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Shafts, Belts, Chains, & Clutches. Shafts – Definition. Generally shafts are members which rotate in order to transmit power or motion. They are usually circular in cross section, and that’s the type we will analyze.

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shafts definition
Shafts – Definition
  • Generally shafts are members which rotate in order to transmit power or motion. They are usually circular in cross section, and that’s the type we will analyze.
  • Shafts do not always rotate themselves, as in the case of an axle – but axles support rotating members.

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elements attached to a shaft
Elements Attached to a Shaft

Shoulders provide axial positioning location, & allow for larger center shaft diameter – where bending stress is highest.

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common shaft materials
Common Shaft Materials
  • Typically shafts are machined or cold-drawn from plain hot-rolled carbon steel. Applications requiring greater strength often specify alloy steels (e.g., 4140).
  • Some corrosion applications call for brass, stainless, Ti, or others.
  • Aluminum is not commonly used (low modulus, low surface hardness).

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shafts for steady torsion
Shafts for Steady Torsion

Often the rotating mass & static load on a shaft are neglected, and the shaft is sized simply to accommodate the transmitted power. In such cases, the engineer typically seeks to limit the maximum shear stress max to some value under the yield stress in shear (Sys), or to limit the twist angle .

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shafts in steady torsion
Shafts in Steady Torsion

Chapter 1 review equations:

kW = FV/1000 = Tn/9600 (1.15)

hp = FV/745.7 = Tn/7121 (1.16)

kW = kilowatts of power

F = tangential force (N)

V = tangential velocity (m/s)

T = torque (N x m)

n = shaft speed (rpm)

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sbs power units
SBS Power Units

Chapter 1 review equation:

hp = FV/33,000 = Tn/63,000 (1.17)

where,

hp = horsepower

F = tangential force (lb.)

V = tangential velocity (ft/min)

T = torque (lb - in.)

n = shaft speed (rpm)

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steady state shaft design
Steady State Shaft Design

Because shafts are in torsion, the shear stress is generally the limiting factor. Recall that

max = Tc/J

where c = radius, and, for a circular shaft,

J = d4/32

As always, use a safety factor of n to arrive at

all = max /n

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limiting the twist angle
Limiting the Twist Angle

In some cases it is desired to limit the twist angle to a certain value. Recall from Chapter 4:

 = TL/GJ (4.9)

L = length

G = shear modulus

 is always in radians (deg. x /180)

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example 9 1
Example 9.1

Design a solid shaft to transmit 500 kW at n = 1200 rpm.

Sys = 300 MPa, and G = 80 GPa. Choose n = 1.5, and limit  to < 4o along the 2m length of the shaft.

Remember, n = rpm, while n = safety factor.  must be in radians in eq. 4.9.

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combined static loads
Combined Static Loads

In the preceding example, we considered only the torsional load. Often the shaft loading is more complicated, and both bending and axial loads must be considered as well. Review from Chapter 3:

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combined static load
Combined Static Load

The axial stress is given by:

x = Mc/I + P/A = 32M/ D3 + 4P/ D2

(M = bending moment, P = axial load)

The torsional stress is given by:

xy = Tc/J = 16T/ D3

(T = Torque, J = polar moment of inertia, c = radius)

(For circular cross sections.)

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maximum shear stress theory
Maximum Shear Stress Theory

Typically the axial load P is small compared to the bending moment M and the torque T, and so it is neglected. (Notice how shear is completely omitted.)

Recall the maximum shear stress criterion from the Sept. 26th lecture:

Sy/n = (x2 + 4 xy 2)1/2(eq. 7.11)

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maximum shear stress theory1
Maximum Shear Stress Theory

Substitute the previous values for x and xyinto eq. 7.11 to obtain:

Sy/n = (32/ D3)[M2 + T2]1/2

This equation, or the related eq. for the maximum energy of distortion theory, is useful for finding either D or n.

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fluctuating loads
Fluctuating Loads

In their support of rotating members, most shafts are subject to fluctuating loads, possibly including a shock component as well. We’ve covered fatigue & impact in previous lectures, and that material is directly applied to the design of shafts.

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slide18
Example 9.2

Find required dia. of shaft using MDET & Soderberg fatigue relation. Surface is ground. Su = 810 MPa, and Sy = 605 MPa. Torque varies by +/- 10%. The fatigue stress [] factor Kf = 1.4. Temp = 500 oC, and n = 2. Survival rate = 50%.

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shock factors
Shock Factors

In shaft design, shock loading is typically accounted for by yet more fudge factors, Ksb (bending shock) and Kst (torsional shock). The values of these factors range from 1.0 to 2.0. The shock factors are applied to their respective stress components.

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critical speeds of shafts
Critical Speeds of Shafts

All structures exhibit one or more natural, or resonant frequencies. When a shaft rotates at speeds equal or close to the natural frequencies, resonance may occur. This is usually to be avoided, although some designs feature resonance.

