QUESTION FIVE
1) Determine the symmetrical and boundary conditions.
(2) Create the finite element model of plane stress and determine the stress and strain states at
points A and B when a=0.2m.
In order to determine the stress and strain states at points A and B, the position of A and B
must be determined on the figure on ansys and the corresponding node is read off the
screen.
From the above diagrams it can be seen that point A corresponds to node 5 and point B
corresponds to the node 4.
Now by plotting the results for our stress and strain, the following is obtained.
STRESS
THE FOLLOWING X,Y,Z VALUES ARE IN GLOBAL COORDINATES
NODE
1
2
3
4
5
SX
SY
0.10002E+07 0.28170E+06
0.10000E+07 38.931
0.84524E+06 -138.55
-7893.3
-0.12807E+07
0.33386E+07 13767.
SZ
0.0000
0.0000
0.0000
0.0000
0.0000
SXY
0.66699
-8.3874
-3.4727
1826.1
314.83
SYZ
0.0000
0.0000
0.0000
0.0000
0.0000
SXZ
0.0000
0.0000
0.0000
0.0000
0.0000
EPELX
EPELY
EPELZ
EPELXY
EPELYZ
0.44237E-05-0.88702E-07-0.18578E-05 0.83777E-11 0.0000
0.48310E-05-0.14491E-05-0.14494E-05-0.10535E-09 0.0000
0.40835E-05-0.12257E-05-0.12248E-05-0.43618E-10 0.0000
0.18180E-05-0.61757E-05 0.18676E-05 0.22937E-07 0.0000
0.16108E-04-0.47720E-05-0.48585E-05 0.39544E-08 0.0000
EPELXZ
0.0000
0.0000
0.0000
0.0000
0.0000
STRAIN
NODE
1
2
3
4
5
From the above results it can be concluded that for this specific model, the stress tensors
are:
 0.3338  10 7
314.83 0 


A
314.83 1.3767  10 4 0  Pa



0
0 0 


 - 7.893  10 4
1.826  10 4 0 


B   1.826  10 4 - 1.2807  10 6 0  Pa



0
0 0 


And the strain tensors for this model at these specific points are:
0
 0.16108 10 4 0.39544  10-8

