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.

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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

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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

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