Frame & Subframe Analysis for Vertical and Lateral Load

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FRAME & SUBFRAME

ANA

L

YSIS

FOR V

ERTICAL &

LATERAL LOAD

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LEARNING OUTCOME :

1

1..00 ccllaassssiiffyy bracedbraced andand un-bracedun-braced frame.frame. 2.

2.00 awawarare e on on ththe e ananalalysysis is of of brbracaced ed coconcncrerete te frframameded building for vertical load.

building for vertical load. 3

3.0.0 ananalalyyzze e uun-n-brbracaceedd ccooncncrerette e frframamed ed bubuiildldiinng fg foorr wind load by

wind load by cantilever method.cantilever method.

By completing this chapter, students shall be able to: By completing this chapter, students shall be able to:

Continuous Beam Design & Detailing

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Introduction to Frame Structure



_{Building frames}

often consist of

girders/beams

girders/beams that

that

are rigidly

connected to

columns.



_{Frames are}

classified as

braced

or

u

n-

n-braced

braced

against

sidesway.

Beam (300x600mm) Beam (300x600mm) Slab 200mm, 250mm & Slab 200mm, 250mm & 300mm thickness 300mm thickness

Varies column size Varies column size (500x500mm, (500x500mm, 750x750mm, 750x750mm, 900x900mm) 900x900mm) Figure 1: 3D Modeling of Figure 1: 3D Modeling of

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Un-braced frames



 _{Sway frames/Sideways}_{Sway frames/Sideways}

uninhibited. uninhibited.



 _{Framed need to be designed}_{Framed need to be designed}

to resist the lateral loads, to resist the lateral loads, since there is no bracing since there is no bracing elements in the building. elements in the building.



 _{Un-braced frame need to}_{Un-braced frame need to}

carry both vertical &

lateral loads.

Continuous Beam Design & Detailing_–_–Frame analysisFrame analysis

Figure b: Rigid Frame/ Un-braced Frame Figure b: Rigid Frame/ Un-braced Frame

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Beam Beam (horizontal (horizontal element) element) Vertical loading Vertical loading Column Column (vertical (vertical element) element) Lateral/Horizontal Lateral/Horizontal loading loading

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Braced

_{Braced frames}



 _{Non-sway frames/Sideways}_{Non-sway frames/Sideways}

inhibited. inhibited.



 _{No lateral/horizontal loads (eg. wind}_{No lateral/horizontal loads (eg. wind}

load) being transferred either to load) being transferred either to columns or beams.

columns or beams.



 _{Diagonal bracing, shear-walls,}_{Diagonal bracing, shear-walls,}

masonry infill walls, lift/elevator masonry infill walls, lift/elevator shafts and staircases core provide shafts and staircases core provide lateral stability to the structural lateral stability to the structural frame of the building.

frame of the building.



 _{Braced frame could take only}_{Braced frame could take only}

vertical loads.

.

Figure a: Braced Frame Figure a: Braced Frame

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Beam (horizontal element) Beam (horizontal element) Column (vertical Column (vertical element) element) Vertical load Vertical load Bracing type X Bracing type X

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

_{Infilled with brickworks, blockwork or}_{Infilled with brickworks, blockwork or}

precast panels. precast panels.



_{It’s subjected to lateral load the infill}_{It’s subjected to lateral load the infill}

behaves as strut along its

behaves as strut along its compressioncompression diagonal to brace the frame.

diagonal to brace the frame.



_{Forces and stresses in the infill and frame.}_{Forces and stresses in the infill and frame.} 

_{Design procedure}_{Design procedure –}_{– design of infill, design of}_{design of infill, design of}

frame, horizontal deflection frame, horizontal deflection

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Figure d: Shear Wall Structure Figure d: Shear Wall Structure



_{Shear wall structures are}_{Shear wall structures are entirely}_entirely

lateral load resistant and

lateral load resistant and much stiffermuch stiffer horizontall

horizontally than y than rigid frame.rigid frame.



_{Restrict openings in walls.}_{Restrict openings in walls.} 

_{Shear wall structure performs well in}_{Shear wall structure performs well in}

earthquake because ductility is important earthquake because ductility is important in design of shear wall.

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Figure e: Wall-frame Structure Figure e: Wall-frame Structure



_{Combination of shear walls and rigid}_{Combination of shear walls and rigid frames.}_frames. 

