frame analysis -...

FRAME ANALYSIS

Dr. Izni Syahrizal bin Ibrahim Faculty of Civil Engineering

Universiti Teknologi Malaysia

Email: iznisyahrizal@utm.my

Introduction

Building Construction

3D Frame: Beam, Column & Slab

3D Frame: Beam & Column ONLY

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

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)

Alternate Span Loaded Adjacent Span Loaded

1.35Gk

1.5Qk 1.5Qk

Load Set 1

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

All Spans Loaded

Alternate Span Loaded

1.35Gk

Load Set 2

Example 1

ACTION ON BEAMS

6 m 6 m 6 m

Level 1

Level 2

Level 3

Level 4

Level 5

Level 6

Level 7

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

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

6 m 6 m 6 m

4 m 0.47

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)

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

wmax = 1.35Gk + 1.5Qk = 57.53 kN/m wmin = 1.35Gk = 30.66 kN/m

6 m 6 m 6 m

Example 2

ANALYSIS OF ONE-LEVEL SUB-FRAME

6 m 6 m 6 m

Level 1

Level 2

Level 3

Level 4

Level 5

Level 6

Level 7

Example 2

Example 2: Analysis of One Level Sub-Frame

6 m 6 m 6 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

Analysis using LOAD SET 1

KAB KBC KCD

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

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

𝐹𝑐,𝑙 =𝐾𝑐,𝑙

Joint B & Joint C: 𝐹𝐵𝐴, 𝐶𝐵 =𝐾𝐴𝐵, 𝐵𝐶

𝐾𝐴𝐵, 𝐵𝐶+𝐾𝐵𝐶, 𝐶𝐷+𝐾𝑐,𝑢+𝐾𝑐,𝑙= 0.32

𝐹𝐵𝐶, 𝐶𝐷 =𝐾𝐵𝐶, 𝐶𝐷

𝐹𝑐,𝑢 =𝐾𝑐,𝑢

𝐹𝑐,𝑙 =𝐾𝑐,𝑙

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

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

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

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

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

wmax = 57.5 kN/m

95.6 200.4

VAB VBA

wmax = 54.4 kN/m

184.5 142.3

VBC VCB

wmin = 30.7 kN/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

Shear Force and Bending Moment Diagram

SFD (kN)

BMD (kNm)

156.2 78.5

82.0 55.4

142.3 184.5

45.1 51.0 44.6

8.5 7.4 8.7 7.6

24.1 21.1

3.8 10.6 3.7

4.3 4.4

2.7 m 3.1 m 3.4 m

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

wmin = 30.7 kN/m

wmax = 57.5 kN/m wmax = 54.4 kN/m

A B C D

SELF STUDY !!! CASE 3: Span 1 & 3 = wmax, Span 2 = wmin

6 m 6 m 6 m

wmax = 57.5 kN/m wmax = 57.5 kN/m wmin = 29.1 kN/m

A B C D

SELF STUDY !!! CASE 4: Span 1 & 3 = wmin, Span 2 = wmax

6 m 6 m 6 m

wmin = 30.7 kN/m wmin = 30.7 kN/m wmax = 54.4 kN/m

A B C D

Shear Force and Bending Moment Envelope

SFD (kN)

BMD (kNm)

82.0, 82.0 55.4

Case 1 Case 2 Case 3 Case 4

45.1 125.9

142.3 95.6

184.5 200.4

168.8 104.5

112.6 112.6

54.1 43.8

153.4 153.4

43.8 54.1

190.0, 190.0

170.3, 170.3

156.2 78.5

170.3, 170.3

190.0, 190.0

161.9, 161.9

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

190.0, 190.0

170.3, 170.3

156.2 78.5

170.3, 170.3

190.0, 190.0

161.9, 161.9

161.9, 161.9 Case 1 Case 2 Case 3 Case 4

Example 3

ANALYSIS OF TWO FREE-JOINT SUB-FRAME

6 m 6 m 6 m

Level 1

Level 2

Level 3

Level 4

Level 5

Level 6

Level 7

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

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

Span 1 Span 2

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

12=54.4 × 62

12= 163.3 𝑘𝑁𝑚

CASE 2: Span 1 = wmax , Span 2 = wmin

6 m 6 m

wmax = 57.5 kN/m

wmin = 29.11 kN/m

Span 1 Span 2

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

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

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

168.8 97.1

30.2 49.0 56.0

SFD (kN)

BMD (kNm)

6 m 6 m 6 m

wmax = 57.5 kN/m

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

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

3500= 4.6 × 105 𝑚𝑚3

𝐾𝑐,𝑙 =1.6×109

4000= 4.0 × 105 𝑚𝑚3

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

CASE 2: Span 2 = wmax, Span 1 & 3 = wmin

6 m 6 m 6 m

wmax = 30.7 kN/m wmin = 30.7 kN/m wmax = 54.4 kN/m

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

12=30.7 × 62

12= 92.0 𝑘𝑁𝑚

12=54.4 × 62

12= 163.3 𝑘𝑁𝑚

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

3500= 4.6 × 105 𝑚𝑚3

𝐾𝑐,𝑙 =1.6×109

4000= 4.0 × 105 𝑚𝑚3

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

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

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

146.6 146.6

17.7 17.7 20.2 20.2

SFD (kN)

