Low speed shaft:
Given: Ft = 1546.155 H, Fr = 567.339 H, Lt = 0.093 m, Lt/2 = 0.0465 m,
1. Determination of reaction in bearings in the horizontal plane:
Rсх*Lт + Ft * Lт/2 = 0
Rсх*0.093+1546.155*0.0465 = 0
Rсх*0.093 = -71.896
Rсх = 71.896/0.093 = 773.075 N
Ft* Lt/2+Rdh* Lt = 0
1546.155*0.0465+ Rdh *0.093 = 0
Rdh = 71.896/0.093 = 773.075 N
Check: ∑Fnх = 0
Rdh + Rсх - Ft = 0 ; 773.075+773.075-1546.155 = 0 ; 0 = 0
M2lev = Rсх * Lт/2 = 773.075 * 0.0465 = 35.947 Nm
M2pr = M2lev = 35.947 Nm
M3lev = Rсх * Lт- Ft* Lт/2 = 71.895-71.895 = 0
2. Determination of reaction in bearings in the vertical plane:
Rsy*Lt + Fr * Lt/2 = 0
Rsy*0.093+567.339*0.0465 = 0
Rsy = 26.381/0.093 = 283.669 N
Fr* Lt/2+Rdu* Lt = 0
567.339*0.0465+ RDN *0.093 = 0
RDN = 26.38/0.093 = 283.669 N
Check: ∑Fnу = 0
Rsy – Fr+ Rdu = 0 ; 283.669 – 567.339+283.669 = 0 ; 0 = 0
We construct diagrams of bending moments.
M2lev = Rsy * Lt/2 = 283.669 * 0.0465 = 13.19 Nm
M2pr = M2lev = 13.19 Nm
M3lev = Rsy * Lt-Fr* Lt/2 = 26.381-26.381 = 0
3. We build diagrams of torques.
Mk = M2 = Ft*d2/2 = 1546.155*184.959/2 = 145.13 Nm
4. Determine the total radial reactions:
Rc = = 823.476 N
Rd = 
= 823.476 N
5. Determine the total bending moments.
M2 = = 38.29 Nm
7. Check calculation of bearings:
7.1 The basic dynamic load rating of a bearing, Cr, is the constant radial load that the bearing can support with a basic life of 10 revolutions of the inner ring.
Cr = 29100 N for a high-speed shaft (Table K27, p. 410), bearing 306.
Cr = 25500 N for a low-speed shaft (Table K27, p. 410), bearing 207.
The required bearing life Lh for gear reducers is Lh ≥ 60,000 hours.
The suitability of bearings is determined by comparing the calculated dynamic load capacity Crp, N with the basic durability L10h, h. with the required Lh, h. according to the conditions Crp ≤ Cr; L10h ≥ Lh.
The calculated dynamic load capacity Crp, N and basic durability L10h, hours are determined by the formulas:
Crp = 
; L10h =
where RE is the equivalent dynamic load, N;
ω – angular velocity of the corresponding shaft, s
M – exponent: M = 3 for ball bearings (p. 128).
7.1.1 Determine the equivalent load RE = V* Rr*Kv*Kt, where
V – rotation coefficient. V = 1 with rotating inner ring of the bearing (page 130).
Rr – radial load of the bearing, N. Rr = R – total reaction of the bearing.
Kv – safety factor. Kv = 1.7 (Table 9.4, p. 133).
Kt – temperature coefficient. Kt = 1 (Table 9.5, p. 135).
High speed shaft: RE = 1*1.7*1323.499*1 = 2249.448 N
Low-speed shaft: RE = 1*1.7*823.746*1 = 1399.909 N
7.1.2 We calculate the dynamic load capacity Crp and durability L10h of bearings:
High speed shaft: Crp =2249.448 
= 2249.448*11.999 = 26991.126 N; 26991.126 ≤ 29100 - the condition is met.
75123.783 ≥ 60000 - the condition is met.
Low speed shaft: Crp = 1399.909 
= 1399.909*7.559 = 10581.912 N; 10581.912 ≤ 25500 - the condition is met.
848550.469 ≥ 60000 - the condition is met.
The verification calculation showed the profitability of the selected bearings.
