Final Project for the class of "Numerical Methods in Mechanical Engineering" - MECH 309. In this project, various engineering problems were analyzed and solved using advanced numerical approximation methods and MATLAB software.

1. Final Project
MECH 309
Numerical Methods in Mechanical Engineering
Date Submitted: April 13th
, 2018
Jai Patel 260586546
Stasik Nemirovsky 260660024

2. 1 Two Dimensional Heat Transfer
Given the conduction heat transfer, partial differential equation:
(1.1.1)
The points A(0,0) , B(0,1) and C(-1,-1) are located on the L-shaped grid below:
Del operator can be expressed as:
∇= (
∂
𝜕𝑥1
, … ,
∂
𝜕𝑥1
) = ∑ 𝑒𝑖⃗⃗⃗𝑛
𝑖=1
∂
𝜕𝑥1
(1.1.2)
Alternatively, in 2D we get:
∇ = (
𝜕
𝜕𝑥
,
𝜕
𝜕𝑦
) (1.1.3)
To approximate ∇(𝐾∇𝑇) at point (𝑋𝑖, 𝑌𝑗) using finite
differences scheme we should first expand the terms. It is
important to note that since 𝐾(𝑋𝑖, 𝑌𝑗) is a known value at
any point on the grid; we should only use finite differences
to approximate 𝑇(𝑋𝑖, 𝑌𝑗), to avoid introducing
unnecessary errors.
Therefore,
∇(𝐾∇𝑇) = ∇ [𝐾
𝜕𝑇
𝜕𝑥
𝑖 + 𝐾
𝜕𝑇
𝜕𝑦
𝑗] (1.1.4)
Using chain rule:
∇(𝐾∇𝑇) = 𝐾
𝜕2 𝑇
𝜕𝑥2 +
𝜕𝐾
𝜕𝑥
∙
𝜕𝑇
𝜕𝑥
+ 𝐾
𝜕2 𝑇
𝜕𝑦2 +
𝜕𝐾
𝜕𝑦
∙
𝜕𝑇
𝜕𝑦
(1.1.5)
Using center finite difference scheme to approximate 𝑇(𝑋𝑖, 𝑌𝑗), we get;
∇(𝐾∇𝑇) = 𝐾 [
𝑇𝑖+1,𝑗−2𝑇𝑖,𝑗+𝑇𝑖−1,𝑗
∆𝑥2 ] +
𝜕𝐾
𝜕𝑥
∙ [
𝑇𝑖+1,𝑗−𝑇𝑖−1,𝑗
2∆𝑥
] + 𝐾 [
𝑇𝑖,𝑗+1−2𝑇𝑖,𝑗+𝑇𝑖,𝑗−1
∆𝑦2 ] +
𝜕𝐾
𝜕𝑦
∙ [
𝑇𝑖,𝑗+1−𝑇𝑖,𝑗−1
2∆𝑦
] (1.1.6)
Given 𝐾(𝑥, 𝑦) = 2 + cos(𝑥 + 𝑦) and 𝐹(𝑥, 𝑦) = exp(−𝑥2
− 𝑦2
), and knowing that the temperature T
vanishes on the boundary of the grid above (𝑇 = 0) we can now solve the boundary value problem with
a system of equations:
−(2 + cos(𝑥𝑖 + 𝑦𝑗)) ∙ [
𝑇𝑖+1,𝑗−2𝑇𝑖,𝑗+𝑇𝑖−1,𝑗
∆𝑥2 ] − (− sin(𝑥𝑖 + 𝑦𝑗)) ∙ [
𝑇𝑖+1,𝑗−𝑇𝑖−1,𝑗
2∆𝑥
] −
(2 + cos(𝑥𝑖 + 𝑦𝑗)) ∙ [
𝑇𝑖,𝑗+1−2𝑇 𝑖,𝑗+𝑇𝑖,𝑗−1
∆𝑦2 ] − (− sin(𝑥𝑖 + 𝑦𝑗)) ∙ [
𝑇𝑖,𝑗+1−𝑇𝑖,𝑗−1
2∆𝑦
] = exp(−𝑥𝑖
2
− 𝑦𝑗
2
) (1.1.7)
Figure 1.1.1 "L" Shaped Grid

