Engineering report
Derivation of stiοΏ½ness matrix for a beam
Nasser M. Abbasi
July 7, 2016
Contents
1 Introduction 1
2 Direct method 32.1 Examples using the direct beam stiοΏ½ness matrix . . . . . . . . . . . . . . . . 11
2.1.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.1.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.1.3 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3 Finite elements or adding more elements 213.1 Example 3 redone with 2 elements . . . . . . . . . . . . . . . . . . . . . . . . 21
4 Generating shear and bending moments diagrams 29
5 Finding the stiοΏ½ness matrix using methods other than direct method 315.1 Virtual work method for derivation of the stiοΏ½ness matrix . . . . . . . . . . 325.2 Potential energy (minimize a functional) method to derive the stiοΏ½ness matrix 33
6 References 35
iii
Contents CONTENTS
iv
Chapter 1
Introduction
A short review for solving the beam problem in 2D is given. The deflection curve, bendingmoment and shear force diagrams are calculated for a beam subject to bending momentand shear force using direct stiοΏ½ness method and then using finite elements method byadding more elements. The problem is solved first by finding the stiοΏ½ness matrix using thedirect method and then using the virtual work method.
1
CHAPTER 1. INTRODUCTION
2
Chapter 2
Direct method
Local contents2.1 Examples using the direct beam stiοΏ½ness matrix . . . . . . . . . . . . . . . . . . 11
3
CHAPTER 2. DIRECT METHOD
Starting with only one element beam which is subject to bending and shear forces. Thereare 4 nodal degrees of freedom. Rotation at the left and right nodes of the beam andtransverse displacements at the left and right nodes. The following diagram shows thesign convention used for external forces. Moments are always positive when anti-clockwisedirection and vertical forces are positive when in the positive π¦ direction.
The two nodes are numbered 1 and 2 from left to right. π1 is the moment at the left node(node 1), π2 is the moment at the right node (node 2). π1 is the vertical force at the leftnode and π2 is the vertical force at the right node.
x
y
1 2
M1 M2
V1V2
The above diagram shows the signs used for the applied forces direction when acting inthe positive sense. Since this is a one dimensional problem, the displacement field (theunknown being solved for) will be a function of one independent variable which is theπ₯ coordinate. The displacement field in the vertical direction is called π£ (π₯). This is thevertical displacement of point π₯ on the beam from the original π₯ β ππ₯ππ . The followingdiagram shows the notation used for the coordinates.
x,u
y,v
Angular displacement at distance π₯ on the beam is found using π (π₯) = ππ£(π₯)ππ₯ . At the left
node, the degrees of freedom or the displacements, are called π£1, π1 and at the right nodethey are called π£2, π2. At an arbitrary location π₯ in the beam, the vertical displacement isπ£ (π₯) and the rotation at that location is π (π₯).
The following diagram shows the displacement field π£ (π₯)
1 21 2
main displacement field quantity we need to find
x dvx
dx
v 1 v 2
vx
x,u
y,v
In the direct method of finding the stiοΏ½ness matrix, the forces at the ends of the beam arefound directly by the use of beam theory. In beam theory the signs are diοΏ½erent from whatis given in the first diagram above. Therefore, the moment and shear forces obtained usingbeam theory (ππ΅ and ππ΅ in the diagram below) will have diοΏ½erent signs when comparedto the external forces. The signs are then adjusted to reflect the convention as shown inthe diagram above using π and π.
4
CHAPTER 2. DIRECT METHOD
For an example, the external moment π1 is opposite in sign to ππ΅1 and the reaction π2is opposite to ππ΅2. To illustrate this more, a diagram with both sign conventions is givenbelow.
Beam theory positive sign directions for bending moments and shear forces
Applied external moments and end forces shown in positive sense directions
M1 M2
V1V2
1 2
MB1 MB2
VB1 VB2
The goal now is to obtain expressions for external loads ππ and π π in the above diagramas function of the displacements at the nodes {π} = {π£1, π1, π£2, π2}π.
In other words, the goal is to obtain an expression of the form οΏ½ποΏ½ = [πΎ] {π} where [πΎ] is thestiοΏ½ness matrix where οΏ½ποΏ½ = {π1,π1, π 2,π2}
π is the nodal forces or load vector, and {π} isthe nodal displacement vector.
In this case [πΎ] will be a 4 Γ 4 matrix and οΏ½ποΏ½ is a 4 Γ 1 vector and {π} is a 4 Γ 1 vector.
Starting with π1. It is in the same direction as the shear force ππ΅1. Since ππ΅1 =πππ΅1ππ₯ then
π1 =πππ΅1ππ₯
Since from beam theory ππ΅1 = βπ (π₯) πΌπ¦ , the above becomes
π1 = βπΌπ¦ππ (π₯)ππ₯
But π (π₯) = πΈπ (π₯) and π (π₯) = βπ¦π where π is radius of curvature, therefore the above becomes
π1 = πΈπΌπππ₯ οΏ½
1ποΏ½
Since 1π =
οΏ½π2π’ππ₯2 οΏ½
οΏ½1+οΏ½ππ’ππ₯ οΏ½
2οΏ½3/2 and for a small angle of deflection ππ’
ππ₯ βͺ 1 then 1π = οΏ½π
2π’ππ₯2
οΏ½, and the
above now becomes
π1 = πΈπΌπ3π’ (π₯)ππ₯3
Before continuing, the following diagram illustrates the above derivation. This comes frombeam theory.
5
CHAPTER 2. DIRECT METHOD
Deflection curve
x
x
Radius of curvature
vx
x dv
dx
1 d 2v
dx2
Now π1 is obtained. π1 is in the opposite sense of the bending moment ππ΅1 hence anegative sign is added giving π1 = βππ΅1. But ππ΅1 = βπ (π₯) πΌ
π¦ therefore
π1 = π (π₯)πΌπ¦
= πΈπ (π₯)πΌπ¦
= πΈ οΏ½βπ¦π οΏ½
πΌπ¦
= βπΈπΌ οΏ½1ποΏ½
= βπΈπΌπ2π€ππ₯2
