Question

The following data refer to two close-coiled helical steel springs.

 

SPRING

FREE HEIGHT

d

D

N

1

55 mm

4,75 mm

30 mm

8

2

48 mm

5,3 mm

44 mm

6

 

 

One spring is placed inside the other, not touching each other, and compressed between a pair of parallel plates until the distance between the plates is 40 mm. Make a neat detailed sketch of the spring and plate assembly. If G = 83 GPa for both springs, calculate the following:

 

5.1 The force applied to the spring assembly. ​​​​​/4/

 

5.2 The shear stress induced in each spring.​​​​​​/3/

 

5.3 The total strain energy stored in the spring assembly.​​​​/3/

 

5.4 The stiffness of each spring.​​​​​​​/3/

 

5.5​The equivalent stiffness of a single spring that can be substituted in place of these springs.​​​​

Answer 1

$G = 83\ GPa = 83×10^9\ Pa\\ G = 83× 10^3\ N/mm²$

(Spring index)

$C_1$ $=D/d = 30/4.75=6.3$

$C_2 =D/d = 44/5.3 = 8.3$

(Pitch)

$p_1=$ free length/N-1 $=55/8-1 =$

$55/7=7.9mm$

$p_2= 48/6-1 = 48/5 = 9.6mm$

$τ = F×D/2×\pi d² = 83×10³×30/2×π(4.75)² =$

$8.8×10⁴\ Nm$