10J X

<br>

closed
A coil helical spring is made of 10 mm diameter steel wire. Spring consists of 10 complete turns
Problem5,2:
with a mean diameter of 120 mm. The spring carries an axial pull of 200 N. Determine the shear stress induced
in the sprina.
neglecting the efect of stress concentration. Determine the deflection in spring, its stifness and strain energy stored by it, if
modulus of rigidity of spring material is 60 kN/mm'.
Solution: Given data :D = 120 mm, d = 10 mm, n = 10, W = 200 N, G = 60 kN/mm = 60 x 10° N/mm.
Procedure : (i) Shear stress induced :
Shear stress induced in the spring (neglecting the stress concentration) is,
8 WD x 200 x 120
88x
= 61.115 N/mm
TIX 103
(iü) Deflection in spring :

=
8 WD'n 8x 200 x (120) x 10
=
Deflection, 8 Ga x
60 10*x 10 46.08 mm
(iit) Stiffness: W 200 =
Stiffness, S =8 46, 08 4.3402 N/mmn
(iv) Strain energy stored :

1 1
= ×
6 =,x 200 x 46.08 = 4608 N.mm =
U
7x
4.608 N.m or J
W

Nunerical Type No. 2: "Design of Hellcal Spring Considering Stress due to Curvature of Wire"
Problem(5.3/: A helical spring is made from a wire of 8 mm diameter and has outside diameter 90 mm. If the
permissible shear stress is 350 N/mm and modulus of rigidity is 84
kN/mm, find the axial load, which the spring can carry
and the deflection per active turn.
(0 Neglecting the effect of curvature. (ü) Considering the effect of curvature.
Solution:Given data:d= 8 mm, D, = 90 mm, t = 350 N/mm“, G =
84 kN/mm = 84 x
10
= D,-d= N/mm.
Procedure: We have, D 90-8 = 82 mm
C = P82
d = 10.25
:
Case () Neglecting the effect of curvature:
K, 1 =
We have, =1+c- 1 +3x1o25 1.048

() Torsional shear stress induced in the spring (neglecting curvature effect) is

given by.
8 WD

8x Wx 82
350 1.0488 X

350 x
x
8 =
W= 1.0488 x
818.26 N
8x 82
 Elements
<br>

i) Deflection per 5.13

active turn is
given by,
8 WD³ Design of Spr
Gd
8 x 818.26 x 82
Case (I): Considering 84 x 10 xg4 =
10.49 mm
We know that,
the effect of curvature:
Wahl's stress
concentration
factor is.
4C-1
K= 4C 0.615 4 x
n 4
– 10.25-1 0.615
Shear stress C
4x 10.25 -
induced in 4
10.25 1,1411
the spring is given
by,
T = Kx 8 WD

350 = 1.1411 x 8xWx 82

TX 8
350 x nx 8
W=
1,1411 x8x82 = 752.073 N
() Deflection per active
turn isgiven by,
8 WD
8x 752.073 x 82
n Gd =
84 x 10x8 9.6417 mm
Probleri 5./: A
helical compression
spring carries a load
token as 8. Assume of 500 N with a deflection
permissible of 25 mm. The spring index may
t= 350 MPa. Modulus of rigidity = 84 kN/mm, 4C-1
be
spring index. Find number Wahl's factor as 0.615
of active turns of spring. cA +
C
where Cis
Solution: Given data: W = 500 N, S = 25 mm,
C= 8,
= 350 MPa =
350 N/mm, G = 84 kN/mm =
Procedure : Wahl's stress concentration 84x 10 N/mm²
factor is,

K = 4C-1 0.615 4 x
8-1 0.615
4C-4*C
C
4x8 -4 =
1.184
Maximum shear stress induced is
given by,
8 WD 8 WC
T = Kx

=
8
WC 8x 500 x8
-(:c-)
d Kx = 1.184x TIX 350
d = 5.87 mm
and D = 5.87 = 46.96 mm
Cxd= 8x
Number of active turns:It is calculated from the deflection formula.
8 WD'n
We have, Gd
x (46.96) xn
=
8x 500
25 84 x 10x (5.87)*
= 6.0190 7 (say)
re
otal number of active turns
n
= n =

ends of bronze material

Problem 5.5: Design a helical compression spring with grounded for valve used in an hydraulic
cuit. Take following values.
(a) Spring stress = 100 N/mm (c) Maximum load = 100 N,

index, C= 12, (b) Safe shear = mm.