π2 is now found. It is in the opposite sense of the shear force ππ΅2, hence a negative sign isadded giving
π2 = βππ΅2
= βπππ΅2ππ₯
Since ππ΅2 = βπ (π₯) πΌπ¦ , the above becomes
π2 =πΌπ¦ππ (π₯)ππ₯
But π (π₯) = πΈπ (π₯) and π (π₯) = βπ¦π where π is radius of curvature. The above becomes
π2 = βπΈπΌπππ₯ οΏ½
1ποΏ½
But 1π =
οΏ½π2π€ππ₯2 οΏ½
οΏ½1+οΏ½ππ€ππ₯ οΏ½
2οΏ½3/2 and for small angle of deflection ππ€
ππ₯ βͺ 1 hence 1π = οΏ½π
2π’ππ₯2
οΏ½ then the above
becomes
π2 = βπΈπΌπ3π’ (π₯)ππ₯3
Finally π2 is in the same direction as ππ΅2 so no sign change is needed. ππ΅2 = βπ (π₯) πΌπ¦ ,
6
CHAPTER 2. DIRECT METHOD
therefore
π2 = βπ (π₯)πΌπ¦
= βπΈπ (π₯)πΌπ¦
= βπΈ οΏ½βπ¦π οΏ½
πΌπ¦
= πΈπΌ οΏ½1ποΏ½
= πΈπΌπ2π’ππ₯2
The following is a summary of what was found so far. Notice that the above expressionsare evaluates at π₯ = 0 and at π₯ = πΏ. Accordingly to obtain the nodal end forces vector οΏ½ποΏ½
οΏ½ποΏ½ =
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π1
π1
π2
π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=
β§βͺβͺβͺβͺβͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβͺβͺβͺβͺβ©
πΈπΌ π3π’(π₯)ππ₯3 οΏ½
π₯=0
βπΈπΌ π2π’ππ₯2 οΏ½π₯=0
βπΈπΌ π3π’(π₯)ππ₯3 οΏ½
π₯=πΏ
πΈπΌ π2π’ππ₯2 οΏ½π₯=πΏ
β«βͺβͺβͺβͺβͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβͺβͺβͺβͺβ
(1)
The RHS of the above is now expressed as function of the nodal displacements π£1, π1, π£2, π2.To do that, the field displacement π£ (π₯) which is the transverse displacement of the beamis assumed to be a polynomial in π₯ of degree 3 or
π£ (π₯) = π0 + π1π₯ + π2π₯2 + π3π₯3
π (π₯) =ππ£ (π₯)ππ₯
= π1 + 2π2π₯ + 3π3π₯2 (A)
Polynomial of degree 3 is used since there are 4 degrees of freedom, and a minimum of 4free parameters is needed. Hence
π£1 = π£ (π₯)|π₯=0 = π0 (2)
And
π1 = π (π₯)|π₯=0 = π1 (3)
Assuming the length of the beam is πΏ, then
π£2 = π£ (π₯)|π₯=πΏ = π0 + π1πΏ + π2πΏ2 + π3πΏ3 (4)
And
π2 = π (π₯)|π₯=πΏ = π1 + 2π2πΏ + 3π3πΏ2 (5)
Equations (2-5) gives
{π} =
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π0π1
π0 + π1πΏ + π2πΏ2 + π3πΏ3
π1 + 2π2πΏ + 3π3πΏ2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=
βββββββββββββββ
1 0 0 00 1 0 01 πΏ πΏ2 πΏ3
0 1 2πΏ 3πΏ2
βββββββββββββββ
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π0π1π2π3
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
Solving for ππ givesβ§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π0π1π2π3
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=
ββββββββββββββββ
1 0 0 00 1 0 0
β 3πΏ2 β 2
πΏ3πΏ2 β 1
πΏ2πΏ3
1πΏ2 β 2
πΏ31πΏ2
ββββββββββββββββ
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=
ββββββββββββββββ
π£1π1
3πΏ2π£2 β
1πΏπ2 β
3πΏ2π£1 β
2πΏπ1
1πΏ2π1 +
1πΏ2π2 +
2πΏ3π£1 β
2πΏ3π£2
ββββββββββββββββ
π£ (π₯), the field displacement function from Eq. (A) can now be written as a function of the
7
CHAPTER 2. DIRECT METHOD
nodal displacements
π£ (π₯) = π0 + π1π₯ + π2π₯2 + π3π₯3
= π£1 + π1π₯ + οΏ½3πΏ2
π£2 β1πΏπ2 β
3πΏ2
π£1 β2πΏπ1οΏ½ π₯2 + οΏ½
1πΏ2
π1 +1πΏ2
π2 +2πΏ3
π£1 β2πΏ3
π£2οΏ½ π₯3
= π£1 + π₯π1 β 2π₯2
πΏπ1 +
π₯3
πΏ2π1 β
π₯2
πΏπ2 +
π₯3
πΏ2π2 β 3
π₯2
πΏ2π£1 + 2
π₯3
πΏ3π£1 + 3
π₯2
πΏ2π£2 β 2
π₯3
πΏ3π£2
Or in matrix form
π£ (π₯) = οΏ½1 β 3 π₯2
πΏ2 + 2 π₯3
πΏ3 π₯ β 2π₯2
πΏ + π₯3
πΏ2 3 π₯2
πΏ2 β 2 π₯3
πΏ3 βπ₯2
πΏ + π₯3
πΏ2οΏ½
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
= οΏ½ 1πΏ3οΏ½πΏ3 β 3πΏπ₯2 + 2π₯3οΏ½ 1
πΏ2οΏ½πΏ2π₯ β 2πΏπ₯2 + π₯3οΏ½ 1
πΏ3οΏ½3πΏπ₯2 β 2π₯3οΏ½ 1
πΏ2οΏ½βπΏπ₯2 + π₯3οΏ½οΏ½
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
The above can be written as
π£ (π₯) = οΏ½π1 (π₯) π2 (π₯) π3 (π₯) π4 (π₯)οΏ½
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
(5A)
π£ (π₯) = [π] {π}
Where ππ are called the shape functions. The shape functions areβββββββββββββββ
π1 (π₯)π2 (π₯)π3 (π₯)π4 (π₯)
βββββββββββββββ
=
ββββββββββββββββ
1πΏ3οΏ½πΏ3 β 3πΏπ₯2 + 2π₯3οΏ½
1πΏ2οΏ½πΏ2π₯ β 2πΏπ₯2 + π₯3οΏ½1πΏ3οΏ½3πΏπ₯2 β 2π₯3οΏ½
1πΏ2οΏ½βπΏπ₯2 + π₯3οΏ½
ββββββββββββββββ
We notice that π1 (0) = 1 and π1 (πΏ) = 0 as expected. Alsoππ2 (π₯)ππ₯
οΏ½π₯=0
=1πΏ2
οΏ½πΏ2 β 4πΏπ₯ + 3π₯2οΏ½οΏ½π₯=0
= 1
Andππ2 (π₯)ππ₯
οΏ½π₯=πΏ
=1πΏ2
οΏ½πΏ2 β 4πΏπ₯ + 3π₯2οΏ½οΏ½π₯=πΏ
= 0
Also π3 (0) = 0 and π3 (πΏ) = 1 andππ4 (π₯)ππ₯
οΏ½π₯=0
=1πΏ2
οΏ½β2πΏπ₯ + 3π₯2οΏ½οΏ½π₯=0
= 0
andππ4 (π₯)ππ₯
οΏ½π₯=πΏ
=1πΏ2
οΏ½β2πΏπ₯ + 3π₯2οΏ½οΏ½π₯=πΏ
= 1
The shape functions have thus been verified. The stiοΏ½ness matrix is now found by substi-tuting Eq. (5A) into Eq. (1), repeated below
οΏ½ποΏ½ =
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π1
π1
π2
π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=
ββββββββββββββββββββββββ
πΈπΌ π3π£(π₯)ππ₯3 οΏ½
π₯=0
βπΈπΌ π2π£ππ₯2 οΏ½π₯=0
βπΈπΌ π3π£(π₯)ππ₯3 οΏ½
π₯=πΏ
πΈπΌ π2π£ππ₯2 οΏ½π₯=πΏ
ββββββββββββββββββββββββ
8
CHAPTER 2. DIRECT METHOD
Hence
οΏ½ποΏ½ =
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π1
π1
π2
π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=
ββββββββββββββββββ
πΈπΌ π3
ππ₯3(π1π£1 + π2π1 + π3π£2 + π4π2)
βπΈπΌ π2
ππ₯2(π1π£1 + π2π1 + π3π£2 + π4π2)
βπΈπΌ π3
ππ₯3(π1π£1 + π2π1 + π3π£2 + π4π2)
πΈπΌ π2
ππ₯2(π1π£1 + π2π1 + π3π£2 + π4π2)
ββββββββββββββββββ
(6)
Butπ3
ππ₯3π1 (π₯) =
1πΏ3
π3
ππ₯3οΏ½πΏ3 β 3πΏπ₯2 + 2π₯3οΏ½
=12πΏ3
Andπ3
ππ₯3π2 (π₯) =
1πΏ2
π3
ππ₯3οΏ½πΏ2π₯ β 2πΏπ₯2 + π₯3οΏ½
=6πΏ2
Andπ3
ππ₯3π3 (π₯) =
1πΏ3
π3
ππ₯3οΏ½3πΏπ₯2 β 2π₯3οΏ½
=β12πΏ3
Andπ3
ππ₯3π4 (π₯) =
1πΏ2
π3
ππ₯3οΏ½βπΏπ₯2 + π₯3οΏ½
=6πΏ2
For the second derivativesπ2
ππ₯2π1 (π₯) =
1πΏ3
π2
ππ₯2οΏ½πΏ3 β 3πΏπ₯2 + 2π₯3οΏ½
=1πΏ3
(12π₯ β 6πΏ)
Andπ2
ππ₯2π2 (π₯) =
1πΏ2
π2
ππ₯2οΏ½πΏ2π₯ β 2πΏπ₯2 + π₯3οΏ½
=1πΏ2
(6π₯ β 4πΏ)
Andπ2
ππ₯2π3 (π₯) =
1πΏ3
π2
ππ₯2οΏ½3πΏπ₯2 β 2π₯3οΏ½
=1πΏ3
(6πΏ β 12π₯)
Andπ2
ππ₯2π4 (π₯) =
1πΏ2
π2
ππ₯2οΏ½βπΏπ₯2 + π₯3οΏ½
