CHAP 4 FINITE ELEMENT ANALYSIS OF
BEAMS AND FRAMES
FINITE ELEMENT ANALYSIS AND DESIGN
Nam-Ho Kim
1
INTRODUCTION
• We learned Direct Stiffness Method in Chapter 2
– Limited to simple elements such as 1D bars
• In Chapter 3, Galerkin Method and Principle of Minimum
Potential Energy can be applied to more complex elements
• we will learn Energy Method to build beam finite element
– Structure is in equilibrium when the potential energy is minimum
• Potential energy: Sum of strain energy and potential of
applied loads
Potential of

=
U
+
V
• Interpolation scheme:
applied loads
v(x)
=
N(x)
 {q}
Strain energy
Beam
Interpolation Nodal
deflection
function
DOF
2
BEAM THEORY
• Euler-Bernoulli Beam Theory
– can carry the transverse load
– slope can change along the span (x-axis)
– Cross-section is symmetric w.r.t. xy-plane
– The y-axis passes through the centroid
– Loads are applied in xy-plane (plane of loading)
y
y
Neutral axis
Plane of loading
x
L
z
A
F
F
3
BEAM THEORY cont.
• Euler-Bernoulli Beam Theory cont.
– Plane sections normal to the beam axis remain plane and normal to
the axis after deformation (no shear stress)
– Transverse deflection (deflection curve) is function of x only: v(x)
– Displacement in x-dir is function of x and y: u(x, y)
dv
u(x,y) = u0 (x) − y
dx
u du0
d2 v
 xx =
=
−y 2
x dx
dx
y
dv
dx
y(dv/dx)
Neutral axis
x
L
q=
F
q = dv/dx
y
v(x)
4
BEAM THEORY cont.
• Euler-Bernoulli Beam Theory cont.
u du0
d2 v
 xx =
=
−y 2
x dx
dx
– Strain along the beam axis: 0 = du0 / dx
– Strain xx varies linearly w.r.t. y; Strain yy = 0
– Curvature: −d2 v / dx 2
– Can assume plane stress in z-dir
basically uniaxial status
d2 v
 xx = E xx = E0 − Ey 2
dx
• Axial force resultant and bending moment
d2 v
P =   xx dA = E0  dA − E 2  ydA
dx A
A
A
d2 v 2
M = −  y xx dA = −E0  ydA + E 2  y dA
dx A
A
A
Moment of inertia I(x)
P = EA0
d2 v
M = EI 2
dx
EA: axial rigidity
EI: flexural rigidity
5
BEAM THEORY cont.
• Beam constitutive relation
– We assume P = 0 (We will consider non-zero P in the frame element)
– Moment-curvature relation:
d2 v
M = EI 2
dx
Moment and curvature is linearly dependent
• Sign convention
+Vy
+M
y
x
+P
+M
+P
+Vy
– Positive directions for applied loads
y
p(x)
x
C1
F1
C2
F2
C3
F3
6
GOVERNING EQUATIONS
• Beam equilibrium equations
dVy


 fy = 0  p(x)dx +  Vy + dx dx  − Vy = 0


dVy
dx
dM 
dx

−M +  M +
dx  − ( pdx )
+ Vy dx = 0
dx
2


– Combining three equations together:
– Fourth-order differential equation
Vy = −
Vy +
dVy
dx
dx
M+
Vy
dx
dM
dx
d4 v
EI 4 = p(x)
dx
p
M
= −p(x)
dM
dx
dx
7
STRESS AND STRAIN
• Bending stress
d2 v
 xx = −Ey 2
dx
 xx (x,y) = −
d2 v
M = EI 2
dx
M(x)y
I
Bending stress
– This is only non-zero stress component for Euler-Bernoulli beam
• Transverse shear strain
 xy =
u v
v v
+
=− +
=0
y x
x x
dv
u(x,y) = u0 (x) − y
dx
– Euler beam predicts zero shear strain (approximation)
VQ

=
– Traditional beam theory says the transverse shear stress is xy
Ib
– However, this shear stress is in general small compared to
the bending stress
8
POTENTIAL ENERGY
• Potential energy
• Strain energy
 =U+ V
– Strain energy density
2
2
2
2




