chapter 3- self assesment truss structure

8/11/2019 Chapter 3- Self Assesment Truss Structure

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Mechatronics Engineering DepartmentFaculty of Engineering

National University of Sciences and Technology

EM 415

SPECIAL TOPICS IN MECHATRONICS

CHAPTER 3 EXAMPLE

Mr. Anas Bin Aqeel

Department of Mechatronics Engineering

National University of Sciences and Technology

Pakistan


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Instructions

This file is based on Chapter 3 Example 2 given in your notes. It is taken from a previous exam paper.

Open the file in Powerpoint and launch slideshow (from menu: Slideshow - from Beginning or click at the bottom right of

the window)

The example is broken down into a series of steps to illustrate how to tackle a question like this. At various points, you will be

given multiple choice questions to work through which will show you how to develop the stiffness matrix for each element,

construct the global stiffness matrix and solve to determine the displacement and stresses of the elements.

Click on the Start button to progress

Start


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Here the node numbers are shown in circles at the nodes and the element numbers in square boxes in

the middle of the element.

Note we have also added the X and Y axis defining the global co-ordinate system.

Stage 1

1 2

34

Continue

1

2 3

A pin jointed frame has been represented using three 1-D

truss elements, Figure 3.9. Use the Finite Element Method

and the elemental stiffness matrix to calculate the

displacements and elemental stresses. The structure is

made from Aluminium with E = 70000 MPa, with the

following cross sectional areas for the rod elements: A12=

300 mm2

, A23= 150 mm2

, A42= 150 mm2

.

Figure 3.9 X

Y


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Before starting, decide on a set of units, and to work consistently

with these all the way through. We could convert everything to SI

units (kg, m, s, N, Pa). In this case, the dimensions are given in

mm, so it makes more sense to use milimetres for length, but

what units should we use for force and stress?

Click on the button A, B or C or D, click clue to get a hint or jumpstraight to the answer

Stage 3: Decide what units to use

Clue

Go straight to answer







300 mm2

, A23= 150 mm2

, A42= 150 mm2

.

Figure 3.9

A

B

C

D

N, GPa

kN, MPa

N, Pa

N, MPa

Units

Force = ?

Length = mm

Stress/E = ?


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Decide on the base units of mass, length and time and then derive the

units for the others. In this case, we have decided to use mm for length,

try the suggestion above for mass and time

Click on the back button to return to the question

Stage 3: Clue

Back







300 mm2

, A23= 150 mm2

, A42= 150 mm2

.

Figure 3.9

Units

Mass = g

Length = mm

Time = ms

Force =

Stress/E =


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Press continue to go proceed

Answer: CORRECT

Continue

The base units are kg, mm and ms, giving the units shown above







300 mm2

, A23= 150 mm2

, A42= 150 mm2

.

Figure 3.9

Units

Mass [M] = g

Length [L]= mm

Time [T] = ms

N

s

mkg

10s

10m10kg223

3-3

force

MPa10N/m

10m

N

area

force 6223

stress


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Answer: INCORRECT







300 mm2

, A23= 150 mm2

, A42= 150 mm2

.

Figure 3.9

Start with the base units of mass, length and time given above

and work out the corresponding units of force and stress

Go back to question

Units

Mass = g

Length = mm

Time = ms

Force =

Stress/E =


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300 mm

2

, A23= 150 mm

2

, A42= 150 mm

2

.

Figure 3.9

Answer: INCORRECT

Start with the base units of mass, length and time given above

and work out the corresponding units of force and stress.

Go back to question

Mass = g

Length = mm

Time = ms

Force =

Stress/E =

Units


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Answer

Continue







300 mm

2

, A23= 150 mm

2

, A42= 150 mm

2

.

Figure 3.9

The base units are kg, mm and ms, giving the units shown above

Units

Mass [M] = g

Length [L]= mm

Time [T] = ms

N

s

mkg

10s

10m10kg223

3-3

force

MPa10N/m

10m

N

area

force 6223

stress


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Next we will consider each element in turn, starting with Element 1.

First sketch the element as shown and then calculate the angle.

What is the angle in this case?

Click on the button A, B or C or D, click clue to get a hint or jumpstraight to the answer

Stage 3: Consider each element in turn

Clue


1 1 2







300 mm

2

, A23= 150 mm

2

, A42= 150 mm

2

.

Figure 3.9

X

Y

A

B

C

D

0

45

90

180

Element 1


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The angle of the element is defined as the angle measured at the node

with the lowest number ANTICLOCKWISE from the global X axis to the

element.

