nonlinear introduction

7/21/2019 Nonlinear Introduction

1/16

Basic Introduction to

Nonlinear AnalysisRonald D. ZiemianBucknell University

The function of a structural engineer isto design not to analyze

Norris and Wilbur1960

Analysis is a means to an endrather than the end itself.

Role of the analysis:

forces, moments and deflections design equations

insight into the behavior of a structure better the understanding, better the design

Limit States Design:Prior to limit of resistance, significant nonlinearresponse, including- geometrical effects (P-, P-)- material effects (yielding, cracking, crushing)

- combined effects

AISC Ch. C: P-, P- (App. 7)App. 1: Inelastic Design

Seismic: Pushover AnalysisOther: Progressive Collapse

Impetus:

Nonlinear Analysis

Limit States Design

Available Software

Research

Education

Nonlinear AnalysisHand methods

Second-order effects i.e. Moment Amplification Factors (B1 and B2 factors)

Material nonlinear effects i.e. plastic analysis (upper and lower bound theories)

Computer Methods (focus of todays lecture) Lots of variations

all use same basic concepts (most important to today)

one approach will be presented (basis for MASTAN2)

Please keep in mind All methods are approximate

Not a substitute, but a complement to good engineering

Lecture Overview

Brief Introduction (done!)Computer Structural Analysis (Review?)

Basis for Material Nonlinear ModelsIncorporating Geometric Nonlinear Behavior

Critical Load Analysis

Overview of MASTAN2 software

Summary and Concluding Remarks


2/16

How does the computer getthese results?

State-of-the-Art Crystal Ball? Not quite.

By applying 2 requirements and 1 translator Two Requirements:

Equilibrium (equations in terms of Fs and Ms, 1 per d.o.f.)

Compatibility (equations in terms of s and s, 1 per d.o.f.)

Translator apples to oranges Constitutive Relationship (i.e. Hookes Law, = E )

Generalized to Force-to-Displacement (i.e. F=k)

Re-write equilibrium eqs. in terms of unknown displacements

# of Equil. Eqs. = # of Unknown Displacements

Equilibrium Equations

A B C

D 40 kips

CDxD

BDxD

ADxD

xD

fff

Fux-

++=

=

40

0:d.o.f.

40 kips

BDxDf

BDyDf

Membe

rBD

BDxDf

BDyDf

MemberCD

CDxDf

CDyDf

CDyDf

CDxDf

ADyDf

Membe

rAD

ADxDf

ADyDf

ADxDf

D

Free BodyDiagram

CDyD

BDyD

ADyD

yD

fff

Fvy-

++=

=

0

0:d.o.f.

Translator: Forces Displacements

ui

vi

vj

uj

i

j

(Deformed)

i

j(Undeformed)

fxi

fxj

fyj

fyi

Big Question:Where do these known stiffness coefficients kscome from? Little Answer:

Function of members material and geometricproperties, including its orientation.

jjiiyj

jjiixj

jjiiyi

jjiixi

vkukvkukf

vkukvkukf

vkukvkukf

vkukvkukf

44434241

34333231

24232221

14131211

+++=

+++=

+++=

+++=

F for all members

BDD

BDBDD

BDBDB

BDBDB

BDBDyD

BD

D

BDBD

D

BDBD

B

BDBD

B

BDBD

xD

BDD

BDBDD

BDBDB

BDBDB

BDBDyB

BDD

BDBDD

BDBDB

BDBDB

BDBDxB

vkukvkukfvkukvkukf

vkukvkukf

vkukvkukf

44434241

34333231

24232221

14131211

+++=+++=

+++=

+++=

Member BD:

ADD

ADADD

ADADA

ADADA

ADADyD

ADD

ADADD

ADADA

ADADA

ADADxD

ADD

ADADD

ADADA

ADADA

ADADyA

ADD

ADADD

ADADA

ADADA

ADADxA

vkukvkukf

vkukvkukf

vkukvkukf

vkukvkukf

44434241

34333231

24232221

14131211

+++=

+++=

+++=

+++=

Member AD:

A B C

DuD

vD

uA

vAuB

vBuC

vC

Member CD:

CDD

CDCDD

CDCDC

CDCDC

CDCDyD

CD

D

CDCD

D

CDCD

C

CDCD

C

CDCD

xD

CDD

CDCDD

CDCDC

CDCDC

CDCDyC

CDD

CDCDD

CDCDC

CDCDC

CDCDxC

vkukvkukfvkukvkukf

vkukvkukf

vkukvkukf

44434241

34333231

24232221

14131211

+++=+++=

+++=

+++=

Substituting into Equil. Eqs.

A B C

D40 kN

CD

xD

BD

xD

AD

xD

xD

fff

Fux-

++==

40

0:d.o.f.

ADD

ADADD

ADADA

ADADA

ADADxD vkukvkukf 34333231 +++=

Member AD:

BDD

BDBDD

BDBDB

BDBDB

BDBDxD vkukvkukf 34333231 +++=

Member BD:

Member CD:CDD

CDCDD

CDCDC

CDCDC

CDCDxD vkukvkukf 34333231 +++=

( )( )( )CD

D

CDCD

D

CDCD

C

CDCD

C

CD

BD

D

BDBD

D

BDBD

B

BDBD

B

BD

AD

D

ADAD

D

ADAD

A

ADAD

A

AD

D

vkukvkuk

vkukvkuk

vkukvkuk

u

34333231

34333231

3433323140

:d.o.f.

+++

++++

++++=

Substituting into Equil. Eqs. (cont.)

A B C

D40 kN

CD

yD

BD

yD

AD

yD

yD

fff

Fvy-

++==

0

0:d.o.f.

( )( )( )CDD

CDCD

D

CDCD

C

CDCD

C

CD

BD

D

BDBD

D

BDBD

B

BDBD

B

BD

AD

D

ADAD

D

ADAD

A

ADAD

A

AD

D

vkukvkuk

vkukvkuk

vkukvkuk

v

44434241

44434241

44434241

0

:d.o.f.

+++

++++

++++=

Member AD:

Member BD:

Member CD:

ADD

ADADD

ADADA

ADADA

ADADyD vkukvkukf 44434241 +++=

BDD

BDBDD

BDBDB

BDBDB

BDBDyD vkukvkukf 44434241 +++=

CDD

CDCDD

CDCDC

CDCDC

CDCDyD vkukvkukf 44434241 +++=


3/16

So, where are we at?

We have two equilibrium equations (1 perd.o.f.) in terms of a lot of displacements:

( )( )( )

( )( )( )CD

D

CDCD

D

CDCD

C

CDCD

C

CD

BD

D

BDBD

D

BDBD

B

BDBD

B

BD

AD

D

ADAD

D

ADAD

A

ADAD

A

AD

D

CD

D

CDCD

D

CDCD

C

CDCD

C

CD

BD

D

BDBD

D

BDBD

B

BDBD

B

BD

AD

D

ADAD

D

ADAD

A

ADAD

A

AD

D

vkukvkuk

vkukvkuk

vkukvkukv

vkukvkuk

vkukvkuk

vkukvkuku

44434241

44434241

44434241

34333231

34333231

34333231

0:

40:

+++

++++

++++=

+++

++++

++++=

What card havent we played yet?