0
A   0.39544  10-8 - 0.4772  10-5


0
0 - 0.4858 10-5

0
 0.1818  10 5 0.22937  10-7

0
B   0.22937  10-7 - 0.6175  10-5


0
0 0.1867  10-5



 Pa





 Pa



(3) Graph the stress xx in line AC and go on to determine the stress concentration factor (a=0.2).
In order to determine a graph for sigma xx along line AC, we must first determine the
amount of nodes along this line and in turn we can calculate the distance between each
node, there are 28 nodes along line AC, and point A starts at 0.2m and point C is at 1m,
hence the distance between each node is (1-0.2)/28 = 2.857cm
These results were all listed on excel and the corresponding stress in the x direction for each
node was identified.
Sigma xx in line AC
4.00E+06
Stress (sigma xx) (Pa)
3.50E+06
3.00E+06
2.50E+06
2.00E+06
1.50E+06
1.00E+06
5.00E+05
0.00E+00
0
0.1
0.2
0.3
0.4
0.5
Distance (m)
0.6
0.7
0.8
0.9
1
node
5
271
269
267
265
263
261
259
257
255
253
251
249
247
245
243
241
239
237
235
233
231
229
227
225
223
221
219
3
Distance (m)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
0.2
0.228571428
0.257142856
0.285714284
0.314285712
0.34285714
0.371428568
0.399999996
0.428571424
0.457142852
0.48571428
0.514285708
0.542857136
0.571428564
0.599999992
0.62857142
0.657142848
0.685714276
0.714285704
0.742857132
0.77142856
0.799999988
0.828571416
0.857142844
0.885714272
0.9142857
0.942857128
0.971428556
0.999999984
stress
(Pa)
3.34E+06
2.52E+06
2.05E+06
1.76E+06
1.58E+06
1.46E+06
1.37E+06
1.31E+06
1.26E+06
1.22E+06
1.19E+06
1.16E+06
1.14E+06
1.12E+06
1.10E+06
1.08E+06
1.07E+06
1.05E+06
1.04E+06
1.02E+06
1.01E+06
9.90E+05
9.74E+05
9.57E+05
9.39E+05
9.17E+05
8.95E+05
8.72E+05
8.45E+05
To determine the stress concentration factor:
The stress concentration factor can be determined by obtaining the maximum stress that
occurs, which is at point A
𝜎𝑚𝑎𝑥
Factor of safety = 𝜎𝑎𝑝𝑙𝑙𝑖𝑒𝑑
=
3.34×106
1×106
=3.34
(4) Investigate the effect of radius a on the stress level (at point A) through the finite element
modelling. Please verify the numerical result against the formulae in Section 8.5 of textbook.
To investigate the effect of of radius a on the stress level at point a, we need to create a
different Ansys models for each of the radius. We are initially given a radius of 0.2m and so
for the analysis we will choose to take 2 different radii, one that 0.1m below given and one
that is 0.1m above the given, hence we will analyse the effect of radius at 0.1, 0.2, 0.3m.
Stress at radius 0.1m
THE FOLLOWING X,Y,Z VALUES ARE IN GLOBAL COORDINATES
NODE
1
2
3
4
5
SX
SY
0.10000E+07 66289.
0.10000E+07 9.4150
0.96458E+06 -33.024
-24595.
-0.10271E+07
0.30135E+07 44579.
SZ
0.0000
0.0000
0.0000
0.0000
0.0000
SXY
-1.2684
-2.1150
-0.83423
3855.7
-4098.9
SYZ
0.0000
0.0000
0.0000
0.0000
0.0000
SXZ
0.0000
0.0000
0.0000
0.0000
0.0000
SYZ
0.0000
0.0000
0.0000
0.0000
0.0000
SXZ
0.0000
0.0000
0.0000
0.0000
0.0000
SYZ
0.0000
0.0000
0.0000
0.0000
0.0000
SXZ
0.0000
0.0000
0.0000
0.0000
0.0000
Stress at radius 0.2m
THE FOLLOWING X,Y,Z VALUES ARE IN GLOBAL COORDINATES
NODE
1
2
3
4
5
SX
SY
0.10002E+07 0.28170E+06
0.10000E+07 38.931
0.84524E+06 -138.55
-7893.3
-0.12807E+07
0.33386E+07 13767.
SZ
0.0000
0.0000
0.0000
0.0000
0.0000
SXY
0.66699
-8.3874
-3.4727
1826.1
314.83
Stress at radius 0.3m
THE FOLLOWING X,Y,Z VALUES ARE IN GLOBAL COORDINATES
NODE
1
2
3
4
5
SX
SY
0.10005E+07 0.70367E+06
0.10001E+07 86.677
0.59650E+06 -327.64
-6735.4
-0.17201E+07
0.38666E+07 10033.
SZ
0.0000
0.0000
0.0000
0.0000
0.0000
SXY
4.8918
-20.133
-5.0809
769.27
647.51
From the above stress tables it can be seen that for all different variations in the radius of
the circle the stress of the x component at point a (node 5) remains almost constant with
