Chapter 9_Frame Analysis(New)

Frame Analysis 185

99 FFRRAAMMEE AANNAALLYYSSIISS

9.1 Introduction

In a typical reinforced concrete building, the structural system is quite complex. The structure is three-

dimensional (3D), comprising floor slabs, beams, columns, and footings, which monolithically

connected and act integrally to resist vertical loads (permanent and variable) and lateral loads (wind

loads, seismic loads). Figure 9.1 shows the example of 3D view of a reinforced concrete building

structure. This structure can be idealized as 3D frame which consist of slabs, beams and columns as

shown in figure 9.2(a). In many cases the slabs are analyzed separately, hence the idealized frame

consist only beams and columns as shown in Figure 9.2(b).

The analysis of 3D frame may represent the real behavior of the structure but it is quite

complicated since the structure is highly indeterminate. The analysis is normally carried out by

computer since manual calculations is unfeasible.

In order to simplify the analysis, the 3D frame structure is generally divided into a series of

independent parallel two dimensional plane frames (2D) along the column lines in the longitudinal and

transverse directions of the building as shown in figure 9.3. These 2D frames may further be simplified

into substitute frames (Sub-frame) or continuous beams as shown in figure 9.4.

Figure 9.1: 3D view of reinforced concrete building structure

186 Frame Analysis

Figure 9.2: 3D frame of building structure

Figure 9.3: 2D frame of building structure

Figure 9.4: Substitute frames

(a) (b)

(a) (b)

Sub-frame:

Roof level

Sub-frame:

Floor level

Frame Analysis 187

9.2 Types of Frames

Reinforced concrete frames which consist of columns and beams can be divided into two types:

i. Braced frames

Frames that not contribute to the overall stability of the structure. None of the lateral actions,

including wind, are transmitted to the columns and beams but carries by bracing members such

as shear walls. This type of frame support vertical actions only.

ii. Unraced frames

Unbraced frames are frames that contribute to the overall stability of the structure. All lateral

actions, including wind, are transmitted to the columns and beams since there are no bracing

members such as shear walls are provided. These types of frames have to support vertical and

lateral actions.

9.3 Method of Analysis

The primary objective of structural analysis is to obtain a set of internal forces and moments

throughout the structure that are in equilibrium with the design loads for the required loading

combinations. The general provisions relating to analysis of the structure are set out in EN 1992-1-1,

Figure 9.5:

Braced frame

Figure 9.6:

Unbraced frame

Shear walls as

bracing members

188 Frame Analysis

Section 5. Generally it will be satisfactory to determine envelopes of forces and moments by linear

elastic analysis of all or parts of the structure and allow for redistribution.

Code of practices permit the use of approximate analysis techniques in which the structure can be

considered as a series of sub-frames. EC 2 does not specifically describe the extent to which various

columns and beams are included in the sub-frames. The methods of sub-frames analysis discussed here

are based on BS 8110.

i. One-level Sub-frame

Each sub-frame may be taken to consist of the beams at one level together with the columns

above and below. The ends of the columns remote from the beams may generally be assumed

to be fixed unless the assumption of a pinned end is clearly more reasonable (for example,

where a foundation detail is considered unable to develop moment restraint)

ii. Two-points Sub-frame

The moments and forces in certain individual beam may be found by considering a simplified

sub-frame consisting only of that beam, the columns attached to the end of that beam and the

beams on either side, is any. The column and beam ends remote from the beam under

consideration may generally be assumed to be fixed unless the assumption of pinned is clearly

more reasonable. The stiffness of the beams on either side of the beam considered should be

taken as half their actual values if they are taken to be fixed at their outer ends.

iii. Continuous beam and one-point sub-frame

The moments and forces in the beams at one level may also be obtained by considering

the beams as a continuous beam over supports providing no restraint to rotation.

Figure 9.9: Continuous beam for analysis of beams

Figure 9.7: Sub-frame for analysis of beams and columns

Figure 9.8: Sub-frame for analysis of individual beam

Kb1 0.5Kb2 0.5Kb3 0.5Kb1 Kb2

Frame Analysis 189

The ultimate moments for column may be calculated by simple moment distribution

procedures, on the assumption that the column and beam ends remote from the junction

under consideration are fixed and that the beams posses half their actual stiffness. The

arrangement of the design ultimate variable loads should be such as to cause the

maximum moment the column.

9.4 Actions and Combination of Actions

The actions on buildings is due to permanent (dead load), variable (imposed, wind, dynamic, seismic

loads) and accidental load. In most cases, multistory buildings for office or residential purpose are

design for dead, imposed and wind loads.

Separate actions or loads must be applied to the structure in appropriate directions and various

types of actions combined with partial safety factors selected to cause the most severe design condition

for the member under consideration. In general the combination of actions discussed in section 6.4.1 of

this module should be investigated.

Example 9.1

The framing plans for a multistorey building are shown in Figure E9.1. The main dimensions,

structural features, loads, materials etc. are also set out in the figure. Analyse subframe 3/A-D, Level 1

to determine shear forces and bending moments of corresponding beams and columns. Use all the three

methods of analysis that discussed in section 9.3 of this module.

Figure 9.10: One-point sub-frames for analysis of columns

0.5Kb 0.5Kb 0.5Kb 0.5Kb

190 Frame Analysis

Data:

Permanent office building (Design life = 50 years).

Location : Near sub-urban

Zone 1 of Malaysia wind

speed mapping

Topography : Flat area slope < 0.05 Buildings around within

one- kilometers radius.

Beams in grid line 1,2,3,12 : 250 x 600 mm.

Beams in grid line A, B, C & D: 250 x 500mm.

Slab thickness: 150 mm.

Columns: 300 x 400 mm.

Imposed load: 4.0 kN/m2.

Finishes, ceiling, services etc.: 0.75 kN/m

2.

Partitions : 0.5 kN/m2

Figure E9.1

2 @

4 m

=

8 m

6 @

3.5

m

=

21 m

4

m

1m

7

@ 5

m

= 3

5 m

2

@ 4

m

=

8 m

6 m 6 m 6 m

Level 1

Level 2

Level 3

Level 4

Level 5

Level 6

Level 7

Frame Analysis 191

9.5 Analysis of Frame for Lateral Loads

Building frames are subjected to lateral loads as well as vertical loads. The necessity for careful

attention to these forces increases for tall buildings. Buildings must not only have sufficient lateral

resistance to prevent failure but also must have sufficient resistance to deflection to prevent damage to

their various parts. Rigid frame buildings are highly indeterminate; their analysis by exact methods

(unless computers are used) is so lengthy as to make the approximate methods very popular. The two

popular approximate method of analysis for lateral loads are portal method and cantilever method.

In the portal method, the frame is theoretically divided into independent portals. The shear in each

storey is assumed to be divided between the bays in proportion to their spans. The shear in each bay is

then divided equally between the columns. The column end moments are the column shear multiplied

by one-half the column height. Beam moments balance the column moments. The external column

only resist axial load which is found by dividing the overturning moment at any level by the width of

the building.

In cantilever method the axial loads in column are assumed to be proportional to the distance from

the centre of gravity of the frame. It is also usual to assume that all the column in a storey are of equal

cross-sectional area and the point of contraflexures are located at the mid-points of all columns and

beams.

It should be emphasized that these approximate methods may give quite inaccurate results for

irregular or high-rise structures. Application of cantilever method is illustrated by example 9.2.

9.6 Calculation of Wind Load

Wind forces are variable loads which act directly on the internal and external surfaces of structures.

The intensity of wind load on a structure is related to the square of the wind velocity and the dimension

of the members that are resisting the wind. Wind velocity is dependent on geographical location, the

height of the structure, the topography of the area and the roughness of the surrounding terrain.

The response of a structure to the variable action of wind can be separated into two components, a

background component and a resonant component. The background component involves static

deflection of the structure under the wind pressure. The resonant component, on the other hand,

involves dynamic vibration of the structure in response to changes in wind pressure. In most structures,

the resonant component is relatively small and structural response to wind forces is usually treated

using static methods of analysis only. Wind creates pressure on the windward side of a building and suction on its other three sides.

Wind also produces suction on flat roofs, on the leeward side of sloping roofs, and even on the

windward side of roofs with a pitch less than 30o (Figure 9.11).

Figure 9.11: Effect of wind on buildings

192 Frame Analysis

Three procedures are specified in MS 1553: 2002 for the calculation of wind pressures on buildings:

the simplified procedure, limited in application to buildings of rectangular in plan and not greater than

15.0 m high; analytical procedure, limited to regular buildings that are not more than 200 m high and

structure with roof spans less than 100 m; and the wind tunnel procedure, used for complex buildings.

(a) Simplified Procedure

The steps of simplified procedure, described in MS 1553 Appendix A, are as follows:

1. Determined the basic wind speed, Vs in accordance with MS 1553 Figure A1 assuming the wind can come from any direction.

2. Determine the terrain/height multiplier, Mz,cat as given in Table A1 MS 1553.

3. Determine the external pressure coefficient, Cp,e for surface of enclose building as given in A2.3 and A2.4

4. Determine the internal pressure coefficient, Cp,i for surface of enclose building which shall be taken as +0.6 or -0.3. The two cases shall be considered to determine the critical load

requirements for the appropriate condition.

5. The design wind pressure, p in Pa, shall be taken as:

p = 0.613(Vs)2 (Mz,cat)

2 (Cpe Cpi)

(b) Analytical Procedure

In the analytical procedure, the design wind pressure, shall be determined using the following equation

as mentioned in Section 2 of MS 1553:

p = 0.613 [Vdes]2 Cfig Cdyn Pa

where

Vdes = design wind speed

= Vsit l

l = importance factor given in Table 3.2 MS 1553

Vsit = site wind speed

= Vs Md Mz,cat Ms Mh

Vs = basic wind speed 33.5 m/s zone I and 32.5 m/s zone II respectively

as given in Figure 3.1 MS 1553 Md = wind directional multiplier = 1.0

Mz,cat = terrain/height multiplier, given in Table 4.1 MS 1553

Ms = shielding multiplier, given in Table 4.3 MS 1553. Shall be taken as

1.0 if the effects of shielding are ignored or not applicable.

Mh = hill shape multiplier. Shall be taken as 1.0 except that for particular

cardinal direction in the local topographic zones.

