2) Frame Analysis

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Moham

Mohammad Soffimad Soffi bin Md. Nobin Md. Nohh

Introduction



 In the design of RC structures based on BS 8110 it has to analyseIn the design of RC structures based on BS 8110 it has to analyse the structure subjected to all probable combinations of loads, the structure subjected to all probable combinations of loads, considering the ultimate limit state.

considering the ultimate limit state.

B B F F C C 3 3₁₁ 7 722 R R e e

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Method of Frame Analysis

1)

1) ComCompleplete sute sub-frb-frameame

3 3₁₁ 7 722 R R e e_i_i

n n f f o o_rr c c e e d d C C o o n n c crr e e t t e e D D e e s si i g g_n_n 2 2

The frame consists of The frame consists of all beams at each level all beams at each level with columns top and with columns top and bottom of beams. bottom of beams. Moments at columns Moments at columns and beams are and beams are

tabulated by analyzing tabulated by analyzing the complete the complete sub-frame.

frame.

Method of Frame Analysis

2)

2) SimSimpliplifiefied sub-frd sub-frameame

B B F F C C 3 3₁₁ 7 722 R R

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Method of Frame Analysis

3)

3) SimplSimplified ified sub-frsub-frame ame at poiat pointnt

3 3₁₁ 7 722 R R e e_i_i

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

columns at top and

bottom, and neighbouring bottom, and neighbouring beams coming into the beams coming into the point.

point.

Analysis of Braced Frame



 A building is saying as bracedA building is saying as braced frame when the horizontal loadings frame when the horizontal loadings are resisting by the shear walls or are resisting by the shear walls or bracing. bracing. B B F F C C 3 3₁₁ 7 722 R R e e

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Analysis of Braced Frame

3 3₇1 1 722

R R e e_i_i

n n f f o o_rr c c e e d d C C o o n n c crr e e t t e e D D e e s si i g g_n_n 2 2 1 1..44GGk k + + 11..66QQkk 11..44GGk k + + 11..66QQkk 11..44GGk k + + 11..66QQkk 11..44GGk k + + 11..66QQkk 1.0Gk 1.0Gk 1.0Gk 1.0Gk 1 1..44GGk k + + 11..66QQkk 11..44GGk k + + 11..66QQkk 1 1..44GGk k + + 11..66QQkk 11..44GGk k + + 11..66QQkk 1 1..00GGkk 11..00GGkk

Example 1.1

A four storey braced building is given in Figure P1.1. Perform the A four storey braced building is given in Figure P1.1. Perform the analysis for ABCD. Given the following data:

analysis for ABCD. Given the following data:

B B F F C C 3 3₁₁ 7 722 R R

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Example 1.1

Solution Solution



 Beam StiffnessBeam Stiffness



 Column StiffnessColumn Stiffness

3 3₁₁ 7 722 R R e e_i_i

Example 1.1

Completed Sub-Frame Analysis Completed Sub-Frame Analysis

B B F F C C 3 3₁₁ 7 722 R R e e

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Example 1.1



 Load Case 1Load Case 1

3 3₁₁ 7 722 R R e e_i_i

All spans loaded with maximum dead plus

All spans loaded with maximum dead plus imposed loadsimposed loads

B B F F C C 3 3₁₁ 7 722 R R

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3 3₁₁ 7 722 R R e e_i_i

Example 1.1



B B F F C C 3 3₁₁ 7 722

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3 3₁₁ 7 722 R R e e_i_i

n n f f o o_rr c c e e d d C C o o n n c crr e e t t e e D D e e s si i g g_n_n 2 2 B B F F C C 3 3₁₁ 7 722 R R

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Example 1.1



3 3₁₁ 7 722 R R e e_i_i

n n f f o o_rr c c e e d d C C o o n n c crr e e t t e e D D e e s si i g g_n_n 2 2 B B F F C C 3 3₁₁ 7 722

