1. A horizontal shaft of 1-in diameter rotates at 500 rpm in a sleeve-type bearing. The coefficient of friction is 0.15. If the reaction between the shaft and the bearing is 800 lbs, compute the horsepower lost in the bearing. (0.4759 hp) Solution:
µ=
F
F = µN = 0.15(800) = 120 lbs.
. .× .. T= F ( ) = = 5 ft. lb HPloss = (
πT ) = π
HPloss = 0.4759 hp
2. A bearing carries a load of 4.450 kN. The shaft diameter is 100 mm, the coefficient of friction is 0.01, and the shaft speed is 400 rpm. Compute the power lost in the bearing. (93.2 watts) Solution:
µ=
F W
F = µ(W) = 0.01(4,450) = 44.5 N
= 2.225 N m ) Plost = 2µ(2.225)(
T = (44.5N)(
Plost = 93.2 watts
3. A 36 mm shaft uses a sleeve bearing that sustains a load of 4,000N. Compute the length of the bearing and the L/D ratio considering that bearing pressure is 1.3 MPa. (2.37) Solution:
Pressure = Load/Area Area =
= 3.077 x 10− m . x
Then: Area = LD
. × A = = 85.47 mm L= D
= . =2.37 D .
4. A 2 7/26 in. journal bearing, 4in. long supports a load of 500 lbs at 600 rpm. If the room temperature is 75̊ F, what viscosity of oil should be used to limit the bearing surface temperature to 140̊ F? Assume no artificial cooling. (12.45centipoise) Solution:
f=
.× DZ =0.002 c
Where:
ΔT+ fPV = k LD
9 f(500)[ π [
[−+] ] (600) ] =
f = 0.008899, thus Z = 12.45 centipoise
5. A support shaft for a new product has been designed with a diameter of 1.6 inches. The force analysis shows that: Radial bearing load Fr = 370 lb Axial bearing load Fa = 130 lb Shaft speed n = 350 rpm Design life specification is 10 years of operation, 50 days/year, 20hr/day Design reliability specification is R = 95% Impact factor If =1.9 Life adjustment factor K r = 0.62 Equivalent dynamic bearing load Pe= 392 lbs Calculate basic dynamic load requirement for R = 95% Solution:
L=t ×n days)(20hrs)(60mins)×350 rev L=10yrs.(501yr 1day 1 hr min L=2.1×10 revs
C = K ×L10 I P
C = . .× × 1.9)( 392 lbs) Cd= 5192 lbs.
6. A radial ball bearing with a specific dynamic capacity of 4500 lbs. is to carry a load of 1500 lbs. The bearing shaft rotates at 1000 rpm. Evaluate the maximum period over which the bearing could be used safely if reliability is 91 percent. (54,133 h rs.) Solution:
The bearing life is given by the relation, L b = k rr b (cd /F)3.33
(1)
where, L b = bearing life, million revolutions k r = reliability factor r b = reliability cd = dynamic capacity factor F = load on the bearing. Now from Figure 1, for 91 percent reliability, k r = 0.92. Substituting values in (1), L b = 0.92 × 91[(4500) / (1500)]3.33 L b = 3248 million revolutions L b = 3248 × 106 revolutions Rev = life(hrs.) ×60rpm life(hrs.) = [(rev) / (60 x rpm)] Substituting values in (2), life(hrs.) = [(3248 × 106) / (60 × 1000)] life(hrs.) = 54,133 hrs.
(2)
7. An SKF Explorer 6309 deep groove ball bearing is to operate at 3 000 r/min under a constant radial load Fr = 10 kN. Oil lubrication is to be used, the oil has an actual kinematic viscosity ν = 20 mm2/s at normal operating temperature. The desired reliabili ty is 90% and it is assumed that the operating conditions are very clean. What is the basic rating life of the bearing? (940) Solution:
The basic rating life for 90% reliability is:
a) From the product table for bearing 6309, C = 55,3 kN. Since the load is purely radial, P = Fr = 10 kN (→ Equivalent dynamic bearing load). Therefore, in million revolutions
or in operating hours
8. What is the rating life in hours of operation (L10h) for deep groove ball bearing 6208 operating at rotational speed n = 650 min-1, with a radial load Fr of 3.2 kN or 326 kgf ? (19,000 hrs.) Solution:
The dynamic equivalent radial load: Pr Fr 3.2kN(326kgf) ＝
＝
Basic dynamic load rating Cr for bearing 6208 given is 29.1 kN (2970 kgf), ball bearing speed factor fn relative to rotational speed n = 650 min-1 from the given fn = 0.37. Thus life factor fh from formula is:
fh
fn
＝
=
0.37×
9. .
