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Engineering report
Derivation of stiļæ½ness matrix for a beam
Nasser M. Abbasi
July 7, 2016
Contents
1 Introduction 1
2 Direct method 32.1 Examples using the direct beam stiļæ½ness matrix . . . . . . . . . . . . . . . . . 13
2.1.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.1.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.1.3 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3 Finite elements or adding more elements 253.1 Example 3 redone with 2 elements . . . . . . . . . . . . . . . . . . . . . . . . . 25
4 Generating shear and bending moments diagrams 35
5 Finding the stiļæ½ness matrix using methods other than direct method 375.1 Virtual work method for derivation of the stiļæ½ness matrix . . . . . . . . . . . 385.2 Potential energy (minimize a functional) method to derive the stiļæ½ness matrix 40
6 References 43
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 moment andshear force using direct stiļæ½ness method and then using finite elements method by addingmore elements. The problem is solved first by finding the stiļæ½ness matrix using the directmethod and then using the virtual work method.
1
CHAPTER 1. INTRODUCTION
2
Chapter 2
Direct method
Local contents2.1 Examples using the direct beam stiļæ½ness matrix . . . . . . . . . . . . . . . . . . . 13
3
CHAPTER 2. DIRECT METHOD
Starting with only one element beam which is subject to bending and shear forces. There are4 nodal degrees of freedom. Rotation at the left and right nodes of the beam and transversedisplacements at the left and right nodes. The following diagram shows the sign conventionused for external forces. Moments are always positive when anti-clockwise direction andvertical forces are positive when in the positive š¦ direction.
The two nodes are numbered 1 and 2 from left to right. š1 is the moment at the left node(node 1), š2 is the moment at the right node (node 2). š1 is the vertical force at the leftnode and š2 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 the š„coordinate. The displacement field in the vertical direction is called š£ (š„). This is the verticaldisplacement of point š„ on the beam from the original š„āšš„šš . The following diagram showsthe notation used for the coordinates.
x,u
y,v
Angular displacement at distance š„ on the beam is found using š (š„) = šš£(š„)šš„ . At the left node,
the degrees of freedom or the displacements, are called š£1, š1 and at the right node they arecalled š£2, š2. At an arbitrary location š„ in the beam, the vertical displacement is š£ (š„) andthe rotation at that location is š (š„).
The following diagram shows the displacement field š£ (š„)
4
CHAPTER 2. DIRECT METHOD
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 stiļæ½ness matrix, the forces at the ends of the beam arefound directly by the use of beam theory. In beam theory the signs are diļæ½erent from whatis given in the first diagram above. Therefore, the moment and shear forces obtained usingbeam theory (ššµ and ššµ in the diagram below) will have diļæ½erent signs when comparedto the external forces. The signs are then adjusted to reflect the convention as shown in thediagram above using š and š.
For an example, the external moment š1 is opposite in sign to ššµ1 and the reaction š2is opposite to ššµ2. 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 šš and š š in the above diagram as
5
CHAPTER 2. DIRECT METHOD
function of the displacements at the nodes {š} = {š£1, š1, š£2, š2}š.
In other words, the goal is to obtain an expression of the form ļ潚ļæ½ = [š¾] {š} where [š¾] is thestiļæ½ness matrix where ļ潚ļæ½ = {š1,š1, š 2,š2}
š is the nodal forces or load vector, and {š} is thenodal displacement vector.
In this case [š¾] will be a 4 Ć 4 matrix and ļ潚ļæ½ is a 4 Ć 1 vector and {š} is a 4 Ć 1 vector.
Starting with š1. It is in the same direction as the shear force ššµ1. Since ššµ1 =šššµ1šš„ then
š1 =šššµ1šš„
Since from beam theory ššµ1 = āš (š„) š¼š¦ , the above becomes
š1 = āš¼š¦šš (š„)šš„
But š (š„) = šøš (š„) and š (š„) = āš¦š where š is radius of curvature, therefore the above becomes
š1 = šøš¼ššš„ ļæ½
1šļæ½
Since 1š =
ļ潚2š¢šš„2 ļæ½
ļæ½1+ļ潚š¢šš„ ļæ½
2ļæ½3/2 and for a small angle of deflection šš¢
šš„ āŖ 1 then 1š = ļ潚
2š¢šš„2
ļæ½, and the above
now becomes
š1 = šøš¼š3š¢ (š„)šš„3
Before continuing, the following diagram illustrates the above derivation. This comes frombeam theory.
Deflection curve
x
x
Radius of curvature
vx
x dv
dx
1 d 2v
dx2
6
CHAPTER 2. DIRECT METHOD
Now š1 is obtained. š1 is in the opposite sense of the bending moment ššµ1 hence anegative sign is added giving š1 = āššµ1. But ššµ1 = āš (š„) š¼
š¦ therefore
š1 = š (š„)š¼š¦
= šøš (š„)š¼š¦
= šø ļæ½āš¦š ļæ½
š¼š¦
= āšøš¼ ļæ½1šļæ½
= āšøš¼š2š¤šš„2
š2 is now found. It is in the opposite sense of the shear force ššµ2, hence a negative sign isadded giving
š2 = āššµ2
= āšššµ2šš„
Since ššµ2 = āš (š„) š¼š¦ , the above becomes
š2 =š¼š¦šš (š„)šš„
But š (š„) = šøš (š„) and š (š„) = āš¦š where š is radius of curvature. The above becomes
š2 = āšøš¼ššš„ ļæ½
1šļæ½
But 1š =
ļ潚2š¤šš„2 ļæ½
ļæ½1+ļ潚š¤šš„ ļæ½
2ļæ½3/2 and for small angle of deflection šš¤
