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Basic- Machine- Design- Situational- Problem-BY- Alcorcon-1
Mechanical Engineering (University of Rizal System)
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Table of Contents
P.A.R.T •
MACHINE DESIGN SHORT PROBLEMS
I would like to dedicate this work to:
My wife gemma Rosa!5lIcorcon
To our children,
Jolin Cristophet and 5?l[e~andr
For their support
.
1. Strength of Materials
1
2. Mechanics
20
3. Machine Shop
40
4. Machine Elements
53
5. Stresses
58
. 6. Shaft
88
7. Keys
119
8. Coupling
130
9. Pressure Vessel
136
10. Bolts and Power Screw
149
11. Flywheel
162
12. Spring
172
13. Gears
187
14. Bearing
211
15. Belts
219
16. Brake
227
17. Clutch
231
18. Machineries
239
P.A R."'T'" ••
SITllATIONAL PROBLEMS
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274
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Strength 0/ MClteriaLs ~ Beams
1
STRENGTH C)F
IIVIA.TE R:J:A.LS
~
PARTl
S;;HO!RT
PRO:BLEM1S
... ~
..
.•... ..
..
' . . . ..
..
...•
'0'
.. ,
",
"" CANTILEVER BEAMS
1. Couple
~
With Fo;rmulas
a
~
Maximum Shear =
Maximum Moment = Mo
8 = maximum slope
~
~
~
~
L
M
e= M L
EI
Y = maximum deflection
~
M L2
y= 2Et
2. Concentrated load at free end
~
~
~ e= 2EI
v
~
>
t
P
P L3
3E I
3. Concentrated load at the mid­span
~
Shear = P
~
Moment =. -~.
PL
2
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~
*-pL
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51 reTlCjth
2
~e=PL2
or Materials ­ Beams
St r enqt l i
or Materials ­ Hewns
P L3
3
~
v> 48E I
8El
2. Uniform load
5P L3
~
y= 48E:1
~ Shear « wL
2
~
4. Uniform distributed load
Moment= w ~
wL
8
3
~ e:;; 24Et
~
4
~Y:; 384El
~
5wL
3. Varying load
~
5. Uniformly varying load
wL
~
2
Moment
~
L
~
~.
RESTRAINED BEAMS
~
1. Concentrated load at mid­span.
~
c SIMPLY SUPPORTED BEAMS
~
1. Concentrated load at mid­span
.J!i~:
~
­­­Lk
2:F
Moment", ~
PL
8
3
PL
P
~
Shear",
A
~ Y"'192EI
2. Uniform distributed load
wL
~
Shear", -
2
~
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t :
~
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4
~
St r en.qtl: of MClteriClls
Beams
,", For Rectangular Beams:
2
Moment:;:;··WL
12
1. SI = flexural stress
SI = ,6M
bh 2
Where: M = maximum moment
3. Uniform varying load
~
5
St icnqt t: (if l\,l(l(crials ­ Beams
2. Ss = shearing stress
3V
Ss = ­­.
2bh
.
Moment
Where: V = maximum shear
2
wL
20
Momem at B = ­­~
L
PROBLEM 1 (Apr. 1999)
A horizontal cantilever beam, 16 ft long is subjected to a load of 500 Ib located to its
center. The
dimension of the beam is 2 x 4 inches respectively. W = 100 Iblft, find
its flexural stress.
C 4510 pSI
A. 3150 psi
D. 5663 psi
B. 2912 psi
Y = deflection at mid­span
~
wl4
y == 768
,
SOLUTION
PROPPED BEAMS
roller at right end.
y = mid­span deflection
~
~
~p
IJ
For a cantilever beam:
1. Concentrated load at mid­span with fixed at left end and
F = total load at the center
•.
L
F = 500 + 100(16) = 2100lbs
Hl-.r~1i
500lbs
~ w =100 'b/..
L = 16 ft
'""""
M = maximum moment
2. Uniform distributed load with fixed at left end and pin at
right end
M
F x (U2)
M
2100(16/2) = 16,800in­lb
S = M_c_= ~2/4(0 8,61
I
= 3150 psi
(2)(4)3
12
~
~
PROBLEM 2
A 12 m simply supported beam with
30 KN load at the center has a maximum
2
deflection of: (EI = 6000 KN.m )
A. 150 mm
B. 160 mm
~
Under simply supported beam table.
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C. 170 mm
D. 180 mm
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6
St r cn at ! of Materials -Beams
SI rcn qtli o] Mat erials ­ Beams
4­
R 1 (10) = (4 x 2)(8) + 10(2)
R 1 = 8.4 KN
y = maximum deflection
3
PL
y = 48E I
where:
P = 30 KN
L = 12 m
2
EI = 6000 KN.m
...­
*30KN
ti.
12 m
10 KN
I'T1
~2rn
:2 kN/rT1l
10 rn
2:M 1 = 0
::6
7
R2 (10) = (4 x 2)(2) + 10(8)
t
R2
R1
R2 = 9.6 KN
By cutting at 10 KN section, and consider the right side of the section:
3
(30)(12)
y = 48(6000)
=0.180m=180mm
MA = R2 (2) = 9.6 (2) = 19.2 KN.m
PROBLEM 3
A 10m simply supported beam with 25 KN load 3 m from the left end has a
2
maximum slope at the left end: (IE = 8000 KN.m )
A. 00056 rad
C. 0.0186 rad
B. 0.0765 rad
D. 0.1823 rad
PROBLEM 5
A 10m simply supported beam with uniform load of 3 KN/m from right end to left end
2
has a maximum deflection of: (EI = 12,000 KN.m )
A. 43.23 mm
C. 54.23 mm
B. 32.55 mm
D. 36.45 mm
SOLUTION
14-" li it­U'
2
125 KN
3m...
2
Pb(L _b )
fk­ ­­­'­­­­­­­­'-
A
6 (E I) L
10m
y = maximum deflection
7rn
KN/nl
5 w L4
384
EI
Y=
~
where:
where:
w = 3 KN/m
L = 10 m
2
EI = 12,000 KN.m
P = 25 KN
b =7 m
L = 10 m
2
EI = 8000 KN.m
25 (7) (10 2 ­ 7 2 )
fk ­ ­':­­­'­­'­,­­­­­­'­
6 (8000) (1 0)
4
5 (3) (10)
= 0.03255 m = 3255 mm
y = 384 (12,000)
0.0186 radians
PROBLEM 4
A 10m simply supported beam has a uniform load of 2 KN/m extended from left end
to 4 m and has a concentrated load of 10 KN, 2 m from the right end. Find the
2
maximum moment at the 10 KN concentrated load. (EI = 10,000 KN.m )
A. 19.20 KN.m
C. 23.45 KN.m
B. 26.34 KN.m
D. 12.34 KN.m
..1-]lIO[-U'
Solving for the reactions at both ends:
LM2 = 0
'W = 3
PROBLEM 6
An 8 m simply supported beam has a uniform load of 2 KN/m from left end to right
end and concentrated load of 10 KN at the center has a maximum deflection of: (EI =
2
5,000 KN.m )
A. 12.34 mm
C. 21.33 mm
D. 34.34 mm
B. 42.66 mm
~1-]'·"Im
Considering the 10 KN concentrated load at the center.
3
PL
yl = 48E I
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St r c-iunl: of MUlerials -Beam s
St r enqt ii o] Materials ~ Beams
8
--
8EI
(10)(8)_ '" 0.021333 m
yl '" 48(5000)
Considering the effect of uniform distributed load of 2 KN/m.
5(2 )(8)4
­­­­­
384(5,000)
25 (14)2
8
'" 0.021333 m = 21.333 mm
8(9000)
iP
=25 KN
L= 14 m
0.068 radians
PROBLEM 9
A 10m cantilever beam has a uniform load of 2.5 KN/m from left to right end. Find
2
•
the maximum deflection of the beam. (EI = 11,000 KN.m )
A. 423. 45 mm
C. 323.45 mm
B. 198.23 mm
D. 284.10 mm
y = y1 + y2
y = 21.333 + 21.333
~
where:
P = 25 KN
L", 14 m
2
EI '" 9,000 KN.m
y1 '" 21.333 mm
5wL4
Y2 '" 384 EI
P L2
()=
3
9
42.66 mm
PROBLEM 7
A 10m cantilever beam has a concentrated load of 10 KN at the free end. Find the
2
maximum deflection of the beam.(EI = 7,000 KN.m )
A. 476.20 mm
C. 544.34 mm
B. 342.34 mm
D. 764.44 mm
EI!!IIl'!13I
W
L4
v SET
>
~+.J' "[.J"
y = maximum deflection
y",
P L3
t
~
3EI
P.10 KN
10 m
where:
w '" 2.5 KN/m
L = 10 m
2
EI '" 11,000 KN.m
4
­ ~L
= 0.2841 m = 284.10 mm
y - 8 (11,000)
where:
P = 10 KN
L = 10 m
2
EI '" 7000 KN.m
PROBLEM 10
A 10m cantilever beam has a uniform load of 2 KN/m from left to right end and with
concentrated load of 8 KN at the center. Find the maximum slope of the beam. (EI =
2
13,000 KN.m )
A. 0.011 rad
C. 0.033 rad
B. 0.022 rad
D. 0.044 rad
10(10)3
Y = ­ ­ ­ ­ = 0.47619 m = 476.20 mm
3(7000)
PROBLEM 8
A 14 m cantilever beam has a concentrated load of 25 KN at the mid­span. Find the
2
maximum slope of the beam. (EI = 9,000 KN.m )
A. 0.00234 rad
C. 1.23 rad
B. 0.068 rad
D. 0.123 rad
EI!!IIl'!13I
Considering the effect of the uniform load of 2 KN/m using beam table,
81
EI!!IIl'!13I
Using beam table, for a cantilever beam with concentrated load at mid­span,
"'~-
6EI
2 (10)3
fl l
'"
6(13.000)
'" 0.0256 rad
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St renqili of MateriClls . Becuns
10
Strength of MCllcliu[s
Considering the effect of concentrated load at the mid-span,
2
fh = P L
8EI
_8 (10)2
8 (13,000)
O2 =
= 81 + 82
8 = 0.002564 + 0.007692 = 0.0333 radians
PROBLEM 11
A 14 m cantilever beam has a load of 16 KN 6 m from the fixed end.
2
maximum slope of the beam. (EI = 6,000 KN.m )
A. 0.183 rad
C. 0863 rad
B. 0.048 rad
D. 0.064 rad
IHN"imD
I
Pa 2
2EI
16 (6)2
2(6,000)
Find the
6m
IP
=16KN
T
.
PROBLEM 13
A 10m simply supported beam has a triangular load from zero at left end to a
maximum of 10 KN/m at the right end. Find the maximum deflection of the beam. (EI
2
= 10,000 KN.m )
C. 84.67 mm
A. 5424 mm
B. 65.10 mm
D. 76.56 mm
where: w = 10 KN/m
L = 10 m
2
EI = 10,000 KN.m
0.048 radians
PL 2
81 = 2EI
8 (12)2
0.054 radians
4
Considering the effect of concentrated load at free end:
2(8,000)
1.5(12)3
6 (8,000)
2.5 W L
Y= 384EI
L = 14 m
Em!!miD
­­­­
6EI
For a triangular load with simply supported beam,
PROBLEM 12
A 12 m cantilever beam has a uniform load of 1.5 KN/m extended from fixed end up
to 4 m and a concentrated load of 8 KN at the free end. Find the maximum slope of
the beam. (EI = 8,000)
A. 0.126 rad
C 0.654 rad
B. 0.234 rad
D. 0.345 rad
81
L
Em!!miD
where:
P = 16 KN
a=6m
2
EI = 6,000 KN.m
8
l
W
8 = 81 + 82
8 = 0.072 + 0.054 = 0.126 radians
For a cantilever beam with a concentrated load at a distance from fixed end:
8
11
= 0.007692 rad
tl 2
8
lkwns
P = 8 KN/m
W =1.5 KN/m
0.072 radians
Considering the effect of uniform load of 1.5 KN/m.
L
=12 m
4
y
2.5(10.li1.QL = 0.06510 m
­384(10,000)
65.10mm
PROBLEM 14A 8 m cantilever beam has a triangular load from left end to a maximum of 12 KN/m
2
at the fixed end. What is the maximum deflection of the beam? (EI = 12,000 KN.m )
A. 136.53 mm
C. 165.34 mm
B. 145.34 mm
D. 15434 mm
.1e ] , I " [ e 1S'
For a triangular load of cantilever beam,
4
W L
y = 30EI
where:
w = 12 KN/m
j
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Sf rCllgt h of Materials - Beams
12
SlrcIII/II, ojl\lul('riuls
L=8m
2
EI = 12,000 KN.m
y=
_2(8)~
13
I'I<OULJ:-:M 17
A vertical load of 400 N acts at the end of a horizontal rectangular cantilever beam 2
m long and 25 mm wide. If the allowable bending stress is 130 Mpa, find the depth of
the beam.
A. 23.45 mm
C. 56.34 mm
B. 38.43 mm
D. 45.34 mm
__
30 (12,000) - 0 13653 m = 136.53 mm
PROBLEM 15
A 12 m restrained beams with two end fixed has a concentrated load of 15 KN at the
2
center. Find the maximum deflection of the beam. (EI = 7,000 KN.m )
A. 8.45 mm
C. 19.28 mm
B. 34.23 mm
D. 24.67 mm
E::1!!iirmI
E::1!!iirmI
For a cantilever beam with load act at the free end:
P = 400 N = 0.40 KN
M=PL
M = (0.40)(2)
M = 0.80 KN.m
For restrained beam with concentrated load at the center,
P = 15 KN
PL3
Y = 192EI
~
where: P = 15 KN
L = 12 m
2
EI = 7,000 KN.m
15(12)3
Y = 192-(7,000)
Beams
L:12m
~
Using the formula of flexural stress,
6M
s= --.2
bh
PROBLEM 16
A 15 m restrained beam with two end fixed has a uniform load of 3 KN/m, find the
2
maximum deflection of the beam. (EI = 9000KN-m )
A. 56.34 mm
C. 34.56 mm
B 23.56 mm
D. 43.94 mm
130000 = 6 (0.80)
,
(0.025) h2
h
PROBLEM 18
A simply supported beam is 50 mm by 200 mm in
25 mm
cross section and 4 m long. If the flexural stress is not to exceed 8.3 Mpa, find the
maximum mid-span concentrated load.
A. 2.77 KN
C. 3.45 KN
B. 6.34 KN
D. 4.34 KN
E::1!!iirmI
SOLUTION
For restrained beam with uniform distributed load,
where: w = 3 KN/m
L = 15 m
2
EI = 9000 KNm
400 N
2m
h = 0.03843 m = 38.43 mm
0.01928 m = 19.28 mm
wL4
Y = 384EI
t. p :::
~
~
For a simply supported beam with load at the center,
M = maximum moment
M = PU4
M=~
~p
4
A
M=P
L = 15 m
4m
­:.a
Using the shearing stress formula for beam
s= §~
3 (15)4
Y = - - - - - - = 0.043945 m = 43.945 mm
384(9000)
bh 2
Where:
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200 mm
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SI r ctiqt.t: of Materials - Beams
]4
Sln'T/fl/l,
b = 0.050 m
h = 0.20 m
2.77 KN
PROBLEM 19
A simple wooden beam, 50 mm wide by 250 mm deep and 7.5 m long has a
maximum deflection of 102 mm under a uniform load "w". E = 12,411 Mpa
a. What is the uniform load, w.
b. What is the slope?
c. What is the maximum moment?
d. What is the maximum shear?
e. What is the flexural stress?
f. What is the shearing stress?
Em!!'.iIrSmI
bh 3
1=12
f.
S =
2.'­". = _3J7.5l~
2bh
= 900 K a
2(0.05)(0.25)
9
(0.050)(0.250 )3 = 0.0000651 m 4
12
4
5w(7.5)
384(12411 x 103 )(0.0000651)
50 mm
2m
6m
~P2m
~
e
PROBLEM 21
A cantilever beam having a span of 6 m, carries a triangular load of 20 KN/m at its
fixed end to zero at the free end of the beam. Determine the deflection at the free
12
2
end of the beam if EI = 50 x 10 N.mm .
A. 17.28 mm
C. 15.62 mm
B. 16.48 mm
D. 12.63 mm
Em!!'.iIrSmI
3
wL
24EI
2 (7.5)3
24 (12411 x 10 3 )(0.0000651)
0.0435 rad
For triangular load of cantilever beam:
w = 20 KN/m = 20 N/mm
wL4
M = maximum moment
y = 30 EI
M=wL 2/8= 2(7.5)2
8
1
a
2mP~
648 (30,760 x 10 )
w = 2 KN/m
d.
P
PROBLEM 20
A simply supported beam having a span of 6 m carries two concentrated loads of 80
KN at its middle thirds. Compute the deflection at the mid­span due to this load. Let
9
EI = 30,670 x 10 N .rnrn".
A. 18 mm
C. 24 mm
B. 20 mm
D. 30 mm
Y= 23(80,000)(6,000)3 = 20 mm
250 mm
c.
6(14)
= ~= 26,880 Kpa = 26.880 Mpa
S = ­­
b h2
(0.05)(0.25)2
3
y = 23PL
648EI
y = 384EI
b.
6M
e.
4-i-]'lin-gl
a. For a uniform loading beam under simply supported:
5wL4
0.102=
]5
Beams
V = 2(7.5L = 7.5 KN
2
6P
._­­ ­ 2
8,300 = (0.05) (0.20)
1=
III j\!1u/eriols
= 14KN.m
Y=
20 (6,000)4
12
30 (50 x 10 )
= 17.28 mm
V = maximum shear = w U2
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"1­]lliit­a
PROBLEM 22
A simply supported rectangular beam 50 mm wide by 100 mm deep carries a uniform
load of 1200 N/m over its entire length. What is the maximum length of the beam If
the flexural stress is limited to 20 Mpa.
A 3.33 m
C. 5.2 m
B. 4.25 m
D. 6.0
m
P = 50 KN
~
PL3
Y=
3E1
t
29.63 = 50,000 (4,000)3
38­'-
1i'­]lI"t­U'
EI = 35.9996 x 10
wL2
Moo ­­8
s,
6M
20 = ­­­­­­
(50)(100)2
9
N.m
2
= 35,999.6 KN.mm 2 = 36,000 KN.m 2
PROBLEM 25
A restrained beam carries a triangular load which increases from zero to 30 KN/m
from A to B having a span of 4 m. Determine the moment at A
A. 12 KN.m
C. 18 KN.m
B. 24 KN.m
D. 16 KN.m
100 mm
6M
bh 2
50 mm
Emut­U'
3
M = 1,666 x 10 N-mm = 1,666 N-m
w:;; 30 KN/m
2
1 666 = 1200 L
,
8
2
30
\p
PROBLEM 23
A simply supported beam having a span of 20 m carries a concentrated load of 10 KN
at a distance of 5 m from the left support and a counter clockwise couple of 6 KN.m
acting 3 m from the right support. The maximum moment due to this load is:
A. 39 KN.m
C. 36 KN.m
B. 43 KN.m
D. 48 KN.m
..i­]'II1t­a
20 R1 = 10 (15) + 6
R1 = 7.8 KN
wL
MA
L = 3.33 m
LMs = 0
17
Strength of MUlerials - Beams
St rcnot u or Materials - Beams
If)
5
110 KN
6 m+
A
tR'
20 m
f
M
A
= EO) (4)2
36'
B'
16 KN.m
I
I
I
I
I
~/
I
L=4 m
PROBLEM 26
A cantilever beam 3 m long carries a concentrated load of 35 KN at its free end The
material is structural steel and the maximum bending stress is not to exceed 125
Mpa. Determine the required diameter of the bar if it is circular
A. 204.5 mm
C. 164.6 mm
B. 188.6 mm
D. 179.2 mm
6 KN-m
P
SOLUTION
~
3m
(\B
M = 7.8 (5) = 39 KN.m
Moo 35 (3)
M = 105 KN.m
PROBLEM 24
A cantilever beam having a span of 4 m caries a concentrated load of 50 KN at the
free end. What is the flexural rigidity if the beam deflects 29.63 mm at its free end.
2
2
A. 36,000 KN.m
C. 24,800 KN.m
2
2
B. 42,000 KN.m
D. 54,200 KN.m
A
s = ~c
6
105 X 10 N.mm
I
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=35 KN
t
lOMoARcPSD|3535879
Strength oiMa u-ii at-,
SIII'II9Ih of M(lterials ­ Beams
IS
M
\
\
fJ8 KN.m
1'1\( lllLEM 29
nd
1
\
I' ,
105xlO h (di2)
125
64
d
Ii/,w,,"
= 20450 mm
PROBLEM 27
A steel beam 2 m In length is simply supported at each end and carnes a
concentrated load of 100 KN acting 05 m from one of the supports. Determine the
maximum bending stress set up in the beam if the cross-section is rectangular, 100
mm wide by 150 mm deep.
C. 97 Mpa
A. 100 Mpa
D. 150 Mpa
B. 120 Mpa
Em!!ImD
A restrained beam having a span of 6 m carnes a uniform IOdd 01 O~ Nzrn throughout
6
Its span. If E = 200,000 Mpa, I = 6080 x 10 rnrn" determine the deflection at the midspan.
C. 036 mm
A 0.14mm
D. 0.48 mm
B. 0.25 mm
Em!!ImD
wL
384 EI
50,000 (6) (6,000)3
y­­­­­
6
\
384 (200,000)(6,080 x 10 )
LMB = 0
2 R1 = 100 (1.5)
0.5 m ~
R, = 75 KN
6
M = 37.5 KN.m = 37.5 x 10 N.mm
6M
bh 2
S = ­­­.­ = -~
6(37.5X 10 6 )
S = 100 N/mm
p = 100 KN
.a­­ 2m- - B
M = 75 (0.50)
100(150)2
2
tR1
iI'50mm
= 100 Mpa
100 mm
PROBLEM 28
A beam made of titanium has a yield point of 850 Mpa. The beam has 25 mm x 50
mm rectangular cross-section and bends about an axis parallel to the 25 mm face. If
the maximum bending stress is 650 Mpa, find the corresponding bending moment.
A. 6.8 KN.m
C. 4.5 KN.m
D. 8.4 KN.m
B. 8.4 KN,m
m:mmmI
6M
S = ---.
bh 2
6M
650= ­ ­ ­ (25)(50)2
M = 6.8 X 10" N.mm
~
4
y
f150mm
25mm
Downloaded by King Nixon Lagrisola (knlagrisola@gmail.com)
0.14 mm
L = 15 m
lOMoARcPSD|3535879
20
Mechunics
21
Mccllullics
Note.
If a body is at rest, then initial velocity IS zero.
If a body is moving at constant speed, then acceleration IS zero.
If a body is completely stop after the trip, then velocity is zero
~
~
~
2. Free Falling Body
Vf=O
a. VI = va ± gt
Fr
,
c. S= Vbl ±%gf
•
•••
•
=
I
I
I
I
I
**
" "
,,
••
•
·
I
I
I
I
I
I
5
_BL:r
tN
-
Va
•
(. If speed is uniform:
S = vt
­
s
If speed varies:
Vo
2
± 2aS
~
•
-~
I
vJ
Note:
If a body is dropped freely, then initial velocity is zero
~
If a body is thrown upward, then it reached to a maximum height where the
velocity is zero.
Vf
-----.
14
9=0
H
I~
c.S= Vat ±% a t;l.
R
~
x
where
a = acceleration
a is + if accelerating
a is - if decelerating
V o = initial velocity
VI = final velocity
S = distance traveled
t = time
I
3. Projectile Motion
­­­­­.
2
r,
I
I
~
1. Rectilinear motion
- motion is along a straight path.
b. Vf
·
g = acceleration due to gravity
g is + if a body is moving down
a is - if a body is moving up
Vo = initial velocity
v, = final velocity
S = distance traveled
t = time
~w
=f N
a. Vr = Vo ± at
2
Vl
where:
F = frictional force
f = coefficient of static friction
N = normal force
"
I
± 2g8
b.
IEImm
2
Vo
,
1''''''-
Components of initial velocity, va
vox
=v; cose
Voy = va sine
For Horizontal Displacement:
"The horizontal component of initial velocity will not change throughout the
flight. vox = Vlx contant
=
l
x = horizontal displacement = vox t
Downloaded by King Nixon Lagrisola (knlagrisola@gmail.com)
(v, coss) t
lOMoARcPSD|3535879
22
"
Mcch anic:«
R = maximum range
Vo
2
23
Mechanics
.
(. For uniform speed:
sln28
9
" a
For Vertical Displacement:
,
y = verticaldisplacemeI1t
vay t
1t':igrS v"Sjn8.t - 112 9 t2
nDN
Where:
" Vfy = final velocity along vertical = Vay • g t.= vasih!:l· gt
w = angular velocity, rad/s, rev/sec
2
a = angular acceleration in rad/s", rev/s
t = time
8 = rad or rev
a = acceleration, m/s"
R = radius of rotation, m
N = speed, rev/s
v = velocity, m/s
"
f., Vt == final resultant velocity ==
Vo
c;.. H = maximum
2nRN
" v
Note:
29
Use + if accelerating
Use - if decelerating
Where:
Vox = initial velocity along horizontal
v oy = initial velocity along vertical
Vlx = final velocity along horizontal
Vfy = final velocity along vertical
VI = final velocity
Vo = initial velocity
t = time of flight
"
Polar Moment of Inertia
Using English units:
Note:
~
The horizontal component velocity is constant at any time of flight.
~
At maximum point the value of 8 = O.
~
If Y is below the reference point, then the sign of y is negative.
Where:
J m = polar moments of inertia of masses, It­lb­sec"
2
p = Ib!ft
L = length, It
2
g = 32.2 ft/sec
J = polar moment of inertia of area with constant cross-section, ft4
4. Circular Motion
Radian ­ is a unit of angular measure equal to 57.30°
Angular speed ­ is the angle through which it turns per unit time.
&zgular acceleration ­ is the rate of change of its angular speed with respect to time.
.. '
FORMULAS:
~/
For accelerating object:
c;. WI :: Wa ± a t
r~
'W?=Wa ± 2 a 6
6 == wo t ± 112 a
r
Jm = P L J
\~ "
Where:
2
J m = kg_m
(, Radius of Gyration
:/
­,
Using 51 units:
..........
Wo ••·•
2
"
e­-
­
WI
Using English Units
-:
".
"~
......... __ .::
Downloaded by King Nixon Lagrisola (knlagrisola@gmail.com)
p = kg/m
3
L = length, m
J = m"
lOMoARcPSD|3535879
24
K" = radius of gyration
Ko ==
g~P
(f
J m := Wk o
9
25
Mechu7Iics
Mechu7Iics
= anqular acceleration, rad per sec?
2
USing 81 Units:
2
T o:= J m a == mko a
Where:
Where:
J m = polar moments of inertia of masses, It-lb-sec"
To := torque, N-m
Ko = ft
g = 32.2 tt/sec"
W := weight, Ib
J,'l := moment of inertia, kg_m
2
a := angular acceleration, rad per sec
Using 51 Units
Kinetic Energy
(. Kinetic Energv of Translating body
2
1
2
KE:= ­mv
2
KE == Wv
29
2
Where:
J m:= kg_m2
p:= kg/m 3
Ko:= m
m := mass, kg
Where:
"
2
Radius of Oscillation
For isosceles triangle:
For circle:
For parabola:
314 of height
5/8 of diameter
5/7 of the height
KE = kinetic energy, Joules
m:= mass, kg
v = velocity, mls
W:= weight, N
g = acceleration due to gravity, rn/s"
(, Kinetic Energy of Rotating body
"
Center of Percussion
KE == 1fz Jln {t)2
Q := distance from axis of rotation to center of percussion
Where:
J m:= moment of inertia, tt-lb-sec''
(. Total Kinetic Energy
Where:
r := the distance form axis of rotation to center of gravity of body
KE := Y2mv2 + Yz. J m {t)2
(,. Force of a Blow:
(., Formulas relating Torque and Angular Acceleration
Average force of blow :=
Using English units:
To == J m a
ws
d
Where:
8 := total height, It
W := weight of driver in Ibs
o := distance in feet which pile is driven
«:.. Linear Impulse and Momentum
Where:
To := torque in pounds-feet
2
J", := moment of inertia, ft-lb-sec
Ko:= radius of gyration, It
W
'
L Inear rnornentum» m v = - v
9
Downloaded by King Nixon Lagrisola (knlagrisola@gmail.com)
()) = angular velocity, radlsec
lOMoARcPSD|3535879
r,
26
\1-
I
Mechanics
Mcrhultics
Linear impulse » F x t
F :=: forcet = time
~
m = mass
27
\, Unbanked curve
v = velocity
W :=: weight
Angular impulse and Momentum
tan (8 + x) :=:
v2
a
==
~-
9
gR
H
AOQularmornentUfTl :=: J m W
Angular impulse :=: To x t
Where:
f :=: coefficient of friction
R :=: radius of curvature
8 = super elevation angle or banking angle
x :=: side thrust friction angle
H = super elevation
Angular impulse :=: Change in angular momentum
TdX f=,J m{4>f -wo)
.
.
~,
.
Fe
\,. Acceleration
Normal acceleration :=: V'/r
Tangential acceleration :=:dvldt
,
Centrifugal Force
j
Fe :=: rnas, :=: rI1 (l/r)
\
I
Ig',
:=:
wv 2
gr
V
.~
•I
W
FR
I
I
I
FR = .m!l
I R
I
I
I
I
I
tan
~'
v2
noitoM~.,- -
~n
_
• • •
FR
H
9
~._
gR
Where:
m = mass, kg
~
Downloaded by King Nixon Lagrisola (knlagrisola@gmail.com)
.
c.
~.-
-;;::._
.....
..
9
• •
FR
2. If the object is decelerating, the direction of reverse effective force is the
same with the direction of motion.
Banked curve
a
9
.
Banking Curves:
f
~a
1. If the object is accelerating, the direction of reverse effective force is
opposite to the direction of motion.
...
v = velocity, m/s
r :=: radius of curvature, m
f = coefficient of friction
a = acceleration, m/s"
m = mass, kg
W = weight, N
g :=: 9.81 m/s" = 32.2 ft/s 2
\,
=w
Direction of Reverse Effective Force
Where:
"
reverse effective force
W:=: weight, N
Motion
i
a = acceleration, rn/s"
lOMoARcPSD|3535879
:2H
PROllLEM I (OCt. 2000)
A car travels with an initial velocity of 10 m/s or 36 km/hr. If it is decelerating at the
rate of 3 rn/s", how far, in meters does it travel before stopping?
A. 17
C. 19
B. 21
D 15
~
From the ground to a maximum height:
V" = 114km/s==114,000m/s
2==V 2
0
Vt==Vo-gt
-2aS
.:
o == 114,000 - 9.81 t
(0)2 == (10)2 - 2 (3) S
..,.
6 m/s
/ /. £"\
,,'-
VI == 0 at maximum height
PROBLEM
Vt
29
Mecltwlics
!Hech ci nie s
i
:_-------------
­­-
.....
~
~,
l
t = 11620.795 sec == 3.228 hrs
S = 16.67 m
Time of object in going up == time of object in going down
PROBLEM 2 (Oct. 2000)
A block weighing 56 Ibs rest on horizontal surface. The force needed to move along
the surface is 20 Ibs. Determine the coefficient of friction.
C 0.36
A. 0.0
B 0.112
D.028
Em:!!imD
So that total time to go up and back.
== 3.228 + 3.228 == 645 hrs
PROBU':M 5 (ME Bd. Oct. 97)
An occupant moves toward the center of a merry go around at 6 rn/s If the merry go
around rotates at 6 rpm. Compute the acceleration component of the occupant
normal to the radius.
A. 6.79 rn/s"
B. 8.29 m/s"
F==fN==fW
C. 7.54 rn/s"
D. 3.77 rn/s"
20 == f (56)
~
f == 0.357
v == 2rcRN
6 == 2 rt R (6/60)
R == 955 m
PROBLEM 3 (Oct. 2000)
A baseball is thrown straight upward with a velocity of 20 m/s. Compute for the time
elapse for the baseball to return. Assume for a zero drag.
A. 1.84
C. 250
B. 2.21
D. 2.04
a == acceleration
a
Em:!!imD
a
VI = Va - g t
v2
R
62
9.55
3.77 m/s
2
o == 20 - 9.81 t
PROBLEM 6 (ME Bd. Apr. 98)
t == 2.038 sec
PROBLEM 4 (Apr. 1999)
A 114 km/s speed is projected vertically how long will it take to go back to its initial
position.
C. 7.23 hrs
A. 645 hrs
B. 1.23 hrs
D. 845 hrs
,\'
,~
A wheel accelerates from rest with a = 5 rad/sec.sec. Compute how many revolutions
are made in 4 seconds.
A. 5.71 rev
C.700rev
B. 6.36 rev
D. 2000 rev
...-r.JX'ur.1~.
1 rev == 2rc rad
,I';,
..
