Beam
Two dominant mathematical models for beams are the shear undeformable (EulerBernoulli) model and the shear-deformable (Timoshenko) model.
In Euler-Bernoulli beam theory, shear deformations are neglected, and plane sections
remain plane and normal to the longitudinal axis. In the Timoshenko beam theory, plane
sections still remain plane but are no longer normal to the longitudinal axis. The
difference between the normal to the longitudinal axis and the plane section rotation is the
shear deformation.
Shear deflection effects are often significant in the lateral deflection of short beams. The
significance decreases as the slenderness ratio (ratio of the beam length to the radius of
gyration of the beam cross-section) increases.
•
For slender beams (L/r > 20) both theories give the same result;
•
For stocky beams (L/r < 10) Timoshenko beam is physically more realistic because it
includes the shear deformations
Since the Timoshenko beam theory is higher order than the Euler-Bernoulli theory, it is
known to be superior in predicting the transient response of the beam. The superiority of
the Timoshenko model is more pronounced for beams with a low aspect ratio. It is shown
that use of an Euler-Bernoulli based controller to suppress beam vibration can lead to
instability caused by the inadvertent excitation of unmodelled modes.
In ANSYS, BEAM188 (or BEAM189) is suitable for analysing slender to moderately
stubby/thick beam structures. This element is based on Timoshenko beam theory which
is a first order shear deformation theory: transverse shear strain is constant through the
cross section; that is, cross sections remain plane and undistorted after deformation.
Shear deformation effects are included. This element is well-suited for linear, large
rotation, and/or large strain nonlinear applications.
Shear deflection effects are activated in the stiffness matrices of ANSYS beam elements
by including a nonzero shear deflection constant (SHEAR_) in the real constant list for
that element type. The shear deflection constant is defined as the ratio of the actual beam
cross-sectional area to the effective area resisting shear deformation. The shear constant
should be equal to or greater than zero. The element shear stiffness decreases with
increasing values of the shear deflection constant. A zero shear deflection constant may
be used to neglect shear deflection. Shear deflection constants for several common
sections are as follows: rectangle (6/5), solid circle (10/9), hollow (thin-walled) circle (2),
hollow (thin-walled) square (12/5). Shear deflection constants for other cross-sections
can be found in structural handbooks.
Finite Element Analysis
Euler Beam Theory
According to classical beam theory (Euler beam theory), the deformation of an arbitrary
point along the beam is defined by its transverse displacement v and its rotation with
respect to the z-axis, φ, as shown in Figure.
y
A
−y
x
B
dx
dv
Figure 1: Deformation of beam
It is derived with reference to Figure 1 that
u dv
φ=
− =
y dx
(1)
where u is the axial displacement.
The curvature, κ, is given by
dφ d 2 v
κ =
=
dx dx 2
(2)
The moment-curvature equation is given by
κ=
M
EI
(3)
The bending moment, M, and shear force, V, are related to the transverse displacement
function as
M = EI
d 2v
dx 2
d 3v
V = EI 3
dx
2
(4a)
(4b)
Beam
Element Stiffness Matrix
A linear Euler beam element is shown in Figure 2. The origin of the x-axis is located at
node 1, and the length of element is L.
v1
1
φ1
x
2
v2
φ2
L
Figure 2: A beam element
According to Euler beam theory; neglecting shear deformation effect, which is valid for
thin beams. Each element has 2 nodes, 2 DOFs per node: deflection and angular
displacement.
A cubic polynomial function is chosen.
(5)
v = a 0 + a1x + a 2 x 2 + a 3 x 3
where a0 – a3 are coefficients to be determined. These four coefficients have no physical
interpretation. We have to define (or solve) these in terms of nodal degrees of freedom:
d1y, φ1, d2y, and φ2, i.e. the deflections and rotations at the two nodes.
The first derivative
dv 1
is
dx
dv
=a1+2a 2 x + 3a 3 x 2
dx
(6)
The boundary conditions at x = 0 are deflection = d1y and bending slope =
L are deflection = d2y and bending slope =
dv
= φ2.
dx
dv
= φ1; at x =
dx
At x = 0,
1
a0 = d1 y
(7)
a1 = φ1
(8)
Note that the first derivative of v is still a continuous function of x.
3
Finite Element Analysis
At x = L,
d 2 y = d1 y + φ1L + a2 L2 + a3 L3
(9)
φ2 = φ1 + 2a2 L + 3a3 L2
(10)
Eqn. 9 – Eqn. 10 × L/3 yields
3
3
2
1
a2 =
− 2 d1 y + 2 d 2 y − φ1 − φ2
L
L
L
L
(11)
Substituting Eqn. 11 into Eqn. 10 yields
a3 =
2
2
1
1
d − 3 d 2 y + 2 φ1 + 2 φ2
3 1y
L
L
L
L
(12)
The deflection function can now be rewritten in terms of x, dyi and φi.
3
3
2
1
d x 2 + 2 d 2 y x 2 − φ1 x 2 − φ2 x 2
2 1y
L
L
L
L
2
2
1
1
+ 3 d1 y x 3 − 3 d 2 y x 3 + 2 φ1 x 3 + 2 φ2 x 3
L
L
L
L
v = d1 y + φ1 x −
(13)
Rearranging Eqn. 13 yields
 3x 2 2 x 3 