Click on the back button to return to the question

Stage 3: Clue

Back







300 mm

2

, A23= 150 mm

2

, A42= 150 mm

2

.

Figure 3.9

Node with lower node number

1 1 2 X

Y

Element 1


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Answer: CORRECT

Continue

The element is aligned with the global X axis, so the angle between the X axis and the element is zero.







300 mm

2

, A23= 150 mm

2

, A42= 150 mm

2

.

Figure 3.9

= 0

1 1 2 X

Y

Element 1


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300 mm2, A23

= 150 mm2, A42

= 150 mm2.

Figure 3.9

1 1 2 X

Y

Element 1

Node with lower node number

Answer: INCORRECT

Remember the angle of the element is defined as the angle

measured at the node with the lowest number ANTICLOCKWISE

from the global X axis to the element.

Go back to question


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Answer

Continue

The element is aligned with the global X axis, so the angle between the X axis and the element is zero.







300 mm2, A23

= 150 mm2, A42

= 150 mm2.

Figure 3.9

= 0

1 1 2 X

Y

Element 1


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We know all of the quantities to work out the stiffness matrix for

element 1. In your exam, you will be given a formula sheet with

the standard formula for a truss finite element given above

Next







300 mm2, A23

= 150 mm2, A42

= 150 mm2.

Figure 3.9

Stage 3: Element 1

cosc

sins

scss-cs-

csccs-c-

s-cs-scs

cs-c-csc

L

EA

22

22

22

22

j

j

i

i

V

U

V

U

UK

Truss Finite Element: For a truss finite element

aligned at angle to global X axis with a linear shape

function. Relative to the global co-ordinate system:


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Clue

Multi-choice








300 mm2, A23

= 150 mm2, A42

= 150 mm2.

Figure 3.9

Calculate the stiffness matrix for element 1

Stage 3: Element 1

cosc

sins

scss-cs-

csccs-c-

s-cs-scs

cs-c-csc

L

EA

22

22

22

22

j

j

i

i

V

U

V

U

UK




= 0

1 1 2 X

YElement 1


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Stage 3: Element 1

A CB

The stiffness matrix is:

0000

042000042000-

0000

042000-042000

1 =][k

0000

042042-

0000

024-042

1 =][k

42000042000-0

0000

420000420000

0000

1 =][k







300 mm2, A23

= 150 mm2, A42

= 150 mm2.

Figure 3.9

cosc

sins

scss-cs-

csccs-c-

s-cs-scs

cs-c-csc

L

EA

22

22

22

22

j

j

i

i

V

U

V

U

UK




= 0

1 1 2 X

YElement 1


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Answer: INCORRECT

Back







300 mm2, A23

= 150 mm2, A42

= 150 mm2.

Figure 3.9

cosc

sins

scss-cs-

csccs-c-

s-cs-scs

cs-c-csc

L

EA

22

22

22

22

j

j

i

i

V

U

V

U

UK




= 0

1 1 2 X

YElement 1

Be careful with the sine and cosine terms, remember alpha = o, so s = 0 and c = 1


23/35

Answer: INCORRECT

Back







300 mm2, A23

= 150 mm2, A42

= 150 mm2.

Figure 3.9

cosc

sins

scss-cs-

csccs-c-

s-cs-scs

cs-c-csc

L

EA

22

22

22

22

j

j

i

i

V

U

V

U

UK




= 0

1 1 2 X

YElement 1

Be careful with the units, remember we are working in mm and MPa, so you

can use the quantities exactly as they appear in the question.


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Press continue to go to next question

Answer:

Continue







300 mm2, A23

= 150 mm2, A42

= 150 mm2.

Figure 3.9

= 0

1 1 2 X

YElement 1

2

2

1

1

1

0000

042000042000-

0000

042000-042000

V

U

V

U

Uk =][

From the question, we have E = 70000 MPa, A12= 300 mm2, and L12= 500mm.

We know the angle alpha = 0, so sin() = 0 and cos() = 1

We can plug all these numbers into the formula for [K] to give the result shown


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We can now repeat the process for element 2.

Next







300 mm2, A23

= 150 mm2, A42

= 150 mm2.

Figure 3.9

Stage 3: Element 2

cosc

sins

scss-cs-

csccs-c-

s-cs-scs

cs-c-csc

L

EA

22

22

22

22

j

j

i

i

V

U

V

U

UK




2

3

2

X

YElement 2


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Clue

Multi-choice








300 mm2, A23

= 150 mm2, A42

= 150 mm2.

