P. Sci. Chapter 11 Motion & Forces. Motion when something changes position.
Engineering reportused for external forces. Moments are always positive when anti-clockwise...
Transcript of Engineering reportused for external forces. Moments are always positive when anti-clockwise...
Engineering report
Derivation of sti๏ฟฝness matrix for a beam
Nasser M. Abbasi
July 7, 2016
Contents
1 Introduction 1
2 Direct method 32.1 Examples using the direct beam sti๏ฟฝness 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 sti๏ฟฝness matrix using methods other than direct method 375.1 Virtual work method for derivation of the sti๏ฟฝness matrix . . . . . . . . . . . 385.2 Potential energy (minimize a functional) method to derive the sti๏ฟฝness 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 sti๏ฟฝness method and then using finite elements method by addingmore elements. The problem is solved first by finding the sti๏ฟฝness 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 sti๏ฟฝness 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 ๐ฆ direction.
The two nodes are numbered 1 and 2 from left to right. ๐1 is the moment at the left node(node 1), ๐2 is the moment at the right node (node 2). ๐1 is the vertical force at the leftnode and ๐2 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 ๐ฅcoordinate. The displacement field in the vertical direction is called ๐ฃ (๐ฅ). This is the verticaldisplacement of point ๐ฅ on the beam from the original ๐ฅโ๐๐ฅ๐๐ . The following diagram showsthe notation used for the coordinates.
x,u
y,v
Angular displacement at distance ๐ฅ on the beam is found using ๐ (๐ฅ) = ๐๐ฃ(๐ฅ)๐๐ฅ . At the left node,
the degrees of freedom or the displacements, are called ๐ฃ1, ๐1 and at the right node they arecalled ๐ฃ2, ๐2. At an arbitrary location ๐ฅ in the beam, the vertical displacement is ๐ฃ (๐ฅ) andthe rotation at that location is ๐ (๐ฅ).
The following diagram shows the displacement field ๐ฃ (๐ฅ)
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 sti๏ฟฝness matrix, the forces at the ends of the beam arefound directly by the use of beam theory. In beam theory the signs are di๏ฟฝerent from whatis given in the first diagram above. Therefore, the moment and shear forces obtained usingbeam theory (๐๐ต and ๐๐ต in the diagram below) will have di๏ฟฝerent signs when comparedto the external forces. The signs are then adjusted to reflect the convention as shown in thediagram above using ๐ and ๐.
For an example, the external moment ๐1 is opposite in sign to ๐๐ต1 and the reaction ๐2is opposite to ๐๐ต2. 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 ๐๐ and ๐ ๐ in the above diagram as
5
CHAPTER 2. DIRECT METHOD
function of the displacements at the nodes {๐} = {๐ฃ1, ๐1, ๐ฃ2, ๐2}๐.
In other words, the goal is to obtain an expression of the form ๏ฟฝ๐๏ฟฝ = [๐พ] {๐} where [๐พ] is thesti๏ฟฝness matrix where ๏ฟฝ๐๏ฟฝ = {๐1,๐1, ๐ 2,๐2}
๐ is the nodal forces or load vector, and {๐} is thenodal displacement vector.
In this case [๐พ] will be a 4 ร 4 matrix and ๏ฟฝ๐๏ฟฝ is a 4 ร 1 vector and {๐} is a 4 ร 1 vector.
Starting with ๐1. It is in the same direction as the shear force ๐๐ต1. Since ๐๐ต1 =๐๐๐ต1๐๐ฅ then
๐1 =๐๐๐ต1๐๐ฅ
Since from beam theory ๐๐ต1 = โ๐ (๐ฅ) ๐ผ๐ฆ , the above becomes
๐1 = โ๐ผ๐ฆ๐๐ (๐ฅ)๐๐ฅ
But ๐ (๐ฅ) = ๐ธ๐ (๐ฅ) and ๐ (๐ฅ) = โ๐ฆ๐ where ๐ is radius of curvature, therefore the above becomes
๐1 = ๐ธ๐ผ๐๐๐ฅ ๏ฟฝ
1๐๏ฟฝ
Since 1๐ =
๏ฟฝ๐2๐ข๐๐ฅ2 ๏ฟฝ
๏ฟฝ1+๏ฟฝ๐๐ข๐๐ฅ ๏ฟฝ
2๏ฟฝ3/2 and for a small angle of deflection ๐๐ข
๐๐ฅ โช 1 then 1๐ = ๏ฟฝ๐
2๐ข๐๐ฅ2
๏ฟฝ, and the above
now becomes
๐1 = ๐ธ๐ผ๐3๐ข (๐ฅ)๐๐ฅ3
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 ๐1 is obtained. ๐1 is in the opposite sense of the bending moment ๐๐ต1 hence anegative sign is added giving ๐1 = โ๐๐ต1. But ๐๐ต1 = โ๐ (๐ฅ) ๐ผ
๐ฆ therefore
๐1 = ๐ (๐ฅ)๐ผ๐ฆ
= ๐ธ๐ (๐ฅ)๐ผ๐ฆ
= ๐ธ ๏ฟฝโ๐ฆ๐ ๏ฟฝ
๐ผ๐ฆ
= โ๐ธ๐ผ ๏ฟฝ1๐๏ฟฝ
= โ๐ธ๐ผ๐2๐ค๐๐ฅ2
๐2 is now found. It is in the opposite sense of the shear force ๐๐ต2, hence a negative sign isadded giving
๐2 = โ๐๐ต2
= โ๐๐๐ต2๐๐ฅ
Since ๐๐ต2 = โ๐ (๐ฅ) ๐ผ๐ฆ , the above becomes
๐2 =๐ผ๐ฆ๐๐ (๐ฅ)๐๐ฅ
But ๐ (๐ฅ) = ๐ธ๐ (๐ฅ) and ๐ (๐ฅ) = โ๐ฆ๐ where ๐ is radius of curvature. The above becomes
๐2 = โ๐ธ๐ผ๐๐๐ฅ ๏ฟฝ
1๐๏ฟฝ
But 1๐ =