_{Usually use in reinforced concrete structure,}_{Usually use in reinforced concrete structure,}

however use of steel in bracing and rigid frames however use of steel in bracing and rigid frames offer advantage in horizontal interaction.

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Figure

Figure g: Framg: Frame -e - tubetube



_{Lateral resistance provided by very}_{Lateral resistance provided by very}

stiff moment resisting frames that form stiff moment resisting frames that form tube around the perimeter of the

tube around the perimeter of the building.

building.



_{Consists of very closely spaced}_{Consists of very closely spaced}

columns (2-4 m apart) joined by deep columns (2-4 m apart) joined by deep spandrel girders.

spandrel girders.



_{Suitable in reinforced concrete or}_{Suitable in reinforced concrete or}

steel construction with building from steel construction with building from 40-100 stories high.

40-100 stories high.



_{Repetitive frame pattern, grid like}_{Repetitive frame pattern, grid like}

façade

façade – – relatively efficient, easyrelatively efficient, easy construction.

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Figure a: Empire State Figure a: Empire State Building

Building

Figure b: Citicorp Building, Figure b: Citicorp Building, New York City

New York City

Figure c: Mercantile Building, St Louis Figure c: Mercantile Building, St Louis

Figure d: Hearst Tower, New York City Figure d: Hearst Tower, New York City

Figure 4(a-d): Braced Frame Examples Figure 4(a-d): Braced Frame Examples

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Lateral Stability in Frame Structure

Figure 5: Elements in Frame Structure

Figure 5: Elements in Frame Structure Figure 6: Frame structure Figure 6: Frame structure provides lateral stabilityprovides lateral stability (Plan View)

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A A B B C C 1 1 22 33 44 column

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B B C C 1 1 22 33 44 SHEAR WALL SHEAR WALL Column

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A A B B C C 1 1 22 33 44 SHEAR WALL SHEAR WALL column

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Approximate Method of

Frame Analys

is

Figure 8

Figure 8a: a: Actual Actual 5-storeys Frame5-storeys Frame Structure

Structure

Figure 8

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A.

A. Monolithic Monolithic Braced Braced Frames Frames Not Not Providing Providing Lateral Lateral StabilityStability

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Simplificati

on

into

sub-frame

The structural frame is divided into sub-fra

The structural frame is divided into sub-frames consisting ofmes consisting of beams at

beams at one levelone level and columns and columns above and above and below that below that level withlevel with ends taken as fixed (or pinned).

ends taken as fixed (or pinned).

A

A BB CC DD

Approximate Method of

Frame Analys

is

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A.

A. Monolithic Monolithic Braced Braced FramesFrames –– con’t con’t

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Simplificati

on for

individual beams & columns

The simplified sub-frame consists of

The simplified sub-frame consists of the beamthe beam to be designed,to be designed, the

the columns attachedcolumns attached to thto the ene ends of ds of the the beambeam andand the beamsthe beams on either side

on either side if any.if any.

The column and beam ends r

The column and beam ends remote from the beam consideredemote from the beam considered are taken as fixed and

are taken as fixed and thethe stiffness of the beams on either sidestiffness of the beams on either side

should be taken as

half

their actual sizetheir actual size..

The moments for design for an individual column may be found The moments for design for an individual column may be found from the same sub-frame analysis provided that its central from the same sub-frame analysis provided that its central beam is the longer of the two beams framing into t

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A.

(2)

(2) Simplification for individual beams & columns Simplification for individual beams & columns

Beam A-B Beam A-B K K KK//22 Beam B-C Beam B-C K/2 K/2 _K_K _K_K//2₂ Beam C-D Beam C-D A A BB CC DD K/2 K/2 KK : :

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A.

A. Monolithic Monolithic Braced Braced FramesFrames –– con’t con’t (3) Continuous Beam Simplification

(3) Continuous Beam Simplification

The beam at the floor considered may be taken as a

The beam at the floor considered may be taken as a continuouscontinuous beam over

beam over supports providing supports providing no restraint no restraint to rotation. to rotation. ThisThis gives more cons

gives more conservatiervativeve desigdesign than the proceduren than the procedures previousls previously.y. (4)Asymmetrically-loadedColumns

(4)Asymmetrically-loadedColumns

This method is to be us

This method is to be used where the beam has been ed where the beam has been analyzed.analyzed. The column moments can be calculated on the assumpti

The column moments can be calculated on the assumption:on:

-- colcolumn aumn and beand beam endm ends remos remote frote from the jm the juncunctiotion undn underer consideration are fixed.

consideration are fixed.