Compatibility Eqs. (consistent deflections)

CD

CC 0vv ==

CD

CC 0uu ==AD

AA 0vv ==

AD

AA 0uu ==

BD

BB 0vv ==

BD

BB 0uu ==

Member-to-Support

A B C

D

D

MemberCDM

embe

rAD

Membe

rAD

Membe

rBD

Membe

rBD

CD

D

BD

D

AD

DD vvvv ===

CD

D

BD

D

AD

DD uuuu ===

Member-to-MemberDv

AD

Dv

ADDu

BD

Dv

BD

Du

CD

Dv

CD

Du

Du

Time for some serious simplifying

Applying Compatibility to Equil. Eqs.:

( )( )( )

( )( )( )CD

D

CDCD

D

CDCD

C

CDCD

C

CD

BD

D

BDBD

D

BDBD

B

BDBD

B

BD

AD

D

ADAD

D

ADAD

A

ADAD

A

AD

D

CD

D

CDCD

D

CDCD

C

CDCD

C

CD

BD

D

BDBD

D

BDBD

B

BDBD

B

BD

AD

D

ADAD

D

ADAD

A

ADAD

A

AD

D

vkukvkuk

vkukvkuk

vkukvkukv

vkukvkuk

vkukvkuk

vkukvkuku

44434241

44434241

44434241

34333231

34333231

34333231

0:

40:

+++

++++

++++=

+++

++++

++++=

Which simplifies to

All = 0

All = uD

All = vD

After simplifying

( ) ( ) DCDBDADDCDBDADDD

CDBDAD

D

CDBDAD

D

vkkkukkkv

vkkkukkku

0:

40:

444444434343

343434333333

+++++=

+++++=

A B C

D40 kN

Solve for Unknown Displacements

Since ks are known, we have2 Equations and 2 Unknowns

? ?

? ?

uD = # and vD = ##

##

#

0

0

0

0

0

0

====

====

==

==

==

==

==

==

CDD

BDD

ADDD

CD

D

BD

D

AD

DD

CD

CC

CD

CC

BD

BB

BD

BB

AD

AA

AD

AA

vvvv

uuuu

vv

uu

vv

uu

vv

uu

With all displacements, solve formember forces

ADD

ADADD

ADADA

ADADA

ADADyD

ADD

ADADD

ADADA

ADADA

ADADxD

ADD

ADADD

ADADA

ADADA

ADADyA

ADD

ADADD

ADADA

ADADA

ADADxA

vkukvkukf

vkukvkukf

vkukvkukf

vkukvkukf

44434241

34333231

24232221

14131211

+++=

+++=

+++=

+++=

Member AD:

Member BD:

BDD

BDBDD

BDBDB

BDBDB

BDBDyD

BDD

BDBDD

BDBDB

BDBDB

BDBDxD

BDD

BDBDD

BDBDB

BDBDB

BDBDyB

BDD

BDBDD

BDBDB

BDBDB

BDBDxB

vkukvkukf

vkukvkukf

vkukvkukf

vkukvkukf

44434241

34333231

24232221

14131211

+++=

+++=

+++=

+++=

Member CD:

CDD

CDCDD

CDCDC

CDCDC

CDCDyD

CDD

CDCDD

CDCDC

CDCDC

CDCDxD

CDD

CDCDD

CDCDC

CDCDC

CDCDyC

CDD

CDCDD

CDCDC

CDCDC

CDCDxC

vkukvkukf

vkukvkukf

vkukvkukf

vkukvkukf

44434241

34333231

24232221

14131211

+++=

+++=

+++=

+++=

Summary of Computer Approach For each d.o.f., write an equilibrium equation:

Fexternal = fmember Re-write (translate) each member force in terms of

its end displacements (Stiffness Eqs.)

fmember = k member member end Substitute Stiffness Eqs. into above Equil. Eqs.

Simplify Equil. Eqs. by applying member-to-memberand member-to-support compatibility conditions

Solve n Equil. Eqs. for the n unknown displacements

Use Stiffness Eqs. to calculate member forces

Apply Equil. Eqs. to solve for reactions


4/16

Lots of Questions

So, this is how most commercial programs such

as SAP2000, RISA, STAAD, etc. get theanswer? Yes! Known as Direct Stiffness Method

So, all such programs will give the same answer? Yes, as long as it is a static 1st-order elastic analysis.

