3. (30 marks)
The bar shown in the figure below is cylindrical in shape. When laying flat on the ground, it has a
length of 2m. When supported from a ceiling mount, the bar wil deflect under its own weight. The
ar is made from HDPE with E=10 MPa and a density of 0.95 g/cm?. By
eaking the bar into 8
finite elements, each with a free length of 250mm, determine the displacement of the bar tip, and
the reaction force at the ceiling support:
a. Due to the self weight of the bar only, and
. Ifa 1 kN load is applied centrally at the tip.
(Note: Matlab may be used to solve equations, please submit a basic code with your submission.
All “process calculations" should be done by hand)
© 100mm
«
"Flat length" = 2m
PowerPoint Presentation
MECH4410 – Mechanics of Solids 2
and FEA
Bar Elements and 2D Transformations
Peter Robinson
Lecturer, School of Engineering
The University of Newcastle
2
Lecture Objectives
• Present the equations used to determine the behaviour of bar
elements
• Derive the stiffness matrix of the bar element using the direct method
• Interpretation of a stiffness matrix
• Define “shape function”
• Express the shape function matrix, N, of a 2-noded bar element
• Express the strain-displacement matrix, B, of a 2-noded bar element
• Derive the stiffness matrix of the bar element using the formal
procedure
• Able to manually solve for axial displacements and stresses for a
simple assembly of bar Elements
• Determine the work equivalent loading for a bar subject to a
distributed load
• Transforming the stiffness of a bar element in 2D space.
3
Bar or Truss Elements
• A 1D element that can only transmit axial force (no bending).
• Consider the bar with a cross-sectional area A, modulus E and initial
length L
• Elastic behaviour – consider analogies with springs.
???? = ?????? = ??
????
????
=
??
??
• Force in the bar:
?? = ????
????
????
= ????
??2 − ??1
??
4
Bar or Truss Elements
• The bar cannot sustain shear force or bending moment.
• Any effect of transverse displacement is ignored.
• No intermediate applied loads (only applied at the ends).
5
Global Stiffness Matrix – Direct Method
• Consider a bar with uniform cross-section of length L as shown
elow:
• At node 1:
??1 + ???? = 0
• At Node 2:
??2 − ???? = 0
• So
????
??
??1 − ??2 = ??1
????
??
??2 − ??1 = ??2
• In matrix form:
????
??
1 −1
−1 1
??1
??2 =
??1
??2
?? ≡ ????
??
, ?? −??−?? ??
6
Interpretation of a Stiffness Matrix
• What do the numbers in a stiffness matrix mean?
A column of k is a vector of nodal loads that must be applied to the
element at its nodes to sustain a deformation state in which the
co
esponding degree of freedom (d.o.f.) has unit value and all other
d.o.f. are zero.
7
Interpretation of a Stiffness Matrix
• What do the numbers in a stiffness matrix mean?
A column of k is a vector of nodal loads that must be applied to the
element at its nodes to sustain a deformation state in which the
co
esponding degree of freedom (d.o.f.) has unit value and all other
d.o.f. are zero.
• An example:
If u1 = 1 and u2 = 0 then
????
??
1 −1
−1 1
1
0 =
??1
??2
,ℎ???????? ??1??2
=
????
??
1
−1 =
??
−??
8
Interpretation of a Stiffness Matrix
• Example from exercises:
• Therefore rows and columns sum to zero: static equili
ium.
9
Global Stiffness Matrix – Formal Method
• Direct method works for simple systems only.
• For complex systems, the stiffness matrix is derived by stating the
work done by external loads is stored in the element as elastic strain
energy.
• General formula is given by K = ∫????????????
B is the strain-displacement matrix
E is the material matrix
dV is the increment of the element volume V
• To obtain B we first need to know the relationship between the
displacement of an a
itrary point on the bar and the displacement of
the nodes - this relationship defines what is happening within a body,
ased on surface movement.
Best to show with an example.
10
Global Stiffness Matrix – Formal Method
• Case 1: bar element subject to axial loading
• Move node 1 by +1 unit to the right, and hold node 2 fixed.
• How far does a point x = L/2 move?
• How far does a point x = 3L/4 move?
