Mechanical Properties of Solids 2

Question 1

what is hooke’s law

Answer 1

stress is directly propoertional strain, within the limit of elasticity.

The limit of elasticity is the limit up to which stress/ strain is a constant. It is only up to a stress.
stress= k x strain
where k is the proportionality constant and is
known as modulus of elasticity

Question 2

what s=is
(i) elastic limit
(ii) permenant set
(iii) ultimate tensile strength

Answer 2

(i) Point B on the curve. It is the maximum stress on whose removal, the body regains its original shape exactly. It is called elastic limit/ yield point. Corresponding stress is called yield stress.

(ii) Permenant set is the plastic deformation created in the body when a stress greater than elasic limit is applied.

(iii) Ultimate tensile strength is the stress corresponding to D on curve. Beyond this point, additional strain is produced
even by a reduced applied force and fracture
occurs at point E.

Question 3

what are ductile and brittle materials

Answer 3

(i) materials for which a large amt of plastic deformation takes place between ultimate tensile strength and fracture point is called ductile.

(ii) Materials for which a small amout of plastic defromation is possible between ultmitae tensile strength and fracture point is calle dbrittle
glass

Question 4

explain regions of stress strain curve

Answer 4

• Frm O- A, hooke’s law is followed by body. when stress is removed it regains original shape and size exactly

-from A-B, the body still regains original shape but hook’s law is not followed exactly. body still shows elastic behaviour. B is called elastic limit

-From b-D, the strain increases rapidly for even a small change in stress. At an point C on the body, there is a permanent plastic deformation called permanenet set. D is ultimate tensile strenght

• ## E is called th efracture point when the body breaks.

Question 5

what are elastomers

Answer 5

although elastic region is very large, the material does not obey Hooke’s law over most of the
region. Secondly, there is no well defined plastic
region. Substances like tissue of aorta, rubber
etc. which can be stretched to cause large strains
are called elastomers

Question 6

what is modulus

Answer 6

the is the ratio of stress to strain

(i) Young’s modulus:
Ratio of tensile stress to longitudinal strain

(i) Bulk Modulus
Ratio of bulk stress to volume strain. Reciprocal of bulk modulus is compressibility( it is the fractional change in volume per unit increase in pressure)

Question 7

what is poisson’s ratio

Answer 7

the ratio of laterial strain to longitudinal strain within the limit of elasticity,
Theoretically poissons ratio lies between -1 and 0.5 but it is never neative and practically it is between 0.2 and 0.4. for most materials

stel= 0.28- 0.30
aluminium - 0.33

Question 8

why are pillars I shaped

Answer 8

When we make a horizontal beam and apply the load in the centre, the beam starts bending. This is called sagging.
The amount by which is bents i given by delta= Wl3/4bd3Y

This bending can be reduced the increasing the depth of the beam. However, a deep beam has a tendency of bending under heavy load. This is called buckling.

Both this phenomena can be prevented by using I shaped beams/girders. The horizontal parts are called flanges while verical parts are called web. The flanges provide large loadbearing surface and enough depth to prevent
bending. This shape reduces the weight of the
beam without sacrificing the strength and hence
reduces the cost.

When used horizontally, the stress is mostly concentrated on the ends and the centre experiences almost no stress.

I shaped girder will not collapse under its own weight

Question 9

why is the maximum height of a mountain 10km

Answer 9

At the bottom of a mountain of height h, the
force per unit area due to the weight of the
mountain is hρg where ρ is the density of the
material of the mountain and g is the
acceleration due to gravity. The material at the
bottom experiences this force in the vertical
direction, and the sides of the mountain are free.
Therefore this is not a case of pressure or bulk
compression. There is a shear component,
approximately hρg itself. Now the elastic limit
for a typical rock is 30 × 107 N m-2. Equating
this to hρg, with ρ = 3 × 103 kg m-3 gives
hρg = 30 × 107 N m-2 . Or
h = 30 × 107 N m-2/(3 × 103 kg m-3 × 10 m s-2)
= 10 km