Chapter 4 – Lateral Force
Resisting Systems
Dr.-Ing. Girma Zerayohannes
Dr.-Ing. Adil Zekaria
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z
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Chapter 4- Lateral Force Resisting
Systems (LFRS)
• 4.1 Introduction
• All structures from the simplest to most
complex must be provided with suitable LFRS
• Simple structures such as isolated elevated
water tanks, sign boards, simple ware houses,
etc.
• More complex structures buildings, bridges,
waterfront structures, ships, etc.
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z
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Chapter 4- Lateral Force Resisting
Systems
• A cantilever column or a pin supported column
with lateral restraint at ground level play the role
of LFRS for the similar structures
• Elaborate LFRS consisting of frames, walls,
combinations of frames and walls, and other
more complex systems are required for the more
complex structures
• In the latter the vertical elements are rigidly
connected with horizontal diaphragms enabling
them to act in unison.
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Chapter 4- Lateral Force Resisting
Systems
• The most commonly used structural systems
are:
• (i) Wall systems
• (ii) Frame systems
• (iii) Mixed wall-frame system
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Chapter 4- Lateral Force Resisting
Systems
•
•
•
•
4.2Wall System
4.2.1 Stable arrangement of walls
(i) There must be at least 3 walls
(ii) The axes of the walls should not intersect
at a point
• (iii) All 3 walls should not be parallel
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Chapter 4- Lateral Force Resisting
Systems
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Chapter 4- Lateral Force Resisting
Systems
• 4.2.2 Avoid high torsion
• Note: eccentric arrangement of wall is the
most frequent cause of collapse during EQ
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Chapter 4- Lateral Force Resisting
Systems
• 4.2.3 Distribution of story shear among the
walls
• 4.2.3.1 Statically determinate wall system
• Note: the story shear and the forces in the
walls are statically equivalent
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Chapter 4- Lateral Force Resisting
Systems
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Chapter 4- Lateral Force Resisting
Systems
• Note: the story shear and the forces in the
walls are statically equivalent
•  V2x = 100 kN
•  V1y+ V3y = 0
•  Torsion exerted by the story shear Vx
•  V2x(5) + V3y(8) – V1y(10) = 0
•  V3y= -500/18 and V1y= 500/18
•  V3y= -27.78 kN and V1y= 27.78 kN
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Chapter 4- Lateral Force Resisting
Systems
• To reduce the magnitudes of the forces,
• (i) reduce the magnitude of the torsional
moment 500 kNm by reducing the distance
b/n the story shear Vx and and the center of
stiffness S that lies on wall axis of Wall 2 and
• (ii) Increasing the lever arm b/n walls 1 and 3,
placing them as far apart from each other as
possible ,i.e., at the periphery
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Chapter 4- Lateral Force Resisting
Systems
• 4.2.3.2 Statically indeterminate wall system
• More than three walls
• Additional compatibility conditions are to be
considered to determine all shear wall forces
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Chapter 4- Lateral Force Resisting
Systems
•
•
•
•
•
Determination of the center of stiffness
In the following:
Iix = Moment of inertia of wall i w.r.t x-Axis
Iiy = Moment of inertia of wall i w.r.t y-Axis
xi , yi = Distance of shear center of wall i from
origin of chosen coordinate system
• xi , yi = Distance of shear center of wall i from
the center of stiffness
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Chapter 4- Lateral Force Resisting
Systems
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Chapter 4- Lateral Force Resisting
Systems
• Goal is to:
• (i) determine the center of stiffness
• (ii) distribute the horizontal force passing through
M
• A shear force Vx through the center of stiffness S
results only in translation in the x-direction and
no rotation.
• This means the same amount of deflection for all
walls connected with each other by means of the
diaphragm
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Chapter 4- Lateral Force Resisting
Systems
•  Vx is distributed according to their stiffness
(rigidity)  according to the moment of
inertias w.r.t the y-axis
V1x 
Vx I1 y
I
iy
; V2 x 
Vx I 2 y
I
; etc
iy
• Note that the resultant of the distributed
forces is equal to Vx and passes through S.
•  S can be determined by determining Vx
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Chapter 4- Lateral Force Resisting
Systems
• Line of action of Vx is yS away from E and given
by:
ys
V ( y ) V ( y )



V
V
ix
i
ix
ix
i
x
• Substitution for Vix in terms of Vx from above
and factoring the constants VV and

simplifying yields:
x
ix
ys
I (y )


I
iy
i
iy
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Chapter 4- Lateral Force Resisting
Systems
• Similarly from consideration of story shear Vy
in the y-direction
xs
I (x )