Figure 3.9

Calculate the stiffness matrix for element 2

Stage 3: Element 2

cosc

sins

scss-cs-

csccs-c-

s-cs-scs

cs-c-csc

L

EA

22

22

22

22

j

j

i

i

V

U

V

U

UK




X

Y

2

3

2Element 2


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Clue

Back







300 mm2, A23

= 150 mm2, A42

= 150 mm2.

Figure 3.9

Start off by drawing the element and working out the angle alpha.

From the question, we have E = 70000 MPa, A23= 150 mm2, and L23= 566mm

(calculated from the Figure).

We can plug all these numbers into the formula shown above for [K]

cosc

sins

scss-cs-

csccs-c-

s-cs-scs

cs-c-csc

L

EA

22

22

22

22

j

j

i

i

V

U

V

U

UK




X

Y

2

3

2Element 2

= ?


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Stage 3: Element 2

A CB

The stiffness matrix is:







300 mm2, A23

= 150 mm2, A42

= 150 mm2.

Figure 3.9

cosc

sins

scss-cs-

csccs-c-

s-cs-scs

cs-c-csc

L

EA

22

22

22

22

j

j

i

i

V

U

V

U

UK




X

Y

2

3

2Element 2

13125131251312513125

13125-131251312513125

13125-131251312513125

1312513125-13125-13125

9281928192819281

9281928192819281

9281-928192819281

92819281-92819281

9281928192819281

9281-928192819281

9281-928192819281

92819281-9281-9281


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Answer: CORRECT

Continue







300 mm2, A23

= 150 mm2, A42

= 150 mm2.

Figure 3.9

X

Y

2

3

2

Element 2

= 135

3

3

2

2

2

9281928192819281

9281-928192819281

9281-928192819281

92819281-9281-9281

V

U

V

U

Uk =][


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Answer: INCORRECT

Back







300 mm2, A23

= 150 mm2, A42

= 150 mm2.

Figure 3.9

Be careful with the angle, remember this is measured at the node with the lower

number, anticlockwise from the global X axis

cosc

sins

scss-cs-

csccs-c-

s-cs-scs

cs-c-csc

L

EA

22

22

22

22

j

j

i

i

V

U

V

U

UK




X

Y

2

3

2Element 2

= ?


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Answer: INCORRECT

Back







300 mm2, A23= 150 mm2, A42= 150 mm

2.

Figure 3.9

Be careful with the term L, this is the length of the element, so in this case,

you will need to work this out using trigonometry from the information given

in the figure.

cosc

sins

scss-cs-

csccs-c-

s-cs-scs

cs-c-csc

L

EA

22

22

22

22

j

j

i

i

V

U

V

U

UK




X

Y

2

3

2Element 2


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Answer:

Continue







300 mm2, A23= 150 mm2, A42= 150 mm

2.

Figure 3.9

From the question, we have E = 70000 MPa, A12= 300 mm2, and L12= 500mm.

We know the angle alpha = 0, so sin() = 0 and cos() = 1

We can plug all these numbers into the formula for [K] to give the result shown

X

Y

2

3

2

Element 2

= 135

3

3

2

2

2

9281928192819281

9281-928192819281

9281-928192819281

92819281-9281-9281

V

U

V

U

Uk =][


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Press continue to show the stiffness matrix for element 3

Stage 3: Element 3

Continue







300 mm2, A23= 150 mm2, A42= 150 mm

2.

Figure 3.9

Finally, we will repeat the procedure for element 3. You should be able to work

this out using the formula given. Once you have written it down, check it against

the answer given next...

cosc

sins

scss-cs-

csccs-c-

s-cs-scs

cs-c-csc

L

EA

22

22

22

22

j

j

i

i

V

U

V

U

UK




2

4

3

X

Y

Element 3


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Press continue

Answer:

Continue







300 mm2, A23= 150 mm2, A42= 150 mm

2.

Figure 3.9

Check you have correctly calculated the matrix for element 3.

Note that next to each element, we have written the corresponding

displacement vector {u}this will help when we are constructing the global

stiffness matrix

Element 3

2

4

3

X

Y

4

4

2

2

3

928192819281-9281-

928192819281-9281-

9281-9281-92819281

9281-9281-92819281

V

U

V

U

Uk =][

= 45


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You have finished the first stage of this question.

We will cover the rest in the lectures in Week

3.

chapter 3- self assesment truss structure

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