Topic 2 Scheme

SOLUTION FIGURE 1 B

C

60

60

A

D

E

Ay

Ey

2m

2m

+

2m

 MA

=

0

35 ( 8 ) - Ey ( 4 )

=

0

=

70 kN

Ey

35kN 2m

(  )

( 1 mark )

+

 Fy

=

0

=

0

Ay

=

- 35 kN

Ay

=

35 kN

Ay + 70 - 35

(  )

( 1 mark )

Joint A +

FAB sin 60 FAB

 Fy

- 35 + FAB sin 60 FAB

FAB cos 60

=

0

=

0

=

40.415 kN

FAE

( Tension )

( 1 mark )

35kN

+

 Fx

=

0

FAE + 40.415 cos 60

=

0

FAE

=

- 20.208 kN

FAE

=

20.208 kN ( Compression )

( 1 mark )

Joint B

+

 Fy

=

0

=

0

FBE

=

- 40.415 kN

FBE

=

40.415 kN

- 40.415 sin 60 - FBE sin 60 FBC

40.415 cos 60

FBE

40.415kN 40.415 sin 60

FBE cos 60

( Compression )

FBE sin 60

( 1 mark )

+

 Fx

=

0

FBC - 40.415 cos 60 - 40.415 cos 60

=

0

FBC

=

40.415 kN ( Tension )

( 1 mark )

Truss has symmetrical load and structure

FAB

=

FDC

=

40.415 kN ( Tension )

( 1 mark )

FAE

=

FDE

=

20.208 kN ( Compression )

( 1 mark )

FBE

=

FCE

=

40.415 kN ( Compression )

( 1 mark )

C

B

1

x

Ax

A

60

60

D

E Ay

Ey

2m

2m

2m

2m

tan 60

=

x 2

x

=

3.464 m

( 1 mark )

+

 MA

=

0

1 ( 3.464 ) - Ey ( 4 )

=

0

Ey

=

0.866

(  )

( 1 mark )

+

 Fy

=

0

Ay + 0.866

=

0

Ay

=

- 0.866

Ay

=

0.866

(  )

( 1 mark )

+

 Fx

=

0

=

0

Ax

=

- 1

Ax

=

1

Ax + 1

(  )

( 1 mark )

FAB sin 60 FAB

Joint A

FAB cos 60 FAE

1

0.866

+

 Fy

- 0.866 + FAB sin 60

=

0

=

0

FAB

=

1 ( Tension )

( 1 mark ) +

 Fx

=

0

FAE + 1 cos 60 - 1

=

0

FAE

=

0.5 ( Tension )

( 1 mark ) Joint B +

 Fy

=

0

=

0

FBE

=

- 1

FBE

=

1

- 1 sin 60 - FBE sin 60 FBC

1 cos 60

FBE cos 60

FBE

1

1 sin 60

( Compression )

FBE sin 60

( 1 mark ) +

 Fx

FBC - 1 cos 60 - 1 cos 60 FBC

=

0

=

0

=

1 ( Tension )

( 1 mark ) Joint D + FDC sin 60

 Fy

FDC sin 60

FDC

FDC

FDC cos 60

=

0

=

0

=

0

( 1 mark )

FDE

+

 Fx FDE

=

0

=

0

( 1 mark )

Joint C + 1 FCE cos 60

1

FCE

FCE sin 60

 Fy

=

0

- FCE sin 60

=

0

FCE

=

0

( 1 mark )

The magnitude of the horizontal displacement at point C due by the external load P0 Member

L P1



( m )

( m )

AB

40.415

1

4

161.66

AE

- 20.208

0.5

4

- 40.416

BC

40.415

1

4

161.66

( kN )

BE

- 40.415

-1

4

161.66

CD

40.415

0

4

0

CE

- 40.415

0

4

0

DE

- 20.208

0

4

0 444.564



( 7 x 0.5 = 3.5 marks ) ( 1 mark )

C

B

3.464m

x

A

60

60

D

E

2m

x

2m

=

 22

=

4m

+

2m

2m

3.4642

( 1 mark )

The magnitude of the horizontal displacement at point C due by the temperature in members AB , BC and CD

Member

P1

c ( / C )

AB

1

11 x 10

AE

0.5

0

BC

1

11 x 10

BE

-1

0

-6

-6

-6

L

t



( m )

(  C )

( m )