7.1.3 Compose a tabular answer:
Main dimensions and operational dimensions of bearings:
8. Structural layout of the drive:
8.1 Design of gears:
Gear:
At the ends of the teeth, chamfers of size f = 1.6 mm are made. Chamfer angle αf on chevron wheels with hardness of working surfaces HB< 350, αф = 45°. Способ получения заготовки – ковка или штамповка.
8.1.1 Installing the wheel on the shaft:
To transmit rotating torque by a gear pair, a keyed connection with a H7/r6 fit is used.
8.1.2 When using chevron wheels as a gear pair, there is no need to take care of the axial fixation of the wheel, however, to prevent axial displacement of the bearings towards the wheel, we install two bushings on both sides of the wheel.
8.2 Shaft design:
The transition section of the shafts between two adjacent steps of different diameters is made with a groove:
8.2.2 On the first and third stages of the low-speed shaft, we use a keyed connection with keys having the following dimensions:
8.3 Design of the gearbox housing:
The body is made of cast iron grade SCh 15. The body is detachable. Consists of a base and a lid. It has a rectangular shape, with smooth outer walls without protruding structural elements. At the top of the housing cover there is an inspection window, closed by a cover with an vent. At the bottom of the base there are two plugs - drain and control.
Thickness of walls and stiffeners δ, mm: δ=1.12 =1.12*3.459=3.8 mm.
To fulfill the condition δ≥6 mm, we take δ = 10 mm.
8.3.1 The gearbox is fastened to the base frame (plate) using four M12 studs. The flange width is 32 mm, the axis coordinate of the hole for the stud is 14 mm. The connection between the cover and the base of the housing is carried out with six M8 screws. The inspection window cover is secured with four M6 screws.
8.4 Check calculation of shafts
8.4.1. We determine the equivalent moment using the formula for shafts:
High speed shaft: Meq = 
= 
= 63.011 (N)
Low speed shaft: Meq = 
= = 150.096 (N)
8.4.2. We determine the calculated equivalent stresses δeq and compare them with the permissible value [δ]u. We choose steel 45 for the drive and driven shafts, for which [δ]u = 50 mPa
d = 42 – diameter of the low-speed shaft in the dangerous section.
Conclusion: the strength of the high-speed and low-speed shafts is ensured.
Lubrication
9.1 For general purpose gearboxes, continuous lubrication with liquid oil using a non-flow crankcase method (dipping) is used. This method is used for gears with peripheral speeds from 0.3 to 12.5 m/sec.
9.2 The choice of oil type depends on the value of the calculated contact stress in the teeth GН and the actual peripheral speed of the wheels U. The oil type is selected according to table 10.29, page 241. In this gearbox at U = 1.161 m/sec, GН = 412, I-G-A-68 grade oil is used.
9.3 For single-stage gearboxes, the oil volume is determined at the rate of 0.4...0.8 liters. per 1 kW of transmitted power. P = 2.2 kW, U = 2.2*0.5 = 1,100 l. The oil volume in the designed gearbox is 1,100 liters. The gearbox is filled with oil through the inspection window. The oil level is controlled using a control plug. The oil is drained through the drain plug.
9.4 Lubrication of bearings:
In designed gearboxes, liquid and plastic lubricants are used to lubricate rolling bearings. Lubricant is filled into the bearing manually with the bearing cover removed. The most common grease for rolling bearings is fatty grease (GOST 1033-79), fatty konstalin UT-1 (GOST 1957-75).
1) We make a calculation of the shaft diagram:
Circumferential force F t = 7945.9 N
Radial force F r = 2966.5 N
Axial force F a = 1811 N
2) Let's draw up a design diagram of the shaft:
Find l 1:
l 1 = V P /2 + (5h10) + v 2T /2, (123)
l 1 = 37/2 + 10 + 63/2 = 60.5 = 60 mm.
Find l 2:
l 2 = in 2T /2 + (5h10) + in 2B + (5h10) + V P /2, (124)
l 2 = 63/2 + 10 + 45 + 10 + 37/2 = 114 mm.
l 3 = 37/2+1.2*70+1.5*60=192 mm (125)
2) F M =vT 3 *250=7915.965 H
3) Ma =F a *d 2T /2 = 221578.5 H; (126)
M A = 0; (127)
Y V (60 + 114)-221578.6-2966.5*60 = 0, (128)
Y A (60+114)+114*2966.5= 221578.6
Check: ?Y=0, (130)
Y A +Y B -F r =0, (131)
670.13+2296.37-2966.5=0 - the condition is met.
