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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Reinforced Concrete Design 2

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.


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.


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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� (i) Vertical loads (1.2Gk + 1.2Qk) for sub-frame +

(ii) Wind load (1.2Wk) for complete frame

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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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� 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 =−=


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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 ===


� 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


� 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 =−=


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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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� 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


� 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 =−=


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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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Documents

2) Frame Analysis