CH 10: Mechanical Springs
Springs are important mechanical elements because of their flexibility and
controllable stiffness. Springs allow controlled application of force or torque; also
they can be used for storing and releasing energy.
In general, springs may be classified as: wire springs, flat springs, and special shaped
springs.
 Wire springs (round or square wires) are helical in shape and can be made to
resist tension, compression, or torsion.
Stresses in Helical Springs
Consider a helical compression spring of mean coil diameter “𝐷”
and wire diameter “𝑑” subjected to compressive force “𝐹”.
 However, the wire is curved and the curvature increases the shear stress and
this is accounted for by another correction factor 𝐾𝐶 and thus the equation
becomes:
  Kc Ks
where 𝐾𝐶 is the “curvature correction factor”.
 Or easier the two correction factors are combined together as a single
correction factor 𝐾𝐵 where:
4𝐶 + 2
𝐾𝐵 = 𝐾𝐶 𝐾𝑆 =
4𝐶 − 3
Thus;
𝜏 = 𝐾𝐵
 If we remove a portion of the spring, the internal reactions will
be a direct shear 𝐹 and a torque 𝑇 = 𝐹𝐷/2 where each will
cause a shear stress, and the maximum shear will occur at the
inner surface of the wire which is equal to:
 max 
8𝐹𝐷
𝜋 𝑑3
Deflection of Helical Springs
Tr F

J
A
The deflection-force relation can be obtained using Castigliano's theorem.
Substituting 𝑇 = 𝐹𝐷/2 , 𝑟 = 𝑑/2 , 𝐽 =

8FD
d 3
𝜋
32
 The total strain energy in the spring wire has two components torsional and
shear.
𝜋
𝑑 4 , 𝐴 = 𝑑 2 gives:
4
8 FD 4 F

d 3 d 2
U
 Defining the spring index which is a measure of coil curvature as:
𝐶 = 𝐷/𝑑
For most springs C ranges from 6 to 12
T 2L F 2L

2GJ 2 AG
Substituting for 𝑇, 𝐴 & 𝐽 and knowing that 𝐿 = 𝜋𝐷𝑁
where 𝑁 = 𝑁𝑎 is the Number of active coils, we get:
We get:

where K s 
2C  1  8FD 
8FD

  Ks
2C  d 3 
d 3
U
2C  1
is called the “Shear stress correction factor”
2C
 Applying Castigliano's theorem to get the deflection “𝑦”;
y
This equation assumes the spring wire to be straight and subjected to torsion and
direct shear.
since 𝐶 = 𝐷/𝑑 we can write:
y
The effect of transverse
shear is neglected
 Knowing that the “spring rate” 𝑘 = 𝐹/𝑦
where 𝐶1′ and 𝐶2′ are elastic constants;
C1' 
Thus,
𝑘=
𝑑4𝐺
8𝐷3 𝑁𝑎
𝑁𝑎 : Number of active coils
Compression Springs
There are four types of ends used for compression springs:
 Plain ends: ends are non-interrupted (same as if the
spring was cut into sections).
 Plain-Ground ends: plain ends that are grinded flat.
 Squared (or closed) ends: ends are squared by
deforming them to zero degree helix angle.
 Squared and Ground ends: ends are grinded after
squaring.
 Table 10-1 gives the dimension formulas (free length, solid length, pitch) and the
number of active coils 𝑁𝑎 for the different types of ends.
The max compression of spring = free length (𝑙𝑜 ) - solid length (𝑙𝑠 )
Stability of compression springs:
Similar to columns, compression springs may buckle if the deflection (i.e., load)
becomes too large.
 The critical value of deflection (i.e., the value causing buckling) is given as:
U 8FD3 N 4 FDN

 2
F
d 4G
d G
  C ' 1\2 
ycr  l0C1' 1  1  22  
  eff  


very small
8FD3 N 
1  8FD3 N
1 

d 4G  2C 2 
d 4G
4 F 2 D 3 N 2 F 2 DN

d 4G
d 2G
2 ( E  G )
E
'
and C 2 
2( E  G )
2G  E
2
and 𝜆𝑒𝑓𝑓 is the effective slenderness ratio;
eff  l0 \ D
where “𝛼” is the end condition constant (see Table 10-2)
 Absolute stability is obtained when (C 2 \  eff )  1 (i.e., 𝑦𝑐𝑟 gives complex
number)
'
2
 C 2'  2eff
Thus,
l0 
D 2( E  G )

2G  E
 For steels, this turns out to be:
Buckling will not occur if
this condition is satisfied
l0  2.63
D

