P h E lPushover Example(Appendix C)(Appendix C)

Derrell Manceaux, P.E.

“PUSHOVER” ANALYSIS of a TWO COLUMN BENT

Derrell Manceaux, P.E.Derrell Manceaux, P.E.

Pushover Concept

To find the total deflection a structure can obtain prior to collapse

Pushover Procedure

Push on a structure until an event occurs, calculate properties and continue pushing until next event occurs, ….. continue this iteration until collapsethis iteration until collapse

Pushover Procedure

1) Calculate member properties 1) Calculate member properties 2) Obtain Moment-Curvature for column with axial loads

(max & min). 3) Determine EQ force required to obtain ultimate moment

Keep pushing untilPush until hinge forms

p p gmechanism develops

First hinge Second hinge

Pushover Procedure

4) Determine additional EQ force required to obtain 4) Determine additional EQ force required to obtain rotational capacity

5) Calculate deflections 6) Continue to push until first rotational capacity is

reached-failure 7) Calculate final deflection

Keep pushing untilrotational capacity is reached

Rotation capacity

Example Problem

Structure Properties

G t Geometry H=15 ft W=32 ft

DL=1200 k

L=20 ft Plastic Hinge

L =20 inch LpA=20 inch LpB=20 inch Dead Load 15

’20

’

DL=1200 kip

32’32’A B

“Moment-Curvature” for Column- (Dead Load Only)

Axial Dead Load PDL=1200 k PDL= 600 k/column

Yield Moment=Ultimate MomentM M 1600 k ft

My Mu

My=Mu=1600 k-ft Yield Curvature

Ø =140x10-6/inch Øy 140x10 /inch Ultimate Curvature

Øu=2350x10-6/inchu

Øy Øu

Iteration 1-Determine EQ force required to obtain Yield Moment for DL only

Apply Unit Load and Take Ratios Apply Unit Load and Take Ratios M=7.5 k-ft with unit load

EQ 1600/7 5 213 3 ki EQ=1600/7.5 = 213.3 kip

My=1600 k-ft

7.5

A B

Revise axial loads for EQ force being applied

EQ x L = PEQ x w 213.3 x 20 = PEQ x 32 PEQ=± 133.33 kip

Pcol A= PDL - PEQ

Pcol A = 600 - 133.3 = 466.7 kcol A

Pcol B= PDL + PEQ

A B

col B DL EQ

Pcol B = 600 + 133.3 = 733 k

Obtain Moment-Curvature for column with revised axial loads

C l “A” Column “A” Axial Load=466 kip

Yield Moment=Ultimate Moment Yield Moment=Ultimate MomentMAy=1265 k-ft

Yield Curvature Yield CurvatureØAy=138 x 10-6/inch

Ultimate Curvature Col B Ultimate Curvature

ØAu=2500 x 10-6/inch EIA= MAy / ØAy

Col A

A Ay Ay1265 x 12/138 x 10-6

=1.1 x 108 k-in2

Øy Øu

Obtain Moment-Curvature for column with revised i l l daxial loads

C l “B” Column “B” Axial Load=733 kip

Yield Moment=Ultimate Moment Yield Moment=Ultimate MomentMBY=1670 k-ft

Yield Curvature Yield CurvatureØBY=143x10-6/inch

Ultimate Curvature Col B Ultimate Curvature

ØBu=2200x10-6/inch EIB= MBy / ØBy

Col A

B By By1670 x 12/143 x 10-6

=1.4 x 108 k-in2

Øy Øu

Iteration 2-Determine EQ force required to obtain First Yield in Column “A”Apply unit load to determine linear ratios (scale Apply unit load to determine linear ratios (scale factor).