Generally the designer tries to keep the speed at least 25% lower than o. But in some cases, the operating speed is higher.

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the rayleigh equation
The Rayleigh Equation

ncr = (1/2)[ (gW)/(W2)]1/2

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example 9 4
Example 9.4

d = 30mm, D = 50mm, E = 210 Gpa. Find ncr

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shaft attachments
Shaft Attachments
  • Many different methods, each with pros and cons of both function, ease of use, and cost: the designer must balance between these factors.
  • Some methods are very weak compared to the shaft (e.g., a set screw), others are stronger than the shaft itself.

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shaft attachments pins
Shaft Attachments: Pins

Straight Tapered Roll

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shaft attachments tapered clamps
Shaft Attachments: Tapered Clamps

www.ringfeder.com

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stresses in keys
Stresses in Keys

Distribution of force is quite complicated. The common assumption is that the torque T is carried by a tangential force F acting on radius r:

T = Fr

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stresses in keys1
Stresses in Keys
  • From T = Fr, both shear and compressive bearing stresses may be calculated from the width and length of the key.
  • The safety factor ranges from n = 2 (ordinary service) to n = 4.5 (shock).
  • The stress concentration factor in the keyway ranges from 2 to 4.

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problem 9 20
Problem 9.20

A 3/8 x 3/8 x 3” long key holds a 3” long hub onto a 1 ½” dia. shaft. The key and the shaft both have an allowable stress of 10 ksi. What is the factor of safety against shear failure if the transmitted torque is 3.5 kip-in.?

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splines
Splines

Splines permit axial motion between matching parts, but transmit torque. Common use is automotive driveshafts – check your R/C car.

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couplings
Couplings
  • In many designs involving shafts, two shafts must be connected co-axially. Couplings are used to make these connections.
  • Couplings are either rigid or flexible. Rigid couplings require very close alignment of the shafts, generally better than .001” per inch of separation.

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rigid couplings sleeves
Rigid Couplings – Sleeves

The simplest type of coupling is the simple sleeve coupling. But this also has the lowest torque capacity.

http://www.grainger.com/Grainger/wwg/start.shtml

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rigid couplings flanged
Rigid Couplings - Flanged

Great web resource:

http://www.powertransmission.com/pages/couplings.htm

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flexible couplings
Flexible Couplings
  • There are many types of flexible couplings as well. Generally something flexible is sandwiched in between, or connected to, rigid flanges attached to each shaft.
  • Alignment is still important! Reaction forces increase with misalignment, and often bearings are not sized

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slide35
Two-piece “Donut” (or toroidal) flexible coupling

http://viva.rexnord.com/content/features.html

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universal joints
Universal Joints
  • U-joints are considered linkages rather than couplings, but serve the same purpose of transmitting rotation.
  • Very large angular displacements may be accommodated.
  • Single joints are not constant-velocity. Almost always, two joints are used. The angles must be equal for uniform velocity.

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slide37
Shafts parallel

but offset

Shafts not parallel but intersecting

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it s not nanotechnology but you could get rich
It’s Not Nanotechnology, But You Could Get Rich!
  • Despite decades of research and 1000s of Ph.D. theses, even “highly engineered” shafts fail all too frequently. Even NASA can’t always get it right.
  • Often the connections are to blame: keys, splines, couplings, and so on. Often fatigue wear failure is the culprit.
  • The world may beat a path to your door!

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purchased vs designed components
Purchased vs. Designed Components
  • We went into shafts in some detail because shafts tend to be custom-designed for each application.
  • Often the components that the engineer puts onto the shaft, however, are purchased. These components can be analyzed as much as one likes, but it’s usually best to work with the manufacturer’s application data.

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belts and chains
Belts and Chains
  • Economical alternatives to gears for transmitting rotary motion between shafts. Can be lighter, smaller, and as efficient as gears.
  • However, usually the life is more limited.
  • Bicycle “gears” (chain drive) versus automotive gears.

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timing belts
Timing Belts
  • Toothed or synchronous belts don’t slip, and therefore transmit torque at a constant ratio: great for applications requiring precise timing, such as driving an automotive camshaft from the crankshaft.
  • Very efficient. More $ than other types of belts.

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flat round and v belts
Flat, Round, and V Belts
  • Flat and round belts work very well. Flat belts must work under higher tension than V belts to transmit the same torque as V belts. Therefore they require more rigid shafts, larger bearings, and so on.
  • V belts create greater friction by wedging into the groove on the pulley or sheave. This greater friction = great torque capacity.

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v belt sheave cross section
V Belt & Sheave Cross Section

The included angle 2 ranges between 34o and 40o.

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flat belts vs v belts
Flat Belts vs. V Belts
  • Flat belt drives can have an efficiency close to 98%, about the same as a gear drive.
  • V belt drive efficiency varies between 70% and 96%, but they can transmit more power for a similar size. (Think of the wedged belt having to come un-wedged.)
  • In low power applications (most industrial uses), the cheaper installed cost wins vs. their greater efficiency: V belts are very common.

