Distribution of Forces in Lateral Load Resisting · PDF fileDistribution of Forces in Lateral...

Distribution of Forces

in Lateral Load Resisting Systems

Many slides from 2009 Myanmar Slides of Profs Jain and Rai 1

Part 2. Horizontal Distributionand Torsion

IITGN Short CourseGregory MacRae

Reinforced Concrete Cast-in-Situ Slabs

• Moment of inertial for bending in its own plane

( Very large quantity!!)

• Practically, floor is infinitely stiff for bending

deformation in its own plane.

12

3tbI

• The slab is subject to horizontal load.

b

t

2Sudhir K Jain

Floor Diaphragm Action

Plan of a one-storey building

with shear walls

t = floor thickness; width of the beam representing floor diaphragm

b = floor width; depth of the beam representing floor diaphragm

L = span of the beam representing floor diaphragm

Springs represent lateral

stiffness walls / frames

k/2k k

b

L L

3Sudhir K Jain

Vertical load analogy for floor diaphragm action

Lateral earthquake force, EL

Beam representing

floor diaphragm

Ibeam = tb3/12

K K/2 K

4

Floor Diaphragm Action

Sudhir K Jain

In Plane Force Out of Plane Force

In Plane Deformation of

Floor

Out of Plane Deformation

of Floor5

In-plane versus out-of-plane deformation of floor

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Floor Deformations

In-Plane Floor Deformation Out of Plane Floor Deformation

6Sudhir K Jain

Rigid-body movements of a rigid floor diaphragm

Rotation about z-axis

Longitudinal

Translation

Translation in x-direction Translation in y-direction

Transverse

Translation

Angle of

rotation

Resultant Translation

Combination of translations and rotation

7Sudhir K Jain

Example 1: Effect of floor diaphragm action

Slab thickness = 150 mm

E = 25,500 N/mm2

k = 2300 103 N/mm

4123

104.612

8000150mmI diaphragm

• Actual Analysis 440 120 440

• Rigid Diaphragm 455 90 455

• Tributary Area 250 500 250

Force in Springs

k k0.2k

20 m 20 m

8m

1000 kNI

E I

1000 kN

k 0.2k k

8Sudhir K Jain

Rigid Diaphragm

Assumption is generally

used for RC floors

Example 2: Centre of Mass

• Given floor plan and lumped masses per unit area

• Locate centre of mass of the floor

Force in Springs

700 kg/m2

1000 kg/m2

8m

5 m

30 m

10 m

Force in Springs

A

X

CM (A)

B

CM (B)

C

CM

CM (C)

z

y

y

9

F

Sudhir K Jain

Example 2: Centre of Mass

• Locate centre of mass of segments A, B, C as:

– CM(A) = (4.0, 7.5); CM (B) = (4.0, 2.5);

– CM(C) = (19.0, 5.0)

• Calculate centre of mass of floor asyx,

mm

ymy

mm

xmx

i

ii

i

ii

1.5

4.14

10Sudhir K Jain

Centre of Stiffness (of a Single-storey Building)

• Point on the floor through which a lateral load should pass in

order to have only rigid body translation (i.e., no rigid body

rotation).

• Use the above definition to locate the centre of stiffness.

• Example:

k1.5k0.5k

30 m

10m

14 m

1.5kA

B

321

1.2k

11Sudhir K Jain

k ∆1.5k ∆0.5k ∆

F

∆

X

Force equilibrium:

Moment equilibrium:

kkkkF 35.15.0

mk

kx

kkkkxF

3.173

52

52305.1145.00.

Example 3. Centre of Stiffness

12Sudhir K Jain

Force equilibrium:

Moment equilibrium:

kkkF 7.22.15.1

mk

ky

kkyF

6.57.2

15

105.102.1.

