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A close-coiled helical spring has a mean coil diameter 10 times the diameter of the wire. A load of 3,5 kN causes a deflection of 95 mm and maximum stress of 400 MPa. Take G = 80 GPa. Calculate: 5.1 The wire diameter, (3) 5.2 The mean coil diameter, (3) 5.3 The number of coils, (3) 5.4 The energy absorbed when subjected to the maximum load, and (3) 5.5 The stiffness of the spring. (3)

Question-AnswerCategory: Strength of MaterialsA close-coiled helical spring has a mean coil diameter 10 times the diameter of the wire. A load of 3,5 kN causes a deflection of 95 mm and maximum stress of 400 MPa. Take G = 80 GPa. Calculate: 5.1 The wire diameter, (3) 5.2 The mean coil diameter, (3) 5.3 The number of coils, (3) 5.4 The energy absorbed when subjected to the maximum load, and (3) 5.5 The stiffness of the spring. (3)
Ayush asked 2 months ago

A close-coiled helical spring has a mean coil diameter 10 times the diameter of the wire. A load of
3,5 kN causes a deflection of 95 mm and maximum stress of 400 MPa.
Take G = 80 GPa.
Calculate:
5.1 The wire diameter, (3)
5.2 The mean coil diameter, (3)
5.3 The number of coils, (3)
5.4 The energy absorbed when subjected to the maximum load, and (3)
5.5 The stiffness of the spring. (3)

1 Answers
Maweto Chito answered 2 months ago

 
 

Q.
Given,
D: lod
P: 3.5 KN
S
95 inm
T
400MPa
GS
80 Gla
5.1.
8 PD
Tid3
400x106 .
8 x 3.5x10x (hod)
n (03)
2
-4
olsa 2.228 xló54.
<?
4G
volume
(400 4106,
nd
4 प
nd
* Tom x
4x box 10
9
3
(400x1063?
Tx(14.997-167)
x nx 149.27410x4.05%
4
Чx Box lo
166.18

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