2x 2 x3 
v = 1 − 2 + 3 d1 y +  x −
+ 2 φ1
L
L 
L
L 


 3x 2 2 x 3 
 x 2 x3 
+  2 − 3 d 2 y +  −
+ 2 φ2
L 
 L
 L L 
(14)
or
v = N1d1 y + N 2φ1 + N 3 d 2 y + N 4φ2
(15)
where Ni are generally termed as shape functions. (Recall that for a tensile element,
x
x
u = N1d1x + N 2 d 2 x , where N1 = 1 − and N 2 = .)
L
L
The curvature can also be rewritten in terms of x, dyi and φi.
dv
6
6
4
2
=−
φ1 2 d1 y x + 2 d 2 y x − φ1 x − φ2 x
dx
L
L
L
L
6
6
3
3
+ 3 d1 y x 2 − 3 d 2 y x 2 + 2 φ1 x 2 + 2 φ2 x 2
L
L
L
L
Rearranging Eqn. 16 yields
4
(16)
Beam
 4 x 3x 2 
dv  6 x 6 x 2 
= − 2 + 3  d1 y + 1 −
+ 2  φ1
dx  L
L 
L
L 

(17)
 6x 6x2 
 2 x 3x 2 
+  2 − 3  d 2 y +  − + 2  φ2
L 
L 
L
 L
or
dv
= N1d1 y + N 2φ1 + N 3 d 2 y + N 4φ2
dx
For deflection
3x 2 2 x 3
1− 2 + 3
L
L
2
2x
x3
x−
+ 2
L
L
2
3x 2 x3
− 3
L2
L
2
x
x3
− + 2
L L
For slope dv/dx
6x 6x2
− 2 + 3
L
L
4 x 3x 2
1−
+ 2
L
L
6x 6x2
−
L2 L3
2 x 3x 2
−
+ 2
L
L
It is seen at any given coordinate,
∑N
i
(18)
x=0
x=L
N1
1
0
N2
0
0
N3
0
1
N4
0
0
x=0
x=L
N1
0
0
N2
1
0
N3
0
0
N4
0
1
= 1 . At node 1 N1 = 1 and N 2 = N 3 = N 4 = 0 , so
that v = d1 y ; similarly for slope N1 = N 3 = N 4 = 0 and N 2 = 1 , so that φ = φ1 .
The element stiffness and equations can be derived using a direct equilibrium approach.
F1 y = V = EI
d 3v(0) EI
= 3 (12d1 y + 6 Lφ1 − 12d 2 y + 6 Lφ2 )
dx 3
L
M 1 = − M = − EI
d 2v(0 ) EI
= 3 6 Ld1 y + 4 L2φ1 − 6 Ld 2 y + 2 L2φ2
2
dx
L
(
(19)
)
(20)
5
Finite Element Analysis
F2 y = −V = − EI
M 2 = M = EI
d 3v(L ) EI
= 3 (− 12d1 y − 6 Lφ1 + 12d 2 y − 6 Lφ2 )
dx 3
L
d 2v(L ) EI
= 3 6 Ld1 y + 2 L2φ1 − 6 Ld 2 y + 4 L2φ2
2
dx
L
(
)
(21)
(22)
or in matrix form,
6 L − 12 6 L   d1 y 
 F1 y 
 12
M 
 6 L 4 L2 − 6 L 2 L2   φ 
 1  EI 
  1 
 = 3
 F2 y  L − 12 − 6 L 12 − 6 L  d 2 y 