=1πΏ2
(6π₯ β 2πΏ)
9
CHAPTER 2. DIRECT METHOD
Hence Eq. (6) becomes
οΏ½ποΏ½ =
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π1
π1
π2
π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=
ββββββββββββββββββββββββ
πΈπΌ π3
ππ₯3(π1π£1 + π2π1 + π3π£2 + π4π2)οΏ½
π₯=0
βπΈπΌ π2
ππ₯2(π1π£1 + π2π1 + π3π£2 + π4π2)οΏ½
π₯=0
βπΈπΌ π3
ππ₯3(π1π£1 + π2π1 + π3π£2 + π4π2)οΏ½
π₯=πΏ
πΈπΌ π2
ππ₯2(π1π£1 + π2π1 + π3π£2 + π4π2)οΏ½
π₯=πΏ
ββββββββββββββββββββββββ
=
βββββββββββββββββββββββββ
πΈπΌ οΏ½ 12πΏ3π£1 +6πΏ2π1 β
12πΏ3π£2 +
6πΏ2π2οΏ½
π₯=0
βπΈπΌ οΏ½ 1πΏ3(12π₯ β 6πΏ) π£1 +
1πΏ2(6π₯ β 4πΏ) π1 +
1πΏ3(6πΏ β 12π₯) π£2 +
1πΏ2(6π₯ β 2πΏ) π2οΏ½
π₯=0
βπΈπΌ οΏ½ 12πΏ3π£1 +6πΏ2π1 β
12πΏ3π£2 +
6πΏ2π2οΏ½
π₯=πΏ
πΈπΌ οΏ½ 1πΏ3(12π₯ β 6πΏ) π£1 +
1πΏ2(6π₯ β 4πΏ) π1 +
1πΏ3(6πΏ β 12π₯) π£2 +
1πΏ2(6π₯ β 2πΏ) π2οΏ½
π₯=πΏ
βββββββββββββββββββββββββ
=πΈπΌπΏ3
βββββββββββββββ
12 6πΏ β12 6πΏβ (12π₯ β 6πΏ)π₯=0 βπΏ (6π₯ β 4πΏ)π₯=0 β (6πΏ β 12π₯)π₯=0 βπΏ (6π₯ β 2πΏ)π₯=0
β12 β6πΏ 12 β6πΏ(12π₯ β 6πΏ)π₯=πΏ πΏ (6π₯ β 4πΏ)π₯=πΏ (6πΏ β 12π₯)π₯=πΏ πΏ (6π₯ β 2πΏ)π₯=πΏ
βββββββββββββββ
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
(7)
Or in matrix form, after evaluating the expressions above for π₯ = πΏ and π₯ = 0 asβ§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π1
π1
π2
π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=πΈπΌπΏ3
ββββββββββββββββ
12 6πΏ β12 6πΏ6πΏ 4πΏ2 β6πΏ 2πΏ2
β 12πΏ3 β 6
πΏ212πΏ3 β 6
πΏ2
(12πΏ β 6πΏ) πΏ (6πΏ β 4πΏ) (6πΏ β 12πΏ) πΏ (6πΏ β 2πΏ)
ββββββββββββββββ
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=πΈπΌπΏ3
βββββββββββββββ
12 6πΏ β12 6πΏ6πΏ 4πΏ2 β6πΏ 2πΏ2
β12 β6πΏ 12 β6πΏ6πΏ 2πΏ2 β6πΏ 4πΏ2
βββββββββββββββ
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
The above now is in the form
οΏ½ποΏ½ = [πΎ] {π}
Hence the stiοΏ½ness matrix is
[πΎ] =πΈπΌπΏ3
βββββββββββββββ
12 6πΏ β12 6πΏ6πΏ 4πΏ2 β6πΏ 2πΏ2
β12 β6πΏ 12 β6πΏ6πΏ 2πΏ2 β6πΏ 4πΏ2
βββββββββββββββ
Knowing the stiοΏ½ness matrix means knowing the nodal displacements {π} when giventhe forces at the nodes. The power of the finite element method now comes after all thenodal displacements π£1, π1, π£2, π2 are calculated by solving οΏ½ποΏ½ = [πΎ] {π}. This is becausethe polynomial π£ (π₯) is now completely determined and hence π£ (π₯) and π (π₯) can now beevaluated for any π₯ along the beam and not just at its end nodes as the case with finitediοΏ½erence method. Eq. 5A above can now be used to find the displacement π£ (π₯) and π (π₯)everywhere.
π£ (π₯) = [π] {π}
π£ (π₯) = οΏ½π1 (π₯) π2 (π₯) π3 (π₯) π4 (π₯)οΏ½
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
To summarise, these are the steps to obtain π£ (π₯)
1. An expression for [πΎ] is found.
2. οΏ½ποΏ½ = [πΎ] {π} is solved for {π}
10
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
3. π£ (π₯) = [π] {π} is calculated by assuming π£ (π₯) is a polynomial. This gives the displace-ment π£ (π₯) to use to evaluate the transverse displacement anywhere on the beam andnot just at the end nodes.
4. π (π₯) = ππ£(π₯)ππ₯ = π
ππ₯ [π] {π} is obtained to evaluate the rotation of the beam any whereand not just at the end nodes.
5. The strain π (π₯) = βπ¦ [π΅] {π} is found, where [π΅] is the gradient matrix [π΅] = π2
ππ₯2 [π].
6. The stress from π = πΈπ = βπΈπ¦ [π΅] {π} is found.
7. The bending moment diagram from π(π₯) = πΈπΌ [π΅] {π} is found.
8. The shear force diagram from π (π₯) = πππ₯π(π₯) is found.
2.1 Examples using the direct beam stiοΏ½ness matrix
The beam stiοΏ½ness matrix is now used to solve few beam problems. Starting with simpleone span beam
2.1.1 Example 1
A one span beam, a cantilever beam of length πΏ, with point load π at the free end
L
P
The first step is to make the free body diagram and show all moments and forces at thenodes
L
P
R
M1M2
π is the given force. π2 = 0 since there is no external moment at the right end. HenceοΏ½ποΏ½ = [πΎ] {π} for this system is
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π π1
βπ0
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=πΈπΌπΏ3
βββββββββββββββ
12 6πΏ β12 6πΏ6πΏ 4πΏ2 β6πΏ 2πΏ2
β12 β6πΏ 12 β6πΏ6πΏ 2πΏ2 β6πΏ 4πΏ2
βββββββββββββββ
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
Now is an important step. The known end displacements from boundary conditions issubstituted into {π}, and the corresponding row and columns from the above system ofequations are removed1. Boundary conditions indicates that there is no rotation on theleft end (since fixed). Hence π£1 = 0 and π1 = 0. Hence the only unknowns are π£2 and π2.Therefore the first and the second rows and columns are removed, giving
β§βͺβͺβ¨βͺβͺβ©βπ0
β«βͺβͺβ¬βͺβͺβ=
πΈπΌπΏ3
ββββββ12 β6πΏβ6πΏ 4πΏ2
ββββββ
β§βͺβͺβ¨βͺβͺβ©π£2π2
β«βͺβͺβ¬βͺβͺβ
1Instead of removing rows/columns for known boundary conditions, we can also just put a 1 on thediagonal of the stiοΏ½ness matrix for that boundary conditions. I will do this example again using this method
11
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
Now the above is solved for
β§βͺβͺβ¨βͺβͺβ©π£2π2
β«βͺβͺβ¬βͺβͺβ. Let πΈ = 30 Γ 106 psi (steel), πΌ = 57 ππ4, πΏ = 144 ππ, and
π = 400 lb, henceοΏ½ οΏ½1 P=400;2 L=144;3 E=30*10^6;4 I=57.1;5 A=(E*I/L^3)*[12 6*L -12 6*L;6 6*L 4*L^2 -6*L 2*L^2;7 -12 -6*L 12 -6*L;8 6*L 2*L^2 -6*L 4*L^29 ];10 load=[P;P*L;-P;0];11 x=A(3:end,3:end)\load(3:end)οΏ½ οΏ½
Λ
which givesx =-0.2324-0.0024
Therefore the vertical displacement at the right end is π£2 = 0.2324 inches (downwards) andπ2 = β0.0024 radians. Now that all nodal displacements are found, the field displacementfunction is completely determined.