BMD (kNm) 89.3

Example 4

CONTINUOUS BEAM + ONE FREE-JOINT SUB-FRAME

Example 4

6 m 6 m 6 m

Level 1

Level 2

Level 3

Level 4

Level 5

Level 6

Level 7

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

Joint A: 𝐹𝐵𝐴 =5.63

5.63+0= 1.00

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

6 m 6 m 6 m

A B C D

wmax = 57.5 kN/m

Fixed End Moment, FEM

12=57.5 × 62

12= 172.6 𝑘𝑁𝑚

12=54.4 × 62

12= 163.3 𝑘𝑁𝑚

12=30.7 × 62

12= 92.0 𝑘𝑁𝑚

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

A B C D

VAB VBA VBC

wmax = 57.5 kN/m

wmax = 54.4 kN/m

217.9 137.2

wmin = 30.7 kN/m

VCD VDC

SELF STUDY !!!

6 m 6 m 6 m

A B C D

wmax = 57.5 kN/m

SELF STUDY !!!

6 m 6 m 6 m

A B C D

wmin = 29.1 kN/m

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

FINALLY DO THE SFD & BMD ENVELOPE !!!

Sub-Frame: Bending Moment in Column

wmax = 57.5 kN/m

0.5KAB

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

3500= 4.6 × 105 𝑚𝑚3

𝐾𝑐,𝑙 =1.6×109

4000= 4.0 × 105 𝑚𝑚3

Column A

Moment in upper column, Mupper

𝑀𝑢𝑝𝑝𝑒𝑟 = 𝐹𝐸𝑀 ×𝐾𝑐,𝑢

𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝐴𝐵= 172.6 ×

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 ×

4.6+4.0+0.5×7.5= 𝟓𝟔. 𝟎 kNm

64.3 56.0

𝑀𝐵𝐴 =𝑤𝐿2

12=57.5 × 62

12= 172.6 𝑘𝑁𝑚

12=29.1 × 62

12= 87.3 𝑘𝑁𝑚

∆𝐹𝐸𝑀 = 𝑀𝐴𝐵 −𝑀𝐵𝐶 = 𝟖𝟓. 𝟑 𝒌𝑵𝒎

6 m 6 m

wmax = 57.5 kN/m

wmin = 29.1 kN/m

0.5KAB 0.5KBC

Column B

𝑀𝑢𝑝𝑝𝑒𝑟 = ∆𝐹𝐸𝑀 ×𝐾𝑐,𝑢

𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝐴𝐵+0.5𝐾𝐵𝐶= 85.3 ×

4.6+4.0+0.5×7.5+0.5×7.5= 𝟐𝟒. 𝟐 kNm

𝑀𝑢𝑝𝑝𝑒𝑟 = ∆𝐹𝐸𝑀 ×𝐾𝑐,𝑙

𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝐴𝐵+0.5𝐾𝐵𝐶= 85.3 ×

4.6+4.0+0.5×7.5+0.5×7.5= 𝟐𝟏. 𝟐 kNm

21.2 24.2

12=29.1 × 62

12= 87.3 𝑘𝑁𝑚

12=57.5 × 62

12= 172.6 𝑘𝑁𝑚

∆𝐹𝐸𝑀 = 𝑀𝐶𝐷 −𝑀𝐵𝐶 = 𝟖𝟓. 𝟑 𝒌𝑵𝒎

6 m 6 m

0.5KBC 0.5KCD

Column C

𝑀𝑢𝑝𝑝𝑒𝑟 = ∆𝐹𝐸𝑀 ×𝐾𝑐,𝑢

𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝐵𝐶+0.5𝐾𝐶𝐷= 85.3 ×

4.6+4.0+0.5×7.5+0.5×7.5= 𝟐𝟒. 𝟐 kNm

𝑀𝑢𝑝𝑝𝑒𝑟 = ∆𝐹𝐸𝑀 ×𝐾𝑐,𝑙

𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝐵𝐶+0.5𝐾𝐶𝐷= 85.3 ×

4.6+4.0+0.5×7.5+0.5×7.5= 𝟐𝟏. 𝟐 kNm

24.2 21.2

wmax = 57.5 kN/m

0.5KCD

𝑀𝐷𝐶 =𝑤𝐿2

12=57.5 × 62

12= 172.6 𝑘𝑁𝑚

Column D

𝑀𝑢𝑝𝑝𝑒𝑟 = 𝐹𝐸𝑀 ×𝐾𝑐,𝑢

𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝐶𝐷= 172.6 ×

4.6+4.0+0.5×7.5= 𝟔𝟒. 𝟑 kNm

𝑀𝑢𝑝𝑝𝑒𝑟 = 𝐹𝐸𝑀 ×𝐾𝑐,𝑙

𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝐶𝐷= 172.6 ×

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

(+) ()