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lOMoARcPSD|3535879
30
31
Mechanics
Me(' han i ('.s
e
H c= w.t + 1/2 (f
w, c= 0 (from rest)
o c= 0 + 1/2 (5/2rr)(4)2
e = 6.36 rev
mv~
i
c=
Fe
mg
R
t
9 R
v c= 343 m/s
PROBLEM 7 (ME Bd. Apr. 98)
What minimum distance can a truck slide on a horizontal asphalt road if it is traveling
at 25 m/s. The coefficient of sliding friction between asphalt and rubber is at 0.6 The
weight of the truck is 8500 kg.
A. 44.9
C. 532
B. 58.5
D. 63.8
Solving for N:
v==2rrRN
343 = 2 rr (1.2) N
V
//--1'_.'l~ -.
/ = 9.81(1.2)
I
A.2m),
,,
,
.
,
,
;'
,
···
'
'
·
"~"
,
...... "
N = 0.455 rev/s X 2rr = 2.86 rad/s
SOLUTION
v = 25 m/s
--.
v=o
--.
I..
;1
I
PROBLEM 10 (ME Bd. Apr. 98)
Compute the speed a satellite to orbit the earth at an elevation of 100 km.
radius is at 6400 km. Assume no change of gravity with the elevation
C. 8740 m/s
A. 6320 m/s
B. 7120 rn/s
D. 7920 rn/s
I
Em!!DtmI
Considering the truck:
FR
-----.
IFH = 0
~
Fr c= FR
W
f W = ---a
g
a c= f g = (0.6)(9.81)
c=
5.887 m/s
To maintain the satellite to orbit the earth,
Fe = r,
2
L,<.-X,,l,
I
2
2
981 (6400,000 + 100,000)
v == 7985 m/s
PROBLEM 11 (ME Bd. Apr. 98)
A liquid full is to be rotated in the vertical plane. What minimum angular velocity in
radians/sec is needed to keep the liquid not spilling if the rotating arm is 12 meters?
A. 2.26
C. 3.16
B. 25.6
D. 2.86
To keep the liquid not spilling out,
gR
v
PROBLEM 9 (ME Bd. Apr. 98)
SOLUTION
mv_ == mg
R
i
After the slide it will stop so that V2 = o.
2
2
V2 = V1 + 2aS
o = (25)2 + 2(5.886)S
S = 532 m
An elevator weighing 2000 bs is moving vertically upward with an acceleration of 3
tt/s". A man standing in said elevator weighs 180 lbs. Compute the tension in the
supporting cable under this condition.
C 1820lbs
T
A. 2180lbs
B. 2191 Ibs
D. 23831bs
t
....{.)'i'iMDI
I Fv c- 0
W == total load
00
~
f
Fe = Fg
Earth's
W = 2000 + 180 = 2180 Ibs
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FRt
lOMoARcPSD|3535879
32
MechcUlics
Mechanics
T
PROBLEM 14 (ME Bd. Apr. 96)
A drop hammer of 1 ton dead weight capacity is propelled downward by a 12 in
diameter cylinder. At 100 psi air pressure, what IS the Impact velocity if the stroke IS
28 inches?
C. 15.8 fps
A. 63.2 fps
o 474 fps
B. 316 fps
W + FH
W
W+ - a
g
T
T
2180
+~-
2180
32.2
3:)
(3) = 23831bs
Em!!ImD
PROBLEM 12 (ME Bd. Apr. 98)
The capsule orbits the earth 180 km above the surface. In what velocity in m/s
necessary for a circular orbit consider the earth radius at 6400 km and at 92 rn/s"?
A. 8864
C. 7780
B. 7016
O. 8058
W = 1 ton = 2000 Ibs
Em!!ImD
Force = Pressure x Area
Solving for the force acting on piston:
Force
= P (2C 0 2 )
4
Fe = F g
Force = 100 [ (rr/4)(12)2]
Force = 11,310 Ibs
~
rnv"
--=mg
R
6400 km
mv 2
--=g
w
R
R = distance from center of
Earth
earth to the capsule
R = (6400 + 180) 1000 = 6,580.000 m
Substitute:
v2
------=9.2
6,580,000
v = 7780 m/s
v} = 0 + 2(214.3)(28/12)
V2 = 31.6 fps
PROBLEM 13 (ME Bd. Apr. 96)
A truck skids to a stop 60 m after the application of the brakes while traveling at 90
Km/hr. What is its acceleration in rn/sec"?
A. -5.21 rn/sec"
C. 6.36
B -7.06
O. 5.76
SOLUTION
V1
90(1000)
= --._.--= 25 rn/s
3600
v/ = V,2 + 2aS
V2 = 0 (stopped)
o = (25)2 + 2a(60)
a = -521 rn/sec"
F = ma
F = (W/g) a
11,310 = (2000/322) a
2
a = 182.1 ft/s
v/ = V,2+ 2aS
v, = 0 (from rest)
a = 1821 + g
2
a = 1821 + 322 = 2143 ft/s
-
v = 90 kph
~
~
......
'4
5 = 60 m
-
v=o
....,
PROBLEM 15 (ME Bd. Apr. 96)
A flywheel rotates at 120 rpm or 12.57 rad/sec slowed down to 102 rpm or 10.68
rad/sec during the punching operation that requires 3/4 second of the punching
portion of the cycle. Compute the angular acceleration of the flywheel in rad/sec".
A. -2.52 rae/sec"
C. 3.15 rae/sec"
0 2.22 rad/sec''
B. -2 75 rae/sec"
...
. :r .~ ji[.l~.
a
ex
u
W2 ="!1
10.68 -12.57
-_.---~----
0.75
-252 rae/sec'
' ,.
\
'~r
Downloaded by King Nixon Lagrisola (knlagrisola@gmail.com)
lOMoARcPSD|3535879
34
Mcciuinics
35
Mechanics
C. 400 m
D. 600 m
A. 350 m
B 500 m
PROBLEM 16
A 2-ton weight is lowered at a constant acceleration of 2 rt/s".
stress?
A 4,344.34 Ib
C. 3,751.55 Ib
B. 5,344.56 Ib
D. 6,356.24Ib
What is the cable
"'.-]",,[-1\-
Solving first the acceleration:
Vr = Vo + at
50=20+a(10)
SOLUTION
Solving for the distance traveled:
2
S =0 Vo t + Y2 a t
S = 20(10) + 1/2 (3)(10)2
S = 350 m
Cable Stress
v =20 m/s
v = 50 m/s
•
•
----+
----+
I.
t = 10 seconds.
~I
When lowering the load:
r
FR = direction of FR is upward since the
direction of load is downward.
W
FR= - a
R
F
g
FR -
Ex20~)
32.2
Em!!ImD
W = 2 tons
(2) = 248.45 Ib
PROBLEM 19
A box sliding on a floor has a frictional resistance of 100 N. If coefficient of friction is
0.3, find the weight of the box.
A. 222.22 N
C. 444.44 N
B. 111.11 N
D. 333.33 N
Fr =0 frictional resistance
l:F v
=0
0
r. = f N
Cable stress = W - FR
Cable stress =0 (2 x 2000) - 248.45
Cable stress =0 3,751.55 Ibs
100 = 0.3 N
N = 333.33 N
PROBLEM 17
The speedometer of a vehicle changes from 20 km/hr to 80 km/hr In 10 seconds.
Find the acceleration for this period in m/s".
A. 1.67
C. 2.34
B. 6.34
0 8.45
W = N =0 333.33 N
PROBLEM 20
An object dropped on a gravity travels 300 m after how many seconds?
A. 5.34 sec
C. 7.82 sec
B. 6.34 sec
D. 10.23 sec
SOLUTION
V1 = 20 (1000/3600)
V1 = 5.55 m/s
V2 = 80 (1000/3600)
V2 =0 22.22 m/s
v =20 kph
v =80 kph
~
•
•
Va
----+
~
----+
t = 10 seconds.
~
S
V2 = V1 + at
22.22 = 5.55 + a(10)
a =0 1.67 rn/s"
=0
=0
0 (dropped)
Vo t
+ '12 g t
2
300 =0 0 + '12 (9.81) t
I
\
Ii
,
PROBLEM 18
A car changes accelerates its speed from 20 m/s to 50 rn/s in 10 sec.
distance traveled during the period of acceleration.
Find the
t =0 7.82 sec
Downloaded by King Nixon Lagrisola (knlagrisola@gmail.com)
5 =300 m
2
lOMoARcPSD|3535879
3G
Mechanics
2
-
v, 2 - 2 g S
s
VI = 0 (at maximum height)
0
2
V =40 m/s
S = 8155 m
­­:­::­­0" -,
IlE!!i.mD
From maximum height to the ground:
'
Va = 0
Vt =
­­-
H = maximum height
PROBLEM 22
A stone is projected upward from the ground travels a maximum height of 50 m. Find
the velocity of stone when it returns to the ground.
A. 31.32 m/s
C. 4323 rn/s
O. 2345 m/s
B. 5434 m/s
2
981
17877 m
R
E:mImD
t
= (40)2 - 2 (9.81) S
(45)2 Sin 2(30)
R
PROBLEM 24
A shot is projected 40° from the horizontal at the rate of 30 m/s will reached the
maximum height of:
C. 32.34 m
A 20.34 m
O. 4534 m
B. 18.95 m
SOLUTION
VI
37
Mechanics
PROBLEM 21
An object IS projected upward at the rate of 40 m/s. How high will the object rise?
C 65.34 m
A. 5434 m
O. 8155 m
B. 7645 m
v/ + 2 g S
'
,
H =
'u
~-_.
2
(~0)2
. 2
sin 0.
2g
.~_,-
sin ~O
2(9.81)
2
H=1895m
,\
,
I
'
'
I
I
•
I
'
'
I
I
'
I
'
'
I
I
S'SOml tv
V? = 02 + 2 (9.81) (50)
H =
V
~
PROBLEM 25
A capsule orbit the earth at the rate of 6000 rn/s. If the radius of earth is 3000 km,
find the height of capsule from the surface of the earth ..
A. 435.66 m
C. 543.23 m
B. 669.72 m
O. 78545 m
.i·"··"·lh"
The capsule will orbit the earth if:
Vf = 31.32 m/s
•
6400 krn
Fe = Fg
PROBLEM 23
An object is projected upward at the rate of 45 m/s at 30° from the horizontal. Find
the maximum range of projectile.
A. 178.77 m
C 18934 m
o 10934m
B. 165.34 m
SOLUTION
~
"r;;.
"
/
/
~<;:
R
R
maximum range
2 sin 28
g
~
30'
,
,
"
(6,000)2 = 9.81 R
,
,
/'7/7/%7777/// ; \
I..
Wv 2
- - ~W
gR
R = distance from center of earth to the
capsule above the earth surface.
2
v = g R
.1
R = 3669724.771 m
R = 3669.724 km
R = r" + H
3669724 = 3000 + H
H = 669.72 m
Downloaded by King Nixon Lagrisola (knlagrisola@gmail.com)
w
Earth
lOMoARcPSD|3535879
38
Mcclwnic!i
PROBLEM 26
A car that travels 100 mph reduces its speed 50 mph in 30 seconds.
distance traveled by the car until it stops.
A 4402.10 m
C. 5434.56 m
B. 3452.34 m
D. 6456.35 m
Find the
v =100 mph
~
e..
I.
V = 50 mph
~
Tan H
Sl
30 sec
~
(40)2
Tan H
8 = 9.26
SOLUTION
-
S2
0
V=O
~I
= 50 (5280/3600) = 73.33 ft/s
VI = va - at
14667 = 73.33 - a (30)
a = 2.44 m/s"
SI
=
Vo t
- Y2 a t
2
S, = 146.67 (30) - Y2 (2.44) (30)2
S, = 3300.10 m
Solving for S2:
VI
2
=
Va
2
-
2a S2
0 2 = (73.33)2 - 2(2.44)S2
S2 = 1102 m
S = total distance traveled
S = 3300.10 + 1102 = 4402.10m
PROBLEM 27
A car traveling at 40 m/s in a radius of 1000 m will have a banking angle of:
0
A. 7.34°
C. 9.26
B. 183.4°
D. 12.3°
SOLUTION
Tan 8
v = 40 m/s BI""'1II
-I iL~
0.1631
..
~u
---
9.81 (1 000)
Va = 100 (5280/3600) = 14667 ftls
VI
39
Mechanic!i
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gR
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,
/
/
.....
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I
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\
I
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~I
0
R=1000m,
\
I
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lOMoARcPSD|3535879
40
Machine Shop Pta ct icc
Mact n n«: Shop Practice
IV~C="':J
3
11'1 E
!:» ..... C» F»
41
m:mmB
Co
Time = Length to be,welcl
Welding rate
(; Volume Consumption of gas. V
V = (Va + Va) L
~
Where:
V = total gas consumption of oxygen and acetylene
Va = volume of oxygen consumed per unit length of seam
Va = volume of acetylene needed per unit length of seam
L = length of seam
Im!!Jm]
(. Cutting Speed, v:
v = 11: 0 N
~
Where:
v = cutting speed
D = diameter of workpiece
N = speed
t.. Time of Milling
"
PH.OBLEM 1 (Oct. 2000)
= Length to be cut
Determine the time in seconds, to saw a rectangular magnesium bar 5 in wide and 2
in thick if the length of cut is 5 in. The power hacksaw does 120 strokes/min and the
feed/stroke is 0.127 mm.
Cutting rate
(; Time»
.~Le_n.-':g,thC0bcu
No. ot teeth x Speed, N x Feed rate
A 189
C. 99
B. 500
D. 20
~1·]'"[B
~
"
Time of Cutting = Length to be cut
Cutting rate
time
Time
= Length to be cut
Cutting rate
'" For a hacksaw, there is only one cut stroke per revolution.
c. Time: .__
Lengthtobe cut
Speed, N x ~trok
rev
x . ~.e F
stroke
Length to be cut
--"- - . - - - - - - -
Cutting rate
For a hacksaw. there is only one cut stroke per revolution.
Feed per stroke = 0.127 mm = 0.005 in
time =
5
--_._--_..- - - - - - - - = 8.33 min
strokes
in
120-.
(0.005---)
min
stroke
500 sec
PROBLEM 2 (Oct. 2000)
In an Oxy-Acetylene manual welding method. to weld a 3 1/2 ft long seam in a 0.375"
thick steel plate at a consumption rate of 9 ft3/ft for oxygen and 7 ft31ft for acetylene.
Compute for the total combined gas consumption in ft3
A 48
C. 56
B 24.5
D. 31.50
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lOMoARcPSD|3535879
42
Machine Shop Practice
SOLUTION
Machine Shop Prcut ic«:
43
Cutting time- 10.42 min
v = total gas consumption
PROBLEM 6 (Oct. 1998)
Compute the cutting speed in fpm of a workpiece with 2 inches diameter and running
at 100 rpm.
A. 52
C. 25
D. 26
B. 62
v = (Vo + Va) L
= 56 ft3
V = (9 + 7) (35)
PROBLEM 3 (Oct. 1998)
Compute for the drill penetration in in/min when a drill turns at 1200 rpm and the feed
of 0.005 inlrev. Material is steel.
A. 1.8
C. 12
B. 3.6
D. 6
ED:!ImI:I
V=ITDN
V = IT (2/12) (100)
52.35 ftlmin
SOLUTION
PROBLEM 7 (ME Bd. Apr. 98)
Drill Penetration = (Feed rate) N
Drill Penetration = (0.005 - ' ~
)(1200 rev)
rev
min
6 in/min
PROBLEM 4 (Oct. 1998)
What is the value in degrees of 1 radian?
C. 57.29°
A. 5749°
B. 57.94°
D. 57.92°
Compute the cutting speed in fpm of a workpiece with 2 inches diameter and turning
at 100 rpm?
A. 72
C. 62
B. 102
D. 52
ED:!ImI:I
v=rrDN
v = IT (2/12)(100)
SOLUTION
1 rad x] 80°
IT rad
57.29°
PROBLEM 8 (Oct. 2000)
PROBLEM 5 (Oct. 1998)
Compute the manual cutting time in minutes, of a steel plate 4 ft by 8 ft by 2.5 cm
thick with a hand cutting speed of 3.8 to 4 mm/sec, cutting lengthwise.
A. 10
C. 0.38
B. 1.81
D. 8.16
-=~.
.
TIme =
Considering the average value:
Length
Determine the average time to cut by automatic oxygen-acetylene (machine)
crosswise a 4 ft x 8 ft x 4 in thick steel plate.
A. 6.85 min
C. 10 min
B. 318 sec
D. 360 sec
Using the average cutting rate of 9 in/min for oxygen-acetylene cutter.
SOLUTION
Cutting speed
= 52 ftlmin
~8+4
2
=3.9 mm/sec
Time
Length to be cut
cutting rate
4(12)
---
9
-r-
.
5333 min
= 5333(60)
320 sec
= 8 ft (12) (25.4) = 2438.40 mm
Length
Cutting time = ·Cuttingspeed
, PROBLEM 9 (ME Bd. Apr. 96)
2438.40 = 625.23 sec
3.9
Calculate the rpm for machining a cast iron workpiece 6 inches in diameter.
lowest cutting speed for cast Iron is 50 fpm.
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The
lOMoARcPSD|3535879
44
Machine Shop Practice
A. 35.3
C. 43.3
l\!Jachine Shop Pra ct icc
C. 33.3
O. 53.32
45
I'~O()LEM
12 (ME Bd. Oct. 95)
How long Will it take to saw a rectangular piece of aluminum plate 8 in wide and 1 1/2
In. thick if the length of the cut is 8 in, the power hacksaw makes 120 rev/min and
average feed per stroke is 0.0060 in?
A.13.11
C.11.11
B 14.01
O. 12.03
SOLUTION
v=nON
50 = rr(6/12)N
E1':'m!immI
N = 31.83 rpm
The answer is: C (nearest)
time
Length of cut
------------
Cutting Rate
PROBLEM 10 (ME Bd. Oct. 95)
With the arc welding rate of 18 in/min, how long will It take to weld 1/2 in thick plate by
3 feet long seam?
A. 3 min
C. 2 min
B. 1.5 min
O. 4 min
11
A1
J1
1/2"
SOLUTION
Length to be weld
time
- ' " - - - - - -
Welding rate
L = J'
time = (3 x 12)/18 = 2 min
V
V
For a hacksaw. there is only 1 cut stroke per revolution.
.
&n
.
time = - - - -.--- -------..- - - . - - = 11.11 min
120 ev x~troke
xO.0060
In
min
rev
stroke
.r.
PROBLEM 13 (ME BD. Oct. 95)
Using oxyacetylene welding method to weld a 3 ft long seam in a 3/8 in thick plate at
a consumption rate of 9 ft31ft of weld for oxygen and 7 felft for acetylene. What is the
total combined gas consumption in ft3?
A. 51
B. 48
C. 45
O. 55
E1':'m!immI
PROBLEM 1 1 (ME Bd. Apr. 95)
How long will it take to mill a 3/4" by 2" long keyway in a 3" diameter shafting with a
24 tooth cutter turning at 100 rpm and 0.005" feed/tooth?
A. 0.136 min
C. 0.196 min
B. 0.166 min
O. 0.106 min
v
total gas consumption of oxygen and acetylene
v
ft3
(9 + 7 ) - - - (3) ft length = 48 ft3
ft length
SOLUTION
PROBLEM 14
time
A grinding wheel rotates 1750 rpm and has a surface speed of 2290 ft/min.
diameter of the wheel in inches.
A. 3 in
C. 4 in
B. 5 in
O. 6 in
Length to be cut
-----_.
- . -
Cutting rate
time
2in
24 te~
x1 o~e_v
rev
time
0.167 min
"-N !Ii it-D
x~.O-
min
tooth
v=rrON
2290 = rr (0) (1750)
0=5 in
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Find the
lOMoARcPSD|3535879
46
PROBLEM 15 (ME Bd. Apr. 96)
To facilitate the milling (roughing) work of a cast steel material using a 1 1/2 inch
diameter cutter choose between the two available stock. Find the
speed of the
cutter in rpm. -high speed steel cutter with a cutting speed of 50 fpm - carbide tipped
cutter with a cutting speed of 200 fpm
A. 389
C. 572
B. 509
O. 412
SOLUTION
47
MUc!liTI<' Shop Practice
Mcicttinc Shop Practice
o = 0.955 ft (12) == 11.461n
PROBLEM 18
A workpiece turning at 20 rad/s has a diameter of 4 in. Find the peripheral speed of
workpiece
C. 5.23 fps
A. 1.23 fps
O. 7.34 fps
B. 3.33 fps
EmmmD
v=nON
High speed steel cutter (slower)
50(12) = n(1.5)N
N
20 ( 1rev)
N
3.183 rev/s
2n
v == nON = IT (4/12) (3.183) = 3.33 fps
N = 127 rpm
PROBLEM 19
Carbide tipped cutter (faster)
Two plates 4 ft x 8 It plate % in thick is to be joined lengthwise. A welding machine to
be used has a welding rate of 20 in/min. How long will it take to join the plates?
.
A. 7.34 min
C. 4.80 min
B. 1.23 min
O. 6.45 min
200(12) = n(1.5)N
N = 509 rpm
EmmmD
PROBLEM 16
A piston reciprocates 100 times per minute (single stroke). The length of the stroke is
6 in. What is the average wristpin velocity in ft/min?
A. 50 ft/min
C. 75 ftimin
B.100ft/min
0.125ft/min
SOLUTION
\
Time
--------._--
Time
8 x 12' = 4.8 min
20
Welding rate
,
v == LN
ft
stroke
.
v == (6/12) _ . - (100) ---.- = 50 ft/rnin
rrun
stroke
4"
Length to be weld
~
(
8"
PROBLEM 17
A workpiece turninq at 300 rpm has a tangential speed of 15 ft/s. Find the diameter. PROBLEM 20
I A 10 It long seam can be weld in 10 min. The welding rate is:
of the workpiece.
A. 8 in/min
C 10 in min
C. 13.23 in
A. 11.46 in
B. 12 in/min
O. 14 in/min
O. 17.34 in
B. 15.23 in
EmmmD
V=nON
EmmmD
.
Length to be weld
Time = . - - - ----.--Welding rate
15 == IT (0) (300/60)
~."fi _
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I
8''''
lOMoARcPSD|3535879
i'vlachirlc ShoJi )'/rl( I ice
48
Machine Shop Prartire
1l4·I l!mm
10(12)
10
--
- - --- -----
Welding Rate
Length to be cut
time =
A seam can be weld in 3 min. If welding rate is 15 in/min, find the length of seam.
A. 20 in
C. 25 in
D. 45 in
B. 30 in
SOLUTION
time
5
0.2
25 sec
PROBLEM 24
Length to be weld
-------~--
Welding rate
L
Length to be weld
It takes 1 min to mill a 3;'/ x 4" keyway of a hub by means of a cutter having 30 teeth
at 120 rpm. Find the equivalent feed per tooth, In/tooth
A 000222
C 0.00333
B 000444
D. 0.00111
EI!!ImD
---------~
15
Length to be weld
= 45 in
time
PROBLEM 22
How long will it take to mill %" by 1.5" long keyway by means of 25 teeth cutter turning
at 120 rpm and 0.008" feed/tooth.
A. 1.23 sec
C. 4.23 sec
D. 375 sec
B 634 sec
SOLUTION
Length to be cut
---.---------_._-
Cutting rate
- - ----
4
,---~
------.
rev
teeth
120 . x 30 -- - x Feed per tooth
min
rev
Feed per tooth
= 0.00111 in/tooth
PROBLEM 25
How long will it take to cut a rectangular plate 4 it wide by means of a saw that has a
speed of 160 rev/min and an average feed per stroke of 0.009 in?
A. 22.22 min
C. 55.55 min
D. 44.44 min
4"
B 3333 min
Length to be cut
time
--
time
------------"
time
,-----'
Cutting rate
time
PROBLEM 21
3
-
12 in/min
Welding rate
Time
-~--
Cutting rate
1.5
120 ~ev
x 25 teeltl x O.OOB
min
4~)
rev
0.0625 min (60)
EI!!ImD
in
tooth
= 3.75 sec
time
Length to be cut
-.---
---
-
Cutting rate
For a hacksaw, there is only one cut stroke per revolution.
PH.OBLEM 23
A 5 in long keyway is 10 be mill at the rate of 0.2 in/min. How long will it take to mill
the keyway?
A. 20 sec
C 25 sec
B. 15 sec
0 30 sec
4 (12)
lime = .'
----.
160 rE3~
x l~trokE3.
x 0.009
in
min
rev
stroke
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= 33.33 minute
_.~
~
lOMoARcPSD|3535879
,~!uchine
50
PROBLEM 26
A 4ft wide x 8 ft long and 1,12 In thick steel plate is be cut into 4 equal parts crosswise
If cutting rate is 0.02 In/stroke and saw speed is 110 rpm, how long will it take to do
the job?
A. 53.45 min
C. 65.45 min
D. 67.45 min
B. 3423 min
4"
SOLUTION
L
L
~
L
total length of cut
L
4 cuts (4 ftl cut)
time
total length of plate to be cut
3 cuts (4 ft / cut) = 12 ft
Length__to be cut
.._--
time
time
--
Cutting rate
4"
= 16 ft
---,
..- - - - - - - - -
Cutting rate
16 (12)
10
19.20 min
USing oxyacetylene welding method to weld 10ft long seam In a 1,12 In thick plate at a
3
3
consumption rate of 10ft /ft oxygen and 8 ft /ft for acetylene. What is the total gas
consumption?
C,160fe
A 150 ft3
o 180 fe
B. 170 ft'!
PHOBLEM 27
1/2"
~
It takes 5 min to cut a plate 20 ft long by means of a saw having a speed of 250 rpm.
What is the average feed per stroke 01 the saw, in/stroke?
A 0.534
C. 0.745
B. 0.192
D 0.342
v
Total gas consumption
~
v
Volume ()f~9
(Length to be weld)
length to be weld
V
(10+8)(10)
V
180 ft'!
Length to be cut
-_.__
- - - - - _ ~ .
Cutting rate
20
20 (12)
-.---- .----.--.----.---
250
rev
stroke
.-- x 1 -- - x Feed per stroke
min
rev
Feed per stroke
EW'M~:sazc
11
For a hacksaw, there is only one cut stroke per revolution.
5=-
8"
Length to be cut
PROBLEM 29
For a hacksaw, there is only one cut stroke per revolution.
12(12)
.
time = -.~_c
= 65.45 min
rev
stroke
In
110- x 1 - --- x 0.02
min
rev
stroke
time
51
r\i!uchinc Shop Prac t icc
Shop Piact icc
PROBLEM 30
A 24 ft2 lateral surface for a cylindrical tank is to be made by joining 2 ft by 2 ft steel
3/lt
plate to form 6 It x 4 ft. The consumption rate for both oxygen and nitrogen is 20 ft
using oxyacetylene welding method. Find total gas consumption needed, fe.
A, 250 fe
C. 440 fe
3
B. 470 ft3
D. 280lt
~
o 192 in/stroke
PROBLEM 28
4"
2"
An oxyacetylene cutter IS use to cut a 4 ft x 8 ft steel plate crosswise and divided the
whole plate into five equal parts. If the cutting rate is 10 in per minute, how long Will It
take to cut the plates.
A 19.20 min
C. 21.23 min
D 1734 min
B. 13.45 min
2"
L = total length to be weld
L = 6+4+4 = 14ft
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~
6"
~
lOMoARcPSD|3535879
52
Total gas consumption
Total gas consumption
Mac n in« Shop Practice
·
..
4.
(20) (14)
280 ft3
PROBLEM 31
.
A 10ft X 20 ft plate is to be divided into 8 pieces 5 ft x 5 ft steel plate by means of an .
oxyacetylene cutter that has a cutting rate of 10 in per minute. How long will it takes
to do the Job?
C. 60 min
A. 50 min
B 70 min
O. 80 min
SOLUTION
10"
L = 20 + 10 + 10 + 10
L = 50 ft
.
Length to be cut
time = - - - . - - Cutting rate
M1ACHINE
ELEIVIEN-r!:»
-
5"
5"
L = total length to be cut
53
Mach inc Elcrncli t s
1. Cylinders Rolling in opposite direction:
20"
50 (12)
10
= 60 min
~
Tangential speed
V2
V1 = V2 = TI 01 N1 = TI 02 N2
(.
Relation of diameter and speed
D1 N1 = D2 N2
to Speed Ratio ==
".
Center Distance
Speed of Driver
Speedof the Driven
R + R
1
2
c
= D 1 +~2
2
2. Cylinders Rolling in the same direction
4L.
Tangential speed
V1 = V2 = TI D1 N1 = TI D2 N2
\.
Relation of diameter and speed
01 N1 = D2 N2
(.
Speed Ratio
Speed of Driver
= Speed of the Driven
I
I
(.
Center Distance
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R z - R1
l:)?=_l:)l
2
101(
.
C
..
I
lOMoARcPSD|3535879
54
Mach.irie Elements
PROBLEM 1 (ME Bd. Apr. 97)
Two parallel shafts connected by pure rolling turn in the same direction having a
speed ratio of 2.75. What is the distance of the shafts if the smaller
cylinder is 22
cm in diameter?
A. 16.67 cm
C. 30.25 cm
B. 25.25 cm
O. 19.25 cm
PROBLEM 3
A 5 ft diameter pulley is turning at 200 rpm. Find the speed of the pulley mid-way of
the center and outer surface of the pulley.
A. 20.45 ft/s
C. 24.34 ft/s
B. 30.45 fils
O. 26.18 fils
E'iI!!ImD
SOLUTION
v == 2 IT Im N
Using the relation of 0 and N:
0 1 N1 == O2 N 2
22(2.75) == O2(1)
O2 == 60.5 cm
r == 5/2 == 2.5 ft
rm == 2.5/2 == 1.25 ft
v == 2 t: (1.25) (200/60)
For cylinders in the same direction:
C == O2 _=-P-l
2
v == 26.18 fils
PROBLEM 4
60.5 - 22
C
_ ..
C
19.25 cm
~-
A wheel is turning at 360 rpm.
the radius of the flywheel?
A. 8ft
B. 2ft
2
PROBLEM 2 (ME Bd. Apr. 97)
A lineshaft runs at 360 rpm. An 18" pulley on the shaft is belt connected to a 12"
pulley on the countershaft.
From a 15" pulley on the countershaft, motion is
transmitted to the machine. Compute/check the required diameter of the pulley on the
machine to give a spindle speed of 660 rpm.
A. 16"
B. 121/4"
C. 101/2"
O. 81/2"
12"
SOLUTION
OL N L
55
Mci ci': inc ElcITICf/( S
== Dc Nc
If tangential speed of the wheel is 15080 ft/s, what is
C. 4 ft
O. 10ft
E'iI!!ImD
v==ITON
150.80 == IT 0 (360/60)
o
8ft
R
8/2
4ft
18(360) == 12N c
Nc == 540 rpm (speed of countershaft)
Dc Nc == OMNM
If center distance is
Lineshaft
Countershaft
Machine
15(540) == OM(660)
SOLUTION
OM == 1227
Say 12 1/4"
Speed ratio == - ~ N
-
3
-1-
2
0 1 N 1 == D 2N 2
O2 == (N,/N 2)DI
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'.
I
C
= 40 in.1"
lOMoARcPSD|3535879
57
Machine Elcmcnt s
5G
Machine Elements
EmmiImI
D2 = DI (3/1)
DA N A = DB NB
DA (6) = DB (4)
DA = 2/3 DB
D2 = 3 D,
For cylinders rolling in opposite direction:
C =~_-+:
40
D,
DB NB = De Ne
DB (4) = De (2)
De = 2 DB
D
Dc
CAe = - A + DB + -2
2
2
3D + D
1
1 _._-,,-
2
20 in (smaller pulley)
PROBLEM 6
Three cylinders A:8:C rolling in external contact has a speed ratio of 4:2:1. If the
diameter of cylinder A is 10 ern, find the center distance between cylinder A and C.
A.40cm
C.45cm
8. 35 cm
D. 50 ern
K-N':JmmI
DA NA
DB NB
10(4)
DB (2)
DB NB
20 (2)
o, (1)
2/3 DB
---- + DB + 2D
--B
2
2
30 = 2.33 DB
DB = 12857 in
DA = 2/3 (12.857) = 8.571 in
Dc = 2 (12.857) = 25.714 in
2
Dc = 40 cm
CAe
=30in
De Ne = Do No
25.714 (2) = Do (1)
Do = 51.428 in
2Do
D
CAO = - A + DB + Dc + - 2
2
8571
51.428
CAO = - - - + 12857 + 25714 + - - -
DB = 20 cm
o, Nc
CAe =
~c
Dc
DA
+ DB+--
2
2
10
40
CAe
-- + 20 + --2
2
CAe
45 cm
PROBLEM 7
Four cylinders A:8:C:D has a speed ratio of 6:4:2:1. If center distance between A and
C is 30 inches, find the center distance A and D.