1
1
1
d
v
1
d
v
U0 =  xx  xx = E( xx )2 = E  − y 2  = Ey 2  2 
2
2
2  dx 
2
 dx 
– Strain energy per unit length
2
2
1 2  d2 v 
1  d2 v 
UL (x) =  U0 (x,y,z)dA =  Ey  2  dA = E  2   y 2dA
2
2  dx  A
 dx 
A
A
1  d2 v 
UL (x) = EI  2 
2  dx 
2
Moment of
inertia
– Strain energy
L
U =  UL (x)dx =
0
2
d v
1
EI
 2  dx

0
2
 dx 
L
2
9
POTENTIAL ENERGY cont.
• Potential energy of applied loads
NF
NC
i=1
i=1
V = −  p(x)v(x)dx −  Fv(x
i
i ) −  Ci
L
0
dv(x i )
dx
• Potential energy
2
NC
NF
L
dv(x i )
1 L d v
 = U + V =  EI  2  dx −  p(x)v(x)dx −  Fv(x
)
−
C

i
i
i
0
2 0  dx 
dx
i=1
i=1
2
– Potential energy is a function of v(x) and slope
– The beam is in equilibrium when  has its minimum value


=0
v
v*
v
10
RAYLEIGH-RITZ METHOD
1. Assume a deflection shape
v(x) = c1f1(x) + c 2 f2 (x)..... + c n fn (x)
–
–
Unknown coefficients ci and known function fi(x)
Deflection curve v(x) must satisfy displacement boundary conditions
2. Obtain potential energy as function of coefficients
(c1,c 2 ,...c n ) = U + V
3. Apply the principle of minimum potential energy to determine
the coefficients
 
=
=
c1 c 2
=

=0
c n
11
EXAMPLE – SIMPLY SUPPORTED BEAM
p0
• Assumed deflection curve
v(x) = C sin
x
L
E,I,L
• Strain energy
2
d v
1
C2EI4
U =  EI  2  dx =
2 0  dx 
4L3
L
2
• Potential energy of applied loads (no reaction forces)
2p L
x
dx = − 0 C
L

0
0
EI4 2 2p0L
• Potential energy  = U + V = 3 C −
C
4L

L
L
V = −  p(x)v(x)dx = −  p0C sin
•
2p0L
d EI4
PMPE:
=
C
−
=0
3
dC
2L


4p0L4
C=
EI5
12
EXAMPLE – SIMPLY SUPPORTED BEAM cont.
• Exact vs. approximate deflection at the center
p0L4
Capprox =
76.5EI
p0L4
Cexact =
76.8EI
• Approximate bending moment and shear force
4p0L2
d2 v
2
x
x
M(x) = EI 2 = −EIC 2 sin
= − 3 sin
dx
L
L

L
4p L
d3 v
3
x
x
Vy (x) = −EI 3 = −EIC 3 cos
= − 20 cos
dx
L
L

L
•

pL
p
1  p0L3
x − 0 x3 + 0 x 4 
Exact solutions v(x) = 
EI  24
12
24 
p0L
p
x + 0 x2
2
2
pL
Vy (x) = 0 − p0 x
2
M(x) = −
13
EXAMPLE – SIMPLY SUPPORTED BEAM cont.
Deflection
0.8
v(x)/v_max
•
1.0
0.6
0.4
v-exact
0.2
v-approx.
0.0
0
0.2
0.4
Bending Moment M(x)
• Bending
moment
0.6
0.8
1
0
0.2
0.4
0.6
0.8
1
-0.04
-0.06
-0.08
-0.10
M_exact
-0.12
M_approx
Error increases
0.00
-0.02
x
x
-0.14
0.6
V_exact
• Shear force
Shear Force V(x)
0.4
V_approx
0.2
0.0
-0.2
-0.4
-0.6
0
0.2
0.4
x
0.6
0.8
1
14
EXAMPLE – CANTILEVERED BEAM
–p0
• Assumed deflection
C
v(x) = a + bx + c1x 2 + c 2 x 3
E,I,L
F
• Need to satisfy BC
v(0) = 0, dv(0) / dx = 0
v(x) = c1x 2 + c 2 x 3
L
• Strain energy U = EI ( 2c1 + 6c 2 x )2 dx
2 0
• Potential of loads
L
V ( c1,c 2 ) = −  ( −p0 ) v(x)dx − Fv(L) − C
0
dv
(L)
dx
 p0L3

