tugas elhing

8/11/2019 Tugas elhing

1/20

For elemen 1

f1x k1 -k1 d1x

f3x -k1 k1 d3x

For elemen 2

f3x k2 -k2 d3x

f2x -k2 k2 d2x

d3x(1) = d3x(2)

F1x = f1x(1)

F2x = f2x(2)

F3x = f3x(1) + f3x(2)

Therefore the force-displacement equations for this spring system are:

F1x = k1d1x - -k1d3x

F2x = -k2d3x + k2d2x

F3x = -k1d1x + k1d3x + k2d3x -

In Matrix,

F1x k1 0 -k1 d1x

F2x 0 k2 -k2 d2x

F3x -k1 -k2 k1+k2 d3x

x

= x

= x

=


2/20

k2d2x


3/20

k(1)

= 1000 x 1 -1

-1 1

k(2)

= 2000 x 1 -1

-1 1

k(3)

= 3000 x 1 -1

-1 1

K = 1000 0 -1000 0 d1x

0 3000 0 -3000 d2x

-1000 0 3000 -2000 d3x

0 -3000 -2000 5000 d4x

Force at node 4 (F4x= 5000lb)

3000 -2000 d3x 0

-2000 5000 d4x 5000

d3x 9.09091E-08 5000

d4x 2000

solving for d3xand d4xgives

d3x = 0.909091 in

d4x = 1.363636 in

F1x 1000 0 -1000 0

F2x 0 S 0 -3000

F3x -1000 0 3000 -2000

F4x 0 -3000 -2000 5000

F1x = -909.091 lb

F2x = #VALUE! lb

F3x = 0 lb

F4x = 5000 lb

For elemen 1

f1x 1000 -1000 0

f3x -1000 1000 0.909091

x

x =

= x

= x

= x

1 3

3 4

4 2


4/20

For elemen 2

f3x 2000 -2000 0.909091

f4x -2000 2000 1.363636

For elemen 3

f4x 3000 -3000 1.363636

f2x -3000 3000 0= x

= x


5/20

k1 = 1000 lb/in

k2 = 2000 lb/in

k3 = 3000 lb/in

P = 5000 lb

F1x

F2x

F3x

F4x

2000 0

3000 5000

0

0

0.909090909

1.363636364

f1x = -909.091

f3x = 909.0909

=

x


6/20

f3x = -909.091

f4x = 909.0909

f4x = 4090.909

f2x = -4090.91


7/20

k = 200 KN/m

= 20 mm = 0.02 m

k(1)

= k(2)

= k(3)

= k(4)

= 200

K = 200 -200 0 0 0

-200 400 -200 0 0

0 -200 400 -200 0

0 0 -200 400 -2000 0 0 -200 200

200 -200 0 0 0 d1x F1x

-200 400 -200 0 0 d2x F2x

0 -200 400 -200 0 d3x F3x

0 0 -200 400 -200 d4x F4x

0 0 0 -200 200 d5x F5x

Applying the boundary condition (d1x=0 and d5x=20mm) and the known force give:

400 -200 0 0 d2x 0

-200 400 -200 0 d3x 0

0 -200 400 -200 d4x 0

0 0 -200 200 0.02 F5x

400 -200 0 d2x 0

-200 400 -200 d3x 0

0 -200 400 d4x 4

d2x 3.125E-08 120000

d3x 80000

d4x 40000

d2x = 0.005 m

d3x = 0.01 m

d4x = 0.015 m

F1x = -1 KN

F5x = 1 KN

x =

x =

=

x =

x


8/20

For elemen 1

f1x = 200 -200 0

f2x = -200 200 0.005

For elemen 2

f2x = 200 -200 0.005

f3x = -200 200 0.01

For elemen 3

f3x = 200 -200 0.01

f4x = -200 200 0.015

For elemen 4

f4x = 200 -200 0.015

f5x = -200 200 0.02x

x

x

x


9/20

1 -1 KN/m

-1 1

determinan= 32000000

Adjoint

80000 40000 0 400 -200 -200

160000 80000 0 -200 400 -200

80000 120000 4-200 -200 400

0 400 0

-200 400 400

0 -200 0

x

x


10/20

f1x = -1 KN

f2x = 1 KN

f2x = -1 KN

f3x = 1 KN

f3x = -1 KN

f4x = 1 KN

f4x = -1 KN

f5x = 1 KN


11/20

0 -200 0 120000 80000

400 400 -200 80000 160000

40000 800000 400 0

400 -200 -200

-200 400 -200

-200 -200 400

=


12/20


13/20

40000

80000

120000


14/20

d1x = d3x = d4x = 0

d2x(1)