2
2 
M 2 
 6 L 2 L − 6 L 4 L   φ2 
(23)
For comparison, the element stiffness matrix for a Timoshenko beam element is given
below.
 F1 y 
 12
M 

EI  6 L
 1
 = 3
 F2 y  L (1 + ϕ ) − 12

M 2 
 6L
6L
(4 + ϕ )L
2
− 6L
(2 − ϕ )L2
− 12
− 6L
12
− 6L
6L
  d1 y 
(2 − ϕ )L   φ1 
 
− 6 L  d 2 y 
(4 + ϕ )L2   φ2 
2
12 EI 12 g
= 2 and γ is Timoshenko shear coefficient (Cowper 1966;
L
γAGL2
Hutchinson 2000).
where ϕ =
6
(24)
Beam
Example 1
Use finite element method to determine the nodal displacements and rotations and the
global and element forces for the beam shown in Figure 3. The beam is divided into 2
elements, fixed at node 1 and has a roller support at node 2. An elastic spring is attached
to node 3. A downward vertical force of P = 50 kN is applied at node 3. Let E = 200
GPa and I = 2×10-4 m4 throughout the beam, and let k = 200 kN/m and L = 2 m.
P
1
(1)
2
L
(2)
L
3
(3)
k
4
Figure 3: A beam supported by a spring
Solution
Step 1: Element stiffness matrix
EI = 200 × 109 × 2 × 10−4 = 4 × 107 (N ⋅ m 2 ) = 40 (MN ⋅ m 2 )
Element 1:
6L
 F1 y 
 12
M 
 6 L 4 L2
 1  EI 
 = 3
 F2 y  L − 12 − 6 L

2
M 2 
 6L 2L
12
 12
 12
16
= 5× 
− 12 − 12

8
 12
− 12 6 L   d1 y 
− 6 L 2 L2   φ1 
 
12 − 6 L  d 2 y 

− 6 L 4 L2   φ2 
− 12 12   d1 y 
− 12 8   φ1 
 
12 − 12 d 2 y 

− 12 16   φ2 
7
Finite Element Analysis
Element 2:
6L
 F2 y 
 12
M 
 6 L 4 L2
 2  EI 
 = 3
 F3 y  L − 12 − 6 L

2
 M 3 
 6L 2L
12
 12
 12
16
= 5× 
− 12 − 12

8
 12
− 12 6 L  d 2 y 
− 6 L 2 L2   φ2 
 
12 − 6 L  d 3 y 

− 6 L 4 L2   φ3 
− 12
− 12
12  d 2 y 
8   φ2 
 
12 − 12 d 3 y 

− 12 16   φ3 
Element 3:
 F4 y 
 1 − 1 d 4 y 
  = k
 
− 1 1  d 3 y 
 F3 y 
 1 − 1 d 4 y 
= 0.2 × 
 
− 1 1  d 3 y 
Step 2: Global stiffness matrix
60
60
0
0
0   d1 y 
− 60
 F1 y   60
 
M  
80
40
0
0
0   φ1 
− 60
 1   60

 F2 y  − 60 − 60 60 + 60 − 60 + 60
60
0  d 2 y 
− 60
  
 
40 − 60 + 60 80 + 80
40
0   φ2 
− 60
M 2  =  60
F   0
0
60 + 0.2 − 60 − 0.2 d 3 y 
− 60
− 60
 3y  
 
0
60
40
80
0   φ3 
− 60
M 3   0

F   0
0
0
0
0
0.2  d 4 y 
− 0.2
 4y  
60 − 60 60
0
0
0   d1 y 
 60
 
 60
80 − 60 40
0
0
0   φ1 

− 60 − 60 120
0
0  d 2 y 
− 60 60
 

=  60
40
0
160 − 60 40
0   φ2 
 0
0
− 60 − 60 60.2 − 60 − 0.2 d 3 y 
 

0
0
60
40
60
80
0
−
  φ3 

 
 0
0
0
0
0
0.2  d 4 y 
− 0.2

8
Beam
Step 3: Boundary conditions and nodal displacements
d1 y = φ1 = d 2 y = d 4 y = 0 and F3 y = − P = −0.05 MN
Thus,
 0   160 − 60 40   φ2 

 
 
