Question 5(2)
NODE
1
2
3
4
5
7
9
11
13
15
SX
SY
SZ
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.
0.10002E+07 0.28043E+06
0.10002E+07 0.27633E+06
0.10002E+07 0.26952E+06
0.10002E+07 0.25913E+06
0.10001E+07 0.24725E+06
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
SXY
0.66699
-8.3874
-3.4727
1826.1
314.83
-12.469
-51.178
-66.515
-132.10
-163.51
SYZ
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
SXZ
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
Where A and B are point 4 and 5 respectively.
3.3386E + 06 3.1483E + 02 0
[𝛔A] = [3.1483E + 02 1.3767E + 04 0]
0
0
0
−7.8933E + 03
[𝛔B] = [ 1.8261E + 03
0
NODE
1
2
3
4
5
7
9
11
13
15
17
19
0
0]
0
1.8261E + 03
−1.2807E + 06
0
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
0.44255E-05-0.94839E-07-0.18560E-05-0.15661E-09 0.0000
0.44314E-05-0.11463E-06-0.18500E-05-0.64282E-09 0.0000
0.44411E-05-0.14747E-06-0.18401E-05-0.83545E-09 0.0000
0.44561E-05-0.19768E-06-0.18250E-05-0.16593E-08 0.0000
0.44732E-05-0.25502E-06-0.18078E-05-0.20538E-08 0.0000
0.44930E-05-0.32123E-06-0.17879E-05-0.22224E-08 0.0000
0.45150E-05-0.39505E-06-0.17657E-05-0.24584E-08 0.0000
1.6108E − 05
[𝜖 A] = [ 3.9544E − 09
0
1.8180E − 06
[𝜖 B] = [ 2.2937E − 08
0
3.9544E − 09
−4.7720E − 06
0
0
0
−4.8585E − 06
2.2937E − 08
−6.1757E − 06
0
0
0
1.8676E − 06
]
]
EPELXZ
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
(3) Graphing the stress sigma xx along the line AC
From the graph below, the stress along the lie Ac decreases from its maximum point
at A. This can be explained by the following factor
-
Higher stress around sharp corners (point A)
Higher stress where notches are present
Sigma xx along the line AC
4,00E+06
Stress, 𝛔xx (MPa)
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
0,6
Distance from Point A, y (m)
0,7
0,8
0,9
Stress concentration factor:
k = 𝛔max/𝛔xx
K= 3338600/1E6
K= 3.3386
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
For a=0.15
NODE
1
2
3
4
5
7
9
11
13
15
17
19
SX
SY
SZ
0.10001E+07 0.15294E+06
0.10000E+07 21.493
0.91731E+06 -74.028
-22086.
-0.11189E+07
0.31647E+07 17882.
0.10001E+07 0.15223E+06
0.10001E+07 0.14997E+06
0.10001E+07 0.14625E+06
0.10001E+07 0.14073E+06
0.10001E+07 0.13429E+06
0.10001E+07 0.12684E+06
0.10000E+07 0.11854E+06
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
SXY
0.23162
-4.6987
-0.49866
3274.2
-409.21
-8.6485
-30.770
-42.807
-74.475
-88.907
-98.307
-110.18
SYZ
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
SXZ
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
For a=0.25
NODE
SX
SY
SZ
0.10003E+07 0.46103E+06
0.10001E+07 59.593
0.74163E+06 -218.55
-6966.2
-0.14659E+07
0.35659E+07 11113.
0.10004E+07 0.45885E+06
0.10002E+07 0.45322E+06
0.10002E+07 0.44303E+06
0.10002E+07 0.42879E+06
0.10001E+07 0.41143E+06
0.10001E+07 0.39118E+06
0.10001E+07 0.36847E+06
0.10001E+07 0.34379E+06
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
SXY
65.991
-13.724
-2.2395
373.23
1012.1
2.6159
65.746
-23.498
-102.49
-150.36
-188.91
-235.54
-277.04
SYZ
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
SXZ
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
SX
SY
SZ
1 0.10005E+10 0.70367E+09
2 0.10001E+10 86677.
3 0.59650E+09-0.32764E+06
4 -0.67354E+07-0.17201E+10
5 0.38666E+10 0.10033E+08
7 0.10005E+10 0.70058E+09
9 0.10005E+10 0.69152E+09
11 0.10004E+10 0.67637E+09
13 0.10004E+10 0.65385E+09
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
SXY
SYZ
4891.8
0.0000
-20133.
0.0000
-5080.9
0.0000
0.76927E+06 0.0000
0.64751E+06 0.0000
-68105.