๏ฟฝ๐2๐ค๐๐ฅ2 ๏ฟฝ
๏ฟฝ1+๏ฟฝ๐๐ค๐๐ฅ ๏ฟฝ
2๏ฟฝ3/2 and for small angle of deflection ๐๐ค
๐๐ฅ โช 1 hence 1๐ = ๏ฟฝ๐
2๐ข๐๐ฅ2
๏ฟฝ then the above
becomes
๐2 = โ๐ธ๐ผ๐3๐ข (๐ฅ)๐๐ฅ3
Finally ๐2 is in the same direction as ๐๐ต2 so no sign change is needed. ๐๐ต2 = โ๐ (๐ฅ) ๐ผ๐ฆ ,
7
CHAPTER 2. DIRECT METHOD
therefore
๐2 = โ๐ (๐ฅ)๐ผ๐ฆ
= โ๐ธ๐ (๐ฅ)๐ผ๐ฆ
= โ๐ธ ๏ฟฝโ๐ฆ๐ ๏ฟฝ
๐ผ๐ฆ
= ๐ธ๐ผ ๏ฟฝ1๐๏ฟฝ
= ๐ธ๐ผ๐2๐ข๐๐ฅ2
The following is a summary of what was found so far. Notice that the above expressions areevaluates at ๐ฅ = 0 and at ๐ฅ = ๐ฟ. Accordingly to obtain the nodal end forces vector ๏ฟฝ๐๏ฟฝ
๏ฟฝ๐๏ฟฝ =
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐1
๐1
๐2
๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=
โงโชโชโชโชโชโชโชโชโชโชโจโชโชโชโชโชโชโชโชโชโชโฉ
๐ธ๐ผ ๐3๐ข(๐ฅ)๐๐ฅ3 ๏ฟฝ
๐ฅ=0
โ๐ธ๐ผ ๐2๐ข๐๐ฅ2 ๏ฟฝ๐ฅ=0
โ๐ธ๐ผ ๐3๐ข(๐ฅ)๐๐ฅ3 ๏ฟฝ
๐ฅ=๐ฟ
๐ธ๐ผ ๐2๐ข๐๐ฅ2 ๏ฟฝ๐ฅ=๐ฟ
โซโชโชโชโชโชโชโชโชโชโชโฌโชโชโชโชโชโชโชโชโชโชโญ
(1)
The RHS of the above is now expressed as function of the nodal displacements ๐ฃ1, ๐1, ๐ฃ2, ๐2.To do that, the field displacement ๐ฃ (๐ฅ) which is the transverse displacement of the beam isassumed to be a polynomial in ๐ฅ of degree 3 or
๐ฃ (๐ฅ) = ๐0 + ๐1๐ฅ + ๐2๐ฅ2 + ๐3๐ฅ3
๐ (๐ฅ) =๐๐ฃ (๐ฅ)๐๐ฅ
= ๐1 + 2๐2๐ฅ + 3๐3๐ฅ2 (A)
Polynomial of degree 3 is used since there are 4 degrees of freedom, and a minimum of 4free parameters is needed. Hence
๐ฃ1 = ๐ฃ (๐ฅ)|๐ฅ=0 = ๐0 (2)
And
๐1 = ๐ (๐ฅ)|๐ฅ=0 = ๐1 (3)
Assuming the length of the beam is ๐ฟ, then
๐ฃ2 = ๐ฃ (๐ฅ)|๐ฅ=๐ฟ = ๐0 + ๐1๐ฟ + ๐2๐ฟ2 + ๐3๐ฟ3 (4)
And
๐2 = ๐ (๐ฅ)|๐ฅ=๐ฟ = ๐1 + 2๐2๐ฟ + 3๐3๐ฟ2 (5)
8
CHAPTER 2. DIRECT METHOD
Equations (2-5) gives
{๐} =
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐0๐1
๐0 + ๐1๐ฟ + ๐2๐ฟ2 + ๐3๐ฟ3
๐1 + 2๐2๐ฟ + 3๐3๐ฟ2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=
โโโโโโโโโโโโโโโ
1 0 0 00 1 0 01 ๐ฟ ๐ฟ2 ๐ฟ3
0 1 2๐ฟ 3๐ฟ2
โโโโโโโโโโโโโโโ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐0๐1๐2๐3
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
Solving for ๐๐ givesโงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐0๐1๐2๐3
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=
โโโโโโโโโโโโโโโโ
1 0 0 00 1 0 0
โ 3๐ฟ2 โ 2
๐ฟ3๐ฟ2 โ 1
๐ฟ2๐ฟ3
1๐ฟ2 โ 2
๐ฟ31๐ฟ2
โโโโโโโโโโโโโโโโ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=
โโโโโโโโโโโโโโโโ
๐ฃ1๐1
3๐ฟ2๐ฃ2 โ
1๐ฟ๐2 โ
3๐ฟ2๐ฃ1 โ
2๐ฟ๐1
1๐ฟ2๐1 +
1๐ฟ2๐2 +
2๐ฟ3๐ฃ1 โ
2๐ฟ3๐ฃ2
โโโโโโโโโโโโโโโโ
๐ฃ (๐ฅ), the field displacement function from Eq. (A) can now be written as a function of thenodal displacements
๐ฃ (๐ฅ) = ๐0 + ๐1๐ฅ + ๐2๐ฅ2 + ๐3๐ฅ3
= ๐ฃ1 + ๐1๐ฅ + ๏ฟฝ3๐ฟ2
๐ฃ2 โ1๐ฟ๐2 โ
3๐ฟ2
๐ฃ1 โ2๐ฟ๐1๏ฟฝ ๐ฅ2 + ๏ฟฝ
1๐ฟ2
๐1 +1๐ฟ2
๐2 +2๐ฟ3
๐ฃ1 โ2๐ฟ3
๐ฃ2๏ฟฝ ๐ฅ3
= ๐ฃ1 + ๐ฅ๐1 โ 2๐ฅ2
๐ฟ๐1 +
๐ฅ3
๐ฟ2๐1 โ
๐ฅ2
๐ฟ๐2 +
๐ฅ3
๐ฟ2๐2 โ 3
๐ฅ2
๐ฟ2๐ฃ1 + 2
๐ฅ3
๐ฟ3๐ฃ1 + 3
๐ฅ2
๐ฟ2๐ฃ2 โ 2
๐ฅ3
๐ฟ3๐ฃ2
Or in matrix form
๐ฃ (๐ฅ) = ๏ฟฝ1 โ 3 ๐ฅ2
๐ฟ2 + 2 ๐ฅ3
๐ฟ3 ๐ฅ โ 2๐ฅ2
๐ฟ + ๐ฅ3
๐ฟ2 3 ๐ฅ2
๐ฟ2 โ 2 ๐ฅ3
๐ฟ3 โ๐ฅ2
๐ฟ + ๐ฅ3
๐ฟ2๏ฟฝ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
= ๏ฟฝ 1๐ฟ3๏ฟฝ๐ฟ3 โ 3๐ฟ๐ฅ2 + 2๐ฅ3๏ฟฝ 1
๐ฟ2๏ฟฝ๐ฟ2๐ฅ โ 2๐ฟ๐ฅ2 + ๐ฅ3๏ฟฝ 1
๐ฟ3๏ฟฝ3๐ฟ๐ฅ2 โ 2๐ฅ3๏ฟฝ 1
๐ฟ2๏ฟฝโ๐ฟ๐ฅ2 + ๐ฅ3๏ฟฝ๏ฟฝ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
The above can be written as
๐ฃ (๐ฅ) = ๏ฟฝ๐1 (๐ฅ) ๐2 (๐ฅ) ๐3 (๐ฅ) ๐4 (๐ฅ)๏ฟฝ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
(5A)
๐ฃ (๐ฅ) = [๐] {๐}
9
CHAPTER 2. DIRECT METHOD
Where ๐๐ are called the shape functions. The shape functions areโโโโโโโโโโโโโโโ
๐1 (๐ฅ)๐2 (๐ฅ)๐3 (๐ฅ)๐4 (๐ฅ)
โโโโโโโโโโโโโโโ
=
โโโโโโโโโโโโโโโโ
1๐ฟ3๏ฟฝ๐ฟ3 โ 3๐ฟ๐ฅ2 + 2๐ฅ3๏ฟฝ
1๐ฟ2๏ฟฝ๐ฟ2๐ฅ โ 2๐ฟ๐ฅ2 + ๐ฅ3๏ฟฝ1๐ฟ3๏ฟฝ3๐ฟ๐ฅ2 โ 2๐ฅ3๏ฟฝ
1๐ฟ2๏ฟฝโ๐ฟ๐ฅ2 + ๐ฅ3๏ฟฝ
โโโโโโโโโโโโโโโโ
We notice that ๐1 (0) = 1 and ๐1 (๐ฟ) = 0 as expected. Also๐๐2 (๐ฅ)๐๐ฅ
๏ฟฝ๐ฅ=0
=1๐ฟ2
๏ฟฝ๐ฟ2 โ 4๐ฟ๐ฅ + 3๐ฅ2๏ฟฝ๏ฟฝ๐ฅ=0
= 1
And๐๐2 (๐ฅ)๐๐ฅ
๏ฟฝ๐ฅ=๐ฟ
=1๐ฟ2
๏ฟฝ๐ฟ2 โ 4๐ฟ๐ฅ + 3๐ฅ2๏ฟฝ๏ฟฝ๐ฅ=๐ฟ
= 0
Also ๐3 (0) = 0 and ๐3 (๐ฟ) = 1 and๐๐4 (๐ฅ)๐๐ฅ
๏ฟฝ๐ฅ=0
=1๐ฟ2
๏ฟฝโ2๐ฟ๐ฅ + 3๐ฅ2๏ฟฝ๏ฟฝ๐ฅ=0
= 0
and๐๐4 (๐ฅ)๐๐ฅ
๏ฟฝ๐ฅ=๐ฟ
=1๐ฟ2
๏ฟฝโ2๐ฟ๐ฅ + 3๐ฅ2๏ฟฝ๏ฟฝ๐ฅ=๐ฟ
= 1
The shape functions have thus been verified. The sti๏ฟฝness matrix is now found by substitutingEq. (5A) into Eq. (1), repeated below
๏ฟฝ๐๏ฟฝ =
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐1
๐1
๐2
๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=
โโโโโโโโโโโโโโโโโโโโโโโโ
๐ธ๐ผ ๐3๐ฃ(๐ฅ)๐๐ฅ3 ๏ฟฝ
๐ฅ=0
โ๐ธ๐ผ ๐2๐ฃ๐๐ฅ2 ๏ฟฝ๐ฅ=0
