Illustrative examples – II

ILLUSTRATIVE EXAMPLES – II

Problem 8: The elongations of a mild steel rod of diameter 12 mm over a gauge length of 300. mm is 0.190 mm when an axial load of 15 kN is applied to it. Find the Young's Modulus of mild steel.

Given Data

Load = P = 15 × 10³

Diameter of rod = 12 mm

Length of rod = l = 300 mm

Elongation = dl = 0.190 mm

Solution

 

Problem 9: A tension test was conducted on a steel rod at Strength of Materials Laboratory and the following observations were made:

Initial Diameter = 14.0 mm

Gauge Length = 100 mm

Yield Point Load = 41.5 kN

Ultimate Load = 69.2 kN

Load at Rupture = 61.6 kN

Diameter at Neck = 11.2 mm

Elongation at Break = 38 mm

Elongation at Load 25.1 kN (P) = 0.08 mm= dl    (AU)

Calculate:

(i) Young's Modulus or Modulus of Elasticity, (ii) Yield Strength, (iii) Ultimate Strength, (iv) Nominal Stress at Rupture, (v) Actual Stress at Rupture, (vi) Percentage Elongation and (vii) Percentage Reduction in cross-sectional Area.

Solution

(i) Young's Modulus or Modulus of Elasticity (E)

 

Problem 10: A tension test was conducted on a 10 mm mild steel rod and the following readings were taken:

Gauge Length = 200 mm = 1

Load at yield point = 21 kN

Maximum load applied = 30 kN

Load at breaking = 26 kN

Diameter at the neck = 8 mm  = d

Distance between the gauge marks after breaking (1 + dl) = 212 mm

Extension over the gauge length for a load of 12.5 kN (P) = 0.15 mm = (dl)

Determine:

(i) Young's Modulus of the rod material, (ii) Yield Stress, (iii) Ultimate Stress, (iv) Nominal Stress at Breaking, (v) Actual Stress at Breaking (vi) Percentage Elongation and (vii) Percentage Reduction in Cross-sectional Area.

Answers

(i) 2.12 × 10⁵ N/mm², (ii) 267.29 N/mm² , (iii) 381.84 N/mm², (iv) 330.91 N/mm² (v) 517.02 N/mm², (vi) 6 %, (vii) 36 %

 

Problem 11: A circular bar of 25 mm diameter and 200 mm long is subjected to a tensile force of 60 kN and the elongation is 0.2 mm. Calculate the modulus of elasticity.

Given Data

Diameter of the bar = d = 25 mm

Length of the bar = l = 200 mm

Tensile load = P = 60 kN = 60,000 N

Elongation = dl = 0.2 mm

To Find

Modulus of Elasticity or Young's Modulus (E)

Solution

Within the elastic limit,  E = Stress / Strain

Area of cross-section, = A π × 25² / 4 = 491.07 m²

Stress = P / A = 60,000 / 491.07 = 122.18 N/mm²

Strain = dl / l = 0.02 / 200 = 0.001

Modulus of Elasticity = E = 122.18 / 0.001 = 122.18kN / mm2 = 1.22 × 10⁵ N/ mm²

 

Problem 12: A square bar of 75 mm side is subjected to an axial pull of 240 kN. The extension of the bar is 0.18 mm over a gauge length of 200 mm. The change in lateral dimension is 0.010 mm. Compute the Young's modules, Poisson's ratio, Rigidity modules and Bulk modules.

Solution

(i) Young's Modulus (E)

 

Problem 13: A hollow steel column has an internal diameter 0.6 times as that of the external. The material attains an ultimate stress of 420 N/mm2. If the working load is 1250 kN and the factor of safety allowed is 4, determine the outer (external) and internal diameters.

Given Data

Ultimate stress = 420 N/mm²

Working load = 1250 kN

Factor of safety = 4

To Find

(1) Outer Diameter (D_o) and (ii) Internal Diameter (D_i)

Solution

 

Problem 14: The modulus of elasticity and modulus of rigidity of a material are 1 x 105 MPa and 0.4 × 10⁵ MPa respectively. Calculate the change in volume of a 100 mm size cube of that material when it is subjected to a uniform pressure of 50 MPa on all its faces. (AU)

Given Data

Modulus of Elasticity = E = 1× 10⁵ MPa

Modulus of Rigidity = G = 0.4 × 10⁵ MPa

Direct Stress = f = 50 MPa

Side of Cube = a = 100mm

To Find

Change in Volume (dv)

Solution

 

Problem 15: A mild steel flat 150 mm wide and 20 mm thick of length 5000 mm carries a tensile load of 300 kN. If E = 2 × 10⁵ N/mm² and 1/m = 0.25. Calculate the change in length and width of the bar.

Given Data

Width of the mild steel = w = 150 mm

Thickness of mild steel flat = t = 20 mm

Length of the mild steel flat = l = 5000 mm

Tensile Load = P = 300 kN = 300 × 10³ N

Young’s Modulus = E = 2 × 10⁵ N/mm²

Poisson’s Ratio = l/m = 0.25

To Find

(i) Change in Length (dl) and (ii) Change in Width (dw)

Solution

(i) Change in Length (dl)

 

Problem 16: A round rod of 20 mm diameter carries an axial tensile force of 60 kN. Find the axial strain and lateral strain of the rod. E = 200 GPa. Poisson's Ratio = 0.25.

Answers

Axial Stress = 191 MPa, Axial Strain = 9.55 × 10^-4, Lateral Strain = 2.39 × 10^-4