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flat belt drive example
Flat Belt Drive Example

Chapter 13, Problem 1

A flat belt 4 in. wide and in. thick operates on pulleys of diameters 5 in. and 15 in. and transmits 10 hp. Determine:

 (a) The required belt tensions.

  (b) The belt length.

  Given: the speed of the small pulley is 1500 rpm, the pulleys are 5 ft apart, the coefficient of friction is 0.30, and the weight of the belt material is 0.04 lb/in.3.

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flat belt drive example1
Flat Belt Drive Example

T = (F1 – F2)r (eq. 13.1)

hp = (F1 – F2)V/33,000 = Tn/63,000 (eq. 13.3)

sin = (r2 – r1)/c (eq. 13.6)

 =  – 2  (eq. 13.7)

Fc = (w/g)V2 (eq. 13.13)

F1 = Fc + [ /(-1)](T1/r1) (eq. 13.20)

 = ef/sin (eq. 13.21)

(Where  = 90o for a flat belt)

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chains
Chains
  • Compared to belts, chains can transmit more power for a given size, and can maintain more precise speed ratios.
  • Like belts, chains may suffer from a shorter life than a gear drive. Flexibility is limited by the link-length, which can cause a non-uniform output at high speeds.

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chains1
Chains
  • Chain drives can be very efficient. Bicycle example.
  • The fact that the user controls the length (with master links) is a plus. However, the sprockets wear out much more frequently than does a belt sheave. Take your pick!

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chain sizes u s
Chain Sizes (U.S.)
  • The “chain number” is nominally the roller-to-roller pitch in 1/80 inch increments.
  • Size 40 chain = ½” pitch ~ bike chain.
  • Size 80 chain = 1” pitch.
  • Size 120 chain = 1 ½” pitch.

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clutches brakes
Clutches & Brakes
  • Both use friction to control rotational power.
  • Clutches are used to couple & decouple rotating members; typically a power source from the rest of a machine. Auto example.
  • Brakes are used to dissipate rotational energy.

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friction materials
Friction Materials
  • Clutches and brakes depend on friction to operate. Typically one surface is metal, either steel or cast iron. The other surface is usually of a composite nature, for example soft metal particles embedded with reinforcing fibers in a bonding matrix.
  • Conflicting requirements of minimal wear, but acceptable f.

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single plate disc clutch
Single Plate Disc Clutch

(for yoke shifter)

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wet clutches
“Wet” Clutches
  • Why on earth would an engineer design a clutch where the plates operate in an oil bath? Isn’t friction the idea?
  • Cooling, smooth operation (no ‘grabbing,’ and reduced wear, that’s why.
  • True that f is reduced and so sizes must be increased – but a worthwhile tradeoff.

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simple band brake
Simple Band Brake

Very similar to a belt drive; torque capacity is T = (F1 – F2)r

Equations 13.42 through 13.45 in text

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differential band brake
Differential Band Brake

The friction force helps to apply the band: therefore it is “self-energizing.”

Can become self-locking:

Fa = (1/a)(cF2 – sF1)

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short shoe drum brakes
Short-Shoe Drum Brakes

If the shoe is short (less than 45o contact angle), a uniform pressure distribution may be assumed which simplifies the analysis in comparison to long-shoe brakes.

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self energizing self locking brakes
Self-Energizing & Self-Locking Brakes

If the rotation is as shown, then

Fa = (Fn/a)(b – fc).

If b <= fc, then the brake is self-locking.

Think of a door stop, that is a self-locking short shoe brake.

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long shoe drum brakes
Long-Shoe Drum Brakes

Cannot assume uniform pressure distribution, so the analysis is more involved.

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internal long shoe drum brakes
Internal Long-Shoe Drum Brakes

Formerly in wide automotive use; being replaced by caliper disc brakes, which offer better cooling capacity (and many other advantages).

How many HP can a set of brakes dissipate?

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slide68
Chapter 13, Problem 19

The band brake shown in Figure P13.18 has a power capacity of 40 kW at 600 rpm. Determine the belt tensions.

Given: = 250, r = 250 mm,

a = 500 mm, and f = 0.4.

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chapter 13 problem 19
Chapter 13, Problem 19

Torque T = (9549 x kW)/n

= (9549 x 40)/600 = 636.6 N-m (eq. 1.15)

F1 = F2e f, (eq. 13.44)

where  is in radians, so

F1 = F2e (.4)(4.363) = 5.727F2

T = (F1 – F2)r (eq. 13.42),

Or, T = (.25)(5.727F2 – F2) = 1.182F2

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chapter 13 problem 191
Chapter 13, Problem 19

Therefore, F2 = 538.6 N, and,

F1 = 3,085 N

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