Distribution of Forces in Lateral Load Resisting Systems


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1 Distribution of Forces in Lateral Load Resisting Systems Part 2. Horizontal Distribution and Torsion IITGN Short Course Gregory MacRae Many slides from 2009 Myanmar Slides of Profs Jain and Rai 1
2 Reinforced Concrete CastinSitu Slabs The slab is subject to horizontal load. b t Moment of inertial for bending in its own plane I 3 tb 12 ( Very large quantity!!) Practically, floor is infinitely stiff for bending deformation in its own plane. 2
3 Floor Diaphragm Action L L b /2 Plan of a onestorey building with shear walls Springs represent lateral stiffness walls / frames t = floor thicness; 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 3
4 Floor Diaphragm Action Lateral earthquae force, EL Beam representing floor diaphragm I beam = tb 3 /12 K K/2 K Vertical load analogy for floor diaphragm action 4
5 Inplane versus outofplane deformation of floor In Plane Force Out of Plane Force In Plane Deformation of Floor Out of Plane Deformation of Floor 5
6 Floor Deformations InPlane Floor Deformation Out of Plane Floor Deformation 6
7 Rigidbody movements of a rigid floor diaphragm Longitudinal Translation Transverse Translation Translation in xdirection Translation in ydirection Angle of rotation Resultant Translation Rotation about zaxis Combination of translations and rotation 7
8 Example 1: Effect of floor diaphragm action 20 m 20 m 0.2 8m Slab thicness = 150 mm E = 25,500 N/mm 2 = N/mm I diaphragm mm I N 1000 N 0.2 E I Force in Springs Actual Analysis Rigid Diaphragm Tributary Area Rigid Diaphragm Assumption is generally used for RC floors 8
9 Example 2: Centre of Mass Given floor plan and lumped masses per unit area Locate centre of mass of the floor 8m 5 m 1000 g/m 2 Force in Springs 700 g/m 2 10 m 30 m y A CM (A) C CM F B Force CM (B) in Springs X CM (C) y z 9
10 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 x, y as x y m x i m m y i i m i i i 14.4 m 5.1 m 10
11 Centre of Stiffness (of a Singlestorey 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: A 10m m B 14 m 11
12 Example 3. Centre of Stiffness X F Force equilibrium: F Moment equilibrium: F. x x m 52 12
13 Example 3. Centre of Stiffness y F Force equilibrium: Moment equilibrium: F F. y y 5.6m
14 Lateral load distribution due to rigid floor diaphragm (Symmetrical case no torsion) Wall stiffness 1 = 2 =0.3 EL Building Plan 3 = F F 1 F 2 F 3 Wall stiffness F EL Definition of lateral stiffness F1 1, F2 2, F3 3 F1 F2 F3 EL 1 2 F1 EL; F2 EL; F EL 14
15 Example 4: 200 N applied along ydirection y 4 10m m x 30 m Locate centre of stiffness : (15m, 5m) Locate centre of mass : (15m, 5m) Hence, no torsion 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 Wall 1, 2, 3 share load proportional to stiffness F N; F N F N 15
16 Eccentric Systems Load at centre of mass = Load at centre of stiffness + Twisting moment about the centre of stiffness e CM CS EL EL e.el CM CS EL EL M = e.el CS CM 16
17 Analysis of force induced by twisting moment (rigid floor diaphragm) r 1 r 3 r 5 r r 5 1 CS r 4 3 r 2 2 r 2 r 1 i = Lateral stiffness of the i th element r i = Perpendicular distance of the i th element from centre of stiffness = Rotation of the floor diaphragm in its own plane 17
18 Analysis of force induced by twisting moment (rigid floor diaphragm) Displacement of i th element, in its own plane, due to rotation about centre stiffness i r i Resisting force in i th element F i i r i Restoring moment by force in i th element M i Fr i i i r 2 i By moment equilibrium M t i r 2 i Force in the i th element r i i Fi 2 iri M t 18
19 Example 5: Load distribution in eccentric system y 17.33m m 10m m m (a) Walls 1.5 x 5.56m 5m CS CM e x =2.33m (b) CS and CM e y =0.56m CM 200N = CM 466Nm = 200Nx 2.33m CS 200N (c) Forces F F 1 F 2 CS F 3 CM 5 200N Nm F F 1 2 CS F 3 CM 5 F 5 Translational Forces (d) Torsional Forces
20 Example 5: Load distribution in eccentric system Analysis for 200 N force acting at centre of stiffness This force is resisted by walls 1, 2, and 3 in proportion to their lateral stiffness. This gives: 200 F F F Analysis for 466 Nm moment acting on the diaphragm at CS: N N The twisting moment of 466 Nm is resisted by all the walls (including walls 4 and 5). N 20
21 Example 5: Load distribution in eccentric system Wall i ri 2 iri i r i i i F1 2 iri r M t 613.3
22 Example 5: Load distribution in eccentric system F 1 F 2 CS F 3 CM 5 200N + F Nm F F 1 2 CS F 3 CM 5 F 5 Translational Forces The total force resisted by the walls (= translational force + torsional force): F 1 = = 79.9 N F 2 = = 34.6 N F 3 = = 85.6 N F 4 = 5.1 N = 5.1 N F 5 = 5.1 N = 5.1 KN Torsional Forces F F F 1 2 CS F 3 CM 5 F 5 Total Forces 22