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A.

(3)

(3) Continuous Beam Simplification Continuous Beam Simplification & Asymmetrically-loaded& Asymmetrically-loaded

Columns Columns A A BB CC DD A A K/2 K/2 B&C B&C K/2 K/2 K/2 K/2 K/2 K/2

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A.

A. Monolithic Monolithic Braced Braced FramesFrames –– con’t con’t Choice of Critical Loading

Choice of Critical Loading Arrangement Arrangement

It will normally be sufficient to con

It will normally be sufficient to consider the following arrangement ofsider the following arrangement of vertical load:

vertical load: (a

(a)) AlAll spl span an loloadaded wed witith thh thee maximum design ultimate loadmaximum design ultimate load (1.35Gk + 1.5Qk).(1.35Gk + 1.5Qk). (a)

(a) AltAlternernatiative ve spaspans lns loadeoaded witd with thh thee maximummaximum design ultimate load (1.35Gkdesign ultimate load (1.35Gk + 1.5Qk) and all other spans loaded with the

+ 1.5Qk) and all other spans loaded with the minimumminimum design ultimate loaddesign ultimate load (1.0Gk).

(1.0Gk).

max

max _max_max _max_max

max

max _max_max

max max min min min

min _min_min

(a) (a)

(b) (b)

3 cases of load arrangement in Frame analysis for

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Approximate Method of

Frame Analys

is

B. Rigid Un

B. Rigid Un--braced Frames Providing Lateral Stabilitybraced Frames Providing Lateral Stability

Where rigid frames provides Where rigid frames provides lateral stability, they must be lateral stability, they must be analyzed for

analyzed for

horizontal &

vertical loads

..

As an alternative to the As an alternative to the

complete 3D frame structure complete 3D frame structure analysis, the Code gives the analysis, the Code gives the following method for sway following method for sway frames of three or more frames of three or more approximately equal bays. approximately equal bays. Continuous Beam Design & Detailing

W W i i nn d d L L o o a a d d Gravity Load Gravity Load

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Approximate Method of

Frame Analys

is

B. Rigid Un

B. Rigid Un--braced Framesbraced Frames -

-The design is based on the

The design is based on the

more severe

of the conditions:of the conditions: 1.

1. elastic elastic analysis analysis oror vertical loadsvertical loads only with maximum design loadonly with maximum design load 1.35Gk + 1.5Qk

1.35Gk + 1.5Qk 2.

2. or or thethe sum of the momentssum of the moments obtained from:obtained from: (a)

(a) elastic elastic analysis analysis ofof subframessubframes as define in section with allas define in section with all the beams loaded with

the beams loaded with 1.21.2Gk + 1.2 QGk + 1.2 Qkk (Ho(Horizrizontontal loal loads aads arere

ignored).

(b)

(b) elastic analelastic analysis of the ysis of the complete frame complete frame assuming points assuming points ofof contra flexure at the centers of all beams and columns for wind contra flexure at the centers of all beams and columns for wind load

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Un

-

braced Frame

Analysis by

Cantilever Method

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Lateral Loads on Building Frames:

Cantilever Method

Introduction

11.. TThhee cantilever methodcantilever method is based on the same action asis based on the same action as

long

cantilever beam

subjected to a transverse load.subjected to a transverse load. 2

2.. TThhee lateral loadslateral loads on frame tend to tip up the frame over, oron frame tend to tip up the frame over, or cause a

cause a

rotation

of the frame about neutral axisof the frame about neutral axis..

(Note:

(Note: the neutral the neutral axis is in axis is in horizontal plane thorizontal plane that passes thhat passes through therough the

columns at each floor level.)

3.

3. To cTo counounterateract tct this his tiptippinping, tg, the ahe axiaxial fol forcerces or s or strestress in ss in thethe columns will be

columns will be TENSILETENSILE on one side and COMPRESSIVEon one side and COMPRESSIVE onon the other side of the neutral axis.

the other side of the neutral axis. 4.

