
4 m
wmax = 57.53 kN/m wmin = 30.66 kN/m
wmax = 57.53 kN/m wmin = 30.66 kN/m wmax = 54.42 kN/m
wmin = 29.11 kN/m
A B C D
Kc,l
Kc,u
Analysis using LOAD SET 1
KAB KBC KCD
Example 2: Analysis of One Level Sub-Frame
Structural Dimension Beam: 250 mm 600 mm Column: 300 mm 400 mm
Moment of Inertia, I = bh3/12
Beam: 250×6003
12= 4.5 × 109 𝑚𝑚4
Column: 300×4003
12= 1.6 × 109 𝑚𝑚4
Stiffness, K = I/L
Beam: 𝐾𝐴𝐵 = 𝐾𝐵𝐶 = 𝐾𝐶𝐷 =4.5×109
6000= 7.5 × 105 𝑚𝑚3
Column: 𝐾𝑐,𝑢 =1.6×109
3500= 4.6 × 105 𝑚𝑚3
𝐾𝑐,𝑙 =1.6×109
4000= 4.0 × 105 𝑚𝑚3
Example 2: Analysis of One Level Sub-Frame
Distribution Factor, F = K/K
Joint A & Joint D: 𝐹𝐴𝐵, 𝐶𝐷 =𝐾𝐴𝐵, 𝐶𝐷
𝐾𝐴𝐵, 𝐶𝐷+𝐾𝑐,𝑢+𝐾𝑐,𝑙= 0.47
𝐹𝑐,𝑢 =𝐾𝑐,𝑢
𝐾𝐴𝐵, 𝐶𝐷+𝐾𝑐,𝑢+𝐾𝑐,𝑙= 0.28
𝐹𝑐,𝑙 =𝐾𝑐,𝑙
𝐾𝐴𝐵, 𝐶𝐷+𝐾𝑐,𝑢+𝐾𝑐,𝑙= 0.25
Joint B & Joint C: 𝐹𝐵𝐴, 𝐶𝐵 =𝐾𝐴𝐵, 𝐵𝐶
𝐾𝐴𝐵, 𝐵𝐶+𝐾𝐵𝐶, 𝐶𝐷+𝐾𝑐,𝑢+𝐾𝑐,𝑙= 0.32
𝐹𝐵𝐶, 𝐶𝐷 =𝐾𝐵𝐶, 𝐶𝐷
𝐾𝐴𝐵, 𝐵𝐶+𝐾𝐵𝐶, 𝐶𝐷+𝐾𝑐,𝑢+𝐾𝑐,𝑙= 0.32
𝐹𝑐,𝑢 =𝐾𝑐,𝑢
𝐾𝐴𝐵, 𝐵𝐶+𝐾𝐵𝐶, 𝐶𝐷+𝐾𝑐,𝑢+𝐾𝑐,𝑙= 0.19
𝐹𝑐,𝑙 =𝐾𝑐,𝑙
𝐾𝐴𝐵, 𝐵𝐶+𝐾𝐵𝐶, 𝐶𝐷+𝐾𝑐,𝑢+𝐾𝑐,𝑙= 0.17
Example 2: Analysis of One Level Sub-Frame
CASE 1: Span 1 & 2 = wmax, Span 3 = wmin
Fixed End Moment
𝑀𝐴𝐵 = 𝑀𝐵𝐴 =𝑤𝐿2
12=57.5 × 62
12= 172.6 𝑘𝑁𝑚
𝑀𝐵𝐶 =𝑤𝐿2
12=54.4 × 62
12= 163.3 𝑘𝑁𝑚
𝑀𝐶𝐷 =𝑤𝐿2
12=30.7 × 62
12= 92.0 𝑘𝑁𝑚
6 m 6 m 6 m
3.5 m
4 m
wmax = 57.5 kN/m
wmin = 30.7 kN/m wmax = 54.4 kN/m
A B C D
Span 1 Span 2 Span 3
Example 2: Analysis of One Level Sub-Frame
51.00 -8.46 -8.71 -24.07
0.18 -0.27 0.29 -0.09
1.31 -0.78 0.38 -1.04
0.42 -0.56 4.45 3.23
49.09 -1.81 -13.82 -26.16
0.28 A B 0.19 C 0.19 D 0.28
0.25 0.47 0.32 0.17 0.32 0.32 0.17 0.32 0.47 0.25
-172.6 172.6 -163.3 163.3 -92.0 92.0
43.0 80.5 -3.0 -1.6 -3.0 -22.7 -12.1 -22.7 -42.9 -22.9
-1.5 40.3 -11.3 -1.5 -21.5 -11.3
0.4 0.7 -9.2 -4.9 -9.2 7.3 3.9 7.3 5.3 2.8
-4.6 0.3 3.7 -4.6 2.6 3.7
1.1 2.1 -1.3 -0.7 -1.3 0.6 0.3 0.6 -1.7 -0.9
-0.6 1.1 0.3 -0.6 -0.9 0.3
0.2 0.3 -0.4 -0.2 -0.4 0.5 0.3 0.5 -0.1 -0.1
44.6 -95.6 200.4 -7.4 -184.5 142.3 -7.6 -125.9 45.1 -21.1
Moment Distribution Method
Example 2: Analysis of One Level Sub-Frame
51.00 -8.46 -8.71 -24.07
0.18 -0.27 0.29 -0.09
1.31 -0.78 0.38 -1.04
0.42 -0.56 4.45 3.23
49.09 -1.81 -13.82 -26.16
0.28 A B 0.19 C 0.19 D 0.28
0.25 0.47 0.32 0.17 0.32 0.32 0.17 0.32 0.47 0.25
-172.6 172.6 -163.3 163.3 -92.0 92.0
43.0 80.5 -3.0 -1.6 -3.0 -22.7 -12.1 -22.7 -42.9 -22.9
-1.5 40.3 -11.3 -1.5 -21.5 -11.3
0.4 0.7 -9.2 -4.9 -9.2 7.3 3.9 7.3 5.3 2.8
-4.6 0.3 3.7 -4.6 2.6 3.7
1.1 2.1 -1.3 -0.7 -1.3 0.6 0.3 0.6 -1.7 -0.9
-0.6 1.1 0.3 -0.6 -0.9 0.3
0.2 0.3 -0.4 -0.2 -0.4 0.5 0.3 0.5 -0.1 -0.1
44.6 -95.6 200.4 -7.4 -184.5 142.3 -7.6 -125.9 45.1 -21.1
Moment in the COLUMN at each joint
Example 2: Analysis of One Level Sub-Frame
51.00 -8.46 -8.71 -24.07
0.18 -0.27 0.29 -0.09
1.31 -0.78 0.38 -1.04
0.42 -0.56 4.45 3.23
49.09 -1.81 -13.82 -26.16
0.28 A B 0.19 C 0.19 D 0.28
0.25 0.47 0.32 0.17 0.32 0.32 0.17 0.32 0.47 0.25
-172.6 172.6 -163.3 163.3 -92.0 92.0
43.0 80.5 -3.0 -1.6 -3.0 -22.7 -12.1 -22.7 -42.9 -22.9
-1.5 40.3 -11.3 -1.5 -21.5 -11.3
0.4 0.7 -9.2 -4.9 -9.2 7.3 3.9 7.3 5.3 2.8
-4.6 0.3 3.7 -4.6 2.6 3.7
1.1 2.1 -1.3 -0.7 -1.3 0.6 0.3 0.6 -1.7 -0.9
-0.6 1.1 0.3 -0.6 -0.9 0.3
0.2 0.3 -0.4 -0.2 -0.4 0.5 0.3 0.5 -0.1 -0.1
44.6 -95.6 200.4 -7.4 -184.5 142.3 -7.6 -125.9 45.1 -21.1
Moment in the BEAM at each joint
Example 2: Analysis of One Level Sub-Frame
6 m
wmax = 57.5 kN/m
95.6 200.4
VAB VBA
wmax = 54.4 kN/m
6 m
184.5 142.3
VBC VCB
wmin = 30.7 kN/m
6 m
125.9 45.1
VCD VDC
Solve for each span using equilibrium method of analysis: VAB = 155.1 kN VBA = 190.0 kN VBC = 170.3 kN VCB = 156.2 kN VCD = 105.5 kN VDC = 78.5 kN
Reaction Calculation at Support
Example 2: Analysis of One Level Sub-Frame
Shear Force and Bending Moment Diagram
SFD (kN)
BMD (kNm)
155.1
190.0
170.3
105.5
156.2 78.5
95.6
113.5
200.4
82.0 55.4
142.3 184.5
125.9
