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JN Reddy
The Finite Element Method
Read: Sections 4.6 and 5.4
Plane (2D) Truss and Frame Elements
• Review of bar finite elementin the local coordinates
• Plane truss element• Review of beam finite
element in the localcoordinates
• Plane frame element• Numerical examples
CONTENTS
JN Reddy
FINITE ELEMENT ANALYSIS OF PLANE TRUSSES AND FRAMES
Trusses and Frames: 2
JN Reddy
REVIEW OF THE BAR ELEMENT
Linear bar element in the element coordinate system
K u Fe e e=
Trusses & Frames: 3
1
2
1 1 11 1 12
K F,e
e ee e e ee
e
QE A f hh Q
− = = + −
he
211 1,e eQ f 2 2,e eQ fex
ey
Horizontal
EeAehe
⎡⎢⎣1 0 −1 00 0 0 0−1 0 1 00 0 0 0
⎤⎥⎦⎧⎪⎨⎪⎩ue1ve1ue2ve2
⎫⎪⎬⎪⎭ =
⎧⎪⎨⎪⎩F e10F e20
⎫⎪⎬⎪⎭K Fe e eD =
he 2
1θe
y
xeF1
eF2 ex
eyev1
eu1
2eu
2
ev
Inclined
JN Reddy
Bar Element in Global Coordinates
EeAehe
⎡⎢⎣1 0 −1 00 0 0 0−1 0 1 00 0 0 0
⎤⎥⎦⎧⎪⎨⎪⎩ue1ve1ue2ve2
⎫⎪⎬⎪⎭ =
⎧⎪⎨⎪⎩F e10F e20
⎫⎪⎬⎪⎭ he 2
1
θe
y
xeF1
eF2 ex
eyev1
eu1
2eu
2
ev
Bar element in the element coordinates
K Fe e eD =
xe = x cos θ + y sin θ
ye = −x sin θ + y cos θxeye
=cos θ sin θ− sin θ cos θ
xy
2
1
θe
y
x
ex
ey( , )e ex y
( , )e ex y P
Transformation relations between the two coordinate systems
Trusses & Frames: 4
●
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Transformation relations between the displacements of the two coordinate systems
he 2
1
θe
y
x
,F e2 ex
eyev1
eu2
ev2
exF1
eu1
eyF1
ev1
eyF2
ev2
exF2
eu, 2
eF1
eu1
1 1
1 1
2 2
2 2
cos sin,
sin cos
cos sinsin cos
e e
e e
e e
e e
u uv v
u uv v
θ θθ θ
θ θθ θ
= − = −
0 0
{∆e} = [T e]{∆e}
⎧⎪⎨⎪⎩ue1ve1ue2ve2
⎫⎪⎬⎪⎭ =
⎡⎢⎣cos θe sin θe 0 0− sin θe cos θe 0 00 0 cos θe sin θe0 0 − sin θe cos θe
⎤⎥⎦⎧⎪⎨⎪⎩ue1ve1ue2ve2
⎫⎪⎬⎪⎭
Bar Element in Global Coordinates
Trusses & Frames: 5
{ } { }[ ]e e eF T F=
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Bar Element in Global Coordinates: Truss
[Ke][T e]{∆e} = [T e]{F e}, [T e]−1 = [T e]T
[T e]T[Ke][T e]{∆e} = {F e} or [Ke]{∆e} = {F e}[Ke] = [T e]T[Ke][T e], {F e} = [T e]T{F e}
K Fe e eD =
Trusses & Frames: 6
[Ke] =EA
h
⎡⎢⎢⎣cos2 θ 1
2 sin 2θ − cos2 θ −12 sin 2θ12 sin 2θ sin2 θ −12 sin 2θ − sin2 θ− cos2 θ −12 sin 2θ cos2 θ 1
2 sin 2θ
−12 sin 2θ − sin2 θ 12 sin 2θ sin2 θ
⎤⎥⎥⎦
{F e} =
⎧⎪⎨⎪⎩F e1F e2F e3F e4
⎫⎪⎬⎪⎭ =
⎧⎪⎨⎪⎩P e1 cos θeP e1 sin θeP e2 cos θeP e2 sin θe
⎫⎪⎬⎪⎭+⎧⎪⎨⎪⎩fe1 cos θefe1 sin θefe2 cos θefe2 sin θe
⎫⎪⎬⎪⎭
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EXAMPLE 1
Given truss
P = 100 kNy
xA B
C2P = 200 kN
L
L
P = 100 kN
x
2P = 200 kN
1
yF2yF1
xF2
xy
x
y
y
xF1
)2,1(1)4,3(21 90q =
2 45q =
)6,5(3
2
Finite element discretization
Trusses & Frames: 7
Element Global Geom. Mater. Orient.
number nodes prop. prop.