A. 68.57 m
C. 5634 m
8. 83.45 m
D. 7634 m
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2
68.57 in
lOMoARcPSD|3535879
58
51 resses
I
5
STRESSES
12. Strain (unit strain) - is the total deformation measured in the direction of the line
of stress.
13. Modulus of Elasticity - is the ratio of stress to the strain.
14. Residual Stress - internal inherent, trapped locked-up body stress that exists
within a material as a result of things other than external loading such as cold
working, heating or cooling, etching, repeated stressing and electroplating.
~
15. Compressive Strength - is the ability of a material to resist being crushed.
16. Bending Strength - is the quality of a material which resists forces from causing
a member to bend or deflect in the direction in which the load is applied.
~
Force
. 2
2
2
1. Stress (S) = -----, lb/in , kg/cm , KN/m
Area
Stress - a total resistance that a material offers to an applied load.
2. Ultimate stress (Su) - is the stress that would cause failure
3. Yield stress(Sy) - maximum stress without causing deformation
4. Allowable stress(Sail)
59
Stresses
~
1. Tensile Stress (S,)
S, = Ft
A
D
For solid circular cross-section: A = 2: 0"
4
= Ultimate stress/Factor of Safety
For hollow circular cross-section: A = ~(Oo
5. Design stress(Sd) - stress used in determining the size of a member.
Based on ultimate stress:
Based on yield stress: ~"i=-c±
where: FS = factor of safety
6. Working stressfSc) = stress actually occurring under operating condition.
7. Endurance limit or fatigue limit = the maximum stress that will not cause failure
when the force is reverse indefinitely.
8. Elastic limit - is the maximum stress to which a standardized test specimen may
10. Strength - is the ability of metal to withstand loads without breaking down.
11. Shear strength - is the ability of metal to withstand forces thus following a
number of twist.
_Oj2)
F
For rectangular cross-section: A "'base x height
2. Compressive Stress(Sc)
Sc =
F
Fe
A
For solid circular cross-section: A = ~
be subjected without a permanent deformation.
9. Proportional Limit - is the stress at which the stress-strain curve deviates from a
straight line.
2
4
4
For hollow circular cross-section: A = ~
02
~D
2
4
(0 0 _Oj2)
For rectangular cross-section: A = base x height
Note: For both compressive and shearing stress, the area is perpendicular to the line
of force.
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Srress es
61
5 I r esses
3. Shearin g Stress( Ss}
A
For single bolt of rivet needed to join to plates together.
Ss
5. Factor of safety( FS}
D
F
a. Based on yield strengt h
F
where:
cross section)
~ircula
(fO~
.il4p~
For single rivet: A
For double riveted joint: A =2(rr74 p2)
F '" applied load
Note: The shearin g area is always parallel
to the line of force.
b. Based on ultimat e strengt h
F
II
where:
B. Shearin g due to punchin g of hole.
Sail ::: allowab le stress
Su = ultimate stress
5 y '" yeild stress
(for punchin g a hole)
(for square hole)
t~
Where: S '" length of side of square
t '" plate thickne ss
6. Torsion al Shear Stress( Ss}
nOs
where:
T '" torque
J '" polar momen t of inertia
Pressur e needed to punch a hole. F:
(focsolipShaft)
F= (fXlx8 0,tons
~£.
c '" distanc e from neutral axis to the farthest fiber
c '" r (for circular cross section)
d = diamete r
d '" hole diameter, in
t '" thicknes s, in
7. Bendin g Stress( Sr}
FROM MACH INERI ES HAND BOOK p. 1149:
W ::: working strength to screwe d up tightly to packed joint
where:
F
P,ojec ted "'ea
..
= DL
+ ""J
g....--L--:a
I
Where: Sj '" working stress, psi
d '" bolt diamete r, in
where: A
F
Sf:= Me
2
W:=iSI (O.55d ;>0;250)
4. Bearing Stress( Sb)
~
16T
o
FROM MACH INERI ES HAND BOOK p. 1924:
Where:
T
~
M '" momen t
c = distanc e of farthest fiber from neutral axis
I '" momen t of inertia about the neutral axis
3/12
(for rectang ular cross section)
I = bh
M
I
Z := section modulu s := - : = -
e
L
1<
D
)j
D
8. Strain and Elonga tion
c
Strain
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y
Sf
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62
Stresses
(. Stress ==
Sa=' Smax=.§min
2
Smax == maximum stress
Smm == minimum stress
A
L
"
(., Poisson's Ratio(u) == is the ratio of lateral unit deformation to axial unit
deformation.
(.
where:
~
Y == elongation due to applied load
L == original length
F == force
A == area
S == stress
U=
where:
,I
(t2 - tl)
e.
E
-
E
2G
G == shear modulus of elasticity
E == modulus of elasticity
F
.
F
AE
E == stram ==-.....
9. Thermal Elongation; Stresses
e
63
Stresses
F
U ==
r
~
v
L
h2
q I"""
-J
V/2
Longitudinal Strain
== By.
Ex
~
Ex == L z -L1
L
1At J4UJ.ij
.
Lateral Strain
-----,,___
~.=
Ez
Ex
E == WZ - W1
y
L1
w1
E == t2 - t 1
z
11
k-V/2
PROBLEM I (Apr. 1999)
A steel rod 30 mm & 800 mm long has an allowance elongation not to exceed 1.5
where:
Y == elongation due to temperature difference
k == coefficient of thermal expansion. m/m-oC
t 1 == initial temperature, "C
b == final temperature, °C
mm, find the allowable load in kN.
A. 278
B. 248
C. 316
D. 236
4-t-l i 'i'(,]:'1
(., Relation between shearing and tensile stress based on theory of failure:
For steel:
~y
Ssrnax
1
(., Variable Stress
E == 30 x 10 6 psi
Y _
206,785,714 kpa
FL
FS
AE
FS == factor of safety
Sy == yield point
Sn == endurance limit
Sm == mean stress
0.0015 ==
where:
Smax +Smin
2
Sa == variable component stress
F(0.800)
- - -
~ (0.03)2(206,785,714)
4
F == 274 KN
1'[{OBLf<:M 2 (Oct. 2000)
Compute the induced/compressive stress, In kpa. of a steel solid shafting of 50 mm
diameter and 800 mm in length that is subjected to an increase of temperature by
80"C.
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64
Stresses
Stresses
A. 181,816
B. 242,816
65
SOLUTION
C. 218.182
D. 196,560
Z == Polar section modulus
SOLUTION
Z == TIS
Where:
For steel:
T == torque
S == stress
6
E == 30 X 10 psi == 206,785,714 kpa
6
G == 12 x 10 psi == 82,714,286 kpa
6
a == thermal elongation == 11.7 x 10- m/mDC
Using thermal elongation formula:
S == K E (t2 - t1)
S == 11.7x10
6(206,785,714)(80)
IT d
3
Simplify:
IT d
TIS
3
_ IT (3)3 == 5.30 in3
16
16 -
193,551.43 Kpa
PROBLEM 5 (Apr. 1999)
The answer is: D (nearest)
PROBLEM 3 (Oct. 2000)
Compute for the load in KN on a 3 cm diameter, 100 cm long steel rod if its maximum
elongation exceed 0.12 cm.
C. 148
A. 178
B. 196
D. 287
Em!!ImD
Compute the allowable load in kN on a 20 mm x 120 cm long steel rod with a
maximum elongation must not exceed 1 mrn.
A. 35
C. 66
B. 55
D 76
~
For steel:
E == 30x10 6psi == 206,785,714kpa
For steel:
6
E == 30 x 10 psi
FL
y == - AE
0.0012 ==
16 T
S
206,785,714 kpa
y ==
~
AE
0001 ==
F (1.0)
.._ - - - - - -~
IT (0.03)2 (206,785,714)
F (1
-J (0020)2(206,785,714)
F == 54.1364 KN
4
F == 175.40 KN
PROBLEM 6 (Oct. 2000)
Determine the load in kw on a 25 mm diameter x 1200 mm long steel shaft if its
maximum elongation exceeds 1 mm.
A. 83
C. 103
B. 125
D 234
The answer is: A (nearest)
PROBLEM 4 (Oct. 2000)
Compute for the polar section modulus of a SAE 1060 shafting having a diameter of 3
inches. Use a factor of safety of 2 and design stress at 800 psi.
A. 4.7
C. 4.2
B. 6.1
D. 5.3
Ei-l"imD
For steel:
E == 30 X 10 6 psi zz: 206785,714 kpa
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ast·m1D(.]tI
FL
Y = _.AE
F(120)
.
0.001 = . - - . - . - - .
IT (0.025)2(206,785,714)
4
F = 84.59 KN
S
Em!!mm
3
to fix a lever to a shaft,
Determ ine the minimu m mean diamete r of a taper pin for use
r is 2 inches and
diamete
shaft
The
in-lbs.
700
of
torque
m
if it is transmi t a maximu
2.
of
safety
of
factor
the materia l allowab le stresse s is 15,000 psi. Use
mm
5.2
C.
A. 7.0 mm
D. 6.2 mm
B. 7.2 mm
3/4"
3'
DI!!iIir+D
3
r
700
F
PROB LEM 8 (ME Bd. Oct. 97)
What pressur e is required for punchin g a hole 2 inches in diamete
plate?
A.10to ns
C. 20 tons
B 30 tons
D. 40 tons
1
s, == ~
r thru a 1/4 in steel
700lbs
A
700
15,000 = .. -~
(IT/4)d 2
d = 0.2437 in = 6.19 mm
SOLU TION
PROB LEM 1 1 (ME Bd. Apr. 97)
a steel plate made of SAE
How many 5/16 inch holes can be punch in one motion in
strength for shear is
ultimate
The
tons.
55
1010 steel, 7/16 Inch thick using a force of
safety.
of
50 ksi and use 2 factor
C. 5
A. 58
D 62
B 37
From Machin eries Handbo ok, 1924:
Pressur e == D x t x 80, tons
D = hole diamete r
~
t == plate thickne ss
Pressur e = 2 x 1/4 x 80 == 40 tons
T
F
Weight = Volume x density
Weight = [(3/4)(3 x 12)(20 x 12)J(0.284)
Weight = 1840.32 Ibs
Solving for the equival ent diamete r for a force of 55 tons.
1/4"
~
PROB LEM 9 (ME Bd. Oct. 97)
28,000
PROB LEM 10 (ME Bd. Oct. 97)
The total weight of steel plates 3/4" x 3' x 20' is:
C. 3676 Ibs
A 1838 Ibs
D. 1848 Ibs
B. 1698 Ibs
Where:
Sn
FS
S == 9,333.3 3 PSI
PROB LEM 7 (ME Bd. Oct. 97)
Using density of steel == 0.284 Ib/in
67
St resse's
Stress es
bb
2"
enduran ce strength is
A link has a load factor of 0.8, the surface is 092 and the
ed to a reversin g
subject
is
it
if
link
the
of
stress
ing
28,000 psi. Compu te the alternat
load. Assume a factor of safety of 3.
C. 9333
A. 8150
D. 7260
B 10.920
Ss
Ss
F
A
F
rtdt
50 X 10 3 =
d == 16 In
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(55 x 2000)
rrd(7/1 6)
•
I¢'-H
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68
Stresses
No. of holes =
_~Ji
PROBLEM 14 (ME Bd. Apr. 96)
= 5.12
(5/16)
Therefore no. of holes is 5 because the punching machine cannot punch entirely the
remaining 0.12 in.
A steel tie rod on bridge must be made withstand a pull of 5000 Ibs. Find the diameter
of the rod assuming a factor of safety of 5 and ultimate stress of 64.000 psi.
C. 0.71
A 0.75
D. 0.79
B. 0.84
SOLUTION
PROBLEM 12 (ME Bd. Apr. 97)
The shaft whose torque vanes from 2000 to 6000 in-lbs has 1 1/2 inches in diameter
and 60,000 psi yield strength. Compute for the shaft mean average stress.
C 5162 psi
A. 6036 psi
B. 6810 psi
D. 5550 psi
SI = F/A
If factor of safety is given: St
Where:
SOLUTION
SOlin
~6!max
nd
=~ ()O
3
16Tmin
nd3
S mean --
n(1.5)3
-~
FS
A
64,000
5000
5
J1:d2
16(2000)
Smax + Smin __
-,~_.
F
4
d = 0.71 in
= 5000 Ibs
3018 psi
= - n(1sT
2
PROBLEM 15 (ME Bd. Apr. 96)
If the weight of 6" diameter by 48" long SAE 1030 shafting is 174.5 kg, then what will
be the weight of chromium SAE 51416 of same size?
A. 305.5 Ibs
C 4264 Ibs
B. 384.61bs
D. 465.1 Ibs
9054 + 3018
2
6036 psi
EE!l!ii tel II
PROBLEM 13 (ME Bd. Apr. 96)
A journal bearing with diameter of 76.2 mm is subjected to a load of 4900 N while
rotating
at 200
rpm. If its coefficient of friction is 0.02 and UD = 2.5, find its projected
.
2
area In mm .
A. 12,090
B. 14,165
C. 13,050
D. 14,516
4900 N
I
SOLUTION
.,
o
D
FS = factor of safety
= 9054 psi
Using variable stress formula equation:
~
Su
FS
_§'l = ~
Using shaft stress formula for pure torsion:
Smax
69
Stresses
Solving for the equivalent no. of holes
= 2.5
Substitute the diameter:
L
- - - = 2.5
76.2
L = 190.5 mm
~L
Projected A rea
The major component of different steels is iron, therefore their densities do not differ
much.
Ib
174.5 kg (2000 ---)
Weight
kg
Weight
38481bs
PROBLEM 16 (ME Bd. Apr. 96)
If the ultimate shear strength of a steel plates is 42,000 psi, what
force is necessary to punch a 0.75 inch diameter hole in a 0.625
inch thick plate?
C. 68,080 Ibs
A 63,0081bs
B. 61,800 Ibs
D. 66,8001bs
EmmmmI
76.2 mm
A = projected area is rectangular
A = 0 xL
A = 76.2(190.5) = 14,516 mm 2
r.;~,
:!~;
SOLUTION:
Force ­r- Stress x Area
Where: Area = Circumference x thickness
~:
..~
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0.625"
8
~O.75"
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70
SI resse»
Stress es
Area == n D t
71
";ON""'.
Force == 42,000 [n: x 0.75 x 0.625J
Solving for diamete r of bushing hole after heating:
y == kL(t2 - t.) == 0.00000 68(1.99 9)(24.5 ) == 0.00033 3"
D == 1.999 + 0.00033 3 == 199933 3"
Force == 61,850 Ibs
PROB LEM 17 (ME Bd. Oct. 95)
What is the modulu s of elasticity if the stress is 4,000
psi and unit a strain
0.00105 ?
6
6
A. 41 .905 x 10
B. 42.300 X 10
C. 41.202 X 10 6
D. 43.101 X 10 6
of
Solving for diamete r of shaft after cooling:
y == -0.0007 3(2) == -0.0014 6"
d == 2 - 0.00146 == 1.99854 0"
Clearan ce == 1.99933 3 - 1.99854 0 == 0.00079 3"
PROB LEM 20 (ME Bd. Oct. 95)
What modulu s of elasticity in tension is required to obtain
a unit deforma tion of
0.00105 m/m from a load produci ng a unit tensile stress of 44,000
psi?
6
A. 42.300 x 106 psi
C 41.202 x 10 psi
6
6
B. 43.101 x 10 psi
D. 41.905 x 10 psi
SOLUT ION
E == Stress/S train
E == 44,000/ 0.00105 == 41.905 x10 6psi
EmmmD
PROB LEM 18 (ME Bd. Oct. 95)
A 2.5 in diamete r by 2 in long journal bearing is to carry a
5500-lb load at 3600 rpm
using SAE 40 lube oil at 200°F through a single hole at
25 psi.
Comput e the
bearing pressure.
A. 1100ps i
C. 900 pSI
B. 1000 psi
D. 950 psi
I
5500lb s
SOLU TION
..
Bearing pressur e is the same as bearing stress.
F
s, == LD
Sb ==
5500
== 1100 psi
2(2.5)
~/
b
I<
Project ed A
rea
. •2"
2.5"
>[
.5 in i
PROB LEM 19 (ME Bd. Oct. 95)
It is a problem of expans ion and shrinka ge of steel materia l
so that the slightly bigger
shafting of 2" diamete r can be inserted/fitted to the slightly
smaller hole of a steel
bushing of 1.999" diamete r with the following
process /materia l/data to apply:
Coefficient of expans ion of carbon steel == 0.00000 68"I""F tempera
ture raised by gas
heating == 24.5°F. Cooling media to use dry ice with boiling point
of -109.3° F (-78SC )
Shrinka ge rate below boiling point is 0.00073 in/in Determ
ine the final clearan ce
betwee n the
expand ed steel bushing hole against the shrinka ge of the steel
shaft.
A. 000079 3"
B. 0.00069 3"
C. 0.00075 0"
D. 0.00080 0"
E == Stress/S train == 44,000/ 0.00105 == 41.905 x 10 6 psi
PROB LEM 21 (ME Bd. Oct. 95)
What force P is required to punch a 1/2 in. hole on a 3/8 in.
thick plate it the ultimate
shear strength of the plate is 42,000 psi?
A. 24.9401 bs
C. 24,6201bs
B. 24.9601 bs
D. 24,7401bs
EmmmD
P == Stress x Area
P == Ssu X (x d t) == 42,000(n: )(1/2)(3/8) == 24,7401 bs
PROB LEM 22 (ME Bd. Apr. 95)
What pressur e is required to punch a hole 2" diamete r through
a 1/4" steel plate?
A 10 tons
C. 20 tons
B. 30 tons
D. 40 tons
SOLU TION
From Machin ery's Handbo ok p. 1924:
P == d x t x 80 tons == 2 x 1/4 x 80 == 40 tons
1'!{OH LEM 23 (ME Bd. Apr. 95)
Comput e the working strength of 1" bolt which is screwed
up tightly in packed joint
when the allowab le working stress is 13.000 pSI
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72
Stress es
A. 3600 psi
B. 3900lb s
73
Stress es
C. 3950lb s
D. 3800lb s
PROB LEM 26 (ME Bd. Apr. 94)
Determine the estimated weight of an A-36 steel plates size 3/16
x 6' x 20'
A.9191 bs
C.1012 1bs
B. 8291bs
D. 7351bs
1 in
SOLU TION
F
F
From Machinery's Handbo ok p 1149:
SOLU TION
W = working strength to screwed up tightly to packed joint
From Faires p. 574,
W = St (0.55d 2 - 0.25d)
Where: St = working stress, psi
d = bolt diameter, in
Density of steel = 0.284 lb/in"
Weight = Volume x Density
W = 13,000 [0.55(1 l- 0.25(1)] = 3900lb s
6'
Weight = (3/16)(6 x 12)(20 x 12)(0.284) = 9201bs.
PROB LEM 24 (ME Bd. Apr. 95)
What is the working strength of a 2" bolt which is screwed
up tightly in a packed
jointly when the allowable working stress 12,000 psi?
A. 20,1201bs
C. 20,1001bs
B. 20,4001bs
D. 20,2001bs
SOLU TION
2
-
E:1!!ImD
F
F
0.25d)
W = 12,000 [0.55(2 l- 0.25(2)]
= 20,400 Ibs
Determine the estimated weight of an A-36 steel plate size W'
x 4' x 8'.
A. 280 kgs
C. 332 kgs
B. 301 kgs
D. 297 kgs
Fb = 14.176
F = total force applied on cylinde r head
F = 10(14.176) = 141. 76 KN
Pressure = F/A
A = area of cylinde r
141.76
Pressure =
2 = 288.8 Kpa
(n/4)(0 .25)
SOLU TION
From Faires p. 574,
density of steel = 0.284 Ib/in3
Weight = volume x density
1/2(4 x 12)(8 x 12)(0.284)
Weight = 654.336 Ibs = 297 kg
Fb = force applied per bolt
s, = Fb / A
A = cross-sectional area of bolt
50,000 =.
Fb
(rcl4)(0 .019)2
PROB LEM 25 (ME Bd. Apr. 94)
L
A 19 mm stud bolts is used to fastened on a 250 mrn diamete
r cylinder head of diesel
engine. If there are 10 stud bolts, determine the pressure inside
the cylinder if bolt
stress is limited to 50 Mpa.
C. 3426 Kpa '
A. 288.8 Kpa
D. 4828 Kpa
B. 2888 Kpa
20mm
From Machin ery's Handbo ok p 1149:
W = St (0.55d
PROB LEM 27
PROB LEM 28
1~"
V
4'
250mm
A column supports a compressive load of 250 KN. Determine
the outside diamete r of
column if inside diamete r is 185 mm and compressive stress of
50 Mpa.
A. 200.62 mm
C. 216.42 mm
B. 201.47 mm
D. 208.41 mm
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75
5tn'ss(·...,
s: resses
Im!!Im3
SOLUTION
F = 250 KN
S =
S 0= F/A
50000 0=
,
A
0=
A
250
A
3
5 X 10- m2
S =
F = 25.132 KN
PROBLEM 31
PROBLEM 29
A steel hal/ow tube is use to carry a tensile load of 500 KN at a stress of 140 Mpa. If
outside diameter is 10 times the tube thickness, find the thickness of the tube.
c. 20.64 mm
A. 11.24 mm
B. 107 mm
O. 22.61 mm
What force is necessary to punch a 30 mm hole in 12.5 mm thick plate if ultimate
shear stress is 410 Mpa?
C. 483 KN
A. 564 KN
O. 983 KN
B. 342 KN
Im!!Im3
Using shearing stress formula:
F
S
SOLUTION
A
S
£:
A
140,000
500
A
A 0= 3.57143 X 10- 3 m 2
F
F
80,000 = ---(7t/4)(0.02)2
Substitute:
3
5 x 10- 0= rr/4 (0 02 - (0. 185)2J
3
6.366 X 10- 0= 0 02 - 0.034225
Do 0= 0.201472 m 0= 201.47 mm
0=
F
F
------2
(7t/4)0
2)
A 0= (rr/4 0 0 - (rr/4 0;2)
2
A 0= rr/4(00 - 0, 2)
S
20mm
F
~
t
Do 0= 10t
Do 0= OJ + 2t
10t 0= D, + 2t
F
ndt
F
410,000 = 7t(0.03)(0.0125)
F 0= 483.0198 KN
PROBLEM 32
Two 30 mm thick plate is fastened by two bolts, 25 mm in diameter. If the plates is
subjected to 50 KN tension, find the bearing stress in bolts.
A. 33,333.33 Kpa
C. 5555.55 Kpa
B. 4444.44 Kpa
O. 555,555 Kpa
0; 0= 8t
A 0= rr/4 (0 02 _ 0,2)
3
3.57143 x 10- 0= 7t/4((10t)2 - (8t)2J
4.5473 x 10-3 0= 36e
t 0= 0.112389 m 0= 11.24 mm
25mm
SOLUTION
50 KN
+-
PROBLEM 30
Using bearing stress formula:
A 20 mm diameter rivet is used to fastened two 25 mm thick plate. If the shearing
stress is 80 Mpa. What tensile force applied each plate to shear the bolt?
A. 26.35 KN
C. 30.41 KN
B. 28.42 KN
O. 25.13 KN
S =
£.
A
A = area = 2 d t ( for two bolts)
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50 KN
---l
lOMoARcPSD|3535879
Stress es
S ==£
S ==
PROB LEM 35
50
2(0.025 )(0.03)
What is the modulu s of elasticit y if stress is 300 Mpa and strain
of 0.00138 ?
A. 217.39 x 103 Mpa
C. 220.61 Mpa
3
B. 200.61 x 10 Mpa
D. 215.21 Mpa
== 33,333. 33 Kpa
PROB LEM 33
SOLU TION
A 2.5 inches shaft is subjecte d to 3 KN-m torque. Find the stress
develop ed.
A. 48.62 Mpa
C. 59.68 Mpa
B. 52.75 Mpa
D. 38.64 Mpa
.
Modulu s of Elasticit y, (E)
E =
SOLUT ION
Using shaft stress formula :
16 T
TC d
S ==
300
0.00138
Stress
= --.-Strain
== 217.39 x 10 3 Mpa
PROB LEM 36
d == 2.5 in x 1/12x1 /3.281 == 0.0634 9m
S
77
Stress es
dt
T
3
=3 KN.m
In a 2.0 m cantilev ered I-beam, a 2 Mton weight is applied at
free end. If the allowab le
stress in beam is 110 Mpa, determi ne the section modulu s.
3
A. 18.54 in 3
C. 26.83 in
3
B. 21.77 in 3
D. 24.28 in
16 (3)
SOLU TION
TC (0.0634 9)3
~
For a cantilev er beam with load at free end:
M == F xL
M == (2 x 1000 x 0.00981 )(2) = 39.24 KN-m
S == 59,680. 68 Kpa == 59.68 Mpa
p= 2 Mtan
t
PROB LEM 34
A shaft when subject ed to pure torsion develop ed a stress of 50
Z == Section modulu s
Z == IIc == MIS == 39.24/1 10,000 == 3.5673 x 10.4 m 3 x
(3937)3
3
Z == 21.77 in
IEm!ImD
PROB LEM 37
Mpa. If polar momen t
of inertia is 6.1359 x 10. m , determi ne the maximu m torque
the shaft could handle.
A. 1.23 KN-m
C. 1.84 KN-m
B. 1.68 KN-m
D. 2.48 KN-m
7
4
A 5 KN force acting at the end of a 3 m cantilever beam. If section
J == polar momen t of inertia
TC d
4
J == - -
beam is 10 ;n 3, what is the stress induced ?
A. 86,285 Kpa
C. 29,684 Kpa
B. 78,318 Kpa
D. 91,535 Kpa
modulu s of the
32
6.1359 xlO. 3 ==
d == 0.05 m
S= 16T
3
TC d
50,000 ==
4
TCd
32
16 T
TC (0.05)3
~
~
For a cantilev er beam with load at free end:
M == maximu m momen t = F x L
M == 5(3) == 15 KN-m
Z == Section modulu s == lie
Z == 10 in 3x 1/(39.37 )3 = 1.6387 x10·4m 3
M
15
S = flexural stress == =
4
Z
1.6387x1 0-
T == 1.227 KN-m
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P=5 KN
t
91,535. 62 Kpa
lOMoARcPSD|3535879
78
PROBLEM 38
A 6 mm steel wire is 5m long and stretches 8 mm under a given load. If modulus of
elasticity is 200 Gpa, find the load applied.
A.7KN
C.9KN
B. 8 KN
D. 10 KN
SOLUTION
Y ==
FL
AE
Em!!IiImI
Y = ~
AE
2
A == 0.5 ern" x 1/(100)2 == 5 X 10'5 m
6mm
4m
E == 176580000 Kpa == 176.58 Gpa
i
F(5)
[(nI4)(0.006)2](200x10 6 )
mm
PROBLEM 41
F == 9.047 KN
PROBLEM 39
A steel wire 10m long, hanging vertically supports a tensile
'f' F
load of 2 KN. Neglecting the weight of wire, determine the required diameter if the
stress is not to exceed 140 Mpa and the total elongation is not to exceed 5 mm.
Assume E == 200 Gpa.
A. 2 mm
C. 4 mm
B.3mm
D.5mm
A 20 m rod is stretches to a strain of 0.001. Determine the deflection of the rod.
A. 20 mm
C. 30 mm
B. 25 mm
D. 35 mm
SOLUTION
Strain == Y/L
0.001 == Y/20
Im!!ImD
.
FL
Y == elongation
== AE
2
0.5 cm
0.001 == (225xO.00981)(4)
(5 x 10-5 )E
5m
0.008 ==
79
Stresses
Stresses
Y == 0.020 m == 20 mm
D
0.005 == ~
2(10)
A(200x10 6 )
A == 2 X 10'5 m2
2
A == n/4 d
2 x'lO'5 ==n/4 d2
d == 0.00504 m == 5.04 mm
PROBLEM 42
A rail having a coefficient of linear expansion of 11.6 x 10'6 m/m-oC increases its
length when heated from 70°F to 113°F. Determine the strain.
A. 2.04 x 10'4
C. 4.21 X 10'4
B. 6.05 X 10'4
D. 2.77 X 10'4
Em!!IiImI
PROBLEM 40
An iron rod 4 m long and 0.5 ern" in cross section stretches 1 mm when a mass of
225 kg is hang on it. Compute the modulus of elasticity of the iron.
A. 176.58 Gpa
C. 160.41 Gpa
B. 169.81 Gpa
D. 180.26 Gpa
Converting to DC difference:
L1Clt.F == 519
t.C/(113-70) == 519
L1C == 23.889°C
Strain == S/E == K (t2 - t.) == 11.6 X 10'6 (23.889) == 2.77 x 10'4
PROBLEM 43
What temperature will the rails just touch if steel railroad is 10m long are laid with
clearance of 3 mm at initial temperature of 15°C? Use k == 11.7 X 10'6 m/m-oC.
A. 35.64°C
C. 45.64°c
B. 40.64°C
D. 50. 64°C
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80
E5t!!ImD
Sf = 15,000 psi
Y=kL(b-t1)
PROBLEM 46
Clearance = Y/2 + Y/2
Clearance = Y = 0.003
6
A1A'Wb
J
V/2
k-V/2
0.003 = 11.7 x 10. (10)(b - 15)
t2 = 40.64°C
For a given material, the modulus of elasticity is 15,000,000 PSI In tension and
6,000,000 psi in shear. Determine the Poisson's ratio for this material.
A. Q4
C. 0.3
B. 0.25
D. 0.5
SOLUTION
Using Poisson's ratio formula:
E
G = --2(1 + u)
3
the steel beam is 8.0 in . How much stress is induced?
A. 10,000 psi
C.12,000psi
B.15,000psi
D.14,200psi
The section modulus for
P'7
~
10ft
M
z= Sf
2G
100
u = ---1
2(40)
u = 0.25
SOLUTION
A 1000 lb force acts at the end of a 10ft cantilever beam.
z = section modulus
G=_E_
2(1 + u)
E
u = --1
A 25 mm shaft is keyed to a 300 mm diameter pulley and transmits 3 kw of power.
The keyed assembly rotates at 1725 rpm. What is the tangential force at the key?
A. 1.33 KN
C. 2.33 KN
D. 8.34 KN
B. 5.23 KN
PROBLEM 45
M = F x L = 1000(10) = 10,000 ft-Ib
M = 120,000 in-Ib
2
PROBLEM 47
6 000 000 = 15,000,000
2(1 + u)
"
2(1 + u) = 2.5
u = 0.25
For a cantilever beam:
2
For a given material, the modulus of elasticity is 100 GN/m in tension and 40 GN/m
in shear. Find the Poisson's ratio.
A. 0.10
C. 0.25
B. 0.50
D. 0.35
U·)!'mmI
PROBLEM 44
SOLUTION
81
Stresses
Stresses
P=2rcTN
3 = 2 rt T (1725/60)
T = 0.01661 KN-m
300mm
But: T = F x r
0.01661 = F (0.025/2)
F = 1.33 KN
01
'
PROBLEM 48
What force is necessary to punch a 1 in hole in a 118 in steel plate if the ultimate
shearing stress is 60,000 psi and the ultimate compressive stress is 80,000 psi?
A. 34,520.34 Ib
C. 12,344.56 Ib
B. 8,345.56 Ib
D. 23,561.90 Ib
E:r!!iIm
Solving for the shearing area during punching:
3 = 120,000
Sf
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82
Stresses
A = re d t = re (1) (1/8) = 0.3926 in
S= ~
83
Stresses
2
PROBLEM 51
A hollow rivet has an outside diameter of 5 mm and an inside diameter of 3 mm. If
2
6
the allowable shearing stress is 400 x 10 N/m , what maximum shearinq force can
the rivet sustain if subjected to double shear?
A. 10 KN
C. 15 KN
B. 20 KN
O. 25 KN
A
60,000= _F_
0.3926
F = 23,562 Ibs
e
EmDImII
~D=
2
A=2 [%(0 -0;2)]
•
1 in
PROBLEM 49
A U-bolt supports a load of 6000 lb. The cross-section of the bolt has a diameter of
F = 400 X 10 N/m
S = F/A
A
2
~ (1 I 2)2] = 0.3927 in
-5
F = 10.05 KN
PROBLEM 52
A vertical load of 400 Newtons acts at the end of a horizontal rectangular cantilever
2
6000lb
PROBLEM 50
A 1 in diameter shaft has a 2 in diameter collar resting on a support. The axial load
on the shaft is 10,000 Ib and the thickness of the collar is 1/2 in. How much shearing
stress is induced?
A. 6,366.20 psi
C. 5,244.56 psi
B. 4,234.56 psi
O. 8,456.34 psi
F
10,000 Ibs
=
SOLUTION
SHAFT· 1" dia.
;~
F
COLLAR· 2" dia.
Ss = reds t
1/2"
Ss = 10,000
re(1)(0.5)
F
F
2
= 6,000/0.3927 = 15,278.84 psi
S, = 6,366.20 psi
)1
2
3
= 400 x 10 KN/m
400 x 10 - 2.51327x10
1/2"
5 mm
F
3-
SOLUTION
S= ~
A = 2 !: [(0.005)2 - (0.003)2]
4
A = 2.51327 X 10-5 m2
6
1/2 inch. How much stress is induced in the sides of the bolt?
