Double Roll Crusher Machine Design

COLLEGE OF ENGINEERING AND ARCHITECTURE MECHANICAL ENGINEERING DEPARTMENT NEGROS ORIENTAL STATE UNIVERISTY MAIN CAMPUS II, BAJUMPANDAN DUMAGUETE CITY

DOUBLE ROLL CRUSHER DESIGN

PRESENTED TO ENGR. ELIJAH SERATE

IN PARTIAL FULFILLMENT OF THE REQUIREMENTS IN ME 421- MACHINE DESIGN 2

BY ROMMEL LITO C. NEPALAR

2012

[DOUBLE ROLL CRUSHER DESIGN]

CONTENTS: INTRODUCTION…………………………………………………………………………………………………………………………3 REVIEW OF RELATED LITERATURE……………………………………………………………………………………………..4 Other Crusher Types…………………………………………………………………………………………………….4 Single Roll Crusher……………………………………………………………………………………………4 Jaw Crusher……………………………………………………………………………………………………..5 Brad-multi Roll Crusher…………………………………………………………………………………….5 Clinker Crusher…………………………………………………………………………………………………6 DRAWING…………………………………………………………………………………………………………………………………..7 Isometric View (a)…………………………………………………………………………………………………………7 Isometric View (b)…………………………………………………………………………………………………………8 Top View……………………………………………………………………………………………………………………….9 Front View……………………………………………………………………………………………………………………10 Side View……………………………………………………………………………………………………………………..10 DESIGN CONSIDERATION………………………………………………………………………………………………………….11 DESIGN CALCULATION………………………………………………………………………………………………………………11 Solving for Radius of the Rolls …………………………………………………………………………11 Solving for Feedrate…………………………………………………………………………………………12 Solving for the Crusher Power…………………………………………………………………………12 Solving for the Force Required………………………………………………………………...……..13 Solving for Torque……………………………………………………………………………………………13 Design Horsepower to drive mechanism including losses………………………………14 Transmitted Power on V-belt………………………………………………………………………….14 DRIVING SYSTEM DESIGN………………………………………………………………………………………………………….14 V-belt Design………………………………………………………………………………………………………………..14 Solving for Design Hp……………………………………………………………………………………….14 Solving for the Diameter of the Drive and Driven Sheave……………………………….15 Solving for Center Distance………………………………………………………………………………15 Solving for V-belt Length………………………………………………………………………………….15 Solving for V-belt Speed…………………………………………………………………………………..16 Solving for kd(small diameter factor)……………………………………………………………….16 Solving for Rated Hp………………………………………………………………………………………. 16 Solving for Adjusted Hp……………………………………………………………………………………16 Solving for the Number of Belts………………………………………………………………………17 Solving for the Tension…………………………………………………………………………………….17 Computing the Arc of Contact………………………………………………………………………….18 Power Transmitted on V-belt…………………………………………………………………………..18 Solving for Thickness of Sheave……………………………………………………………………….18 Computing the weight of the Motor Sheave…………………………………………………..19 Computing the weight of the Driven Sheave…………………………………………………..19 SHAFT DESIGN…………………………………………………………………………………………………………………………..19 Shaft 1………………………………………………………………………………………………………………………… 19 Computing the Angle of Wrap………………………………………………………………………….20 Solving for Forces on Belt at Section A…………………………………………………………….20 Forces at Section A due to Weight of the Sheave, W2……………………………………..21

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Total Load at Section A…………………………………………………………………………………….21 Solving for Weight of the Roll………………………………………………………………………….21 Solving for the Force Required to Crush the Coal (Fr)…………………………………….22 Solving for the Vertical Component of the Bearing ………………………………………..23 Solving for the Horizontal Component of the Bearing …………………………………….24 Data for the Loading at Shaft 1………………………………………………………………………..26 Data for the Moment at Shaft 1……………………………………………………………………….26 Bearing Selection for Shaft 1……………………………………………………………………………27 Design for Key for the Pulley at Shaft 1…………………………………………………………………………28 Solving for Maximum Torque…………………………………………………………………………..29 Solving for the Length of Key……………………………………………………………………………30 Bolt Design for the Bearings on Shaft 1…………………………………………………………………………30 Solving for the Size of the Bolt…………………………………………………………………………31 Shaft 2………………………………………………………………………………………………………………………….31 Solving for the Vertical Component of the Bearing ……………………………………… 32 Solving for the Horizontal Component of the Bearing………………………………………33 Data for the Loading at Shaft 2………………………………………………………………………..33 Data for the Moment at Shaft 2……………………………………………………………………..34 Bearing Selection for Shaft 2…………………………………………………………………………..35 Bearing Housing Design on Shaft 2………………………………………………………………….35 Bearing Housing Design on Shaft 2………………………………………………………………….36 Spring Design……………………………………………………………………………………………………………….37 Solving for Spring Index………………………………………………………………………………….37 Solving for Stress Factor………………………………………………………………………………….37 Stress Because of the Load……………………………………………………………………………..38 Solving the Number of Active Coils…………………………………………………………………38 Solving for Solid Height……………………………………………………………………………………38 Scale of Spring (k)……………………………………………………………………………………………39 Force to Compress the Spring to Solid Height……………………………………….........39 Permissible Solid Stress………………………………………………………………………………….39 Solving for Pitch………………………………………………………………………………………………39 Spring Mounting……………………………………………………………………………………………..40 Bolt Design for Mounting the Crusher…………………………………………………………………………40 Solving for the Maximum Tensile Force………………………………………………………….41 Solving for Fm......................................................................................................42 Solving for Fr ……………………………………………………………………………………………………43 DESIGN SUMMARY……………………………………………………………………………………………………………………44 LIST OF REFERENCE……………………………………………………………………………………………………………………46 APPENDIX………………………………………………………………………………………………………………………………….47

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INTRODUCTION Double roll crushers consist of two adjacent rolls placed parallel to each other and rotated in opposite directions. These are typically used in situations in which fines are to be minimized. They are widely employed on friable materials such as coal, lime, limestone, petroleum coke, and chemicals. As the two rolls rotate toward each other, the material is pulled down into the crushing zone where it is grabbed and compressed by the rolls. Product size is determined by the size of the gap between the rolls, and this gap can be changed to vary product size or to compensate for wear. Since both rolls rotate at the same speed, there is no relative motion between the two roll surfaces, and crushing is primarily accomplished by compression. Compression crushing is extremely efficient, as energy is only used to crush those particles larger than the gap between the rolls. Fines are reduced because already sized material passes freely through the crusher with no further reduction. Protection from uncrushables is provided by means of a retractable roll assembly. It retracts instantly when an uncrushable is encountered, then reverts to its original position once the uncrushable has cleared the crushing chamber with no stoppage of the crusher.

