1. Chapter 4 – Lateral Force Resisting Systems
Dr.-Ing. Girma Zerayohannes
Dr.-Ing. Adil Zekaria
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

2. 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

3. 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.
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

4. Chapter 4- Lateral Force Resisting Systems
The most commonly used structural systems are:
(i) Wall systems
(ii) Frame systems
(iii) Mixed wall-frame system
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

5. 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
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

6. Chapter 4- Lateral Force Resisting Systems
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7. 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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8. Chapter 4- Lateral Force Resisting Systems
4.2.3 Distribution of story shear among the walls
Statically determinate wall system
Note: the story shear and the forces in the walls are statically equivalent
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

9. Chapter 4- Lateral Force Resisting Systems
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

10. 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= kN and V1y= kN
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

11. 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
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

12. Chapter 4- Lateral Force Resisting Systems
Statically indeterminate wall system
More than three walls
Additional compatibility conditions are to be considered to determine all shear wall forces
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

13. 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
= Distance of shear center of wall i from the center of stiffness
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

14. Chapter 4- Lateral Force Resisting Systems
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15. 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
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

16. 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
Note that the resultant of the distributed forces is equal to Vx and passes through S.
 S can be determined by determining Vx
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

17. Chapter 4- Lateral Force Resisting Systems
Line of action of Vx is yS away from E and given by:
Substitution for Vix in terms of Vx from above and factoring the constants and simplifying yields:
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

18. Chapter 4- Lateral Force Resisting Systems
Similarly from consideration of story shear Vy in the y-direction
As an example determine the center of stiffness of the statically indeterminate wall system shown in the previous slide
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

19. 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
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

20. 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).
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

21. 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
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

22. 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)
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

23. Chapter 4- Lateral Force Resisting Systems
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
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

24. 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 
Where k = the proportionality constant that has to be determined so that we can determine the wall forces resulting from torsion
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

25. 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: 
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

26. Chapter 4- Lateral Force Resisting Systems
Substituting the vale of k in wall force eqns as a result of torsion above:
Note that V1y  0 for wall 1
Thus the total force in wall i resulting from Vx and TS will be:
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

27. Chapter 4- Lateral Force Resisting Systems
Note that the 1st term in the expression for Viy is  0 because Vy = 0
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

28. 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
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

29. 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
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

30. 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 5
500 4
400 3
300
-0.4a 2
200
-0.1a 1
100
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

31. 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!!
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

32. Chapter 4- Lateral Force Resisting Systems
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

33. 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
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

34. 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
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

35. 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?
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

36. 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?
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

37. Chapter 4- Lateral Force Resisting Systems
Project the 3-D frame analysis
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

38. 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
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

39. Dr.-Ing. Girma Z. Dr.-Ing. Adil Z
Deformation pattern of a dual system
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

40. Dr.-Ing. Girma Z. Dr.-Ing. Adil Z
Typical deflection, moment and shear diagrams
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41. Chapter 4- Lateral Force Resisting Systems
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42. Chapter 4- Lateral Force Resisting Systems
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

43. Chapter 4- Lateral Force Resisting Systems
See analysis model
Dr.-Ing. Girma Z. Dr.-Ing. Adil Z