Transcript Introduction to Technical Mathematics

01 - Applied Forces
01 – Applied Forces
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The intent of this presentation is to present enough information to provide the reader with a
fundamental knowledge of mechanical applied forces used within Michelin and to better
understand basic system and equipment operations.
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01 - Applied Forces
Module 1 – Levers
Module 2 – Torque
Module 3 – Sling Angles
01 – Applied Forces
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Module 1: Levers
Module 1
Levers
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Module 1: Levers
Categories of Levers
Class 1 Lever
A lever is considered a simple machine. In its most elementary form a lever is a rigid bar that rotates
around a pivot point. This pivot point is called the fulcrum. An object being moved by a lever is called
the load. The closer the load is to the fulcrum, the easier it is to move.
F2
F1
D1
D2
Fulcrum
In the above figure, at one end of the lever we have the effort force F1; at the other end the resistance
force F2. The fulcrum is between F1 and F2 resulting in a class 1 lever. Some other examples of class
1 levers are seesaws, claw hammers, pliers, scissors, and oars.
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Module 1: Levers
Categories of Levers
Class 1 Lever
Both F1 and F2 tend to turn the lever around the fulcrum in opposite directions. The tendency to revolve
the lever is called Moment of Force or Torque. The moment of force is equal to the force multiplied by its
distance from the fulcrum.
Moment of Effort = F1 x D1
Moment of Resistance = F2 x D2
F2
F1
D1
D2
Fulcrum
The lever is balanced or in equilibrium when;
The moment of effort = the moment of resistance
F1 x D1 = F2 x D2
The mechanical advantage (MA) of a lever is the ratio of the output force to the input force F2/F1. If we
want a mechanical advantage in force, D1 must be greater than D2. When using levers, if we reduce
effort, we increase effort distance; and if effort required increases, we decrease effort distance.
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Module 1: Levers
Categories of Levers
Class 2 Lever
In a class 2 lever the effort force, F1 is located to one end of the bar, the resistance force, F2 is located
between F1 and the fulcrum, and the fulcrum is located on the end of the bar opposite to F1.
F2
D2
D1
F1
Some examples of class 2 levers are wheelbarrows and nutcrackers.
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Module 1: Levers
Categories of Levers
Class 3 Lever
In a class 3 lever the effort force, F1 is located between the resistance force, F2 and the fulcrum. Notice
that for this class of lever, the effort force is greater than the resistance force. This is different from the
first-class and second-class levers. However, also notice that the effort force moves through a shorter
distance than the resistance force.
F2
D2
D1
F1
Some examples of class 3 levers are human arm, tweezers and boom crane (cherry picker).
Note: The distance is always measured to the fulcrum.
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Module 1: Levers
Categories of Levers
Example:
How much effort F1 is needed to hold a 750 lb. Crate in the shown position?
F1 x D1 = F2 x D2
F1 x 6yds. = 750lbs. x 1yd.
6yds.(F1) = 750yd-lbs.
F1 = 750yd-lbs.
6yds.
F1 = 125lbs.
6 yds.
1 yd.
F1
F2 = 750 lbs.
MA = F2/F1
MA = 750lbs./125lbs.
MA = 6
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Module 1: Levers
Solve the following exercises showing all calculations.
1).
F2 = 450lbs.
5yds.
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F1 = ?
7yds.
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Module 1: Levers
Solve the following exercises showing all calculations.
2).
460kg.
800mm
300mm
F1 = ?
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Module 1: Levers
Solve the following exercises showing all calculations.
3).
800lb
10ft.
3ft.
F1 = ?
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Module 1: Levers
Solve the following exercises showing all calculations.
F1 = ?
4).
Dia. 1ft.
14"
300lbs.
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Module 1: Levers
This Concludes
Module 1
Levers
01 – Applied Forces
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Module 2: Torque
Module 2
Torque
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Module 2: Torque
Units of Torque
The symbol for torque is τ. The concept of torque, also called a moment of force or couple, originated
with the work of Archimedes on levers. A force applied to a lever, multiplied by its distance from the
lever's fulcrum, is the torque applied at that point.
F2
F1
D1
D2
Fulcrum
Torque is a force that tends to rotate or turn things around a pivot point. Basically, torque can be thought
of as rotational force or a turning or twisting force. You generate a torque any time you apply a force
using a wrench. Tightening the lug nuts on your wheels is a good example. When you use a wrench, you
apply a force to the handle. This force creates a torque at the center of the lug nut, which tends to turn
the lug nut.
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Module 2: Torque
Units of Torque
The force applied to a lever, multiplied by its distance from the lever's fulcrum, is torque. As a formula,
you can define Torque as :
F
D
Torque = Force x Distance
Another way of expressing the above equation is; torque is the product of the magnitude of force and the
perpendicular distance from the force to the axis of rotation (i.e. the pivot point).
