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MECG1022 Design Project Report

Design of an Overhanging Countershaft

Date: 11/29/2018 Name: Byron Andr´es Mac´ıas Balda Desing Project No: 1 Instructor: Professor Carlos Gabriel Helguero

Mechanical Engineering Faculty ESPOL

´ Indice 1. Objectives and Introduction 1.1. Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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2. Design criteria and assumptions

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3. Design procedures and descriptions with abridged derivations 3.1. Step 1: V-Belts. . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2. Step 2: Gears . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3. Step 3: Countershaft . . . . . . . . . . . . . . . . . . . . . . . 3.4. Step 4: Wedges . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5. Step 5: Bearing . . . . . . . . . . . . . . . . . . . . . . . . . .

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4. Results and discussions

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5. Conclusions and recommendations

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6. Appendix 6.1. V-Belt . . . . 6.2. Gears . . . . 6.3. Countershaft 6.4. Wedger . . . 6.5. Ball Bearing

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V-Belt . . . . . . . . . . . . . . Charges in the Countershaft . Shear strength X . . . . . . . . Momentum in X . . . . . . . . Shear strength Y . . . . . . . . Momentum in Y . . . . . . . . Torque delivered by the motor

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´ Indice de figuras 1. 2. 3. 4. 5. 6. 7.

1. 1.1.

Objectives and Introduction Objectives

The objectives of this project, is the design of a countershaft that rotates with a nominal power of 10.5 hp, other elements such as belts and gears are found along the countershaft, must design the gears and select the right type of belt.

1.2.

Introduction

The countershaft is a component whose purpose is to transmit mechanical energy through its rotation, is subject to the force of torsion and bending. The countershaft is also an element subject to a product of repeated load of the other components, such as gears, belts, which carry these entails to the fault mechanical for fatigue.The cross section of a countershaft is usually circular although occasionally it may have another type of section such as corrugated or polygonal, the circular diameter of the countershaft is determined based on the efforts it supports, the material used and the conditions of manufacture and use that are will give. The other part of the design is the V-belts selection, knowing that the rated power of the countershaft is 10.5 hp and the angular speed of the engine is 2000 rpm, a suitable diameter was selected for the engine, as well as a step length and type of band; this in order that the number of bands to use is not too high in addition to the time of life of the belt this in a suitable range. After obtaining the angular velocity of the countershaft, the gears were determined and designed, selecting a number of teeth for the gear and also determining the appropriate material so that the safety factors for the bending effort and the surface stress are at least greater than one. The gears are common components in the countershaft and serve to transmit a circular movement through the contact of the gear.

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2.

Design criteria and assumptions

The main criteria in the design were to obtain values of safety factors suitable for each element, in addition that the highest is for the countershaft with the purpose that the last component that fails due to fatigue is this. Among the assumptions that were made were: For gears : • Number of gear teeth 35. • Material used AGMA maximum grade brinell hardness equal to 380. • Reliability of 90 % • Factor Ka in uniform impact of driving machine and uniform driven machine • Cycle life N = 5,54 × 108 For V-Belts: • Diameter on the motor pulley is seven inches. • V-Belt B180 Countershaft: • Material AISI 1050 with Sut = 100Kpsi Bearing • Ball bearing #6310 with internal diameter d=1.96 in

3.

Design procedures and descriptions with abridged derivations

The total system design is composed of the following component design: VBelts, Gears, Countershaft, Wedge, Bearing.

3.1.

Step 1: V-Belts.

Started in the design of the V-Belts to determine the angular velocity of the countershaft, a type B belt was selected for power ranges between 1-25 hp with width 0.65 in. Then the type of belt was determined by testing the different available step length values, where the minimum diameter for the smaller pulley is 5.4 in., The selection focused on obtaining the least amount of belts possible to use.

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V-Belt Selected: • B180 • d= 7 in • # Belt: 3

3.2.

Step 2: Gears

Having the angular velocity of the countershaft, a number of gear teeth of the 35 was selected, knowing the nominal power of 10.5 hp, the torque of the pinion was obtained, as well as the tangential and radial loads. Then the bending efforts, contact effort, the design factors from the selected material and the conditions in which the gears will be applied were determined.

3.3.

Step 3: Countershaft

To design the countershaft the section was first determined along the axis where shoulders would be placed, to diminish the concentration of efforts, rounded shoulders were preferred. Then, given the safety factor of 1.85, the diameter of the contershaft and the diameter of the shoulder were obtained, where the gear was placed using a wedge, as the value of bending moments and torques is the maximum winged of the gears, it was assumed that the The same man can be used on the side of the V-Belts, assumptions of the factors Kb , Kt , Kts are made in a moment as it is solved in Appendix.

