Que: Two tapering bars of the same material are subjected to tensile load P. The larger diameter of each of the bar id D. The diameter of the bar A at its small end is D/2 and bar B is D/3. If both the bar are the same length, then what is the ratio of elongation of bar A to bar B. (Engg Services-2006)

(a) 3 : 2
(b) 2 : 3
(c) 4 : 9
(d) 1 : 3

Ans: (b)
Explanation: For a tapering rod,
ΔL = 4WL / π d1 d2 E

ΔL for A bar = 4WL / (π D D/2 E) = 8WL / (π D² E)

ΔL for B bar = 4WL / (π D D/3 E) = 12WL / (π D² E)

So, Ratio of ΔL for A bar to ΔL for B bar = 8 / 12 =2 / 3

Que: Two identical circular rod of the same diameter and same length are subjected to same magnitude of axial tensile force. One of the rod is made out of mild steel having the modulus of elasticity of 206 Gpa. The other rod is made out of cast iron having modulus of elasticity of 100 Gpa. Both the material are homogeneous and isotropic and the axial force causes the same amount of uniform stress in them. if the stresses developed are within the proportional limit of respective materials, then which of the following statement are correct (Gate-2007)

(a) Both rods elongate by the same amount
(b) Mild steel rod elongates more than Cast iron rod
(c) Cast iron rod elongates more than Mild steel
(d) As the stresses are equal, strain are also equal in both the rods

Ans: (c)
Explanation: Stress are equal in both materials
Length of the both rods are same
From equation of Modulus of elasticity = stress / strain  = (force * length) / (area * change in length)
here force, length and area remain same so modulus of elasticity are directly proportional to change in length
So
For Mild steel  206 ≈ 1 /  change in length
change in length for mild steel ≈ 1 / 206 ≈ 0.0048

For Cast iron  100 ≈ 1 /  change in length
change in length for Cast iron ≈ 1 / 100 ≈ 0.01
so (elongation) change in length are higher in Cast iron

Que: The total area under the stress-strain curve of a mild steel specimen tested upto failure under tension is a measure of its (Gate-2002)

(a) Breaking Strength
(b) Toughness
(c) Hardness
(d) Stiffness

Ans: (b)
Explanation: Breaking strength is the maximum stress that a material can withstand while being stretched or pulled before failing or breaking so its not related with strain only stress are considered,
                    Hardness is the resistance of a material to localized deformation so hardness is propositional to accepted load before any deformation at here strain = change in length / original length are not consider.
                  Toughness is the ability of a material to absorb energy and plastically deform before fracturing and absorb energy per unit volume calculated by area under the stress-strain curve.

Engineering Mechanics example (Gate-2005)

Que: A time variation of the position of a particle in rectilinear motion is given by  x = 2t³ + t² + 2t  If v is the velocity and a is the acceleration of particle in constant units, the motion started with (Gate-2005)

(a) v = 0, a = 0

(b) v = 0, a = 2

(c) v = 2, a = 0

(d) v = 2, a = 2

Ans: (d)

Explanation: 

x = 2t³ + t² + 2t  So,

Velocity v = dx/dt = 6t² + 2t +2

Acceleration  a = 12t + 2

At the start of motion  t = 0, Then

v =  0 + 0 + 2 = 2

a = 0 + 2 = 2

Que: a car moving with uniform acceleration covers 450 m in a 5 second interval, and covers 700 m in the next 5 second interval. the acceleration of the car is (Gate-98)

(a) 7 m/s²
(b) 10 m/s²
(c) 25 m/s²
(d) 50 m/s<sup>2

Ans: (b)
Explanation: Here 
u = initial velocity
v = final velocity
a = acceleration
t = time
s = distance

using the relation</sup> s = ut + 1/2 at² and v=u + at, for the first 5 second interval,we have

450 = 5u + 25/2 a   ___(i)

v = u + 5a

for the next 5 second interval (v=u + 5a) will be the initial velocity, then

700 = (u + 5a) 5 + 25/2 a  

700 = 5u + 25a + 25/2 a   ___(ii)

From equations (i) and (ii), we get

700-450 = (5u +25a + 25/2 a) – (5u + 25/2 a)

250 = 25a

a = 250/25

a =10 m/s²

Que: Which one of the following is correct in respect of Poisson's ratio (µ) limits for an isotropic solid? (Engg Services-2004)

(a) - ∞ ≤ µ ≤ ∞

(b) ¼ ≤ µ ≤ 1/3

(c) -1 ≤ µ ≤ ½

(d) - ½ ≤ µ ≤ 1/2

Ans: (b)

Explanation: Poisson's ratio always lies between 0.25 and 0.33 for most of the engineering materials.

Que: A solid uniform bar of diameter d and length l is hanging vertically from its free end. The elongation of the bar due to self weight is (Engg Services-2005)

(a) Proportional to l and inversely Proportional to to d²
(b) Proportional to l² and inversely proportional to d²
(c) Proportional to l but independent of d
(d) Proportional to l² but independent of d

Ans: (a)
Explanation: Extension due to self weight (W) is
Here E = Modulus of elasticity
d = Diameter of rod
L = Length of rod
ΔL = Elongation

From equation of modulus of elasticity of self weighted rod = E = (W x L) / 2(π/4 d² x ΔL)
ΔL = (W x L) / 2(π/4 d² x E)
From above equation concluded that Elongation (ΔL) are directly Proportional to l and inversely Proportional to to d²