Finite Element Method: Truss Element Introduction

àThe finite element method (FEM) is a technique for analyzing the behavior of engineered structures subjected to a variety of loads. FEM is the most widely applied computer simulation method in engineering. àThe basic idea is to divide a complicated structure into small and manageable pieces (discretization) and solve the algebraic equation.

Applications of FEM in Engineering

Simulation

Experiment

Engineered system Vehicle Tire Cellular phone Li-ion battery system Predicted measure Safety Cornering force Reliability Thermal behavior

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Available Commercial FEM Software

• ANSYS (General purpose) • PATRAN (Pre/Post Processor) • I-DEAS (Complete CAD/CAM/CAE package) • NASTRAN (General purpose) • HyperMesh (Pre/Post Processor) • COSNOS (General purpose) • ABAQUS (Nonlinear and dynamic analyses) • Dyna-3D (Crash/impact analysis)

Types of Finite Elements

1-D (Line) Element 2-D (Plane) Element 3-D (Solid) Element

(Spring, truss, beam, pipe, etc.) (Membrane, plate, shell, etc.) (3-D fields – temperature, displacement, stress, flow velocity, etc.)

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element Truss Element in 2D Space

x, y : global coordination system uˆ : deformation

xˆ, yˆ : local coordination system L : length

dˆ : displacement fˆ : force

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Linear Static Analysis

Most structural analysis problems can be treated as linear static problems, based on the following assumptions. 1. Small deformations (loading pattern is not changed due to the deformed shape) 2. Static loads (the load is applied to the structure in slow or steady fashion) 3. Elastic materials (no plasticity or failures)

Hooke’s Law and Deformation Equation

du$ e = s = Ee dx$

Equilibrium d F du$I As = T = Constant G AE J = 0 x dx$ H dx$K

Assumptions $ $ 1. Truss cannot support shear force. f1y = 0 f 2 y = 0 2. Ignore the effect of lateral deformation

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Direct Stiffness Method

DSM is an approach to calculate a stiffness matrix for a system by directly superposing the stiffness matrixes of all elements. DSM is beneficial to get the stiffness matrix of relatively simple structures consist of several trusses or beams.

Step 1: Determination of element type Step 2: Determination of deformation

Assume a linear deformation function as F d$ - d$ I u$ = a + a x$ = 2 x 1x x$ + d$ 1 2 HG L KJ 1x

In vector form

Rd$ U u$ = N N 1x 1 2 Sd$ V T 2 x W x$ x$ N = 1- N = 1 L 2 L

N1, N2: shape function

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Step 3: Strain and stress calculation Step 4: Derivation of element stiffness matrix

Deformation rate - strain F d$ - d$ I T = As = AE 2 x 1x x G L J du$ d$ - d$ H K e = = 2 x 1x x $ AE dx L f$ = -T = d$ - d$ 1x L e 1x 2 x j Stress - strain AE f$ = T = d$ - d$ 2 x L e 2 x 1x j s x = Ee x

R f$ U AE 1 -1 Rd$ U 1x = L O 1x S f$ V L M-1 1 PSd$ V T 2 x W N QT 2 x W

AE 1 -1 k$ = L O L NM-1 1 QP

NOTE: In a truss element, stiffness (spring constant, k) is equivalent to AE/L

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Step 5: Constitution of global stiffness matrix Step 6: Calculation of nodal displacement

Construct a global stiffness matrix in a • Use boundary conditions global coordinate system • Solve the system of linear algebraic equations F = K d to calculate the N beg K = K = å k nodal deformation in a truss structure e=1

N F = lFq = å f (e) e=1

Step 7: Calculation of stress and strain in an element

Compute a strain and stress at any point within an element

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Example (Truss element in 1D space) Example>

A structure consists of three beams (see below figure). Find (a) the global stiffness matrix in global coordinate system, (b) the displacements at the node 2 and 3, (c) the reaction forces at the node 2 and 3. 3000lb loading acts in the positive x- direction at node 2. The length of the elements is 30in. Young’s modulus for the elements 1 and 2: E = 30×106 psi Cross-sectional area for the elements 1 and 2: A = 1in2 Young’s modulus for the elements 3: E = 15×106 psi Cross-sectional area for the elements 3: A = 2in2