šš„ āŖ 1 hence 1š = ļ潚
2š¢šš„2
ļæ½ then the above
becomes
š2 = āšøš¼š3š¢ (š„)šš„3
Finally š2 is in the same direction as ššµ2 so no sign change is needed. ššµ2 = āš (š„) š¼š¦ ,
7
CHAPTER 2. DIRECT METHOD
therefore
š2 = āš (š„)š¼š¦
= āšøš (š„)š¼š¦
= āšø ļæ½āš¦š ļæ½
š¼š¦
= šøš¼ ļæ½1šļæ½
= šøš¼š2š¢šš„2
The following is a summary of what was found so far. Notice that the above expressions areevaluates at š„ = 0 and at š„ = šæ. Accordingly to obtain the nodal end forces vector ļ潚ļæ½
ļ潚ļæ½ =
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š1
š1
š2
š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=
ā§āŖāŖāŖāŖāŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖāŖāŖāŖāŖā©
šøš¼ š3š¢(š„)šš„3 ļæ½
š„=0
āšøš¼ š2š¢šš„2 ļ潚„=0
āšøš¼ š3š¢(š„)šš„3 ļæ½
š„=šæ
šøš¼ š2š¢šš„2 ļ潚„=šæ
ā«āŖāŖāŖāŖāŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖāŖāŖāŖāŖā
(1)
The RHS of the above is now expressed as function of the nodal displacements š£1, š1, š£2, š2.To do that, the field displacement š£ (š„) which is the transverse displacement of the beam isassumed to be a polynomial in š„ of degree 3 or
š£ (š„) = š0 + š1š„ + š2š„2 + š3š„3
š (š„) =šš£ (š„)šš„
= š1 + 2š2š„ + 3š3š„2 (A)
Polynomial of degree 3 is used since there are 4 degrees of freedom, and a minimum of 4free parameters is needed. Hence
š£1 = š£ (š„)|š„=0 = š0 (2)
And
š1 = š (š„)|š„=0 = š1 (3)
Assuming the length of the beam is šæ, then
š£2 = š£ (š„)|š„=šæ = š0 + š1šæ + š2šæ2 + š3šæ3 (4)
And
š2 = š (š„)|š„=šæ = š1 + 2š2šæ + 3š3šæ2 (5)
8
CHAPTER 2. DIRECT METHOD
Equations (2-5) gives
{š} =
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š0š1
š0 + š1šæ + š2šæ2 + š3šæ3
š1 + 2š2šæ + 3š3šæ2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=
āāāāāāāāāāāāāāā
1 0 0 00 1 0 01 šæ šæ2 šæ3
0 1 2šæ 3šæ2
āāāāāāāāāāāāāāā
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š0š1š2š3
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
Solving for šš givesā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š0š1š2š3
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=
āāāāāāāāāāāāāāāā
1 0 0 00 1 0 0
ā 3šæ2 ā 2
šæ3šæ2 ā 1
šæ2šæ3
1šæ2 ā 2
šæ31šæ2
āāāāāāāāāāāāāāāā
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=
āāāāāāāāāāāāāāāā
š£1š1
3šæ2š£2 ā
1šæš2 ā
3šæ2š£1 ā
2šæš1
1šæ2š1 +
1šæ2š2 +
2šæ3š£1 ā
2šæ3š£2
āāāāāāāāāāāāāāāā
š£ (š„), the field displacement function from Eq. (A) can now be written as a function of thenodal displacements
š£ (š„) = š0 + š1š„ + š2š„2 + š3š„3
= š£1 + š1š„ + ļæ½3šæ2
š£2 ā1šæš2 ā
3šæ2
š£1 ā2šæš1ļæ½ š„2 + ļæ½
1šæ2
š1 +1šæ2
š2 +2šæ3
š£1 ā2šæ3
š£2ļæ½ š„3
= š£1 + š„š1 ā 2š„2
šæš1 +
š„3
šæ2š1 ā
š„2
šæš2 +
š„3
šæ2š2 ā 3
š„2
šæ2š£1 + 2
š„3
šæ3š£1 + 3
š„2
šæ2š£2 ā 2
š„3
šæ3š£2
Or in matrix form
š£ (š„) = ļæ½1 ā 3 š„2
šæ2 + 2 š„3
šæ3 š„ ā 2š„2
šæ + š„3
šæ2 3 š„2
šæ2 ā 2 š„3
šæ3 āš„2
šæ + š„3
šæ2ļæ½
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
= ļæ½ 1šæ3ļ潚æ3 ā 3šæš„2 + 2š„3ļæ½ 1
šæ2ļ潚æ2š„ ā 2šæš„2 + š„3ļæ½ 1
šæ3ļæ½3šæš„2 ā 2š„3ļæ½ 1
šæ2ļæ½āšæš„2 + š„3ļæ½ļæ½
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
The above can be written as
š£ (š„) = ļ潚1 (š„) š2 (š„) š3 (š„) š4 (š„)ļæ½
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
(5A)
š£ (š„) = [š] {š}
9
CHAPTER 2. DIRECT METHOD
Where šš are called the shape functions. The shape functions areāāāāāāāāāāāāāāā
š1 (š„)š2 (š„)š3 (š„)š4 (š„)
āāāāāāāāāāāāāāā
=
āāāāāāāāāāāāāāāā
1šæ3ļ潚æ3 ā 3šæš„2 + 2š„3ļæ½
1šæ2ļ潚æ2š„ ā 2šæš„2 + š„3ļæ½1šæ3ļæ½3šæš„2 ā 2š„3ļæ½
1šæ2ļæ½āšæš„2 + š„3ļæ½
āāāāāāāāāāāāāāāā
We notice that š1 (0) = 1 and š1 (šæ) = 0 as expected. Alsošš2 (š„)šš„
ļ潚„=0
=1šæ2
ļ潚æ2 ā 4šæš„ + 3š„2ļæ½ļ潚„=0
= 1
Andšš2 (š„)šš„
ļ潚„=šæ
=1šæ2
ļ潚æ2 ā 4šæš„ + 3š„2ļæ½ļ潚„=šæ
= 0
Also š3 (0) = 0 and š3 (šæ) = 1 andšš4 (š„)šš„
ļ潚„=0
=1šæ2
ļæ½ā2šæš„ + 3š„2ļæ½ļ潚„=0
= 0
andšš4 (š„)šš„
ļ潚„=šæ
=1šæ2
ļæ½ā2šæš„ + 3š„2ļæ½ļ潚„=šæ
= 1
The shape functions have thus been verified. The stiļæ½ness matrix is now found by substitutingEq. (5A) into Eq. (1), repeated below
ļ潚ļæ½ =
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š1
š1
š2
š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=
āāāāāāāāāāāāāāāāāāāāāāāā
šøš¼ š3š£(š„)šš„3 ļæ½
š„=0
āšøš¼ š2š£šš„2 ļ潚„=0
āšøš¼ š3š£(š„)šš„3 ļæ½
š„=šæ
šøš¼ š2š£šš„2 ļ潚„=šæ
āāāāāāāāāāāāāāāāāāāāāāāā
Hence
ļ潚ļæ½ =
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š1
š1
š2
š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=
āāāāāāāāāāāāāāāāāā
šøš¼ š3
šš„3(š1š£1 + š2š1 + š3š£2 + š4š2)
āšøš¼ š2
šš„2(š1š£1 + š2š1 + š3š£2 + š4š2)
āšøš¼ š3
šš„3(š1š£1 + š2š1 + š3š£2 + š4š2)
šøš¼ š2
šš„2(š1š£1 + š2š1 + š3š£2 + š4š2)
āāāāāāāāāāāāāāāāāā
(6)
Butš3
šš„3š1 (š„) =
1šæ3
š3
šš„3ļ潚æ3 ā 3šæš„2 + 2š„3ļæ½
=12šæ3
10
CHAPTER 2. DIRECT METHOD
Andš3
šš„3š2 (š„) =
1šæ2
š3
šš„3ļ潚æ2š„ ā 2šæš„2 + š„3ļæ½
=6šæ2
Andš3
šš„3š3 (š„) =
1šæ3
š3
šš„3ļæ½3šæš„2 ā 2š„3ļæ½
=ā12šæ3
Andš3
šš„3š4 (š„) =
1šæ2
š3
šš„3ļæ½āšæš„2 + š„3ļæ½
=6šæ2
For the second derivativesš2
šš„2š1 (š„) =
1šæ3
š2
šš„2ļ潚æ3 ā 3šæš„2 + 2š„3ļæ½
=1šæ3
(12š„ ā 6šæ)
Andš2
šš„2š2 (š„) =
1šæ2
š2
šš„2ļ潚æ2š„ ā 2šæš„2 + š„3ļæ½
=1šæ2
(6š„ ā 4šæ)
Andš2
šš„2š3 (š„) =
1šæ3
š2
šš„2ļæ½3šæš„2 ā 2š„3ļæ½
=1šæ3
(6šæ ā 12š„)
Andš2
šš„2š4 (š„) =
1šæ2
š2
šš„2ļæ½āšæš„2 + š„3ļæ½
=1šæ2
(6š„ ā 2šæ)
11
CHAPTER 2. DIRECT METHOD
