Back to Study material
FEM

Unit-5

Analysis of structures


  • Definition: A truss is a structure that consists of

1)     All straight members

2)     Connected together with pin joints

3)     Connected only at the ends of the members

4)     And all external forces (loads & reactions) must be applied only at the joints.

  • The analysis of trusses is usually based on the following simplifying assumptions:

1)     The centroidal axis of each member coincides with the line connecting the centers of the adjacent members and the members only carry axial force.

2)     All members are connected only at their ends by frictionless hinges in plane trusses.

  • Trusses Using FEA In this section, we will apply basic finite element techniques to solve general two dimensional truss problems. The technique is a little more complex than that originally used to solve truss problems, but it allows us to solve problems involving statically indeterminate structures.

 


Stiffness matrix of a truss element

  • The truss may be statically determinate or indeterminate. All members are subjected to only direct stresses (tensile or compressive). Joint displacements are selected as unknown variables.
  • Since there is no bending of the members we have to ensure only displacement continuity (C continuity) and there is no need to worry about slope continuity (C continuity).
  • Here we select two noded bar element for the formulation of stiffness matrix of truss element. Since the members are subjected to only axial forces, the displacements are only in the axial directions of the members.
  • Therefore, the nodal displacement vector for the bar element is

                                                     {}=     

Where, and are the displacements in axial direction of the element. The stiffness matrix of a bar element is

                          

[]=

 

 

Problems-

Example 1-

Analyze the truss as shown in fig. cross sectional area of members are AB= 1000 , BC=800 ,CA= 800 take E

C:\Users\Ssd\Desktop\Untit 5 1.png

 

Fig.1

Ans-

C:\Users\Ssd\Desktop\Unit 5 2.png

 

Fig.2

 

Step-1 degrees of freedom

Discretization –

Elements

Nodes

Displacements mm

Boundary conditions

1

AB

2

BC

=0

3

CA

-

Assume x-axis horizontal through point (& vertical through point A). the co-ordinates of node A(0,1.5) , B(4,1.5) & C(2,0) . take E in Gpa.

Member

L

l

m

AE/l (KN/mm)

AB

4

0

4

1

0

50

BC

-2

-1.5

2.5

-0.8

-0.6

64

CA

-2

1.5

2.5

-0.8

0.6

64

 

Step-2 element stiffness matrices

Stiffness matrix of element AB .

Stiffness matrix of element BC

Stiffness matrix of element CA

Step 3- global stiffness matrix

Total Dof are 06, size of stiffness matrix 6*6

Step -4 reduced stiffness matrix

Since eliminate corresponding rows & columns from global stiffness matrix

Step -5 equation of equilibrium

 

 

Example -2

Figure shows a plane , truss with three members cross- sectional area of all members 500 young modulus is 200 KN/. Determine deflection at loaded joint .

 

Solution  

Step 1 – degree of freedom ;-

Discretization

Element

Nodes

Displacement (mm)

Boundary cond

1

AD

2

BD

3

CD

Assume origin support A(0,0). The co-ordinates of other nodes&  B(1000,0), C(2000,0) and D(1500,1000)

Member

L

l

m

AE/L (KN/M)

AD

1500

1000

1802.8

0.832

0.555

88.75

BD

500

1000

1118

0.447

0.894

143.112

CD

-500

1000

1118

-0.477

0.894

143.112

 

Step -2 element stiffness matrices

Stiffness matrix of element AD

 

Step – 4 equations of equilibrium

 

 

Example 3

For the truss as shown in fig. using finite element method, determines deflections at loaded joints. The joint B is subjected to 50 KN horizontal force towards left & 80 KN force vertically downward. take cross sectional area of all the members 1000 young modulus is 200 Gpa.

C:\Users\Ssd\Desktop\UNIT 5 4.png

 

                     Fig.4

Ans-

Step -1 degree of freedom (06)

Discretization

Element

Nodes

Displacement (mm)

Boundary cond

1

AB

2

DB

3

CB

 

Assume origin point B. the co-ordinates points are

Members

L

l

m

AE/L (kN/mm)

AB

4000

-3000

5000

0.8

-0.6

40

DB

4000

3000

5000

0.8

0.6

40

CB

-4000

30000

5000

-0.8

0.6

40

 

Step -2  stiffness  matrix of element AB .

Stiffness matrix at element DB.

