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Linear Spring Elements
CHAPTER
LINEAR SPRING ELEMENTS
3
Linear springs are the simplest elements. In this chapter we use them to present the main concept of an element equation derivation, and to explain how the local stiffness matrices can be assembled in order to compose the global stiffness matrix of the structure, how the boundary conditions can be specified and adopted, and how the displacement field of the structural system can be derived. Even though linear springs are elements carrying axial forces in only one direction, the steps for the finite element (FE) formulation of the problem is similar even for more complicated cases.
3.1 THE ELEMENT EQUATION As already mentioned, the first step for the FE formulation of a structure is the derivation of the element equation fr g ¼ ½kfd g
(3.1)
correlating the loads {r} acting on the nodes of an element with the corresponding nodal displacements {d}. More specifically, the main target of this step is the derivation of the element’s stiffness matrix [k] correlating the above column matrices {r} and {d}. Figure 3.1a shows a system of linear springs before loading. As Figure 3.1b indicates, a consequence of the action of axial loads at the end of the spring members are axial displacements. To derive the element equation, the method of direct equilibrium will be adopted.
3.1.1 THE MECHANICAL BEHAVIOR OF THE MATERIAL It is well known that the equation describing the mechanical behavior of a spring is given by Hook’s law: F ¼ kðd2x d1x Þ
(3.2)
where k is the spring constant, ðd2x d1x Þ is the spring’s elongation, and F is the axial force causing spring’s elongation. Essentials of the Finite Element Method. http://dx.doi.org/10.1016/B978-0-12-802386-0.00003-7 Copyright © 2015 Elsevier Inc. All rights reserved.
41
42
CHAPTER 3 LINEAR SPRING ELEMENTS
L
1
2 d2x
(a)
d1x
r2x
r1x
(b) FIGURE 3.1
(a) System of springs before loading. (b) Deformation of a system of springs due to action of axial forces on its nodes.
3.1.2 THE PRINCIPLE OF DIRECT EQUILIBRIUM In order to correlate the force F (that causes the spring’s elongation) to nodal forces, the equilibrium equation will be applied in the two pieces of the spring 1-2 (see Figure 3.2). Therefore: r1x + F ¼ 0
(3.3)
F + r2x ¼ 0
(3.4)
The Equations (3.3) and (3.4) can be written in the following matrix form:
r1x r2x
¼
F F
(3.5)
Using Equation (3.2), Equation (3.5) yields
r1x r2x
¼
kðd2x d1x Þ kðd2x d1x Þ
(3.6)
or Node 2
Node 1 r1x
FIGURE 3.2 Equilibrium of the two pieces of the spring.
F
F
r2x
3.2 THE STIFFNESS MATRIX OF A SYSTEM OF SPRINGS
r1x r2x
¼
k k k k
d1x d2x
43
(3.7)
Therefore, Equation (3.7) is the element’s equation, and the matrix
½k ¼
k k k k
(3.8)
is the stiffness matrix of the spring element.
3.2 THE STIFFNESS MATRIX OF A SYSTEM OF SPRINGS The procedure for deriving the stiffness matrix for a system of springs will be demonstrated using the example of Figure 3.3. The main steps to be followed are: (a) to derive the stiffness matrix for each spring element, (b) to expand the stiffness matrices to the degrees of freedom of the whole structure, and (c) to assembly the expanded stiffness matrices in order to derive the stiffness matrix of the structure.
3.2.1 DERIVATION OF ELEMENT MATRICES Taking into account Equation (3.7), the following element equations for the corresponding elements can be derived: Element 1, nodes 1 and 2
Element 2, nodes 2 and 3
Element 3, nodes 3 and 4
Node 1
k1
r1x r2x
r2x r3x
r3x r4x
¼
¼
¼
k1 k1 k1 k1
k2 k2 k2 k2
k3 k3 k3 k3
d1x d2x
d2x d3x
d3x d4x
k2 Node 2
(3.9)
(3.10)
(3.11)
k3
Node 4
Node 3 N
Element 1
Element 2
FIGURE 3.3 A simple structural system composed of three spring elements.
Element 3
44
CHAPTER 3 LINEAR SPRING ELEMENTS
3.2.2 EXPANSION OF ELEMENT EQUATIONS TO THE DEGREES OF FREEDOM OF THE STRUCTURE Since the structure contains four nodes, the above element equations (Equations 3.9–3.11) should be expanded to four degrees of freedom, that is: Element 1, nodes 1 and 2 8 9 2 k1 k1 r1x > > > < > = 6 k r2x 1 k1 ¼6 4 0 0 r3x > > > > : ; 0 0 r4x
Element 2, nodes 2 and 3
Element 3, nodes 3 and 4
0 0 0 0
38 9 d1x > 0 > > < > = 07 d 7 2x 0 5> d > > : 3x > ; 0 d4x
(3.12)
8 9 2 0 0 0 r1x > > > < > = 6 0 k r2x k 2 2 ¼6 > r3x > > 4 0 k2 k2 > : ; 0 0 0 r4x
38 9 d1x > 0 > > < > = 07 d 7 2x 0 5> d > 3x > : > ; 0 d4x
(3.13)
38 9 d1x > 0 0 0 > > < > = 0 0 0 7 d 7 2x 5 0 k3 k3 > d > > : 3x > ; 0 k3 k3 d4x
(3.14)
8 9 2 r1x > 0 > > < > = 6 0 r2x ¼6 > r3x > > 40 > : ; r4x 0
3.2.3 ASSEMBLY OF ELEMENT EQUATIONS Superposition (addition) of the element Equations (3.12)–(3.14) yields:
8 9 2 38 9 d1x > k1 R1x > k1 0 0 > > > > > < = 6 < > = k1 k1 + k2 k2 R2x 0 7 d2x 7 6 ¼4 0 k2 k2 + k3 k3 5> R > d > > > > : 3x > ; : 3x > ; R4x 0 0 k3 k3 d4x
(3.15)
It should be noted that the quantities Rix (i ¼ 1,2,3,4) are the resultants of the nodal forces.