Cfig. = aerodynamic shape factor

= Cp,e KaKcKlKp for external pressure

Cp,e = external pressure coefficient given Tables 5.2(a) and Table 5.2(b)

Frame Analysis 193

MS 1553 for windward and leeward walls respectively for rectangular

enclosed buildings.

Ka, Kc, Kl, and Kp are area reduction factor, combination factor, local pressure factor

and porous cladding reduction factor respectively. All shall be taken as 1.0 in most

cases.

Cdyn = dynamic response factor. Shall be taken as 1.0 unless the structure

is wind sensitive

Example 9.2

For the multistorey building in Example 9.1,

(a) Calculate the wind load on the building.

(b) Calculate the bending moments for all beams and columns, due to wind load from (a).

(c) Analyse the subframe consisting of Beam 3/A-D, Level 1 with the columns above and below

them, subjected to vertical load only.

(d) Sketch the bending moment diagrams for each loading (b) & (c) and the combined loadings.

194 Frame Analysis

EXAMPLES

Example 9.1(a) : Actions on beam

Example 9.1(b) : Analysis of one level sub-frame

Example 9.1(c) : Analysis of two-point sub-frame

Example 9.1(d) : Continuous beam + one-point sub-frame

Example 9.2(a) : Calculation of wind load

Example 9.2(b) : Lateral load analysis : Cantilever method

Example 9.2(c) : Analysis of one level sub-frame - Vertical load only

Frame Analysis 195

Example 9.1(a) : Action on beams page 1/2

Ref. Calculations Output

SPECIFICATION

Loading:

Finishes, ceiling, services etc. = 0.75 kN/m2

Density of concrete = 25 kN/m3

Imposed load = 4.00 kN/m2

Partition = 0.50 kN/m2

Dimension:

Slab thickness, h = 150 mm

Beam size, b x h = 250 x 600 mm

Column size, b x h = 300 x 400 mm

Loading distribution:

4.0m

0.47 0.45 0.47

0.42 0.39 0.42

5.0m

6.0m 6.0m 6.0m

Slab 2-3/A-B : L y/L x = 6.0 / 4.0 = 1.50 Case 2

Slab 3-4/A-B : L y/L x = 6.0 / 5.0 = 1.20 Case 2

Slab 2-3/B-C : L y/L x = 6.0 / 4.0 = 1.50 Case 1

Slab 3-4/B-C : L y/L x = 6.0 / 5.0 = 1.20 Case 1

Slab 2-3/C-D : L y/L x = 6.0 / 4.0 = 1.50 Case 2

Slab 3-4/C-D : L y/L x = 6.0 / 5.0 = 1.20 Case 2

ACTIONS

Loads on slab, n kN/m2 :

Slab selfweight = 0.15 x 25 = 3.75 kN/m2

Finishes, ceiling etc. = 0.75 kN/m2

Characteristic permanent load, g k = 4.50 kN/m2

Imposed load = 4.00 kN/m2

Partition = 0.50 kN/m2

Characteristic variable load, q k = 4.50 kN/m2

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2

3

4

A B C D

196 Frame Analysis

Example 9.1(a) : Action on beams page 2/2


Loads on beam, w kN/m :

6.0 m 6.0 m 6.0 m

Span 1 (A-B) : w 1

Perm. load from slab = 0.47 x 4.50 x 4.0 = 8.46 kN/m


Beam selfweight = 0.45 x 0.25 x 25 = 2.81 kN/m

Characteristic permanent load, G k = 20.72 kN/m

Variable load fr.slab = 0.47 x 4.50 x 4.0 = 8.46 kN/m

Variable load fr. slab = 0.42 x 4.50 x 5.0 = 9.45 kN/m

Chacracteristic variable load, Q k = 17.91 kN/m

Design load, 1.35G k + 1.5Q k = 54.84 kN/m

1.35G k = 27.98 kN/m

Span 2 (B-C) : w 2




Characteristic permanent load, g k = 19.69 kN/m





1.35G k = 26.58 kN/m

Span 3 (C-D) : w 3




Characteristic permanent load, g k = 20.72 kN/m

Variable load from slab = 0.47 x 4.50 x 4.0 = 8.46 kN/m

Variable load from slab = 0.42 x 4.50 x 5.0 = 9.45 kN/m



1.35G k = 27.98 kN/m

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A BC D

w1 w2 w3

Frame Analysis 197

Example 9.1(b) : Analysis of one level sub-frame page 1/10


Data:

Actions:

w 1 : kN/m

w 1 kN/m w 2 kN/m w 3 kN/m 3.5 m 1.35Gk = 28.0

1.35Gk+1.5Qk= 54.8

w 2 :

4.0 m 1.35Gk = 26.6

6.0 m 6.0 m 6.0 m 1.35Gk+1.5Qk= 51.9

w 3 :

Size : 1.35Gk = 28.0

Beam : b x h = 250 x 600 mm 1.35Gk+1.5Qk= 54.8

Column : b x h = 300 x 400 mm

Momen of Inertia : I = bh3/12

Beam : I = 250 x 600 3/12 = 4.5 x 10

9 mm

4

Column: I = 300 x 400 3/12 = 1.6 x 10

9 mm

4

Stiffness : K = I /L

Column: K cu= 1.6 x 109 / 3500 = 4.6 x 10

5 mm

3

K cl = 1.6 x 109 / 4000 = 4.0 x 10

5 mm

3

Beam: K AB = 4.5 x 109 / 6000 = 7.5 x 10

5 mm

3

K BC = 4.5 x 109 / 6000 = 7.5 x 10

5 mm

3

K CD = 4.5 x 109 / 6000 = 7.5 x 10

5 mm

3

Distribution factor : F = K / K

Joint A: F AB = K AB/(K AB + K cu + K cl) = 0.47

F cu = K cu/(K AB + K cu + K cl) = 0.28

F cl = K cl/(K AB + K cu + K cl) = 0.25

Joint B: F BA = K AB/(K AB+K BC+K cu+K cl) = 0.32

F BC = K BC/(K AB+K BC+K cu+K cl) = 0.32

F cu = Kcu/(K AB+K BC+K cu+K cl) = 0.19

F cl = K cl/(K AB+K BC+K cu+K cl) = 0.17

Joint C: F CB = K BC/(K BC+K CD+K cu+K cl) = 0.32

F CD = K CD/(K BC+K CD+K cu+K cl) = 0.32

F cu = K cu/(K BC+K CD+K cu+K cl) = 0.19

F cl = K cl/(K BC+K CD+K cu+K cl) = 0.17

Joint D: F DC = K CD/(K CD+K cu+K cl) = 0.47

F cu = K cu/(K CD+K cu+K cl) = 0.28

F cl = K cl/(K CD+K cu+K cl) = 0.25

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A B C D

198 Frame Analysis



Case 1:: Span 1 & 2 design permanent & variable loads 1.35G k + 1.5Q k

Span 3 design permanent loads 1.35G k

Fixed end moment :

-M AB = M BA = w 1L 12/12

54.8 kN/m 51.9 kN/m 28.0 kN/m 3.5m = 54.8 x 6.02/12

= 164.5 kNm

-M BC = M CB = w 2L 22/12

4.0m = 51.9 x 6.02/12

6.0 m 6.0 m 6.0 m = 155.7 kNm

-M CD = M DC = w1L 12/12

= 28.0 x 6.02/12

= 83.9 kNm

Moment distribution

48.58 -7.90 -9.26 -21.65

0.17 -0.24 0.26 -0.07

1.22 -0.71 0.32 -0.95

0.40 -5.23 4.07 3.25

46.80 -1.72 -13.91 -23.87

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

-164.5 164.5 -155.7 155.7 -83.9 83.9

40.9 76.8 -2.8 -1.5 -2.8 -22.8 -12.2 -22.8 -39.2 -20.9

-1.4 38.4 -11.4 -1.4 -19.6 -11.4

0.4 0.7 -8.6 -4.6 -8.6 6.7 3.6 6.7 5.3 2.8

-4.3 0.3 3.3 -4.3 2.7 3.3

1.1 2.0 -1.2 -0.6 -1.2 0.5 0.3 0.5 -1.6 -0.8

-0.6 1.0 0.3 -0.6 -0.8 0.3

0.1 0.3 -0.4 -0.2 -0.4 0.4 0.2 0.4 -0.1 -0.1

42.5 -91.1 191.3 -6.9 -176.5 134.2 -8.1 -116.8 40.6 -18.9

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A B C D

Frame Analysis 199



Shear force :

91.1 54.8 191.3 176.5 51.9 134.2 116.8 28.0 40.6

6.0 6.0 6.0

V AB V BA V BC V CB V CD V DC

M @ B = 0

6.0 V AB - (54.8 x 6.0 x 3.0) + 191.3 - 91.1 = 0

V AB = ( 987.13 - 191.3 +91.1 ) / 6.0 = 147.8 kN

V BA = ( 54.8 x 6.0) - 147.8 = 181.2 kN

M @ C = 0

6.0 V BC - (51.9 x 6.0 x 3.0) + 134.2 - 176.5 = 0

V BC = ( 934.03 - 134.2 +176.5 ) / 6.0 = 162.7 kN

V CB = ( 51.9 x 6.0) - 162.7 = 148.6 kN

M @ D = 0

6.0 V CD - (28.0 x 6.0 x 3.0) - 116.8 + 40.6 = 0

V CD = ( 503.56 +116.8 - 40.6 ) / 6.0 = 96.6 kN

V DC = ( 28.0 x 6.0) - 96.6 = 71.2 kN

Shear force and bending moment diagrams

147.8 162.7 96.6

2.7 3.1 3.5

181.2 148.6 71.2

191.3 134.2

91.1 176.5 116.8

42.5 48.6 40.6

7.9 6.9 9.3 8.1 21.6 18.9

50.1

78.7

108.1

21.3 3.5 4.1 9.5

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200 Frame Analysis





Fixed end moment :

