Matrix analysis for structure example

8/9/2019 Matrix analysis for structure example

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Name: Engr. Gerion Jay C. Babuyo Page No.: 1/13

Proffesor: Jerry B. Maratas ce, m.eng(struct.) m.asep

Subject: MCE 121S - Matrix Analysis of Structures Date: January 15, 2015

Assignment No.: 2

PROBLEM NO.1.1:

1.) Determine the structural stiffness matrix for the entire structure by using the basic definition of stiffness2.) Determine the structural stiffness matrix by combining the individual element stiffness.3.) If the left end of the structure is fixed, solve for the nodal displacements and member forces if F 2 = 10 kips, F 3 = -5

kips, and F 4 = 8 kips. Use the partitioning method and the method of removing rows and columns from the originalstructural stiffness matrix E = 29 x 10 6 psi for all members.

F2 F3 F4

Areas for all members = 2 sq. in.

Member 1 Member 2 Member 3

Solution:

1.) Note that the structure has 4 degrees of freedom with 4 translations at each node; therefore we will have a 4 X 4structural stiffness matrix.

= eq.1 To obtain the structural stiffness matrix and to find the forces that is on the node, we introduce a unit displacement( ) at each node one at a time.

At node 1: 1 = 1; 2 = 0; 3 = 0; 4 = 0

We have; F1 F2 F3 F4

F1 = k11 = k1 1 = k1 k1 1 k1 1 F2 = k21 = -k1 1 = -k1 F1 F2 F3 = k31 = 0 k1 1 k1 1 F4 = k41 = 0

At node 2: 1 = 0; 2 = 1; 3 = 0; 4 = 0


F1 = k12 = -k1 2 = -k1 k1 2 k1 2 k2 2 k2 2 F2 = k22 = k1 2 + k2 2 = k1 + k2 F1 F2 F3 F3 = k32 = -k2 2 = -k2 k1 2 k1 2 k2 2 k2 2 F4 = k42 = 0

At node 3: 1 = 0; 2 = 0; 3 = 1; 4 = 0


F1 = k13 = 0 k2 3 k2 3 k3 3 k3 3 F2 = k23 = -k2 3 = -k2 F2 F3 F4 F3 = k33 = k2 3 + k3 3 = k2 + k3 k2 3 k2 3 k3 3 k3 3 F4 = k43 = -k3 3 = -k3

20 in. 15 in. 10 in.

1 2 3 4

1 2

1 2 3 4

1 2 3

2 3 4

1 2 3 4


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Assignment No.: 2

At node 4: 1 = 0; 2 = 0; 3 = 0; 4 = 1


F1 = k14 = 0 k3 4 k3 4 F2 = k24 = 0 F3 F4 F3 = k34 = -k3 4 = -k3 k3 4 k3 4 F4 = k44 = k3 4 = k3

Substitute all the elements from each node to equation 1 ( eq.1 ) and our structural stiffness matrix will become,

= 0 0 0

00 0 k i = ; k 1 = ; k 2 = ; k 3 =

= [120 120 0 0120 760 115 000 1150 16 110110 110 ]

2.) The individual stiffness matrix of a single element can be obtained using the definition of the stiffness but we are allfamiliar that for a one dimensional bar (1D bar) the individual stiffness matrix is just equal to;

= 1 11 1 The individual element stiffness are;Stiffness for Element 1: = 120 120120 120 Stiffness for Element 2: = 115 115115 115 Stiffness for Element 2: = 110 110110 110

To determine the structural stiffness matrix using individual stiffness matrix, it is necessary to assign global

displacement coordinates. In this would problem the global displacement coordinates are;

Therefore, the structural stiffness matrix will be;

= [120 120 0 0120 760 115 000 1150 16 110110 110 ]

1 2 3 4

3 4

1 2

1

2

2 3

2

3

3 4

4

3

1 2 3 4

1

2

3

4

1 2

1

2

2 3

2

3

3 4

3

4


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Assignment No.: 2

3.) Given that the left end of the structure is to be fixed and all the forces are known;

10 k -5k 8k

Areas for all members = 2 sq. in.


We already obtained the structural stiffness matrix of the structure above in no.2 which is;

= [

120 120 0 0120 760 115 000 1150 16 110110 110 ]

The force-displacement equation will become

F = k ; Note that the displacement in the fixed-end support is equal to zero

1058 =

[120 120 0 0120 760 115 000 1150 16 110110 110 ]

[0234]

We can reduce the matrix into a 3 x 3 matrix by eliminate the 1 st column and row due to u1 = 0;

1058 = [760 115 01150 16 110110 110 ] 234

To find the displacement on each node, we need to find the inverse of the 3 x 3 structural stiffness matrixes andmultiply it with the forces given in the left side. In this problem we use the Gauss-Jordan row reduction method tofind the inverse that is;

[A][I] = 760 115 01150 16 110110

110

1 0 00 1 00 0 1

= [1 47 000 970 110110 110 ]607 0 047 1 00 0 1

= [1 04900 1 790145 ][

1009 409 0409 709 049 79 1]

20 in. 15 in. 10 in.

