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SEISMIC LOADS LATERAL LOAD FLOW FRAMES and SHEAR WALLS.

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Presentation on theme: "SEISMIC LOADS LATERAL LOAD FLOW FRAMES and SHEAR WALLS."— Presentation transcript:

1 SEISMIC LOADS LATERAL LOAD FLOW FRAMES and SHEAR WALLS

2 SEISMIC LOAD

3

4 At N, W : Ss = 1.50 S 1 = 0.60 Determine Spectral Response Parameters at design location

5 Site Class : D Ss > 1.25 Fa = 1.0 S 1 > 0.5 Fv = 1.5 Determine Site Coefficients Determine Design Spectral Acceleration Parameters S MS = (1.0)(1.5) = 1.5 S DS = (2/3)(1.5) = 1.0

6 Cs = S DS /(R/I) =1.0/(R/I) Class II : I = 1.0 Ordinary Moment Resisting Frame : R = 3.5 V = 1.0/3.5 W 0.3 W

7 Seismic Load is generated by the inertia of the mass of the structure : V BASE Redistributed (based on relative height and weight) to each level as a ‘Point Load’ at the center of mass of the structure or element in question : F X V BASE W x h x  (w h) ( V BASE ) (Cs)(W)V BASE = F x =

8 Total Seismic Loading : V BASE = 0.3 W W = W roof + W second

9 W roof

10 W second flr

11 W = W roof + W second flr

12 V BASE

13 Redistribute Total Seismic Load to each level based on relative height and weight F x = F roof F second flr V BASE (w x )(h x )  (w h)

14 F x = V BASE (w x )(h x )  (w h) In order to solve the equivalent lateral force distribution equation, we suggest you break it up into a spreadsheet layout Floorw h(w)(h) (w)(h)/  (w)(h)VbaseFx Roof166.67k 30ft5000k-ft k68.75k 2 nd 200k 15ft3000k-ft k41.25k  (366.67k)  (8000k-ft)  (110k) Vbase = 0.3W = 0.3(166.67k+200k) = 0.3(366.67k) = 110k

15 Load Flow to Lateral Resisting System : Distribution based on Relative Rigidity Assume Relative Rigidity : Single Bay MF : Rel Rigidity = Bay MF : Rel Rigidity = Bay MF : Rel Rigidity = 3

16 Distribution based on Relative Rigidity :  R = = 4 P x = ( R x /  R ) (P total ) P MF1 = 1/4 P total

17 Lateral Load Flow diaphragm > collectors/drags > frames

18 STRUCTURAL DIAPHRAGM A structural diaphragm is a horizontal structural system used to transfer lateral loads to shear walls or frames primarily through in- plane shear stress Basically, combined with vertical shear walls or frames IT ACTS LIKE A LARGE I-BEAM

19 STRUCTURAL DIAPHRAGM Flexible or Semi-flexible Type: Plywood Metal Decking

20 STRUCTURAL DIAPHRAGM Rigid Diaphragm Type: Reinforced Concrete Slab Concrete-filled Metal Deck composite Slab Braced/horizontal truss

21 STRUCTURAL DIAPHRAGM Rigid Diaphragm: Almost no deflection Can transmit loads through torsion Flexible Diaphragm: Deflects horizontally Cannot transmit loads through torsion

22 COLLECTORS and DRAGS

23 COLLECTORS and DRAG STRUTS A beam element or line of reinforcement that carries or “collects” loads from a diaphragm and carries them axially to shear walls or frames. A drag strut or collector behaves like a column.

24 Lateral Load Flow diaphragm > collectors/drags > frames DIAPHRAGM COLLECTOR FRAME

25 Lateral Load Flow diaphragm > collectors/drags > frames DIAPHRAGM COLLECTOR FRAME LATERAL LOAD

26 Lateral Load Flow diaphragm > collectors/drags > frames DIAPHRAGM COLLECTOR FRAME LATERAL LOAD

27 DIAPHRAGM COLLECTOR FRAME LATERAL LOAD FRAME COLLECTOR

28 LATERAL FORCE RESISTING SYSTEMS: MOMENT Resisting frames Diagonally BRACED frames SHEAR walls

29 INSTABILITY OF THE FRAME Pinned connections cannot resist rotation. This is not a structure but rather a mechanism.

30 STABILIZE THE FRAME FIX ONE OR MORE OF THE BASES

31 FIX ONE OR MORE OF THE CORNERS STABILIZE THE FRAME

32 ADD A DIAGONAL BRACE

33 RELATIVE STIFFNESS OF FRAMES AND WALLS LOW DEFLECTION HIGH STIFFNESS ATTRACTS MORE LOAD HIGH DEFLECTION LOW STIFFNESS ATTRACTS LESS LOAD

34 BRACED FRAMES

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36 SHEAR WALLS

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41 MOMENT FRAMES

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43 INDETERMINATE STRUCTURES SOLVE BY “PORTAL FRAME METHOD”

44 MOMENT FRAMES SOLVE BY “PORTAL FRAME METHOD” PINNED BASE =4 UNKNOWNS, 3 EQUATIONS, STATICALLY INDETERMINATE TO FIRST DEGREE

45 MOMENT FRAMES SOLVE BY “PORTAL FRAME METHOD” FIXED BASE =6 UNKNOWNS, 3 AVAILABLE EQUATIONS OF EQUILIBRIUM STATICALLY INDETERMINATE TO THE 3 RD DEGREE


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