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1. Analysis of a Simply Supported Beam
Zeinab El-Sayegh
ID: 7126247
Department of Engineering and Computer Science
Presented in a Partial Fulfillment of the Requirement For The
MECH 6441
Concordia University
Montreal, Quebec, Canada
December , 2014

2. Analysis of a Simply Supported Beam
by
Zeinab El-Sayegh
ID: 7126247
December 2014
A Project Submitted to the Graduate Faculty
of The University of Concordia in Partial Fulfillment
of the MECH 6441
Requirements for the Degree
Master of Engineering
Mechanical Engineering
2014

3. c 2014
Zeinab El-Sayegh
All Rights Reserved

4. 4
Abstract
This project will deal with the analysis of a simply supported beam that consist
of two materials with applied complex loading. Calculation of the Airy stress
function, stress tensor, principal stresses, maximum shear stress, and octahedral
shear stress. Similar calculation will be done for the strain components. In
addition study on the displacements, change in length will be done. Finally
design parametric study on the maximum of the maximum principal stress and
minimum of the minimum principal stress in the beam with respect to the aspect
ratio of the beam and forces will be performed.
Index words: simply supported beam, complex loading, Airy
stress/strain, stress/strain tensor, principal
stresses/strains, maximum shear stress/strains,
octahedral shear stress/strains, parametric design.

5. Contents
1 Introduction and Literature Review 11
1.1 Simply Supported beam . . . . . . . . . . . . . . . . . . . . . . 11
1.1.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.1.2 Material properties . . . . . . . . . . . . . . . . . . . . . 12
1.1.3 Model Used . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.2 Stress Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.2.1 Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.2.2 Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.2.3 Airy Stress/strain . . . . . . . . . . . . . . . . . . . . . . 13
1.2.4 Stress/strain Tensor . . . . . . . . . . . . . . . . . . . . . 14
1.2.5 Principle Stresses/strains . . . . . . . . . . . . . . . . . . 14
1.2.6 Maximum Shear Stress/strains . . . . . . . . . . . . . . . 14
1.2.7 Octahedral Shear Stress/strains . . . . . . . . . . . . . . 14
1.2.8 Displacement . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.3 Load Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2 Stress Calculation 17
2.1 Airy Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.2 Stress Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.3 Principal Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.4 Maximum Shear Stress . . . . . . . . . . . . . . . . . . . . . . . 25
2.5 Octahedral Shear Stress . . . . . . . . . . . . . . . . . . . . . . 25
2.6 Maximum of all stresses . . . . . . . . . . . . . . . . . . . . . . . 26
3 Strain Calculation 27
3.1 Strain Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.2 Principal Strains . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.3 Maximum Shear Strain . . . . . . . . . . . . . . . . . . . . . . . 29
3.4 Maximum of all Strains . . . . . . . . . . . . . . . . . . . . . . . 29
3.5 Geometric Calculations . . . . . . . . . . . . . . . . . . . . . . . 31
3.6 Displacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
3.7 Length Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
5

6. 6 CONTENTS
4 Complement Study 35
4.1 Parametric Study . . . . . . . . . . . . . . . . . . . . . . . . . . 35
4.2 Bending Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
4.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
A MATLAB Coding 39
A.1 Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
A.2 strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

7. List of Figures
1.1 Typical Simply Supported Beam [1] . . . . . . . . . . . . . . . . 11
1.2 The beam model used in this study . . . . . . . . . . . . . . . . . 12
1.3 The Model used with Load . . . . . . . . . . . . . . . . . . . . . 15
2.1 Model with Located Co-Ordinate System . . . . . . . . . . . . . 18
2.2 The variation of the principle stress as a function of the position 26
3.1 Variation of the maximum principle strain as a function of and y 30
3.2 Variation of the minimum principle strain as a function of x and y 30
3.3 The variation of x displacement as a function of x and y . . . . . 32
3.4 The variation of y displacement as a function of x and y . . . . . 32
4.1 Bending Simply Supported Beam . . . . . . . . . . . . . . . . . 36
4.2 Bending Beam with couple moments . . . . . . . . . . . . . . . . 36
7

8. 8 LIST OF FIGURES

9. List of Tables
1.1 Steel and Aluminum Material Properties [2] . . . . . . . . . . . . 12
9

10. 10 LIST OF TABLES

11. Chapter 1
Introduction and Literature
Review
This chapter includes an introduction about the project content. Definition of
simply supported beam will be introduced, stress and strain definitions will be
proposed, and the methodology used to solve this project will be suggested.
1.1 Simply Supported beam
In this section it is required to define the simply supported beam term, highlight
on few properties of this structure and produce a model of the beam with the
required complex loading.
1.1.1 Definition
A beam is a structural element that is capable of withstanding load primarily by
resisting bending [1]. For a simply supported beam, the structure is supported
at each end which are free to rotate and have no moment resistance, and the
load is distributed in some way along its length.The simply supported beam
shown in figure 1.1 shows the typical shape of a simply supported beam with
out any force. The simply supported beam has on one side a hinged support
H
Figure 1.1: Typical Simply Supported Beam [1]
11

12. 12 CHAPTER 1. INTRODUCTION AND LITERATURE REVIEW
and on the other side a roller support. If the load was vertical and their is no
moment then both supports produce only vertical forces, however in the general
case the hinged support is able to produce vertical and horizontal forces.
1.1.2 Material properties
In this project we assume that the beam of length L, depth 2c and unit width.
Has the top half of the beam made of aluminum, and bottom half made of
steel. A small search is made to conduct the material properties which is shown
in table 1.1, typical properties of shown are assumed at room temperature (
25 C ), we also assume that the materials is stainless steel 304 and aluminum
wrought alloy 6061-T6 [2] These properties are based on the assumptions made,
Table 1.1: Steel and Aluminum Material Properties [2]
Properties Stainless Steel Aluminum
Density ( Mg/m3
) 7.86 2.71
Modulus of Elasticity (GPa) 193 68.9
Modulus of Rigidity ( GPa) 75 26
Poisson’s Ratio 0.27 0.35
Yield Strength of tension (MPa) 207 225
and according the reference used. Other values may be conducted for different
assumptions and using different references.
1.1.3 Model Used
In this part a new model of the beam will be defined according to the material
properties , and the material assumptions. Figure 1.2 shows the new model
with the two surfaces of aluminum and steel over each others.
Figure 1.2: The beam model used in this study