โ๐ธ๐ผ ๐3๐ฃ(๐ฅ)๐๐ฅ3 ๏ฟฝ
๐ฅ=๐ฟ
๐ธ๐ผ ๐2๐ฃ๐๐ฅ2 ๏ฟฝ๐ฅ=๐ฟ
โโโโโโโโโโโโโโโโโโโโโโโโ
Hence
๏ฟฝ๐๏ฟฝ =
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐1
๐1
๐2
๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=
โโโโโโโโโโโโโโโโโโ
๐ธ๐ผ ๐3
๐๐ฅ3(๐1๐ฃ1 + ๐2๐1 + ๐3๐ฃ2 + ๐4๐2)
โ๐ธ๐ผ ๐2
๐๐ฅ2(๐1๐ฃ1 + ๐2๐1 + ๐3๐ฃ2 + ๐4๐2)
โ๐ธ๐ผ ๐3
๐๐ฅ3(๐1๐ฃ1 + ๐2๐1 + ๐3๐ฃ2 + ๐4๐2)
๐ธ๐ผ ๐2
๐๐ฅ2(๐1๐ฃ1 + ๐2๐1 + ๐3๐ฃ2 + ๐4๐2)
โโโโโโโโโโโโโโโโโโ
(6)
But๐3
๐๐ฅ3๐1 (๐ฅ) =
1๐ฟ3
๐3
๐๐ฅ3๏ฟฝ๐ฟ3 โ 3๐ฟ๐ฅ2 + 2๐ฅ3๏ฟฝ
=12๐ฟ3
10
CHAPTER 2. DIRECT METHOD
And๐3
๐๐ฅ3๐2 (๐ฅ) =
1๐ฟ2
๐3
๐๐ฅ3๏ฟฝ๐ฟ2๐ฅ โ 2๐ฟ๐ฅ2 + ๐ฅ3๏ฟฝ
=6๐ฟ2
And๐3
๐๐ฅ3๐3 (๐ฅ) =
1๐ฟ3
๐3
๐๐ฅ3๏ฟฝ3๐ฟ๐ฅ2 โ 2๐ฅ3๏ฟฝ
=โ12๐ฟ3
And๐3
๐๐ฅ3๐4 (๐ฅ) =
1๐ฟ2
๐3
๐๐ฅ3๏ฟฝโ๐ฟ๐ฅ2 + ๐ฅ3๏ฟฝ
=6๐ฟ2
For the second derivatives๐2
๐๐ฅ2๐1 (๐ฅ) =
1๐ฟ3
๐2
๐๐ฅ2๏ฟฝ๐ฟ3 โ 3๐ฟ๐ฅ2 + 2๐ฅ3๏ฟฝ
=1๐ฟ3
(12๐ฅ โ 6๐ฟ)
And๐2
๐๐ฅ2๐2 (๐ฅ) =
1๐ฟ2
๐2
๐๐ฅ2๏ฟฝ๐ฟ2๐ฅ โ 2๐ฟ๐ฅ2 + ๐ฅ3๏ฟฝ
=1๐ฟ2
(6๐ฅ โ 4๐ฟ)
And๐2
๐๐ฅ2๐3 (๐ฅ) =
1๐ฟ3
๐2
๐๐ฅ2๏ฟฝ3๐ฟ๐ฅ2 โ 2๐ฅ3๏ฟฝ
=1๐ฟ3
(6๐ฟ โ 12๐ฅ)
And๐2
๐๐ฅ2๐4 (๐ฅ) =
1๐ฟ2
๐2
๐๐ฅ2๏ฟฝโ๐ฟ๐ฅ2 + ๐ฅ3๏ฟฝ
=1๐ฟ2
(6๐ฅ โ 2๐ฟ)
11
CHAPTER 2. DIRECT METHOD
Hence Eq. (6) becomes
๏ฟฝ๐๏ฟฝ =
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐1
๐1
๐2
๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=
โโโโโโโโโโโโโโโโโโโโโโโโ
๐ธ๐ผ ๐3
๐๐ฅ3(๐1๐ฃ1 + ๐2๐1 + ๐3๐ฃ2 + ๐4๐2)๏ฟฝ
๐ฅ=0
โ๐ธ๐ผ ๐2
๐๐ฅ2(๐1๐ฃ1 + ๐2๐1 + ๐3๐ฃ2 + ๐4๐2)๏ฟฝ
๐ฅ=0
โ๐ธ๐ผ ๐3
๐๐ฅ3(๐1๐ฃ1 + ๐2๐1 + ๐3๐ฃ2 + ๐4๐2)๏ฟฝ
๐ฅ=๐ฟ
๐ธ๐ผ ๐2
๐๐ฅ2(๐1๐ฃ1 + ๐2๐1 + ๐3๐ฃ2 + ๐4๐2)๏ฟฝ
๐ฅ=๐ฟ
โโโโโโโโโโโโโโโโโโโโโโโโ
=
โโโโโโโโโโโโโโโโโโโโโโโโโ
๐ธ๐ผ ๏ฟฝ 12๐ฟ3๐ฃ1 +6๐ฟ2๐1 โ
12๐ฟ3๐ฃ2 +
6๐ฟ2๐2๏ฟฝ
๐ฅ=0
โ๐ธ๐ผ ๏ฟฝ 1๐ฟ3(12๐ฅ โ 6๐ฟ) ๐ฃ1 +
1๐ฟ2(6๐ฅ โ 4๐ฟ) ๐1 +
1๐ฟ3(6๐ฟ โ 12๐ฅ) ๐ฃ2 +
1๐ฟ2(6๐ฅ โ 2๐ฟ) ๐2๏ฟฝ
๐ฅ=0
โ๐ธ๐ผ ๏ฟฝ 12๐ฟ3๐ฃ1 +6๐ฟ2๐1 โ
12๐ฟ3๐ฃ2 +
6๐ฟ2๐2๏ฟฝ
๐ฅ=๐ฟ
๐ธ๐ผ ๏ฟฝ 1๐ฟ3(12๐ฅ โ 6๐ฟ) ๐ฃ1 +
1๐ฟ2(6๐ฅ โ 4๐ฟ) ๐1 +
1๐ฟ3(6๐ฟ โ 12๐ฅ) ๐ฃ2 +
1๐ฟ2(6๐ฅ โ 2๐ฟ) ๐2๏ฟฝ
๐ฅ=๐ฟ
โโโโโโโโโโโโโโโโโโโโโโโโโ
=๐ธ๐ผ๐ฟ3
โโโโโโโโโโโโโโโ
12 6๐ฟ โ12 6๐ฟโ (12๐ฅ โ 6๐ฟ)๐ฅ=0 โ๐ฟ (6๐ฅ โ 4๐ฟ)๐ฅ=0 โ (6๐ฟ โ 12๐ฅ)๐ฅ=0 โ๐ฟ (6๐ฅ โ 2๐ฟ)๐ฅ=0
โ12 โ6๐ฟ 12 โ6๐ฟ(12๐ฅ โ 6๐ฟ)๐ฅ=๐ฟ ๐ฟ (6๐ฅ โ 4๐ฟ)๐ฅ=๐ฟ (6๐ฟ โ 12๐ฅ)๐ฅ=๐ฟ ๐ฟ (6๐ฅ โ 2๐ฟ)๐ฅ=๐ฟ
โโโโโโโโโโโโโโโ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
(7)
Or in matrix form, after evaluating the expressions above for ๐ฅ = ๐ฟ and ๐ฅ = 0 asโงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐1
๐1
๐2
๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=๐ธ๐ผ๐ฟ3
โโโโโโโโโโโโโโโโ
12 6๐ฟ โ12 6๐ฟ6๐ฟ 4๐ฟ2 โ6๐ฟ 2๐ฟ2
โ 12๐ฟ3 โ 6
๐ฟ212๐ฟ3 โ 6
๐ฟ2
(12๐ฟ โ 6๐ฟ) ๐ฟ (6๐ฟ โ 4๐ฟ) (6๐ฟ โ 12๐ฟ) ๐ฟ (6๐ฟ โ 2๐ฟ)
โโโโโโโโโโโโโโโโ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=๐ธ๐ผ๐ฟ3
โโโโโโโโโโโโโโโ
12 6๐ฟ โ12 6๐ฟ6๐ฟ 4๐ฟ2 โ6๐ฟ 2๐ฟ2
โ12 โ6๐ฟ 12 โ6๐ฟ6๐ฟ 2๐ฟ2 โ6๐ฟ 4๐ฟ2
โโโโโโโโโโโโโโโ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
The above now is in the form
๏ฟฝ๐๏ฟฝ = [๐พ] {๐}
Hence the sti๏ฟฝness matrix is
[๐พ] =๐ธ๐ผ๐ฟ3
โโโโโโโโโโโโโโโ
12 6๐ฟ โ12 6๐ฟ6๐ฟ 4๐ฟ2 โ6๐ฟ 2๐ฟ2
โ12 โ6๐ฟ 12 โ6๐ฟ6๐ฟ 2๐ฟ2 โ6๐ฟ 4๐ฟ2
โโโโโโโโโโโโโโโ
Knowing the sti๏ฟฝness matrix means knowing the nodal displacements {๐} when given theforces at the nodes. The power of the finite element method now comes after all the nodaldisplacements ๐ฃ1, ๐1, ๐ฃ2, ๐2 are calculated by solving ๏ฟฝ๐๏ฟฝ = [๐พ] {๐}. This is because the polyno-
12
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
mial ๐ฃ (๐ฅ) is now completely determined and hence ๐ฃ (๐ฅ) and ๐ (๐ฅ) can now be evaluated forany ๐ฅ along the beam and not just at its end nodes as the case with finite di๏ฟฝerence method.Eq. 5A above can now be used to find the displacement ๐ฃ (๐ฅ) and ๐ (๐ฅ) everywhere.
๐ฃ (๐ฅ) = [๐] {๐}
๐ฃ (๐ฅ) = ๏ฟฝ๐1 (๐ฅ) ๐2 (๐ฅ) ๐3 (๐ฅ) ๐4 (๐ฅ)๏ฟฝ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
To summarise, these are the steps to obtain ๐ฃ (๐ฅ)
1. An expression for [๐พ] is found.
2. ๏ฟฝ๐๏ฟฝ = [๐พ] {๐} is solved for {๐}
3. ๐ฃ (๐ฅ) = [๐] {๐} is calculated by assuming ๐ฃ (๐ฅ) is a polynomial. This gives the displace-ment ๐ฃ (๐ฅ) to use to evaluate the transverse displacement anywhere on the beam andnot just at the end nodes.
4. ๐ (๐ฅ) = ๐๐ฃ(๐ฅ)๐๐ฅ = ๐
๐๐ฅ [๐] {๐} is obtained to evaluate the rotation of the beam any where andnot just at the end nodes.
5. The strain ๐ (๐ฅ) = โ๐ฆ [๐ต] {๐} is found, where [๐ต] is the gradient matrix [๐ต] = ๐2
๐๐ฅ2 [๐].
