11Apr 21, 2023

Last course Interpretations and properties of the

stiffness matrix (cont’d) The DSM for plane and space trusses

2

Today

One-dimensional bar element

3

Example

See Example 3.8 from D.L. Logan (2007).

E=1.2 x 106 psi for all elements.

See the Matlab code, Lect5_Truss2.m for the solution.

Answer: σ1=-945 psi; σ2=1440 psi; σ3=-2850 psi

4

Homework 4 Given a space truss structure as

shown here. A=1.53e-3 m2; E= 75 GPa Determine the nodal

displacements, support reactions, and the element internal forces (use Matlab).

Use SAP2000 and compare your results.

Instead of use truss model in SAP2000, use rigid frame model and compare the results.Source: http://www.andrew.cmu.edu/course/24-ansys/htm/s4_truss.htm

5

Truss Structures in SAP2000

In SAP2000, there is no special truss element To do analysis of a truss structure, use Frame

element If the truss is an ideal truss (see the truss

structure assumptions), you may either: Set the geometric Section properties j, i33, and i22 all

to zero, or Release both bending rotations, R2 and R3, at both

ends and release the torsional rotation, R1, at either end.

6

One-dimensional Bar Element– Principal-of-Virtual Work Approach

7

From direct to virtual work approach So far we have derived the spring element and the bar

element by directly applied the physical laws (direct approach), i.e. Hooke’s law and force equilibrium law.

More complex finite elements, however, may not be able to be formulated using the direct approach.

So we turn to alternative approach for deriving the element stiffness equation.

There are several methods for obtaining the stiffness equation, but in this course we are using the principle of virtual work.

8

Basic concepts from the theory of elasticity Consider a 1D bar as shown here.

Assume that the bar has: Length L Cross-sectional area A Modulus of elasticity E Coefficient of thermal expansion α

x, u

b=b(x)

L

X=0 X=L

P

9

A, E and α may vary along the bar, i.e. A=A(x), E=E(x), and α= α(x)

The bar is subjected to: Distributed axial loal b=b(x), in the unit of [force]/[length],

Note that this load is in the category of body force because it acts at every point in the body

Concentrated force P at the right end, and Temperature change T 0C (may also vary along the bar).

The loads and temperature change causes the bar deforms, which can be described completely by an axial displacement field,

x, u

b=b(x)

L

X=0 X=L

P

u=u(x)

10

In the theory of elasticity, there are three basic set of equations, i.e. Strain-displacement equations Stress-strain equations Equations of equilibrium

To obtain strain-displacement equations, consider the free body diagram of the bar segment between sections at distances x and x+Δx:

u(x): displacement of the section at distance x u(x+Δx): displacement of the section at distance x+Δx

x x+Δx

u(x) u(x+Δx)

Δx

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The average strain in the bar segment is

If Δx approaches zero, then the average strain becomes the strain at any point in the section located at distance x, i.e.

( ) ( )u x x u x

x

u(x) u(x+Δx)

Δx

Δ 0

( ) ( )( ) lim ( ) ,x

x

u x x u x dx u x u

x d x

12

Thus, the strain-displacement equation for the bar is

The strain, ε, at each point of the bar consists of two parts, i.e. The strain due to thermal expansion, The strain due to external loads,

If the bar is not restrained, then the thermal strain will not induce any thermal stress.

The strain that induces stress is

( )d u

xd x

0 T P

0 T

…(1)

13

The stress-strain equation is given by Hooke’s law, viz.

Now, consider the force equilibrium of the bar segment:

: average axial force along the bar segment

Force equilibrium:

( )E T

b

σ(x) σ(x+Δx)

Δx

b

( ) ( ) ( ) ( ) 0x x A x x x A x b x

…(2)

14

N(x): internal normal (axial) force in the bar at distance x If Δx approaches zero, then

( ) ( ) ( ) ( ) 0x x A x x x A x b x

( ) ( ) 0N x x N x b x

( ) ( )0

N x x N xb

x

0

( ) ( )lim ( ) 0x

N x x N xb x

x

σ(x) σ(x+Δx)

Δx

b

15

Thus, the equation of equilibrium for the bar is

Or shortly,

We can obtain the governing equation for the 1D bar using Eqs. (1), (2) and (3). Multiplying Eq. (2) by A results in

( )( ) 0

dN xb x

dx

, 0xN b …(3)

( ) ( ( ) )N x EA x T …(4)

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Substituting Eq. (1) into Eq. (4), we obtain

Now substituting this equation into Eq. (3)

If E, A, α and T constant along the bar, then the equation simply becomes

( ) ( ( ) )N x EA x T …(4)

( )( ) ( )

du xN x EA T

dx …(5)

( )( ( )) ( ) 0

d du xEA T b x

dx dx

2

2

( )( ) 0

d u xEA b x

dx

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In conclusion, the governing equation for the bar is

Given the external loads and displacement boundary condition, utilizing this 2nd order differential equation we can obtain the displacement field, the strain field and the stress field.

x, u

b=b(x)

L

X=0 X=L

P

( )( ( )) ( ) 0

d du xEA T b x

dx dx

1818Apr 21, 2023

Recall the example of bar structure

Find the exact solutions for the displacement, strain, and stress along the bar and compare to the finite element solutions.