13. 1.2. STRESS ANALYSIS 13
This new model only defines the new material combination and loading is
not applied yet, next step will be to define the appropriate loading on the beam.
1.2 Stress Analysis
In this section definitions of related types of stress and strains will be defined,
summarized demonstration to indicate the relation between the definitions and
the related problem.
1.2.1 Stress
Stress is a physical quantity that expresses the internal forces that neighboring
particles of a continuous material exert on each other [3]. So basically stress is
directly related to the force, it can be said that the stress is the force per unit
area. In this area one may also define the shear stress which is the he component
of stress coplanar with a material cross section. Shear stress is derived from the
force vector component parallel to the cross section [3].
1.2.2 Strain
In a similar manner as in section 1.2.1 a definition of strain may be implemented.
physically strain is defined as deformation of a solid due to stress, so whenever
we have stress their will be strain. A similar concept of the shear stress is
found in the strain which is called engineering shear strain and it represents the
measure of the change in angle along a certain axis.
1.2.3 Airy Stress/strain
This function can be used as a scaler potential to find the stress/strain tensor,
it satisfies the equilibrium in the absence of body forces,but it should be noted
that it can only be used for two dimensional problems ex. plane stress/ plane
strain. for the case of the airy stress function the governing equation is:
▽φ4
+ (1 − υ)▽2
ϕ = 0 (1.1)
for the case of equilibrium the equation becomes:
▽φ4
= 0 (1.2)
while for the case of airy strain function, the governing equation becomes:
▽φ4
+
(1 − 2υ)
1 − υ
▽2
ϕ = 0 (1.3)
In these equations function φ(x, y) is the stress/strain function and ϕ is the
potential term.

14. 14 CHAPTER 1. INTRODUCTION AND LITERATURE REVIEW
1.2.4 Stress/strain Tensor
The stress/strain tensor is defined from nine elements which is the matrix form
of the stresses over axises. In the case of two dimensional study the tensor size
reduces to be come 2*2 instead of 3*3.
1.2.5 Principle Stresses/strains
A principle plane is usually any plane in which the shear stresses are zero. To be
able to find the principle stress/strain their is some calculations that needs to
be done, calculation of three invariants from the the stress/strain tensor should
be done as defined below: For the case of stress [4]:
I1 = σxx + σyy + σzz (1.4)
I2 = σxxσyy + σyyσzz + σxxσzz − τxy
2
− τyz
2
− τxz
2
(1.5)
I3 = σxxσyyσzz − σxxτyz
2
− σyyτxz
2
− σzzτxy
2
− 2τxyτxzτyz (1.6)
these invariants will be plugged in a third degree equation and solved to find
three stresses, the characteristic equation is a second name of the third degree
equation which is defined as [4]:
σ3
− I1σ2
+ I2σ − I3 = 0 (1.7)
Similar analysis can be retrieved for principle strains calculation.
1.2.6 Maximum Shear Stress/strains
The maximum shear stress/strain is calculated from the maximum principle
shear/strain by summing the smallest and biggest one and dividing the summa-
tion by two.
1.2.7 Octahedral Shear Stress/strains
This section includes the study of the stress/strains on a plane equally inclined
with respect to the co-ordinate axes. The octahedral plane is a plane that makes
the same angle with the three principle directions. The stress and shear over
the plane can be calculated as[4]:
σoct =
I1
3
(1.8)
τoct
2
=
1
3
q
(σ1 − σ2)
2
+ (σ2 − σ3)
2
+ (σ3 − σ1)
2
(1.9)
1.2.8 Displacement
The displacement is related to the strain since the strain it self is a ratio of
length, to be able to determine the displacement at any point two displacement
functions u (x,y) and v(x,y) should be determined from the geometry of the
structure.

15. 1.3. LOAD ANALYSIS 15
1.3 Load Analysis
In this section an overview of the load structure will be performed. It is assumed
that the load applied to the top surface is downward distributed loading of
trapezoidal distribution with intensity F1 N/m at left end and F2 N/m at right
end, with F1 = 0.7F2. Figure 1.3 shows the load applied.
Figure 1.3: The Model used with Load

16. 16 CHAPTER 1. INTRODUCTION AND LITERATURE REVIEW

17. Chapter 2
Stress Calculation
In this chapter solution for the stress constants listed in chapter one will be
solved. The solution will be on two ways , solution by hand with analytical
results and solution with MATLAB that can facilitate the solving of high order
equations. A study for the stress/strain and displacement will be conducted
according to the defined parameters in chapter one. Certain study of the stress
over the materials will be done using hand calculations and software calculations.
Some assumptions are taken into consideration in this section and dimensions
are given as:
L = 4m,
c = 0.25m,
w = 1m,
f1 = 3.5KN/m,
f2 = 5KN/m,
These assumptions satisfy the problem condition of F1 = 0.7F2 and are appli-
cable to real life problems.
2.1 Airy Stress
The strategy used in this section is to take a general form of the solution and
start eliminating some constants according to the boundary conditions. Lets
start by defining an axis co-ordinate to start identifying the boundary condi-
tions.
17

18. 18 CHAPTER 2. STRESS CALCULATION
Figure 2.1: Model with Located Co-Ordinate System
Figure 2.1 shows the co-ordinate system used, where the origin is one the
geometric center of the beam. All parameters are kept parameters and no
value is assumed till now, figure 2.1 shows the dimensions of the beam. Now
we can write the boundary conditions which are classified to strong and weak
conditions. The strong boundary conditions are:
y = ±b; τxy = 0 (2.1)
y = −b; σy = 0 (2.2)
y = b; σy = −f1 − 0.3
f2
L
x (2.3)
These boundary conditions should be satisfied in the strong sense. We shall
now require three linearly independent weak boundary conditions on the ends
when x = ±L/2, but since our model is not symmetric we should define each
condition alone, for x = L/2
Fx(L/2) =
Z b
−b
σx(L/2, y)dy = 0 (2.4)
Fy(L/2) =
Z b
−b
τxy(L/2, y)dy = FB (2.5)
M(L/2) =
Z b
−b
σx(L/2, y)ydy = 0 (2.6)