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3 3₁₁ 7 722 R R e e_i_i

Example 1.1

Simplified sub-frame Simplified sub-frame 

B B F F C C 3 3₁₁ 7 722 R R

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Example 1.1

3 3₇1 1 722 R R e e_i_i

Example 1.1



B B F F C C 3 3₁₁ 7 722

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Example 1.1

3 3₇1 1 722 R R e e_i_i

Example 1.1



B B F F C C 3 3₁₁ 7 722 R R

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Example 1.1

3 3₇1 1 722 R R e e_i_i

Example 1.1

Simplified Sub-Frame at Point Simplified Sub-Frame at Point

B B F F C C 3 3₁₁ 7 722

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Example 1.1



 Point B & CPoint B & C

3 3₁₁ 7 722 R R e e_i_i

Example 1.2



 Figure P1.2 shows the four spans sub frame. Given:Figure P1.2 shows the four spans sub frame. Given: U

UDDL L ((aalll l ssppaannss)):: CCoonncceennttrraatteed d llooaad d ((ssppaan n BBCC)): :

B B F F C C 3 3₁₁ 7 722 R R

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Example 1.2

  SolutionSolution UDL UDL Max

Max = 1.= 1.4(204(20) + ) + 1.61.6(15(15) = 5) = 52 kN2 kN/m/m Min Min = 1= 1.0(2.0(20) = 0) = 20 k20 kN/N/mm Concentrated Load Concentrated Load Ma Max = x = 1.1.44(3(300) + ) + 1.1.6(6(1515) = ) = 66 66 kNkN MiMin = n = 1.1.0(0(3030) = ) = 30 30 kNkN Load Case 1 Load Case 1 3 3₁₁ 7 722 R R e e_i_i

Example 1.2

Load Case 2 Load Case 2 B B F F C C 3 3₁₁ 7 722

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Ana

Analysis of Unb

lysis of Unbrace

raced

d Fram

Frame

e



 For unbracedFor unbraced frame, the greatest oframe, the greatest of the following f the following moments andmoments and shearing forces are to be taken for design purposes:

shearing forces are to be taken for design purposes:



Three cases loading arrangements as braced sub-frame (max =Three cases loading arrangements as braced sub-frame (max = 1.4Gk + 1.6Qk, min = 1.0Gk) 1.4Gk + 1.6Qk, min = 1.0Gk) 3 3₁₁ 7 722 R R e e_i_i

Ana

Analysis of Unb

lysis of Unbrace

raced

d Fram

Frame

e



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

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

B B F F C C 3 3₁₁ 7 722 R R

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Ana

Analysis of Unb

lysis of Unbrace

raced

d Fram

Frame

e

Analysis of Horizontal Load Using Portal Method Analysis of Horizontal Load Using Portal Method



 The following assumptions have to be made:The following assumptions have to be made:



Loads applied at beam-column junction.Loads applied at beam-column junction.



Total horizontal shear at any level is carried by columns at theTotal horizontal shear at any level is carried by columns at the points of

points of contraflexurecontraflexure immediately below immediately below the level.the level.



The points of The points of contraflexurecontraflexure occur at the occur at the mid-heights of columnsmid-heights of columns and at

and at midspamidspansns of beams.of beams.



Each bay acts as a separate portal and the horizontal load isEach bay acts as a separate portal and the horizontal load is divided between bays in proportion to span.

divided between bays in proportion to span.

3 3₁₁ 7 722 R R e e_i_i

Example 1.3



 Draw the bending moment diagram of the 5 storey building frameDraw the bending moment diagram of the 5 storey building frame

B B F F C C 3 3₁₁ 7 722

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Example 1.3

  SolutionSolution 3 3₁₁ 7 722 R R e e_i_i

Example 1.3

Analysis of horizontal load Analysis of horizontal load

Roof

Roof floor floor = = (1.2 (1.2 x x 3) 3) x x (3.5/2) (3.5/2) = = 6.30 6.30 kNkN 3

3rdrd_{and 4}_{and 4}thth _Floor_Floor _{= (1.2}₌_{(1.2 x}_{x 3)}_{3) x}_{x [(3.5/2)}_{[(3.5/2) +(3.5/2)]}_{+(3.5/2)] =}_{= 12.6}_{12.6 kN}_kN

B B F F C C 3 3₁₁ 7 722 R R e e

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Example 1.3 (Roof Floor)



 Axial Force in column:Axial Force in column:



 Shear force in beamShear force in beam

3 3₁₁ 7 722 R R e e_i_i

Example 1.3 (Roof Floor)