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Module 2: Torque
Units of Torque
The common values used to establish torque are force and distance. The common units used to express
force are; pounds (lbs) in the U.S., newtons (N) in the ISO system. Sometimes kilograms (kg) are also
used in the place of newtons as a more practical unit. The common units used to express distance are;
inches (in) or feet (ft) in the U.S., meters (m) in the ISO system.
Common examples of those values are listed below;
U.S. examples
T=FxD
T=FxD
lbs-in = lbs x in ; lbs-ft = lbs x ft
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ISO examples
T=FxD
T=FxD
Nm = N x m; kgm = kg x m
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Module 2: Torque
Torque and Speed Relations
In machine design it is often necessary to incorporate a power transmission device between an energy source and the
desired output motion. Examples of power transmission devices may include (but are not limited to); gear trains,
friction drives, timing belts, flat belts, chain & sprocket drives, levers, and screw drives.
The Power Transmission device will often include a ratio or mechanical advantage aspect. A ratio can increase the
output torque and decrease output speed of a mechanism, or decrease output torque and increase output speed, but
not both in the same direction. A classical example is the gears on a multi-speed bicycle.
Choosing a low gear that allows you to pedal easily up hill, but with a lower bicycle speed. Conversely a high gear
provides a higher bicycle speed, but more torque is required to turn the crank arm of the pedal. This tradeoff is
fundamentally due to the law of energy conservation and is the key concept of mechanical advantage. With a given
power source you can either achieve high speed (rpm) output or high force/torque output but not both.
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Module 2: Torque
Torque and Speed Relations
To understand the relationship between speed and torque better you have to back-up to a point of understanding the
basics behind the transfer of rotation within a power transmission device. A pair of gears, in mesh, would be the
simplest approach to explain the principles. Although we are using a pair of gears in mesh, these basic principles still
apply to virtually all power transmission devices.
When two gears are in mesh a ratio is formed. That ratio is determined by the distances from the center of the gear to
the point of contact, or in other words the radius of the gear. For instance, in a device with two gears, if one gear is
twice the diameter of the other, the ratio would be 2:1.
Speed (RPM)
Force
Speed (RPM)
This ratio would be consistent in the fact that it applies to most aspects
of the gear train. The diameters would be a 2:1 ratio, the circumference
would be 2:1, the number of teeth (if it is a gear train) would be 2:1, as
well as the radii as illustrated in figure below.
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Gear 2
Gear 1
Torque
Torque
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Radius
Radius
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Module 2: Torque
Torque and Speed Relations
For the purposes of this illustration we’ll assume the fact that in the
figure at right, Gear 2 has a 2:1 ratio with Gear 1. In this figure, you can
see that the radius of Gear 1 is smaller than Gear 2. Establishing the fact
that the smaller gear is spinning twice as fast as the larger gear, because
the circumference has to travel twice as far for every revolution of the
larger gear. The same is true for the larger gear, it is spinning at half the
speed of the smaller gear, because the circumference only travels half as
far for every revolution of the smaller gear.
Speed (RPM)
Force
Speed (RPM)
Gear 2
Gear 1
Torque
Torque
Radius
Radius
The fact that one gear is spinning faster or the fact that one gear is spinning half as fast as the other is because of the
ratio between the gears; the gear ratio. If both gears had a 1:1 ratio, all aspects would be equal and they would rotate
at the same speed.
The same would be true when a comparison of speed and torque is used. Speed (rpm) represents how fast the gears
are spinning, where torque represents how much force the rotating gear is capable of supplying at the center of the
axle or shaft. Remember, torque is a product of force and distance.
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Module 2: Torque
Torque and Speed Relations
Using the same previous example in figure above, an example of 2:1 ratio between the gears, the speed (rpm) will be
cut in half, but the torque will be doubled from Gear 1 to Gear 2. Inversely the speed (rpm) will double, but the torque
will be cut in half from Gear 2 to Gear 1. Simply because the amount of force transmitted from one gear to another
stays the same no matter how many gears are utilized. When the force is applied at the point of tooth contact between
both gears (in mesh) the force is then transmitted thru the radius and is multiplied by the distance to the center of the
axle or shaft. Much like the results of the action of a lever shown in figure below.
F2
F1
Thus if Gear 1 is the input and
Gear 2 is the output:
The output speed (rpm) will be
half the input speed (rpm).
The output torque will be twice that
of the input.
D1
D2
Fulcrum
Conversely, if Gear 2 is the
input and Gear 1 is the output:
The output speed (rpm) will be
twice the input speed (rpm).