 p0L4

2
= c1 
− FL − 2CL  + c 2 
− FL3 − 3CL2 
 3

 4

15
EXAMPLE – CANTILEVERED BEAM cont.
• Derivatives of U:
U
= 2EI ( 2c1 + 6c 2 x ) dx = EI 4Lc1 + 6L2c 2
c1
0
L
(
)
U
= 6EI ( 2c1 + 6c 2 x ) xdx = EI 6L2c1 + 12L3c 2
c 2
0
L
• PMPE:  = 0
c1

=0
c 2
(
)
p0L3
EI 4Lc1 + 6L c 2 = −
+ FL2 + 2CL
3
p0L4
2
3
EI 6L c1 + 12L c 2 = −
+ FL3 + 3CL2
4
(
2
(
)
)
• Solve for c1 and c2: c1 = 23.75  10−3 , c 2 = −8.417  10 −3
• Deflection curve: v(x) = 10 −3 ( 23.75x 2 − 8.417x 3 )
• Exact solution: v(x) =
(
1
5400x 2 − 800x 3 − 300x 4
24EI
)
16
EXAMPLE – CANTILEVERED BEAM cont.
Deflection
v(x)/v_max
•
0.0
0.0
0.0
v-exact
0.0
v-approx.
0.0
0
0.2
0.4
x
0.6
0.8
1
Error increases
• Bending
moment
Bending Moment M(x)
500.00
M_exact
400.00
M_approx
300.00
200.00
100.00
0.00
-100.00
0
0.2
0.4
x
0.6
0.8
1
• Shear force
Shear Force V(x)
600.0
500.0
400.0
300.0
V_exact
V_approx
200.0
0
0.2
0.4
x
0.6
0.8
1
17
FINITE ELEMENT INTERPOLATION
• Rayleigh-Ritz method approximate solution in the entire beam
– Difficult to find approx solution that satisfies displacement BC
• Finite element approximates solution in an element
– Make it easy to satisfy displacement BC using interpolation technique
• Beam element
– Divide the beam using a set of elements
– Elements are connected to other elements at nodes
– Concentrated forces and couples can only be applied at nodes
– Consider two-node bean element
– Positive directions for forces and couples
F2
F1
– Constant or linearly
distributed load
C1
x
C2
p(x)
18
FINITE ELEMENT INTERPOLATION cont.
• Nodal DOF of beam element
– Each node has deflection v and slope q
– Positive directions of DOFs
– Vector of nodal DOFs {q} = {v1 q1 v 2
q2 } T
• Scaling parameter s
– Length L of the beam is scaled to 1 using scaling parameter s
v2
v1
x − x1
s=
,
L
dx = Lds,
1
ds = dx,
L
ds 1
=
dx L
q1
q2
x
L
x1
s=0
x2
s=1
• Will write deflection curve v(s) in terms of s
19
FINITE ELEMENT INTERPOLATION cont.
• Deflection interpolation
– Interpolate the deflection v(s) in terms of four nodal DOFs
– Use cubic function: v(s) = a0 + a1s + a2s2 + a3 s3
– Relation to the slope:
dv dv ds 1
q=
=
= (a1 + 2a2s + 3a3 s2 )
dx ds dx L
– Apply four conditions:
dv(0)
dv(1)
v(0) = v1
= q1
v(1) = v 2
= q2
dx
dx
– Express four coefficients in terms of nodal DOFs
v1 = v(0) = a0
dv
1
q1 =
(0) = a1
dx
L
v 2 = v(1) = a0 + a1 + a2 + a3
dv
1
q2 =
(1) = (a1 + 2a2 + 3a3 )
dx
L
a0 = v 1
a1 = Lq1
a2 = −3v1 − 2Lq1 + 3v 2 − Lq2
a3 = 2v1 + Lq1 − 2v 2 + Lq2
20
FINITE ELEMENT INTERPOLATION cont.
• Deflection interpolation cont.
v(s) = (1 − 3s2 + 2s3 )v1 + L(s − 2s2 + s3 )q1 + (3s 2 − 2s3 )v 2 + L( −s 2 + s3 )q2
 v1 
q 
 
