The Finite Element Method for the Analysis of Non-Linear and ...

The Finite Element Method for the Analysis of 

Non-Linear and Dynamic Systems 

Prof. Dr. Eleni Chatzi 

Lecture 6 - 5 November, 2012 

Institute of Structural Engineering Method of Finite Elements II 1

2D Axisymmetric, Plane Strain and Plane Stress Elements 

The nodal point coordinates determine the spatial configuration of the 

element at time 0 and t using: 

n∑ 

n∑ 

0 x 1 (r) = N 0 k x k 1 

0 x 2 (r) = N 0 k x k 2 

and t x 1 (r) = 

k=1 

n∑ 

N t k x k 1 

k=1 

t x 2 (r) = 

k=1 

n∑ 

N t k x k 2 

k=1 

Isoparametric Elements ⇒ t u i (r) = ∑ n 

k=1 N t k uk i , i=1,2 



Shape Functions 

n = 4 nodes 

N1 a = 1 4 (1 + r)(1 + s), N 2 a = 1 4 (1 − r)(1 + s), N 3 a = 1 (1 − r)(1 − s), 

4 

N4 a = 1 (1 + r)(1 − s) 

4 

n = 8 nodes 

N b 5 = 1 2 (1 − r2 )(1 + s), N b 6 = 1 2 (1 − s2 )(1 − r), N b 7 = 1 2 (1 − r2 )(1 − s), 

N b 8 = 1 2 (1 − s2 )(1 + r) 

N b 1 = N a 1 − 1 2 N b 5 − 1 2 N b 8 , N b 2 = N a 2 − 1 2 N b 6 − 1 2 N b 7 , N b 3 = N a 3 − 1 2 N b 6 − 1 2 N b 7 , 

N4 b = N 4 a − 1 2 N 7 b − 1 2 N 8 

b 

n = 9 nodes 

N9 c = (1 − r2 )(1 − s 2 ), 

Ni c = N i b − 1 4 N 9 c, i=1,2,3,4 

Nj c = N j b − 1 2 N 9 c, i=5,6,7,8 









Example 

Establish the Total Lagrangian Strain - Displacement matrices for 

the following element (assuming large displacements/large strains): 



Example 

The nodal coordinates at time 

t are: 

t 

u 1 2=0.5 

t 

u 1 1= 1 

t u 1 1 = 1 

t u 2 1 = 0 

t u 3 1 = 0 

t u 4 1 = 1 

t u 1 2 = 0.5 

t u 2 2 = 0.5 

t u 3 2 = 0 

t u 4 2 = 0 

The Jacobian for a 2D system is defined as: 

009 Example 6.18 

7 

⎡ 

⎤ ⎡ ⎤ 

0 J = 

⎢ 

⎣ 

∂ t x 1 

∂r 

∂ t x 1 

∂s 

∂ t x 2 

∂r 

∂ t x 2 

∂s 

⎥ 

⎦ = 

⎢ 

⎣ 

3 

2 

0 

0 

3 

2 

⎥ 

⎦ 



Example - Linear Strain Displacement Matrix component t 0B L0 

Using the 4 node 2D element shape functions and the relevant 

matrix we obtain: 



Example - Nonlinear Strain Displacement Matrix component t 0B L1 

We first need to calculate the following terms: 



Example - Linear Strain Displacement Matrix component t 0B L1 



Example - Nonlinear Strain Displacement Matrix component t 0B NL 

Using the 4 node 2D element shape functions and the relevant matrix we 

obtain: 


The Beam Element 

*The section on the Beam Element is taken from Prof. H. Waisman’s 

notes of the FEM II course - CEEM Department, Columbia University 

F-16 Aeroelastic Structural Model 

Exterior 

model 

95% are 

shell 

elements 

FEM model: 

150000 Nodes 

Internal structure 

zoom. Some Brick 

and tetrahedral 

elements 

http://www.colorado.edu/engineering/CAS/Felippa.d/FelippaHome.d/Home.html 


Beam Elements 

Two main beam theories: 

Euler-Bernoulli theory (Engineering beam theory) -slender beams 

Timoshenko theory thick beams 

Euler - Bernoulli Beam 


Beam Elements 

Euler Bernoulli Beam Assumptions - Kirchhoff Assumptions 

Normals remain straight (they do not bend) 

Normals remain unstretched (they keep the same length) 

Normals remain normal (they always make a right angle to the neutral 

plane) 


Beam Elements - Strong Form 

υ´ υ´ 

υ 



Equilibrium 

– distributed load per unit length 

– shear force 

Combining the equations 



(1) 

Free end with applied load 

(2) 

(S) 

(3) 

Simple support 

(4) 

(5) 