A. 12,734.45 psi
C. 13,345.56 psi
B. 14,567.23 psi
O. 15,278.12 psi
A = 2[
k
0
beam 2m long and 25 mm wide. If the allowable bending stress is 130 MN/m , find
the depth of the beam in mm.
A. 23.44 mm
C. 30.23 rnrn
B. 38.43 mm
O. 43.23 mm
SOLUTION
~
tP=400 N
2m
Moment = 400(2) = 800 N.m = 0.80 KN.m
3]
z = MIS = [0.80/130 x 10 (1000)3
Z = 6153.85 rnrn"
bh 2
z=6
6153.85 = (25)~
6
h = 38.43 mm
.1/2"
1"
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h
25mm
lOMoARcPSD|3535879
84
Stresses
A simply supported timber beam is 50 mm by 200 mm in cross-section and 4 m long.
If the fiber stress is not to exceed 8.3 Mpa and the beam weight is neglected, find the
maximum mid-span concentrated load that the beam can support if the 200 mm
dimension is vertically oriented.
A. 2,766.67 N
C. 2,345.56 N
B. 3,233.45 N
D. 1,234.45 N
~p
PROBLEM 55
A.1Mm~
For rectangular beam:
Z = section modulus of rectangular beam
B. 173 rnrn"
D. 190 mm
200mm
Based on tensile stress:
6
(50)(200)2
-'---'--'----'--
S =
SOmm
6
4
z = 333,333.33 rnrn" = 3.33 x 10. m
3
2
~
2m
A
104 = 18,000
A
2
A '" 173 mm
M=Sz
4
6)(3.333
Moo (8.3 x 10
10. ) = 2,766.67 N.m
For simply supported beam:
M= ~L-.
C.1~m
SOLUTION
2
Z= - -
Z =
F
210000 = - ,
0.00144
F = 302.4 KN = 302,400 N
A steel hanger of 2 m length is to carry an axial load of 18 KN. If the tensile stress is
restricted to 104 Mpa and the elongation caused by this load is restricted to 1 mm,
what is the minimum cross-sectional area the member can have? E = 200,000 Mpa
SOLUTION
bh
85
Stresses
PROBLEM 53
Based on the given deflection:
4
y
276667 = P(4)
,.
4
1 =
P = 2,766.67 N
F = 18 KN
FL
AE
18,000 (2,000)
A (200,000)
A = 180 rnrn"
PROBLEM 54
Therefore use the higher area: A = 180 rnrn"
A lap joint consists of steel plate 250 mm by 18 mm in thickness is connected by 4-20
mm diameter rivets. Compute the bearing capacity of the rivet connection if the
allowable bearing stress is 210 Mpa.
A. 302,400 N
C. 420,340 N
B. 360,200 N
D. 460,240 N
PROBLEM 56
SOLUTION
F
Sb = A
A = bearing area
A = 4(dxt) = 4 (0.020 x 0.018)
2
A = 0.00144 m
F
A short hollow steel cylinder with a wall thickness of 38 mm is to carry a compression
load, applied uniformly on the end, of 7,800 KN. If the allowable working stress in
steel in compression is 138 Mpa, then the minimum outside diameter of the cylinder
required to safely support the load is:
A. 512 mm
C. 396 mm
B. 482 mm
D. 460 mm
Em!!imD
s =
F
A
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86
Stresses
138
SOLUTION
A
A = 56,521.74 rnrn"
A=Ao-A,
2 = ~ [D 2 ~ D 2J
A = ~ D 2 _ ~ D1
4
4°
4
01
Dj = Do- 2t
D j = DO - 2(38)
S = KE (t2 - t1)
6
S = (6.5 x 10 6 inl'F) (30 x 10 psi)( 130) = 25,350 psi
PROBLEM 59
= Do - 76
TC
2
A steel rod with a cross-sectional area of 160 rnrn" is to be attached between two
fixed points 1.25 m apart. If the rod is too short by 0.25 mm, find the stress applied to
put the rod back to fitness. E = 200,000 Mpa
A. 40 Mpa
C. 50 Mpa
B. 30 Mpa
D. 60 Mpa
2
56,521.74 = -[Do -(D o-76) J
4
2
2
79,965.71 = D0 - D0 + 152D o - 5,776
Do = 511.5 mm
38m~
k
~
Do
)1
PROBLEM 57
A 25 mm diameter steel bar is loaded in double shear until failure, the ultimate load is
found to be 446 KN. If the allowable stress is to be based on a factor of safety 3 what
must be the diameter of a pin designed for an allowable load of 26 KN in single shear
A. 14.79 mm
C. 13.24 mm
be?
B. 15.64 mm
D. 12.63 mm
SOLUTION
L = 1.25 - 0.00025 = 1.24975 m
y =
S~
E
0.00025 = S (
SOLUTION
1.24975 )
200,000,000
S = 40,008 Kpa = 40.008 Mpa
Fu
Su = -
A
For double shear:
446,000
Su =
2[~(5)]
Sail
Sail
87
Stresses
7,800,000
454.52 Mpa
.4
Su _ 454.52 = 151.51 Mpa
FS 3
F
A
151.51 = 26,000
~d2
4
d=14.79mm
PROBLEM 58
A steel bar, initially free of stress, is held between rigid supports. Determine the
stress in the bar if temperature drops 130°F. K = 6.5 x 10'6 inl'F, E = 30 x 106 psi
A. 25,350 psi
C. 24896 psi
B. 26,234 psi
D. 23,654 psi
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.(
1.25 m
J
lOMoARcPSD|3535879
88
89
Shafting
Shafting
e Shaft power
6
Sha~t
A. Power Formula in 81 unit:
TN
T
where:
P == power, KW
T = torque, KN-m
N == speed, rps
B. Power in English
mmmmm!
where:
P == power, Hp
T == torque, ft-lb
N == speed, rpm
1. Shaft - is a rotating member that is used to transmit power.
F
2. Axle - a stationary member carrying rotating wheels, pulleys.
(; Torque, T
R
3. Machine shaft - shaft which is an integral part of the machine.
4. Line shaft - transmission shaft driven by prime mover.
5. Counter shaft - transmission shaft intermediate between the line shaft and
the driven machine.
where:
F == applied force
R == radius = 0/2
"
Shaft
Stresses in shaft when sUbjected to pure torsion(Ss)
T
6. Spindles - is a short axles and shafts.
.&EiiU&i:Q:a:UiUU,W
7. Transmission shaft - is a shaft used to transmit power between the source and
the machines absorbing the power, and include countershafts, line shafts, head
shafts, and all factory shafting.
~
A. For solid shaft:
W
OTO R
• ,~1%
%4~
Machine Shaft
Main Shaft
0
I'
Counter Shaft
I,
T]
~1,
I.
.. ~
-J
~
...
~
B. For hollow shaft:
,~
Driven Machine 1
Driven MaChine 2
S5
where:
Do == outside diameter
0,
== inside diameter
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•
tOlE
t
t
lOMoARcPSD|3535879
90
"
91
Shafting
Shafting
J
= polar moment of inertia
c
T
d
= is the distance of the farthest fiber from neutral axis
b. For hollow shaft:
= torque
= diameter of shaft
Torsional deflection(8), rad
where:
•
L = length of shaft
J = polar moment of inertia
Strength of shaft with assumed allowable stresses
(PSME Code p. 18)
a. For Main Shafts:
b. For Line Shafts:
G = modulus of rigidity in shear = 11.5 x 10 psi for steel
6
c. For short shafts:
FROM MACHINERIES HANDBOOK:
where:
For solid shaft:
P = power, Hp
N = speed, rpm
o = diameter, inch
For hollow shaft:
•
(From Machineries Handbook)
iii Stress in shaft when subjected to Torsion and Bending
loads:
•
F
a. For solid shaft:
'a.iIJ21t!~l
~
Shaft diameter for 0.08 degrees per foot of length of shaft
deflection.
T
Where:
where:
T = torque
M = moment
5 s = maximum shear stress
51 = maximum tensile or compressive stress
o = diameter, in
T = torque, in-Ib
P = horsepower
N = speed, rpm
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92
Sh (~ t
ino
!cJ:1
Shu!tin.t}
For Sl units:
2. Rectangular
D = 2.26 tIT 'or D = 125.70
~.
bd(b 2 +d2 )
12
J =
d~I
Where:
o :::: diameter, mm
P = power, watts
N = speed, rpm
T = torque, N-mm
•
bd
b
I.
tol
3 +1.6 <l.
d
Shaft deflection of 1 degree for a length of 20 times its
diameter.
D=0.10 Wf·. orD=4.0
~
Where:
b -= bose
d = height (shortest side)
3. Circle
Where"
ltD
J= -
P = power, hp
D = diameter, in
speed, rpm
torque, in-Ib
=
•
Linear deflection of shafting
3.
Shafting subjected to no bending action of pulleys except its own weight
}
3
4
Zp =
ltD
16
32
Where: 0
N
T~
L =6.95
2
-::=
diameter
4. Hollow tube
.
J = -"-(D 4_d4 )
r
32
frJ2
Z =
p
D4 4
-"-r
-d
16
D
\.
b. Shafting subjected to bending action of pulleys, etc
L=5.2 ~
Where: L = shaft length, ft
= shaft diameter, In
D
= outside diameter
d
=
J
Inner diameter
5. Hexagonal
o
J = 1.0825 s' = 0.12F'
POLAR MOMENT OF INERTIA & POLAR SECTION MODULUS
J :::: polar moment of inertia
S
Zp - polar section modulus
1. Square
J = 0.1667a
Where:
4
z, = 0.206 a'
Zp
~Io
= 0.20F'
5
F
= length 01 side
6. Triangle
J = 0.036
a = length of side
s'
Z, = 0.05 s'
S
length of side
s
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5
lOMoARcPSD|3535879
n4
PROBLEM I (Oct. 2000)
A 0.75x10
Compute for the ~wislng
1200 rpm.
A. 1166
B. 915
B.1.0X10
moment in m-fb developed when the shaft delivers 20 hp at
2rrT(1200)
33,000
87.53 ft-Ib (12) ~
0.9549 KN-m (1000)(1000)
~
6
0.9549 x 10 Nmm
PROBLEM 5 (Ocl. 2000)
1050.42In-lb
D!!D1':D
, A 102 mm diameter shaft is driven at 3000 rpm by a 300 hp pnme mover. The shalt
, drives a 121.9 cm diameter chain sprocket having 85% output efficiency. Compute
the torque In in-Ib developed in
the shalt.
A. 5.600
C 8,150
B. 7.100
D. 6.300
mm:mmI
p
c
.2itT~
..:r L
33.000
(3.1x10
--
JG
6)(1400)
- - 4- . _ - - -
[r (~ o)
0.003774 rad (180
e/rr)
](80.000)
300 ~
T
c
.2.rrTJ3000)
33,000
525.211 h-lb (12) ~
6,302.53 in-lb
0.216 degree
PIWBLEM 6 (Oct. 1998)
PROBLEM:> (Oct. 2000)
Compute for the torsional deflection in degrees. of a 110 mm diameter, 1.4 m long
shaft subjected to a twist moment of 3 x 1o" N.mm
The torsional modulus of
elasticity IS 83000 N/mm'.
A. 0.27
C. 0.20
B. 031
D. 024
Em!!ImD
TL
JG
~
2 IT (T) (200/60)
The answer is: B (nearest)
Find the torsional deflection, in degrees, of a solid steel shaft, 110 mm 00 x 1.4 m
fJ
long subject to 3.1 x 10 N-mm torque. The torsional modulus of elasncrty is 80,000
N/mm-mm.
A. 0.221
C. 0.28
B. 0.195
D 0.21
8
~
T
PROBLEM 2 (Oct. 2000)
8
2rrTN
20 ~
20 ~
~
~
P
2rrTN
33,000
9
C. 1.20 X 10
D. 2.10 X 10"
5
O. 945
P ~
8 ~
5
5
&!!I1t_gl
C. 1050
E!!Dl':D
T ~
95
ShuJll nq
ShaJting
A 3 in diameter short shaft carrying pulleys close to the bearings transmit how much
horsepower if the shaft makes 280 rpm?
A 991 hp
C. 199 hp
B. 919 hp
D. 260 hp
Em:!!JmD
For short shaft:
~
6
(3 x 10 )(1400)
[rr
(~0)4]83.
0.00352 rad (180
e/rr)
P ~
D3N
-
38
Q)3J.280)
38
19895 hp
PI,OBLEM 7 (Oct. 1998)
0.202 degree
What power would a spindle 55 mm In diameter transmit at 480 rpm stress allowed
2
'R OELEM 4 (Oct. 2000)
'ind the torsional moment (in Newton-rum), developed when the shaft delivers 20 kw
t 200 rpm
lor short shalt is 59 N/mm ?
A. 98.68 kw
B. 96.88 kw
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C 68.98 kw
D. 68.89 kw
lOMoARcPSD|3535879
96
1,927.390 N.mm = 1.92739 KN.m
2 rt T N ~ 2 n (1.92739)(480/60) ~
T
P
SOLUTION
o ~ 55 mm ~ 0.055 m
S ~
59 N/mm' ~
59 Mpa ~
s ~
For shaft:
E'ROBLEM 10 (ME Bd. Apr. 98)
A small countershaft is 1 1/2 inch in diameter and has an allowable stress of 8500 psi.
Find the horsepower delivered by the shaft at a speed of 15.7 rad/s.
C 13.31
A. 7.20
B.l.4
O. 14.72
_16T
n (0.055)3
EE!!i it.q+
T = 1.927 KN-m
p ~ 2 rt T N ~
96.88 kw
No correct answer In the given choices.
59,000 kpa
16T
)"[03
59,000 ~
g7
Shaft uvq
Shq{t ing
2 rt (1.927) (480/60) ~
96.88 kw
s ~
16T
rrd
PH.OBLEM 8 (Oct. 1998)
A hollow shaft has an inner diameter of 0.035 m and an outer diameter of 0.06 m.
Compute the torque jf the shear stress is not to exceed 120 Mpa In N-m.
A. 4,400 N-m
C. 4,500 N-m
B. 4,300 N-m
O. 4,200 N-m
~
5.63278 in-Ibs
N :::: 15.7 !ad x~q. ec
sec
N
SOLUTION
16T
3
,(1.5)
~
8,500
T
3
~
min
469.4 It-Ibs
x __ ~r
2n:rad
150 rev/min
p ~
~TN
~
33,000
2,( 469.4 )(150)
33,000
~
1339 h
P
For hollow shaft
s ~
PROBLICM 11 (ME Bd. Apr. 97)
... Determine the torque received by the motor shaft funning at 4250 rpm, transmitting
11 Hp, through a 10 in diameter. 20° involute gear. The shaft is supported by ball
bearings at both ends and the gear is fixed at the middle of 8 in shaft length.
C. 132 ft-Ib
A. 163 in-Ib
B. 167 ft-Ib
0 138 ft-lb
_16TO o _
4
4
n(Oo -0, )
120,000 ~
T ~
__16 (T)JO 06)_
rt [(0.06)4 - (0.035)4 J
4.50 KN-m ~
4500 N-m
IEiI!!Iim
PROBLEM 9 (ME Bd. Apr. 97)
r What power would a spindle
55 mm in diameter transmit at 480 rpm. Stress allowed
z
for short shaft is 59 N/mm
A. 45.12 kw
B. 50.61 kw
45 Hp
O. 39.21 kw
s _ 16T
nd 3
16T
"(55)3
2,HJ
33,000
2nT(4250)
11
T
EI!!Dm
59
C
p ~
~
0
-----
33.000
13.594 tl-Ib
~
163.128 in-lbs
E'HOlll.b:M 12 (ME Bd. Apr. 98)
A 16 It lineshatl has no bending action except Its own weight. What power in Hp can
the shaft deliver at a speed of 200 rpm. Consider that the torsional deflection Will not
exceed 0.08/fllenglh.
C. 244
A. 13.2
1'16
B 158
o
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9A
Shafting
Shafting
SOLUTION
, PROBLEM 14 (ME Bd. Apr. 98)
From Machineries Handbook, 24th Ed.
Compute the diameter of a solid shalt transmitting 75 Hp at 1800 rpm. The nature of
the load and the type of service is such that the allowable S, based on pure torsion is
6000 psi.
p.267
For line shaft with no bending action except its own weight.
0 2 /3 := ~-­
where:
_~
02i3 =
8.95
0 = shaft diameter, in
L ee shaft length, ft
A. 1 7{7"
C. 1 5/16"
B. 21/16"
D. 31/8"
SOLUTION
8.95
P = 2"TN
75(33,000) ~ 2 n T (1800)
T = 218.838lt-lbs
T = 2626.06 in-Ibs
S _ 16T
- nd 3
o = 2.39 in
4
'1
"0
N
1--,
\ 4.6 )
Hp =
99
N = speed, rpm
4
Hp
2.39 ) 200
( 4.6
6,000 = 16(2626.06)
n(d)3
Hp
14.58 hp
d = 1.306 in
Therefore use = 1 5/6 in (standard size)
PROBLEM 13 (ME Bd. Apr. 98)
The torsional deflection of a SAE 1040 steel shaft is 0.8' in a length or 1/2 m. The
shear stress is 69 Mpa. Compute the diameter of the shaft in millimeter. Steel
modulus of elasticity is 79.300 Mpa or N/mm 2 .
A. 51
C. 75
B. 50
O. 62
am:mD
The shalt is subjected to a steady load of 36,000 Ibs at a shear stress of 10,000 psi.
Compute the diameter of the said shalt in inches.
A. 1718
C. 3
B.21M
D.23M
SOLUTION
T = F x r = 36,000 (dI2) = , 8,000 d in-lbs
S
16T
S = 16T
nd3
=ncf3
= ~­
69,000
/ PROBLEM 15 (ME Bd. Apr. 98)
n (d)3
T = 13.548 d
3
KN.m
10,000 = 16(18,000d)
n(d)3
TL
8 =-
d = 3.027 in
JG
0.8
Ll
X
1t
180 0
d = 0.062 m
~548d3)(1@_
PROBLEM 16 (ME Bd. Apr. 97)
4
(nd )(79.300.000)
32
62 mm
Determine the torque received by the motor shaft running at 4,250 rpm, transrnittlnq
11 hp through a 10 in diameter 20° involute gear. The shalt is supported by ball
bearings at both ends and the gear is fixed at the shalt length.
A. 163 in-Ib
C. 167 in-Ib
B. 132in-lb
D.138in-lb
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100
ShaJtiflq
Shafting
torsional sheanng stress IS 28 N/mm"'.
A. 218
C. 312
B. 232
D. 380
SOLUTION
P = nTN
m!!Ji1:mI
11(33,000) = ZnT(4250)
T = 13.594 It-Ib (12) = 163.128 in-los
PROBLEM 17 (ME Bd. Apr. 96)
Compute the maximum unn shear in a 3 inches diameter steel shafting that transmits
24000 in-Ib of torque at 99 rpm.
A. 4530 psi
C. 4250 psi
B. 3860 psi
D. 4930 psi
IImI!!imD
P = 2nT N
750 = 2nT(1500/60)
T = 4.775 KN-m
S ~
16T
nd 3
s.
16(4775)
379,982 Kpa
'n(004)3
380 Mpa
PROBLEM 20 (ME Bd. Oct. 95)
A hollow shaft has an inner diameter of 0.035 m and outer diameter of 0.06 m.
Compute for the torque if the shear stress is not to exceed 120 Mpa In N-m.
A. 4500
C. 4100
B. 4300
D. 4150
s = 16T
nd 3
Ss =
16(24,000)
n(3)3
= 4527 psi
SOLUTION
PROBLEM 18 (ME Bd. Apr. 96)
Compute the linear shalt diameter to transmit 12 HP at 180 rpm with torsional
deflection of 0.08 degrees per foot length.
A. 3 in
C. 5 em
B. 2.35 in
D. 62 mm
s.
16TDo
-2 2-
n(D o
·0,)
120,000.000
SOLUTION
16(T)(0.06)
n[(006)4 ~(0 35)4i
T = 4500 N-m
P = 2nTN
12(33,000) = 2nT(180)
T = 350.14 It-Ibs = 4,201.68 in-los
L=IIt=12in
J = l'td'/32
G = 12,000,000 psi for steel
PROBLEM 21 (ME Bd. Oct. 95)
Design the size of solid steel shalt to be used .or a 500 Hp, 250 rpm application If the
allowable torsional detrection is 1 and the allowable stress is 10,000 psi and modulus
6
01 rigidity IS 13 x 10 pSI.
A. 5" dia.
C. 4-5/8" dia.
D. 4-3/4" dia.
B. 4-7/8" dra.
Q
e = TL
JG
n
0.08' x 180 0
4,201.68(12)
m:m:mmI
4
(nd )(12,000,000)
32
d
~
1 (J 1
2.35 in
PROBLEM 19 (ME Bd. Apr. 96)
Compute the nominal shear stress at the surface in Mpa for a 40 mm diameter shalt
that transmits 750 K!N at 1500 rpm. Axial and bending loads are assumed negligible.
Solving for the shaft diameter 0 based on stress:
P = Zrt TN
500(33.000) ~ 2 rt T (250)
T ~ 10.504.226 It-Ibs
T = 126,051 m-lbs
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102
IO.~
SfW/llill/
Sh({!1 fJlq
Emmit-w'
s = 16T
3
n:d
10.000 c.
16(126.051)
p -
For short shaft-
38
nid)3
Where
d = 4"
(51! 25.4)N
120 = . _
.
38
N = 330 rpm
v = ,D N = IT (0 55 x 3 28)(330)
I)
JG
x Jl!180
('~d2
-
-----
PROBLEM 24 (ME Bd. Oct. 95)
A hollow shaft nas an Inner diameter of 0035 m and outer diameter of 0.06 m
Determine the polar moment of mertta of the hollow sf-a-t.
C 125x0bm~
A 152x0hm~'
)(13.000,000)
B1152x10- 6 m
4-7/8" (standard)
PROBLEM 22 (ME Bd. Oct. 95)
A 2-m solid shaft IS dnven by a 36-',n, gear and transmits power at 120 rpm.
II
J
J
mmm.mI
D.1125x10
b
m
1
4
,
-d, )
32
IT [(006)' - (0035)4]
32
6
m':
j"[d3
PROBLEM 25 (ME Bd. Apr. 95)
Compute the speed of the gear mounted all. a 52.5 mm diameter shalt receiving
power from a driVing motor With 250 Hp
A 2182 rpm
C 2071 rpm
8 2282 rpm
D. 234 f rpm
15T
_.
n(2)3
T = 18850 m-Ibs
Where
T:(d o
J = 1 125 x 10
15T
12.000 =
p =
4
~
allowable shearing stress IS 12 kSI, what horsepower can be transmitted?
A 29 89
C. 35.89
B 38 89
D. 34 89
s
1870 fpm
--
d = 484"
Therefore use d
D shaft diameter, In
N = speed. rpm
P = power. hp
Solving for the shaft diameter based on torsional deflection:
The shaft length IS not given The common practice for torsional deflection IS per 20d
length. (From machineries Handbook)
TL
126.051(20d)
D"N
1.570.8 tt-lbs
2n.~
550
T = torque. f1-lb
N = speed, rps
p = 2'(1570 8)(120! 50)
550
~
p
80
3589 Hp
250
PROBLEM 2:; (ME Bd. Oct. 95)
A short 61 mm diameter shaft transmits 120 Hp Compute the hnear speed of a pulley
55 em diameter mounted on the shaft.
A. 1795 lpm
C. 1765 fpm
8. 1856 fpm
D 2106 torn
D 3N
N
(52.5 25.4)3 N
80
2255 rpm
1he: answer IS B (nearest)
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10"
Shuftinq
S}lU(lllln
PROBLEM 26 (ME Bd. Apr. 95)
A solid cylindrical shaft 48.2 em long IS used for a transmission of mechanical power
at a rate of 37 KW running at 1760 rpm The S~ IS 8.13Mpa. Calculate the diameter.
A. 30 mm
C. 35 mm
B. 40 mm
D 50 mm
f'I<UIJLEM 28 (ME ae, Apr. 94)
A 3" diameter short shaft carrying 2 inches pulleys close to the bearings transmit how
much power If the shaft makes 280 rpm
A 199 Hp
C. 198 Hp
B. 200 Hp
D. 210 Hp
~
SOLUTION
For short shafts
D 3N
(3)3(280)
P~2r TN
37 ~ 2nT(1760/60)
T ~ 0.20 KN-m
S ~
P =
16T
rrd3
8130 =
=
38
38
199 Hp
PROBLE:vI 29 (ME Bd. Apr. 97)
A step shaft made of SAE 1117 steel with an ultimate strength of 69.7 ksi The notch
sensitivity lactor IS 0.00435 IS the constant dependent upon the ultimate strength.
check the radius stress raiser
C 0250
A. 0.192
D. 0775
B. 0386
16(42.017)
nd 1
d ~
105
0.050 m = 50 mm
PROBLEM 27 (ME Bd. Apr. 95)
A hollow shaft with outside diameter of 14 em and wall thickness of 0.8 em transmits
200 KW at 400 rpm What must be the angular deflection of the
shaft If the length IS
5 meters? The material of the shaft IS C4140 steel
C 1 83 deg
A 0.84 dog
B. 1.14 deg
D 1.50 deg
~
From Faires p 113
q =
1
- -- 1 , (a! r)
0983 =
SOLUTION
1
1 I 0.00435
= 0.25
P = 2" T N
200 = Zrr T(400/60)
T = 477465 KN-m
PROHLEM 30
D, = D.
2 t
D = 014 - 2(0008) = 0.124 m
A shafl IS used to transmit 200 KW at 300 rpm by means of a 200 mm diameter
sprocket Determine the force tangent to the sprocket
A 60 44 KN
C 60 88 KN
B. 60 33 KN
D 6366 KN
J = rr/32 (D.' - D,') = rr/32 [(0.14)' - (0.124)"] = 1.45042 x 10" m"
From Vallance p 31. Table 2-6 for C4140
steel.
G ~ 12000.000 pSI = 12.000.000 (101325/147) = 82714.286 Kpa
II
c
TL
~
P=2TITN
200 " 2 II T 1300/60)
T ~ 6.366 KN-m
JG
H
477465(5)
145042,10 '(82.714286)
00199 rad (180"
IT rad)
1.14 deg
T :::;- F x r
6366 = F, (0 20/2)
F = 63.66 KN
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lOb
Shullilln
Find the diameter of a steel shaft which will be used to operate a 110 t0N motor
rotating at 5 rps If torsional stress IS 90 Mpa.
A. 60.2 mm
C. 38.3 mm
B. 583 mm
O. 46.2 rr-m
E·)!''ir.g+
33,000
2nT(1400)
50
T
P=2nTN
110 = 2 n T (5)
T = 3.501 KN-m
16T
rrd
S
--------
33,000
18757 It-Ib x 12in/lt
16T
,
16(2250905)
500
- ---- -----
rId
J
P[,OIlLEM :14
d = 0.058298 m = 58.298 mm
A hollow shaft that has a 100 mm outside diameter and 80 mm msrdc diameter IS
used lo transmit 100 KW at 600 rpm Determine the shaft stress
A 1373 Mpa
C. 19.86 Mpa
B 1682 Mpa
D. 17.21 Mpa
PROBLEM 32
What IS the speed of 63 42 mm shaft that transmits 75 KW if stress is not to exceed
26 Mpa?
A. 550 rpm
C. 650 rpm
B 660 rpm
D. 700 rpm
E.:.t!!immI
S = 16T
...- r.lI.fkl .~
P=2,TN
100 = 2, T 1600/60)
T = 15915 KN-m
4 --
il:(Do -0
S
16T
---n(0.06342)3
16(1.5915)1010)
16TO,
S
J
26 000 =
,
3
d = 2.84 In
16(35()1)
'd
2250905 m-lb
r::d J
J
90000 =
nd
2"TN
P =
Em!!i:mD
S
107
S/wjl illy
PROBLEM :J]
4
)
;[(010)"
13728.73 Kpa
(008)']
1
13.73 Mpa
P],Ol3LSM 35
T = 1.3022 KN-m
P=2rrTN
75 = 2 rt (1 3022)(N)
No 9.166 rps x 60 = 550 rpm
Determine the length of the shortest 2 mm diameter bronze wire which can be twisted
through two complete turns Without exceertnq a shearing stress of 70 Mpa. G =
35,000 Mpa
C. 4.836 mm
A. 6,283 mm
O. 7.263 mm
B 5,484 mm
PROBLEM 33
A steel shalt transrruts 50 Hp at 1400 rpm. If allowable stress is 500 PSI, find the shalt
diameter
A. 3.58 In
C 1 65 In
B. 2.84 In
D 2.54 In
E.:.t!!immI
Ss
70 =
16 T
TId:)
16 T
rt (2)3
T = 109.96 N mm
A = 2 turns = 2 rev = 2 (2TI) - 4"
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I (),~
,'-jlllllllll!)
4)1 =
--
IT (2)4
32
L
--
SOLUTION
(35000)
'
D
lit x 1,3.281
_ 0.30478 m
6.283 mm
T = F(D!2)
PI,OBLEM 36
d
A hollow shaft developed a torque of 5 KN-m and shaft stress IS 40 Mpa If outside
diameter of shaft IS 100 mrn. determine the shaft inner diameter.
A 68 43 mm
C. 58 38 mm
B. 6328 mm
D. 77.64 mm
J
8
ell
Em!!ImD
OiL
s
lOq
~1 (l r !ln
10996 L
1§TD o
4
n(D o
-
ell
D:')
= 5(0.30478/2)
= 0.7619 KN-m
2,nxI/12xl/3.281 = 005079m
7
rrd'i32 = n(O 05079)";32 = 6.533 x 10 m"
TL
JG
T
JG
0.7619
(6533;; 07)(83xl 0 6 )
0.014043 rad/rn x 180 'ITrad
0.805 -rn
16(5)(0 10)
40.000
n[(O 1())4 ]1),D(-~
I'HO!3LEM 39
What IS the minimum diameter of a steel shaft which will be used to operate a 14 KN6
m torque and will not be twist more than 3 In a length of 6 m? (G = 83 x 10 Kpa)
B. 118.45 rnm
A 10064 mm
B 96.80 mm
C 120.72 mm
(0 1)'
D,' c 6 3662 x 10"
D, ~ 0.07764 m = 77.64 mm
l'I,OBUcM :17
What is the polar moment of Inertia of a solid shaft that has a torque ot 1.5 KN-m and
a stress of 25 Mpa?
A. 246 x 10 6 m'l
C 3.46 X 10' m'
B 2.02 x 10- 0 m'l
D 1.24 x 10'[ m 1
EJ!!,it.'%I
e = torsional deflection
(J =
TL
---
JG
Em:!!i:ir:mI
3' (IT/180') -
s
14(6)
------------
16T
nd 3
J(83Xl 0 6 )
ITd 4
5
J = 1.93286 X 10 m'
25.000 = 16(1,5)
ITd
d
_
1.93286 x 10" =
0.0673556 m = 673556 mm
4
J
3
rrd._.
32
4
d = 011845m
= rr(0.06755) = 2.02 x 10" m'
32
red 11
32
= 118.45mm
32
['I,OIlLEM 40
PROBLEM :lH
A force tangent to a foot diameter pulley IS 5 KN and IS mounted on a 2 Inches shaft.
Determine the torsional deflection If G = 83 x 1Or, Kpa
C 0.768 '1m
A 0805 -rn
B 0.654 -rn
D 0.938 .m
A solid shaft 5 m long IS stressed to 60 Mpa when twisted through 4
diameter If G . .:. 83 Gpa
A. 95.46 mm
C 101 32 mm
B 90.42 mm
D. 103.54 rnm
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Find the shaft
lOMoARcPSD|3535879
j
In
Shu!1 inn
IlE!!mmI
Em:!!ii%.:D
s = 16T
For rnam shaft:
3
i'Td
Solving for T in terms of d:
T =
p
3S
rrd
4
(50/25.4)(3001
80
~
2860 Hp
PHOBLEM '+:'\
3
rrd (60.000)
11780.97 d
16
J
___
A 1 5 In diameter short shaft IS used to transmit 44,4 Hp Determine the shah speed
C 400 rps
A 500 rps
0 450 rpm
B. 833 rps
Eqn 1
TL
H
o3 N
80
16
T
II1
.SIi(~lt!nq
JG
x n:/180'
E:.'l!!IiI:.mI
TIS)
J
IITd ](83x106)
32
T = 113774606 d'
SOLUTION:
_ Eqn. 2
Equate 1 and?
/"
113774.606d' = 1178097,r
d = 0.10354 m = 10354 mm
¢'
1
(::I: :;;
r f1 q.:. Cj-l
c.
! Ii:; ;:;d Le.'