45.1 51.0 44.6
22.3
25.5
8.5 7.4 8.7 7.6
24.1 21.1
12.1
3.8 10.6 3.7
4.3 4.4
2.7 m 3.1 m 3.4 m
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 = wmax, Span 1 = wmin
6 m 6 m 6 m
3.5 m
4 m
wmin = 30.7 kN/m
wmax = 57.5 kN/m wmax = 54.4 kN/m
A B C D
Span 1 Span 2 Span 3
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 = wmax, Span 2 = wmin
6 m 6 m 6 m
3.5 m
4 m
wmax = 57.5 kN/m wmax = 57.5 kN/m wmin = 29.1 kN/m
A B C D
Span 1 Span 2 Span 3
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 = wmin, Span 2 = wmax
6 m 6 m 6 m
3.5 m
4 m
wmin = 30.7 kN/m wmin = 30.7 kN/m wmax = 54.4 kN/m
A B C D
Span 1 Span 2 Span 3
Example 2: Analysis of One Level Sub-Frame
Shear Force and Bending Moment Envelope
SFD (kN)
BMD (kNm)
95.6
113.5
200.4
82.0, 82.0 55.4
142.3
184.5
125.9
45.1
Case 1 Case 2 Case 3 Case 4
45.1 125.9
142.3 95.6
55.4
113.5
184.5 200.4
123.0
168.8 104.5
18.4
112.6 112.6
168.8
123.0
104.5
54.1 43.8
130.9
91.5
153.4 153.4
43.8 54.1
130.9
155.1
190.0, 190.0
170.3, 170.3
105.5
156.2 78.5
78.5
105.5
156.2
170.3, 170.3
190.0, 190.0
155.1
161.9, 161.9
183.2
87.3
87.3
183.2
161.9, 161.9
Example 2: Analysis of One Level Sub-Frame
Shear Force and Bending Moment Envelope
SFD (kN)
BMD (kNm) 55.9
48.9 30.0 26.2 30.0 26.2
55.9 48.9
155.1
190.0, 190.0
170.3, 170.3
105.5
156.2 78.5
78.5
105.5
156.2
170.3, 170.3
190.0, 190.0
155.1
161.9, 161.9
183.2
87.3
87.3
183.2
161.9, 161.9 Case 1 Case 2 Case 3 Case 4
Example 3
ANALYSIS OF TWO FREE-JOINT SUB-FRAME
2@
4 m
= 8
m
2@
4 m
= 8
m
7@
5 m
= 3
5 m
6 m 6 m 6 m
6 m 6 m 6 m
Level 1
Level 2
Level 3
Level 4
Level 5
Level 6
Level 7
6@
3.5
m =
21
m
4 m
1 m
Example 3
CASE 1: Span 1 & 2 = wmax
Stiffness, K = I/L
Beam: 𝐾𝐴𝐵 =4.5×109
6000= 7.5 × 105 𝑚𝑚3
𝐾𝐵𝐶 = 0.5 ×4.5×109
6000= 3.75 × 105 𝑚𝑚3
Column: 𝐾𝑐,𝑢 =1.6×109
3500= 4.6 × 105 𝑚𝑚3
𝐾𝑐,𝑙 =1.6×109
4000= 4.0 × 105 𝑚𝑚3
Example 3: Analysis of Two Free-Joint Sub-Frame
6 m 6 m
3.5 m
4 m
wmax = 57.5 kN/m wmax = 54.4 kN/m
A B
Span 1 Span 2
C
Span
AB
Example 3: Analysis of Two Free-Joint Sub-Frame
Distribution Factor, F = K/K
Joint A: 𝐹𝐴𝐵 =7.5
7.5+4.6+4.0= 0.47
𝐹𝑐,𝑢 =4.6
7.5+4.6+4.0= 0.28
𝐹𝑐,𝑙 =4.0
7.5+4.6+4.0= 0.25
Joint B: 𝐹𝐵𝐴 =7.5
7.5+3.75+4.6+4.0= 0.38
𝐹𝐵𝐶 =3.75
7.5+3.75+4.6+4.0= 0.19
𝐹𝑐,𝑢 =4.6
7.5+3.75+4.6+4.0= 0.23
𝐹𝑐,𝑙 =4.0
7.5+3.75+4.6+4.0= 0.20
51.0 -11.9
0.0 -0.4
2.2 -0.1
0.5 -9.3
48.3 -2.1
0.28 A B 0.23
0.25 0.47 0.38 0.20 0.19
-172.6 172.6 -163.3
43.1 81.1 -3.5 -1.9 -1.8
-1.8 40.6 0.0
0.4 0.8 -15.4 -8.1 -7.7
-7.7 0.4 0.0
1.9 3.6 -0.2 -0.1 -0.1
-0.1 1.8 0.0
0.0 0.0 -0.7 -0.4 -0.3
45.4 -96.7 195.6 -10.5 -173.2
Moment Distribution Method Fixed End Moment
𝑀𝐴𝐵 =𝑤𝐿2
12=57.5 × 62
12= 172.6 𝑘𝑁𝑚
𝑀𝐵𝐶 =𝑤𝐿2
12=54.4 × 62
12= 163.3 𝑘𝑁𝑚
Span
AB
CASE 2: Span 1 = wmax , Span 2 = wmin
6 m 6 m
3.5 m
4 m
wmax = 57.5 kN/m
wmin = 29.11 kN/m
A B
Span 1 Span 2
Example 3: Analysis of Two Free-Joint Sub-Frame
55.2 -30.2
0.2 -0.4
2.2 -0.9
4.5 -9.3
48.3 -19.6
0.28 A B 0.23
0.25 0.47 0.38 0.20 0.19
-172.6 172.6 -87.3
43.1 81.1 -32.4 -17.1 -16.2
-16.2 40.5 0.0
4.0 7.6 -15.4 -8.1 -7.7
-7.7 3.8 0.0
1.9 3.6 -1.4 -0.8 -0.7
-0.7 1.8 0.0
0.2 0.3 -0.7 -0.4 -0.3
49.2 -105.1 168.8 -26.4 -112.2
Moment Distribution Method
C
Span
AB
Reaction Calculation at Support
Example 3: Analysis of Two Free-Joint Sub-Frame
6 m
wmax = 57.5 kN/m
96.7 195.6
VAB VBA
Solve for each span using equilibrium method of analysis: VAB = 156.2 kN VBA = 189.0 kN
CASE 1
wmax = 57.5 kN/m
6 m
105.1 168.8
VAB VBA
Solve for each span using equilibrium method of analysis: VAB = 162.0 kN VBA = 183.2 kN
CASE 2
Span
AB
Shear Force and Bending Moment Diagram
Example 3: Analysis of Two Free-Joint Sub-Frame
156.2
189.0
183.2
162.0
105.1
168.8 97.1
195.4
30.2 49.0 56.0
26.4
SFD (kN)
BMD (kNm)
114.9
122.9
Span
