A cylindrical steel rod 0.50 m long and 1 cm in radius is subjected to a tensile force of 1 ｘ 10⁴ N. (a) What is the tensile stress? (b) What is a tensile strain? (c) By what amount does the rod stretch? Answer ((a) 0.31 ｘ 10⁸ N m⁻² (b) 1.6 ｘ 10⁻⁴ (c) 0.8 ｘ 10⁻⁴ m) 



Given Data:

Length of steel rod = L =  0.50 m 

Radius steel rod = r = 1 cm = 0.01 m

Force = F = 1ｘ10⁴ N

Young's modulus of steel = Y = 20 ｘ10¹⁰ N m⁻²


To Find:

(a). Tensile Stress = ?

(b). Tensile Strain = ?

(c). Change in Length = ΔL = ?



Solution:

(a). Tensile Stress = ?

Tensile stress is defined as:

Tensile Stress = `\frac {F}{A}`

where [ A = ㄫ r²] So

Tensile Stress = `\frac {F}{ㄫ r²}`  

by putting values


Tensile Stress = `\frac {1ｘ10⁴ N}{3.1416 ｘ(0.01 m)²}`

Tensile Stress = `\frac {1ｘ10⁴ N}{3.1416 ｘ0.0001 m²}`

Tensile Stress = `\frac {1ｘ10⁴ N}{3.1416 ｘ10⁻⁴ m²}`

Tensile Stress = 0.318ｘ10⁸ N m⁻² -----------Ans.(1)



(b). Tensile Strain = ?

According to Young's Modulus:

Y = Tensile Stress / Tensile Strain

or

Tensile Strain = Tensile Stress / Y


Tensile Strain =  `frac {0.318ｘ10⁸ N m⁻²}{20 ｘ10¹⁰ N m⁻²}`

Tensile Strain =  0.0159ｘ10⁻²

or

Tensile Strain = 1.59ｘ10⁻⁴ ---------------Ans. (2)



(c). Change in Length = ΔL = ?

Tensile strain is defined as:

Tensile strain = `\frac {ΔL}{L}`

or

ΔL = Tensile strain ｘ L

ΔL = 1.59ｘ10⁻⁴ ｘ 0.5 m

ΔL = 0.795ｘ10⁻⁴  m

or

ΔL = 0.8ｘ10⁻⁴  m ----------------Ans. (3)



Similar Questions:

A 1.25 cm diameter cylinder is subjected to a load of 2500 kg. Calculate the stress on the bar in mega Pascal. (Ans: 200 MPa) 



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