As per the given question,
For the section BC,
The diameter of the brass =50mm
Length =200mm
For the section AB and CD,
diameter =20mm
and length =100mm
force is applied at D, then elongation in the total length
Δl=0.15mm
Δl=2Δl1+Δl2
⇒0.15mm=2Δl1+Δl2
We know that,
Y=AΔlFl
Δl=AYFl
Now, substituting the values,
0.15mm=π252YF×200mm+2×π102YF×100mm
the diagram mentioning in the question is not available, so situation and values are not clear.
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