Step 3- reduced stiffness matrix

Step 4- equation of equilibrium

 

 


Steps for the solution of Indeterminate plane frames using finite element method:

1. Divide the frame into number of elements (Take one member as one element)

2. Identify total degrees of freedom (Three D.O.F. at each node, two displacements and rotation)

3. Determine stiffness matrices of all elements ([K]1, [K]2……)

4. Assemble the global stiffness matrix [K] 

5. Impose the boundary conditions and determine reduced stiffness matrix.

6. Determine element nodal load vector [q] (Restrained structure)

7. Determine equivalent load vector[f]

8. Apply equation of equilibrium [K]{∆}={f}, and determine unknown joint displacements.

9. Apply equation [K]{∆}+[q]={f} to determine reactions and moments

Problems-

Example 1-

Analysis the portal frame as shown in fig. using finite element methos take EI constant. Neglect axial deformation

C:\Users\Ssd\Desktop\UnIT 5 5.png

 

Fig.5

Step -1 total DoF =12

(three DOf at each node, two displacement & one rotation)

No. of elements 03 (AB,BC,CD)

Discretization

Element

Nodes

Displacement

Boundary conditions

1

AB

1,2,3,4,5,,6,

1=2=3=zero

2

BC

4,5,6,7,8,9

5=8=zero,4=7

3

DC

10,11,12,7,8,9

10=11=12=zero

 

Step 2- element stiffness matrix

Element stiffness matrix from AB (column member)

Imposing boundary conditions 5=8=0

Element stiffness matrix for DC

Imposing boundary conditions 10=12=0

Step -3 reduced stiffness matrix

Since horizontal sway at B& C are same (4=7) we can modify the above stiffness matrix as

 

Step -4 element nodal load vector

 

 

C:\Users\Ssd\Desktop\UnIT 5 6.png

                                   Fig.6

 

 

Reduced element nodal load vector

Step -5 equivalent load vector

{f}= {q}+ joint forces

Step -6 equation of equilibrium

Step -7 moment calculations

Member AB

=

Member BC

Member DC

 

 

Example – 2

Analyze the rigid frame by using finite element method. Take EI constant . neglect axial deformation

 

 

C:\Users\Ssd\Desktop\UNIT 5 7.png

 

                                          Fig.7

 

Step -1 total DOf =12

(three dof at each node , two displacements & one rotation)

Element

Nodes

Displacement

Boundary conditions

1

AB

1,2,3,4,5,,6,

1=2=3=zero

2

BC

4,5,6,7,8,9

5=8=zero,4=7

3

DC

10,11,12,7,8,9

10=11=12=zero

Step -2 element stiffness matrix for column AB

 

Element stiffness matrix of beam Bc

Element stiffnes matrix of beam DC.

  Imposing boundary condition

1=2=3=5=8=10=11=12=0

Step -3 reduced stiffness matrix

Since horizontal sway at B &C are same (4=7) , we can modify the above stiffness matrix as

Step -4 element nodal load vector

C:\Users\Ssd\Desktop\UnIT 5 8.png                               

                                                                   Fig.8

 

 

 

 

 

Reduced element nodal load vector

 

Step -5 equivalent load vector

Step -6 equation of equilibrium

Step -7 moment calculations

Member AB

Member BC

Member DC

 

 

 


Example 1

For the axisymmetric element shown in figure, determine the element stresses

Let E=210

C:\Users\Ssd\Desktop\UnIT 5 9.png

                                                             Fig.9

Poisson’s Ratio=0.25

The total displacements are

u1=0.05mm ;    W1=0.03mm

u2=0.02mm;  W2=0.02mm

u3=0mm;W3=0mm

 

To Find Element Stresses & Stiffness Matrix

 

 

Solution:-

 

We know that

&

 

=1250mm

Substitute all this values in {B}

                                         Put =E =

We know that stiffness matrix

 

Note ; matrix form [k]= (6*6)

To find stresses

Note – matrix form = 4*1

 

 

Example 2 :-

A long hollow cylinder of inside diameter 80mm & outside diameter 120mm is firmly filled in a hole of another rigid cylinder over it full length as shown in fig. the cylinder is then subjected to an internal pressure of 2 MPa. By using two elements on the 10mm length shown . find the displacement at the inner radius . take E=200 Gpa ,u=0.3

C:\Users\Ssd\Desktop\UNIT 5 10.png

Fig 10

Solution -

Stiffness matrix for element 1

For the both elements 

The stiffness matrices for element

 

 

References-

(1)  Finite Element Analysis -C.S. Krishnanmoorthy, Tata McGraw Hill Publishing Co. Ltd, New Delhi,

(2)  The Finite Element method -ZIENKIEWICZ.O.C.Tata McGraw Hill Pub. New Delhi, 2000

(3)  Finite Element Methods by C R Alaval , PHI

(4)  Finite Elements in Engineering:- Chandrupatta, et. AI. Prentice Hall of India Pvt. Ltd.,

 

 

 

 

 


Index
Notes
Highlighted
Underlined
:
Browse by Topics
:
Notes
Highlighted
Underlined