Wait a minuteIs this the basic analysisprocedure for the finite element method? Yes! Bit more tricky to get ks, s, and s

Two Big Questions

Where do those stiffness coefficients

come from? You mean the ones that relate member endforces to member end displacements?

Yeah, those ks !

What happens when we go static nonlinearor even dynamic? Same basic procedure, but apply loads in

increments and perform a series of analyses.Then, sum incremental results.

< Much more to come on this! >

Important PointsThe only opportunity for most computer

analysis software to model the actualbehavior of the structure is through themember stiffness terms.

So, to include first-order effects

second-order effects material nonlinear behavior

Must modify member stiffness!!!

Lets review member stiffness

Stiffness Coefficients, ks

Lets start with high school physics Extension Spring Experiment

After:

F

F

k

1

F = k

Before:Force

Displacement

After:

M

k

1

M = k

M

Stiffness Coefficients, ks (cont.)

More advanced high school physics lab Rotational Spring Experiment

Before:Moment

Rotation

How about real structural members?Axial force member

F = k(A,L,E)

Stiffness k function of: Geometry: Area and Length (A,k & L, k)

Material: Elastic Modulus (E,k)

After:F

F

k1

Before:


5/16

How about real members? (cont.)Flexural members

Before:

Stiffness k function of: Geometry: Moment of Inertia & Length (I,k& L, k) Material: Elastic Modulus (E,k)

Before:

After:

F

After:M

F = k(I,L,E)M = k(I,L,E)

Other factor impacting stiffnessOrientation of member

consider axial force member:

kv=EA/Lkh=0

VerticalMember kh=EA/L

kv=0

HorizontalMember

Orientation of axial force member

Important Point: Less vertical a member,the less stiffness to resist vertical loads.

kv=(sin2)EA/L

kh=(cos2)EA/L

Summary: Three Perspectives

Reality: What you see

F1

Three Perspectives (cont.)What you see on your computer screen:

DL

WL

Collection of elementsconnected by sharingcommon nodes

Three PerspectivesWhat your computer actually sees:

Assemblage ofequivalent springs {F} = [K]{}

k

DLWL


6/16

Analysis Review: Key Points

Reviewed the Direct Stiffness Method Equilibrium Translator F() Compatibility

Response of structure controlled bystiffness of members (a.k.a. springs)

First-order elastic stiffness of memberfunction of: Material Property (E)

Geometric Properties (A, I, L, and orientation)

Time to go nonlinearlets begin with material nonlinear

Material Nonlinear (Inelastic)

Best place to start is with a tensile test

= P/A

= /L

E

1

yield Perfectly Plastic (E = 0)

Ela

stic

Normal Stress: Structural MembersFor typical structural steel members

(L/d>10), elastic/inelastic behaviorcontrolled by normal stresses s actingalong the length axis of the member.

Normal stress produced by: Axial force (P/A)

Major and/or minor axis flexure (Mc/I)

Combination of above effects (i.e. P/A + Mc/I) Warping (not today!)

We will assume elastic-perfectly-plasticmaterial (often done for steel)

E=0 k=0

Post-Yield:

= y

Inelastic Behavior: Axial ForceOriginally:

P=P/A=0

Yield:

Py

Py=Ay

= yPlastic Hingeat P = Py or

when P/Py = 1.0

k=EA/L1

Elastic:

P

< y

Inelastic Behavior:Flexure

Mp=Zy

M

My=Sy

MA

A

k=4EI/L

1EIelastic

-y +y

M < My

Section A-A

Inelastic Behavior:Flexure (cont.)

MA

A

Mp=Zy

k=4EI/L

1

M

My=Sy

Section A-A

EIelastic

-y +y

M=Sy=My


7/16

Inelastic Behavior:Flexure (cont.)