Hence Eq. (6) becomes
ļ潚ļæ½ =
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š1
š1
š2
š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=
āāāāāāāāāāāāāāāāāāāāāāāā
šøš¼ š3
šš„3(š1š£1 + š2š1 + š3š£2 + š4š2)ļæ½
š„=0
āšøš¼ š2
šš„2(š1š£1 + š2š1 + š3š£2 + š4š2)ļæ½
š„=0
āšøš¼ š3
šš„3(š1š£1 + š2š1 + š3š£2 + š4š2)ļæ½
š„=šæ
šøš¼ š2
šš„2(š1š£1 + š2š1 + š3š£2 + š4š2)ļæ½
š„=šæ
āāāāāāāāāāāāāāāāāāāāāāāā
=
āāāāāāāāāāāāāāāāāāāāāāāāā
šøš¼ ļæ½ 12šæ3š£1 +6šæ2š1 ā
12šæ3š£2 +
6šæ2š2ļæ½
š„=0
āšøš¼ ļæ½ 1šæ3(12š„ ā 6šæ) š£1 +
1šæ2(6š„ ā 4šæ) š1 +
1šæ3(6šæ ā 12š„) š£2 +
1šæ2(6š„ ā 2šæ) š2ļæ½
š„=0
āšøš¼ ļæ½ 12šæ3š£1 +6šæ2š1 ā
12šæ3š£2 +
6šæ2š2ļæ½
š„=šæ
šøš¼ ļæ½ 1šæ3(12š„ ā 6šæ) š£1 +
1šæ2(6š„ ā 4šæ) š1 +
1šæ3(6šæ ā 12š„) š£2 +
1šæ2(6š„ ā 2šæ) š2ļæ½
š„=šæ
āāāāāāāāāāāāāāāāāāāāāāāāā
=šøš¼šæ3
āāāāāāāāāāāāāāā
12 6šæ ā12 6šæā (12š„ ā 6šæ)š„=0 āšæ (6š„ ā 4šæ)š„=0 ā (6šæ ā 12š„)š„=0 āšæ (6š„ ā 2šæ)š„=0
ā12 ā6šæ 12 ā6šæ(12š„ ā 6šæ)š„=šæ šæ (6š„ ā 4šæ)š„=šæ (6šæ ā 12š„)š„=šæ šæ (6š„ ā 2šæ)š„=šæ
āāāāāāāāāāāāāāā
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
(7)
Or in matrix form, after evaluating the expressions above for š„ = šæ and š„ = 0 asā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š1
š1
š2
š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=šøš¼šæ3
āāāāāāāāāāāāāāāā
12 6šæ ā12 6šæ6šæ 4šæ2 ā6šæ 2šæ2
ā 12šæ3 ā 6
šæ212šæ3 ā 6
šæ2
(12šæ ā 6šæ) šæ (6šæ ā 4šæ) (6šæ ā 12šæ) šæ (6šæ ā 2šæ)
āāāāāāāāāāāāāāāā
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=šøš¼šæ3
āāāāāāāāāāāāāāā
12 6šæ ā12 6šæ6šæ 4šæ2 ā6šæ 2šæ2
ā12 ā6šæ 12 ā6šæ6šæ 2šæ2 ā6šæ 4šæ2
āāāāāāāāāāāāāāā
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
The above now is in the form
ļ潚ļæ½ = [š¾] {š}
Hence the stiļæ½ness matrix is
[š¾] =šøš¼šæ3
āāāāāāāāāāāāāāā
12 6šæ ā12 6šæ6šæ 4šæ2 ā6šæ 2šæ2
ā12 ā6šæ 12 ā6šæ6šæ 2šæ2 ā6šæ 4šæ2
āāāāāāāāāāāāāāā
Knowing the stiļæ½ness matrix means knowing the nodal displacements {š} when given theforces at the nodes. The power of the finite element method now comes after all the nodaldisplacements š£1, š1, š£2, š2 are calculated by solving ļ潚ļæ½ = [š¾] {š}. This is because the polyno-
12
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
mial š£ (š„) is now completely determined and hence š£ (š„) and š (š„) can now be evaluated forany š„ along the beam and not just at its end nodes as the case with finite diļæ½erence method.Eq. 5A above can now be used to find the displacement š£ (š„) and š (š„) everywhere.
š£ (š„) = [š] {š}
š£ (š„) = ļ潚1 (š„) š2 (š„) š3 (š„) š4 (š„)ļæ½
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
To summarise, these are the steps to obtain š£ (š„)
1. An expression for [š¾] is found.
2. ļ潚ļæ½ = [š¾] {š} is solved for {š}
3. š£ (š„) = [š] {š} is calculated by assuming š£ (š„) is a polynomial. This gives the displace-ment š£ (š„) to use to evaluate the transverse displacement anywhere on the beam andnot just at the end nodes.
4. š (š„) = šš£(š„)šš„ = š
šš„ [š] {š} is obtained to evaluate the rotation of the beam any where andnot just at the end nodes.
5. The strain š (š„) = āš¦ [šµ] {š} is found, where [šµ] is the gradient matrix [šµ] = š2
šš„2 [š].
6. The stress from š = šøš = āšøš¦ [šµ] {š} is found.
7. The bending moment diagram from š(š„) = šøš¼ [šµ] {š} is found.
8. The shear force diagram from š (š„) = ššš„š(š„) is found.
2.1 Examples using the direct beam stiļæ½ness matrix
The beam stiļæ½ness 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 šæ, with point load š at the free end
L
P
13
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
The first step is to make the free body diagram and show all moments and forces at thenodes
L
P
R
M1M2
š is the given force. š2 = 0 since there is no external moment at the right end. Henceļ潚ļæ½ = [š¾] {š} for this system is
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š š1
āš0
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=šøš¼šæ3
āāāāāāāāāāāāāāā
12 6šæ ā12 6šæ6šæ 4šæ2 ā6šæ 2šæ2
ā12 ā6šæ 12 ā6šæ6šæ 2šæ2 ā6šæ 4šæ2
āāāāāāāāāāāāāāā
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
Now is an important step. The known end displacements from boundary conditions issubstituted into {š}, 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 š£1 = 0 and š1 = 0. Hence the only unknowns are š£2 and š2.Therefore the first and the second rows and columns are removed, giving
ā§āŖāŖāØāŖāŖā©āš0
ā«āŖāŖā¬āŖāŖā=
šøš¼šæ3
āāāāāā12 ā6šæā6šæ 4šæ2
āāāāāā
ā§āŖāŖāØāŖāŖā©š£2š2
ā«āŖāŖā¬āŖāŖā
Now the above is solved for
ā§āŖāŖāØāŖāŖā©š£2š2
ā«āŖāŖā¬āŖāŖā. Let šø = 30 Ć 106 psi (steel), š¼ = 57 šš4, šæ = 144 šš, and
š = 400 lb, henceļæ½ ļæ½1 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)ļæ½ ļæ½
1Instead of removing rows/columns for known boundary conditions, we can also just put a 1 on the diagonalof the stiļæ½ness matrix for that boundary conditions. I will do this example again using this method
14
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
Ė
which givesx =-0.2324-0.0024
Therefore the vertical displacement at the right end is š£2 = 0.2324 inches (downwards) andš2 = ā0.0024 radians. Now that all nodal displacements are found, the field displacementfunction is completely determined.