{π} =
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
00
β0.2324β0.0024
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
From Eq. 5A
π£ (π₯) = [π] {π}
= οΏ½π1 (π₯) π2 (π₯) π3 (π₯) π4 (π₯)οΏ½
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
00
β0.2324β0.0024
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
= οΏ½ 1πΏ3οΏ½πΏ3 β 3πΏπ₯2 + 2π₯3οΏ½ 1
πΏ2οΏ½πΏ2π₯ β 2πΏπ₯2 + π₯3οΏ½ 1
πΏ3οΏ½3πΏπ₯2 β 2π₯3οΏ½ 1
πΏ2οΏ½βπΏπ₯2 + π₯3οΏ½οΏ½
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
00
β0.2324β0.0024
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=0.002 4πΏ2
οΏ½πΏπ₯2 β π₯3οΏ½ +0.232 4πΏ3
οΏ½2π₯3 β 3πΏπ₯2οΏ½
But πΏ = 144 inches, and the above becomes
π£ (π₯) = 3.992 0 Γ 10β8π₯3 β 1.695 6 Γ 10β5π₯2
To verify, let π₯ = 144 in the above
π£ (π₯ = 144) = 3.992 0 Γ 10β81443 β 1.695 6 Γ 10β51442
= β0.232 40
The following is a plot of the deflection curve for the beamοΏ½ οΏ½1 v=@(x) 3.992*10^-8*x.^3-1.6956*10^-5*x.^22 x=0:0.1:144;3 plot(x,v(x),'r-','LineWidth',2);4 ylim([-0.8 0.3]);5 title('beam deflection curve');6 xlabel('x inch'); ylabel('deflection inch');7 gridοΏ½ οΏ½
12
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
Λ
Now instead of removing rows/columns for the known boundary conditions, a 1 is put onthe diagonal. Starting again with
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π π1
βπ0
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=πΈπΌπΏ3
βββββββββββββββ
12 6πΏ β12 6πΏ6πΏ 4πΏ2 β6πΏ 2πΏ2
β12 β6πΏ 12 β6πΏ6πΏ 2πΏ2 β6πΏ 4πΏ2
βββββββββββββββ
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
Since π£1 = 0 and π1 = 0, thenβ§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
00βπ0
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=πΈπΌπΏ3
βββββββββββββββ
1 0 0 00 1 0 00 0 12 β6πΏ0 0 β6πΏ 4πΏ2
βββββββββββββββ
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
00π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
The above system is now solved as before. πΈ = 30Γ 106 psi, πΌ = 57 ππ4, πΏ = 144 in, π = 400 lbοΏ½ οΏ½1 P=400;2 L=144;3 E=30*10^6;4 I=57.1;5 A=(E*I/L^3)*[12 6*L -12 6*L;6 6*L 4*L^2 -6*L 2*L^2;7 -12 -6*L 12 -6*L;8 6*L 2*L^2 -6*L 4*L^29 ];10 load=[0;0;-P;0]; %put zeros for known B.C.11 A(:,1)=0; A(1,:)=0; A(1,1)=1; %put 1 on diagonal12 A(:,2)=0; A(2,:)=0; A(2,2)=1; %put 1 on diagonal13 AοΏ½ οΏ½
ΛA =1.0e+07 *0.0000 0 0 00 0.0000 0 00 0 0.0007 -0.04960 0 -0.0496 4.7583οΏ½ οΏ½
1 sol=A\load %SOLVEοΏ½ οΏ½Λsol =00-0.2324-0.0024
The same solution is obtained as before, but without the need to remove rows/column from
13
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
the stiοΏ½ness matrix. This method might be easier for programming than the first methodof removing rows/columns.
The rest now is the same as was done earlier and will not be repeated.
2.1.2 Example 2
This is the same example as above, but the vertical load π is now placed in the middle ofthe beam
L
P
In using stiοΏ½ness method, all loads must be on the nodes. The vector οΏ½ποΏ½ is the nodal forcesvector. Hence equivalent nodal loads are found for the load in the middle of the beam.The equivalent loading is the following
L
P/2P/2 PL/8PL/8
Therefore, the problem is as if it was the following problem
L
P/2P/2
PL/8PL/8
Now that equivalent loading is in place, we continue as before. Making a free body diagramshowing all loads (including reaction forces)
L
P/2
P/2
PL/8PL/8
M1
R
14
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
The stiοΏ½ness equation is now written as
οΏ½ποΏ½ = [πΎ] {π}β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π β π/2π1 β ππΏ/8
βπ/2ππΏ/8
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=πΈπΌπΏ3
βββββββββββββββ
12 6πΏ β12 6πΏ6πΏ 4πΏ2 β6πΏ 2πΏ2
β12 β6πΏ 12 β6πΏ6πΏ 2πΏ2 β6πΏ 4πΏ2
βββββββββββββββ
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
There is no need to determine π and π1 at this point since these rows will be removed dueto boundary conditions π£1 = 0 and π1 = 0 and hence those quantities are not needed tosolve the equations. Note that the rows and columns are removed for the known boundarydisplacements before solving οΏ½ποΏ½ = [πΎ] {π}. Hence, after removing the first two rows andcolumns, the above system simplifies to
β§βͺβͺβ¨βͺβͺβ©βπ/2ππΏ/8
β«βͺβͺβ¬βͺβͺβ=
πΈπΌπΏ3
ββββββ12 β6πΏβ6πΏ 4πΏ2
ββββββ
β§βͺβͺβ¨βͺβͺβ©π£2π2
β«βͺβͺβ¬βͺβͺβ
The above is now solved for
β§βͺβͺβ¨βͺβͺβ©π£2π2
β«βͺβͺβ¬βͺβͺβusing the same numerical values for π, πΈ, πΌ, πΏ as in the
first exampleοΏ½ οΏ½1 P=400;2 L=144;3 E=30*10^6;4 I=57.1;5 A=(E*I/L^3)*[12 6*L -12 6*L;6 6*L 4*L^2 -6*L 2*L^2;7 -12 -6*L 12 -6*L;8 6*L 2*L^2 -6*L 4*L^29 ];10 load=[-P/2;P*L/8];11 x=A(3:end,3:end)\loadοΏ½ οΏ½
Λx =-0.072630472854641-0.000605253940455
Therefore
{π} =
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
00
β0.072630472854641β0.000605253940455
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
This is enough to obtain π£ (π₯) as before. Now the reactions π and π1 can be determined ifneeded. Going back to the full οΏ½ποΏ½ = [πΎ] {π}, results in
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π β π/2π1 β ππΏ/8
βπ/2ππΏ/8
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=πΈπΌπΏ3
βββββββββββββββ
12 6πΏ β12 6πΏ6πΏ 4πΏ2 β6πΏ 2πΏ2
β12 β6πΏ 12 β6πΏ6πΏ 2πΏ2 β6πΏ 4πΏ2
βββββββββββββββ
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
00
β0.072630472854641β0.000605253940455
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=πΈπΌπΏ3
βββββββββββββββ
0.871 57 β 3.631 5 Γ 10β3πΏ0.435 78πΏ β 1.210 5 Γ 10β3πΏ2
3.631 5 Γ 10β3πΏ β 0.871 570.435 78πΏ β 2.421 Γ 10β3πΏ2
βββββββββββββββ
Hence the first equation becomes
π β π/2 =πΈπΌπΏ3
οΏ½0.871 57 β 3.631 5 Γ 10β3πΏοΏ½
and since πΈ = 30 Γ 106 psi (steel) and πΌ = 57 ππ4and πΏ = 144 in and π = 400 lb, then
15
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
π =οΏ½30Γ106οΏ½57
1443οΏ½0.871 57 β 3.631 5 Γ 10β3 (144)οΏ½ + 400/2 . Therefore
π = 400 lb
and π1 β ππΏ/8 = πΈπΌπΏ3οΏ½0.435 78πΏ β 1.210 5 Γ 10β3πΏ2οΏ½ + ππΏ/8 , hence
π1 = 28 762 lb-ft
Now that all nodal reactions are found, the displacement field is found and the deflectioncurve can be plotted.