3.2.4 DERIVATION OF THE FIELD VALUES Equation (3.15) can be written in an abbreviated notation, that is, fRg ¼ ½kfdg or ½kfdg ½I fRg ¼ fO41 g. An alternative formulation of the latter matrix equation is:
fd g ½ ½k ½I fRg
¼ fO41 g
(3.16)
where [k] is the global stiffness matrix, [I] is the 4 4 unit matrix, {O4x1} is a 4 1 vector containing zeros, and the vectors {d}, {R} are fd g ¼ ½ d1x d2x d3x d4x T and fRg ¼ ½ R1x R2x R3x R4x T . Therefore, Equation (3.16) expresses the following system of equations:
3.2 THE STIFFNESS MATRIX OF A SYSTEM OF SPRINGS
2
k1 k1 0 0 1 0 0 6 k1 k1 + k2 k2 0 0 1 0 6 4 0 k2 k2 + k3 k3 0 0 1 0 0 k3 k3 0 0 0
8 9 d1x > > > > > > d2x > > > 8 9 > > 3> > > d3x > > 0 > 0> > > > > > < < > = = 0 7 d 0 4x 7 ¼ 5 0 > > R1x > >0> > > > : > ; > 1 > R2x > 0 > > > > > > > > > R > > : 3x > ; R4x
45
(3.17)
Boundary Conditions The above matrix equation represents an algebraic system of four equations with eight unknowns. For the above system to be solved, four more equations should be added. The missing equations can be specified by the following boundary conditions: d1x ¼ 0
(3.18)
R2x ¼ 0
(3.19)
R3x ¼ 0
(3.20)
R4x ¼ N
(3.21)
The above boundary conditions can be formulated in a matrix form as follows: 2
1 60 6 40 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 1 0 0
0 0 1 0
8 9 d1x > > > > > > d2x > > > 8 9 > > 3> > > > 0 > d 0> > > > 3x > > > < = < > = 07 7 d4x ¼ 0 0 5> R > > 0> > > > 1x > > : > ; > 1 > R N > > 2x > > > > > > > R > > : 3x > ; R4x
(3.22)
Final Solution Therefore, superposition of Equations (3.17) and (3.22) yields an algebraic system of eight equations with eight unknowns, providing the nodal displacements and forces: ⎡ k1 ⎢ −k1 ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ 1 ⎢ ⎢ 0 ⎢ 0 ⎢ ⎣⎢ 0
−k1 k1 + k2 k k −k2 0 0 0 0 0
0 k −k2 k + k3 k k2 −k3 k 0 0 0 0
0 0 −k3 k k k3 0 0 0 0
−1 0 0 0 ⎤ 0 −1 0 0 ⎥⎥ 0 0 −1 0 ⎥ ⎥ 0 0 0 −1 ⎥ 0 0 0 0⎥ ⎥ 0 1 0 0⎥ 0 0 1 0⎥ ⎥ 0 0 0 1 ⎦⎥
⎧ d1x ⎫ ⎧ 0 ⎫ ⎪d ⎪ ⎪ 0 ⎪ ⎪ 2x ⎪ ⎪ ⎪ ⎪ d3x ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d4x ⎪ ⎪ 0 ⎪ ⎨ ⎬=⎨ ⎬ ⎪ R1x ⎪ ⎪ 0 ⎪ ⎪ R2x ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ R3x ⎪ ⎪ 0 ⎪ ⎪R ⎪ ⎪ N ⎪ ⎩ 4x ⎭ ⎩ ⎭
(3.23)
Knowing the nodal displacements d1x, d2x, d3x, and d4x, the internal forces r1x, r2x, r3x, and r4x can now be obtained using Equations (3.9)–(3.11).
46
CHAPTER 3 LINEAR SPRING ELEMENTS
EXAMPLE 3.1 Determine the field values (nodal displacements and internal forces) for the following structural system. Data k ¼ 40,000 N/m, F ¼ 5000 N Element 2
k 3 k 2
1
F
Element 1 4 k Element 3
Step 1: Derivation of Element Equations
r1x r2x r2x r3x r2x r4x
¼
¼
¼
k k k k k k k k k k k k
Step 2: Expansion of the Element Equations to Element 1, nodes 1 and 2 8 9 2 k r1x > > > = 6 < > k r2x 6 ¼ > 4 0 > r3x > > : ; 0 r4x
d1x d2x d2x d3x d2x d4x
for element 1
(3.24)
for element 2
(3.25)
for element 3
(3.26)
the Degrees of Freedom of the Structure 38 9 k 0 0 > d1x > > = < > k 0 07 7 d2x (3.27) 0 0 0 5> > > d3x > : ; 0 0 0 d4x
3.2 THE STIFFNESS MATRIX OF A SYSTEM OF SPRINGS
Element 2, nodes 2 and 3
Element 3, nodes 2 and 4
47
8 9 2 0 r1x > > > = 6 < > 0 r2x ¼6 4 0 r > > > ; : 3x > 0 r4x
38 9 0 0 0 > d1x > > = < > k k 0 7 d 7 2x k k 0 5> d > > ; : 3x > 0 0 0 d4x
(3.28)
8 9 2 r1x > 0 > > = 6 < > 0 r2x ¼6 4 0 r > > > ; : 3x > 0 r4x
0 k 0 k
38 9 0 > d1x > > = < > k 7 d 7 2x 0 5> d > > ; : 3x > k d4x
(3.29)
0 0 0 0
Step 3: Superposition of Elements Equation and Derivation of the Global Matrix Addition of Equations (3.27)–(3.29) yields 9 2 8 38 9 k k 0 0 > R1x > > > > > > d1x > = 6 < 7< d2x = k 3k k k R2x 6 7 ¼ R > 4 0 k k 0 5> d > > > > ; ; : 3x > : 3x > 0 k 0 k R4x d4x
(3.30)
Equation (3.30) can be written in the following abbreviated notation: fR g ¼ ½ k fd g
(3.31)
½kfd g ½I fRg ¼ fO41 g
(3.32)
or
or
½ ½k ½I
fd g f Rg
¼ fO41 g
(3.33)
