Question

A cantilever of steel fixed horizontally is subjected to a load of 225gm at its free end.The geometric moment of inertia of the cantilever is 4.5×10^-11m^4.If the length of cantilever and yougs modulus of steel are 1m and 200×10^9pa respectively.calculate the depression at the loaded end.​

Answer 1

Length of cantilever, L=50cm=0.5m

Deflection at the end w(L)=15mm

Find: Deflection w(x) at 30cm from rigid end.

Depression at a point in a cantilever beam with load at one end is given by,w(x)=

6EI

Px

2

(3L−x)

We have,

w(L)

w(x)

=

3EI

PL

3

6EI

Px

2

(3L−x)

w(0.5m)

w(0.3m)

=

3EI

P(0.5)

3

6EI

P(0.3)

2

(3(0.5)−0.3)

This gives, w(0.3m)=6.48mm

$Length of cantilever, L=50cm=0.5m</h3><h3>Deflection at the end w(L)=15mm</h3><h3>Find: Deflection w(x) at 30cm from rigid end.</h3><h3>Depression at a point in a cantilever beam with load at one end is given by,w(x)=6EIPx2(3L−x)</h3><h3>We have,</h3><h3>w(L)w(x)=3EIPL36EIPx2(3L−x)</h3><h3></h3><h3>w(0.5m)w(0.3m)=3EIP(0.5)36EIP(0.3)2(3(0.5)−0.3)</h3><h3></h3><h3>This gives, w(0.3m)=6.48mm</h3><h3></h3><h3>$

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