3.36
＝
Therefore, with fh = 3.36 from the rated life, L10h, is approximately 19,000 hours.
9. A punch press has a no. 308 spherical roller bearing. The main shaft runs at 1600 rpm and the bearing experiences a radial load of 2000 lbs. and a thrust of 1400 lbs. Under steady loading conditions and for outer bearing ring rotation evaluate
(a) bearing life for 90% reliability (b) bearing life for 96% reliability. Solution:
From table 9 – 4 (see prep), for a 40mm (8 × 5) bore size bearing, the values of basic static load rating C(o)1 = 12,900 lbs. and of basic dynamic load rating C1 = 13,700 lbs. In order to calculate the equivalent load Pe, the values of x and y are to be evaluated from table 9 – 3 The value of factor [(iFA) / (C(o)1)] where i=1 FA = thrust on the bearing is [(iFA) / (C(o)1)] = [(1 × 1400) / (12,900)] = 0.108 and value of factor [(FA) / (VFR )] Where: V = 1.2 for outer ring rotation FR = the radial load on the bearing [(FA) / (VFR )] = [(1400) / (1.2 × 2000)] = 0.5833 From table 9 – 3 (see prep), for [(iF A) / (C(o)1)] = 0.108, e1 = 0.30. Since [(FA) / (VFR )] > e1, so x = 0.56, y = 1.45. The equivalent load for the bearing is expressed by, Pe = xVFR + yFA Substituting values in (1), Pe = 0.56 × 1.2 × 2000 + 1.45 × 1400 Pe = 3374 lbs. Also, Pe = VFR = 1.2 × 2000 = 2400 lbs. Since 3374 > 2400 thus Pe = 3374 lbs. (a) The life of the bearing in revolutions is given by, L b = (C1 / Pe)3 = [(13,700) / (3374)]3 or, L b = 66.95 million revolutions In hours the life is, L b = [(66.95 × 106) / (n × 60)] where, n = rpm speed of shaft. L b = [(66.95 × 106) / (1600 × 60)] L b = 697.4 hours (b) From Figure 9 – 3 (see prep), for a 96% reliability, life adjustment reliability factor R r = 0.55.
(1)
Thus for 96% reliability, Life of bearing in rev = 0.55 × 66.95 × 106 = 36.8225 × 106 rev. Life of bearing in hrs. = 0.55 × 697.4 = 383.62 hrs. Thus we see that as the reliability is increased from 90% to 96% the bearing life decreases from 697.4 hrs. to 383.6 hrs.
10. A sleeve bearing has an outside diameter of 1.5in and a length of 2 in., the wall thickness is 3/16 in. The bearing is subjected to a radial load of 450 lb. Find the bearing pressure.(150psi) Solution:
Solving for the bearing pressure:
P=
F = = 150 psi D .
11. A rotating tube mill in a cement mill consists of a horizontal cylinder 7 ft. and 6 in. in diameter. This mill weighs 50 tons when loaded and is supported on 16-in rollers, two at each end, carried on 5-in journals. The rollers are located 30 deg on either side of the vertical centerline. The coefficient of rolling friction si 0.0125 and the coefficient of journal friction are 0.06. The mill revolves at 20 rpm. Determine the tangential vel. of the roller. (471 fpm.) Solution:
V = the tangential velocity of the roller = tangential vel. of the mill
V = πDN V = π(7.5)(20) V = 471 fpm.
12. A spinning cylinder mill in a cement mill consists of a horizontal cylinder 7 ft. and 6 in. in diameter. This mill weighs 50 tons when loaded and is supported on 16-in rollers, two at each end, carried on 5-in journals. The rollers are located 30 deg on either side of the vertical centerline. The coefficient of rolling friction si 0.0125 and the coefficient of journal friction are 0.06. The mill revolves at 20 rpm. Find the horsepower to overcome the rolling friction. (10.3hp) Solution:
F = tangential force =
co
F = 57, 735 lbs. P = the hp to overcome rolling friction
FV , ., P= , P=
P = 10.3 hp
13. The specific dynamic capacity of a rolling bearing is 7500 lb. If subjected to an equivalent radial load of 1000 lbs., 1600 rpm. What is the expected life in millions of revolutions? Solution:
L10 = the life in millions of revolutions for wheel bearing
L10
=
L10 = 825.79 million revolutions.
