2) Frame Analysis
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Transcript of 2) Frame Analysis
FRAME ANALYSISBy:
Mohammad Soffi bin Md. Noh
Introduction
� In the design of RC structures based on BS 8110 it has to analyse the structure subjected to all probable combinations of loads, considering the ultimate limit state.
� Once the bending moment, shear force etc. were obtained, reinforcements can be designed according to the standard.
� Generally, three dimensional wide frame analysis is the most accurate method to analyse the frame building.
� 3-D frame is complex and need to be carried out using relevant computer software. Clause 3.2.1.1 BS 8110: Part 1: 1997 states that the analysis may be simplified appropriately sub-frame.
� Hence there are 3 levels sub-frames:� Complete sub-frame
� Simplified sub-frame
� Simplified sub-frame at point
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Method of Frame Analysis
1) Complete sub-frame
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The frame consists of all beams at each level with columns top and bottom of beams.
Moments at columns and beams are tabulated by analyzing the complete sub-frame.
Method of Frame Analysis
2) Simplified sub-frame
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The frame consists of a selected beam with columns and neighbouringbeams at both sides of selected beam.
Method of Frame Analysis
3) Simplified sub-frame at point
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The frame consists of a selected point or node with columns at top andbottom, and neighbouring beams coming into the point.
Analysis of Braced Frame
� A building is saying as braced frame when the horizontal loadings are resisting by the shear walls or bracing.
� The analysis of braced frame is only considered for the vertical loads which are dead and imposed load.
� For the combination of dead load and imposed load, the following loading patterns are considered:� All spans loaded with maximum
dead plus imposed loads� Alternate spans loaded with
maximum dead load and imposed load and all other spans loaded with minimum dead load
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Analysis of Braced Frame
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1.4Gk + 1.6Qk 1.4Gk + 1.6Qk 1.4Gk + 1.6Qk 1.4Gk + 1.6Qk
1.0Gk1.0Gk1.4Gk + 1.6Qk 1.4Gk + 1.6Qk
1.4Gk + 1.6Qk 1.4Gk + 1.6Qk1.0Gk 1.0Gk
Example 1.1
A four storey braced building is given in Figure P1.1. Perform the analysis for ABCD. Given the following data:
All columns = 350 mm x 300 mm Gk = 25 kN/m
All beams = 300 mm x 600 mm Qk = 10 kN/m
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Example 1.1
Solution� Beam Stiffness
� Column Stiffness
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Example 1.1
Completed Sub-Frame Analysis
Max load = 1.4(25) + 1.6(10) = 51 kN/m
Min Load = 1.0(25) = 25 kN/m
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Example 1.1
� Load Case 1
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All spans loaded with maximum dead plus imposed loads
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Example 1.1
� Load Case 2
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Example 1.1
� Load Case 3
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Example 1.1
Simplified sub-frame� Load Case 1
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Example 1.1
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Example 1.1
� Load Case 2
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Example 1.1
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Example 1.1
� Load Case 3
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Example 1.1
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Example 1.1
Simplified Sub-Frame at Point� Point A & D
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Example 1.1
� Point B & C
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Example 1.2
� Figure P1.2 shows the four spans sub frame. Given:
UDL (all spans): Concentrated load (span BC):
Gk = 20 kN/m Gk = 30 kNQk = 15 kN/m Qk = 15 kN
Sketch the loadings arrangement.
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Example 1.2
� SolutionUDLMax = 1.4(20) + 1.6(15) = 52 kN/m Min = 1.0(20) = 20 kN/mConcentrated LoadMax = 1.4(30) + 1.6(15) = 66 kN Min = 1.0(30) = 30 kN
Load Case 1
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Example 1.2
Load Case 2
Load Case 3
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Analysis of Unbraced Frame
� For unbraced frame, the greatest of the following moments and shearing forces are to be taken for design purposes:
� Three cases loading arrangements as braced sub-frame (max = 1.4Gk + 1.6Qk, min = 1.0Gk)
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Analysis of Unbraced Frame
� (i) Vertical loads (1.2Gk + 1.2Qk) for sub-frame +
(ii) Wind load (1.2Wk) for complete frame
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Analysis of Unbraced Frame
Analysis of Horizontal Load Using Portal Method
� The following assumptions have to be made:� Loads applied at beam-column junction.