-M AB = M BA = w 1L 12/12

28.0 kN/m 51.9 kN/m 54.8 kN/m 3.5m = 28.0 x 6.02/12

= 83.9 kNm

-M BC = M CB = w 2L 22/12

4.0m = 51.9 x 6.02/12

6.0 m 6.0 m 6.0 m = 155.7 kNm

-M CD = M DC = w1L 12/12

= 54.8 x 6.02/12

= 164.5 kNm

Moment distribution

21.65 9.26 7.90 -48.58

0.07 -0.26 0.24 -0.17

0.95 -0.32 0.71 -1.22

-3.25 -4.07 5.23 -0.40

23.87 13.91 1.72 -46.80

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

-83.9 83.9 -155.7 155.7 -164.5 164.5

20.9 39.2 22.8 12.2 22.8 2.8 1.5 2.8 -76.8 -40.9

11.4 19.6 1.4 11.4 -38.4 1.4

-2.8 -5.3 -6.7 -3.6 -6.7 8.6 4.6 8.6 -0.7 -0.4

-3.3 -2.7 4.3 -3.3 -0.3 4.3

0.8 1.6 -0.5 -0.3 -0.5 1.2 0.6 1.2 -2.0 -1.1

-0.3 0.8 0.6 -0.3 -1.0 0.6

0.1 0.1 -0.4 -0.2 -0.4 0.4 0.2 0.4 -0.3 -0.1

18.9 -40.6 116.8 8.1 -134.2 176.5 6.9 -191.3 91.1 -42.5

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A B C D

Frame Analysis 201



Shear force :

40.6 28.0 116.8 134.2 51.9 176.5 191.3 54.8 91.1

6.0 6.0 6.0


M @ B = 0

6.0 V AB - (28.0 x 6.0 x 3.0) + 116.8 - 40.6 = 0

V AB = ( 503.56 - 116.8 +40.6 ) / 6.0 = 71.2 kN

V BA = ( 28.0 x 6.0) - 71.2 = 96.6 kN

M @ C = 0

6.0 V BC - (51.9 x 6.0 x 3.0) + 176.5 - 134.2 = 0

V BC = ( 934.03 - 176.5 +134.2 ) / 6.0 = 148.6 kN

V CB = ( 51.9 x 6.0) - 148.6 = 162.7 kN

M @ D = 0

6.0 V CD - (54.8 x 6.0 x 3.0) - 191.3 +91.1 = 0

V CD = ( 987.13 +191.3 - 91.1 ) / 6.0 = 181.2 kN

V DC = ( 54.8 x 6.0) - 181.2 = 147.8 kN


71.2 148.6 181.2

2.5 2.9 3.3

96.6 162.7 147.8

176.5 191.3

40.6 116.8 134.2 91.1

18.9 21.6 48.6

8.1 9.3 6.9 7.9 42.5

50.1

78.7

108.1

9.5 4.6 4.0 21.3

msy '09

202 Frame Analysis





Fixed end moment :

-M AB = M BA = w 1L 12/12

54.8 kN/m 26.6 kN/m 54.8 kN/m 3.5m = 54.8 x 6.02/12

= 164.5 kNm

-M BC = M CB = w 2L 22/12

4.0m = 26.6 x 6.02/12

6.0 m 6.0 m 6.0 m = 79.7 kNm

-M CD = M DC = w1L 12/12

= 54.8 x 6.02/12

= 164.5 kNm

Moment distribution

53.50 -29.44 29.44 -53.50

0.52 -0.73 0.73 -0.52

2.35 -2.21 2.21 -2.35

3.84 -10.06 10.06 -3.84

46.80 -16.44 16.44 -46.80

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

-164.5 164.5 -79.7 79.7 -164.5 164.5

40.9 76.8 -27.0 -14.4 -27.0 27.0 14.4 27.0 -76.8 -40.9

-13.5 38.4 13.5 -13.5 -38.4 13.5

3.4 6.3 -16.5 -8.8 -16.5 16.5 8.8 16.5 -6.3 -3.4

-8.3 3.1 8.3 -8.3 -3.1 8.3

2.1 3.9 -3.6 -1.9 -3.6 3.6 1.9 3.6 -3.9 -2.1

-1.8 1.9 1.8 -1.8 -1.9 1.8

0.5 0.8 -1.2 -0.6 -1.2 1.2 0.6 1.2 -0.8 -0.5

46.8 -100.3 159.7 -25.8 -104.5 104.5 25.8 -159.7 100.3 -46.8

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A B C D

Frame Analysis 203



Shear force :

100.3 54.8 159.7 104.5 26.6 104.5 159.7 54.8 100.3

6.0 6.0 6.0


M @ B = 0

6.0 V AB - (54.8 x 6.0 x 3.0) + 159.7 - 100.3 = 0

V AB = ( 987.13 - 159.7 +100.3 ) / 6.0 = 154.6 kN

V BA = ( 54.8 x 6.0) - 154.6 = 174.4 kN

M @ C = 0

6.0 V BC - (26.6 x 6.0 x 3.0) + 104.5 - 104.5 = 0

V BC = ( 478.41 - 104.5 +104.5 ) / 6.0 = 79.7 kN

V CB = ( 26.6 x 6.0) - 79.7 = 79.7 kN

M @ D = 0

6.0 V CD - (54.8 x 6.0 x 3.0) - 159.7 +100.3 = 0

V CD = ( 987.13 +159.7 - 100.3 ) / 6.0 = 174.4 kN

V DC = ( 54.8 x 6.0) - 174.4 = 154.6 kN


154.6 79.7 174.4

2.8 3.0 3.2

174.4 79.7 154.6

159.7

100.3 159.7 104.5 104.5 100.3

46.8 53.5 53.5

29.4 25.8 29.4 25.8 46.8

15.1

117.7 117.7

23.4 12.9 12.9 23.4

msy '09

204 Frame Analysis



Case 4:: Span 2 design permanent & variable loads 1.35G k + 1.5Q k

Span 1 & 3 design permanent loads 1.35G k

Fixed end moment :

-M AB = M BA = w 1L 12/12

28.0 kN/m 51.9 kN/m 28.0 kN/m 3.5m = 28.0 x 6.02/12

= 83.9 kNm

-M BC = M CB = w 2L 22/12

4.0m = 51.9 x 6.02/12

6.0 m 6.0 m 6.0 m = 155.7 kNm

-M CD = M DC = w1L 12/12

= 28.0 x 6.02/12

= 83.9 kNm

Moment distribution

20.93 12.58 -12.58 -20.93

-0.06 -0.02 0.02 0.06

0.37 0.26 -0.26 -0.37

-3.25 -1.58 1.58 3.25

23.87 13.91 -13.91 -23.87

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

-83.9 83.9 -155.7 155.7 -83.9 83.9

20.9 39.2 22.8 12.2 22.8 -22.8 -12.2 -22.8 -39.2 -20.9

11.4 19.6 -11.4 11.4 -19.6 -11.4

-2.8 -5.3 -2.6 -1.4 -2.6 2.6 1.4 2.6 5.3 2.8

-1.3 -2.7 1.3 -1.3 2.7 1.3

0.3 0.6 0.4 0.2 0.4 -0.4 -0.2 -0.4 -0.6 -0.3

0.2 0.3 -0.2 0.2 -0.3 -0.2

-0.1 -0.1 0.0 0.0 0.0 0.0 0.0 0.0 0.1 0.1

18.3 -39.3 121.8 11.0 -145.4 145.4 -11.0 -121.8 39.3 -18.3

msy '11

A B C D

Frame Analysis 205



Shear force :

39.3 28.0 121.8 145.4 51.9 145.4 121.8 28.0 39.3

6.0 6.0 6.0


M @ B = 0

6.0 V AB - (28.0 x 6.0 x 3.0) + 121.8 - 39.3 = 0

V AB = ( 503.56 - 121.8 +39.3 ) / 6.0 = 70.2 kN

V BA = ( 28.0 x 6.0) - 70.2 = 97.7 kN

M @ C = 0

6.0 V BC - (51.9 x 6.0 x 3.0) + 145.4 - 145.4 = 0

V BC = ( 934.03 - 145.4 +145.4 ) / 6.0 = 155.7 kN

V CB = ( 51.9 x 6.0) - 155.7 = 155.7 kN

M @ D = 0

6.0 V CD - (28.0 x 6.0 x 3.0) - 121.8 +39.3 = 0

V CD = ( 503.56 +121.8 - 39.3 ) / 6.0 = 97.7 kN

V DC = ( 28.0 x 6.0) - 97.7 = 70.2 kN


70.2 155.7 97.7

2.5 3.0 3.5

97.7 155.7 70.2

145.4 145.4

39.3 121.8 121.8 39.3

18.3 20.9 20.9

12.6 11.0 12.6 11.0 18.3

48.8 48.8

88.1

9.2 5.5 5.5 9.2

msy '09

206 Frame Analysis



Shear Force Envelop

154.6 162.7 181.2

181.2 162.7 154.6

Bending Moment Envelop

191.3 176.5 176.5 191.3

100.3 100.3

46.8 53.5

29.4 25.8 29.4 25.8 46.8

48.8 15.1 48.8

88.1

117.7 117.7

23.4 12.9 5.5 5.5 12.9 23.4

msy '09

Frame Analysis 207

Example 9.1(c) : Analysis of two free-joint sub-frame page 1/6


Beam A-B (Beam C-D similar)


: Span 2 design permanent & variable loads 1.35G k + 1.5Q k

Fixed end moment :

-M AB = M BA = w 1L 12/12

4.6 54.84 kN/m 51.89 kN/m = 54.8 x 6.02/12

= 164.5 kNm

7.5 3.75 -M BC = M CB = w 2L 22/12

4.0 = 51.9 x 6.02/12

6.0 m 6.0 m = 155.7 kNm

Distribution factor:

Joint A: F AB = 7.5 / (7.5 + 4.6 + 4.0) = 0.47

F cu = 4.6 / (7.5 + 4.6 + 4.0) = 0.28

F cl = 4.0 / (7.5 + 4.6 + 4.0) = 0.25

Joint B: F BA = 7.5 / (7.5 + 3.75 + 4.0 + 4.6) = 0.38

F BC = 3.8 / (7.5 + 3.75 + 4.0 + 4.6) = 0.19

F cu = 4.0 / (7.5 + 3.8 + 4.0 + 4.6) = 0.20

F cl = 4.6 / (7.5 + 3.8 + 4.0 + 4.6) = 0.23

Moment distribution

49.36 -9.95

0.02 -0.34

2.07 -0.08

0.48 -7.75

46.80 -1.79

0.28 A B 0.20

0.25 0.47 0.38 0.23 0.19

-164.5 164.5 -155.7

40.9 76.8 -3.3 -2.0 -1.7

-1.7 38.4 0.0

0.4 0.8 -14.5 -8.9 -7.3

-7.3 0.4 0.0

1.8 3.4 -0.1 -0.1 -0.1

-0.1 1.7 0.0

0.0 0.0 -0.6 -0.4 -0.3

43.2 -92.6 186.3 -11.4 -165.0

msy '11

A B

208 Frame Analysis




: Span 2 design permanent loads 1.35G k

Fixed end moment :