- If ends are ground and squared (α=0.5) it becomes:
l0  5.26D
Spring Materials
Springs are manufactured using hot (or cold) working processes depending on size
of the wire and spring index.
 A variety of materials may be used for making springs, Table 10-3 gives
description of the most commonly used steels.
 Spring materials may be compared based on their tensile strength. However, the
tensile strength for wires depends on the wire diameter, and the strengthdiameter relation is:
“Ultimate” tensile strength
S ut  A / d m
𝐴 & 𝑚 are material constants
 Table 10 - 4 gives the material constants for different wire materials.
 The diameters for standard gage wires are found in Table A-28.
 However, springs are subjected to shear not tension and we need to consider
yield strength not ultimate strength.
An approximate relation between shear yield strength 𝑆𝑦𝑠 and ultimate tensile
0.35 S ut  S ys  0.52 S ut
strength 𝑆𝑢𝑡 is:
 Thus, the free length of the spring is found as:
𝑙𝑜 = 𝑙𝑠 + (1 + 𝜉)𝑦𝑚𝑎𝑥
 If the spring runs over-a-rod or in-a-hole, we can we use their diameters as
governing value for the coil ID or OD (with some clearance included).
 However, in cases where there are no constraints on coil diameter, we can solve
for it by setting the shear yield strength (with design factor considered) to be
equal to the maximum shear stress at solid length.
S ys
 Table 10-6 gives a better approximation of the relation between 𝑆𝑦𝑠 and 𝑆𝑢𝑡 for
different materials.
 Table 10-5 gives the elastic constants 𝐸 & 𝐺 for different spring materials.
( ns ) d
 KB
8Fs D
d 3
substituting and solving for 𝐶 gives:
See Example 10-1 from text
2
C
Only for As-wound
 2   
2  
3
 
  4
4
 4 
Design of Helical Compression Springs for Static Service
where ,  
Make “a priori” decisions (if no specific requirements are given):
 Material: HD steel should be the first choice since it has the lowest relative
cost.
 Function: maximum load and spring stiffness or maximum displacement.
 Type of ends: squared ends should be the first choice since it gives good
stability and has low cost.
 Manufacturing: as-wound should be the first choice since it has the lowest
cost.
 Safety: use a design factor at solid length of at least 1.2
S ys
(ns ) d
and  
8(1   ) Fmax
d 2
and thus: 𝐷 = 𝐶𝑑
 The recommended ranges for spring index and number of active turns are:
4  C  12
&
3  N a  15
 In addition, a “figure of merit” that depends on the material cost and weight
can be used to select between the different feasible designs.
(ns ) d  1.2
fom  (relative _ material _ cos t )
 Working range: to ensure linearity we should avoid closing the spring to its
solid length under the maximum load. why?
 2d 2 Nt D
where  is the material weight density
4

“fom”: the closer
to zero the better
It is recommended to confine the operating range of the spring to the central
75% of its possible compression distance (i.e., between F = 0 and F = Fs)
Design Strategy:
Thus we can write:
Fs  (1   ) Fmax
where Fs is the force needed to compress the spring to solid length, and  is
the fractional overrun to closure.
- It is recommended that   0.15
 Use the given constraints and recommended ranges of the parameters to
identify the feasible designs, then use the “fom” to choose between the
feasible designs.
 See the design flow chart (fig. 10-3) in text.
 Consider all constraints and make a priori decisions.
 Make the wire diameter as your design variable and choose initial wire
diameter.
 Compute the other parameters: D, C, OD or ID, Na, 𝑙𝑠 , 𝑙𝑜 , (𝑙𝑜 )cr , ns , fom.
 Choose other wire diameters and recalculate the other parameters and
tabulate the data.
 Maximum shear stress will occur at point “B”
 B  (K ) B
8 FD
d 3
Where (𝐾)𝐵 is correction factor for curvature and direct shear.
See Example 10-2 from text
(K ) B 
Extension Springs
Extension springs are used to carry tensile loading; they require some means to
transfer the tensile load to the body of the spring, such as threaded plug or swivel
hook.
4C 2  1
4C 2  4
&
C2 
2r2
d
 Extension springs usually have “initial tension” 𝐹𝑖 , and thus the
load-deflection relation is:
F  Fi  ky
 Due to the initial tension in the spring, there is torsional shear stress always
present in the coil.
 The preferred “range” of the “uncorrected” initial shear stress is:
i 
 Extension springs are made such that the body coils are touching each other and
the spring usually has pre-tension.
 Stresses in the body are handled the same as
compression springs.
 Maximum tensile stress will occur at point “A”
on the inner surface of the ring:
16 FD 4 F

d 3 d 2
where 𝑁𝑏 is the number of body coils
l0  2( D  d )  ( Nb  1)d  (2C  1  Nb )d
Where (𝐾)𝐴 is the bending stress correction factor for curvature.
4C12  C1  1
4C1 (C1  1)
G
E
 The free length of an extension spring with
ordinary twisted end loops is found as:
Bending stress Axial stress
(K ) A 
where 𝐶 is the spring index
 When determining the stiffness of the spring “𝑘”, the deformation of the hooks
needs to be accounted for and thus an “equivalent” active number of turns is
used:
Na  Nb 
 Also, stresses in the hook need to be considered.
 A  (K ) A
231
C 3