Mp=Zy

k=4EI/L

1

M

My=Sy

Section A-A -y +y

E=0

EI


8/16

Inelastic Behavior:Combination P & M

Mp=Zy

k=4EI/L

1

M

MA

AP

-y

+y

M < Mp

= Mc/I + P/A

+y

-y

Plastic

Elas

tic

Inelastic Behavior:Combination P & Mfor Plastic Hinge

MA

AP

+y

-y

+y

-y

+y

-y= +

M < Mp

M/Mp < 1

P < Py

P/Py < 1

Fully yielded

section when:

Elastic

YieldSurface

Plastic Hinge Criterion:P/Py

MA

AP

M/Mp

1.0

1.0

-y P=PyM=0

+y

-y P


9/16

NoAxial Force

Second-Order Effects

A.K.A. Geometric Nonlinear BehaviorEquilibrium Equations

Reality: Should be formulated on deformedshape

Difficulty: Deformed shape (deformations) isa function of the member forces, which are inturn a function of the deformations(Chicken n Egg)

Remedy: Perform a series of analyses withloads applied in small increments and updategeometry after each load increment.

Truss is susceptible to2nd-Order effects,luckily is oftenquite small.

Equilibrium Equations Formulated on

Undeformed Shape Formulated on

Deformed Shape

HP P

H

Different reactions and member forces.

k=3EI/L3

1

H

Equilibrium Equations

H

P

M=HL

k=3EI/L3

1

H

keff < k1

Effectivelateralstiffnessis reduced!

Formulated onUndeformed Shape

H

P

M=HL+P

Formulated onDeformed Shape

Focus on Lateral Stiffness Formulated on Undeformed Shape: Linear Response

Before:

kspringP

H

After:

klateral = kspring1

H

Lateral Stiffnessis slope of H-response curve


10/16

Focus on Lateral Stiffness (cont.) Formulated on Deformed Shape: Nonlinear Response

Before:

kspring P

H

After:

klateral < kspring1

H kspring1

Effective lateralstiffness is reduced

Focus on Lateral Stiffness (cont.) Equilibrium Formulated on Deformed Shape

PH kspring

R=kspring

Mo = 0 RL = HL + P

R = H + P/L

H = klateral with klateral = kspring P/L

Lateral Stiffness (slope of response curve)

Pt. o

L L

Lets start by assuming L L,

kspring = H + P/L

H = kspring P/L

H = (kspring P/L)

Some thoughts here

This simple analysis becomes less accurate as/L becomes large (i.e. /L >> 1/5) Remedy: Perform an incremental analysis and update

geometry after each load incrementhence, limit/L in each step to some small amount

Keep in mind serviceability limits are oftensomething like /L < 1/400

Most importantly, klateral = kspring P/L takes onthe form:

k2nd-Order El. = k1st-Order El. + kgGeometric Stiffness

Geometric Stiffness

Effective lateral stiffness of a member: decreases as a member is compressed

kg is negative for compressive P

backpacker example

increases when subjected to tension kg is positive for tensile P guitar string example

Employing geometric stiffness approach Other methods exist (i.e. stability functions)

How about real members? (recall)Flexural members subjected to axial force

F

Stiffness k function of: Geometry: Moment of Inertia & Length (I,k& L, k)

Material: Elastic Modulus (E ,k)

Axial Force: Compressive (P, k)

M

P

P

F = k(I,L,E,P) M = k(I,L,E,P)

Closer look at stiffness terms

Flexural members subjected to axial force

M = k(I,L,E,P) withk = 4EI/L 2PL/15

M

P

F = k(I,L,E,P) withk = 12EI/L3 6P/5L

F

P

Again, basic form:

k2nd-Order El. = k1st-Order El. + kg


11/16

Geometric Nonlinear AnalysisEmploy Direct Stiffness Method applying loads

in increments: Solve Equil. Eqs. {dF} = [K]{d}

At start of increment, modify member stiffnessto account for presence of member forces (suchas axial force): k = kelastic + kg with kg = geometric stiffness

At end of increment, update model of structuralgeometry to include displacements

Continue to accumulate results of load increments(i = i-1 + d and fi = fi-1 + df) until all of load isapplied or elastic instability is detected.