3. From inspection, we know the grid spacing is ∆𝑥 = ∆𝑦 = 0.25, we get:
16(−2 − cos(𝑥𝑖 + 𝑦𝑗)) ∙ [𝑇𝑖+1,𝑗 − 2𝑇𝑖,𝑗 + 𝑇𝑖−1,𝑗] + 2 sin( 𝑥𝑖 + 𝑦𝑗) ∙ [𝑇𝑖+1,𝑗 − 𝑇𝑖−1,𝑗] +
16(−2 − cos(𝑥𝑖 + 𝑦𝑗)) ∙ [𝑇𝑖,𝑗+1 − 2𝑇𝑖,𝑗 + 𝑇𝑖,𝑗−1] + 2 sin( 𝑥𝑖 + 𝑦𝑗) ∙ [𝑇𝑖,𝑗+1 − 𝑇𝑖,𝑗−1] =
exp(−𝑥𝑖
2
− 𝑦𝑗
2
) (1.1.8)
After collecting the terms, we get a system of 𝑛 equations corresponding to the amount of nodes in a 𝑛𝑥𝑛
grid:
16(−2 − cos(𝑥𝑖 + 𝑦𝑗)) ∙ [𝑇𝑖+1,𝑗 + 𝑇𝑖,𝑗+1 + 𝑇𝑖,𝑗−1 + 𝑇𝑖−1,𝑗 − 4𝑇𝑖,𝑗]
+ 2 sin( 𝑥𝑖 + 𝑦𝑗) ∙ [𝑇𝑖+1,𝑗 + 𝑇𝑖,𝑗+1 + 𝑇𝑖,𝑗−1 + 𝑇𝑖−1,𝑗] = exp(−𝑥𝑖
2
− 𝑦𝑗
2
) (1.1.9)
Here Delaunay is a useful tool as it is used as a finite element method in order to numerically approximate
the solution for the partial differential equations. Delaunay triangulation is used to generate mesh made up
of triangles of the temperature field of the plate[1]
. Trisurf is then used as a tool in order to display the
triangles that defines in the “m” by 3 matrix as a surface. Trisurf defines the triangular face and index’s the
vector that contains the x and y coordinate as well has the z coordinate that represents the temperature[2]
.
Figure 1.1.2 represents the temperature field of the heat generation on the L shaped plate where the
temperature along the boundary vanishes
(ie:0) for n=25. It can be concluded that the
temperature at the center of the plate is the
largest, as you move away from the center the
temperature at each node decreases.
Figure 2 displays the temperatures along line
B-C for 9x9 L-shaped grid. From inspection of
the plot we notice that the max temperature
along B-C is approximately in the mid-line
which is consistent with the temperature field
in the grid, as line B-C crosses nearby the max
temperature of the grid.
Figure 1.1.2: Temperature Plot of Heat Generation of an L shaped plate,
where the temperature on boundary of the plate vanishes where n=25

4. Order of convergence refers to the convergence of the finite different approximation to the exact solution
of the derivative, as ℎ → 0. For illustration purposes we choose the following example:
𝑓(𝑥) = 𝑒𝑥𝑝(−𝑥)𝑠𝑖𝑛(𝑥^2 /2) (1.1.10)
Using forward Finite Differences, we want to see how “fast” the derivative of the function, evaluated at
𝑥 = 2, converges the exact derivative. The exact derivative is:
𝑓′(2) = −0.23569874791
Forward finite difference approximation is known to have an error
of order 𝑂(ℎ). From the table attached, it is obvious that ℎ (grid
spacing) is proportional to the error. Therefore, we conclude that the
order of convergence of first derivative Forward Finite Difference
Scheme is 𝑂(ℎ). Similarly, using central finite difference, the
developed error is of the order 𝑂(ℎ2
). Since we have first order
Central finite difference terms in the developed scheme of the heat
equation, we conclude that the order of convergence of our
developed scheme is 𝑂(ℎ2
)[3]
.
New boundary conditions are introduced; ∇𝑇 ∙ 𝑛 = 0 along A-B.
Since 𝑛 is the normal to line A-B, the above condition can be simplified:
[
𝜕𝑇
𝜕𝑥
+
𝜕𝑇
𝜕𝑦
] ∙ 𝑛 𝑥 = 0
𝑑𝑇
𝑑𝑥
= 0 (1.1.11)
From the above equation it can be concluded that the temperature on the boundary points along A-B must
be equal to the corresponding nodes to their left, as there cannot be a difference in temperatures between
those (dT/dx =0).
Table 1.1: Error of Forward Finite Difference Vs
Grid Spacing (h)
Figure 2: Temperature field along the line B-C where the
temperature along the boundary vanishes, n=9