 6.9 4 

e 0.105C
6.5  MPa

&
C1 
2r1
d
Usually r1  D \ 2
D: mean coil diameter
 Table 10-7 gives the maximum allowable shear stress for extension springs (as
percentage of the tensile strength).
See Example 10-6 from text
(e, f)
CH 11: Rolling-Contact Bearings
Also called “antifriction bearings” or “rolling bearings”.
➢ The starting friction is about twice the running friction.
Different from journal bearings in that the load is transferred by elements in rolling
contact rather than sliding.
With rolling bearings, we do not design the bearing but rather we select a bearing
according to our design requirements (the bearings are already designed).
Bearings are designed to take radial load or
thrust load, or combination of both.
• Nomenclature of ball bearings;
▪ Four main parts: inner ring, outer ring, balls
(or rollers) & separator (retainer).
➢ How balls are inserted in the grooves?
• Some types of ball bearings: see fig. 11-2.
(c)
(d, e)
(f, h)
(g)
(i, j)
Deep groove bearing: takes radial and some thrust load.
Filling notch bearing: has more balls i.e. takes more radial load, but less
thrust.
Angular contact bearing: more thrust.
Shielded & Sealed bearings: protection against dirt.
Self-aligning bearings: withstands more misalignment.
Double row bearing: takes twice the load of single row, but less parts
and space than two bearings.
Thrust bearings: thrust load only.
• Some types of roller bearings: see fig. 11-3.
(a)
(b)
(c)
(d)
Straight roller bearing: takes higher radial load than ball bearing (more
contact area), but needs perfect geometry & does not take thrust load.
Spherical-roller thrust bearing: useful for heavy loads & misalignment
(contact area increases with load).
Thrust: thrust load only.
➢ Why rollers are tapered?
Needle bearing: useful when radial space is limited.
➢ For example:
SKF rates for 106 revolutions
• Bearing life is a measure of the “Number of revolutions of the inner ring (outer
ring is fixed)” or “Number of hours of use (at a standard speed)” until the first
evidence of fatigue.
• According to ABMA, “Rating life” or minimum life or “𝐿10 ” life or “𝐵10 ” life is the
number of revolutions (or hours at fixed speed) that 90% of a group of bearings
will achieve or exceed before failure criterion develops.
➢ Median or average life refers to 50th percentile life of a group of bearings. It
can be up to 4 or 5 times the 𝐿10 life.
Bearing Load-Life Relation at Rated Reliability
When identical groups of bearings are tested until life-failure
criterion at different loads, the data can be plotted as:
• Thus, we can write:
where,
a=3
FL1\ a = Const.
= FL1 / a
L in revolutions
(1)
for Ball bearings
Obtained from testing
a = 10 / 3 for Roller bearings
From eqn. (1) we can write:
1/ a
F1 L1
= F2 L2
1/ a
• Manufacturers rate their bearings for a fixed number of revolutions at a certain
radial load called the “catalog load rating” 𝐶10 .
xD : Non-dimensional life measure where:
• To choose a bearing from the catalog, we can replace 𝐹1 and 𝐿1 with catalog
values 𝐶10 and 𝐿10 :
1/ a
Bearing Life
RD : Desired reliability.
x0 , & b : Weibull parameters.
Timken rates for 90×106 revolutions
C10 L10
• Other types:
▪ Instrument bearings; high precision, made of stainless steel.
▪ Non precision; no separator, made of sheet metal.
▪ Ball bushings; permit sliding & some rotation.
When a bearing is in operation, contact stresses occur on
the inner ring, rolling elements and outer ring.
If the bearing is clean, lubricated, sealed against dust, and
operates at reasonable temperature, then metal fatigue
(surface pitting or flaking) will be the only cause of failure.
Bearing Types
(a)
(b)
Tapered-roller bearings: take both radial & thrust loads (higher loads
than ball bearings).
L 60 D n D
xD =
=
L10 60 R n R
Note that if 𝑅𝐷 = 0.9 is used,
this will give the same result
as the previous equation.
Also, 𝑎𝑓 can be included in
the previous equation.
The typical values of the Weibull distribution parameters for SKF ball bearings are
𝑥𝑜 = 0.02, 𝜃 = 4.459 & 𝑏 = 1.483 where 𝑥𝑜 and 𝜃 are in million revolutions.
or
C10 ( R nR 60)1 / a = FD ( D nD 60)1 / a
Catalog load rating (KN or lb)
Q: why would we need a reliability higher than 0.9?
Desired speed (rev/min)
Desired life (hours)
Desired load (KN or lb)
Rating life (hours)
Rating speed (rev/min)
A: take for example a gearbox having six bearings each with 0.9 reliability.
The total reliability will be:
(0.9) 6 = 0.53 Only!
See Example 11-3 from text
Solving for 𝐶10 gives:
1/ a
  n 60 
C10 = FD  D D 
  R nR 60 
• The ABMA identifies the boundary dimensions of bearings using a 2-digit number
called the “dimension-series code” where the first digit refers to the width and
the second refers to the height.
➢ Note that a variety of bearings sizes that may have the same bore.
See Example 11-1 from text
Relating Load, Life and Reliability
The catalog gives load rating for 0.9 reliability “𝐶10 ”
Q: what if we desire a higher reliability?
A: Since bearing life is a random variable that follows a Weibull distribution,
the catalog load rating “𝐶10 ” can be found as:


xD
C10  a f FD 
1/ b 
 x0 + ( − x0 )(1 − RD ) 
1/ a
For 𝑅 ≥ 0.9
where,
a f : Application factor to compensate for non-steady load.
❖ Table 11-2 lists the dimensions and load ratings C10 and Co for two types of the
02-series ball bearings (from the SKF catalog).
➢ The Co is called the “static load rating” which is the maximum radial load a
bearing can withstand while it is not rotating.
▪ Co value depends on the number and dimensions of the balls or rollers in
the bearing.
✓ Why is the Co value smaller than the C10 value,
and would that cause a problem?
✓ What is the importance of the fillet radius and
shoulder diameter? (see the figure)
❖ Table 11-3 lists the dimensions and load ratings for some cylindrical-roller
bearings (from the SKF catalog)
✓ Why the shoulder diameter is not listed?
❖ To assist the designers in bearing selection, bearing manufacturers give some
recommendations on bearing life (see Table 11-4) and load application factor
(see Table 11-5).
Variable Loading
Bearing loads are frequently variable, it can be:
▪ Piecewise constant loading in cyclic pattern.
▪ Continuously variable loading in repeatable
pattern.
▪ Random.
• Let us consider the piecewise constant pattern,
eqn. (1) can be written as:
F a L = Constant = K
▪ If the bearing runs at load level F1 until point A, then the partial damage can
be measured as:
D = F1al A
Combined Radial and Thrust Loading
Ball bearings are capable of resisting both radial and thrust loading.
• Let 𝐹𝑎 & 𝐹𝑟 be the axial (thrust) and radial loads and take “𝐹𝑒 ” as the “equivalent
radial load” (i.e., it will do the same damage as both).
▪ Also, define a “rotation factor”, 𝑉, such that:
𝑉=1
when inner ring rotates
𝑉 = 1.2
when outer ring rotates
Note that if the bearing is
subjected to radial load only, the
rotation factor 𝑉 can be included
directly in the equation used for
calculating 𝐶10 by multiplying the
design load 𝐹𝐷 with 𝑉.
Fe = X iVFr + Yi Fa
i = 1 when Fa / VFr  e
i = 2 when Fa / VFr  e
(1)
where,
why?
where,
❖ Table 11-1 gives the values of X 1 , X 2 , Y1 , Y2
▪ “ e ” depends on Fa / C o (calculate the value of Fa / C o then take the
corresponding e value).
Note that 𝐶𝑜 needs to be known (i.e., a bearing must be selected) to find 𝐹𝑒 .
Thus, an iterative solution is needed when the bearing is loaded by a
combined radial and thrust loads, as will be seen later in Example 11-7.
See Example 11-4 from text

The damage done by loads Fe1 , Fe 2 & Fe3 is,
D = Fea1l1 + Fea2l2 + Fea3l3
▪ From testing, it was found that 𝐹𝑒 can be represented as:
Feq =  f i (a fi Fei ) a
▪ Consider the piecewise constant loading pattern
shown.

Fei : is equivalent radial load for combined
radial-thrust loads.
𝑙𝑖 : is the number of revolutions.
▪ The equivalent steady load “𝐹𝑒𝑞 ” when run for l1 + l2 + l3 revolutions, will do
the same damage:
D = Feq (l1 + l2 + l3 )
(2)
▪ Equating eqns. (1) and (2), and solving for 𝐹𝑒𝑞 we get,
a
1/ a
 F al + F a l + Fea3l3 
a 1/ a
Feq =  e1 1 e 2 2
 =  f i Fei
l1 + l2 + l3