Clamped support 


Beam Elements - Strong Form to Weak Form 

Multiply Eqns. Multiply (1), Eq. (4) (1), (5) by (4) w(5) and by integrate w and integrate over theover domain the domain 

First integration by parts 

First integration by parts 

First integration First integration by parts by parts 

Second integration by parts gives 





Beam Elements - Strong Form to Weak Form 

Arrive at the weak form 

(W) 

Note: 

1. The spaces are C 1 continuous, i.e. the derivative must also be 

continuous 

Note: 

1. The spaces are C 1 continuous, i.e. the derivative must also be 

continuous 

2. The left side is symmetric in w and υ (bi-linear form: 

a(υ, w)=a(w, υ)) this will lead to a symmetric Stiffness Matrix 

2. The left side is symmetric in w and v (bi-linear form: a(v,w)=a(w,v) 

this will lead to symmetric stiffness matrix 


Beam Elements - FE Formulation 

Physical domain 

Natural domain 

Element 

displacement 

vector 

Element 

force 

vector 


Beam Elements - Shape Functions 

Hermite 

Hermite 

Polynomials 

Polynomials 

Note: The choice of a cubic polynomial is related to the homogeneous 

strong form of the problem EIυ ′′′′ = 0. 


Beam Elements - Shape Functions 

The displacement is approximated by 

However note : 

From coordinate transformation (mapping) 


Beam Elements - Galerkin 

Finally, the weight functions and trial solutions are 

Finally, the weight functions and trial solutions 

Where the shape functions are 


Beam Elements - FE Matrices 

From 

From 

the weak 

the 

form, 

weak 

we 

form, 

had 

we had 

The second The second derivative derivative becomes becomes 

and 


Beam Elements - FE Matrices 

Stiffness matrix 

Force vector 

Assuming constant distributed force 


Beam Elements - Example 

Consider a clamped-free beam 

with EI = 10 4 Nm 2 

s = −20N 

m = 20Nm 

Consider a clamped-free beam 

Pre-processing 



For element (1) For element (2) 

[1] [2] [3] [4] [3] [4] [5] [6] 

Assembly into a Global Stiffness Matrix 



Boundary force matrix 

Element (1) has no boundary on Γ S or Γ M 

For element (2) we have 

[3] 

[4] 

[5] 

[6] 

Assembly to global boundary force vector 



Body force vector 

(distributed 

loads) 

(Point loads) 

For element (1) Given: 

For element (2) Given: 



The global force vector 

-9 

-15.3 

-4 

15.3 

-20 

20 

=KNOWN 

Post-processing 


Timoshenko Beam 

In a Timoshenko Beam, a plane normal to 

the beam axis before deformation doesnt 

remain normal after deformation (short and 

thick beams, sandwich composite beams). 

i.e. θ ≠ dv 

dx : 

γ = −θ + dv Transverse Shear strain 

dx 

ε = −y dθ Normal strain 

dx 

The strain energy of an element may be written as 

G - shear modulus 

b - widthh 

µ - correction factor 

h - height 



Chapter 9: THE TL TIMOSHENKO PLANE BEAM ELEMENT 9–8 

2-node C 1 (cubic) element 

for Euler-Bernoulli beam model: 

plane sections remain plane and 

normal to deformed longitudinal axis 

2-node C 0 linear-displacement-and-rotations 

element for Timoshenko beam model: 

plane sections remain plane but not 

normal to deformed longitudinal axis 

(a) 

(b) 

C 1 element 

with same DOFs 

Figure 9.6. Sketch of the kinematics of two-node beam finite element models based on 

(a) Euler-Bernoulli beam theory, and (b) Timoshenko beam theory. These 

models are called C 1 and C 0 beams, respectively, in the FEM literature. 



Substituting the relations into U 



I - moment of inertia 

I -moment - A -of of cross-sectional inertia, A-cross-sectional area area 

A 

- cross-sectional area 

Galerkins 

Galerkin’s 

Approximation 

Approximation 

(Linear 

(Linear 

shape 

shape 

functions) 

functions) 

Galerkin’s Approximation (Linear shape functions) 

Element stiffness 





Exact integration 

Shear Locking 

Example: Cantilever Beam, 

20 elements, different 

span L to depth h 

Reduced 1-point quadrature integration 

Ratio of tip displacement for 

Timoshenko beam theory and 

Euler-Bernoulli theory 



The Shear locking Phenomenon 

This occurs when significant bending is present. It is owed to the use 

of linear shape functions that cannot accurately model the curvature 

that is present in the actual material under bending condition. 

Instead, a shear stress is introduced which causes the element to 

reach equilibrium conditions for smaller displacements that the real 

ones. 

The element therefore appears stiffer. 