where [k] is the global stiffness matrix, [I] is the 4 4 unit matrix, fO41 g is a 4 1 vector containing zeros, and the vectors {d}, {R} are 8 9 d > > > = < 1x > d2x (3.34) fd g ¼ d > > > ; : 3x > d4x 9 8 R1x > > > > = < R2x (3.35) f Rg ¼ R > > > ; : 3x > R4x
Continued
48
CHAPTER 3 LINEAR SPRING ELEMENTS
EXAMPLE 3.1—CONT’D Step 4: Boundary Conditions (a) Boundary conditions for nodal displacements
Due to the supports on the nodes 1, 3, and 4, the boundary conditions regarding displacements are: 9 d1x ¼ 0 = (3.36) d3x ¼ 0 ; d4x ¼ 0 The above boundary conditions can be written in the following matrix form: 2 38 9 8 9 1 0 0 0 > > >0> > > > d1x > 6 0 0 1 0 7 < d2x = < 0 = 6 7 ¼ 4 0 0 0 1 5 > d3x > > 0 > > > ; > : > ; : 0 0 0 0 0 d4x
(3.37)
or in an abbreviated notation ½BCd fdg ¼ fDOg or
fd g ½ ½BCd ½O4x4 f Rg where
2
1 60 ½BCd ¼ 6 40 0 2 0 60 ½O4x4 ¼ 6 40 0
(3.38)
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 0
¼ fDOg 3 0 07 7 15 0 3 0 07 7 05 0
fDOg ¼ ½ 0 0 0 0 T
(3.39)
(3.40)
(3.41)
(3.42)
(b) Boundary conditions for nodal forces
Taking into account the force acting on node 2, the following boundary condition can be used: R2x ¼ F The above boundary condition can be expressed in the following matrix form: 9 8 9 2 38 0 0 0 0 > R1x > > >0> > > > 6 0 0 0 0 7< R2x = < 0 = 6 7 4 0 0 0 0 5> R3x > ¼ > 0 > > > ; : > ; > : F 0 1 0 0 R4x
(3.43)
(3.44)
3.2 THE STIFFNESS MATRIX OF A SYSTEM OF SPRINGS
49
or in an abbreviated notation ½BCRfRg ¼ fROg or
½ ½O44 ½BCR
where
fd g f Rg
2
0 60 ½BCR ¼ 6 40 0
0 0 0 1
(3.45)
¼ fROg
0 0 0 0
(3.46)
3 0 07 7 05 0
(3.47)
fROg ¼ ½ 0 0 0 F T
(3.48)
It should be noted that since the first three rows are occupied by the boundary conditions for displacements, the boundary condition for the forces should be placed in the fourth line. (c) Matrix equation of boundary conditions
Adding Equations (3.39) and (3.46), the following matrix equation of boundary conditions can be obtained: fd g ½ ½BCd ½BCR ¼ fDOg + fROg (3.49) f Rg Step 5: Equation of Structure Combining Equations (3.33) and (3.49), the following equation of the structure, containing both stiffness matrix and boundary conditions, can be obtained: ½k ½I fO41 g fd g (3.50) ¼ ½BCd ½BCR fRg fDOg + fROg The above system expresses the following set of algebraic equations:
⎡ k ⎢ −k ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ 1 ⎢ ⎢ 0 ⎢ 0 ⎢ ⎣⎢ 0
−k
0
0
−1
0
0
3k
−k
−k
0
−1
0
−k
k
0
0
0
−1
−k
0
k
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
⎤ ⎥ ⎥ 0 ⎥ ⎥ −1 ⎥ 0 ⎥ ⎥ 0 ⎥ 0 ⎥ ⎥ 0 ⎦⎥ 0 0
⎧ d1x ⎫ ⎪d ⎪ ⎪ 2x ⎪ ⎪ d3x ⎪ ⎪ ⎪ ⎪ d4x ⎪ ⎨ ⎬= ⎪R1x ⎪ ⎪R2x ⎪ ⎪ ⎪ ⎪R3x ⎪ ⎪R ⎪ ⎩ 4x ⎭
⎧0 ⎫ ⎪0 ⎪ ⎪ ⎪ ⎪0 ⎪ ⎪ ⎪ ⎪0 ⎪ ⎨ ⎬ ⎪0 ⎪ ⎪0 ⎪ ⎪ ⎪ ⎪0 ⎪ ⎪F ⎪ ⎩ ⎭
(3.51)
Continued
50
CHAPTER 3 LINEAR SPRING ELEMENTS
EXAMPLE 3.1—CONT’D Step 6: Solution of the Algebraic System Taking into account that k ¼ 40,000 N/m and F ¼ 5000 N, the following results can be obtained: 9 9 8 8 0 d1x > > > > > > > > > > > > > > 0:041 > d2x > > > > > > > > > > > > > > > > d3x > 0 > > > > > > > > = = < < 0 d4x ¼ 1666:67 > R1x > > > > > > > > > > > > > > 5000 > R > > > > 2x > > > > > > > > > > > > > > > > 1666:67 R 3x > > > > ; ; : : 1666:67 R4x
(3.52)
Step 7: Internal Forces Using the above values for the nodal displacements, the internal forces can now be obtained by Equations (3.27)–(3.29): Element 1, nodes 1 and 2 9 8 9 8 r1x > 1666:67 > > > > > > > = < = < 1666:67 r2x ¼ 0 r > > > > > > ; : ; > : 3x > 0 r4x
(3.53)
9 8 9 8 r1x > 0 > > > > > > > = < = < 1666:67 r2x ¼ 1666:67 > r > > > > > ; : ; > : 3x > 0 r4x
(3.54)
9 8 9 8 r1x > > 0 > > > > = = > < < > 1666:67 r2x ¼ 0 > > > > r3x > > > ; : : ; > 1666:67 r4x
(3.55)
Element 2, nodes 2 and 3
Element 3, nodes 2 and 4
3.2 THE STIFFNESS MATRIX OF A SYSTEM OF SPRINGS
51
EXAMPLE 3.2: SOLUTION OF A SIMPLE STRUCTURE COMPOSED OF SPRINGS BY CALFEM/MATLAB
Element 2 Node 1 Element 1
k1
k2
1
F
Node 2 2 k3
k4
k5