v(s) = [N1(s) N2 (s) N3 (s) N4 (s)]  1 
v 2 

 q2 

• Shape functions
N1(s) = 1 − 3s2 + 2s3
N2 (s) = L(s − 2s2 + s3 )
N3 (s) = 3s2 − 2s3
N4 (s) = L( −s2 + s3 )
– Hermite polynomials
– Interpolation property
v(s) = N {q}
1.0
N1
0.8
N3
0.6
0.4
N2/L
0.2
0.0
0.0
-0.2
0.2
0.4
0.6
N4/L
0.8
1.0
21
FINITE ELEMENT INTERPOLATION cont.
• Properties of interpolation
– Deflection is a cubic polynomial (discuss accuracy and limitation)
– Interpolation is valid within an element, not outside of the element
– Adjacent elements have continuous deflection and slope
• Approximation of curvature
– Curvature is second derivative and related to strain and stress  v1 
q 
d2 v 1 d2 v 1
 1
=
=
[
−
6
+
12s,
L(
−
4
+
6s),
6
−
12s,
L(
−
2
+
6s)]
 
dx 2 L2 ds2 L2
v 2 
 q2 
2
dv 1
= 2 B  {q}
2
B: strain-displacement vector
dx
L
41
14
– B is linear function of s and, thus, the strain and stress
– Alternative expression: d2 v 1 T
= 2 q  {BT }
2
dx
L 14 41
– If the given problem is linearly varying curvature, the approximation is
accurate; if higher-order variation of curvature, then it is approximate 22
FINITE ELEMENT INTERPOLATION cont.
• Approximation of bending moment and shear force
d2 v EI
M(s) = EI 2 = 2 B  {q}
dx
L
Linear
dM
d3 v EI
Vy = −
= −EI 3 = 3 [ −12 −6L 12 −6L]{q}
dx
dx
L
Constant
– Stress is proportional to M(s); M(s) is linear; stress is linear, too
– Maximum stress always occurs at the node
– Bending moment and shear force are not continuous between adjacent
elements
23
EXAMPLE – INTERPOLATION
v1
• Cantilevered beam
• Given nodal DOFs
{q} = {0, 0, − 0.1, − 0.2} T
•
•
•
•
v2
q1
q2
L
Deflection and slope at x = 0.5L
Parameter s = 0.5 at x = 0.5L
1
L
1
L
Shape functions: N1( 21 ) = , N2 ( 21 ) = , N3 ( 21 ) = , N4 ( 21 ) = −
2
8
2
8
Deflection at s = 0.5:
v( 21 ) = N1( 21 )v1 + N2 ( 21 )q1 + N3 ( 21 )v 2 + N4 ( 21 )q2
=
v
Lq
1
L
1
L
 0 +  0 +  v 2 −  q2 = 2 − 2 = −0.025
2
8
2
8
2
8
• Slope at s = 0.5:
dN3
dN
dN4 
dv 1 dv 1  dN1
=
=  v1
+ q1 2 + v 2
+ q2
dx L ds L  ds
ds
ds
ds 
(
)
(
)
1
1
= v1 ( −6s + 6s2 ) + q1 1 − 4s + 3s2 + v 2 (6s − 6s 2 ) + q2 −2s + 3s 2 = −0.1
L
L
24
EXAMPLE
• A beam finite element with length L
L3
L2
v1 = 0, q1 = 0, v 2 =
, q2 =
3EI
2EI
• Calculate v(s)
L
F
v(s) = N1(s)v1 + N2 (s)q1 + N3 (s)v 2 + N4 (s)q2
v(s) = (3s2 − 2s3 )v 2 + L( −s2 + s3 )q2
• Bending moment
d2 v EI d2 v EI
M(s) = EI 2 = 2 2 = 2 (6 − 12s)v 2 + L( −2 + 6s)q2 
dx
L ds
L
EI 
L3
L2 
= 2 (6 − 12s)
+ L( −2 + 6s)
L 
3EI
2EI 
= L(1 − s) = (L − x)
Bending moment cause by unit force at the tip
25
FINITE ELEMENT EQUATION FOR BEAM
• Finite element equation using PMPE
– A beam is divided by NEL elements with constant sections
• Strain energy
– Sum of each element’s strain energy
LT
NEL
x(2 )
NEL
0
e =1
x1
e =1
e
U =  UL (x)dx =   (e) UL (x)dx =  U( )
e
– Strain energy of element (e)
2
2
1  d2 v 
EI 1 1  d2 v 
(e)
U = EI (e)  2  dx = 3   2  ds
x1 2 dx
L 0 2  ds 