19. 2.1. AIRY STRESS 19
for the second side were x = −L/2 the conditions are:
Fx(−L/2) =
Z b
−b
σx(−L/2, y)dy = 0 (2.7)
Fy(−L/2) =
Z b
−b
τxy(−L/2, y)dy = FA (2.8)
M(−L/2) =
Z b
−b
σx(−L/2, y)ydy = 0 (2.9)
It is noted that Fa and FB can be found from drawing a free body diagram and
applying newtons second law. The parametric values that is obtained are:
FA =
f2 + 2f1
6
L = 8 (2.10)
FB =
2f2 + f1
6
L = 9 (2.11)
From the model we have the normal traction is not normal and it varies with
x1
(n = 1) , so it demands a polynomial of order (n + 5 = 6) , the general form
of such a polynomial is:
φ = ax2
+ bxy + cy2
+ dx3
+ ex2
y + fxy2
+ gy3
+hx4
+ ix3
y + jx2
y2
+ kxy3
+ ly4
+ mx5
+nx4
y + ox3
y2
+ px2
y3
+ qxy4
+ ry5
+ sx6
+tx5
y + ux4
y2
+ vx3
y3
+ wx2
y4
+ zxy5
+ aay6
(2.12)
The equation looks big but non need to rewrite it again. This equation will
be solved by hand and using MATLAB software. Starting with the by hand
solution, equation 2.12 should at the beginning satisfy equation 1.2 then the
conditions resulting are:
24h + 8j + 24l + 120mx + 24ny + 24ox + 24py + 24qx
+120ry + 360aay2
+ 360sx2
+ 48ux2
+ 24uy2
+24wx2
+ 48wy2
+ 120txy + 72vxy + 120zxy = 0 (2.13)
(120m + 24o + 24q)x + (360s + 48u + 24w)x2
+(24n + 24p + 120r)y + (360aa + 24u + 48w)y2
+(120t + 72v + 120z)xy + (24h + 8j + 24l) = 0 (2.14)
The conditions for the relation to be true are:
120m + 24o + 24q = 0 (2.15)
360s + 48u + 24w = 0 (2.16)
24n + 24p + 120r = 0 (2.17)
360aa + 24u + 48w = 0 (2.18)
120t + 72v + 120z = 0 (2.19)
24h + 8j + 24l = 0 (2.20)

20. 20 CHAPTER 2. STRESS CALCULATION
From the first condition 6 equations were established. Now to apply the bound-
ary condition one must first get an explicit form of the stress and shears,
σx =
∂2
φ
∂y2
= 2ux4
+ 6vx3
y + 2ox3
+ 12wx2
y2
+ 6px2
y + 2jx2
+ 20zxy3
+ 12qxy2
+ 6kxy + 2fx + 30aay4
+ 20ry3
+ 12ly2
+ 6gy + 2c (2.21)
σy =
∂2
φ
∂x2
= 30sx4
+ 20tx3
y + 20mx3
+ 12ux2
y2
+ 12nx2
y + 12hx2
+ 6vxy3
+ 6oxy2
+ 6ixy
+ 6dx + 2wy4
+ 2py3
+ 2jy2
+ 2ey + 2a (2.22)
τxy = −
∂2
φ
∂x∂y
= −5tx4
− 8ux3
y − 4nx3
− 9vx2
y2
− 6ox2
y − 3ix2
− 8wxy3
− 6pxy2
− 4jxy − 2ex − 5zy4
− 4qy3
− 3ky2
− 2fy − b (2.23)
Now solving the boundary conditions one by one to eliminate and solve for the
constants, starting with equation 2.1:
y = ±b; τxy = 0
τxy(x, −1/2) = f − b − (3k)/4 + q/2 − (5z)/16 − 2ex + 2jx
− (3px)/2 + wx − 3ix2
− 4nx3
+ 3ox2
− 5tx4
+ 4ux3
− (9vx2
)/4 (2.24)
= (f − b − (3k)/4 + q/2 − (5z)/16) + x(−2e + 2j − 3p/2 + w)
+ x2
(3i + 3o − 9v/4) + x3
(−4n + 4u) + x4
(−5t) (2.25)
τxy(x, 1/2) = −b − f − (3k)/4 − q/2 − (5z)/16 − 2ex − 2jx
− (3px)/2 − wx − 3ix2
− 4nx3
− 3ox2
− 5tx4
− 4ux3
− (9vx2
)/4 (2.26)
= −b − f − (3k)/4 − q/2 − (5z)/16 + x(−2e − 2j − 3p/2 − w)
+ x2
(−3i − 3o − 9v/4) + x3
(−4n − 4u)x4
(−5t) (2.27)

