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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
S. Socrate 2016
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
S. Socrate 2016
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
S. Socrate 2016
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=
S. Socrate 2016