4. A load of 2.5kN is lo be lifted by a steel wire. What should be the minimum diameter
of the wire if the stress in the wire is not to exceed 100 N/mm 2.Also determine the
work done if the length of the steel wire is 5m. Take E = 2 x 105 N/mm².
Saluti
Answers
Step-by-step explanation:
4. A load of 2.5kN is lo be lifted by a steel wire. What should be the minimum diameter
of the wire if the stress in the wire is not to exceed 100 N/mm 2.Also determine the
work done if the length of the steel wire is 5m. Take E = 2 x 105 N/mm².
Saluti
Solution:
Given data:
Load, P = 2.5KN = 2500N
Stress induced in the wire, S(sigma) = 100 N/mm^2
Length of wire, L = 5m.
Young's modulus of elasticity, E = 2 × 10^5 N/mm^2
To find:
- minimum diameter of the wire, d = ?
- work done, w = ?
Calculation:
In first step
We know that
Stress = Load/ Area
100 = 2500 / πr^2
r^2. = 2500 / 100 × π
r^2 = 7.96
r. = √7.96
r. = 2.90 mm
If diameter is equal to twice of radius, then
d = 2r
= 2 × 2.90
= 5.8 mm
Hence the diameter of the wire is 5.8 mm
In second step
work done = load × distance
= 2500 × 5
= 12500 N m Ans.