F1

607 R1 115 R1 + R2 R3 47 R2 + R1709 R2 110 R2 + R3


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Assignment No.: 2

= [1 0 00 1 00 0 1][20 20 2020 35 3520 35 45]

The inverse of the matrix now is located on the right side. Finding the displacement will be obtained thru the productof the forces and the inverse of the matrix.

[20 20 2020 35 3520 35 45][

1058] =

[234]

[201020 5208201035 5358201035 5458] =

[234]

[260305385]

=

[234]

[4.4828 105.2586 106.6379 10]

=

[234]

Using the stiffness element for member 1 and expressing into force-displacement equation we have,

= 120 120120 120 10 = 29 102120 120120 120 04.4828 10

F1 =

29 102 x

4.4828 10

F 1 = -13k

The member forces can be obtained using the individual element stiffness matrix

Member 1:

= 29 102120 120120 120 04.4828 10 = 1313

79 R3 +R2 45 R349 R3 +R1


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Assignment No.: 2

Member 2:

= 29 102115 115115 115 4.4828 10

5.2586 10 = 33 Member 3:

= 29 102110 110110 110 5.2586 10

6.6379 10 = 88 The member forces can also be found using partitioning method;

The partitions below show the forces in each member;

Therefore the member forces and the nodal displacements are:

= 1313 =0 . =6.6379 . = 33 =4.4828 10 . = 88

=5.2586 10 .

20 in. 15 in. 10 in.

-13 k 10 k -5k 8k

-13 k 10 k -5k 8k

Node 1 Node 2 Node 3 Node 4

13 k 3k

13 k 10 k 5k

8k


13 k 10 k

3k

Member 1 Member 2Member 1

13 k

13 k

13 k 8k Member 1 Member 2 Member 3

3k 8k

10 k 5k


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Assignment No.: 2

PROBLEM NO.1.3:

Solve for nodal displacement and member end forces for the structure in the figure below. Also construct the axialforce diagrams.

E = 30 x 10 6 psi

Member 1 Member 2

Solution:

Notice that the loads are distributed all over the length, thus we need to express this into nodal loads such that;

(a) (b) (c)

Isolate a section in member 1 at Figure (b)

= 2 Note: = 0 (at support) 0 = | 0 = 10 F 1 100; F 1 = 10

k

Isolate a section in member 2 at Figure (b)

= 3 10 Note: = 0 (at support) 0 = { 10} 0 = 1015 15; F 2 = 32.5 k Our equivalent nodal forces will become;

F 3 = (2(10) + 3(15)) (10 + 32.5) = 32.5k

The force-displacement with stiffness element matrix for each member is;

Member 1: = 110 110110 110 0 + Member 2: = 215 215215 215 +

1 in. 2 2 in. 215 k

10 in. 15 in.

2 k/in. 3 k/in.

15 k =

2 k/in. 3 k/in.

2 k/in. 3 k/in.

15 k + F1

F2

15 k

F3

F1

F2 F3

2 k/in.

F1

x

2x

P(x)

3 k/in.

10 k 3x

P xF2

x

R

10 k 15 k

32.5 k 22.5 k

1 21

2

2 32

3


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Assignment No.: 2

Combining the individual element stiffness matrix we have the structural stiffness matrix which can be reduced into a2 x 2 matrix.

1032.57.5 = [

110 110 0110 730 2150 215 215 ]023 32.57.5 = 730 215215 215

Finding the inverse of the matrix to obtain the displacement

= . .; det. [k] = = ; Adj. [k] = 215 215215 730 = 75

215 215215 730 =

10 1010 17.5

The displacements are;

10 1010 17.532.57.5 = ; = 400 456.25 = 0.0133 .0.0152 . ; u1 = 0 in .The member forces can be obtained using individual stiffness matrix.

Member 1: =

110 110

110 110 00.0133

+ 1010

= 5030

Member 2: = 215 215215 215 0.01330.0152 + 22.522.5 = 3015

Therefore the member forces and the nodal displacements are:

= 5030 =0 . =0.0152 . = 3015 =0.0133 .

1

1

2 3

2

3

15 k

3 k/in. 2 k/in.

50 k50 k

30 k 30 k 15 k

50 k (+)

(-)

(+)30 k

15k

Member 1 Member 2


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Assignment No.: 2

PROBLEM NO.1.5:

Determine the equivalent nodal forces for the loads shown in the figures below.