(e) Deflection 15
of rigidity = 4 x10 N/mm,
FModulus coils, (v) Stiffoere of sorina.
rind out 0 Diametor of snrina wire, (i) Diameter oT Sprig cot, () vumber of spring
<br>

15
Problem S.6): A an automobile suspension system. The spring has stifres
closed coid helical spring is used for
squared and grounded ends. Load on the spring causes a total deflection of 9 mm. By taking permissible shecr
stress of 400 MPa, fnd: () Wire diameter of spring. (i) Length of spring.
Take spring index= 6 and G= 80 x10 N/mm'.
Solution: Given data:S = 85 N/mm, S = 9 mm, t = 400 MPa = 400 N/mm,C= 6, G = 80 × 10 N/mm
Procedure : () Wire diameter of spring :

We have, Spring stiffness, S =

W = =
85 x 9
=
Sx 765 N
Wahl's stress concentration factor is,

= 4C-1 0.615 4
x6-1 0.615
K + =
4C– 4 C 4x6-4 6 1.2525
Maximum shear stress induced is given by,
8 WD 8 WC
= Kx
t Kx
Id=
= 8 WC 8x 765x6
d' Kx = 1.2525 x = 36.5843
TT
X 400
=
d 6.0497 mm
=
and D
=
Cxd = 6x6.0497 =
36.29 mm
Diameter of spring wire 6.0497 mm and Mean diameter
of spring coil = 36.29 mm
(i) Length of spring :Deflction isgiven by,
8 WD'n

Gxd
8xGxd* 9x 80x10° x (6.0497)
n 8 WD =
3.29
8× 765 >x (36.29) 4 (say)

Total number of active turns (n) = 4.

<br>

pesign of Machine Elements 5.15 Design of Springs

Given squared and grounded ends, therefore, total number of turns is given by,
= n + 4 +
n' 2= 2
=6
Solid length, =
L n'xd=6x 6.0497 = 36.2982 mm
Free length
n'xd + ômax + 0.15 x imax = (6 x 6.0497) + 9 + (0.15 x 9) = 46.648 mm
=
L4

5.7/Find the maximum

Problem shear stress and deflection induced in a helical spring, has to absorb
1000 N-m energy. Use the data as, D = 100 mm, d = 20 mm, n = 30, G = 84
kN/mm.
i it
:
Solution: Given data U= 1000 N.m 1000 103 N-mm, D = 100 mm,
= x

d= 20 mm, n = 30, G = 84 KN/mm' = 84 ×

103 N/mm²
Procedure: () Maximum shear stress induced:
We D 100
have, Spring index =
C=ã =20 =5
Wahl's stress concentration factor is,

4C-1 0.615 4x5-1 0.615

K= 4C-4 C
4x5 -4 5
=
1.3105
From the expression of strain energy stored in a spring, we have,

U= 4K' G

where, V= Volume of wire = Cross-sectional area of spring wire x Length of wire

4K2 xd)x
G^4
(xon)

Ux4x K²xG 1000 x 10³ x 4 x (1.3105)2 x 84 x 103

= 1,94,886,93ipate
|x (rDn) (x(202)x (x 100 x 30)

=
441.46 N/mm²
(ii) Deflection:
8x Kx Wx D
We have,
x

W =
TXTX dB 441.46 IX (20)3 = 10582.88 N
8xKx D 8x1.3105 × 100

Deflection is given by,

8WD³nn 8x 10582.88 x 100 x 30
=
Gx d4
84 x 103x (20)4
188.98 mm
<br>

mmn
= 22.69
n'-1 24-1
Problen 5.112 Design a helical compression spring for maximum load of 800 N with a deflection of 25 mm. Spring
index is 5 and
Wahl's correction factor is l.3. Maximum permissible shear stress for spring wire is 400 MPa and modulus of
igidity is 84 KN/mm.
= 800 N, C= 5, 8 = 25 mm,
Solution: Given data : t= 400 N/mm²,
W

= 84 x 10° N/mm², K = 1.3

G=84 kN/mm
Procedure: Design of helical spring:
(1) Mean diameter of spring coil Maximum shear stress induced
: is given by,