x(2 )
e
y
C1
p(x)
x
1
F1
x 1(1)
C2
C3
C4
2
3
4
F2
x 2(1) = x 1(2 )
F3
x 2(2 ) = x 1(3 )
F4
x 2(3 ) = x 1(4 )
C5
5
F5
x 2(4 )
26
FE EQUATION FOR BEAM cont.
• Strain energy cont.
– Approximate curvature in terms of nodal DOFs
2
 d2 v 
 d2 v   d2 v 
T
(e) T
(e)
=
=
{
q
}
B
B
{
q
}




 2
 2  2 
   
ds
ds
ds
1

4
4

1
41
14





– Approximate element strain energy in terms of nodal DOFs
1 ( e ) T  EI 1
T

(e)
U = {q }  3  B  B  ds 
2
L 0

(e)
{q( ) } =
e
1 (e) T (e) (e)
{q } [k ]{q }
2
• Stiffness matrix of a beam element
 −6 + 12s 
L( −4 + 6s)
1
EI
e
  −6 + 12s L( −4 + 6s) 6 − 12s L( −2 + 6s) ds
[k ( ) ] = 3  
0
L  6 − 12s 


L( −2 + 6s)
27
FE EQUATION FOR BEAM cont.
• Stiffness matrix of a beam element
6L −12 6L 
 12
 6L 4L2 −6L 2L2 
EI
e

[k ( ) ] = 3 
L  −12 −6L 12 −6L 

2
2 
 6L 2L −6L 4L 
Symmetric, positive semi-definite
Proportional to EI
Inversely proportional to L
• Strain energy cont.
NEL
U = U
(e)
e =1
1 NEL ( e ) T ( e ) ( e )
=  {q } [k ]{q }
2 e=1
– Assembly
U=
1
{Q s } T [K s ]{Q s }
2
28
EXAMPLE – ASSEMBLY
y
x
2EI
• Two elements
3 • Global DOFs
EI
2
1
2L
F2
v1
{Q s } T = {v1 q1 v 2 q2 v 3 q3 }
L
q1
v2
q2
−3 3L  v1
 3 3L

2
2 
EI  3L 4L −3L 2L  q1
(1)
[k ] = 3
L  −3 −3L 3
−3L  v 2

2
2 
3L
2L
−
3L
4L

 q2
F3
v2
q2
v3
q3
6L −12 6L  v 2
 12

2
2 
EI  6L 4L −6L 2L  q2
( 2)
[k ] = 3
L  −12 −6L 12 −6L  v 3

2
2 
6L
2L
−
6L
4L

 q3
−3 3L
0
0 
 3 3L
 3L 4L2 −3L 2L2
0
0 


3L −12 6L 
EI  −3 −3L 15
[K s ] = 3 

L  3L 2L2 3L 8L2 −6L 2L2 
0
0 −12 −6L 12 −6L 

2
2 
0
0
6L
2L
−
6L
4L


29
FE EQUATION FOR BEAM cont.
• Potential energy of applied loads
 F1 
– Concentrated forces and couples
C 
 1 
ND
T
V
=
−
v
q
v
......
q
=
−
{
Q
}
{Fs }
F


