¡Descarga Máquinas térmicas Pequeños depósitos endurecidos que se forman en los riñones y suelen dol y más Apuntes en PDF de Máquinas Térmicas solo en Docsity!
SOLUTION (10.1)
Known: A special C-clamp uses a 0.5-inch diameter Acme thread and a collar of
0.625-inch effective mean diameter.
Find: Estimate the force required at the end of a 5-in. handle to develop a 200 lb
clamping force.
Schematic and Given Data:
6 in.
1/2 in. Acme thread
d = 5/8 in. c
Assumptions:
- Coefficients of running friction are estimated as 0.15 for both the collar and the
screw.
- The screw has a single thread.
Analysis:
- From section 10.3.1, and considering that service conditions may be conducive to
relatively high friction, estimate f = f c
≈ 0.15 (for running friction).
- From Table 10.3, p = 0.1 in., and with a single thread, L = 0.1 in.
- From Fig. 10.4(a),
α = 14.
o
and d
m
= d −
p
= 0.5 − 0.05 = 0.45 in.
- From Eq. (10.1),
λ = tan
L
πd
m
= tan
π
o
- From Eq. (10.6),
α
n
= tan
(tan α cos λ) = tan
(tan 14.
o
cos 4.
o
o
(Note: with λ ≈ 4
o
, it is obvious that α
n
≈ α and well within the accuracy of
assumed friction coefficients)
- From Eq. (10.4),
T =
Wd
m
f !d
m
n
!d
m
cos "
n
f L
W f
c
d
c
T =
(150)(0.45)
2
(0.15)!(0. 45) + 0. 1(cos 14. 47˚)
!(0. 45)(cos 14. 47˚) " (0. 15)(0. 1)
(150)(0.15)(0.625)
2
T = 7.70 + 7.03 = 14.73 lb in. Use T ≈ 15 lb in.
At the end of a 6-in. handle, the clamping force required ≈ 15/6 = 2.5 lb ■
SOLUTION (10.2)
Known: A double-threaded Acme screw of known major diameter is used in a jack
having a plain thrust collar of known mean diameter. Coefficients of running friction
are estimated as 0.09 for the collar and 0.12 for the screw.
Find:
(a) Determine the pitch, lead, thread depth, mean pitch diameter, and helix angle of
the screw.
(b) Estimate the starting torque for raising and for lowering a 4000 N load.
(c) If the screw is lifting a 4000 N load, determine the efficiency of the jack.
Schematic and Given Data:
Load
4,000 N
f
d
c
f
c
d m
Double-threaded Acme
screw
d = 1 in.
d
c
=
f = 0.
f = 0.
c
50 mm
Assumptions:
- The starting friction is about 1/3 higher than running friction.
- The screw is not exposed to vibration.
Analysis:
- From Table 10.3, there are 5 threads per inch.
p = 1/5 = 0.2 in. = 0.0051 m ■
Because of the double-threaded screw,
L = 2p = 0.4 in. = 0.0102 m ■
T = 7.00 + 9 = 16.00 N•m
- From Eq. (10.4), the friction free torque for raising the load is
T =
[ ]
(0.0102)(cos 14.36˚)
π (0.02286)cos 14.36˚
= 6.49 N•m
- Efficiency = 6.49/16.00 = 40.6% ■
- Work input to the screw during one revolution = 2πT = 2π(16.00) = 100.5 N•m
- Work output during one revolution = WL = (4000)(2)(0.0051) = 40.8 N•m
- Efficiency = Work out/Work in = 40.8/100.5 = 40.6%
Comments:
- For a double threaded screw the work output during one revolution is WL where L
= 2p.
- If a small thrust bearing were used so that the collar friction could be neglected,
the efficiency would increase to 6.49/7.00 = 92.7%.
SOLUTION (10.3)
Known: A square-threaded, single thread power screw is used to raise a known load.
The screw has a mean diameter of 1 in. and four threads per inch. The collar mean
diameter is 1.5 in. The coefficient of friction is estimated as 0.1 for both the thread and
the collar.
Find:
(a) Determine the major diameter of the screw.
(b) Estimate the screw torque required to raise the load.