=
SWD 8WC (c
Kxd3= Kx
KX rd2

8WC 8x 800 x 5
=
d2 = Kx TXT 1.3 x 33.0909
TX 400
<br>

5.18 Design
ot Sçrirçs
Design of Machine Elements
= mm
5.75 mm and D Cxd=
= = 5x5.75 28.75
d =
= = mnm and Mean diameter of spring coil = D 28.75
Diameter of spring wire d 5.75
:
(2) Number of turns It is
calculated from the deflection formula.
8WD³n
8 =
Gx d4
x (5.75)* =
SxGxd4 25 x 84 x 10 15.09 16 (say)
n 8WD
x
8x 800 (28.75)3
Total number of active turns = n= 16
Assuming squared and grounded ends, total number of turns is given by.
n' = n + 2 = 16 + 2 = 18
(3) Solid length: Ls = n'xd = 18 x 5.75 = 103.5 mm
(4) Free length: Le = + ômax + 0.15 x omax
n'xd
4 = (18x 5.75) + 25 + (0.15 x 25) = 132.25 mm
Free length 132.25
(5) Pitch of the coil:p mm
n'-1 18-1 =7.7794

Numeical Type No. 3 : "Design of Helical SpringOperating between a Load Range"

Problem5.12: helical valve spring is to be designed for an operating load range of approximately 135
A

T
defection of the spring for the load range is 7.5 mm. Assume spring index of 10. Pemissible shear stress for the materizl
N

the spring = 480 MPa and its modulus of rigidity = 80 KN/mm. Design the spring.
t
Take Wahl's factor 4C-1 Q615
=ctC being the spring index.
: C = 10,
Solution: Given data t = 480 MPa, G = 80 kN/mm = 80 x 10 N/mm. Load range is 135 N, for wnic
deflection of spring is 7.5 mm. Therefore, we can write, W = 135 N,
for which, corresponding §= 7.5 mm.
Procedure:
Wahl's correction factor, K= 4C-1 0.615 (4x 10) -1 0.615
+ = 1.1448
4C-4 T C (4x 10) -4 10
Designof helical spring :
(1) Mean diameter of spring coil :Maximum
shear stress induced is given by,
8WC

Kx 8WC. 1.1448 x 8x 135 x 10

= 10.37
TX 480
d = 3.22 mm
We knOw, oL D 10 x3.22 = 32.2 mm
Thus. diameter of spring wire, = Cxd=
d 3.22 mm, and Mean diameter
of helical spring coil, D= 32.2 mm
(2) Number of turns : It is calculated from deflection
formula.
8wc'n
8 =
G.d
=
Sx Gxd 7.5 x 80 x 10° x
3.22
n =
8WC' 8x 135 >x (10)
For safe desian, minimum number (say) 1.7884
of coils or turns of spring should
Assuming squared and grounded ends, total be taken as A
number of turns is aiven by
n' =n +2 + 2 = 6 = 4

(3) Solid length: L: n'xd=6 3.22 = 19.32 mm x

(4) Free length : L= Ls + Omax + 0.15 x Omax = 19.32 + 7.5 +

(0.15 x
7.5) = 27.945 mm
<br>

Design of Machine Elements

5.19
Design of Springs
: Free length 27.945
(5) Pitch of coil p= =
n'-1 6-1 5.589 mm
Problen 5.13/ Design a close coiled helical compression spring
axial deflection of spring for load range is mm. for service load ranging from 2250 N to
6 Assume a spring index of 5. Permissible 2750 N. The
shear stress
and modulus of rgidity is
G= 84 kN/mm. Take design stress 25% excess of permissible stress intensity is 420 N/mm
intermittent operation. for severe condition and
Solution: Given data : W; = 2250 N,
= 2750 N, C
W,
=5,
Permissible shear stress = 420
N/mm², G = 84 x 103 N/mm?
Deflection for the load range
(W,- W) = (2750- 2250) = (500 N) is, & = 6 mm.
Procedure: Wahl's stress concentration factor is,

K=
4C-1 + 0.615 4 x5-1 0.615
=
4C-4 C 4
x5 -4 1.3105
For severe conditions and intermittent
operations, take
Design stress = 1.25 x Permissible stress
= 1.25 >x
420 = 525 N/mm
(1) Diameter of spring wire: Maximum stress shear induced due to maximum load (Wa) is given by,
8W,D 8W,C
(e-)
=
T Kx = Kx

8W; C x
d2 = Kx = 1.3105 x 8x 2750 5
TX 525
Thus, d = 9.35 mm and D = Cxd =5x9.35 = 46.75 mm
Diameter of spring wire, d = 9.35 mm
And, Mean diameter of spring coil, D = 46.75 mm
(2) Number of turns : It is calculated from the deflection formula.
8WD n
Gx d4

