Solutions for Problems in "A First Course in the Finite Element Method" (6th Edition) by Daryl Logan

1
Chapter 1
1.1. A finite element is a small body or unit interconnected to other units to model a larger
structure or system.
1.2. Discretization means dividing the body (system) into an equivalent system of finite elements
with associated nodes and elements.
1.3. The modern development of the finite element method began in 1941 with the work of
Hrennikoff in the field of structural engineering.
1.4. The direct stiffness method was introduced in 1941 by Hrennikoff. However, it was not
commonly known as the direct stiffness method until 1956.
1.5. A matrix is a rectangular array of quantities arranged in rows and columns that is often used
to aid in expressing and solving a system of algebraic equations.
1.6. As computer developed it made possible to solve thousands of equations in a matter of
minutes.
1.7. The following are the general steps of the finite element method.
Step 1
Divide the body into an equivalent system of finite elements with associated
nodes and choose the most appropriate element type.
Step 2
Choose a displacement function within each element.
Step 3
Relate the stresses to the strains through the stress/strain law—generally called
the constitutive law.
Step 4
Derive the element stiffness matrix and equations. Use the direct equilibrium
method, a work or energy method, or a method of weighted residuals to relate the
nodal forces to nodal displacements.
Step 5
Assemble the element equations to obtain the global or total equations and
introduce boundary conditions.
Step 6
Solve for the unknown degrees of freedom (or generalized displacements).
Step 7
Solve for the element strains and stresses.
Step 8
Interpret and analyze the results for use in the design/analysis process.
1.8. The displacement method assumes displacements of the nodes as the unknowns of the
problem. The problem is formulated such that a set of simultaneous equations is solved for
nodal displacements.
1.9. Four common types of elements are: simple line elements, simple two-dimensional elements,
simple three-dimensional elements, and simple axisymmetric elements.
1.10 Three common methods used to derive the element stiffness matrix and equations are
(1) direct equilibrium method
(2) work or energy methods
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2
(3) methods of weighted residuals
1.11. The term ‘degrees of freedom’ refers to rotations and displacements that are associated with
each node.
1.12. Five typical areas where the finite element is applied are as follows.
(1) Structural/stress analysis
(2) Heat transfer analysis
(3) Fluid flow analysis
(4) Electric or magnetic potential distribution analysis
(5) Biomechanical engineering
1.13. Five advantages of the finite element method are the ability to
(1) Model irregularly shaped bodies quite easily
(2) Handle general load conditions without difficulty
(3) Model bodies composed of several different materials because element equations are
evaluated individually
(4) Handle unlimited numbers and kinds of boundary conditions
(5) Vary the size of the elements to make it possible to use small elements where necessary

3
Chapter 2
2.1
(a)
[k(1)
] =
1 1
1 1
0 – 0
0 0 0 0
– 0 0
0 0 0 0
k k
k k
[k(2)
] =
2 2
2 2
0 0 0 0
0 0 0 0
0 0 –
0 0 –
k k
k k
[k3
(3)
] = 3 3
3 3
0 0 0 0
0 0 –
0 0 0 0
0 – 0
k k
k k
[K] = [k(1)] + [k(2)] + [k(3)]
[K] =
1 1
3 3
1 1 2 2
3 2 2 3
0 – 0
0 0 –
– 0 –
0 – –
k k
k k
k k k k
k k k k
(b) Nodes 1 and 2 are fixed so u1 = 0 and u2 = 0 and [K] becomes
[K] =
1 2 2
2 2 3
–
–
k k k
k k k
{F} = [K] {d}
3
4
x
x
F
F
=
1 2 2
2 2 3
–
–
k k k
k k k
3
4
u
u

