Stress Distributions Numerical-Finite Element Method-Assignment Solution, Exercises of Mathematical Methods for Numerical Analysis and Optimization

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ASSIGNMENT

NUMERICAL

PROBLEM # 1.

A, E, I constant P

L

P

L

W 2

W 1

w x

x

NUMERICAL PROBLEM # 1.

STATEMENT

Using a one- beam element idealization, find the stress distribution under a load of P

for the uniform cantilever beam shown in the figure 1.15 (Hint: Use the displacement

model of Equation 1.78, the strain displacement relation given in equation 9.25 and the

stress-strain relation  xx  E  xx , where E is the Young’s Modulus.)

Solution:

Step I: Idealization

The one-beam element idealization as given in the statement can be justified by considering

the fixed and free end of beam as end nodes. As there are two nodes, the axial and angular

displacement of node 1 is denoted by W 1 and W 2 while those of node 2 are denoted by W 3 and

W 4 respectively.

Beam subjected to

load

And

2 2

2

w

x

W 1

(e) . 6

l

(12 x -6 l )

2 +W 3

(e) . 6

l

(6 l -12 x )

2 +W 2

(e) 2 . 4

l

(6 x -4 l )

2 +W 4

(e) . 4

l

(6 x – 2 l )

2

-2 W 1

(e) W 3

(e) 6

l

(12 x -6 l )

2 -2 W 3

(e) W 2

(e) 5

l

( 72x

2 +24 l

2 -84 lx )+ 2 W 2

(e) W 4

(e) 4

l

(36 x

2 -36 l x +8 l

2 )

+2 W 4

(e) W 1

(e) 5

l

( 72x

2

• 60lx+ 12 l

2 )+ 2 W

(e) W 2

(e) 5

l

( 72x

2 +24 l

2 -84 lx )+ 2 W 3

(e) W 4

(e) 5

l

( 60 lx -12 l

2

• 72x

2 )

So after substituting the value of

2

2

w

x

in equation 1.79 and integrating it, we get

w x =

EI[ 3

l

(6 W 1

(e) +2W 2

(e) l

2 +6W 3

(e) +2 W 4

(e) l

2

• 6W 1

(e) W 2

(e) l -12W 1

(e) W 3

(e)

• 6W 1

(e) W 4

(e) l -6W 2

(e) W 3

(e) l

+2W 2

(e) W 4

(e) l

2 -6W 3

(e) W 4

(e) )]

This can be written in the matrix form as

w x =

. 3

2 EI

l

 

( ) ( ) ( ) ( ) 1 2 3 4

e e e e W W W W

( ) 1 2 2 ( ) 2 ( ) 3 2 2 ( ) 4

e

e

e

e

l l W

l l l l (^) W

l l (^) W

l l l l (^) W

 ^  ^ 

 ^ 

 ^ 

Comparing the above equation with equation 1.79, we get

( ) e W = Vector of unknown displacement for element e

( ) e W =

( ) 1 ( ) 2 ( ) 3 ( ) 4

e

e

e

e

W

W

W

W

For element 1

W 1

(e)

1

2

3

4

W

W

W

W

And stiffness matrix of an element =

( ) e  k   

2 2 ( ) 3

2 2

e

l l

EI l^ l^ l^ l K l l l

l l l l

  ^ 

………………………… (A)

Thus A can be used to find stiffness matrix of element 1

Step IV: Assembly of element stiffness matrices and element load vectors

Since there is only one element, the assembly stiffness matrix is given as

W 1 = W 1

(1) W 2 = W 2

(2) W 3 = W 3

(3) W 4 = W 4

(4)

6 3 6 3

2 2 3 2 3

6 3 6 3 ~

2 2 3 3 2

l l

l l l l

K

l l

l l l l

  

 

  ^  

   ^ 

  ^ ^ ^    

 

 (^)    

(1)

(^1 )

(3)

(^2 )

(3)

3 3 (4)

4 4

W (^) W

W (^) W

W W

W W









After deleting the rows and the columns corresponding to the degrees of freedom W 1 , W2.

Since W 1 =W 2 =0 are the boundary conditions, and putting l =L, we obtain the stiffness matrix

[k] of beam as

( )

(^3 )

e

EI L

L L L

K

 ^ 

And the load vector is given as

3

4

P

P

P

Since the vertical load P is applied at node 2 ( in the direction of W 3 ) is given by positive

sign i.e. P and the rotational load ( bending moment at node 2 in the direction of W 4 ) is 0,

thus load vector of beam corresponding to degrees of freedom W 3 and W 4 can be expressed

as

Step VI: Element Stresses

By using equation 9.25 i.e.

2 ( ) ( ) 2

e e xx

w B W y x

And we know that

( ) e (^) ( ) xx (^) xx

 E^ e

So, we get

2 ( ) ( ) 2

e^ e xx

w E B W y x

Since there is only one element, so

2 (1) (1) xx 2

w E B W y x



Where

2

2

w

x

W 1

(e) . ( )

e l

(12 x -6 l

( e) ) +W 2

(e) . ( )

e l

(6 x -4 l

(e) ) +W 3

(e) . ( )

e l

(6 l

(e) -12 x ) +W 4

(e) . ( )

e l

(6 x – 2 l

(e) )

Thus the stresses in the element can be determined

(1)  (^) xx (. ) x y   yE [W1. 3

L

(12 x -6L) +W 2. 2

L

(6 x -4L)+W 3. 3

L

(6L-12 x )+W 4. 2

L

(6 x – 2L)]

Note:

Since the values of W1, W 2 , W3, W 4 and L are not mentioned in the given numerical problem.

Therefore the above equation represents a general solution for given problem.

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