11
Global Stiffness Matrix – Formal Method
• Case 1: bar element subject to axial loading
• Move node 1 by +1 unit to the right, and hold node 2 fixed.
• How far does a point x = L/2 move? Ans: ½ unit
• How far does a point x = 3L/4 move? Ans: ¼ unit
12
• This allows a relation to be
developed, showing the
displacement along the bar, with
espect to movement of node 1:
?? =
?? − ??
?? ??1
This is called a ‘shape function’
Global Stiffness Matrix – Formal Method
• Case 1: bar element subject to axial loading
• Move node 1 by +1 unit to the right, and hold node 2 fixed.
13
Global Stiffness Matrix – Formal Method
• Case 2: bar element subject to axial loading
• Move node 2 by +1 unit to the right, and hold node 1 fixed.
• How far does a point x = L/2 move?
• How far does a point x = 3L/4 move?
14
Global Stiffness Matrix – Formal Method
• Case 2: bar element subject to axial loading
• Move node 2 by +1 unit to the right, and hold node 1 fixed.
• How far does a point x = L/2 move? Ans: ½ unit
• How far does a point x = 3L/4 move? Ans: ¾ unit
15
• This allows a relation to be
developed, showing the
displacement along the bar, with
espect to movement of node 2:
?? =
??
?? ??2
This is called a ‘shape function’
Global Stiffness Matrix – Formal Method
• Case 2: bar element subject to axial loading
• Move node 2 by +1 unit to the right, and hold node 1 fixed.
16
Global Stiffness Matrix – Formal Method
• Combining gives:
?? =
?? − ??
??
??1 +
??
??
??2
?? = ?? − ??
??
??
??
??1
??2
?? = ?? � ??
• ?? is a shape function matrix
• ?? is a vector of element d.o.f
17
Shape Functions
?? =
?? − ??
??
??1 +
??
??
??2
• This is the sum of two shape functions:
??1 =
?? − ??
??
, ??2=
??
??
• The shape function defines how displacement (u) varies with position
(x), when a unity displacement is applied at one end, and all other
degrees of freedom are fixed.
• Now – how do calculate strains, and stresses? We create B, the
strain-displacement matrix.
18
Strain Displacement Matrix
Axial strain, ????, is the gradient of axial displacement:
?? = ?? � ??
???? =
????
????
=
????
????
?? = ????
And for stress:
?? = ?????? = ??????
If we know the shape functions:
?? =
?? − ??
??
??
??
?? = −
1
??
1
??
This is the B matrix for a bar-type element.
B matrix is the
derivative of the
shape function
matrix.
19
Strain Displacement Matrix
• For a constant cross-sectional area:
???? = ??????
• Therefore:
?? = �????????????
?? = �
0
??
− �1 ??
�1 ??
?? − �1 ?? �
1
?? ??????
?? =
????
??
1 −1
−1 1
• Same stiffness matrix as the direct method.
• Proof that it works for a bar element.
20
Example 5.1
Determine the stresses in the two-bar assembly shown in the figure below.
21
Example 5.1
Stiffness matrices for each element:
??1 =
2????
??
1 −1
−1 1 and ??2 =
????
??
1 −1
−1 1
Global matrix:
?? =
????
??
2 −2 0
−2 3 −1
0 −1 1
22
Example 5.1
Knowing ???? = ??
????
??
2 −2 0
−2 3 −1
0 −1 1
??1
??2
??3
=
??1
??2
??3
Boundary conditions:
??1 = ??3 = 0
??2 = ??
????
??
2 −2 0
−2 3 −1
0 −1 1
0
??2
0
=
??1
??
??3
23
Example 5.1
????
??
2 −2 0
−2 3 −1
0 −1 1
0
??2
0
=
??1
??
??3
Solving the 2nd line yields:
3??????2
??
= ?? ⇒ ??2 =
????
3????
24
Example 5.1
What about the stresses?
Solve for ??1 and ??3 and divide by the respective, ‘A’, or:
??1 = ????1 = ?????? = ??
−1
??
1
??
??1
??2
= ??
??2 − ??1
??
=
??
??
????
3????
− 0 =
??
3??
??2 = ????2 = ?????? = ??
−1
??
1
??
??2
??3
= ??
??3 − ??2
?? =
??