21. 2.1. AIRY STRESS 21
Quitting the equations to zero a set of equations can be determined:
f − b − (3k)/4 + q/2 − (5z)/16 = 0 (2.28)
−2e + 2j − 3p/2 + w = 0 (2.29)
3i + 3o − 9v/4 = 0 (2.30)
−4n + 4u = 0 (2.31)
−5t = 0 (2.32)
−b − f − (3k)/4 − q/2 − (5z)/16 = 0 (2.33)
−2e − 2j − 3p/2 − w = 0 (2.34)
−3i + 3o − 9v/4 = 0 (2.35)
4n + 4u = 0 (2.36)
(2.37)
One can conclude that, t = 0 , n = u = 0. Writing equation 2.2 :
y = −b; σy = 0
= 2a − e + j/2 − p/4 + w/8 + 6dx − 3ix
+ (3ox)/2 − (3vx)/4 + 12hx2
+ 20mx3
− 6nx2
+ 30sx4
− 10tx3
+ 3ux2
(2.38)
= (2a − e + j/2 − p/4 + w/8) + x(6d − 3i + 3o/2 − 3v/4)
+ x2
(12h − 6n + 3u) + x3
(20m − 10t) + 30sx4
(2.39)
The conditions becomes:
2a − e + j/2 − p/4 + w/8 = 0 (2.40)
6d − 3i + 3o/2 − 3v/4 = 0 (2.41)
12h − 6n + 3u = 12h = 0 (2.42)
20m − 10t = 20m = 0 (2.43)
s = 0 (2.44)
Three more parameters are reviled s = m = h = 0. Similarly with equation 2.3
:
y = b; σy = −f1 − 0.3
f2
L
(x + L/2) (2.45)
= 2a + e + j/2 + p/4 + w/8 + 6dx + 3ix + (3ox)/2
+ (3vx)/4 + 12hx2
+ 20mx3
+ 6nx2
+ 30sx4
+ 10tx3
+ 3ux2
(2.46)
= 2a + e + j/2 + p/4 + w/8 + 6dx + 3ix + (3ox)/2 + (3vx)/4 (2.47)
= (2a + e + j/2 + p/4 + w/8) + x(6d + 3i + 3o/2 + 3v/4) (2.48)
Setting the equations to its correspondence,
2a + e + j/2 + p/4 + w/8 = −4.25 (2.49)
6d + 3i + 3o/2 + 3v/4 = −0.375 (2.50)

22. 22 CHAPTER 2. STRESS CALCULATION
The above 22 equations represent the dominant equations that should be sat-
isfied, taking into consideration that the initial number of unknowns where 25,
we still have 3 more to compose, but lets do a small evaluation of the equations
we have using the MATLAB coding, it is shown that w = aa = 0. The left
equations are:
7.5r − 6l − 2e = 0 (2.51)
2a + e − 1.5l − 1.25r + 4.25 = 0 (2.52)
6d + 1.5q + 0.375 = 0 (2.53)
Now we should solve the weak conditions to find the remaining parameters
it is only required to find one weak condition for example equation 2.4, the
parameters become:
• a = −1.0625
• b = 34.5
• c = −1
• d = −0.0156
• e = −3.1875
• f = −0.03125
• g = 92
• h = 0
• i = −0.0625
• j = 0
• k = −46
• l = 0
• m = 0
• n = 0
• o = −0.0625
• p = −4.25
• q = 0.0625
• r = −0.85
• s = 0
• t = 0
• u = 0
• v = 0
• w = 0
• z = 0
• aa = 0
For more details about the MATLAB coding and the use of boundary con-
ditions, please refer to appendix A.1. Now the airy function, stresses, and shear

23. 2.2. STRESS TENSOR 23
becomes:
φ(x, y) = −0.0625x3
y2
− 0.0625x3
y − 0.0156x3
+ 4.25x2
y3
− 3.1875x2
y
− 1.0625x2
+ 0.0625xy4
− 46xy3
− 0.03125xy2
+ 34.5xy − 0.85y5
+ 92y3
− y2
(2.54)
σx(x, y) = −0.125x3
+ 25.5x2
y + 0.75xy2
− 276xy − 0.0625x
− 17y3
+ 552y − 2 (2.55)
σy(x, y) = 8.5y3
− 6.375y − 0.375xy − 0.375xy2
− 0.0936x − 2.125 (2.56)
τxy(x, y) = 0.375x2
y + 0.1875x2
− 25.5xy2
+ 6.375x − 0.25y3
+ 138y2
+ 0.0625y − 34.5 (2.57)
2.2 Stress Tensor
The Tensor is in two dimensional form of:
σx τxy
τxy σy
The stress tensor can be found from the Airy stress values, using equations 2.55
, 2.56, and 2.57 . It is noted that the tensor depends on the position of the
particle relevant to (x ,y) ,










−0.125x3
+ 25.5x2
y + 0.75xy2
0.375x2
y + 0.1875x2
− 25.5xy2
+ 6.375x
−276xy − 0.0625x − 17y3
+ 552y − 2 −0.25y3
+ 138y2
+ 0.0625y − 34.5
0.375x2
y + 0.1875x2
− 25.5xy2
+ 6.375x 8.5y3
− 6.375y − 0.375xy
−0.25y3
+ 138y2
+ 0.0625y − 34.5 −0.375xy2
− 0.0936x − 2.125










If the matrix seems small please refer to the indicated equations. The stress
tensor is not affected with the material properties due to the fact that the stress
deals with the force and area, but does not deal with the body properties it self.