∆

F 1.5k ∆

y 1.2k ∆

Example 3. Centre of Stiffness

13Sudhir K Jain

ELkkk

kFEL

kkk

kFEL

kkk

kF

321

33

321

22

321

11 ;;

Building Plan

k 2=0.3kWall stiffness

k 1=kk 3=k

EL

Wall stiffness

Definition of lateral stiffness

Fk

F

ELFFF

kFkFkF

321

332211 ,,

Lateral load distribution due to rigid floor diaphragm

(Symmetrical case – no torsion)

F3F2F1

EL

14Sudhir K Jain

k 1.2k0.5k

30 m

10m

9m

4

k

y

23

5

1

x

kNkkk

kF

kNkkk

kFkN

kkk

kF

9.882.15.0

2002.1

0.372.15.0

2005.0;1.74

2.15.0

200

3

21

Example 4: 200 kN applied along y-direction

• Locate centre of stiffness : (15m, 5m)

• Locate centre of mass : (15m, 5m)

• Hence, no torsion

• Wall 1, 2, 3 share load proportional to stiffness

15Sudhir K Jain

k The centre of stiffness

(CS) is at the centre

of the building. If the

centre of mass is also

here then the building

undergoes translation

but no torsion

Load at centre of mass = Load at centre of stiffness + Twisting

moment about the centre of stiffness

• CM

• CS

CM

ELe.EL

EL

ELCS

M = e.EL

ELCS

CM

Eccentric Systems

e

16Sudhir K Jain

r3

r2

r1

r5r1 r3

r5

r2

1

2

3

5

4

r4

ki = Lateral stiffness of the ith element

ri = Perpendicular distance of the ith element from centre of stiffness

= Rotation of the floor diaphragm in its own plane

Analysis of force induced by twisting moment (rigid floor diaphragm)

17Sudhir K Jain

CS

• Displacement of ith element, in its own plane,

due to rotation about centre stiffness

• Resisting force in ith element

• Restoring moment by force in ith element

• By moment equilibrium

• Force in the ith element

ii r

iii rkF

2

iiiii rkrFM

2

iit rkM

t

ii

iii M

rk

rkF

2

Analysis of force induced by twisting moment (rigid floor diaphragm)

18Sudhir K Jain

Example 5: Load distribution in eccentric system

k 1.5k0.5k

30 m

10m

14m

4

1.2k

y

2 3

5

1

x

1.5k

(a) (b)

17.33m

15m

5m5.56m

ex=2.33m

ey=0.56m

200kN 200kN

1 34

200kN

•CS•

CM5

F3

21 34

•

CM5

F1F2 F3

2

F4

F5

•CS466kNm

466kNm = 200kNx 2.33m

•CS

(c) Forces

(d)

F1F2

=• CM •CM

•

CM

•CS

Sudhir K JainTranslational Forces Torsional Forces

Walls CS and CM

Analysis for 200 kN force acting at centre of stiffness

This force is resisted by walls 1, 2, and 3 in proportion to their lateral stiffness. This gives:

Analysis for 466 kN-m moment acting on the diaphragm at CS:

The twisting moment of 466 kN-m is resisted by all the walls (including walls 4 and 5).

kNkkk

kF 7.66

5.15.0

2001

kNkkk

kF 3.33

5.15.0

2005.02

kNkkk

kF 100

5.15.0

2005.13


20Sudhir K Jain

1 k 17.33 17.33k 300.3k 13.2

2 0.5k 3.33 1.67k 5.5k 1.3

3 1.5k 12.67 19.00k 240.8k 14.4

4 1.5k 4.44 6.66k 29.6k 5.1

5 1.2k 5.56 6.67k 37.1k 5.1

t

ii

ii Mrk

rkF

21

k3.613

ikWallir iirk 2

iirk


Sudhir K Jain

The total force resisted by the walls (= translational force + torsional force):F1 = 66.7+13.2 = 79.9 kN

F2 = 33.3+1.3 = 34.6 kN

F3 = 100 – 14.4 = 85.6 kN

F4 = 5.1 kN = 5.1 kN

F5 = 5.1 kN = 5.1 KN


22Sudhir K Jain

1 34

•

CM5

F1F2 F3

2

F4

F5

•CS

Total Forces

Translational Forces Torsional Forces

1 34

200kN

•CS•

CM5

F3

21 34

•

CM5

F1F2 F3

2

F4

F5

•CS466kNm

+F1F2

(a) Without Torsion

All frames must follow through the same displacements at each level.