minor variations throughout, this shows that no matter what the radius will be the x
component of the stress will remain the same throughout at point A
Now to verify the numerical results against the formulae in section 8.5
𝜎𝜃𝜃 = 𝜎𝑥𝑥 =
𝑇
𝑎2
𝑇
3𝑎4
[1 + 2 ] − [1 + 4 ] 𝑐𝑜𝑠2𝜃
2
𝑟
2
𝑟
Now for that numerical equation, if the point on a model corresponds to the outer edge of the
radius, for example in our case point A corresponds to 0.2m which is the same as the radius of the
circle. Hence in this case a will be equal to r and so as a result all values of a/r will be equal to 1. This
validation should in turn apply to all values of A and r, and this has been proven through our ansys
analysis as all the variations in the radius still produce an almost identical stress at point A in the x
component.
Further proof of the equation:
=
𝑇
𝑎2
3𝑎4
[1 + 2 + 1 + 4 ] 𝑠𝑖𝑛𝑐𝑒 𝜃 = 90°
2
𝑟
𝑟
𝑇
0.12
3 × 0.14
[1 +
+
1
+
]
2
0.12
0.14
=
=
1 × 106
[1 + 1 + 1 + 3]
2
= 3 × 106
This result should be the same for a radius of 0.1m, 0.2m and 0.3m.
(5) If change the structure from the thin plate to a thick prismatic block with a long through hole (the
figure below becomes a cross-sectional view), compare the stress and strain states at A and B via
finite element modelling?
In order to change the plate from a thin plate to a thick plate, the thickness in ansys must be
changed to a selected variable, hence in this case thickness was changed from 0.1m to a thickness of
2m, the reason that 2m was selected was simple because the width and length are 1m each and so
in order to obtain a prismatic block the thickness must be greater than the length and width.
Ansys model obtained after meshing:
From the above model the stress and strain at both points A (node 5) and point B (node 4) were
analysed and compared to previous results.
STRESS (FOR THICK PLATE THICKNESS = 2m)
THE FOLLOWING X,Y,Z VALUES ARE IN GLOBAL COORDINATES
NODE
1
2
3
4
5
SX
SY
0.10002E+07 0.28170E+06
0.10000E+07 38.931
0.84524E+06 -138.55
-7893.3
-0.12807E+07
0.33386E+07 13767.
SZ
0.0000
0.0000
0.0000
0.0000
0.0000
SXY
0.66699
-8.3874
-3.4727
1826.1
314.83
SYZ
0.0000
0.0000
0.0000
0.0000
0.0000
SXZ
0.0000
0.0000
0.0000
0.0000
0.0000
EPELYZ
0.0000
0.0000
0.0000
0.0000
0.0000
EPELXZ
0.0000
0.0000
0.0000
0.0000
0.0000
SYZ
0.0000
0.0000
0.0000
0.0000
0.0000
SXZ
0.0000
0.0000
0.0000
0.0000
0.0000
STRAIN (FOR THICK PLATE THICKNESS = 2m)
THE FOLLOWING X,Y,Z VALUES ARE IN GLOBAL COORDINATES
NODE
1
2
3
4
5
EPELX
EPELY
EPELZ
EPELXY
0.44237E-05-0.88702E-07-0.18578E-05 0.83777E-11
0.48310E-05-0.14491E-05-0.14494E-05-0.10535E-09
0.40835E-05-0.12257E-05-0.12248E-05-0.43618E-10
0.18180E-05-0.61757E-05 0.18676E-05 0.22937E-07
0.16108E-04-0.47720E-05-0.48585E-05 0.39544E-08
STRESS (FOR THIN PLATE THICKNESS = 0.1m)
THE FOLLOWING X,Y,Z VALUES ARE IN GLOBAL COORDINATES
NODE
1
2
3
4
5
SX
SY
0.10002E+07 0.28170E+06
0.10000E+07 38.931
0.84524E+06 -138.55
-7893.3
-0.12807E+07
0.33386E+07 13767.
SZ
0.0000
0.0000
0.0000
0.0000
0.0000
SXY
0.66699
-8.3874
-3.4727
1826.1
314.83
STRAIN (FOR THIN PLATE THICKNESS = 0.1m)
NODE
1
2
3
4
5
EPELX
EPELY
EPELZ
EPELXY
EPELYZ
0.44237E-05-0.88702E-07-0.18578E-05 0.83777E-11 0.0000
0.48310E-05-0.14491E-05-0.14494E-05-0.10535E-09 0.0000
0.40835E-05-0.12257E-05-0.12248E-05-0.43618E-10 0.0000
0.18180E-05-0.61757E-05 0.18676E-05 0.22937E-07 0.0000
0.16108E-04-0.47720E-05-0.48585E-05 0.39544E-08 0.0000
EPELXZ
0.0000
0.0000
0.0000
0.0000
0.0000
From the above results it can be clearly seen that a change in the thickness of the plate will
have absolutely no effect on the points A and B, as the stress and strain of all components at
points A and B for different thicknesses are exactly the same.
Stress Tensors for 2m thickness.
 0.3338  107
314.83 0 