{š} =
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
00
ā0.2324ā0.0024
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
From Eq. 5A
š£ (š„) = [š] {š}
= ļ潚1 (š„) š2 (š„) š3 (š„) š4 (š„)ļæ½
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
00
ā0.2324ā0.0024
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
= ļæ½ 1šæ3ļ潚æ3 ā 3šæš„2 + 2š„3ļæ½ 1
šæ2ļ潚æ2š„ ā 2šæš„2 + š„3ļæ½ 1
šæ3ļæ½3šæš„2 ā 2š„3ļæ½ 1
šæ2ļæ½āšæš„2 + š„3ļæ½ļæ½
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
00
ā0.2324ā0.0024
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=0.002 4šæ2
ļ潚暄2 ā š„3ļæ½ +0.232 4šæ3
ļæ½2š„3 ā 3šæš„2ļæ½
But šæ = 144 inches, and the above becomes
š£ (š„) = 3.992 0 Ć 10ā8š„3 ā 1.695 6 Ć 10ā5š„2
To verify, let š„ = 144 in the above
š£ (š„ = 144) = 3.992 0 Ć 10ā81443 ā 1.695 6 Ć 10ā51442
= ā0.232 40
The following is a plot of the deflection curve for the beamļæ½ ļæ½1 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 gridļæ½ ļæ½
15
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
Ė
Now instead of removing rows/columns for the known boundary conditions, a 1 is put onthe diagonal. Starting again with
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š š1
āš0
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=šøš¼šæ3
āāāāāāāāāāāāāāā
12 6šæ ā12 6šæ6šæ 4šæ2 ā6šæ 2šæ2
ā12 ā6šæ 12 ā6šæ6šæ 2šæ2 ā6šæ 4šæ2
āāāāāāāāāāāāāāā
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
Since š£1 = 0 and š1 = 0, thenā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
00āš0
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=šøš¼šæ3
āāāāāāāāāāāāāāā
1 0 0 00 1 0 00 0 12 ā6šæ0 0 ā6šæ 4šæ2
āāāāāāāāāāāāāāā
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
00š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
The above system is now solved as before. šø = 30 Ć 106 psi, š¼ = 57 šš4, šæ = 144 in, š = 400 lbļæ½ ļæ½1 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 Aļæ½ ļæ½
ĖA =
16
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
1.0e+07 *0.0000 0 0 00 0.0000 0 00 0 0.0007 -0.04960 0 -0.0496 4.7583ļæ½ ļæ½
1 sol=A\load %SOLVEļæ½ ļæ½Ėsol =00-0.2324-0.0024
The same solution is obtained as before, but without the need to remove rows/column fromthe stiļæ½ness matrix. This method might be easier for programming than the first method ofremoving 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 š is now placed in the middle ofthe beam
L
P
In using stiļæ½ness method, all loads must be on the nodes. The vector ļ潚ļæ½ is the nodal forcesvector. Hence equivalent nodal loads are found for the load in the middle of the beam. Theequivalent loading is the following
L
P/2P/2 PL/8PL/8
17
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
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
The stiļæ½ness equation is now written as
ļ潚ļæ½ = [š¾] {š}ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š ā š/2š1 ā ššæ/8
āš/2ššæ/8
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=šøš¼šæ3
āāāāāāāāāāāāāāā
12 6šæ ā12 6šæ6šæ 4šæ2 ā6šæ 2šæ2
ā12 ā6šæ 12 ā6šæ6šæ 2šæ2 ā6šæ 4šæ2
āāāāāāāāāāāāāāā
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
There is no need to determine š and š1 at this point since these rows will be removed dueto boundary conditions š£1 = 0 and š1 = 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 ļ潚ļæ½ = [š¾] {š}. Hence, after removing the first two rows andcolumns, the above system simplifies to
ā§āŖāŖāØāŖāŖā©āš/2ššæ/8
ā«āŖāŖā¬āŖāŖā=
šøš¼šæ3
āāāāāā12 ā6šæā6šæ 4šæ2
āāāāāā
ā§āŖāŖāØāŖāŖā©š£2š2
ā«āŖāŖā¬āŖāŖā
The above is now solved for
ā§āŖāŖāØāŖāŖā©š£2š2
ā«āŖāŖā¬āŖāŖāusing the same numerical values for š, šø, š¼, šæ as in the first
exampleļæ½ ļæ½1 P=400;2 L=144;
18
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
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)\loadļæ½ ļæ½
Ėx =-0.072630472854641-0.000605253940455
Therefore
{š} =
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
00
ā0.072630472854641ā0.000605253940455
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
This is enough to obtain š£ (š„) as before. Now the reactions š and š1 can be determined ifneeded. Going back to the full ļ潚ļæ½ = [š¾] {š}, results in
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š ā š/2š1 ā ššæ/8
āš/2ššæ/8
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=šøš¼šæ3
āāāāāāāāāāāāāāā
12 6šæ ā12 6šæ6šæ 4šæ2 ā6šæ 2šæ2
ā12 ā6šæ 12 ā6šæ6šæ 2šæ2 ā6šæ 4šæ2
āāāāāāāāāāāāāāā
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
00
ā0.072630472854641ā0.000605253940455
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=šøš¼šæ3
āāāāāāāāāāāāāāā
0.871 57 ā 3.631 5 Ć 10ā3šæ0.435 78šæ ā 1.210 5 Ć 10ā3šæ2
3.631 5 Ć 10ā3šæ ā 0.871 570.435 78šæ ā 2.421 Ć 10ā3šæ2
āāāāāāāāāāāāāāā
Hence the first equation becomes
š ā š/2 =šøš¼šæ3
ļæ½0.871 57 ā 3.631 5 Ć 10ā3šæļæ½
and since šø = 30 Ć 106 psi (steel) and š¼ = 57 šš4and šæ = 144 in and š = 400 lb, then
š =ļæ½30Ć106ļæ½57
1443ļæ½0.871 57 ā 3.631 5 Ć 10ā3 (144)ļæ½ + 400/2 . Therefore
š = 400 lb
and š1 ā ššæ/8 = šøš¼šæ3ļæ½0.435 78šæ ā 1.210 5 Ć 10ā3šæ2ļæ½ + ššæ/8 , hence
š1 = 28 762 lb-ft
Now that all nodal reactions are found, the displacement field is found and the deflection
19
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
curve can be plotted.
