Stiffness Matrices 2-Finite Element Method-Assignment Solution, Exercises of Mathematical Methods for Numerical Analysis and Optimization

This assignment solution was submitted to Amar Sharma for Finite Element Method course at Aligarh Muslim University. It includes: Stiffness, Matrix, Planar, Frame, Element, Local, Coordinate, System, Young, Modulus

Typology: Exercises

2011/2012

Uploaded on 07/08/2012

ramu
ramu 🇮🇳

4.4

(57)

135 documents

1 / 3

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
ASSIGNMENT # 07
Title: Problem # 6.7
PROBLEM # 6.7
The stiffness matrix of a planar frame element in the local coordinate system is given by
E
Where E is the Young's modulus, A is the area of cross section, I is the moment of
inertia, and L is the length. Using this. generate the stiffness matrices of the three
elements shown in Figure 6.7 in the local coordinate system and indicate the respective
local degrees of freedom.
docsity.com
pf3

Partial preview of the text

Download Stiffness Matrices 2-Finite Element Method-Assignment Solution and more Exercises Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity!

ASSIGNMENT # 07

Title: Problem # 6.

PROBLEM # 6.

The stiffness matrix of a planar frame element in the local coordinate system is given by

E

Where E is the Young's modulus, A is the area of cross section, I is the moment of inertia, and L is the length. Using this. generate the stiffness matrices of the three elements shown in Figure 6.7 in the local coordinate system and indicate the respective local degrees of freedom.

SOLUTION:

As each node has 3 degree of freedoms, therefore each local stiffness matrix is of order 6×6. And as there are 3 elements thus the global stiffness matrix is of order 12×12. The stiffness matrix of a planar frame element in the local coordinate system is given , using this we can get the local stiffness matrices for respective elements. Each node has 3 DOFs as u 1 , v 1 , 1.

For Element 1:

A=6 in^2 L=36 in E=30×10^6 psi I= 2 in^4

For Element 2:

A=6 in^2 L=60 in E=30×10^6 psi I= 2 in^4