Stiffness Matrix of a Tapered Bar-Finite

Q. 5.23

Derive the stiffness matrix of a tapered bar with linearly varying area of cross section using a direct approach.

SOLUTION:-

Divide the continuum into two finite elements as shown

The field variable is the deflection u. The element equations can be expressed in matrix

form as

where [k] is called the stiffness or characteristic matrix, ff is the vector of nodal displacements, and P is the vector of nodal forces of the element. We shall derive the element stiffness matrix from the basic definition of the stiffness coefficient. For an element we can write

where [k] is called the stiffness or characteristic matrix, u is the vector of nodal displacements, and P is the vector of nodal forces of the element. We shall derive the element stiffness matrix from the basic definition of the stiffness coefficient.

docsity.com

The stiffness influence coefficient kij is defined as the force needed at node i (in the direction of ,ui) to produce a unit displacement at node j (uj = 1) while all other nodes are restrained. k11 = force at node 1 to produce a unit displacement at node 1 with node 2 restrained It means u1= 1 , u2=0 as shown in following figure. From equilibrium k21 = -k11

Similarly

k22 = force at node 2 to produce a unit displacement at node 2 with node 1 restrained It means u1=0 , u2=1 as shown in following figure. From equilibrium k12 = -k22

We can find k as

Force = stress × area = strain × young modulus × area

For unit displacement F = k

As from above discussion

docsity.com

Similarly

As from above discussion

So

As area of element is changing linearly so we can take average nodal area for an element

For element 1

For element 2

docsity.com