MN Solved Example 3

8/9/2019 MN Solved Example 3

1/10

Technical University Gheorghe Asachi of Iasi Faculty of Civil Engineering, Department of Structural Mechanics

Solved example for Numerical Methods Combined Shear and Bending

Page 1 of 10

Consider the following stepped beam:

a) Determine the free nodal displacements;b) Determine the reactive forces;c) Determine the bending moments and the shear forces at the nodes;

d) Plot the shear force and bending moment diagrams.

Given data:

82

3 4

kNq 10

m

L 4 m

kNE 2.1 10

m

I 1.29 10 m

Step 1:

Input data (units are kN and m):

ORIGIN 1

q 1

E 2.1 108

I 1.2910 3

L4

2

L

q LQ

2 q

EIy

2q LM

8

2EI y

1 2 3

x

y

D1 = v 1 D2 = 1

D3 = v 2 D4 = 2

D5 = v 3 D6 = 3

R 1

R 2

R 3

0.5L


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Page 2 of 10

Step 2:

Write the finite element force-displ acement relations:

Element 1:

1 1

1 1 1 12 2

1 1 1 y1 1 1 1 1 1n 3

2 1 1 212 2

2 1 1 1 1 2

F 12 6L 12 6L v0

EIM 6L 4L 6L 2L0F fn k1 d

F 12 6L 12 6L v0 L

M 6L 2L 6L 4L0

EIyE I

2 E I

Qq L

1

2

Mq L

1 2

8

i 1

k1

EIyi

12

6 Li

12

6 Li

6 Li

4 Li 2

6 Li

2 Li

2

12

6 Li

12

6 Li

6 Li

2 Li 2

6 Li

4 Li

2

Li

3

k1

5.079 104

1.016 105

5.079 104

1.016 105

1.016 105

2.709 105

1.016 105

1.355 105

5.079 104

1.016 105

5.079 10 4

1.016 105

1.016 105

1.355 105

1.016 105

2.709 105

fn1

0

0

0

0


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Page 3 of 10

Element 2:

2

22

2 22 22 2

2 2 2 y2 22 2 2 2n 3

3 32 222 2

3 32 2 2 22

q L4

F v12 6L 12 6Lq L2EIM 6L 4L 6L 2L48F fn k2 d

F v12 6L 12 6Lq L L4M 6L 2L 6L 4L

q L48

fn2

10

3.333

10

3.333

i 2

k2

EIyi

12

6 Li

12

6 Li

6 Li

4 Li

2

6 Li

2 Li

2

12

6 Li

12

6 Li

6 Li

2 Li

2

6 Li

4 Li

2

Li

3

k2

8.127 105

8.127 105

8.127 105

8.127 105

8.127 105

1.084 106

8.127 105

5.418 105

8.127 105

8.127 105

8.127 105

8.127 105

8.127 105

5.418 105

8.127 105

1.084 106

fn2

q L2

2

q L2 2

12

q L2

2

q L2 2

12


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Page 4 of 10

Step 3:

Determine the assembled structural stiffness matrix and write the structural force-displacement relation:

1

11 22 22 2

n 1 22 2

3

1 1

2 2

3 3 32

0 0

0 0vq L q L

00 04 4Q v v Qq L q LF f K D K 0M M48 48

vq L q L

4 4q L

R R

R R

R q

48

R

1

5

62

2

3

4

D

K

D

D

DL

4

D

8

D

ii 1 6 jj 1 6 k1exp ii jj 0

i 1 4 j 1 4 k1exp i j k1i j

k1exp

5.079 104

1.016 105

5.079 104

1.016 105

0

0

1.016 105

2.709 105

1.016 105

1.355 105

0

0

5.079 104

1.016 105

5.079 104

1.016 105

0

0

1.016 105

1.355 105

1.016 105

2.709 105

0

0

0

0

0

00

0

0

0

0

00

0

k2expii jj

0 k2expi 2 j 2

k2i j

k2exp

0

0

0

0

0

0

0

0

0

0

0

0

0

0

8.127 105

8.127 105

8.127 105

8.127 105

0

0

8.127 105

1.084 106

8.127 105

5.418 105

0

0

8.127 105

8.127 105

8.127 105

8.127 105

0

0

8.127 105

5.418 105

8.127 105

1.084 106

K k1exp k2exp


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Page 6 of 10

1

25

r rr rn r 6

1

2

r 3

n n nr nn n2

3

42

R

R

FR DD

D

D

0

q LD4Dq L

f K K DD48 K2F f K K 0 0

Q q L

4Mq L

48

K1 stack submatrix K 1 1 1 6( ) submatrix K 5 6 1 6( ) submatrix K 2 4 1 6( )( )