= d2x(2)

= d2x(3)

= d2x

k(1)

= 1000 -1000 k(2)

= 2000

-1000 1000 -2000

K = 1000 -1000 0 0

-1000 6000 -2000 -3000

0 -2000 2000 0

0 -3000 0 3000

Applying the boundary conditions (d1x=d3x=d4x=0) and the known forces (F2x=P) gives:

P = (k1+k2+k3)d2x

d2x = 0.833333

Solving the forces gives:

F1x = -833.333

F3x = -1666.67

F4x = -2500

1 2 2


15/20

P = 5000 lb

k1 = 1000 lb/in

k2 = 2000 lb/ink3 = 3000 lb/in

-2000 k(3)

= 3000 -3000

2000 -3000 3000

3 2 4


16/20

The Potental energy :

p for elemen 1 :

p(1)

= 1/2 k1 x (d3x-d1x)2

- f1xd1x -

p for elemen 2 :

p(2)

= 1/2 k2 x (d4x-d3x)2

- f3xd3x -

p for elemen 3 :

p

(3)

= 1/2 k3 x (d2x

-d4x

)

2

- f2x

d2x

-

Minimizing the total potential energy p :

p/ d1x = 0 = -k1d3x + k1d1x -

p/ d2x = 0 = k3d2x - k3d4x -

p/ d3x = 0 = k1d3x - k1d1x -

p/ d4x = 0 = k2d4x - k2d3x -

in Matrix form

1000 0 1000 0 d1x f1x

0 3000 0 -3000 d2x f2x-1000 0 3000 -2000 d3x f3x(1) + f3x(2)

0 -3000 -2000 5000 d4x f4x(1) + f4x(2)

where, f1x = F1x 1000 0 1000

f2x = F2x 0 3000 0

f3x(1) + f3x(2 = F3x -1000 0 3000

f4x(1) + f4x(2 = F4x 0 -3000 -2000

x =


17/20

k1 = 1000 lb/in

k2 = 2000 lb/in

k3 = 3000 lb/in

P = 5000 lb

f3xd3x

f4xd4x

f4xd4x

f1x(1)

f2x(3)

k2d4x + k2d3x - f3x(1)

- f3x(2)

k3d2x + k3d4x - f4x(2)

- f4x(3)

0 d1x F1x

-3000 d2x F2x

-2000 d3x F3x

5000 d4x F4x

x =


18/20

For elemen 1 and 2

k(1)

= k(2)

= 1 -1

-1 1

For elemen 3

k

(3)

= 1 -1-1 1

K = 1 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 1

F1x 1 -1 0 0

F2x -1 2 -1 0

F3x 0 -1 2 -1

F4x 0 0 -1 1

The boundary condition are:

d1x = d4x = 0

Applying the boundary condition and knon forces gives :

3000 2 -1 d2x

0 -1 2 d3x

d2x = 0.002 ind3x = 0.001 in

The global nodal forces are calculated as:

F1x 1 -1 0 0

F2x -1 2 -1 0

F3x 0 -1 2 -1

F4x 0 0 -1 1

= 1000000 x

= 1000000 x

= 1000000 x

1000000

1000000

x

x

1000000 x

x

d1x d2x d3x d4x


19/20

For elemen 1 and 2

A = 1 in2

E = 30000000 psi

30 in

F2x = 3000 lb

For elemen 3

A = 2 in2

E = 15000000 psi

d1x

d2x

d3x

d4x 2000000 -1000000

-1000000 2000000

Matrix invers = 3.33333E-13 2000000 1000000

1000000 2000000

0 -2000

0.002 3000

0.001 0

0 -1000

= lbs

x

x


20/20

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