Fig. 1.5a

= +

Solution:

We need two equations to solve the problem, one compatibility equation and an equilibrium equation. Note that thedisplacement of the members are to be equal which we can utilize it as an equation to solve the nodal forces.

= ; = ; = ; = = ; = = ; =; = = + ; = ; =

Fig. 1.5b

= +

Solution:

We know that; = (for a triangular loading)=

;

=

= ;0=

= ; = = ; = ; =

R1

F

b a L

F

b a L

Equivalent Nodal Force R2R1 R2R1

Equivalent Nodal Force R2R1

Equivalent Nodal Force = =

Equivalent Nodal Force R2 R2R1

Equivalent Nodal Force R2R1

Equivalent Nodal Force

L

kx kx

R1

kx

p0(x) x (in.)

P (x)

= =


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Assignment No.: 2

Fig. 1.5c

= +

Solution:

The total force can be obtained by integrating the whole area;

=

sin =

cos = 1 1

=

Thus;

= ; = = cos1 0= sin 0= 0 0 ; 0= ; = =

;

= ;

=

Fig. 1.5d

= +

Solution:


= = = = ; = ; = 0= ; 0= ; = = ; =

R1Equivalent Nodal Force

R2 R2R1Equivalent Nodal Force

R2R1


R1

p0(x) x (in.)

P (x)

= =

L

sin sin

sin



R2R1

Equivalent Nodal Force =312 =34

L

kx kx

R1

p0(x) x (in.)

P (x)

kx


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Assignment No.: 2

Fig. 1.5e

= +

Solution:


=

=

=

= ; = ; = 0= ; 0= ; = = ; = Fig. 1.5f

= +

Solution:

We know that; = ( for a trapezoidal loading )= ; = 1 2 ; =21 ; =21

=1 2212 ;

=2 21 22 12

0= 212 233 133 ( further simplification yields ) 2 =1 2 ; =1 2 = ; =121 2 1 2 =1 2



R2R1

Equivalent Nodal Force =420 =45

L

R1

p0(x) x (in.)

P (x)

kx

kx

kx



R2R1


= =

L

R1

p0(x) x (in.)

P (x)


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Assignment No.: 2

PROBLEM NO.1.11:

Member 1 in the structure shown in the figure below undergoes a temperature increase of 50 oF. Using E = 29 x 10 6 psi and = 6.5 x 10 -6 in/in/ oF, find the nodal displacement and member forces. Draw a free body diagram of eachmember.

Node 1 Node 2 Node 3

Solution:

Due to increase in temperature a nodal force (compression) is developed in each node of member 1 which isequivalent to

= thus the forces are;

=29 1026.5 1050=18.85 The force-displacement equations for each member are;For Member 1: For Member 2:

= 29 1015 1515 15 0 = 29 10110 110110 110 Combining individual stiffness and simplifying we have;

18.8518.858 =

29 10[15 15 015 310 1100 110 110 ]0 ;

18.858 =

29 10310 110110 110

Finding the inverse of the 2 x 2 matrix we have;

Adj. [k] = 110 110110 310 ; det. [k] = - = [k]-1 = .110 110110 310 =50110 110110 310 = 5 55 15

The displacements are now;

5 55 1518.858=(29 103)23 ; = 1.8707 10

8.8794 10 The member forces can be obtained using individual stiffness matrix;

For Member 1:

= 29 1015 1515 15 01.8707 103 + 18.8518.85 = = 88

2 in. 2 1 in. 28k

Member 1 Member 2

10 in. 10 in.

18.85 k 18.85 k

8k


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Assignment No.: 2

For Member 2:

= 29 10110 110110 110 1.8707 10

8.8794 10 + 00 = = 88 The diagram below illustrates the forces acting on its member;

Therefore, the nodal displacements and the member forces are;

= 88 =0 . = 8.8794 104 . = 88 =1.8707 103 .

PROBLEM NO.1.13:

Determine the static and kinematic indeterminacy for the following structures

Fig. 1.13a Solution:

This truss has 4 unknown reactions, 12 degrees of freedom and 4 restraintsavailable on its supports, also 3 equilibrium equations can be utilize in thisstructure thus it s;

Static Indeterminacy = Reactions Static Eq . = 4 3 = 1o

Kinematic Indeterminacy = Df Restraints = 12 4 = 8o

Fig. 1.13b Solution:

This 1D bar has only 1 unknown reaction, 5 degrees of freedom and 1restraint available on its support, also only 1 equilibrium equation can beutilize in this structure thus its;

Static Indeterminacy = Reactions Static Eq . = 1 1 = 0o

Kinematic Indeterminacy = Df Restraints = 5 1 = 4o

18.85 k 18.85 k

Node 1 Node 2 Node 3

8k

8k Member 1 Member 28k 8k 8k

8k


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Matrix analysis for structure example

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