4

15

6.6 x 10-4

4

0

0

4

15

6.6 x 10-4

4

0

0

CD

0

11 x 10

4

15

0

CE

0

0

4

0

0

DE

0

0

4

0



0 1.32 x 10-3

( 3 x 0.5 = 1.5 marks ) ( 1 mark )

The magnitude of the horizontal displacement at point C due by the external load and temperature in members AB , BC and CD , CH

=

1

 P0 P1 L

+

 P1 c L t

[ 444.564 ]

+

1.32 x 10-3

+

1.32 x 10-3

AE

CH

=

1 AE

 CH

=

444.564

(  )

AE

( 2 marks )

SOLUTION FIGURE 2 B

C

D

H

G

E

1

 6m

Ax

A

F

Ay

Fy 3m

+

 MA

1 ( 6 ) - Fy ( 9 ) Fy

3m

=

0

=

0

=

0.667

3m

(  )

( 1 mark )

+

 Fy

=

0

Ay + 0.667

=

0

Ay

=

- 0.667

Ay

=

0.667

(  )

( 1 mark )

+

 Fx

=

0

=

0

Ax

=

- 1

Ax

=

1

Ax + 1

(  )

( 1 mark )

tan 

=

6 3



=

63.435

( 1 mark )

Joint A + FAB

 Fy

=

0

FAB - 0.667

=

0

=

0.667

FAB

( Tension )

FAH

1

( 1 mark ) 0.667

+

 Fx

FAH - 1 FAH

=

0

=

0

=

1 ( Tension )

( 2 marks ) Joint B +

0.667 FBH sin 63.435

=

0

=

0

FBH

=

- 0.746

FBH

=

0.746

- 0.667 - FBH sin 63.435

FBC

FBH

 Fy

FBH cos 63.435

( Compression )

( 1 mark ) +

 Fx

FBC - 0.746 cos 63.435 FBC

=

0

=

0

=

0.334 ( Tension )

( 1 mark ) Joint H +

 Fy

FHC - 0.746 sin 63.435 0.746 sin 63.435

FHC

FHC

0.746

=

0

=

0

=

0.667 ( Tension )

0.746 cos 63.435 1

FHG

( 2 marks ) +

 Fx

=

0

FHG - 1 + 0.746 cos 63.435

=

0

FHG

=

0.666 ( Tension )

( 1 mark )

Joint C +

0.334

=

0

=

0

FCG

=

- 0.746

FCG

=

0.746

- 0.667 - FCG sin 63.435

FCD

FCG

 Fy

FCG cos 63.435

0.667

( Compression )

FCG sin 63.435

+

 Fx

FCD - 0.334 - 0.746 cos 63.435 FCD

=

0

=

0

=

0.668 ( Tension )

Joint G +

FGD - 0.746 sin 63.435

0.746 sin 63.435 0.746

FGD

FGD

=

0

=

0

=

0.667 ( Tension )

0.746 cos 63.435 0.666

 Fy

FGF

+

 Fx

=

0

FGF - 0.666 + 0.746 cos 63.435 =

0

FGF

=

0.332 ( Tension )

Joint D +

FDF cos 63.435

FDF

=

0

=

0

FDF

=

- 0.746

FDF

=

0.746

- 0.667 - FDF sin 63.435

FDE

0.668

 Fy

0.667

( Compression )

FDF sin 63.435

+

 Fx

FDE - 0.668 - 0.746 cos 63.435 FDE

=

0

=

0

=

1.002 ( Tension )

Joint E +

1.002

 Fy FEF

=

0

=

0

FEF

x B

C

D

H

FEDG

E

1

63.435 6m

1

A

FFD

0.667 3m

3m FFD cos 63.435 FFG

EF

1

FFD sin 63.4350.667 x 6m

3m

63.435

F 0.667

3m

( 2 marks )

+

 MF

1 ( 6 ) - FED ( 6 ) FED

=

0

=

0

=

1 ( Tension )

( 2 marks )

+

 Fy

=

0

=

0

FFD

=

- 0.746

FFD

=

0.746

FFD sin 63.435 + 0.667

( Compression )

( 2 marks ) +

 Fx

0.746 cos 63.435 - 1 + 1 - FFG FFG

=

0

=

0

=

0.334 ( Tension )

( 2 marks )

The horizontal displacement at joint E if the members of AH and CH have occur elongation during the preparation 



( mm )

( mm )

0

0

1

1.5

1.5

BC

0.334

0

0

BH

- 0.746

0

0

CD

0.668

0

0

CG

- 0.746

0

0

CH

0.667

1.5

1.001

DE

1

0

0

DF

- 0.746

0

0

DG

0.667

0

0

Member

P1

AB

0.667

AH

EF

0

0

0

FG

0.334

0

0

GH

0.666

0



0 2.501

( 2 x 1 = 2 marks )