For short shaft
03 N
p =
38
(1.5)3 N
44.4
38
N = 500 rpm x 1160
PROBLEM .+ I
What IS the diameter of a line shaft that transmit 150 KW at 15 rps?
A 228 In
C 1.62 in
0 2.04 In
B. 3.54 In
Em:!!ii%.:D
8.33 IpS
PR01J!.EM 4-+
A 3 in solid shalt IS desired to replaced a hollow shaft havnq 4 In outside diameter
Consider the strength to be the same. determine the mside diameter of hollow shaft
A 2.5 In
C 3.48111
B 3 0 In
0 4.0 In
For line shaft.
p =
E:.'l!!IiI:.mI
O'N
53.5
150'0.746 =
For 50110 shaft
03(15x60)
16T
S = 15T
53.5
nd 3
D , 2.28 in
PHOBLEM 42
A main shaft has 50 mm diameter IS running at 300 rpm What power that could be
delivered by the shaft?
A 3040 Hp
C. 32 50 Hp
B. 2860 Hp
O. 16.42 Hp
0.1886T
,,(3)3
For hollow shaft
16TO r
- 4"' --- 4
s
"(0,,
01886 T =
0
1
16T(4)
rrl4 J 0')
(41',0,'=108
[J, = 348 In
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)
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112
Sh(~{t
SlluJrll1()
P[{OULEM 45
A motor IS used to drive a centrifugal pump thai discharges 3000 h.rnin at a head 01
10 m. The pump efficiency IS 6SC'o and running at 550 rpm. Find the torsional stress of
shaft If shaft diameter IS 35 mm.
C 1285 Mpa
A. 1385 Mpa
B 11 85 Mpa
D 1487 Mpa
IlE:!!IiI:mI
Brakl Power
Solving for the power output of the pump"
Q = 3000 li/rrun = 3 m'!/min
P ~ wQ h
P ~ 9 81 (3160)( 10)
P ~ 4905 KW
16(0.1252)
n(0.035)
]
14876.63 Kpa
535
D = 2 In
['!,OllU':M 47
An 800 mm diameter circular saw blade IS driven by a 1800 rpm motor with belt
speed ratio of 1 8. Find the panphcra! speed of the blade
C. 13243 IVsec
A 137.43 ft/sec
D. 13521 It/sec
B 14065 ft/sec
NI / N;. _ 1.8
1800 ; N. = 1 8
N.· = 1000 rpm
V • IT D rl
,(08)11000/60) = 4188 m/sec x 3 281
137.43 It/sec
[,1'OBl.!':"1 4H
14.876 Mpa
A machine shaft IS supported on bearings 1 m apart IS to transmit 190 KW at 300 rpm
while subjected to bending load 01 500 kg at the center If maximum shearing stress IS
40 Mpa. determine the shaft diameter
A. 100 mill
C. 94 mm
B. 90 mm
0 98 mm
Hali,,-p'
PROBLEM 4(;
A circular saw blade has a circular speed of 25 m/sec and 500 mm diameter IS driven
Find the required diameter of the
by means of belt that has an effiCiency of 95%
driven shaft If power transmitted IS 50 hp.
A. 1 In
C. 3 In
B 2 In
D. 4 in
E'.l!!iil.:'mI
V =rrDN
25 = rt (0.5) N
N = 954.93 rpm
For line shaft:
P =
D:'(954931
IIE:!!iml3
Solving for tho power Input of the pump:
Brake power = 4.905/068 = 7213 KW
P=2nTN
7.213 = 2, T(550/60)
T = 0.1252 KN·m
s-
150(095)
I L;
iun
For Simply supported beam with load at center.
PL
(500x000981}(1)
M
4
4
1.226 KN m
M
I
Sol\/Ing lor the torque developed"
P=2nTN
190 = 2 rt T (300·60)
T
604i88KNm
For shaft subjected to torsion and bending load.
Based on shearing.
S
D3N
p ~
16
_
"'
')
\ M" - T"
;-rd\
53.5
J
~
0 ()OO
P lei = D N
53,5
16
')
" ,'(1 226)" T (6.04788)
:r:cl
d
009227 rr; = 9227 mm
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<'
500 kg
~
PI
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II"
PROHLSM 49
A 100 mm diameter shaft \5 subjected to a torque of 6 KN-m and a bendinq moment
of 2.5 KN-m Find the maximum bending stress developed.
A. 4584 Mpa
C 50.2B Mpa
B. 6025 Mpa
D. 5546 Mpa
I!m!!iiI:mI
For line shan
p
For shaft subjected to torsion and bending IOrHL
16 .
-- I'M~\"2
Ted'
.
Based on compression,
16
--
?- -
')
n(0.1)
0 3 (900 )
PROBLEM 52 (ME Bd. Oct. 1996)
I
A centnlugal pump IS dvectty coupled to a molar The pump rating IS 3600 liters per
minute against a total head of 8 meters of water. Pump efftcrency IS 65"~,
at a speed
of SSO rpm Calculate the torsional stress Induced on the 40 mm diameter motor
shaft.
C. 10.01085 kpa
A 11,193.45 kpa
O. 16,34567 kpa
B. 13467.34 kpa
J
40,836.62 Kpa
200
53.5
D = 2 28 In or 2 7/16 III (standard)
-', --,
__ T.:' I
S, ='- ",125+\1 2 5 ;- - (6;- ,
S,
D"N
53.5
SOLUTION
31 =
I J5
:-.,!Ui!llfU!
ShU/II/H;
4584 Mpa
PROBLl';M E)O
A round steel shaft transmits 1/2 Hp at 1800 rpm. Tile shear modulus of elastiCity is
12.000,000 PSI The torsional deflection IS not to exceed 1 deg In a Jer-otn equal to 20
times the diameter Find the shaft diameter.
A 1.04 In
C a 864 In
B. a 257 In
D 2 045 In
E'.l!!iiI:mI
p
,
1,2 =
,
33000
T
S
= 1.4589 f1-lb = 175071n-lb
H =
o - 3,000 Ills
S
TL
JG
("xrr/180'
( IT
32
d
-.=
3 rn?s
WOh
981(3)18)
e
0.65
7,24 kw
E = 65%
P",2rrTN
7.24
2 n T (550'60)
Tea 1208 KN-m
16T
2nT(1800)
-
1.t}!IJ'"
Im:!!mm
p,
2nTN
33.000
=
Brake Power
rrdJ
16(0 1258)
n(O 04)3
~
10.01085kpa
17507(20d)
PROBLEM 5;,
d')(12000000)
A round steel shaft transmits 0.375 kw at 1800 rpm. Tho shear rnodu.us of elasticuy
IS 80 Gpa The tors.onal deflection is not 10 exceed 1 degree In a length equal to 20
rnaroe'c-s Fmd the shaft diameter
C 12344mm
A 6621 mm
16300 mm
3 8342 mm
025727,n
o
PHOBLEM 51 (ME Bd, Oct. 1996)
A line shaft IS to transmit 200 Hn at 900 rpm. Find the diameter of the shaft.
A. 23/16 m
C 315/16 m
S. 27/16 In
0 13/16 In
E.l!!imD
P
2,1 N
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j)h
i nq
Sl](~f
.'-,}wfl inq
0.375 = 2 rrT (180060)
T = 0 001989 KN-m
11=
SOLUTION
TL
m'I,JI ~, -:.: 1/2 m",I'J
m=wV
wV 1 1 i2wV 2
JG
x-n
0 001989(20d)
180
6
(80xl0 )
n:d4
-_
V, = 1/2 V,
32
d = 00066206 m = 6.62 mm
(rr/4)(D,' - D,')L = 1/2 (ni4)(d')L
,..,
'")
2
0,," - 0' " 112 d
(1.5)' - Dr' = 1/2 (11"
D. = 1 .322 In
PROBLEM 54
A round steel shaft rotates at 200 rpm and IS subjected to a torque of 225 N.m and a
bending moment of 340 Nrn The allowable sheanng stress IS 40 MN/n/ the
allowable tensile IS 53,3 MNim;! Find the diameter
A 23.45 mm
C 28.56 mm
0 4149 mm
B. 3445 mm
E:.7:!!mmI
Based on sbeacr.o16 r '--2
S. = _ \ M
nd' _
40 x 10 =
If it
rotates at 1725 rpm and shear modulus of elasticity IS 12.000,000 pSI. what is the
torsional deflection In degrees per foot of length?
C 3.21'
A 3.82"
B. 256
D. 423"
n:d J
7000=
16T
.
"(I! 4)3
T = 1.789 fI-lb
16
2
2
---, - M + vM t T
nd<l
53 3 x 10' =
A 1/4 In diameter shaft IS designed with a working stress 01 7000 pSI In shear
S = 16T
16 '
-- -, -- r;
'l,(0.34 t ,(0.225t
- _
j
nd
,
Based on compression.
'I
I'ROBU:M ,,6
IIE!!Im3
, T,l
d = 0.0373 m = 37 3 mrn
51 =
117
16 10.34 ,
;r:d 3
J=
\(0::34l (0.225)2
r
d = 0 04149 m = 41 49 mm
Therefore choose the larger diameter,
nd 4
rr(1 4)4
_ 3 835 x 10' ff'
=
32
32
L·lf1=12In
J/nrad
(1789xI2,112)
TL
= 00559 rad x 180
8=
JG
3835xl0·'(12.000.000)
.
d ·,41 49 mm
II = 3 21
PFOl-lLEM C,7
PROBLEM 55
A 1 In diameter shaft IS to be replaced with a hollow shaft of the same rnatenal
weighing half as much, but equally strong In torsion The outside diameter of the
hollow shaft IS to be 1 1/2 in Find the inside diameter.
C I 653 III
A. 1322 In
B. 2 123 In
0 3 123 In
A hollow shaft has an outside diameter of 4 In and an Inside diameter of 3 In. Find
the marnctcr of a solid shaft with equal strength In torsion
A2121n
B323,n
C >1?3 In
[) 3 S2 In
IIE!!Im3
S, "1 =- SI"...,.....
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lOMoARcPSD|3535879
I] H
J'. ( ·!I -,
....... llfl/llllq
16T
IT d
3
11 q
16TO"
:::::
--1
--.-1
7T(Oo
°
D )
0,/ - 0,'1-:::. rj)
(4)'. (3)' 0 d' (4)
do 352 In
I'ROIILEM 58
GmD
A 1;4 In steel shaft transmits 30 In-Ib of torque The effective length of the shaft IS 12
in: the modulus of etasncuv In shear IS 12.000,000 pSI. Find the angular deflection in
degrees.
A. 234"
C, 448'
B 1,23'
0, 634"
~
mm:mD
J =
H :::::
ITd'
'~(TI
-'
32
TL
(30)(12)
JG
1. Key - a machine member employed at the mtartacc of a pair of mating male and
terna'c Circular cross-sectional members to prevent relative angular motion
between these mating members
>1
~
. )4
4 . = 3835 x 10ft
32
3835xl0" (12000000)
= 0.078 rad x 180"/mad
00448'
2. Keyway
a groove ill the shaft and mating member to which the key fits.
3. Splines - permanent keys made integral With the shaft and fitting Into keyways
broached Into the mating
PHOl3LEM SCI
A 1 1/2 In Monel shaft IS used In a torsional application. Based on operating speed of
100 rpm and safety factor of 12. How much horsepower can it transmit? Assume the
shearrng stress IS 3/4 of the ultimate tensile stress which is 100.000 pSI
A 657 hp
C 234 hp
B. 345 hp
0. 5.23 hp
IIE!!imD
has a square cross-section with tlalf of ItS depth sunk In the shaft
4. Square key
ano half in the hub.
has a reqular cross section With a smaller om-cison placed In the
5. Flat key
radial direction with half sunk In the shaft and half In the hub and IS used where
the weakening of the shaft by the keyway IS serious
6. Round key - has Circular cross section
S" = 3'4 S,
S" = 34 (100,000) 0 75,000 pSI
7. Barth key
S _ SSLJ
,- FS
consists 01 one-half of a Circular disk fitting into a rectangular
8. Woodruff key
keyway In the female member and a semi-circular keyway In the male member
S,075,000'12
6.250 pSI
S,
0
S,
0
,
9. GibMhead taper key - is a flat key With a special qrb-he ad to facilitate easy
dnvrnq and removal of the key
16T
7Td 3
6250=
10. Saddle key
16T
-',
T -4141 751n·lb 0345.141!·lb
P = ~"TN_
= 2IT(34514)(100)
33,000
IS a flat key used Without a 'Keyway In the shaft
11. Kennedy keys
are tapered SqU,Hf' keys With the diagonal dimension In a
circumferential direction.
"(1.5r
33,000
IS a square key With bottom two corners beveled.
= 6 57 hp
is one which has a uqhl [II into one member and a loose Sliding lit
12. Feather key
In thr' 1Tl,1tltl(j member thus allowmq tile hub to move along the shaft but prevents
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121
{(I'II_",
ILO
rotation on the shalt
Types of keys:
1. square key
2 flat key
3 found key
4. barth key
5 woodruff key
6 oro-nead key
7. saddle key
8 kennedy key
9. leather key
3. Compressive Stress (So) of key
Ih/
Fe
So
(h/2)L
h
//
rmmmm
Compressive area - ~
4. Shearing Stress (5,) of key
s, = ~
Key
where
F
h
r
w
wL
L -= tenqth of key
W -= width of key
h - height of key
w
Shaft
5. Relation of key and shaft for the same material:
1. Power of key:
A. P
21t T N, KW
B P
21tTN H
33000' P
2. Force transmitted, F
F
where
d
T
T
d/2
shaft diameter
w
D
4
L = 12 D
6. Force tangent to pulley rim
l?
F'
0/
T=T'
F.r=F'.R
~
F
R
_ force tanqent to the key
force tangent to pulley rim
radn.s of pulley
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R
T'
r
2
lOMoARcPSD|3535879
122
tc: c-~ I ':>
PROBLE:vl I (ME Bd. Oct. 97)
A 76.2 rnrn In diameter shafting of SAE 1040 grade, cold rolled, navinq a yield point of
50 kSI and with a 3/4 x 5 Inches key Compute the minimum Yield pomt In the 'Key In
order to transmit the torque of the shaft. The tactor of satety to use IS 2 and S
0.50 S/
A 39120 kSI
C 27920 kSI
B 42.130ksl
D. 47.120 kSI
7788
f-
For shearing ot key
For Key
wL
S,
28844.4
00111(L)
)
54mm
2 F
h L
3/4"
" 25,000 pSI
2(28.844.41
90,000.000
S = 16T
o 015875(Lj
L ~ 0.040377 m = 40377 em
Therefore choose the longer length of key
L c 433 em
rrd3
T
L
1.11 em
(
FS
50,000
25,000 =
<~; .
For compression of key'
~
3 In
Sy
2
1
'1
;.<:
P' ,
L -c 00433 m = 4.33 em
0.75"'
d = 762 mm
S"
1.5875 em
F
S.
60,000000
S,.
28,8444 N
(0054/2)
SOLUTION
w -= 3/;" ~
L = 5"
1'23
16T
n( 3)'
132,535 94 m-lb
T = F r
I'H013LEM 3 (Ocl. 1998)
132,535.94 = F(3/2)
F = 88,357.29Ibs
For sheanng of key
F
88,357.29
A 76.2 mm cramercr shattmq of SAE 1040 grade, cold rolled, havno a Yield oom' or
50 kSI and With a 3/4 x 3/4 x 5 In key Compute the minimum Yield pou-t In the key In
order transmit the torque of the shaft The factor of safely to usc IS 2 and S,c, .;: 05
S,
wL
075(5)
S,
23,561 94 psi
Sy = S, x FS
S, = 23,561 94 x 2 = 47,120 PSI
4720 kSI
For Key
A keyed sprocket deliver a torque of 778.8 N m thru the shaft of 54 rnrn 00. The '",ey
thickness IS 1.5875 em and the Width IS 1.11 em. Compute the length of the same
key. The permissible stress value of 60 Mpa for shear and 90 Mpa for tension.
A 39.12 em
C 52.22 em
B. 4.82 em
0 432 em
F
o 4721 ksi
Em!!Ii[-':W
PHOflLEM 2 (ME Bd. Ocl. 97)
Em!!iiIm
C. 4712 kSI
A 47 52 kSI
B. 4725 kSI
w ::::: ~': " =-- 0 75"'
- c-:
L d c:: 76.2 rnm = 3 In
.,
S,
S"
S =
T
FS
16T
Tl"d 3
25,000
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50.000
2
16T
c
IT(3)3
c
25,000 PSI
3"
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124
T = 132.535.94 in-Ib
T = Fr
132,535.94 = F(3/2)
F = 88,357.29Ibs
For shearing of key:
F
88,357.29
Ss
wL
= --0.75(5)
PROBLEM 5
A metric M4 x 15 square key is used with a 16 mm shaft. If the allowable shearing
stress is 50 Mpa. How much torque can the assembly handle? The rotational speed
of the shaft is 600 rpm.
C. 24 N-m
A. 20 N-m
D. 34 N-m
B. 12 N-m
23,561.94 psi
Sy = Ss x FS
S, = 23,561.94x2 = 47,120 psi
SOLUTION
47.20 ksi
PROBLEM 4 (ME Bd. Apr. 98)
A keyed gears delivers a torque of 912.4 N-m thru its shaft of 635 mm outside
diameter. If the key has thickness of 15.875 mm and width of 11.1125, find the length
of the key. Assume the permissible stress values of 61.2 Mpa for shear and tension
at 99.8 Mpa.
A. 47.42 mm
C. 42.22 mm
B. 39.72 mm
D 46.92 mm
An M4 x 15 square key, means w = 4 mm,
h = 4 mm and L = 15 mm
4mm
4mm
For shearing of key:
F
Ss= -
wL
Ss = 50 Mpa or 50 N/mm
2
F
50 =
(4)(15)
F = 3000 N
T = F (d/2) = 3000(0.016/2) = 24 N-m
Em!!iImII
T = Fxr
912.4 = F (0.0635/2)
F = 28,737 N
Based on shearing of key:
F
Ss = - wL
61200 000
"
125
K('l/S
=~37
0.0111125L
L = 0.04225 m = 42.25 mm
Based on compressive of key:
Sc = 2F
hL
99,800,000
L = 0.03628 m
2(28,737)
~ . _ - -
0.015875L
= 36.28 mm
Therefore use the longer length to be safe:
L = 42.25 mm
15.875m _
11.11 em
~'
E
I
16 mm
)1
PROBLEM 6
A flat key is to be designed for 75 mm diameter shaft which will transmit 150 'r<YV at
400 rpm. If allowable shearing stress is 200 Mpa and key width is 15 mm, determine
the length of key.
C. 33.75 mm
A. 30.65 mm
D. 32.85 mm
B 31.83 mm
Em!!iImII
P = 2rrTN
150 = 2rrT(400/60)
T = 3.581 KN-m
T = Fxr
3.581 = F x (0.075/2)
F = 95.493 KN
I
IE
I
For shearing of key:
F
So =
Lw
200,000 =
_9~3
L(0.015)
L = 0.03183 m = 3183 rnrn
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)
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126
127
J,! '1/',
PROBLEM 7
A rectangular key is used in a pulley to transmit 100 KW at 1000 rpm on 50 mm shaft
diameter. Determine the force required to remove the shaft from the hub if friction is
0.4.
A. 33.24 KN
C. 28.35 KN
B. 36.85 KN
D. 30.55 KN
1\ <+3 KN
B. 48 KN
C. 45 KN
D. 46 KN
K=I-15·UW..'
For line shaft:
D 3N
P = -53.5
SOLUTION
3
15/0.746 =
P=2nTN
100 = 2nT(1000/60)
T = 0.9549 KN-m
T = Fxr
0.9549 = F x (0.050/2)
F = 38.197 KN
.0 (600 )
53.5
o = 1.2148 in = 0.03085 m
L = 1 in = 0.025398 m
w = 0/4 = 0.03085/4 = 0.007714 m
F
Ss =
Lw
The friction on key experience both upper half on one side and lower half on the other
side.
F
230,000
= (0.001'714)(0.025398)
F = 45.061 KN
Force required to remove the shaft from the hub
P = 2 f F = 2(0.4)(38.197) = 30.55 KN
PROBLEM 10
PROBLEM 8
A 7/16 in height x 3 in length flat key is keyed to a 2 inches diameter shaft. Determine
the torque in the key if bearing stress allowable is 25 Ksi.
A. 16,406.25 in-Ib
C. 17.42 in-lb
B. 15,248.56 in-Ib
0 246.75 in-lb
A square key is to be used in a 40 mm diameter shaft and that will developed a 2 KNm torque. If bearing stress of the key is 400 Mpa, determine the cross sectional
dimension of square key to be used if key length is 30 mm.
2
C. 446.80 rnrn"
A. 324.80 mm
D. 277.77 mrn"
B. 246.80 rnrn"
&-i-"'5it-D
"W"'mE'
w = 7/16 = 0.4375 in
F
Sc = -~
7/16"
(h / 2)L
T = Fxr
2 = F x (0.04/2)
F = 100 KN
w
w
F
25000 = -----,
3 (0.4375/2]
Sc =
F = 16,406.25 \bs
F
---
(h / 2)L
400000 =
100
,
(0.030/ 2)(h)
T = Fxr
h = 0.01666 m = 16.6667 mm
w = h = 16.667 mm
A '" 16.667 x 16.667 = 277.77 rnrn"
T = 16,406.25 x (2/2) '" 16,406.25 in-Ibs
!(
,
PROBLEM 9
A rectangular key is used in a pulley connected to a line shaft at 15 KW and 600 rpm.
If shearing of key is 230 Mpa, determine the force acling on key length if key width
one-fourth of shaft diameter and key length is 1 inch
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40 mm
)
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KellS
12R
EmmmD
PROBLEM II
A 100 KN force is acting on a key that has a length of 4 times its height. If bearing
stress of key is 400 Mpa, determine the height of key.
A. 1054 mm
C. 12.65 mm
B.1118mm
D.15.25mm
SOLUTION
s, '"
For the same shaft and key material:
w '" 1/4 0
w = 1/4 (0.12)
w :" 003 m '" h
F
Sc'" - - (h /2)L
F
(h/2)L
F
410,000 = ­ (0.14)(0.03)
H
w
F '" 1722 KN
Fxr'" F'xR
1722(0.12) '" F' (1.2)
F' '" 172.20 KN
L '" 4 h
100
400,000 '" (4h)(h /2)
h '" 0.01118 m
h '" 11.18 mm
PROBLEM 12
A one meter pulley is fastened to a 100 mm shaft by means of 25 mm square key and
150 mm long. What force tangent to pulley rim will shear the key if shear stress is 200
Mpa?
A. 65 KN
C. 75 KN
B. 70 KN
D. 80 KN
EiD!Im'D
F
Ss '" Lw
25mm
F
25mm
200,000 '" 0.025(0.15)
F = 750 KN
Torque on key = Torque on pulley
r, '" Tp
F x r = F' x R
750(0.100) = F' (1)
F' '" 75 KN
'(
I
100 mm
)
PROBLEM 13
A 1.2 m pulley is fastened to a 120 mm shaft by means of a square key with 140 mm
long. What force tangent to pulley rim will crush the key if bearing stress is 410 Mpa.
Assume key and shaft are of the same material.
A. 16042 KN
C. 18042 KN
B. 16525 KN
D. 172.20 KN
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130
Couplif/S}
131
Coupling
2. Total force transmitted(F)
8
Coupling
3. Force transmitted per bolt(F b )
~
where:
n = no. of bolts
Dc = bolt circle diameter
Flange
---71 t~
~
•
1. Coupling - is a mechanical device
which is used to connect length of
shafts permanently.
2. Rigid couplings - couplings that do
not allow angular, or rotational
flexibility and used with collinear
shafts.
Shaft
4. Shearing of bolts(Ss)
•
Shear Area
5. Compressive stress on bolts and flange(Sc)
Flange
3. Flange coupling - type of rigid coupling which consist of two halves of flanges
connected by each other by bolts.
4. Sleeve or collar couplings - rigid coupling which is a cylindrical collar pressed
over the ends of two co-linear shafts.
where:
t = thickness of flange
d = bolt diameter
5. Flexible couplings - coupling which allows angularity to take care of
misalignment of the shafts
A. P = 2. n TN, KW
B P =
21tTN
33,000,HP
~i
--+--
mmmm
1. Power transmitted
I
-4 l~t
d
Compressed Area
PROBLEM 1 (ME Bd. Oct. 97)
A flanged bolt coupling has ten (10) steel 25.4 mm diameter bolts evenly tighten
around a 415 mm bolt circle. Determine the torque capacity of the connection if the
2
allowable shearing stress in the bolt is 50 MN/m .
C. 46.15 KN.m
A. 5995 KNm
B. 52.6 KN.m
D. 43.8 KN.m
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132
Ccn ipl i iu]
(nI4)d 2
P = 2rcTN
60 = 2 rt T (180/60)
T = 3183 KN-m
where: Fb = force transmitted per bolt
50,000
Fb
= . . - .~
-1
SOLUTION
For shearing stress in bolts
s, = -~
133
COlLpiing
Em!!J:mD
-'-
(n 14)(0.0254)2
Torque on coupling:
T = F (OJ2)
3.183 = F (0.180/2)
F = 353677 KN (total force)
Fb =' 25335 KN
F = total force on bolts = n x F b
F = 10(25.335) = 25335 KN
T = F (-~)
o
2
= 253.35(0.415/2) = 52.57 KN.m
F
Fb = force transmitted per bolt
Fb
PROBLEM 2
A flange bolt coupling consist of eight 20 mm diameter steel bolts space evenly
Sc = dt
around a bolt circle 300 mm in diameter. If the coupling is subjected to a torque of
15.1 KN-m, determine the maximum shearing stress in the bolts?
A. 40 Mpa
C. 70 Mpa
B 38 Mpa
O. 29 Mpa
FS
Fy
Sc
4.421
(0016)(0.019)
.148
-+---
14542
-
n
353677
= -
8
= 14,542.67 Kpa
= 4.421 KN
14.542 Mpa
= 30.8
PROBLEM 4
A turbine is connected to a generator by means of flange coupling that has a bolt
~i·l![H
T =F x r
15.1 = F (0.300/2)
20 mm
Lomm
F = 10067 KN (total force)
L
Fb = force per bolt
Fb = Fin =' 100.67/8 = 1258 KN
For shearing of bolt
Ss ~
.~ rc d2
4
12.58
rc
2
- (0020)
4
-
-_.~
40043.38 Kpa = 4004 Mpa
PROBLEM :3
A flange coupling having 180 mm bolt circle and 19 mm thick uses 8 bolts, 16 mm
diameter to connect two shafts. It is use to transmit 60 KW at 180 rpm. Determine the
factor of safety in bearing if yield point in compression is 448 Mpa
A. 156
C 30.8 /
B 185
o. 25.4
circle diameter of 500 mm. The generator output is 40 MW, 3600 rpm and 90%
etticiency. If there are 16 bolts, determine the force acting on each bolt
A. 26.41 KN
C. 35.62 KN
B 29.47 KN
O. 3261 KN
Emmir.m
40000
Brake Power = .i.:»:
= 44444.44 KW
0.9
P = 2rcTN
44,444.44 = 2 rt T (3600/60)
T = 117.89 KN-m
o co. 500 mm = 0.50 m
T = F (0/2)
117.89 = F(0.50/2)
F = 471.57 KN
F b = Fin
where: n = no. of bolts
71.57
F I , = ---- = 29.47 KN
16
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1:34
( '0 I I l ' II 1111
COll/J[i,,,/
PH.OBLEM 5
.,:r]I{·K~
A 75 mm diameter shaft is transmitting 300 KW ill 600 rpm A solid coupling has 6
bolts each 18 mm In diameter. Find the required bolt circle diameter if shearing stress
in bolts is 27.5 Mpa.
A. 22740 mm
C 25450 mm
B. 233.60 mm
D. 27260 mm
T
P=2rrTN
300 = 2 1l T (600/60)
T = 47746 KN-m
= - ~
(rr / 4)d
4=:.~
18 mm
2
27 500 =
f},__
,
rr/4(0.018)2
Fb = 6.9978 KN
F = 6(6.9978) = 41.9868 KN
•
s,
Fb
dt
T = F (0/2)
15 = F(O.25/2)
F = 120 KN
SOLUTION
S,
75mm
Dc
Solving for t based on compressive on key:
15
15000 =
,
(0.025)t
t = 0.040 m = 40 mm
PROBLEM 6
The total force of 125 KN is required of flange coupling that has 5 bolts has also
shearing stress of 30 Mpa for bolt. Determine the required diameter of bolt.
A. 62.54 KN
C 45.62 KN
B. 54.21 KN
O. 32.57 KN
--i·'!!ImD
Ss
125/5 = 25 KN
Fb
­­­­­­,­-
(rt /4)d 2
25
30.000 = ­ ­ (rr/4)d 2
d = 00325 m
25 mm
Fb = F = 120 = 15 KN
n
8
T = F(D c/2)
4.774 = 41.9868(Oc/2)
Dc = 0.22740 m = 22740 mm
Fb
135
= 32.57 mm
PROBLEM 7
A flange coupling with bolt circle diameter of 250 mm uses 8 bolts, 25 mm diameter.
the torque transmitted is 15 KN-m and compressive stress of bolt is 15 Mpa.
Determine the required flange thickness.
A. 25 mm
C. 35 mm
B. 30 mm
D. 40 mm
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250mm
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136
,9
I :I
Pressure \', ,.,>.,>/.[
Pressure V, 'S.'>, '/
~rs;u
"~s.el
rnnxrrnum pressure at the bottom
w h == (SG x ww) h
I'
P
-<-t
h
For Spherical Pressure Vessel
1<
)j
Di
(.. If efficiency of joint is not considered:
S =!,]J~
4t
,
S== POi
41 e
(.. OJ :::: Do -2t
For Cylindrical Pressure Vessel
where:
I
,
,
,,
,,
(. If efficiency of Joint is considered:
(, If the ratio of wall thickness to the inside diameter ( ~
) is less than 0.07 then
OJ
.
the cylinder is considered as thin-wall.
,
,
I
I
t~
,,
,
,,,
~t
),
D'I
,
k
,
~
.
!
e:::: joint efficiency
t
(.. If efficiency of Joint is not considered:
S, == tangential stress or hoop stress
If the ratio of t/D, is greater than 007, then the vessel is considered as thick-wall.
S ",PDi
2t
SL
longitudinal stress
Sl
PDf
4t
"
2te
l1-~+tS 'lQ 2
t ::::
Ie
Dr
~
s,
When the vessel is subjected to an internal and external pressures:
S == tangential stress
t == thickness of the wall
For Maximum Internal Stress:
2
S
1/
"
For a cylindrical vessel filled with fluid:
2.
~
2-
fo
Do :::: outside diameter
2
_ Pi(fo Hi ) ~2Por
" OJ '" 0 0 - 2t
where: t:::: wall thickness
tangential stress
internal pressure
SI
P,
4t e
where:
e == joint efficiency
0, == inside diameter
J
1 == wall thickness
D == inside diameter
(.
PDt
SI-Ii
where:
<. If efficiency of joint is considered:
PO
SI == ------1...
Using Lame's equation for internal pressure:
~
- rj
,,
~
For Maximum External Stress:
SIO ~
Or{bP~ir 2
2
•... ':'./
2
ro 2 .-. r,2
I
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2
+ rj )
.t<
" r.
:<
.
:I
): ~
)1I
I
r
I
o
'
t
lOMoARcPSD|3535879
Pressure \ ·(·ss,·/
l0lJ
where:
St, = maximum internal tangential
stress
SIO = maximum external tangential
stress
P, = internal pressure
Po = external pressure
r0 = outside radius
r, = inside radius
PROBLEM 1 (Oct. 2000)
Find the thickness of a metal needed in the semi-spherical end of cylindrical
vessel 0.70 meter in diameter subjected to an internal pressure of 2.7 N/mm 2
The material is mild carbon steel and tensile stress is 69 N/mm.
A. 5.2 mm
C. 7.53 mm
B. 6.02 mm
O. 6.84 mm
l' 1« 1I\L1': M :\ (Oct. 1998)
Deterrrune the bursting pressure of steel with diameter of 10 inches and make of
1/4 111 thick plate. The joint efficiency is at 70% and the tensile strength is 60 ksi.
C. 2.400 psi
A 4.020 psi
B. 4.002 pSI
D. 4,200 psi
SOLUTION
Using spherical vessel formula for thin wall:
Considenng the efficiency,
8
PO,
4te
Using spherical vessel formula for thin wall:
S
60 ksi = 60,000 psi
Two semi-spherical ends is considered as sphere.
2
P = 2.7 N/mm = 2.7 Mpa = 2700 Kpa
2
S = 69 N/mm = 69 Mpa = 69,000 Kpa
60,000
SOLUTION
S =
~Oi
P(10)_
= .£2~)
= 4(11'4)(0.70)
P = 4200 psi
4t
69,000
139
Pressure \',',.,,-1
PROBLEM 4 (ME Bd. Oct. 97)
Determine the safe wall thickness of a 30 inches steel with internal pressure of
7.82 Mpa. The yeild stress of material is at 275.48 Mpa. The factor of safety to
use is 2.0.