where f i is the fraction of the total revolutions run under Fei


▪ Also, we can include the application factor for each segment:
1/ a
See Example 11-5 from text
Selection of Ball and Straight Roller Bearings
See Example 11-7 from text
Selection of Tapered Roller Bearings
Tapered roller bearings are more complicated than ball
and straight roller bearings.
• The four components of a tapered roller bearing are:
cone (inner ring), cup (outer ring), tapered rollers and
cage (retainer).
• The assembled bearing consists of two separate parts:
1. The cone assembly (cone, rollers and cage).
2. The cup.
• Tapered roller bearing can carry radial or thrust loads
or any combination of the two.
• Even if the bearing is under radial load only, because of the cup surface
inclination, a thrust reaction will be induced, and it will try to separate the cone
and cup assemblies.
▪ One way to overcome this problem is to use two tapered roller bearings in
opposite orientations: “direct or indirect mounting” see fig. 11-14.
❖ Fig. 11-15 shows a catalog page for tapered roller bearing from Timken Company
[90×106 rev. life].
❖ The Weibull distribution parameters used for the Timken tapered roller bearings
are found in Table 11-6 [Manufacturer 1].
• To determine the equivalent design
load for each bearing, first we need to
identify the bearing that carries the
external thrust load (if any is present),
and label that bearing as Bearing A
while the other one will be named
Bearing B.
▪ Then, the equivalent design loads
for each of the two bearings can
be calculated as:
➢ If
𝐹𝑖𝐴 ≤ (𝐹𝑖𝐵 + 𝐹𝑎𝑒 )
➔
𝐹𝑒𝐴 = 0.4𝐹𝑟𝐴 + 𝐾𝐴 (𝐹𝑖𝐵 + 𝐹𝑎𝑒 )
𝐹𝑒𝐵 = 𝐹𝑟𝐵
➢ If
𝐹𝑖𝐴 > (𝐹𝑖𝐵 + 𝐹𝑎𝑒 )
➔
𝐹𝑒𝐵 = 0.4𝐹𝑟𝐵 + 𝐾𝐵 (𝐹𝑖𝐴 − 𝐹𝑎𝑒 )
𝐹𝑒𝐴 = 𝐹𝑟𝐴
where,
𝐹𝑒𝐴 & 𝐹𝑒𝐵 : are the equivalent radial loads for bearings A & B.
𝐹𝑟𝐴 & 𝐹𝑟𝐵 : are the direct radial loads acting on bearings A & B.
𝐹𝑖𝐴 & 𝐹𝑖𝐵 : are the induced axial loads on bearings A & B.
𝐹𝑎𝑒 : is the external axial load.
See Example 11-8 from text
▪ The induced axial component can be found as:
Fi =
where,
0.47Fr
K
K = 0.389cot
Design Assessment for Selected Rolling-Contact Bearings
and  is half the cup angle.
➢ Before a particular bearing is selected, an estimated value of 𝐾 = 1.5 is used.
When we design a machine, each component (e.g., gears, shafts, bearings, etc.) is
designed separately. However, the components interact and influence each other.
It is always a good check to do a design assessment after all elements have been
designed (or selected) to make sure that all elements will perform as they are
assumed to do.
• For example, if the machine has several bearings we can do design assessment to
check the reliability of each of them and the total reliability for all.
CH 17: Flexible Mechanical Elements
Flexible mechanical elements (belts, chains, ropes) are used in conveying systems and
to transmit power over long distances (instead of using shafts and gears).
• The use of flexible elements simplifies the design and reduces cost.
• Also, since these elements are elastic and usually long, they play a role in
absorbing shock loads and reducing vibrations.
• For ball and roller bearings, solving for the reliability, we get:
• Disadvantage, they have shorter life than gears, shafts, etc.
  a f FD  a

