SWINBURNE UNIVERSITY OF TECHNOLOGY (SARAWAK CAMPUS)
FACULTY OF ENGINEERING AND INDUSTRIAL SCIENCE
HES5320 Solid Mechanics Semester 2, 2011
Group Assignment
Lecturer: Dr. Saad A. Mutasher
By
Group No. 2
Stephen Bong Pi Yiing (4209168)
Ngui Yong Zit (4201205)
Ling Wang Soon (4203364)
Due Date: 5 pm, 28th
October 2011 (Friday)
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 2 of 30
Question 1
For plate shown in Figure 1, use solidwork simulation to calculate the maximum principle stresses and
their locations. The materials of plate is Alloy steel (E = 210 GPa, ν = 0.29). Use the option Design
Scenario in solidwork to study the effect of hole diameter on principle stress. The diameters of hole are
(20, 25, 30, 40, 50, 60, 70, 80, 90, 100) mm. Plot separately the graph of principle stress vs. hole diameter.
Vertical axis of graphs should be stress and horizontal axis hole diameter. Discuss the results.
Figure 1
Solutions
Fig. Q1A and Fig. Q1B below shows the results of finite element analysis (FEA) simulation by using
SolidWorks and the plot of principle stress vs. hole diameter.
Fig. Q1A: Locations of principle stresses obtained using SolidWorks FEA Simulation when the hole’s
diameter is 50 mm.
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 3 of 30
Fig. Q1B: Plot of principle stress vs. hole diameter.
Hole Diameter (mm) Principal Stresses (MPa)
20 4.9915
25 4.7507
30 4.9517
40 5.3035
50 5.8161
60 6.2621
70 6.3928
80 6.8952
90 7.1401
100 7.2091
Table 1: Variation of principle stresses with their respective hole diameter
Discussion: According to the definition of normal stress, dA
dF
A
F
A=
∆
∆=
→∆ 0limσ , which is a measurement of
the amount of internal forces contained in a deformable materials. It can be also defined by the internal
force per unit area (P. P., Benham; R. J., Crawford & C. G., Armstrong, 1996, pp. 43). Mohr’s circle is an
alternative which represents all possible states of normal and shear stress on any plane through a stressed
point. From the Mohr’s circle, the plane causes the material to experience zero shear stress is termed
principal planes and the normal stresses acting on them are termed principal stresses which always
denoted as (σ1 and σ2, or, σmax and σmin) (P. P., Benham; R. J., Crawford & C. G., Armstrong, 1996, pp.
298 & pp. 301). Based on the results obtained from the SolidWorks FEA simulation as shown in Fig.
Q1A above, the magnitude of the principal stresses are 7.7 MPa and 0.4 MPa respectively. Apart from
that, the stress is maximum at the necking of the plate and minimum at the top and bottom edges of the
hole. When a load of 5 kN is applied at the free end of the plate, the plate experience uniform stress
distribution and the deformation only takes place in tangential direction of the applied load. When a hole
4.5
5
5.5
6
6.5
7
7.5
20 30 40 50 60 70 80 90 100
Pri
nci
pal
Str
ess,
σ (
MP
a)
Hole Diameter (mm)
Principal Stress, σ (MPa) vs. Hole Diameter (mm)
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 4 of 30
is drilled with an offset of 0.25 m from the fixed end, the distribution of stresses deviates when the load of
5 kN is applied at the free end. The deviation of colours in the legend beside the simulation as shown in
Fig. Q1A above indicates the magnitude of principle stresses with respective to their locations. In order to
study the effect of principle stresses resulted by various hole diameters, the top edge of the hole has been
selected as a reference point. According to the definition of normal stress, A
F=σ where F is the applied
load and A is the area normal to the applied force, the upsurge in hole diameters will results the reduction
in area which consequence the increase in principle stress as shown in Fig. Q1B above.
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 5 of 30
Question 2
A solid plate of 400 mm diameter and 20 mm thickness is acted upon by a uniform distributed pressure of
1000 kN/m2 acting upward.
(a) Calculate the central deflection use solidwork simulation and compare the result with analytical
solution, check the effect of mesh size.
(b) Use solidworksimulation; sketch the distribution of deflection of the plate under the load and the
radial and tangential stresses along the radius of plate. Then compare the results with the analytical
solution.