4. The The cancantiltileveever mer methothod is d is thetherefrefore aore approppropripriate iate if thf the fre frame ame isis tall & slender

tall & slender, or has columns with, or has columns with different cross-sectionaldifferent cross-sectional

area.

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Lateral Loads on Building Frames:

Cantilever Method

Neutral axis Neutral axis

Assumption in Cantilever Method for

Fixed-Supported Frame

11.. AA

hinge

is placed at the center of eachis placed at the center of each girders & columns (zero moment).

girders & columns (zero moment).

2.

2. ThThe ae axixiaal sl strtreess ss in in a ca cololumumn in iss proportional to its

proportional to its distance from thedistance from the centroid

centroid of the cross-sectional areas ofof the cross-sectional areas of the columns at a given floor level.

the columns at a given floor level.

T T e e n n s s _i_i l l e e C C o o m m p p r r e e s s s s _i_i v v e e Z Z Z Z c c o o_n_n t t cont cont Note: o

Note: ohingehinge

(zero M) (zero M) Continuous Beam Design & Detailing

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Lateral Loads on Building Frames:

Cantilever Method

Procedures in Frame Analysis by

Procedures in Frame Analysis by Cantilever Method:Cantilever Method:

11.. PPllaacce e aa

hinge

at the center of each girders & columns .at the center of each girders & columns . 2

2.. DDeetteerrmmiinne e tthhee

centroid

of the columns’ cross sectional areas by :of the columns’ cross sectional areas by :

3.

3. AnaAnalyzlyze e a a secsectiotion n throthrough ugh the the hinhinge ge at at the the top top stostory. ry. DetDetermermineine the

the

axial force in each column

. . (Tips: (Tips: the the force force isis proportional to its distance from the centroid

proportional to its distance from the centroid of the columns’ crossof the columns’ cross--sectional areas.)

sectional areas.) 4

4.. DDeetteerrmmiinne e tthhee

remaining hinge forces

(V&N).(V&N). 5.

5. ReRepepeat at ststep ep 3-3-4 f4 for or lolower wer lelevevel ol of tf the he frframame.e. 6.

6. DrDraw aw the the bebendndining mg momomenent dt diaiagragram om of tf the he frframame.e. _ _ _ _ _x_{x A}_A x x A A     

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Example 1:

Determine (approximately)

Determine (approximately) thethe reactions at the basereactions at the base of the columns ofof the columns of the frame sh

the frame shown in Figure own in Figure below. below. The columns The columns are assumed to are assumed to havehave equal

equal cross-sectional cross-sectional areas. areas. Use Use thethe cantilever methodcantilever method of analysis.of analysis. 30kN 30kN 15kN 15kN 6m 6m 4m 4m 4m 4m A A B B C C D D E E F F

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Example 1 (cont)

(i).

(i). Placed hPlaced hinge at inge at the midpoint the midpoint of columns of columns and girdersand girders

& determine the centroid of the columns’ cross sectional areas. & determine the centroid of the columns’ cross sectional areas.

30kN 30kN 15kN 15kN 6m 6m 4m 4m 4m 4m G G H H I I L L K K J J A A B B C C D D E E F F _ _ _ _ ₀₀₍_{( )}_{) 6}₆₍_{( ))} 3 3 2 2 x x AA _A_A _A_A x x mm A A AA           6m 6m Centroid, Centroid, x =3m x =3m Plan View Plan View

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Example 1 (cont)

(ii).

(ii). Analyze Analyze level level at at top top storey.storey.

2m 2m 3 3mm 33mm 3P 3P = 10kN= 10kN 3P3P = 10kN= 10kN O O a. a. Determine Determine PP 10 10 0 0 3 3 ((33) ) 3 3 ((33) ) 3300((22) ) 00 3 3 o o M M      P P   P P       P P    b.

b. Determine Determine remaining hinge remaining hinge forcesforces

I I_y_y I I_x_x H H_x_x 0 0 1010 y y yy F F     I I  kN kN   0 0 1100((33) ) 2 2 0 0 1155 I I x x xx M M __{ }_{ }_ __H H __{ }_{ }H H _ kN kN



+ + + + 0 0 1515 F F     I I  kN kN   10kN 10kN 15kN 15kN K K_x_x 30kN 30kN 15kN 15kN 6m 6m 4m 4m 4m 4m G G H H I I L L K K J J A A B B C C D D E E F F

kN kN K K F F _x_x __₀₀__ _x_x __₁₅₁₅



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(ii).