24. 24 CHAPTER 2. STRESS CALCULATION
2.3 Principal Stresses
Principle stresses are evaluated from the stress tensor using the methodology
indicated in section 1.2.5. starting by the stress invariants,
I1 = −0.125x3
+ 25.5x2
y + 0.375xy2
− 276.375xy − 0.1561x
− 8.5y3
+ 545.625y − 4.125 (2.58)
I2 = −0.094x4
y2
− 0.094x4
y − 0.0235x4
+ 8.5x3
y3
− 6.37x3
y − 2.125x3
− 433.6x2
y4
− 0.1875x2
y3
+ 214x2
y2
− 2.48x2
y − 27.7x2
+ 4698xy4
− 25.5xy3
− 1967xy2
+ 535xy
+ 440x − 144.6y6
+ 69y5
− 14243y4
− 2.875y3
+ 6003y2
− 1155.94 − 1186 (2.59)
I3 = 0 (2.60)
The over all equation is from the third degree in the form :
σ3
− I1σ2
+ I2σ − I3 = 0
Solving with the obtained values of invariants and taking into consideration that
I3 is zero which leads one of the σ to be zero,and makes the equation easier to
solve buy quadratic equation method to obtain the two other σ,
σ1 = 4272.8y − 0.078x − 138.2xy
+
v
u
u
u
u
u
u
u
u
u
u
u
u
t
0.0078x6
− 3.875x5
y + 325.26x4
y2
+ 34.73x4
y
+0.066x4
− 6.375x3
y3
− 7047x3
y2
− 59.4x3
y
+4.76x3
+ 650x2
y4
− 1101x2
y3
+ 51676x2
y2
− 57.1x2
y
+55.4x2
− 3.1875xy5
− 7047xy4
+ 614.5xy3
−146864xy2
− 15.49xy − 879.7x + 325.25y6
− 138y5
+ 23849y4
+65.8y3
+ 136847y2
+ 61.17y + 2380
+ 0.1875xy2
+ 12.75x2
y − 0.0625x3
− 4.25y3
− 2.0625 (2.61)
σ2 = 0 (2.62)
σ3 = 4272.8y − 0.078x − 138.2xy
−
v
u
u
u
u
u
u
u
u
u
u
u
u
t
0.0078x6
− 3.875x5
y + 325.26x4
y2
+ 34.73x4
y
+0.066x4
− 6.375x3
y3
− 7047x3
y2
− 59.4x3
y
+4.76x3
+ 650x2
y4
− 1101x2
y3
+ 51676x2
y2
− 57.1x2
y
+55.4x2
− 3.1875xy5
− 7047xy4
+614.5xy3
− 146864xy2
− 15.49xy − 879.7x + 325.25y6
−138y5
+ 23849y4
+ 65.8y3
+ 136847y2
+ 61.17y + 2380
+ 0.1875xy2
+ 12.75x2
y − 0.0625x3
− 4.25y3
− 2.0625 (2.63)
(2.64)

25. 2.4. MAXIMUM SHEAR STRESS 25
It is noted that these principle stresses are obtained using MATLAB coding for
more details about the coding please refer to Appendix A.1 , and they are as
a function of the position point p(x,y). For example, taking the center point
which is the point of origin p(0,0) the principle stresses become:
σ1 =
√
304705/16 − 33/16 = 32.437KPa
σ2 = 0
σ3 = −
√
304705/16 − 33/16 = −36.56KPa
And similarly at every point of the model we are able to find the localized
principle stresses.
2.4 Maximum Shear Stress
The maximum shear stress is obtained from the equation:
τmax =
σ1 + σ3
2
τmax = −0.0625x3
+ 12.75x2
y + 0.1875xy2
− 138.2xy
− 0.078x − 4.25y3
+ 273y − 2.0625 (2.65)
where σ1, and σ3 are the minimum and maximum principle stresses , then the
maximum shear stress becomes. As indicated before this is a function of the
point position (x,y).
2.5 Octahedral Shear Stress
The octahedral shear stress is calculated as indicated in equations 1.8, and 1.9,
σoct =
I1
3
σoct = −0.0417x3
+ 8.5x2
y + 0.125xy2
− 92.125xy − 0.052x
− 2.83y3
+ 181.875y − 1.375 (2.66)
τoct
2
=
1
3
q
(σ1 − σ2)
2
+ (σ2 − σ3)
2
+ (σ3 − σ1)
2
τoct
2
=
v
u
u
u
u
u
u
u
u
u
u
u
u
t
0.01x6
− 4.25x5
y + 433.625x4
y2
+ 46.25x4
y
+0.073x4
− 2.83x3
y3
− 9396.8x3
y2
−83.5x3
y + 4.9x3
+ 578.3x2
y4
− 137.8x2
y3
+ 69045x2
y2
−77.8x2
y + 55.4x2
− 4.25xy5
− 6264.5xy4
+ 683xy3
− 197130xy2
+336xy − 879.5x + 112.4y6
− 13.67y5
+22303y4
+ 77.5y3
+ 186465y2
− 689y + 2383

26. 26 CHAPTER 2. STRESS CALCULATION
2.6 Maximum of all stresses
As illustrated before the stresses are as a function of the position point p(x,y).
In this section an analysis of the stresses will be performed to identify the
maximum stresses and their location. Starting with the principle stresses and
plotting the variation of the principle stress as a function of the x and y is shown
in figure 2.2.
Figure 2.2: The variation of the principle stress as a function of the position
From the graph it is shown that the maximum principle stress occurs happens
at the point x = −2, and y = 0.5.The principle stresses at this point are:
σ1 = 596.13KPa
σ2 = 0
σ3 = −3.5040KPa
One may also check for the lowest principle stress and it happens to be σmin =
−602.1250, and this point which is p(-2,-0.5) the other principle stress is σ =
−3∗10−4
. The maximum octahedral stress will happen for the first case because
the difference is bigger then,
σoct = 198.7082KPa
τoct
2
= 283.4972KPa

27. Chapter 3
Strain Calculation
In this chapter the strain will be calculated according to the known equation
σ = Eǫ, but the strain change along the y axis will depend on the material
property due to the combination of the materials ( aluminum, steel ).
3.1 Strain Tensor
The strain tensor will be divided to two parts for the section −0.5 ≤ y ≤ 0 in
this part E will be for steel, and the second part 0 ≤ y ≤ 0.5, in this part E will
be for Aluminum. Using Hooke’s Law σ = Eǫ it can be known that:
if − 0.5 ≤ y ≤ 0
ǫst = 1.45 ∗ 10−5
σ (3.1)
if0 ≤ y ≤ 0.5
ǫal = 5.18 ∗ 10−6
σ (3.2)
It can be also written in the form of a matrix, for −0.5 ≤ y ≤ 0 ,










5.18 ∗ 10−6
(−0.125x3
+ 25.5x2
y + 0.75xy2
5.18 ∗ 10−6
(0.375x2
y + 0.1875x2
− 25.5xy2
−276xy − 0.0625x − 17y3
+ 552y − 2) +6.375x − 0.25y3
+ 138y2
+ 0.0625y − 34.5)
5.18 ∗ 10−6
(0.375x2
y + 0.1875x2
− 25.5xy2
5.18 ∗ 10−6
(8.5y3
− 6.375y − 0.375xy
+6.375x − 0.25y3
+ 138y2
+ 0.0625y − 34.5) −0.375xy2
− 0.0936x − 2.125)