-M AB = M BA = w 1L 12/12

4.6 54.84 kN/m 26.58 kN/m = 54.8 x 6.02/12

= 164.5 kNm

7.5 3.75 -M BC = M CB = w 2L 22/12

4.0 = 26.6 x 6.02/12

6.0 m 6.0 m = 79.7 kNm


Joint A: F AB = 7.5 / (7.5 + 4.6 + 4.0) = 0.47

F cu = 4.6 / (7.5 + 4.6 + 4.0) = 0.28

F cl = 4.0 / (7.5 + 4.6 + 4.0) = 0.25

Joint B: F BA = 7.5 / (7.5 + 3.75 + 4.0 + 4.6) = 0.38

F BC = 3.8 / (7.5 + 3.75 + 4.0 + 4.6) = 0.19

F cu = 4.0 / (7.5 + 3.8 + 4.0 + 4.6) = 0.20

F cl = 4.6 / (7.5 + 3.8 + 4.0 + 4.6) = 0.23

Moment distribution

53.63 -25.95

0.20 -0.34

2.07 -0.76

4.56 -7.75

46.80 -17.11

0.28 A B 0.20

0.25 0.47 0.38 0.23 0.19

-164.5 164.5 -79.7

40.9 76.8 -32.1 -19.6 -16.0

-16.0 38.4 0.0

4.0 7.5 -14.5 -8.9 -7.3

-7.3 3.7 0.0

1.8 3.4 -1.4 -0.9 -0.7

-0.7 1.7 0.0

0.2 0.3 -0.6 -0.4 -0.3

46.9 -100.6 159.7 -29.7 -104.1

msy '11

A B

Frame Analysis 209



Shear force :

Case 1

92.6 54.8 186.3

6.0

V AB V BA

M @ B = 0

6.0 V AB - (54.8 x 6.0 x 3.0) + 186.3 - 92.6 = 0

V AB = ( 987.13 - 186.3 +92.6 ) / 6.0 = 148.9 kN

V BA = ( 54.8 x 6.0) - 148.9 = 180.2 kN

Case 2

100.6 54.8 159.7

6.0

V AB V BA

M @ B = 0

6.0 V AB - (54.8 x 6.0 x 3.0) + 159.7 - 100.6 = 0

V AB = ( 987.13 - 159.7 +100.6 ) / 6.0 = 154.7 kN

V BA = ( 54.8 x 6.0) - 154.7 = 174.4 kN

Shear force and bending moment diagram

154.7

148.9

2.7 174.4

2.8 180.2

100.6 186.3

92.6 159.7

46.9 53.6

26.0 29.7

109.6

117.6

23.5 14.8

msy '11

210 Frame Analysis



Beam B-C

Case 1:: Span 1 and 2 design permanent & variable loads 1.35G k + 1.5Q k

: Span 3 design permanent load 1.35G k

Fixed end moment :

4.6 -M AB = M BA = w 1L 12/12

54.84 kN/m 51.89 kN/m 27.98 kN/m = 54.8 x 6.02/12

= 164.5 kNm

3.75 7.50 3.75 -M BC = M CB = w 2L 22/12

4.0 = 51.9 x 6.02/12

= 155.7 kNm

6.0 m 6.0 m 6.0 m -M CD = M DC = w 3L 32/12

= 28.0 x 6.02/12

= 83.9 kNm


Joint B: F BA = 3.75 / (3.75 + 4.57 + 4.0 + 7.5) = 0.19

F BC = 7.50 / (3.75 + 4.57 + 4.0 + 7.5) = 0.38

F cu = 4.57 / (3.75 + 4.57 + 4.0 + 7.5) = 0.23

F cl = 4.00 / (3.75 + 4.57 + 4.0 + 7.5) = 0.20

Joint C: F CB = 7.50 / (3.75 + 7.50 + 4.0 + 4.6) = 0.38

F CD = 3.75 / (3.75 + 7.50 + 4.0 + 4.6) = 0.19

F cu = 4.57 / (3.75 + 7.50 + 4.0 + 4.6) = 0.23

F cl = 4.00 / (3.75 + 7.50 + 4.0 + 4.6) = 0.20

Moment distribution

1.13 -16.74

0.11 0.01

-0.07 -0.59

3.13 0.39

-2.04 -16.55

B 0.23 C 0.23

0.19 0.20 0.38 0.38 0.20 0.19

164.5 -155.7 155.7 -83.9

-1.7 -1.8 -3.3 -27.1 -14.5 -13.6

0.0 -13.6 -1.7 0.0

2.6 2.7 5.1 0.6 0.3 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

165.4 1.0 -167.6 129.0 -14.6 -97.7

msy '11

A

B C

D

Frame Analysis 211




: Span 1 and 3 design permanent load 1.35G k

Fixed end moment :

4.6 -M AB = M BA = w 1L 12/12

27.98 kN/m 51.89 kN/m 27.98 kN/m = 28.0 x 6.02/12

= 83.9 kNm

3.75 7.50 3.75 -M BC = M CB = w 2L 22/12

4.0 = 51.9 x 6.02/12

= 155.7 kNm

6.0 m 6.0 m 6.0 m -M CD = M DC = w 3L 32/12

= 28.0 x 6.02/12

= 83.9 kNm


Joint B: F BA = 3.75 / (3.75 + 4.57 + 4.0 + 7.5) = 0.19

F BC = 7.50 / (3.75 + 4.57 + 4.0 + 7.5) = 0.38

F cu = 4.57 / (3.75 + 4.57 + 4.0 + 7.5) = 0.23

F cl = 4.00 / (3.75 + 4.57 + 4.0 + 7.5) = 0.20

Joint C: F CB = 7.50 / (3.75 + 7.50 + 4.0 + 4.6) = 0.38

F CD = 3.75 / (3.75 + 7.50 + 4.0 + 4.6) = 0.19

F cu = 4.57 / (3.75 + 7.50 + 4.0 + 4.6) = 0.23

F cl = 4.00 / (3.75 + 7.50 + 4.0 + 4.6) = 0.20

Moment distribution

20.38 -20.38

0.11 -0.11

0.59 -0.59

3.13 -3.13

16.55 -16.55

B 0.23 C 0.23

0.19 0.20 0.38 0.38 0.20 0.19

83.9 -155.7 155.7 -83.9

13.6 14.5 27.1 -27.1 -14.5 -13.6

0.0 -13.6 13.6 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

100.6 17.8 -138.9 138.9 -17.8 -100.6

msy '11

A

B C

D

212 Frame Analysis



Shear force :

Case 1

167.6 51.9 129.0

6.0

V BC V CB

M @ C = 0

6.0 V BC - (51.9 x 6.0 x 3.0) + 129.0 - 167.6 = 0

V BC = ( 934.03 - 129.0 +167.6 ) / 6.0 = 162.1 kN

V CB = ( 51.9 x 6.0) - 162.1 = 149.3 kN

Case 2

138.9 51.9 138.9

6.0

V BC V CB

M @ C = 0

6.0 V BC - (51.9 x 6.0 x 3.0) + 138.9 - 138.9 = 0

V AB = ( 934.03 - 138.9 +138.9 ) / 6.0 = 155.7 kN

V BA = ( 51.9 x 6.0) - 155.7 = 155.7 kN

Shear force and bending moment diagram

162.1

155.7

3.1 149.3

3.0 155.7

167.6 138.9

138.9 129.0

20.4 20.4

17.8 17.8

85.6

94.6

8.9 8.9

msy '11

Frame Analysis 213

Example 9.1(d) : Continuous beam + one free-joint sub-frame page 1/9


Continuous Beam :

Bending moment and shear force in beam.

6.0 m 6.0 m 6.0 m

Moment of Inertia :

I = bh3/12 = 250 x 600

3/12 = 4.50 x 10

9mm

4

Stiffness :

AB : k AB = 0.75 I /L = 3.38 x 109 / 6000 = 5.63 x 10

5 mm

3

BC : k BC = I /L = 4.50 x 109 / 6000 = 7.5 x 10

5 mm

3

CD: k CD = 0.75 I /L = 3.38 x 109 / 6000 = 5.63 x 10

5 mm

3


Joint A: F AB = k BA/(k AB+ 0 )

= 5.63 / (5.63 + 0.00) = 1.00

Joint B: F BA = k AB/(k AB+ k BC)

= 5.63 / (5.63 + 7.50) = 0.43

F BC = k BC/(k AB+ k BC)

= 7.50 / (5.63 + 7.50) = 0.57

Joint C: F BC = k BC/(k BC+ k CD)

= 7.50 / (7.5 + 5.63) = 0.57

F CD = k CD/(k BC+ k CD)

= 5.63 / (7.5 + 5.63) = 0.43

Joint D: F DC = k CD/(k CD+ 0 )

= 5.63 / (5.63 + 0.00) = 1.00

msy '11

A BC D

214 Frame Analysis





54.8 kN/m 51.9 kN/m 28.0 kN/m

6.0 m 6.0 m 6.0 m

Fixed End Moment :

-M AB = M BA = w 1 L 12/12 = 54.84 x 6.0

2 / 12 = 164.5 kNm

-M BC = M CB = w 2 L 22/12 = 51.89 x 6.0

2 / 12 = 155.7 kNm

-M CD = M DC = w 3 L 32/12 = 27.98 x 6.0

2 / 12 = 83.9 kNm

Moment Distribution :

0.00 1.00 0.43 0.57 0.57 0.43 1.00 0.00

-164.5 164.5 -155.7 155.7 -83.9 83.9

164.5 -3.8 -5.1 -41.0 -30.7 -83.9

82.3 -20.5 -2.5 -42.0

-26.5 -35.3 25.4 19.1

12.7 -17.6

-5.4 -7.3 10.1 7.6

5.0 -3.6

-2.2 -2.9 2.1 1.6

1.0 -1.4

-0.4 -0.6 0.8 0.6

0.0 208.5 -208.5 127.8 -127.8 0.0

Shear force :

