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Loading Conditions on each Section (x) Applied loading only along the axis (x) of the bar. The only internal resultant at any sections ┴ x is the axial force N(x)

Find N(x)along the bar (axial force diagram) by cutting the bar at each x and imposing x-equilibrium.

For the example shown, equilibrium at x gives: for x<xB :Σ Fx = 0 = – N(x) +FC +FB ! N(x) = FC + FB

for x>xB :Σ Fx = 0 = – N(x)+FC ! N(x) = FC

And the entire axial force diagram is:

For distributed loading fx(x) (with fx (x) in N/m + along x), obtain the force by integrating fx(x) along the bar. For the bar shown:

dxxfxL

xx∫= )()N(

The differential relationship between the distributed load fx (x) and the axial force N(x) is

)(xfdxxd

x−=)N(

This can be directly obtained from ΣFx=0 on a dx slice of the bar

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Structural response Elongation: Displacement field : with ux(x0)=u0 determined by Boundary conditions (e.g., u0 =0 at support)

Section deformation Section at x has displacement ux(x) Section at x+dx has displacement ux(x+dx)= ux+dux Local measure of deformation at section x : (change in length)/(original length) à

dxxdxdxdu L

a

Lx ∫∫ ==

00

)(εδ

∫∫ +=+=x

xa

x

x

xxx dxxudx

dxduxuxu

00

)()()( 00 ε

cross sections stay flat ! Same εa at all points of section

)()( xdxdux x

a =ε

Kinematics constraint (geometry of deformation) Cross sections ┴ x : stay flat, translate by ux(x)

dxdux

Strain !" section deformation

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dAzyx(xA

n∫= ),,() σN

Section equilibrium The Axial Force N(x) at section x is obtained by integrating the contributions of each elemental area dA, which carries a normal stress σn

Constitutive Properties If the material is linear elastic, and the modulus of elemental area dA is E, the stress can be obtained as: )(),,(),,( xzyxEzyx an εσ ⋅=

Section Response

dAzyxExEA

xEA(xx

dxduxEA

dxdudAzyxE

dxdu(x

dAdxduzyxEdAxzyxEdAzyx(x

Aeff

eff

xinverteff

x

A

x

A

x

Aa

An

∫

∫

∫∫∫

=

=⎯⎯ →←==

===

),,()()(

)()())()()(),,()

),,()(),,(),,()

NN

N εσ

Constant over cross section

Effective Section Stiffness:

If only 1 material, E(x)! (EA)eff=E(x)A(x);

If 2 materials (E1, E2) ! (EA)eff= E1 A1 + E2 A2 S. Socrate 2016

Special case: homogeneous bar (modulus E) ; constant cross section A ; constant axial force

! !

;

material theof Modulus sYoung' :

strain axial :

stress normal :

bar theof stiffness axial :

/1

,

δδ

εσ

δε

σ

δδ

KK

geometrymaterial

a

n

a

n

LAEPP

EAL

LAEK

E

L

A

PK

==

=

=

=

=

==

"#$

N

N

Equilibrium (x)

ΣFx=0 à N =P (constant along bar)

N

δ

1 K

εa

1 E

σn

structure

material

0

0

0

δ = L - L0 : elongation of the bar

L

0

0

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ΣFx=0 à N

σ n ( x , y ,z )= E ( x , y ,z )⋅εa ( x )

N (x )= σ n ( x , y ,z )A∫ dA

Uniaxial Loading

Sections ┴ x: stay flat, translate along x by ux(x)

dux ( x ) / dx

εa ( x )=dux ( x )dx

ux ( x )= ux ( x0 )+duxdxdx '

x0

x∫ = u0 + ε( x ')dx '

x0

x∫

δ =duxdxdx

0L∫ = ε( x )dx

0L∫

Internal Resultant obtained by imposing Kinematic assumption

Stress ↔ Resultant

Deformation at section x

Strain↔ Stress

Section Deformation ↔ Strain

Resultant ↔ Deformation

Effective Section Stiffness …++=

= ∫

2211)()(

),,()()(

AEAExEA

dAzyxExEA

eff

xA

effx

Displacement field and Elongation

Synoptic Table

)()()(

xEA(xx

dxdu

eff

x )N=

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Solution Procedure

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Solution Procedure

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