AB
Example 3: Analysis of Two Free-Joint Sub-Frame
CASE 1: Span 1 & 2 = wmax, Span 3 = wmin
6 m 6 m 6 m
3.5 m
4 m
wmax = 57.5 kN/m
wmin = 30.7 kN/m wmax = 54.4 kN/m
B C
Span 1 Span 2 Span 3
A D
Span
BC
Example 3: Analysis of Two Free-Joint Sub-Frame
Distribution Factor, F = K/K
Joint B: 𝐹𝐵𝐴 =3.75
3.75+7.5+4.6+4.0= 0.19
𝐹𝐵𝐶 =7.5
3.75+7.5+4.6+4.0= 0.38
𝐹𝑐,𝑢 =4.6
3.75+7.5+4.6+4.0= 0.23
𝐹𝑐,𝑙 =4.0
3.75+7.5+4.6+4.0= 0.20
Joint C: 𝐹𝐶𝐵 =7.5
3.75+7.5+4.6+4.0= 0.38
𝐹𝐶𝐷 =3.75
3.75+7.5+4.6+4.0= 0.19
𝐹𝑐,𝑢 =4.6
3.75+7.5+4.6+4.0= 0.23
𝐹𝑐,𝑙 =4.0
3.75+7.5+4.6+4.0= 0.20
Fixed End Moment
𝑀𝐴𝐵 = 𝑀𝐵𝐴 =𝑤𝐿2
12=57.5 × 62
12= 172.6 𝑘𝑁𝑚
𝑀𝐵𝐶 = 𝑀𝐶𝐵 =𝑤𝐿2
12=54.4 × 62
12= 163.3 𝑘𝑁𝑚
𝑀𝐶𝐷 = 𝑀𝐷𝐶 =𝑤𝐿2
12=30.7 × 62
12= 92.0 𝑘𝑁𝑚
Stiffness, K = I/L
Beam: 𝐾𝐴𝐵 = 𝐾𝐵𝐶 = 0.5 ×4.5×109
6000= 3.75 × 105 𝑚𝑚3
𝐾𝐵𝐶 =4.5×109
6000= 7.5 × 105 𝑚𝑚3
Column: 𝐾𝑐,𝑢 =1.6×109
3500= 4.6 × 105 𝑚𝑚3
𝐾𝑐,𝑙 =1.6×109
4000= 4.0 × 105 𝑚𝑚3
Span
BC
1.00 -16.61
0.11 0.01
-0.08 -0.59
3.11 0.41
-2.15 -16.44
B 0.23 C 0.23
0.19 0.20 0.38 0.38 0.20 0.19
172.6 -163.3 163.3 -92.0
-1.8 -1.9 -3.5 -27.0 -14.4 -13.5
0.0 -13.5 -1.8 0.0
2.6 2.7 5.1 0.7 0.4 0.3
0.0 0.3 2.6 0.0
-0.1 -0.1 -0.1 -1.0 -0.5 -0.5
0.0 -0.5 -0.1 0.0
0.1 0.1 0.2 0.0 0.0 0.0
173.4 0.9 -175.3 136.7 -14.5 -105.6
Moment Distribution Method
Example 3: Analysis of Two Free-Joint Sub-Frame
Span
BC
Example 3: Analysis of Two Free-Joint Sub-Frame
CASE 2: Span 2 = wmax, Span 1 & 3 = wmin
6 m 6 m 6 m
3.5 m
4 m
wmax = 30.7 kN/m wmin = 30.7 kN/m wmax = 54.4 kN/m
B C
Span 1 Span 2 Span 3
A D
Span
BC
Example 3: Analysis of Two Free-Joint Sub-Frame
Distribution Factor, F = K/K
Joint B: 𝐹𝐵𝐴 =3.75
3.75+7.5+4.6+4.0= 0.19
𝐹𝐵𝐶 =7.5
3.75+7.5+4.6+4.0= 0.38
𝐹𝑐,𝑢 =4.6
3.75+7.5+4.6+4.0= 0.23
𝐹𝑐,𝑙 =4.0
3.75+7.5+4.6+4.0= 0.20
Joint C: 𝐹𝐶𝐵 =7.5
3.75+7.5+4.6+4.0= 0.38
𝐹𝐶𝐷 =3.75
3.75+7.5+4.6+4.0= 0.19
𝐹𝑐,𝑢 =4.6
3.75+7.5+4.6+4.0= 0.23
𝐹𝑐,𝑙 =4.0
3.75+7.5+4.6+4.0= 0.20
Fixed End Moment
𝑀𝐴𝐵 = 𝑀𝐵𝐴 =𝑤𝐿2
12=30.7 × 62
12= 92.0 𝑘𝑁𝑚
𝑀𝐵𝐶 = 𝑀𝐶𝐵 =𝑤𝐿2
12=54.4 × 62
12= 163.3 𝑘𝑁𝑚
𝑀𝐶𝐷 = 𝑀𝐷𝐶 =𝑤𝐿2
12=30.7 × 62
12= 92.0 𝑘𝑁𝑚
Stiffness, K = I/L
Beam: 𝐾𝐴𝐵 = 𝐾𝐵𝐶 = 0.5 ×4.5×109
6000= 3.75 × 105 𝑚𝑚3
𝐾𝐵𝐶 =4.5×109
6000= 7.5 × 105 𝑚𝑚3
Column: 𝐾𝑐,𝑢 =1.6×109
3500= 4.6 × 105 𝑚𝑚3
𝐾𝑐,𝑙 =1.6×109
4000= 4.0 × 105 𝑚𝑚3
Span
BC
20.25 -20.25
0.11 0.11
0.59 -0.59
3.11 -3.11
16.44 -16.44
B 0.23 C 0.23
0.19 0.20 0.38 0.38 0.20 0.19
92.0 -163.3 163.3 -92.0
13.5 14.4 27.0 -27.0 -14.4 -13.5
0.0 -13.5 13.5 0.0
2.6 2.7 5.1 -5.1 -2.7 -2.6
0.0 -2.6 2.6 0.0
0.5 0.5 1.0 -1.0 -0.5 -0.5
0.0 -0.5 0.5 0.0
0.1 0.1 0.2 -0.2 -0.1 -0.1
108.6 17.7 -146.6 146.6 -14.5 -108.6
Moment Distribution Method
Example 3: Analysis of Two Free-Joint Sub-Frame
Span
BC
Reaction Calculation at Support
Example 3: Analysis of Two Free-Joint Sub-Frame
6 m
wmax = 54.4 kN/m
175.3 136.7
VBC VCB
Solve for each span using equilibrium method of analysis: VBC = 169.7 kN VCB = 156.8 kN
CASE 1
wmax = 54.4 kN/m
6 m
146.6 146.6
VBC VCB
Solve for each span using equilibrium method of analysis: VCB = 163.3 kN VCB = 163.3 kN
CASE 2
Span
BC
Shear Force and Bending Moment Diagram
Example 3: Analysis of Two Free-Joint Sub-Frame
163.3
163.3
156.8
169.7
175.3
146.6 146.6
195.4
17.7 17.7 20.2 20.2
SFD (kN)
BMD (kNm) 89.3
98.3
Span
BC
Example 4
CONTINUOUS BEAM + ONE FREE-JOINT SUB-FRAME
Example 4
2@
4 m
= 8
m
2@
4 m
= 8
m
7@
5 m
= 3
5 m
6 m 6 m 6 m
6 m 6 m 6 m
Level 1
Level 2
Level 3
Level 4
Level 5
Level 6
Level 7
6@
3.5
m =
21
m
4 m
1 m
Example 4: Continuous Beam + One Free-Joint Sub-Frame
6 m 6 m 6 m
A B C D
Moment of Inertia
𝐼𝑏𝑒𝑎𝑚 =𝑏ℎ3
12=250 × 6003
12= 4.5 × 109 𝑚𝑚4
Stiffness, K = I/L
𝐾𝐴𝐵 = 𝐾𝐶𝐷 = 0.75 ×𝐼𝑏𝑒𝑎𝑚𝐿
= 0.75 ×4.5 × 109
6000= 5.63 × 105 𝑚𝑚3
𝐾𝐵𝐶 =𝐼𝑏𝑒𝑎𝑚𝐿
=4.5 × 109
6000= 7.5 × 105 𝑚𝑚3
Distribution Factor, F = K/K
Joint A: 𝐹𝐵𝐴 =5.63
5.63+0= 1.00