6. The stress from ๐ = ๐ธ๐ = โ๐ธ๐ฆ [๐ต] {๐} is found.
7. The bending moment diagram from ๐(๐ฅ) = ๐ธ๐ผ [๐ต] {๐} is found.
8. The shear force diagram from ๐ (๐ฅ) = ๐๐๐ฅ๐(๐ฅ) is found.
2.1 Examples using the direct beam sti๏ฟฝness matrix
The beam sti๏ฟฝness 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 ๐ฟ, with point load ๐ 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
๐ is the given force. ๐2 = 0 since there is no external moment at the right end. Hence๏ฟฝ๐๏ฟฝ = [๐พ] {๐} for this system is
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ ๐1
โ๐0
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=๐ธ๐ผ๐ฟ3
โโโโโโโโโโโโโโโ
12 6๐ฟ โ12 6๐ฟ6๐ฟ 4๐ฟ2 โ6๐ฟ 2๐ฟ2
โ12 โ6๐ฟ 12 โ6๐ฟ6๐ฟ 2๐ฟ2 โ6๐ฟ 4๐ฟ2
โโโโโโโโโโโโโโโ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
Now is an important step. The known end displacements from boundary conditions issubstituted into {๐}, 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 ๐ฃ1 = 0 and ๐1 = 0. Hence the only unknowns are ๐ฃ2 and ๐2.Therefore the first and the second rows and columns are removed, giving
โงโชโชโจโชโชโฉโ๐0
โซโชโชโฌโชโชโญ=
๐ธ๐ผ๐ฟ3
โโโโโโ12 โ6๐ฟโ6๐ฟ 4๐ฟ2
โโโโโโ
โงโชโชโจโชโชโฉ๐ฃ2๐2
โซโชโชโฌโชโชโญ
Now the above is solved for
โงโชโชโจโชโชโฉ๐ฃ2๐2
โซโชโชโฌโชโชโญ. Let ๐ธ = 30 ร 106 psi (steel), ๐ผ = 57 ๐๐4, ๐ฟ = 144 ๐๐, and
๐ = 400 lb, hence๏ฟฝ ๏ฟฝ1 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)๏ฟฝ ๏ฟฝ
1Instead of removing rows/columns for known boundary conditions, we can also just put a 1 on the diagonalof the sti๏ฟฝness 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
ห
which givesx =-0.2324-0.0024
Therefore the vertical displacement at the right end is ๐ฃ2 = 0.2324 inches (downwards) and๐2 = โ0.0024 radians. Now that all nodal displacements are found, the field displacementfunction is completely determined.
{๐} =
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
00
โ0.2324โ0.0024
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
From Eq. 5A
๐ฃ (๐ฅ) = [๐] {๐}
= ๏ฟฝ๐1 (๐ฅ) ๐2 (๐ฅ) ๐3 (๐ฅ) ๐4 (๐ฅ)๏ฟฝ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
00
โ0.2324โ0.0024
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
= ๏ฟฝ 1๐ฟ3๏ฟฝ๐ฟ3 โ 3๐ฟ๐ฅ2 + 2๐ฅ3๏ฟฝ 1
๐ฟ2๏ฟฝ๐ฟ2๐ฅ โ 2๐ฟ๐ฅ2 + ๐ฅ3๏ฟฝ 1
๐ฟ3๏ฟฝ3๐ฟ๐ฅ2 โ 2๐ฅ3๏ฟฝ 1
๐ฟ2๏ฟฝโ๐ฟ๐ฅ2 + ๐ฅ3๏ฟฝ๏ฟฝ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
00
โ0.2324โ0.0024
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=0.002 4๐ฟ2
๏ฟฝ๐ฟ๐ฅ2 โ ๐ฅ3๏ฟฝ +0.232 4๐ฟ3
๏ฟฝ2๐ฅ3 โ 3๐ฟ๐ฅ2๏ฟฝ
But ๐ฟ = 144 inches, and the above becomes
๐ฃ (๐ฅ) = 3.992 0 ร 10โ8๐ฅ3 โ 1.695 6 ร 10โ5๐ฅ2
To verify, let ๐ฅ = 144 in the above
๐ฃ (๐ฅ = 144) = 3.992 0 ร 10โ81443 โ 1.695 6 ร 10โ51442
= โ0.232 40
The following is a plot of the deflection curve for the beam๏ฟฝ ๏ฟฝ1 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 grid๏ฟฝ ๏ฟฝ
15
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
ห
Now instead of removing rows/columns for the known boundary conditions, a 1 is put onthe diagonal. Starting again with
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ ๐1
โ๐0
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=๐ธ๐ผ๐ฟ3
โโโโโโโโโโโโโโโ
12 6๐ฟ โ12 6๐ฟ6๐ฟ 4๐ฟ2 โ6๐ฟ 2๐ฟ2
โ12 โ6๐ฟ 12 โ6๐ฟ6๐ฟ 2๐ฟ2 โ6๐ฟ 4๐ฟ2
โโโโโโโโโโโโโโโ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
Since ๐ฃ1 = 0 and ๐1 = 0, thenโงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
00โ๐0
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=๐ธ๐ผ๐ฟ3
โโโโโโโโโโโโโโโ
1 0 0 00 1 0 00 0 12 โ6๐ฟ0 0 โ6๐ฟ 4๐ฟ2
โโโโโโโโโโโโโโโ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
00๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
The above system is now solved as before. ๐ธ = 30 ร 106 psi, ๐ผ = 57 ๐๐4, ๐ฟ = 144 in, ๐ = 400 lb๏ฟฝ ๏ฟฝ1 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 A๏ฟฝ ๏ฟฝ
หA =
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 4.7583๏ฟฝ ๏ฟฝ
1 sol=A\load %SOLVE๏ฟฝ ๏ฟฝหsol =00-0.2324-0.0024
The same solution is obtained as before, but without the need to remove rows/column fromthe sti๏ฟฝness 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 ๐ is now placed in the middle ofthe beam
L
P
In using sti๏ฟฝness method, all loads must be on the nodes. The vector ๏ฟฝ๐๏ฟฝ 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 sti๏ฟฝness equation is now written as
๏ฟฝ๐๏ฟฝ = [๐พ] {๐}โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ โ ๐/2๐1 โ ๐๐ฟ/8
โ๐/2๐๐ฟ/8
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=๐ธ๐ผ๐ฟ3
โโโโโโโโโโโโโโโ
12 6๐ฟ โ12 6๐ฟ6๐ฟ 4๐ฟ2 โ6๐ฟ 2๐ฟ2
โ12 โ6๐ฟ 12 โ6๐ฟ6๐ฟ 2๐ฟ2 โ6๐ฟ 4๐ฟ2
โโโโโโโโโโโโโโโ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
There is no need to determine ๐ and ๐1 at this point since these rows will be removed dueto boundary conditions ๐ฃ1 = 0 and ๐1 = 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 ๏ฟฝ๐๏ฟฝ = [๐พ] {๐}. Hence, after removing the first two rows andcolumns, the above system simplifies to
โงโชโชโจโชโชโฉโ๐/2๐๐ฟ/8
โซโชโชโฌโชโชโญ=
๐ธ๐ผ๐ฟ3
โโโโโโ12 โ6๐ฟโ6๐ฟ 4๐ฟ2
โโโโโโ
โงโชโชโจโชโชโฉ๐ฃ2๐2
โซโชโชโฌโชโชโญ
The above is now solved for
โงโชโชโจโชโชโฉ๐ฃ2๐2
โซโชโชโฌโชโชโญusing the same numerical values for ๐, ๐ธ, ๐ผ, ๐ฟ as in the first
example๏ฟฝ ๏ฟฝ1 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)\load๏ฟฝ ๏ฟฝ
หx =-0.072630472854641-0.000605253940455
Therefore
{๐} =
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
00
โ0.072630472854641โ0.000605253940455
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
This is enough to obtain ๐ฃ (๐ฅ) as before. Now the reactions ๐ and ๐1 can be determined ifneeded. Going back to the full ๏ฟฝ๐๏ฟฝ = [๐พ] {๐}, results in
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ โ ๐/2๐1 โ ๐๐ฟ/8
โ๐/2๐๐ฟ/8
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=๐ธ๐ผ๐ฟ3
โโโโโโโโโโโโโโโ
12 6๐ฟ โ12 6๐ฟ6๐ฟ 4๐ฟ2 โ6๐ฟ 2๐ฟ2
โ12 โ6๐ฟ 12 โ6๐ฟ6๐ฟ 2๐ฟ2 โ6๐ฟ 4๐ฟ2
โโโโโโโโโโโโโโโ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
00
โ0.072630472854641โ0.000605253940455
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=๐ธ๐ผ๐ฟ3
โโโโโโโโโโโโโโโ
0.871 57 โ 3.631 5 ร 10โ3๐ฟ0.435 78๐ฟ โ 1.210 5 ร 10โ3๐ฟ2
3.631 5 ร 10โ3๐ฟ โ 0.871 570.435 78๐ฟ โ 2.421 ร 10โ3๐ฟ2
โโโโโโโโโโโโโโโ
Hence the first equation becomes
๐ โ ๐/2 =๐ธ๐ผ๐ฟ3
๏ฟฝ0.871 57 โ 3.631 5 ร 10โ3๐ฟ๏ฟฝ
and since ๐ธ = 30 ร 106 psi (steel) and ๐ผ = 57 ๐๐4and ๐ฟ = 144 in and ๐ = 400 lb, then
๐ =๏ฟฝ30ร106๏ฟฝ57
1443๏ฟฝ0.871 57 โ 3.631 5 ร 10โ3 (144)๏ฟฝ + 400/2 . Therefore
๐ = 400 lb
and ๐1 โ ๐๐ฟ/8 = ๐ธ๐ผ๐ฟ3๏ฟฝ0.435 78๐ฟ โ 1.210 5 ร 10โ3๐ฟ2๏ฟฝ + ๐๐ฟ/8 , hence
๐1 = 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.
