SEISMIC LOADS

LATERAL LOAD FLOW

FRAMES and SHEAR WALLS

SEISMIC LOAD

At 37.80 N , -122.37 W :

Ss = 1.50

S1 = 0.60

Determine Spectral Response Parameters at design location

Site Class : D

Ss > 1.25

Fa = 1.0

S1 > 0.5

Fv = 1.5

Determine Site Coefficients

Determine Design Spectral Acceleration Parameters

SMS = (1.0)(1.5) = 1.5

SDS = (2/3)(1.5) = 1.0

Cs = SDS /(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

Seismic Load is generated by the inertia of the mass of the structure : VBASE

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 : FX

VBASE Wx hx

(w h)

( VBASE )

(Cs)(W)VBASE =

Fx =

Total Seismic Loading :

VBASE = 0.3 W

W = Wroof + Wsecond

Wroof

Wsecond flr

W = Wroof + Wsecond flr

VBASE

Redistribute Total Seismic Load to each level based on relative height and weight

Fx =

Froof

Fsecond flr

VBASE (wx)(hx)

(w h)

Fx =VBASE (wx)(hx)

(w h)

In order to solve the equivalent lateral force distribution equation, we suggest you break it up into a spreadsheet layout

Floor w h (w)(h) (w)(h)/(w)(h) Vbase Fx

Roof 166.67k 30ft 5000k-ft 0.625 110k68.75k

2nd 200k 15ft 3000k-ft 0.375 110k 41.25k

(366.67k) (8000k-ft) (110k)

Vbase = 0.3W = 0.3(166.67k+200k) = 0.3(366.67k) = 110k

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

Distribution based on Relative Rigidity :

R = 1+1+1+1 = 4

Px = ( Rx / R ) (Ptotal)

PMF1 = 1/4 Ptotal

Lateral Load Flow

diaphragm > collectors/drags > frames

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

STRUCTURAL DIAPHRAGM

Flexible or Semi-flexible Type:

PlywoodMetal Decking

STRUCTURAL DIAPHRAGM

Rigid Diaphragm Type:

Reinforced Concrete SlabConcrete-filled Metal Deck composite SlabBraced/horizontal truss

STRUCTURAL DIAPHRAGM

Rigid Diaphragm:

Almost no deflectionCan transmit loads through torsion

Flexible Diaphragm:

Deflects horizontallyCannot transmit loads through torsion

COLLECTORS and DRAGS

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.

Lateral Load Flow

diaphragm > collectors/drags > frames

DIAPHRAGM

COLLECTOR

COLLECTOR

FRAME

FRAME

Lateral Load Flow

diaphragm > collectors/drags > frames

DIAPHRAGM

COLLECTOR

COLLECTOR

FRAME

FRAME

LATERAL

LOAD

Lateral Load Flow

diaphragm > collectors/drags > frames

DIAPHRAGM

COLLECTOR

COLLECTOR

FRAME

FRAME

LATERAL

LOAD

DIAPHRAGM

COLLECTOR

COLLECTOR

FRAME

FRAME

LATERAL

LOAD

FRAMECOLLECTOR

COLLECTOR

LATERAL FORCE RESISTING SYSTEMS:

MOMENT Resisting frames

Diagonally BRACED frames

SHEAR walls

INSTABILITY OF THE FRAME

Pinned connectionscannot resist rotation.

This is not a structurebut rather a mechanism.

STABILIZE THE FRAME

FIX ONE OR MORE OF THE BASES

FIX ONE OR MORE OF THE CORNERS

STABILIZE THE FRAME

STABILIZE THE FRAME

ADD A DIAGONAL BRACE

RELATIVE STIFFNESS OF FRAMES AND WALLS

LOW DEFLECTION

HIGH STIFFNESS

ATTRACTS MORE LOAD

HIGH DEFLECTION

LOW STIFFNESS

ATTRACTS LESS LOAD

BRACED FRAMES

BRACED FRAMES

SHEAR WALLS

SHEAR WALLS

SHEAR WALLS

SHEAR WALLS

SHEAR WALLS

MOMENT FRAMES

MOMENT FRAMES

MOMENT FRAMES

INDETERMINATE STRUCTURES

SOLVE BY “PORTAL FRAME METHOD”

MOMENT FRAMES

SOLVE BY “PORTAL FRAME METHOD”

PINNED BASE =4 UNKNOWNS, 3 EQUATIONS, STATICALLY INDETERMINATE TO FIRST DEGREE

MOMENT FRAMES

SOLVE BY “PORTAL FRAME METHOD”

FIXED BASE =6 UNKNOWNS, 3 AVAILABLE EQUATIONS OF EQUILIBRIUMSTATICALLY INDETERMINATE TO THE 3RD DEGREE