− 0.05 = − 60 60.2 − 60 d 3 y 
 0   40 − 60 80   φ 
 3 

 
Solving the equations we obtain
−3
 φ2   − 1.22 × 10 

  
−3
d 3 y  = − 5.7 × 10 (m)
 φ   − 3.66 × 10−3 
 3 

9
Finite Element Analysis
Step 4: Element forces
Element 1:
60 − 60 60   d1 y 
 F1 y   60
 M   60
80 − 60 40   φ1 
 1 
=
 
 
 F2 y  − 60 − 60 60 − 60 d 2 y 

M 2   60
40 − 60 80   φ2 
60 − 60 60  
0

 60



 60
80 − 60 40  
0


=


− 60 − 60 60 − 60 
0



−3 
40 − 60 80  − 1.22 × 10 
 60
 60   − 7.32 × 10 −2 (MN)   − 73.2 (kN) 
 

 40  
−2
 − 4.88 × 10 (MN ⋅ m) − 48.8 (kN ⋅ m)

=
=
= −1.22 × 10 −3 × 

 
 
−2
− 60  7.32 × 10 (MN)   73.2 (kN) 
−2
 80  − 9.76 × 10 (MN ⋅ m) − 97.6 (kN ⋅ m)
Element 2:
60 − 60 60  d 2 y 
 F2 y   60
M   60
80 − 60 40   φ2 
 2 
=
 
 
 F3 y  − 60 − 60 60 − 60 d 3 y 

 M 3   60
40 − 60 80   φ3 
0
60 − 60 60  

 60


 60
−3 
80 − 60 40   − 1.22 × 10 
=


− 60 − 60 60 − 60  − 5.7 × 10 −3 


40 − 60 80  − 3.66 × 10 −3 
 60
− 4.92 × 10 −2 (MN)  49.2 (kN) 
 


−2
9.8 × 10 (MN ⋅ m)  98 (kN ⋅ m) 
=
=

 
−2
− 4.92 × 10 (MN) − 49.2 (kN)

 
0
0

Element 3:
 F4 y 
 1 − 1 d 4 y 
  = 0.2 × 
 
− 1 1  d 3 y 
 F3 y 
0

 1 − 1 
= 0.2 × 


−3 
− 1 1  − 5.7 × 10 
 1.14 × 10−3 
 1.14 
(MN) = 
=
 (kN)
−3 
− 1.14
− 1.14 × 10 
10
Beam
Example 2
A cantilever beam subjected to a vertical load, P = 100 kN, at its free end is shown in
Figure 4. The length of the beam, L, is 2 m, and the cross-section is a square of side
dimension, b = 100 mm. The elastic modulus is 200 GPa and the Poisson’s ratio is 0.3.
Use one element to determine the deflection at the free end assuming both Euler and
Timoshenko beam elements, respectively.
P
1
(1)
2
L
Figure 4: A cantilever beam
Solution
When Euler beam element is used, the element force-displacement relationship is
6 L − 12 6 L   d1 y 
 F1 y 
 12
M 
 6 L 4 L2 − 6 L 2 L2   φ 
 1  EI 
  1 
 = 3
 F2 y  L − 12 − 6 L 12 − 6 L  d 2 y 

2
2 
M 2 
 6 L 2 L − 6 L 4 L   φ2 
When Timoshenko beam element is used, the element force-displacement relationship is
 F1 y 
 12
M 

EI  6 L
 1
 = 3
 F2 y  L (1 + ϕ ) − 12


 6L

M 2 
6L
(4 + ϕ )L
2
− 6L
(2 − ϕ )L2
− 12
− 6L
12
− 6L
6L
  d1 y 
(2 − ϕ )L   φ1 
 
− 6 L  d 2 y 
(4 + ϕ )L2   φ2 
2
After applying boundary conditions d1 y = φ1 = 0 and F2 y = − P , the reduced global forcedisplacement relationship is
Euler beam element
− P  EI  12 − 6 L  d 2 y 
 = 3 

2 
 0  L  − 6 L 4 L   φ2 
Timoshenko beam element
11
Finite Element Analysis
− P 
EI  12
 = 3