Element 3 Node 3 Element 4
Element 5
3
Data k1 ¼ 1300N=m k2 ¼ 1400N=m k3 ¼ 1500N=m k4 ¼ 1600N=m k5 ¼ 1700N=m F ¼ 8000N
Continued
52
CHAPTER 3 LINEAR SPRING ELEMENTS
EXAMPLE 3.2: SOLUTION OF A SIMPLE STRUCTURE COMPOSED OF SPRINGS BY CALFEM/MATLAB—CONT’D Step 1: Topology Matrix Step 1: Topology matrix Elements
Edof = [1 1 2 The element 3 is connected to the degrees of freedom 2 and 3
212 323 423 5 2 3]
Step 2: Loads Step 2: Loads Matrix 3×1. The number of rows is equal to the number of degrees of freedom
f = zeros(3,1); f(2) = 8000;
The force acting to the degree of freedom 2 is equal to 8000
Step 3: Element Matrices k1¼13000;ep1¼k1;Ke1¼spring1e(ep1); k2¼14000;ep2¼k2;Ke2¼spring1e(ep2); k3¼15000;ep3¼k3;Ke3¼spring1e(ep3); k4¼16000;ep4¼k4;Ke4¼spring1e(ep4); k5¼17000;ep5¼k5;Ke5¼spring1e(ep5);
3.2 THE STIFFNESS MATRIX OF A SYSTEM OF SPRINGS
53
Step 4: Assembly of the Element Stiffness Matrices K¼zeros(3,3); K¼assem(Edof(1,:),K,Ke1); K¼assem(Edof(2,:),K,Ke2); K¼assem(Edof(3,:),K,Ke3); K¼assem(Edof(4,:),K,Ke4); K¼assem(Edof(5,:),K,Ke5); Step 5: Boundary Conditions Step 5: Boundary conditions On the node 1 the displacement is 0
bc=[10;30];
On the node 3 the displacement is 0
Step 6: Solution of the System of Equations Step 6: Solution of the system of equations Output of the solution is the vector a containing the displacements on each degree of freedom and the vector r containing the reactions
[a,r]=solveq(K,f,bc)
Results of the Step 6 a5 0 0.1067 0 r5 1.0e+03 * -2.8800 0.0000 -5.1200 Continued
54
CHAPTER 3 LINEAR SPRING ELEMENTS
EXAMPLE 3.2: SOLUTION OF A SIMPLE STRUCTURE COMPOSED OF SPRINGS BY CALFEM/MATLAB—CONT’D Step 7: Computation of the Nodal Displacements of Each Element ed1¼extract(Edof(1,:),a) ed2¼extract(Edof(2,:),a) ed3¼extract(Edof(3,:),a) ed4¼extract(Edof(4,:),a) ed5¼extract(Edof(5,:),a) Results of Step 7 ed1 5 0 0.1067 ed2 5 0 0.1067 ed3 5 0.1067 0 ed4 5 0.1067 0 ed5 5 0.1067 0 Step 8: Computation of the Spring Forces es1¼spring1s(ep1,ed1) es2¼spring1s(ep2,ed2) es3¼spring1s(ep3,ed3) es4¼spring1s(ep4,ed4) es5¼spring1s(ep5,ed5) Results of Step 8 es1 5 1.3867e+03 es2 5 1.4933e+03 es3 5 -1600 es4 5 -1.7067e+03 es5 5 -1.8133e+03 The CALFEM/MATLAB Computer Code >>Edof¼[1 1 2;2 1 2;3 2 3;4 2 3;5 2 3]; f¼zeros(3,1);f(2)¼8000;
REFERENCES
55
k1¼13000;ep1¼k1;Ke1¼spring1e(ep1); k2¼14000;ep2¼k2;Ke2¼spring1e(ep2); k3¼15000;ep3¼k3;Ke3¼spring1e(ep3); k4¼16000;ep4¼k4;Ke4¼spring1e(ep4); k5¼17000;ep5¼k5;Ke5¼spring1e(ep5); K¼zeros(3,3); K¼assem(Edof(1,:),K,Ke1); K¼assem(Edof(2,:),K,Ke2); K¼assem(Edof(3,:),K,Ke3); K¼assem(Edof(4,:),K,Ke4); K¼assem(Edof(5,:),K,Ke5); bc¼[1 0;3 0]; [a,r]¼solveq(K,f,bc) ed1¼extract(Edof(1,:),a) ed2¼extract(Edof(2,:),a) ed3¼extract(Edof(3,:),a) ed4¼extract(Edof(4,:),a) ed5¼extract(Edof(5,:),a) es1¼spring1s(ep1,ed1) es2¼spring1s(ep2,ed2) es3¼spring1s(ep3,ed3) es4¼spring1s(ep4,ed4) es5¼spring1s(ep5,ed5)
REFERENCES [1] Austrell P-E, Dahlblom O, Lindemann J, Olsson A, Olsson K-G, Persson K, et al. CALFEM a finite element toolbox, Version 3.4., Division of Structural Mechanics, Lund University; 2004. [2] Cook RD, Malkus DS, Plesha ME, Witt RJ. Concepts and applications of finite element analysis. Hoboken: John Wiley & Sons; 2002. [3] Bhatti MA. Fundamental finite element analysis and applications. Hoboken: John Wiley & Sons; 2005. [4] Logan DL. A first course in the finite element method. Boston, MA: Gengage Learning; 2012. [5] Oden JT, Becker EB, Carey GF. Finite elements: an introduction, volume I. New Jersey: Prentice Hall; 1981. [6] Fish J, Belytschko T. A first course in finite elements. New York: Wiley; 2007. [7] Zienkiewicz OC, Taylor RL, Fox DD. The finite element method for solid and structural mechanics. 7th ed. Oxford: Butterworth-Heinemann; 2013. [8] Deif AS. Advanced matrix theory for scientists and engineers. 2nd ed. London: Abacus Press; 1991.