V = − (Fv
+
C
q


ND 
s
2
 1 1 2
i i
i i)
i=1




CND 
– Distributed load (Work-equivalent nodal forces)
NEL
x(2 )
NEL
e =1
x1
e =1
e
x(2 )
e
V = −   (e) p(x)v(x)dx =  V (e)
1
V (e) =  (e) p(x)v(x)dx = L(e)  p(s)v(s)ds
x1
0
1
V
(e)
=L
(e)
 p(s) ( v N + q N + v N + q N ) ds
1 1
1
2
2
3
2
4
0
 (e) 1

 (e) 1

 (e) 1

 (e) 1

= v1  L  p(s)N1ds  + q1  L  p(s)N2ds  + v 2  L  p(s)N3ds  + q2  L  p(s)N4ds 
0
0
0
0








= v1F1(e) + q1C1(e) + v 2F2(e) + q2C(e)
2
30
EXAMPLE – WORK-EQUIVALENT NODAL FORCES
• Uniformly distributed load
pL
F1 = pL  N1(s)ds = pL  (1 − 3s + 2s )ds =
0
0
2
1
1
pL2
2
2
3
C1 = pL  N2 (s)ds = pL  (s − 2s + s )ds =
0
0
12
1
1
pL
F2 = pL  N3 (s)ds = pL  (3s2 − 2s3 )ds =
0
0
2
1
1
pL2
2
2
3
C2 = pL  N4 (s)ds = pL  ( −s + s )ds = −
0
0
12
1
 pL
T
{F} = 
2
1
pL2
12
pL
2
2
3
pL2 
−

12 
p
pL/2
pL2/12
Equivalent
pL/2
pL2/12
31
FE EQUATION FOR BEAM cont.
• Finite element equation for beam
6L −12 6L   v1   pL / 2   F1 
 12
  2

  
2
2 
EI  6L 4L −6L 2L   q1   pL / 12   C1 
 =
+ 
3
L  −12 −6L 12 −6L  v 2   pL / 2   F2 

2
2 
2



q
6L
2L
−
6L
4L
−
pL
/ 12  C2 

 2 
– One beam element has four variables
– When there is no distributed load, p = 0
– Applying boundary conditions is identical to truss element
– At each DOF, either displacement (v or q) or force (F or C) must be
known, not both
– Use standard procedure for assembly, BC, and solution
32
PRINCIPLE OF MINIMUM POTENTIAL ENERGY
• Potential energy (quadratic form)
 =U+ V =
1
{Q s } T [K s ]{Q s } − {Q s } T {Fs }
2
• PMPE
– Potential energy has its minimum when
[K s ]{Q s } = {Fs }
[Ks] is symmetric & PSD
• Applying BC
– The same procedure with truss elements (striking-the-rows and
striking-he-columns)
[K]{Q } = {F}
[K] is symmetric & PD
• Solve for unknown nodal DOFs {Q}
33
BENDING MOMENT & SHEAR FORCE
• Bending moment
d2 v EI d2 v EI
M(s) = EI 2 = 2 2 = 2 B  {q}
dx
L ds
L
– Linearly varying along the beam span
• Shear force
 v1 
q 
dM
d3 v
EI d3 v EI
 
Vy (s) = −
= −EI 3 = − 3 3 = 3 [ − 12 −6L 12 −6L]  1 
dx
dx
L ds
L
v 2 
 q2 
– Constant
– When true moment is not linear and true shear is not constant, many
elements should be used to approximate it
My
• Bending stress  x = −
I
• Shear stress for rectangular section
1.5Vy 
4y 2 
 xy (y) =
1 − 2 
bh 
h 
34
y
EXAMPLE – CLAMPED-CLAMPED BEAM
• Determine deflection &
slope at x = 0.5, 1.0, 1.5 m
• Element stiffness matrices
v1
q1
v2
x
1
2
1m
q2
6 −12 6  v1
 12
 6
4 −6 2  q1
(1)

[k ] = 1000 
 −12 −6 12 −6  v 2


6
2
−
6
4

 q2
3
1m
F2 = 240 N
v2
q2
v3
q3
6 −12 6  v 2
 12
 6
4
−6
2  q2
(2)