Joint B: 𝐹𝐵𝐴 =5.63
5.63+7.5= 0.43
𝐹𝐵𝐶 =7.5
5.63+7.5= 0.57
Joint C: 𝐹𝐵𝐶 =7.5
5.63+7.5= 0.57
𝐹𝐶𝐷 =5.63
7.5+5.63= 0.43
Joint D: 𝐹𝐷𝐶 =5.63
5.63+0= 1.00
Load Set 1
Example 4: Continuous Beam + One Free-Joint Sub-Frame
CASE 1: Span 1 & 2 = wmax, Span 3 = wmin
6 m 6 m 6 m
A B C D
wmax = 57.5 kN/m
wmin = 30.7 kN/m wmax = 54.4 kN/m
Span 1 Span 2 Span 3
Fixed End Moment, FEM
𝑀𝐴𝐵 = 𝑀𝐵𝐴 =𝑤𝐿2
12=57.5 × 62
12= 172.6 𝑘𝑁𝑚
𝑀𝐵𝐶 = 𝑀𝐶𝐵 =𝑤𝐿2
12=54.4 × 62
12= 163.3 𝑘𝑁𝑚
𝑀𝐶𝐷 = 𝑀𝐷𝐶 =𝑤𝐿2
12=30.7 × 62
12= 92.0 𝑘𝑁𝑚
Example 4: Continuous Beam + One Free-Joint Sub-Frame
0.00 1.00 0.43 0.57 0.57 0.43 1.00 0.00
-172.6 172.6 -163.3 163.3 -92.0 92.0
172.6 -4.0 -5.3 -40.7 -30.5 -92.0
86.3 -20.4 -2.7 -46.0
-28.3 -37.7 27.8 20.9
13.9 -18.8
-6.0 -7.9 10.8 8.1
5.4 -4.0
-2.3 -3.1 2.3 1.7
1.1 -1.5
-0.5 -0.6 0.9 0.7
0.00 217.9 -217.9 137.2 -137.2 0.00
Moment Distribution Method
A B C D
Example 4: Continuous Beam + One Free-Joint Sub-Frame
VAB VBA VBC
6 m
wmax = 57.5 kN/m
217.9
wmax = 54.4 kN/m
6 m
217.9 137.2
VCB
wmin = 30.7 kN/m
6 m
137.2
VCD VDC
Solve for each span using equilibrium method of analysis: VAB = 136.3 kN VBA = 208.9 kN VBC = 176.7 kN VCB = 149.8 kN VCD = 114.9 kN VDC = 69.1 kN
Reaction Calculation at Support
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 = wmax, Span 1 = wmin
6 m 6 m 6 m
A B C D
wmax = 57.5 kN/m
wmin = 30.7 kN/m wmax = 54.4 kN/m
Span 1 Span 2 Span 3
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 = wmax, Span 2 = wmin
6 m 6 m 6 m
A B C D
wmax = 57.5 kN/m wmax = 57.5 kN/m
wmin = 29.1 kN/m
Span 1 Span 2 Span 3
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 = wmax, Span 1 & 3 = wmin
6 m 6 m 6 m
A B C D
wmin = 30.7 kN/m wmin = 30.7 kN/m
wmax = 54.4 kN/m
Span 1 Span 2 Span 3
FINALLY DO THE SFD & BMD ENVELOPE !!!
Example 4: Continuous Beam + One Free-Joint Sub-Frame
Sub-Frame: Bending Moment in Column
6 m
3.5 m
4 m
wmax = 57.5 kN/m
Kc,l
Kc,u
0.5KAB
Fixed End Moment, FEM
𝑀𝐴𝐵 =𝑤𝐿2
12=57.5 × 62
12= 172.6 𝑘𝑁𝑚
Moment of Inertia, I = bh3/12
Beam: 250×6003
12= 4.5 × 109 𝑚𝑚4
Column: 300×4003
12= 1.6 × 109 𝑚𝑚4
Stiffness, K = I/L
Beam: 𝐾𝐴𝐵 = 𝐾𝐵𝐶 = 𝐾𝐶𝐷 =4.5×109
6000= 7.5 × 105 𝑚𝑚3
Column: 𝐾𝑐,𝑢 =1.6×109
3500= 4.6 × 105 𝑚𝑚3
𝐾𝑐,𝑙 =1.6×109
4000= 4.0 × 105 𝑚𝑚3
Column A
Example 4: Continuous Beam + One Free-Joint Sub-Frame
Sub-Frame: Bending Moment in Column
Moment in upper column, Mupper
𝑀𝑢𝑝𝑝𝑒𝑟 = 𝐹𝐸𝑀 ×𝐾𝑐,𝑢
𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝐴𝐵= 172.6 ×
4.6
4.6+4.0+0.5×7.5= 𝟔𝟒. 𝟑 kNm
Moment in lower column, Mlower
𝑀𝑢𝑝𝑝𝑒𝑟 = 𝐹𝐸𝑀 ×𝐾𝑐,𝑙
𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝐴𝐵= 172.6 ×
4.0
4.6+4.0+0.5×7.5= 𝟓𝟔. 𝟎 kNm
64.3 56.0
Example 4: Continuous Beam + One Free-Joint Sub-Frame
Sub-Frame: Bending Moment in Column
Fixed End Moment, FEM
𝑀𝐵𝐴 =𝑤𝐿2
12=57.5 × 62
12= 172.6 𝑘𝑁𝑚
𝑀𝐵𝐶 =𝑤𝐿2
12=29.1 × 62
12= 87.3 𝑘𝑁𝑚
∆𝐹𝐸𝑀 = 𝑀𝐴𝐵 −𝑀𝐵𝐶 = 𝟖𝟓. 𝟑 𝒌𝑵𝒎
6 m 6 m
3.5 m
4 m
wmax = 57.5 kN/m
wmin = 29.1 kN/m
Kc,l
Kc,u
0.5KAB 0.5KBC
Column B
Example 4: Continuous Beam + One Free-Joint Sub-Frame
Sub-Frame: Bending Moment in Column
Moment in upper column, Mupper
𝑀𝑢𝑝𝑝𝑒𝑟 = ∆𝐹𝐸𝑀 ×𝐾𝑐,𝑢
𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝐴𝐵+0.5𝐾𝐵𝐶= 85.3 ×
4.6
4.6+4.0+0.5×7.5+0.5×7.5= 𝟐𝟒. 𝟐 kNm
Moment in lower column, Mlower
𝑀𝑢𝑝𝑝𝑒𝑟 = ∆𝐹𝐸𝑀 ×𝐾𝑐,𝑙
𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝐴𝐵+0.5𝐾𝐵𝐶= 85.3 ×
4.0
4.6+4.0+0.5×7.5+0.5×7.5= 𝟐𝟏. 𝟐 kNm
21.2 24.2
Example 4: Continuous Beam + One Free-Joint Sub-Frame
Sub-Frame: Bending Moment in Column
Fixed End Moment, FEM
𝑀𝐵𝐶 =𝑤𝐿2
12=29.1 × 62
12= 87.3 𝑘𝑁𝑚
𝑀𝐶𝐷 =𝑤𝐿2
12=57.5 × 62
12= 172.6 𝑘𝑁𝑚
∆𝐹𝐸𝑀 = 𝑀𝐶𝐷 −𝑀𝐵𝐶 = 𝟖𝟓. 𝟑 𝒌𝑵𝒎
6 m 6 m
3.5 m
4 m
wmin = 29.1 kN/m wmax = 57.5 kN/m
Kc,l
Kc,u
0.5KBC 0.5KCD
Column C
Example 4: Continuous Beam + One Free-Joint Sub-Frame