 0  L (1 + ϕ ) − 6 L
− 6 L  d 2 y 
(4 + ϕ )L2   φ2 
Euler beam element
PL3
d2 y = −
3EI
PL2
φ2 = −
2 EI
Timoshenko beam element
PL3 (4 + ϕ )
d2 y = −
12 EI
2
PL
φ2 = −
2 EI
Since I =
b4
E
5
, A = b2 , G =
, and for square cross-sections, γ = ,
2(1 + ν )
12
6
ϕ=
12 EI
2(1 + ν )b 2 2 × (1 + 0.3) × 0.12
=
=
= 7.8 × 10−3
5 2
γAGL2
γL2
×2
6
Thus, these two types of beam elements give similar results.
12
Beam
Example 3
A cantilever beam subjected to a vertical load, P, at its free end is shown in Figure 5. Use
two Euler beam elements to determine:
a)
Nodal deflections and rotations;
b) Element forces and moments;
c)
Reactions.
P
1
(1)
(2)
2
3
2L
Figure 5: A cantilever beam
Use these symbols in the derivation required.
E: Young’s modulus,
I: second moment of area,
L: length
Solution
The force-displacement relationship of element 1 is
6𝐿𝐿 ⎧𝑑𝑑1𝑦𝑦 ⎫
𝜙𝜙1
2𝐿𝐿2
�
−6𝐿𝐿 ⎨𝑑𝑑2𝑦𝑦 ⎬
4𝐿𝐿2 ⎩ 𝜙𝜙2 ⎭
𝐹𝐹1𝑦𝑦
12
𝐸𝐸𝐸𝐸 6𝐿𝐿
𝑀𝑀1
� �= 3�
𝐹𝐹2𝑦𝑦
𝐿𝐿 −12
6𝐿𝐿
𝑀𝑀2
6𝐿𝐿
4𝐿𝐿2
−6𝐿𝐿
2𝐿𝐿2
−12
−6𝐿𝐿
12
−6𝐿𝐿
𝐹𝐹2𝑦𝑦
12
⎧ ⎫ 𝐸𝐸𝐸𝐸
𝑀𝑀2
6𝐿𝐿
= 3�
𝐹𝐹
−12
𝐿𝐿
3𝑦𝑦
⎨ ⎬
6𝐿𝐿
⎩ 𝑀𝑀3 ⎭
6𝐿𝐿
4𝐿𝐿2
−6𝐿𝐿
2𝐿𝐿2
−12 6𝐿𝐿 ⎧𝑑𝑑2𝑦𝑦 ⎫
𝜙𝜙2
−6𝐿𝐿 2𝐿𝐿2
�
12 −6𝐿𝐿 ⎨𝑑𝑑3𝑦𝑦 ⎬