(c) If collar friction is eliminated, determine the minimum value of thread coefficient
of friction needed to prevent the screw from overhauling.
Schematic and Given Data:
Load
25,000 lb
d
m
= 1 in.
d c
= 1. 5 in.
4 threads/in.
f = f = 0.
c
f
d c
f
c
d m
Assumption: The screw is not exposed to vibration.
Analysis:
- From Fig. 10.4(c),
d = d
m
p
2
= 1 +
- 25
2
= 1. 125 in.
- From Eq. (10.4a),
T =
Wd
m
2
f !d
m
!d
m
" f L
W f
c
d
c
2
(25000)(1)
2 [ ]
(0. 1) π
(1)
- 25
π
(1) −
(0. 1)(0. 25)
(25000)(0. 1)(1. 5)
2
T = 2263 lb in. + 1875 lb in. = 4138 lb in. ■
- From Eq. (10.7a), the screw is self-locking if
f!
L
"d
m
f ≥ 0.08 ■
Therefore, the minimum value of thread coefficient of friction needed to prevent
the screw from overhauling is 0.08.
SOLUTION (10.4)
Known: A double-threaded Acme stub screw of known major diameter is used in a
jack having a plain thrust collar of known mean diameter. Coefficients of running
friction are estimated as 0.10 for the collar and 0.11 for the screw.
Find:
(a) Determine the pitch, lead, thread depth, mean pitch diameter, and helix angle of
the screw.
(b) Estimate the starting torque for raising and for lowering a 5000 lb load.
(c) If the screw is lifting a 5000 lb load at the rate of 4 ft/min, determine the screw
rpm. Also determine the efficiency of the jack under this steady-state condition.
(d) Determine if the screw will overhaul if a ball thrust bearing (of negligible friction)
were used in place of the plain thrust collar.
T = 799.9 + 640.1 = 1440 lb in. to raise the load ■
From Eq. (10.5),
T =
Wd
m
2
f
π d
m
− Lcos
α
n
π d
m
cos
α
n
f L
W f
c
d
c
2
=
3500(1.925)
2
[ ]
0
. 147
π (1.925)
0.5cos 14.45˚
π (1.925)cos 14.45˚
− 0.147(0.5)
3500(0.133)(2.75)
2
T = 230.0 + 640.1 = 870.1 lb in. to lower the load ■
4(12) in. /min
- 5 in. /rev
= 96 rpm ■
- From Eq. (10.4), with f c
= 0.1, f = 0.
T =
[ ]
π (1.925)
0.5cos 14.45˚
π (1.925)cos 14.45˚
T = 667.1 + 481.3 = 1148.4 lb in.
- From Eq. (1.3),
W
in
Tn
(1148. 4/12)
5252
= 1.75 hp
W
out
(3500)
33000
= 0.424 hp
Therefore, efficiency =
- From Eq. (10.7), the screw is self-locking if
f ≥
L cos!
n
"d
m
- 5(cos 14. 45˚)
Thus, if f = 0.11, the screw is self-locking and not overhauling. ■
SOLUTION (10.7)
Known: A square-threaded, single thread power screw is used to raise a known load.
The screw has a mean diameter of 1 in. and four threads per inch. The collar mean
diameter is 1.75 in. The coefficient of friction is estimated as 0.1 for both the thread
and the collar.
Find:
(a) Determine the major diameter of the screw.
(b) Estimate the screw torque required to raise the load.
(c) If collar friction is eliminated, determine the minimum value of thread coefficient
of friction needed to prevent the screw from overhauling.
Schematic and Given Data:
Load
13,750 lb
f
d
c
f
c
d m
d
m
= 1 in.
d
c
=
4 threads/in.
f = f = 0.
c
1.75 in.
Assumption: The screw is not exposed to vibration.
Analysis:
- From Fig. 10.4(c),
d = d
m
p
2
= 1 +
- 25
2
= 1. 125 in.