(c) Load Case 2 – For the same plate, assume a different loading case as follows – at the centre a Point
Load, F, of 2000 N acting downwards plus a constant pressure, p, of 1000 kN/m2 acting upwards
over the entire plate (i.e. in the opposite direction to F). Perform an analysis and note the resulting
plate profile. A Graph (to scale) comparing FEA deflection with theoretical deflection (Vertical axis
of graphs should be ‘w’ deflection and horizontal axis diametral location.). Discuss the results.
The material of plate is Alloy steel (E = 210 GPa, ν = 0.29). Assume the plate is clamp at the ends.
Solutions
(a) Simulation
Fig. Q2A: Contour plot of central deflection of the plate caused by uniform distributed pressure of
1000 kPa obtained by using SolidWorks FEA Simulation
Based on Fig. Q2A above, the maximum deflection of the plate when a uniform distributed pressure
of 1000 kPa (1 MPa) is acted upward on the plate is 1.636 × 10-1
mm or 0.1636 mm.
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 6 of 30
Analytical Solution
d = 400 mm = 0.4 m, a = 0.2 m, h = 20 mm = 0.02 m, p = 1000 kN/m2.
For Circular Plate, Fixed Outer Edge (Clamp at the ends or clamped periphery), Loaded Uniformly
(see Appendix), the maximum deflection is given by
D
paw
64
4
max =
where)1(12
2
3
ν−=
EhD is the flexural rigidity.
( )[ ]mm 0.1636=
×
−×=
−=
−==
m) Pa)(0.02 10210(16
29.01m) Pa)(0.2 101000(3
16
)1(3
)1(1264
64 9
243
3
24
3
2
44
maxEh
pa
Eh
pa
D
paw
ν
ν
FEA (SolidWorks Simulation)
The analytical deflection with variable radius are calculated by using Microsoft Excel. The plots of
deflections obtained by SolidWorks simulation and analytical solutions are shown in Fig. Q2B
below:
Fig. Q2B: Plot of deflection (mm) vs. radius (mm)
0.00E+00
2.00E-02
4.00E-02
6.00E-02
8.00E-02
1.00E-01
1.20E-01
1.40E-01
1.60E-01
1.80E-01
0 50 100 150 200
Def
lect
ion (
mm
)
Radius (mm)
Deflection (mm) vs. Radius (mm)
SolidWorks FEA Simulation
Analytical Solution
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 7 of 30
Check with mesh size:
Shell
Mesh Size: 5 mm, Maximum Deflection = 0.16356 mm
Mesh Size: 10 mm, Maximum Deflection = 0.16359 mm
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 8 of 30
Mesh Size: 15 mm, Maximum Deflection = 0.16361 mm
Mesh Size: 20 mm, Maximum Deflection = 0.16371 mm
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 9 of 30
Mesh Size: 25 mm, Maximum Deflection = 0.16376
Mesh Size: 30 mm, Maximum Deflection = 0.164 mm
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 10 of 30
Mesh Size: 35 mm, Maximum Deflection = 0.16397
Mesh Size = 40 mm, Maximum Deflection = 0.16437 mm
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 11 of 30
Solid
Mesh Size = 5 mm
Mesh Size = 10 mm
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 12 of 30
Mesh Size = 15 mm
Mesh Size = 20 mm
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 13 of 30
Mesh Size = 25 mm
Mesh Size: 30 mm
The maximum deflections of the circular plate with their respectively mesh sizes are tabulated in
Table 2 and a plot of deflection vs. mesh size are shown in Fig. Q2B below:
Mesh Size
(mm)
Maximum Deflection (mm)
Thin Shell Solid
5 0.16356 0.16944
10 0.16359 0.16902
15 0.16361 0.16847
20 0.16371 0.16733
25 0.16376 0.16593
30 0.164 0.16488
35 0.16397 0.16332
40 0.16437 0.16236
Table 2: Maximum deflection with their respective mesh sizes
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 14 of 30
Fig. Q2C: Plot of maximum deflections (mm) vs. mesh size (mm)
(b) Simulation
Fig. Q2C: Distribution of deflection of the plate under the load and the radial stresses along the
radius of the plate obtained by SolidWorks FEA Simulation
0.162
0.163
0.164
0.165
0.166
0.167
0.168
0.169
0.17
5 10 15 20 25 30 35 40
Def
lect
ion (
mm
)
Mesh Size (mm)
Deflection (mm) vs. Mesh Size (mm)
Thin Shell
Solid
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 15 of 30
Fig. Q2D: Distribution of deflection of the plate under the load and the tangential stresses along the
radius of the plate obtained by SolidWorks FEA Simulation
Analytical Solution
The bending stresses (radial stress, σr, and tangential stress, σθ) are given by 3
12
h
zM r
r =σ and
3
12
h
zMθθσ = where ( ) ( )[ ]22
3116
rap
M r νν +−+= and ( ) ( )[ ]22311
16ra
pM ννθ +−+= are the
moment-slope relationships ((P. P., Benham; R. J., Crawford & C. G., Armstrong, 1996, pp. 447).