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element node # Step 4: Derivation of element element # 1 2 stiffness matrix 2 3 6 (1) (2) (1)(30 ´10 ) L 1 -1O 6 L 1 -1O k = k = M P = 10 M P AE 1 -1 30 N-1 1 Q N-1 1 Q k$ = L O L M-1 1 P 3 4 N Q (2)(15´106 ) 1 -1 1 -1 k (3) = L O = 106 L O 30 NM-1 1 QP NM-1 1 QP

node # d1x d2 x d3 x d4 x Step 5: Constitution of global 1 -1 0 0 stiffness matrix L O M-1 1+1 -1 0 P K = 106 M P Answer of (a) M 0 -1 1+1 -1P N 0 0 -1 1 Q

RF1x U L 1 -1 0 0 ORd1x U |F | M-1 2 -1 0 P|d | | 2 x | = 106 M P| 2 x | SF V 0 -1 2 -1 Sd V | 3x | M P| 3x | F M P d T| 4x W| N 0 0 -1 1 QT| 4x W|

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Step 6: Calculation of nodal Using the boundary conditions d1x = d4 x = 0 displacement 3000 2 -1 d R U = 106 L OR 2 x U S V M PSd V T 0 W N-1 2 QT 3x W

So, the displacements for this system are

d2 x = 0.002in. d3x = 0.001in. Answer of (b)

Substituting the displacements to the equation, the load can be obtained as follows. Answer of (c)

Step 7: Calculation of stress and F s = strain in an element x A

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Selection of Deformation Function

1. The deformation function needs to be continuous within an element. 2. The deformation function needs to provide continuity among elements.

1-D linear deformation function satisfies (1) and (2). à compatibility

3. The deformation function needs to express the displacement of a rigid body and the constant strain in element.

u$ = a1 + a2 x$ : 1-D linear deformation function à completeness

a1 considers the motion of a rigid body. $ $ a2 xˆ considers the constant strain ( e x = d u d x = a 2 )

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element Truss Element in 2D Space

Vector transformation in 2-D

Displacement vector d in global coordinate and local coordinate systems. $ $ $ $ d = dxi + dy j = dx i + dy j

Global Local

Relation between two coordination systems

$ d Rdx U C S R x U | | = L O C = cosq S = sinq Eq. (1) Sd$ V M-S CPSd V T| y W| N QT y W

L C SO : Transformation Matrix NM-S CQP

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Stiffness matrix in global coordinate system

The stiffness matrix of truss element in local coordinate system

R f$ U AE 1 -1 Rd$ U 1 = L O 1x f$ = k$d$ : Step 4 on page 6 S f$ V L M-1 1 PSd$ V T 2 W N QT 2 x W

The stiffness matrix of truss element in global coordinate system

R f1x U Rd1x U | | | | : What we want to f1y d1y | | = k| | f = kd construct in 2-D space S f V Sd V | 2 x | | 2 x | f d T| 2 y W| T| 2 y W|

Let’s calculate the relation between k$ and k .

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Transform the loading in a same way,

* f$ = T f Eq. (3)

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

From Eq.(2) and (3)

$ $ $ C S 0 0 f = kd * L O where, T = NM0 0 C SQP T * f = k$T * d

We needs the inverse matrix of T*; however, because T* is not the square matrix, it cannot be immediately transformed.

Invite ˆ ˆ and ˆ ˆ ( f1y, d 1 y ) ( f2y, d 2 y )

$ d Rd1x U L C S 0 0OR 1x U |d$ | M-S C 0 0P|d | | 1y | = M P| 1y | $ Sd$ V 0 0 C S Sd V d = Td | 2 x | M P| 2 x | d$ M 0 0 -S CP d | 2 y | N QT| 2 y W| T W Transformation Matrix

In a same way,

fˆ = T f

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Expanded 4×4 matrix k$

R f$ U 1 0 -1 0 Rd$ U | 1x | L O| 1x | f$ AE M 0 0 0 0P d$ | 1y | = M P| 1y | S f$ V L -1 0 1 0 Sd$ V | 2 x | M P| 2 x | | f$ | M 0 0 0 0P|d$ | T 2 y W N QT 2 y W

The stiffness matrix of truss elements in global coordinate system

T f = k$ T d f = T -1 k$ T d = T T k$ T d

T is an orthogonal matrix (TT=T-1).

k = T T k$ T

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

So, éCS 0 0 ù ê ú 2 2 -SC 0 0 L C CS -C -CSO T = ê ú 2 2 ê0 0 CS ú AE M S -CS -S P ê ú k = M P T $ ë0 0 -SC û L M C2 CS P k = T k T MSymmetry S2 P Eq. (4) é1 0- 1 0 ù N Q ê ú AE 0 0 0 0 kˆ = ê ú L ê-1 0 1 0 ú ê ú ë0 0 0 0 û Assemble the stiffness matrix for the whole system

N N å k (e) = K å f (e) = F e=1 e=1

The relation between the nodal loading and displacement in global coordinate system

F = K d

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

ExampleExample (Truss element in 2D space)

Three truss elements are assembled and 10,000lb loading is applied at node 1 as shown in figure. Find the displacements at node 1. For all elements, Young’s modulus: E = 30 ´106 psi , cross-sectional area: A = 2in.2 .