Sub-Frame: Bending Moment in Column
Moment in upper column, Mupper
𝑀𝑢𝑝𝑝𝑒𝑟 = ∆𝐹𝐸𝑀 ×𝐾𝑐,𝑢
𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝐵𝐶+0.5𝐾𝐶𝐷= 85.3 ×
4.6
4.6+4.0+0.5×7.5+0.5×7.5= 𝟐𝟒. 𝟐 kNm
Moment in lower column, Mlower
𝑀𝑢𝑝𝑝𝑒𝑟 = ∆𝐹𝐸𝑀 ×𝐾𝑐,𝑙
𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝐵𝐶+0.5𝐾𝐶𝐷= 85.3 ×
4.0
4.6+4.0+0.5×7.5+0.5×7.5= 𝟐𝟏. 𝟐 kNm
24.2 21.2
Example 4: Continuous Beam + One Free-Joint Sub-Frame
Sub-Frame: Bending Moment in Column
6 m
3.5 m
4 m
wmax = 57.5 kN/m
Kc,l
Kc,u
0.5KCD
Fixed End Moment, FEM
𝑀𝐷𝐶 =𝑤𝐿2
12=57.5 × 62
12= 172.6 𝑘𝑁𝑚
Column D
Example 4: Continuous Beam + One Free-Joint Sub-Frame
Sub-Frame: Bending Moment in Column
Moment in upper column, Mupper
𝑀𝑢𝑝𝑝𝑒𝑟 = 𝐹𝐸𝑀 ×𝐾𝑐,𝑢
𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝐶𝐷= 172.6 ×
4.6
4.6+4.0+0.5×7.5= 𝟔𝟒. 𝟑 kNm
Moment in lower column, Mlower
𝑀𝑢𝑝𝑝𝑒𝑟 = 𝐹𝐸𝑀 ×𝐾𝑐,𝑙
𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝐶𝐷= 172.6 ×
4.0
4.6+4.0+0.5×7.5= 𝟓𝟔. 𝟎 kNm
56.0 64.3
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
Wind Action
• Static deflection of the structure under wind pressure
Background Component
• Dynamic vibration of the structure under wind pressure
• Relatively small
• Normally treated using static method of analysis
Resonant Component
Wind Action
Separation Zone
Windward Wake Leeward
Wind Separation point
Stagnation point
Plan View Side View
15
(+) ()
() ()
Effect of Wind on Buildings
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
Wind Action: Simplified Method
Appendix A, MS 1553: 2002 1. Determined basic wind speed, Vs (Figure A1, MS 1553: 2002) 2. Determine the terrain/height multiplier, Mz,cat (Table A1, MS
1553: 2002) 3. Determine external pressure coefficient, Cp,e for surface of enclose
building (A2.3 and A2.4, MS 1553: 2002) 4. Determine internal pressure coefficient, Cp,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 = 𝟎. 𝟔𝟏𝟑 𝑽𝒔𝟐 𝑴𝒛,𝒄𝒂𝒕
𝟐𝑪𝒑,𝒆 − 𝑪𝒑,𝒊
Wind Action: Simplified Method
Figure A1, MS 1553: 2002
Wind Action: Simplified Method
Wind Action: Simplified Method
Wind Action: Simplified Method
Wind Action: Analytical Procedure
Section 2, MS 1553: 2002
𝑽𝒅𝒆𝒔 = 𝑽𝒔𝒊𝒕𝒍
𝑪𝒇𝒊𝒈 = 𝑪𝒑,𝒆𝑲𝒂𝑲𝒄𝑲𝒍𝑲𝒑
𝑽𝒔𝒊𝒕 = 𝑽𝒔𝑴𝒅𝑴𝒛,𝒄𝒂𝒕𝑴𝒔𝑴𝒉
𝑪𝒅𝒚𝒏 = 𝟏. 𝟎 𝒊𝒇
> 𝟏 𝑯𝒆𝒓𝒕𝒛 (𝒖𝒏𝒍𝒆𝒔𝒔 𝒕𝒉𝒆 𝒔𝒕𝒓𝒖𝒄𝒕𝒖𝒓𝒆 𝒊𝒔 𝒘𝒊𝒏𝒅 𝒔𝒆𝒏𝒔𝒊𝒕𝒊𝒗𝒆)
𝒑 = 𝟎. 𝟔𝟏𝟑 𝑽𝒅𝒆𝒔𝟐𝑪𝒇𝒊𝒈𝑪𝒅𝒚𝒏
If others, see Next
Design wind pressure (in Pa):
Wind Action: Analytical Procedure
Vdes = Design wind speed l = Importance factor (Table 3.2, MS 1553: 2002) Vs = Basic wind speed (33.5 m/s: Zone 1 & 32.5 m/s: Zone 2) Md = Wind directional multiplier = 1.0 Mz, cat = Terrain/Height multiplier (Table 4.1, MS 1553: 2002) Ms = Shielding multiplier (Table 4.3, MS 1553: 2002) = 1.0 if effects of shielding is ignored or not applicable Mh = Hill shape multiplier = 1.0 except for particular cardinal direction in the local topographic zones Cfig = Aerodynamic shape factor Cp,e = External pressure coefficient (Table 5.2(a) & 5.2(b), MS 1553: 2002) Cdyn = Dynamic response factor Ka = Area reduction factor = 1.0 (in most cases) Kc = Combination factor = 1.0 (in most cases) Kl = Local pressure factor = 1.0 (in most cases) Kp = Porous cladding reduction factor = 1.0 (in most cases)
Wind Action: Analytical Procedure
Section 2, MS 1553: 2002
𝑪𝒅𝒚𝒏 =
𝟏 + 𝟐𝒍𝒉 𝒈𝒗𝟐𝑩𝒔 +
𝒈𝑹𝟐𝑺𝑬𝒕𝜻
𝟏 + 𝟐𝒈𝒗𝒍𝒉
𝑩𝒔 =𝟏
𝟏 +𝟑𝟔 𝒉 − 𝒔 𝟐 + 𝟔𝟒𝒃𝒔𝒉
𝟐
𝑳𝒉
𝑺 =𝟏
𝟏 +𝟑. 𝟓𝒏𝒂𝒉 𝟏 + 𝒈𝒗𝒍𝒉
𝑽𝒅𝒆𝒔𝟏 +
𝟒𝒏𝒂𝒃𝟎𝒉 𝟏 + 𝒈𝒗𝒍𝒉𝑽𝒅𝒆𝒔
𝑬𝒕 =𝟎. 𝟒𝟕𝑵
𝟐 + 𝑵 𝟓/𝟔
𝑵 =𝒏𝒂𝑳𝒉 𝟏 + 𝒈𝒗𝒍𝒉
𝑽𝒅𝒆𝒔
Wind Action: Analytical Procedure
Wind Action: Analytical Procedure
Wind Action: Analytical Procedure
Wind Action: Analytical Procedure
Wind Action: Analytical Procedure
Example 5
WIND LOAD CALCULATION
Example 5: Wind Load Calculation
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, na = nc: 0.2 Hertz Mode Shape: Linear, k: 1.0 Average Building Density: 160 kg/m3