0
P
=
1 2 2
2 2 3
–
–
k k k
k k k
3
4
u
u
{F} = [K] {d} [K]–1 {F} = [K]–1 [K] {d}

4
 K]–1 {F} = {d}
Using the adjoint method to find [K–1]
C11 = k2 + k3 C21 = (– 1)3
(– k2)
C12 = (– 1)1 + 2
(– k2) = k2 C22 = k1 + k2
[C] = 2 3 2
2 1 2
k k k
k k k
and CT
= 2 3 2
2 1 2
k k k
k k k
det [K] = | [K] | = (k1 + k2) (k2 + k3) – ( – k2) (– k2)
 | [K] | = (k1 + k2) (k2 + k3) – k2
2
[K –1
] =
[ ]
det
T
C
K
[K –1] =
2 3 2
2 1 2
2
1 2 2 3 2
( )( ) –
k k k
k k k
k k k k k
=
2 3 2
2 1 2
1 2 1 3 2 3
k k k
k k k
k k k k k k
3
4
u
u
=
2 3 2
2 1 2
1 2 1 3 2 3
0
k k k
k k k P
k k k k k k
 u3 = 2
1 2 1 3 2 3
k P
k k k k k k
  u4 = 1 2
1 2 1 3 2 3
( )
k k P
k k k k k k
(c) In order to find the reaction forces we go back to the global matrix F = [K]{d}
1
2
3
4
x
x
x
x
F
F
F
F
=
1
1 1
2
3 3
1 1 2 2 3
3 2 2 3 4
0 0
0 0
0
0
u
k k
u
k k
k k k k u
k k k k u
F1x = – k1 u3 = – k1
2
1 2 1 3 2 3
k P
k k k k k k
 F1x = 1 2
1 2 1 3 2 3
k k P
k k k k k k
F2x = – k3 u4 = – k3 1 2
1 2 1 3 2 3
( )
k k P
k k k k k k
 F2x = 3 1 2
1 2 1 3 2 3
( )
k k k P
k k k k k k
2.2

5
k1 = k2 = k3 = 1000
lb
in.
(1) (2) (2) (3)
[k(1)] =
(1)
(2)
k k
k k
; [k(2)] =
(2)
(3)
k k
k k
By the method of superposition the global stiffness matrix is constructed.
(1) (2) (3)
[K] =
0 (1)
(2)
0 (3)
k k
k k k k
k k
 [K] =
0
2
0
k k
k k k
k k
Node 1 is fixed  u1 = 0 and u3 = 
{F} = [K] {d}
1
2
3
?
0
?
x
x
x
F
F
F
=
0
2
0
k k
k k k
k k
1
2
3
0
?
u
u
u

 
 

 
 
 
 

3
0
x
F
=
2
2
3 2
0 2
2
x
k u k
k k u
F k u k
k k
  u2 =
2k
k
=
2
=
1 in.
2
 u2 = 0.5
F3x = – k (0.5) + k (1)
F3x = (– 1000
lb
in.
) (0.5) + (1000
lb
in.
) (1)
F3x = 500 lbs
Internal forces
Element (1)
(1)
1
(2)
2
x
x
f
f
=
1
2
0
0.5
u
k k
k k u
 (1)
1x
f = (– 1000
lb
in.
) (0.5)  (1)
1x
f = – 500 lb
(1)
2x
f = (1000
lb
in.
) (0.5)  (1)
2x
f = 500 lb
Element (2)

6
(2)
2
(2)
3
x
x
f
f
=
2 2
3 3
0.5 0.5
1 1
u u
k k k k
u u
k k k k

(2)
2
(2)
3
– 500 lb
500 lb
x
x
f
f
2.3
(a) [k(1)
] = [k(2)
] = [k(3)
] = [k(4)
] =
k k
k k
By the method of superposition we construct the global [K] and knowing {F} = [K] {d}
we have
1
2
3
4
5
?
0
0
?
x
x
x
x
x
F
F
F P
F
F
=
0 0 0
2 0 0
0 2 0
0 0 2
0 0 0
k k
k k k
k k k
k k k
k k
1
2
3
4
5
0
0
u
u
u
u
u
(b)
0
0
P =
2 2 3
3 2 3 4
3 4
4
2 0 0 2
2 2
0 2 0 2
u
k k ku ku
k k k u P ku ku ku
k k ku ku
u
 u2 = 3
2
u
; u4 = 3
2
u
Substituting in the second equation above
P = – k u2 + 2k u3 – k u4
 P = – k 3
2
u
+ 2k u3 – k 3
2
u
 P = ku3
 u3 =
P
k
u2 =
2
P
k
; u4 =
2
P
k
(c) In order to find the reactions at the fixed nodes 1 and 5 we go back to the global
equation {F} = [K] {d}
F1x = – ku2 = –
2
P
k
k
 F1x =
2
P
F5x = – ku4 = –
2
P
k
k
 F5x =
2
P
Check
Fx = 0  F1x + F5x + P = 0
(1)
(2)
(3)