In case of 0 ≤ y ≤ 0.5 the stress tensor will be ,










1.45 ∗ 10−5
(−0.125x3
+ 25.5x2
y + 0.75xy2
1.45 ∗ 10−5
(0.375x2
y + 0.1875x2
− 25.5xy2
−276xy − 0.0625x − 17y3
+ 552y − 2) +6.375x − 0.25y3
+ 138y2
+ 0.0625y − 34.5)
1.45 ∗ 10−5
(0.375x2
y + 0.1875x2
− 25.5xy2
1.45 ∗ 10−5
(8.5y3
− 6.375y − 0.375xy
+6.375x − 0.25y3
+ 138y2
+ 0.0625y − 34.5) −0.375xy2
− 0.0936x − 2.125)










27

28. 28 CHAPTER 3. STRAIN CALCULATION
3.2 Principal Strains
The principle stresses will be evaluated under the two conditions also, in this
can we will get ǫst and ǫAl, for the case of steel, −0.5 ≤ y ≤ 0
ǫ1 = 1582y − 0.45x − 801xy
+ 0.5
v
u
u
u
u
u
u
u
u
u
u
u
u
u
u
u
t
0.5x6
− 214x5
y + 21884x4
y2
+ 2337x4
∗ y
+4.5x4
− 429x3
y3
− 474160x3
y2
− 3999x3
y
+320x3
+ 43766x2
y4
− 6947.7x2
y3
+ 3476795x2
y2
−3840.8x2
y + 3727x2
− 214xy5
− 474160xy4
+41352xy3
− 988xy2
− 1042xy − 59189x
+21882y6
− 9284y5
+ 1604585y4
+ 4428y3
+9207088y2
+ 4115.6y + 160160
+ 1.08xy2
+ 73.95x2
y − 0.3625x3
− 24.65y3
− 11.96 (3.3)
ǫ2 = 0 (3.4)
ǫ3 = 1582y − 0.45x − 801xy
− 0.5
v
u
u
u
u
u
u
u
u
u
u
u
u
u
u
u
t
0.5x6
− 214x5
y + 21884x4
y2
+ 2337x4
∗ y
+4.5x4
− 429x3
y3
− 474160x3
y2
− 3999x3
y
+320x3
+ 43766x2
y4
− 6947.7x2
y3
+ 3476795x2
y2
−3840.8x2
y + 3727x2
− 214xy5
− 474160xy4
+41352xy3
− 988xy2
− 1042xy − 59189x
+21882y6
− 9284y5
+ 1604585y4
+ 4428y3
+9207088y2
+ 4115.6y + 160160
+ 1.08xy2
+ 73.95x2
y − 0.3625x3
− 24.65y3
− 11.96 (3.5)
for the case of Aluminum 0 ≤ y ≤ 0.5, the principle stresses become
ǫ1 = 3955.78y − 1.13x − 2003xy + 2.72xy2
+ 184.9x2
y − 0.9x3
− 61.6y3
+ 0.5
v
u
u
u
u
u
u
u
u
u
u
u
u
t
3.28x6
− 1340x5
y + 136774x4
y2
+ 14605x4
y + 28x4
− 2681x3
y3
−2963500x3
y2
− 24995x3
y + 2004x3
+ 273538x2
y4
− 4342x2
y3
+21729973x2
y2
− 24005x2
y + 2329x2
− 1340xy5
−2963500xy4
+ 258451xy3
− 61756298xy2
− 6515xy
−369933x + 136767y6
− 58029y5
+10028662y4
+ 27674y3
+ 57544301y2
+ 25722y + 1001003
− 29.9 (3.6)
ǫ2 = 0 (3.7)

29. 3.3. MAXIMUM SHEAR STRAIN 29
ǫ3 = 955.78y − 1.13x − 2003xy + 2.72xy2
+ 184.9x2
y − 0.9x3
− 61.6y3
− 0.5
v
u
u
u
u
u
u
u
u
u
u
u
u
t
3.28x6
− 1340x5
y + 136774x4
y2
+ 14605x4
y + 28x4
− 2681x3
y3
−2963500x3
y2
− 24995x3
y + 2004x3
+ 273538x2
y4
− 4342x2
y3
+21729973x2
y2
− 24005x2
y + 2329x2
− 1340xy5
−2963500xy4
+ 258451xy3
− 61756298xy2
− 6515xy
−369933x + 136767y6
− 58029y5
+10028662y4
+ 27674y3
+ 57544301y2
+ 25722y + 1001003
− 29.9 (3.8)
It is noted that the strains are measured in the micro units.
3.3 Maximum Shear Strain
The maximum shear strain is obtained from the equation,
γmax = ǫ1 + ǫ3 (3.9)
In the Aluminum the maximum shear strain becomes,
γmax = −1.8x3
+ 369x2
y + 5.4xy2
− 4007xy − 2.26345x − 123y3
+ 7911y − 59.8
(3.10)
For the case of steel the maximum shear strain becomes,
γmax = −0.725x3
+147.9x2
y+2.175xy2
−1603xy−0.90538x−49.3y3
+3164y−24
(3.11)
3.4 Maximum of all Strains
In a similar manner as listed in section 2.6, the variation of the maximum
principle stress is plotted as a function of x and y as shown in figure 3.1.

30. 30 CHAPTER 3. STRAIN CALCULATION
Figure 3.1: Variation of the maximum principle strain as a function of and y
The plot shown contains both the aluminum and the steel it can be noticed
that at the point (0,y) their is discontinuity due to the joining of both materials
at this point. It also can be noticed that the maximum principle strain occurs
at the point p(-2,0.5) where the maximum strain happens to be ǫmax = 8694.6
, in this case the minimum strain is ǫmin = −50.8. Now plotting the minimum
principle strain which is ǫ3 for example is shown in figure 3.2,
Figure 3.2: Variation of the minimum principle strain as a function of x and y
for the minimum case the values are at p(-2,-0.5) where ǫmin = −3492.325 ,
in this case the maximum strain is ǫmin = −0.00174.