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REVIEW OF RELATED LITERATURE OTHER CRUSHER TYPES Single Roll Crusher Single Roll Crushers are typically used as primary crushers. A single roll crusher has a roll assembly consisting of a roll shaft and a fabricated roll shell with integral fixed teeth. In the single roll crusher, three different methods of reduction occur: impact, shear and compression. Entering the crusher through the feed hopper, the feed material is struck by the teeth of the revolving roll. While some breakage occurs here by impact, the rotation of the roll carries the material into the crushing chamber formed between the breaker plate and the roll itself. As the turning roll compresses the material against the stationary breaker plate, the teeth on the roll shear the material. Sized material falls directly out through the discharge end of the crusher which is completely open. There are no screen bars, and consequently there is no recrushing of the sized materials, a factor that helps to reduce power demand while minimizing product fines. The clearance between the breaker plate and the roll determines the product size. This clearance is adjustable from outside the machine by a shim arrangement. Adding or removing shims causes the plate to pivot about its top hinge, moving it into or away from the roll. For protection against uncrushable debris, the breaker plate assembly is secured with an automatic release device. As pressure from the uncrushable is exerted against the plate, the device allows the entire breaker plate assembly to move away from the roll instantly. The uncrushable drops clear of the machine by gravity, and the breaker plate assembly immediately returns to its normal crushing position.

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Jaw Crusher For crushing of hard, abrasive materials, Jaw Crushers are often preferred, since this type of machine will crush virtually any mineral. Jaw Crushers differ substantially from other types of crushers. There is no rotary motion in the crushing cycle, and all crushing is done by compression of the feed material between two massive jaws, which in effect are a type of breaker plate. Jaw plates can be either smooth or corrugated. While one jaw is fixed, the other jaw pivots about a top hinge. This moving jaw is shaped to move firmly and squarely against the material, at 250 to 400 strokes per minute. There is no rubbing or grinding, only compression, which produces a generally cubical product with minimum fines. The moving jaw is so balanced that fully 95% of the drive motor power is used for crushing, while only 5% of the power is needed to move the jaw itself. As a result of this high mechanical efficiency, smaller motors may be used, keeping power costs down. Behind the stationary jaw are shims, used to compensate for plate wear and to adjust the closed side setting. For protection from uncrushables, there is also an automatic drive disengagement feature that acts instantaneously on the moveable jaw assembly.

Brad-multi Roll Crusher The Brad-Multi-Roll Crusher was introduced to produce material with output smaller than 1/4" (6mm) while producing a minimum of extreme fines. This produces a product having a very steep gradation curve, making it ideally suited for preparation of fuel and sorbents for fluid bed boilers and for other

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applications that require such a gradation. This machine is capable of handling moist materials when operated in open circuit. Drying is usually recommended when closed-circuit operation is needed to meet a specific gradation curve.

The Brad-Multi-Roll Crusher is well suited to abrasive materials. To compensate for any reduction in charge level resulting from wear, additional charge can simply be added. This greatly simplifies maintenance, and provides a machine with exceptionally high availability. For some closed-circuit applications, integral screening can be accomplished. The great compactness of this design means that a smaller building is required, while minimizing the need for peripheral equipment. Optional sound housings can be provided to control noise emissions. When compared with other machines that require air classification to achieve specified product sizes, users of the Brad-Multi-Roll Crusher will enjoy significant and continuing savings in power costs.

Clinker Crusher The Clinker Crusher is used for handling of bottom ash, though it can readily be configured to handle other materials. It is now widely employed as a direct replacement for the most common makes and sizes of clinker grinders. Its efficient design improves maintainability and increases component life, thereby reducing both operation and maintenance costs. For example, the cast, high chrome roll segments are reversible to maximize their wear life. These segments can also be changed out with the unit in place by means of a large door in the rear of the frame. For retrofit, no changes to foundations are needed under normal circumstances.

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DRAWING:

Isometric View (a)

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Isometric View (b)

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Top View

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Front View

Side View

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DESIGN CONSIDERATION: Roll Crusher Type: Double Roll Crusher Material to be crushed: Bituminous coal, Broken Maximum Feed Size: 2 3/8 in. Distance Between rolls: 1 ½ in. Operating Condition: dry condition Further specifications and design considerations were made along the design process.

DESIGN CALCULATION: Solving for Radius of the Rolls: (assuming all surfaces are smooth and particle to be crushed is spherical) To solve for the radius of the rolls, it is convenient to assume that the particle to be crushed is spherical and roll surfaces are smooth. The figure below shows a spherical particle about to enter the crushing zone of a roll crusher. The nip angle is defined as the angle that is tangent to the roll surfaces at the points of contact between the rolls and the particle. Usually the nip angle is between 20˚ and 30˚ but in some large roll crushers it is up to 40˚. For the design I choose 20˚ nip angle.

Where: D/2

R= radius of the roll L= distance between rolls, 1 ½ in. ϴ d= diameter of the feed, 2 3/8 in. ϴ= Nip angle, 20˚ (

)

R

L

R= 6.067001948 in. D= 12.1340039

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For the roll diameter, the calculated diameter is not available in the market so I use D = 15” available at http://www.gundlachcrushers.com/crushers/roll-crushers-coal-salts-limeminerals.cfm with a chisel tooth profile.