Wind
Win
d
Side Elevation Plan View
b = 18 m
b =
18
m
d = 51 m
h = 25 m
Example 5: Wind Load Calculation
Using Analytical Procedure in Section 2, MS 1553: 2002
Design wind pressure, 𝒑 = 𝟎. 𝟔𝟏𝟑 𝑽𝒅𝒆𝒔𝟐𝑪𝒇𝒊𝒈𝑪𝒅𝒚𝒏
(1) Design wind speed, 𝑽𝒅𝒆𝒔 = 𝑽𝒔𝒊𝒕𝒍 = 𝟑𝟐. 𝟓 × 𝟏. 𝟏𝟓 = 𝟑𝟕. 𝟑𝟕 𝒎/𝒔 where 𝑉𝑠𝑖𝑡 = 𝑉𝑠𝑀𝑑𝑀𝑧,𝑐𝑎𝑡𝑀𝑠𝑀ℎ = 33.5 × 1.00 × 0.97 × 1.00 × 1.00 = 32.5 𝑚/𝑠 Vs = Basic wind speed (Figure A.1) = 33.5 m/s Md = Wind directional multiplier = 1.00 Mz,cat = Terrain/Height multiplier (Table 4.1) = 0.97 by interpolation Ms = Shielding multiplier = 1.00 Mh = Hill shape multiplier = 1.00 l = Importance factor (Table 3.2) = 1.15
Example 5: Wind Load Calculation
25 m
Example 5: Wind Load Calculation
Example 5: Wind Load Calculation
(2) Aerodynamic shape factor, 𝑪𝒇𝒊𝒈 = 𝑪𝒑,𝒆𝑲𝒂𝑲𝒄𝑲𝒍𝑲𝒑
𝑪𝒇𝒊𝒈 𝑾 = 𝑪𝒑,𝒆𝑲𝒂𝑲𝒄𝑲𝒍𝑲𝒑 = 𝟎. 𝟖𝟎
𝑪𝒇𝒊𝒈 𝑳 = 𝑪𝒑,𝒆𝑲𝒂𝑲𝒄𝑲𝒍𝑲𝒑 = −𝟎. 𝟐𝟓
where Cp,e = External pressure coefficient Windward wall = 0.80 (for varying z) Leeward wall = -0.25 by interpolation (for d/b = 51 m/18 m = 2.8) Ka = Area reduction factor = 1.00 Kc = Combination factor = 1.00 Kl = Local pressure factor = 1.00 Kp = Porous cladding reduction factor = 1.00
Example 5: Wind Load Calculation
2.8
Example 5: Wind Load Calculation
(3) Dynamic response factor,𝑪𝒅𝒚𝒏 =
𝟏+𝟐𝒍𝒉 𝒈𝒗𝟐𝑩𝒔+𝒈𝑹𝟐𝑺𝑬𝒕𝝃
𝟏+𝟐𝒈𝒗𝒍𝒉= 𝟏. 𝟑𝟐𝟕
where lh = Turbulence intensity at z = h (Table 6.1) = 0.205 by interpolation gv = Peak factor = 3.7 given
Bs = Background factor = 1
1+36 ℎ−𝑠 2+64𝑏𝑠ℎ
2
𝐿ℎ
= 0.7431
S = Size reduction factor = 1
1+3.5𝑛𝑎ℎ 1+𝑔𝑣𝑙ℎ
𝑉𝑑𝑒𝑠1+
4𝑛𝑎𝑏0ℎ 1+𝑔𝑣𝑙ℎ𝑉𝑑𝑒𝑠
= 0.188
Et = Spectrum of turbulence = 0.47𝑁
(2+𝑁)5/6= 0.623
= Ratio of structural damping to critical damping (Table 6.2) = 0.05
gR = Peak factor for resonance response = 2 log𝑒 600𝑛𝑎 = 3.09
h = 25 m s = 0 m
𝐿ℎ = 1000ℎ
10
0.25
= 1257 𝑚
bsh = 51 m
na = 0.2 Hz given b0h = bsh = 51 m
𝑁 =𝑛𝑎𝐿ℎ 1 + 𝑔𝑣𝑙ℎ
𝑉𝑑𝑒𝑠= 11.83
Example 5: Wind Load Calculation
25 m
Example 5: Wind Load Calculation
Example 5: Wind Load Calculation
Level Height (m) Mz,cat (Table
4.1)
𝒑 = 𝟎. 𝟔𝟏𝟑 𝑽𝒅𝒆𝒔𝟐𝑪𝒇𝒊𝒈𝑪𝒅𝒚𝒏 (N/m2)
Windward Leeward Total, ptotal
Vdes,z (m/s) pW Vdes,h(m/s) pL pW - pL
7 (Roof) 25.0 0.970 37.4 910 37.4 -284 1194
6 21.5 0.950 36.6 872 37.4 -284 1156
5 18.0 0.920 35.4 815 37.4 -284 1099
4 14.5 0.890 34.3 766 37.4 -284 1050
3 11.0 0.840 32.4 683 37.4 -284 967
2 7.5 0.790 30.4 601 37.4 -284 885
1 4.0 0.750 28.9 543 37.4 -284 827
Example 5: Wind Load Calculation
0.910
Total (kN/m2)
1.156
pL (kN/m2)
0.872
0.766
0.601
1.050
0.885
0.248
0.248
0.248
0.248
pW (kN/m2)
Level 7 (Roof)
Level 6
Level 4
Ground
Level 2
1.194
Frame Analysis for Lateral Actions
• 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
Portal 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
Cantilever Method
Frame Stability
Pinned beam-to-column connections
Frame is not stable when subjected to lateral forces
Vertical load La
tera
l lo
ad
Frame Stability – Unbraced
Pinned base
Rigid beam-to-column connection
Unstable Frame
Rigid base
Frame Stability – Braced
Unstable Frame
Pinned connections
Rigid core
Stable Frame
Example 6
LATERAL LOAD ANALYSIS – CANTILEVER METHOD
Example 6: Lateral Load Analysis (Cantilever Method)
Design wind load for each floor level of sub-frame 3 = 1.2wk: where wk = Total wind pressure, ptotal Contact area Roof: 1.2 1.194 kN/m2 4.50 m 1.75 m = 11.3 kN Level 6: 1.2 1.156 kN/m2 4.50 m 3.50 m = 21.8 kN Level 5: 1.2 1.099 kN/m2 4.50 m 3.50 m = 20.8 kN Level 4: 1.2 1.050 kN/m2 4.50 m 3.50 m = 19.8 kN Level 3: 1.2 0.967 kN/m2 4.50 m 3.50 m = 18.3 kN Level 2: 1.2 0.885 kN/m2 4.50 m 3.50 m = 16.7 kN Level 1: 1.2 0.827 kN/m2 4.50 m 3.75 m = 16.7 kN
5 m 4 m
Example 6: Lateral Load Analysis (Cantilever Method)
Roof
Level 6
Level 5