D = 152855.1152 MN, p = 1000 kPa, h = 20 mm, ν = 0.29, E = 210 GPa.
The maximum radial and tangential stresses occur at r = a and z = h/2 = 0.02 m/2 = 0.01 m
( ) ( )[ ] ( ) ( )[ ]
( ) MPa 75−=−=−⋅×
=
+−+=+−+⋅⋅===
=
Pa 00000075m 2.02m) 02.0(8
Pa) 101000(3
318
331
162
1212
2
2
3
22
2
22
33
2
max , aah
pra
ph
hh
zM
ar
hzr
r υυυυσ
( ) ( )[ ] ( ) ( )[ ]
MPa 21.75−=−=×
−=−=
+−+=+−+⋅⋅===
=
Pa 00002175m) 02.0(4
m) m)(0.2 29.0Pa)( 101000(3
4
3
3118
3311
162
1212
2
23
2
2
22
2
22
33
2
max ,
h
ap
aah
pra
ph
hh
zM
ar
hz
υ
υυυυσθ
θ
The variation of bending stresses (radial and tangential) with respect to the deviation of radius are
calculated by using Microsoft Excel and the plots of variation of radial stresses and tangential
stresses with respect to the deviation of radius for both the SolidWorks FEA Simulation and
analytical solution are shown in Fig. Q2E and Fig. Q2F below.
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 16 of 30
Fig. Q2E: Plot of radial stresses (MPa) vs. radius (mm)
Fig. Q2F: Plot of tangential stresses (MPa) vs. radius (mm)
-50
-40
-30
-20
-10
0
10
20
30
40
50
60
70
80
0 50 100 150 200Rad
ial
Str
ess,
σr (M
Pa)
Radius (mm)
Radial Stress, σr (MPa) vs. Radius (mm)
SolidWorks FEA Simulation
Analytical Solution
-50
-40
-30
-20
-10
0
10
20
30
0 50 100 150 200
Tan
gen
tial
Str
ess,
σθ (
MP
a)
Radius (mm)
Tangential Stress, σθ (MPa) vs. Radius (mm)
SolidWorks FEA Simulation
Analytical Solution
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 17 of 30
(c) Simulation
Fig. Q2G: Deflection profile of the circular plates with uniform distribution pressure of 1 MPa
(upwards) and a point load of 2000 N (downwards). The maximum deflection is 0.1532 mm.
Analytical Solution
The maximum deflection of the circular plate which subjected to a uniform pressure distribution of 1
MPa (upwards) and a point load of 2000 N (downwards) can be determined by using the method of
superposition.
From part (a), the maximum deflection caused by the uniform pressure distribution is wmax, pressure =
0.1636 mm.
For Circular Plate, Fixed Outer Edge (clamped periphery), and Loaded by Central Concentrated
Force (Point Load) (see Appendix), the deflections with respect to diametral deviation and maximum
deflection are given by
−+
= 222
ln216
raa
rr
D
Fw
π and
D
Faw
π16
2
max =
mm 0.1532=
−=
+=
+=
mm 0104.0mm 1636.0
) N 1152.152855(16
)m .20N)( 2000(mm 1636.0
2
LoadPoint max,onDistributi Pressure Uniformmax,2 Case Loading max,
π
www
The deflections of the circular plate with respect to the diametral deviation are calculated by using
Microsoft Excel. Fig. Q2H below shows the plot of deflection (mm) vs. diametral location (mm).
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 18 of 30
Fig. Q2H: Plot of deflection (mm) vs. diametral location (mm)
Discussion: Since the uniform distributed pressure acting upwards in much greater than the
magnitude of point load which acting downwards, therefore, it can be concluded that the overall
deflected profile is in the positive y-direction which is also proven by using either the analytical
solution (method of superposition) and SolidWorks FEA Simulation. Apart from that, since the point
load is acting downwards at the centre of the circular plate and the distribution of pressure is uniform
over the entire plate, thus, the plot of deflection vs. diametral location as shown in Fig. Q2H above is
like a bell curve due to its symmetrical loadings property. Since the circular plate is clamped at its
periphery, therefore, there has no deflection at its end (r = a = 200 mm). Based on Fig. Q2H, there
has not much deviations between the results obtained by using analytical solution and SolidWorks
FEA Simulation. Besides, as shown in Fig. Q2E and Fig. Q2F, the deviations of results between the
two methodologies is extremely small as well. Thus, SolidWorks FEA Simulation can be used for
practical experiments and modeling as it provides good approximations.