Solving for Feedrate: Q= π60DWωLρB (t/h) Where: Q= crusher capacity D= diameter of roll, m (15 in= 0.381 m) W= width of the roll, m (50 in = 1.27 m) ω= roll speed, 130 rpm L= distance between rolls, m (1 ½ in = 0.0381 m) ρB = bulk density of feed material, t/m3 (.833 t/m3) For the ρB of coal, refer to http://wiki.answers.com/Q/What_is_the_density_of_coal. Q= π (60)(0.381)(1.27)(130)(0.0381)(.833) Q= 376.3077383 t/h

Solving for the Crusher Power From: http://www.gundlachcrushers.com/crushers, with data from a 2000 Series Roll Crusher I can get the power required for the roll. For a 2000 Series Roll Crusher     

Single-stage and two-stage models 15" diameter rolls (380 mm) Roll face up to 60" (1525 mm) wide

Motor Power, 20 kw (26.80965147 hp) Capacity up to 400 tph

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From the Design Hp Formula: Design Hp =Transmitted Hp x Nsf Where: Nsf = 1.4+0.2 (Table 17.7 by Faires, p. 460) Design Hp = 26.80965147 Hp Transmitted Hp = 26.80965147 / 1.6 = 16.75603217 Hp

Comparing the power and the feedrate for commercial crusher. P1/C1 = P2/C2 Where: P1 and C1 = Power and Capacity for the crusher to be design P2 and C2 = Power and Capacity for a commercial crusher = P1 = 15.76356142 hp

Solving for Torque P= 2πTN T= Where: P= 15.76356142 hp N= 130 rpm T=

x

T= 636.8615984 lb-ft

Solving for the Force Required F=T/r F=

x

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F = 1018.978557 lb

Design horsepower to drive mechanism including losses According to Morse, p.452, losses in V-belt is 5%.

Transmitted power on V-belt Transmitted power on V-Belt= Required power to drive the crusher/0.95 = 15.76356142 / 0.95 = 16.59322255 hp

DRIVING SYSTEM DESIGN V-Belt Design Design Consideration: Motor type: squirrel cage motor Type of service: continuous (16 hr/day) service V-belt material: Leather belt From: http://www.engineeringtoolbox.com/electrical-motors-hp-torque-rpm-d_1503.html, Power versus torque and motor velocity in electric motors Table. Nearest capable of driving is: Motor Power = 20 hp RPM= 500 rpm

Solving for Design Hp Design Hp =Transmitted Hp x Nsf Where: Nsf = 1.4+0.2 (Table 17.7 by Faires, p. 460) Design Hp= 16.59322255 x (1.4+0.2) = 26.54915608 Hp (From figure 17.14 Belt Selection from Horsepower and Speed by Faires, p.457, I choose section D)

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Solving for the diameter of the drive and driven sheave N2D2=N1D1 Where: N1= Motor speed, 500 rpm N2= Crusher speed, 130 rpm D2= diameter of the driven sheave D1= diameter of the drive sheave From: Table 17.3, Standard V-belt Lengths; Horsepower Constants (Faires, p. 458) If D2= 48 in. and D1= 13 in. N2= (13x500)/48 N2= 135.4166667 rpm

If D2=58 in. and D1= 16 in. N2= (16x500)/58 N2= 137.9310345 rpm If D2=58 in and D1= 14 in. N2= (14x500)/ 58 N2 = 120.6896552 rpm I choose D2=48 in. and D1= 13 in. which is nearest to 130 rpm.

Solving for center distance (Eq. by Faires, p, 457) C= [(D2+D1)/2]+D1 C= [(48+13)/2]+13

or C =D2 whichever is longer

C= 43.5 in. I choose C = 48 in. which is longer Note: The center distance is made adjustable so that the belts can be mounted into the grooves without harmful stretching and so that initial tension can be maintained (Faires, p. 457).

Solving for belt length (Eq. by Faires, p. 446 ) L = *(π/2)(D2 - D1)] + 2C + [(D2 - D1)2 / 4C] L = *(π/2)(48 - 13)] + 2(48) + [(48 - 13)2 / 4(48)] L=157.3580798 in.

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From table 17.3, Standard V-belt length (by Faires p. 458) at section D, select D158 with pitch length equal to 161.3 in.

Solving for V-belt speed Vm = πD1N1 / 12 = π(13)(500)/12 Vm = 1701.696021 ft/min

Solving for kd (small diameter factor) From: table 17.4, by Faires, p.459 D2/D1 = (48/13)= 3.692307692 Kd= 1.14

Solving for Rated Hp From equation by Faires, p.456 * (

+

)

Where : Vm= belt speed, 1701.696021 ft/min Kd = 1.14 Rated hp constants: a= 18.788 c= 137.7 e= 0.0848 Rated hp constants (from table 17.3, by Faires, p. 458) *

(

)

+

Rated Hp = 9.505200567 Hp

Solving for Adjusted Hp Horsepower rating must be corrected for length of belt (Kl) and arc of contact (K ). (Faires, p.457) Adjusted Hp = kϴxkl (Rated Hp)

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Where: Kl= Length correction factor, 0.92 from table 17.6 Length Correction Factors by Faires, (p.459) kϴ= arc of contact factor From table 17.5, Arc of Contact factors (Faires, p.459) (D2 – D1) / C = (48 – 13) / 48 = 0.729166666 By interpolation, 0.70 0.729166666 0.80

0.89 Kθ 0.87

Kθ = 0.884166668 Therefore, Adjusted Hp = 0.92 x 0.884166668 x 9.505200567 = 7.731846993 Hp

Solving for Number of Belts Number of Belts = Design Hp/ Adjusted Hp = 26.54915608 / 7.731846993 = 3.433740489 Say 4 belts

Use 4 D158 leather V-belts with pitch length equal to 161.3 in. from section D, b x t = 1 ¼ in., 3/4 in., figure 17.14, Belt Section from Horsepower and Speed by Faires, p. 457.