−6𝐿𝐿 4𝐿𝐿2 ⎩ 𝜙𝜙3 ⎭
The force-displacement relationship of element 2 is
13
Finite Element Analysis
The global force-displacement relationship is
𝐹𝐹1𝑦𝑦
12
⎧ ⎫
⎡ 6𝐿𝐿
𝑀𝑀1
⎪ ⎪
𝐸𝐸𝐸𝐸 ⎢−12
𝐹𝐹2𝑦𝑦
= 3⎢
⎨ 𝑀𝑀2 ⎬ 𝐿𝐿 ⎢ 6𝐿𝐿
⎢ 0
⎪𝐹𝐹3𝑦𝑦 ⎪
⎣ 0
⎩ 𝑀𝑀3 ⎭
6𝐿𝐿
4𝐿𝐿2
−6𝐿𝐿
2𝐿𝐿2
0
0
12
6𝐿𝐿
⎡ 6𝐿𝐿
4𝐿𝐿2
⎢
𝐸𝐸𝐸𝐸 −12 −6𝐿𝐿
= 3⎢
2𝐿𝐿2
𝐿𝐿 ⎢ 6𝐿𝐿
⎢ 0
0
⎣ 0
0
−12
−6𝐿𝐿
12 + 12
−6𝐿𝐿 + 6𝐿𝐿
−12
6𝐿𝐿
−12
−6𝐿𝐿
24
0
−12
6𝐿𝐿
The boundary conditions are 𝑑𝑑1𝑦𝑦 = 𝜙𝜙1 = 0.
6𝐿𝐿
2𝐿𝐿2
−6𝐿𝐿 + 6𝐿𝐿
4𝐿𝐿2 + 4𝐿𝐿2
−6𝐿𝐿
2𝐿𝐿2
6𝐿𝐿
2𝐿𝐿2
0
8𝐿𝐿2
−6𝐿𝐿
2𝐿𝐿2
0
0
−12
−6𝐿𝐿
12
−6𝐿𝐿
0
0
−12
−6𝐿𝐿
12
−6𝐿𝐿
0 ⎧𝑑𝑑1𝑦𝑦 ⎫
𝜙𝜙1
0 ⎤⎪
⎪
⎥ ⎪𝑑𝑑 ⎪
6𝐿𝐿 ⎥ 2𝑦𝑦
2𝐿𝐿2 ⎥ ⎨ 𝜙𝜙2 ⎬
−6𝐿𝐿⎥ ⎪
⎪𝑑𝑑3𝑦𝑦 ⎪
⎪
4𝐿𝐿2 ⎦ ⎩ 𝜙𝜙3 ⎭
0 ⎧𝑑𝑑1𝑦𝑦 ⎫
𝜙𝜙1
0 ⎤⎪
⎪
⎥ ⎪𝑑𝑑 ⎪
6𝐿𝐿 ⎥ 2𝑦𝑦
2𝐿𝐿2 ⎥ ⎨ 𝜙𝜙2 ⎬
−6𝐿𝐿⎥ ⎪
⎪𝑑𝑑3𝑦𝑦 ⎪
⎪
4𝐿𝐿2 ⎦ ⎩ 𝜙𝜙3 ⎭
The reduced global force-displacement relationship is
0
24
0
𝐸𝐸𝐸𝐸 0
8𝐿𝐿2
0
� �= 3�
−𝑃𝑃
𝐿𝐿 −12 −6𝐿𝐿
0
6𝐿𝐿
2𝐿𝐿2
0
4 0
⎧
⎫
⎪ 0 ⎪
0 4𝐿𝐿
𝑃𝑃𝐿𝐿3 = �
2 𝐿𝐿
⎨
⎬
⎪6𝐸𝐸𝐸𝐸 ⎪
3 𝐿𝐿
⎩ 0 ⎭
𝑃𝑃𝐿𝐿3
⎧−
⎫
2
⎪ 6𝐸𝐸𝐸𝐸 ⎪
0
0
=�
3
2
⎨ 𝑃𝑃𝐿𝐿 ⎬
3
⎪ 6𝐸𝐸𝐸𝐸 ⎪
⎩ 0 ⎭
𝑑𝑑2𝑦𝑦 =
−2 𝐿𝐿 ⎧𝑑𝑑2𝑦𝑦 ⎫
𝜙𝜙2
−3 𝐿𝐿
�
−2 𝐿𝐿 ⎨𝑑𝑑3𝑦𝑦 ⎬
−3 2𝐿𝐿 ⎩ 𝜙𝜙3 ⎭
0
−3
−2
−3
0 ⎧𝑑𝑑2𝑦𝑦 ⎫
𝜙𝜙2
𝐿𝐿
�
𝑑𝑑
𝐿𝐿 ⎨ 3𝑦𝑦 ⎬
2𝐿𝐿 ⎩ 𝜙𝜙3 ⎭
𝐿𝐿𝜙𝜙2
𝑃𝑃𝐿𝐿3
−
2
12𝐸𝐸𝐸𝐸
0
⎡𝐿𝐿
⎧ 𝑃𝑃𝐿𝐿3 ⎫
⎪
⎪ ⎢