3.4.

Step 4: Wedges

Knowing the diameter of the shoulder and the thickness of the gear, at this point the calculations for the wedge are obtained, since the wedge is designed where there are greater bending moments, the same design can be used to place on the end pulley of the countershaft .

3.5.

Step 5: Bearing

The bearings were selected according to the diameter of the countershaft and the contershaft was modified so that the bearings were adjusted, ball bearings were used and the number of millions of revolutions was calculated using the axial force P = 600 N and the magnitude of the reactions as the radial force in the bearing.

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4.

Results and discussions

The results obtained give adequate safety values, such as a high life time for each component, and it is possible that the material used for the countershaft allows the ultimate fatigue failure to be countershaft, which is what is recommended for each element. a different type of material was selected less for the wedge which the same material as the countershaft.

5.

Conclusions and recommendations

In conclusion the components of the countershaf will fail first by fatigue in a period of time high by the materials that were selected for this design, to achieve this it is advisable to use a software like excel or matlab to test with different materials such as for gears, to go by varying the diameter of the pulley in the motor, the length of the passage that will be used and for the conuntershaft. Materials that result in small safety factors should not be used.

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6. 6.1.

Appendix V-Belt

d= 7 in D=15.748 in C=72.90 in N=2000 rpm V-Belt: B180

../../../Desktop/vbelt.png

Figura 1: V-Belt

Lp = 181,8 Hnom = 10,5hp    r 2   π π 2 C = 0,25  Lp − (D + d) + Lp − (D + d) − 2(D − d)  2 2 C = 72,90in

πdN 12 V = 3665,19f t/min V =

D −d 2C ◦ θ = 86,56 , 3,02rad

θ = π − 2sin−1

 Kc = 0,965 → Fc = Kc

V 1000

2

Fc = 12,96N

K1 = 0,143543 + 0,007648θ − 0,000015052θ 2 −→ K1 = 0,69 7

K2 = 1,2 Htab = 4,83 Ha = K1 K2 Htab Ha = 4hp

Ks = 1,1, nd = 1 −→ Hd = Hnom Ks nd Hnom = 11,55hp

NB ≥

Hd = 2,85 −→ 3 Ha

4F =

d 63025 H Nb

N (d/2)

4F = 34,66

F1 = Fc +

4exp(0,5123θ) exp(0,5123θ) − 1

F1 = 56,99N

F2 = F1 − 4F F2 = 22,33N

nf s =

Ha Nb Hnom Ks

nf s = 1,04

Kb = 576 Kb d K T2 = F1 + b D T1 = 139,27lb, T2 = 93,56lb T1 = F1 +

8

K = 1193, b = 10,928  ! !−b −1   K −b K + Np =   T1 T2  Np = 1,54 × 1010

Lp

t=

720V t = 352188,712

6.2.

Gears

Nω = N ∗ Dd ←→ 889,002rpm Np = 24 Ng = 35 pd = 10 φ = 20◦ HP ST C Tp =

P Nω

Tp = 744lb − in

2Pd Tp

Wt =

Nω

Wt = 620,327lb Wr = Wt tan(φ) Wr = 1387,77lb

dp = dg =

Nω = 2,4in pd Ng pd

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= 3,5in

Vt =

dp Nω π

12 V = 558,57f t/min Qv = 7 Flexion effort σb =

Wt pd Ka Km Ks KB KI FJ Kv Jg = 0,376 Jp = 0,352

F=

12 = 1,2in −→ Km = 1,6 pd Ka = 1

A Kv = √ A + Vt

!B −→ B = 0,73, A = 65,12 Kv = 0,79

Kf = Kb = KI = 1

σbp = 2973,3psi σbg = 27844,8psi

Sf b =

KL S 0 KT KR f b

AGMA GRADE 2 HB=380 Sf b 0 = 6235 + 174HB − 0,126HB2 Sf b 0 = 54160,6

60min N = 889,002rpm 1h 



! 2080h (5years)(1turn) turn − year 10

N = 5,54 × 108

KL = 1,3558N −0,0178 KL = 0,947 460 + T◦F 620 ◦ = 75 −→ 167◦ F

KT = T◦C

Kt = 1,01

KR = 0,85

Sf b = 59743,8 np = ng =

Sf b σbp

Sf b σbg

=2

= 2,145

Contact effort r σc = C p

Wt Ca Cm Cs Cf FId Cv Ca = 1,25 Cm = 1,6

Cv = 0,76 ν = 0,28, E = 30 × 106 Kpsi −→ Cp = 2276 ρp = 0,35, ρg = 0,65 −→ I = 0,089 dp = 2,4in

σc = 162459psi

Sf c =

CL CH S 0 CT CR f c 11

AGMA GRADE 2 HB=380 Sf c 0 = 2700 + 364HB −→ Sf c 0 = 165320psi CR = 1 CT = 0,86 CH = 1 CL = 1,4488N −0,023 −→ CL = 0,91 Sf c = 178237psi ! Sf c 2 nsc = σc nsc = 1,16

6.3.