4) Determine the support reactions in the horizontal plane:
Due to the inevitable misalignment of the shaft connection, the low-speed shaft is loaded with additional force F M - the force of the couplings.
For two-stage gearbox:
F M = 250vТ 2Т 2) =7915.96 H, (132)
We direct the force F M so that it increases the stress and deformation from the force F t (in the worst case).
Equilibrium condition for a point
V: ?M V =0, (133)
X A (l 1 +l 2) - F t l 2 - F M l 3 =0 (133)
Let us write down the equilibrium condition for the point
A: ?M A =0, (134)
X B (l 1 +l 2)+F t l 1 -F M (l 1 +l 2 +l 3)=0, (135)
Check: ?Х=0, (136)
X A + F t +X B - F M =0,
10.75+7945.9+15.55-7915.965=0 - the condition is met.
5) We construct a diagram of bending moments from forces F g and F a
M Reference =670.13*60 =40207.8 Nm;
M Left =Y A l 1 +Fa d 2T /2=40207.8+221578.6 =261786.4 Nm;
M B =Y A (l 1 +l 2)+ Fa d 2T /2-F r l 2 =0 (Check!)
6) We construct a diagram of the bending moments due to the force Ft.
M C. =-X A l 1 =-10.75·60=-644.4·N·m;
M B =Х A (l 1 +l 2)+Ft l 2 =-1870.5+353.4=-1517.1 N m;
M D =-X A (l 1 +l 2 +l 3)+ F t (l 1 +l 2)+X B l 3,
M D = -10.75*366+3.1*306+15.55*192=0 (Check!)
The diagram of bending moments is presented in Appendix A.
7) We build a total diagram of bending moments
The ordinates of the total diagram of bending moments from the combined action of these forces are found using the formula:
M B = -1517.1 N m;
The total diagram of bending moments is in Appendix A.
8) We build a diagram of torques:
T = F t d 2t /2, (138)
T = 7945.97525/2 = 2085817.12 Nm
Torque diagram in Appendix A.
9) Determine the total reactions of the supports:
The most loaded is support B, where the radial force = 8458b51 N acts.
5.2 Drawing up design diagrams for a low-speed shaft and determining reactions in the supports
From previous calculations we have:
L 1 = 69 (mm)
Support reactions:
1. in the XDZ plane:
∑М 1 = 0; R X 2 ∙ 2 l 1 - F t ∙ l 1 = 0; R X 2 =F t /2 = 17833/2 = 8916.5 N
∑М 2 = 0; - R X 1 ∙ 2 l 1 - F t ∙ l 1 = 0; R X 1 =F t /2 = 17833/2 = 8916.5 N
Check: ∑X= 0; R X 1 + R X 2 - F t = 0; 0 = 0
2. in the YOZ plane:
∑М 1 = 0; F r ∙ l 1 + F a ∙ d 2 /2 – R y 2 ∙ 2 l 1 = 0; V
R y 2 = (F r ∙ l 1 + F a ∙ d 2 /2)/ 2 l 1 ;N
R y 2 = (F r ∙ 69+ F a ∙ d 2 /2)/ 2 ∙ 69 = 9314.7 N
∑М 2 = 0; - R y 1 ∙ 2 l 1 + F a ∙ d 2 /2 – F r ∙ l 1 = 0;
R y 1 = (F a ∙ d 2 /2 - F r ∙ l 1)/ 2 l 1 ;N
R y 1 = (F a ∙ 524/2 - F r ∙ 69)/ 2 ∙ 69 = 2691.7 N
Check: ∑Y= 0; - R y 1 + R y 2 – F r = 0; 0 = 0
Total support reactions:
P r 1 = √ R 2 X 1 + R 2 Y 1 ;H
P r 1 = √ 8916.5 2 + 2691.7 2 = 9313.9 N
P r 2 = √ R 2 X 2 + R 2 Y 2 ;H
P r 2 = √ 8916.5 2 + 9314.7 2 = 12894.5 N
We select bearings according to the more loaded support Z.
We accept 219 light series radial ball bearings:
D = 170 mm; d = 95 mm; B = 32 mm; C = 108 kN; C 0 = 95.6 kN.