54.8 208.5 208.5 51.9 127.8 127.8 28.0

6.0 6.0 6.0

V AB V BA V BC V CB V CD V DC M @ B = 0

6.0 V AB - (54.8 x 6.0 x 3.0) + 208.5 = 0

V AB = ( 987.13 - 208.5 ) / 6.0 = 129.8 kN

V BA = ( 54.8 x 6.0) - 129.8 = 199.3 kN

M @ C = 0

6.0 V BC - (51.9 x 6.0 x 3.0) + 127.8 - 208.5 = 0

V BC = ( 934.03 - 127.8 +208.5 )/6.0 = 169.1 kN

V CB = ( 51.9 x 6.0) - 169.1 = 142.2 kN

M @ D = 0

6.0 V CD - (28.0 x 6.0 x 3.0) - 127.8 = 0

V CD = ( 503.56 +127.8 ) / 6.0 = 105.2 kN

V DC = ( 28.0 x 6.0) - 105.2 = 62.6 kN

msy '11

A B C D

Frame Analysis 215





28.0 kN/m 51.9 kN/m 54.8 kN/m

6.0 m 6.0 m 6.0 m

Fixed End Moment :

-M AB = M BA = w 1 L 12/12 = 27.98 x 6.0

2 / 12 = 83.9 kNm

-M BC = M CB = w 2 L 22/12 = 51.89 x 6.0

2 / 12 = 155.7 kNm

-M CD = M DC = w 3 L 32/12 = 54.84 x 6.0

2 / 12 = 164.5 kNm


0.00 1.00 0.43 0.57 0.57 0.43 1.00 0.00

-83.9 83.9 -155.7 155.7 -164.5 164.5

83.9 30.7 41.0 5.1 3.8 -164.5

42.0 2.5 20.5 -82.3

-19.1 -25.4 35.3 26.5

17.6 -12.7

-7.6 -10.1 7.3 5.4

3.6 -5.0

-1.6 -2.1 2.9 2.2

1.4 -1.0

-0.6 -0.8 0.6 0.4

0.0 127.8 -127.8 208.5 -208.5 0.0

Shear force :

28.0 127.8 127.8 51.9 208.5 208.5 54.8

6.0 6.0 6.0


6.0 V AB - (28.0 x 6.0 x 3.0) + 127.8 = 0

V AB = ( 503.56 - 127.8 ) / 6.0 = 62.6 kN

V BA = ( 28.0 x 6.0) - 62.6 = 105.2 kN

M @ C = 0

6.0 V BC - (51.9 x 6.0 x 3.0) + 208.5 - 127.8 = 0

V BC = ( 934.03 - 208.5 +127.8 )/6.0 = 142.2 kN

V CB = ( 51.9 x 6.0) - 142.2 = 169.1 kN

M @ D = 0

6.0 V CD - (54.8 x 6.0 x 3.0) - 208.5 = 0

V CD = ( 987.13 +208.5 ) / 6.0 = 199.3 kN

V DC = ( 54.8 x 6.0) - 199.3 = 129.8 kN

msy '11

A B C D

216 Frame Analysis





54.8 kN/m 26.6 kN/m 54.8 kN/m

6.0 m 6.0 m 6.0 m

Fixed End Moment :

-M AB = M BA = w 1 L 12/12 = 54.84 x 6.0

2 / 12 = 164.5 kNm

-M BC = M CB = w 2 L 22/12 = 26.58 x 6.0

2 / 12 = 79.7 kNm

-M CD = M DC = w 3 L 32/12 = 54.84 x 6.0

2 / 12 = 164.5 kNm


0.00 1.00 0.43 0.57 0.57 0.43 1.00 0.00

-164.5 164.5 -79.7 79.7 -164.5 164.5

164.5 -36.3 -48.4 48.4 36.3 -164.5

82.3 24.2 -24.2 -82.3

-45.6 -60.8 60.8 45.6

30.4 -30.4

-13.0 -17.4 17.4 13.0

8.7 -8.7

-3.7 -5.0 5.0 3.7

2.5 -2.5

-1.1 -1.4 1.4 1.1

0.0 147.0 -147.0 147.0 -147.0 0.0

Shear force :

54.8 147.0 147.0 26.6 147.0 147.0 54.8

6.0 6.0 6.0


6.0 V AB - (54.8 x 6.0 x 3.0) + 147.0 = 0

V AB = ( 987.13 - 147.0 ) / 6.0 = 140.0 kN

V BA = ( 54.8 x 6.0) - 140.0 = 189.0 kN

M @ C = 0

6.0 V BC - (26.6 x 6.0 x 3.0) + 147.0 - 147.0 = 0

V BC = ( 478.41 - 147.0 +147.0 )/6.0 = 79.7 kN

V CB = ( 26.6 x 6.0) - 79.7 = 79.7 kN

M @ D = 0

6.0 V CD - (54.8 x 6.0 x 3.0) - 147.0 = 0

V CD = ( 987.13 +147.0 ) / 6.0 = 189.0 kN

V DC = ( 54.8 x 6.0) - 189.0 = 140.0 kN

msy '11

A B C D

Frame Analysis 217




Span 1 & 3 design permanent loads 1.35G k

28.0 kN/m 51.9 kN/m 28.0 kN/m

6.0 m 6.0 m 6.0 m

Fixed End Moment :

-M AB = M BA = w 1 L 12/12 = 27.98 x 6.0

2 / 12 = 83.9 kNm

-M BC = M CB = w 2 L 22/12 = 51.89 x 6.0

2 / 12 = 155.7 kNm

-M CD = M DC = w 3 L 32/12 = 27.98 x 6.0

2 / 12 = 83.9 kNm


0.00 1.00 0.43 0.57 0.57 0.43 1.00 0.00

-83.9 83.9 -155.7 155.7 -83.9 83.9

83.9 30.7 41.0 -41.0 -30.7 -83.9

42.0 -20.5 20.5 -42.0

-9.2 -12.3 12.3 9.2

6.1 -6.1

-2.6 -3.5 3.5 2.6

1.8 -1.8

-0.8 -1.0 1.0 0.8

0.5 -0.5

-0.2 -0.3 0.3 0.2

0.0 143.8 -143.8 143.8 -143.8 0.0

Shear force :

28.0 143.8 143.8 51.9 143.8 143.8 28.0

6.0 6.0 6.0


6.0 V AB - (28.0 x 6.0 x 3.0) + 143.8 = 0

V AB = ( 503.56 - 143.8 ) / 6.0 = 60.0 kN

V BA = ( 28.0 x 6.0) - 60.0 = 107.9 kN

M @ C = 0

6.0 V BC - (51.9 x 6.0 x 3.0) + 143.8 - 143.8 = 0

V BC = ( 934.03 - 143.8 +143.8 )/6.0 = 155.7 kN

V CB = ( 51.9 x 6.0) - 155.7 = 155.7 kN

M @ D = 0

6.0 V CD - (28.0 x 6.0 x 3.0) - 143.8 = 0

V CD = ( 503.56 +143.8 ) / 6.0 = 107.9 kN

V DC = ( 28.0 x 6.0) - 107.9 = 60.0 kN

msy '11

A B C D

218 Frame Analysis



Shear Force Diagram

140.0 169.1 199.3

129.8 155.7 189.0

62.6 142.2 107.9

60.0 79.7 105.2

105.2 79.7 60.0

107.9 142.2 62.6

189.0 155.7 129.8

x 4 199.3 169.1 140.0

Distance Mid-span moment

Span 1 Span 1

x 1 = 129.8 / 54.8 = 2.4 m M 1 = 129.8x 2.4 / 2 = 153.6 kNm

x 2 = 62.6 / 28.0 = 2.2 m M 2 = 62.6x 2.2 / 2 = 70.1 kNm

x 3 = 140.0 / 54.8 = 2.6 m M 3 = 140.0x 2.6 / 2 = 178.8 kNm

x 4 = 60.0 / 28.0 = 2.1 m M 4 = 60.0x 2.1 / 2 = 64.2 kNm

Span 2 Span 2

x 1 = 169.1 / 51.9 = 3.3 m M 1 = (169.1 x3.3/2) -208.5 = 67.1 kNm

x 2 = 142.2 / 51.9 = 2.7 m M 2 = (142.2 x2.7/2) -127.8 = 67.1 kNm

x 3 = 79.7 / 26.6 = 3.0 m M 3 = (79.7 x3.0/2) -147.0 -= 27.4 kNm

x 4 = 155.7 / 51.9 = 3.0 m M 4 = (155.7 x3.0/2) -143.8 = 89.7 kNm

Span 3 Span 3

x 1 = 105.2 / 28.0 = 3.8 m M 1 = (105.2 x3.8/2) - 127.8 =70.1 kNm

x 2 = 199.3 / 54.8 = 3.6 m M 2 = (199.3 x3.6/2) - 208.5 =153.6 kNm

x 3 = 189.0 / 54.8 = 3.4 m M 3 = (189.0 x3.4/2) - 147.0 =178.8 kNm

x 4 = 107.9 / 28.0 = 3.9 m M 4 = (107.9 x3.9/2) - 143.8 =64.2 kNm

Bending Moment Diagram

Case 1

208.5 208.5 Case 2

Case 3

Case 4

127.8 127.8

-27.4

64.2 64.2

89.7

178.8 178.8

msy '11

x2

x1

x3

Frame Analysis 219



Sub-frame: Bending Moment in Columns

Momen of Inertia : I = bh3/12

Column: I = 300 x 4003 /12 = 1.6 x 10

9 mm

4

Beam : I = 250 x 6003 /12 = 4.5 x 10

9 mm

4

Stiffness : K = I /L

Columnlower : K cl = 1.6 x 109 / 3500 = 4.6 x 10

5 mm

3

Columnupper : K cu = 1.6 x 109 / 4000 = 4.0 x 10

5 mm

3

Beam: K AB = 4.5 x 109 / 6000 = 7.5 x 10

5 mm

3

K BC = 4.5 x 109 / 6000 = 7.5 x 10

5 mm

3

K CD = 4.5 x 109 / 6000 = 7.5 x 10

5 mm

3

Column A

FEM = wL2/12

54.8 kN/m = 54.8 x 6.02

/12

= 164.52 kNm

6.0m

Moment in upper column,

M = FEM x K cu /(K cu + K cl + K b/2)

