# Stress, strain, Young’s modulus – problems and solutions

1. A nylon string has a diameter of 2 mm, pulled by a force of 100 N. Determine the stress!

Known :

Force (F) = 100 N

Diameter (d) = 2 mm = 0.002 m

Radius (r) = 1 mm = 0.001 m

Wanted : The stress

Solution :

Area :

A = π r2

A = (3.14)(0.001 m)2 = 0.00000314 m2

A = 3.14 x 10-6 m2

The stress : Read :  Young’s double-slit experiment - problems and solutions

2. A cord has original length of 100 cm is pulled by a force. The change in length of the cord is 2 mm. Determine the strain!

Known :

Original length (l0) = 100 cm = 1 m

The change in length (Δl) = 2 mm = 0.002 m

Wanted : The strain

Solution :

The strain : Read :  Applications of Bernoulli's principle

3. A string 4 mm in diameter has original length 2 m. The string is pulled by a force of 200 N. If the final length of the spring is 2.02 m, determine : (a) stress (b) strain (c) Young’s modulus

Known :

Diameter (d) = 4 mm = 0.004 m

Radius (r) = 2 mm = 0.002 m

Area (A) = π r2 = (3.14)(0.002 m)2

Area (A) = 0.00001256 m2 = 12.56 x 10-6 m2

Force (F) = 200 N

Original length of spring (l0) = 2 m

The change in length (Δl) = 2.02 – 2 = 0.02 m

Wanted : (a) The stress (b) The strain c) Young’s modulus

Solution :

(a) The stress (b) The Strain (c) Young’s modulus Read :  Motion on the rough inclined plane with the friction force - application of Newton's law of motion problems and solutions

4. A string has a diameter of 1 cm and the original length of 2 m. The string is pulled by a force of 200 N. Determine the change in length of the string! Young’s modulus of the string = 5 x 109 N/m2

Known :

Young’s modulus (E) = 5 x 109 N/m2

Original length (l0) = 2 m

Force (F) = 200 N

Diameter (d) = 1 cm = 0.01 m

Radius (r) = 0.5 cm = 0.005 m = 5 x 10-3 m

Area (A) = π r2 = (3.14)(5 x 10-3 m)2 = (3.14)(25 x 10-6 m2)

Area (A) = 78.5 x 10-6 m2 = 7.85 x 10-5 m2

Wanted : The change in length (Δl)

Solution :

Young’s modulus formula : The change in length : Read :  Diffraction grating – problems and solutions

5. A concrete has a height of 5 meters and has unit area of 3 m3 supports a mass of 30,000 kg. Determine (a) The stress (b) The strain (c) The change in height! Acceleration due to gravity (g) = 10 m/s2. Young’s modulus of concrete = 20 x 109 N/m2

Known :

Young’s modulus of concrete = 20 x 109 N/m2

Initial height (l0) = 5 meters

Unit area (A) = 3 m2

Weight (w) = m g = (30,000)(10) = 300,000 N

Wanted : (a) The stress (b) The strain (c) The change in height!

Solution :

(a) The stress (b) The Strain (c) The change in height Read :  Resolve initial velocity into horizontal and vertical components of projectile motion