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Data for stiffness matrix calculation

d1x d1y d2 x d2 y L0 0 0 0 O 6 M P 1 c30 ´10 ha2f 0 1 0 -1 Step 4: Derivation of element stiffness ka f = M P matrix (from Eq. (4)) 120 M0 0 0 0 P N0 -1 0 1 Q

d d d d 1x 1y 3 x 3 y d1x d1y d4 x d4 y L 0.5 0.5 -0.5 -0.5O L 1 0 -1 0O 30 ´106 2 M 0.5 0.5 -0.5 -0.5P 6 M P a2f c ha f 3 c30 ´10 hb2g 0 0 0 0 k = M P kb g = M P 120 ´ 2 M-0.5 -0.5 0.5 0.5 P -1 0 1 0 M P 120 M P N-0.5 -0.5 0.5 0.5 Q N 0 0 0 0Q

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Step 5: Constitution of global stiffness matrix

d1x d1y d2 x d2 y d3 x d3 y d4 x d4 y ML 1.354 0.354 0 0 -0.354 -0.354 -1 0PO M 0.354 1.354 0 -1 -0.354 -0.354 0 0P M 0 0 0 0 0 0 0 0P M 0 -1 0 1 0 0 0 0P K = a500,000fM P M-0.354 -0.354 0 0 0.354 0.354 0 0P M-0.354 -0.354 0 0 0.354 0.354 0 0P M -1 0 0 0 0 0 1 0P NM 0 0 0 0 0 0 0 0QP

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Step 6: Calculation of nodal displacement

Using boundary condition at node 2,3, and 4.

0 R U L 1.354 0.354 0 0 -0.354 -0.354 -1 0O R d1x U |-10,000| M 0.354 1.354 0 -1 -0.354 -0.354 0 0P | d | | | M P | 1y | | F2 x | M 0 0 0 0 0 0 0 0P |d2 x = 0| | F | M 0 -1 0 1 0 0 0 0P | | | 2 y | |d2 y = 0| S V = a500,000fM P ´S V F3x M-0.354 -0.354 0 0 0.354 0.354 0 0P d = 0 | | | 3x | F M-0.354 -0.354 0 0 0.354 0.354 0 0P d = 0 | 3y | M P | 3y | | F | M -1 0 0 0 0 0 1 0P |d = 0| | 4x | M P | 4 x | | F | 0 0 0 0 0 0 0 0 |d = 0| T 4y W NM QP T 4 y W

Calculate unknown displacement

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

F Step 7: Calculation of stress and strain in an element ( s = ) A

-2 Rd1x = 0.414 ´10 U (1) F2 y 6 | -2 | s = 30 ´10 d = -1.59 ´10 A s a1f = 0 -1 0 1 | 1y | = 3965psi 120 S d = 0 V | 2 x | | d = 0 | T 2 y W -2 Rd1x = 0.414 ´10 U 6 | -2 | a2f 30 ´10 L- 2 - 2 2 2 O| dy = -1.59 ´10 | s = (2) 2F3x 120 2 M 2 2 2 2 PS d = 0 V s = N Q| 3x | A | d = 0 | T 3y W = 1471psi and

-2 Rd1x = 0.414 ´10 U 6 | -2 | a3f 30 ´10 d1y = -1.59 ´10 s = -1 0 1 0 | | = -1035psi F 120 S d = 0 V s (3) = 4x | 4x | A | d = 0 | T 4y W

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Step 8: Confirmation the loading equilibrium at node 1

2 2 2 Fx = 0 a1471psif 2in. - a1035psif 2in. = 0 å c h 2 c h

2 2 2 Fy = 0 a3695psif 2in. + a1471psif 2in. -10,000 = 0 å c h c h 2

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element Truss Element in 3D Space

