SOLVED NUMERICAL PROBLEMS-STRENGTH OF MATERIAL

SOLVED NUMERICAL PROBLEMS

STRENGTH OF MATERIALS

Q. Find the ratio of second moment of area about the centroid axis and about the base of a triangle. Given   base=200 mm and altitude = 600 mm.

A.

Second moment of area is moment of inertia ‘I’

(i) About the centroid axis,    I centroid axis  = bd3/36, say= I1

(ii) About the base,  I base            = bd3/12, say = I2

I1/I2 = (bd3/36)/( bd3/36) = 1/3

NOTE: Dimensions given are superfluous.

Q. What is total elongation of a bar of uniform cross-section under its own weight and an external load equal to the weight of the bar.

A.

Elongation due to external load ‘W’ =δL1= WL/AE

Elongation due to the weight of the bare ‘W’ =δL2= WL/2AE

Total elongation =δL1 +δL2 =(3/2) WL/AE

The ratio δL2/ δL1 = ½

Q. A beam of square section with side ‘a’. Its diagonal is horizontal. Find the section modulus of the beam.

A.

There will be one triangle above and another below the horizontal diagonal. The moment of inertia of one triangle about the base is bd3/12= a4/12. Therefore the total moment of inertia will be 2 x a4/12.

Length of the diagonal =

Height of the triangle = [a2 – (a/ )2]0.5 =a/

Therefore section modulus , Z = I/ymax = (2 x a4/12) / a/  = a3/3

Q.  The deflection of free end of a cantilever of rectangular cross-section due to certain loading is 0.8 cm. Find the deflection of the free end if the depth doubles. Keep the width same.

A.

Deflection δ at free end = WL3/3EI=0.8

I = bd3/12 with depth as ‘d’ and width as ‘b’

I=b(2d)3/12 = 4 bd3/12  with depth as 2d and width as b

Deflection with depth as 2d will be =0.8/4=0.2 cm

Q. Find the ratio of torques of solid shaft of diameter ‘D’ and a hollow shaft with diameters ‘D’ and ‘d’. Given allowable shear stress ‘z’.

A.

Torsional moment is torque’ T ’.

Tsolid = (πD3/16) z

Thollow = π(D4—d4)/16 D) z

Ratio Tsolid / Thollow = [(πD3/16) z]/[ π(D4—d4)/16 D) z]= D4/(D4—d4)

Tsolid / Thollow = D4/(D4—d4)

Q.  Bending moment of M and a torque T act on a shaft. What is ratio of maximum bending stress to the maximum shear stress?

A.

Equivalent bending moment is

Meq = (1/2) [M+(M2 +T2)0.5]

Maximum bending stress σ = (Meq/I) (D/2)

Equivalent torque is

Teq = (M2 +T2)0.5

Maximum shear stress z = (Teq/ 2I) (D/2)

Therefore ratio σ/z = 2 Meq/Teq

Q. Write the relationships between uniformly distributed load ‘w’, shear force ‘V’ and bending moment ‘M’.

A.

(a) The relation between UDL ‘w’ and shear force ‘V’ is

dV/dx = — w

(b)The relation between shear force ‘V’ and bending moment ‘M’ is

dM/dx = Vx

Q. The Euler load for a column is 1000 kN and the crushing load is 1500 kN. Find the Rankine load.

A.

We know 1/PEu +1/Pc = 1/PR

1/1000 + 1/1500 = 1/PR

PR = 600 kN

Q. Explain the difference between Neutral layer and neutral axis.

A.

Neutral Layer: It is a plane (area) under zero stress and under zero strain. It passes through the neutral axis of the beam.

Neutral axis: It is a line where neutral fiber cuts the cross section of the beam. It   does not have any stress or no strain.