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
-200 -150 -100 -50 0 50 100 150 200
Def
lect
ion,
w (
mm
)
Diametral Location (mm)
Deflection, w (mm) vs. Diametral Location (mm)
SolidWorks FEA
SimulationAnalytical Solution
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 19 of 30
Question 3
For the truss shown in Figure 2
(a) Use method of joints or method of sections to calculate the forces in each member.
(b) Use finite element method to calculate the displacement in nodes 3 and 7.
(c) Calculate the forces in each element (member) and compare your results with part (a).
Figure 2
Solutions
(a) Support Reactions:
( ) ( ) ( )
( )↓=−=⇒=−+−=
=⇒=−+−=
∑∑
kN 50kN 50
kN 250
121
221
0kN 501kN 50 ;0
0m 9kN 150m 6m 3kN 50 ;0
RRRF
RRM
y
Method of Joints
Joint 7
(C) kN 150
(T) kN 212.13
=
=−
=
=
=+−
=
−
−−
−
−
∑
∑
75
7675
76
76
02
2
0
02
2kN 150
0
F
FF
F
F
F
F
x
y
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 20 of 30
Joint 6
(T) kN 150
(C) kN 150
=
=−
=
=
=−
=
−
−−
−
−−
∑
∑
63
6376
65
7665
02
2
0
02
2
0
F
FF
F
F
FF
F
x
y
Joint 5
(C) kN 50
(C) kN 141.42
=
=−+
=
=
=−+−
=
−
−−
−
−
∑
∑
54
5453
53
53
0kN 1502
2
0
02
2kN 250kN 150
0
F
FF
F
F
F
F
x
y
Joint 3
(T) kN 50
(T) kN 100
=
=−+−
=
=
=−
=
−
−−
−
−−
∑
∑
32
5332
43
4353
02
2kN 150
0
02
2
0
F
FF
F
F
FF
F
x
y
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 21 of 30
Joint 2
(T) kN 50
(C) kN 70.71
=
=++−
=
=
=−
=
−
−−
−
−−
∑
∑
21
4221
42
4232
02
2
0
02
2
0
F
FF
F
F
FF
F
y
x
By taking the sum of forces in x-direction at Joint 4 gives F1-4 = 0.
(b) Member (Element) Numbering
A = 2500 mm2 = 2500 × 10
-6 m
2 E = 200 GPa = 2 × 10
11 Pa
EA = 5 × 108 N
Member Length (m) θ (deg.) θcos=c θsin=s LAEk =
(MN/m)
(1) 3 90 0 1 166.67
(2) 3 0 1 0 166.67
(3) 3 90 0 1 166.67
(4) 3 0 1 0 166.67
(5) 3 90 0 1 166.67
(6) 23 135 22− 22 117.85
(7) 3 0 1 0 166.67
(8) 3 0 1 0 166.67
(9) 3 0 1 0 166.67
(10) 23 135 22− 22 117.85
(11) 23 135 22− 22 117.85
The general stiffness matrix for the element in global co-ordinates is given by:
[ ]
−−
−−
−−
−−
=
22
22
22
22
scsscs
csccsc
scsscs
csccsc
K G
e
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 22 of 30
The stiffness matrix for each element in global co-ordinates will be:
[ ]
−
−=
67.166067.1660
0000
67.166067.1660
0000
1 G
eK [ ]
−
−
=
0000
067.166067.166
0000
067.166067.166
2 G
eK
[ ]
−
−=
67.166067.1660
0000
67.166067.1660
0000
3 G
eK [ ]
−
−
=
0000
067.166067.166
0000
067.166067.166
4 G
eK
[ ]
−
−=
67.166067.1660
0000
67.166067.1660
0000
5 G
eK [ ]
−−
−−
−−
−−
=
93.5893.5893.5893.58
93.5893.5893.5893.58
93.5893.5893.5893.58
93.5893.5893.5893.58
6 G
eK
[ ]
−
−
=
0000
067.166067.166
0000
067.166067.166
7 G
eK [ ]
−
−
=
0000
067.166067.166
0000
067.166067.166
8 G
eK
[ ]
−
−
=
0000
067.166067.166
0000
067.166067.166
9 G
eK [ ]
−−
−−
−−
−−
=
93.5893.5893.5893.58
93.5893.5893.5893.58
93.5893.5893.5893.58
93.5893.5893.5893.58
10 G
eK
[ ]
−−
−−
−−
−−
=
93.5893.5893.5893.58
93.5893.5893.5893.58
93.5893.5893.5893.58
93.5893.5893.5893.58
11 G
eK