8xGx d4 6x 84 x
10³ x (9.35)4
n= 500 x (46.75)3
=
9.42 10 (say)
8WD3 8x
Total number of active turns = n = 10
Assuming squared and grounded ends, total number of turns is given by.
+ 2 = 10 + 2 = 12
n'=,n
(3) Maximumdeflection : For load range of 500 N, deflection is 6 mm. Therefore, maximum defletion for maximum
load of 2750 N can be calculated as follows :

mm
For 500 N 8= 6

=
2750 x 6
= 33 mm
For 2750 N Ômax
500
L =
n'xd= 12
x
9.35 = 112.2 mm
(4) Solid length:
Free length : =
L+ Ômax + 0.15 omax = 112.2 + 33 + (0.15 x 33)
=
150.15 mm
(5) 4
Free length 150.15 =
(6) Pitch of coil: p = 13.65 mm
n'-1
12
-1
<br>

Problen5.164Design a helical spring for a spring loaded safety valve for the following conditions.
pressure
Operating
=Pi =1N/mm
Maximum pressure, when the valve blows off freely = p, = 1.075 N/mm
Maximum lift of valve (when the pressure is 1.075 N/mm') =
8= 6 mm.
Diameter of valve seat = D, = 100 mm.
Maximunm shear stress = T= 400 N/mm.
Modulus of rigidity = G= 86 x10 N/mm²
Spring index = C= 5.5.

Solution: Given data :

p = 1
N/mm, p: = 1.075
N/mm, 8 = 6 mm, D, = 100 mm, t =
400 N/mm,
G= 86 x 10
N/mm, C = 5.5.

Procedure:
(1) Diameter of spring wire and diameter of spring coil :
Initial tensile force acting on the spring is,

2
= 7853.98 N

Wi = D, xp1 =7x100x1
4
Maximum tensile force is,

W: = x Dx x
4
P.
4
100 x 1.075 = 8443.03 N
- = N
Load range = W,- W, = 8443.03 7853.98 589.05
4C-1 + 0.615 4x 5.5-1 + 0.615 = 1.2785
Wahl's factor, K= 4C- 4 c 4x 5.5–4 5.5

8 W;C
= Kx
(: W; is maximum load acting on spring)
We have, Tmax
<br>

Design of Machine Elements 5.22 Design

of Springs

400 = 8x 8443.03 x 5.5

1.2785x
d= 19.44 mm
Mean diameter, D = Cxd= 5.5x 19.44 = 106.92 mm
(2) Number of coils :
For load range, Wrange = 589.05 =
6 mm
N õrange

For maximum load, = 8443.03 8443.03 x6 =

N Šmax = 85.99 mm
W,

589.05

We have, Smax
8 Wmax
XCx
G. d

8 x 8443.03 x 5.5xn
85.99
86 x 10x 19.44
Active number of coils = n = 12.79 14 (say)
Total Number of coils, n' =n+ 2 = 14 +
2= 16 [Assuming squared and grounded ends
(3) Solid length, L = 19.44 = 311.04 mm
n'xd= 16 ×

(4) Free length, Lt =

n'x d + 8max + 0.15 Smay = (16 x 19.44) + 85.99 + (0.15 x 85.99) = 409.92 mm
(5) Pitch of coil: p = L 409,92
n'-1 16 -1 =27.328 mm
<br>

Numerical Type No. 6 : "Design of HelicalSprings Arranged in Series or Parallel"

Problem A helical spring of stiffness 12 N/mm is placed on the top of other spring having stiffness 8 N/mm. Find
the force required to give a total deflection of 50 mm.
Solution: Given data : S, = 12 N/mm, S = 8 N/mm, &= mm
50
Procedure: Since one spring is placed on the top of second spring, it means that, springs are
connected in series.
We have,
S1
S 12
S = 4.8 N/mm
Also, =
W
S

W =
Sx§ = 4.8 x 50 = 240 N
Problem5.22): A closed coil helical spring of 12 active coils
has a spring stifness of K. lt is cut into two
5 and 7 turns. Determine the spring stiffness springs having
of resultant springs.
Solution: Given data: n, 5, n, = 7, n = 12
=
Procedure: We have, axial deflection,
8WD n
Gx d4 ... (1)
By definition, Stiffness (S) = K= [Stiffness is termed as
'K' in problem
W
K
(8 WD'n)
G d
From equation (1)
<br>

Design of Machine Elements Design of Springs

5.29

18 WL
Stress in full length leaves: OF =
bt (2n, + 3n)
12 WL
Stress in graduated leaves: +
bt (2n, 3n)
18 WL =
12 WL
We can write, bt? (2n, + 3n) 1.5 X
ht (2ng
t 3h)
O; = 1.5 x oG
nn tho ctrecs induCed in araduated ea