š£ (š„) = [š] {š}
š£ (š„) = ļ潚1 (š„) š2 (š„) š3 (š„) š4 (š„)ļæ½
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
= ļ潚1 (š„) š2 (š„) š3 (š„) š4 (š„)ļæ½
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
00
ā0.072630472854641ā0.000605253940455
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
= ļæ½ 1šæ3ļæ½3šæš„2 ā 2š„3ļæ½ 1
šæ2ļæ½āšæš„2 + š„3ļæ½ļæ½
āāāāāāā0.072630472854641ā0.000605253940455
āāāāāā
=6.052 5 Ć 10ā4
šæ2ļ潚暄2 ā š„3ļæ½ +
0.072 63šæ3
ļæ½2š„3 ā 3šæš„2ļæ½
Since šæ = 144 inches, the above becomes
š£ (š„) =6.052 5 Ć 10ā4
(144)2ļæ½(144) š„2 ā š„3ļæ½ +
0.072 63(144)3
ļæ½2š„3 ā 3 (144) š„2ļæ½
= 1.945 9 Ć 10ā8š„3 ā 6.304 7 Ć 10ā6š„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 right end,and the vertical load š now at distance š from the left end and at distance š from the rightend, and a uniform distributed load of density š lb/in is on the beam.
20
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
Pm (lb/in
a b
M1
Using the following values: š = 0.625šæ, š = 0.375šæ, šø = 30 Ć 106 psi (steel), š¼ = 57 šš4, šæ =144 in, š = 1000 lb, š = 200 lb/in.
In the above, the left end reaction forces are shown as š 1 and moment reaction as š1 andthe reaction at the right end as š 2. Starting by finding equivalent loads for the point loadš and equivalent loads for for the uniform distributed load š. All external loads must betransferred to the nodes for the stiļæ½ness 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
21
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
Using free body diagram, with all the loads on it gives the following diagram (In this diagramš is the reaction moment and š 1, š 2 are the reaction forces)
Now that all loads are on the nodes, the stiļæ½ness equation is applied
ļ潚ļæ½ = [š¾] {š}ā§āŖāŖāŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖāŖāŖā©
š 1 āšš2(šæ+2š)
šæ3 ā ššæ2
šā ššš2
šæ2 ā ššæ2
12
š 2 āšš2(šæ+2š)
šæ3 ā ššæ2
šš2ššæ2 + ššæ2
12
ā«āŖāŖāŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖāŖāŖā
=šøš¼šæ3
āāāāāāāāāāāāāāā
12 6šæ ā12 6šæ6šæ 4šæ2 ā6šæ 2šæ2
ā12 ā6šæ 12 ā6šæ6šæ 2šæ2 ā6šæ 4šæ2
āāāāāāāāāāāāāāā
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
Boundary conditions are now applied. š£1 = 0, š1 = 0,š£2 = 0. Therefore the first, second andthird rows/columns are removed giving
šš2ššæ2
+ššæ2
12=
šøš¼šæ3
4šæ2š2
Hence
š2 = ļ潚š2ššæ2
+ššæ2
12 ļæ½ ļ潚æ4šøš¼ļæ½
Substituting numerical values for the above as given at the top of the problem results in
š2 =āāāāā(1000) (0.625 (144))2 (0.375 (144))
(144)2+(200) (144)2
12
āāāāā
āāāāāā
1444 ļæ½30 Ć 106ļæ½ (57)
āāāāāā
= 7.719 9 Ć 10ā3ššš
22
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
Hence the field displacement š¢ (š„) is now found
š£ (š„) = [š] {š}
š£ (š„) = ļ潚1 (š„) š2 (š„) š3 (š„) š4 (š„)ļæ½
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
= ļ潚1 (š„) š2 (š„) š3 (š„) š4 (š„)ļæ½
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
000
7.719 9 Ć 10ā3
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=1šæ2
ļæ½āšæš„2 + š„3ļæ½ ļæ½7.719 9 Ć 10ā3ļæ½
= 3.722 9 Ć 10ā7š„3 ā 5.361 Ć 10ā5š„2
And a plot of the deflection curve isļæ½ ļæ½1 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');8 gridļæ½ ļæ½
Ė
23
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
24
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 š£ (š„) becomesmore accurate (By comparing the the result with the exact solution based on using the beam4th order diļæ½erential 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 stiļæ½ness matrix for thewhole beam (including both elements) will be 6 by 6. For each element however, the samestiļæ½ness matrix will be used as above and that will remain as before 4 by 4.
25
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 stiļæ½ness matrix for each element is found and then the global stiļæ½ness matrix is assem-bled. Then ļ潚ššššššļæ½ = ļ潚¾ššššššļæ½ ļ潚ššššššļæ½ is solved as before. The first step is to move all loads tothe nodes as was done before. This is done for each element. The formulas for equivalentloads remain the same, but now šæ becomes šæ/2. The following diagram show the equivalentloading for š
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 š is
26
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
Now the above two diagrams are put together to show all equivalent loads with the originalreaction forces to obtain the following diagram
