Set-3 Final by Jahangir (27-36)R.p65 - SIA GROUP

S.34 Spectrum ALL-IN-ONE Journal for Engineering Students, 2013
as,
Total torque required to turn the hand wheel is given
d ⎛ D 60 ⎞
T = P × + µ cWR
⎜Q R = = = 30 mm ⎟
2 ⎝ 2 2 ⎠
= 2327 .935×
50
+ (0.18×
10×
10
3 30)
2
×
T = 112.198 × 10 3 N-mm
Torque applied to hand wheel is given as,
D1
T = 2P 1 ×
2
D 1
112.198 × 10 3 = 2×
100×
2
∴
D 1
=
2×
112.198×
10
2×
100
D 1
= 1121.98 ~ 1122 mm
D 1
= 1122 mm or 1.122 m
∴ Diameter of hand wheel is 1.122 m.
Q7. (a) Explain the following terms of the
spring,
(i) Free length
(ii) Solid height
(iii) Spring rate
(iv) Active and inactive coils
(v) Spring index and
(vi) Stress factor.
Answer :
April/May-12, Set-3, Q7(a)
(i) Free Length
It is defined as the total length of helical compressed
spring when it is unloaded. When the spring is loaded or
compressed completely then it is given as,
l f
= Solid length + Maximum deflection + Clearance
between adjacent coils.
Clearance between adjacent coil is usually taken as
0.15 × Maximum deflection.
i.e., 0.15 (δ max
)
∴ l f
= n 1
× d + δ max
+ 0.15 (δ max
)
(ii) Solid Height
When a helical spring is compressed axially along
the helical axis, till the coils comes in contact with each other
then solid length is given as the product of total number of
coils and the diameter of wire. It is denoted by l s
.
= n 1
× d
l s
3
(iii) Spring Rate
It is defined as the ratio of applied load to deflection
of spring. It is also known as stiffness or spring constant. It
is denoted by ‘k’ and given as,
k = δ
W
Where,
W = Applied load in Newtons
δ = Deflection of spring in mm.
(iv) Active and Inactive Coils
It is defined as the coils which are actively
participating in spring action when the load is applied. In
squared and grounded ends, the bottom most does not
participate in spring action and hence, it is called as inactive
coil. The number of inactive turns increases from the bottom
side as the load increases. The coils other than inactive
coils are known as active number of coils.
(v) Spring Index
Spring index is defined as the ratio between mean diameter
of coil and diameter of wire. It is represented by ‘C’.
(vi)
Mathematically, C =
D
Mean diameter of coil
Diameter of wire
d
=
D
d
Figure
Stress Factor
It is also known as Wahl’s factor.
For remaining answer refer Unit-VII, Q5.
(b) Design a helical compression spring for a
maximum load of 1000 N for a deflection
of 25 mm using the value of spring index
as 5. The maximum permissible shear
stress for spring wire is 420 MPa and
modulus of rigidity is 84 kN/mm 2 .
4C -1 0.615
Take Wahl’s factor = +
4C - 4 C ,
where C = Spring index.
April/May-12, Set-3, Q7(b)
B.Tech. III-Year II-Sem.
( JNTU-Anantapur)