0x Tx L

03A A 0A A

1919Apr 21, 2023

For this problem, p(x)=0 and T=0. Thus, the governing equation can be written as

The area of a cross section at a distance x:

Boundary conditions:

x L

duEA F

dx

0T

2( ) ( 3)A x A x

L

T0 x L

00

xu

Displacement/geometrical/essential boundary condition

x ; uFH ( )( ( ) ) 0

d du xEA x

dx dx

Force/mechanical/natural boundary condition

2020Apr 21, 2023

The solution of the governing equation is

The strain field is

The stress field is

T T

0 T

3( ) ln

2 3 2

FL Lu x

EA L x

T

0 T

( )( )

3 2

Ldu x Fx

dx EA L x

T

0 T

( ) ( )3 2

LFx E x

A L x

Displacement field

21

Recall the finite element solutions

2 T

3 0

3/15

8 /15

D FLD EA

1 11 1 2 1

T 0

2

0.5 5x x

d d FE E

L A

2 22 2 2 1

T 0

2

0.5 3x x

d d FE E

L A

2222Apr 21, 2023

Medan Perpindahan

0

0.1

0.2

0.3

0.4

0.5

0.6

0 0.2 0.4 0.6 0.8 1

X (L T)

u (

FL

T/E

A0)

Pendekatan

Eksak

2323Apr 21, 2023

Medan Tegangan

0.00

0.20

0.40

0.60

0.80

1.00

1.20

0 0.2 0.4 0.6 0.8 1

X (L T)

( F

/A0 ) Pendekatan

Eksak

24

Strain energy

Energy The capacity or ability to do work.

Work A product of a force and a distance in the direction

where the force moves. Elastic strain energy (or elastic deformation

energy) Energy stored in a solid body in the deformed state. The capacity of internal forces (or stresses) to do

work through deformation (strains) in the solid body.

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Strain energy for a bar

A bar has only one stress component along its axis (uniaxial stress condition)

Consider an infinitesimal element subjected to normal stress σx only

x xN dy dz

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Due to Nx, the element extends Δx

Assume that the material is elastic-linear

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The work done by Nx

Thus, the strain energy of the infinitesimal element is

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The strain energy per unit volume is then

The strain energy per unit volume is called strain energy density, U0

U0 is equal to the area under the stress-strain curve and above the strain axis.

The strain energy for the whole solid body:12 x x

V

U dV

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The strain energy for the bar:

x, u

b=b(x)

L

X=0 X=L

P

120

( ) ( ) L

U x x Adx

30

Problem example Consider the following two bars

Suppose that the bars absorb the same amount of energy corresponding to the axial load

Neglecting stress concentration, compare the axial loads acting on bar 1, P1, and bar 2, P2

1 2

31

Solution Bar 1

Bar 2

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Solution (cont’d)

33

Homework 5 (due date: FEM Mid-sem Exam Time)1. Show how to obtain that the solutions for the

displacement and strain fields of the tapered bar.

T T

0 T

3( ) ln

2 3 2

FL Lu x

EA L x

T

0 T

( )( )

3 2

Ldu x Fx

dx EA L x

x ; uFH

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2. Compute the displacement and stress at the right end of the bar, i.e. u(LT) and σ(LT) by modeling the bar using finite elements. The area of each element is taken to be equal to the section area at the midpoint of the element.

a. Use one element, with L1=LT

b. Use two elements, with L1=L2=LT/2

c. Use three elements, with L1=L2=L3=LT/3

d. Use four elements, with the length of each element LT/4

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3. Examine the convergence of the finite element solutions in Problem #2 by:

a. Computing the relative error (%) for each solution, both for the displacement and the stress.

b. Draw convergence graphs for the displacement and the stress, with the abscissa is the number of element and the ordinate is the solutions, including the exact solution. Give your comment regarding the finite element solutions.

4. Prove that the exact strain energy in the bar is

T

0

ln 3

4

FLU

EA

36

The principle of virtual work A virtual (imaginary) displacement is an imaginary and

very small change in the configuration of a system The principle of virtual work (or more precisely the

principle of virtual displacement) is stated as follows:

If a body in equilibrium is subjected to arbitrary virtual displacements associated with a compatible deformation of the body, the virtual work of external forces on the body is equal to the virtual strain energy of the internal stresses

Compatible (admissible) displacements are those that satisfy the boundary conditions and ensure that no discontinuities, such as voids or overlaps

37

The principle of the virtual work

δU: the virtual strain energy of the internal stresses

δW: the virtual work of external forces on the body

X, u

Z, w

w

δw

δw is the virtual displacement

38

The virtual strain energy for a bar

δε : virtual strain, which is the strain associated with the virtual displacement, i.e.

The external virtual work:

0( ) ( )

LU x x Adx

( ) ( )d

x u xdx

x, u

b=b(x)

L

X=0 X=L

P

0( ) ( ) ( )

LW u x b x dx P u L

39

Notice that

Why is it not equal to the definition of actual strain energy?

120

( ) ( ) L

U x x dV

Discussion

40

Formulation of bar element Consider the two-node bar element as follows

b(x) : body force along the bar element in the unit of force per unit length (e.g. kN/m)

u=u(x) : displacement field within the element, i.e. displacement as a function of point x within the element

0 x L

41

Displacements at nodes 1 and 2:

Forces at nodes 1 and 2:

The first and fundamental step in the finite element formulation is to assume the displacement field within the element in terms of its nodal displacements

42

Here it is assumed that the displacement field is a linear function

Let us now express the assumed function in terms of nodal displacements d1 and d2

43

As a result,

44

Written in matrix form

Thus we can write

45

Where

N is also called the matrix of interpolation functions, because it interpolates the displacement field u=u(x) from the nodal displacements

L x x

L L

N

1

2

d

d

d

Matrix of shape functions

Vector of nodal displacements