(b) With Torsion

All frame displacements at each level must be compatible with level translational and torsionaldisplacements.

Multistorey Frames

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F4

None of the floors undergoany rotation as forces passthrough the CS (i.e. CS =CM)

(CS)i

F3

(CS)i

F2

(CS)iF1

Multistorey Frames

Important:

• First calculate lateral load at different floors for the entire building

• Then distribute to different frames/walls as per floor diaphragm behavior

Do Not

• Calculate seismic design force directly for individual frames of the building

24Sudhir K Jain

Multistorey Frames

Plan of a building with space frame: this may be thought of as four 2-bay frames in the y-direction, and three 3-bay frames in the x-direction

A B C D1

2

3

Planx

y

25Sudhir K Jain

A

B

C

1 2 3

Plan

• Frames 1&3 same

• Frame spacing same

The requirement is:(a) Displacement in frames 1, 2 & 3 are equal at floor 1.

(b) Displacement in frames 1, 2 & 3 are equal at floor 2.

(c) Displacement in frames 1, 2 & 3 are equal at floor 3.

Design force in y-direction

on the entire building

1000

400

100

• Missing column

• Symmetric system

26Sudhir K Jain

Example 6: 3-storey symmetric building

36

52

91

9

8

710

11

12 15

14

13

18 21

2017

1916

29

23

28

Impose the conditions

This will ensure proper load distribution.

1613107

1714118

1815129

1000

400

100

27Sudhir K Jain

Frame A Frame B Frame C


Simple calculate the member forces

100

400

1000

Imaginary rigid

links

to ensure floor diaphragm action

Think of the translational problem as:

28Sudhir K Jain


2-D Frame with Rigid Lines

Direction of

Earthquake force

A B C D E

1

2

3

4

5(a)

Frame 1 Frame 2 Frame 3 Frame 4

Link bars

29Sudhir K Jain

Approximate Lateral Load Distribution

• Exact distribution requires computer analysis

• How do we carry out approximate hand calculations for

buildings up to 4 stories without torsion?

– Assume that all 2-D frames have same displacement profile

(shape only) for lateral loads

– Now match roof displacement only

– If assumption is exactly valid, analysis will still be exact

30Sudhir K Jain

Example 7. Approximate Distribution, No Torsion

A B C

PlanDesign force in y-direction

on the entire building

1000

400

100

301000

400

100

A

151000

400

100

B

301000

400

100

C

31Sudhir K Jain

unitsk

unitsk

unitsk

C

B

A

5030

1500

10015

1500

5030

1500

unitski 2005010050

fffk

kf

fffk

kf

fffk

kf

C

C

BB

AA

25.0200

50

50.0200

100

25.0200

50 500

200

50

Frame B

250

100

25

Building Frames A and C

1000

400

100

32Sudhir K Jain


Evaluate , , such that roof displacement is same

+ + = 1.0

A B C

Plan Entire Building

1000

400

100

. 1000

. 400

. 100

. 1000

. 400

. 100

. 1000

. 400

. 100

A B C

EQ

(No Torsion Case)

33Sudhir K Jain


• Illustration

– Parts of building in double height

– Symmetric1000

700

400

50

Portion in double

height Plan Elevation

1 2 3 4

1000

700

400

50

34Sudhir K Jain


Further Simplification

• For load distribution, relative lateral stiffness is

needed

1 2 31 1 1

1

1

1k

2

2

1k

3

3

1k

Relative terms only are required

35Sudhir K Jain

Approximate Lateral Load Stiffness of Frames

• Number of approximate methods, e.g.,

– McLeod’s Method

– Computer methods for analysing frames

• Caution

– Do not believe in storey stiffness as

– This assumes beams are infinitely rigid!

– Never happens

!!L

EI3

12

36Sudhir K Jain

Torsion in Multistory Buildings

• Centre of stiffness at different floors

• Number of definitions

• Depends on usage

• Implementation

37Sudhir K Jain

None of the floors undergoany rotation

(CS)i

F3

(CS)i

F2

(CS)iF1

F3

F2

F1

Design lateral load profile

Centre of Stiffness for multistory buildings

38Sudhir K Jain

Torsion in Multistory Buildings . . .