0.0000
-0.12718E+06 0.0000
-0.17385E+06 0.0000
-0.30026E+06 0.0000
SXZ
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
1
2
3
4
5
7
9
11
13
15
17
19
21
For a=0.3
NODE
15
17
19
0.10003E+10 0.62738E+09
0.10003E+10 0.59655E+09
0.10002E+10 0.56194E+09
0.0000
0.0000
0.0000
-0.36070E+06
-0.41291E+06
-0.45815E+06
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
(Please refer to graph on the next page) Using the formula in Section 8.5 , where maximum stress
value occurs at a=r and 𝜃 = 90°. Upon changing our a values (1.5,2.5 and 3) it is evident that our
stress values at point a remain the same. Mathematically this is because our a and r values are
equal. However from the graph we can see that stress xx, will not remain constant, and does not
correspond to formulae calculations. This formula is applicable for small values of a , with respect to
the length of the rectangle. In our scenario, this makes the formulae not applicable.
𝜎𝑥𝑥 =
𝑇
𝑎2
𝑇
3𝑎 4
+
−
+
[1
]
[1
] 𝑐𝑜𝑠2𝜃
2
𝑟2
2
𝑟4
𝜎𝑚𝑎𝑥 =
𝑇
𝑇
𝑎2
𝑇
3𝑎4
[1 + 2 ] − [1 + 4 ] 𝑐𝑜𝑠180
2
𝑟
2
𝑟
𝑎2
= [1 + 2 + 1 + 3]
2
𝑎
𝑇
2
= [1 + 1 + 1 + 3]
= 3𝑇 𝑊ℎ𝑒𝑟𝑒 𝑇 = 1𝐸6
=3Mpa
Therefore:
The stress level at point A for radii a =0.15, 0.25 and 0.3 and the following results plotted at Point A
(Max stress)
R vs. Max Stress
8
Stress, 𝛔xx, (MPa)
7
6
5
4
3
2
1
0
0
0,1
0,2
0,3
0,4
0,5
0,6
Radius, a (m)
(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?
At thickness T : T= 2
NODE
1
2
3
4
5
7
9
11
13
15
17
19
21
23
25
SX
SY
SZ
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.
0.10002E+07 0.28043E+06
0.10002E+07 0.27633E+06
0.10002E+07 0.26952E+06
0.10002E+07 0.25913E+06
0.10001E+07 0.24725E+06
0.10001E+07 0.23354E+06
0.10001E+07 0.21825E+06
0.10001E+07 0.20177E+06
0.10000E+07 0.18450E+06
0.10000E+07 0.16680E+06
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
SXY
0.66699
-8.3874
-3.4727
1826.1
314.83
-12.469
-51.178
-66.515
-132.10
-163.51
-176.94
-195.73
-208.62
-212.61
-214.79
SYZ
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
Therefore yielding
0.33386E + 07
[𝛔A] = [ 3.1483E + 02
0
−7.8933E + 03
[𝛔B] = [ 1.8261E + 03
0
3.1483E + 02
1.3767E + 04
0
1.8261E + 03
−1.2807E + 06
0
0
0]
0
0
0]
0
SXZ
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
NODE
1
2
3
4
5
7
9
11
13
15
17
19
21
23
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
0.44255E-05-0.94839E-07-0.18560E-05-0.15661E-09 0.0000
0.44314E-05-0.11463E-06-0.18500E-05-0.64282E-09 0.0000
0.44411E-05-0.14747E-06-0.18401E-05-0.83545E-09 0.0000
0.44561E-05-0.19768E-06-0.18250E-05-0.16593E-08 0.0000
0.44732E-05-0.25502E-06-0.18078E-05-0.20538E-08 0.0000
0.44930E-05-0.32123E-06-0.17879E-05-0.22224E-08 0.0000
0.45150E-05-0.39505E-06-0.17657E-05-0.24584E-08 0.0000
0.45388E-05-0.47464E-06-0.17418E-05-0.26204E-08 0.0000
0.45637E-05-0.55804E-06-0.17167E-05-0.26705E-08 0.0000
EPELXZ
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
Therefore yielding
1.6108E − 05
[𝜖 A] = [ 3.9544E − 09
0
1.8180E − 06
[𝜖 B] = [ 2.2937E − 08
0
3.9544E − 09
−4.7720E − 06
0
0
0
−4.8585E − 06
2.2937E − 08
−6.1757E − 06
0
0
0
1.8676E − 06
]
]
Therefore our data suggests that whether we have a prismatic block or a thin plate , our stress and
strain values remain unaltered
Question 6 (b) : Calculate the principal stresses at point A and B for plane stress (t=0.1m) and
plane strain problems, respectively.