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
27
3.1. Example 3 redone with 2 elements CHAPTER 3. FINITE ELEMENTS OR . . .
ļ潚ļæ½ = [š¾] {š} for each element is now constructed. Starting with the first elementāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāā
š 1 āšļæ½ šæ2 ļæ½
2
š1 āšļæ½ šæ2 ļæ½
2
12
āšš2ļæ½ šæ2+2šļæ½
(šæ/2)3ā
šļæ½ šæ2 ļæ½
2
šļæ½ šæ2 ļæ½2
12 āšļæ½ šæ2 ļæ½
2
12 ā ššš2
ļæ½ šæ2 ļæ½2
āāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāā
=šøš¼
ļ潚æ2ļæ½3
āāāāāāāāāāāāāāāāāāāāāāā
12 6 ļ潚æ2ļæ½ ā12 6 ļ潚æ2ļæ½
6 ļ潚æ2ļæ½ 4 ļ潚æ2ļæ½2
ā6 ļ潚æ2ļæ½ 2 ļ潚æ2ļæ½2
ā12 ā6 ļ潚æ2ļæ½ 12 ā6 ļ潚æ2ļæ½
6 ļ潚æ2ļæ½ 2 ļ潚æ2ļæ½2
ā6 ļ潚æ2ļæ½ 4 ļ潚æ2ļæ½2
āāāāāāāāāāāāāāāāāāāāāāā
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
And for the second elementāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāā
āšš2ļæ½ šæ2+2šļæ½
(šæ/2)3ā
šļæ½ šæ2 ļæ½
2
šļæ½ šæ2 ļæ½2
12 āšļæ½ šæ2 ļæ½
2
12 ā ššš2
ļæ½ šæ2 ļæ½2
š 3 āšš2ļæ½ šæ2+2šļæ½
ļæ½ šæ2 ļæ½3 ā
šļæ½ šæ2 ļæ½
2
šļæ½ šæ2 ļæ½2
12 + šš2š
ļæ½ šæ2 ļæ½2
āāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāā
=šøš¼
ļ潚æ2ļæ½3
āāāāāāāāāāāāāāāāāāāāāāā
12 6 ļ潚æ2ļæ½ ā12 6 ļ潚æ2ļæ½
6 ļ潚æ2ļæ½ 4 ļ潚æ2ļæ½2
ā6 ļ潚æ2ļæ½ 2 ļ潚æ2ļæ½2
ā12 ā6 ļ潚æ2ļæ½ 12 ā6 ļ潚æ2ļæ½
6 ļ潚æ2ļæ½ 2 ļ潚æ2ļæ½2
ā6 ļ潚æ2ļæ½ 4 ļ潚æ2ļæ½2
āāāāāāāāāāāāāāāāāāāāāāā
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£2š2
š£3š3
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
The 2 systems above are assembled to obtain the global stiļæ½ness matrix equation givingāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāā
š 1 āšļæ½ šæ2 ļæ½
2
š1 āšļæ½ šæ2 ļæ½
2
12
ā2šš2ļæ½ šæ2+2šļæ½
ļæ½ šæ2 ļæ½3 ā 2
šļæ½ šæ2 ļæ½
2
ā2ššš2
ļæ½ šæ2 ļæ½2
š 3 āšš2ļæ½ šæ2+2šļæ½
ļæ½ šæ2 ļæ½3 ā
šļæ½ šæ2 ļæ½
2
šļæ½ šæ2 ļæ½2
12 + šš2š
ļæ½ šæ2 ļæ½2
āāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāā
=šøš¼
ļ潚æ2ļæ½3
āāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāā
12 6 ļ潚æ2ļæ½ ā12 6 ļ潚æ2ļæ½ 0 0
6 ļ潚æ2ļæ½ 4 ļ潚æ2ļæ½2
ā6 ļ潚æ2ļæ½ 2 ļ潚æ2ļæ½2
0 0
ā12 ā6 ļ潚æ2ļæ½ 12 + 12 ā6 ļ潚æ2ļæ½ + 6 ļ潚æ2ļæ½ ā12 6 ļ潚æ2ļæ½
6 ļ潚æ2ļæ½ 2 ļ潚æ2ļæ½2
ā6 ļ潚æ2ļæ½ + 6 ļ潚æ2ļæ½ 4 ļ潚æ2ļæ½2+ 4 ļ潚æ2ļæ½
2ā6 ļ潚æ2ļæ½ 2 ļ潚æ2ļæ½
2
0 0 ā12 ā6 ļ潚æ2ļæ½ 12 ā6 ļ潚æ2ļæ½
0 0 6 ļ潚æ2ļæ½ 2 ļ潚æ2ļæ½2
ā6 ļ潚æ2ļæ½ 4 ļ潚æ2ļæ½2
āāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāā
ā§āŖāŖāŖāŖāŖāŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
š£3š3
ā«āŖāŖāŖāŖāŖāŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖāŖāŖāŖāŖāŖā
The boundary conditions are now applied giving š£1 = 0, š1 = 0 since the first node is fixed,and š£3 = 0. And putting 1 on the diagonal of the stiļæ½ness matrix corresponding to these
28
3.1. Example 3 redone with 2 elements CHAPTER 3. FINITE ELEMENTS OR . . .
known boundary conditions results ināāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāā
00
ā2šš2ļæ½ šæ2+2šļæ½
(šæ/2)3ā 2š(šæ/2)2
ā2 ššš2
(šæ/2)2
0šļæ½ šæ2 ļæ½
2
12 + šš2š
ļæ½ šæ2 ļæ½2
āāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāā
=šøš¼
ļ潚æ2ļæ½3
āāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāā
1 0 0 0 0 00 1 0 0 0 0
0 0 24 0 0 6 ļ潚æ2ļæ½
0 0 0 8 ļ潚æ2ļæ½2
0 2 ļ潚æ2ļæ½2
0 0 0 0 1 0
0 0 6 ļ潚æ2ļæ½ 2 ļ潚æ2ļæ½2
0 4 ļ潚æ2ļæ½2
āāāāāāāāāāāāāāāāāāāāāāāāāāāāāāāā
āāāāāāāāāāāāāāāāāāāāāāāāā
00š£2š2
0š3
āāāāāāāāāāāāāāāāāāāāāāāāā
As was mentioned earlier, another method would be to remove the rows/columns whichresults in ā
āāāāāāāāāāāāāāāāāāāāāāāāā
ā2šš2ļæ½ šæ2+2šļæ½
(šæ/2)3ā 2
šļæ½ šæ2 ļæ½
2
ā2ššš2
ļæ½ šæ2 ļæ½2
šļæ½ šæ2 ļæ½2
12 + šš2š
ļæ½ šæ2 ļæ½2
āāāāāāāāāāāāāāāāāāāāāāāāāā
=šøš¼
ļ潚æ2ļæ½3
āāāāāāāāāāāāāāāāā
24 0 6 ļ潚æ2ļæ½
0 8 ļ潚æ2ļæ½2
2 ļ潚æ2ļæ½2
6 ļ潚æ2ļæ½ 2 ļ潚æ2ļæ½2
4 ļ潚æ2ļæ½2
āāāāāāāāāāāāāāāāā
āāāāāāāāāā
š£2š2
š3
āāāāāāāāāā
Giving the same solution. There are 3 unknowns to solve for. Once these unknowns aresolved for, š£ (š„) for the first element and for the second element are fully determined. Thefollowing code displays the deflection curve for the above beamļæ½ ļæ½
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;19 -2*P*a*b^2/(L/2)^2;20 0;21 P*a^2*b/(L/2)^2+(m*(L/2)^2)/12]
29
3.1. Example 3 redone with 2 elements CHAPTER 3. FINITE ELEMENTS OR . . .
22 sol=A\loadļæ½ ļæ½ĖA =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 š£2 = ā0.2735 in (downwards displacement) and š2 = ā0.0019radians and š3 = 0.0076 radians. š£ (š„) polynomial is now found for each element
š£šššš1 (š„) = ļ潚1 (š„) š2 (š„) š3 (š„) š4 (š„)ļæ½
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£1š1
š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
ļ潚1 (š„) š2 (š„) š3 (š„) š4 (š„)ļæ½
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
00š£2š2
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
=āāāāā
1
ļæ½ šæ2 ļæ½3 ļæ½3 ļæ½
šæ2ļæ½ š„2 ā 2š„3ļæ½ 1
ļæ½ šæ2 ļæ½2 ļæ½ā ļæ½
šæ2ļæ½ š„2 + š„3ļæ½
āāāāā
āāāāāāā0.2735ā0.0019
āāāāāā
=2.188šæ3 ļæ½2š„3 ā
32šæš„2ļæ½ ā
0.007 6šæ2 ļ潚„3 ā
12šæš„2ļæ½
The above polynomial is the transverse deflection of the beam for the region 0 ā¤ š„ ā¤ šæ/2.