[k ] = 1000 
 −12 −6 12 −6  v 3


2
−6
4  q3
 6
6 −12 6
0
0   v1   F1 
 12
 
 6
4 −6 2
0
0   q1   C1 


 −12 −6 24 0 −12 6  v 2  240 
1000 

  = 
0
6
2
0
8
−
6
2
q


 2 
 0
0 −12 −6 12 −6  v 3   F3 


  
C
0
0
6
2
−
6
4
q

  3   3 
35
EXAMPLE – CLAMPED-CLAMPED BEAM cont.
• Applying BC
24 0  v 2  240 
1000 
 =


q
0
8
0

 2 

• At x = 0.5
v 2 = 0.01
q2 = 0.0
s = 0.5 and use element 1
v( 21 ) = v1N1( 21 ) + q1N2 ( 21 ) + v 2N3 ( 21 ) + q2N4 ( 21 ) = 0.01 N3 ( 21 ) = 0.005m
q( 21 ) =
dN3
1
v
= 0.015rad
(1) 2
L
ds s= 1
2
• At x = 1.0
either s = 1 (element 1) or s = 0 (element 2)
v(1) = v 2N3 (1) = 0.01 N3 (1) = 0.01m
q(1) =
dN3
1
v
= 0.0rad
2
(1)
L
ds s=1
v(0) = v 2N1(0) = 0.01 N1(0) = 0.01m
q(0) =
dN1
1
v
= 0.0rad
2
(2)
L
ds s=0
Will this solution be accurate or approximate?
36
EXAMPLE – CANTILEVERED BEAM
• One beam element
• No assembly required
• Element stiffness
6 −12 6  v1
 12
 6
4 −6 2  q1

[K s ] = 1000 
 −12 −6 12 −6  v 2


6
2
−
6
4

 q2
p0 = 120 N/m
EI = 1000 N-m2
C = –50 N-m
L = 1m
• Work-equivalent nodal forces
 1 − 3s2 + 2s3 
 F1e 
 1/ 2   60 


C 

 

2
3
1 (s − 2s + s )L 
 1e 
 L / 12   10 

 = p0L 0 
 ds = p0L 
=

2
3
F
1/
2
60
3s
−
2s
 2e 



 

2
3
C2e 
 ( −s + s )L 
−L / 12 −10 
37
EXAMPLE – CANTILEVERED BEAM cont.
• FE matrix equation
6 −12 6   v1   F1 + 60 
 12
 6
4 −6 2   q1   C1 + 10 
  = 
1000 

 −12 −6 12 −6  v 2   60 

  
6
2
−
6
4

  q2  −10 − 50 
• Applying BC
12 −6  v 2   60 
1000 
 =


q
−
6
4
−
60

 2 

v 2 = −0.01m
q2 = −0.03 rad
• Deflection curve: v(s) = −0.01N3 (s) − 0.03N4 (s) = −0.01s3
• Exact solution: v(x) = 0.005(x 4 − 4x 3 + x 2 )
38
EXAMPLE – CANTILEVERED BEAM cont.
• Support reaction (From assembled matrix equation)
1000 ( −12v 2 + 6q2 ) = F1 + 60
1000 ( −6v 2 + 2q2 ) = C1 + 10
F1 = −120N
C1 = −10N m
• Bending moment
EI
B  {q}