√ ◦1 2 3 A, h1 = L E θ1 = 90
◦
2 1 3 A, h2 = 2L E θ2 = 45
JN Reddy
EXAMPLE 1 (continued)
P = 100 kN
x
2P = 200 kN
1
yF2yF1
xF2
xy
x
y
y
xF1
)2,1(1)4,3(21 90q =
)6,5(3
2
[K3] =EA
L
⎡⎢⎣0.3536 0.3536 −0.3536 −0.35360.3536 0.3536 −0.3536 −0.3536−0.3536 −0.3536 0.3536 0.3536−0.3536 −0.3536 0.3536 0.3536
⎤⎥⎦
The element stiffness matrices are [1/(2√2) = 0.3536]
Trusses & Frames: 8
2[ ]K =
2 45q = [K2] =EA
L
⎡⎢⎣0 0 0 00 1 0 −10 0 0 00 −1 0 1
⎤⎥⎦1[ ]K =
JN Reddy
EXAMPLE 1 (continued)
Assembled stiffness coefficients
Trusses & Frames: 9
Connectivity array for 2 DoF per node truss
3 4 5 61 2 5 6é ùê úê úë û
1 2 3 4
[B] =(1)(2)
(2) (2)11 11 12 12 13
(2) (2)14 15 13 16 14
(2)22 22 23 24
(2) (2)25 23 26 24
, , 0,0, , ,
, 0, 0,, ,
K K K K KK K K K KK K K KK K K K
= = =
= = =
= = =
= =
(1) (1) (1)33 11 34 12 34 12
(1) (1) (1)35 13 36 14 34 12
(1) (1) (1)35 13 36 14 44 22
(1) (1) (1) (2)45 23 46 24 55 33 33
(1) (2) (1) (2)56 34 34 66 44 44
, , ,, , ,, , ,, , ,
, .
K K K K K KK K K K K KK K K K K KK K K K K K KK K K K K K
= = =
= = =
= = =
= = = +
= + = +
P = 100 kN
x
2P = 200 kN
1
yF2yF1
xF2
xy
x
y
y
xF1
1 1 2( , )2 3 4( , )
1 90q =
3 5 6( , )
2
)2,1(1
2 3 4( , )
)2,1(1
2 3 4( , )
2 45q =
JN Reddy
Assembled system of equations of the truss
EXAMPLE 1 (continued)
The displacement continuity conditions are
u11 = u31 = U1, v
11 = v
31 = V1
u12 = u21 = U2, v
12 = v
21 = V2
u22 = u32 = U3, v
22 = v
32 = V3
Trusses & Frames: 10
EA
L
⎡⎢⎢⎢⎢⎢⎢⎢⎣
1.3536 0.3536 −1.0 0.0 |− 0.3536 −0.35360.3536 0.0 0.0 |− 0.3536 −0.3536
1.0 0.0 | 0.0 0.01.0 | 0.0 −1.0
symm. −−− −−− −−− |−−−− −−−−| 0.3536 0.3536| 1.3536
⎤
⎦
11
11
22
22
33
33
x
y
x
y
x
y
FUFVFUFV
FUFV
ì üì ü ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ïï ï ï ï=í ý í ýï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ïî þ ï ïî þ
0 0.
0 0.
0.