Solving for the Tension F1

F2

F1 / F2 = efθ

Eq. 1

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Where: F1 = total tension on the tight side, lb F2 = total tension on the slack side, lb f = coefficient of friction, 0.25 from Table 16-15 Coefficient of Friction for Belts, for leather on cast iron pulley θ = arc of contact, rad

Computing the arc of contact θ = , 180 – 2sin -1 [ (D2 – D1) / 2C + - (π / 180) = { 180 – 2sin-1 [ (48 – 13) / 2(48) + - (π / 180) = 2.395222057 rad From eq.1 F1 / F2 = efθ F1 = e(.25) (2.395222057) F2 F1 = 1.819943605 F2

Eq.2

Power Transmitted on V-belt From equation 16.7 (Daughtie and Vallance, p.383) Hp = [ (F1 – F2) / 550 ] x Vm Where: Hp= transmitted Hp, 16.59322255 hp F1 =total tension on the tight side, lb F 2= total tension on the slack side, lb νm = belt speed, 28.36153368 ft/sec (F1-F2)= 550(16.59322255)/28.36153368 (F1-F2)= 321.7827023

Eq. 3

Equating equation 2 and 3 1.819943605 F2 – F2 = 321.7827023 F2 = 392.4449198 lbs. F1 = 1.819943605 (392.4449198) F1 = 714.2276221 lbs.

Solving for thickness of the sheave T = width of the belt x number of belt

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= (1 ¼ )x 4 T = 5 in.

Computing the weight of the motor sheave W1 = (π/4) D12tρ Where: D1 = diameter of the small sheave, 13 in. t = thickness of small sheave, 5 in. ρ = density of pulley material used, 0.256 lb/in³ for ordinary cast iron from Table 2-1, General Properties of commonly used materials (Doughtie and Vallance, p.11) W1 = (π/4)(13)2(5)(0.256) W1= 169.8973307 lbs.

Computing the weight of the Driven sheave W2 = (π/4) D22tρ Where: D2 = diameter of the driven sheave, 48 in. t = thickness of sheave, 5 in. ρ = density of pulley material used, 0.256 lb/in³ for ordinary cast iron from Table 2-1, General Properties of commonly used materials (Doughtie and Vallance, p.11) W2 = (π/4)(48)2(5)(0.256) W2 = 2316.233432 lbs

SHAFT DESIGN Shaft 1 Design Consideration: Material Used: AISI C1117 Normalized Carbon Steel Tensile strength: 63,700 psi Yield strength: 35,000 psi From Table AT 8 Typical Properties of Steel (Faires, p.578) Shaft diameter: 4 in

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8"

36"

36"



Fx1

F1

Fy1

F2



Fy2

Fr

Ft

Fx2

Components of shaft 1

Computing the Angle of Wrap Cos α = (D2-D1)/2C from eq. by Spotts, p.293 Where: D₂ = diameter of big sheave, 48 in D₁ = diameter of small sheave, 13 in C = centre distance, 48in Therefore, α= cos-1 α = 68.61805742˚

Solving for Forces on Belt at Section A F1 = 714.2276221 lbs. F2 = 392.4449198 lbs. Fx1 = F1sinα = 714.2276221 (sin 68.61805742˚) Fx1 = 665.0678856 lbs

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Fy1 = F1cosα = 714.2276221 (cos 68.61805742˚) Fy1 = 260.3954756 lbs (downward) Fx2 = F2sinα = 392.4449198 (sin 68.61805742˚) Fx2 = 365.4449198 lbs Fy2 = F2cosα = 392.4449198 (cos 68.61805742˚) Fy2 = 143.0788706 lbs (upward) Horizontal Component, Fx= Fx1 + Fx2 Fx = 665.0678856 + 365.4449198 Fx = 1030.512805 lb (to the right) Vertical Component, Fy = Fy1 - Fy2 Fy = 260.3954756 – 143.0788706 Fy = 117.316605 lbs (downward)

Forces at Section A Due to Weight of the Sheave, W2 W2 = 2316.233432 lbs

Total Load at section A FHA = 1030.512805 lbs FVA = 117.316605 + 2316.233432 FVA = 2433.550037 lbs.

Solving for the weight of the roll Roll material: high carbon steel w= ρ v where: ρ = density of the roll, 0.283244983 lb/in3 (Material Science and Engineering 4th Edition, V. Raghavan, p.396) v= roll volume v= (π/4)(D-d)2w where: D= diameter of the roll, 15 in. d= shaft diameter, 4 in. w= roll width, 50 in. v= (π/4)(15-4)2(50) v= 4751.658889 in3 therefore,

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w = (4751.658889) (.283244983) w = 1345.883541 lbs.

Solving for the Force Required to Crush the Coal (Fr) ς = F/A where: ς = compressive strength of coal, 500 psi, (Table 8.2, Handbook of Coal Analysis by J.G. Speight) A= area of the coal to be crushed Solving for area: e= effective length of the roll, 60%

19/16 x w 3/4

1 1/2

X = √( ) X = 0.920682491 W = 2(0.920682491) W = 1.841364983 Therefore: A = (1.841364983)(50)(0.6) A = 55.24094949 in2 Fr = 500 x 55.24094949 Fr= 27620.47475 lbs. Ft = force required to drive the crusher, 1018.978557 lb Total Roll Load FVC = w - Ft FVC = 1345.883541- 1018.978557 FVC = 326.904984 lbs (downward) FHC = Fr = 27620.47475 lbs

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Solving for Vertical Component of the Bearing Fva

Fvc

8"

36"

36"

Rvb

Rvd

FVA = 2433.550037 lbs FVC = 326.904984 lbs ΣMB = 0 72 Rvd = 8 (2433.550037) - 36(326.904984) Rvd = 106.9419566 lbs ΣY=0 Rvb = Fva + Fvc + Rvd Rvb = 2433.550037 + 326.904984 + 106.9419566 Rvb = 2867.39678 lbs 2433.550037 lbs

6.53809968 lb/in A

B

8"

C

11"

D

E

11"

50"

2867.396978 lbs

106.9419566 lbs

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Shear Diagram 433.846941 lbs 106.941957 lbs

0

2433.550037 lbs

Moment Diagram

0

1176.386495 in-lb 14696.08395 in-lb

19468.4003 in-lb

Solving for Horizontal Component of the Bearing FHA

FHC

8"

36"

36"

RHB

RHD

FHC = 27620.47475 lbs. (left) FHA = 1030.512805 lbs. (right) ΣMB = 0 72 RHD = 36(27620.47475) + 8(1030.512805)

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RHD = 13924.7388 lbs ΣY = 0 RHB + FHC = FHA + RHD RHB = 1030.512805 + 13924.7388 – 27620.47475 RHB = -12665.22315 lbs Therefore: RHB = 12665.22315 lbs (upward) 1030.512805 lbs 552.409495 lb/in

A

B

8"

C

D

11"

E

0

11"

50"

12665.22315 lbs

13924.7388 lbs

Shear Diagram 13695.73596 lbs

1030.512805 lbs

X = 24.79326224 in

13924.7388 lbs

Moment Diagram 328678.1846 in-lb

158897.198 in-lb

153172.1267 in-lb

8244.10244 in-lb

0

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Resultant moment MB = √ MB = 21141.99223 in-lb Mc = √ MC = 159575.3565 in-lb MD= √ MD= 153176.6461 in-lb MF= √ MF= 328678.1846 in-lb Therefore, maximum moment occurs at section F.