14. The weight on each wheel of a motor driven travelling crane is 6000 lb. The wheels are 2ft. in diameter and are carried on journals 4 in., in diameter and 8 in. long. The coefficient of rolling friction is 0.02 and sliding friction 0.5. Determine the horsepower of the motor required to drive a four-wheel crane travelling 500 ft/min. Solution:
. = 7.27 hp , .. = 3. 03 hp ,
Hprolling friction =
Hpsliding friction =
Hptotal = 7.27 + 3.03 = 10.3 hp
15. A sleeve bearing has an outside diameter of 0.75 in and a length of 2.5 in., the wall thickness is 3/16 in. The bearing is subjected to a radial load of 900 lbs. Determine the bearing pressure for the given parameters.( psi)
480
Solution:
Solving for the bearing pressure:
P=
F = 9 = 480psi D ..
16. It is given that a bearing transmits a load of 7.350 kN. The diameter of shaft is 100 mm, the coefficient of friction and the shaft speed are 0.01 and 650 rpm, respectively. Calculate the power loss in the bearing. (250.15 watts) Solution:
µ=
F W
F = µ(W) = 0.01(4,450) = 44.5 N
= 3.675 N m ) Plost = 2µ(3.675 N)( T = (73.5N)(
Plost = 250.15 watts
17. Determine the optimum model number for a cylindrical roller bearing operating at the rotational speed N = 450 min^-1, with a radial load F r of 200 kN (20,400kgf), and which must have a life ( L10h) of over 20,000 hours. Given:
The life factor f h = 3.02 ( L10h at 20,000) and the speed factor f n = 0.46 (n = 450 min-1). Solution:
Cr =
ℎ
Pr =
. × 200 .
Cr = 1,313 kN -- The smallest bearing that fulfills all the requirements is
NU2336.
18. Find the projected area and hence the length of a journal-bearing with a 3 in. diameter journal acted upon by a load of 2200 lbs. The shaft speed is 200 fpm, viscosity of lubricating oil is 32 centipoise and the coefficient of friction for the bearing is 0.01. Solution:
The speed of the shaft in rpm is given by, ns = [v / (0.262d)] where, ns = speed of shaft, rpm v = speed of shaft, fpm d = journal diameter, in. (Note: Journal is that portion of shaft that is covered inside the bearing.) Substituting values in (1), ns = [(200) / (0.262 × 3)] From Fig. 9 – 4 (see prep), for f = 0.01, the value of factor [(μns) / P] = 25. Therefore, 25 = [(μns) / P] Where:
μ = viscosity of lubricating oil, cp
P = bearing pressure, psi. Substituting values of known quantities in (2), 25 = [(32 × 254.45) / P] = P = 325.7 psi The projected bearing area, A b, is given by, A b = [(total load on the bearing, lb.) / (bearing pressure, psi)] (4) Substituting values in (4), A b = [(2200) / (325.7)] in. 2 = A b = 6.75 sq. – in. A b = L b × d where, L b = length of bearing. L b = (A b / d) = [(6.75) / 3] = 2.25 in.
∵ ∵
(1)
(2)
19. Evaluate the length of a journal bearing to support a shaft-pulley arrangement shown in Figure 1. The shaft is 5 ft. long and has a diameter of 1(1/2) in. The pulley, having a concentrated load of 1000 lbs., is located at a distance of 2 ft. from the right end of the shaft. The design stress value of the shaft is 150 psi and its weight is neglected.
Solution:
In Figure 1, R L and R R , the reaction provided by the bearing, are to be evaluated first in order to calculate the bearing length. Considering all moment of forces about left support point A and for equilibrium,
∑MA = 0 Thus,
(1)
1000 × 3 – 5 × R R = 0
or, R R = [(300) / 5] = 600 lbs. Similarly, considering moment of forces about right support point B and for equilibrium: ∑MB = 0 Thus, R L × 5 – 2 × 1000 = 0 or, R L = [(2000) / 5] = 400 lbs Since R R > R L, thus a value of R = 600 lbs. will be used to find the bearing length. Therefore the bearing length l b is given by, l b = [R / (Sd s)] where, S = design stress value ds = shaft diameter, Substituting values in (3), l b = [(600) / (150 × 1.5)] = 2.7 in. Thus a bearing with length of 2.7 in. will be suitable.
