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matrix algebra example problem solving, Exercises of Mathematics

Mathematics

matrix algebra solutions to a problem about a frame using matrix algebra

Typology: Exercises

2022/2023

Uploaded on 12/14/2023

fatimae-roque
fatimae-roque 🇵🇭

1 document

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bg1
Lbm1 6:= Abm 0.075:=
Lbm2 6:= Ebm 30 106
:=
Lbm3 6:= Ibm 480 10 6-
:=
Lbm4 9:=
Lbm5 Lbm4
2Lbm3
2
+:=
STIFFNESS MATRIX OF EACH MEMBER
K1
Ebm Ibm
Lbm1
3
Abm Lbm1
2
Ibm
cos 90deg( )2
12 sin 90deg( ) 2
+
Abm Lbm1
2
Ibm
12-
cos 90deg( )sin 90deg( )
6-Lbm1
sin 90deg( )
Abm Lbm1
2
Ibm
cos 90deg( )2
12 sin 90deg( ) 2
+
-
Abm Lbm1
2
Ibm
12-
cos 90deg( )sin 90deg( )
-
6 Lbm1
sin 90deg( )
( )
-
Abm Lbm1
2
Ibm
12-
cos 90deg( )sin 90deg( )
Abm Lbm1
2
Ibm
sin 90deg( )2
12 cos 90deg( ) 2
+
6 Lbm1
cos 90deg( )
Abm Lbm1
2
Ibm
12-
cos 90deg( )sin 90deg( )
-
Abm Lbm1
2
Ibm
sin 90deg( )2
12 cos 90deg( ) 2
+
-
6 Lbm1
cos 90deg( )
6-Lbm1
sin 90deg( )
6 Lbm1
cos 90deg( )
4 Lbm1
2
6 Lbm1
sin 90deg( )
6 Lbm1
cos 90deg( )
( )
-
2 Lbm1
2
Abm Lbm1
2
Ibm
cos 90deg( )2
12 sin 90deg( ) 2
+
-
Abm Lbm1
2
Ibm
12-
-cos 90deg( )sin 90deg( )
6 Lbm1
sin 90deg( )
Abm Lbm1
2
Ibm
cos 90deg( )2
12 sin 90deg( ) 2
+
Abm Lbm1
2
Ibm
12-
cos 90deg( )sin 90deg( )
6 Lbm1
sin 90deg( )
Abm Lbm1
2
Ibm
12-
-cos 90deg( )sin 90deg( )
Abm Lbm1
2
Ibm
sin 90deg( )2
12 cos 90deg( ) 2
+
-
6-Lbm1
cos 90deg( )
Abm Lbm1
2
Ibm
12-
cos 90deg( )sin 90deg( )
Abm Lbm1
2
Ibm
sin 90deg( )2
12 cos 90deg( ) 2
+
6-Lbm1
cos 90deg( )
6-Lbm1
sin 90deg( )
6 Lbm1
cos 90deg( )
2 Lbm1
2
6 Lbm1
sin 90deg( )
6-Lbm1
cos 90deg( )
4 Lbm1
2
:=
K1
800
0
2400-
800-
0-
2400-
0
375000
0
0-
375000-
0
2400-
0
9600
2400
0-
4800
800-
0-
2400
800
0
2400
0-
375000-
0-
0
375000
0-
2400-
0
4800
2400
0-
9600
=
pf3
pf4
pf5

Partial preview of the text

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L

bm

:= 6 A

bm

L

bm

:= 6 E

bm

6 := 

L

bm

:= 6 I

bm

  • 6 := 

L

bm

L

bm

L

bm

2 L bm

2 := +

STIFFNESS MATRIX OF EACH MEMBER

K

E

bm

I

bm

L

bm

3

A

bm

L

bm

2 

I

bm

cos 90deg( )

2  12 sin 90deg( )

2

A

bm

L

bm

2 

I

bm

 cos 90deg( )sin 90deg( )

- 6 L

bm

 sin 90deg( )

A

bm

L

bm

2 

I

bm

cos 90deg( )