π£ (π₯) = [π] {π}
π£ (π₯) = οΏ½π1 (π₯) π2 (π₯) π3 (π₯) π4 (π₯)οΏ½
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
= οΏ½π1 (π₯) π2 (π₯) π3 (π₯) π4 (π₯)οΏ½
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
00
β0.072630472854641β0.000605253940455
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
= οΏ½ 1πΏ3οΏ½3πΏπ₯2 β 2π₯3οΏ½ 1
πΏ2οΏ½βπΏπ₯2 + π₯3οΏ½οΏ½
βββββββ0.072630472854641β0.000605253940455
ββββββ
=6.052 5 Γ 10β4
πΏ2οΏ½πΏπ₯2 β π₯3οΏ½ +
0.072 63πΏ3
οΏ½2π₯3 β 3πΏπ₯2οΏ½
Since πΏ = 144 inches, the above becomes
π£ (π₯) =6.052 5 Γ 10β4
(144)2οΏ½(144) π₯2 β π₯3οΏ½ +
0.072 63(144)3
οΏ½2π₯3 β 3 (144) π₯2οΏ½
= 1.945 9 Γ 10β8π₯3 β 6.304 7 Γ 10β6π₯2
The following is the plot
2.1.3 Example 3
Assuming the beam is fixed on the left end as above, but simply supported on the rightend, and the vertical load π now at distance π from the left end and at distance π from theright end, and a uniform distributed load of density π lb/in is on the beam.
16
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
Pm (lb/in
a b
M1
Using the following values: π = 0.625πΏ, π = 0.375πΏ, πΈ = 30 Γ 106 psi (steel), πΌ = 57 ππ4,πΏ = 144 in, π = 1000 lb, π = 200 lb/in.
In the above, the left end reaction forces are shown as π 1 and moment reaction as π1 andthe reaction at the right end as π 2. Starting by finding equivalent loads for the point loadπ and equivalent loads for for the uniform distributed load π. All external loads must betransferred to the nodes for the stiοΏ½ness method to work. Equivalent load for the abovepoint load is
Pb2L2a
L3Pa2L2b
L3
a b
Pab2
L2
Pa2b
L2
Equivalent load for the uniform distributed loading is
mL2
mL2
12mL2
mL2
12
Using free body diagram, with all the loads on it gives the following diagram (In thisdiagram π is the reaction moment and π 1, π 2 are the reaction forces)
17
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
Now that all loads are on the nodes, the stiοΏ½ness equation is applied
οΏ½ποΏ½ = [πΎ] {π}β§βͺβͺβͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβͺβͺβ©
π 1 βππ2(πΏ+2π)
πΏ3 β ππΏ2
πβ πππ2
πΏ2 β ππΏ2
12
π 2 βππ2(πΏ+2π)
πΏ3 β ππΏ2
ππ2ππΏ2 + ππΏ2
12
β«βͺβͺβͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβͺβͺβ
=πΈπΌπΏ3
βββββββββββββββ
12 6πΏ β12 6πΏ6πΏ 4πΏ2 β6πΏ 2πΏ2
β12 β6πΏ 12 β6πΏ6πΏ 2πΏ2 β6πΏ 4πΏ2
βββββββββββββββ
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
Boundary conditions are now applied. π£1 = 0, π1 = 0,π£2 = 0. Therefore the first, second andthird rows/columns are removed giving
ππ2ππΏ2
+ππΏ2
12=
πΈπΌπΏ3
4πΏ2π2
Hence
π2 = οΏ½ππ2ππΏ2
+ππΏ2
12 οΏ½ οΏ½πΏ4πΈπΌοΏ½
Substituting numerical values for the above as given at the top of the problem results in
π2 =βββββ(1000) (0.625 (144))2 (0.375 (144))
(144)2+(200) (144)2
12
βββββ
ββββββ
1444 οΏ½30 Γ 106οΏ½ (57)
ββββββ
= 7.719 9 Γ 10β3πππ
Hence the field displacement π’ (π₯) is now found
π£ (π₯) = [π] {π}
π£ (π₯) = οΏ½π1 (π₯) π2 (π₯) π3 (π₯) π4 (π₯)οΏ½
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
= οΏ½π1 (π₯) π2 (π₯) π3 (π₯) π4 (π₯)οΏ½
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
000
7.719 9 Γ 10β3
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=1πΏ2
οΏ½βπΏπ₯2 + π₯3οΏ½ οΏ½7.719 9 Γ 10β3οΏ½
= 3.722 9 Γ 10β7π₯3 β 5.361 Γ 10β5π₯2
And a plot of the deflection curve isοΏ½ οΏ½1 clear all; close all;2 v=@(x) 3.7229*10^-7*x.^3-5.361*10^-5*x.^23 x=0:0.1:144;4 plot(x,v(x),'r-','LineWidth',2);5 ylim([-0.5 0.3]);6 title('beam deflection curve');7 xlabel('x inch'); ylabel('deflection inch');
18
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
8 gridοΏ½ οΏ½Λ
19
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
20
Chapter 3
Finite elements or adding moreelements
Finite elements generated displacements are smaller in value than the actual analyticalvalues. To improve the accuracy, more elements are added. To add more elements, thebeam is divided into 2,3,4 and more beam elements. To show how this works, example 3above is solved again using two elements. It is found that displacement field π£ (π₯) becomesmore accurate (By comparing the the result with the exact solution based on using thebeam 4th order diοΏ½erential equation. It is found to be almost the same with only 2 elements)
3.1 Example 3 redone with 2 elements
The first step is to divide the beam into two elements and number the degrees of freedomand global nodes as follows
Pm (lb/inP
L
L/2 L/2
m (lb/in)Divide into 2 elements
There is 6 total degrees of freedom. two at each node. Hence the stiοΏ½ness matrix for thewhole beam (including both elements) will be 6 by 6. For each element however, the samestiοΏ½ness matrix will be used as above and that will remain as before 4 by 4.
21
3.1. Example 3 redone with 2 elements CHAPTER 3. FINITE ELEMENTS OR . . .
1 2 3
1 2 3
v 1 v 2 v 3
The stiοΏ½ness matrix for each element is found and then the global stiοΏ½ness matrix isassembled. Then οΏ½ππππππποΏ½ = οΏ½πΎπππππποΏ½ οΏ½ππππππποΏ½ is solved as before. The first step is to move allloads to the nodes as was done before. This is done for each element. The formulas forequivalent loads remain the same, but now πΏ becomes πΏ/2. The following diagram showthe equivalent loading for π
1 2 3
a b
L/2 L/2
Pb2L/22a
L/23Pa2L/22b
L/23Pab2
L/22Pa2b
L/22
a 0.125L
b 0.375L
Equivalent loading of point load P after moving to nodes
of second element
The equivalent loading for distributed load π is
1 2 3
L/2 L/2
mL/22
mL/22
mL/22
mL/22
12
mL/22
12
mL/22
12
mL/22
12
Now the above two diagrams are put together to show all equivalent loads with the originalreaction forces to obtain the following diagram
22
3.1. Example 3 redone with 2 elements CHAPTER 3. FINITE ELEMENTS OR . . .
1 2 3
L/2 L/2
mL/22
mL/22
mL/22
mL/22
12
mL/22
12
mL/22
12
mL/22
12
Pb2L/22a
L/23
Pa2L/22b
L/23
Pab2
L/22
Pa2b
L/22
R1R3
M1
Nodal loading for 2 elements beam after performing equivalent loading
οΏ½ποΏ½ = [πΎ] {π} for each element is now constructed. Starting with the first elementβββββββββββββββββββββββββββββββββ
