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A text reference detailing the procedure to conduct a frame analysis due to Gk, Qk and Wk
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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
msy '11
2
3
4
A B C D
196 Frame Analysis
Example 9.1(a) : Action on beams page 2/2
Ref. Calculations Output
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
Perm. load from slab = 0.42 x 4.50 x 5.0 = 9.45 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
Perm. load from slab = 0.45 x 4.50 x 4.0 = 8.10 kN/m
Perm. load from slab = 0.39 x 4.50 x 5.0 = 8.78 kN/m
Beam selfweight = 0.45 x 0.25 x 25 = 2.81 kN/m
Characteristic permanent load, g k = 19.69 kN/m
Variable load fr. slab = 0.45 x 4.50 x 4.0 = 8.10 kN/m
Variable load fr. slab = 0.39 x 4.50 x 5.0 = 8.78 kN/m
Chacracteristic variable load, Q k = 16.88 kN/m
Design load, 1.35G k + 1.5Q k = 51.89 kN/m
1.35G k = 26.58 kN/m
Span 3 (C-D) : w 3
Perm. load from slab = 0.47 x 4.50 x 4.0 = 8.46 kN/m
Perm. load from slab = 0.42 x 4.50 x 5.0 = 9.45 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 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
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
msy '11
A BC D
w1 w2 w3
Frame Analysis 197
Example 9.1(b) : Analysis of one level sub-frame page 1/10
Ref. Calculations Output
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
msy '11
A B C D
198 Frame Analysis
Example 9.1(b) : Analysis of one level sub-frame page 2/10
Ref. Calculations Output
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
msy '11
A B C D
Frame Analysis 199
Example 9.1(b) : Analysis of one level sub-frame page 3/10
Ref. Calculations Output
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
msy '11
200 Frame Analysis
Example 9.1(b) : Analysis of one level sub-frame page 4/10
Ref. Calculations Output
Case 2:: Span 2 & 3 design permanent & variable loads 1.35G k + 1.5Q k
Span 1 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 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
msy '11
A B C D
Frame Analysis 201
Example 9.1(b) : Analysis of one level sub-frame page 5/10
Ref. Calculations Output
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
V AB V BA V BC V CB V CD V DC
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
Shear force and bending moment diagrams
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
Example 9.1(b) : Analysis of one level sub-frame page 6/10
Ref. Calculations Output
Case 3:: Span 1 & 3 design permanent & variable loads 1.35G k + 1.5Q k
Span 2 design permanent loads 1.35G k
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
msy '11
A B C D
Frame Analysis 203
Example 9.1(b) : Analysis of one level sub-frame page 7/10
Ref. Calculations Output
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
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) + 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
Shear force and bending moment diagrams
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
Example 9.1(b) : Analysis of one level sub-frame page 8/10
Ref. Calculations Output
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
Example 9.1(b) : Analysis of one level sub-frame page 9/10
Ref. Calculations Output
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
V AB V BA V BC V CB V CD V DC
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
Shear force and bending moment diagrams
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
Example 9.1(b) : Analysis of one level sub-frame page 10/10
Ref. Calculations Output
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
Ref. Calculations Output
Beam A-B (Beam C-D similar)
Case 1:: Span 1 design permanent & variable loads 1.35G k + 1.5Q k
: 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
Example 9.1(c) : Analysis of two free-joint sub-frame page 2/6
Ref. Calculations Output
Case 2:: Span 1 design permanent & variable loads 1.35G k + 1.5Q k
: 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
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
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
Example 9.1(c) : Analysis of two free-joint sub-frame page 3/6
Ref. Calculations Output
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
Example 9.1(c) : Analysis of two free-joint sub-frame page 4/6
Ref. Calculations Output
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
Distribution factor:
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
Example 9.1(c) : Analysis of two free-joint sub-frame page 5/6
Ref. Calculations Output
Case 2:: Span 2 design permanent & variable loads 1.35G k + 1.5Q k
: 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
Distribution factor:
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
Example 9.1(c) : Analysis of two free-joint sub-frame page 6/6
Ref. Calculations Output
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
Ref. Calculations Output
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
Distribution factor:
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
Example 9.1(d) : Continuous beam + one free-joint sub-frame page 2/9
Ref. Calculations Output
Case 1:: Span 1 & 2 design permanent & variable loads 1.35G k + 1.5Q k
Span 3 design permanent loads 1.35G k
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
Example 9.1(d) : Continuous beam + one free-joint sub-frame page 3/9
Ref. Calculations Output
Case 2:: Span 2 & 3 design permanent & variable loads 1.35G k + 1.5Q k
Span 1 design permanent loads 1.35G k
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
Moment Distribution :
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
V AB V BA V BC V CB V CD V DC M @ B = 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
Example 9.1(d) : Continuous beam + one free-joint sub-frame page 4/9
Ref. Calculations Output
Case 3:: Span 1 & 3 design permanent & variable loads 1.35G k + 1.5Q k
Span 2 design permanent loads 1.35G k
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
Moment Distribution :
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
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) + 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