๐ฃ (๐ฅ) = [๐] {๐}
๐ฃ (๐ฅ) = ๏ฟฝ๐1 (๐ฅ) ๐2 (๐ฅ) ๐3 (๐ฅ) ๐4 (๐ฅ)๏ฟฝ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
= ๏ฟฝ๐1 (๐ฅ) ๐2 (๐ฅ) ๐3 (๐ฅ) ๐4 (๐ฅ)๏ฟฝ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
00
โ0.072630472854641โ0.000605253940455
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
= ๏ฟฝ 1๐ฟ3๏ฟฝ3๐ฟ๐ฅ2 โ 2๐ฅ3๏ฟฝ 1
๐ฟ2๏ฟฝโ๐ฟ๐ฅ2 + ๐ฅ3๏ฟฝ๏ฟฝ
โโโโโโโ0.072630472854641โ0.000605253940455
โโโโโโ
=6.052 5 ร 10โ4
๐ฟ2๏ฟฝ๐ฟ๐ฅ2 โ ๐ฅ3๏ฟฝ +
0.072 63๐ฟ3
๏ฟฝ2๐ฅ3 โ 3๐ฟ๐ฅ2๏ฟฝ
Since ๐ฟ = 144 inches, the above becomes
๐ฃ (๐ฅ) =6.052 5 ร 10โ4
(144)2๏ฟฝ(144) ๐ฅ2 โ ๐ฅ3๏ฟฝ +
0.072 63(144)3
๏ฟฝ2๐ฅ3 โ 3 (144) ๐ฅ2๏ฟฝ
= 1.945 9 ร 10โ8๐ฅ3 โ 6.304 7 ร 10โ6๐ฅ2
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 ๐ now at distance ๐ from the left end and at distance ๐ from the rightend, and a uniform distributed load of density ๐ 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: ๐ = 0.625๐ฟ, ๐ = 0.375๐ฟ, ๐ธ = 30 ร 106 psi (steel), ๐ผ = 57 ๐๐4, ๐ฟ =144 in, ๐ = 1000 lb, ๐ = 200 lb/in.
In the above, the left end reaction forces are shown as ๐ 1 and moment reaction as ๐1 andthe reaction at the right end as ๐ 2. Starting by finding equivalent loads for the point load๐ and equivalent loads for for the uniform distributed load ๐. All external loads must betransferred to the nodes for the sti๏ฟฝness 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 diagram๐ is the reaction moment and ๐ 1, ๐ 2 are the reaction forces)
Now that all loads are on the nodes, the sti๏ฟฝness equation is applied
๏ฟฝ๐๏ฟฝ = [๐พ] {๐}โงโชโชโชโชโชโชโชโชโจโชโชโชโชโชโชโชโชโฉ
๐ 1 โ๐๐2(๐ฟ+2๐)
๐ฟ3 โ ๐๐ฟ2
๐โ ๐๐๐2
๐ฟ2 โ ๐๐ฟ2
12
๐ 2 โ๐๐2(๐ฟ+2๐)
๐ฟ3 โ ๐๐ฟ2
๐๐2๐๐ฟ2 + ๐๐ฟ2
12
โซโชโชโชโชโชโชโชโชโฌโชโชโชโชโชโชโชโชโญ
=๐ธ๐ผ๐ฟ3
โโโโโโโโโโโโโโโ
12 6๐ฟ โ12 6๐ฟ6๐ฟ 4๐ฟ2 โ6๐ฟ 2๐ฟ2
โ12 โ6๐ฟ 12 โ6๐ฟ6๐ฟ 2๐ฟ2 โ6๐ฟ 4๐ฟ2
โโโโโโโโโโโโโโโ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
Boundary conditions are now applied. ๐ฃ1 = 0, ๐1 = 0,๐ฃ2 = 0. Therefore the first, second andthird rows/columns are removed giving
๐๐2๐๐ฟ2
+๐๐ฟ2
12=
๐ธ๐ผ๐ฟ3
4๐ฟ2๐2
Hence
๐2 = ๏ฟฝ๐๐2๐๐ฟ2
+๐๐ฟ2
12 ๏ฟฝ ๏ฟฝ๐ฟ4๐ธ๐ผ๏ฟฝ
Substituting numerical values for the above as given at the top of the problem results in
๐2 =โโโโโ(1000) (0.625 (144))2 (0.375 (144))
(144)2+(200) (144)2
12
โโโโโ
โโโโโโ
1444 ๏ฟฝ30 ร 106๏ฟฝ (57)
โโโโโโ
= 7.719 9 ร 10โ3๐๐๐
22
2.1. Examples using the direct beam . . . CHAPTER 2. DIRECT METHOD
Hence the field displacement ๐ข (๐ฅ) is now found
๐ฃ (๐ฅ) = [๐] {๐}
๐ฃ (๐ฅ) = ๏ฟฝ๐1 (๐ฅ) ๐2 (๐ฅ) ๐3 (๐ฅ) ๐4 (๐ฅ)๏ฟฝ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
= ๏ฟฝ๐1 (๐ฅ) ๐2 (๐ฅ) ๐3 (๐ฅ) ๐4 (๐ฅ)๏ฟฝ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
000
7.719 9 ร 10โ3
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=1๐ฟ2
๏ฟฝโ๐ฟ๐ฅ2 + ๐ฅ3๏ฟฝ ๏ฟฝ7.719 9 ร 10โ3๏ฟฝ
= 3.722 9 ร 10โ7๐ฅ3 โ 5.361 ร 10โ5๐ฅ2
And a plot of the deflection curve is๏ฟฝ ๏ฟฝ1 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 grid๏ฟฝ ๏ฟฝ
ห
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 ๐ฃ (๐ฅ) becomesmore accurate (By comparing the the result with the exact solution based on using the beam4th order di๏ฟฝerential 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 sti๏ฟฝness matrix for thewhole beam (including both elements) will be 6 by 6. For each element however, the samesti๏ฟฝness 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 sti๏ฟฝness matrix for each element is found and then the global sti๏ฟฝness matrix is assem-bled. Then ๏ฟฝ๐๐๐๐๐๐๐๏ฟฝ = ๏ฟฝ๐พ๐๐๐๐๐๐๏ฟฝ ๏ฟฝ๐๐๐๐๐๐๐๏ฟฝ 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 ๐ฟ becomes ๐ฟ/2. The following diagram show the equivalentloading for ๐
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 ๐ 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 . . .