C. 21.6 mm
A. 3/4 in
D. 5/8 111
B. 23.4 cm
(0.7)_
4 (tj---
0.00685 mm = 6.85 m
E1I!DI':D
PROBLEM 2 (Oct. 2000)
A cylindrical tank with 10 inches inside diameter contains oxygen gas at 2500
psi. Calculate the required wall thickness in millimeter under stress of 28000 psi.
A. 10.54
C. 10.24
B. 11.34
O. 12.44
SOLUTION
8
=~Oi
28,000
8, =
~
For cylindrical pressurized tank:
= ~50l
PD~
= PO,
FS
2t
!<
30in
~
PROBLEM 5 (ME Bd. Oct. 97)
2t
t = 0.446 in (25.4)
1,
P = 7.82 Mpa
2t
275.48 = 7.82@~)
2
2t
t = 0.8516 in = 21.6 mm
2t
~(.1
Using thin-wall formula:
= 11.34 mm
Determine the bursting steam pressure of a steel shell with diameter of 10 inches
and made of 114 inch thick steel plate. The joint efficiency is at 70% and the
tensile strength is 60 ksi.
C 42.8 ksi
A. 4200 psi
D 8500 psi
B. 105 kSI
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140
Pressure \lessel
Pr c ss urc \Ie sscl
Ei3!Iim3I
St =
Steel shell usually spherical:
Considering the efficiency,
S
PO
-- ---'2t
Based on operating pressure:
180(60)
9500 = : - - - - -
POi
2t
4tTj
60,000
t
P(1Q)--
=:
141
4(0:25 )(0 70)
=:
0.568"
Based on pop-out pressure:
P =: 4200 psi
9500 c;
200(60)
2t
PROBLEM 6 (ME Bd. Apr. 98)
Compute the safe wall thickness of a 76.2 cm diameter steel tank. The tank is
subjected to 7.33 Mpa pressure and the steel material has Yield stress of 215.4
Mpa. The factor of safety to use is 3.
A. 1 1/2 in
C. 4.09 cm
B. 3.89 cm
D. 369 cm
t = 0_631"
A standard plate thickness of 5/8" (0.625") would be safe
bursting pressure ustnq a wall thickness of 5/8" (0.625"):
S, =:
POi
2t
P(60)
2(-0.625)
65,000 =
SOLUTION
P = 1354 psi
~jt
Usually steel tank is cylindrical,
Sy
POi
FS
2t
215.4
7.33(76.2)
2t
3
PROBLEM 8 (ME Bd. Oct. 95)
P = 7.33Mpa
--'-~-._--
=:
)1
!<
Determine the thickness of a steel air receiver with 30 inches diameter and
pressure load of 120 pSI, design stress of 8000 psi.
A. 1/4 in
C. 5/8 in
D. 1/2 in
C 3/8 In
76.2 em
SOLUTiON
3.89 cm
For cylindrical tank:
PO
St = ----,
2t
PROBLEM 7 (ME Bd. Oct. 95)
A steel cylindrical air receiver with 5 feet diameter and pressure load of 180 psi,
design stress of 9500 psi maximum. The pressure vessel is to be provided with 1
1/2" diameter drain valve Installed at the bottom of the vessel and safety
pressure relief valve installed either at the top most or at the side with prop-out
rating of 200 psi. assume a 100% weld joint efficiency. The lap welding tensile
strength is 65,000 psi. determine the bursting pressure of this air receiver.
C. 1454 psi
A. 1154 psi
B. 1354 psi
0 1254 pSI
Ei3!Iim3I
0, = 5 It =60 in
_.-
Solving for the
8000
120(30)
---_.
2t
t =: 0.225.
Therefore: Use 1/4" standard thickness
(Kent's p. 6-03)
PROBLEM 9 (ME Bd. Oct 93)
A compression ring is to be used at the junction of a conical head and shell.
Determine the required area of the compression ring if the pressure is 50 psi and
the stress is 13,750 pSI. Assume an efficiency of the JOint IS 80%.
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142
Pressure Vessel
143
PreSSlUT Vessel
Em!!iirf3
IED!DImI
20mm
'I- ~t
k
S =
Using thin wall cylinder formula:
--Do
S
>I
PD,
2t
PD,
!<
2te
140,000 = P(0.50)
P = 11,200 Kpa
t = 0.00227 D,
Do
Do
Do
D, + 2t
D, + 2(0.00227 D,)
1.00454 D,
Area
rr/4 [( 1.00454 D,2 - D,2 J
2
2
Area = 0.00715 D, in
11.20 Mpa
PROBLEM 12
A water tank 10m x 12 m is completely filled with water. Determine the minimum
thickness of the plate if stress is limited to 50 Mpa.
A. 11.77 mm
C. 13.55 mm
B. 12.66 mm
D. 12.44 mm
KJe1!ii«ml
PROBLEM 10 (ME Bd. Apr. 94)
A cylindrical tank with 10" inside diameter contains oxygen gas at 2500 psi.
Calculate the required wall thickness in (mm) under stress of 28,000 psi.
A. 11.44 mm
C. 11.34 mm
B. 10.6 mm
D. 10.3 mm
SOLUTION
t~·
PD,
SI = .-.
2t
~t
Solving for maximum pressure located at the
bottom of the tank. For cylindrical pressure
vessel
P
wh=9.81(12)
117.72 Kpa
P = 2500 psi
S = ~Di
2(1)
t = 0.4464 in = 11.34 mm
)j
2(0.02)'
13.750 =J'0D;
2t(0.80)
28 000 = ~50J.1Cl)_
,
500mm
1<
10in
>I
1<
>I
10 m
2t
50.000 = 2~7.(10)
2t
t = 0.011772 m = 11.77 mm
PROBLEM 11
PROBLEM 13
Determine the internal pressure of a cylindrical tank 500 mm internal diameter,
20 mm thick and 3 m long if stress is limited to 140 Mpa.
C 1120 Mpa
A. 1006 Mpa
B. 10.52 Mpa
D. 12.88 Mpa
A spherical shell of 1.8 m outside diameter and 1.725 m inside diameter contains
helium at a pressure of 10.4 Mpa. Compute the stress in the shell.
A. 124.08 Mpa
C. 96.48 Mpa
B. 119.06 Mpa
D. 88.46 Mpa
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]44
]45
Pressure Vessel
Pressure Vessel
=~§
.i·'K:m:mI
PO,
FS
~20
Do -0,
2
2t
= __ P (2)
4.5
2 (0.015)
P = 14 Mpa
1.8 -1.725
t = - - - - - - = 0.0375 m
t~
2
S = ~i
OJ = 1.725 m
1 = 119.60 Mpa
10.4 (1.72_5
PROBLEM 14
A 600 mm diameter spherical steel pressure vessel has a uniform 8 mm wall
thickness and an internal gage pressure of 10 Mpa. If the ultimate stress of steel
is 420 Mpa, determine the factor of safety with respect to tensile of failure.
A. 230
B. 1.89
16
t)ROBL~M
Oo=1.8m
4(0.0375)
4t
~t
C. 2.69
D. 148
The cylinder of a hydraulic press is made up of thick walled cylinder having an
inside diameter of 300 rnrn. It is subjected to an internal pressure of 40 Mpa
Determine the thickness of the cylinder without exceeding a shearing stress of 80
fvlpa.
C. 48.92 mm
A. 62.13 mm
D. 7240 mm
B. 5861 mm
EJ'immi
Using thick wall formula: (Lame's Equation)
SOLUTION
Solving first the inside diameter:
OJ = Do - 2t
0, = 600 - 2(8) = 584 mm = 0.584 m
t.
8mm
l
~ J~+ -1J
= Q~?
U~:r
=
P =40 Mpa
-J
1
= 10980 mm
= 010980 m
~8m
~t
I(
Using the spherical vessel formula:
S =
P~j
10(0584)
4 (0.008)
4t
FS
?"'S
420
18250
PROBLEM 17
= 18250 Mpa
The internal pressure of a 400 mm inside diameter cylindrical tank is 10 Mpa and
tank thickness IS 25 mm. Determine the stress developed if joint efficiency is
95%.
C 86.75 Mpa
A 8010 Mpa
D. 7842 Mpa
B. 84.21 Mpa
2.30
PROBLEM ]5
Determine the largest internal pressure which maybe applied to a cylindrical tank
2 m in diameter and 15 mm wall thickness, if the ultimate stress of steel is 420
Mpa and a factor of safety of 45.
A. 14 Mpa
C. 18 Mpa
I
B. 2.6 Mpa
D. 22 Mpa
-J.~
SOLUTION
~
~ 25m
S
PO,
15 mm
S
p
10,000(040)
------_._"-----~-
2(0.025)(0.95)
P OJ
2t
1<
2m
10 Mpa
2tT]
84.21 Mpa
S
:;
300 mmn
)1
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84.210.526 Kpa
I(
400 mmn
:;
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146
Pressure Vcssd
Pressure Vessel
PROBLEM 18
A spherical tank 15 mm thick has an internal pressure of 5 Mpa. The joint
efficiency is 96% and stress is limited to 46,875 Kpa. Find the tank internal
radius.
A. 540 mm
C. 270 mm
B. 200 mm
O. 300 mm
SOLUTION
IID!!JmD
Using thick-wall formula:
r iTs! + PI) - 1l
2l\i(St- Pi) J
t= Q
0.050 =
.0. 30 r ((30,000 + Pi) -1l
l~ (30,000 - p;)
2
S = ~
I' ,,:
4111
46 875 =
,
.~
5,000LCJ,L_
4(0.015)(096)
I
0, = 0.540 m = 540 mm
r, = 540/2 = 270 mm
I(
I
1333 =
I I
)! ~5m
.
,~
I(30,~_l
l \i (30,000 - Pi)
0.333 =
I
Di
147
I
IIlJ£:+.~03(
~L
(30,000-Pi )
J
l
1J
J
Squaring both sides:
000
1.333 =
+ p;
3000 -PI
:3
PROBLEM 19
A cylindrical tank has an inside diameter of 5 in and is subjected to an internal
pressure of 500 psi. If maximum stress is 1200 psi, determine the required
thickness.
A. 1.0 in
C 1.4 in
B. 1.2 In
O. 1.6 in
30,000 + P, = 53333.33 - 1.7777P,
P, = 8400 Kpa = 8.4 Mpa
PROBLEM 21
A round vertical steel tank has an Inside diameter of 3 m and 6 m height. It
3
SOLUTION
contains gasoline with a density of 750 kg/m If the allowable tensile stress is 25
Mpa, find the minimum thickness required
A 2.65 mm
C. 3.65 mm
B. 2.85 mm
O. 1.82 mm
To check if the problem is thin or thick wall:
S = POI
2t
1200 = .?g~(5)
wa·"'U(m11
2t
Solving for maximum pressure at the bottom of tank.
t = 1.04166 in
tID, = 104166/5
tiD, = 0208> 007
P
Therefore, use thick-wall formula:
S
By Lame's Equation for thick vessel
t=
l
l
~ I~: +-~? -1J ~= J~-ci:
~cif
w h = (750 x 9.81 Ii 000)(6) = 44.145 Kpa
POI
2t
25.000 =
-1]
1395 in
4·~@1
2t
t = 2.648 x 10. m = 2.648 mm
3
!'\-{013LEM 22
PROBLEM 20
A thick wall cylinder has 50 mm thick and internal die-meter of 300 mrn. If stress
is limited to 30 Mpa, determine the maximum internal pressure
A 8.1 Mpa
C 8.3 Mpa
O. 8.4 Mpa
B 82 Mpa
A cylinder having an internal diameter of 16 in and external diameter of 26 inches
IS subjected to 1500 psi external pressure and internal pressure of 9.000 psi.
Determine the hoop stress on outer surface of cylinder.
A. 8,742.65 psi
C. 9.400.62 psi
0 9,471.43 psi
B 7,642.85 psi
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148
Pressure Vessel
noll s & Power Screw
J49
SOLUTION
r, = 16/2 = 8 in
r o = 26/2 = 13 in
2
2
2
2Pj r, - Po( r0 + r, )
Sto
= --- ----..
2
2
ro
SIO
s;
-+1
-+
-
r,
2(9000)(8)2 -1500(13 2 ... 8 2)
----_ -._---_._-._..
13 2 .. 8 2
7642.85 psi
1500 psi
-+1
-+
-+
I<
k
16in
26in
~
>!
)1
m:mm:tmtJ
PROBLEM 23
A cylindrical tank has a mean diameter of 40 cm and thickness of 10 ern. It is
subjected to an external pressure of 140 kq/crn" and maximum internal
tangential stress is 900 kq/crn". Determine the maximum internal tangential
stress is 900 kq/crn". Determine the maximum internal pressure of the tank.
2
C. 1246.5 g/cm
A. 2947.5 kg/cm
2
D. 1942.6 kq/crn"
B. 34265 kg/cm
Em!!JirmI
r, = 40 - (10/2) = 35 cm
= 40 + (10/2) = 45 cm
2
2
2
Stl = fJ-,~r
2Po o
Io
-
........ 140
r
~
Bolts and Screws are threaded fasteners which are used to hold together
machine members which require easy dismantling.
,. Pitch(p) = is the axial distance between adjacent threads.
~.
Lead '
(,
Minor diameter - IS the smallest diameter of the threads.
~,
Pitch diameter - is the mean of major and minor diameters.
<. Major diameter - is the outside diameter of the threads and is the nominal
diameter
r0 2 - r,
SI'
SI'
900( 45 2 + 35 2) - 2(140)( 45)2
- --452 ' _ 352 _.-.2947.5 kq/crn"
!S the axial distance a thread advances in one revolution.
1<
t t t
Stress area - is the area of an imaginary Circle whose diameter is the mean of
the pitch and minor diameter.
)1
Stress area =
40 em
IT
l
4.
2
..o., +D,'
-_. i
2
)
Ie. Common types of bolts and screws:
1. machine bolts
2. stud bolt
3. eye bolt
4. U-bolt
I..
5. stove bolt
6. cap screw
7. set screw
Types 01 threads:
~.;
UNC(Unified National Course) - for general use, except where other types are
t,Ynrr.mended.
, UNF(IJ'lified National fine), frequently useo in automotive and aircraft work and
Wllt:lf; a line adjustment is requllf:cJ
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lOMoARcPSD|3535879
150
130/{ s
&. Power Screw
3. UNEF(Unified National Extra Fine) ~
very fine adjustment is required
151
Bolts & Power Screw
used in aeronautical equipment and where
o
~
nominal diameter
F, = Initial tension
Common forms of threads:
1. Acme thread
2. Sellers square thread
Co
«':
1. Formulas from Vallance:
"
Sw = C(A r ) 0 4 18
Depth of tap
L = p (for single thread)
L = 2p (for double thread)
L = 3p (for triple th read)
= 1.5 0 (for cast iron)
'" Linear velocity = (rotational speed)(Lead)
v = NL
1.25 0 (for steel)
D
'" Initial torque = 0.2 Fa 0
where:
Sw = permissible working stress, psi
Fa = applied load, Ib
A = stress area, in 2
C = 5,000 (for carbon steel)
C = 15,000 (for alloy steel)
o = nominal diameter
2. Formulas from Faires:
Sy(A s )1/ 2
"
Sd = ~-­
6
"
€.-
,
Iv
(. Lead (L)
(. Fa = C(A) 1418
"
x
Pitch(p) ~IS the axial distance a thread
advances in one revolution.
1
p =
No. of threads per inch
Sy(A s )3 / 2
Fe = ------.-6
Depth of tap = 1.5 0 (for cast iron)
= 0 (for steel of wrought iron)
c
Lead angle(x)
"
rrO m
Torque applied to turn the screw, T
For square thread:
7
T
= yv¥~
l1t~ni
~
For ACME thread:
Depth of Tap
T
= -~. WO m lcos<1>tanx+f]
------.--------2
cos <1> - f tan x
where:
Om = mean diameter of screw
W = weight or load
f = coefficient of friction of th reads
0
<1J = 14.4
Initial Torque = C 0 F,
where:
Sd = design tensile stress
Fe = tensile load
As = stress area
Sy = yield stress
C = 0.20 for as received
= 0 15 for lubricated
Lead
tan x =
'" Torque required to overcome collar friction, Te
r. = fs_~J:-S
= fe W o,
2
2
where
0, = (Do + 0,)/2 = rll + r,
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GT
lOMoARcPSD|3535879
152
a-ieJ!,ureg,
Dc = mean diameter of collar
r 0 = outside radius of collar
r, = inside radius of collar
fe = coefficient of friction of collar
From Machineries Handbook:
W '" working strength of bolt
2
W '" S, (055d . 0.25d)
W '" 12,000[0.55(2)2 . 0.25(2)]
W '" 20,400 Ibs '" 2040 klb
(, Total torque to operate the screw, TT
TT = T +- Te
(, Output power of screw(P o )
PROBLEM 3 (ME Bd. Oct. 97)
Po = Weight x linear velocity of screw
'" W xv
C
Power input of screw(Pi)
MOTOR
Power Screw
P, '" 2 IT TT N
C
1.5:~
Holl s <v, Power Screw
Bolts &. Patne-r .screw
An eyebolt is lifting a block weighing 350 Ibs. The eyebolt is of SAE C 1040 material
with Su '" 67 ksi and Sy = 55 ksi, what is the stress area ( In inches square) of the bolt
if it is under the unified coarse series thread?
A. 0.1341
C. 0.0991
O. 0.1043
B. 0.1134
Em!!ImD
Efficiency of power screwte).
a. For square thread:
e
e
From Faires p. 150:
Fe = applied load on bolt
Power Output/Power Input
tan x(1 - f tan
S A 3/2
fO
tan x + f + (c_c )(1 ~ f tan x)
Om
b. For ACME thread:
tan x(cos o - f sin x)
--_._---.._ . _ - - - - - - - - - - - - - - -..e '"
f 0
.
tan x cos o + f cos x + (c_c_ )(cos<jl - f sin x)
Om
Fe
= .X __s __
6
213
I
_ (. 6 Fe
A s - 1-- I
\ SY )
Stress Area
= r~(350)
t
1,55,000 I
3
0.1134 in
2
PROBLEM 1 (Oct. 1998)
Compute the working strength of 1 In bolt is screwed up tightly in packaged joint when
the allowable stress is 13,000 psi.
A. 3,6001bs
C. 3,8001bs
B. 3,7001bs
D. 3,9001bs
EmI!BtmI
2
F '" S(0.55d - 0.25d) '" 13,000[0.55(1)2 - 0.25(1)}
F '"
PROBLEM 4 (ME Bd. Apr. 97)
If the pitch of a screw is 2/9 find the thread per inch.
A. 0.34
B 4.5
C. 54
O. 17
E'!!Jrm
\<
3,9001bs
Pitch
No of thread per inch
1
PROBLEM 2 (ME Bd. Oct. 97)
What is the working strength of a 2 inches bolt which is screwed up tightly in a packed
joint when the allowabl, working stress is 12,000 psi.
A. 204 klb
C 234 klb
O. 18 klb
B 224 klb
2/9 = - . - - - - - No. of thread per inch
No. of thread per inch
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= 45
~r
1 in
>\
lOMoARcPSD|3535879
]54
Bolts & Pot ncr Screw
Boll s &. Power Screw
PROBLEM 5 (ME Bd. Exam.)
Find the horsepower lost when a collar is loaded with 1000 Ib rotates at 25 rpm, and
has a coefficient of friction 0.15. The outside diameter of the collar is 4 in and the
inside diameter is 2 in.
A. 0.045 hp
C. 0.089 hp
B. 0.89 hp
D. 0.56 hp
Em!!iiImI
-
= 015(~O)2+
2(12)
= 18.75ft-lb
2
0.00515 KN-m
P = 2nTN
1000
1
P = 2n(0.00515)(--)(----)
60
0.746
P = 072 Hp
P = 2(n)(18.75)(25) = 008925 h
33,000
P
PROBLEM 6 (ME Bd. Apr. 98)
Compute how many 3/8 inch diameter set screws required to transmit 3 Hp at a shaft
speed of 1000 rpm. The shaft diameter is 1 inch.
A. 1 1/2
C. 3
B. 2
D. 1
Em!!iiImI
PROBLEM 8
Determine the permissible working stress of a UNC bolts that has a stress area of
2
0.606 in if material used is carbon steel.
A. 4055.5 psi
C. 4675.5 psi
B. 5244.5 psi
D. 4186.5 psi
IEm!!imD
From Machineries Handbook, 24th Ed. p. 1452
Sw = C (Ar)o 418
_ DNd 2 / 3
-,,-50-
p -
where:
d
ro + rj
T = fW(-----)
T = 0.15(100)(000981)(O.05CJ..:+-.Q:02?.1
~
2
)
Solving for the power lost from collar:
P = 2 IT TN/ 33,000
3
SOLUTION
T = f W rj
rl = mean radius of collar
T c -- -fcW(r2-~ o +r,)
H
155
at 1000 rpm and the coefficient of friction between the collar and the pivot surface is
0.15.
C. 0.5 Hp
A 08 Hp
D. 1.2 Hp
B. 0.3 Hp
C = 5000 (for carbon steel)
D = shaft diameter, in
d = set screw diameter, in
N = speed, rpm
Sw = 5000(0606)°418 = 4055.49 psi
PROBLEM 9
1(1 000)d 2 /3
50
0.4383 in
Number of set screws
0.4383
3/8
2
The stress area of NC bolt is 0.763 in , if material used is carbon steel, determine the
applied load on the bolt.
C. 4675.5 psi
A. 3407.141bs
D. 4186.5 psi
B. 5244.5 psi
1.17
(say 2)
IEm!!imD
PROBLEM 7 (ME Bd. Apr. 95)
What is the frictional Hp acting on a collar loaded with 100 kg weight? The collar has
an outside diameter of 100 mm and an internal diameter of 40 mm. The collar rotates
Fa = C (Ar)l 418
C = 5,000 (for carbon steel)
Fa = 5,000(0763)1"18 = 3407.141bs
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noll." 'v, Power SCrelL!
156
Boll.'-; ,"x. Power Screw
PROBLEM 10
A 12 cm x 16 cm air compressor has 5 bolts on cylinder head with yield stress of 440
Mpa. If the bolt stress area is 0,13 inc', determine the maximum pressure inside the
cylinder.
A. 546.71 psi
C 742.78 psi
B. 671.96 psi
D. 840.60 psi
SOLUTION
Sy = 440,000 x (14.7/101325)
Sy = 63834.196 psi
Fe
~
v = linear speed In raising the screw
v = NL
8 = N(8 mm x 1/25.4 x 1/12)
N = 3048 rpm
Solvinq for the power input:
P, = 2 n T N
P, = 211:(0,050)(304.8/60)
P, = 1.5959 KW x 1/0.746
P, = 2.139 Hp
«':
Sy(A s )3 / 2
l~7
IBfPm
-----
6
PROBLEM 13
(63,834.19)(013)3/2
,,-_ -'-'
---6- ,,--
F
Fe = 498.67 Ibs
F = 498.67(5) = 249337 Ibs
Pressure = F/A
2493,37
Pressure = --,.-------(n:/4)(12/2.54)2
Fe
12cm
A single square thread power screw has a lead of 6 mm and mean diameter of 34
mm If it is used to lift a load of 26 KN and coefficient of friction of thread is 0.15,
determine the torque required to turn the screw.
A. 919 N-m
C. 72.6 N-m
B. 65.8 N-m
D. 865 N-m
671.97 psi
~i-l!wmI
PROBLEM 11
The cylinder head of ammonia compressor has core gasket area of 80 cm 2 and
2
flange pressure of 90 kg/cm . Determine the torque applied on the bolt if nominal
diameter of bolt used is 3/4 inch and there 5 bolts.
A. 47628 in-Ibs
C, 696.28 in-lbs
B. 586.28 in-lbs
0, 666.26 in-lbs
Solving for the lead angle of screw:
L
6
tan x =
= - - - = 0.05617
n:D m
n:(34)
Solvinq for the torque required to turn the screw:
T =
E:1!!imD
WD...'11...I· (tan..x..:+.fL
2
L 1-
I
f tan x J
Total initial tension = 90(80) = 7200 kg x 2.205 Ibs = 158761bs
Initial tension per bolt = 15876/5 = 3175.2 Ibs
T
26(0.034) I 0.05617 \- 0,15
-------1
--..--..-----.-....
Solving for the initial torque applied per bolt:
T = 0.2 F, D = 0.2(3175.2)(3/4) = 476.28 in-Ibs
T
= 0.0919 KN-m = 91.90 N-m
2
16mm
L1-0.15(0.05617)
PROBLEM 14
PROBLEM 12
The total torque required to turn the power screw is 50 N-m. If linear speed of screw
is 8 ft/min and lead of 8 mm, determine the Hp input of the power screw.
A.282Hp
C2.14Hp
B 2.54 Hp
D 238 Hp
An ACME thread power screw that has a mean diameter of 25 mm and pitch of 5 mm
IS used to lift a load of 500 kg. If friction on threads is 0.1, determine the torque
needed to turn the screw.
C. 13.10 N-m
A. 1030 N-m
D. 14.10 N-m
B. 1263 N-m
.....I N·~ .J r
tan x ~
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L
5
«o.,
n(25)
0.0636
lOMoARcPSD|3535879
158
Bolts &. Power Screll'
Bolts &. Power Screw
WOrn lcos(j)tanx+f)l
- - - - - _ _._2
cose-t tan x
T
-"'-_.-
For double square thread: L =0 2P
2(10)
tan x =
- - - - = 0.039788
;c(80x2)
..
For ACME thread, <jl =0 14.5 0
W =0 500 x 0.00981 = 4.905 KN
W Om = 4.905 (0.025) =0 0.122625 KN.m
T =0
(0.12)[COS45~63+J
T = WOrn rJt~nx+!l
2 L 1 - I tan x
= 0.01030 KN-m
2
cos 14.5°-0.1 0(0.0636)
I
(12.80) 0039788 + 0.13 I
2
11-0.13(0.039788)
=0
1.0923 KN-m
Solving lor the total torque:
PROBLEM 15
The torque required to overcome collar friction of a 100 mm mean diameter collar
power screw is 50 N-m and collar friction of 0.15. Determine the weight lifted by the
screw.
A 458.57 kg
C 487.57 kg
B. 478.57 kg
D. 679.57 kg
SOLUTION
TT
=0
T + T,
But: r, =0 0.20 TT
TT =0 T + 0.20TT
0.80TT =0 1.0923
TT = 1.3653 KN-m
=0
1365.30 N-m
PROBLEM 17
The root diameter of a double square thread power screw is 0.55 in. The screw has a
pitch of 0.2 in. Determine the major diameter.
A. 0.524 in
C. 0842 in
B. 0.750 in
D. 0961 in
Torque required to overcome collar friction:
=0
Om =0 2r
W Om =0 80(008 x 2) =0 12.8 KN-m
T =0
T =0 10.30 N-m
Tc
159
!,"-W (r 0 +rJ
2
Taking the relation between Om and r.
Do +0;
Om = - -
=0
2
2ro +2r,
----
2
SOLUTION
r0 + r,
For double square thread:
L =0 2p = 2(0.2)
r o+ r, =0 0.100
Solving for the weight:
50 =
For square thread:
Do = 0, + L/2
Do = 0.55 + 0.4/2
(0.15)JWL~ 9_01
2
W =0 6666.667 N x 1/9.81
=0
= 0.4
67957 kg
=0
0.750 in
PROBLEM 18
A power screw consumes 6 Hp in raising a 2800 Ib weight at the rate of 30 ft/min.
PROBLEM 16
A double square thread power screw has a mean radius 01 80 mm and a pitch 01 10
mm is use to lift a load of 80 KN. If friction of screw is 0.13 and collar torque is 20% of
input torque, determine the Input torque required.
C 1246.30 N-m
A. 83076 N-m
B. 83576 N-m
0 136530 N-m
Em!!ImD
Solvinq for the horsepower output:
SOLUTION
tan x
Determine the efficiency of the screw.
A. 12.5%
C. 42.42%
B. 16.8%
D. 66.62%
L
nD rn
Hp,
=0
'N x vei()c;,ty ,hp
-3~O
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160
Bolts & Power Screw
Bolts &. Power Screw
Hp.,
0.70 ==
30x2800
Hp.,
33,000
2.545 Hp
Solving for the efficiency of the screw:
By using quadratic formula:
Po
Screw Efficiency
])xnat(01. - [x_~at
tanx+0.10+0
2x
0.1tanx + 0.07 == tanx - 0.1 Otan
2x·
0.1tan
0.3tanx + 0.07 == 0
-------.--
Pi
Screw Efficiency == 2.545/6 == 42.42%
tan x ==
_.~--
- (-0.3) ± J(-0.3)2~
tanx == 0.255
x == tan' 0.255
PROBLEM 19
A square thread power screw has a pitch diameter of 1.5" and a lead of 1". Neglecting
collar friction, determine the coefficient of friction for threads if screw efficiency is
63.62%
A. 0.113
C. 0.146
B. 0.121
0.151
o
II:DmimDI
Solving for the lead angle:
L
1
tan x ==
..--- == == 0.2122
nOm
n(1.5)
Using the formula of efficiency:
tan x(1 - f tan x)
e == . _ - -----tan x + f + (fcOc 10 m )(1- f tan x)
Since collar friction is negligible (fc == 0), then the quantitytt, Dc IDm)(1 . ftanx) == 0
0.6362 ==
..?1l
0.2122[1-_fJ.9 212
0.2122 + f + 0
0.2122 + f == 0.333 - 0.070f
f==0.113
PROBLEM 20
A square thread screw has an efficiency of 70% when friction of threads is 0.10 and
collar friction is negligible.
A. 1430°
B. 1037 0
4(0.1 0-)W07)
2(0.10)
C. 12.43°
o 16.45°
SOLUTION
Using efficiency formula'
tanx(1 ftanx)
e == ---- .. _ . _ ~ tanx I f I (f, 0, IOm)(1-ftanx)
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14.30°
161
lOMoARcPSD|3535879
I ():2
1"11/11'11 t ' 'I
I fi:{
FI.1jU l l W I '/
2. Total weight of Flywheel(W)
11 Flyvvheel·
W := Wr + Wah
where
Wr =- weight of the tlvw-reel rim
W"h - weight of arm an the hub
3. Weight of the flywheel rim(W,)
W r =- Vxw
W, = (rr D b t)(w)
Om
where
b =- Width of the flywheel nm
\ =- thickness of flywheel rim
w =- density of llywheel material
W"I'>
Om
b
4. Energy required to punch a metal(E)
E=1/2Ftp
Shaft
E = 1/2(8," x A)l,
b
l.
Flvwheel
a rotating energy reservoir which absorbs energy from a power
source dunnq a portion ef the operattnq cycle and delivcrs that stored energy
as useful Work our.nq the other portion of the cycle.
(.
1
2
3
4
5
6
Flywheel applications:
Punch press
Shears
Interna! combustion engines
Compressors
Reciprocating pumps
Steam engines
where:
A -=0 shearing area
(for Circular hole)
A =- IT d t p
F _ average force needed to punch a t-ore
S~
=- uittrnate shear stress
" _ thickness of the plate
d =- hole o.ar-ieter
D; .,-- mea-t diameter of flywhee!
I
...
tI.
d
5. Power needed to punch a hole(P)
~nergy.
Time needed to punch a hole
P =
1. Kinetic energy released by the flywheel(KE).
W
2
2
KE = -(v, ··V2 )
2g
6. Kinetic energy released by the fl'f/'lheel =- Energy needed to punch a
hole
where
7. Coefficient of fluctualion(C;)
maximum speoo = It 0 N.
v. = minimum speed = It 0 N2
W := total weight 01 the tlvwheel
9 =- acceleration due \0 gra',Ilty =- 9.81 Ill/SCC
V'
-
C,
~-l,.
wnere
v2
v
v;=.v 1+v-;;.
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2
C,
2(v, -v 2 )
v, +v2
-
lOMoARcPSD|3535879
I f31
Flll/I'he!'!
j()~
Fllfll'!!t'd
PRO[JLE:Y1 I (ME Bd. Oct 95)
A cast Iron flywheel IS rotated at a speed of 1200 rpm and bavmq a mean rcldluc, o! I
foot. If the weight of the rim IS 30 Ibs, what IS tile centrifugal force? Use laclG! C - -11
A 14.8001bs
C 7 ml
B 14.860 Ibs
D 14 760 IrJS
IIE!!Ii1ElII
Solvmo for Hl(' hl1, \I V,.clqr1l
W _ 120'.W­1
W
12<1181181­97417kg
PROBLEM :1
A press IS used to punch 10 holes per minute of 30 mm diameter hole trorn 25 mrr
thick plate. Determine the power needed to punch a hole If ultimate shear stress is
400 Mpa
C 3.58 KW
A. 2.68 KW
D 096 KW
B 1.96KW
Solving fer the tangential speed.
v = IT D N = n(1 x 2)(1200,601
v = 125 664 tt/scc
E:I!!hir·W+
USing centrifugal force formula
Wv 2
F.