2 
L
EI
= 2 ( −6 + 12s)v1 + L( −4 + 6s)q1 + (6 − 12s)v 2 + L( −2 + 6s)q2 
L
= 1000[ −0.01(6 − 12s) − 0.03( −2 + 6s)]
M(s) =
= −60s N m
• Shear force
EI
Vy = − 3 12v1 + 6Lq1 − 12v 2 + 6Lq2 
L
= −1000[ −12  ( −0.01) + 6( −0.03)]
= 60N
39
EXAMPLE – CANTILEVERED BEAM cont.
• Comparisons
0.000
0.000
FEM
Exact
-0.002
FEM
Exact
-0.005
-0.010
q
v
-0.004
-0.006
-0.020
-0.008
-0.025
Deflection
Slope
-0.030
-0.010
0
0.2
0.4
x
0.6
0.8
0
1
10
0.2
0.4
x
0.6
0.8
1
0
FEM
Exact
0
FEM
Exact
-20
-10
-40
Vy
-20
M
-0.015
-30
-60
-80
-40
-100
Bending moment
-50
-60
Shear force
-120
0
0.2
0.4
x
0.6
0.8
1
0
0.2
0.4
x
0.6
0.8
1
40
PLANE FRAME ELEMENT
• Beam
– Vertical deflection and slope. No axial deformation
• Frame structure
– Can carry axial force, transverse shear force, and bending moment
(Beam + Truss)
• Assumption
– Axial and bending effects
are uncoupled
– Reasonable when deformation
u2
is small
• 3 DOFs per node
u1
2
2
u1
v1
q2
p
q2
{ui , v i , qi }
u2
q1
F
v2
• Need coordinate transformation like plane truss
v2
v1
1
q1
1
u2
3
v2
q2
3
4
u1
v1
q1 41
PLANE FRAME ELEMENT cont.
• Element-fixed local coordinates x − y
• Local DOFs {u,v, q}
Local forces {fx , fy , c}
• Transformation between local and global coord.
 fx1   cos f sin f
f  
 y1   − sin f cos f
 c1   0
0
 =
0
fx2   0
fy2   0
0
  
0
 c 2   0
0
0
1
0
0
0
0   fx1 
 
0
0
0   fy1 

0
0
0   c1 
 
cos f sin f 0   fx2  Local coordinates
v2
− sin f cos f 0   fy2 
 
0
0
1  c 2 
y
x
0
0
v1
{ f } = [T ]{ f }
{q} = [T ]{q}
y
u1
f
u2
q2
2
q1 1
x
Global coordinates
42
PLANE FRAME ELEMENT cont.
• Axial deformation (in local coord.)
EA  1 −1  u1   fx1 
 = 


L  −1 1  u2  fx2 
• Beam bending
6L −12 6L   v1   fy1 
 12
 6L 4L2 −6L 2L2   q   
EI 
  1  =  c1 
L3  −12 −6L 12 −6L   v 2  fy2 

2
2 
6L
2L
−
6L
4L

  q2   c 2 
• Basically, it is equivalent to overlapping a beam with a bar
• A frame element has 6 DOFs
43
PLANE FRAME ELEMENT cont.
• Element matrix equation (local coord.)
0
 a1
 0
12a2

 0
6La2

0
 −a1
 0 −12a2

6La2
 0
0
−a1
0
6La2
0
−12a2
4L2a2
0
−6La2
0
a1
0
−6La2
0
12a2
2L2a2
0
−6La2
 u1   fx1 
  f 

6La2  v1   y1 

2
2L a 2   q1   c1 
  =  
0  u2   fx2 
−6La2   v 2   fy2 
   
2
4L a 2   q2   c 
 2
0
EA
L
EI
a2 = 3
L
a1 =
[k ]{q} = { f }
• Element matrix equation (global coord.)
[k ][T ]{q} = [T ]{f }
[T ]T [k ][T ]{q} = {f }
[k ]{q} = {f }
[k ] = [T ]T [k ][T ]
• Same procedure for assembly and applying BC
44
PLANE FRAME ELEMENT cont.
• Calculation of element forces
– Element forces can only be calculated in the local coordinate
– Extract element DOFs {q} from the global DOFs {Qs}
– Transform the element DOFs to the local coordinate {q} = [T ]{q}
– Then, use 1D bar and beam formulas for element forces
AE
( u2 − u1 )
L
EI
– Bending moment M(s) = 2 B  {q}
L
– Axial force P =
– Shear force Vy (s) =
• Other method:
EI
[ −12 −6L 12 −6L] q
L3
 − Vy1 
6L −12 6L   v1 
 12


 6L 4L2 −6L 2L2   q 
−
M
EI


 1
1


=


 
3
+
V


L
−
12
−
6L
12
−
6L
 y2 
v 2 

2
2 
M 
6L
2L
−
6L
4L

  q2 
 2 
45