= 164.5 x 4.0 / (4.0 + 4.6 + 3.8) 53.4

= 53.4 kNm 61.0

Moment in lowercColumn,

M = FEM x K cl /(K cu + K cl + K b/2)

= 164.5 x 4.6 / (4.0 + 4.6 + 3.8)

= 61.0 kNm

msy '11

Kcl

Kcu

0.5KAB

220 Frame Analysis



Column B

FEM 1 = wL2/12

54.8 kN/m 26.6 kN/m = 54.8 x 6.02 /12

= 164.5 kNm

FEM 2 = wL2/12

= 26.6 x 6.02 /12

6.0 m 6.0 m = 79.7 kNm

M = FEM 1 - FEM 2 = 164.5 - 79.7 = 84.8 kNm


M = M x K cu /(K cu + K cl + 0.5K AB + 0.5K BC)

= 84.8 x 4.0 / (4.0 + 4.6 + 3.75 + 3.75)

= 21.1 kNm 21.1

24.1

Moment in lower column,

M = M x K cl /(K cu + K cl + 0.5K AB + 0.5K BC)

= 84.8 x 4.6 / (4.0 + 4.6 + 3.75 + 3.75)

= 24.1 kNm

Column C

FEM 1 = wL2/12

26.6 kN/m 54.8 kN/m = 26.6 x 6.02 /12

= 79.7 kNm

FEM 2 = wL2/12

= 54.8 x 6.02 /12

6.0 m 6.0 m = 164.5 kNm

M = FEM 1 - FEM 2 = 164.5 - 79.7 = 84.8 kNm


M = M x K cu /(K cu + K cl + 0.5K BC + 0.5K CD)

= 84.8 x 4.0 / (4.0 + 4.6 + 3.75 + 3.75)

= 21.1 kNm 21.1

24.1

Moment in lower column,

M = M x K cl /(K cu + K cl + 0.5K BC + 0.5K CD)

= 84.8 x 4.6 / (4.0 + 4.6 + 3.75 + 3.75)

= 24.1 kNm

msy '11

Kcl

Kcu

0.5KAB 0.5KBC

Kcl

Kcu

0.5KAB 0.5KBC

Frame Analysis 221



Column D

FEM = wL2/12

54.8 kN/m = 54.8 x 6.02

/12

= 164.52 kNm

6.0 m


M = FEM x K cu /(K cu + K cl + 0.5K CD)

= 164.5 x 4.0 / (4.0 + 4.6 + 3.8) 53.4

= 53.4 kNm 61.0

Moment in lowercColumn,

M = FEM x K cl /(K cu + K cl + 0.5K CD)

= 164.5 x 4.6 / (4.0 + 4.6 + 3.8)

= 61.0 kNm

msy '11

Kcl

Kcu

0.5KCD

222 Frame Analysis

Example 9.2(a) : Calculation of wind load page 1/2


Specification

Building dimensions,

Height, h = 30 m

Width, b = 18 m

Length, d = 51 m

Building usage & location,

Permanent office

Sub-urban, Zone 1

Flat area, building around

Calculation of wind load based on MS 1553:2002

Design wind pressure

p = 0.613 [V des]2 C fig Cdyn Pa 2.8333

V des = design wind speed

= V sit l

Table 3.2 l = importance factor = 1.0

V sit = V s M d M z,cat M s M h

Fig. 3.1 V s = basic wind speed = 33.5 m/s

2.2 M d = wind directional multiplier = 1.00

Table 4.1 M z,cat= terrain/height multiplier = 1.00

4.3.1 M s = shielding multiplier = 1.00

4.4 M h = hill shape multiplier = 1.00

V sit = 33.5 x 1.00 x 1.00 x 1.00 x 1.00

= 33.50 m/s

V des = 33.50 x 1.0 = 33.50 m/s

C fig. = aerodynamic shape factor

= C p,e K aK cK lK p for external pressure

Table 5.2(a) C p,e = external pressure coefficient =

Table 5.2(b) : windward wall = 0.70

: leeward wall = -0.25

5.4.2 K a = area reduction factor = 1.00

5.4.3 K c = combination factor = 1.00

5.4.4 K l = local pressure factor = 1.00

5.4.5 K p = porous cladding reduction factor = 1.00

6.1 C fig = (0.70 - -0.25) x 1.00 x 1.00 x 1.00 x 1.00

= = 0.95

Cdyn = dynamic response factor = 1.00

msy '11

h

b

Side elevation

d

b

Plan

Wind

Wind

Frame Analysis 223

Example 9.2(a) : Calculation of wind load page 2/2


Design wind pressure

p = 0.613 x 33.5 x 0.95 x 1.00

= 653.54 N/m2

= 0.654 kN/m2

Characteristic wind load for

each sub-frame,

= 0.65 +(4.0 x 5.00) /2

= 2.94 kN/m height

msy '11

4m5m

2

4m5m

224 Frame Analysis

Example 9.2(b) : Lateral load analysis : Cantilever method page 1/10


Characteristic wind load, W k = 2.94 kN/m height

Design wind load, W d = 1.20 W k = 3.53 kN/m height

Point load for each floor level :

Roof level : 3.5 (1.00 + 1.75) = 9.70 kN

Level 2 - 6 : 4.5 (3.5) =3.5 (1.75 + 1.75) = 12.35 kN

Level 1 : 3.5 (1.75 + 2.00) = 13.23 kN

9.7 1.0 * Assumptions,

1.Point of contraflexure

K 3.5 are located at the

12.3 mid-points of all

columns and beams.

L 3.5 2.The direct axial loads

12.3 in the columns are in

propotion to their

M 3.5 distances from the

12.3 centre of gravity of

the frame.

N 3.5 3. All the column in a

12.3 storey are of equal

cross-sectional area.

P 3.5

12.3

Q 3.5

13.2

R 4.0

6 m 6 m 6 m

x

Centre of gravity of the structure,

x = (6.0N + 12N + 18N) / 4 N = 9 m

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 :

msy '11

A B C D

Frame Analysis 225



Level K and above

9.7

1.75m

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,

M @ K = 0

(9.7 x 1.75) + (1.0P x 6.0) - (1.0P x 12.0) 3.0P = 0.85 kN

-(3.0P x 18.0) = 0 1.0P = 0.28 kN

60.0P = 17.0 kN 1.0P = 0.28 kN

P = 17.0 / 60.0 = 0.28 kN 3.0P = 0.85 kN

Shear force in beams and columns,

F 1 F 2 F 3

1.75m

H 1 H 2 H 3 H 4

6 m 6 m 6 m

0.85 kN 0.28 kN 0.28 kN 0.85 kN

Consider sub-frame to the left of F 1 :