By expanding these stiffness matrices, the stiffness matrix for the whole structure will be:
[ ]
−−
−−−
−−−
−−−
−−−
−−−−
−−−
−−−−
−−−
−−−−
−−−
−−−
−
−
=
93.5893.5893.5893.580000000000
93.586.22593.5893.58067.16600000000
93.5893.586.22593.5867.166000000000
93.5893.5893.586.2250000067.1660000
0067.16606.22593.580093.5893.580000
067.1660093.5827.392067.16693.5893.580000
0000006.22593.5867.166093.5893.5800
0000067.16693.5827.3920093.5893.58067.166
000093.5893.5867.16606.22593.580000
00067.16693.5893.580093.5827.392067.16600
00000093.5893.58006.22593.5867.1660
00000093.5893.58067.16693.586.22500
000000000067.166067.1660
000000067.1660000067.166
G
eS
K
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 23 of 30
−−
−−−
−−−
−−−
−−−
−−−−
−−−
−−−−
−−−
−−−−
−−−
−−−
−
−
=
7
7
6
6
5
5
4
4
3
3
2
2
1
1
7
7
6
6
5
5
4
4
3
3
2
2
1
1
93.5893.5893.5893.580000000000
93.586.22593.5893.58067.16600000000
93.5893.586.22593.5867.166000000000
93.5893.5893.586.2250000067.1660000
0067.16606.22593.580093.5893.580000
067.1660093.5827.392067.16693.5893.580000
0000006.22593.5867.166093.5893.5800
0000067.16693.5827.3920093.5893.58067.166
000093.5893.5867.16606.22593.580000
00067.16693.5893.580093.5827.392067.16600
00000093.5893.58006.22593.5867.1660
00000093.5893.58067.16693.586.22500
000000000067.166067.1660
000000067.1660000067.166
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
F
F
F
F
F
F
F
F
F
F
F
F
F
F
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
Boundary Conditions: 0 and 0 ,0 511 === YXY δδδ
By applying the boundary conditions the equations can be reduced to:
−−
−−
−−
−−−
−−−
−−−
−−
−−−
−−
−−−
−
=
7
7
6
6
4
4
3
3
2
2
1
7
7
6
6
4
4
3
3
2
2
1
93.5893.5893.5893.580000000
93.586.22593.5893.580000000
93.5893.586.22593.580000000
93.5893.5893.586.22500067.166000
00006.22593.5867.166093.5893.580
000093.5827.3920093.5893.5867.166
000067.16606.22593.58000
00067.1660093.5827.392067.1660
000093.5893.58006.22593.580
000093.5893.58067.16693.586.2250
0000067.166000067.166
Y
X
Y
X
Y
X
Y
X
Y
X
X
Y
X
Y
X
Y
X
Y
X
Y
X
X
F
F
F
F
F
F
F
F
F
F
F
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
Substitute the magnitude of the external forces, the equations become:
−−
−−
−−
−−−
−−−
−−−
−−
−−−
−−
−−−
−
=
−
−
7
7
6
6
4
4
3
3
2
2
1
93.5893.5893.5893.580000000
93.586.22593.5893.580000000
93.5893.586.22593.580000000
93.5893.5893.586.22500067.166000
00006.22593.5867.166093.5893.580
000093.5827.3920093.5893.5867.166
000067.16606.22593.58000
00067.1660093.5827.392067.1660
000093.5893.58006.22593.580
000093.5893.58067.16693.586.2250
0000067.166000067.166
150
0
0
0
50
0
0
0
0
0
0
Y
X
Y
X
Y
X
Y
X
Y
X
X
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
By using T1-84, the displacement is given by:
−
−
−
−
=
−
−
−−
−−
−−
−−−
−−−
−−−
−−
−−−
−−
−−−
−
=
−
2711.7
8999.0
8999.0
9257.2
2738.0
3006.0
3268.0
0254.2
3006.0
7251.1
3006.0
150
0
0
0
50
0
0
0
0
0
0
93.5893.5893.5893.580000000
93.586.22593.5893.580000000
93.5893.586.22593.580000000
93.5893.5893.586.22500067.166000
00006.22593.5867.166093.5893.580
000093.5827.3920093.5893.5867.166