Example 9.1(d) : Continuous beam + one free-joint sub-frame page 5/9
Ref. Calculations Output
Case 4:: Span 2 design permanent & variable loads 1.35G k + 1.5Q k
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
Moment Distribution :
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
V AB V BA V BC V CB V CD V DC M @ B = 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
Example 9.1(d) : Continuous beam + one free-joint sub-frame page 6/9
Ref. Calculations Output
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
Example 9.1(d) : Continuous beam + one free-joint sub-frame page 7/9
Ref. Calculations Output
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
Example 9.1(d) : Continuous beam + one free-joint sub-frame page 8/9
Ref. Calculations Output
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
Moment in upper column,
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
Moment in upper column,
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
Example 9.1(d) : Continuous beam + one free-joint sub-frame page 9/9
Ref. Calculations Output
Column D
FEM = wL2/12
54.8 kN/m = 54.8 x 6.02
/12
= 164.52 kNm
6.0 m
Moment in upper column,
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
Ref. Calculations Output
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
Ref. Calculations Output
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
Ref. Calculations Output
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
Example 9.2(b) : Lateral load analysis : Cantilever method page 2/10
Ref. Calculations Output
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
Consider sub-frame to the left of F 2 :
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
Consider sub-frame to the left of F 3 :
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
Example 9.2(b) : Lateral load analysis : Cantilever method page 3/10
Ref. Calculations Output
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
Axial force in columns,
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
Shear force in beams and columns,
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
Consider sub-frame to the left of F 1 :
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
Consider sub-frame to the left of F 2 :
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
Consider sub-frame to the left of F 3 :
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
Example 9.2(b) : Lateral load analysis : Cantilever method page 4/10
Ref. Calculations Output
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
Axial force in columns,
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
Shear force in beams and columns,
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
Consider sub-frame to the left of F 1 :
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
Consider sub-frame to the left of F 2 :
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
Consider sub-frame to the left of F 3 :
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
Example 9.2(b) : Lateral load analysis : Cantilever method page 5/10
Ref. Calculations Output
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
Axial force in columns,
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
Shear force in beams and columns,
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
Consider sub-frame to the left of F 1 :
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
Consider sub-frame to the left of F 2 :
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
Consider sub-frame to the left of F 3 :
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
Example 9.2(b) : Lateral load analysis : Cantilever method page 6/10
Ref. Calculations Output
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
Axial force in columns,
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
Shear force in beams and columns,
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
Consider sub-frame to the left of F 1 :
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
Consider sub-frame to the left of F 2 :
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
Consider sub-frame to the left of F 3 :
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
Example 9.2(b) : Lateral load analysis : Cantilever method page 7/10
Ref. Calculations Output
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
Axial force in columns,
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
Shear force in beams and columns,
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
Consider sub-frame to the left of F 1 :
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
Consider sub-frame to the left of F 2 :
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
Consider sub-frame to the left of F 3 :
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
Example 9.2(b) : Lateral load analysis : Cantilever method page 8/10
Ref. Calculations Output
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
Axial force in columns,
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
Shear force in beams and columns,
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
Consider sub-frame to the left of F 1 :
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
Consider sub-frame to the left of F 2 :
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
Consider sub-frame to the left of F 3 :
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
Example 9.2(b) : Lateral load analysis : Cantilever method page 9/10
Ref. Calculations Output
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
Example 9.2(b) : Lateral load analysis : Cantilever method page 10/10
Ref. Calculations Output
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
Ref. Calculations Output
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
Ref. Calculations Output
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
V AB V BA V BC V CB V CD V DC
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
Shear force and bending moment diagrams
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
Ref. Calculations Output
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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