Illustrative examples - I

ILLUSTRATIVE EXAMPLES - I

Problem 1: A steel rod of 16 mm diameter and 300 mm long is subjected to a tensile load of 30 kN. Determine the intensity of stress and strain if the elongation of the rod due to the load is 0.21 mm.

Given Data

Diameter of the rod  = d = 16 mm

Length of the rod = 1 = 300 mm

Tensile Load = p = 30 kN = 30 × 10³ N

Elongation of the rod = dl = 0.21 mm

To Find

(i) Tensile Stress and (ii) Tensile Strain

Solution

 

Problem 2: A steel wire of length 10 m and diameter 5 mm is used to hang a load at its bottom. The stress and strain in the wire are found to be 140 N/mm2 and 0.0007 respectively. Determine the load it carries and the elongation of wire.

Given Data

Diameter of the wire = d = 5 mm

Length of the wire = l = 10m = 10,000 mm

Tensile Stress = f1 = 140 N/mm² (Since the load is hung by the wire.)

Tensile Strain = 0.0007

To Find

(i) Tensile Load P and (ii) Elongation of the wire dl

 

Solution

 

Problem 3: A rectangular wooden column of length 3 meters and size 300 mm x 200 mm carries an axial load of 300 kN. The column is found to be shortened by 1.5 mm under the load. Find the stress and strain in the column and state their nature. (AU)

Given Data

Length of the column = 1 = 3 m 3000 mm

Area of the Column = A = 300 mm × 200 mm = 60,000 mm²

Axial Load P = 300 kN = 300 × 1000 N

Shortening of Column = dl = 1.5 mm

To Find

(i) Nature of load, (ii) Stress and (iii) Strain

Solution

(i) Nature of load: Since the load causes shortening of the member, the stress and strain produced in the column are compressive in nature.

(ii) Compressive Stress = fc = Compressive Load / Area of Cross-section = P / A

300 × 1000 / 60,000 N/mm² = 5 N/mm²

(iii) Compressive Strain = ec = Decrease in length / Original length = dl / l

= 1.5 / 3000 = 0.0005

 

Problem 4: An axial tensile load of 10 kN is applied on a steel rod. Find the diameter of the rod if the stress is not to exceed 10 MN/m².

Answer: Diameter of the rod = d = 35.68 mm [1 Mega Newton = 10⁶ N]

 

Problem 5: Three rivets of diameter 30 mm are provided to carry a total shear force of 210 kN. Determine the stress developed in the rivets.

Given Data

Shear carried by 3 rivets = 210 KN

Shear carried by 1 rivet (P) = (210/3) = 70 kN = 70 × 10³ N

Diameter of rivet (d) = 30 mm

Solution

Area of cross-section of one rivet = A = π × 30² / 4 = 706.9 mm²

Shear Stress = q = P / A = 70 × 30³ / 706.9 = 99.02 N/mm²

 

Problem 6: Find the force required to punch a 16 mm diameter hole in a plate of 20 mm thickness, if the shear strength of the material of the plate is 340 MN/m². (AU)

Solution

 

Problem 7: A truss member is connected to a tie bar by means of four bolts. The allowable shear stress in the bolts is 110 N/mm². Compute the minimum diameter of bolts, if the maximum load in the bar is 100 kN.

Solution

Shear Stress = q = P / A =  110 N/mm²

Total Area of cross-section of 4 bolts = A = P / q

= 100 × 10³ = / 110 = 909.09 mm²

Area of cross-section of 1 bolt = (909.09 / 4) = 227.27 mm²

Diameter of bolt = d = 17.3 mm

Hence, adopt 18 mm diameter bolts.