F v = 0 : F 1 - 0.85 = 0 F 1 = 0.85 kN

M @F 1 = 0: (H 1 x 1.75) - (0.85 x 3.00) = 0 H 1 = 1.46 kN


F v = 0 : F2 - 4.7P - 1.3P = 0 F 2 - 0.85 - 0.28 = 0 F 2 = 1.13 kN

M @F 2 = 0 : (H 1 + H 2) x 1.75 - (0.85 x 9.0) H 2 = 3.40 kN

- (0.28 x 3.0) = 0


F v = 0 : F2 - 4.7P - 1.3P = 0 F 3 - 0.85 - 0.28 + 0.28 = 0 F 3 = 0.85 kN

M @F 2 = 0 : (H 1 + H 2 + H3) x1.75 - (0.85 x 15.00) H 3 = 3.40 kN

- (0.28 x 9.0) + (0.28 x 3.0) = 0

F H = 0 : 14.8 - H1 - H2 - H3 - H4 = 0 9.7 - H 1 - H2 - H 3 - H 4 = 0 H 4 = 1.46 kN

msy '11

K

K

226 Frame Analysis



Level L and above

9.7

3.5m

12.3

1.75m

H 1 H 2 H 3 H 4

6 m 6 m 6 m

3.0P 1.0P 1.0P 3.0P


M @ L = 0

(9.7 x 5.25) + (12.3 x 1.75) + (1.0P x 6.0) 3.0P = 3.63 kN

- (1.0P x 12.0) -(3.0P x 18.0) = 0 1.0P = 1.21 kN

60.0P = 72.5 kN 1.0P = 1.21 kN

P = 72.5 / 60.0 = 1.21 kN 3.0P = 3.63 kN


0.85 kN 0.28 kN 0.28 kN 0.85 kN

1.46 F 1 3.40 F 2 3.40 F 3 1.46

1.75

1.75

H 1 H 2 H 3 H 4

6 m 6 m 6 m

3.63 kN 1.21 kN 1.21 kN 3.63 kN


F v = 0 : F 1 - 3.63 + 0.85 = 0 F 1 = 2.78 kN

M @F 1 = 0: (H 1 + 1.46) x 1.75 - (2.78 x 3.00) = 0 H 1 = 3.31 kN


F v = 0 : F2 - 4.7P - 1.3P = 0 F 2 - 2.78 - 0.93 = 0 F 2 = 3.70 kN

M @F 2 = 0 : (H 1 + H 2) x 1.75 + (4.85 x 1.75) H 2 = 7.72 kN

- (2.78 x 9.0) - (0.93 x 3.0) = 0


F v = 0 : F2 - 4.7P - 1.3P = 0 F 3 - 2.78 - 0.93 + 0.93 = 0 F 3 = 2.78 kN

M @F 2 = 0 : (H 1 + H 2 + H3) x1.75 + (8.25 x 1.75) H 3 = 7.72 kN

- (2.78 x 15.00) - (0.93 x 9.0) + (0.93 x 3.0) = 0

F H = 0 : 14.8 - H1 - H2 - H3 - H4 = 0 22.1 - H 1 - H2 - H 3 - H 4 = 0 H 4 = 3.31 kN

msy '11

L

Frame Analysis 227



Level M and above

9.7

12.3 3.5m

12.3 3.5m

1.75m

H 1 H 2 H 3 H 4

6 m 6 m 6 m

3.0P 1.0P 1.0P 3.0P


M @ M = 0

(9.7 x 8.75) + (12.3 x 5.25) + (12.3 x 1.75) + 3.0P = 8.57 kN

(1.0P x 6.0) - (1.0P x 12.0) -(3.0P x 18.0) = 0 1.0P = 2.86 kN

60.0P = 171.3 kN 1.0P = 2.86 kN

P = 171.3 / 60.0 = 2.86 kN 3.0P = 8.57 kN


3.63 kN 1.21 kN 1.21 kN 3.63 kN

3.31 F 1 7.72 F 2 7.72 F 3 3.31

1.75

1.75

H 1 H 2 H 3 H 4

6 m 6 m 6 m

8.57 kN 2.86 kN 2.86 kN 8.57 kN


F v = 0 : F 1 - 8.57 + 3.63 = 0 F 1 = 4.94 kN

M @F 1 = 0: (H 1 + 3.31) x 1.75 - (4.94 x 3.00) = 0 H 1 = 5.16 kN


F v = 0 : F2 - 4.7P - 1.3P = 0 F 2 - 4.94 - 1.65 = 0 F 2 = 6.59 kN

M @F 2 = 0 : (H 1 + H 2) x 1.75 + (11.03 x 1.75) H 2 = 12.04 kN

- (4.94 x 9.0) - (1.65 x 3.0) = 0


F v = 0 : F2 - 4.7P - 1.3P = 0 F 3 - 4.94 - 1.65 + 1.65 = 0 F 3 = 4.94 kN

M @F 2 = 0 : (H 1 + H 2 + H3) x1.75 + (18.74 x 1.75) H 3 = 12.04 kN

- (4.94 x 15.00) - (1.65 x 9.0) + (1.65 x 3.0) = 0

F H = 0 : 14.8 - H1 - H2 - H3 - H4 = 0 34.4 - H 1 - H2 - H 3 - H 4 = 0 H 4 = 5.16 kN

msy '11

LM

228 Frame Analysis



Level N and above

9.7

12.3 3.5m

12.3 3.5m

12.3 3.5m

1.75m

H 1 H 2 H 3 H 4

6 m 6 m 6 m

3.0P 1.0P 1.0P 3.0P


M @ M = 0

(9.7 x 12.3)+ (12.3 x 8.75) + (12.3 x 5.25) + (12.3 x 1.75) + 3.0P = 15.67 kN

(1.0P x 6.0) - (1.0P x 12.0) -(3.0P x 18.0) = 0 1.0P = 5.22 kN

60.0P = 313.3 kN 1.0P = 5.22 kN

P = 313.3 / 60.0 = 5.22 kN 3.0P = 15.67 kN


8.57 kN 2.86 kN 2.86 kN 8.57 kN

5.16 F 1 12.04 F 2 12.04 F 3 5.16

1.75

1.75

H 1 H 2 H 3 H 4

6 m 6 m 6 m

15.67 kN 5.22 kN 5.22 kN 15.67 kN


F v = 0 : F 1 - 15.67 +8.57 = 0 F 1 = 7.10 kN

M @F 1 = 0: (H 1 + 5.16) x 1.75 - (7.10 x 3.00) = 0 H 1 = 7.01 kN


F v = 0 : F2 - 4.7P - 1.3P = 0 F 2 - 7.10 - 2.37 = 0 F 2 = 9.47 kN

M @F 2 = 0 : (H 1 + H 2) x 1.75 + (17.20 x 1.75) H 2 = 16.36 kN

- (7.10 x 9.0) - (2.37 x 3.0) = 0


F v = 0 : F2 - 4.7P - 1.3P = 0 F 3 - 7.10 - 2.37 + 2.37 = 0 F 3 = 7.10 kN

M @F 2 = 0 : (H 1 + H 2 + H3) x1.75 + (29.24 x 1.75) H 3 = 16.36 kN

- (7.10 x 15.00) - (2.37 x 9.0) + (2.37 x 3.0) = 0

F H = 0 : 14.8 - H1 - H2 - H3 - H4 = 0 46.7 - H 1 - H2 - H 3 - H 4 = 0 H 4 = 7.01 kN

msy '11

LN

Frame Analysis 229



Level P and above

9.7

12.3 3.5m

12.3 3.5m

12.3 3.5m

12.3 3.5m

1.75m

H 1 H 2 H 3 H 4

6 m 6 m 6 m

3.0P 1.0P 1.0P 3.0P


M @ P = 0

(9.7 x 15.8)+ (12.3 x 12.3)+ (12.3 x 8.75) + (12.3 x 5.25) + 3.0P = 24.9 kN

(12.3 x 1.75)+ (1.0P x 6.0) - (1.0P x 12.0) -(3.0P x 18.0) 1.0P = 8.3 kN

= 0 1.0P = 8.3 kN

60.0P = 498.6 kN 3.0P = 24.9 kN

P = 498.6 / 60.0 = 8.31 kN


15.67 kN 5.22 kN 5.22 kN 15.67 kN

7.01 F 1 16.36 F 2 16.36 F 3 7.01

1.75

1.75

H 1 H 2 H 3 H 4

6 m 6 m 6 m

24.93 kN 8.31 kN 8.31 kN 24.93 kN


F v = 0 : F 1 - 24.93 +15.67 = 0 F 1 = 9.3 kN

M @F 1 = 0: (H 1 + 7.01) x 1.75 - (9.3x 3.00) = 0 H 1 = 8.9 kN


F v = 0 : F2 - 4.7P - 1.3P = 0 F 2 - 9.26 - 3.09 = 0 F 2 = 12.3 kN

M @F 2 = 0 : (H 1 + H 2) x 1.75 + (23.37 x 1.75) H 2 = 20.7 kN

- (9.3 x 9.0) - (3.09 x 3.0) = 0


F v = 0 : F2 - 4.7P - 1.3P = 0 F 3 - 9.26 - 3.09 + 3.09 = 0 F 3 = 9.3 kN

M @F 2 = 0 : (H 1 + H 2 + H3) x1.75 + (39.73 x 1.75) H 3 = 20.7 kN

- (9.3 x 15.00) - (3.09 x 9.0) + (3.09 x 3.0) = 0

F H = 0 : 14.8 - H1 - H2 - H3 - H4 = 0 59.1 - H 1 - H2 - H 3 - H 4 = 0 H 4 = 8.9 kN

msy '11

LP

230 Frame Analysis



Level Q and above

9.7

12.3 3.5m

12.3 3.5m

12.3 3.5m

12.3 3.5m

12.3 3.5m

1.75m

H 1 H 2 H 3 H 4

6 m 6 m 6 m

3.0P 1.0P 1.0P 3.0P


M @ P = 0

(9.7 x 19.3)+ (12.3 x 15.8)+ (12.3 x 12.3) + (12.3 x 8.75) + 3.0P = 36.3 kN

(12.3 x 5.25)+ (12.3 x 1.75)+ (1.0P x 6.0) - (1.0P x 12.0) 1.0P = 12.1 kN

-(3.0P x 18.0) = 0 1.0P = 12.1 kN

60.0P = 727.0 kN 3.0P = 36.3 kN

P = 727.0 / 60.0 = 12.12 kN


24.9 kN 8.3 kN 8.3 kN 24.9 kN

8.9 F 1 20.7 F 2 20.7 F 3 8.9

1.75

1.75

H 1 H 2 H 3 H 4

6 m 6 m 6 m

36.3 kN 12.1 kN 12.1 kN 36.3 kN


F v = 0 : F 1 - 36.35 +24.93 = 0 F 1 = 11.4 kN

M @F 1 = 0: (H 1 + 8.9) x 1.75 - (11.4x 3.00) = 0 H 1 = 10.7 kN


F v = 0 : F2 - 4.7P - 1.3P = 0 F 2 - 11.42 - 3.81 = 0 F 2 = 15.2 kN

M @F 2 = 0 : (H 1 + H 2) x 1.75 + (29.55 x 1.75) H 2 = 25.0 kN

- (11.4 x 9.0) - (3.81 x 3.0) = 0


F v = 0 : F2 - 4.7P - 1.3P = 0 F 3 - 11.42 - 3.81 + 3.81 = 0 F 3 = 11.4 kN

M @F 2 = 0 : (H 1 + H 2 + H3) x1.75 + (50.23 x 1.75) H 3 = 25.0 kN

- (11.4 x 15.00) - (3.81 x 9.0) + (3.81 x 3.0) = 0

F H = 0 : 14.8 - H1 - H2 - H3 - H4 = 0 71.4 - H 1 - H2 - H 3 - H 4 = 0 H 4 = 10.7 kN

msy '11

LQ

Frame Analysis 231



Level R and above

9.7

12.3 3.5m

12.3 3.5m

12.3 3.5m

12.3 3.5m

12.3 3.5m

13.2 3.5m

2.0m

H 1 H 2 H 3 H 4

6 m 6 m 6 m

3.0P 1.0P 1.0P 3.0P


M @ P = 0

(9.7 x 23.0)+ (12.3 x 19.5)+ (12.3 x 16.0) + (12.3 x 12.5) + 3.0P = 51.1 kN

(12.3 x 9.00)+ (12.3 x 5.50)+ (13.2 x 2.00)+ (1.0P x 6.0) 1.0P = 17.0 kN

- (1.0P x 12.0) -(3.0P x 18.0) = 0 1.0P = 17.0 kN

60.0P = 1021.4 kN 3.0P = 51.1 kN

P = 1021 / 60.0 = 17.02 kN


36.3 kN 12.1 kN 12.1 kN 36.3 kN

10.7 F 1 25.0 F 2 25.0 F 3 10.7

1.75

2.00

H 1 H 2 H 3 H 4

6 m 6 m 6 m

51.1 kN 17.0 kN 17.0 kN 51.1 kN


F v = 0 : F 1 - 51.07 +36.35 = 0 F 1 = 14.7 kN

M @F 1 = 0: (H 1x 2.00) + (10.7 x 1.75) - (14.7x 3.0)=0 H 1 = 12.7 kN


F v = 0 : F2 - 4.7P - 1.3P = 0 F 2 - 14.72 - 4.91 = 0 F 2 = 19.6 kN

M @F 2 = 0 : (H 1 + H 2) x 2.00 + (35.72 x 1.75) H 2 = 29.6 kN

- (14.7 x 9.0) - (4.91 x 3.0) = 0


F v = 0 : F2 - 4.7P - 1.3P = 0 F 3 - 14.72 - 4.91 + 4.91 = 0 F 3 = 14.7 kN

M @F 2 = 0 : (H 1 + H 2 + H3) x2.00 + (60.73 x 1.75) H 3 = 29.6 kN

- (14.7 x 15.00) - (4.91 x 9.0) + (4.91 x 3.0) = 0

F H = 0 : 14.8 - H1 - H2 - H3 - H4 = 0 84.7 - H 1 - H2 - H 3 - H 4 = 0 H 4 = 12.7 kN

msy '11

LR

232 Frame Analysis



Shear Force in Beams and Columns

0.85 1.13 0.85

1.46 3.40 3.40 1.46

2.78 3.70 2.78 3.5m

3.31 7.72 7.72 3.31

4.94 6.59 4.94 3.5m

5.16 12.04 12.04 5.16

7.10 9.47 7.10 3.5m

7.01 16.36 16.36 7.01