Data for the Loading at Shaft 1 Section A B C D

Load Vertical 2433.550037 lbs 2867.396978 lbs 326.904984 lbs 106.9419566 lbs

Horizontal 1030.512805 lbs 12665.22315 lbs 27620.47475 lbs 13924.7388lbs

Resultant 2642.749028 lbs 12985.75538 lbs 27622.40924 lbs 13925.14945 lbs

Data for the Moment at Shaft 1 Section B C D F

Moment Vertical 13468.4003 in-lb 14696.08395 in-lb 1176.38695 in-lb 0

Horizontal 8244.10244 in-lb 158897.198 in-lb 153172.1287 in-lb 328678.1846 in-lb

Resultant 21141.99223 in-lb 159575.3565 in-lb 153176.6461 in -lb 328678.1846 in -lb

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Bearing Selection for Shaft 1 The maximum actual radial load of the bearing is at section D Fr = 13925.14945 lbs. From eq. 9-17 by Doughtie and Vallance, p.209 Fc = (KaKl) KoKpKsKtFr When a rolling bearing turns while receiving a load, a lot of stress is repeatedly placed on the small contact surface of the bearing rings and rolling elements, and the bearing must maintain high precision while rotating. That means bearing materials must satisfy the following demands.      

Must be hard. Rolling fatigue life must be long. Wear must be slight. Must be shock-resistant. Dimensions must not vary largely with the passing of time. Must be economical and easy to machine.

Design Consideration: Material Used for the bearing: High Carbon Chrome Bearing Steel The expected life of operation for the bearings is 10 years for 16 hrs/day operation. Shaft diameter may be reduced to accommodate the bearing. Where: Fc = catalog rating of bearing, lb (Table 9-7 & 9-8) Fr= actual radial load on the bearing, 4643.7678 lbs Ka= application factor taking into account the amount of shock (Table 9-4), 1.0 for uniform and steady load Kl =

, life factor

Ha = desired life of bearing, hours of use Hc= catalog rated life, 10,000 hrs. Krel= reliability factor (Table 9-3), 1.0 Ha = 10(365)(16) Ha = 58400 hrs. Hc =10,000 hrs. Kl =

= 1.800822669

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Ko= oscillation factor, 1.0 for constant rotational speeds of the races Kp = preloading factor, 1.0 for non-preloaded ball bearings Ks =

, speed factor

Na = rotational speed of bearing, 135.4166667 rpm Nc = catalog rated rotational speed, 500 rpm Kr = rotational factor, 1.0 for bearing with fixed outer races and rotating inner races Ks =

= 0.646994673

Kt = thrust factor, 1.0 for no thrust-load component Therefore, FC = (1.0 x 1.800822669 ) x 1.0 x 1.0 x 0.646994673 x 1.0 x 13925.14945 FC = 16224.50736 lbs From Table 9-7 Typical radial capacity Fc of ball bearings (Doughtie and Vallance, p.212). I select SAE 419 Two-row angular type ball bearing with radial capacity (Fc) of 20300 lbs and shaft diameter be reduced to 3.7402 in. From Table 8 outside diameter is equal to 9.8425 in and width 2.1654 in.

Design for Key for the Pulley on Shaft 1 Design consideration of Key: Material used: AISI C1117, Normalized Carbon Steel, same as the material used in shafting Su = 63,700 psi Sy = 35,000 psi From Table AT 19 Key Dimension (Faires, p.594) For shaft diameter of 3.7402 in b= 7/8”

t = 5/8 “

MACHINE DESIGN 2

28

2012


Shearing Stress, Ss Ss= Where: Sy = yield stress, 35,000 psi Fs = factor of safety, 3.0 from Table 1.1 Factors of Safety for repeated, one direction, gradual mild shock (Faires, p.20) Therefore, Ss =

= 5833.3333333 psi

Compressive Stress, Sc Sc= Where: Sy = yield stress, 35,000 psi Fs = factor of safety, 3.0 from Table 1.1 Factors of Safety for repeated, one direction, gradual mild shock (Faires, p.20) Therefore, Sc =

= 11,666.66667 psi

Solving for the Maximum Torque, T Hp = Where: Hp = transmitted power, 16.59322255 hp T = maximum torque N = rpm of the shaft, 135.4166667 rpm Therefore, T= T= T = 643.5654046 ft-lb or 7722.784855 in-lb Then, F= F=

MACHINE DESIGN 2

29

2012


F = 4129.610638 lbs.

Solving for the Length of the Key a. Based on the bearing stress of the shaft and since the shaft and the key have the same material it is also equal to the bearing stress of the key. Sc= 11,666.66667 = L = 1.132693203 in. b. Based on the shearing stress of the key Ss = 5833.3333333 = L = 0.809066574 in. Therefore, I use L =1.132693203 in. and

in x in Key.

Bolt Design for the Bearings on Shaft 1 Bearing housing design is a four bolt flange housing design.

Design Condition: Material Used: AISI C1117, Normalized Carbon Steel Sy : 35,000 psi Su : 63,700 psi From Table AT 8 Typical Properties of Steel (Faires, p.578) Number of bolts: 4 bolts Fs = 3.0, from Table 1.1 Factors of Safety, based on yield strength of steel, ductile metals (Faires, p.20)

MACHINE DESIGN 2

30


2012

To compute for the size of bolt I will use the resultant force that will cause shearing (F = 27622.40924 lbs) divided by the number of bolts.

