S. Socrate 2013 K. Qian

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 à N(x) = FC + FB

for x>xB :Σ Fx = 0 = – N(x)+FC à 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

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 à 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 ßà section deformation

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)à (EA)eff=E(x)A(x);

If 2 materials (E1, E2) à (EA)eff= E1 A1 + E2 A2

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

Solution Procedures:

1)  Force Method for Statically Indeterminate (SI) Bars in Axial Loading

•  Remove redundant support à Replace with redundant reaction

•  Solve the companion Statically Determinate (SD) problem

•  Obtain the displacement at the redundant support in terms of the redundant reaction

•  Impose zero displacement at the redundant support and obtain the redundant reaction

•  Back-substitute the redundant reaction in the solution to the companion SD probem to find the solution to the SI problem.

2)  Method of Joints for Statically Determinate trusses

•  Determine the free and constrained DOFs of the joints in the truss

•  Draw FBDs of the joints for the free DOFs. Draw the unknown axial forces, N , positive, i.e., coming out of the joints.

•  Impose equilibrium of the joints along the free DOFs (ΣFx=0 and/or ΣFy=0). Obtain all the axial forces in the bars.

•  Draw FBDs of the joints for the constrained DOFs. Now the axial forces are known: draw them the way they act (pushing the joint if N is compressive, pulling if N is tensile) indicating their magnitude in the FBDs. Draw the cartesian components of the unknown reactions positive ( along x and y)

•  Impose equiibrium of the joints along the constrained DOFs (ΣFx=0 and/or ΣFy=0). Obtain the cartesian components of the reactions at the supports. Check global equilibrium!

3.  Geometry of deformation in trusses

1.  For displacing joint, J , with displacement vector uJ: {uJx , uJ

y}, subjected to unknown load PJ: {PJ

x ,PJy}:

2.  Draw each bar, e.g., AJ, connected to joint J

•  for each bar obtain the cartesian components of the unit vector along the bar pointing toward the moving joint, J (e.g., for AJ : rAJ: {r AJ

x , r AJy})

•  obtain the elongation of each bar by scalar product between the vectors e.g., for AJ :

δAJ=rAJ�uJ=r AJx uJ

x+ r AJy uJ

y

•  obtain the axial forces in each bar due to the displacement of joint J , e.g.,

N AJ=KAJδAJ

3.  Draw the FBD of the moving joint. Draw the cartesian components of the unknown external load PJ and impose equilibrium along x and y à obtain {PJ

x ,PJy}

4.  Draw FBDs of each support pin (e.g., A) along the constrained DOFs. Draw the cartesian components of the unknown reactionàobtain {RA

x ,RAy}

5.  If more than one joint moves: use SUPERPOSITIONàsolve problem for each moving joint and superpose the solutions.