Board Exam answer key | Strength of Materials - Unit - 3 - 2/3 marks

Subject : Strength of Materials                  Scheme : M

UNIT – 3, GEOMETRICAL PROPERTIES OF SECTIONS AND THIN SHELLS.

Board Exam Answer key 2/3 marks 

1.     Define center of gravity

·   Center of gravity is defined as the point through which the total weight is acting.

·        The center of gravity is same for different positions of the body.

2.     Define moment of inertia

·   Moment of inertia of any section about an axis is the second moment of inertia, represents the internal resistance of the body to resist rotation about that axis

·      Moment of inertia is the product of area and square of the distance from the reference axis to centroid.

·        MOI = Area x (distance)^2

3.     Distinguish between thin and thick cylinders.

Cylindrical shell is having the thickness of the wall is less than 1/10^th to 1/15^th of its diameter is known as thin cylinder

Eg. Boilers, water pipes.

Cylindrical shell is having the thickness of the wall is more than 1/10^th to 1/15^th of its diameter is known as thick cylinder

Eg. Gas cylinders, Gun barrels

4.    Define centroid.

·     The point at which the point through the total area of lamina or figure is acting.

·      Centroid is always acting at the center point of lamina

5.    State parallel axis theorem.

    Parallel axis theorem states that The moment of inertia of a body about an axis parallel to the body passing through its centre is the sum of moment of inertia of a body about the axis passing through the middle and product of the mass of the body times the square of the distance between the 2 axes

I=Ic+Mh^2

6.    A boiler 2.8m diameter is subjected to a steam pressure of 0.68N/mm^2. Find the hoop stress and longitudinal stress, if the thickness of the boiler plate is 10mm.

Hoop stress, f1 = pd/2t =   (.68 x 2800) / (2 x 10) = 95.2 N/mm^2

Longitudinal stress, f2 = pd/4t = (.68x2800) / (4 x 10) = 47.6 N/mm^2

7.    State perpendicular axis theorem.

    Perpendicular axis theorem states that the moment of inertia of a planar body about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with the perpendicular axis and lying in the plane of the body.

Iz = Ix + Iy

8.    Derive moment of inertia for rectangular area.

The moment of inertia of a rectangle with respect to an axis passing through its centroid, is given by the following expression:

Ixx = bh^3 / 12

where b is the rectangle

           h is the height

9. Distinguish between centre of gravity and centroid.

Center of gravity

·   Center of gravity is defined as the point through which the total weight is acting.

·        The center of gravity is same for different positions of the body.

Centroid :

·        Centroid is defined as the point through which the total area is acting

·        Centroid is always acting at the center of lamina

 

10. List out the stresses, induced in thin cylindrical shells.

1.     Hoop stress

2.     Longitudinal stress

3.     Radial stress

11. What is centeroidal axis and axis of reference?

    Centeroidal axis

        A line passing through the centroid of that plane area is known as centroidal axis.

Axis of reference

            The center of gravity of a plane area or a body is always calculated with reference to some assumed axis is known as axis of reference

12. Define thin cylindrical shell.

    Cylindrical shell is having the thickness of the wall is less than 1/10^th to 1/15^th of its diameter is known as thin cylinder

Eg. Boilers, water pipes.

13.  A steel penstock of 1.5m diameter and 15mm thick is subjected to an internal pressure of 15bar. Calculate the hoop stress and longitudinal stress at the bottom of the penstock.

Circumferential or hoops stress = pd/2t = (1.5x1500) / (2x15) = 75N/mm^2

Longitudinal Stress = pd/4t = (1.5 x1500) / (4x15) = 37.5 N/mm^2.