 − x0 
 xD 
  C10 

R =1− 

 − x0




b
Belts
For 𝑅 ≥ 0.9
• There are four basic types of belts (Table 17-1):
▪
▪
▪
▪
See Examples 11-9 & 11-10 from text
Flat belts ~ crowned pulleys.
Round belts ~ grooved pulleys.
V-belts ~ grooved pulleys.
Timing belts ~ toothed pulleys.
• Characteristics of belt drives:
▪ Pulley axis must be separated by certain minimum distance. Why?
▪ Can be used for long centers distance.
▪ Except for timing belts, there is some slipping between belt and pulley, thus
angular velocity ratio is not constant or equal to the ratio of pulley diameters.
▪ A tension pulley can be used to maintain tension in the belt.
• There are two main configurations for belt drives; open and crossed (Fig 17-1)
where the direction of rotation will be reversed for the crossed belt drive.
• The figure shows reversing and non-reversing belt drives. Always there is one
loose side and this depends on the driver pulley and the direction of rotation.
❖ Fig. (17-3) shows flat belt drive for out-of-plane pulleys.
❖ Fig. (17-4) shows how clutching action can be obtained by shifting the belt from
loose to a tight pulley.
❖ Fig. (17-5) shows two types of variable-speed belt drives.
Flat and Round Belt Drives
where
Flat belt drivers produce very little noise and they absorb more vibration from the
system than V-belts.
Also, flat belts drives have high efficiency of about 98 % (same as for gears) compared
to 70-96 % for V-belts.
• For open belt drives, the contact angles are:
 d =  − 2 sin −1
D−d
2C
 D =  + 2 sin −1
D−d
2C
Larger contact-angle
for the large pulley
w 2
V
g
where g = 9.81 m/s2, w : is weight per unit length, V=πDn
▪ The initial tension can be expressed as:
▪ And the length of the belt is:
4C 2 − ( D − d ) 2 +
Fi =
1
( D D + d d )
2
• For crossed belt drives, the contact angle is the same for both pulleys:
 =  + 2 sin −1
D+d
2C
4C 2 − ( D + d ) 2 +
1
( D + d )
2
Note that  is the smallest
value of the contact angle
F1 − Fc
= e f
F2 − Fc
Minimum value of Fi needed to transmit
T e f + 1
a certain value of torque without slipping
D e f − 1
➢ This equation shows that if Fi is zero; then T is zero (i.e. there is no
transmitted torque).
Fi =
▪ Tight side tension:
F1 = Fi + Fc + F 
▪ Substituting in F1 & F2 equations we get:
= Fi + Fc + T / D
2 𝑒 𝑓
𝐹1 = 𝐹𝑐 + 𝐹𝑖 𝑓
▪ Loose side tension:
= Fi + Fc − T / D
(1)
▪ The belting equation relates the possible belt tension values with the
coefficient of friction and it is defined as:
▪ Substituting in eqn.(1) we find the relation between Fi and T
• Force Analysis:
F2 = Fi + Fc − F 
F1 + F2
− Fc
2
where f: coefficient of friction, : contact angle.
▪ And the belt length is:
L=
It also can be written as:
Fc =
d : diameter of smaller pulley
C : centers distance
L=
▪ The total transmitted force is the difference between F1 & F2
2T
F1 − F2 =
D
▪ The centrifugal tension Fc can be found as:
Fc = m r22
where  is the angular velocity & m is the mass per unit length.
D : diameter of larger pulley
where:
Fi : initial tension, Fc : hoop tension due to centrifugal force,
and F : tension due to transmitted torque.
𝑒
Note that “D” refers to the
diameter of the driver pulley
smallest value of the contact
angle
𝐹2 = 𝐹𝑐 + 𝐹𝑖
+1
2
𝑒 𝑓 + 1
Can be used to find the F1 & F2 values when
the belt is on the verge of slipping or to find
F1 & F2 for small Fi values where slipping is
occurring (note that the kinetic coefficient
of friction should be used in such case)
• Steps for analyzing flat belts include:
➢ Plotting F1 & F2 vs. Fi we can see
that the initial tension needs to be
sufficient so that the difference
between F1 & F2 curves is 2T/D.
1. Find  for the smallest pulley from geometry (find ef if needed).
FC =
2. From belt material and speed find FC.
❖ Table 17-2 gives the manufacturers
specifications for the allowable tension
for each type of belts.
w 2
V
g
3. Find the transmitted torque.
• When a belt is selected, the tension in the tight side is set to be equal to the max
allowable tension for that belt type.
➢ However, severity of flexing at the pulley, and the belt speed affect the belt
life, thus they need to be accounted for.
T = H d n = ( H nom K S nd ) n
4. From torque T, find the transmitted load.
( F1 ) a − F2 = 2T D
5. From belt material, drive geometry & speed, find ( F1 ) a .
▪ Therefore the max allowable tension is found as:
( F1 ) a = bFa C P Cv
(F1 )a=bFa C P CV
6. Find F2
where:
Note that F2 must
be larger than zero
F2 = ( F1 ) a − ((F1 ) a − F2 )
Fa : allowable tension per unit width for a specific belt material (kN/m)
(Table 17-2)
7. From ( F1 ) a , F2 & Fc find Fi .
b : belt width (m)
Fi =
CP : pulley correction factor (for the severity of flexing), it is found from
(Table 17-4) for the small pulley diameter. Use CP=1 for urethane belts
( F1 ) a + F2
− FC
2
8. Check if the friction of the belt material is sufficient to transmit the torque.
𝑓 > 𝑓′
CV : velocity correction factor. (For velocities other than 3 m/s), it is found
from Fig. 17-9 for leather belts. For polyamide or urethane belts use CV=1
where
f=
• The transmitted horsepower can be found as:
1