Solving the Size of the Bolt Based on Shearing stress Ss = Where: Sy = yield stress, 35,000 psi Fs= factor of safety, 3.0 Ss= Where: F = maximum shear force, 27622.40924 lbs A= cross sectional area of the bolt Therefore, =

(

)

=

(

)

d= 1.227714667 in. From Table 6-1 Unified & American National threads, coarse, fine, and extra-fine series (Doughtie and Vallance, p.130) I select 1 in-11 UNC bolt with Ar = 0.969 in2 and minor diameter of 1.0747 in. The bolt being screwed into the post is the threaded part.

Shaft 2 Design Consideration: Material Used: AISI C1117 Normalized Carbon Steel Tensile strength: 63,700 psi Yield strength: 35,000 psi From Table AT 8 Typical Properties of Steel (Faires, p.578) Shaft diameter: 4 in.

MACHINE DESIGN 2

31

2012


36"

36"

Fr

w = 1345.883541 lbs Fr = 27620.47475 lbs Total Load FvB = 1345.883541 lbs FHB = 27620.47475 lbs

Solving for the Vertical Component of the Bearing FVB

36"

36"

RvC

RVA

RVA = RVC = 1345.883541/ 2 RVA = RVC = 672.417705 lbs

MACHINE DESIGN 2

32

2012


26.91767082 lbs

B

A

C

11"

D 11"

50"

672.417705 lbs

672.417705 lbs


o X= 25 in

672.417705 lbs


7396.594755 in-lb

7396.594755 in-lb

o

Solving for the Horizontal Component of the bearing FHB

36"

RHA

36"

RHC

FHB = 27620.47475 lbs

MACHINE DESIGN 2

33

2012


RHA = RHC = 27620.47475 / 2 RHA = RHC = 13810.23738 lbs 552.409495 lb/in

A

B

C

11"

50"

D 11"

13810.23738 lbs

13810.23738 lbs


0

X = 25 in 13810.23738 lbs


151912.6112 in-lb 151912.6112 in-lb

0

Resultant Moment ME = √ ME = 324925.0439 in-lb MB = MC =√ MB = MC = 152092.574 in-lb maximum moment occurs at section E.

Data for the Loading at Shaft 2 Section A B C

Load Vertical 672.417705 lbs 1345.883541 lbs 672.417705 lbs

Horizontal 13810.23738 lbs 27620.47475 lbs 13810.23738 lbs

Resultant 13826.59763 lbs 27653.24625 lbs 13826.59763 lbs

MACHINE DESIGN 2

34

2012


Data for the Moment at Shaft 2 Section E B or C

Moment Vertical 15801.81607 in-lb 7396.594755 in-lb

Horizontal 324540.578 in-lb 151912.6112

Resultant 324925.0439 in-lb 152092.574 in-lb

Bearing Selection for Shaft 2 The maximum actual radial load of the bearing is at section A or C. Fr = 13826.59763 lbs. From eq. 9-17 by Doughtie and Vallance, p.209 Fc = (KaKl) KoKpKsKtFr When a rolling bearing turns while receiving a load, a lot of stress is repeatedly placed on the small contact surface of the bearing rings and rolling elements, and the bearing must maintain high precision while rotating. That means bearing materials must satisfy the following demands.      

Must be hard. Rolling fatigue life must be long. Wear must be slight. Must be shock-resistant. Dimensions must not vary largely with the passing of time. Must be economical and easy to machine.

Design Consideration: Material Used for the bearing: High Carbon Chrome Bearing Steel The expected life of operation for the bearings is 10 years for 16 hrs/day operation. Shaft diameter may be reduced to accommodate the bearing. Where: Fc = catalog rating of bearing, lb (Table 9-7 & 9-8) Fr= actual radial load on the bearing, 1794.885353 lbs Ka= application factor taking into account the amount of shock (Table 9-4), 1.0 for uniform and steady load Kl =

, life factor

MACHINE DESIGN 2

35

2012


Ha = desired life of bearing, hours of use Hc= catalog rated life, 10,000 hrs. Krel= reliability factor (Table 9-3), 1.0 Ha = 10(365)(16) Ha = 58400 hrs. Hc =10,000 hrs. Kl = = 1.800822669 Ko= oscillation factor, 1.0 for constant rotational speeds of the races Kp = preloading factor, 1.0 for non-preloaded ball bearings Ks =

Ks =

, speed factor Na = rotational speed, 135.4166667 rpm Nc = catalog rated rotational speed, 500 rpm Kr = rotational factor, 1.0 for bearing with fixed outer races and rotating inner races = 0.646994673

Kt = thrust factor, 1.0 for no thrust-load component Therefore, FC = (1.0 x 1.800822669 ) x 1.0 x 1.0 x 0.646994673 x 1.0 x 13826.59763 FC =16109.71837 lbs From Table 9-7 Typical radial capacity Fc of ball bearings (Doughtie and Vallance, p.212). I select SAE 419 Two-row angular type ball bearing with radial capacity (Fc) of 20300 lbs and shaft diameter be reduced to 3.7402 in. From Table 8 outside diameter is equal to 9.8425 in and width 2.1654 in.

Bearing Housing Design on Shaft 2

Material Used: AISI C1117, Normalized Carbon Steel

MACHINE DESIGN 2

36

2012


SPRING DESIGN

Design Consideration: Load: F = 27620.47475 lbs (based on horizontal reaction for shearing) Number of Spring: 8 Load = 27620.47475 / 8 Load = 3452.559344 lbs Spring material: Hard drawn Wire, squared and ground ends Type of service: average service From Table 9 Century Spring Catalog, I use 4053 spring with wire diameter, Dw = 1.00 in with maximum suggested load = 4200 lbs Free Length = 10 in Dm = 3 in Deflection = .4 in

Solving for Spring Index C = Dm/Dw C=3/1 C=3

Solving for Stress Factor K= K=

+ +

MACHINE DESIGN 2

37

2012


K= 1.58

Stress Because of the Load Ss = k Ss = 1.58 Ss = 38.67346463 ksi Design stress Ssd = 0.324 Su Where: Su = 140/ Dw0.19 Ssd = 47.304/Dw 0.19

(.85)

Ssd = 40.2084/10.19 Ssd = 40.2084 ksi Ssd > Ss (Applicable)

Solving the Number of Active Coils Nc = Nc =

(

)

Nc = 6.168263649 or 6 active coils

Solving for Solid Height At Table At 16 Approximate Free Length and Solid Heights (Faires, p.589) Solid Height = DwNc + 2Dw = (1)(6) + 2 (1) = 8 in

MACHINE DESIGN 2

38

2012


Scale of Spring (k) k= F/δ k= 3452.559344/.4 k= 8631.39836 lb/in

Force to Compress the Spring to Solid Height Fc = k (free length – solid height) Fc = (8631.39836)(10 – 8) Fc = 17262.79672 lbs Solving For Solid Stress Ssolid stress = (Ss/F)(Fc) Ssolid stress = (38.67346463/3452.559344)(17262.79672) Ssolid stress = 193.3673232 ksi

Permissible Solid Stress Ss = Where: Q = 70 X= 0.19 Ss = Ss = 70 ksi. The spring would take a permanent set if compressed to solid height.