๏ฟฝ๐๏ฟฝ = [๐พ] {๐} for each element is now constructed. Starting with the first elementโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
๐ 1 โ๐๏ฟฝ ๐ฟ2 ๏ฟฝ
2
๐1 โ๐๏ฟฝ ๐ฟ2 ๏ฟฝ
2
12
โ๐๐2๏ฟฝ ๐ฟ2+2๐๏ฟฝ
(๐ฟ/2)3โ
๐๏ฟฝ ๐ฟ2 ๏ฟฝ
2
๐๏ฟฝ ๐ฟ2 ๏ฟฝ2
12 โ๐๏ฟฝ ๐ฟ2 ๏ฟฝ
2
12 โ ๐๐๐2
๏ฟฝ ๐ฟ2 ๏ฟฝ2
โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
=๐ธ๐ผ
๏ฟฝ๐ฟ2๏ฟฝ3
โโโโโโโโโโโโโโโโโโโโโโโ
12 6 ๏ฟฝ๐ฟ2๏ฟฝ โ12 6 ๏ฟฝ๐ฟ2๏ฟฝ
6 ๏ฟฝ๐ฟ2๏ฟฝ 4 ๏ฟฝ๐ฟ2๏ฟฝ2
โ6 ๏ฟฝ๐ฟ2๏ฟฝ 2 ๏ฟฝ๐ฟ2๏ฟฝ2
โ12 โ6 ๏ฟฝ๐ฟ2๏ฟฝ 12 โ6 ๏ฟฝ๐ฟ2๏ฟฝ
6 ๏ฟฝ๐ฟ2๏ฟฝ 2 ๏ฟฝ๐ฟ2๏ฟฝ2
โ6 ๏ฟฝ๐ฟ2๏ฟฝ 4 ๏ฟฝ๐ฟ2๏ฟฝ2
โโโโโโโโโโโโโโโโโโโโโโโ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
And for the second elementโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
โ๐๐2๏ฟฝ ๐ฟ2+2๐๏ฟฝ
(๐ฟ/2)3โ
๐๏ฟฝ ๐ฟ2 ๏ฟฝ
2
๐๏ฟฝ ๐ฟ2 ๏ฟฝ2
12 โ๐๏ฟฝ ๐ฟ2 ๏ฟฝ
2
12 โ ๐๐๐2
๏ฟฝ ๐ฟ2 ๏ฟฝ2
๐ 3 โ๐๐2๏ฟฝ ๐ฟ2+2๐๏ฟฝ
๏ฟฝ ๐ฟ2 ๏ฟฝ3 โ
๐๏ฟฝ ๐ฟ2 ๏ฟฝ
2
๐๏ฟฝ ๐ฟ2 ๏ฟฝ2
12 + ๐๐2๐
๏ฟฝ ๐ฟ2 ๏ฟฝ2
โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
=๐ธ๐ผ
๏ฟฝ๐ฟ2๏ฟฝ3
โโโโโโโโโโโโโโโโโโโโโโโ
12 6 ๏ฟฝ๐ฟ2๏ฟฝ โ12 6 ๏ฟฝ๐ฟ2๏ฟฝ
6 ๏ฟฝ๐ฟ2๏ฟฝ 4 ๏ฟฝ๐ฟ2๏ฟฝ2
โ6 ๏ฟฝ๐ฟ2๏ฟฝ 2 ๏ฟฝ๐ฟ2๏ฟฝ2
โ12 โ6 ๏ฟฝ๐ฟ2๏ฟฝ 12 โ6 ๏ฟฝ๐ฟ2๏ฟฝ
6 ๏ฟฝ๐ฟ2๏ฟฝ 2 ๏ฟฝ๐ฟ2๏ฟฝ2
โ6 ๏ฟฝ๐ฟ2๏ฟฝ 4 ๏ฟฝ๐ฟ2๏ฟฝ2
โโโโโโโโโโโโโโโโโโโโโโโ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ2๐2
๐ฃ3๐3
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
The 2 systems above are assembled to obtain the global sti๏ฟฝness matrix equation givingโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
๐ 1 โ๐๏ฟฝ ๐ฟ2 ๏ฟฝ
2
๐1 โ๐๏ฟฝ ๐ฟ2 ๏ฟฝ
2
12
โ2๐๐2๏ฟฝ ๐ฟ2+2๐๏ฟฝ
๏ฟฝ ๐ฟ2 ๏ฟฝ3 โ 2
๐๏ฟฝ ๐ฟ2 ๏ฟฝ
2
โ2๐๐๐2
๏ฟฝ ๐ฟ2 ๏ฟฝ2
๐ 3 โ๐๐2๏ฟฝ ๐ฟ2+2๐๏ฟฝ
๏ฟฝ ๐ฟ2 ๏ฟฝ3 โ
๐๏ฟฝ ๐ฟ2 ๏ฟฝ
2
๐๏ฟฝ ๐ฟ2 ๏ฟฝ2
12 + ๐๐2๐
๏ฟฝ ๐ฟ2 ๏ฟฝ2
โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
=๐ธ๐ผ
๏ฟฝ๐ฟ2๏ฟฝ3
โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
12 6 ๏ฟฝ๐ฟ2๏ฟฝ โ12 6 ๏ฟฝ๐ฟ2๏ฟฝ 0 0
6 ๏ฟฝ๐ฟ2๏ฟฝ 4 ๏ฟฝ๐ฟ2๏ฟฝ2
โ6 ๏ฟฝ๐ฟ2๏ฟฝ 2 ๏ฟฝ๐ฟ2๏ฟฝ2
0 0
โ12 โ6 ๏ฟฝ๐ฟ2๏ฟฝ 12 + 12 โ6 ๏ฟฝ๐ฟ2๏ฟฝ + 6 ๏ฟฝ๐ฟ2๏ฟฝ โ12 6 ๏ฟฝ๐ฟ2๏ฟฝ
6 ๏ฟฝ๐ฟ2๏ฟฝ 2 ๏ฟฝ๐ฟ2๏ฟฝ2
โ6 ๏ฟฝ๐ฟ2๏ฟฝ + 6 ๏ฟฝ๐ฟ2๏ฟฝ 4 ๏ฟฝ๐ฟ2๏ฟฝ2+ 4 ๏ฟฝ๐ฟ2๏ฟฝ
2โ6 ๏ฟฝ๐ฟ2๏ฟฝ 2 ๏ฟฝ๐ฟ2๏ฟฝ
2
0 0 โ12 โ6 ๏ฟฝ๐ฟ2๏ฟฝ 12 โ6 ๏ฟฝ๐ฟ2๏ฟฝ
0 0 6 ๏ฟฝ๐ฟ2๏ฟฝ 2 ๏ฟฝ๐ฟ2๏ฟฝ2
โ6 ๏ฟฝ๐ฟ2๏ฟฝ 4 ๏ฟฝ๐ฟ2๏ฟฝ2
โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
โงโชโชโชโชโชโชโชโชโชโชโชโจโชโชโชโชโชโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
๐ฃ3๐3
โซโชโชโชโชโชโชโชโชโชโชโชโฌโชโชโชโชโชโชโชโชโชโชโชโญ
The boundary conditions are now applied giving ๐ฃ1 = 0, ๐1 = 0 since the first node is fixed,and ๐ฃ3 = 0. And putting 1 on the diagonal of the sti๏ฟฝness matrix corresponding to these
28
3.1. Example 3 redone with 2 elements CHAPTER 3. FINITE ELEMENTS OR . . .
known boundary conditions results inโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
00
โ2๐๐2๏ฟฝ ๐ฟ2+2๐๏ฟฝ
(๐ฟ/2)3โ 2๐(๐ฟ/2)2
โ2 ๐๐๐2
(๐ฟ/2)2
0๐๏ฟฝ ๐ฟ2 ๏ฟฝ
2
12 + ๐๐2๐
๏ฟฝ ๐ฟ2 ๏ฟฝ2
โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
=๐ธ๐ผ
๏ฟฝ๐ฟ2๏ฟฝ3
โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
1 0 0 0 0 00 1 0 0 0 0
0 0 24 0 0 6 ๏ฟฝ๐ฟ2๏ฟฝ
0 0 0 8 ๏ฟฝ๐ฟ2๏ฟฝ2
0 2 ๏ฟฝ๐ฟ2๏ฟฝ2
0 0 0 0 1 0
0 0 6 ๏ฟฝ๐ฟ2๏ฟฝ 2 ๏ฟฝ๐ฟ2๏ฟฝ2
0 4 ๏ฟฝ๐ฟ2๏ฟฝ2
โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
โโโโโโโโโโโโโโโโโโโโโโโโโ
00๐ฃ2๐2
0๐3
โโโโโโโโโโโโโโโโโโโโโโโโโ
As was mentioned earlier, another method would be to remove the rows/columns whichresults in โ
โโโโโโโโโโโโโโโโโโโโโโโโโ
โ2๐๐2๏ฟฝ ๐ฟ2+2๐๏ฟฝ
(๐ฟ/2)3โ 2
๐๏ฟฝ ๐ฟ2 ๏ฟฝ
2
โ2๐๐๐2
๏ฟฝ ๐ฟ2 ๏ฟฝ2
๐๏ฟฝ ๐ฟ2 ๏ฟฝ2
12 + ๐๐2๐
๏ฟฝ ๐ฟ2 ๏ฟฝ2
โโโโโโโโโโโโโโโโโโโโโโโโโโ
=๐ธ๐ผ
๏ฟฝ๐ฟ2๏ฟฝ3
โโโโโโโโโโโโโโโโโ
24 0 6 ๏ฟฝ๐ฟ2๏ฟฝ
0 8 ๏ฟฝ๐ฟ2๏ฟฝ2
2 ๏ฟฝ๐ฟ2๏ฟฝ2
6 ๏ฟฝ๐ฟ2๏ฟฝ 2 ๏ฟฝ๐ฟ2๏ฟฝ2
4 ๏ฟฝ๐ฟ2๏ฟฝ2
โโโโโโโโโโโโโโโโโ
โโโโโโโโโโ
๐ฃ2๐2
๐3
โโโโโโโโโโ
Giving the same solution. There are 3 unknowns to solve for. Once these unknowns aresolved for, ๐ฃ (๐ฅ) for the first element and for the second element are fully determined. Thefollowing code displays the deflection curve for the above beam๏ฟฝ ๏ฟฝ
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\load๏ฟฝ ๏ฟฝหA =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 ๐ฃ2 = โ0.2735 in (downwards displacement) and ๐2 = โ0.0019radians and ๐3 = 0.0076 radians. ๐ฃ (๐ฅ) polynomial is now found for each element
๐ฃ๐๐๐๐1 (๐ฅ) = ๏ฟฝ๐1 (๐ฅ) ๐2 (๐ฅ) ๐3 (๐ฅ) ๐4 (๐ฅ)๏ฟฝ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ1๐1
๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
๏ฟฝ๐1 (๐ฅ) ๐2 (๐ฅ) ๐3 (๐ฅ) ๐4 (๐ฅ)๏ฟฝ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
00๐ฃ2๐2
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
=โโโโโ
1
๏ฟฝ ๐ฟ2 ๏ฟฝ3 ๏ฟฝ3 ๏ฟฝ
๐ฟ2๏ฟฝ ๐ฅ2 โ 2๐ฅ3๏ฟฝ 1
๏ฟฝ ๐ฟ2 ๏ฟฝ2 ๏ฟฝโ ๏ฟฝ
๐ฟ2๏ฟฝ ๐ฅ2 + ๐ฅ3๏ฟฝ
โโโโโ
โโโโโโโ0.2735โ0.0019
โโโโโโ
=2.188๐ฟ3 ๏ฟฝ2๐ฅ3 โ
32๐ฟ๐ฅ2๏ฟฝ โ
0.007 6๐ฟ2 ๏ฟฝ๐ฅ3 โ
12๐ฟ๐ฅ2๏ฟฝ
The above polynomial is the transverse deflection of the beam for the region 0 โค ๐ฅ โค ๐ฟ/2.
