m4l26 Lesson 26 The Direct Stiffness Method: Temperature Changes and Fabrication Errors in Truss Analysis

7/31/2019 m4l26 Lesson 26 The Direct Stiffness Method: Temperature Changes and Fabrication Errors in Truss Analysis

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Version 2 CE IIT, Kharagpur

Module4

Analysis of Statically

IndeterminateStructures by the DirectStiffness Method


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Lesson26

The Direct StiffnessMethod: Temperature

Changes and

Fabrication Errors in Truss Analysis


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Instructional ObjectivesAfter reading this chapter the student will be able to1. Compute stresses developed in the truss members due to temperature

changes.

2. Compute stresses developed in truss members due to fabrication members.3. Compute reactions in plane truss due to temperature changes and fabricationerrors.

26.1 IntroductionIn the last four lessons, the direct stiffness method as applied to the trussanalysis was discussed. Assembly of member stiffness matrices, imposition ofboundary conditions, and the problem of inclined supports were discussed. Dueto the change in temperature the truss members either expand or shrink.

However, in the case of statically indeterminate trusses, the length of themembers is prevented from either expansion or contraction. Thus, the stressesare developed in the members due to changes in temperature. Similarly the errorin fabricating truss members also produces additional stresses in the trusses.Both these effects can be easily accounted for in the stiffness analysis.

26.2 Temperature Effects and Fabrication Errors


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The force displacement equation for the entire truss may be written as,

{ } [ ]{ } { }t puk p )(+= (26.5)

where , { } p is the vector of external joint loads applied on the truss and ( ){ }t p is thevector of joint loads developed in the truss due to change intemperature/fabrication error of one or more members. As pointed out earlier. inthe truss analysis, some joint displacements are known due to boundaryconditions and some joint loads are known as they are appliedexternally.Thus,one could partition the above equation as,

[ ] [ ][ ] [ ]

{ }{ }

( )( )

+

=

t u

t k

k

u

u

k

p

p

u

u

k k

k k

p

p

2221

1211(26.6)

where subscript u is used to denote unknown quantities and subscript k is usedto denote known quantities of forces and displacements. Expanding equation(26.6),

{ } [ ]{ } [ ]{ } ( )t k k uk puk uk p ++= 1211 (26.7a){ } [ ]{ } [ ]{ } ( ){ }21 22u u k u t p k u k u p= + + (26.7b)

If the known displacement vector { } { }0=k u then using equation (26.2a) theunknown displacements can be calculated as

{ } [ ] { } ( ){ }( )t k k u p pk u =1

11 (26.8a)

If { } 0k u then{ } [ ] { } [ ]{ } ( ){ }( )t k k k uu puk pk u =

12

1(26.8b)

After evaluating unknown displacements, the unknown force vectors arecalculated using equation (26.7b).After evaluating displacements, the memberforces in the local coordinate system for each member are evaluated by,

{ } [ ][ ]{ } { }t puT k p += (26.9a)or

( )( )

+

=

t

t

p

p

v

u

v

u

L AE

p

p'2

'1

2

2

1

1

'2

'1

sincos00

00sincos

11

11


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Expanding the above equation, yields

{ } { } L AE

vu

v

u

L

AE p +

=

2

2

1

1

1 sincossincos (26.10a)

And,

{ } { }

1

12

2

2

cos sin cos sin

u

v AE p AE L

u L

v

=

(26.10b)

Few problems are solved to illustrate the application of the above procedure tocalculate thermal effects /fabrication errors in the truss analysis:-

Example 26.1

Analyze the truss shown in Fig.26.2a, if the temperature of the member (2) israised by C

o

40 .The sectional areas of members in square centimeters areshown in the figure. Assume 25 / 102 mm N E = and 1/ 75, 000 = per C o .


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The numbering of joints and members are shown in Fig.26.2b. The possibleglobal displacement degrees of freedom are also shown in the figure. Note thatlower numbers are used to indicate unconstrained degrees of freedom. From thefigure it is obvious that the displacements 0876543 ====== uuuuuu due toboundary conditions.The temperature of the member (2) has been raised by C

o

40 . Thus,

T L L =

( )( ) 3102627.2402375000

1 == L m (1)

The forces in member (2) due to rise in temperature in global coordinate systemcan be calculated using equation (26.4b).Thus,

( )( )( )

=

sin

cos

sin

cos)(

2

1

6

5

L AE

p

p

p

p

t

t

t

t

(2)

For member (2),

242 102020 mcm A == ando

45=


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( )( )( )( )

5

6 4 11 3 3

1

2

1

21

220 10 2 10 2.2627 10 /10 1

21

2

t

t

t

t

p

p

p

p

=

(3)

( )( )( )( )

5

6

1

2

1

1150.82 1

1

t

t

t

t

p

p

kN p

p

=

(4)