Solving for Pitch At Table At 16 Approximate Free Length and Solid Heights (Faires, p.589) Free length = PNc + 2Dw 10 = P (6) + 2 (1) P = 1.33333333 in

Solving for Pitch Angle Pitch Angle, λ = tan-1 (1.33333333/2π) λ = 11.98081357˚

MACHINE DESIGN 2

39

2012


Spring Mounting

Bolt design for Mounting the Crusher Design Condition: Material Used: AISI C1117, Normalized Carbon Steel Sy : 35,000 psi Su : 63,700 psi

MACHINE DESIGN 2

40

2012


Solving for the Maximum Tensile Force

D= 1.5 in

1030.512805 lbs

4

D=1.5 in

3

8 in

ϴ 19 in

69 in

B r Fr Fd

1

Fm

2

58 in

TOP VIEW

1030.512805 lbs

25 in

69 in

1.5 in

T1 = T2 T3 = T4 = (1.5/70.5) F1

MACHINE DESIGN 2

41

2012


ΣMo = 0 1030.512805 x 25 = 2 (70.5) T1 + 2 (1.5) T3 1030.512805 x 25 = 2 (70.5) T1 + 2 (1.5) (1.5/70.5) T1

T1 = 182.6323599 lbs

Max т = √(

)

Where: ϴx = TA/ Ar ϴy = 0 = Fr / n Ar Max т = N = number of bolts Ar = stress area Fs = 3.0, from Table 1.1 Factors of Safety, based on yield strength of steel, ductile metals (Faires, p.20)

Solving for Fm r=√ r = 45.06939094 in 1030.512805 (19) = 4 (45.06939094) Fm Fm = 108.6088745 lbs Fm Components: tanϴ = 29/34.5

MACHINE DESIGN 2

42


2012 ϴ = 40.46222749˚

Fmy = 108.6088745 sin 40.46222749˚ Fmy = 70.4813605 lbs Fmx = 108.6088745 cos 40.46222749˚ Fmx = 82.63331921 lbs

Solving for Fr Fr = √ Fr = 1104.090756 lbs

= √(

5833.3333333 =

)

+

Ar = 0.263546125 in2 From Table 6-1 Unified & American National threads, coarse, fine, and extra-fine series (Doughtie and lVallance, p.130) I select ¾ in-10 UNC bolt with Ar = 0.334 in2 and minor diameter of 0.6273 in.

MACHINE DESIGN 2

43

2012


DESIGN SUMMARY Elements Motor V-belt

Sheave of V-belt

Shaft 1

Bearing on Shaft 1

Key at Shaft 1

Bolt on Bearing @ Shaft 1

Shaft 2

Bearing on Shaft 2

Specifications 20 Hp 500 rpm C = 48 in L = 161.3 in Vm = 1701.696021 ft/min N = 4 leather belts b x t = 1 ¼ x ¾ in D1 = 13 in D2 = 48 in T = 5 in C = 48 in N1 = 500 rpm N2 = 135.4166667 rpm AISI C117 Normalized Carbon Steel D = 4 in L = 80 in High Carbon Chrome Bearing Steel Ha = 58400 hrs. SAE 419 Two-row Angular Type Ball Bearing Bore = 3.7402 in AISI C117 Normalized Carbon Steel b = 7/8 in t = 5/8 in L= 1.132693203 in AISI C117 Normalized Carbon Steel Ar = 0.969 in2 N= 4 bolts 1 ¼ in-11 UNC bolt AISI C117 Normalized Carbon Steel D = 4 in L = 72 in High Carbon Chrome Bearing Steel Ha = 58400 hrs. SAE 419 Two-row Angular Type Ball Bearing Bore = 3.7402 in

Loadings

HpTransmitted = 16.59322255

W1 = 169.8973307 lbs W2 = 2316.233432 lbs

@ section F, Mmax = 328678.1846 in-lb T = 7722.784855 in-lb Fr = 13925.14945 lbs

T = 7722.784855 in-lb F = 4129.610638 lbs

F = 27622.40924 lbs

@ section E Mmax = 324925.0439 in-lb

Fr = 13826.59763 lbs

MACHINE DESIGN 2

44

2012 Spring

Bolt on Crusher

[DOUBLE ROLL CRUSHER DESIGN] Hard drawn wire Squared and Ground ends Dw = 1 in Free length = 10 in Dm = 3 in Deflection = 0.4 in Nc = 6 active coils P = 1.3333333 in λ = 11.980811357˚ C = 3 in Solid height = 8 in AISI C117 Normalized Carbon Steel N= 4 bolts Ar = 0.334 in2 select ¾ in-10 UNC bolt

F = 3452.559344 lbs. Ss = 38.67346463 ksi Ssd = 40.2084 ksi. Fc = 17262.79672 lbs. k= 8631.39836 lb/in Ssolid stress = 193.3673232 ksi Spermissible = 70 ksi.

T1 = 182.6323599 lbs F = 1030.512805 lbs Fr = 1104.030756 lbs Fm = 108.6088745 lbs

MACHINE DESIGN 2

45

2012


LIST OF REFERENCES Books : Doughtie, V.L., &Vallance, A. (1978).Design of Machine Members(4thed.).New York: McGrawHill, INC. Faires, V.M. (1969). Design of Machine Elements (4thed.). New York: MacMillan Company. Morse, F. (1953).Power Plant Engineering. Philippines: Litton Educational Pubishing, INC. Spotts, M.F. (1991). Design of Machine Elements (6thed.). Singapore: Simon & Schuster (Asia) Pte Ltd.