2  12 sin 90deg( )

2

A

bm

L

bm

2 

I

bm

 cos 90deg ( )sin 90deg( )

6 L

bm

 sin 90deg( )

A

bm

L

bm

2 

I

bm

 cos 90deg( )sin 90deg( )

A

bm

L

bm

2 

I

bm

sin 90deg( )

2  12 cos 90deg( )

2

6 L

bm

 cos 90deg( )

A

bm

L

bm

2 

I

bm

 cos 90deg( )sin 90deg( )

A

bm

L

bm

2 

I

bm

sin 90deg( )

2  12 cos 90deg( )

2

6 L

bm

 cos 90deg( )

- 6 L

bm

 sin 90deg( )

6 L

bm

 cos 90deg( )

4 L

bm

2 

6 L

bm

 sin 90deg( )

6 L

bm

 cos 90deg( )

2 L

bm

2 

A

bm

L

bm

2 

I

bm

cos 90deg( )

2  12 sin 90deg( )

2

A

bm

L

bm

2 

I

bm

  •  cos 90deg( )sin 90deg( )

6 L

bm

 sin 90deg( )

A

bm

L

bm

2 

I

bm

cos 90deg( )

2  12 sin 90deg( )

2

A

bm

L

bm

2 

I

bm

 cos 90deg ( )sin 90deg( )

6 L

bm

 sin 90deg( )

A

bm

L

bm

2 

I

bm

  •  cos 90deg( )sin 90deg( )

A

bm

L

bm

2 

I

bm

sin 90deg( )

2  12 cos 90deg( )

2

- 6 L

bm

 cos 90deg( )

A

bm

L

bm

2 

I

bm

 cos 90deg( )sin 90deg( )

A

bm

L

bm

2 

I

bm

sin 90deg( )

2  12 cos 90deg( )

2

- 6 L

bm

 cos 90deg( )

- 6 L

bm

 sin 90deg( )

6 L

bm

 cos 90deg( )

2 L

bm

2 

6 L

bm

 sin 90deg( )

- 6 L

bm

 cos 90deg( )

4 L

bm

2 

K

k 1

E

bm

I

bm

L

bm

3

A

bm

L

bm

2 

I

bm

A

bm

L

bm

2 

I

bm

6.L

bm

6.L

bm

6.L

bm

4L

bm

2

- 6. L

bm

2.L

bm

2

A

bm

L

bm

2 

I

bm

A

bm

L

bm

2 

I

bm

- 6. L

bm

- 6. L

bm

6.L

bm

2L

bm

2

- 6. L

bm

4.L

bm

2

k 2

k 1

k 3

k 1

cosθ

L

bm

sinθ

L

bm

L

bm

T

cosθ

  • sinθ

sinθ

cosθ

cosθ

  • sinθ

sinθ

cosθ

T

T

T

T

K

1_

T

T k 1

 T

K

K

K

K

K

FIXED-END MOMENTS EACH MEMBER

FEM

A

:= 0 FEM

V

12 L

bm

:= =64.9 FEM

M

12 L

bm

2 

Ff 5

FEM

A

cosθ 5

 FEM

V

sinθ 5

FEM

A

sinθ 5

 FEM

V

cosθ 5

FEM

M

FEM

A

cosθ 5

 FEM

V

sinθ 5

FEM

A

sinθ 5

 FEM

V

cosθ 5

FEM

M

KS

START OF LOAD-DISPLACEMENT ANALYSIS IN THE GLOBAL AXES

LOADS FROM

FIXED-END MOMENT, Ff 5

LOADS FROM

NODES

Pf

:= 5 Pj

P Pj - Pf

:= = 5 d KS

  • 1 P

START OF EACH MEMBER EVALUATION

MEMBER 1

d 1

:= T

= u 1

T

d 1

:= = Qf 1

:= 0 Q

k 1

u 1

 Qf 1

:= = F

T

T Q 1

MEMBER 2

d 2

:= T

= u 2

T

d 2

4 Qf 2

:= 0 Q

k 2

u 2

 Qf 2

:= = F

T

T Q 2