ansys plane stress diagram:
NODE
S1
1
2
3
4
5
S2
S3
SINT
SEQV
0.0000
-0.39327E+07-0.36125E+08 0.36125E+08 0.34328E+08
0.0000
-0.10005E+07-0.50442E+08 0.50442E+08 0.49950E+08
0.0000
-0.99401E+06-0.41477E+08 0.41477E+08 0.40989E+08
0.40623E+08 0.29613E+06 0.0000
0.40623E+08 0.40476E+08
0.0000
-0.52431E+06-0.13091E+09 0.13091E+09 0.13064E+09
principal stresses at point A
Stress 1 = 0 Pa
Stress 2 = -0.52431E6 Pa
Stress 3 = -0.13091E9 Pa
NODE
1
2
3
4
5
principal stresses at point B :
Stress 1 = 0.4062E8 Pa
Stress 2 = 0.29613E6 Pa
Stress 3 = 0 Pa
S1
S2
S3
SINT
SEQV
-0.51788E+07-0.51539E+08-0.63797E+08 0.58619E+08 0.53552E+08
-0.10008E+07-0.68719E+08-0.72123E+08 0.71122E+08 0.69483E+08
-0.99134E+06-0.59163E+08-0.64828E+08 0.63837E+08 0.61201E+08
0.59344E+08 0.39498E+06-0.28860E+08 0.88204E+08 0.77817E+08
-0.73477E+06-0.10322E+09-0.18740E+09 0.18667E+09 0.16192E+09
principal stresses : point A
Stress 1 = -0.73477E6 Pa
Stress 2 = -0.10322E9 Pa
Stress 3 = -0.18740E9 Pa
principal stresses : point B
Stress 1 = 0.59344E8 Pa
Stress 2 = 0.39498E6 Pa
Stress 3 = -0.28860E8 Pa
PLANE STRESS
PLANE STRAIN
(c ) Plot the relation between delta T (-60 – 60 degrees Celsius) and these principal stresses.
Refer to appendix for Ansys data
For Plane Stress:
After obtaining the data from ANSYS we can pot the realtion between the changes in
temperature for the Plane stress and Plane strain problem for point A and point B.
Plane Stress at point A
Plane Stress : point A
Temp
(oC)
-60
Sigma1
(Pa)
3.93E+08
Sigma 2
(Pa)
1.61E+06
Sigma 3
(Pa)
0
-40
2.62E+08
1.07E+06
0
-20
1.31E+08
5.40E+05
0
0
1.08E+05
8044.5
0
20
0
-5.24E+05
-1.31E+08
40
0
-1.06E+06
-2.62E+08
60
0
-1.59E+06
-3.93E+08
5,00E+08
4,00E+08
3,00E+08
2,00E+08
For Plain Stress at
point A Sigma1 (Pa)
1,00E+08
-100
0,00E+00
-50
0
-1,00E+08
-2,00E+08
-3,00E+08
-4,00E+08
-5,00E+08
50
100
For Plain Stress at
point A Sigma 2 (Pa)
For Plain Stress at
point A Sigma 3 (Pa)
Plane Stress at point B
Temp
(oC)
-60
-40
-20
0
20
40
60
Plane Stress : point B
Sigma1
Sigma 2
(Pa)
(Pa)
0
8.92E+05
0
5.95E+05
0
2.98E+05
0
-890.27
4.06E+07 2.96E+05
8.41E+07 5.93E+05
1.28E+08 8.90E+05
Sigma 3
(Pa)
-1.33E+08
-8.97E+07
-4.63E+07
-2.82E+06
0
0
0
1,50E+08
1,00E+08
For Plain Stress at
point B Sigma1 (Pa)
5,00E+07
-100
0,00E+00
-50
0
50
100
For Plain Stress at
point B Sigma 2 (Pa)
For Plain Stress at
point B Sigma 3 (Pa)
-5,00E+07
-1,00E+08
-1,50E+08
For plane Strain
plane Strain : point A
Temp
(oC)
-60
Sigma1
(Pa)
5.61E+08
Sigma 2
(Pa)
3.09E+08
Sigma 3
(Pa)
2.23E+06
-40
3.74E+08
2.06E+08
1.49E+06
-20
1.87E+08
1.03E+08
7.48E+05
0
6.54E+03
-73648
2.52E+05
20
-
-1.03E+08
-1.87E+07
7.35E+05
40
60
1.48E+06
2.22E+06
-2.06E+06
-3.75E+08
-3.10E+08 -5.62E+08
8,00E+08
6,00E+08
4,00E+08
For Plain Strain at
point A Sigma1 (Pa)
2,00E+08
-100
0,00E+00
-50
0
-2,00E+08
50
100
For Plain Strain at
point A Sigma 2 (Pa)
For Plain Strain at
point A Sigma 3 (Pa)
-4,00E+08
-6,00E+08
-8,00E+08
At the point B
Plane
strain
Temp
(oC)
-60
Point B
Sigma1
(Pa)
8.33E+07
Sigma 2
(Pa)
-1.18E+06
Sigma 3
(Pa)
-1.89E+08
-40
5.53E+07
-7.89E+05
-1.27E+08
-20
2.72E+07
3.94E+05
-6.47E+07
0
4.24E+02
-8.09E+05
-2.70E+06
20
5.93E+07
3.95E+05
-2.89E+07
40
1.21E+08
7.90E+05
-5.69E+07
60
1.83E+08
1.19E+06
-8.50E+07
2,50E+08
2,00E+08
1,50E+08
1,00E+08
For Plain Strain at
point B Sigma1 (Pa)
5,00E+07
-100
0,00E+00
-50
0
-5,00E+07
-1,00E+08
50
100
For Plain Strain at
point B Sigma 2 (Pa)
For Plain Strain at
point B Sigma 3 (Pa)
Discussion of graphs:
Trends:
At point A for Plane stress and Plane Strain:
Between:
Temps (-60° -20°): increase in principal stresses
Temps (20°- 60°) : decrease in principal stresses
At point B for Plane Stress and Plane strain
Between:
temps (-60° -20°): decreases in principal stresses
Temps (20°- 60°): increases in principal stresses
Comparison:
From the tabulated data, we can see that in our plane stress data,
At point A
Stress 3 is zero from (-60° -0°) and stress 1 between (20° -60°)
At point B
Stress 1 is zero from (-60° -0°) and stress 3 between (20° -60°)
This is due to no stresses acting in the 3rd Plane
Discussion:
d) Calculate these principal stresses again if changing the materials from mild steel to
aluminium and copper.