Countershaft

../../../Desktop/arbolcargas.png

Figura 2: Charges in the Countershaft

WT = 620,327lb Wr = 1387,44lb P = 600lb F1 + F2 = 79,32lb W = 225,4lb For P X + Mox = 0 −1387,77(9,84) − Rrx (20,46) = 0 Rrx = −667,432lb

+&

X

Fx = 0

Rox + 1387,77 − 667,432 = 0 Rox = −720,338lb For Y +

X

12

Moy

−620,327(9,84) + Rry (20,46) − 79,32(30,97 + 32,27 + 31,62) − W (31,62) = 0 Ry = 1014,44lb

+↑

X

Fy = 0

Roy + Rry − 79,32(3) − 620,327 − 225,4 = 0 Roy = 69,247lb

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../../../Desktop/Cortantex.png Figura 3: Shear strength X ../../../Desktop/momentox.png Figura 4: Momentum in X ../../../Desktop/cortantey.png Figura 5: Shear strength Y ../../../Desktop/momentumy.png Figura 6: Momentum in Y

../../../Desktop/torca.png

Figura 7: Torque delivered by the motor

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Shoulder to 10.57 in to the origen Ma = 6627,53lb − in Tm = 744,39 Ka = 0,79 Kb = 0,9 −→ supposed Kd = 1,01 Ke = Kc = 1 Kf = 1,7, Kf s = 1,5 −→ supposed n = 1,85

Sut] = 100KP si −→ AISI1050 Se = 0,5 ∗ 0,79 ∗ 0,9 ∗ 1,019 ∗ Sut Se = 35,90KP si   h i 1/3 2 1/2   2(K M )  3(K T )  fs m  16n  f a   + d =   Se Sut   π  d = 1,84in Typical shoulder relationship D = 1,2(d) = 2,2in r/d= 0,1 −→ r = 0,184 Looking at concentration tables Kt = 1,6, Kts = 1,45 qt = 0,87, qts = 0,9 Kf = 1,52, Kf s = 1,405 Then the safety factor value is recalculated Ka = 0,82, Kd = 1,01, Kb = 0,82 Se = 37,88KP si

σa 0 =

32Kf Ma πd 3

= 16924,1KP si

σm 0 = 1481 1 σ 0 σ 0 = a + m nf s Se Sut nf s = 2,16 15

6.4.

Wedger

The wedge is centered on the gear side. F = 1,2in D = 2,2in −→ Wnominal = 0,5in Ta = 6627,57lb − in Tm = 744,39lb − in Se = 37,38KP si Sut = 100KP si

2Ta = 6025,03lb D 2T Fa = m = 676,718lb D Fa =

Fa = 10041,7P si w∗F F τm = m = 1127,86P si w∗F τa =

q σa 0 = 3(τa )2 = 17392,7P si q σm 0 = 3(τm )2 = 1953,51P si

nf s =

σa 0 Se

1 + σSmut0

nf s = 2,06 Pressure effort σcontac =

Fm + Fa = 11169,6P si Acontac

Sy = 84KP si −→ ns = ns = 7,52

16

Sy σcontac

6.5.

Ball Bearing ForRo

√

Fr = 720,3382 + 69,2472 = 723,659lb Fa = P = 600lb Bearing −→ #6310, d = 1,96in C = 10600lb Co = 8150lb Fa = 0,073 −→ e = 0,27 Co V = 1,

Fa = 0,83 > e VF r

X = 0,56, Y = 1,61 P = XCFr + Y Fa = 1371,25  3 C L10 = = 461,92millionsof revolution P ForRR Fr = 1214,31lblb Fa = P = 600lb Bearing −→ #6310, d = 1,96in C = 10600lb Co = 8150lb Fa = 0,073 −→ e = 0,27 Co V = 1,

Fa = 0,49 > e VF r

X = 0,56, Y = 1,61 P = XCFr + Y Fa = 1646,01  3 C = 267,06millionsof revolution L10 = P

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