Question

a)Explain the terms plastic deformation and elastic deformation. b)Plot a graph of stress-strain of a metal. State and explain important features of a graph. c) A copper wire 200cm long and 1.22mm in diameter is fixed horizontally below the two supports 200cm apart. Find the mass in grams of the load which , when suspended as the midpoint of the wire, produces a sang of 2cm at that point.(young modulus for copper in 1.2×10^11 N/m^3)​

Answer 1

a) Plastic deformation is a permanent deformation that occurs in a material when it is subjected to a stress beyond its yield point, causing a rearrangement of its internal structure. The material will not return to its original shape after the stress is removed.

Elastic deformation, on the other hand, is a temporary deformation that occurs in a material when it is subjected to a stress within its elastic limit, causing a change in its shape. The material will return to its original shape once the stress is removed.

b) The stress-strain graph of a metal typically shows a linear elastic region, a yield point, a plastic deformation region, and a fracture point. The important features of the graph include:

• Elastic region: The region where the material deforms elastically and returns to its original shape once the stress is removed. The slope of the line in this region is known as the modulus of elasticity or Young's modulus.
• Yield point: The point at which the material begins to deform plastically and permanently.
• Plastic deformation region: The region where the material undergoes plastic deformation, and the stress increases with strain.
• Fracture point: The point at which the material fails and fractures.

c) The formula to find the mass of the load is: m = (πr^2Lρg)/2

where:

r = radius of the wire = 1.22/2 x 10^-3 m

L = length of the wire = 200 cm = 2 m

ρ = density of copper = 8.96 x 10^3 kg/m^3

g = acceleration due to gravity = 9.81 m/s^2

First, we need to find the elongation of the wire when it is suspended by the load. The elongation is given by: ΔL = FL/AY

where:

F = force exerted by the load

A = cross-sectional area of the wire = πr^2

Y = Young's modulus of copper = 1.2 x 10^11 N/m^2

The sag of the wire is 2 cm, which means the elongation is 4 cm. Converting to meters, ΔL = 0.04 m.

Substituting the values into the formula for the mass of the load:

m = (πr^2Lρg)/2 = (π x (1.22/2 x 10^-3)^2 x 2 x 8.96 x 10^3 x 9.81 x 0.04)/2 = 10.9 g (rounded to one decimal place)

Therefore, the mass of the load that produces a sag of 2 cm is 10.9 grams