� Total horizontal shear at any level is carried by columns at thepoints of contraflexure immediately below the level.
� The points of contraflexure occur at the mid-heights of columns and at midspans of beams.
� Each bay acts as a separate portal and the horizontal load is divided between bays in proportion to span.
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Example 1.3
� Draw the bending moment diagram of the 5 storey building frame subjected to 3 kN/m wind load as shown in Figure P1.3.
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Example 1.3
� Solution
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Example 1.3
Analysis of horizontal loadRoof floor = (1.2 x 3) x (3.5/2) = 6.30 kN
3rd and 4th Floor = (1.2 x 3) x [(3.5/2) +(3.5/2)] = 12.6 kN2nd Floor = (1.2 x 3) x [(3.5/2) + (4/2)] = 13.5 kN
1st Floor = (1.2 x 3) x [(4/2) + (4/2)] = 14.4 kN
Ground Floor = (1.2 x 3) x (4/2) = 7.2 kN
Ratio of axial force in columnAxial Force of external column : Axial force of internal column
N1 : N2
8 : 2
4P : 1P
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Example 1.3 (Roof Floor)
� Axial Force in column:
� Shear force in beam
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Example 1.3 (Roof Floor)
� Horizontal Force in Column:
ΣMF1 = 0 ΣMF2 = 0
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0)3(65.0)75.1(1 =−H
kNH 11.175.1
95.11 ==
0)2(16.0)8(65.0)75.1)(( 21 =−−+ HH
kNH 04.211.115.32 =−=
Example 1.3 (Roof Floor)
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Example 1.3 (4th Floor)
� Axial force in Column:
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41 24.34 NkNPN ===
32 81.0 NkNPN ===
Example 1.3 (4th Floor)
� Shear force in beamΣFy = 0
ΣFy = 0
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024.365.01 =−+F
kNF 24.316.065.081.024.32 =−−+=
kNF 59.265.024.31 =−=
081.024.316.065.02 =−−++F
Example 1.3 (4th Floor)
� Horizontal force in columnΣ MF1 = 0
� ΣMF2 = 0
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0)3(24.3)3(65.0)75.1(11.1)75.1(1 =−++H
kNH 33.31 =
)8)(65.024.3()75.1)(04.211.1()75.1)(( 21 −−+++ HH
0)2)(16.081.0( =−−kNHH 43.9)( 21 =+
kNH 10.633.343.92 =−=
Example 1.3 (4th Floor)
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Example 1.3 (3rd Floor)
� Axial force in Column:ΣMs = 0P(6) – P(10) – 4P(16) + 6.3(8.75) + 12.6(5.25) + 12.6(1.75) = 0P = 2.11 kN N1 = 4P = 8.44 kN = N4 N2 = 1P = 2.11 kN = N3
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Example 1.3 (3rd Floor)
� Shear force in beam� ΣFy = 0
� ΣFy = 0
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044.824.31 =−+F
kNF 5.681.024.311.244.82 =−−+=
kNF 20.524.344.81 =−=
011.244.881.024.32 =−−++F
Example 1.3 (3rd Floor)
� Horizontal force in columnΣ MF1 = 0
� ΣMF2 = 0
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0)3(44.8)3(24.3)75.1(33.3)75.1(1 =−++H
kNH 58.51 =
)8)(24.344.8()75.1)(10.633.3()75.1)(( 21 −−+++ HH
0)2)(81.011.2( =−−kNHH 83.15)( 21 =+
kNH 25.1058.583.152 =−=
Example 1.3 (3rd Floor)
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Example 1.3 (2nd Floor)
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Example 1.3 (1st Floor)
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Simplified Approach
� 1.4Gk + 1.6Qk
� 1.4(25) + 1.6(10) = 51 kN/m
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Simplified Approach
� 1.2Gk + 1.2Qk
� 1.2(25) + 1.2(10) = 42 kN/m
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Result of Portal Analysis at 3rd Floor
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Vertical Load + Horizontal Load
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31.5327.03
16.07
13.67 13.67
16.07 31.53
27.03
1.2Gk + 1.2Qk
1.2Wk
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