Comparison: 1st- and 2nd-Order Analysis Results

Moments increase by ~10%

2nd-Order Inelastic AnalysisEmploy Direct Stiffness Method applying loads

in increments: Solve Equil. Eqs. {dF} = [K]{d}At start of increment, modify member stiffness

to account for presence of member forces andany yielding:

k = kelastic + kgeometric + kplasticAt end of increment, update model of structural

geometry to include displacementsContinue to accumulate results of load increments

(i = i-1 + d and fi = fi-1 + df) until all of load isapplied or inelastic instability is detected.

Critical Load Analysis (Basics)Definition: Critical or buckling load is the

load at which equilibrium may be satisfiedby more than one deformed shape.

Big Q: How does computer software calculate this?

P Solution

#1 P

Solution

#2

P


12/16

Critical Load Analysis (Background)

Elastic stiffness of a member k = kel + kg kel is f(A or I, L, and E)

kg is f(P,L), also note directly proportional to PElastic stiffness of structure [K] = k

[K] = [Kel] + [Kg] [Kg] directly proportional to applied force

i.e. Double applied forces, hence, double internal forcedistribution and double [Kg]

To the computer, buckling will occur whenour equilibrium equations {F} = [K]{} permitnon-unique solutions, e.g. det[K] = 0.

Example

Demonstrate computational

method for calculating theelastic critical load (bucklingload) for the structuralsystem shown.

A, IL, E

PRigid Beam

Example: Key Stiffness Terms

vert

A, IL, E

HP

Lateral Stiffness:

klateral

H = klaterallatklateral = 12EI/L3 6P/5L

kvertical

Vertical Stiffness:

P = kverticalvert

latPRigid BeamH

Example: Solution

1. Apply reference load, and use 1st-order elastic analysis to obtaininternal force distribution.

A, IL, E

PRigid Beam

klateral = 12EI/L3 6P/5L

klateral = 0 when P = 10EI/L2

(Ptheory=9.87EI/L2)Pcr = P = 10EI/L2

Pcr

2. Determine load factor at whichsystem stiffness degrades topermit buckling.

P = 400 kipsW1265

100 kips

=9.4

=1.7

Thoughts on Critical Load AnalysisComputer analysis for a large system:

First, apply reference and perform analysis Solve equilibrium eqs. {Fref} = [K]{}

With displacements solve for member forces

Second, assemble [Kel

] and [Kg

] based on {Fref

}

Finally, determine load factor causing instability;computationally this means find load factor at which[K]=[Kel]+[Kg] becomes singular Determine at which det([Kel]+[Kg) = 0

Eigenvalue problem: Eigenvalues = Critical Load Factors, sEigenvectors = Buckling modes

Accuracy increases with more elements percom ression members (2 often ade uate)


13/16

Basic Introduction Complete

Acquire nonlinear analysis software Commercial programs Educational software (i.e. MASTAN2)

Where do I go from here? (Learning to drive)

Review the slides (Read the drivers manual) Acquire nonlinear software (Borrow a friends car)

Work lots of examples (Go for a drive, scary at first) Apply nonlinear analysis in design (NASCAR? not quite)

MASTAN2:

-Educational software-GUI commercial programs

-Limited # of pre- and post-processing optionsto reduce learning curve

-Suite of linear and nonlinear 2D and 3Danalysis routines

-Available with textbookor online at no cost

www.mastan2.com orwww.aisc.org [Steel Tools]

MASTAN2

1st-Order Elastic: [Ke]{}={F}

Levels of Analysis:

2nd-Order Elastic: [Ke + Kg]{d}={dF}

1st-Order Inelastic: [Ke + Kp]{d}={dF}

2nd-Order Inelastic: [Ke + Kg + Kp]{d}={dF}

Critical Load: [Ke + Kg]{d}={0}

Yield Surface:

Function of P, Mmajor, and Mminor

1

st

-Order

Elastic

Lateral displacement,

2

nd

-Order

Elastic

Elastic Stability Limit

Hes

Lateral

load,

H

Inelastic Limit Load

st

-Order

Inelastic

Hp

H= P

H= P

P

P

Elastic Critical Load

Inelastic Critical Load

He

Hi

2

nd

-Order

Inelastic

Inelastic Stability

Limit

His

Actual

MASTAN2

1

st

-Order

Elastic

u

in.)