The Beam Element in Large Displacements 

Assumptions: 

Plane sections initially normal to the neutral axis remain plane 

The longitudinal (ε ηη) and the two shear stresses(ε ηξ , ε ηζ ) are the only non 

zero ones. 

A general 3D beam element would then be: 



Along the lines of the truss element formulation we need to express the 

coordinates of a random point within the beam element. Using (r, s, t) as 

the Cartesian coordinates at a point within an element with N nodal 

points, this is written: 

t x i = 

N∑ 

N t k x k i + t 2 

k=1 

N∑ 

k=1 

a k N k t V k 

ti + s 2 

N∑ 

k=1 

b k N k t V k 

si, i = 1, 2, 3 

Vectors V s and V t define the orientation of the cross-section for the beam: 

They are normal to the axis of the beam and to each other. The values a 

and b define the size of the cross section of the beam. 

The relative displacement components would be: 

t u i = t x i − 0 x i 

u i = t + ∆ t x i − t x i 

are the incremental components 



This leads to the following formulations 

t 

u i = 

N∑ 

N kt u k i + t 2 

k=1 

N∑ 

k=1 

a k N k ( t V k 

ti − 0 V k ti) + s 2 

N∑ 

k=1 

b k N k ( t V k si − 0 V k 

si) 

and u i = 

where 

N∑ 

N k u k i + t 2 

k=1 

N∑ 

k=1 

a k N k V k 

ti + s 2 

N∑ 

k=1 

Vti k = t + ∆ t Vti k − t Vti 

k 

Vsi k = t + ∆ t Vsi k − t Vsi 

k 

b k N k V k 

si, i = 1, 2, 3 



We need to express the components V k 

ti , ZV k 

si of the vectors Vk t , V k s in 

terms of the nodal rotational degrees of freedom per node k: 

θ T k = [ θ k x θ k y θ k z 

This in linear analysis would be written as: 

V k t = θ k × t V k t 

V k s = θ k × t V k s 

but in the case of large displacements a second order Taylor approximation 

needs to be used: 

V k t = θ k × t V k t + 1 2 θ k × (θ k × t V k t ) 

V k s = θ k × t V k s + 1 2 θ k × (θ k × t V k s) 

] 



The finite element equations in this case are: 

t K 

⎡ 

⎢ 

⎣ 

. 

u k 

θ k 

. 

⎤ 

⎥ 

⎦ = t + ∆t R − t F (1) 

Having solved Eqn (1) for bfu k , θ k , we obtain the approximations for the 

nodal point displacement and direction vectors: 

t + ∆t u k = t u k + u 

∫ k 

t + ∆t V k t = t Vt k + dθ k × τ Vt 

k 

V 

∫ k 

t + ∆t V k s = t Vs k + dθ k × τ Vs 

k 

V k 

The above formulation corresponds to an iteration of the Newton-Raphson 

method 



Example: 

Assume the following 2-node beam element. Evaluate the coordinate and 

displacement interpolations and derivatives required for the setup of the 

strain displacement matrices in the TL and UL formulation. 



Example: 

First we need to calculate the nodal coordinates at times t and 0 ( t x 1 , t x 2 ), 

( 0 x 1 , 0 x 2 ). Using the corresponding geometries and the following formula: 

t x i = 

N∑ 

N t k x k i + t 2 

k=1 

N∑ 

k=1 

a k N k t V k 

ti + s 2 

N∑ 

k=1 

where N k are linear shape functions, we obtain: 

b k N k t V k 

si, i = 1, 2, 3 



Example: 

Hence, the displacement components are at time t given by the following 

formula: 

N∑ 

t u i = N t k u k i + t N∑ 

a k N k ( t Vti k − 0 Vti) k + s 2 

2 

k=1 

k=1 

N∑ 

k=1 

b k N k ( t V k si − 0 V k 

si), i = 1, 2 

hence: 



Example: 

For the calculation of the incremental displacements we will need to 

calculate the rotational components V s t (since V k t = 0). Using the 

following formula: 

V k s = θ k × t V k s + 1 2 θ k × (θ k × t V k s) 

we get: 

θ k × t V k s = 

∣ 

e 1 e 2 e 3 

0 0 θ 

−sin( t θ k ) cos( t θ k ) 0 

⎡ 

∣ = ⎣ 

−θ k cos( t θ k ) 

−θ k sin( t θ k ) 

0 

⎤ 

⎦ 

and θ k × (θ k × t V k s) = 

∣ 

e 1 e 2 e 3 

0 0 θ 

−θ k cos( t θ k ) −θ k sin( t θ k ) 0 

⎡ 

∣ = ⎣ 

θ 2 k sin(t θ k ) 

−θ 2 k cos(t θ k ) 

0 

⎤ 

⎦ 



Example: 

Then, the incremental displacements are given by 

and u i = 

N∑ 

N k u k i + t 2 

k=1 

N∑ 

k=1 

a k N k V k 

ti + s 2 

N∑ 

k=1 

b k N k V k 

si, i = 1, 2 

hence: 



Example: 

Finally, the required derivatives for the TL and UL formulation will be: 

where we assumed t L = 0 L = L. 