Determine moment diagram with unit load Determine moment diagram with unit load. EQ1=MAY/7.5=1265/7.5= 168.67 kip

7.5

Calculate deflection at first yield

3 Δ1= EQ1 x (h x 12)3

3 x (EIA + EIB) Δ1

Δ1 = 168.67 x (15 x 12)3 = 1.31 in3 x (1 1 + 1 4) x 1083 x (1.1 + 1.4) x 108

K 3 EIΔ

Basic stiffness formula for deflection Fix-Pin condition:

A BK=3 EIΔL3

Keep Pushing until Second Hinge FormsKeep Pushing until Second Hinge FormsDetermine additional EQ force required to obtain yield moment in second columnobtain yield moment in second column Reserve capacity in column “B”= Mreserve B=MBY- MAY=1670-1265=405 k-ft

A B

Determine additional EQ force required toDetermine additional EQ force required to obtain yield moment in second column

Place pin in col “A” and apply unit load Determine EQ2 with ratios: Determine EQ2 with ratios: EQ2=Mreserve B/15 = 405/15 = 27 kip

M=15M=0

A B

Calculate additional deflection for second hinge in Col ‘B’

EIB = 1.4 x 108 k-in2

Δ EQ x (h x 12)3 Δ2 Δ2= EQ2 x (h x 12)3

3 x EIB

2

Δ2 = 27 x (15 x 12)3 = 0.37 in3 x (1.4 x 108)

A B

Failure is assumed when a plastic hinge reaches rotational capacity

Column ‘A’-Rotational Capacity

Yield Curvature ØAy=138 x 10-6/inch

Col B

Col A

Ultimate Curvature ØAu=2500 x 10-6/in θpA=LpA x (ØAu - ØAy)pA pA Au Ay

θpA= 20 x (2500-138) x 10-6

= 0.047

Øy Øu

Failure is assumed when a plastic hinge reaches rotational capacity

Øy ØuØy Øu

Failure is assumed when a plastic hinge reaches rotational capacity

Failure is assumed when a plastic hinge reaches rotational capacity

Column ‘B’-Rotational Capacity

Yield Curvature- ØBy=143 x 10-6/inch

Ultimate Curvature- ØBu=2200 x 10-6/in

θpB=LpB x (ØBu - ØBy) θpB= 20 x (2200-143) x 10-6

Col B

Col AØy Øu θpB 20 x (2200 143) x 10 = 0.041

Col AØy Øu

Following the formation of the second hinge, the rotational capacity of Column ‘A’ was reduced: θA Reduced= Δ2/ (h x 12-0.5LpA) θA Reduced= 0.37/ (15 x 12-10)= 0.0021A Reduced

And the remaining rotational capacity in column ‘A’ is:

θA Remaining = θpA - θA Reduced

controls

A Remaining pA A Reduced

θA Remaining = 0.047-0.0021 = 0.045 > θpB = 0.041

C ti t h til fi t t ti lContinue to push until first rotational capacity is reached:Δ3= θpB (h x 12-0.5LPB) = 0.041(15 x 12-10) = 6.97 in

Total Deflection

Δtotal= ∑ Δn = 1.31 + 0.375 + 6.97 = 8.66 inch

Push until hinge formsKeep pushing until

t ti l it i h dPush until hinge forms rotational capacity is reached

First hinge Rotation capacity

Keep pushing untilmechanism developsmechanism develops

Second hinge

Total EQ Force Required for FailureTotal EQ Force Required for Failure

Push until hinge formsKeep pushing until

t ti l it i h d

EQtotal= ∑ EQn = 168.7 + 27 + 0 = 195.7 kip

Push until hinge forms rotational capacity is reached

First hinge Rotation capacity

Keep pushing untilmechanism developsmechanism develops

Second hinge

Ductility Factor – “R”

R Δ / Δ R = μ = Δtotal/ Δyield

R = μ = 8.66 / 1.31 = 6.6 Δyield Δtotal

Note:AASHTO “R” factor for multi column bent is 5

Summary

Understand concept and procedure of performing a Pushover AnalysisT fi d th t t l d fl ti t t bt i i t To find the total deflection a structure can obtain prior to collapse

Describe the “Pushover” Procedure

Push until hinge forms

Keep pushing untilrotational capacity is reachedg

First hinge Rotation capacity

Keep pushing untilmechanism developsmechanism develops

Second hinge