HES5320 Solid Mechanics, Semester 2, 2011, Group Assignment

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


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


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


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


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


Check with mesh size:

Shell

Mesh Size: 5 mm, Maximum Deflection = 0.16356 mm


Group Assignment




Group Assignment


Mesh Size: 25 mm, Maximum Deflection = 0.16376


Group Assignment


Mesh Size: 35 mm, Maximum Deflection = 0.16397

Mesh Size = 40 mm, Maximum Deflection = 0.16437 mm

Group Assignment


Solid

Mesh Size = 5 mm

Mesh Size = 10 mm

Group Assignment


Mesh Size = 15 mm

Mesh Size = 20 mm

Group Assignment


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


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


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


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)


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)


Analytical Solution

Group Assignment


(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


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


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


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


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


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


−−

−−−

−−−

−−−

−−−

−−−−

−−−

−−−−

−−−

−−−−

−−−

−−−

−

−

=

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


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


( ) ( )[ ]( )

( ) ( )[ ]( ) (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


APPENDIX

Group Assignment


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


Group Assignment


Group Assignment


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.

Documents

HES5320 Solid Mechanics, Semester 2, 2011, Group Assignment