- From Eq. (10.4a),
T =
Wd
m
2
f !d
m
!d
m
" f L
W f
c
d
c
2
[ ]
π (1)
π(1) − (0.1)(0.25)
T = 1245 lb in. + 1203 lb in. = 2448 lb in. ■
- From Eq. (10.7a), the screw is self-locking if
f ≥
L
π d
m
π (1)
- From Eq. (10.1),
! = tan
" 1 L
#d
m
= tan
˚
- For starting, increase the coefficients of friction by 1/3:
f
c
= 0.133, f = 0.
- From Eq. (10.6),
α n
= tan
(tan α cos λ) = tan
(tan 14.
˚
cos 4.
˚
˚
- From Eq. (10.4),
T =
Wd
m
f !d
m
n
!d
m
cos "
n
f L
W f
c
d
c
5000(1. 925)
2
- 147 !(1. 925) + 0. 5cos 14. 45˚
!(1. 925)cos 14. 45˚ " 0. 147(0. 5)
T = 1142.7 + 831.3 = 1974 lb in. to raise the load ■
- From Eq. (10.5),
T =
Wd
m
2
f !d
m
" Lcos #
n
!d
m
cos #
n
W f
c
d
c
2
5000(1. 925)
2
- 147 !(1. 925) " 0. 5cos 14. 45˚
!(1. 925)cos 14. 45˚ + 0. 147(0. 5)
T = 328.5 + 831.3 = 1160 lb in. to lower the load ■
4(12) in. /min
- 5 in. /rev
= 96 rpm ■
- From Eq. (10.4), with f c
= 0.1, f = 0.
T =
- 11 !(1. 925) + 0. 5cos 14. 45˚
!(1. 925)cos 14. 45˚ " 0. 11(0. 5)
T = 953 + 625 = 1578 lb in.
10 From Eq. (1.3),
W
in
Tn
(1578/12)
5252
= 2.40 hp
W
out
(5000)
33000
= 0.606 hp
- Therefore, efficiency =
- From Eq. (10.7), the screw is self-locking if
f ≥
L cos!
n
"d
m
- 5(cos 14. 45˚)
Thus, if f = 0.11, the screw is self-locking and not overhauling. ■
SOLUTION (10.9)
Known: A jack uses a single square-thread screw to raise a known load. The major
diameter and pitch of the screw and the thrust collar mean diameter are known.
Running friction coefficients are estimated.
Find:
(a) Determine the thread depth and helix angle.
(b) Estimate the starting torque for raising and lowering the load.
(c) Estimate the efficiency of the jack for raising the load.
(d) Estimate the power required to drive the screw at a constant 1 revolution per
second.
Schematic and Given Data:
Load
50 kN
Single square-thread screw
d = 36 mm
p = 6 mm
d = 80 mm
f = 0.
f = 0.
c
c
f
d
c
f c
d m
p
Assumption: The starting friction is about 1/3 higher than running friction.
Analysis:
- From Fig. 10.4(c),
Thread depth = p/2 = 6/2 = 3 mm ■
- Work output during one revolution
= W• p = (50,000)(0.006) = 300 N•m
- Efficiency =
Work
out
Work
in
- Check:
Torque during load raising with f = f c
T =
(50, 000)(0. 033)
2
( )
π
T = 47.8 N•m
Efficiency =
T
(with zero friction)
T
(actual)
- Check (partial):
Torque during load raising if collar friction is eliminated = 173 N•m
Efficiency (screw only) =
- From Eq. (1.2),
W =
nT
(60)(413)
9549
= 2.6 kW ■
SOLUTION (10.10)
Known: An ordinary C-clamp uses a 1/2 in. Acme thread and a collar of 5/8 in. mean
diameter.
Find: Estimate the force required at the end of a 5-in. handle to develop a 200 lb
clamping force.
Schematic and Given Data:
5 in.
1/2 in. Acme thread
d = 5/8 in.
c
Assumptions:
- Coefficients of running friction are estimated as 0.15 for both the collar and the
screw.
- The screw has a single thread.
Analysis:
- From section 10.3.1, and considering that service conditions may be conducive to
relatively high friction, estimate f = f c
≈ 0.15 (for running friction).
- From Table 10.3, p = 0.1 in., and with a single thread, L = 0.1 in.
- From Fig. 10.4(a),
d
m
= d −
p
= 0.5 − 0.05 = 0.45 in.