ln
( F1 ) a − FC
F2 − FC
Minimum friction needed to transmit
the torque without slipping
Alternatively, the comparison can be made between the
calculated Fi and the minimum required value of Fi
H = ( F1 − F2 )V = Tn
➢ However, when designing a belt drive, a design factor nd must be included to
account for unquantifiable effects. Also, a correction factor KS is used to
account for load deviations from the nominal value (i.e., over loads).
n fs = H a H nom K S
9. Find the factor of safety
See Example 17-1 from text
▪ Thus the design horsepower is:
H d = H nom K S nd
V-Belts
K1: contact angle correction factor (Table 17-13).
Note: the contact angles for V-belts are found using the same
equations used for flat belts.
The cross-sectional dimensions of V-belts are standardized. Each
letter designates a certain cross-section (see Table 17-9).
• A V-belt can be specified by the cross-section letter followed
by the inside circumference length.
❖ Table 17-10 gives the standard lengths for V-belts.
▪ However, calculations involving the belt length are usually based on the pitch
length of standard belts.
❖ Table 17-11 gives the quantity to be added to the inside length.
Example: Pitch length of C-1500 belt is: 1500 + 72 = 1572 mm.
• The standard angle for the V-belts cross-section is 40˚ but the sheave (pulley with
groove) has slightly smaller angle. This causes the belt to wedge itself inside the
groove and thus increases the friction.
• The operating speed for V-belts needs to be high, and the recommended speed
range is from 5 to 25 m/s. Best performance is obtained at speed of 20 m/s.
K2: belt length correction factor (Table 17-14).
▪ The design horsepower is found as:
➢
Based on the innercircumference length
Power that needs to
be transmitted from
the power source to
the driven machine
H d = H nom K S nd
where,
Hnom: nominal horsepower of the power source.
KS: service factor for overloads (Table 17-15).
nd: design factor of safety.
▪ The number of belts needed to transmit the design horsepower is found as:
Hd
Ha
Nb 
• For V-belts, the pitch length LP, and center-to-center distance are found as:
where Nb is an integer
• The belting equation for V-belts is the same equation used for flat belts. The
effective coefficient of friction for Gates Rubber Company belts is 0.5123
LP = 2C +  ( D + d ) / 2 + ( D − d ) 2 /(4C )
and
2







C = 0.25 LP − ( D + d ) +  LP − ( D + d ) − 2( D − d ) 2 
2
2






➢ While there are no limitations on the center-to-center distance for flat belts,
for V-belts the center-to-center distance should not exceed “3(D+d)” because
the excessive vibrations of the loose side will shorten the belt life. why?
➢ Also, the centers distance should not be less than D.
Thus,
F1 − Fc
= e 0.5123
F2 − Fc
▪ Where the centrifugal tension Fc is found as:
2
V 
Fc = K C 