π 1 βποΏ½ πΏ2 οΏ½
2
π1 βποΏ½ πΏ2 οΏ½
2
12
βππ2οΏ½ πΏ2+2ποΏ½
(πΏ/2)3β
ποΏ½ πΏ2 οΏ½
2
ποΏ½ πΏ2 οΏ½2
12 βποΏ½ πΏ2 οΏ½
2
12 β πππ2
οΏ½ πΏ2 οΏ½2
βββββββββββββββββββββββββββββββββ
=πΈπΌ
οΏ½πΏ2οΏ½3
βββββββββββββββββββββββ
12 6 οΏ½πΏ2οΏ½ β12 6 οΏ½πΏ2οΏ½
6 οΏ½πΏ2οΏ½ 4 οΏ½πΏ2οΏ½2
β6 οΏ½πΏ2οΏ½ 2 οΏ½πΏ2οΏ½2
β12 β6 οΏ½πΏ2οΏ½ 12 β6 οΏ½πΏ2οΏ½
6 οΏ½πΏ2οΏ½ 2 οΏ½πΏ2οΏ½2
β6 οΏ½πΏ2οΏ½ 4 οΏ½πΏ2οΏ½2
βββββββββββββββββββββββ
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
And for the second elementββββββββββββββββββββββββββββββββββββββββ
βππ2οΏ½ πΏ2+2ποΏ½
(πΏ/2)3β
ποΏ½ πΏ2 οΏ½
2
ποΏ½ πΏ2 οΏ½2
12 βποΏ½ πΏ2 οΏ½
2
12 β πππ2
οΏ½ πΏ2 οΏ½2
π 3 βππ2οΏ½ πΏ2+2ποΏ½
οΏ½ πΏ2 οΏ½3 β
ποΏ½ πΏ2 οΏ½
2
ποΏ½ πΏ2 οΏ½2
12 + ππ2π
οΏ½ πΏ2 οΏ½2
ββββββββββββββββββββββββββββββββββββββββ
=πΈπΌ
οΏ½πΏ2οΏ½3
βββββββββββββββββββββββ
12 6 οΏ½πΏ2οΏ½ β12 6 οΏ½πΏ2οΏ½
6 οΏ½πΏ2οΏ½ 4 οΏ½πΏ2οΏ½2
β6 οΏ½πΏ2οΏ½ 2 οΏ½πΏ2οΏ½2
β12 β6 οΏ½πΏ2οΏ½ 12 β6 οΏ½πΏ2οΏ½
6 οΏ½πΏ2οΏ½ 2 οΏ½πΏ2οΏ½2
β6 οΏ½πΏ2οΏ½ 4 οΏ½πΏ2οΏ½2
βββββββββββββββββββββββ
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£2π2
π£3π3
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
23
3.1. Example 3 redone with 2 elements CHAPTER 3. FINITE ELEMENTS OR . . .
The 2 systems above are assembled to obtain the global stiοΏ½ness matrix equation givingββββββββββββββββββββββββββββββββββββββββββββββββββββββββ
π 1 βποΏ½ πΏ2 οΏ½
2
π1 βποΏ½ πΏ2 οΏ½
2
12
β2ππ2οΏ½ πΏ2+2ποΏ½
οΏ½ πΏ2 οΏ½3 β 2
ποΏ½ πΏ2 οΏ½
2
β2πππ2
οΏ½ πΏ2 οΏ½2
π 3 βππ2οΏ½ πΏ2+2ποΏ½
οΏ½ πΏ2 οΏ½3 β
ποΏ½ πΏ2 οΏ½
2
ποΏ½ πΏ2 οΏ½2
12 + ππ2π
οΏ½ πΏ2 οΏ½2
ββββββββββββββββββββββββββββββββββββββββββββββββββββββββ
=πΈπΌ
οΏ½πΏ2οΏ½3
βββββββββββββββββββββββββββββββββββββ
12 6 οΏ½πΏ2οΏ½ β12 6 οΏ½πΏ2οΏ½ 0 0
6 οΏ½πΏ2οΏ½ 4 οΏ½πΏ2οΏ½2
β6 οΏ½πΏ2οΏ½ 2 οΏ½πΏ2οΏ½2
0 0
β12 β6 οΏ½πΏ2οΏ½ 12 + 12 β6 οΏ½πΏ2οΏ½ + 6 οΏ½πΏ2οΏ½ β12 6 οΏ½πΏ2οΏ½
6 οΏ½πΏ2οΏ½ 2 οΏ½πΏ2οΏ½2
β6 οΏ½πΏ2οΏ½ + 6 οΏ½πΏ2οΏ½ 4 οΏ½πΏ2οΏ½2+ 4 οΏ½πΏ2οΏ½
2β6 οΏ½πΏ2οΏ½ 2 οΏ½πΏ2οΏ½
2
0 0 β12 β6 οΏ½πΏ2οΏ½ 12 β6 οΏ½πΏ2οΏ½
0 0 6 οΏ½πΏ2οΏ½ 2 οΏ½πΏ2οΏ½2
β6 οΏ½πΏ2οΏ½ 4 οΏ½πΏ2οΏ½2
βββββββββββββββββββββββββββββββββββββ
β§βͺβͺβͺβͺβͺβͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
π£3π3
β«βͺβͺβͺβͺβͺβͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβͺβͺβͺβͺβͺβ
The boundary conditions are now applied giving π£1 = 0, π1 = 0 since the first node is fixed,and π£3 = 0. And putting 1 on the diagonal of the stiοΏ½ness matrix corresponding to theseknown boundary conditions results in
ββββββββββββββββββββββββββββββββββββββ
00
β2ππ2οΏ½ πΏ2+2ποΏ½
(πΏ/2)3β 2π(πΏ/2)2
β2 πππ2
(πΏ/2)2
0ποΏ½ πΏ2 οΏ½
2
12 + ππ2π
οΏ½ πΏ2 οΏ½2
ββββββββββββββββββββββββββββββββββββββ
=πΈπΌ
οΏ½πΏ2οΏ½3
ββββββββββββββββββββββββββββββββ
1 0 0 0 0 00 1 0 0 0 0
0 0 24 0 0 6 οΏ½πΏ2οΏ½
0 0 0 8 οΏ½πΏ2οΏ½2
0 2 οΏ½πΏ2οΏ½2
0 0 0 0 1 0
0 0 6 οΏ½πΏ2οΏ½ 2 οΏ½πΏ2οΏ½2
0 4 οΏ½πΏ2οΏ½2
ββββββββββββββββββββββββββββββββ
βββββββββββββββββββββββββ
00π£2π2
0π3
βββββββββββββββββββββββββ
As was mentioned earlier, another method would be to remove the rows/columns whichresults in β
βββββββββββββββββββββββββ
β2ππ2οΏ½ πΏ2+2ποΏ½
(πΏ/2)3β 2
ποΏ½ πΏ2 οΏ½
2
β2πππ2
οΏ½ πΏ2 οΏ½2
ποΏ½ πΏ2 οΏ½2
12 + ππ2π
οΏ½ πΏ2 οΏ½2
ββββββββββββββββββββββββββ
=πΈπΌ
οΏ½πΏ2οΏ½3
βββββββββββββββββ
24 0 6 οΏ½πΏ2οΏ½
0 8 οΏ½πΏ2οΏ½2
2 οΏ½πΏ2οΏ½2
6 οΏ½πΏ2οΏ½ 2 οΏ½πΏ2οΏ½2
4 οΏ½πΏ2οΏ½2
βββββββββββββββββ
ββββββββββ
π£2π2
π3
ββββββββββ
Giving the same solution. There are 3 unknowns to solve for. Once these unknowns aresolved for, π£ (π₯) for the first element and for the second element are fully determined. Thefollowing code displays the deflection curve for the above beamοΏ½ οΏ½
1 clear all; close all;2 P=400;3 L=144;4 E=30*10^6;5 I=57.1;6 m=200;7 a=0.125*L;8 b=0.375*L;9 A=E*I/(L/2)^3*[1 0 0 0 0 0;10 0 1 0 0 0 0;11 0 0 24 0 0 6*L/2;12 0 0 0 8*(L/2)^2 0 2*(L/2)^2;13 0 0 0 0 1 0;14 0 0 6*L/2 2*(L/2)^2 0 4*(L/2)^2];15 A16 load = [0;17 0;18 -(m*L/2) - 2*P*b^2*(L/2+2*a)/(L/2)^3;
24
3.1. Example 3 redone with 2 elements CHAPTER 3. FINITE ELEMENTS OR . . .
19 -2*P*a*b^2/(L/2)^2;20 0;21 P*a^2*b/(L/2)^2+(m*(L/2)^2)/12]22 sol=A\loadοΏ½ οΏ½
ΛA =1.0e+08 *0.0000 0 0 0 0 00 0.0000 0 0 0 00 0 0.0011 0 0 0.01980 0 0 1.9033 0 0.47580 0 0 0 0.0000 00 0 0.0198 0.4758 0 0.9517load =00-15075-8100087750sol =00-0.2735-0.001900.0076
The above solution gives π£2 = β0.2735 in (downwards displacement) and π2 = β0.0019radians and π3 = 0.0076 radians. π£ (π₯) polynomial is now found for each element
π£ππππ1 (π₯) = οΏ½π1 (π₯) π2 (π₯) π3 (π₯) π4 (π₯)οΏ½
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£1π1
π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
οΏ½π1 (π₯) π2 (π₯) π3 (π₯) π4 (π₯)οΏ½
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
00π£2π2
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
=βββββ
1
οΏ½ πΏ2 οΏ½3 οΏ½3 οΏ½
πΏ2οΏ½ π₯2 β 2π₯3οΏ½ 1
οΏ½ πΏ2 οΏ½2 οΏ½β οΏ½
πΏ2οΏ½ π₯2 + π₯3οΏ½
βββββ
βββββββ0.2735β0.0019
ββββββ
=2.188πΏ3 οΏ½2π₯3 β
32πΏπ₯2οΏ½ β
0.007 6πΏ2 οΏ½π₯3 β
12πΏπ₯2οΏ½
The above polynomial is the transverse deflection of the beam for the region 0 β€ π₯ β€ πΏ/2.