α = 14.
o
- From Eq. (10.1),
λ = tan
L
πd
m
= tan
π(0. 45)
o
- From Eq. (10.6),
α
n
= tan
(tan α cos λ) = tan
(tan 14.
o
cos 4.
o
o
(Note: with λ ≈ 4
o
, it is obvious that α n
≈ α and well within the accuracy of
assumed friction coefficients)
Analysis:
- From Table 10.3,
p = L = 1/(1.75) = 0.571 in.
- From Fig. 10.4(d),
d
m
= d!
p
2
= 3!
- 571
2
= 2. 71 in.
and α = 2.
o
- Since cos α = 0.999, use Eq. (10.4a):
T =
Wd
m
2
f !d
m
!d
m
" f L
W f
c
d
c
2
T = 11,851 lb in. = 988 lb ft (during load raising)
- To raise the gate 36 in./min with L = 0.571 in.
requires 36/0.571 = 63.05 ≈ 63 rpm ■
- From Fig. (1.3),
W =
(63. 05)(988)
5252
= 11.9 hp say 12 hp
Therefore, 12 hp are required to drive each screw. ■
- Check:
Work output per gate = 52,000 lb (3 ft/min)
= 156,000 lb ft/min = 4.73 hp
! = tan
- 1 L
"d
m
= tan
o
From Fig. 10.8, Efficiency ≈ 40 %
Thus, W required
= 11.83 hp ■
SOLUTION (10.23)
Known: The bolt shown is made from cold drawn steel. The load fluctuates
continuously between 0 and 8000 lb.
Find:
(a) The minimum required value of initial load to prevent loss of compression of the
plates.
(b) The minimum force in the plates for the fluctuating load when the preload is 8500
lb.
Schematic and Given Data:
F
e
= 0 to 8 , 000 lb
k
c
= 6 k
b
F
e
F
e
Assumption: The bolt, nut, and plate materials do not yield.
Analysis:
- Compression of the plates is lost when F c
= 0 when maximum load is applied.
From Eq. (10.13)
F
i
= F
c
k
c
k
b
c
F
e
= 0 +
6k
b
k
b
b
8, 000 =
6
7
(8, 000) = 6,857 lb ■
- Minimum force in plates occurs when fluctuating load is maximum.
From Eq. (10.13),
F
c
= F
i
k
c
k
b
c
F
e
= 8, 500 -
6k
b
k
b
b
8, 000 = 8, 500 -
6
7
(8, 000) = 1,643 lb ■
0
20,
40,
60,
54,
50,
34,
Total force (N)
Time
F
b
F
c
F
e
10,
30,
SOLUTION (10.26)
Known: The cylinder head of a piston-type air compressor is held in place by ten bolts.
Total joint stiffness is four times total bolt stiffness. Each bolt is tightened to an initial
tension of 5000 N. The total external force acting to separate the joint fluctuates
between 10,000 and 20,000 N.
Find: Draw a graph (plotting force vs. time) showing three or four external load
fluctuations, and draw corresponding curves showing the fluctuations in total bolt load
and total joint clamping force.
Schematic and Given Data:
F = 10,000 to 20,000 N e
Assumption: The bolt size and material are such that the bolt load remains within the
elastic range.
Analysis:
- Using Eq. (10.13) for F e
= 20,000 N,
F
b
= F
i
( )
k
b
k
b
c
F
e
= [(5000)(10)] +
= 54,000 N
F
c
= F
i
( )
k
c
k
c
k
b
F
e
= 34,000 N
- For F e
= 10,000 N,
F
b
(10,000) = 52,000 N
F
c
(10,000) = 42,000 N
0
20
40
60
54
50
34
Total force (kN)
Time
F
b
F
c
F
e
10
42
52
30
SOLUTION (10.27)
Known: Two parts of a machine are held together by bolts that are initially tightened
to provide a total initial clamping force of 10,000 N. The elasticities are such that k c
2k b
Find:
(a) Determine the external separating force that would cause the clamping force to be
reduced to 1000 N.
(b) If this separating force is repeatedly applied and removed, determine values of
mean and alternating force acting on the bolts.
