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postnew [5]
4 years ago
7

A cylindrical specimen of some metal alloy having an elastic modulus of 108 GPa and an original cross-sectional diameter of 3.7

mm will experience only elastic deformation when a tensile load of 1890 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.45 mm.
Physics
1 answer:
kaheart [24]4 years ago
5 0

Answer:

L= 276.4 mm

Explanation:

Given that

E= 180 GPa

d= 3.7 mm

F= 1890 N

ΔL= 0.45 mm

We know that ,elongation due to load F in a cylindrical bar is given as follows

\Delta L =\dfrac{FL}{AE}

L=\dfrac{\Delta L\times AE}{F}

Now by putting the values in the above equation we get

L=\dfrac{0.45\times 10^{-3}\times \dfrac{\pi}{4}\times (3.7\times 10^{-3})^2\times 108\times 10^9}{1890}\ m

L=0.2764 m

L= 276.4 mm

Therefore the length of the specimen will be 276.4 mm

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