2
P
+
2
P
+ P = 0
 0 = 0
2.4
(a) [k(1)] = [k(2)] = [k(3)] = [k(4)] =
k k
k k
By the method of superposition the global [K] is constructed.
Also {F} = [K] {d} and u1 = 0 and u5 = 
1
2
3
4
5
?
0
0
0
?
x
x
x
x
x
F
F
F
F
F
=
0 0 0
2 0 0
0 2 0
0 0 2
0 0 0
k k
k k k
k k k
k k k
k k
1
2
3
4
5
0
?
?
?
u
u
u
u
u
(b) 0 = 2k u2 – k u3 (1)
0 = – ku2 + 2k u3 – k u4 (2)
0 = – k u3 + 2k u4 – k  (3)
From (2)
u3 = 2 u2
From (3)
u4 = 2
2
2
u
Substituting in Equation (2)
– k (u2) + 2k (2u2) – k 2
2
2
u
 
 
 
 
 – u2 + 4 u2 – u2 –
2
= 0  u2 =
4
u3 = 2
4
 u3 =
2
u4 = 4
2
2
 u4 =
3
4
(c) Going back to the global equation

8
{F} = [K]{d}
F1x = – k u2 =
4
k  F1x =
4
k
F5x = – k u4 + k  = – k
3
4
+ k 
 F5x =
4
k
2.5
u1 u2 u2 u4
[k (1)
] =
1 1
1 1
; [k (2)
] =
2 2
2 2
u2 u4 u2 u4
[k (3)
] =
3 3
3 3
; [k (4)
] =
4 4
4 4
u4 u3
[k (5)
] =
5 5
5 5
Assembling global [K] using direct stiffness method
[K] =
1 1 0 0
1 1 2 3 4 0 2 3 4
0 0 5 5
0 2 3 4 5 2 3 4 5
Simplifying
[K] =
1 1 0 0
1 10 0 9 kip
0 0 5 5 in.
0 9 5 14
2.6 Now apply + 3 kip at node 2 in spring assemblage of P 2.5.
 F2x = 3 kip
[K]{d} = {F}

9
[K] from P 2.5
1 1 0 0
1 10 0 9
0 0 5 5
0 9 5 14
1
2
3
4
0
0
u
u
u
u
=
1
3
3
0
F
F
 
 
 
 
 
 
 
(A)
where u1 = 0, u3 = 0 as nodes 1 and 3 are fixed.
Using Equations (1) and (3) of (A)
2
4
10 9
9 14
u
u
=
3
0
 
 
 
Solving
u2 = 0.712 in., u4 = 0.458 in.
2.7
f1x = C, f2x = – C
f = – k = – k(u2 – u1)
 f1x = – k(u2 – u1)
f2x = – (– k) (u2 – u1)
1
2
x
x
f
f
=
k –k
–k k
1
2
u
u
 [K] =
k –k
–k k
same as for
tensile element
2.8
k1 = 1000
1 1
1 1
; k2 = 1000
1 1
1 1
So

10
[K] = 1000
1 1 0
1 2 1
0 1 1
{F} = [K] {d} 

1
2
3
?
0
500
F
F
F

 
 

 

 
 
=
1 1 0
10 1 2 1
00
0 1 1
 
 
 
 

 
 
1
2
3
0
?
?
u
u
u
 0 = 2000 u2 – 1000 u3 (1)
500 = – 1000 u2 + 1000 u3 (2)
From (1)
u2 =
1000
2000
u3  u2 = 0.5 u3 (3)
Substituting (3) into (2)
 500 = – 1000 (0.5 u3) + 1000 u3
 500 = 500 u3
 u3 = 1 in.
 u2 = (0.5) (1 in.)  u2 = 0.5 in.
Element 1–2
(1)
1
(1)
2
x
x
f
f
= 1000
(1)
1
(1)
2
500lb
1 1 0 in.
1 1 0.5 in. 500lb
x
x
f
f
 

  

 
 

   
Element 2–3
(2)
2
(2)
3
x
x
f
f
= 1000
(2)
2
(2)
3
500 lb
1 1 0.5 in.
1 1 1 in. 500 lb
x
x
f
f
 
  
 

 
 