(2)
(3)
20. A 36 in. diameter pulley revolving at 500 rpm is mounted on a shaft driven by an electric motor as shown in Figure 1 below. The distance between the centers of the pulley and the bearing is 12 in. and the bearing friction coefficient is 0.02. The pulley is belt driven and the tensions on the tight slack sides are 400 lbs. and 200 lbs., respectively. Design a suitable journal bearing to support the arrangement.
Solution:
The journal (portion of shaft covered by bearing) is subjected to torsional, bending and shear stresses due to the belt tensions. These stresses will determine the size (length) of the journal bearing. The twisting moment induced in the journal due to net force (belt pull) acting on it is given by: T = (P1 – P2) (d p / 2) (1) Wherein:
P1 = belt tension on tight side P2 = belt tension on slack side (P1 – P2) = net belt tension = St d p = pulley diameter.
Table: Design data for journal bearing Type of Bearing Electric motors Pumps Automobiles: Main bearing Crankpin Wrist pin Air compressors: Main bearing Crankpin Wrist pin
Allowable Pressure, l b/d b Range p b psi 100-200 1.5-2.0 80-100 1.5-2.0 500-600 1000-1200 1200-1800
0.5-1.0 0.5-1.0 0.8-1.2
125-250 300-600 500-1000
1.0-2.0 1.0-1.7 1.5-2.0
St will produce a moment of force about the bearing center point. Substituting values in (1), T = (400 – 200) [(36) / 2] lb. – in. T = 3600 lbs. – in. The bending moment is given by, M = (P1 + P2)b Where: b = distance between centers of bearing and pulley. Substituting values in (2), M = (400 + 200) × 12 lbs. – in. M = 7200 lbs. – in. The equivalent moment, ME, due to T and M is given by, ME = (M/2)[1 + √{1 + (T/M)2}] Substituting values of M and T in (2), ME = [(7200) / 2] [1 + √{1 + {(3600) / (7200)}2}] ME = 7625 lbs. – in. The relation between the equivalent moment, ME, allowable bearing pressure 'p b' and moment of inertia for the journal 'I j' is given by, ME = [(p b × I j) / r] Where: I j = [(πd b4) / (64)] (for circular cross-section) (I j / r) = [(πd b4) / {(64) × (d b / 2)}] (I j / r) = [(πd b3) / (32)] Substituting expression for (I j / r) from (5) in (4), ME = p b × [(πd b3) / (32)] Rearranging (6) for d b, d b = [(32 × ME) / (π × P)](1/3) From the table, for an electric motor bearing, the allowable pressure is 150 psi (average value). Substituting values of ME and p b in (7), d b = [(32 × 7625) / (π × 150)](1/3) or, d b = 8 in. From the table, for an electric motor bearing, (l b / d b) = 1.75 (average value) where, l b = bearing length Thus, l b = 1.75 × d b or, I b = 1.75 × 8 = 14 in. The energy loss due to friction is given by, hf = f b × W × [(πd b × n p) / (12)] [(ft. – lb) / (min.)] (8) where, f b = bearing friction coefficient W = total bearing load, lb. and W = (P1 + P2)lb. n p = pulley speed, rpm. Substituting values in (8), hf = 0.02 × 600 × [(π × 8 × 500) / (12)] [(ft.– lb) / (min.)]
(2)
(3)
∴
∴ ∴
(4)
(5) (6) (7)
hf = 12566.371 [(ft. – lb) / (min.)] Since, 1 btu = 778[(ft. – lb) / (min.)] hf = [(12566.371) / (778)] [(btu) / (min.)] hf = 16.152 [(btu) / (min.)] The heat lost per unit of projected bearing area is: hf = [(16.152) / (l b × d b) [(btu) / (min. – in.2)] Substituting values of l b and d b in (9), hf = [(16.152) / (14 × 8)] [(btu) / (min. – in.2)] hf = 0.114 [(btu) / (min. – in.2)]
(9)
From Figure 9 – 11 (see prep), for a heat loss of 0.144 [(btu) / (min. – in.2)], the rise in temperature is about 40°F. This value is acceptable and the bearing designed is okay.
