Download - 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

BFC 3172

Reinforced Concrete Design 2

Method of Frame Analysis

1) Complete sub-frame

BFC 3172


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

BFC 3172


The frame consists of a selected beam with columns and neighbouringbeams at both sides of selected beam.


3) Simplified sub-frame at point

BFC 3172


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

BFC 3172


Analysis of Braced Frame

BFC 3172


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

BFC 3172


Example 1.1

Solution� Beam Stiffness

� Column Stiffness

BFC 3172


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

BFC 3172


Example 1.1

� Load Case 1

BFC 3172


All spans loaded with maximum dead plus imposed loads

BFC 3172


BFC 3172


Example 1.1

� Load Case 2

BFC 3172


BFC 3172


BFC 3172


Example 1.1

� Load Case 3

BFC 3172


BFC 3172


BFC 3172


Example 1.1

Simplified sub-frame� Load Case 1

BFC 3172


Example 1.1

BFC 3172


Example 1.1

� Load Case 2

BFC 3172


Example 1.1

BFC 3172


Example 1.1

� Load Case 3

BFC 3172


Example 1.1

BFC 3172


Example 1.1

Simplified Sub-Frame at Point� Point A & D

BFC 3172


Example 1.1

� Point B & C

BFC 3172


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.

BFC 3172


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

BFC 3172


Example 1.2

Load Case 2

Load Case 3

BFC 3172


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)

BFC 3172



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

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

BFC 3172



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.

BFC 3172


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.

BFC 3172


Example 1.3

� Solution

BFC 3172


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

BFC 3172


Example 1.3 (Roof Floor)

� Axial Force in column:

� Shear force in beam

BFC 3172



� Horizontal Force in Column:

ΣMF1 = 0 ΣMF2 = 0

BFC 3172


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 =−=


BFC 3172


Example 1.3 (4th Floor)

� Axial force in Column:

BFC 3172


41 24.34 NkNPN ===

32 81.0 NkNPN ===


� Shear force in beamΣFy = 0

ΣFy = 0

BFC 3172


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

BFC 3172


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 =−=


BFC 3172


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

BFC 3172



� Shear force in beam� ΣFy = 0

� ΣFy = 0

BFC 3172


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

BFC 3172


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 =−=


BFC 3172


Example 1.3 (2nd Floor)

BFC 3172


Example 1.3 (1st Floor)

BFC 3172


BFC 3172


Simplified Approach

� 1.4Gk + 1.6Qk

� 1.4(25) + 1.6(10) = 51 kN/m

BFC 3172


Simplified Approach

� 1.2Gk + 1.2Qk

� 1.2(25) + 1.2(10) = 42 kN/m

BFC 3172


Result of Portal Analysis at 3rd Floor

BFC 3172


Vertical Load + Horizontal Load

BFC 3172


31.5327.03

16.07

13.67 13.67

16.07 31.53

27.03

1.2Gk + 1.2Qk

1.2Wk

BFC 3172