LINEAR SPRING ELEMENTS
3
Linear springs are the simplest elements. In this chapter we use them to present the main concept of an element equation derivation, and to explain how the local stiffness matrices can be assembled in order to compose the global stiffness matrix of the structure, how the boundary conditions can be specified and adopted, and how the displacement field of the structural system can be derived. Even though linear springs are elements carrying axial forces in only one direction, the steps for the finite element (FE) formulation of the problem is similar even for more complicated cases.
3.1 THE ELEMENT EQUATION As already mentioned, the first step for the FE formulation of a structure is the derivation of the element equation fr g ¼ ½kfd g
(3.1)
correlating the loads {r} acting on the nodes of an element with the corresponding nodal displacements {d}. More specifically, the main target of this step is the derivation of the element’s stiffness matrix [k] correlating the above column matrices {r} and {d}. Figure 3.1a shows a system of linear springs before loading. As Figure 3.1b indicates, a consequence of the action of axial loads at the end of the spring members are axial displacements. To derive the element equation, the method of direct equilibrium will be adopted.
3.1.1 THE MECHANICAL BEHAVIOR OF THE MATERIAL It is well known that the equation describing the mechanical behavior of a spring is given by Hook’s law: F ¼ kðd2x d1x Þ
(3.2)
where k is the spring constant, ðd2x d1x Þ is the spring’s elongation, and F is the axial force causing spring’s elongation. Essentials of the Finite Element Method. http://dx.doi.org/10.1016/B978-0-12-802386-0.00003-7 Copyright © 2015 Elsevier Inc. All rights reserved.
41
42
CHAPTER 3 LINEAR SPRING ELEMENTS
L
1
2 d2x
(a)
d1x
r2x
r1x
(b) FIGURE 3.1
(a) System of springs before loading. (b) Deformation of a system of springs due to action of axial forces on its nodes.
3.1.2 THE PRINCIPLE OF DIRECT EQUILIBRIUM In order to correlate the force F (that causes the spring’s elongation) to nodal forces, the equilibrium equation will be applied in the two pieces of the spring 1-2 (see Figure 3.2). Therefore: r1x + F ¼ 0
(3.3)
F + r2x ¼ 0
(3.4)
The Equations (3.3) and (3.4) can be written in the following matrix form:
r1x r2x
¼
F F
(3.5)
Using Equation (3.2), Equation (3.5) yields
r1x r2x
¼
kðd2x d1x Þ kðd2x d1x Þ
(3.6)
or Node 2
Node 1 r1x
FIGURE 3.2 Equilibrium of the two pieces of the spring.
F
F
r2x
3.2 THE STIFFNESS MATRIX OF A SYSTEM OF SPRINGS
r1x r2x
¼
k k k k
d1x d2x
43
(3.7)
Therefore, Equation (3.7) is the element’s equation, and the matrix
½k ¼
k k k k
(3.8)
is the stiffness matrix of the spring element.
3.2 THE STIFFNESS MATRIX OF A SYSTEM OF SPRINGS The procedure for deriving the stiffness matrix for a system of springs will be demonstrated using the example of Figure 3.3. The main steps to be followed are: (a) to derive the stiffness matrix for each spring element, (b) to expand the stiffness matrices to the degrees of freedom of the whole structure, and (c) to assembly the expanded stiffness matrices in order to derive the stiffness matrix of the structure.
3.2.1 DERIVATION OF ELEMENT MATRICES Taking into account Equation (3.7), the following element equations for the corresponding elements can be derived: Element 1, nodes 1 and 2
Element 2, nodes 2 and 3
Element 3, nodes 3 and 4
Node 1
k1
r1x r2x
r2x r3x
r3x r4x
¼
¼
¼
k1 k1 k1 k1
k2 k2 k2 k2
k3 k3 k3 k3
d1x d2x
d2x d3x
d3x d4x
k2 Node 2
(3.9)
(3.10)
(3.11)
k3
Node 4
Node 3 N
Element 1
Element 2
FIGURE 3.3 A simple structural system composed of three spring elements.
Element 3
44
CHAPTER 3 LINEAR SPRING ELEMENTS
3.2.2 EXPANSION OF ELEMENT EQUATIONS TO THE DEGREES OF FREEDOM OF THE STRUCTURE Since the structure contains four nodes, the above element equations (Equations 3.9–3.11) should be expanded to four degrees of freedom, that is: Element 1, nodes 1 and 2 8 9 2 k1 k1 r1x > > > < > = 6 k r2x 1 k1 ¼6 4 0 0 r3x > > > > : ; 0 0 r4x
Element 2, nodes 2 and 3
Element 3, nodes 3 and 4
0 0 0 0
38 9 d1x > 0 > > < > = 07 d 7 2x 0 5> d > > : 3x > ; 0 d4x
(3.12)
8 9 2 0 0 0 r1x > > > < > = 6 0 k r2x k 2 2 ¼6 > r3x > > 4 0 k2 k2 > : ; 0 0 0 r4x
38 9 d1x > 0 > > < > = 07 d 7 2x 0 5> d > 3x > : > ; 0 d4x
(3.13)
38 9 d1x > 0 0 0 > > < > = 0 0 0 7 d 7 2x 5 0 k3 k3 > d > > : 3x > ; 0 k3 k3 d4x
(3.14)
8 9 2 r1x > 0 > > < > = 6 0 r2x ¼6 > r3x > > 40 > : ; r4x 0
3.2.3 ASSEMBLY OF ELEMENT EQUATIONS Superposition (addition) of the element Equations (3.12)–(3.14) yields:
8 9 2 38 9 d1x > k1 R1x > k1 0 0 > > > > > < = 6 < > = k1 k1 + k2 k2 R2x 0 7 d2x 7 6 ¼4 0 k2 k2 + k3 k3 5> R > d > > > > : 3x > ; : 3x > ; R4x 0 0 k3 k3 d4x
(3.15)
It should be noted that the quantities Rix (i ¼ 1,2,3,4) are the resultants of the nodal forces.