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

LCx 0 O MC 0 P M y P MCz 0 P A E 1 -1 LCx Cy Cz 0 0 0 O k = L O M 0 C P L M-1 1 PM 0 0 0 C C C P M x P N QN x y z Q M 0 Cy P M 0 C P N z Q

2 2 L Cx CxCy CxCz -Cx -CxCy -CxCz O M C2 C C -C C -C2 -C C P M y y z x y y y z P A E M C2 -C C -C C -C2 P k = z x z y z z L M C2 C C C C P M x x y x z P M C2 C C P M y y z P Symmetry C2 NM z QP

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

ExampleExample (Truss element in 3D space) Solve the 3-D truss problem as shown in below figure. The elastic moduli for all elements are E =1.2 ´106 psi . 1000lb loading is applied in the negative z-direction at node 1. Node 2, 3, and 4 are supported on sockets with balls (x,y, and z direction movements are constrained). The movement along the y direction is constrained by a roller at Node 1.

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Step 4: Derivation of element stiffness matrix

For element 3 -72.0 C = = - 0.833 x 86.5

Cy = 0

-48.5 C = = - 0.550 z 86.5

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Submatrix l ML0.69 0 0.46PO l = M 0 0 0 P NM0.46 0 0.30QP Stiffness matrix of the element k

d1x d1y d1z d4 x d4 y d4 z (0.187)(1.2 ´106 ) l -l k (3) = L O 86.5 NM -l l QP

For element 1

(1) L =80.5in. Cx = - 0.89 Cy = 0.45 Cz = 0

L 0.79 -0.40 0O d d d d d d 1x 1y 1z 2 x 2 y 2 z M P 6 l = -0.40 0.20 0 (1) (0.320)(1.2 ´10 ) L l -l O M P k = M P NM 0 0 0QP 80.5 N -l l Q

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

For element 2

(2) L =108in. Cx = - 0.667 Cy = 0.33 Cz = 0.667

L 0.45 -0.22 -0.45O d1x d1y d1z d3 x d3 y d3z 6 M P (2) (0.729)(1.2 ´10 ) L l -l O l = M-0.22 0.11 0.45P k = M P NM-0.45 0.45 0.45QP 108 N -l l Q

Step 5, 6, 7

Using the boundary conditions d1y = 0, d2 x = d2 y = d2z = 0, d3x = d3y = d3z = 0, d4x = d4y = d4z = 0

d1x d1z 9000 -2450 K = L O NM-2450 4450QP Stiffness matrix

R 0 U L 9000 -2450ORd1x U = Stiffness equation in global coordinate system S V M PSd V T-1000W N-2450 4550QT 1z W

d1x = - 0.072in. Calculated displacement (The negative signs mean that the

d1z = - 0.264in. displacements are in the negative x and z direction)

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element Inclined or Skewed and Support

' To be convenient, apply the boundary condition d 3 y to the local coordinate x'- y'

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Displacement vector transformation at node 3:

Rd3¢x U cosa sina Rd3x U = L O Sd¢ V M-sina cosaPSd V T 3y W N QT 3y W or L cosa sina O ld3¢q= t3 ld3q t3 = NM-sina cosaQP

Displacement vector transformation at all nodes:

kd¢p= T1 kdp T kdp= T1 kd¢p

Rd1x U Rd1¢x U |d | |d ¢ | | 1y | L I 0 0 O| 1y | |d2 x | T |d2¢x | | |= T d ¢ = M 0 I 0 P| | Sd V 1 l q M PSd ¢ V | 2 y | M 0 0 t T P| 2 y | N 3 Q |d3x | |d3¢x | |d | |d ¢ | T| 3y W| T| 3y W|

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Loading vector transformation :

In local coordinate system l f3¢q= t3 l f3q

In global coordinate system

R f1x U R f1x U | f | | f | | 1y | L I 0 0 O| 1y | f f | 2 x | M P| 2 x | k f ¢p= T1 k f p or = 0 I 0 S f V M PS f V | 2 y | 0 0 t | 2 y | NM 3 QP | f 3¢x | | f 3x | | f ¢ | | f | T| 3y W| T| 3y W|

Note: f1x = f '1x , f1y = f '1y , f 2 x = f '2 x , f 2 y = f '2 y

d1x = d1¢x , d1y = d1¢y , d2 x = d2¢x , d2 y = d2¢y

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Transformation of the finite element formulation in global coordinates::

Boundary condition: d1x = 0, d1y = 0, d3¢y = 0 F2 y = 0, F3¢x = 0 F2 x

Calculation of displacement and loading: unknown displacements d 2 x , d 2 y , d 3 ¢ x reaction forces F 1 x , F 1 y and that of inclined roller F3¢y

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

ExampleExample (Truss element in inclined support)

Find displacements and reacting forces for the plane trusses in the figure. E = 210 GPa and

A= 6.00 ´10-4 m2 for nodes 1 and 2, and A= 6 2 ´10-4 m2 for node 3.

