For the simple bar shown in figure determine the displacement, strain, stress caused due to self weight. Given length of the bar is 0.5m, Cross Section Area of the bar is 0.1m2, ρ=7848 kg/m3 and Young’s Modulus 2x1011 N/m2 .

A,

L,

E

ρ

Analytical Solution:Displacement in the bar at any section ux = xρgl

2Ex7848x9.81x0.5

2x2x1011=

Displacement in the bar at the free end x=0.5 is = = 4.812x10-87848x9.81x0.52

2x2x1011

PROBLEM

Finite Element Analysis Solution:Discretize the given structure with two 1-D bar element of equal length le= 0.25 m

Q1

Q2

12

3

Q2

12Elemental Stiffness Matrix for a 1-D Bar Element of uniform

cross section is given by EAle

[ke] = +1 -1-1 +1

Then the Elemental Stiffness Matrix Element 1 is given by[k1] = 2x1011x0.1

0.25

Then the Elemental Stiffness Matrix Element 2 is given by[k2] =

[k1] = 1010 +8 -8-8 +8Then we have [k2] = 1010 +8 -8

-8 +8and

Then the Overall Stiffness Matrix 2 is given by

2x1011x0.10.25

+1 -1-1 +1

2

3

2 3

+1 -1-1 +1

1

2

1 2

[Ko] = 10108 -8 0-8 8+8 -80 -8 8

1

2

3

1 2 3

=1058 -8 0-8 16 -80 -8 8

1

2

3

1 2 3

Characteristic equation for the over all problem is given by [Ko] {Q} = {Fo}

Apply the boundary conditions. Since at node 1 the bar is fixed hence Q1=0, Then by elimination approach eliminating the first row and column in the characteristic equation we have

10108 -8 0-8 16 -80 -8 8

1

2

3

1 2 3Q1Q2Q3

12 1

962.36

The elemental load vector due to body force is given by ρΑgle2[fb] =

11

Then the Elemental load vector in the element 1 is given by

Similarly Elemental load vector in the element 2 is given by 2

3= 962.36 1

1

Then the Overall Nodal Force Vector due to self weight will be {Fo}=12 1

962.361

2

3

7848x9.81x0.1x0.252

[fb1] =

11

1

2= 962.36 1

1

For the Element 1 q1 = Q1 = 0 and q2 = Q2 = 3.61x10-8 mm

10108 -8 0-8 16 -80 -8 8

Q1Q2Q3

12 1

962.36 1010 16 -8-8 8

Q2Q3

21962.36Then we

Solving we get Q2 = 3.61x10-8mm and Q3 = 4.81x10-8mm Solution for the displacements are Q1= 0, Q2 = 3.61x10-8mm and Q3 = 4.81x10-8mm DETERMINATION OF OTHER UNKNOWNS

ξ= -1 ξ=+1ξ=01 2

q2q1ξ

Displacement in an element is given by {U} = [N]{q}In which [N] = [ N1 N2 ] is the Shape function matrix

{q} = q1 is unknown nodal displacement vectorq2

in which N1=(1-ξ)

2 N2=(1+ξ)

2 are the shape functions of 1-D Bar Element.and

Thus U(-1) = 0 at node 1 and U(+1) = 3.61x10-8 mm at node 2

Thus {U} = N1q1 + N2q2 = (1-ξ)2

(1+ξ)2(0) + (3.61x10-8)

2.5x10-3(1+ξ)

U(ξ)=1.805x10-8(1+ξ)

Displacement any where with in the element is given by U(ξ)=1.805x10-8(1+ξ)

=

Strain {ε} = [B] {q} But [B] = 1le [ -1 1]

{ε} = 10.25

[ -1 1]

Thus {ε} = 1le [ -1 1] q1

q20

3.61x10-8

Stress {σ} = [D] {ε} = E ε = (2x1011)x(1.444x10-7) = 2.888x104 N/mm2 (Tensile)

= 1.444x10-7 (Tensile)

For the Element 2 q1 = Q2 = 3.61x10-8mm and q2 = Q3 = 4.61x10-8 mm

=

ξ= -1 ξ=+1ξ=01 2

q2q1ξ

Displacement in an element is given by {U} = [N]{q}In which [N] = [ N1 N2 ] is the Shape function matrix

{q} = q1 is unknown nodal displacement vectorq2

in which N1=(1-ξ)

2 N2=(1+ξ)

2 are the shape functions of 1-D Bar Element.and

Thus U(-1) = 3.61x10-8 mm at node 1 and U(+1) = 4.81x10-8 mm at node 2

Thus {U} = N1q1 + N2q2 = (1-ξ)2

(1+ξ)2(3.61x10-6) + (4.81x10-8) 10-8(4.21+0.6ξ)

Thus the displacement any where with in the element is given by U(ξ)=10-8(4.21+0.6ξ)

Strain {ε} = [B] {q} But [B] = 1le [ -1 1]

{ε} = 10.25

[ -1 1] Thus {ε} = 1

le [ -1 1] q1q23.61x10-8

4.81x10-8

Stress {σ} = [D] {ε} = E ε = (2x1011)x(4.8x10-8) = 9600 N/mm2 (Tensile)

= 4.8x10-8 (Tensile)

Reaction forces are determined from the eliminated row of the characteristic equation of the system i.e.

Thus 1011[8 -8 0]Q1Q2Q3

= R1 Thus R1=1011[8 -8 0]0

3.61x10-8 4.81x10-8

R1 = -28.88x103 N

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3

1

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3 4

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1

1

3

2

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1

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5

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Exit the Menu4

1

2,3

Enter the coordinates and apply until all the coordinates are given

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1

3

Pick the points where the lines are to be created

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1

Pick the line using the curser till all are

completed

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3

1

2

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3

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Pick the required line for discretizing

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Observe the details which has to be decided for the element type, Material constant and Real constant

1

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After creating the mesh. To check go to plotctrls >Number>highlight the node number and element numbering and then press ok. You can observe the elements and node are numbered.

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Pick the nodes where the boundary

conditions are to be specified

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2

1

Apply the boundary condition and then exit

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After exiting save the file

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Close the file2

3

1

After solution is done go to Postprocessor to view the results

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1

COMPARISON

SOLUTION FOR THE DISPLACEMENTSFEM Manual Calculated Solution ANSY SolutionQ1= 0 Q1= 0Q2 = 3.61x10-8mm Q2 = 3.61x10-8mmQ3 = 4.81x10-8mm Q3 = 4.81x10-8mm