JN Reddy
EXAMPLE 1 (continued)
F 11 + F31 = F
1x , F
12 + F
32 = F
1y
F 13 + F21 = F
2x , F
14 + F
22 = F
2y
F 23 + F33 = F
3x , F
24 + F
34 = F
3y
{∆} =
⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
U1V1U2V2U3V3
⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭, {F} =
⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
F 11 + F31
F 12 + F32
F 13 + F21
F 14 + F22
F 23 + F33
F 24 + F34
⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭=
⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
F 1xF 1yF 2xF 2yF 3xF 3y
⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭Boundary conditions
U1 = V1 = U2 = V2 = 0, F 3x = P, F3y = −2P
where the global forces and displacements are
100 kN
x
− 200 kN
1
yF2yF1
xF2
xy
x
y
y
xF1
)2,1(1)4,3(21 90q =
2 45q =
)6,5(3
2
1U
1V2V
2U
3U
3V
Trusses & Frames: 11
JN Reddy
EXAMPLE 1 (continued)
EA
L
0.3536 0.35360.3536 1.3536
U3V3
=P−2P⎧⎪⎨⎪⎩
F 1xF 1yF 2xF 2y
⎫⎪⎬⎪⎭ =EA
L
⎡⎢⎣−0.3536 −0.3536−0.3536 −0.35360.0 0.00.0 −1.0
⎤⎥⎦⎧⎨⎩U3V3
⎫⎬⎭U3 = (3 + 2
√2)PL
EA= 5.828
PL
EA, V3 = −3PL
EA
F 1x = −P, F 1y = −P, F 2x = 0.0, F 2y = 3P
Solution100 kN
x
− 200 kN
1
yF2yF1
xF2
xy
x
y
y
xF1
)2,1(1)4,3(21 90q =
2 45q =
)6,5(3
2
1U
1V2U2V
3U
3U
Trusses & Frames: 12
(P = 105 N, EA = 108 N)
(m)
(N)1 1 2 21 2 1 23 2( ) ( ) ( ) ( );F F P F F P=- = =- =-
JN Reddy
σe = − Pe1
Ae=P e2Ae
P e1P e2
=AeEehe
1 −1−1 1
ue1ue2⎧⎪⎨⎪⎩
ue1ve1ue2ve2
⎫⎪⎬⎪⎭ =
⎡⎢⎣cos θe sin θe 0 0− sin θe cos θe 0 00 0 cos θe sin θe0 0 − sin θe cos θe
⎤⎥⎦⎧⎪⎨⎪⎩ue1ve1ue2ve2
⎫⎪⎬⎪⎭
Post-computation of member displacements and stresses
EXAMPLE 1 (continued)
100 kN
x
− 200 kN
1
yF2yF1
xF2
xy
x
y
y
xF1
)2,1(1)4,3(21 90q =
2 45q =
)6,5(3
2
1U
1V2U2V
3U
3U
Trusses & Frames: 13
1 1 2 2 1 2 1 21 1 1 1 2 2 3 2 2 3
30 3 2 2; ( ) ,PL PLu v u v u u U v v VAE AE
= = = = = = = + = = =-
We have
JN Reddy
Post-computation of member displacements and stresses
EXAMPLE 1 (continued)
Trusses & Frames: 14
( )
12 3 1 3 1 3
22 3 2 3 2 3 3
1 1 2 21 2 1 2
1 2
3
12
3 2
3 2( ) ( )
cos sin
cos sin
,
,
PLu U V VA
u U V U V
P P P P P P
P PA A
q q
q q
s s
= + = =-
= + = +
=- = =- =-
=- =
JN Reddy
PLANE FRAME STRUCTURES
he1 2
eee fQF 111 +=
eee qQF 233 +=
eee qQF 122 +=
eee qQF 466 +=eee fQF 244 +=
eee qQF 355 +=
1 2
eS1 eS2
ew1ew2
eu1eu2
he
Displ. degrees of freedom in the element coordinates
Force degrees of freedom in the element coordinates
2
1 α
z
xeu1
eS1
ew1
ew2 eu2 exeS2
ez
Displ. degrees of freedom in the local coordinates
eu1
eS1
ew1
ew2
eu2
eS2 2
1α
z
x
Displ. degrees of freedom in the global coordinates
Trusses & Frames: 15
JN Reddy
Frame Element in Global Coordinates
[Ke][T e]{∆e} = [T e]{F e}, [T e]−1 = [T e]T
[T e]T[Ke][T e]{∆e} = {F e} or [Ke]{∆e} = {F e}[Ke] = [T e]T[Ke][T e], {F e} = [T e]T{F e}
Te e eD D=
1 1 2 2
1 1 2 2
1 1 2 2
0 00 0
0 0 1 0 0 1
cos sin cos sinsin cos , sin cos
e e e e
e e e e
e e e e
u u u uw w w w