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

Figure A1, MS 1553: 2002

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):

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)

Section 2, MS 1553: 2002

𝑪𝒅𝒚𝒏 =

𝟏 + 𝟐𝒍𝒉 𝒈𝒗𝟐𝑩𝒔 +

𝒈𝑹𝟐𝑺𝑬𝒕𝜻

𝟏 + 𝟐𝒈𝒗𝒍𝒉

𝑩𝒔 =𝟏

𝟏 +𝟑𝟔 𝒉 − 𝒔 𝟐 + 𝟔𝟒𝒃𝒔𝒉

𝑳𝒉

𝑺 =𝟏

𝟏 +𝟑. 𝟓𝒏𝒂𝒉 𝟏 + 𝒈𝒗𝒍𝒉

𝑽𝒅𝒆𝒔𝟏 +

𝟒𝒏𝒂𝒃𝟎𝒉 𝟏 + 𝒈𝒗𝒍𝒉𝑽𝒅𝒆𝒔

𝑬𝒕 =𝟎. 𝟒𝟕𝑵

𝟐 + 𝑵 𝟓/𝟔

𝑵 =𝒏𝒂𝑳𝒉 𝟏 + 𝒈𝒗𝒍𝒉

𝑽𝒅𝒆𝒔

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

Side Elevation Plan View

b = 18 m

d = 51 m

h = 25 m

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

(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

(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

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

= 1257 𝑚

bsh = 51 m

na = 0.2 Hz given b0h = bsh = 51 m

𝑁 =𝑛𝑎𝐿ℎ 1 + 𝑔𝑣𝑙ℎ

𝑉𝑑𝑒𝑠= 11.83

Level Height (m) Mz,cat (Table

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

Total (kN/m2)

pL (kN/m2)

pW (kN/m2)

Level 7 (Roof)

Level 6

Level 4

Ground

Level 2

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

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

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

6 m 6 m 6 m

A B C D x = 9 m

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

A B C D

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

LEVEL K AND ABOVE

Level K and Above

11.3 kN

6 m 6 m 6 m

1.75 m

1.0P 1.0P

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

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

11.3 kN

6 m 6 m 6 m

1.75 m

0.99 kN

0.33 kN

H3 H2 H1

F1 F2 F3

0.99 kN 0.33 kN

LEVEL L AND ABOVE

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

L H4 H3 H2 H1

1.75 m

6 m 6 m 6 m 4.86 kN

21.8 + 11.3 = 33.1 kN

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

0.33 kN

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

6 m 6 m 6 m 4.86 kN

21.8 + 11.3 = 33.1 kN

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

0.33 kN

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

LEVEL M AND ABOVE

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

21.8 kN

H4 H3 H2 H1

20.8 kN

M 1.75 m

6 m 6 m 6 m 12.48 kN

21.8 + 11.3 + 20.8 = 53.9 kN

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

1.62 kN

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

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

21.8 + 11.3 + 20.8 = 53.9 kN

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

1.62 kN

4.86 kN

LEVEL N AND ABOVE

LEVEL P AND ABOVE

LEVEL Q AND ABOVE

LEVEL R AND ABOVE

Do the Following Level:

SHEAR FORCE IN BEAMS & COLUMNS

6 m 6 m 6 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

BENDING MOMENTS IN BEAMS & COLUMNS

2.98 6.93 2.94

11.61 15.48 11.61 8.63 20.16 20.16 8.98

30.48 22.86 14.23 33.13

2.97 2.97

6.93 6.93

8.63 20.16

20.16 8.98

14.23 33.13 33.23 13.74

Continues for Level N, P, Q, R

11.61 15.48

Example 7

ANALYSIS OF ONE LEVEL SUB-FRAME – VERTICAL LOAD

Example 7: Analysis of One Level Sub-Frame – Vertical Load Only

Level 6

Level 5

Level 4

Level 3

Level 2

Level 1

6 m 6 m 6 m

A B C D

CASE 1: 1.2Gk + 1.2Qk

Fixed End Moment

12=48.7 × 62

12= 146.2 𝑘𝑁𝑚

12=46.1 × 62

12= 138.4 𝑘𝑁𝑚

12=48.7 × 62

12= 146.2 𝑘𝑁𝑚

6 m 6 m 6 m

48.7 kN/m 46.1 kN/m

A B C D

48.7 kN/m

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

48.7 kN/m

82.2 165.6

VAB VBA

46.1 kN/m

146.9 146.9

VBC VCB

48.7 kN/m

165.6 82.2

VCD VDC

SFD (kN)

BMD (kNm)

138.4 160.1

156.2 132.3

60.6 97.5

146.9 146.9

82.2 43.8

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

Bending Moment due to Vertical Load (1.2Gk + 1.2Qk)

60.6 97.5

146.9 146.9

82.2 43.8

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

frame analysis -...

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