Level 4
Level 3
Level 2
Level 1
11.3 kN
21.8 kN
20.8 kN
19.8 kN
18.3 kN
16.7 kN
16.7 kN
6 m 6 m 6 m
4 m
3
.5 m
3
.5 m
3
.5 m
3
.5 m
3
.5 m
3
.5 m
A B C D x = 9 m
K
L
M
N
P
Q
R
Example 6: Lateral Load Analysis (Cantilever Method)
Roof
Level 6
Level 5
Level 4
Level 3
Level 2
Level 1
11.3 kN
21.8 kN
20.8 kN
19.8 kN
18.3 kN
16.7 kN
16.7 N
6 m 6 m 6 m
4 m
3
.5 m
3
.5 m
3
.5 m
3
.5 m
3
.5 m
3
.5 m
A B C D
K
L
M
N
P
Q
R
Column A B C D
Distance from Centroid 9.00 3.00 3.00 9.00
Column Axial Load 3.0P 1.0P 1.0P 3.0P
Example 6: Lateral Load Analysis (Cantilever Method)
LEVEL K AND ABOVE
Example 6: Lateral Load Analysis (Cantilever Method)
Level K and Above
H4
11.3 kN
6 m 6 m 6 m
1.75 m
3.0P
K
1.0P 1.0P
3.0P
H3 H2 H1
Axial Force in Columns 𝑀@𝐾 = 0 11.3 × 1.75 + 1.0𝑃 × 6 − 1.0𝑃 × 12 − 3.0𝑃 × 18 = 0 P = 0.33 kN
Example 6: Lateral Load Analysis (Cantilever Method)
Shear Force in Beams & Columns Consider sub-frame to the left of F1: 𝐹𝑣 = 0: F1 – 0.99 = 0 F1 = 0.99 kN 𝑀@𝐹1 = 0: 𝐻1 × 1.75 − 0.99 × 3 = 0 H1 = 1.70 kN Consider sub-frame to the left of F2: 𝐹𝑣 = 0: F2 – 0.99 – 0.33= 0 F2 = 1.32 kN 𝑀@𝐹2 = 0: 𝐻1 + 𝐻2 × 1.75 − 0.99 × 9 − 0.33 × 3 = 0 H2 = 3.96 kN Consider sub-frame to the left of F3: 𝐹𝑣 = 0: F3 – 0.99 – 0.33 + 0.33= 0 F3 = 0.99 kN 𝑀@𝐹3 = 0: 𝐻1 + 𝐻2 + 𝐻3 × 1.75 − 0.99 × 15 − 0.33 × 9 + 0.33 × 3 = 0 H3 = 3.96 kN 𝐹𝐻 = 0: 11.3 – H1 – H2 – H3 – H4 = 0 H4 = 1.68 kN
H4
11.3 kN
6 m 6 m 6 m
1.75 m
0.99 kN
K
0.33 kN
H3 H2 H1
F1 F2 F3
0.99 kN 0.33 kN
Example 6: Lateral Load Analysis (Cantilever Method)
LEVEL L AND ABOVE
Example 6: Lateral Load Analysis (Cantilever Method)
Level L and Above
Axial Force in Columns 𝑀@𝐿 = 0 11.3 × 5.25 + 21.8 × 1.75 + 1.0𝑃 × 6 − 1.0𝑃 × 12 − 3.0𝑃 × 18 = 0 P = 1.62 kN
6 m 6 m 6 m 3.0P 1.0P
1.0P 3.0P
11.3 kN
21.8 kN
3.5
m
K
L H4 H3 H2 H1
1.75 m
Example 6: Lateral Load Analysis (Cantilever Method)
6 m 6 m 6 m 4.86 kN
1.70
21.8 + 11.3 = 33.1 kN
K
L H4 H3 H2 H1
1.75 m
1.62 kN 1.62 kN 4.86 kN
F1 F2 F3 1.75 m
0.99 kN
3.96
0.33 kN
3.96
0.33 kN
1.68
0.99 kN
Shear Force in Beams & Columns Consider sub-frame to the left of F1: 𝐹𝑣 = 0: F1 – 4.86 + 0.99 = 0 F1 = 3.87 kN 𝑀@𝐹1 = 0: 𝐻1 + 1.70 × 1.75 − 4.86 − 0.99 × 3 = 0 H1 = 4.93 kN Consider sub-frame to the left of F2: 𝐹𝑣 = 0: F2 – 4.86 + 0.99 – 1.62 + 0.33 = 0 F2 = 5.16 kN 𝑀@𝐹2 = 0: 𝐻1 + 𝐻2 × 1.75 + 1.70 + 3.96 × 1.75 − 4.86 − 0.99 × 9 − 1.62 − 0.33 × 3 = 0
H2 = 11.52 kN
Example 6: Lateral Load Analysis (Cantilever Method)
6 m 6 m 6 m 4.86 kN
1.70
21.8 + 11.3 = 33.1 kN
K
L H4 H3 H2 H1
1.75 m
1.62 kN 1.62 kN 4.86 kN
F1 F2 F3 1.75 m
0.99 kN
3.96
0.33 kN
3.96
0.33 kN
1.68
0.99 kN
Shear Force in Beams & Columns Consider sub-frame to the left of F3: 𝐹𝑣 = 0: F3 – 4.86 + 0.99 – 1.62 + 0.33 + 1.62 – 0.33 = 0 F3 = 3.87 kN 𝑀@𝐹3 = 0: 𝐻1 + 𝐻2 + 𝐻3 × 1.75 + 1.70 + 3.96 + 3.96 × 1.75 − 4.86 − 0.99 × 15 − 1.62 − 0.33 × 9
+ 1.62 − 0.33 × 3 = 0 H3 = 11.52 kN 𝐹𝐻 = 0: 33.1 – H1 – H2 – H3 – H4 = 0 H4 = 5.13 kN
Example 6: Lateral Load Analysis (Cantilever Method)
LEVEL M AND ABOVE
Example 6: Lateral Load Analysis (Cantilever Method)
Level M and Above
Axial Force in Columns 𝑀@𝐿 = 0
11.3 × 8.75 + 21.8 × 5.25 + 20.8 × 1.75 + 1.0𝑃 × 6 − 1.0𝑃 × 12 − 3.0𝑃 × 18 = 0 P = 4.16 kN
11.3 kN
6 m 6 m 6 m 3.0P 1.0P 1.0P
3.0P
21.8 kN
K
L
H4 H3 H2 H1
3.5 m
20.8 kN
3.5 m
M 1.75 m
Example 6: Lateral Load Analysis (Cantilever Method)
6 m 6 m 6 m 12.48 kN
4.93
21.8 + 11.3 + 20.8 = 53.9 kN
L
M H4 H3 H2 H1
1.75 m
4.16 kN 4.16 kN 12.48 kN
F1 F2 F3 1.75 m
4.86 kN
11.52
1.62 kN
11.52
1.62 kN
5.13
4.86 kN
Shear Force in Beams & Columns Consider sub-frame to the left of F1: 𝐹𝑣 = 0: F1 – 12.48 + 4.86 = 0 F1 = 7.62 kN 𝑀@𝐹1 = 0: 𝐻1 + 4.93 × 1.75 − 12.48 − 4.86 × 3 = 0 H1 = 8.13 kN Consider sub-frame to the left of F2: 𝐹𝑣 = 0: F2 – 12.48 + 4.86 – 4.16 + 1.62 = 0 F2 = 10.16 kN 𝑀@𝐹2 = 0: 𝐻1 + 𝐻2 × 1.75 + 4.93 + 11.52 × 1.75 − 12.48 − 4.86 × 9 − 4.16 − 1.62 × 3 = 0