Lei
gl
30(125 664)2
F
32.2(1)
I'ROU ~M
c
14.7241bs
2 (ME Bd. Apr. 94)
W'h
~·:JX(fi[.1
V·
V,
Vi'
KE
2rrR,N c 20(1016)',200,'60)
21 279 rr-sec
2rr(1 0161(200,'601(1 ­ 0 0871
19,427 rn.soc
W"
­_. (V t
"
Sorv.nq lor the welghl o! lhe r.rn t aserr on e'lergy reqlmer"­'en(:
W-~
\"/!~
,212791=
.0=
(J
30 mm
I'RC)BIYM 4
The kmetrc energy needed to punch a hole IS 5 KJ wt-a: IS the maximum ttuck.iess
of a hole that can be punch If hole diameter IS 25 mrn ard ultimate shearinq stress of
plate is 420 Mpa
A 12.61 rn­n
C 1741 mm
D 19.62mm
B 1468m"
E
E
E
5
:
3400lO 90)
981
25 mmC:
Em!!iit.':+
V ­> ~)
2g
energy requrred to shear a plate
E _ 1'2 (S. x A)t"
A _ IT d I,. = rt (0 03)(0 025)
A
0.002356 m'
E c 1'21400000 x 0 002356)(0 0251
E = 1178 KN­m
11.78
Energy
1 963 KW
P ower --=
Time
6
E
A flywheel for a punching press must be capable of furnishing 3400 N-m cl energ:1
dunnq the 25°,) revolution while tile hole IS being punched The flywheel maxumtrspeed IS 200 rpm and the speed decreases 8.7 c,c our.nq the load stroke The mear
radius of the run IS 1016 mm diameter and contributes 90~-o
of t"le energy
requirements Approximate total weight cf the llywheel 10 be 1.20 times that 01 tile
rim Find the weight of the flywheel.
A 975 kg
C
B. 652 kg
D
v,
1 min '­'­ 60 sec
t = time required to punch one hole
I = 60/10 c 6 sec
energy reqUired to shear a plate
1 /2 (S",>( Al tr_
1'2 IS, lie d I, Itr
1i2 j420.000!(n: x 0.025 X tl,)l"
'T'
17.41 mm
t­, ­ 0.01 /~l1
11] 42TI?J
tpL:.::.:
2,'981,1"
.<;-25 rnrn
81151 "'9
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n 1111'Il( '( ,I
1 ijl;
During iJ punching process o! 350 kg flywheel the speed vanes from 200 rpm to 180
rpm with 1 m mean diameter Deterrr.r-e the ktnettc onerqy released by the flywheel
A 364 KJ
C 451 KJ
B 6.28 KJ
0 5.62 KJ
C. 60 mm
o 55 rr-rn
A. 45 r­im
B 30 mm
Em!!ii[.g+
Sheared Area
E _ 1!2(S??xA)t
SOLUTION
v·
v.
II 7
PI 1111'/1<',,1
I'f,UIlIX;V1 r,
10 = 1/2 (420000)(A)(0 020)
A c 0.0023809 m
r: 0 N 1
ell )1200,601
"0 N2
"(1)1180,601
1047 m/sec
9.425 m/sec
20 mm
Let
­ length of side cf a square hole
A = shear area
A c 14x) t
0.0023809 = 4x(0 02)
x c 002976 m = 29.76 mm
'I
KE =-- kinetic energy released by Ihe flywheel
_
W
,
2
(v.: v.;")
Kt: -=2g
009811 110
7?
KE = 1350xO
-I
4)
219.81)
KE = 364 KJ
I,~
'Q
0C'
4L.-:))
j
PR013LfcM "
A 1 m mean diameter flywheel. SOD kg weight changes Its speed from 220 rom to 200
rpm during shearing process. What average lorce is needed to shear a 30 rnrn th nk
f'r,OHLFM fi
pta:e
A plate 200 mm wide and 25 mm thick ,...../Ilh strength cl 410 Mpa IS to be shear along
Its width During shear.nq process an 800 nom mean diameter flywheel changes its
spe od Iro-n 200 rpm to 180 rpm Deter-ru-e the wc.qht 01 flywheel to be usc
A. 326584 kg
C 3821.72 kg
B 3452.71 kg
0 3845.97 kg
V,
,,0 N,
"ON
8377 m'snc
75398 rr/sec
E - 25625 KJ
25625 ­=
If'
".
~
it
W
,
(VI""
2g
KE =
Shear Area
29
234 KN
1082 KN
(1 )(220/60) ­::: 11 519 m/sec
10472 m/sec
W = 500 x 0.00981 = 4905 KN
E = 1,'2 IS., .<1.,;1): I
E - 1,'2 (410,000)10 2 x 0 025)10 025)
KE
o
v: = :1(1)(200/60)
.'10.8)1200/60)
'10.8)1180/60)
w ('I.',
C
EI::I!!im'I3I
v,
EI!!:im'mI
v:
A. 384 Kr-;
B. 653 Kr­.
2,
v'),1 =
,
4905 [ 1
.
­ \ 1.5191"
2(981)
1,1 0.427 J~]
KE = 5756 KJ
E =
1 FI
2
~25m
5.756 = 1,2 (FIIO 03)
c c 383.75 Kr-;
v, ')
-
.
200mm
PROBLEM Y
[1'8 37lt - (7.5398)
219811 .
The enercv required to puncf a ho.e IS 3 KJ frorr a II/wheel mean diameter of 800
mrn that slows Clown trom 33 rps to 3 rps during ouicunq. If Ii\'ei;)ht ,=,1 arm and hub
account 10"" of rm weight, cetetrrune the rim Vie 9hl
A 421 08 kg
C 48268 kg
0 41668 kg
B "5740 kg
W ­ 37729 KN x 1:0 00981 = 38"597 kJ
f'ROHLEM 7
The energy requ,'red 10 punch a square role from 20 mil thick plate IS 10 r(J It the
uiti'late strength 01 pla:e is 420 Mpa. detorrn.ne the maxnr­um sides of square that
can be punchec.
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I fl,'-<,
FIL/Lt'lU'd
A 32-1~rnPl
B 24261 rrrn
I.E!!m:D
v·
v
,1(0.80il3 3), 8293 m/scc
n(O 80)(3) = 75398 rrsec
EE!!IiS3
Sorvmo for the weight of the Hywheal:
W
KE =
-(v,
2g
?
b = 51
W, = (n D b II W
2
-v,)
~
VV
1 (j~)
Fl!JlI'llcd
C 286.76 mm
D. 29878 mm
450 = ;r(2 x 045)[51)11)(7200)
1 = 006649 m = 6649 mm
?
Dc = 900 mm
--;[(8293)"-(7539811
3 = ..
2(9.81)
W
44358 KN x 1 '000981
51
5(66491
33245 mm
b
b
503 147 kg
b
Solving for the nrn weight:
W = VV + VV d ,
VV ~
+ 10 ooVV,
vv,
k
,I
PHOBLlc:vI 12
503.147 = W, + 010W
W = 4574 kg
The power required 10 shear a plate 50 mm thick for 10 seconds IS 5 KW. If ultimate
strength of plate material IS 420 Mpa How vade should the plate be?
C 90?8 mrn
A 8042 mm
B 85 65 mm
D 95 23 mm
PHOBLE:vI 10
A flywheel has a total weight of 500 kg and the weight of arm and hub IS 8°'0 of total
weight. It has a mean diameter of 1 m and width of 300 mm Deterrruno the thickness
of the fiy\vheel if density of material IS 7000 kgl"/
A. 60.65 mm
C 6972 mm
D 75 42 mel
B. 6548 mm
.,.'S!imD
Power x time
51101 = 50 KJ
E
E
I.E!!m:D
E = 1·2 (S" x AI t
50 = 1'2 1420.000)[A)10 05)
A = 00047619 m
Solving for the nrn werqht
VV ::;- W r
-....
Shear Area
w
W,,'I
\V = W r + 8°'"W
500 = W. + 008[500)
VI = 460 kg
Dc = 1 m
A = wt
0.0047619 = w(0.051
w = 0095238 m
IN
Solving for the rim thtcxnes s .
Wc(oDbl)w
460 = c (1110 3)(11[70001
1 = 0.06972 m
I ~_ 69 72 rnrr
----'
95,23 m­e
l'EOlJLE:Yl I:,
k
)1
A 900 m-n mean diameter flywneel has a 'N,d:h Cit 350 m>, 7S rr.m thick <.nd dens ty
the cr-c-qv relcasoo if .ts
of 7100 kq.rn''. \Jeg'ectInQ :he weight of arm nne hub, fl~d
speed cranges from 280 rpm to 250 rpm
C 548 KJ
A. 7.89 KJ
B. 9 30 KJ
D. 6 71 KJ
PROBLEM I I
and mean rartus of 450 mm If rim Width IS 5
A flywr.eel has a rim weight of 450 kC~
times the rim thickness ana materia! density IS 720C kq/m'. determine the Width of
Ilywheel
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lOMoARcPSD|3535879
170
FIIJ/I"! II '('1
Fi'l" ,Ii "f'!
~
C 19866 Ibs
D 20077 Ibs
A. 190 66 Ibs
B. 195661bs
Solving for welghi of rim
W. = J( 0 b t w
W,
= ,,(09)(0.35)(0.075117100) c
Em!!ImmI
52696 kg
w = W,+W.I'r.
Dc;:: 900 rnm
W = W, + 0 = 526 96Kg
v, = ,,(09)(280/60) = 13195 m/sec
V2 = ;t10 91(250/601 = 11,781 m/sec
80
W
'
29(V," - v :
IE
W = 526 96 x 0 00981 = 5 169 KN
'5169'
KE ,- \
.J [113195)' -111 781)!J
2(981)
2
~
Y1 - v;;
2
v, + Vi "- 160
160 - V,
V2
USing kinetic energy torrnura
KE =
v2
V;
v.;
>1
Cr
2(V1
V2)
VI - V2
b = 350 mm
21v,
009
(160
-----
v,ll
160
v- = 83.6 ft.sec
v' ~ 160 - 836 c 764fusec
930 KJ
W
::.'
2
--(v, -v 2 )
2q
PROIlLEM 14
KE =
A flywheel welghmg 1000 kg has a rao.us of gyration of 1.5 m The normal operaunq
speed IS 160 rpm and ecett.ccn: of Huctuauor, IS 0.08, determine the energy released
by the flywheel
A 31.70 KJ
C 4170 KJ
B 3670 KJ
D 4670 KJ
3500 =--l(83.6)'
2(322)
W _c 19566 Ibs
IIE!!'.'iiI:D
v-
c
,,(15 x 2)(160/601
2(v 1 - V?)
V 1
v:
v,l
2(25132
008
25 132 rrvsec
25 132 + VI'
25132 + v- = 628318
V; = 23 20 rnsec
W = 1000 kg (0 009811
W,
2
IV, - - v" - )
KE
2~
981 KN
9_8 1
1
) 11251321'
2(9.81) ,
(232/]
25v_
c
KE
4670 KJ
1'!,OElLEM 1f>
A 5 11 mean diameter llywheel has to absorb 3500 ft-Ib of energy and maintain a
coettrcrent 01 uuctuanc- of 0.09 11 mean speed IS 80 n/soc. find the weight of
flywheel.
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W
I
I
176.4)2 J
17\
lOMoARcPSD|3535879
-
I -"',
,
In
.'­iJHIIl!J
1. Stress of coil spring(S,)
12
Spring
S, ~
8KF~m
ltd
2 Stress factor(K)
K ~
~
4C­1 + O~
4C­4
C
3 Spring Index (e)
C ~
Om
d
Uses of spring:
Where
1
2.
3
4
To absorb energy or shock loads, as In automobile shock absorbers
To mamlam contact between machine members. as In valves and clutches
To act as a source of energy. as In clocks
To serve as measuring device, as In spring scales
Om = D, ­ d
CL~
Om = D: + d
4 Deflection(y)
3
Types of springs:
y ~
a
b
Helical compression spring
Helical tenson
c. Torsion
d. Spira!
e. Leaf spring
BFC n
Gd
t
SL
Om
where:
F = axtal load
D·'1 = mean diameter
d = wire diameter
n = no. of active or effective coils
G =- modulus of rigidity
:+- d
Dc>
Materials used in spring:
a Oil-tempered spring wire
b. mUSIC wire
c. hard drawn spring wire
d carbor steei
e. chrome-vanadium steel
f. chrome-silicon steel
9 stainless steel
5.
Deflection at solid force(y,)
y. ~ Free length - Solid length
6
Spring rate(K)
K ~ Fly ~ F, I Y1
K
Tabu!ated Data of springs:
Types of coil end
Plam
GrourJd
Squared
Squared & Gro.mc
Actual no, of coil
N
N
n -+- 2
n • 2
y,
Solid Length
(n ./ 11d
Free Length
np + d
nd
np
in ~
np + 3d
np + 2d
3) d
(n -. 2\ d
7
F,/ Y2
F2 -F,
Y2 ­ Y,
Impact load on spring:
W(h+y) ~
F
-y
2
where
F = maximum force acting on the spnrr;
y
dcllecuon on spring
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T_
t
constant
I
)!
lOMoARcPSD|3535879
17,1
W ::- weight of the object
h ~ height of the object
t -= tfucxncas of plates
n,] = no of gradualed leaves
nl
no of full length leaves
y = dallecuon of spring
PHOIlLEM I (Oct, 1999)
8 For series connected spring:
A body weighing 1000 Ibs Ieus from a height of 6 in and strikes a 2000 lb/m spring
The deformation of the spring IS
C 3
A 2
D 5
B. 4
Y == total elongation
SOLUTION
Y :;: Yj + Y2 + Y3
F,
F
F
K1
F, ~ F, ~ Fe
K2
y
2 +--3
--+--K
K2 K3
F
total load
~
USing the formula of Impact load on spring'
F ~ 2000 lb/m tV) = 2000 V Ibs
(2000 V)
1000 (6 + V) ~ " .
V
2
6 + Y ==
l-V6~0
K3
y
~
£L~5-F3
K
K2
*_
K3
""r'--'"
K,
K2
If the springs are of the same material
F, ~ F, ~ F 3 = F/3
F
v
18FL
,
m:m:mD
SolVing for Call mean diameter
0" ~ 0" d
0", ~ 9256,0.9525 c 8.3035 cm
Leaf springs:
8, =
K3
Pl{OBLF:M 2 (ME ac. Oct, 97)
Compute the defl€ctlon of an 18 COils helical spring having a load of 100 kg The
modulus 01 ciasncity in shear of spring IS 96.62 Gpa. 00 of 9256 cm and With Wife
diameter of 9.525 mm. The spnng IS squared and ground ends
A. 9 cm
C. 11 em
B 101 cm
14cm
°
F == F1 + F2 + F3
11
(V - 3) tV + 2) ~ 0
V ~ 3 In: V = ,2 In (neglected)
Therefore' y = 3 In
F
Y = total elongation
Y == Y1 == Y2 == YJ
l
By factoring.
For parallel connected spring:
9
IT)
,'-,/)) II If/
C =
-"~
2
bt (2n g + 3n, )
12FL3
bt 3 (2n g + 3n,)
where
51 ~
flexural stress
F
L
b
load at the supports
distance of force to produce maximum moments
width of plates
~
d
.. ~
.8.::l0 35 ~
8.717
0.9525
For square and ground ends
Actual no of coils = n + 2
Where n = no. of active coils
18 = n + 2
n ::::: 16 coils
Solvmq for the deflection'
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I
9.525 mm ~
I.
I
I
\
Ie
I
0<--
0,
9.256 em
)
r-I
I
I
)1
,
lOMoARcPSD|3535879
lill
y
y
t77
,'-,J> / 1/ I [I
G~J
96 62x10"I0 0095251
00903, m -= 9.037 em
k -
IEm!!Im3
101 6 - 9 525
rj
k
K2 = 0.64 kg/mm
Y = total deflection
y =
~Yl+y~,:
1(1(1
+
Y
114(]
11111
11,(,4
K,
11111
+
F,
F3
K2
K:J
KJ = 0.64 kg/mm
= 5625 mm
IU,4
100 kg
}
D,
F
Y =
Compute the maximum ceuecto-i of a 20 coils helical sprr-q hEWing a load of 75 kqs.
The spring is a square/ground ends w.th modulus of elasticity In shear of 79 84 Gpa.
ou.stde diameter of 101 6 rnrn. wtre drameter of 9.525 mm
C 15 i7 mm
A 140.7 rnm
B 112 7 rnrn
D 126 7 rum
D
K, = 0.4 kg/mm
Y
FH()l--H,Ei\'i :1 (ME Bd. Oct. 97)
D
spring constant
F
k
8(IUOxO 00981,(8 717}'(161
BFC'n
+75 kg
For square Ci'ld g:ound ends
Actual no. of uctve coils = n + 2
?O ::: n +- 2
n = 18 co.ls (acuve)
A high alloy spring aavmo squared and ground ends a'ld has a total of 16 Calls and
modulus of elashcity In shear of 85 Gpa
~
92075 mm
l'I,OULEM 5 (ME Bd, Apr, 96)
Compute the Wahl factor
The spnng
outside diameter IS 9 66 em Wire diameter IS 0,65 cm
C. 1.10
A, 1058
B
1 185
D
12
E'I!!iiI':.mI
I
9.525 mm
C - Spring index
C
0111
d
92075
9,525
9,67
-7l i<,I l
I
I
I(
0,
)
f
101.6 mm
Solving for the deflection:
y c
8FC"n
8175xO 00981)19671 3(18)
Gd
79 84x1 0 610009525)
=- 012597 m
125.97 rr­m
Whaal Facor
c:-
Whaal Factor =
I'J{OULEM 4 (ME Bd, Oct 97)
A three extenSion call springs are hooked In scncs tl..·at support a single weight of 100
kg The first spring ,'S rated at 0 400 kq.r-rm and the other 2 lower springs is rated at
O 64 kq/mrn Compute the total (jeflectlon.
C. 156 mrn
A 563 mm
B 268 mm
D. 250 mm
IIE:!!i:lm
Solvinq for spring Index'
mean diameter
D,,, = D, - d ~ 9 66 - 0 65
D,­ = 901 em
C = D,Jd = 901065 = 13,86
Dr"
4C 1
0,615
4C4
C
4(1386)-1
o 61 5
4(n86) 4
13 86
I
0.65 em
­7l
I
I
I
I
I,
I
0,
I(
9.66 em
)
r-I
I
I
)1
,
1 1023
I'1,OULlcM (i (ME Bd, Oct 95)
A COil spring With 5 em outsde diameter is required to work under the load of 190 N
the ends are to
The Wire diameter IS 5 mm. the spnng IS to have 6 act ve coils and
be closed and ground. Detsrrnme the total number of Calls The modulus of rigidity IS
80 Gpa and the mean radius IS to be 23 mm. With 7 mm pitch of the SPWlg_
A 6.5 COils
C. 7,5 ous
B 8 5 Calls
0 9 5 Calls
For sor-es connected spring.
F = F, = F
F, = 100 kg
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lOMoARcPSD|3535879
17R
Sprilln
IIE!!immI
I"" lIILEM 'l
A spnnq has a rate 01 50 Ib with a spring Index of 8 If stress Induced IS 90 000 ~:;'
determine the wire diameter.
A. 0058 In
C 0452 In
B. 0.828 In
D. 01157 In
For square and ground ends
Total no of coils = n + 2 .:.;. 6 + 2 ~
8 coils or 8,5 calls
PHOHLF:M 7 (ME Bd. Oct. 95)
A helical spring having square and ground ends a total of 18 cons and its matenal has
modulus of elasticity In shear of 78,910 Gpa. If the spring has an outside diameter of
1042 em and wire diameter of a 625 em, compute the maximum deflection that can
be produced In the spring due to a load of 50 kgs
A. 302 mm
B 342 mm
C 490 mm
D 322 mm
Em!IimI
Solving for K'
K
0
y
y
C
418) 1
0615
4(81 4
8
1184
0
USing stress formula
S
0
For square and ground ends
Actual no. of coils = n + 2
18=11+2
n = 18 - 2 = 16 active calls
y
0_615
CoD,. d
8 c D, d
D",
8 d
Dm 0 D, - dolO 42 - 0625
0"1 = 9795 em
C
D";d
9 795/0 625
15672
8FC
4C-1
4C-4
0
K
IIE!!immI
0
l ,'q
8KFD",
'.
nd'
8(1 184)(50)(8d)
90.000
d
:-Ed J
0 1157 In
0
Jn
Gd
O.625cm ~
8(50 xO 00981)(15 672)3(16)
--
----
0
PHOllLI::M 10
I
I
I
I
I
(7891x 10 6 )(000625)
0490 m
1
(
490 mm
I(
D,
10.42 em
)r-I
I
)i,
Determine the maximum shearing stress of a helical spring composed ot 20 turns of
20 rnrn diameter wire on a mean radius of 80 rnrn when the spring IS supporting a
load 012 KN Go 83,000 Gpa
C 120.60 Mpa
A 10820 Mpa
B
['HOBLF:M 8
A extension coli spnng IS to be elonqate 5 In under a load of 50 lb. What IS u-e spring
rate?
A s Jb-mn
C 151b/mln
B 10 lb/mm
o 20lb/mm
Em!iiriD
F.y
Spnng Rate
5015
0
10 lb.m!n
D
8868 Mpa
1 ~2KN
IE!!imD
s
8KFD",
r::d3
C
D,,;d c
4C1
K -
Spnnq Rate
9862 Mpa
K
s
s
4C
-~
(80 x 2)/20 _ 8
0.615
4
C
4(8)
1
0.615
4(8)
4
8
1 184
I
8(1184112)(008x2)
"1002)' ­­120 603 Kpa ­
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120.603 Mpa
20 mm ~
__ I
1
I
1
1
I(
i(
D,
)1
I
I
Do
) ,I
lOMoARcPSD|3535879
J so
A helical spring IS made by wrapping steel wire 20 mm diameter around a forming
cylinder 150
mm ITl diameter.
Compute the elonqauon of the spnng without
exceeding a shearing stress of 140 Mpa If It IS composed of 18 turns. Let G = 83,000
Mpa
A. 9642 mm
C 121 36 mm
B 1006 mm
0 89 62 mm
PI'OfJ\X:vJ 1:1
A spring sustam 200 H-Ib o! energy wrtf deflection of 3 tn
Assume that the coil
diameter IS 7 times the wire diameter and allowable stress of 100,000 pSI. deterrrune
the wire diameter.
C.05681n
A. 0.416 In
D. 0672 In
B. 0 321 In
Em!!hit.g+
SOLUTION
200
F
D». c 0, + d = 150 + 20 " 170 mm
C = D,,!d = 170/20 = 8.5
K
4C1
4C
K
tHJ
,t..;pr-III!/
l'kOULEM I 1
K
C
4(8.5) . 1
0.615
4(8.5)
8.5
s ._. 8KFD
r: d
140,000
4
1.172
~F
}
Il1
3
"(0 02)3
7
~
d
0.615
4C 4
Sc ~
=
C
I
= 1 2128
8(12128)(800)(7d)
nd
0
0615
3
100,000
d
4(7) ~ 1
4(7) ~ 4
8KFD r ,
rtd
-~
8(1.1721(F)(0 1701
~
7d
Om
d
4C ~1
C
0.615
4
= 800lbs
3 12
J
0416 in
F = 2207 KN
3
y
y
8FC n
8 (2.207) (8.5) l (18)
Gd
(83,000.000) (0.02)
P](OBLEM 14
I
20 mm -?>lI
11757 mm
I
I
,
I
_
I
IE
150 mm
0
)1
I
I
.I
~ I
A weight of 100 Ibs stru.es a COil from a height of 18 Inches and deflects the spring of
6 Inches Find the average force acting on the spring.
A. 600lb
C. 800lb
B 700 Ib
D. 900 Ib
PHORLEY! 12
IIE!!ii1:mI
It IS found that a load of 50 Ibs on extension COil spnnq deflects 8 5 In What load will
the spring deflects 25 In?
A 10 64 Ib
C 13 48 Ib
B. 12 48 Ib
0 14 70 Ib
USing the formula of Impact load on
spring.
\
4-i·,,··it·H'
0
Wlh + y)
Tall
F
~y
2
The spring rate of spring is constant:
K,
100(18 + 61 =
K:,
FI / y,
F2 / Y:I
50'85
Fe' '2 5
F-
F
800 Ibs
14 70 tbs
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~
2
(6)
18;n
lOMoARcPSD|3535879
IR2
u·n
.'i[!1 ifill
PHOBLEM 15
Three corl spring are hooked In senes and support a w8rglll 01 70 kg One spring has
a spring rate of 0,209 kg/min and the other two have sprinq rates of 0.643 kq/mm.
Find the deflection
A 346.71 mm
C 55265 mm
B 389.30 mm
D. 416.58 mm
].tijlI T.r- ~
.~
C 3250 rnm
D 3460 mm
A 27 60 rnm
B. 2980 rnrn
EI:'I!!iil':.lD
1
D" ~ D, 'd
D", = 117 ' 13
OTT - 104 mm
C :::: 0",1 d
y = lola] defleclion
y = y + y? + v>
C = 104/13
C " 8
For series connection of spring.
F, = F i · = F s = 70 kg
K
FlY
Y = F/K
y =
y
y
y
I
13 m ~
Gd
I
K1 = 0.209 kg/mm
+ F:,iK J
F; K· + F~/K;:
700 209 + 70/0643 + 70/0 643
55265 mm
Y =
PROBLEM 16
8(1 11(8)3(75)
I
6)(0.013)
IE
(80xl0
70 kg
SOLUTION
I
Ty, NK~=F
2s mm
v, = y, + 0025
IIE!!1:mmI
::
i~~
For parallel springs'
y, = y = YJ = y.,
i=
!-
F1
F,
F2
y,
F, i k, ::: 9010 717
y,
125,52 mm
F', = F, = 360/4
Y?::: YJ
F2
:~
i :!FJ
I
Do = 117 mm
))
PHOHLE:vI 1H
A concentric helical spnng IS use to support a load of 90 KN. The Inner spnng has a
rate of 495 8 KN/m and outside spring IS 126.5 KN/m If initially the Inner spring IS 25
mm shorter than the outer spring. find the percent load car-reo by Inner sprrnq
A. 3465""
C. 6825""
B 5586""
D. 768P"
Four compression coil spring In parallel support
a load of 360 kg. Each spring has a gradient of 0.717 kg/mm. Find the deflection.
C 138.52 mm
A 125.52 mm
D. 145.52 mm
B. 132.52 mm
:~
-,
I
Y = 003249 m
y _ 32.49 mm
K2 = 0.643 kg/mm
KJ = 0.643 kg/mm
=
8Fc ln
:~
I !"~
F4
90 kg
+360 kg
K
Fly
F, - v. K, c 495 8 Y
126,5 YL
F,
F, + F L :.:: 90
1
YI
~-
495.8 v, t 1265 YL = 90
4958y + 126,5(y, + 0025) = 90
495 Sv- t 126 5 v: + 3 1625 c 90
v, = 01395 m
F. = 4958(0 13951 = 69 185 KN
69185
'J" Load Carried
=: 76 8r,0
y, = y
90
PROBLEM 17
A force of 1 1 KN IS acting on a 75 active coils With wire diameter of 13 mm. The
outside diameter of corlrs 117 rum and G ::: 80 GN/m 2 Find the coil deflection.
PROl3LEM
I~)
How long a wire IS needed 10 make a helical spring havmq a mean diameter of 1 Inch
If there are 8 active coils?
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lOMoARcPSD|3535879
A
B
25 13 In
2665 In
C 30211n
D. 3234 In
IIiE!!DIiD
L
K
4C 1
4C ,1
S
stress at solid length
S -
8KFD r·)
0.615
4(8) 1
o 615
C
4(8)4
8
1 184
rrrJ
wire length
S,
L
lR5
Sprlllfj
Ik-\
8(1184)(15)(010)
,,0.0125) 3
Circumference x No of calls
L = cD(n)
'11)(8)- 2513,n
FI<OBLEM 20
IIiE!!DIiD
For square and ground ends
Actual no or coils = n + 2
where: n = no. of effective coil
12 = n+2
n = 10 calls
o 10053 In"
Solvmq for mass
w
mN
m = V w
(0.10053)(0282)
Free Length
Free Length
Free Length
o 0283 Ib
~i t·W'
For square and ground ends:
Solid Length = (n + 2)d ~ (10 + 2)(12 5)
150 mm
y, = solid length deflection
y = Free Length
SOlid length
F, ~ force at solid length = k Y'>
F,. ~ 150(0100) = 15 KN
C = D-! d = 100/12.5 = 8
250 - 150
100 mm
1
20 mm
np ·t 2d
10(20) + 2(125(
225 mm
Solid Length c_ (n + 2)d
Solid Length = (10 + 2)(12 5) = 150 mm
y. = FL - SL c 225 - 150 = 75 mm
PROBLEM 21
A square and ground ends spring has a free length 01 250 rnm. There are 10 active
calls with Wife diameter of 12.5 mm If the spring rate is 150 KN,irn and mean
diameter IS 100 rnm. determine the solid stress.
A 2.3155-\ Mpa
C 7.8427 Mpa
B 7.6548 Mpa
D. 8.432.9 Mpa
= 2.315S4Mpa
A squared and ground ends spring has a pitch of 20 111m wire diameter of 12.5 mm.
If there are 12 actual number of coils. lind the deflection when the spring IS
compressed to Its solid length
A, 78 mm
C 77 mm
o 79mm
B. 75 mm
~'it.]:
V : .: Volume of spring
V c (,/4 de) L = 1'/4 (0 081"](20)
2.315.544Kpa
22
1>J«()3Lt:~l
A 008" diameter sprnq has a length of 20 in if density of spring is 0 282 Ib,',ln:;
determine the mass of spring
A. 00395 Ib
C 00485 Ib
B. 0.02831b
D. 0.06861b
c
l'I~CJL-:M
2:;
A spring with plain ends has 15 active calls. diameter or 6 mm and Pll!,)) of 10 mm If
spring rate IS 100 KN/m. determine the solid force
A. 4 KN
C 6 KN
B 5 KN
D. 7 KN
IIE!!ImD
For plain end type of spring'
Solid Length = (n + l)d
SOlid Length = (15 + 1)(6)
SOlid Length =- 96 mm
Free Length = np + d = 15(tO) + 6
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lOMoARcPSD|3535879
IHG
Le( I/o..,
I ,~7
156 rum
v. = FL, SL = 156, 96
13 (3.E3Clr
60 mrn
F, = kyo = 100(00601 = 6KN
PRO[JLE\1 24
A spring has a spring rate of 30 KN.im. If wire dIameter IS 10 rnrn with mean diameter
of 70 rnrn. determine the number of active cors G --= 80 G~/m2.
A. 654
C. 842
B. 782
D 972
Em!!ImD
y =
engaging teeth
~
8C'n
I. Addendum
height of tooth above pitch Circle: or the distance between the pitch
circle and the top of the tooth
Gd
C = D, / d
70/10
7
2
3
v = 8C n
F
(.. Gears - are machine elements that transmit motion by means of successively
Gd
the circle that bounds the outer ends 01 the teeth
Arc of action
arc of the pitch Circle through which a tooth travels from the first
pornt of contact With the mating tooth to the prtcn POint
8(7)'J n
1/30
Addendum Circle
180xH/'J(O 0101
4.
n : : : 9 72 corls
Arc 01 approach - are of the Circle through which a tooth travels from the point of
contact Wittl the mating tooth to the pitch POint
;-l. AXial plane - In a pair of gears It IS the plane trial contains the two axes. In a
PH OLlLEM 2"
single gear, It may be any plane containing the a xts and the giver, pomt
A spring has a diameter 0125 mm and 1,2 active coils if a load of 10 KN IS apoheo it
deflects 75 mm Determine the mean diameter of the spring If G = 80 Glv.m".
C 134.65mm
A 12465mm
B 129.65 rum
D 14065 rnrc
IE!!immI
y
7.
T
IF = 10 KN
75 m
1
8C'n
GeJ
Backlash
the amount by w-uch trc wrrith of ttle tooth space exceeds the
thickness of the engaging tooth on the prtcr, circles.
8. Base Circle
•
q.
8(10IiC)J(12)
0075 =
b, Arc of recess
arc of the pitch Circle through wtuch a tooth travels from Its
contact with the mating tooth at the prtch pomt to the point where tho contact
ceases.
the circle from which an Involute tooth ',$ generated or developed.