Properties for aluminium
E = 70 GPa
v = 0.35
Properties for copper
E = 115 GPa
v = 0.34
APPENDIX:
FOR PLANE STRESS AND PLANE STRAIN Q6
Temp of -60 oC
4
5
0.0000
-0.89200E+06-0.13314E+09 0.13314E+09 0.13269E+09
0.39315E+09 0.16052E+07 0.0000
0.39315E+09 0.39235E+09
Temp of -40 oC
4
5
0.0000
-0.59497E+06-0.89698E+08 0.89698E+08 0.89402E+08
0.26214E+09 0.10728E+07 0.0000
0.26214E+09 0.26160E+09
Temp of -20 oC
4
5
0.0000
-0.29794E+06-0.46258E+08 0.46258E+08 0.46109E+08
0.13112E+09 0.54044E+06 0.0000
0.13112E+09 0.13085E+09
For a temperature of 60oC, we get the following stresses:
4
5
0.12750E+09 0.89019E+06 0.0000
0.12750E+09 0.12706E+09
0.0000
-0.15891E+07-0.39294E+09 0.39294E+09 0.39214E+09
Temp of 40oC
4
5
0.84063E+08 0.59316E+06 0.0000
0.84063E+08 0.83768E+08
0.0000
-0.10567E+07-0.26192E+09 0.26192E+09 0.26139E+09
Temp of 20oC
4
5
0.40623E+08 0.29613E+06 0.0000
0.40623E+08 0.40476E+08
0.0000
-0.52431E+06-0.13091E+09 0.13091E+09 0.13064E+09
Temp of 0oC
4
5
0.0000
-890.27
0.10810E+06 8044.5
-0.28175E+07 0.28175E+07 0.28170E+07
0.0000
0.10810E+06 0.10431E+06
PLANE STRAIN
Temp = -60 oC
4
5
0.83344E+08-0.11833E+07-0.18882E+09 0.27217E+09 0.24128E+09
0.56120E+09 0.30937E+09 0.22304E+07 0.55897E+09 0.48487E+09
temp = -40 oC
4
5
0.55293E+08-0.78872E+06-0.12678E+09 0.18207E+09 0.16151E+09
0.37405E+09 0.20623E+09 0.14891E+07 0.37256E+09 0.32317E+09
temp = -20 oC
4
5
0.27242E+08-0.39416E+06-0.64740E+08 0.91981E+08 0.81746E+08
0.18690E+09 0.10308E+09 0.74783E+06 0.18615E+09 0.16148E+09
temp = 0 oC
4
5
423.96
6540.3
-0.80919E+06-0.26977E+07 0.26981E+07 0.23981E+07
-73648.
-0.25203E+06 0.25858E+06 0.22925E+06
temp = 20 oC
4 0.59344E+08 0.39498E+06-0.28860E+08 0.88204E+08 0.77817E+08
5 -0.73477E+06-0.10322E+09-0.18740E+09 0.18667E+09 0.16192E+09
temp = 40 oC
4 0.12139E+09 0.78954E+06-0.56911E+08 0.17830E+09 0.15758E+09
5 -0.14761E+07-0.20637E+09-0.37455E+09 0.37308E+09 0.32361E+09
Temp = 60 oC
4
5
0.18343E+09 0.11841E+07-0.84962E+08 0.26839E+09 0.23735E+09
0.39315E+09 0.16052E+07 0.0000
0.39315E+09 0.39235E+09