A
314.83 1.3767  104 0  Pa



0
0 0 


 - 7.893  10 4
1.826  10 4 0 


B   1.826  10 4 - 1.2807  10 6 0  Pa



0
0 0 


Strain Tensors for 2m thickness.
0
 0.16108 10 4 0.39544  10-8

0
A   0.39544  10-8 - 0.4772  10-5


0
0 - 0.4858 10-5

0
 0.1818  10 5 0.22937  10-7

0
B   0.22937  10-7 - 0.6175  10-5


0
0 0.1867  10-5

QUESTION SIX
(a) Define the boundary condition


 Pa





 Pa



b) Calculate the principal stresses at point A and B for plane stress (t=0.1m) and
plane strain problems, respectively.
The principal stresses for the ansys plane stress diagram is shown below and the principal
stresses at points A and B are taken from nodes 4 and 5.
NODE
1
2
3
4
5
S1
S2
S3
0.0000
-0.39327E+07-0.36125E+08
0.0000
-0.10005E+07-0.50442E+08
0.0000
-0.99401E+06-0.41477E+08
0.40623E+08 0.29613E+06 0.0000
0.0000
-0.52431E+06-0.13091E+09
SINT
0.36125E+08
0.50442E+08
0.41477E+08
0.40623E+08
0.13091E+09
SEQV
0.34328E+08
0.49950E+08
0.40989E+08
0.40476E+08
0.13064E+09
Hence the principal stresses at point A are:
Stress 1 = 0 Pa
Stress 2 = -0.52431E6 Pa
Stress 3 = -0.13091E9 Pa
And the principal stresses at point B are:
Stress 1 = 0.4062E8 Pa
Stress 2 = 0.29613E6 Pa
Stress 3 = 0 Pa
Similarly on ansys the diagram, we can obtain a plane strain diagram and repeat the same
process to obtain the principal stresses at the points A and B. The diagram is on the
following page along with the principal stresses.
NODE
S1
S2
S3
SINT
SEQV
1
2
3
4
5
-0.51788E+07-0.51539E+08-0.63797E+08
-0.10008E+07-0.68719E+08-0.72123E+08
-0.99134E+06-0.59163E+08-0.64828E+08
0.59344E+08 0.39498E+06-0.28860E+08
-0.73477E+06-0.10322E+09-0.18740E+09
0.58619E+08
0.71122E+08
0.63837E+08
0.88204E+08
0.18667E+09
0.53552E+08
0.69483E+08
0.61201E+08
0.77817E+08
0.16192E+09
Hence the principal stresses at point A are:
Stress 1 = -0.73477E6 Pa
Stress 2 = -0.10322E9 Pa
Stress 3 = -0.18740E9 Pa
And the principal stresses at point B are:
Stress 1 = 0.59344E8 Pa
Stress 2 = 0.39498E6 Pa
Stress 3 = -0.28860E8 Pa
c) Plot the relation between delta T (-60 – 60 degrees celcius) and these principal
stresses.
In order to complete this part of the question, there is a large number of variables that must be
obtained and these must in turn be compared and graphed. The principal stresses at points A and B
must be obtained for the plain stress and plain strain for different temperatures ranging from -60 to
60.
Plain Stress
For a temperature of 60oC, we get the following stresses:
NODE
S1
S2
S3
SINT
SEQV
1
0.0000
-0.97105E+07-0.10760E+09 0.10760E+09 0.10309E+09
2
0.0000
-0.10017E+07-0.15097E+09 0.15097E+09 0.15047E+09
3
0.0000
-0.98170E+06-0.12349E+09 0.12349E+09 0.12300E+09
4 0.12750E+09 0.89019E+06 0.0000
0.12750E+09 0.12706E+09
5
0.0000
-0.15891E+07-0.39294E+09 0.39294E+09 0.39214E+09
Temp of 40oC
NODE
S1
S2
S3
1
0.0000
-0.68216E+07-0.71863E+08
2
0.0000
-0.10011E+07-0.10071E+09
3
0.0000
-0.98786E+06-0.82481E+08
4 0.84063E+08 0.59316E+06 0.0000