000067.16606.22593.58000
00067.1660093.5827.392067.1660
000093.5893.58006.22593.580
000093.5893.58067.16693.586.2250
0000067.166000067.1661
7
7
6
6
4
4
3
3
2
2
1
Y
X
Y
X
Y
X
Y
X
Y
X
X
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 24 of 30
Therefore, the displacements at nodes 3 and 7 are:
mm 7.2711mm 0.8999mm 0.3268mm 2.0254 −=−=== 7733 and , , YXYX δδδδ
Direction: δx3 – Right
δy3 – Upwards
δx7 – Left
δy7 – Downwards
(c) The forces in each member is given by:
( ) ( )[ ]( ) [ ]
( ) ( )[ ]( ) [ ]
( ) ( )[ ]( ) [ ]
( ) ( )[ ]( ) [ ]
( ) ( )[ ]( ) [ ]
( ) ( )[ ]( )
( ) ( )[ ]( ) [ ]
( ) ( )[ ]( ) [ ]
( ) ( )[ ]( ) [ ] (C) kN 150
(C) kN 50
0
(C) kN 212.99
(C) kN 149.99
(T) kN 150.05
(C) kN 100.01
(T) kN 50.05
(T) kN 50.10
=
−
−=
=
=
−−=
−−=
=
−
−=
−=
=
−
−
−
−−=
−−=
=
−=
=
=
−
−=
−−=
=
−
−=
−−=
=
−=
−−=
=
=
−=
kN 2711.7
8999.00167.166
kN 2738.0
3006.00167.166
kN
2738.0
3006.0
3006.0
01167.166
kN
2711.7
8999.0
8999.0
9257.2
2
2
2
2
2
2
2
285.117
kN 8999.0
9257.21067.166
kN
8999.0
9257.2
3268.0
0254.2
101067.166
kN
2738.0
3006.0
3268.0
0254.2
101067.166
kN
3268.0
0254.2
3006.0
7251.1
010167.166
kN
3006.0
7251.1
3006.0
10067.166
7
7
899
4
4
888
4
4
1
177
7
7
6
6
666
6
6
555
6
6
3
3
444
4
4
3
3
333
3
3
2
2
222
2
2
1
111
Y
X
Y
X
Y
X
X
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
X
sckf
sckf
scckf
scsckf
sckf
scsckf
scsckf
scsckf
scckf
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 25 of 30
( ) ( )[ ]( )
( ) ( )[ ]( ) (T) kN 142.13
(T) kN 70
=
−=
−−=
=
−
−−=
−−=
kN 3268.0
0254.2
2
2
2
285.117
kN
2738.0
3006.0
3006.0
7251.1
2
2
2
2
2
2
2
285.117
3
3
111111
4
4
2
2
101010
Y
X
Y
X
Y
X
sckf
scsckf
δ
δ
δ
δ
δ
δ
Support Reactions:
−
−
=
−
−
−
−
−−−
−−−−
−
=
0843250
21320
101050
2711.7
8999.0
8999.0
987.2
0
0
2738.0
3006.0
3268.0
0254.2
3006.0
7251.1
0
3006.0
0067.16606.22593.580093.5893.580000
067.1660093.5827.392067.16693.5893.580000
000000000067.166067.1660
5
5
1
.
.
.
R
R
R
Y
X
Y
Thus, R1Y = -50.1010 kN and R5Y = 250.0848 kN
Based on the calculations above, the difference between the results obtained by using finite element
analysis and method of joints is extremely small, and therefore, it can be concluded that finite
element analysis is a powerful methodology for modeling a complex structure.
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 26 of 30
APPENDIX
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 27 of 30
TABULATED SOLUTIONS OF CIRCULAR PLATES
(Ref.: Budynas, R. G., 1999, Advanced Strength and Applied Stress Analysis, pp. 337-340, 2nd
edn.,
McGraw-Hill, Singapore)
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 28 of 30
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 29 of 30
Group Assignment
HES5320 Solid Mechanics, Semester 2, 2011 Group No. 2 Page 30 of 30
References:
Budynas, R. G., 1999, Advanced Strength and Applied Stress Analysis, pp. 337 – 340, WCB McGraw-
Hill, Singapore.
Benham, P. P.; Crawford, R. J.; Armstrong, C. G., 1996, Mechanics of Engineering Materials, Pearson
Longman, China.