31. 3.5. GEOMETRIC CALCULATIONS 31
3.5 Geometric Calculations
In his section a calculation of the displacements ux(x, y) and uy(x, y) for any
point will be determined. In addition, the change in length of any one of the
diagonals of the beam and change in any one side length of the beam will be
calculated.
3.6 Displacement
From the formula point of view the the displacement along x which is defined
as ux and the displacement along y which is defined by uy are defined as:
ux(x, y) =
Z
ǫxx∂x (3.12)
uy(x, y) =
Z
ǫyy∂y (3.13)
it is noted that when integrating along y direction their are two different values
for ǫ due to the fact that their are two different materials, so along x direction
their will be two different displacement for two different conditions, similarly
along y direction. The displacements become: for steel −0.5 ≤ y ≤ 0 ,
ux(x, y) = −0.18125x4
+ 49.3x3
y + 2.175x2
y2
− 800x2
y
− 0.18125x2
− 98.6xy3
+ 3201.6xy − 11.6x
− 197.2y3
− 8.7y2
+ 9999y − 19.575 (3.14)
uy(x, y) = 12.325y4
− 12.325y − 0.54288xy − 1.0875xy2
− 0.725xy3
− 18.4875y2
− 0.1x − 2.311 (3.15)
for aluminum 0 ≤ y ≤ 0.5 ,
ux(x, y) = −0.453125x4
+ 123.25x3
y + 5.4375x2
y2
− 2001x2
y
− 0.453125x2
− 246.5xy3
+ 8004xy − 29x
− 493y3
− 21.75y2
+ 24998y − 48.9375 (3.16)
uy(x, y) = 30.8125y4
− 1.3572xy − 2.71875xy2
− 1.8125xy3
− 46.21875y2
− 30.8125y (3.17)
In this point one can plot the variation of the displacement as a function of
co-ordinate system by combining equation 3.14 and 3.16 for the boundary con-
ditions to get the x displacement, and equations 3.15 and 3.17 also over the
boundary conditions to get the y displacement.

32. 32 CHAPTER 3. STRAIN CALCULATION
Figure 3.3: The variation of x displacement as a function of x and y
Figure 3.4: The variation of y displacement as a function of x and y
3.7 Length Change
The length change depend directly on u(x,y) and ǫ, to find the change in length
at a certain point either along x or along y :
ǫyy =
δLy
Ly
(3.18)
ǫxx =
δLx
Lx
(3.19)
L it self should be horizontal or vertical, if L is a combination then another
formula should be written that includes an angle orientation, to get back to the

33. 3.7. LENGTH CHANGE 33
simple case of L it also will depend on the position to predefine the elasticity
value then, for steel
δLy = Ly(49y3
− 37y − 2.2xy − 2.2xy2
− 0.543x − 12.3) (3.20)
δLx = Lx(−0.7x3
+ 148x2
∗ y + 4.4xy2
− 1600xy
− 0.36x − 99y3
+ 3201y − 11.6) (3.21)
for aluminum,
δLy = Ly(123.25y3
− 92.4y − 5.44xy − 5.44xy2
− 1.36x − 30.8) (3.22)
δLx = Lx(−1.8x3
+ 369x2
y + 11xy2
− 4002xy
− 0.9x − 246y3
+ 8004y − 29) (3.23)
So to find for example the length change of a horizontal side we look at the x
component and then look at its y position to determine if it is in the steel side
or aluminum.

34. 34 CHAPTER 3. STRAIN CALCULATION

35. Chapter 4
Complement Study
Two main topics will be discussed in this chapter the bending of the beam, and
parametric study on some ratios.
4.1 Parametric Study
In this chapter a design parametric study will be conducted for the maximum
of the maximum principal stress and minimum of the minimum principal stress
in the beam with respect to the aspect ratio of the beam and F2. It is noted
that F2 is related to F1 and the aspect ratio is related to the length of the side
which is actually in this case the length to the width as ( 4 : 1 ) or in general (
L : 1 ).
As the force F2 increase the force F1 will increase with it too, due to the relation
F1 = 0.7F2 in general case the maximum of the maximum principle stress will
be located at the P ( -2 , 0.5 ) and the minimum of the minimum principle
stresses will be located the point P ( -2 , -0.5 ) this is the general case for what
ever the force vary, due to the fact that it will have the same distribution of the
trapezoidal shape.
The aspect ratio will affect the deflection and the displacement, as the beam gets
longer the displacement and increase and the beam will start bending more and
more on its center point which is the deepest point form the pin and roller. As
the beam gets longer the probability of bending gets larger and larger especially
if the width remains one unit or gets smaller.
Considering the relation between the aspect ratio and the principle stresses, the
maximum of the maximum and the minimum of the minimum will always be
located on the extreme points, even when the aspect ratio changes these stresses
will be located at the extreme of the new dimensions. This is due to the point
of application of the forces and the type of the load applied which is constant
in this case as trapezoidal. The location where the pin and roller are connected
also play an important role in determining the position of the stresses, but in
the scope of this project they are fix connected at the extreme of the beam. The
35

36. 36 CHAPTER 4. COMPLEMENT STUDY
aspect ratio can also have a significant effect on the break and fatigue of the
system, as the aspect ratio gets bigger in a sense that the difference between the
length and the width gets larger this will increase the probability of the failure
of the system.
4.2 Bending Beam
Assuming that the beam undergoes a bending moment with the same load force
applied, then the beam will become as shown in figure 4.1 ,
Figure 4.1: Bending Simply Supported Beam
In this case we assume that the load is presented by moments and that the
beam is symmetrical and only have couple moments are the free ends, these
assumptions lead to the drawing shown in figure 4.2
Figure 4.2: Bending Beam with couple moments
Now and using the missioned assumptions it will be easy to solve the prob-