30
3.1. Example 3 redone with 2 elements CHAPTER 3. FINITE ELEMENTS OR . . .
๐ฃ (๐ฅ) for the second element is found in similar way
๐ฃ๐๐๐๐2 (๐ฅ) = ๏ฟฝ๐1 (๐ฅ) ๐2 (๐ฅ) ๐3 (๐ฅ) ๐4 (๐ฅ)๏ฟฝ
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
๐ฃ2๐2
0๐3
โซโชโชโชโชโชโชโฌโชโชโชโชโชโชโญ
= ๏ฟฝ๐1 (๐ฅ) ๐2 (๐ฅ) ๐3 (๐ฅ) ๐4 (๐ฅ)๏ฟฝ
โโโโโโโโโโโโโโโ
โ0.2735โ0.0019
00.0076
โโโโโโโโโโโโโโโ
= ๏ฟฝ๐1 (๐ฅ) ๐2 (๐ฅ) ๐4 (๐ฅ)๏ฟฝ
โโโโโโโโโโ
โ0.2735โ0.00190.0076
โโโโโโโโโโ
=โโโโโโ
1
๏ฟฝ ๐ฟ2 ๏ฟฝ3 ๏ฟฝ๏ฟฝ
๐ฟ2๏ฟฝ3โ 3 ๏ฟฝ๐ฟ2๏ฟฝ ๐ฅ
2 + 2๐ฅ3๏ฟฝ1
๏ฟฝ ๐ฟ2 ๏ฟฝ2 ๏ฟฝ๏ฟฝ
๐ฟ2๏ฟฝ2๐ฅ โ 2 ๏ฟฝ๐ฟ2๏ฟฝ ๐ฅ
2 + ๐ฅ3๏ฟฝ1
๏ฟฝ ๐ฟ2 ๏ฟฝ2 ๏ฟฝโ ๏ฟฝ
๐ฟ2๏ฟฝ ๐ฅ2 + ๐ฅ3๏ฟฝ
โโโโโโ
โโโโโโโโโโ
โ0.2735โ0.00190.0076
โโโโโโโโโโ
Hence
๐ฃ๐๐๐๐2 (๐ฅ) =0.030 4๐ฟ2 ๏ฟฝ๐ฅ3 โ
12๐ฟ๐ฅ2๏ฟฝ โ
0.007 6๐ฟ2 ๏ฟฝ
14๐ฟ2๐ฅ โ ๐ฟ๐ฅ2 + ๐ฅ3๏ฟฝ โ
2.188๐ฟ3 ๏ฟฝ
18๐ฟ3 โ
32๐ฟ๐ฅ2 + 2๐ฅ3๏ฟฝ
Which is valid for ๐ฟ/2 โค ๐ฅ โค ๐ฟ. 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 di๏ฟฝerence 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 ๐ค lbs per unit length isgiven by
๐ฃ (๐ฅ) = โ๐ค๐ฅ2
48๐ธ๐ผ๏ฟฝ3๐ฟ2 โ 5๐ฟ๐ฅ + 2๐ฅ2๏ฟฝ
While the analytical deflection for the same beam but when there is a point load P at distance๐ from the left end is given by
๐ฃ (๐ฅ) = โ๐(โจ๐ฟ โ ๐โฉ3 (3๐ฟ โ ๐ฅ)๐ฅ2 + ๐ฟ2(3(๐ฟ โ ๐)(๐ฟ โ ๐ฅ)๐ฅ2 + 2๐ฟ โจ๐ฅ โ ๐โฉ3))
Therefore, the analytical expression for deflection is given by the sum of the above expres-sions, giving
๐ฃ (๐ฅ) = โ๐ค๐ฅ2
48๐ธ๐ผ๏ฟฝ3๐ฟ2 โ 5๐ฟ๐ฅ + 2๐ฅ2๏ฟฝ โ ๐(โจ๐ฟ โ ๐โฉ3 (3๐ฟ โ ๐ฅ)๐ฅ2 + ๐ฟ2(3(๐ฟ โ ๐)(๐ฟ โ ๐ฅ)๐ฅ2 + 2๐ฟ โจ๐ฅ โ ๐โฉ3))
Where โจ๐ฅ โ ๐โฉmeans it is zero when ๐ฅโ๐ is negative. In other words โจ๐ฅ โ ๐โฉ = (๐ฅ โ ๐)๐๐๐๐ก๐๐ก๐๐ (๐ฅ โ ๐)
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 function๐ฃ (๐ฅ) 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 ๐(๐ฅ) = โ๐ธ๐ผ๐2๐ฃ(๐ฅ)๐๐ฅ2 and shear
force is given by ๐ (๐ฅ) = ๐๐๐๐ฅ = โ๐ธ๐ผ๐
3๐ฃ(๐ฅ)๐๐ฅ3 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
๐ฃ (๐ฅ) =2.188๐ฟ3
๏ฟฝ2๐ฅ3 โ 32๐ฟ๐ฅ
2๏ฟฝ โ 0.007 6๐ฟ2
๏ฟฝ๐ฅ3 โ 12๐ฟ๐ฅ
2๏ฟฝ0.030 4๐ฟ2
๏ฟฝ๐ฅ3 โ 12๐ฟ๐ฅ
2๏ฟฝ โ 0.007 6๐ฟ2
๏ฟฝ14๐ฟ
2๐ฅ โ ๐ฟ๐ฅ2 + ๐ฅ3๏ฟฝ โ 2.188๐ฟ3
๏ฟฝ18๐ฟ
3 โ 32๐ฟ๐ฅ
2 + 2๐ฅ3๏ฟฝ
โซโชโชโชโฌโชโชโชโญ
0 โค ๐ฅ โค ๐ฟ/2๐ฟ/2 โค ๐ฅ โค ๐ฟ
Hence
๐(๐ฅ) = โ๐ธ๐ผโ0.0076(6๐ฅโ๐ฟ)
๐ฟ2 + 2.188(12๐ฅโ3๐ฟ)๐ฟ3
โ0.0076(6๐ฅโ2๐ฟ)๐ฟ2 + 0.0304(6๐ฅโ๐ฟ)
๐ฟ2 โ 2.188(12๐ฅโ3๐ฟ)๐ฟ3
โซโชโชโฌโชโชโญ
0 โค ๐ฅ โค ๐ฟ/2๐ฟ/2 โค ๐ฅ โค ๐ฟ
using ๐ธ = 30 ร 106 psi and ๐ผ = 57 in4 and ๐ฟ = 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 ๐ฃ (๐ฅ) obtained using the finite elements methodis a third degree polynomial and after di๏ฟฝerentiating twice to obtain the bending moment
(๐(๐ฅ) = โ๐ธ๐ผ๐2๐ฃ๐๐ฅ2 ) the result becomes a linear function in ๐ฅ while in the analytical solution
case, when the load is distributed, the solution ๐ฃ (๐ฅ) is a fourth degree polynomial. Hencethe bending moment will be quadratic function in ๐ฅ 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 sti๏ฟฝness matrix usingmethods other than direct method
Local contents5.1 Virtual work method for derivation of the sti๏ฟฝness matrix . . . . . . . . . . . . . 385.2 Potential energy (minimize a functional) method to derive the sti๏ฟฝness matrix . 40
37
5.1. Virtual work method for derivation of . . . CHAPTER 5. FINDING THE . . .
There are three main methods to obtain the sti๏ฟฝness matrix
1. Variational method (minimizing a functional). This functional is the potential energyof the structure and loads.
2. Weighted residual. Requires the di๏ฟฝerential 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 sti๏ฟฝnessmatrix
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 ๐ = {๐ข, ๐ฃ, ๐ค} and assuming the external loads are given by ๏ฟฝ๐๏ฟฝ acting on the nodes,
hence these point loads will do work given by {๐ฟ๐}๐ ๏ฟฝ๐๏ฟฝ on that unit volume where {๐} 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 {๐ฟ๐}๐ {๐} in that same unit volume.