Example: 

Also, the Jacobian will be: 

And the associated displacement derivatives are: 

These will be used to derive the strain-displacement matrices B and finally the 

stiffness matrix K 


Special Considerations-Geometric Stiffness 

An alternate approach to the Large Displacement Problem for practical 

considerations (namely truss, beam problems) 

A cable, when subjected to a large tension force, has an increased lateral stiffness. 

If a long rod is subjected to a large compressive force, and is on the verge of 

buckling, we know that the lateral stiffness of the rod has been reduced 

significantly and a small lateral load may cause the rod to buckle. This general 

type of behavior is caused by a change in the “geometric stiffness of the 

structure. This stiffness is a function of the load in the structural member and 

can be either positive or negative. 



Cable Element 

The fundamental equations for the geometric stiffness for a rod or a cable are 

very simple to derive. Consider the horizontal cable, of length L with an initial 

tension T. If the cable is subjected to lateral displacements, v i and v j , at both 

DYNAMIC ANALYSIS OF STRUCTURES 

ends, as shown, then additional forces, F i and F j , must be developed for the 

cable element to be in equilibrium in its displaced position. 

T 

Fi 

´ 

Deformed Position 

F j 

T 

v i 

´ 

i 

L 

´ 

´ 

j 

v j 

T 

T 

Note that we haveFigure assumed 11.1. all forces Forces andActing displacements on a Cable are positive Element in the up 

direction. We have also made the assumption that the displacements are small 

Taking and do not moments changeabout the tension point j in in the the cable. deformed position, the following equilibrium 

equation can be written: 



Taking moments about point j in the deformed position, the following 

equilibrium equation can be written: 

F i = T (vi − vj) 

L 

And from vertical equilibrium the following equation is apparent: 

F i = −F j 

Combining the above the lateral forces can be expressed in terms of the lateral 

displacements by the following matrix equation: 

[ ] 

Fi 

= T [ ] [ ] 

1 −1 vi 

or symbolically, F 

F j L −1 1 v G = K Gv 

j 

Note that the 2 × 2 geometric stiffness, K G , matrix is not a function of the mechanical 

properties of the cable and is only a function of the elements length and the force in the 

element. Hence, the term “geometric or “stress stiffness matrix is introduced as opposed 

to the “mechanical stiffness matrix which is based on the physical properties. The 

geometric stiffness exists in all structures; however, it only becomes important if it is 

large compared to the mechanical stiffness of the structural system. 



Beam Element 

In the case of a beam element with bending properties in which the deformed 

shape is assumed to be a cubic function due to the rotations φ i and φ j at the 

ends, additional moments M i and M j are developed. The force-displacement 

relationship is given by the following equation: 

⎡ 

⎢ 

⎣ 

F i 

M i 

F j 

M j 

⎤ 

⎥ 

⎦ = 

T 

30L 

⎡ 

⎢ 

⎣ 

⎤ 

36 3L −36 3L 

3L 4L 2 −3L −L 2 

⎥ 

−36 −3L 36 −3L ⎦ 

3L −L 2 −3L 4L 2 

⎡ 

⎢ 

⎣ 

v i 

φ i 

v j 

φ j 

⎤ 

⎥ 

⎦ 

or FG = KGv 

The well-known elastic force deformation relationship, for a prismatic beam 

without shearing deformations, is 

⎡ ⎤ ⎡ 

⎤ ⎡ ⎤ 

F i 

12 6L −12 6L v i 

⎢ M i 

⎥ 

⎣ F j 

⎦ = EI 

⎢ 6L 4L 2 −6L −2L 2 

⎥ ⎢ φ i 

⎥ 

L 3 ⎣ −12 −6L 12 −6L ⎦ ⎣ v j 

⎦ or FE = KEv 

M j 

−6L −2L 2 −6L 4L 2 φ j 

Therefore, the total forces acting on the beam element will be: 

F T = F E + F G = [K E + K G]v = K T v 


Special Considerations - Geometric Stiffness 

Conclusion - Geometric Stiffness 

In the case of constant (dead) loads where T is usually constant the 

calculation of the large displacement effect for cable and beam 

elements is performed by the addition in the FEM code of an 

appropriate extra “geometric” stiffness term.

The Finite Element Method for the Analysis of Non-Linear and ...

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