30
3.1. Example 3 redone with 2 elements CHAPTER 3. FINITE ELEMENTS OR . . .
š£ (š„) for the second element is found in similar way
š£šššš2 (š„) = ļ潚1 (š„) š2 (š„) š3 (š„) š4 (š„)ļæ½
ā§āŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖā©
š£2š2
0š3
ā«āŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖā
= ļ潚1 (š„) š2 (š„) š3 (š„) š4 (š„)ļæ½
āāāāāāāāāāāāāāā
ā0.2735ā0.0019
00.0076
āāāāāāāāāāāāāāā
= ļ潚1 (š„) š2 (š„) š4 (š„)ļæ½
āāāāāāāāāā
ā0.2735ā0.00190.0076
āāāāāāāāāā
=āāāāāā
1
ļæ½ šæ2 ļæ½3 ļæ½ļæ½
šæ2ļæ½3ā 3 ļ潚æ2ļæ½ š„
2 + 2š„3ļæ½1
ļæ½ šæ2 ļæ½2 ļæ½ļæ½
šæ2ļæ½2š„ ā 2 ļ潚æ2ļæ½ š„
2 + š„3ļæ½1
ļæ½ šæ2 ļæ½2 ļæ½ā ļæ½
šæ2ļæ½ š„2 + š„3ļæ½
āāāāāā
āāāāāāāāāā
ā0.2735ā0.00190.0076
āāāāāāāāāā
Hence
š£šššš2 (š„) =0.030 4šæ2 ļ潚„3 ā
12šæš„2ļæ½ ā
0.007 6šæ2 ļæ½
14šæ2š„ ā šæš„2 + š„3ļæ½ ā
2.188šæ3 ļæ½
18šæ3 ā
32šæš„2 + 2š„3ļæ½
Which is valid for šæ/2 ā¤ š„ ā¤ šæ. The following is a plot of the deflection curve using the above2 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 with theanalytical 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 diļæ½erence in the result
31
3.1. Example 3 redone with 2 elements CHAPTER 3. FINITE ELEMENTS OR . . .
more clearly.
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 š¤ lbs per unit length isgiven by
š£ (š„) = āš¤š„2
48šøš¼ļæ½3šæ2 ā 5šæš„ + 2š„2ļæ½
While the analytical deflection for the same beam but when there is a point load P at distanceš from the left end is given by
š£ (š„) = āš(āØšæ ā šā©3 (3šæ ā š„)š„2 + šæ2(3(šæ ā š)(šæ ā š„)š„2 + 2šæ āØš„ ā šā©3))
Therefore, the analytical expression for deflection is given by the sum of the above expres-sions, giving
š£ (š„) = āš¤š„2
48šøš¼ļæ½3šæ2 ā 5šæš„ + 2š„2ļæ½ ā š(āØšæ ā šā©3 (3šæ ā š„)š„2 + šæ2(3(šæ ā š)(šæ ā š„)š„2 + 2šæ āØš„ ā šā©3))
Where āØš„ ā šā©means it is zero when š„āš is negative. In other words āØš„ ā šā© = (š„ ā š)šššš”šš”šš (š„ ā š)
The following diagram is a plot of the analytical deflection with the two elements deflectioncalculated using Finite elements above.
32
3.1. Example 3 redone with 2 elements CHAPTER 3. FINITE ELEMENTS OR . . .
In the above, the blue dashed curve is the analytical solution, and the red curve is the finiteelements solution using 2 elements. It can be seen that the finite element solution for thedeflection is now in a very good agreement with the analytical solution.
33
3.1. Example 3 redone with 2 elements CHAPTER 3. FINITE ELEMENTS OR . . .
34
Chapter 4
Generating shear and bendingmoments diagrams
After solving the problem using finite elements and obtaining the field displacement functionš£ (š„) 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 š(š„) = āšøš¼š2š£(š„)šš„2 and shear
force is given by š (š„) = šššš„ = āšøš¼š
3š£(š„)šš„3 then these diagrams are now readily plotted as shown
below for example three above using the result from the finite elements with 2 elements.Recalling from above that
š£ (š„) =2.188šæ3
ļæ½2š„3 ā 32šæš„
2ļæ½ ā 0.007 6šæ2
ļ潚„3 ā 12šæš„
2ļæ½0.030 4šæ2
ļ潚„3 ā 12šæš„
2ļæ½ ā 0.007 6šæ2
ļæ½14šæ
2š„ ā šæš„2 + š„3ļæ½ ā 2.188šæ3
ļæ½18šæ
3 ā 32šæš„
2 + 2š„3ļæ½
ā«āŖāŖāŖā¬āŖāŖāŖā
0 ā¤ š„ ā¤ šæ/2šæ/2 ā¤ š„ ā¤ šæ
Hence
š(š„) = āšøš¼ā0.0076(6š„āšæ)
šæ2 + 2.188(12š„ā3šæ)šæ3
ā0.0076(6š„ā2šæ)šæ2 + 0.0304(6š„āšæ)
šæ2 ā 2.188(12š„ā3šæ)šæ3
ā«āŖāŖā¬āŖāŖā
0 ā¤ š„ ā¤ šæ/2šæ/2 ā¤ š„ ā¤ šæ
using šø = 30 Ć 106 psi and š¼ = 57 in4 and šæ = 144 in, the bending moment diagram plot is
35
CHAPTER 4. GENERATING SHEAR . . .
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 š£ (š„) obtained using the finite elements methodis a third degree polynomial and after diļæ½erentiating twice to obtain the bending moment
(š(š„) = āšøš¼š2š£šš„2 ) the result becomes a linear function in š„ while in the analytical solution
case, when the load is distributed, the solution š£ (š„) is a fourth degree polynomial. Hencethe bending moment will be quadratic function in š„ 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.
36
Chapter 5
Finding the stiļæ½ness matrix usingmethods other than direct method
Local contents5.1 Virtual work method for derivation of the stiļæ½ness matrix . . . . . . . . . . . . . 385.2 Potential energy (minimize a functional) method to derive the stiļæ½ness matrix . 40
37
5.1. Virtual work method for derivation of . . . CHAPTER 5. FINDING THE . . .
There are three main methods to obtain the stiļæ½ness matrix
1. Variational method (minimizing a functional). This functional is the potential energyof the structure and loads.
2. Weighted residual. Requires the diļæ½erential equation as a starting point. Approximatedin weighted average. Galerkin weighted residual method is the most common methodfor 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 stiļæ½nessmatrix
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 š = {š¢, š£, š¤} and assuming the external loads are given by ļ潚ļæ½ acting on the nodes,
hence these point loads will do work given by {šæš}š ļ潚ļæ½ on that unit volume where {š} 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 {šæš}š {š} in that same unit volume.