5.3 Checking bearing life
Let us determine the ratio F a /C 0
F a /C 0 = 3162/95600 = 0.033
According to the table, the ratio F a /C 0 corresponds to e = 0.25
Let us determine the ratio F a /VF r
V – coefficient for rotation of the inner ring
F a /VF r = 3162/6623 = 0.47
Let's determine the equivalent load
Р = (x ∙ V ∙ F r + YF a) ∙ K σ ∙ K T ; N
K σ – safety factor
K T – temperature coefficient
P = (0.56 ∙ 1 ∙ 6623+ 1.78 3162) ∙ 1.8∙1= 16807 N
Let us determine the design durability in million revolutions.
L = (C/P) 3 million vol.
L = (108000/16807) 3 million vol.
Let's determine the estimated durability in hours
L h 1 = L ∙ 10 6 /60 ∙ n 3 ; h
L h 1 = 265 ∙ 10 6 /60 ∙ 2866 = 154 ∙ 10 3 h
L h 1 ≥ 10 ∙ 10 3
154 ∙10 3 ≥ 10 ∙10 3
5.4 Assessing the suitability of selected bearings
Suitability assessment of selected bearings
154 ∙10 3 ≥ 17987,2
154000 ≥ 17987,2
6. Design of transmission elements
6.1 Design selection
Gear wheel – forged, shape – flat
The gear is made integral with the shaft
6.2 Dimensions
1. gear
Its dimensions are defined above
Its dimensions are defined above
Let's determine the hub diameter:
d st = 1.6 ∙ d k; mm
d st = 1.6 ∙ 120 = 192 mm
We accept d st = 200 mm
Let's determine the length of the hub:
l st = (1.2 ÷1.5) ∙ d k; mm
l st = (1.2 ÷1.5) ∙ 120 = 144 ÷180 mm
Because l st ≤ b 2, take l st = 95 mm
Let's determine the thickness of the rim:
δ 0 = (2.5 ÷ 4) ∙m; mm
δ 0 = (2.5 ÷ 4) ∙5 = 12.5 ÷ 20 mm
We accept δ 0 = 16 mm
Let's determine the thickness of the disk:
C = 0.3 ∙ b 2 ; mm
C = 0.3 ∙ 95 = 28.5 mm
We accept C = 30 mm
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Figure 4 Shaft calculation algorithm diagram
Initial data for calculation: T - force acting on the shaft; Fr, Ft, Fx - torques. Since there are no elements on the design shaft that cause axial force Fx = 0, Ft = 20806, Fr = -20806, T = 4383.
Definitions of support reactions
Calculation of reaction of supports
The reactions of the shaft supports are shown in Figure 5.
Figure 5 Diagrams of the traction sprocket shaft
Left support reaction.
where l1,l2,l3,l4 is the distance between the shaft structural elements, l1 = 100, l2 = 630, l3=100, l4=110, = 20806 H.
where = -20806 N.
Right support reaction.
Determine bending moments for the calculated shaft
Horizontal plane Mi, from the axis: for coupling Mi(m) = 0, left support Mi(l) = 0, for left sprocket Mi(l) = - 2039 N*m, for right sprocket Mi(pz) = -2081 N *m, for the right support Mi(n) = -42 N*m. Diagrams of these forces are shown in Figure 5.
Vertical plane Mi, from the axis: for the coupling Mi(m) = 0, left support Mi(l) = 0, for the left sprocket Mi(lz) = 0, for the right sprocket Mi(pz) = 0,
for the right support Mi(n) = 0. Diagrams of these forces are shown in Figure 5.
Mi given: for coupling Mi(m) = 4383 N*m, left support Mi(l) = 4383 N*m, for left sprocket Mi(l) = 4383 N*m, for right sprocket Mi(pz) = 3022 N *m, for the right support Mi(n) = 42 N*m. Diagrams of these forces are shown in Figure 5.
The total bending moment is equal to: for the coupling T(m) = 4383 N*m, left support T(l) = 4383 N*m, for the left sprocket T(l) = 4383 N*m, for the right sprocket T(pz) = 2192 N*m, for the right support T(p) = 0 N*m. Diagrams of these forces are shown in Figure 5.
We select the material for the shaft according to the given loads: Steel 45 GOST 1050-88.