Mechanical Properties of Solids Test 1

The  law states that the strain in a solid is proportional to the applied stress within the elastic limit of that solid body i.e. stress = k x strain.

A body is said to be plastic when its coefficient of restitution or reformation is zero that means that whatever hits it loses its all kinetic energy once the body gets deformed does not reform.

A body is said to be perfectly elastic when its coefficient of restitution is 1 or we observe total  reformation in the body. That means that whatever hits it doesn't lose any of its kinetic energy or once the body gets totally reformed after a collision.

Stress is the force per unit area experienced by the body and strength is the ability to withstand the stress. When stress becomes greater than strength, accidents happen.

Beyond the elastic limit, if still some force is applied to the body, at first the body deforms completely and when it can’t deform more, it starts to break into pieces. Now when the plastic range is small, on less extent of applied external force the body would break. And we know such objects are called brittle.

Tensile stress (or tension) is the stress state leading to expansion; that is, the length of a material tends to increase in the tensile direction. The volume of the material stays constant. When equal and opposite forces are applied on a body, then the stress due to this force is called tensile stress.

Modulus of elasticity= stress/strain =(F/A)/(ΔL/L) So, for same stress Modulus of elasticity ∝( L/ΔL)and ΔL for rubber is more as compared to steel so Modulus of elasticity for rubber will be less as they are inversely proportional and also ΔL/L is less for steel

strain energy=1/2×stress × strain Work done by a force on a wire
W =2LAy(ΔL)²/2L
=1/2(yALΔ/L)ΔL
=1/2​(yΔL/L)(ΔL/L)(AL)
=1/2(Stress)(Strain)(Volume)
(Work)/(volume)=1/2(stress)(strain)

Since mud or putty have no gross tendency to regain their previous shape & they get permanently deformed, they are close to ideal plastics.

To answer this question you should know the definitions of tensile, compressive and shear stress.
Tensile stress causes change (increases the length of cylinder) in the length of the object, compressive strength changes the volume of the object (it can be applied from all sides of the object), shear stress is applied parallel to the surface of an object (in case of the cylinder shear stress will be parallel to circular surface) we can apply all the three stresses in case of cylinder hence option D is correct.

Using Hooke ‘s Law we get

Stress directly proportional to stress = Load/Area=F/pie*r*r

And rp:rq=2:1

When both the wires are under the same stress,strain produced will be the same.
 

When both the wires are under the same stress,strain produced will be the same.

2.when both the wires are loaded by same weight then

Strain p/strain q=(rq)²/(rp)²=¼

At the yield point, stress is not proportional to strain i.e: hooke's law is not obeyed. Hence the elastic behaviour ends and plastic behaviour begins.

Stress is referred to as the applied strain over a unit area, and we know strain is the force acting along the line, which means strain and force have the same dimensions. Thus stress and pressure have the same dimensions too.

In hooke's law, the constant of proportionality defines  the modulus of elasticity which is also defined as the ratio of internal reforming force / external deforming force.

Springs deform permanently because of elastic fatigueness. The elasticity of the material of spring is lost and it deforms permanently. 

The area under the stress-strain curve represents the mechanical energy per unit volume consumed by the material. This is true in the elastic range of the graph where the energy is reversibly sorted within the material. Area under the stress strain curve depicts the energy absorbed by the material prior to failure.

The last and final point of the strain stress curve is a point where the objects get broken into pieces due to the external force applied and hence it is called fracture.

Elastomers are popular in vascular engineering applications, as they offer the ability to design implants that match the compliance of native tissue.
Substances like tissue of aorta can be stretched to cause large strain.

When a soild is compressed or elongated, it's P.E increases. Potential energy increases because work has to be done against force of repulsion during compression and against force of attraction during elongation.