The requirement on design eccentricity can be fulfilled by applying

earthquake force away from centre of mass by a distance 0.5 times the

calculated eccentricity, such that eccentricity between centre of stiffness and

the load becomes 1.5 times the calculated.

CM = centre of mass

CS = centre of stiffness

1.5ex

ELy

.CM

.CS1.5ey

ELx.CM

.CS

ey.CM

.CS

ex

39Sudhir K Jain

Torsion in Multistory Buildings

jsj

jsjdj

be

bee

Q1

Q2

Q3

Q4

40

• Typical building code specifies design eccentricity in terms of

• Static eccentricity esj

• Accidental eccentricity bj

is typically 1.5

is typically 0.05 to 1.0

(5% to 10% of plan dimension bj)

Sudhir K Jain

jb

Torsion in Multistory Buildings . . .

41

Goel and Chopra

(ASCE, Vol.119; No:10)

Sudhir K Jain

CM CS

Can Conduct Analyses Directly Using Computer

Program with Rigid Diaphragm (e.g. ETABS)

• For buildings generally uniform with height

• Centre of stiffness for different floors on the same vertical line

• Treatment similar to that for single storey building

• Example: Earthquake force in X-direction

6m

3m

4m

4.5m 4.5m 4.5m 4.5m

333

287

141

48

All columns 400x400

Exterior Beams 250x600

Interior Beams 300x450

Approximation in Torsion Calculations

42Sudhir K Jain

Earthquake force in X-direction

6m

3m

4m

4.5m 4.5m 4.5m 4.5m

333

287

141

48

All columns 400x400

Exterior Beams 250x600

Interior Beams 300x450

43Sudhir K Jain

Example 9. Approx. Analysis Torsion

A

B

C

D

1 2 3 4 5 1 2 3 4 5

Eccentricity e = 6.5 – 6.1 = 0.4m

Design eccentricity = 1.5e = 0.6m (Dynamic eccentricity)

Design force profile V acting at CM

= Force profile V at CS

+ Twisting moment profile (Mt = 0.6m x V)

m.

,

,,,y

m/kN,kk

m/kN,kk

Pk

CB

DA

16

22053

1322016739010439010

39010

22016

P

44Sudhir K Jain


• Force profile V at CS

– Frames A, D =

– Frames B, C =

• Twisting moment profile Mt

V.V,

,

V.V,

,

195022053

39010

305022053

22016

t

jj

ii Mrk

rk2

45Sudhir K Jain


Frame

(x 103)

(kN/m)

ri

(m)(x 103)

(kN)(x 103)

Fd

(torsion)

Fd

(Direct)

FTO

(Total)

A 16.22 6.9 111.92 772.23 0.0293 0.0170V 0.304V 0.321V

B 10.39 0.9 9.35 8.42 0.0024 0.0014V 0.195V 0.196V

C 10.39 -2.1 -21.82 45.82 -0.0057 -0.0034V 0.195V 0.195V

D 16.22 -6.1 -98.94 603.55 -0.0259 -0.0156V 0.304V 0.288V

1 12.70 -9.0 -114.30 1028.70 -0.0299 -0.0180V . . . . . . . .

2 8.18 -4.5 -36.80 165.65 -0.0096 -0.0057V . . . . . . . .

4 8.18 4.5 36.80 165.65 0.0096 0.0057V . . . . . . . .

5 12.70 9.0 114.30 1028.70 0.0299 0.0018V . . . . . . . .

2

ii

ii

rk

rk2

ii rkiirkik

46Sudhir K Jain


From frame

analysis with

point load at

roof

Dist from

CS

Level

Total Level

Design Force

(kN)

Design Force For Frame

A

(kN)

B

(kN)

C

(kN)

D

(kN)

4 333 106.90 65.27 64.94 101.20

3 287 92.13 56.25 55.97 87.25

2 141 45.26 27.64 27.50 42.86

1 48 15.40 9.41 9.36 14.60

47Sudhir K Jain


From beginning

of example

Thank you!!

48

Distribution of Forces in Lateral Load Resisting · PDF fileDistribution of Forces in Lateral...

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