 2.4 
KC: accounts for mass of the belt (Table 17-16).
• The transmitted force per belt (F1 - F2) is found as:
• Power:
❖ Table 17-12 gives the power rating for each belt cross-section (according to
sheave pitch diameter and belt speed).
▪ The allowable horsepower per-belt, Ha is found as:
H a = K1 K 2 H tab
where,
The power that can be
transmitted by each belt
from Table 17-12
𝐹1 − 𝐹2 =
𝐻𝑑 ⁄𝑁𝑏
𝜔(𝑑/2)
𝐻 ⁄𝑁
or 𝐹1 − 𝐹2 = 𝜋𝑑𝑛 𝑑𝑏 𝑤ℎ𝑒𝑟𝑒 𝑛: (𝑟𝑒𝑣⁄𝑠)
where 𝜔 (rad/s) & d are for the driver pulley.
e f
e f − 1
▪ Thus, F1 can be found as:
F1 = Fc + (F1 − F2 )
▪ Then F2 can be found from:
F2 = F1 − (F1 − F2 )
Fi =
And Fi is found as:
F1 + F2
− Fc
2
Note: K & b values given in Table 17-17 are valid only for the indicated
range. Thus, if NP is found to be larger than 109 it is reported as NP=109,
and the life time “t” (in hours) is found using NP=109. Also, if it is found
to be less than 108, the belt life is considered to be short and
inappropriate.
• In flat-belt force analysis, the
added tension caused by bending
the belt over the pulley is not
accounted for explicitly (since belt
thickness is not that large).
However, in V-belts the effect of
flexural stress is more pronounced,
and thus it affects the durability
(life) of the belt. The figure shows
the two tension peaks T1 & T2
resulting from belt flexure.
• Steps for analyzing V-belts include:
➢ Find V, LP, C,  and e0.5123
➢ Find Hd, Ha then the number of belts Nb
➢ Find Fc, ∆F, F1, F2 & Fi
➢ Find T1, T2 and then belt life NP & t
See Example 17-4 from text
▪ The values of the tension peaks are found as:
Wire Ropes
𝐾𝑏
𝑇1 = 𝐹1 + (𝐹𝑏 )1 = 𝐹1 +
𝑑
𝑇2 = 𝐹1 + (𝐹𝑏 )2 = 𝐹1 +
Wire ropes are made out of steel wires and are used in many
applications (such as hoisting, haulage, aircraft, etc.).
𝐾𝑏
𝐷
Where,
(Fb)1 & (Fb)2 are the added components of tension due to the flexure of
the belt on the smaller and larger pulleys.
❖ K b is used to account for belt flexure, and it is found from Table 17-16;
d & D are the diameters for the small and large pulleys in (meter).
• The life of V-belts is defined as the number of passes the belt can do (NP), and it is
found as:
𝐾 −𝑏
𝐾
𝑁𝑃 = [( ) + ( )
𝑇1
𝑇2
−𝑏 −1
]
Then, life time (in hours) is found as:
𝑁𝑃 𝐿𝑃
3600𝑉
Er: modules of elasticity of the rope
dw: diameter of the wire
D: sheave diameter
✓ This equation shows the importance of using large diameter sheaves (where
it reduces the stress developed in the outer wires).
✓ The recommended 𝐷⁄𝑑𝑤 ratio is 400 and up.
▪ Tension in the rope that causes the same stress caused by bending is called the
“equivalent bending load”, Fb , and it is found as:
𝐸𝑟 𝑑𝑤 𝐴𝑚
𝐹𝑏 = 𝜎𝐴𝑚 =
𝐷
where,
Am: is the area of the metal in the rope. It is usually: Am=0.38d2
(or from Table 17-27)
• Wire ropes are selected according to two considerations:
▪ Static considerations: the ability of the rope to carry the loads.
▪ Wear life (fatigue) considerations: the ability of the rope to live for a certain
number of loading cycles.
• Static considerations:
▪ Since the rope will bend when it passes over the sheave, and this increases the
tension in the rope, the equivalent bending load 𝐹𝑏 needs to be subtracted
from the ultimate tensile load 𝐹𝑢 .
Thus, the static factor of safety is found as:
𝐹 −𝐹𝑏
𝑛𝑠 = 𝑢
𝐹𝑡
❖ Table 17-25 gives the minimum rope factors of safety for different
applications.
• Fatigue considerations:
▪ The amount of wear that occurs in ropes depends on the bearing pressure on
the rope caused by the sheave and the number of bends (number of the passes
of the rope over the sheave) of the rope during operation.
▪ The "allowable fatigue tension" for a rope is found as:
𝐹𝑓 =
(𝑃⁄𝑆𝑢 ) 𝑆𝑢 𝑑 𝐷
2
Maximum tension that can be supported by
the rope for a certain number of bends.
Where,
▪ First step is to determine the maximum tension caused in the rope by the loads
(this includes the dead weight and the tension caused by acceleration and shock
loads).
➢ For example, the tension caused in a hoisting rope due to load
and acceleration/deceleration is:
W: total weight of the load
Su : ultimate tensile strength of the wires.
d : diameter of the rope & D : diameter of the sheave.
o Approximate ranges of the ultimate strength of wires Su for different
wire materials are listed below:
m: number of ropes supporting the load
w: weight per unit length of the rope
l: suspended length of the rope
a: maximum experienced acceleration/deceleration
𝑔: gravity acceleration
▪ Then, to find the static factor of safety, the "ultimate tensile load" of the rope
is found as:
Fu = strength of the rope × nominal area of the rope
(P/Su) : bearing pressure to ultimate strength ratio.
✓ It is found according to the specified life from Fig 17-21.
➢ It should be noted that Su is the ultimate strength of the wires, NOT the
strength of the rope (it is usually not listed in the tables, but it can be
determined using hardness test).
𝑊
𝑎
𝐹𝑡 = ( + 𝑤𝑙) (1 + )
𝑚
𝑔
where,
• Ropes are designated by size and configuration, for example: 25-mm
6×7 haulage rope means: diameter is 25 mm and has 6 strands each
having 7 wires.
❖ Table 17-24 lists some of the standard ropes along with their
properties. Also see Table 17-27.
• When a rope passes around a sheave, bending stress develops (especially in the
outer wires) due to flexing.
▪ Using mechanics principles, the stress in one of the wires of the rope can be
found as:
𝑑𝑤
𝜎 = 𝐸𝑟
𝐷
where,
❖ where K & b are found from Table 17-17.
𝑡=
• There are two basic ways of winding of wire
ropes:
▪ Regular lay: wires and strands are
twisted in opposite directions (do not
kink or untwist).
▪ Lang lay: wires and strands are twisted
in the same direction (more resistance
to wear and fatigue).
Maximum load that
can be supported
Improved plow steel (monitor)
Plow steel
Mild plow steel
▪ Thus, the fatigue factor of safety can be found as:
𝑛𝑓 =
𝐹𝑓 − 𝐹𝑏
𝐹𝑡
1655 < Su < 1930 MPa
1448 < Su < 1665 MPa
1241 < Su < 1448 MPa