 Horizontal Force in Column:Horizontal Force in Column:

B B F F C C 3 3₁₁ 7 722

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Example 1.3 (Roof Floor)

3 3₇1 1 722

R R e e_i_i

Example 1.3 (4

th th

_Floor)

B B F F C C 3 3₁₁ 7 722 R R e e_i_i

n n f f o o

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Example 1.3 (4

th th

_Floor)



Σ ΣFF_y_y= 0= 0 Σ ΣFFyy= 0= 0 3 3₁₁ 7 722 R R e e_i_i

n n f f o o_rr c c e e d d C C o o n n c crr e e t t e e D D e e s si i g g_n_n 2 2 0 0 24 24 .. 3 3 65 65 .. 0 0 1 1++ −− == F F kN kN F F ₃₃_..₂₄₂₄ ₀₀_..₈₁₈₁ ₀₀_..₆₅₆₅ ₀₀_..₁₆₁₆ ₃₃_..₂₄₂₄ 2 2 == ++ −− −− == kN kN F F ₃₃_..₂₄₂₄ ₀₀_..₆₅₆₅ ₂₂_..₅₉₅₉ 1 1 == −− == 0 0 81 81 .. 0 0 24 24 .. 3 3 16 16 .. 0 0 65 65 .. 0 0 2 2++ ++ −− −− == F F

Example 1.3 (4

th th

_Floor)



 Horizontal force in columnHorizontal force in column

B B F F C C 3 3₁₁ 7 722

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Example 1.3 (4

th th

_Floor)

3 3₁₁ 7 722

R R e e_i_i

Example 1.3 (3

rdrd

_Floor)

B B F F C C 3 3₁₁ 7 722 R R e e_i_i

n n f f o o

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Example 1.3 (3

rdrd

Floor)



  ΣΣFF_y_y= 0= 0   ΣΣFF_y_y= 0= 0 3 3₁₁ 7 722 R R e e_i_i

n n f f o o_rr c c e e d d C C o o n n c crr e e t t e e D D e e s si i g g_n_n 2 2 0 0 44 44 .. 8 8 24 24 .. 3 3 1 1++ −− == F F kN kN F F ₈₈_..₄₄₄₄ ₂₂_..₁₁₁₁ ₃₃_..₂₄₂₄ ₀₀_..₈₁₈₁ ₆₆_..₅₅ 2 2 == ++ −− −− == kN kN F F ₈₈_..₄₄₄₄ ₃₃_..₂₄₂₄ ₅₅_..₂₀₂₀ 1 1== −− == 0 0 11 11 .. 2 2 44 44 .. 8 8 81 81 .. 0 0 24 24 .. 3 3 2 2 ++ ++ −− −− == F F

Example 1.3 (3

rdrd

Floor)



 Horizontal force in columnHorizontal force in column

B B F F C C 3 3₁₁ 7 722

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Example 1.3 (3

rdrd

Floor)

3 3₇1 1 722

R R e e_i_i

Example 1.3 (2

ndnd

Floor)

B B F F C C 3 3₁₁ 7 722 R R

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Example 1.3 (1

stst

Floor)

3 3₇1 1 722

R R e e_i_i

n n f f o o_rr c c e e d d C C o o n n c crr e e t t e e D D e e s si i g g_n_n 2 2 B B F F C C 3 3₁₁ 7 722

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Simplified Approach

  1.4Gk + 1.6Qk1.4Gk + 1.6Qk   1.4(25) + 1.6(10) = 51 kN/m1.4(25) + 1.6(10) = 51 kN/m 3 3₁₁ 7 722 R R e e_i_i

Simplified Approach

  1.2Gk + 1.2Qk1.2Gk + 1.2Qk   1.2(25) + 1.2(10) = 42 kN/m1.2(25) + 1.2(10) = 42 kN/m B B F F C C 3 3₁₁ 7 722 R R

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Result of Portal Analysis at 3

rdrd

_Floor

3 3₇1 1 722

R R e e_i_i

Vertical Load + Horizontal Load

B B F F C C 3 3₁₁ 7 722 1.2Gk + 1.2Qk 1.2Gk + 1.2Qk

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3 3₁₁ 7 722 R R e e_i_i