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Published byHunter Avary Modified about 1 year ago

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SEISMIC LOADS LATERAL LOAD FLOW FRAMES and SHEAR WALLS

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SEISMIC LOAD

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At 37.80 N, -122.37 W : Ss = 1.50 S 1 = 0.60 Determine Spectral Response Parameters at design location

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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

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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

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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 =

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

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W roof

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W second flr

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W = W roof + W second flr

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V BASE

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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)

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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 0.625110k68.75k 2 nd 200k 15ft3000k-ft 0.375110k41.25k (366.67k) (8000k-ft) (110k) Vbase = 0.3W = 0.3(166.67k+200k) = 0.3(366.67k) = 110k

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Load Flow to Lateral Resisting System : Distribution based on Relative Rigidity Assume Relative Rigidity : Single Bay MF : Rel Rigidity = 1 2 - Bay MF : Rel Rigidity = 2 3 - Bay MF : Rel Rigidity = 3

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

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Lateral Load Flow diaphragm > collectors/drags > frames

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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

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

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

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

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COLLECTORS and DRAGS

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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.

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

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

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

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DIAPHRAGM COLLECTOR FRAME LATERAL LOAD FRAME COLLECTOR

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

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

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STABILIZE THE FRAME FIX ONE OR MORE OF THE BASES

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FIX ONE OR MORE OF THE CORNERS STABILIZE THE FRAME

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ADD A DIAGONAL BRACE

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

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BRACED FRAMES

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SHEAR WALLS

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MOMENT FRAMES

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INDETERMINATE STRUCTURES SOLVE BY “PORTAL FRAME METHOD”

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

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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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