5. From Figure 1.1.4, it is obvious that the max temperature on the grid is now higher than before. In addition,
the new boundary conditions can be clearly seen through the flat faces along boundary A-B. However,
Radiation is now introduced as well, the new PDE reads:
(1.1.12)
The new formulation cannot be solved the same
way as before, by using regular matrix form and
using Matlab “” command. Instead, a system of
non-linear equations is to be constructed and
solved using Matlab’s built-in function Fsolve.
Figure 1.1.5 illustrates the new temperature field
with the addition of radiation. The new plot has
as strong resemblance on Figure 1.1.2, however,
that is only due to that fact that the values of
temperature were very small. If the temperature
at each node had larger values, the plot would
With boundary conditions T=0, same as in the
part 4, the new solution seems quite similar to
the previous solution. In fact, since temperature
values are small, 𝑇4
becomes a very small
quantity and therefore has minimal influence on
the new solution. Although, the difference is
Figure 1.1.5 Temperature field of heat generation on L shaped plate
where the temperature on the boundary vanishes with the addition of
radiation
Figure 1.1.4: Temperature field of an L shaped plate where the temperature along
AB does not vanish

6. not visible to the naked eye, the temperatures across the grid are slightly lower than in the previous
solution.
Introducing time into the governing equation allows us to explore how the temperature in the L-shaped grid
changes over a specified time interval. This not only allows us to solve for the steady state solution, but
also tells us the amount of time it will take to get to that steady-state temperature. T (temperature) is now
not just a function of the position on the grid (x,y) but also of time (x,y,t). In a way, instead of just solving
for the temperature in the grid once, we are now going to do it n times, where n represents the amount of
time-steps in the predefined time interval. For example, let the time interval be 5 seconds and time-steps be
0.1 seconds - solving for these setting will result in 50 different temperature fields, each corresponding to
specific time step. Adding all these temperature fields into a sequence of images will result in a dynamic
temperature field plot within a specific time interval. If large enough time interval is defined, it will appear
that after some time the temperature field converges to the steady-state solution.
2.1 Buckling of a tree
Buckling occurs due to axial compressive forces acting on a structure.
Moreover, buckling can occur at stresses that are lower than the maximum
stresses in order to reach failure. Therefore, it is crucial when designing
structures that require slender columns that they are able to handle buckling
stress as well. When mass is applied the to the top of a tree it is assumed it
has a massless truck, at a certain length the tree will begin to buckle as seen
in figure 2.1.1. Intuitively, as the critical length Lc to increase as a function
of EI, where E is Young’s Modulus and I as the second moment of area.
Young’s modulus represents the object’s resistance to deform, and I
represents an object’s efficiency to resist bending due to an applied load;
EI is also known as the flexure rigidity of a material. Lc can be concluded
to decrease has a function of mass (m). A larger mass will result in a large
force applied which will ultimately increase the bending moment the base
of the tree as the length increases.
Assuming small strains and small displacement, the flexure displacement
field w(x) is governed by the following Ordinary Differential Equation:
∀𝑥 ∈ [0, 𝐿], 𝑤4(𝑥) + 𝑘2
𝑤′′(𝑥) = 0 (2.1.1)
The general solution of the ODE is the following:
𝑤4
+ 𝑘2
𝑤2
= 0 (2.1.2)
𝑟2(𝑟2
+ 𝑟2) = 0 (2.1.3)
𝑟1,2 = 0 (2.1.4)
𝑟3,4 = ±𝑖𝑘 (2.1.5)
𝑤(𝑥) = 𝑐1 + 𝑐2 𝑥 + 𝑐3 cos(𝑘𝑥) + 𝑐4sin(𝑘𝑥) (2.1.6)
In order to define a unique solution for the general solution it requires to 4 boundary equations to solve the
4 unknowns, 2 boundary conditions at the fixed end and 2 boundary conditions at the free end. The 4
boundary conditions can be given as the following:
𝑤(0) = 0 (2.1.7)
Figure 2.1.1: Simplified Tree Under
Localized Foliage Weight

