TABLES IN VALLANCE: TABLE 13-1, p316, TABLE 13-2, P317
TABLES IN FAIRES: TABLE AT 17, p590
 Springs
A spring is an elastic object that stores mechanical energy.
Springs are typically made of spring steel. There are many spring
designs. In everyday use, the term often refers to coil springs.
When a spring is compressed or stretched from its resting position,
it exerts an opposing force approximately proportional to its
change in length (this approximation breaks down for larger
deflections). The rate or spring constant of a spring is the change
in the force it exerts, divided by the change in deflection of the
spring. That is, it is the gradient of the force versus
deflection curve. An extension or compression spring's rate is
expressed in units of force divided by distance, for example lbf/in
or N/m. A torsion spring is a spring that works by twisting; when it
is twisted about its axis by an angle, it produces
a torque proportional to the angle. A torsion spring's rate is in units
of torque divided by angle, such as N·m/rad or ft·lbf/degree. The
inverse of spring rate is compliance, that is: if a spring has a rate of
10 N/mm, it has a compliance of 0.1 mm/N. The stiffness (or rate)
of springs in parallel is additive, as is the compliance of springs in
series.
TABLES IN VALLANCE: TABLE 13-1, p316, TABLE 13-2, P317
TABLES IN FAIRES: TABLE AT 17, p590
TABLES IN VALLANCE: TABLE 13-1, p316, TABLE 13-2, P317
TABLES IN FAIRES: TABLE AT 17, p590
TABLES IN VALLANCE: TABLE 13-1, p316, TABLE 13-2, P317
TABLES IN FAIRES: TABLE AT 17, p590
TABLES IN VALLANCE: TABLE 13-1, p316, TABLE 13-2, P317
TABLES IN FAIRES: TABLE AT 17, p590
Formula:
𝜏=
16𝑃𝑅
𝜋𝑑 3
𝑑
(1 + 4𝑅)
𝑠𝑆=𝐾8𝐹𝐷𝑚
𝜋𝑑3
𝐾=
C=
Y=
4𝐶−1
4𝐶−4
𝐷𝑚
𝑑
+
0.615
𝐶
(spring factor)
(spring index)
8𝐹𝐶 3 𝑛
𝐶𝑑
Where:
Ss = Torsional shear stress
F = axial load
Dm = mean spring diameter (Do+Di)/2
Do = outside diameter of spring
Di= inside diameter of spring
D= wire diameter
Y=deflection
n= effective no of coils
G=modulus of elasticity
K= spring factor
C= spring index(ranges from 4 to 12, optimum value is 9).
TABLES IN VALLANCE: TABLE 13-1, p316, TABLE 13-2, P317
TABLES IN FAIRES: TABLE AT 17, p590
1. The helical spring shown is axially loaded with a compression
force P equal to 5 kN. The mean diameter of the spring is 100
mm and the wire used is 10 mm as indicated in the figure.
o What is the shear stress at A?
o What is the shear stress at B?
𝑇 = 0.05𝑃 = 0.05(5)
𝑇 = 0.25𝑘𝑁
𝑇1 =
𝑃
𝐴
=
5(1000)
𝜋(52 )
𝑇1 = 63.66 𝑀𝑃𝑎
𝑇2 =
𝑇𝑟
𝐽
=
0.25(1000)(5)
1
2 (5)
𝑇2 = 1273.24 𝑀𝑃𝑎
𝑇𝐴 = 𝑇1 + 𝑇2
𝑇𝐴 = 63.66 + 1273.24
𝑇𝐴 = 1336.90 𝑀𝑃𝑎
𝑇𝐵 = 𝑇2 − 𝑇1
𝑇𝐵 = 1273.24 − 63.66
𝑇𝐵 = 1209.58 𝑀𝑃𝑎
TABLES IN VALLANCE: TABLE 13-1, p316, TABLE 13-2, P317
TABLES IN FAIRES: TABLE AT 17, p590
2. A helical spring is fabricated by wrapping wire 3/4 in. in
diameter around a forming cylinder 8 in. in diameter. Compute
the number of turns required to permit an elongation of 4 in.
without exceeding a shearing stress of 18 ksi. Use Eq. (3-9) and
G = 12 × 106 psi.
Solution:
𝑇𝑚𝑎𝑥 =
16𝑃𝑅
𝑑
+
(1
)
𝜋𝑑 3
4𝑅
18000 =
16𝑃(4)
3
3
𝜋 (4) [1 + 4]
𝑃 = 356.07𝑙𝑏
𝛿=
4=
64𝑃𝑅𝑛
𝐺𝑑
64(356.07)(4)𝑛
3
(12𝑥106 ) ( )
4
𝑛 = 10.41
TABLES IN VALLANCE: TABLE 13-1, p316, TABLE 13-2, P317
TABLES IN FAIRES: TABLE AT 17, p590
3. Compute the maximum shearing stress developed in a
phosphor bronze spring having mean diameter of 200 mm and
consisting of 24 turns of 20-mm diameter wire when the spring
is stretched 100 mm.
Solution:
𝛿=
64𝑃𝑅𝑛
𝐺𝑑
Where:
𝛿 = 100 𝑚𝑚; 𝑅 = 100 𝑚𝑚
𝑑 = 20 𝑚𝑚; 𝑛 = 24 𝑡𝑢𝑟𝑛𝑠
𝐺 = 42000 𝑀𝑃𝑎
100 =
64𝑃(100)24
42000(20)
𝑃 = 437.5 𝑁
𝑇𝑚𝑎𝑥 = 16(437.5)(100)/𝜋(203 )[4(10) −
Where:
M = 2R/d = 2(100)/20 = 10
𝑇𝑚𝑎𝑥 =
16(437.5)(100)
𝜋(20)
𝑇𝑚𝑎𝑥 = 31.89 𝑀𝑃𝑎
[4(10) −
1
4(10)
−4
1
−4
4(10)
TABLES IN VALLANCE: TABLE 13-1, p316, TABLE 13-2, P317
TABLES IN FAIRES: TABLE AT 17, p590
4. alloy spring having squared and ground and has a total 16 coils and
modulus of elasticity shear of 85 GPa. Compute the Wahl factor. The
spring outside diameter is 9.66 cm and wire diameter is 0.65 cm.
Solution :
𝐷𝑚 = 𝑚𝑒𝑎𝑛 𝑑𝑖𝑎𝑚𝑡𝑒𝑟 = 𝐷𝑜 − 𝑑 = 9.66 − 0.65 = 9.01 𝑐𝑚
C = spring index =
Wahl Factor =
𝐷𝑚
𝑑
4𝐶−1
4𝐶−4
=
=
9.01
0.65
= 13.86
4(13.86)−1
4(13.86)−4
= 1.058
5. alloy spring having squared and ground and has a total 16 coils and
modulus of elasticity shear of 85 GPa. Compute the Wahl factor. The
spring outside diameter is 9.66 cm and wire diameter is 0.65 cm.
Solution :
𝐷𝑚 = 𝑚𝑒𝑎𝑛 𝑑𝑖𝑎𝑚𝑡𝑒𝑟 = 𝐷𝑜 − 𝑑 = 9.66 − 0.65 = 9.01 𝑐𝑚
C = spring index =
Wahl Factor =
𝐷𝑚
4𝐶−1
4𝐶−4
𝑑
=
=
9.01
0.65
= 13.86
4(13.86)−1
4(13.86)−4
= 1.058