This area is the Strain energy per
unit volume
Hence, for a unit volume
{๐ฟ๐}๐ ๏ฟฝ๐๏ฟฝ = {๐ฟ๐}๐ {๐}
And for the whole volume
{๐ฟ๐}๐ ๏ฟฝ๐๏ฟฝ = ๏ฟฝ๐{๐ฟ๐}๐ {๐} ๐๐ (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
๐ = [๐] {๐}
Therefore
๐ฟ๐ = [๐] {๐ฟ๐}
{๐ฟ๐}๐ = {๐ฟ๐}๐ [๐]๐ (2)
Since {๐} = ๐ {๐} then {๐} = ๐ [๐] {๐} = [๐ต] {๐} where ๐ต is the strain displacement matrix[๐ต] = ๐ [๐], hence
{๐} = [๐ต] {๐}{๐ฟ๐} = [๐ต] {๐ฟ๐}
{๐ฟ๐}๐ = {๐ฟ๐}๐ [๐ต]๐ (3)
Now from the stress-strain relation {๐} = [๐ธ] {๐}, hence
{๐} = [๐ธ] [๐ต] {๐} (4)
Substituting Eqs. (2,3,4) into (1) results in
{๐ฟ๐}๐ ๏ฟฝ๐๏ฟฝ โ๏ฟฝ๐{๐ฟ๐}๐ [๐ต]๐ [๐ธ] [๐ต] {๐} ๐๐ = 0
Since {๐ฟ๐} and {๐} do not depend on the integration variables they can be moved outsidethe integral, giving
{๐ฟ๐}๐ ๏ฟฝ๏ฟฝ๐๏ฟฝ โ {๐}๏ฟฝ๐[๐ต]๐ [๐ธ] [๐ต] ๐๐๏ฟฝ = 0
Since the above is true for any admissible ๐ฟ๐ then the only condition is that
๏ฟฝ๐๏ฟฝ = {๐}๏ฟฝ๐[๐ต]๐ [๐ธ] [๐ต] ๐๐
This is in the form ๐ = ๐พฮ, therefore
[๐พ] = ๏ฟฝ๐[๐ต]๐ [๐ธ] [๐ต] ๐๐
knowing [๐ต] allows finding [๐] by integrating over the volume. For the beam element though,
๐ = ๐ฃ (๐ฅ) the transverse displacement. This means [๐ต] = ๐2
๐๐ฅ2 [๐]. Recalling from the abovethat for the beam element,
[๐] = ๏ฟฝ 1๐ฟ3๏ฟฝ๐ฟ3 โ 3๐ฟ๐ฅ2 + 2๐ฅ3๏ฟฝ 1
๐ฟ2๏ฟฝ๐ฟ2๐ฅ โ 2๐ฟ๐ฅ2 + ๐ฅ3๏ฟฝ 1
๐ฟ3๏ฟฝ3๐ฟ๐ฅ2 โ 2๐ฅ3๏ฟฝ 1
๐ฟ2๏ฟฝโ๐ฟ๐ฅ2 + ๐ฅ3๏ฟฝ๏ฟฝ
Hence
[๐ต] =๐2
๐๐ฅ2[๐] = ๏ฟฝ 1
๐ฟ3(โ6๐ฟ + 12๐ฅ) 1
๐ฟ2(โ4๐ฟ + 6๐ฅ) 1
๐ฟ3(6๐ฟ โ 12๐ฅ) 1
๐ฟ2(โ2๐ฟ + 6๐ฅ)๏ฟฝ
39
5.2. Potential energy (minimize a . . . CHAPTER 5. FINDING THE . . .
Hence
[๐พ] = ๏ฟฝ๐[๐ต]๐ [๐ธ] [๐ต] ๐๐
= ๏ฟฝ๐
โงโชโชโชโชโชโชโชโจโชโชโชโชโชโชโชโฉ
1๐ฟ3(โ6๐ฟ + 12๐ฅ)
1๐ฟ2(โ4๐ฟ + 6๐ฅ)
1๐ฟ3(6๐ฟ โ 12๐ฅ)
1๐ฟ2(โ2๐ฟ + 6๐ฅ)
โซโชโชโชโชโชโชโชโฌโชโชโชโชโชโชโชโญ
๐ธ ๏ฟฝ 1๐ฟ3(โ6๐ฟ + 12๐ฅ) 1
๐ฟ2(โ4๐ฟ + 6๐ฅ) 1
๐ฟ3(6๐ฟ โ 12๐ฅ) 1
๐ฟ2(โ2๐ฟ + 6๐ฅ)๏ฟฝ ๐๐
= ๐ธ๐ผ๏ฟฝ๐ฟ
0
โงโชโชโชโชโชโชโชโจโชโชโชโชโชโชโชโฉ
1๐ฟ3(โ6๐ฟ + 12๐ฅ)
1๐ฟ2(โ4๐ฟ + 6๐ฅ)
1๐ฟ3(6๐ฟ โ 12๐ฅ)
1๐ฟ2(โ2๐ฟ + 6๐ฅ)
โซโชโชโชโชโชโชโชโฌโชโชโชโชโชโชโชโญ
๏ฟฝ 1๐ฟ3(โ6๐ฟ + 12๐ฅ) 1
๐ฟ2(โ4๐ฟ + 6๐ฅ) 1
๐ฟ3(6๐ฟ โ 12๐ฅ) 1
๐ฟ2(โ2๐ฟ + 6๐ฅ)๏ฟฝ ๐๐ฅ
= ๐ธ๐ผ๏ฟฝ๐ฟ
0
โโโโโโโโโโโโโโโโ
1๐ฟ6(6๐ฟ โ 12๐ฅ)2 1
๐ฟ5(4๐ฟ โ 6๐ฅ) (6๐ฟ โ 12๐ฅ) โ 1
๐ฟ6(6๐ฟ โ 12๐ฅ)2 1
๐ฟ5(2๐ฟ โ 6๐ฅ) (6๐ฟ โ 12๐ฅ)
1๐ฟ5(4๐ฟ โ 6๐ฅ) (6๐ฟ โ 12๐ฅ) 1
๐ฟ4(4๐ฟ โ 6๐ฅ)2 โ 1
๐ฟ5(4๐ฟ โ 6๐ฅ) (6๐ฟ โ 12๐ฅ) 1
๐ฟ4(2๐ฟ โ 6๐ฅ) (4๐ฟ โ 6๐ฅ)
โ 1๐ฟ6(6๐ฟ โ 12๐ฅ)2 โ 1
๐ฟ5(4๐ฟ โ 6๐ฅ) (6๐ฟ โ 12๐ฅ) 1
๐ฟ6(6๐ฟ โ 12๐ฅ)2 โ 1
๐ฟ5(2๐ฟ โ 6๐ฅ) (6๐ฟ โ 12๐ฅ)
1๐ฟ5(2๐ฟ โ 6๐ฅ) (6๐ฟ โ 12๐ฅ) 1
๐ฟ4(2๐ฟ โ 6๐ฅ) (4๐ฟ โ 6๐ฅ) โ 1
๐ฟ5(2๐ฟ โ 6๐ฅ) (6๐ฟ โ 12๐ฅ) 1
๐ฟ4(2๐ฟ โ 6๐ฅ)2
โโโโโโโโโโโโโโโโ
๐๐ฅ
= ๐ธ๐ผ
โโโโโโโโโโโโโโโโ
12๐ฟ3
6๐ฟ2 โ 12
๐ฟ36๐ฟ2
6๐ฟ2
4๐ฟ โ 6
๐ฟ22๐ฟ
โ 12๐ฟ3 โ 6
๐ฟ212๐ฟ3 โ 6
๐ฟ26๐ฟ2
2๐ฟ โ 6
๐ฟ24๐ฟ
โโโโโโโโโโโโโโโโ
=๐ธ๐ผ๐ฟ3
โโโโโโโโโโโโโโโโ
12 6๐ฟ โ12 6๐ฟ6๐ฟ 4๐ฟ2 โ6๐ฟ 2๐ฟ2
โ12 โ6๐ฟ 12 โ6๐ฟ6๐ฟ 2๐ฟ2 โ6๐ฟ 4๐ฟ2
โโโโโโโโโโโโโโโโ
Which is the sti๏ฟฝness matrix found earlier.
5.2 Potential energy (minimize a functional) method toderive the sti๏ฟฝness 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 sti๏ฟฝness matrix.
Let the system total potential energy by called ฮ and let the total internal energy in thesystem be ๐ and let the work done by external loads acting on the nodes be ฮฉ, then
ฮ = ๐ โฮฉ
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 . . .
12โซ๐{๐}๐ {๐} ๐๐ and the work done by external loads is {๐}๐ ๏ฟฝ๐๏ฟฝ, hence
ฮ =12 ๏ฟฝ๐
{๐}๐ {๐} ๐๐ โ {๐}๐ ๏ฟฝ๐๏ฟฝ (1)
Now the rest follows as before. Assuming that displacement can be written as a function ofthe nodal displacements {๐}, hence
๐ = [๐] {๐}
Since {๐} = ๐ {๐} then {๐} = ๐ [๐] {๐} = [๐ต] {๐} where ๐ต is the strain displacement matrix[๐ต] = ๐ [๐], hence
{๐} = [๐ต] {๐}
Now from the stress-strain relation {๐} = [๐ธ] {๐}, hence
{๐}๐ = {๐}๐ [๐ต]๐ [๐ธ] (4)
Substituting Eqs. (2,3,4) into (1) results in
ฮ =12 ๏ฟฝ๐
{๐}๐ [๐ต]๐ [๐ธ] [๐ต] {๐} ๐๐ โ {๐}๐ ๏ฟฝ๐๏ฟฝ
Setting ๐ฮ ๐{๐} = 0 gives
0 = {๐}๏ฟฝ๐[๐ต]๐ [๐ธ] [๐ต] ๐๐ โ ๏ฟฝ๐๏ฟฝ
Which is on the form ๐ = [๐พ]๐ท which means that
[๐พ] = ๏ฟฝ๐[๐ต]๐ [๐ธ] [๐ต] ๐๐
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
43