9.3 12.3 9.3 3.5m

8.9 20.7 20.7 8.9

11.4 15.2 11.4 3.5m

10.7 25.0 25.0 10.7

14.7 19.6 14.7 3.5m

12.7 29.6 29.6 12.7 4.0m

A B C D

6 m 6 m 6 m

Unless otherwise stated all units in kN

msy '11

Frame Analysis 233



Bending Moments in Beams and Columns

2.5 2.5 5.9 3.4 5.9 2.5 2.5

2.5 3.4 2.5

2.5 5.8 8.3 13.5 11.1 13.5 8.3 5.8

8.3 5.9 11.1 5.9 8.3 2.5

5.8 9.0 14.8 21.1 19.8 21.1 14.8 9.0

14.8 13.5 19.8 13.5 14.8 5.8

9.0 12.3 21.3 28.6 28.4 28.6 21.3 12.3

21.3 21.1 28.4 21.1 21.3 9.0

15.5 27.8 36.2 37.0 36.2 27.8 15.5

12.3 27.8 28.6 37.0 28.6 27.8 12.3

18.8 34.3 43.8 45.7 43.8 34.3 18.8

15.5 34.3 36.2 45.7 36.2 34.3 15.5

25.4 44.2 59.3 58.9 59.3 44.2 25.4

18.8 44.2 43.8 58.9 43.8 44.2 18.8

25.4 59.3 59.3 25.4

Unless otherwise stated all units in kNm

msy '11

234 Frame Analysis

Example 9.2(c) : Analysis of one level sub-frame - Vertical load only page 1/2


Case 1:: All spans design permanent & variable loads 1.2G k + 1.2Q k

Fixed end moment :

-M AB = M BA = w 1L 12/12

46.4 kN/m 43.9 kN/m 46.4 kN/m 3.5m = 46.4 x 6.02/12

= 139.1 kNm

-M BC = M CB = w 2L 22/12

4.0m = 43.9 x 6.02/12

6.0 m 6.0 m 6.0 m = 131.6 kNm

-M CD = M DC = w1L 12/12

= 46.4 x 6.02/12

= 139.1 kNm

Moment distribution

41.67 -9.48 9.48 -41.67

0.25 -0.42 0.42 -0.25

1.52 -1.09 1.09 -1.52

0.34 -6.52 6.52 -0.34

39.56 -1.45 1.45 -39.56

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

-139.1 139.1 -131.6 131.6 -139.1 139.1

34.6 64.9 -2.4 -1.3 -2.4 2.4 1.3 2.4 -64.9 -34.6

-1.2 32.5 1.2 -1.2 -32.5 1.2

0.3 0.6 -10.7 -5.7 -10.7 10.7 5.7 10.7 -0.6 -0.3

-5.4 0.3 5.4 -5.4 -0.3 5.4

1.3 2.5 -1.8 -1.0 -1.8 1.8 1.0 1.8 -2.5 -1.3

-0.9 1.2 0.9 -0.9 -1.2 0.9

0.2 0.4 -0.7 -0.4 -0.7 0.7 0.4 0.7 -0.4 -0.2

36.5 -78.1 157.5 -8.3 -139.7 139.7 8.3 -157.5 78.1 -36.5

msy '11

A B C D

Frame Analysis 235

Example 9.2(c) : Analysis of one level sub-frame - Vertical load only page 2/2


Shear force :

78.1 46.4 157.5 139.7 43.9 139.7 157.5 46.4 78.1

6.0 6.0 6.0


M @ B = 0

6.0 V AB - (46.4 x 6.0 x 3.0) + 157.5 - 78.1 = 0

V AB = ( 834.46 - 157.5 +78.1 ) / 6.0 = 125.8 kN

V BA = ( 46.4 x 6.0) - 125.8 = 152.3 kN

M @ C = 0

6.0 V BC - (43.9 x 6.0 x 3.0) + 139.7 - 139.7 = 0

V BC = ( 789.75 - 139.7 +139.7 ) / 6.0 = 131.6 kN

V CB = ( 43.9 x 6.0) - 131.6 = 131.6 kN

M @ D = 0

6.0 V CD - (46.4 x 6.0 x 3.0) - 157.5 + 78.1 = 0

V CD = ( 834.46 +157.5 - 78.1 ) / 6.0 = 152.3 kN

V DC = ( 46.4 x 6.0) - 152.3 = 125.8 kN


125.8 131.6 152.3

2.7 3.0 3.3

152.3 131.6 125.8

157.5 139.7 157.5

78.1 139.7 78.1

41.7

36.5 9.5 8.3 9.5 8.3 41.7 36.5

57.7

92.7 92.7

18.2 4.1 4.1 18.2

msy '11

236 Frame Analysis

Example 9.2(d) :Combination of vertical load and lateral load page 1/1


Bending moment due to wind load 1.2 W k

18.8 43.8 43.8 18.8

25.4 44.2 59.3 58.9 59.3 44.2 25.4

18.8 44.2 43.8 58.9 43.8 44.2 18.8

25.4 59.3 59.3 25.4

Bending moment due to vertical load 1.2 (G k + Q k)

157.5 157.5

78.1 139.7 139.7 78.1

41.7

36.5 9.5 8.3 9.5 8.3 41.7 36.5

57.7

92.7 92.7

18.2 4.1 4.1 18.2

Combination of vertical and wind load 1.2 (G k + Q k + Wk)

201.7 198.6

34.0 80.9 113.4 122.3

22.9

11.1 53.2 67.6 53.2 67.6 60.4 61.9

57.7

96.9 88.5

7.2 63.4 63.4 43.6

Frame Analysis 237

Problems

9.1 The typical arrangements of an eight storey concrete office building are as shown in Figure P9.1.

The building, 21 m wide by 40 m long is situated on an exposed site in the outskirts of a large

town.

(a) If the building is to be designed as a braced structure using shear wall as the bracing

elements, show on the plan the suitable positions of these walls that should be provided to

resist the horizontal forces in both longitudinal and transverse directions.

(b) Briefly outline the advantages and/or disadvantages of the braced building against the

unbraced building in terms of member sizing, forces, costs and design parameters with

reference to the given size of building.

9.2 Part of the plan of a four storey reinforced concrete building and the front view of its substitute

frame are shown in Figure P9.2. The lateral forces on the building in both directions are to be

resisted by shear walls. The transverse beams with a cross-section of 300 x 700 mm, are

continuous over two spans of 8.75 m each. The continuous beams in the longitudinal direction of

the building have a cross-section of 250 x 600 mm and a span of 5.0 m. The columns, with a cross-

section 300 x 400 mm, are 4.8 m high for the first lift, and 3.6 m for the successive lifts, and for

the purpose of analysis the bottom end of the first lift of the column may be assumed fixed to the

ground beams / footings. The overall thickness of the slab is 175 mm. Other relevant dimensions

and detailing are also shown in the figure. The building is to be designed to carry the vertical loads

as follows:

Characteristic live load = 4.0 kN/m2

Characteristic dead load from finishes = 1.5 kN/m2

and services only

(a) Analyse Beam 6/A-C, Level 1 using the continuous beam and subframe methods, with loading arrangements suggested in EC2, and draw their respective bending moment envelopes (for

beams only) showing all the important values. (The maximum span moments should be

determined from shear force diagrams).

(b) Comment on the two sets of the bending moment values obtained in (a). What effect will these bending moment envelopes have on the amount and detailing of the reinforcement to be

provided in Beam 6/A-C.

4

3 @ 7 m

= 21 m

8 @ 5 m = 40

m

3

2

1

A

B

C

D

E

F

G

H

J

Figure P9.1

238 Frame Analysis

Figure P9.2

7

6

5

4

A B C

8750 8750

36

00

36

00

48

00

36

00

8750 8750

Level 2

Level 1

Ground

300 x 700

Roof

300 x 700

300 x 700

50

00

50

00

50

00

25

0 x

600

25

0 x

600

25

0 x

600

Level 3

A

A

Section A-A

70

0

300

17

5

All units in mm

Frame Analysis 239

9.3 Figure P9.3 shows a substitute frame of a seven storey reinforced concrete building which is to be

built without any bracing members. The dimensions and section properties of all beams and

columns are given in the same figure. The first floor beams supported a characteristic dead load

(including selfweight) and a characteristic live load of 20 kN/m and 15 kN/m respectively. The

characteristic horizontal load due to wind pressure is 4.0 kN/m applied through the whole height of

the frame. The lower end of the first lift of the column may be assumed to be fixed to the

foundation.

(a) Using Cantilever method analyse the frame considering the horizontal load only and draw the bending moment diagram for all columns and beams.

(b) Analyse the sub-frame which consists of the first floor beams and all columns associated with the beams considering the vertical loads only.

(c) Draw the bending moment diagram of the sub-frame mentioned in (b) for the combination of vertical and horizontal loads.

(d) Explain the significance and the usage of the bending moments drawn in (c) for the purpose of structural design.

Figure P9.3

Beam:

250 x 600 mm

Ixx = 45.0 x 108 mm

4

Kxx = 7.50 x 105 mm

3

Column:

300 x 400 mm

Ixx = 16.0 x 108 mm

4

K4.0m = 4.00 x 105 mm

3

K3.4m = 4.71 x 105 mm

3

Unless otherwise

stated all units

are in mm

6000 6000

3400

3400

3400

3400

3400

3400

4000

4.0

kN

/m

400

240 Frame Analysis

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