(ii). Analyze Analyze level level BE.BE.

a.

a. Determine Determine PP

b.

b. Determine Determine remaining hinge remaining hinge forcesforces

+ + 30kN 30kN 15kN 15kN 6m 6m 4m 4m 4m 4m G G H H I I L L K K J J A A B B C C D D E E F F 3 3mm 33mm O O 2m2m 4m 4m 3P 3P = 35kN= 35kN 3P 3P = 35kN= 35kN 3 3 35 35 0 0 )) 2 2 (( 15 15 )) 6 6 (( 30 30 )) 3 3 (( 3 3 )) 3 3 (( 3 3 0 0          



M M P P P P P P o o G G_x_x J J_x_x J J_y_y 15kN 15kN 10kN 10kN + + kN kN J J F

F _y_y 00 _y_y 2525



kN kN G G G G M M x x x x B B 5 5 .. 22 22 0 0 )) 3 3 (( 25 25 )) 2 2 (( )) 2 2 (( 15 15 0 0              



0 0 77..55 x x xx F F __{ }_J J __ kN kN



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b.

b. Determine Determine remaining hinge remaining hinge forces forces (cont)(cont)

kN kN L L F F x x x x  00  2222..55



Example 1 (cont)

10kN 10kN L L_x_x 7.5kN 7.5kN 25kN 25kN 35kN 35kN 15kN 15kN (iii).

(iii). Analyze Analyze support support level.level.

22.5kN 22.5kN 35kN 35kN 22.5kN 22.5kN 35kN 35kN 45kNm 45kNm 22.5kN 22.5kN 35kN 35kN 22.5kN 22.5kN 35kN 35kN 45kNm 45kNm 30kN 30kN 15kN 15kN 6m 6m 4m 4m 4m 4m G G H H I I L L K K J J A A B B C C D D E E F F

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(iv).

(iv). Shear Shear (kN) (kN) in in Beams Beams and and ColumnsColumns

6m 6m 4m 4m 4m 4m 10 10 25 25 15 15 22.5 22.5 15 15 22.5 22.5

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To calculate moment at:

Beam , M = F x beam span x 0.5

Column, M = H x storey height x 0.5

Examples:

Beam ,

Beam , M =

M = 10KN

10KN x 6

x 6 m x

m x 0.5 =

0.5 = 30KN

30KN

Column, M = 22.5KN x 4m x0.5 = 45KN

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(iv).

(iv). BMD (kBMD (kNm) in Nm) in Building FraBuilding Frame that me that subjected subjected to lateral to lateral loadload

6m 6m 4m 4m 4m 4m 10 10 25 25 15 15 22.5 22.5 15 15 22.5 22.5 45 45 ₄₅₄₅ 45 45 45 45 30 30 30 30 30 30 ₃₀₃₀ 75 75 75 75 30 30 30 30

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Exercise 1

A A E E II B B F F J J C C G G K K D D H H L L 10kN 10kN 20kN 20kN 4m 4m 4m 4m 4m 4m 4m 4m 5m 5m A=24(10

A=24(10-3-3_)m_)m22 _A=16(10_A=16(10-3-3_)m_)m22 _A=16(10_A=16(10-3-3_)m_)m22 _A=24(10_A=24(10-3-3_)m_)m22 Area Column:

Area Column:

Problem: Problem:

Draw the moment diagram for girder

Draw the moment diagram for girder IJKLIJKL of the of the building frame. building frame. Use the Use the cantilevercantilever method o

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A A _B_B _C_C _D_D E E F F GG HH II J J K K LL M M NN OO PP 20 kN/m 20 kN/m 3 3....22mm 44mm 77....55mm 3.5 m 3.5 m 3.5 m 3.5 m 4.0 m 4.0 m

The concrete building frame shown in Figure below is subjected to

The concrete building frame shown in Figure below is subjected to an ultimatean ultimate

uniformly distributed horizontal load of

uniformly distributed horizontal load of 20 kN/m. All columns have typical20 kN/m. All columns have typical

dimensions, each of 250mm x 200mm. By using cantilever method, calculate the dimensions, each of 250mm x 200mm. By using cantilever method, calculate the bending moment in beam BC.