Solution Homework1

8/6/2019 Solution Homework1

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ES128: Homework 1Solutions

Problem 1For the spring system given in Figure 1,

a.

Number the elements and nodes;b. Assemble the global stiffness and force matrix;c. Partition the system and solve for the nodal displacements;d. Compute the reaction forces.

SolutionWe assign nodes and elements numbers as in the figure below

It follows that the element stiffness matrices are given by

[1] [5]

]5[]1[

1111

3)1(

= k ;

[2] [4]

]4[]2[

1111)2(

= k ;

[4] [5]

]5[]4[

1111

2)3(

= k ;

Figure 1

(El 1)

(El 2)

(El 3)

(El 4)1

2

3

4

5


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2

[5] [3]

]3[]5[

1111)4(

= k .

The global stiffness matrix is given by

[1] [2] [3] [4] [5]

]5[]4[]3[]2[]1[

621032301010100

0101030003

= k .

The displacement and force nodal vectors are

=

5

4

000

d

uu

, =

50

0

f 3

2

1

r

r

r

.

To impose the prescribed boundary conditions, we partition the system as

=F

E

F

E

FFFE

EFEE

f f

dd

K K K K ,

where =

100

010

003

K EE k ;

=

100130

K EF k ;

=62

23K FF k ; =

000

d E ;

=5

4Fd u

u; =

3

2

1

Ef

r

r

r

; and =500

f F .

Since [d E = 0 0 0], FFFF f dK = , so that

==7143.10

1429.71f K d F

1-FFF k

.


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3

Therefore, k u /1429.74 = and ./7143.105 ku =

Since =Ed [0 0 0], EFEF f dK = , so that

==

7143.10

1429.7

1429.32

dK f FEFE .

Therefore, 1429.321 =r (N), 1429.72 =r (N) and 7143.103 =r (N).

Problem 2

Figure 2 shows a two-member plane truss supported by a linearly elastic spring. Thetruss members are of a solid circular cross section having d=20 mm and E=80Gpa. Thelinear spring has stiffness constant 50 N/mm.

a. Assemble the system global stiffness matrix and calculate the globaldisplacements of the unconstrained node;

b. Compute the reaction forces and check the equilibrium conditions;c. Check the energy balance. Is the strain energy in balance with the mechanical

work of the applied force?d. Compute the strain and stress in each bar.

Figure 2


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SolutionWe assign nodes and elements numbers as in the figure below

For element 1

)/(100265.55

)1020(4

10806

239

11 m N

l EAk =

==

;

5/3sin 1 = 5/4cos 1 = [1x] [1y] [4x] [4y]

]4[]4[]1[]1[

8095.14127.28095.14127.24127.2217.34127.2217.38095.14127.28095.14127.24127.2217.34127.2217.3

106)1(

y x y x

=

For element 2

)/(102832.64

)1020(4

10806

239

22 m N

l EAk =

==

;

0sin 2 = 1cos 2 =

[2x] [2y] [4x] [4y]

]4[]4[]2[]2[

000002832.602832.6000002832.602832.6

10 6)2(

y x y x

=

(El 1)

(El 2)

(El 3)

1

2

3

4


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5

For element 3)/(1050 33 m N k = ;

1sin 3 = ;0cos 3 = ;

[3x] [3y] [4x] [4y]

]4[]4[]3[]3[

05.0005.000000

05.0005.000000

10 6)3(

y x y x

=

The global stiffness matrix is[1x] [1y] [2x] [2y][3x] [3y] [4x] [4y]

]4[]4[]3[]3[]2[]2[]1[]1[

8595.14127.205.00008095.14127.24127.25002.90002832.64127.2217.3

05.0005.000000000000000000000002832.60002832.6008095.14127.200008095.14127.24127.2217.300004127.2217.3

10 6

y x y x y x y x

= .

The displacement and force matrices are

=

y

x

uu

4

4

000000

d ;

=

=

3

33

3

2

2

1

1

03

33

3

2

2

1

1

104907.11

106418.9

50sin1015

50cos1015

f

y

x

y

x

y

x

o y

x

y

x

y

x

r

r

r

r

r

r

r

r

r

r

r

r

.

To impose the prescribed boundary conditions, we partition the system as

=F

E

F

E

FFTEF

EFEE

f

f

d

d

K K

K K ,


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where

=

05.000000000000

000000

0002832.600

00008095,14127.2

00004127.2217.3

10K 6EE ;

=

05.00000002832.68095.14127.24127.2217.3

10K 6EF ;

=8595.14127.2

4127.25002.910K 6FF ; =

000000

d E ;

=y

x

uu

4

4Fd ; =

y

x

y

x

y

x

r r r r r r

3

3

2

2

1

1

Er ;

=3

3

F 104907.11106418.9

f .

Since [d E = 0 0 0], FFFF f dK = , so that

)(1804.11

8543.310f K d 3F

1-FF m==

= x u 4 3.8543(mm)=yu 4 11.1804(mm).

Since [d E = 0 0 0],

==

5590.0

0

0

2175.24

9317.10

5757.14

10dK f 3FEFE (N).

= x r 1 14.5757(kN);=

yr 1 -10.9317(kN);= x r 2 -24.2175 (kN);

02 =yr ;


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03 = x r ;=yr 3 -0.5590 (kN).

The resultant of the reaction forces in x direction is 14.5757+(-24.2175)+9.6418=0.

The resultant of the reaction forces in y direction is (-10.9317)+(-0.5590)+11.4907=0.

The strain for element 1 is

433

1 10249.75

101804.115/3108543.35/4

==

The stress for Element 1 isMPa E 58)10249.7(1080 49111 ===

The strain for element 2 is

43

1 106357.94

108543.3

==

The stress for Element 2 isMPa E 1.77)106357.9(1080 49222 ===

The strain energy for Element 1 is2

111 421 d G = 1l =33.0201 (J).

The strain energy for Element 2 is2

222 42

1 d G = 2l =46.671 (J).

The strain energy for the spring is243 2

1 ykuG = =3.1250 (J).

The strain energy of the system is 82.816 (J)

If the work done by the external force is computed assuming the force had remainedconstant from the initial state to the final state we obtainMechanical work of the applied force = ( y y x x u f u f 4444 + )=2x82.816 (J)

In the linear theory of elasticity Clapeyron's theorem states that the potential energy of deformation of a body, which is in equilibrium under a given load, is equal to half thework done by the external forces computed assuming these forces had remainedconstant from the initial state to the final state


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In reality the forces increased slowly from the initial state (force=0) to the final state(force= final force). If the work done by the external force is computed assuming theforce increased from the initial state to the final state we obtainMechanical work of the applied force = 0.5( y y x x u f u f 4444 + )=82.816 (J)

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Solution Homework1