5
0.0000
-0.10567E+07-0.26192E+09
SINT
0.71863E+08
0.10071E+09
0.82481E+08
0.84063E+08
0.26192E+09
SEQV
0.68706E+08
0.10021E+09
0.81992E+08
0.83768E+08
0.26139E+09
Temp of 20oC
NODE
S1
S2
S3
SINT
SEQV
1
2
3
4
5
0.0000
-0.39327E+07-0.36125E+08
0.0000
-0.10005E+07-0.50442E+08
0.0000
-0.99401E+06-0.41477E+08
0.40623E+08 0.29613E+06 0.0000
0.0000
-0.52431E+06-0.13091E+09
0.36125E+08
0.50442E+08
0.41477E+08
0.40623E+08
0.13091E+09
0.34328E+08
0.49950E+08
0.40989E+08
0.40476E+08
0.13064E+09
SINT
0.10438E+07
0.99996E+06
0.10002E+07
0.28175E+07
0.10810E+06
SEQV
0.91402E+06
0.92404E+06
0.86658E+06
0.28170E+07
0.10431E+06
SINT
0.35351E+08
0.51087E+08
0.41537E+08
0.46258E+08
0.13112E+09
SEQV
0.34466E+08
0.50594E+08
0.41043E+08
0.46109E+08
0.13085E+09
SINT
0.71089E+08
0.10135E+09
0.82547E+08
0.89698E+08
0.26214E+09
SEQV
0.68844E+08
0.10086E+09
0.82046E+08
0.89402E+08
0.26160E+09
SINT
0.10683E+09
0.15161E+09
0.12356E+09
0.13314E+09
0.39315E+09
SEQV
0.10323E+09
0.15112E+09
0.12305E+09
0.13269E+09
0.39235E+09
SINT
0.17704E+09
0.21473E+09
0.19256E+09
0.26839E+09
0.55948E+09
SEQV
0.16180E+09
0.20974E+09
0.18455E+09
0.23735E+09
0.48531E+09
SINT
0.11783E+09
0.14293E+09
0.12820E+09
0.17830E+09
0.37308E+09
SEQV
0.10768E+09
0.13961E+09
0.12288E+09
0.15758E+09
0.32361E+09
Temp of 0oC
NODE
S1
S2
S3
1
0.0000
-0.38687E+06-0.10438E+07
2
0.0000
-0.17762E+06-0.99996E+06
3
0.0000
-0.47321E+06-0.10002E+07
4
0.0000
-890.27
-0.28175E+07
5 0.10810E+06 8044.5
0.0000
Temp of -20 oC
NODE
S1
S2
S3
1 0.35351E+08 0.18451E+07 0.0000
2 0.50087E+08 0.0000
-0.99937E+06
3 0.40531E+08 0.0000
-0.10063E+07
4
0.0000
-0.29794E+06-0.46258E+08
5 0.13112E+09 0.54044E+06 0.0000
Temp of -40 oC
NODE
S1
S2
S3
1 0.71089E+08 0.47340E+07 0.0000
2 0.10035E+09 0.0000
-0.99878E+06
3 0.81535E+08 0.0000
-0.10125E+07
4
0.0000
-0.59497E+06-0.89698E+08
5 0.26214E+09 0.10728E+07 0.0000
Temp of -60 oC
NODE
S1
S2
S3
1 0.10683E+09 0.76229E+07 0.0000
2 0.15062E+09 0.0000
-0.99820E+06
3 0.12254E+09 0.0000
-0.10186E+07
4
0.0000
-0.89200E+06-0.13314E+09
5 0.39315E+09 0.16052E+07 0.0000
Plane strain:
At a temp of 60 oC
NODE
S1
S2
S3
1 -0.13433E+08-0.15365E+09-0.19047E+09
2 -0.10025E+07-0.20537E+09-0.21574E+09
3 -0.97373E+06-0.17632E+09-0.19353E+09
4 0.18343E+09 0.11841E+07-0.84962E+08
5 -0.22174E+07-0.30952E+09-0.56170E+09
At a temp of 40 oC
NODE
S1
S2
S3
1 -0.93058E+07-0.10259E+09-0.12713E+09
2 -0.10017E+07-0.13704E+09-0.14393E+09
3 -0.98253E+06-0.11774E+09-0.12918E+09
4 0.12139E+09 0.78954E+06-0.56911E+08
5 -0.14761E+07-0.20637E+09-0.37455E+09
At a temp of 20 oC
NODE
S1
S2
S3
1 -0.51788E+07-0.51539E+08-0.63797E+08
2 -0.10008E+07-0.68719E+08-0.72123E+08
3 -0.99134E+06-0.59163E+08-0.64828E+08
4 0.59344E+08 0.39498E+06-0.28860E+08
5 -0.73477E+06-0.10322E+09-0.18740E+09
SINT
0.58619E+08
0.71122E+08
0.63837E+08
0.88204E+08
0.18667E+09
SEQV
0.53552E+08
0.69483E+08
0.61201E+08
0.77817E+08
0.16192E+09
SINT
0.59071E+06
0.68420E+06
0.52433E+06