Load

Factor

First

hinge

Second hinge

8783 1

st

-Order)

8783 2

nd

-Order)

14

1831

1665

1720

Mechanism Moments at Limit in. kips)

Planar Frame:

2

nd

-Order

Elastic

2

nd

-Order

Inelastic

1

st

-Order

Inelastic

E= 29,000 ksi

y = 36 ksi

Summary and Conclusions

Provided an introduction to nonlinear analysis Review of direct stiffness method

Material nonlinear analysis (Inelastic hinge)

Geometric nonlinear analysis (2nd-Order) 2nd-Order inelastic analysis (combine above)

Critical load analysis (eigenvalue analysis)

Nonlinearthink modifying member stiffness!Overview and availability of MASTAN2Now, its your turn to take it for a spin


14/16

Appendix

Several examples to try out

Solutions by MASTAN2Need a reference text with many

examples? see Matrix Structural Analysis,2nd Ed., by McGuire, Gallagher, and Ziemian(Wiley, 2000)

See tutorial that comes with MASTAN2OK, time to jump in and start driving

Multi-story

Frame:

Demonstrate:

2

nd

-Order Inelastic Analysis

Non-proportional loading

Gravity Load

a

t

e

r

a

l

o

a

d

E= 29,000 ksi

y = 36 ksi

P

0.5P

P

P

P

P

P

P

P

50 kips 25 kips25 kips

MASTAN2

Model:

Gravity Load

2nd-Order

Inelastic

Analysis:

P = 30 kips

2nd-Order

Inelastic

Analysis:

P = 37.7 kips

2nd-Order

Inelastic

Analysis:

Limit State:

P

lim

= 40.3 kips


15/16

2nd-Order

Inelastic

Analysis:

Post-limit State:

P = 40 kips

1st

Plastic Hinge (P=30 kips)

2nd-Order

Inelastic

Analysis:

Strength Limit State (P=40.3 kips)

All members:

A = 9.34810-2 in2

I = 6.954 10-4 in4

E = 29,000 ksi

P = 100 lbs

20

3 @ 20

3P

3P

Truss Hoff et al.):

Demonstrate:


2

nd

-Order Elastic

Experimental P

limit

=220 lbs)

MASTAN2

Model:

Elastic Critical Load: P

cr

= 210.7 lbs 2

nd

-Order Elastic: P

lim

= 210 lbs


16/16

Response

Curves:


2

nd

-Order Elastic

1

st

-Order

Elastic

P = 210 lbs

Beam-Column:

Demonstrate:

Elastic Critical Load Analysis

1. Flexural Buckling

=0.0)

2. Torsional Flexural Buckling

=0.04)

L = 24

PM =

PL

M = PL

W2476E = 29,000 ksi

MASTAN2

Model:


= 0.0)

Isometric View Elevation View Plan View

Elastic Critical Load Analysis = 0.04)

Isometric View Elevation View Plan View

v

mi

in.)

Load

Factor

Suspension System:

2

nd

-Order

Inelastic

1

st

-Order

Inelastic

A = 5.40 in2

y = 150 ksi

A = 50 in2

I = 20,000 in4

Z = 1,000 in3

y = 50 ksi

1.5

1

0.5

2

0

0 50 100 150 25000

1

st

-Order Inelastic

2

nd

-Order Inelastic

Hinge Formation

nonlinear introduction

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