In the next step, write stiffness matrix of each member in global coordinatesystem and assemble them to obtain global stiffness matrix

Element (1): 240 1015,3,0 m Am L === ,nodal points 4-1

4 11'

3

1 0 1 0

0 0 0 015 10 2 101 0 1 03 10

0 0 0 0

k

=

(5)

Member (2): 241020,23,45 m Am L === o , nodal points 3-1

[ ]

=

5.05.05.05.05.05.05.05.0

5.05.05.05.0

5.05.05.05.0

23

1021020 1142k (6)

Member (3): ,30,1015,90 24 m Lm A === o nodal points 2-1


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4 113

3 3

0 0 0 0

0 1 0 115 10 2 100 0 0 03 x 10 10

0 1 0 1

k

=

(7)

The global stiffness matrix is of the order 88 ,assembling the three memberstiffness matrices, one gets

[ ]

=

00000000

010000000100

0014.4714.470014.4714.47

0014.4714.470014.4714.47

000010001000

00000000

0014.4714.47100014.14714.47

010014.4714.470014.4714.147

10 3k (8)

Writing the load displacement equation for the truss

8

7

6

5

4

3

2

1

p

p

p

p

p

p

p

p

=

00000000

010000000100

0014.4714.470014.4714.47

0014.4714.470014.4714.47000010001000

00000000

0014.4714.47100014.14714.47

010014.4714.470014.4714.147

10 3

8

7

6

5

4

3

2

1

u

u

u

uu

u

u

u

+

0

0

1

1

0

0

1

1

640

(9)

In the present case, the displacements 1u and 2u are not known. All otherdisplacements are zero. Also 021 == p p (as no joint loads are applied).Thus,


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8

7

6

5

4

3

2

1

p

p

p

p

p

p

p

p

=

=

00000000

0100000001000014.4714.470014.4714.47

0014.4714.470014.4714.47

000010001000

00000000

0014.4714.47100014.14714.47

010014.4714.470014.4714.147

8

7

6

5

4

3

2

1

u

u

u

u

u

u

u

u

+

0

01

1

0

0

1

1

640 (10)

Thus unknown displacements are

11

32

147.14 47.14 0 11( 150.82 )

47.14 147.14 0 110

u

u

=

(11)

41

42

7.763 107.763 10

u mu m

= =

Now reactions are calculated as

+

+

=

00

1

1

0

0

640

00

0

0

0

0

00000001000000

0014.4714.4700

0014.4714.4700

00001000

000000

000100

14.4714.47

14.4714.47

1000

00

102

13

8

7

6

5

4

3

u

u

p p

p

p

p

p

3

4

5

6

7

8

0

77.63

77.63

77.63

77.63

0

p

p

p

p

p

p

=

kN (12)


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The support reactions are shown in Fig.26.2c.The member forces can be easilycalculated from reactions. The member end forces can also be calculated byusing equation (26.10a) and (26.10b). For example, for member (1),

o

0=

10010 3'2 = p [ ]0101 44

0

0

7.763 10

7.763 10

(13)

= 77.763 kN. Thus the member (1) is in tension.

Member (2)o

45=

281.9410 3'2 = p [-0.707 -0.707 0.707 0.707]

3

3

102942.3

102942.3

0

0

kN.78.1092 = p

Thus member (2) is in compression


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Example 26.2

Analyze the truss shown in Fig.26.3a, if the member BC is made 0.01m too shortbefore placing it in the truss. Assume AE =300 kN for all members.


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14/20


+

=

0

00

0

75.0

0

75.0

0

162.00934.00000162.0094.0

0934.0487.0000433.0094.0054.000162.0094.000162.0094.0

00094.0487.00433.0094.0054.0

000025.0025.00

0433.00433.00866.000

162.0094.0162.0094.025.00575.00

094.0054.0094.0054.0000108.0

8

7

6

5

4

3

2

1

8

7

6

5

4

3

2

1

u

u

u

u

u

u

u

u

AE

p

p

p

p

p

p

p

p

(2)

Note that 0876543 ====== uuuuuu Thus, solving

=

75.0

0

0

0

575.00

0108.011

2

1

AE u

u(3)

01 =u and, mu 32 103478.4

= (4)

Reactions are calculated as,

+

=

00

0

0

75.0

0

162.0094.0094.0054.0

162.0094.0

094.0054.0

25.00

00

2

1

8

7

6

5

4

3

u

u AE

p p

p

p

p

p

(5)

=

211.0

123.0

211.0

123.0

424.0

0

8

7

6

5

4

3

p

p

p

p

p

p

(6)

The reactions and member forces are shown in Fig.26.3c. The member forcescan also be calculated by equation (26.10a) and (26.10b). For example, formember (2),

o

90=


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4300'

2 = p [0 -1 0 1] L L AE

u

u

u

u

2

1

4

3

( ) ( )4

01.0300103478.4

4300 3 =

kN424.04239.0 = (7)

Example 26.3

Evaluate the member forces of truss shown in Fig.26.4a.The temperature of themember BC is raised by C o40 and member BD is raised by C o50 .Assume

AE=300KN for all members and75000

1= per o C.