7. 𝑤′(0) = 0 (2.1.8)
𝑤(3)(𝐿) + 𝑘2
𝑤′(𝐿) = 0 (2.1.9)
for the forth boundary condition we can consider the moment acting at the end of the tree which is given
as:
𝑤′′(𝐿) = 0 (2.1.10)
With the defined boundary conditions, the solution can be determined. However, it can be noted by 2.1.6,
that the trivial solution is when 𝑤(0) = 0. This is trivial since the tree will not have an displacement at the
fixed end. Applying the boundary conditions, can be found by solving a linear system of equations which
can be written in the form:
𝑀𝑦 = 0 (2.1.11)
𝑤(0) = 𝑐1 + 𝑐3 = 0 (2.1.12)
𝑤′(0) = 𝑐2 − 𝑐3 𝑘𝑠𝑖𝑛(𝑘𝑥) + 𝑐4kcos(𝑘𝑥) = 0 (2.1.13)
𝑤3(𝐿) + 𝑘2
𝑤′(𝐿) = 𝑐3 𝑘3
sin(𝑘𝑥) − 𝑐4 𝑘3
sin(𝑘𝑥) + 𝑘2
𝑐2 − 𝑐3 𝑘3
sin(𝑘𝑥) (2.1.14)
+𝑐4 𝑘3
cos(𝑘𝑥) = 0
𝑤′′(𝐿) = −𝑐3 𝑘2
cos(𝐾𝐿) − 𝑐4 𝑘2
sin(𝐾𝐿) = 0 (2.1.15)
(
1 0 1 0
0 1 0 𝑘
0 𝑘2
0 0
0 0 −𝑘2
cos(𝑘𝐿) −𝑘2
sin(𝐾𝐿)
) (
𝑐1
𝑐2
𝑐3
𝑐4
) = (
0
0
0
0
) (2.1.16)
Applying the boundary conditions, the general solution can be obtained as the following:
𝑤(𝑥) = 𝑐1 − 𝑐1 ∗ cos(𝑘𝑥) (2.1.17)
Taking the determinant of M:
𝑘5
cos(𝑘𝐿) = 0 𝑘 ≠ 0 (2.1.18)
cos(𝑘𝐿) = 0 (2.1.19)
L =
𝜋(2𝑛−1)
2𝑘
𝑛 = 1,2,3 … 𝑛 (2.1.20)
n represents the modes; we will only consider the first mode as it is the most critical for buckling
columns.
𝑓𝑜𝑟 𝑛 = 1 𝑎𝑛𝑑 𝑘2
=
𝑚𝑔
𝐸𝐼
(2.1.21)
𝐿2
=
𝜋2 𝐸𝐼
4𝑚𝑔
(2.1.22)
It is important to note that since the determinant of the matrix M is zero, it implies that the matrix is
noninvertible. As a result, there are infinitely many solutions for the general solution 2.1.17.

8. The second moment of area of a circular section with respect to any axis of symmetry can be obtained by
defining in the plane (y,z) which I can be found as a function of r by:
𝐼 = ∬ 𝑦2
𝑑𝑧𝑑𝑦 (2.1.24)
𝐼 =
𝜋𝑟2
4
(2.1.25)
Given the values for 𝜌 =
1000𝑘𝑔
𝑚3 , 𝐸 = 15 𝐺𝑃𝑎 and r=15cm. The mass is can be assumed to the the mass
of the truck which is obtained by:
𝑚 = 𝜌 ∗ 𝑉 (2.1.26)
where V is the volume of a cylinder.
𝑚 = 𝜌 ∗ 𝜋 ∗ 𝑟2
∗ 𝐿 (2.1.27)
finally, the critical length can be obtained by equation 2.1.22 and 2.1.27:
𝐿 𝑐 = √ 𝜋∗15∗109∗(𝜋∗
0.154
4
)
4∗1000∗9.81
3
= 27.69 𝑚 (2.1.28)
As seen in figure 2.1.2 at the critical length was found to be
27.69 m where the tree has a maximum deflection, whereas
at the fixed end there in no deflection. A street lamp was
taken as an everyday object that experiences buckling. In
order to calculate the critical length of a street lamp, new
physical properties must be changed. The elastic modulus of
a street lamp was taken as 200 GPa and the density as 7870
kg/m3 [4]
. In figure 2.1.3 the critical length of a street lamp
from a load applied at the top is 33m. For street lamp with
the same diameter of the tree, the critical length has
increased. Even though the the mass of the street lamp
increases by almost a factor of 8 which should decrease the
length; the elastic modulus had increase by a factor of 13
which should increase the height. Since the increase of the
elastic modulus was larger than the increase in mass, resulted
in an increase in critical length. In figure 2.1.4 the radius of
a street lamp was taken to be smaller as a more accurate
representation of its critical length. The new critical length
of a street lamp was found be to 52.39 m. As the radius
decreases, the volume and therefore the mass decreases in relation resulting in an increase in length to the
point of maximum deflection. The deflection of the tree is determined by the constant c1 in equation 2.2.17,
as previously explained there are infinitely many solutions; therefore, to increase the deflection c1 should
be increased and vice versa.
Figure2. 3.2 Deflection Vs the Critical Length of a Buckling Tree