37. 4.3. RESULTS 37
lem, the radial and tangential stresses are defined as:
σr =
4M
tb2N
1 −
a2
b2
ln
r
a
−
1 −
a2
r2
ln
b
a
(4.1)
σθ =
4M
tb2N
1 −
a2
b2
1 + ln
r
a
−
1 +
a2
r2
ln
b
a
(4.2)
for the purpose of this section we will assume that a = 100mm, b = 150mm
and that r = 50mm. Taking into consideration that,
M = moment (4.3)
N =
1 −
a2
b2
2
− 4
a2
b2
ln2 b
a
(4.4)
for these values we get that:
σr = 332KPa (4.5)
σθ = −743KPa (4.6)
on the other hand we know that the moment is,
M = (3.5 ∗ 4) ∗ 2 + 1.5 ∗ 2 ∗ 8/3
= 36KN.m
knowing the inertial of the system to be, Iy = 0.0415m4
, Iz = 0.0104m4
, and
Iyz = 0m4
then the stress along x direction will be,
σx =
Mc
I
(4.7)
=
36 ∗ 0.5
0.0415
= 433.7KPa
4.3 Results
From the bending moments analysis we found that the maximum stress that
can be found is σ = 433.7KPa, it is useful to compare this result with a similar
one from the airy stress functions that were derived, the maximum σ of all the
beam is calculate to be σ = 596KPa the error then is 27%. This error is due to
the shearing stress which is not taken into consideration in the bending beam
theory in addition to the assumption of point force and not load force.

38. 38 CHAPTER 4. COMPLEMENT STUDY

39. Appendix A
MATLAB Coding
MATLAB coding for the stress/ strain analysis is listed below.
A.1 Stress
syms x y;
syms sigmax(x,y)sigmay(x,y)taux(x,y)tauxy(x,y)
p(x,y) phi(x,y)eq(x,y);
t=0;
u=0;
n=0;
s=0;
m=0;
h=0;
w=0;
aa=0;
d=-0.0156;
q=0.0625;
l=0;
a=-1.0625;
z=0;
p=4.25;
o=-0.0625;
jj=0;
b=34.5;
v=0;
ii=-0.0625;
e=-3.1875;
g=92;
c=-1;
k=-46;
39

40. 40 APPENDIX A. MATLAB CODING
r=-0.85;
f=-0.03125;
theta=a*x^2+b*x*y +c*y^2+d*x^3+e*x^2*y+f*x*y^2+g*y^3
+h*x^4+ii*x^3*y+jj*x^2*y^2 +k*x*y^3+l*y^4+m*x^5
+n*x^4*y +o*x^3*y^2+p*x^2*y^3+q*x*y^4+r*y^5
+s*x^6 +t*x^5*y+u*x^4*y^2+v*x^3*y^3
+w*x^2*y^4+z*x*y^5+aa*y^6;
phi(x,y)=diff(theta,x,4)+diff(theta,y,4)
+2*diff(diff(theta,x,2),y,2)
sigmax(x,y)=diff(theta,y,2)
sigmay(x,y)=diff(theta,x,2)
taux(x,y)=diff(theta,x)
tauxy(x,y)=diff(taux,y)
tauxy(x,y)=-tauxy(x,y)
int(sigmax(x,y),y,-0.5,0.5)
Ione(x,y)=sigmax(x,y)+sigmay(x,y)
Itwo(x,y)=sigmax(x,y)*sigmay(x,y)-(tauxy(x,y))^2
delta=expand(Ione^2-4*Itwo);
sigmaone(x,y)=simplify(0.5*Ione+0.5*sqrt(delta))
sigmatwo(x,y)=expand(0.5*Ione-0.5*sqrt(delta))
taumax=simplify(0.5*(sigmaone(x,y)+sigmatwo(x,y)))
tauoct=expand((1/3)*sqrt((sigmaone(x,y))^2
+(sigmatwo(x,y))^2+(sigmatwo(x,y)-sigmaone(x,y))^2))
%sigmatwo(-2,-0.5)
%sigmaone(-2,-0.5)
%Ione(-2,0.5)
A.2 strain
syms x y strainx(x,y) strainy(x,y) strainxy(x,y);
syms Ione(x,y)Itwo(x,y)strainone(x,y)straintwo(x,y)
% aluminum
%E=1.45*10^-5;
%steel
%E=5.8*10^-6;
strainx(x,y)=vpa(E*10^6*(- x^3/8 + (51*x^2*y)/2
+ (3*x*y^2)/4 - 276*x*y - x/16 - 17*y^3 + 552*y - 2))
strainy(x,y)=vpa(E*10^6*((17*y^3)/2 - (51*y)/8 -
(3*x*y)/8 - (3*x*y^2)/8 - (117*x)/1250 - 17/8))
strainxy(x,y)=E*10^6*((3*x^2*y)/8 + (3*x^2)/16
- (51*x*y^2)/2 + (51*x)/8 - y^3/4 + 138*y^2 + y/16 - 69/2);
Ione(x,y)=strainx(x,y)+strainy(x,y);
Itwo(x,y)=strainx(x,y)*strainy(x,y)-(strainxy(x,y))^2;
delta=expand(Ione^2-4*Itwo);
strainone(x,y)=vpa(expand(0.5*Ione+0.5*sqrt(delta)));

41. A.2. STRAIN 41
straintwo(x,y)=vpa(expand(0.5*Ione-0.5*sqrt(delta)));
strainmax= vpa(strainone(x,y)+straintwo(x,y));
vpa(expand(int(strainx(x,y),x,-2,x)));
vpa(expand(int(strainy(x,y),y,0,y)));

42. 42 APPENDIX A. MATLAB CODING

43. Bibliography
[1] William Dwight Whitney and Benjamin E. Smith. Beam. The century
dictionary and cyclopedia. vol, 1. new york: Century co., 1901. 487 edition.
[2] R.C. Hibbeler. Mechanics of Materials. Seventh edition edition.
[3] Jacob Lubliner. Plasticity Theory. revised edition edition.
[4] Dr. RajamohanGanesan. Stress Analysis in Mechanical Design.
43