25
3.1. Example 3 redone with 2 elements CHAPTER 3. FINITE ELEMENTS OR . . .
π£ (π₯) for the second element is found in similar way
π£ππππ2 (π₯) = οΏ½π1 (π₯) π2 (π₯) π3 (π₯) π4 (π₯)οΏ½
β§βͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβ©
π£2π2
0π3
β«βͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβ
= οΏ½π1 (π₯) π2 (π₯) π3 (π₯) π4 (π₯)οΏ½
βββββββββββββββ
β0.2735β0.0019
00.0076
βββββββββββββββ
= οΏ½π1 (π₯) π2 (π₯) π4 (π₯)οΏ½
ββββββββββ
β0.2735β0.00190.0076
ββββββββββ
=ββββββ
1
οΏ½ πΏ2 οΏ½3 οΏ½οΏ½
πΏ2οΏ½3β 3 οΏ½πΏ2οΏ½ π₯
2 + 2π₯3οΏ½1
οΏ½ πΏ2 οΏ½2 οΏ½οΏ½
πΏ2οΏ½2π₯ β 2 οΏ½πΏ2οΏ½ π₯
2 + π₯3οΏ½1
οΏ½ πΏ2 οΏ½2 οΏ½β οΏ½
πΏ2οΏ½ π₯2 + π₯3οΏ½
ββββββ
ββββββββββ
β0.2735β0.00190.0076
ββββββββββ
Hence
π£ππππ2 (π₯) =0.030 4πΏ2 οΏ½π₯3 β
12πΏπ₯2οΏ½ β
0.007 6πΏ2 οΏ½
14πΏ2π₯ β πΏπ₯2 + π₯3οΏ½ β
2.188πΏ3 οΏ½
18πΏ3 β
32πΏπ₯2 + 2π₯3οΏ½
Which is valid for πΏ/2 β€ π₯ β€ πΏ. The following is a plot of the deflection curve using theabove 2 equations
When the above plot is compared to the case with one element, the deflection is seen to belarger now. Comparing the above to the analytical solution shows that the deflection nowis almost exactly the same as the analytical solution. Hence by using only two elementsinstead of one element, the solution has become more accurate and almost agrees withthe analytical solution.
The following diagram shows the deflection curve of problem three above when using oneelement and two elements on the same plot to help illustrate the diοΏ½erence in the resultmore clearly.
26
3.1. Example 3 redone with 2 elements CHAPTER 3. FINITE ELEMENTS OR . . .
The analytical deflection for the beam in problem three above (fixed on the left and simplysupported at the right end) when there is uniformly loaded with π€ lbs per unit length isgiven by
π£ (π₯) = βπ€π₯2
48πΈπΌοΏ½3πΏ2 β 5πΏπ₯ + 2π₯2οΏ½
While the analytical deflection for the same beam but when there is a point load P atdistance π from the left end is given by
π£ (π₯) = βπ(β¨πΏ β πβ©3 (3πΏ β π₯)π₯2 + πΏ2(3(πΏ β π)(πΏ β π₯)π₯2 + 2πΏ β¨π₯ β πβ©3))
Therefore, the analytical expression for deflection is given by the sum of the above expres-sions, giving
π£ (π₯) = βπ€π₯2
48πΈπΌοΏ½3πΏ2 β 5πΏπ₯ + 2π₯2οΏ½ β π(β¨πΏ β πβ©3 (3πΏ β π₯)π₯2 + πΏ2(3(πΏ β π)(πΏ β π₯)π₯2 + 2πΏ β¨π₯ β πβ©3))
Where β¨π₯ β πβ©means it is zero when π₯βπ is negative. In other words β¨π₯ β πβ© = (π₯ β π)ππππ‘ππ‘ππ (π₯ β π)
The following diagram is a plot of the analytical deflection with the two elements deflectioncalculated using Finite elements above.
In the above, the blue dashed curve is the analytical solution, and the red curve is thefinite elements solution using 2 elements. It can be seen that the finite element solutionfor the deflection is now in a very good agreement with the analytical solution.
27
3.1. Example 3 redone with 2 elements CHAPTER 3. FINITE ELEMENTS OR . . .
28
Chapter 4
Generating shear and bendingmoments diagrams
After solving the problem using finite elements and obtaining the field displacement func-tion π£ (π₯) as was shown in the above examples, the shear force and bending moments along
the beam can be calculated. Since the bending moment is given by π(π₯) = βπΈπΌπ2π£(π₯)ππ₯2 and
shear force is given by π (π₯) = ππππ₯ = βπΈπΌπ
3π£(π₯)ππ₯3 then these diagrams are now readily plotted
as shown below for example three above using the result from the finite elements with 2elements. Recalling from above that
π£ (π₯) =2.188πΏ3
οΏ½2π₯3 β 32πΏπ₯
2οΏ½ β 0.007 6πΏ2
οΏ½π₯3 β 12πΏπ₯
2οΏ½0.030 4πΏ2
οΏ½π₯3 β 12πΏπ₯
2οΏ½ β 0.007 6πΏ2
οΏ½14πΏ
2π₯ β πΏπ₯2 + π₯3οΏ½ β 2.188πΏ3
οΏ½18πΏ
3 β 32πΏπ₯
2 + 2π₯3οΏ½
β«βͺβͺβͺβ¬βͺβͺβͺβ
0 β€ π₯ β€ πΏ/2πΏ/2 β€ π₯ β€ πΏ
Hence
π(π₯) = βπΈπΌβ0.0076(6π₯βπΏ)
πΏ2 + 2.188(12π₯β3πΏ)πΏ3
β0.0076(6π₯β2πΏ)πΏ2 + 0.0304(6π₯βπΏ)
πΏ2 β 2.188(12π₯β3πΏ)πΏ3
β«βͺβͺβ¬βͺβͺβ
0 β€ π₯ β€ πΏ/2πΏ/2 β€ π₯ β€ πΏ
using πΈ = 30 Γ 106 psi and πΌ = 57 in4 and πΏ = 144 in, the bending moment diagram plot is
The bending moment diagram clearly does not agree with the bending moment diagramthat can be generated from the analytical solution given below (generated using my otherprogram which solves this problem analytically)
The reason for this is because the solution π£ (π₯) obtained using the finite elements methodis a third degree polynomial and after diοΏ½erentiating twice to obtain the bending moment
29
CHAPTER 4. GENERATING SHEAR . . .
(π(π₯) = βπΈπΌπ2π£ππ₯2 ) the result becomes a linear function in π₯ while in the analytical solution
case, when the load is distributed, the solution π£ (π₯) is a fourth degree polynomial. Hencethe bending moment will be quadratic function in π₯ in the analytical case.
Therefore, in order to obtain good approximation for the bending moment and shear forcediagrams using finite elements, more elements will be needed.
30
Chapter 5
Finding the stiοΏ½ness matrix usingmethods other than direct method
Local contents5.1 Virtual work method for derivation of the stiοΏ½ness matrix . . . . . . . . . . . . 325.2 Potential energy (minimize a functional) method to derive the stiοΏ½ness matrix 33
31
5.1. Virtual work method for derivation . . . CHAPTER 5. FINDING THE . . .
There are three main methods to obtain the stiοΏ½ness matrix
1. Variational method (minimizing a functional). This functional is the potential energyof the structure and loads.
2. Weighted residual. Requires the diοΏ½erential equation as a starting point. Approxi-mated in weighted average. Galerkin weighted residual method is the most commonmethod for implementation.
3. Virtual work method. Making the virtual work zero for an arbitrary allowed displace-ment.
5.1 Virtual work method for derivation of the stiοΏ½nessmatrix
In virtual work method, a small displacement is assumed to occur. Looking at small volumeelement, the amount of work done by external loads to cause the small displacement is setequal to amount of increased internal strain energy. Assuming the field of displacement isgiven by π = {π’, π£, π€} and assuming the external loads are given by οΏ½ποΏ½ acting on the nodes,
hence these point loads will do work given by {πΏπ}π οΏ½ποΏ½ on that unit volume where {π} is thenodal displacements. In all these derivations, only loads acting directly on the nodes areconsidered for now. In other words, body forces and traction forces are not considered inorder to simplify the derivations.