I
ix
i
ix
• As an example determine the center of
stiffness of the statically indeterminate wall
system shown in the previous slide
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Chapter 4- Lateral Force Resisting
Systems
•
•
•
•
•
Solution: Let t = wall thickness
I1y = I6x = t(2a)3/12 ;
I2x = I3x = I4y = I5y = t(a)3/12,
I1x = I2y = I3y = I4x = I5x = I6  0
 xS = 2.0a ; yS = 1.2a (check as an
assignment)
• It is shown as S in the floor plan
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Chapter 4- Lateral Force Resisting
Systems
• Story shear distribution among the walls
• Step 1: External horizontal force H acts
generally eccentric to assumed origin of axis
(E). In the case of EQ it passes through the
mass center
• Step2 : Story shear determined and made to
pass through the origin of chosen axis (E).
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Chapter 4- Lateral Force Resisting
Systems
• The eccentricity of the external loading H,
shown as yH causes torsion
• Step 3: The statically equivalent actions Vx and
Tsx = Tex+ Vx yS are made to act at the center
of stiffness. (Note that only in the upper most
story is, Vx = H)
• The story shear Vx at the center of stiffness
result in a uniform translation of the in plane
rigid slabs in the x-direction
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Chapter 4- Lateral Force Resisting
Systems
• The story shear is distributed according of
their moments of inertias as discussed before
• The torsional moment TS results in rotation of
the slabs about the center of stiffness. It will
be absorbed by all the walls. (Observe the role
of the diaphragms in distribution the loads to
the walls. W/o the slabs this is not possible)
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Chapter 4- Lateral Force Resisting
Systems
Ts  (V1x  y1  V2 x  y 2  )  (V1 y  x1  V2 y  x 2  )
• Some of these terms are negligible because of
negligible bending stiffness and V1x ,V2x ,…,V1y
, V2y , … are a result of torsion Ts
• Observe that, the deflection components of
the walls are proportional to y1 , y2 ,  , x1 , x2 , 
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Chapter 4- Lateral Force Resisting
Systems
• Thus the shear forces V1x ,V2x ,…,V1y ,V2y , …
in the walls as a result of torsional moment TS
are proportional to the moment of inertia and
the lever arm and therefore their product
•  V1x  k  I1 y  y1   ;V1 y  k  I1x  x1  
• Where k = the proportionality constant that
has to be determined so that we can
determine the wall forces resulting from
torsion
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Chapter 4- Lateral Force Resisting
Systems
• Invoking statical equivalency between story
torsion TS and and the sum of the torsional
moments exerted by the wall forces w.r.t. the
center of stiffness S:
Ts  k I1 y y12  k I 2 y y22    k I1x x12  k I 2 x x22  

Ts  k   I iy y  I ix x
• 
k
 I
2
i
2
i
Ts
iy
yi2  I ix xi2


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Chapter 4- Lateral Force Resisting
Systems
• Substituting the vale of k in wall force eqns as
a result of torsion above:
   I
 T  I x   I
V1x  Ts  I1 y y1
V1 y
s
1x
1

x , 
iy
yi2  I ix xi2 , 
iy
yi2  I ix
2
i
• Note that V1y  0 for wall 1
• Thus the total force in wall i resulting from
Vx and TS will be:
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Chapter 4- Lateral Force Resisting
Systems
V1x 
V1 y 
Vx I iy
I

iy
V y I ix
 I ix


Ts  I1 y y1
 I
iy
yi2  I ix xi2

Ts  I1x x1
 I
iy


y  I ix x
2
i
2
i

;

• Note that the 1st term in the expression for Viy
is  0 because Vy = 0
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Chapter 4- Lateral Force Resisting
Systems
• Example I: For the Statically indeterminate
walls distribute the story shear if:
• Hx = 100kN; yH = 0.2a; a = 6.0m; t = 0.2m, and
the floor is the upper most floor in a building
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Chapter 4- Lateral Force Resisting
Systems
• Example 2:
• Distribute the story shear in the ground floor
of a ten story building with plan and system of
walls similar to the statically indeterminate
example, if the lateral forces in the x- direction
are as shown in the Table below.
• Determine the external forces acting on wall 1
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Chapter 4- Lateral Force Resisting
Systems
Story
Force (kN)
Center of mass
10
1000
-0.2a
9
900
-0.3a
8
800
0.2a
7
700
0.1a
6
600
-0.2a
5
500
-0.3a
4
400
-0.3a
3
300
-0.4a
2
200
-0.1a
1
100
0.1a
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Chapter 4- Lateral Force Resisting
Systems
• 4.3 Frame system
• Regarding stable arrangement and avoiding high
torsion  same as in walls
• Disadvantage  frames are flexible  not
suitable for medium high to high rise buildings if
used alone
• Example of unstable LFRS in the form of only 2
frames whose axes are parallel to each other and
that was actually constructed in Addis collapsed
completely!!
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Chapter 4- Lateral Force Resisting
Systems
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Chapter 4- Lateral Force Resisting
Systems
• 4.3.1 Lateral force distribution between the
frames
• Hand calculation using what are known as the
D-values of columns were common practice
• Same was also instructed in structural design
courses
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Chapter 4- Lateral Force Resisting
Systems
• While the instruction helps add insight about
the response of frames under lateral loads,
the procedure is rather involved and also
outdated and serves no practical purpose in
present day design offices
• The reason is the ease with which 3-D frames
are modeled and analyzed with modern day
software and computers for all kinds of load
combinations
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Chapter 4- Lateral Force Resisting
Systems
• Thus we will look at a simple 3-D frame
example under lateral loading and use the
results to answer some questions such as:
• Is the stiffness of the LFRS equal to the sum of
the stiffnesses of the individual frames in each
direction?
• Are the rigidities of outer columns less than
those the interior columns and thus take less
share of the story shear?
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Chapter 4- Lateral Force Resisting
Systems
• Is the sum of the shear forces equal to the
story shear?
• How does eccentricity affect the distribution
of the story shear?
• How do we account for accidental
eccentricity?
• Can you show an individual frame with its
share of externally applied lateral load?
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Chapter 4- Lateral Force Resisting
Systems
• Project the 3-D frame analysis
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Chapter 4- Lateral Force Resisting
Systems
• 4.3 Mixed wall-frame system (Dual system)
• Response under lateral load is not anymore
that of a cantilever wall or frame because of
the interaction b/n the two.
• Reliable solutions can be found by modeling
the building as a plane structure consisting of
the frames and the walls connected by rigid
links
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• Deformation pattern of a dual system
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• Typical deflection, moment and shear diagrams
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Chapter 4- Lateral Force Resisting
Systems
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Chapter 4- Lateral Force Resisting
Systems
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Chapter 4- Lateral Force Resisting
Systems
• See analysis model
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