Internet: Engineering Tool Box, electrical motors - Hp, torque and rpm. Retrieved July 5,2012 from http://www.engineeringtoolbox.com/electrical-motors-hp-torque-rpm-d_1503.html Handbook of Coal Analysis. Retrieved July 5, 2012 from http://info.com/Handbookofcoalanalysis Roll Crusher. Retrieved August 17, 2012 from http://www.gundlachcrushers.com/crushers/rollcrushers-coal-salts-lime-minerals.cfm What is the density of Coal.Retrieved August 17, 2012 from http://wiki.answers.com/Q/What BL Precision Bearings Radial Ball Bearings Catalog. Retrieve September 12, 2012 from http://www.qbcbearings.com/BuyRFQ/ThrustB_Bearing_B_CSBW.htm Century Compression Spring Catalog. Retrieved Sept 12, 2012 http://www.centuryspring.com

MACHINE DESIGN 2

46

2012


APPENDIX: Motor Velocity (rpm) Power

3450

2000

1000

500

Torque

hp

Kw

(in lbf)

(ft lbf) (Nm)

(in lbf)

(ft lbf) (Nm)

(in lbf)

(ft lbf) (Nm)

(in lbf)

(ft lbf) (Nm)

1

0.75

18

1.5

2.1

32

2.6

3.6

63

5.3

7.1

126

10.5

14.2

1.5

1.1

27

2.3

3.1

47

3.9

5.3

95

7.9

10.7

189

15.8

21.4

2

1.5

37

3.0

4.1

63

5.3

7.1

126

10.5

14.2

252

21.0

28.5

3

2.2

55

4.6

6.2

95

7.9

10.7

189

15.8

21.4

378

31.5

42.7

5

3.7

91

7.6

10

158

13.1

18

315

26.3

36

630

52.5

71

7.5

5.6

137

11

15

236

20

27

473

39

53

945

79

107

10

7.5

183

15

21

315

26

36

630

53

71

1260 105

142

15

11

274

23

31

473

39

53

945

79

107

1891 158

214

20

15

365

30

41

630

53

71

1260 105

142

2521 210

285

25

19

457

38

52

788

66

89

1576 131

178

3151 263

356

30

22

548

46

62

945

79

107

1891 158

214

3781 315

427

MACHINE DESIGN 2

47

2012


40

30

731

61

83

1260 105

142

2521 210

285

5042 420

570

50

37

913

76

103

1576 131

178

3151 263

356

6302 525

712

60

45

1096 91

124

1891 158

214

3781 315

427

7563 630

855

70

52

1279 107

145

2206 184

249

4412 368

499

8823 735

997

80

60

1461 122

165

2521 210

285

5042 420

570

10084 840

1140

90

67

1644 137

186

2836 236

321

5672 473

641

11344 945

1282

100

75

1827 152

207

3151 263

356

6302 525

712

12605 1050 1425

125

93

2283 190

258

3939 328

445

7878 657

891

15756 1313 1781

150

112

2740 228

310

4727 394

534

9454 788

1069 18907 1576 2137

175

131

3197 266

361

5515 460

623

11029 919

1247 22058 1838 2494

200

149

3654 304

413

6302 525

712

12605 1050 1425 25210 2101 2850

225

168

4110 343

465

7090 591

801

14180 1182 1603 28361 2363 3206

250

187

4567 381

516

7878 657

891

15756 1313 1781 31512 2626 3562

275

205

5024 419

568

8666 722

980

17332 1444 1959 34663 2889 3918

300

224

5480 457

620

9454 788

1069 18907 1576 2137 37814 3151 4275

350

261

6394 533

723

11029 919

1247 22058 1838 2494 44117 3676 4987

400

298

7307 609

826

12605 1050 1425 25210 2101 2850 50419 4202 5699

MACHINE DESIGN 2

48

2012


450

336

8221 685

929

14180 1182 1603 28361 2363 3206 56722 4727 6412

550

410

10047 837

1136 17332 1444 1959 34663 2889 3918 69326 5777 7837

600

448

10961 913

1239 18907 1576 2137 37814 3151 4275 75629 6302 8549

Table 1 Power versus torque and motor velocity in electric motors

Table 8.2 Variation Of Compressive Strength by Rank

Figure 17.14 Belt Selection from Horsepower and Speed

MACHINE DESIGN 2

49

2012


Table 17.7 Service Factors

MACHINE DESIGN 2

50

2012


Table 17.5 Arc-of contact factors

Table 17.4 factor

small

diameter

Table 17.3 Standard V-belt Lengths; Horsepower Constants

MACHINE DESIGN 2

51

2012


Table 16-5 Coefficient of Friction For Belts

Table 17.6 Length Correction Factors

MACHINE DESIGN 2

52

2012


Table AT8 Typical Properties of Steel-Various Sizes and Conditions

MACHINE DESIGN 2

53

2012


Table 9-3 Typical Reliability Factors for rolling element bearings

Table 9-4 Typical Values of application factor for roller and ball bearings

MACHINE DESIGN 2

54

2012


Table 9-2 SAE and International Standard dimensions for ball and Roller Bearings

MACHINE DESIGN 2

55

2012


Table 8 Ball Bearing specification

MACHINE DESIGN 2

56

2012


Table 9-7 Typical Radial Capacity for Ball bearings

Table 9-4 Typical Values of Application factor for roller and ball bearings

MACHINE DESIGN 2

57

2012


Table AT 19 Key Dimensions

MACHINE DESIGN 2

58

2012


Table 6-1 Uniform and American National Threads, coarse, fine and extra-fine.

MACHINE DESIGN 2

59

2012


Table 1.1 Factors of safety

Table 9 Century Spring Catalog

MACHINE DESIGN 2

60

2012


Figure AF 15 Stress Factor

MACHINE DESIGN 2

61

2012


Table AT 17 Mechanical Properties of Wire for Coil Springs

MACHINE DESIGN 2

62

2012


Table AT 16 Approximate Free Lengths and Solid Heights

MACHINE DESIGN 2

63

Double Roll Crusher Machine Design

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