3.2.4 DERIVATION OF THE FIELD VALUES Equation (3.15) can be written in an abbreviated notation, that is, fRg ¼ ½kfdg or ½kfdg ½I fRg ¼ fO41 g. An alternative formulation of the latter matrix equation is:
fd g ½ ½k ½I fRg
¼ fO41 g
(3.16)
where [k] is the global stiffness matrix, [I] is the 4 4 unit matrix, {O4x1} is a 4 1 vector containing zeros, and the vectors {d}, {R} are fd g ¼ ½ d1x d2x d3x d4x T and fRg ¼ ½ R1x R2x R3x R4x T . Therefore, Equation (3.16) expresses the following system of equations:
3.2 THE STIFFNESS MATRIX OF A SYSTEM OF SPRINGS
2
k1 k1 0 0 1 0 0 6 k1 k1 + k2 k2 0 0 1 0 6 4 0 k2 k2 + k3 k3 0 0 1 0 0 k3 k3 0 0 0
8 9 d1x > > > > > > d2x > > > 8 9 > > 3> > > d3x > > 0 > 0> > > > > > < < > = = 0 7 d 0 4x 7 ¼ 5 0 > > R1x > >0> > > > : > ; > 1 > R2x > 0 > > > > > > > > > R > > : 3x > ; R4x
45
(3.17)
Boundary Conditions The above matrix equation represents an algebraic system of four equations with eight unknowns. For the above system to be solved, four more equations should be added. The missing equations can be specified by the following boundary conditions: d1x ¼ 0
(3.18)
R2x ¼ 0
(3.19)
R3x ¼ 0
(3.20)
R4x ¼ N
(3.21)
The above boundary conditions can be formulated in a matrix form as follows: 2
1 60 6 40 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 1 0 0
0 0 1 0
8 9 d1x > > > > > > d2x > > > 8 9 > > 3> > > > 0 > d 0> > > > 3x > > > < = < > = 07 7 d4x ¼ 0 0 5> R > > 0> > > > 1x > > : > ; > 1 > R N > > 2x > > > > > > > R > > : 3x > ; R4x
(3.22)
Final Solution Therefore, superposition of Equations (3.17) and (3.22) yields an algebraic system of eight equations with eight unknowns, providing the nodal displacements and forces: ⎡ k1 ⎢ −k1 ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ 1 ⎢ ⎢ 0 ⎢ 0 ⎢ ⎣⎢ 0
−k1 k1 + k2 k k −k2 0 0 0 0 0
0 k −k2 k + k3 k k2 −k3 k 0 0 0 0
0 0 −k3 k k k3 0 0 0 0
−1 0 0 0 ⎤ 0 −1 0 0 ⎥⎥ 0 0 −1 0 ⎥ ⎥ 0 0 0 −1 ⎥ 0 0 0 0⎥ ⎥ 0 1 0 0⎥ 0 0 1 0⎥ ⎥ 0 0 0 1 ⎦⎥
⎧ d1x ⎫ ⎧ 0 ⎫ ⎪d ⎪ ⎪ 0 ⎪ ⎪ 2x ⎪ ⎪ ⎪ ⎪ d3x ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d4x ⎪ ⎪ 0 ⎪ ⎨ ⎬=⎨ ⎬ ⎪ R1x ⎪ ⎪ 0 ⎪ ⎪ R2x ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ R3x ⎪ ⎪ 0 ⎪ ⎪R ⎪ ⎪ N ⎪ ⎩ 4x ⎭ ⎩ ⎭
(3.23)
Knowing the nodal displacements d1x, d2x, d3x, and d4x, the internal forces r1x, r2x, r3x, and r4x can now be obtained using Equations (3.9)–(3.11).
46
CHAPTER 3 LINEAR SPRING ELEMENTS
EXAMPLE 3.1 Determine the field values (nodal displacements and internal forces) for the following structural system. Data k ¼ 40,000 N/m, F ¼ 5000 N Element 2
k 3 k 2
1
F
Element 1 4 k Element 3
Step 1: Derivation of Element Equations
r1x r2x r2x r3x r2x r4x
¼
¼
¼
k k k k k k k k k k k k
Step 2: Expansion of the Element Equations to Element 1, nodes 1 and 2 8 9 2 k r1x > > > = 6 < > k r2x 6 ¼ > 4 0 > r3x > > : ; 0 r4x
d1x d2x d2x d3x d2x d4x
for element 1
(3.24)
for element 2
(3.25)
for element 3
(3.26)
the Degrees of Freedom of the Structure 38 9 k 0 0 > d1x > > = < > k 0 07 7 d2x (3.27) 0 0 0 5> > > d3x > : ; 0 0 0 d4x
3.2 THE STIFFNESS MATRIX OF A SYSTEM OF SPRINGS
Element 2, nodes 2 and 3
Element 3, nodes 2 and 4
47
8 9 2 0 r1x > > > = 6 < > 0 r2x ¼6 4 0 r > > > ; : 3x > 0 r4x
38 9 0 0 0 > d1x > > = < > k k 0 7 d 7 2x k k 0 5> d > > ; : 3x > 0 0 0 d4x
(3.28)
8 9 2 r1x > 0 > > = 6 < > 0 r2x ¼6 4 0 r > > > ; : 3x > 0 r4x
0 k 0 k
38 9 0 > d1x > > = < > k 7 d 7 2x 0 5> d > > ; : 3x > k d4x
(3.29)
0 0 0 0
Step 3: Superposition of Elements Equation and Derivation of the Global Matrix Addition of Equations (3.27)–(3.29) yields 9 2 8 38 9 k k 0 0 > R1x > > > > > > d1x > = 6 < 7< d2x = k 3k k k R2x 6 7 ¼ R > 4 0 k k 0 5> d > > > > ; ; : 3x > : 3x > 0 k 0 k R4x d4x
(3.30)
Equation (3.30) can be written in the following abbreviated notation: fR g ¼ ½ k fd g
(3.31)
½kfd g ½I fRg ¼ fO41 g
(3.32)
or
or
½ ½k ½I
fd g f Rg
¼ fO41 g
(3.33)
