2) Frame Analysis

May 7, 2017 | Author: Azuanuddin Samad | Category: N/A

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FRAME ANALYSIS By: Mohammad Mohammad Soffi bin Md. Noh 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 Par t 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 

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

Method of Frame Analysis 1) Comple Complete te sub-fr sub-frame ame 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 subframe.

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

Method of Frame Analysis 2) Simpli Simplifie fied d sub-frame sub-frame The frame consists of a selected beam with columns and neighbouring beams at both sides of selected beam.

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Method of Frame Analysis

B F C 3 1 7 2

3) Simplified Simplified sub-frame sub-frame at point point

The frame consists of a selected point or node with columns at top and bottom, 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

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

Analysis of Braced Frame

1.4Gk + 1.6Qk

1.4Gk + 1.6Qk

1.4Gk + 1.6Qk

1.4Gk + 1.6Qk

1.4Gk + 1.6Qk

1.0Gk

1.4Gk + 1.6Qk

1.0Gk

1.0Gk

1.4Gk + 1.6Qk

1.0Gk

1.4Gk + 1.6Qk

Example 1.1 A four storey braced building is given in Figure P1.1. Perform the analysis for ABCD. Given the following data: All colu column mns s = 350 350 mm x 300 300 mm Gk = 25 kN/ kN/m All beams = 300 m m x 6 00 m m Qk = 10 kN/m

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.1 Solution  Beam Stiffness



Column Stiffness

Example 1.1

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

Completed Sub-Frame Analysis

Max load = 1.4(25) + 1.6(10) = 51 kN/m Min Load = 1.0(25) = 25 kN/m

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.1 

B F C 3 1 7 2

Load Case 1

All spans loaded with maximum dead plus imposed loads

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.1 

B F C 3 1 7 2

Load Case 2 R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.1 

B F C 3 1 7 2

Load Case 3 R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.1 Simplified sub-frame  Load Case 1

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.1

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.1 

B F C 3 1 7 2

Load Case 2 R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.1

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.1 

B F C 3 1 7 2

Load Case 3 R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.1

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.1 Simplified Sub-Frame at Point  Point A & D

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.1 

B F C 3 1 7 2

Point B & C R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

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 kN Qk = 15 kN/m Qk = 15 kN Sketch the loadings arrangement.

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

Example 1.2 

Solution UDL Max = 1.4(20 1.4(20)) + 1.6 1.6(15 (15)) = 52 52 kN/m kN/m Concentrated Load Max Ma x = 1.4 1.4(30 (30) + 1.6( 1.6(15 15)) = 66 kN Load Case 1

Example 1.2

Min = 1.0(2 1.0(20) 0) = 20 kN/ kN/m m Min Min = 1.0( 1.0(30 30)) = 30 kN

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

Load Case 2

Load Case 3

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Analysis Analysis of Unbrace Unbraced d Frame Frame 

For unbraced frame, the greatest of 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)

Analysis Analysis of Unbrace Unbraced d Frame Frame 

(i) Vertical loads (1.2Gk + 1.2Qk) for sub-frame + (ii) Wind load (1.2Wk) for complete frame

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Analysis Analysis of Unbrace Unbraced d Frame Frame

B F C 3 1 7 2

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 the points of contraflexure immediately below the level.  The points of contraflexure occur at the mid-heights of columns and at midspans midspans of beams.  Each bay acts as a separate portal and the horizontal load is divided between bays in proportion to span. 

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. P1.3.

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

Example 1.3 

Solution R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

Example 1.3 Analysis of horizontal load Roof 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 kN nd 2 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 column Axial Force of external column : Axial force of internal column N1 : N2 8 : 2 4P : 1P

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.3 (Roof Floor)



Axial Force in column:



Shear force in beam

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.3 (Roof Floor) 

B F C 3 1 7 2

B F C 3 1 7 2

Horizontal Force in Column: ΣMF1

=0

ΣMF2

H 1 (1.75) − 0.65(3) H 1

=

1.95 1.75

=

=

1.11kN

0

=0

( H 1 + H 2 )(1.75) − 0.65(8) − 0.16(2) = 0 H 2

=

3.15 − 1.11 = 2.04kN

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.3 (Roof Floor)

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.3 (4 th Floor)



Axial force in Column:

N 1 N 2

= =

4 P = 3.24kN = N 4 P

=

0.81kN = N 3

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.3 (4 th Floor) 

Shear force in beam ΣFy = 0 F 1

+ 0.65 − 3.24 =

F 1

=

0

3.24 − 0.65 = 2.59kN

ΣFy

=0

F 2

+ 0.65 + 0.16 − 3.24 − 0.81 =

F 2

=

0

3.24 + 0.81 − 0.65 − 0.16 = 3.24kN

Example 1.3 (4 th Floor) 

Horizontal force in column Σ MF1 = 0 H 1 (1.75) + 1.11(1.75) + 0.65(3) − 3.24(3) = 0 H 1

=

3.33kN

 ΣMF2 =

0

( H 1 + H 2 )(1.75) + (1.11 + 2.04)(1.75) − (3.24 − 0.65)(8) −

(0.81 − 0.16)(2) = 0

( H 1 + H 2 ) = 9.43kN H 2

=

9.43 − 3.33 = 6.10kN

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.3 (4 th Floor)

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.3 (3rd Floor)



Axial force in Column: =0

ΣMs

P(6) – P(10) – 4P(16) + 6.3(8.75 6.3(8.75)) + 12.6(5.25) 12.6(5.25) + 12.6(1.75) 12.6(1.75) = 0 P = 2.11 kN N1 = 4P = 8.44 kN = N4 N2 = 1P = 2.11 2.11 kN kN = N3

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.3 (3rd Floor)  

Shear force in beam ΣFy = 0 F 1

+ 3.24 − 8.44 =

0

F 1

= 8.44 − 3.24 =

5.20kN

 ΣFy F 2 F 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

=0

+ 3.24 + 0.81 − 8.44 − 2.11 = = 8.44 +

0

2.11 − 3.24 − 0.81 = 6.5kN

Example 1.3 (3rd Floor) 

Horizontal force in column Σ MF1 = 0 H 1 (1.75) + 3.33(1.75) + 3.24(3) − 8.44(3) H 1

=

=

0

5.58kN

 ΣMF2 =

0

( H 1 + H 2 )(1.75) + (3.33 + 6.10)(1.75) − (8.44 − 3.24)(8) −

(2.11 − 0.81)(2) = 0

( H 1 + H 2 ) = 15.83kN H 2

=

15.83 − 5.58 = 10.25kN

B F C 3 1 7 2

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.3 (3rd Floor)

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.3 (2nd Floor)

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Example 1.3 (1st Floor)

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

Simplified Approach  

1.4Gk + 1.6Qk 1.4(25) + 1.6(10) = 51 kN/m

Simplified Approach  

1.2Gk + 1.2Qk 1.2(25) + 1.2(10) = 42 kN/m

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

Result of Portal Analysis at 3 rd Floor

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

Vertical Load + Horizontal Load 1.2Gk + 1.2Qk

27.03

13.67

31.53 16.07

1.2Wk

27.03

13.67 16.07

31.53

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

B F C 3 1 7 2

R e i   n f   o r c e d C o n c r e t e D e s i   g n 2

2) Frame Analysis

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