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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Dr.-Ing. Girma Z. Dr.-Ing. Adil Z Typical deflection, moment and shear diagrams Dr.-Ing. Girma Z. Dr.-Ing. Adil Z

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

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

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