?? 0 −
????
3???? = −
??
3??
25
Uniformly Distributed Loads on a Bar Element
• Distributed, or varying loads along the bar length – convert UDL into
equivalent nodal loads.
• If uniform – half the total load at each node
• If linear, quadratic (or complex) – not so easy.
• Calculate the ‘work equivalent’ nodal loads – look at the work done by the
loads:
• For a distributed load of ?? (N/m):
?? = �
0
??
????????, ??ℎ??????
?? = ????
26
Uniformly Distributed Loads on a Bar Element
This work, is now equivalent to loads at the ends, multiplied by displacement:
?? = ???? = �
0
??
??????????, where ?? are nodal forces
If ?? is constant, we know N = ??−????
??
?? therefore:
???? ????
????
????
= ??�
0
??
?? − ??
??
??
??
????
????
????
???? ????
????
????
= ?? ?? −
??2
2??
??2
2?? 0
?? ????
????
= ????
2
????
2
????
????
Force is equal at each end…!
27
Uniformly Distributed Loads on a Bar Element
What if there are multiple nodes in a bar?
28
Uniformly Distributed Loads on a Bar Element
What if the distributed force is not constant – ie ?? = ?? ?? ?
• Use the same technique:
?? = ∫0
?? ?? ?? ?????? where ?? = ????
we know N = ??−????
??
?? therefore:
???? ????
????
????
= �
0
??
??(??) ?? − ??
??
??
??
????
????
????
Or:
???? = �
??
??
?? ??
?? − ??
?? ????, ????= �
??
??
?? ??
??
?? ????,
If the force is linear – ie ?? = ????
???? = ??∫0
?? ??−??
??
?????? =?? ??
2
6
and ???? = ??∫0
?? ??
??
?????? =?? ??
2
3
Pi/Pj split = 1:2
29
How is this used in FEA?
• FEA packages assign node numbers in a mesh to minimise sparcity (and
andwidth) in a stiffness matrix.
• Aims to solve:
Kd = F
• Solve for unknown displacements by:
• Direct method – some form of gaussian elimination. Easiest to
solve for multiple load cases. Computational time proportional
to nb2, n≡order of K, b ≡bandwidth.
• Iterative method
• What about 2D space?
30
Bar Elements in 2D space
• Examples up until now have focused on 1-D bar elements (or series of).
• What if bars are in 2-D space, and not in line? What is K?
• Transform from 2-D space to 1-D space by transformation matrix
• Never fear – this is done in the software.
31
Bar Elements in 2D space
• Skipping some maths (you can thank me later), the element stiffness matrix,
K, is given by:
?? = ????????? = ????
??
??2 ???? −??2 −????
???? ??2 −???? −??2
−??2 −???? ??2 ????
−???? −??2 ???? ??2
for DOF ?? =
??1
??1
??2
??2
• Where k’ is the 1-D stiffness matrix:
??′ =
????
??
1 −1
−1 1
If ∅ = 90° ???? 0°, this matrix
educes to the 1-D version
32
Example 5.2
• The figure shows a simple planar pin-jointed truss. Determine the
displacement at node 2 and the stress in each bar.
• No bending stresses (pin joints)
• Global matrix will be 6 x 6 (3 u’s, 3 v’s)
33
Example 5.2
• The figure shows a simple planar pin-jointed truss. Determine the
displacement at node 2 and the stress in each bar.
• In local coordinates, we know:
??′ =
????
??
1 −1
−1 1
• As each bar is misaligned, we need global
coordinates.
• For bar 1 - ∅ = 45°: sin 45° = cos 45° = �1 2
• Therefore ??2 = ??2 = ???? = ⁄1 2
• For bar 1:
−−
−−
−−
−−
=
22
22
22
22
scsscs
csccsc
scsscs
csccsc
L
AEk
−−
−−
−−
−−
=
1111
1111
1111
1111
2
2211
L
AEk
vuvu
34
Example 5.2
• The figure shows a simple planar pin-jointed truss. Determine the
displacement at node 2 and the stress in each bar.
• For bar 2 - ∅ = 135°: sin 135° = −cos 135° = �1 2
• Therefore ??2 = ??2 = ⁄1 2 ,