H2 = 18.93 kN
Example 6: Lateral Load Analysis (Cantilever Method)
Shear Force in Beams & Columns Consider sub-frame to the left of F3: 𝐹𝑣 = 0: F3 – 12.48 + 4.86 – 4.16 + 1.62 + 4.16 – 1.62 = 0 F3 = 7.62 kN 𝑀@𝐹3 = 0: 𝐻1 + 𝐻2 + 𝐻3 × 1.75 + 4.93 + 11.52 + 11.52 × 1.75 − 12.48 − 4.86 × 15 − 4.16 − 1.62 × 9
+ 4.16 − 1.62 × 3 = 0 H3 = 18.99 kN 𝐹𝐻 = 0: 53.9 – H1 – H2 – H3 – H4 = 0 H4 = 7.85 kN
6 m 6 m 6 m 12.48 kN
4.93
21.8 + 11.3 + 20.8 = 53.9 kN
L
M H4 H3 H2 H1
1.75 m
4.16 kN 4.16 kN 12.48 kN
F1 F2 F3 1.75 m
4.86 kN
11.52
1.62 kN
11.52
1.62 kN
5.13
4.86 kN
Example 6: Lateral Load Analysis (Cantilever Method)
LEVEL N AND ABOVE
LEVEL P AND ABOVE
LEVEL Q AND ABOVE
LEVEL R AND ABOVE
Do the Following Level:
Example 6: Lateral Load Analysis (Cantilever Method)
SHEAR FORCE IN BEAMS & COLUMNS
Example 6: Lateral Load Analysis (Cantilever Method)
6 m 6 m 6 m
3.5 m
3.5 m
3.5 m
K
L M
0.99 1.32 0.99
1.70 3.96 3.96 1.68
3.87 5.16 3.87
4.93 11.52 11.52 5.13
7.62 10.16 7.62
8.13 18.93 18.99 7.85
Example 6: Lateral Load Analysis (Cantilever Method)
BENDING MOMENTS IN BEAMS & COLUMNS
Example 6: Lateral Load Analysis (Cantilever Method)
2.97
3.96
2.98 6.93 2.94
11.61 15.48 11.61 8.63 20.16 20.16 8.98
22.86
30.48 22.86 14.23 33.13
18.60
13.74
2.97 2.97
3.96
2.98
6.93
6.93 6.93
8.63 20.16
22.86
20.16 8.98
33.23
14.23 33.13 33.23 13.74
Continues for Level N, P, Q, R
2.97
11.61 15.48
2.94
Example 7
ANALYSIS OF ONE LEVEL SUB-FRAME – VERTICAL LOAD
ONLY
Example 7: Analysis of One Level Sub-Frame – Vertical Load Only
Roof
Level 6
Level 5
Level 4
Level 3
Level 2
Level 1
6 m 6 m 6 m
4 m
3
.5 m
3
.5 m
3
.5 m
3
.5 m
3
.5 m
3
.5 m
A B C D
CASE 1: 1.2Gk + 1.2Qk
Fixed End Moment
𝑀𝐴𝐵 = 𝑀𝐵𝐴 =𝑤𝐿2
12=48.7 × 62
12= 146.2 𝑘𝑁𝑚
𝑀𝐵𝐶 =𝑤𝐿2
12=46.1 × 62
12= 138.4 𝑘𝑁𝑚
𝑀𝐶𝐷 =𝑤𝐿2
12=48.7 × 62
12= 146.2 𝑘𝑁𝑚
6 m 6 m 6 m
3.5 m
4 m
48.7 kN/m 46.1 kN/m
A B C D
Span 1 Span 2 Span 3
48.7 kN/m
Example 7: Analysis of One Level Sub-Frame – Vertical Load Only
43.82 -9.97 -9.97 -43.82
0.27 -0.44 0.44 -0.27
1.60 -1.15 1.15 -1.60
0.36 -6.86 6.86 -0.32
41.60 -1.53 1.53 -41.60
0.28 A B 0.19 C 0.19 D 0.28
0.25 0.47 0.32 0.17 0.32 0.32 0.17 0.32 0.47 0.25
-146.2 146.2 -138.4 138.4 -146.2 146.2
36.4 68.2 -2.5 -1.3 -2.5 2.5 1.3 2.5 -68.2 -36.4
-1.3 34.1 1.3 -1.3 -34.1 1.3
0.3 0.6 -11.3 -6.0 -11.3 11.3 6.0 11.3 -0.6 -0.3
-5.6 0.3 5.6 -5.6 -0.3 5.6
1.4 2.6 -1.9 -1.0 -1.9 1.9 1.0 1.9 -2.6 -1.4
-0.9 1.3 0.3 -0.9 -1.3 0.9
0.2 0.4 -0.7 -0.4 -0.7 0.7 0.4 0.7 -0.4 -0.2
38.3 -82.2 165.6 -8.7 -146.9 146.9 8.7 -165.6 82.2 -38.3
Moment Distribution Method
Example 7: Analysis of One Level Sub-Frame – Vertical Load Only
6 m
48.7 kN/m
82.2 165.6
VAB VBA
46.1 kN/m
6 m
146.9 146.9
VBC VCB
48.7 kN/m
6 m
165.6 82.2
VCD VDC
Solve for each span using equilibrium method of analysis: VAB = 132.3 kN VBA = 160.1 kN VBC = 138.4 kN VCB = 138.4 kN VCD = 160.1 kN VDC = 132.3 kN
Reaction Calculation at Support
Example 7: Analysis of One Level Sub-Frame – Vertical Load Only
Shear Force and Bending Moment Diagram
SFD (kN)
BMD (kNm)
132.3
160.1
138.4 160.1
156.2 132.3
82.2
97.5
165.6
60.6 97.5
146.9 146.9
165.6
82.2 43.8
38.3
19.2
10.0 8.7 10.0 8.7
43.8 38.3
4.4 19.2 4.4
2.7 m 3.0 m 3.3 m
Example 7: Analysis of One Level Sub-Frame – Vertical Load Only
Bending Moment due to Vertical Load (1.2Gk + 1.2Qk)
82.2
97.5
165.6
60.6 97.5
146.9 146.9
165.6
82.2 43.8
38.3
19.2
10.0 8.7 10.0 8.7
43.8 38.3
4.4 19.2 4.4
+ Bending Moment due to Wind Load, 1.2Wk (from your exercise)
= Combination of Vertical & Wind Load
Example 7: Analysis of One Level Sub-Frame – Vertical Load Only