The increase of strain energy is {πΏπ}π {π} in that same unit volume.
This area is the Strain energy per
unit volume
Hence, for a unit volume
{πΏπ}π οΏ½ποΏ½ = {πΏπ}π {π}
And for the whole volume
{πΏπ}π οΏ½ποΏ½ = οΏ½π{πΏπ}π {π} ππ (1)
Assuming that displacement can be written as a function of the nodal displacements ofthe element results in
π = [π] {π}
Therefore
πΏπ = [π] {πΏπ}
{πΏπ}π = {πΏπ}π [π]π (2)
Since {π} = π {π} then {π} = π [π] {π} = [π΅] {π} where π΅ is the strain displacement matrix[π΅] = π [π], hence
{π} = [π΅] {π}{πΏπ} = [π΅] {πΏπ}
{πΏπ}π = {πΏπ}π [π΅]π (3)
Now from the stress-strain relation {π} = [πΈ] {π}, hence
{π} = [πΈ] [π΅] {π} (4)
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5.2. Potential energy (minimize a . . . CHAPTER 5. FINDING THE . . .
Substituting Eqs. (2,3,4) into (1) results in
{πΏπ}π οΏ½ποΏ½ βοΏ½π{πΏπ}π [π΅]π [πΈ] [π΅] {π} ππ = 0
Since {πΏπ} and {π} do not depend on the integration variables they can be moved outsidethe integral, giving
{πΏπ}π οΏ½οΏ½ποΏ½ β {π}οΏ½π[π΅]π [πΈ] [π΅] πποΏ½ = 0
Since the above is true for any admissible πΏπ then the only condition is that
οΏ½ποΏ½ = {π}οΏ½π[π΅]π [πΈ] [π΅] ππ
This is in the form π = πΎΞ, therefore
[πΎ] = οΏ½π[π΅]π [πΈ] [π΅] ππ
knowing [π΅] allows finding [π] by integrating over the volume. For the beam element though,
π = π£ (π₯) the transverse displacement. This means [π΅] = π2
ππ₯2 [π]. Recalling from the abovethat for the beam element,
[π] = οΏ½ 1πΏ3οΏ½πΏ3 β 3πΏπ₯2 + 2π₯3οΏ½ 1
πΏ2οΏ½πΏ2π₯ β 2πΏπ₯2 + π₯3οΏ½ 1
πΏ3οΏ½3πΏπ₯2 β 2π₯3οΏ½ 1
πΏ2οΏ½βπΏπ₯2 + π₯3οΏ½οΏ½
Hence
[π΅] =π2
ππ₯2[π] = οΏ½ 1
πΏ3(β6πΏ + 12π₯) 1
πΏ2(β4πΏ + 6π₯) 1
πΏ3(6πΏ β 12π₯) 1
πΏ2(β2πΏ + 6π₯)οΏ½
Hence
[πΎ] = οΏ½π[π΅]π [πΈ] [π΅] ππ
= οΏ½π
β§βͺβͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβͺβ©
1πΏ3(β6πΏ + 12π₯)
1πΏ2(β4πΏ + 6π₯)
1πΏ3(6πΏ β 12π₯)
1πΏ2(β2πΏ + 6π₯)
β«βͺβͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβͺβ
πΈ οΏ½ 1πΏ3(β6πΏ + 12π₯) 1
πΏ2(β4πΏ + 6π₯) 1
πΏ3(6πΏ β 12π₯) 1
πΏ2(β2πΏ + 6π₯)οΏ½ ππ
= πΈπΌοΏ½πΏ
0
β§βͺβͺβͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβͺβͺβ©
1πΏ3(β6πΏ + 12π₯)
1πΏ2(β4πΏ + 6π₯)
1πΏ3(6πΏ β 12π₯)
1πΏ2(β2πΏ + 6π₯)
β«βͺβͺβͺβͺβͺβͺβͺβ¬βͺβͺβͺβͺβͺβͺβͺβ
οΏ½ 1πΏ3(β6πΏ + 12π₯) 1
πΏ2(β4πΏ + 6π₯) 1
πΏ3(6πΏ β 12π₯) 1
πΏ2(β2πΏ + 6π₯)οΏ½ ππ₯
= πΈπΌοΏ½πΏ
0
ββββββββββββββββ
1πΏ6(6πΏ β 12π₯)2 1
πΏ5(4πΏ β 6π₯) (6πΏ β 12π₯) β 1
πΏ6(6πΏ β 12π₯)2 1
πΏ5(2πΏ β 6π₯) (6πΏ β 12π₯)
1πΏ5(4πΏ β 6π₯) (6πΏ β 12π₯) 1
πΏ4(4πΏ β 6π₯)2 β 1
πΏ5(4πΏ β 6π₯) (6πΏ β 12π₯) 1
πΏ4(2πΏ β 6π₯) (4πΏ β 6π₯)
β 1πΏ6(6πΏ β 12π₯)2 β 1
πΏ5(4πΏ β 6π₯) (6πΏ β 12π₯) 1
πΏ6(6πΏ β 12π₯)2 β 1
πΏ5(2πΏ β 6π₯) (6πΏ β 12π₯)
1πΏ5(2πΏ β 6π₯) (6πΏ β 12π₯) 1
πΏ4(2πΏ β 6π₯) (4πΏ β 6π₯) β 1
πΏ5(2πΏ β 6π₯) (6πΏ β 12π₯) 1
πΏ4(2πΏ β 6π₯)2
ββββββββββββββββ
ππ₯
= πΈπΌ
ββββββββββββββββ
12πΏ3
6πΏ2 β 12
πΏ36πΏ2
6πΏ2
4πΏ β 6
πΏ22πΏ
β 12πΏ3 β 6
πΏ212πΏ3 β 6
πΏ26πΏ2
2πΏ β 6
πΏ24πΏ
ββββββββββββββββ
=πΈπΌπΏ3
ββββββββββββββββ
12 6πΏ β12 6πΏ6πΏ 4πΏ2 β6πΏ 2πΏ2
β12 β6πΏ 12 β6πΏ6πΏ 2πΏ2 β6πΏ 4πΏ2
ββββββββββββββββ
Which is the stiοΏ½ness matrix found earlier.
5.2 Potential energy (minimize a functional) method toderive the stiοΏ½ness matrix
This method is very similar to the first method actually. It all comes down to finding afunctional, which is the potential energy of the system, and minimizing this with respectto the nodal displacements. The result gives the stiοΏ½ness matrix.
33
5.2. Potential energy (minimize a . . . CHAPTER 5. FINDING THE . . .
Let the system total potential energy by called Ξ and let the total internal energy in thesystem be π and let the work done by external loads acting on the nodes be Ξ©, then
Ξ = π βΞ©
Work done by external loads have a negative sign since they are an external agent to thesystem and work is being done onto the system. The internal strain energy is given by12β«π{π}π {π} ππ and the work done by external loads is {π}π οΏ½ποΏ½, hence
Ξ =12 οΏ½π
{π}π {π} ππ β {π}π οΏ½ποΏ½ (1)
Now the rest follows as before. Assuming that displacement can be written as a functionof the nodal displacements {π}, hence
π = [π] {π}
Since {π} = π {π} then {π} = π [π] {π} = [π΅] {π} where π΅ is the strain displacement matrix[π΅] = π [π], hence
{π} = [π΅] {π}
Now from the stress-strain relation {π} = [πΈ] {π}, hence
{π}π = {π}π [π΅]π [πΈ] (4)
Substituting Eqs. (2,3,4) into (1) results in
Ξ =12 οΏ½π
{π}π [π΅]π [πΈ] [π΅] {π} ππ β {π}π οΏ½ποΏ½
Setting πΞ π{π} = 0 gives
0 = {π}οΏ½π[π΅]π [πΈ] [π΅] ππ β οΏ½ποΏ½
Which is on the form π = [πΎ]π· which means that
[πΎ] = οΏ½π[π΅]π [πΈ] [π΅] ππ
As was found by the virtual work method.
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Chapter 6
References
1. A first course in Finite element method, 3rd edition, by Daryl L. Logan
2. Matrix analysis of framed structures, 2nd edition, by William Weaver, James Gere
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