0.26981E+07
0.25858E+06
SEQV
0.57907E+06
0.64834E+06
0.47887E+06
0.23981E+07
0.22925E+06
SINT
0.59800E+08
0.72490E+08
0.64886E+08
0.91981E+08
0.18615E+09
SEQV
0.54695E+08
0.70777E+08
0.62152E+08
0.81746E+08
0.16148E+09
SINT
0.11901E+09
0.14430E+09
0.12925E+09
0.18207E+09
0.37256E+09
SEQV
0.10882E+09
0.14091E+09
0.12383E+09
0.16151E+09
0.32317E+09
SINT
0.17822E+09
0.21610E+09
0.19361E+09
0.27217E+09
0.55897E+09
SEQV
0.16294E+09
0.21104E+09
0.18551E+09
0.24128E+09
0.48487E+09
At a temp of 0 oC
NODE
S1
S2
S3
1 -0.46106E+06-0.48509E+06-0.10518E+07
2 -0.31576E+06-0.39472E+06-0.99996E+06
3 -0.47582E+06-0.58591E+06-0.10001E+07
4
423.96
-0.80919E+06-0.26977E+07
5
6540.3
-73648.
-0.25203E+06
At a temp of -20 oC
NODE
S1
S2
S3
1 0.62875E+08 0.50569E+08 0.30752E+07
2 0.71491E+08 0.67930E+08-0.99911E+06
3 0.63877E+08 0.57991E+08-0.10090E+07
4 0.27242E+08-0.39416E+06-0.64740E+08
5 0.18690E+09 0.10308E+09 0.74783E+06
At a temp of -40 oC
NODE
S1
S2
S3
1 0.12621E+09 0.10162E+09 0.72023E+07
2 0.14330E+09 0.13625E+09-0.99826E+06
3 0.12823E+09 0.11657E+09-0.10178E+07
4 0.55293E+08-0.78872E+06-0.12678E+09
5 0.37405E+09 0.20623E+09 0.14891E+07
At a temp of -60 oC
NODE
1
2
3
4
5
S1
S2
S3
0.18955E+09 0.15268E+09 0.11329E+08
0.21510E+09 0.20458E+09-0.99741E+06
0.19258E+09 0.17515E+09-0.10266E+07
0.83344E+08-0.11833E+07-0.18882E+09
0.56120E+09 0.30937E+09 0.22304E+07
Hence from the above data we can obtain the following results:
The Principal stresses at point A for different temperature changes, for plain stress.
For Plain Stress at point A
Temp (oC)
S1 (Pa)
S2 (Pa)
S3 (Pa)
60
0
1.59E+06 3.93E+08
40
0
1.06E+06 2.62E+08
20
0
5.24E+05 1.31E+08
0
1.08E+05
8044.5
0
-20
1.31E+08 5.40E+05
0
-40
-60
2.62E+08
3.93E+08
1.07E+06
1.61E+06
0
0
500000000
400000000
Principal Stress Pa
300000000
200000000
For Plain Stress at point A S1
(Pa)
100000000
0
For Plain Stress at point A S2
(Pa)
-1E+08
-2E+08
For Plain Stress at point A S3
(Pa)
-3E+08
-4E+08
-80
-60
-40
-5E+08
-20
0
20
40
60
80
Temperature change
The Principal stresses at point B for different temperature changes, for plain stress
For Plain Stress at point B
Temp (oC)
S1 (Pa)
S2 (Pa)
60
1.28E+08 8.90E+05
40
8.41E+07 5.93E+05
20
4.06E+07 2.96E+05
0
0.00E+00 -890.27
-20
0.00E+00 2.98E+05
-40
0.00E+00 5.95E+05
-60
0.00E+00 8.92E+05
1.50E+08
S3 (Pa)
0.00E+00
0.00E+00
0.00E+00
2.82E+06
4.63E+07
8.97E+07
1.33E+08
Principal Stress Pa
1.00E+08
-80
5.00E+07
-60
-40
0.00E+00
-20
0
For Plain Stress at point B S1
(Pa)
20
40
-5.00E+07
60
80
For Plain Stress at point B S2
(Pa)
For Plain Stress at point B S3
(Pa)
-1.00E+08
-1.50E+08
Temperature (degrees celcius)
The Principal stresses at point A for different temperature changes, for plain strain.
For Plain Strain at point A
Temp (oC)
S1 (Pa)
S2 (Pa)
60
2.22E+06 3.10E+08
40
1.48E+06 2.06E+06
20
7.35E+05 1.03E+08
0
6.54E+03
-73648
-20
1.87E+08 1.03E+08
-40
3.74E+08 2.06E+08
-60
5.61E+08 3.09E+08
S3 (Pa)
5.62E+08
3.75E+08