   
F1x = 500 [1 –1 0] 1
0
0.5 in. 500 lb
1 in.
x
F
 
    
 
 
 
2.9
(1) (2)
[k(1)
] =
5000 5000
5000 5000

 
 

 
(2) (3)

11
[k(2)] =
5000 5000
5000 5000

 
 

 
(3) (4)
[k(3)] =
5000 5000
5000 5000

 
 

 
(1) (2) (3) (4)
[K] =
5000 5000 0 0
5000 10000 5000 0
0 5000 10000 5000
0 0 5000 5000

 
 
 
 
 
 
 

 
1
2
3
4
?
1000
0
4000
x
x
x
x
F
F
F
F
=
5000 5000 0 0
5000 10000 5000 0
0 5000 10000 5000
0 0 5000 5000

 
 
 
 
 
 
 

 
1
2
3
4
0
u
u
u
u
 u1 = 0 in.
u2 = 0.6 in.
u3 = 1.4 in.
u4 = 2.2 in.
Reactions
F1x = [5000 – 5000 0 0]
1
2
3
4
0
0.6
1.4
2.2
u
u
u
u

 
 

 
 

 
 

 
F1x = – 3000 lb
Element forces
Element (1)
(1)
1
(1)
2
x
x
f
f
=
5000 5000 0
5000 5000 0.6
  
 
 
 

  

(1)
1
(1)
2
3000lb
3000lb
x
x
f
f
Element (2)
(2)
2
(2)
3
x
x
f
f
=
5000 5000 0.6
5000 5000 1.4
  
 
 
 

  

(2)
2
(2)
3
4000lb
4000lb
x
x
f
f
Element (3)
(3) (3)
3 3
(3) (3)
4 4
x x
x x
f f
f f
=
5000 5000 1.4
5000 5000 2.2
  
 
 
 

  

(3)
3
(3)
4
4000lb
4000lb
x
x
f
f
2.10

12
[k(1)] =
1000 1000
1000 1000
[k(2)] =
500 500
500 500
[k(3)] =
500 500
500 500
{F} = [K] {d}
1
2
3
4
?
– 8000
?
?
x
x
x
x
F
F
F
F

 
 

 
 

 
 

 
=
1
2
3
4
0
1000 1000 0 0
?
1000 2000 500 500
0
0 500 500 0
0 500 0 500 0
u
u
u
u
 u2 =
8000
2000

= – 4 in.
Reactions
1
2
3
4
x
x
x
x
F
F
F
F
=
0
1000 1000 0 0
1000 2000 500 500 4
0 500 500 0 0
0 500 0 500 0
  
 
 
 
   
 
   

   
   

   

1
2
3
4
x
x
x
x
F
F
F
F
=
4000
8000
2000
2000
 
 

 
 
 
 
 
lb
Element (1)
(1)
1
(1)
2
x
x
f
f
=
1000 1000 0
1000 1000 – 4
  
 
 
 

  

(1)
1
(1)
2
x
x
f
f
=
4000
4000
 
 

 
lb
Element (2)
(2)
2
(2)
3
x
x
f
f
=
500 500 4
500 500 0
 
 
 
 
 

  

(2)
2
(2)
3
x
x
f
f
=
– 2000
2000
 
 
 
lb
Element (3)

13
(3)
2
(3)
4
x
x
f
f
=
500 500 4
500 500 0
 
 
 
 
 

  

(3)
2
(3)
4
x
x
f
f
2000
2000

 
 
 
lb
2.11
[k(1)
] =
1000 1000
1000 1000

 
 

 
; [k(2)
] =
3000 3000
3000 3000

 
 

 
{F} = [K] {d}
1
2
3
?
0
?
x
x
x
F
F
F
=
1000 1000 0
1000 4000 3000
0 3000 3000

 
 
 
 

 
 
1
2
3
0
?
0.02 m
u
u
u
 u2 = 0.015 m
Reactions
F1x = (– 1000) (0.015)  F1x = – 15 N
Element (1)
1
2
x
x
f
f
 
 
 
=
1000 1000
1000 1000

 
 

 
0
0.015
 
 
 
 1
2
x
x
f
f
 
 
 

15
15

 
 
 
N
Element (2)
2
3
x
x
f
f
 
 
 
=
3000 –3000
–3000 3000
 
 
 
0.015
0.02
 
 
 