K2 augment submatrix K1 1 6 1 1( ) submatrix K1 1 6 5 6( ) submatrixK1 1 6 2 4( )( )

K2

5.079 104

0

0

1.016 105

5.079 104

1.016 105

0

8.127 105

8.127 105

0

8.127 105

8.127 105

0

8.127 105

1.084 106

0

8.127 105

5.418 105

1.016 105

0

0

2.709 105

1.016 105

1.355 105

5.079 104

8.127 105

8.127 105

1.016 105

8.635 105

7.111 105

1.016 105

8.127 105

5.418 105

1.355 105

7.111 105

1.355 106

Krr submatrix K2 1 3 1 3( )

Krr

5.079 104

0

0

0

8.127 105

8.127 105

0

8.127 105

1.084 106

Krn submatrix K2 1 3 4 6( )

Krn

1.016 10 5

0

0

5.079 104

8.127 105

8.127 105

1.016 10 5

8.127 105

5.418 105

Knr submatrix K2 4 6 1 3( )


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Page 7 of 10

Step 5:

Determine the free nodal displacements and the reactive forces according to the support conditions:

r rr r rnr rr rn r

n n nr nn n n nr

r r

n nn

n

n r

f K D K f K K D

F f K

F DF

K F f K D DK D

Since r 0

D 0

0

r nr rn

n n nn n

f K

F f K

F D

D

1

rn nn n n r

1

nn

r

nn n

K K F f f

K

F

D F f

Knr

1.016 105

5.079 104

1.016 105

0

8.127 105

8.127 105

0

8.127 105

5.418 105

Knn submatrix K2 4 6 4 6( )

Knn

2.709 105

1.016 105

1.355 105

1.016 105

8.635 105

7.111 105

1.355 105

7.111 105

1.355 106

fn submatrix fnexp 2 4 1 1( )

fn

0

10

3.333

fr stack submatrix fnexp 1 1 1 1( ) submatrix fnexp 5 6 1 1( )( )

fr

0

10

3.333

Fn

0

Q

M

Fn

0

20

20


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Page 8 of 10

or:

Checking up the support reactions:

Dn lsolve Knn Fn fn( )

Dn

1.995 10 5

4.324 10 5

7.471 10 6

Dn Knn 1

Fn fn( )

Dn

1.995 10 5

4.324 10 5

7.471 10 6

Fr fr Krn Dn

Fr

0.929

39.071

34.429

L

q LQ 2

q

EIy

2q LM

8

2EIy

1 2 3

x

y

R 1 = 0.929 kN

R 2 = 39.071 kN

R 3 = 34.429 kNm

0.5L


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Page 9 of 10

Step 6:

Determine the bending moments and the shear forces at the nodes:

Node 1:

Node 2:

Node 3:

IN SUMMARY:

a)

5

6n 3

4

521.995 10

4.324 10

7.471 1

D

D D

D0

b) 1

r 2

3

0.929

39.071

3

R

F R

4.429R

Fr 1

Q q L2 Fr

2 7.105 10

15

Fr 1

L1

L2 M Q L2 q L2

L2

2 Fr

3 0

Vz1 Fr 1 Vz1 0.929

My1 0

Vz2left Fr 1 Vz2left 0.929

My2left Fr 1

L1 My2left 3.714

Vz2right Fr 1

Q Vz2right 19.071

My2right Fr 1

L1

M My2right 23.714

Vz3 Fr 2 Vz3 39.071

My3 Fr 3 My3 34.429


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Page 10 of 10

d)

L

q LQ

2

q

EIy

2q LM

8

2EIy

1 2 3

x

y

R 1 = 0.929 kN

R 2 = 39.071 kN

R 3 = 34.429 kNm

0.5L

0.929 kN

V z

M y

+

-

-

+

19.071 kN 39.071 kN

3.714 kNm

34.429 kNm

23.714 kNm

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