6𝐸𝐸𝐸𝐸 = ⎢𝐿𝐿
⎨ 𝑃𝑃𝐿𝐿3 ⎬ ⎢
⎪
⎪ ⎢
⎩10𝐸𝐸𝐸𝐸 ⎭
⎣𝐿𝐿
14
−𝐿𝐿
4𝐿𝐿
𝐿𝐿
𝐿𝐿
−12 6𝐿𝐿 ⎧𝑑𝑑2𝑦𝑦 ⎫
𝜙𝜙2
−6𝐿𝐿 2𝐿𝐿2
�
12 −6𝐿𝐿 ⎨𝑑𝑑3𝑦𝑦 ⎬
−6𝐿𝐿 4𝐿𝐿2 ⎩ 𝜙𝜙3 ⎭
−
3
4
−1
−
6
5
𝐿𝐿
⎤
4 ⎥ 𝜙𝜙
2
𝐿𝐿 ⎥
�𝑑𝑑3𝑦𝑦 �
2 ⎥ 𝜙𝜙
3
4𝐿𝐿⎥
5⎦
Beam
0
⎡𝐿𝐿
⎧ 𝑃𝑃𝐿𝐿3 ⎫
⎪
⎪ ⎢
6𝐸𝐸𝐸𝐸 = ⎢0
⎨ 𝑃𝑃𝐿𝐿3 ⎬ ⎢
⎪
⎪ ⎢
⎩10𝐸𝐸𝐸𝐸 ⎭
⎣0
3
4
1
−
4
9
−
20
−
𝐿𝐿
⎤
4 ⎥ 𝜙𝜙
2
𝐿𝐿 ⎥
�𝑑𝑑3𝑦𝑦 �
4 ⎥ 𝜙𝜙
3
11𝐿𝐿⎥
20 ⎦
2𝑃𝑃𝐿𝐿3
⎧
⎫
𝐿𝐿 𝑑𝑑3𝑦𝑦
3𝐸𝐸𝐸𝐸 = �−1
��
�
3
−9 11𝐿𝐿 𝜙𝜙3
⎨2𝑃𝑃𝐿𝐿 ⎬
⎩ 𝐸𝐸𝐸𝐸 ⎭
2𝑃𝑃𝐿𝐿3
⎫
−1 𝐿𝐿 𝑑𝑑3𝑦𝑦
3𝐸𝐸𝐸𝐸
��
�
3 =� 0
2𝐿𝐿 𝜙𝜙3
⎨ 4𝑃𝑃𝐿𝐿 ⎬
⎩− 𝐸𝐸𝐸𝐸 ⎭
⎧
2𝑃𝑃𝐿𝐿2
𝜙𝜙3 = −
𝐸𝐸𝐸𝐸
𝑑𝑑3𝑦𝑦
8𝑃𝑃𝐿𝐿3
=−
3𝐸𝐸𝐸𝐸
𝜙𝜙2 = −
The deflections and rotations are
3𝑃𝑃𝐿𝐿2
2𝐸𝐸𝐸𝐸
𝑑𝑑2𝑦𝑦 = −
5𝑃𝑃𝐿𝐿3
6𝐸𝐸𝐸𝐸
5𝑃𝑃𝐿𝐿3
⎧−
⎫
6𝐸𝐸𝐸𝐸
⎪
⎪
⎪ 3𝑃𝑃𝐿𝐿2 ⎪
𝑑𝑑2𝑦𝑦
⎪
⎧
⎫ ⎪−
𝜙𝜙2
2𝐸𝐸𝐸𝐸
=
3
⎨𝑑𝑑3𝑦𝑦 ⎬ ⎨ 8𝑃𝑃𝐿𝐿 ⎬
−
⎩ 𝜙𝜙3 ⎭ ⎪ 3𝐸𝐸𝐸𝐸 ⎪
⎪
⎪
⎪ 2𝑃𝑃𝐿𝐿2 ⎪
⎩− 𝐸𝐸𝐸𝐸 ⎭
15
Finite Element Analysis
b)
The element force/moment of element 1 is
𝐹𝐹1𝑦𝑦
12
𝐸𝐸𝐸𝐸 6𝐿𝐿
𝑀𝑀1
� �= 3�
𝐹𝐹2𝑦𝑦
𝐿𝐿 −12
6𝐿𝐿
𝑀𝑀2
6𝐿𝐿
4𝐿𝐿2
−6𝐿𝐿
2𝐿𝐿2
−12
−6𝐿𝐿
12
−6𝐿𝐿
6𝐿𝐿 ⎧𝑑𝑑1𝑦𝑦 ⎫
𝜙𝜙1
2𝐿𝐿2
�
−6𝐿𝐿 ⎨𝑑𝑑2𝑦𝑦 ⎬
4𝐿𝐿2 ⎩ 𝜙𝜙2 ⎭
0
⎧ 0 ⎫
−12 6𝐿𝐿 ⎪
5𝑃𝑃𝐿𝐿3 ⎪
−6𝐿𝐿 2𝐿𝐿2
� −
12 −6𝐿𝐿 ⎨ 6𝐸𝐸𝐸𝐸 ⎬
−6𝐿𝐿 4𝐿𝐿2 ⎪ 3𝑃𝑃𝐿𝐿2 ⎪
⎩− 2𝐸𝐸𝐸𝐸 ⎭
12
6𝐿𝐿
𝐸𝐸𝐸𝐸 6𝐿𝐿
4𝐿𝐿2
= 3�
𝐿𝐿 −12 −6𝐿𝐿
6𝐿𝐿
2𝐿𝐿2
𝑃𝑃
2𝑃𝑃𝑃𝑃
�
=�
−𝑃𝑃
−𝑃𝑃𝑃𝑃
The element force/moment of element 2 is