This area is the Strain energy per
unit volume
Hence, for a unit volume
{šæš}š ļ潚ļæ½ = {šæš}š {š}
And for the whole volume
{šæš}š ļ潚ļæ½ = ļ潚{šæš}š {š} šš (1)
Assuming that displacement can be written as a function of the nodal displacements of the
38
5.1. Virtual work method for derivation of . . . CHAPTER 5. FINDING THE . . .
element results in
š = [š] {š}
Therefore
šæš = [š] {šæš}
{šæš}š = {šæš}š [š]š (2)
Since {š} = š {š} then {š} = š [š] {š} = [šµ] {š} where šµ is the strain displacement matrix[šµ] = š [š], hence
{š} = [šµ] {š}{šæš} = [šµ] {šæš}
{šæš}š = {šæš}š [šµ]š (3)
Now from the stress-strain relation {š} = [šø] {š}, hence
{š} = [šø] [šµ] {š} (4)
Substituting Eqs. (2,3,4) into (1) results in
{šæš}š ļ潚ļæ½ āļ潚{šæš}š [šµ]š [šø] [šµ] {š} šš = 0
Since {šæš} and {š} do not depend on the integration variables they can be moved outsidethe integral, giving
{šæš}š ļæ½ļ潚ļæ½ ā {š}ļ潚[šµ]š [šø] [šµ] ššļæ½ = 0
Since the above is true for any admissible šæš then the only condition is that
ļ潚ļæ½ = {š}ļ潚[šµ]š [šø] [šµ] šš
This is in the form š = š¾Ī, therefore
[š¾] = ļ潚[šµ]š [šø] [šµ] šš
knowing [šµ] allows finding [š] by integrating over the volume. For the beam element though,
š = š£ (š„) the transverse displacement. This means [šµ] = š2
šš„2 [š]. Recalling from the abovethat for the beam element,
[š] = ļæ½ 1šæ3ļ潚æ3 ā 3šæš„2 + 2š„3ļæ½ 1
šæ2ļ潚æ2š„ ā 2šæš„2 + š„3ļæ½ 1
šæ3ļæ½3šæš„2 ā 2š„3ļæ½ 1
šæ2ļæ½āšæš„2 + š„3ļæ½ļæ½
Hence
[šµ] =š2
šš„2[š] = ļæ½ 1
šæ3(ā6šæ + 12š„) 1
šæ2(ā4šæ + 6š„) 1
šæ3(6šæ ā 12š„) 1
šæ2(ā2šæ + 6š„)ļæ½
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5.2. Potential energy (minimize a . . . CHAPTER 5. FINDING THE . . .
Hence
[š¾] = ļ潚[šµ]š [šø] [šµ] šš
= ļ潚
ā§āŖāŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖāŖā©
1šæ3(ā6šæ + 12š„)
1šæ2(ā4šæ + 6š„)
1šæ3(6šæ ā 12š„)
1šæ2(ā2šæ + 6š„)
ā«āŖāŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖāŖā
šø ļæ½ 1šæ3(ā6šæ + 12š„) 1
šæ2(ā4šæ + 6š„) 1
šæ3(6šæ ā 12š„) 1
šæ2(ā2šæ + 6š„)ļæ½ šš
= šøš¼ļ潚æ
0
ā§āŖāŖāŖāŖāŖāŖāŖāØāŖāŖāŖāŖāŖāŖāŖā©
1šæ3(ā6šæ + 12š„)
1šæ2(ā4šæ + 6š„)
1šæ3(6šæ ā 12š„)
1šæ2(ā2šæ + 6š„)
ā«āŖāŖāŖāŖāŖāŖāŖā¬āŖāŖāŖāŖāŖāŖāŖā
ļæ½ 1šæ3(ā6šæ + 12š„) 1
šæ2(ā4šæ + 6š„) 1
šæ3(6šæ ā 12š„) 1
šæ2(ā2šæ + 6š„)ļæ½ šš„
= šøš¼ļ潚æ
0
āāāāāāāāāāāāāāāā
1šæ6(6šæ ā 12š„)2 1
šæ5(4šæ ā 6š„) (6šæ ā 12š„) ā 1
šæ6(6šæ ā 12š„)2 1
šæ5(2šæ ā 6š„) (6šæ ā 12š„)
1šæ5(4šæ ā 6š„) (6šæ ā 12š„) 1
šæ4(4šæ ā 6š„)2 ā 1
šæ5(4šæ ā 6š„) (6šæ ā 12š„) 1
šæ4(2šæ ā 6š„) (4šæ ā 6š„)
ā 1šæ6(6šæ ā 12š„)2 ā 1
šæ5(4šæ ā 6š„) (6šæ ā 12š„) 1
šæ6(6šæ ā 12š„)2 ā 1
šæ5(2šæ ā 6š„) (6šæ ā 12š„)
1šæ5(2šæ ā 6š„) (6šæ ā 12š„) 1
šæ4(2šæ ā 6š„) (4šæ ā 6š„) ā 1
šæ5(2šæ ā 6š„) (6šæ ā 12š„) 1
šæ4(2šæ ā 6š„)2
āāāāāāāāāāāāāāāā
šš„
= šøš¼
āāāāāāāāāāāāāāāā
12šæ3
6šæ2 ā 12
šæ36šæ2
6šæ2
4šæ ā 6
šæ22šæ
ā 12šæ3 ā 6
šæ212šæ3 ā 6
šæ26šæ2
2šæ ā 6
šæ24šæ
āāāāāāāāāāāāāāāā
=šøš¼šæ3
āāāāāāāāāāāāāāāā
12 6šæ ā12 6šæ6šæ 4šæ2 ā6šæ 2šæ2
ā12 ā6šæ 12 ā6šæ6šæ 2šæ2 ā6šæ 4šæ2
āāāāāāāāāāāāāāāā
Which is the stiļæ½ness matrix found earlier.
5.2 Potential energy (minimize a functional) method toderive the stiļæ½ness 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 respect tothe nodal displacements. The result gives the stiļæ½ness matrix.
Let the system total potential energy by called Ī and let the total internal energy in thesystem be š and let the work done by external loads acting on the nodes be Ī©, then
Ī = š āĪ©
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 by
40
5.2. Potential energy (minimize a . . . CHAPTER 5. FINDING THE . . .
12ā«š{š}š {š} šš and the work done by external loads is {š}š ļ潚ļæ½, hence
Ī =12 ļ潚
{š}š {š} šš ā {š}š ļ潚ļæ½ (1)
Now the rest follows as before. Assuming that displacement can be written as a function ofthe nodal displacements {š}, hence
š = [š] {š}
Since {š} = š {š} then {š} = š [š] {š} = [šµ] {š} where šµ is the strain displacement matrix[šµ] = š [š], hence
{š} = [šµ] {š}
Now from the stress-strain relation {š} = [šø] {š}, hence
{š}š = {š}š [šµ]š [šø] (4)
Substituting Eqs. (2,3,4) into (1) results in
Ī =12 ļ潚
{š}š [šµ]š [šø] [šµ] {š} šš ā {š}š ļ潚ļæ½
Setting šĪ š{š} = 0 gives
0 = {š}ļ潚[šµ]š [šø] [šµ] šš ā ļ潚ļæ½
Which is on the form š = [š¾]š· which means that
[š¾] = ļ潚[šµ]š [šø] [šµ] šš
As was found by the virtual work method.
41
5.2. Potential energy (minimize a . . . CHAPTER 5. FINDING THE . . .
42
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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