The output torque will be half
that of the input.
With that known, we can establish the relationship between torque and speed. Torque is inversely proportional to
speed (rpm). In other words, there is a trade-off between how much torque a motor delivers, and how fast the shaft
spins.
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Module 2: Torque
Torque and Speed Relations
The relationship of torque and speed (rpm) is the ratio that is established between the common power transmission
devices, such as a gear train, chain & sprocket drive, belt drive system, or friction drive system.
Speed (RPM)
Force
Speed (RPM)
Gear 2
Gear 1
Torque
Torque
Radius
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Module 2: Torque
Calculation Exercises
F
D
1). With a F = 180 lbs (800.68 N) and D = 10 in (0.25 m) find the torque.
2). If a torque of 325 lb-ft (440.64 N-m) is required at a distance of D = 15 in (0.38 m), what would the F be?
3). If F = 150 lbs (667.23 N) and the torque must be 900 lb-ft (1220.24 N-m), what would the D be equal to?
4). Torque = 1100 lbs-ft (1491.4 N-m) and D = 1 ft (0.305 m), find the F.
5). Torque = 300 lb-ft (406.75N-m), and F = 135 lbs (600.51 N), find the D.
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Module 2: Torque
Calculation Exercises
Metric Example:
We have a motor capable of producing 10 N-m of torque. Attached to the motor is a gear with a PCD of 200 driving a
gear with a PCD of 400. What is the torque exerted on the shaft of the driven gear?
Given:
PCD gear #1 = 200
PCD gear #2 = 400
Motor torque = 10 N-m
Find: Torque on shaft of gear #2
Step 1 : Find the force (F1) at the edge of gear #1.
T= F1 x D
F1 
T
D
F1 
10 N  m
0.1m
F1 = 100 N
Step 2 : Find the torque on gear/shaft #2.
T = F1 x D
T = 100 N x 0.2 m
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Module 2: Torque
Calculation Exercises
F
American Example:
Given: PCD gear #1 = 1.97 in.
PCD gear #2 = 5.91 in.
PCD gear #3 = 3.54 in.
Motor torque = 36.88 Lb-ft
Step 1 - Solve F1:
F1 
Step 2 - Solve T2:
Step 3 - Solve F2:
Step 4 - Solve T3:
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T1
D1
2
Gear # 2
Find: F1 ; F2
Gear # 3
Gear # 1
Torque on shaft of gear # 2
Torque on shaft of gear # 3
36.88Lb  ft
0.082 ft
T2 = F1 x D2
F2 
T2
D2
T3 = F2 x D3
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F1
= 449.76 Lbs.
Motor
T2 = 449.76 Lbs x 0.246 ft
110.64Lb  ft
0.246 ft
T2 = 110.64 Lb-ft
= 449.76 Lbs.
T3 = 449.76 Lbs. x 0.147 ft
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F3
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T3 = 66.11 Lb-ft
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Module 2: Torque
Calculation Exercises
American Example:
F
2
Gear # 2
Gear # 3
Conclusion: F1 = F2
The input force on a gear equals the output force on a gear.
The torque transmitted to the shaft through the gear is directly
proportional to the PCD of the gear.
Gear # 1
What would F3 be? ……………
F1
F3
Motor
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Module 2: Torque
Calculation Exercises
F
1).
2
Gear # 2
Gear # 1
Gear # 3
F1
F3
Motor
Given:PCD gear #1 = 7.87 in.(200mm)
PCD gear #2 = 11.81 in.(300mm)
PCD gear #3 = 3.94 in.(100 mm)
Motor torque = 11.06 Lb-ft (15 N-m)
Find:
F1___________
F2___________
F3___________
Torque on shaft of gear # 2_______
Torque on shaft of gear # 3_______
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Module 2: Torque
Calculation Exercises
F1
Gear # 2
2).
Gear # 1
Gear # 3
L
F2
F3
(Load)
Given: PCD gear #1 = 11.81 in.(300mm)
PCD gear #2 = 9.84 in.(250mm)
PCD gear #3 = 3.54 in.(90mm)
F3 (Load) = 992.08 Lbsf (450 kgf)
L (Length) = 15.75 in. (400mm)
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Torque on shaft 1 __________
Torque on shaft 2 __________
Torque on shaft 3 __________
F2 _______
F1 to hold load stationary _______
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Module 2: Torque
Calculation Exercises
Gear 1
3).
Gear 2
Sprocket B
Sprocket A
Motor
Given:PCD gear #1 = 14.53 in.(369mm)
PCD gear #2 = 7.80 in.(198mm)
PCD Sprocket B = 3.94 in.(100mm)
PCD Sprocket A = 8.19 in.(208mm)
Motor output = 14.75 Lb-ft (20 N-m)
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Find:
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Torque on shaft of Sprocket A
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Module 2: Torque
Calculation Exercises
4). A 12mm screw has a recommended torque value of 9.7 kg-m and will snap at 5 times this value. If a 185 lb.
mechanic hangs on the end of a wrench 14 inches long will the screw break?