α α α αα α α α
θ θ θ θ
= − = −
1 1
1 1
1 1
2 2
2 2
2 2
00
0 0 100
0 0 1
cos sinsin cos
cos sinsin cos
e e
e e
e e
e e
e e
e e
u uw w
u uw w
α αα α
θ θα αα α
θ θ
− =
−
0
0
Trusses & Frames: 16
JN Reddy Trusses & Frames: 17
Frame Element in Global Coordinates (the Euler-Bernoulli beam frame element)
JN Reddy
AN EXAMPLE OF A FRAME STRUCTURE
Given structure
a
b
F
A B
C P
Finite element discretization
1 2 3 4 5 64 5 6 7 8 9é ùê úê úë û
1 2 3 4 5 6 (1)
(2) [B] =
Connectivity array for 3 DoF per node
ab
F
A B
C
2
1
1 2
3
1 2
2
1(1,2,3) (4,5,6)
(7,8,9)P
• •
•
(1,2,3)1
(4,5,6)1
(4,5,6)2
(1,2,3)2
x
z
0 0 01 20 , 90 or 270a a= = −
JN Reddy 2-D Problems: 19
(1) (1) (1)11 11 12 12 16 16, , ... ,K K K K K K= = =
(1) (1) (1)22 22 23 23 26 26, , ... ,K K K K K K= = =
(1) (2) (1) (2)44 44 11 45 45 12
(1) (2) (1) (2)46 46 13 55 55 22
(1) (2) (1) (2)56 56 23 66 66 33
(2) (2) (2)47 14 48 15 49 16
(2) (2) (2)57 24 58 25 59 26
(2) (2)67 24 68 35 69
, ,, ,, ,
, , ,, , ,, ,
K K K K K KK K K K K KK K K K K KK K K K K KK K K K K KK K K K K
= + = +
= + = +
= + = +
= = =
= = =
= = = (2)26
(2) (2) (2)77 44 78 45 79 46
(2) (2) (2)88 55 89 56 99 66
,, , ,, , .
KK K K K K KK K K K K K
= = =
= = =
Global stiffness coefficients in terms of element stiffness coefficients
(1) (1) (1) (1)33 33 34 34 35 35 36 36, , ,K K K K K K K K= = = =
ab
F
2
1
1 2
2
1(1,2,3) (4,5,6)
(7,8,9)P
• •
•
(1,2,3)1
(4,5,6)1
(4,5,6)2
(1,2,3)2
ASSEMBLY OF ELEMENT MATRICES
1 2 3 4 5 64 5 6 7 8 9é ùê úê úë û
1 2 3 4 5 6 (1)
(2) [B] =
JN Reddy 20
EXAMPLE 2 (from the textbook)
The y-axis is into the plane of the paper, and is measured counterclockwise from x-axis to x-axis
A = 10 in.2 , I = 10 in.4 , E = 30×106 psi
1°
°
11
12 qQ +
2
11
11 fQ +
12
14 fQ +
13
15 qQ +
12
13 qQ +
14
16 qQ +
x
z1
1
22Q 2
21Q
24Q
25Q
23Q
26Q
2P4P
xz2
x
z
( )1 32 4tana −=
4P
2P
144 in.
72 in.
11
2
3
B
43
lb/in.72P
1 90 ,a =
2a
in.144
in.1082
1a
a
44 45 46 4 4
54 55 56 5 5
64 65 66 6 6
K K K U FK K K U FK K K U F
=
● (1,2,3)2(4,5,6)1
(4,5,6)
(1) (2) (1) (2)44 44 11 45 45 12
(1) (2) (1) (2)46 46 13 55 55 22
(1) (2) (1) (2)56 56 23 66 66 33
(1) (2)4 4 1
(1) (2)5 5 2
(1) (2)6 6 3
, ,, ,, ,
( 2 ) ( 2 ) 4P24 72 48
K K K K K KK K K K K KK K K K K KF F F PF F F P PF F F P P P
= + = +
= + = +
= + = +
= + =
= + = - + - =-
= + = - =-
The rest of the calculations canbe found on pp.278-281 of the Book.
JN Reddy
SUMMARY
In this lecture we have covered the following topics:
• Plane truss element in the local and global coordinates• Transformation of element equations from element
coordinates to global coordinates• Examples, illustrating assembly, application of
boundary conditions, and calculation of elementforces and stresses
• Plane frame element in the local and globalcoordinates
• Transformation of element equations from elementcoordinates to global coordinates
Trusses & Frames: 21
Recommended