9. 2.2 Density Load:
To find the critical length of a more realistic problem, gravity is assumed to act uniformly distributed load
long the height of the tree; this can be identified as self buckling column and seen in figure 2.2.1.
The governing ordinary differential equation for this model is stated as:
𝑤(4)
(𝑥) + 𝑞𝑥𝑤′′(𝑥) + 𝑞𝑤′(𝑥) = 0 (2.2.1)
The fourth order differential equation can suggest the need of 4 boundary conditions. Which are similar to
the previously section, however, in this case the x axis has been changed for simplicity. The four boundary
conditions are as the following:
𝑤(𝐿) = 0 𝑤′(𝐿) = 0 𝑤′′′(0) = 0 𝑤′′(0) = 0 (2.2.2)
Figure 2.2.1: Deflection of a tree due to
Distributed load along its length
Figure 4.1.3: Deflection of Street lamp with radius R=0.15m with
critical length as 33m
Figure2.1.4: Deflection of Street lamp with radius R=0.075m
with critical length as 52.39m

10. Before the general solution is obtained the ODE must be rewritten with z(r). Therefore, the ODE is first
integrated once and 𝑤′(𝑥) = 𝜃(𝑥) = √ 𝑥𝑧(𝑥) is set applied to achieve the following:
0 = −
1
4𝑥
(−
3
2
)
𝑧′
(𝑥) +
1
√ 𝑥
𝑧′(𝑥) + √ 𝑥𝑧′′(𝑥) (2.2.3)
Then the following is substituted 𝑥3
= 𝑟2
into equation 2.2.3 to get:
𝑟
1
3 𝑧′′(𝑥) + 𝑟−
1
3 𝑧′(𝑥) + 𝑞𝑟𝑧(𝑥) −
1
4𝑟
𝑧 = 0 (2.2.4)
in order to get an equation in terms of z(r), a change of variables must be applied.
𝑑𝑧
𝑑𝑥
=
3
2
𝑟
1
3 (
𝑑𝑧
𝑑𝑟
) (2.2.5)
𝑑2 𝑧
𝑑𝑥2 =
9
4
𝑟
2
3 (
𝑑2 𝑧
𝑑𝑟2) +
3
4
𝑟−
1
3 (
𝑑𝑧
𝑑𝑟
) (2.2.6)
Substituting 2.25 and 2.2.6 into 2.2.4 the final ODE is achieved in terms of z(r) as:
𝑟2 𝑑2 𝑧
𝑑𝑟2 + 𝑟
𝑑𝑧
𝑑𝑟
+ (
4
9
𝑞𝑟2
−
1
9
) 𝑧(𝑟) = 0 (2.2.7)
The general solution to the 2.2.7 is said to be:
𝑧(𝑥(𝑟)) = 𝐴𝐽 𝑛(𝜅𝑟) + 𝐵𝐽−𝑛(𝜅𝑟) (2.2.8)
with 𝑞 =
𝜌𝑆𝑔
𝐸𝐼
, 𝜅2
=
4
9
𝑞 and 𝑛 =
1
3
where the Bessel function of the first kind is given:
𝐽 𝑛(𝑦) =
𝑦 𝑛
√ 𝜋 Γ(n+
1
2
)2 𝑛
∫ cos(𝑦𝑐𝑜𝑠(𝜙)
𝜋
0
sin2𝑛
𝜙 𝑑𝜙 (2.2.9)
at n= -1/3
Equation 2.2.8 can be rewritten in terms of 𝜃, 𝑥, 𝐴 and B as the following:
𝜃(𝑥) = √ 𝑥 (𝐴𝐽 𝑛(𝜅𝑥3/2
) + 𝐵𝐽−𝑛(𝜅𝑥3/2
)) (2.2.10)
Finally using the general solution to the fourth order differential equation, the critical length required for
the tree to self buckle can be determined using numerical approximation.
In order to solve the Bessel function, the integral must be approximated using numerical approximation.
However, the integral within the Bessel function cannot be evaluated at the bounds as it goes in infinity.
Therefore, the recurrence relationship for Bessel function must be implemented as the following:
𝐽 𝑛−1 + 𝐽 𝑛+1 = (
2𝑛
𝑦
) 𝐽 𝑛 (2.2.11)
Equation 2.2.11 is rearranged in order to evaluate the integral as:

11. 𝐽 𝑛−1 = (
2𝑛
𝑦
) 𝐽 𝑛 − 𝐽 𝑛+1 (2.2.12)
now we can replace n by 2/3 to achieve the Bessel function of the first kind as:
𝐽−
1
3
(𝑦) = (
4
3𝑦
) 𝐽2
3
(𝑦) − 𝐽5
3
(𝑦) (2.2.13)
The gamma function was evaluated with n=2/3, well as the integral of cos(𝑦𝑐𝑜𝑠(𝜙) sin2𝑛
𝜙 using
Simpson’s method. Once again the Bessel function is evaluated with n=5/3. Finally, the Bessel function at
n=-1/3 can be computed. In order to find the critical length, the first root of the Bessel function when
n=-1/3 must be obtained. Using secant method, two bounds are chosen in order to find the first root, which
was obtained to be 1.8667. To find the critical length we can use the following relationship:
𝑦 = 𝜅 ∗ 𝑟 = 𝜅 ∗ 𝑥
3
2 (2.2.14)
where
𝜅2
=
4
9
𝜌𝑆𝑔
𝐸𝐼
=
4∗1000∗(𝜋∗0.152)∗9.81
9∗(15∗109)∗
𝜋∗0.154
4
= 0.007188 (2.2.15)
and
1.867 = 0.007188 ∗ 𝑥
2
3 (2.2.16)
𝐿 𝑐 = 𝑥 = 40.698 𝑚
In comparison to the results obtain in the previous section,
the critical length has increased. This is due to the uniformly
distributed load along the length of the tree, opposed to a
mass acting at a localized point. Buckling can be explained
by Euler’s theory of buckling. As deflections are caused by
the bending moments along the beam. When there is a
localized point force the bending moment is very large.
When the load is distributed it results in smaller moments,
therefore, smaller deflections and ultimately a larger critical
length. Figure 2.2.2 illustrate the relationship between the
critical length with the deflection of the tree for a distributed
load. It is important to note that the large deflection is due
to the constant B in equation 2.2.10 is set to 1. Since the
general solution can have infinitely many solutions, the
constant B can be reduced to allow for smaller deflections.
In order to numerically check that the numerical
approximation of the solution obtained is correct, it was
found in literature that the critical length of the beam was
found to be:
𝑙 𝑐
3
=
9𝐸𝐼
4𝑞1
𝐽1
3
2
(2.2.17)
Where 𝐽1
3
2
is said to be 1.866 and 𝑞 = 𝜌𝑔𝑆 [5]
. Therefore, using 2.2.17, the numerical solution for the critical
length is 40.69 m, which is exactly the same value obtained from our numerical approximation.
Figure 2.2.2: Deflection of tree due to uniform distributed Load along its
length, where the constant "B" was set to B=1

12. References
[1] Shewchuk, J. R. (2018, April 5). Delaunay Mesh Generation. Lecture presented at Department of
Electrical Engineering and Computer Sciences University of California at Berkeley. Retrieved from
https://people.eecs.berkeley.edu/~jrs/meshpapers/delnotes.pdf
[2] Delaunay. (n.d.). Retrieved April 2, 2018, from
https://www.mathworks.com/help/matlab/ref/trisurf.html
[3] Urroz, G. E. (n.d.). Numerical Solution to Ordinary Differential Equations. Retrieved April 4,
2018, from
http://ocw.usu.edu/Civil_and_Environmental_Engineering/Numerical_Methods_in_Civil_Engineering/O
DEsMatlab.pdf
[4]“AISI 1020 Low Carbon/Low Tensile Steel.” AZoM.com, 11 June 2013,
www.azom.com/article.aspx?ArticleID=6114.
[5]M. (2018, April 10). Structural Mechanics Lecture 10. Lecture. Retrieved from
https://ocw.mit.edu/courses/mechanical-engineering/2-080j-structural-mechanics-fall-2013/course-
notes/MIT2_080JF13_Lecture10.pdf