where [k] is the global stiffness matrix, [I] is the 4 4 unit matrix, fO41 g is a 4 1 vector containing zeros, and the vectors {d}, {R} are 8 9 d > > > = < 1x > d2x (3.34) fd g ¼ d > > > ; : 3x > d4x 9 8 R1x > > > > = < R2x (3.35) f Rg ¼ R > > > ; : 3x > R4x
Continued
48
CHAPTER 3 LINEAR SPRING ELEMENTS
EXAMPLE 3.1—CONT’D Step 4: Boundary Conditions (a) Boundary conditions for nodal displacements
Due to the supports on the nodes 1, 3, and 4, the boundary conditions regarding displacements are: 9 d1x ¼ 0 = (3.36) d3x ¼ 0 ; d4x ¼ 0 The above boundary conditions can be written in the following matrix form: 2 38 9 8 9 1 0 0 0 > > >0> > > > d1x > 6 0 0 1 0 7 < d2x = < 0 = 6 7 ¼ 4 0 0 0 1 5 > d3x > > 0 > > > ; > : > ; : 0 0 0 0 0 d4x
(3.37)
or in an abbreviated notation ½BCd fdg ¼ fDOg or
fd g ½ ½BCd ½O4x4 f Rg where
2
1 60 ½BCd ¼ 6 40 0 2 0 60 ½O4x4 ¼ 6 40 0
(3.38)
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 0
¼ fDOg 3 0 07 7 15 0 3 0 07 7 05 0
fDOg ¼ ½ 0 0 0 0 T
(3.39)
(3.40)
(3.41)
(3.42)
(b) Boundary conditions for nodal forces
Taking into account the force acting on node 2, the following boundary condition can be used: R2x ¼ F The above boundary condition can be expressed in the following matrix form: 9 8 9 2 38 0 0 0 0 > R1x > > >0> > > > 6 0 0 0 0 7< R2x = < 0 = 6 7 4 0 0 0 0 5> R3x > ¼ > 0 > > > ; : > ; > : F 0 1 0 0 R4x
(3.43)
(3.44)
3.2 THE STIFFNESS MATRIX OF A SYSTEM OF SPRINGS
49
or in an abbreviated notation ½BCRfRg ¼ fROg or
½ ½O44 ½BCR
where
fd g f Rg
2
0 60 ½BCR ¼ 6 40 0
0 0 0 1
(3.45)
¼ fROg
0 0 0 0
(3.46)
3 0 07 7 05 0
(3.47)
fROg ¼ ½ 0 0 0 F T
(3.48)
It should be noted that since the first three rows are occupied by the boundary conditions for displacements, the boundary condition for the forces should be placed in the fourth line. (c) Matrix equation of boundary conditions
Adding Equations (3.39) and (3.46), the following matrix equation of boundary conditions can be obtained: fd g ½ ½BCd ½BCR ¼ fDOg + fROg (3.49) f Rg Step 5: Equation of Structure Combining Equations (3.33) and (3.49), the following equation of the structure, containing both stiffness matrix and boundary conditions, can be obtained: ½k ½I fO41 g fd g (3.50) ¼ ½BCd ½BCR fRg fDOg + fROg The above system expresses the following set of algebraic equations:
⎡ k ⎢ −k ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ 1 ⎢ ⎢ 0 ⎢ 0 ⎢ ⎣⎢ 0
−k
0
0
−1
0
0
3k
−k
−k
0
−1
0
−k
k
0
0
0
−1
−k
0
k
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
⎤ ⎥ ⎥ 0 ⎥ ⎥ −1 ⎥ 0 ⎥ ⎥ 0 ⎥ 0 ⎥ ⎥ 0 ⎦⎥ 0 0
⎧ d1x ⎫ ⎪d ⎪ ⎪ 2x ⎪ ⎪ d3x ⎪ ⎪ ⎪ ⎪ d4x ⎪ ⎨ ⎬= ⎪R1x ⎪ ⎪R2x ⎪ ⎪ ⎪ ⎪R3x ⎪ ⎪R ⎪ ⎩ 4x ⎭
⎧0 ⎫ ⎪0 ⎪ ⎪ ⎪ ⎪0 ⎪ ⎪ ⎪ ⎪0 ⎪ ⎨ ⎬ ⎪0 ⎪ ⎪0 ⎪ ⎪ ⎪ ⎪0 ⎪ ⎪F ⎪ ⎩ ⎭
(3.51)
Continued
50
CHAPTER 3 LINEAR SPRING ELEMENTS
EXAMPLE 3.1—CONT’D Step 6: Solution of the Algebraic System Taking into account that k ¼ 40,000 N/m and F ¼ 5000 N, the following results can be obtained: 9 9 8 8 0 d1x > > > > > > > > > > > > > > 0:041 > d2x > > > > > > > > > > > > > > > > d3x > 0 > > > > > > > > = = < < 0 d4x ¼ 1666:67 > R1x > > > > > > > > > > > > > > 5000 > R > > > > 2x > > > > > > > > > > > > > > > > 1666:67 R 3x > > > > ; ; : : 1666:67 R4x
(3.52)
Step 7: Internal Forces Using the above values for the nodal displacements, the internal forces can now be obtained by Equations (3.27)–(3.29): Element 1, nodes 1 and 2 9 8 9 8 r1x > 1666:67 > > > > > > > = < = < 1666:67 r2x ¼ 0 r > > > > > > ; : ; > : 3x > 0 r4x
(3.53)
9 8 9 8 r1x > 0 > > > > > > > = < = < 1666:67 r2x ¼ 1666:67 > r > > > > > ; : ; > : 3x > 0 r4x
(3.54)
9 8 9 8 r1x > > 0 > > > > = = > < < > 1666:67 r2x ¼ 0 > > > > r3x > > > ; : : ; > 1666:67 r4x
(3.55)
Element 2, nodes 2 and 3
Element 3, nodes 2 and 4
3.2 THE STIFFNESS MATRIX OF A SYSTEM OF SPRINGS
51
EXAMPLE 3.2: SOLUTION OF A SIMPLE STRUCTURE COMPOSED OF SPRINGS BY CALFEM/MATLAB
Element 2 Node 1 Element 1
k1
k2
1
F
Node 2 2 k3
k4
k5
