Sample Problems on Difference Between Stress and Strain

Problem 1: A 2 mm diameter nylon thread is tugged by a force of 100 N. Determine the level of stress.

Solution:

Given,

Force= 100N

diameter (d)= 2mm= 0.002 m

radius (r)= diameter(d)/2

              = 0.002/2

              = 0.001 m

To Find: Stress

Area (A)= πr²

                 = (3.14)(0.001)²

Area (A)= 3.14 ×10^-6m²

Stress(σ)= F/A

             = 100 N/3.14×10^-6m²

                 = 31.5×10⁶ N/m²

So, the stress is 31.5×10⁶ N/m²

Problem 2: A force pulls a cord with an original length of 100 cm. The cord length has changed by 2 mm. Identify the strain

Solution:

Given,

Original length(L)= 100 cm= 1m

Change in length(δl)= 2 mm= 0.002 m

To Find: Strain

Strain(ϵ)= Change in length(δl)/Original length(L)

             = 0.002 m/1 m

             = 0.002

So, the stain is 0.002

Problem 3: A concrete slab with a 3 m³ unit area can sustain a mass of 30000 kg. Determine stress. Acceleration due to gravity(g) is 10 m/s².

Solution:

Given,

Unit Area(A) = 3 m³.

Weight = 30000 kg

Acceleration due to gravity(g) = 10m/s²

Force = mg

         = (30000)(10)

         = 300000 N

To Find: Stress

Stress(σ) = F/A

          = 300000 N/ 3 m³

          = 1×10⁵ N/m²

So, the stress is 1×10⁵ N/m²

Problem 4: Original length is 100 cm and the change in length is 5 mm. Determine the strain.

Solution:

Given,

Original length(L) = 100 cm = 1 m

Change in length( δl) = 5 mm = 0.005 m

To Find: Strain

Strain(ϵ) = Change in length(δl)/ Original length(L)

          = 0.005 m/1 m

          = 0.005

So, the strain is 0.005

Problem 5: The initial length of a 4 mm diameter string was 2 m. A force of 200 N is applied to the string. If the spring’s final length is 2.02 m. Determine stress and strain.

Solution:

Given,

Force(F) = 200N

Original length(L) = 2 m

Change in length(δl) = 2.02 m – 2m = 0.02 m

Diameter(d) = 4 mm = 0.004 m 

Radius(r) = diameter/ 2

           = 0.004/2

           = 0.002 m

Area(A) = πr²

             = (3.14)(0.002)²

             = 12.56×10^-6 m²

 To Find: Stress and Strain

Stress(σ) = F/A

              = 200 N/12.56×10^-6 m²

                  = 15.92×10⁶ N/m²

Strain(ϵ) = Change in length(δl)/Original Length(L)

              = 0.02 m/2m

              = 0.01

So, the stress is 15.92×10⁶ N/m² and the strain is 0.01.

Stress and Strain is an important concept in the field of Material Science and Metallurgy to get into insights about the strength of the material. When a material is put under pressure or has a load applied to it, it develops stress is developed. When a solid is put under stress, it has the ability to deform. This deformation is called Strain. The stress is the pressure per unit area of the material, and the resulting strain is the deformation that occurs as a result of this stress. Strain and stress are strongly intertwined because strain occurs solely as a result of stress.