Page 86 - CITS - WCS - Mechanical

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Load required to compress the spring by 10 mm
= spring stiffness x deflection
= 20 (N/mm) x 10(mm)
Load required = 200 N

3 Helical spring is loaded with a force of 400 Newton and it is compressed by 18 mm. What would be
the load required to compress it to 6 mm?
Given force = 400 Newton
Deflection = 18 mm
Spring Stiffness = Force / Compressed length

= 400 /18
= 22.22 Newton / mm
Load required = 22.22 × 6
= 133.32 N

4 Calculate the tensile strain when a force of 3.2 KN is applied to a bar of original length 2.8 m extends
the bar by 0.5 mm.
Force F = 3.2 KN
Original length L = 280 cm
Increased length(D ) = 0.5 mm = 0.05 cm

Tensile strain = ?
Dl
Strain =
L
0.05
=
280
= 0.0001786
Example
1 A steel rod of 10 mm diameter and 175 mm long is subjected to a tensile load of 15 kN. If E = 2 x10
5
N/mm , calculate the change in length.
2
Tensile load = 15 kN = 15000 N
22
Area of cross section = (πr ) = × 5 × 5 mm = 78.57
2
2
7
15000 N
\ Stress = = 191 N/mm 2
0.785× 100 mm 2

Stress
Young’s modulus E =
Strain
191 N/mm 2
E = 2 × 10 N/mm =
2
5
Strain
\ Strain = 191
2× 10 5
175 × 191
Change in length = mm
2 × 10 5
= 0.167 mm.

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CITS : WCS - Mechanical - Exercise 7 CITS : WCS - Mechanical - Exercise 7