Element 3 Node 3 Element 4
Element 5
3
Data k1 ¼ 1300N=m k2 ¼ 1400N=m k3 ¼ 1500N=m k4 ¼ 1600N=m k5 ¼ 1700N=m F ¼ 8000N
Continued
52
CHAPTER 3 LINEAR SPRING ELEMENTS
EXAMPLE 3.2: SOLUTION OF A SIMPLE STRUCTURE COMPOSED OF SPRINGS BY CALFEM/MATLAB—CONT’D Step 1: Topology Matrix Step 1: Topology matrix Elements
Edof = [1 1 2 The element 3 is connected to the degrees of freedom 2 and 3
212 323 423 5 2 3]
Step 2: Loads Step 2: Loads Matrix 3×1. The number of rows is equal to the number of degrees of freedom
f = zeros(3,1); f(2) = 8000;
The force acting to the degree of freedom 2 is equal to 8000
Step 3: Element Matrices k1¼13000;ep1¼k1;Ke1¼spring1e(ep1); k2¼14000;ep2¼k2;Ke2¼spring1e(ep2); k3¼15000;ep3¼k3;Ke3¼spring1e(ep3); k4¼16000;ep4¼k4;Ke4¼spring1e(ep4); k5¼17000;ep5¼k5;Ke5¼spring1e(ep5);
3.2 THE STIFFNESS MATRIX OF A SYSTEM OF SPRINGS
53
Step 4: Assembly of the Element Stiffness Matrices K¼zeros(3,3); K¼assem(Edof(1,:),K,Ke1); K¼assem(Edof(2,:),K,Ke2); K¼assem(Edof(3,:),K,Ke3); K¼assem(Edof(4,:),K,Ke4); K¼assem(Edof(5,:),K,Ke5); Step 5: Boundary Conditions Step 5: Boundary conditions On the node 1 the displacement is 0
bc=[10;30];
On the node 3 the displacement is 0
Step 6: Solution of the System of Equations Step 6: Solution of the system of equations Output of the solution is the vector a containing the displacements on each degree of freedom and the vector r containing the reactions
[a,r]=solveq(K,f,bc)
Results of the Step 6 a5 0 0.1067 0 r5 1.0e+03 * -2.8800 0.0000 -5.1200 Continued
54
CHAPTER 3 LINEAR SPRING ELEMENTS
EXAMPLE 3.2: SOLUTION OF A SIMPLE STRUCTURE COMPOSED OF SPRINGS BY CALFEM/MATLAB—CONT’D Step 7: Computation of the Nodal Displacements of Each Element ed1¼extract(Edof(1,:),a) ed2¼extract(Edof(2,:),a) ed3¼extract(Edof(3,:),a) ed4¼extract(Edof(4,:),a) ed5¼extract(Edof(5,:),a) Results of Step 7 ed1 5 0 0.1067 ed2 5 0 0.1067 ed3 5 0.1067 0 ed4 5 0.1067 0 ed5 5 0.1067 0 Step 8: Computation of the Spring Forces es1¼spring1s(ep1,ed1) es2¼spring1s(ep2,ed2) es3¼spring1s(ep3,ed3) es4¼spring1s(ep4,ed4) es5¼spring1s(ep5,ed5) Results of Step 8 es1 5 1.3867e+03 es2 5 1.4933e+03 es3 5 -1600 es4 5 -1.7067e+03 es5 5 -1.8133e+03 The CALFEM/MATLAB Computer Code >>Edof¼[1 1 2;2 1 2;3 2 3;4 2 3;5 2 3]; f¼zeros(3,1);f(2)¼8000;
REFERENCES
55
k1¼13000;ep1¼k1;Ke1¼spring1e(ep1); k2¼14000;ep2¼k2;Ke2¼spring1e(ep2); k3¼15000;ep3¼k3;Ke3¼spring1e(ep3); k4¼16000;ep4¼k4;Ke4¼spring1e(ep4); k5¼17000;ep5¼k5;Ke5¼spring1e(ep5); K¼zeros(3,3); K¼assem(Edof(1,:),K,Ke1); K¼assem(Edof(2,:),K,Ke2); K¼assem(Edof(3,:),K,Ke3); K¼assem(Edof(4,:),K,Ke4); K¼assem(Edof(5,:),K,Ke5); bc¼[1 0;3 0]; [a,r]¼solveq(K,f,bc) ed1¼extract(Edof(1,:),a) ed2¼extract(Edof(2,:),a) ed3¼extract(Edof(3,:),a) ed4¼extract(Edof(4,:),a) ed5¼extract(Edof(5,:),a) es1¼spring1s(ep1,ed1) es2¼spring1s(ep2,ed2) es3¼spring1s(ep3,ed3) es4¼spring1s(ep4,ed4) es5¼spring1s(ep5,ed5)
REFERENCES [1] Austrell P-E, Dahlblom O, Lindemann J, Olsson A, Olsson K-G, Persson K, et al. CALFEM a finite element toolbox, Version 3.4., Division of Structural Mechanics, Lund University; 2004. [2] Cook RD, Malkus DS, Plesha ME, Witt RJ. Concepts and applications of finite element analysis. Hoboken: John Wiley & Sons; 2002. [3] Bhatti MA. Fundamental finite element analysis and applications. Hoboken: John Wiley & Sons; 2005. [4] Logan DL. A first course in the finite element method. Boston, MA: Gengage Learning; 2012. [5] Oden JT, Becker EB, Carey GF. Finite elements: an introduction, volume I. New Jersey: Prentice Hall; 1981. [6] Fish J, Belytschko T. A first course in finite elements. New York: Wiley; 2007. [7] Zienkiewicz OC, Taylor RL, Fox DD. The finite element method for solid and structural mechanics. 7th ed. Oxford: Butterworth-Heinemann; 2013. [8] Deif AS. Advanced matrix theory for scientists and engineers. 2nd ed. London: Abacus Press; 1991.