1.87E+07
2.52E+05
7.48E+05
1.49E+06
2.23E+06
8.00E+08
6.00E+08
Principal Stress Pa
4.00E+08
-80
2.00E+08
For Plain Strain at point A S1 (Pa)
-60
-40
0.00E+00
-20
0
-2.00E+08
20
40
60
80
For Plain Strain at point A S2 (Pa)
For Plain Strain at point A S3 (Pa)
-4.00E+08
-6.00E+08
-8.00E+08
Temperature
The Principal stresses at point B for different temperature changes, for plain strain.
For Plain Strain at point B
Temp (oC)
S1 (Pa)
S2 (Pa)
60
1.83E+08 1.19E+06
40
1.21E+08 7.90E+05
20
5.93E+07 3.95E+05
0
4.24E+02 8.09E+05
-20
2.72E+07 3.94E+05
-40
5.53E+07 7.89E+05
-60
8.33E+07 1.18E+06
S3 (Pa)
8.50E+07
5.69E+07
2.89E+07
2.70E+06
6.47E+07
1.27E+08
1.89E+08
2.50E+08
2.00E+08
Principal Stress Pa
1.50E+08
-80
1.00E+08
For Plain Strain at point B S1
(Pa)
5.00E+07
-60
-40
0.00E+00
-20
0
-5.00E+07
-1.00E+08
-1.50E+08
20
40
60
80
For Plain Strain at point B S2
(Pa)
For Plain Strain at point B S3
(Pa)
-2.00E+08
-2.50E+08
Temperature
d) Calculate these principal stresses again if changing the materials from mild steel to
aluminium and copper.
In order to complete this part of the question we must first determine the material
properties of both Copper and Aluminium, once these have been obtained we can then go
on to ansys and change the material properties and obtain listed results of the principal
stresses for both plane strain and plain stress for aluminium and Copper.
Copper
E = 115 GPa
v = 0.34
Principal stresses for plane stress of Copper
NODE
1
2
3
4
5
S1
S2
S3
0.0000
-0.26512E+07-0.20272E+08
0.0000
-0.10003E+07-0.28145E+08
0.0000
-0.99674E+06-0.23288E+08
0.21352E+08 0.16125E+06 0.0000
0.0000
-0.28605E+06-0.72788E+08
SINT
0.20272E+08
0.28145E+08
0.23288E+08
0.21352E+08
0.72788E+08
SEQV
0.19085E+08
0.27659E+08
0.22806E+08
0.21272E+08
0.72646E+08
SINT
0.34103E+08
0.41719E+08
0.37293E+08
0.48310E+08
0.11034E+09
SEQV
0.31211E+08
0.40818E+08
0.35839E+08
0.42915E+08
0.95908E+08
SINT
0.12510E+08
0.17229E+08
0.14383E+08
0.11918E+08
0.44335E+08
SEQV
0.11631E+08
0.16751E+08
0.13911E+08
0.11869E+08
0.44250E+08
Principal stresses for plane strain of Copper
NODE
1
2
3
4
5
S1
S2
S3
-0.34889E+07-0.30634E+08-0.37592E+08
-0.10005E+07-0.40855E+08-0.42719E+08
-0.99494E+06-0.35177E+08-0.38288E+08
0.33931E+08 0.21854E+06-0.14379E+08
-0.42162E+06-0.63792E+08-0.11076E+09
Aluminium
E = 70 GPa
v = 0.35
Principal stresses for plane strain of Aluminium
NODE
1
2
3
4
5
S1
S2
S3
0.0000
-0.20238E+07-0.12510E+08
0.0000
-0.10002E+07-0.17229E+08
0.0000
-0.99808E+06-0.14383E+08
0.11918E+08 97534.
0.0000
0.0000
-0.17093E+06-0.44335E+08
Principal stresses for plane strain of Aluminium
NODE
1
2
3
4
5
S1
S2
S3
SINT
SEQV
-0.25615E+07-0.19162E+08-0.23423E+08 0.20862E+08 0.19091E+08
-0.10003E+07-0.25475E+08-0.26584E+08 0.25584E+08 0.25048E+08
-0.99692E+06-0.22015E+08-0.23874E+08 0.22877E+08 0.22006E+08
0.19992E+08 0.13287E+06-0.87764E+07 0.28768E+08 0.25508E+08
-0.25688E+06-0.39958E+08-0.68710E+08 0.68453E+08 0.59534E+08