( 1 mark )

The horizontal displacement at joint E if the members of DE , FG and DF have occur an increase temperature c

L

t



( / C )

( mm )

( C )

( mm )

0.667

0

6 000

0

0

1

0

3 000

0

0

BC

0.334

0

3 000

0

0

BH

- 0.746

0

6 708

0

0

CD

0.668

0

3 000

0

0

CG

- 0.746

0

6 708

0

0

CH

0.667

0

0

Member

P1

AB AH

6 000

0

-6

DE

1

10 x 10

3 000

16

0.48

DF

- 0.746

10 x 10-6

6 708

16

- 0.801

DG

0.667

0

6 000

0

0

EF

0

0

6 000

0

0

3 000

16

0.160

3 000

0

0

FG

0.334

GH

0.666

10 x 10

-6

0

- 0.161



( 3 x 1 = 3 marks ) ( 1 mark )

B

6m

D

H

G

E

x

A

3m

x

C

=

 32 + 6 2

=

6.708 m

3m

F

3m

( 1 mark )

The horizontal displacement at joint E if the members of AH and CH have occur a 1.5 mm elongation during the preparation and the members of DE , FG and DF occur an increase temperature of 16C ,

EH

=

 P1 (  )

EH

=

2.501

 EH

=

2.34 mm

+

-

 P1 c L t

0.161

(  )

( 2 marks )

SOLUTION FIGURE 3

F

3m

D

E 

3m

 A

B

Ay

C Cy

40kN 2m

+

2m

2m

 MA

=

0

40 ( 4 ) - Cy ( 8 )

=

0

=

20 kN

Cy

2m

(  )

( 0.5 mark )

+

 Fy

Ay + 20 - 40 Ay

=

0

=

0

=

20 kN

(  )

( 0.5 mark )

tan 

=

3 2



=

56.310

( 0.5 mark )



=

180 – ( 56.310 + 56.310 )



=

67.38

Joint A + FAD sin 56.310

 Fy

=

0

20 + FAD sin 56.310

=

0

FAD

=

- 24.037 kN

FAD

=

24.037 kN

FAD FAD cos 56.310 FAB

( Compression ) 20kN

( 1 mark ) +

 Fx

FAB - 24.037 cos 56.310 FAB

=

0

=

0

=

13.333 kN ( Tension )

( 1 mark ) Joint B + FBD sin 56.310 FBE sin 56.310 FBD

=

0

FBD sin 56.310 + FBE sin 56.310 - 40

=

0

FBD

=

40

FBE

FBD cos 56.310

 Fy

13.333kN

( 0.5 mark ) +

 Fx

=

0

13.333 - 13.333 + FBE cos 56.310 - FBD cos 56.310

=

0

=

FBD

40kN

FBE

( Equation 1 )

FBE cos 56.310

13.333kN

-

FBE

( Equation 2 )

( 0.5 mark ) Substitute equation 2 into equation 1 FBD

=

40 - FBE

FBD

=

40 - FBD

2FBD

=

40

FBD

=

20 kN ( Tension )

( 1 mark ) Substitute FBD

=

20 kN into equation 2

FBE

=

FBD

FBE

=

20 kN ( Tension )

Joint D

+

 Fy

=

0

FDF sin 56.310 + 24.037 sin 56.310 - 20 sin 56.310

=

0

FDF

=

- 4.037 kN

FDF

=

4.037 kN

FDF sin 56.310 FDF FDF cos 56.310

( Compression )

FDE

24.037 cos 56.310

( 1 mark )

20 cos 56.310 20kN

24.037kN 24.037 sin 56.310

+

20 sin 56.310

 Fx

=

0

=

0

FDE

=

- 22.188 kN

FDE

=

22.188 kN

24.037 cos 56.310 + 20 cos 56.310 + FDE - 4.037 cos 56.310

( Compression )

( 1 mark )

Truss has symmetrical load and structure

FAB

=

FCB

=

13.333 kN ( Tension )

( 0.5 mark )

FAD

=

FCE

=

24.037 kN ( Compression )

( 0.5 mark )

FBD

=

FBE

=

20 kN ( Tension )

( 0.5 mark ) F FDF

=

FEF

=

4.037 kN

3m

( Compression )

D

( 0.5 mark )