𝐹𝐹2𝑦𝑦
24
⎧ ⎫ 𝐸𝐸𝐸𝐸
𝑀𝑀2
0
= 3�
⎨𝐹𝐹3𝑦𝑦 ⎬ 𝐿𝐿 −12
6𝐿𝐿
⎩ 𝑀𝑀3 ⎭
12
6𝐿𝐿
𝐸𝐸𝐸𝐸 6𝐿𝐿
4𝐿𝐿2
= 3�
𝐿𝐿 −12 −6𝐿𝐿
6𝐿𝐿
2𝐿𝐿2
0
8𝐿𝐿2
−6𝐿𝐿
2𝐿𝐿2
−12 6𝐿𝐿 ⎧𝑑𝑑2𝑦𝑦 ⎫
𝜙𝜙2
−6𝐿𝐿 2𝐿𝐿2
�
12 −6𝐿𝐿 ⎨𝑑𝑑3𝑦𝑦 ⎬
−6𝐿𝐿 4𝐿𝐿2 ⎩ 𝜙𝜙3 ⎭
5𝑃𝑃𝐿𝐿3
⎧−
⎫
6𝐸𝐸𝐸𝐸
⎪
⎪
3𝑃𝑃𝐿𝐿2 ⎪
−12 6𝐿𝐿 ⎪
⎪−
⎪
−6𝐿𝐿 2𝐿𝐿2
2𝐸𝐸𝐸𝐸
�
12 −6𝐿𝐿 ⎨ 8𝑃𝑃𝐿𝐿3 ⎬
−6𝐿𝐿 4𝐿𝐿2 ⎪− 3𝐸𝐸𝐸𝐸 ⎪
⎪
⎪
⎪ 2𝑃𝑃𝐿𝐿2 ⎪
⎩− 𝐸𝐸𝐸𝐸 ⎭
−10𝑃𝑃 − 9𝑃𝑃 + 32𝑃𝑃 − 12𝑃𝑃
𝑃𝑃
−5𝑃𝑃𝐿𝐿 − 6𝑃𝑃𝑃𝑃 + 16𝑃𝑃𝑃𝑃 − 4𝑃𝑃𝑃𝑃
𝑃𝑃𝑃𝑃
�=� �
=�
10𝑃𝑃 + 9𝑃𝑃 − 32𝑃𝑃 + 12𝑃𝑃
−𝑃𝑃
−5𝑃𝑃𝐿𝐿 − 3𝑃𝑃𝑃𝑃 + 16𝑃𝑃𝑃𝑃 − 8𝑃𝑃𝑃𝑃
0
16
Beam
c)
The reactions at node 1 can be found using the global force-displacement relationship.
𝐸𝐸𝐸𝐸 12
𝐹𝐹1𝑦𝑦
� �= 3�
𝑀𝑀1
𝐿𝐿 6𝐿𝐿
=
𝐸𝐸𝐸𝐸 12 6𝐿𝐿
�
𝐿𝐿3 6𝐿𝐿 4𝐿𝐿2
6𝐿𝐿
4𝐿𝐿2
−12
−6𝐿𝐿
−12
−6𝐿𝐿
6𝐿𝐿
2𝐿𝐿2
6𝐿𝐿
2𝐿𝐿2
𝑑𝑑1𝑦𝑦
⎧ ⎫
𝜙𝜙1
⎪
⎪ ⎪
⎪
0 0 𝑑𝑑2𝑦𝑦
�
0 0 ⎨ 𝜙𝜙2 ⎬
⎪
⎪𝑑𝑑3𝑦𝑦 ⎪
⎪
⎩ 𝜙𝜙3 ⎭
0
⎧ 0 ⎫
⎪ 5𝑃𝑃𝐿𝐿3 ⎪
⎪−
⎪
⎪ 6𝐸𝐸𝐸𝐸 ⎪
⎪
⎪
3𝑃𝑃𝐿𝐿2
0 0
𝑃𝑃
�
=�
� −
2𝑃𝑃𝑃𝑃
0 0 ⎨ 2𝐸𝐸𝐸𝐸 ⎬
3
⎪ 8𝑃𝑃𝐿𝐿 ⎪
−
⎪ 3𝐸𝐸𝐸𝐸 ⎪
⎪
⎪
⎪ 2𝑃𝑃𝐿𝐿2 ⎪
⎩− 𝐸𝐸𝐸𝐸 ⎭
References
Cowper, G. R. 1966. The shear coefficient in Timoshenko’s beam theory. Journal of
Applied Mechanics 33 (2):335-340.
Hutchinson, J. R. 2000. Shear coefficients for Timoshenko beam theory. Journal of
Applied Mechanics 68 (1):87-92.
17