CHALLENGE PROBLEM:
5). A mechanic needs to torque a mounting bolt (18mm, quality 8.8) to 225 lb-ft. to secure the assembly to the base.
Using an 18 in. long torque wrench, a 4X torque multiplier, and a 1 ft. extension handle, can the mechanic torque this
bolt to the correct value?
NOTES:
The mechanic can generate 75 N-m of torque on a 39 in. torque wrench
The extension handle covers 152.4 mm of the torque wrench handle
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Module 2: Torque
This Concludes
Module 2
Torque
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Module 3: Sling Angles
Module 3
Sling Angles
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Module 3: Sling Angles
Slings
A 1200 pound load is being lifted by wire ropes. The top rope represents a point upon which forces act
in a static system.
Forces F2 and F3 are angular forces that realize both horizontal and vertical forces. We call these
resultant forces.
F1
A
F2
F3
1200 lbs.
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Module 3: Sling Angles
Slings
Laws of physics say that the total vertical downward forces must equal the total of vertical upward forces.
In this example:
F1 = F2 + F3
F1 = 1200 lbs.
F2 = 600 lbs.
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F3 = 600 lbs.
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Module 3: Sling Angles
Slings
Let’s look closer at F3 resultant. Laws of physics also say that any resultant (angular) force can be
divided into horizontal and vertical components. Therefore, f3 resultant can be divided into F3 vertical
and F3 horizontal.
F3 Horizontal
F3 Resultant
F3 Vertical
Finding F3 resultant now becomes a simple matter of solving a right triangle.
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Module 3: Sling Angles
Slings
If we know the vertical force of one sling (F3 vertical) to be 600 pounds and angle A to be 80 degrees,
the following is true.
40 deg.
40 deg.
F1
1/2 Angle A
600 lb.
600 lb.
A
F2
F3
F3v
F3r
1200 lb.
Therefore:
Cosine
Cosine 40 deg.
F3r =
=
600lb
cos.40deg.
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angle A /2 =
F3v
F3r
600lb.
F3r
F3r = 783.24 lbs.
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Module 3: Sling Angles
Slings
What happens if we change angle A to 160 degrees?
Cosine 80 deg.
=
600lb.
F3r
F3r
=
600lb
cos.80deg.
F3r
= 3455.27 lbs.
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Module 3: Sling Angles
Slings
1) Determine the tension in each sling.
30 deg.
4300 lbs.
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Classification : D3
Conservation :
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Module 3: Sling Angles
Slings
2) Determine the tension in each sling.
60 deg.
6800 lbs.
01 – Applied Forces
Presentation : IMS – Tech Managers Conference
Author : IMS Stafff
Author : IMS Staff
Creation date : 5 Nov 2012
Creation date : 08 March 2012
Classification : D3
Classification : D3
Conservation :
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Module 3: Sling Angles
Slings
3) Determine the tension in each wire supporting the traffic light.
165 deg.
F = 750 lbs.
01 – Applied Forces
Presentation : IMS – Tech Managers Conference
Author : IMS Stafff
Author : IMS Staff
Creation date : 5 Nov 2012
Creation date : 08 March 2012
Classification : D3
Classification : D3
Conservation :
Page : ‹#›
Module 3: Sling Angles
Slings
4) Determine the included angle between the ropes. The tension on each rope is 900 lbs.
400 kg.
01 – Applied Forces
Presentation : IMS – Tech Managers Conference
Author : IMS Stafff
Author : IMS Staff
Creation date : 5 Nov 2012
Creation date : 08 March 2012
Classification : D3
Classification : D3
Conservation :
Page : ‹#›
Module 3: Sling Angles
Slings
4 ft.
5) Determine the tension in each of the 4 slings supporting the load.
3 ft.
35000 lbs.
12 ft.
01 – Applied Forces
Presentation : IMS – Tech Managers Conference
Author : IMS Stafff
Author : IMS Staff
Creation date : 5 Nov 2012
Creation date : 08 March 2012
Classification : D3
Classification : D3
Conservation :
Page : ‹#›
Module 3: Sling Angles
This Concludes
Module 3
Sling Angles
01 – Applied Forces
Presentation : IMS – Tech Managers Conference
Author : IMS Stafff
Author : IMS Staff
Creation date : 5 Nov 2012
Creation date : 08 March 2012
Classification : D3
Classification : D3
Conservation :
Page : ‹#›