E

3m

56.310 A

B

Ay

C Cy

1 2m

2m

2m

2m

+

 MA

1 ( 4 ) - Cy ( 8 ) Cy

=

0

=

0

=

0.5

(  )

( 0.5 mark )

+

 Fy

Ay + 0.5 - 1 Ay

=

0

=

0

=

0.5

(  )

( 0.5 mark )

Joint A + FAD sin 56.310

 Fy

=

0

0.5 + FAD sin 56.310

=

0

FAD

=

- 0.601

FAD

=

0.601

FAD FAD cos 56.310 FAB

( Compression ) 0.5

( 1 mark ) +

 Fx

FAB - 0.601 cos 56.310 FAB

=

0

=

0

=

0.333 ( Tension )

FBD sin 56.310 FBE sin 56.310

Joint B

FBD

( 1 mark ) FBE

FBD cos 56.310

FBE cos 56.310 0.333

0.333

1

+

 Fy

FBD sin 56.310 + FBE sin 56.310 - 1

=

0

=

0

FBD

=

1

-

FBE

( Equation 1 )

( 0.5 mark ) +

 Fx

0.333 - 0.333 + FBE cos 56.310 - FBD cos 56.310 FBE

=

0

=

0

=

FBD ( Equation 2 )

( 0.5 mark ) Substitute equation 2 into equation 1 FBD

=

1 - FBE

FBD

=

1 - FBD

2FBD

=

1

FBD

=

0.5 ( Tension )

( 1 mark ) Substitute FBD

=

0.5 into equation 2

FBE

=

FBD

FBE

=

0.5 ( Tension )

Joint D +

 Fy

=

0

=

0

FDF

=

- 0.101

FDF

=

0.101

FDF sin 56.310 + 0.601 sin 56.310 - 0.5 sin 56.310

FDF sin 56.310 FDF

( Compression )

FDF cos 56.310 FDE

0.601 cos 56.310

( 1 mark )  Fx

=

0

0.601 cos 56.310 + 0.5 cos 56.310 + FDE - 0.101 cos 56.310

=

0

0.5 sin 56.310

FDE

=

- 0.555

FDE

=

0.555

0.5 cos 56.310 0.601

+

0.5

0.601 sin 56.310

( Compression )

( 1 mark ) Truss has symmetrical load and structure

FAB

=

FCB

=

0.333 ( Tension )

( 0.5 mark )

FAD

=

FCE

=

0.601 ( Compression )

( 0.5 mark )

FBD

=

FBE

=

0.5 ( Tension )

( 0.5 mark )

FDF

=

FEF

=

0.101 ( Compression )

( 0.5 mark )

The vertical displacement at joint B due to applied load P0 Member

P1

( kN )

L

A

E



( mm )

( mm2 )

( kN / mm2 )

( mm )

AB

13.333

0.333

4 000

2 500

200

0.036

AD

- 24.037

- 0.601

3 606

2 500

200

0.104

BD

20

0.5

3 606

2 500

200

0.072

BE

20

0.5

3 606

2 500

200

0.072

BC

13.333

0.333

4 000

2 500

200

0.036

CE

- 24.037

- 0.601

3 606

2 500

200

0.104

DF

- 4.037

- 0.101

3 606

2 500

200

0.003

DE

- 22.188

- 0.555

4 000

2 500

200

0.099

EF

- 4.037

- 0.101

3 606

2 500

200

0.003 0.529



( 9 x 0.5 = 4.5 marks ) ( 1 mark )

C

B

3m

x

A

D

E

2m

2m

x

=

 22 + 3 2

x

=

3.606 m

2m

2m

( 0.5 mark )

E

=

200 GPa

E

=

200 x 109 N m2

E

=

1 kN

1 m2

1000 N

10002 mm2

200 kN / mm2

( 0.5 mark )

The vertical displacement at joint B due to elongation of the members of the BD and BE 



( mm )

( mm )

0

0

Member

P1

AB

0.333

AD

- 0.601

0

0

BD

0.5

0.5

0.25

BE

0.5

0.5

0.25

BC

0.333

0

0

CE

- 0.601

0

0

DF

- 0.101

0

0

DE

- 0.555

0

0

EF

- 0.101

0

0 0.5



( 2 x 1 = 2 marks ) ( 1 mark ) The vertical displacement at joint B due to applied load and elongation of the members of the BD and BE , BV

=

 P0 P1 L +

 P1 (  )

AE

BV

=

0.529

+

 BV

=

1.029 mm

0.5

(  )

( 2 marks )