Transcript Lecture Slides-wk06

SEISMIC LOADS
LATERAL LOAD FLOW
FRAMES and SHEAR WALLS
SEISMIC LOAD
Determine Spectral Response Parameters at design location
At 37.80 N , -122.37 W :
Ss = 1.50
S1 = 0.60
Determine Site Coefficients
Site Class : D
Ss > 1.25
Fa = 1.0
S1 > 0.5
Fv = 1.5
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
VBASE = (Cs)(W)
( 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
Fx = VBASE Wx hx
S(w h)
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
Froof
Fsecond flr
VBASE (wx)(hx)
Fx =
S (w h)
VBASE (wx)(hx)
Fx =
S (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)
Roof
166.67k
30ft
5000k-ft
2nd
200k
15ft
3000k-ft
S (366.67k)
(w)(h)/S(w)(h)
Vbase
Fx
0.625
110k
68.75k
0.375
110k
41.25k
S(8000k-ft)
Vbase = 0.3W = 0.3(166.67k+200k) = 0.3(366.67k) = 110k
S (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 :
SR = 1+1+1+1 = 4
Px = ( Rx / SR ) (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 inplane shear stress
Basically, combined with vertical shear walls or frames IT ACTS
LIKE A LARGE I-BEAM
STRUCTURAL DIAPHRAGM
Flexible or Semi-flexible Type:
Plywood
Metal Decking
STRUCTURAL DIAPHRAGM
Rigid Diaphragm Type:
Reinforced Concrete Slab
Concrete-filled Metal Deck composite Slab
Braced/horizontal truss
STRUCTURAL DIAPHRAGM
Rigid Diaphragm:
Flexible Diaphragm:
Almost no deflection
Can transmit loads
through torsion
Deflects horizontally
Cannot 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.
COLLECTOR
FRAME
DIAPHRAGM
COLLECTOR
FRAME
Lateral Load Flow
diaphragm > collectors/drags > frames
COLLECTOR
LATERAL
FRAME
DIAPHRAGM
LOAD
COLLECTOR
FRAME
Lateral Load Flow
diaphragm > collectors/drags > frames
COLLECTOR
LATERAL
FRAME
DIAPHRAGM
LOAD
COLLECTOR
FRAME
Lateral Load Flow
diaphragm > collectors/drags > frames
LATERAL
COLLECTOR
FRAME
LOAD
COLLECTOR
DIAPHRAGM
COLLECTOR
COLLECTOR
FRAME
FRAME
LATERAL FORCE RESISTING
SYSTEMS:
MOMENT Resisting frames
Diagonally BRACED frames
SHEAR walls
INSTABILITY OF THE FRAME
Pinned connections
cannot resist rotation.
This is not a structure
but rather a mechanism
STABILIZE THE FRAME
FIX ONE OR MORE OF THE BASES
STABILIZE THE FRAME
FIX ONE OR MORE OF THE CORNERS
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
PINNED BASE =4 UNKNOWNS, 3
EQUATIONS, STATICALLY
INDETERMINATE TO FIRST DEGREE
SOLVE BY “PORTAL FRAME METHOD”
MOMENT FRAMES
FIXED BASE =6 UNKNOWNS, 3 AVAILABLE EQUATIONS OF EQUILIBRIUM
STATICALLY INDETERMINATE TO THE 3RD DEGREE
SOLVE BY “PORTAL FRAME METHOD”