23 Multistorey Frames (a) Without Torsion All frames must follow through the same displacements at each level. F 4 F 3 F 2 (CS) i (CS) i (b) With Torsion All frame displacements at each level must be compatible with level translational and torsional displacements. F 1 (CS) i None of the floors undergo any rotation as forces pass through the CS (i.e. CS = CM)
24 Important: Multistorey Frames 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 24
25 Multistorey Frames A B C D 1 y x Plan 2 3 Plan of a building with space frame: this may be thought of as four 2bay frames in the ydirection, and three 3bay frames in the xdirection 25
26 Example 6: 3storey symmetric building Missing column Symmetric system A B C Plan Frames 1&3 same Frame spacing same Design force in ydirection on the entire building 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. 26
27 Example 6: 3storey symmetric building Frame A Frame B Frame C Impose the conditions This will ensure proper load distribution. 27
28 Example 6: 3storey symmetric building Thin of the translational problem as: Imaginary rigid lins to ensure floor diaphragm action Simple calculate the member forces 28
29 2D Frame with Rigid Lines A B C D E 1 Direction of Earthquae force (a) Lin bars Frame 1 Frame 2 Frame 3 Frame
30 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 2D 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 30
31 Example 7. Approximate Distribution, No Torsion A B C Plan Design force in ydirection on the entire building A B C 31
32 Example 7. Approximate Distribution, No Torsion A B C units 100 units 50 units i units Building Frames A and C f f A B A B f f f f 0.25f 0.50 f f C C f f 0.25f Frame B 32
33 Example 7. Approximate Distribution, No Torsion (No Torsion Case) Plan A B C EQ Entire Building A B C Evaluate,, such that roof displacement is same + + =
34 Example 8. Approximate Distribution, No Torsion Illustration Parts of building in double height Symmetric Portion in double height Plan Elevation
35 Further Simplification For load distribution, relative lateral stiffness is needed Relative terms only are required 35
36 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 12EI!! 3 L 36
37 Torsion in Multistory Buildings Centre of stiffness at different floors Number of definitions Depends on usage Implementation 37
38 Centre of Stiffness for multistory buildings (CS) i F 3 F 3 F 2 (CS) i F 2 F 1 F 1 (CS) i None of the floors undergo any rotation Design lateral load profile 38
39 Torsion in Multistory Buildings... e x.cs.cm e y CM = centre of mass CS = centre of stiffness.cs.cm 1.5e y EL x.cm.cs 1.5e x EL y The requirement on design eccentricity can be fulfilled by applying earthquae 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. 39
40 Torsion in Multistory Buildings Typical building code specifies design eccentricity in terms of Static eccentricity e sj Accidental eccentricity b j Q1 b j e dj e e sj sj b b j j is typically 1.5 is typically 0.05 to 1.0 (5% to 10% of plan dimension b j ) Q2 Q3 Q4 40
41 Torsion in Multistory Buildings... CM CS Goel and Chopra (ASCE, Vol.119; No:10) Can Conduct Analyses Directly Using Computer Program with Rigid Diaphragm (e.g. ETABS) 41
42 Approximation in Torsion Calculations 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: Earthquae force in Xdirection 6m 3m 4m All columns 400x400 Exterior Beams 250x600 Interior Beams 300x m 4.5m 4.5m 4.5m 42
43 Example 9. Approx. Analysis Torsion Earthquae force in Xdirection 6m 3m 4m A B 141 C 48 D 4.5m 4.5m 4.5m 4.5m All columns 400x400 Exterior Beams 250x600 Interior Beams 300x450 43
44 Example 9. Approx. Analysis Torsion P P y A B D C 16, 220N / m 10390, N / m 10390, , , 6. 1m 16, Eccentricity e = = 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 (M t = 0.6m x V) 44
45 Example 9. Approx. Analysis Torsion Force profile V at CS Frames A, D = Frames B, C = 16, 220 V 53, , 390 V 53, V V Twisting moment profile Mt i r i j r 2 j M t 45
46 Example 9. Approx. Analysis Torsion Frame i (x 10 3 ) (N/m) r i (m) i r i (x 10 3 ) (N) i r i 2 (x 10 3 ) Fd (torsion) Fd (Direct) FTO (Total) A V 0.304V 0.321V B V 0.195V 0.196V C V 0.195V 0.195V D V 0.304V 0.288V V V V V i r i i r i 2 From frame analysis with point load at roof Dist from CS 46
47 Example 9. Approx. Analysis Torsion Level Total Level Design Force (N) A (N) Design Force For Frame B (N) C (N) D (N) From beginning of example 47
48 Than you!! 48
4B2. 2. The stiffness of the floor and roof diaphragms. 3. The relative flexural and shear stiffness of the shear walls and of connections.
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