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Element 1

cosq = 0 sinq =1

d1x d1y d2 x d2 y L 0 0 0 0O (6.0 ´10-4 )(210 ´109 ) M 1 0 -1P k (1) = M P 1m M 0 0P NSymmetry 1Q

Element 2

cosq =1 sinq = 0

d2 x d2 y d3 x d3 y L 1 0 -1 0O (6.0 ´10-4 )(210 ´109 ) M 0 0 0P k (2) = M P 1m M 1 0P NSymmetry 0Q

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Element 3

2 2 cosq = sinq = 2 2

d1x d1y d3x d3 y L 0.5 0.5 -0.5 -0.5O (6 2 ´10-4 ) (210´109 ) M 0.5 -0.5 -0.5P k (3) = M P 2 M 0.5 0.5P NSymmetry 0.5Q

Calculation of the matrix K in global coordinates using direct stiffness method:

ML 0.5 0.5 0 0 - 0.5 -0.5PO M 1.5 0 -1 - 0.5 -0.5P M 1 0 -1 0 P K =1260 ´105 N m M 1 0 0 P M P M 1.5 0.5P NSymmetry 0.5Q

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Matrix T 1 , which transforms the displacement at node 3 in global coordinates to the local coordinates x¢ - y¢:

ML1 0 0 0 0 0 PO M0 1 0 0 0 0 P M0 0 1 0 0 0 P T = 1 M0 0 0 1 0 0 P M P M0 0 0 0 2 2 2 2P N0 0 0 0 - 2 2 2 2Q

* T Calculation of K = T 1 K T 1 : ML 0.5 0.5 0 0 -0.5 -0.5 PO M 0.5 1.5 0 -1 -0.5 -0.5 P M 0 0 1 0 -1 0 P T K =1260 ´105 1 M 0 -1 0 1 0 0 P M P M-0.707 -0.707 -0.707 0 1.414 0.707P N 0 0 0.707 0 -0.707 0 Q And,

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

d1x d1y d2 x d2 y d3¢ x d3¢ y ML 0.5 0.5 0 0 -0.707 0 PO M 0.5 1.5 0 -1 -0.707 0 P M 0 0 1 0 -0.707 0.707 P T K T T =1260 ´105 1 1 M 0 -1 0 1 0 0 P M P M-0.707 -0.707 -0.707 0 1.500 -0.500P N 0 0 0.707 0 -0.500 0.500 Q

Substituting the boundary conditions d1x = d1y = d2 y = d3¢y = 0 to the above equation, F =1000 kN 1 -0.707 d R 2 x U=(1260 ´105 ) L OR 2 x U S F¢ = 0 V M PSd¢ V. T 3x W N-0.707 1.50 QT 3x W

Calculating displacements,

d2 x =11.91mm

d3¢x = 5.613mm

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

Calculating loadings,

F1x = - 500 kN

F1y = - 500 kN

F2 y = 0

F3¢y = 707kN

Free body diagram of truss structure

■

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Truss Element

1. Solution of finite element coincides with the exact solution at nodes. The reason why it

coincides with the exact solution at nodes is because it calculates the element nodal

loading by the energy equivalent based on the linearly assumed displacement form at each

element.

2. Although the nodal value for displacement coincides with the exact solution, the values

between nodes are very inaccuracy in the case of using few elements due to using linear

displacement function at each element. However, when the number of elements increases,

the solution of finite element converges on the exact solution.

3. Stress is derived from the slope of displacement curve as s = E e = E(du dx) . Axial

stress is constant at each element, for u of each element in the solution of finite

element is a linear function. The much more elements are needed for the modeling of the

first derivative of the displacement function like the modeling of axial stress.

4. The most approximate value of the stress appears not at nodes, but at the central point.

It is because the derivative of displacement is calculated more accurate between nodes

than at nodes.

5. Stress is not continuous over the element boundaries. Therefore the equilibrium is not

satisfied over the element boundaries. Also, the equilibrium at each element is usually

not satisfied.

Mechanics and Design SNU School of Mechanical and Aerospace Engineering