 2
3
x
x
f
f
 
 
 
=
15
15

 
 
 
N
2.12
[k(1)
] = [k(3)
] = 10000
1 1
1 1
[k(2)] = 10000
3 3
3 3

 
 

 
{F} = [K] {d}

14
1
2
3
4
?
450 N
0
?
x
x
x
x
F
F
F
F

 
 

 
 

 
 

 
= 10000
1 1 0 0
1 4 3 0
0 3 4 1
0 0 1 1

 
 
 
 
 
 
 

 
1
2
3
4
0
?
?
0
u
u
u
u
0 = – 3 u2 + 4 u3  u2 =
4
3
u3  u2 = 1.33 u3
450 N = 40000 (1.33 u3) – 30000 u3
 450 N = (23200
N
m
) u3  u3 = 1.93  10–2
m
 u2 = 1.5 (1.94  10–2)  u2 = 2.57  10–2 m
Element (1)
1
2
x
x
f
f
 
 
 
= 10000
1 1
1 1 2
0
2.57 10
 
 

 

(1)
1
(1)
2
257 N
257 N
x
x
f
f
 

Element (2)
2
3
x
x
f
f
 
 
 
= 30000
1 1
1 1
2
2
2.57 10
1.93 10


 

 
 

 
 

(2)
2
(2)
3
193 N
193 N
x
x
f
f

 
Element (3)
3 3
4 4
x x
x x
f f
f f
   
   
   
= 10000
1 1
1 1
2
1.93 10
0

 

 
 

(3)
3
(3)
4
193 N
193 N
x
x
f
f

 
Reactions
{F1x} = (10000
N
m
) [1 – 1] 2
0
2.57 10
 
 

 
F1x = – 257 N
{F4x} = (10000
N
m
) [–1 1]
2
1.93 10
0

 

 
 
 F4x = – 193 N
2.13
[k(1)] = [k(2)] = [k(3)] = [k(4)] = 60
1 1
1 1
{F} = [K]{d}

15
1
2
3
4
5
?
0
5 kN
0
?
x
x
x
x
x
F
F
F
F
F

 
 

 
 

 
 

 

 
 
= 60
1 1 0 0 0
1 2 1 0 0
0 1 2 1 0
0 0 1 2 1
0 0 0 1 1
1
2
3
4
5
0
?
?
?
0
u
u
u
u
u
2 3 2 3
3 4 4 3
0 2 – 0.5
0 – 2 0.5
u u u u
u u u u
 u2 = u4
 5 kN = – 60 u2 + 120 (2 u2) – 60 u2
  5 = 120 u2  u2 = 0.042 m
 u4 = 0.042 m
 u3 = 2(0.042)  u3 = 0.084 m
Element (1)
1
2
x
x
f
f
 
 
 
= 60
1 1
1 1
0
0.042
 
 
 

(1)
1
(1)
2
2.5 kN
2.5 kN
x
x
f
f
 

Element (2)
2
3
x
x
f
f
 
 
 
= 60
1 1 1 1
1 1 1 1
0.042
0.084
 
 
 

(2)
2
(2)
3
2.5 kN
2.5 kN
x
x
f
f
 

Element (3)
3 3
4 4
x x
x x
f f
f f
= 60
1 1 1 1
1 1 1 1
0.084
0.042
 
 
 

(3) (3)
3 3
(3) (3)
4 4
2.5 kN 2.5 kN
2.5 kN 2.5 kN
x x
x x
f f
f f
Element (4)
4 4
5 5
x x
x x
f f
f f
   
   
   
= 60
1 1 1 1
1 1 1 1
0.042
0
 
 
 

(4) (4)
4 4
(4) (4)
5 5
2.5 kN 2.5 kN
2.5 kN 2.5 kN
x x
x x
f f
f f
 
   
F1x = 60 [1 –1]
0
0.042
 
 
 
F1x = – 2.5 kN
F5x = 60 [–1 1]
0.042
0
 
 
 
F5x = – 2.5 kN
2.14
[k(1)
] = [k(2)
] = 4000
1 1
1 1
{F} = [K]d}

Solutions for Problems in "A First Course in the Finite Element Method" (6th Edition) by Daryl Logan

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Solutions for Problems in "A First Course in the Finite Element Method" (6th Edition) by Daryl Logan