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Solution

For this problem assembled stiffness matrix is available in Fig.26.4b.The jointsand members are numbered as shown in Fig.26.4b. In the given problem4321 ,,, uuuu and 5u represent unconstrained degrees of freedom. Due to support

conditions, 0876 === uuu .

The temperature of the member (2) is raised byo

50 C.Thus,

mT L L 32 10333.350575000

1 === (1)

The forces are developed in member (2), as it was prevented from expansion.

( )( )( )( )

=

sin

cos

sin

cos

10333.3300 3

2

1

8

7

f

f

f

f

p

p

p

p


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=

1

0

1

0

(2)

The displacement of the member (5) was raised by C o40 . Thus,

mT L L 35 10771.34025000,751 ===

The forces developed in member (5) as it was not allowed to expand is

( )( )

( )( )

=

707.0

707.0

707.0

707.0

10771.3300 3

8

7

6

5

t

t

t

t

p

p

p

p

=

1

1

1

1

8.0 (3)

The global force vector due to thermal load is

( )( )( )( )( )( )( )( )

=

1

0

8.0

8.0

0

0

8.18.0

8

7

6

5

4

3

21

t

t

t

t

t

t

t

t

p

p

p

p

p

p

p

p

(4)

Writing the load-displacement relation for the entire truss is given below.


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+

=

1

08.0

8.0

0

0

8.1

8.0

271.0071.000071.0071.02.00

071.0271.002.0071.0071.00000271.0071.02.00071.0071.0

02.0071.0271.000071.0071.0

071.0071.02.00129.0071.000

071.0071.000071.0271.0020.0

2.00071.0071.000271.0071.0

00071.0071.0020.0071.0271.0

8

7

6

5

4

3

2

1

8

7

6

5

4

3

2

1

u

u

u

u

u

u

u

u

AE

p

p

p

p

p

p

p

p

(5)

In the above problem 087654321 ======== p p p p p p p p and0876 === uuu .

Thus solving for unknown displacements,

=

8.0

0

08.1

8.0

0

0

00

0

271.000071.0071.0

0129.0071.000

0071.0271.0020.0071.000271.0071.0

071.002.0071.0271.0

1

5

4

3

2

1

AE

u

u

u

u

u

(5)

Solving equation (5), the unknown displacements are calculated as

mu

umumumu

0013.0

0,0005.0,0020.0,0013.0

5

4321

=====

(6)

Now, reactions are computed as,

+

=

1

0

8.0

0071.0071.02.00

2.0071.0071.000

071.02.00071.0071.0

5

4

3

2

1

8

7

6

u

u

u

u

u

p

p

p

(7)

All reactions are zero as truss is externally determinate and hence change intemperature does not induce any reaction. Now member forces are calculated by

using equation (26.10b)Member (1): L=5m, o0=


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5'2

AE p = [-1 0 1 0]

2

1

4

3

u

u

u

u

(8)

='2 p 0.1080 Kn

Member 2: L=5m, o90= ,nodal points 4-1

5'2

AE p = [0 -1 0 1] 5

2

1

8

7

10771.3300

u

u

u

u

(9)

=0.1087 kN

Member (3): L=5m, o0= ,nodal points 3-4

5300'

2 = p [-1 0 1 0]

8

7

6

5

u

u

u

u

(10)

=0.0780kN

Member (4 ): ,5,90 m L ==o

nodal points 3-2

5300'

2 = p [0 -1 0 1]

4

3

6

5

u

u

u

u

=0 (11)

Member (5): 25,45 == Lo ,nodal points 3-1

25300'2 = p [-0.707 -0.707 0.707 0.707] 3

2

1

6

5

10333.3300

u

uu

u

(12)

=-0.8619 kN

Member (6) : 25,135 == Lo ,nodal points 4-2


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25

300'2 = p [0.707 -0.707 -0.707 0.707]

4

3

8

7

u

u

u

u

= 0.0150 kN. (13)

SummaryIn the last four lessons, the direct stiffness method as applied to the trussanalysis was discussed. Assembly of member stiffness matrices, imposition ofboundary conditions, and the problem of inclined supports were discussed. Dueto the change in temperature the truss members either expand or shrink.However, in the case of statically indeterminate trusses, the length of themembers is prevented from either expansion or contraction. Thus, the stressesare developed in the members due to changes in temperature. Similarly theerrors in fabricating truss members also produce additional stresses in thetrusses. In this lesson, these effects are accounted for in the stiffness analysis. Acouple of problems are solved.

Documents

m4l26 Lesson 26 The Direct Stiffness Method: Temperature Changes and Fabrication Errors in Truss Analysis