Base hehx angle - the angle, at the base cylinder If an mvolute gear. that the
tooth makes With the gear axis
80xl0"10 025)
C
5.386
C
D"
5386
o
D,
25
0 11 :=. 13465 mm
10. Base pitch
of actron
25 m ~
I
I I. Non-tnt hase pitch - IS the base pitch In the normal plane
,
,
,
-,,
,
,
k
In an Involute gear It IS the pitch on the base Circle or along UlC lmc
Do
)
I,
1 Z. AXlat base pitch
IS the base pitch tn the axtal plane
the distance between the parallel axes 01 spur gears and
I J. Center distance
parflltpl hehcal gears. or between tt-o crossed axes of helical gears and worm
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IHH
(;('111 .'>
(;('(1 ,-,..,
gears.
14. Central plane ~ In a worm gear this IS a plane perpendicular to tt-e gear aXIS
and contains the common perpendicular of the gear and thp worm axrs
1:1. Chordal Thickness
arc
length of the chord sub tended by the Circular thickness
] h. Chordal addendum
the height from the top of the tooth to the chord subtending
the circular-thickness arc
17. Circular pitch - length of the arc 01 the pitch Circle be/ween the centers or other
corresponding POints of the adjacent teeth
I~
Normal circular pitch
1q
Circular thickness
the lenglh of the arc between UW two sides of a gear tooth.
en the pitch Circles unless otherwise specified
~()
:2 J
IS the
Circular pitch In the normal plane
Clearance - tile amount by which the dedendum exceeds the addendum of the
rnatmp tooth.
Cer-t-ai dlameler - the smallest diameter on a gear tooth With which the mating
gear makes
')'1
Cor-tact ratio
:l:J.
Cyclo«l
the ratio of the arc of action to the circular pitch
:12.
:j
lH:J
HIe concave portion of the tooth profrle where It JOins the bottom of
Fd,lel curve
the tooth space The approximate radius of ttus curve IS called the fillet radius
I. Flank of tooth - that surface which IS between the Pitch eucre and the bottom
land
:~-l.
Helix angle - the angle that a helical gear tooth makes the gear axis
3:1.
Internal diameter - the diameter of a Circle comcrd.nq With the tops of the teeth
on an Internal gear.
:H-j. Internal gear
a gear With teeth on the Inner cylindrical' sur-nee
J7. Involute - the curved formed by the path of a pomt on a snalghl line called n18
generatrlx, as It rolls along a convex base curve
:1.'-;. Top land
:~q.
IS the top surface of the tooth
Bottom land - IS the surface of the gear between the fillets of adjacorit teeth
-l-O. Lead - the distance a helical gear or worm would thread along Its (DOS one
revotuton of it were tree to move axially
4- 1
t.me of action
the path of contact In Involute gears It IS a strarqht line passing
through the pitch POint and the tangent to the base cuc.cs.
·12
Module
:l.l, Dedendum - the depth of tooth space below the pitch Circle or the radial
dimensron between the pitch Circle and the bottom of the tooth space
1:).
Outside diameter - the diameter olthe outside Circle
~.)
Drarnetral pitch - the ratio of the number of teeth to the number of millimeters of
pitch oiarneter.
·1-I. Pitch
the distance between Similar, equally spaced tooth surfaces. In a given
direction and alonq a given curve or line.
l(;.
Normal drarnetral prtch - IS the dtametral pitch calculated In the normal plane
and IS equal to the pitch divded by the cosine of hehx angle
the curved formed by the path of d pomt on a Circle as II rolls along a
stra-qturne
L '{.
Effective face Width - that portion of the face Width that actually COmes Into
contact wtlh matmq teeth as occasronalty one member o~ a pair of gears may
have a greater face Width than the other
:2g . Efucrencv - ttlP actual torque ratio ct a gear set divided by Its gear ratio
29.
External gear
:~()
Face 01 tooth - that surface of the tooth which IS between the pitch eucre 10 \'le
top of the tooth
:~
1
Face Width
a gear With teeth on the outer cylindrical surface.
the length of the teeth In c))(1<11 plane
IS the ratio of pitch ciar-ieter In millimeter to the numoer of teeth mm
--1::;
Pitch diameter
·1 t>.
Pitch Circle - a Circle the radius of which is equal to the distance from the gear
axrs to the pitch pomt
th8 diameter of the pitch Circle
4-7. Pressure angle
the angle between the tooth profile and a radical line at Its
Pilch po.n: In Involute teeth the angle between the line of action and the line
tangent to the pitch Circle
-+~.
Roll angle
the angle subtended at the center of the base Circle from teeth
onqm uf an Involute to the pont of tangency of the generatnx from any pomt on
the ~dme
4~)
Involute.
Tip rl ' lle, 1 an arbitrary modrrc.uon of d tuoth profile whereby a small amount 01
mdtelldlls rernoved near the trp of the qear tooth.
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]90
(;('(n'·;
Gears
50. Tooth thickness - the width of tooth measured alonq the pitch circle.
5. Module
5]. Tooth space - the space between the teeth measured along the pitch circle.
52. Whole depth - the total depth of a tooth space, equal to addendum plus
dedendum, also equal to working depth plus clearance.
\9 l
M
M
module
o
M
T
25.4
OP
where:
D '" pitch diameter, mm
.
~
6.
When two gears turning in opposite direction:
C '" center distance
7.
When two gears turning in the same direction..
N2
1. Speed and diameter relation:
C
0 1 Ni'=: D2 Nz
° 2.;.D
2
1
where:
2. Speed and no. of teeth relation:
D '" pitch diameter
T '" number of teeth
N '" speed
T1N, = TzNz
T2
3. Pc = circular pitch
8. Pitch line velocity, V
V=TTON
C = center distance
T 1 = no. of teeth of pinion
T2 = no. of teeth of gear
4. DP = diametral pitch
9.
Dynamic forces on meshing gears:
A. Power transmitted
p= 21tTN, 'r0N
where:
T '" torque, KN-m
N '" speed, rps
HP _ 2rrTN
- 33,000,HP
where:
D = pitch diameter, in
T = no. of teeth
where:
T '" torque, ft-Ibs
N = speed, rpm
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192
D.
r,
1. Pn ==
Ft x r1
where:
3. Fa == F1 tan\fl
N
cos 3 'II
\fI = helix angle
P = diametral pitch
P n = normal diametral pitch
4> == pressure angle
<l>n = normal pressure angle
F1 = transmitted or tangential load
Fa = axial load or end thrust
N == actual number of teeth
N v = virtual number of teeth
f = face width
where:
T 2 = torque developed on driven gear
r2 = radius of driven gear
D. Total load, Tangential load and separation load relation
Fn =
2. tan$n == tane tan'P
cos \fI
where:
C.T2==F t x r2
t:«
p
4. Nv
F[ = tangential force
T 1 = torque developed on driving gear
rl = radius of driving gear
Fn = vF, + Fr
193
Gears
Gears
2
l
case
5. Strength of Helical Gears
where:
Fn = total load or tooth pressure between teeth
Fr = resisting load or separation load
e = pressure angle
F -
SwfY
78
­;­,­­.­.\ - P-78+-JV
6. Dynamic Load on Helical Gears
10. GEAR TOOTH PROPORTION TABLE
Fd== Ft +
200
'GeM-Partsm14ij2~·+
~ :;=
i Working depth
~t1.olesJPh
, Clear.?nce..
r~?fi;et-·
~_
=rl~57_. ±2/P
t.Q~-w:_
rTo!.QJ~lckn_es
Where: P = diametral pitch
~57/P
~7P-n
__
.~
---·_--·---1
~/P 2/P
-
1570~_
.__
O.25/P
-=--
1 .5708/P
N == number of teeth
.
O.Sy + eCf cos 2 'l' + Ft ) 1/2
v == pitch line velocity == nON
,
2.25/P
-l~2)/P·=1
J
where:
Qi(J5.VCfOS~'l+FI)
._.J
_.
.
1
~
~
Worm gears are used where high speed ratios(10:1) and above are desired.
1. Worm Gear Nomenclature
P = linear pitch
~
P
~
ltD
T
where: T = no. of teeth
2.
Helical gear nomenclature:
= distance between adjacent threads
Lead
= the distance from any point on one thread to the corresponding point on
the next turn of the same thread.
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194
Gears
Lead
Gears
= p (for single thread)
PROBLEM 1 (Oct. 1998)
= 2p ( for double thread)
= 3p (for triple thread)
where:
x = lead angle
195
= the angle between the tangent to the pitch helix and the plane of
rotation.
pressure angle
<pn = normal pressure angle
Compute the tooth thickness of 14 1/2 deg. spur gear with diameter pitch of 5.
A. 0.34116in
C.041416in
B.0.31146in
D.0.31461in
IEm!!Dm
<1l =
15708
Tooth thickness
2. v ::; velocity
Tooth thickness
Note: No. of threads on the worm is equal to 1 for single threaded and 2 for double
threaded
Tooth thickness
DP
15708
5
0.31416in
PROBLEM 2 (ME Bd. Oct. 97)
Strength of Worm Gear:
A 36 tooth pinion with a turning speed at 300 rpm drives 120 tooth gear of 14 1/2
degrees involute full depth pressure angle. What would be the speed of the driven
gear.
A. 1000 rpm
C 90 rpm
D.140rpm
B 100 rpm
The worm gear is weaker than the worm, therefore the design for strength is based
on the worm gear.
IEm!!Dm
3. Tan x::; Lead
nO
1.
F =. SPty(120t)' :••...J.
l1200 + V :
t
w
where:
F1 = tangential pitch line load on the gear
Sw = safe stress, Table 12-2
P = circular pitch
f = face width
Y = form factor, Table 11-2
V = pitch line velocity of the gear
Using speed and teeth relation formula:
T, N,
= T2 N2
36(300) = 120(N 2)
N2 = 90 rpm
36 T
120 T
Efficiency of Worm gear
PROBLEM 3 (ME Bd. Oct. 97)
tan x(cos <Pn - f tan x)
A triple thread worm has a pitch diameter of 3 inches
The wheel has 25 teeth and a
pitch diameter of 5 inches. Material tor both the worm and the wheel is at phosphor
bronze. Compute the helix angle (tan a).
C 040
A. 0.20
B. 0.30
D. 14
cos<Pn tan x + f
where:
coefficient of friction
~
IEm!!Dm
~
Bevel gears - are used to connect intersecting shafts, usually at right angle.
Pc
ci rcular pitch
Pc
nD
Number of teeth and Speed relation
T, N, = T2 N2
Lead
Lead
T
_11.(5t = 0.6283 in
- 25
3 P (tor triple thread)
3 x 0.6283 = 18849 In
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196
Gears
tan ,1 = helix angle
L
1.8849
tan <X = -- = '-'.ITD
rr(3)
SOLUTION
= 0.20
From Vallance, p. 282
tan <Dn = tane cos x
tan14S = tano cos45'
¢ = 20.1"
PROBLEM 4 (ME Bd. Oct. 97)
Find the tooth thickness on the tooth circle of a 20 degree full depth involute tooth
having a diametral pitch of 3, circular pitch of 1.0472 and whole depth of tooth at 0.60.
A. 10,7 mm
C. 101 mm
B. 13.2 mm
D. 7.9 mm
SOLUTION
Using the table of gear tooth proportions:
1.5708
Tooth thickness
Tooth thickness
Tooth thickness
PROBLEM 7 (ME Bd. Apr. 96)
Two idlers of 28T and 26T are introduced between the 24T pinion with a turning
speed of 400 rpm driving a final 96T gear. What would be the final
speed of the
driven gear and its direction relative to the driving gear rotation?
A 120 rpm and opposite direction
B. 80 rpm and same direction
C. 100 rpm and opposite direction
D. 100 rpm and same direction
DP
1.5708
3
0.5236 in
197
Gears
~
~
13.29 mm
PROBLEM 5 (ME Bd. Apr. 98)
A pair of gear/pinion of 42 tooth and 18 tooth with a diametral pitch of 0.7874
teeth/cm and the addendum is 0.80/p and the addendum l/p. The gear pressure
angle is 20° Compute the center distance of the pair of gears in meters.
A.0.5026m
C.0.3516m
B. 0.3426 m
D, 0.4013 m
E'I!!DI:D
N4
96 T
Since they are tangent to each other, then
T 1N 1 = TzNz = T 3N3 = T4N4
T1Nl = T4N4
24(400) = 96(N4)
For gears turning in opposite direction:
_ T +T
C - .g-..-p
2 (DP)
N4 = 100 rpm (opposite direction)
T = no. of teeth
42 +18
C = --,.--- = 38.1 cm = 0381 m
2(0.7879)
PROBLEM 8 (ME Bd. Apr. 96)
PROBLEM 6 (ME Bd. Apr. 96)
A helical gear having a 14 1/2° normal pressure angle and transverse diametral pitch
of 23622 per cm. The helix angle is at 45' and has 8 teeth.
Compute the
transverse pressure angle in degrees.
A. 22.2'
C. 19.3°
D. 20.1°
B. 189
A spur pinion rotates at 1800 rpm and transmits to a mating gear 30 HP, The pitch
diameter is 4" and the pressure angle is 14 1/2. Determine the tangential load in Ibs.
A. 495
C. 525
B,535
D 475
&N!!mmI
0
Solving for the torque developed:
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198
Ft
C. 0.7825
O. 0.8085
A. 08095
B. 0.8035
E1!l!1i(.):.
Solving for tangential force:
Force = T/r
Force = 1050.4/2
Force = 525 Ibs
Using D and N relation:
0, N, = 02 N2
O2
Pc
Nj
N2
01 (--)
PROBLEM 9 (ME Bd. Oct. 95)
Compute the pitch angle of a bevel gear given the pinion's number of teeth of 14 and
42 teeth on the gear.
A. 18.40
C. 28.4"
B. 33.4 0
O. 38.4 0
~
02 = 40,
72 T
1<
)1
In opposite direction:
OJ +0 2
C = - "......2
OJ +40 j
From Machinery's Handbook p 844:
0,
4092 in
T
Pitch Angle = tan -1 ---"
Tg
Pc
_1t.~
Pc
0.7854 in.
= tan' 14/42 = 18.4
18 T
For gears tl'~ni g
10.23 =
Pitch angle
199
Gears
Gears
P = 2n:TN
30(33,000) = 2n:T(1800)
T = 87.535 ft-Ibs
T = 1050.4 in-Ibs
T
= ~
n:(4.092)/18
0
PROBLEM 10 (ME Bd. Oct. 95)
Compute for the tooth thickness of 14 1/2 0 spur gear with diametral pitch = 5.
C. 31831
A. 0.3979
B. 003141
0 0.31416
PR, 3LEM 12 (ME Bd. Oct. 95)
The tooth thickness of a gear is 0.5 inch and its circular pitch is 1.0 inch.
the dedendurn of the gear.
A. 0.3183
C 1.250
B. 0.3979
O. 0.1114
Calculate
SOLUTION
SOLUTION
Using the relation of diametral pitch and circular pitch: Pc (OP) =: n:
OP =: n/P,
OP =: n:I1
OP =: n:
Oedendum =: 1.25/DP
Oedendum =: 1.25/n:
Oedendum = 0.3979"
Using the table of gear tooth proportions:
Tooth thickness
1.5708/DP
Tooth thickness
1.5708/5
Tooth thickness = 0.31416"
PROBLEM 1 1 (ME Bd. Apr. 96)
Compute the circular pitch of a pair of gears having a ratio of 4 and a center distance
of 10.23 Each gear has 72 teeth and pinion has 18 teeth.
PROBLEM 13 (ME Bd. Apr. 95)
An internal gear is set up with a 5 in diameter pinion and center distance of 18 inches.
Find the diameter of the internal gear
A. 36"
8 215"
C 26"
O. 41"
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200
1'~()I\.i:!
I ()
Find the tli:;1, incc between centers of a pair of gears, one of which has 12 teeth and
the other :J I teeth The diametral pitch is 7,
A. 3.0 In
C. 4.0 in
B. 3.5 In
D. 4.5 in
SOLUTION
For internal gear:
18
SOLUTION
D2 .' D,
C=
:.lot
Gears
(;{'(If,~
2
For gear turning in opposite direction:
= D 2. :- 5
C
2
= D,­­­­-+ D2
2
D 2 = 41"
DP, =
D,
PROBLEM 14 (ME Bd. Apr. 95)
18in
The minimum whole depth of spur gear of 14-1/2 deq. Involute type with
pitch of 24 and circular pitch of 0,1309:
A 0,09000
B, 0,08900
C, 008987
D. 0.08975
diameter
T,
­D,
= 12/7 = 1.7143 in
D2 = 37/7
= 5.2857 in
Solving for center distance:
_ 17143+5.2857
C
- - - - ~ - - -
2
12 T
37 T
I(
)1
3.5 in
SOLUTION
PROBLEM 17
From Vallance p 262, Table 11-1:
h =~7
h
Determine the pitch diameter of a gear with 28 teeth, 4 diametral pitch.
A. 7 in
C. 9 in
B.8in
D.10in
Pd
2.157
24
0.08987
Em!!JmD
PROBLEM 15 (ME Bd. Apr. 94)
The minimum clearance allowed for meshing spur gears with a circular pitch of
0.1571 and diametral pitch of 20. The spur gear have 25 teeth.
A 0.007855
B. 0.007558
C. 0008578
D. 0007585
Em!!JmD
=T/D
D = TIDP
D
= 28/4 = 7 in
PROBLEM 18
From gear tooth proportions table:
0.1571
Clearance = .
DP
Clearance
DP
01571
20
0007855
Two parallel shafts have an angular velocity ratio of 3 to 1 are connected by gears.
the largest of which has 36 teeth. Find the number of teeth of smaller gear.
A 10
C. 12
B. 11
D. 13
Em!!JmD
T, N,
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T2 N2
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202
( ;( )( II c,
Gears
N 1 / N 2 0= T 2 / T 1
3/1 = 361T,
T 1 = 12 teeth
N, N = 1.6
1 6 Dl
1 6D 1 = 5.2
UlII
[1
D1 j
0,
PROBLEM 19
The parallel shafts have a center distance of 15 in. One of the shaft carries a 40tooth, 2 diametral pitch gear which drives a gear on the other shaft at a speed of 150
rpm. How fast IS the 40­tooth turning?
A 600 rpm
C 150 rpm
B. 300 rpm
D. 75 rpm
Em!!lmD
D, +D 2
­­­­­­-
2
T 1 / DP,
40/2
20 in
20+D 2
D1
D,
To convert module to DP, use this relation:
1<
:J
DP
2
D2
25.~
M
or M = 25.4
DP
25.4
3.90769
6.5
10 in
Using the formula of Speed and Diameter relation:
0 1 N1 0= D2 N2
20(N 1) = 10(150)
N, = 75 rpm
PROBLEM 20
A pair of meshing gears has a diametral pitch of 10, a center distance of 2.6 inches,
and velocity ratio of 1.6. Determine the number of teeth of smaller gear.
A. 10
C. 30
B. 20
D. 40
E:m!IImI
For gear turning in opposite direction:
[)~E2
2
2(2.6) = D, + D2
D, + D2 = 52
D, N,
OP =
0=
15
C =
2
A spur gear 20 degrees full­depth involute teeth has an outside diameter of 195 mm
and a module of 6.5. Determine the number of teeth.
A. 20
C. 28
B. 25
D. 41
For gear turning In opposite direction:
0=
­t­
DP,
T , / D1
10 = T 1 /2
T, = 20 teeth
PROBLEM 21
E:m!IImI
C
0=
:20;1
D2 N2
D,(N, / N2) = D2
. di
N+2
O utside iarneter = ­­­­DP
Where: N = no. of teeth
N+2
195/25.4
390769
N + 2 == 30
N == 28 teeth
PROBLEM 22
What is the pitch diameter of a 40­tooth gear having a circular pitch of 1.5708 in?
A. 20 in
C 30 in
B. 25 in
D. 35 in
E:m!IImI
Pc :::: ­­ltD
T
1.5708 ==
D == 20 in
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ltD
40
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204
Gi'urs
Gears
PROBLEM 23
How many revolutions per minute is a spur gear turning if it has 28 teeth, a circular
pitch of 0.7854 in and a pitch line velocity of 12 It/sec?
A. 378.44 rpm
C. 954.66 rpm
B. 643.82 rpm
D. 392.88 rpm
Pc = ~
M =
DP
25.4
M =
Em!!imD
205
25.4
------
8
M = 3.175
T
0.7854 =
PROBLEM 26
reD
28
D = 7 in
V=reDN
12 = re(7/12)N
N = 6.548 rps x 60
The diametral pitch of a gear having a module of 23 is:
A. 9
C. 10
B. 11
D. 12
EiI!!imD
392.88 rpm
Use this formula to convert module to diametral pitch
P~BLEM
24
How many revolutions per minute is a spur gear turning If it has a module of 2, 40
teeth, and pitch line velocity of 2500 rum/sec?
A. 59683 rpm
C. 476.85 rpm
B. 386.83 rpm
D 312.83 rpm
SOLUTION
M = module
D
M = T
D
...
2 =
40
D = 80 mm
V=reDN
2500 = rt (80)(N)
N = 9947 revlsec (60) = 596.83 rpm
PROBLEM 25
The module of a gear having a diametral pitch of 8 is:
C. 4.23
A. 5.234
B. 3175
D 2.34
SOLUTION
Use this formula to convert diametral pitch to module.
DP
= ~4
25.4
DP
DP
M
2.3
= 11
PROBLEM 27
A gear turning 300 rpm has a diametral pitch of 8. If there are 40 teeth on the gear,
find the pitch line velocity of gear.
A. 654 fps
C. 7.45 rps
B 834 fps
D 9.45 fps
EiI!!imD
2'
DP =
8 =
<1:.~
D
D
D = 5 in
v = re D N = re (5/12)(300/60) = 6.54 ft/s
PH.ODLEM 28
Two gears meshing each other have 18 teeth and 30 teeth. If circular pitch is 023 in.
Find the center distance between gears
C 323 in
A. 523 in
D. 1.76 in
B. 2.34 in
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2 Of)
Gears
G('(/1 S
EmiGr·U'
SOLUTION
Solving for the diametral pitch of the smaller gear:
1(0
Pel
M -
= ---
T
1( (0,)
0.23 = ­ ­ 18
0, = 1.317in
0,
T,M
200
= 20 M 60 M
­'­­"
I­
-
2
M = 5
+ 2.196
1.317
­­_.,­­-
2
2
176 in
PROBLEM 31
PROBLEM 29
A gear with pitch line velocity of 6.126 m/s when turning at 600 rpm. If module is 3,
find the number of teeth on the gear.
A. 60
C. 65
B. 50
0 70
Em!!ImD
A gear has a tooth thickness of 0.1308 in. Find the addendum of the gear at 145
involute.
C.3.12mm
A. 2.12 mm
O. 5.12 mm
B.4.12mm
0
SOLUTION
.
1.57
Toot h thickness = ­ OP
V=1(ON
6 126 = 1( (0) (600/60)
0= 0195m=195mm
3 =
T
T M
C =01+02
2
30
02 = 2.196 in
M =
o ~
02 = T2 M
O2 = 60 M
~'21l
c = 0 1 + O2
o
0, = 20 M
Solving for the diametral pitch of the larger gear:
1(0
P c2 =
­.
T
023 =
:.!()/
0.1308 = :!.57
OP
OP = 12
o
T
195
Addendum
T
T = 65 teeth
1
OP
1/12 = 0.0833 in (25.4) = 2.12 mm
PROBLEM 32
PROBLEM 30
Two gears meshing each other has a center distance of 200 mm. The pinion has 20
teeth and gear has 60 teeth. Find the module of meshing gears.
A. 2
C. 3
B. 4
O. 5
A gear has an addendum of 0.10 in. What is the dedendum of the gear at 20°
pressure involute.
C. 0345 in
A. 0.037 in
O. 0235 in
B. 0.125 in
Em!!ImD
Addendum
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0.80
OP
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201-\
(;('urs
0.10
(;('(1
rs
209
0.3183 = F1 (0.2/2)
F1 = 318 KN
0.80
DP
DP = 8
tans
Fr
= ..-Ft
1
8
Dedendum
..£r
tan 14.5° =
0.125 in
318
F r = 0.823 KN
PROBLEM 33
200 mm
A gear is use to transmit 12 kw at 450 rpm with 26 teeth. If the diametral pitch of the
gear is 10, find the force tangent to the gear.
A.1234KN
C15.42KN
B. 1424 KN
D. 1645 KN
SOLUTION
PROBLEM 35
The pressure angle of the 12 in diameter gear is 20 degrees. The total load of the
gear is 5 KN and is turning at 750 rpm. Find the power delivered by the gear.
A 43.23 KN
C. 5623 KN
B 6423 KN
D. 3423 KN
SOLUTION
P = 2ITTN
F
Cos 8 = ­-t
FN
12 = 2 11 T (450/60)
T = 0.5093 KN-m
~
Cos 20° =
o = 26/10 = 2.6 in
5
F1 = 4.698 KN
T = Fxr
T = Ft x r
r
2.6
0.5093 = F l2"(39.37)
J
T = 4.698
12(3~ 7)
J
T = 0.7159
F = 1542KN
P = 2 IT T N = 2 IT (07159) (750/60) = 56.23 kw
PROBLEM 34
A gear is use to transmit 20 kw at 600 rpm to a driven gear. The pinion has a pitch
diameter of 200 mm. If pressure angle is 14.5 degrees, find the load that tends to
separate the two gears.
A. 0.234 KN
C. 0763 KN
B. 0534 KN
D. 0.823 KN
PROBLEM :36
Three gears meshing each other has a driving power of 30 kw with 900 rpm. The
speed ratio is 1:3:5 and each meshing gear has an efficiency of 96%. Find the torque
developed of the driven gear, KN.m.
C. 5.60
A. 440
B. 7.60
D. 8.20
SOLUTION
SOLUTION
P=2ITTN
20 = 2 IT T (600/60)
T = 03183 KN-m
T =F x r
12in
Po = power output of driven gear
Po = 30 (0.96) (0.96) = 27648 kw
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210
Gears
No
No
speed of driven gear
900
- - = 60 rpm
3 (5)
To = torque output
Po = 2 IT TN
27648 = 2 IT T (60/60)
T = 4.40 KN-m
Bear, uti
211
14 Bearing
~
1. Bearing - a machine member which supports, guide or control the motion of
another.
2. Lubricant - any substance that will form a film between the two suriaces of a
bearing.
3. Babbit - a tin or lead base alloy which is used as bearing material.
4. Sliding Bearing = type of bearing where essentially sliding friction exists.
5. Ball Bearing - type of rolling-element bearing which uses spherical balls rolling
elements.
6. Roller Bearing - type of rolling element bearing which uses cylindrical rollers as
rolling elements.
Classification of Bearings according to load applications:
1. Radial bearing (journal bearing): supports radial load
2. Trust bearing: carries a load collinear to the axis
3. Guide bearing: primarily guides the motion of machine member without specific
regard to the direction of load application.
4. Viscosity = resistance to flow or the property which resist shearing of the
lubricant.
5. Absolute viscosity - viscosity which is determined by direct measurement of
shear resistance.
6. Kinematics viscosity - absolute viscosity divided by the specific gravity.
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212
Rearing
213
lkuril/(/
Units of Viscosity:
r ~- Journal radius, in
L = axial length of bearing, in
n, = journal speed, rps
C, = radial clearance, in
dyne - sec
cm2
Sliding Bearings:
Cd = diametral clearance
Heat dissipation in journal bearings (Vallance p 240)
D­d
H ChLD
778
C. = radial clearance
where:
H = heat dissipated in Btu/min
C h = heat dissipation coefficient, of projected area. It-Ibs/rrun-in"
L = length of bearing, in
D = diameter of bearing, in
Diametral Clearance Ratio
where:
Ball and Roller Bearings
p = unit loading or bearing pressure
p = F/LD
e = eccentricity
e = radial distance between center of bearing and the displaced center of
the journal
D = diameter (bore) of the bearing
d = diameter of the journal
L = axial length of the journal inside the bearing
F = radial load
"
Bearing Sizes and Designation
Example of bearing designation:
SAE or IS 314 is 300 series, NO.14
"
Tabulated data on ball and roller bearings Vallance: Table 9-2 p 206
"
Bearing Capacity based on stresses (Vallance p 205)
where:
Fr
u = viscosity in reyns
n = speed in rps
p = unit loading, psi
fb
D
coefficient of friction
bearing diameter
(.. Petroff's equation for frictional torque (Faires p 302)
Tf:s~unr:.l, p
nLD
K2
5
(for roller bearings)
where:
(,. Frictional torque in bearings (Vallance p 231)
where:
Ttl = frictional torque
F = radial load
Fr
Fr = total radial load, Ibs
n = number of balls or rollers
D = ball diameter or roller diameter, in
L = length of rollers, in
k 1 = 550 for unhardened steel
700 for hardened carbon steel
1000 for hardened alloy steel on flat races
1500 for hardened carbon steel
2000 for hardened alloy steel on grooved races
k2 = 7000 for hardened carbon steel
10,000 for hardened alloy steel
, ,g3<:
s
Cr
(, Radial Load Catalog Capacities of Ball and Roller Bearings (Vallance: pp
207­213)
where:
T, = frictional torque, in-Ib
u = viscosity, reyns (Fig. AF 16, P 595)
(. Tabulated catalog capacities of ball and roller bearings:
Table 9-7, P 212 and Table 9-8, p 213
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214
Bcaiin q
215
Bell t i T/9
PROBU:M 2
ks kt Fr
A 2 in horizontal shaft rotates in a sleeve type bearing
The coefficient of friction IS
o 10 and the shaft applies a load of 500 lb. Find the frictional resistance.
where:
Fe = Catalog rating of bearing, Ib (tables 9-7 and 9-8)
Fr = actual radial load on the bearing, Ib
H, = desired life of bearing, hrs of use
He = catalog rated life of bearing, hr
ka = application factor takinq into account the amount of shock (Table 9-4)
A. 20 Ibs
B. 40 Ibs
C. 30 Ibs
D. 50 Ibs
Em!!iit-H'
F = frictional resistance
F=fN
But: N = W
F = 0.10(500) = 50 Ibs
where:
ko = oscillation factor
= 1.0 for constant rotational speed of the races
= 0.67 for sinusoidal oscillations of the races
kp = preloading factor
= 1.0 for non-preloaded ball bearings and straight roller bearings
k- = rotational factor
= 1.0 for bearings with fixed outer races and rotating inner races
krel = reliability factor, Table 9-3
.. ~KTNa·
3.~
ks=.
.
N
'
C
the
"~
~,;:._
.
PROBLEM 3
A bearing sustains a load of 4450 N. The shaft diameter is 100 mm, the coefficient of
sliding friction is 001, and the shaft speed is 400 rpm. Find the horsepower lost in
the bearing.
A. 9320 watts
C. 83.45 watts
B 76.34 watts
D. 45.23 watts
..
SOLUTION
kt = thrust factor
1.0 if there is no thrust-load component
Solving for frictional force:
F=fN=fW
F = 0.01 (4450) = 44.50 N
PROBLEM 1
A 1 in horizontal shaft rotates at 500 rpm in a sleeve-type bearing. The coefficient of
friction is 0.15. Calculate the horsepower lost in the bearing if the reaction between
the shaft and the bearing is 800 lb.
A. 0.2634 hp
C. 0.0925 hp
B. 1.2344 hp
D. 0.4759 hp
Solving for torque due to friction:
T = F (r) = (44.50)(0.05)
= 2.225 N.m
Solvinq for the power lost:
P = 2 IT TN = 2 IT (2.225)(400/60) = 93.20 watts
SOLUTION
PROBLEM 4
Solving for the frictional force on bearing:
F = frictional force
F = f N = 0.15 (800) = 120 Ib
r = 1/2 = 0.5 in = 0.041666 ft
A 50 mm diameter shaft supported by two sleeve bearings carries a load of 13.3 MN.
The shaft rotates at 150 rpm. If the coefficient of sliding friction between the shaft and
bearings is 0.1, how much power is lost in friction?
A. 261 kw
C.345kw
B. 643 kw
D. 108 kw
T = torque developed due to frictional force
T = F x r = 120 (0.04166) = 5 tt-lb
Solving for the hp lost due to friction:
&miit-H'
P = 2 IT TN = 27t(5)(500) = 0.4759 h
33,000
P
Load for one bearing = 13.3/2
Load for one bearing = 665 MN
6
Fr = frictional resistance = (0 10)(665 x 10 )
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216
217
[3eu t: i WI
Bearing
Usmg the beanng stress formula:
2
S = PIA = 2000/1000 = 2 N/mm or Mpa
2
Fr = 6.65 X 10 KN
Solving for the tangential speed:
v = IT 0 N = IT (0.050)( 150/60)
PROBL~M
= 0.3927 m/s
Power Lost = F r v
2)(0.3927)
Power Lost = (6.65 x 10
= 261.14 kw (per bearing)
Solving for the power lost of 2 bearings:
Total power lost = 2(261.14) = 522.30 kw
7
A 22 mm diameter shaft is supported by sleeve bearing at a distance of 0.50 m. A
load of 2.2 KN is applied at 0.2 m from the left end. The sleeve bearings have an LID
ratio of 1.5. Find the maximum bearing pressure.
A. 1.82 Mpa
C 2 12 Mpa
B. 3.45 Mpa
O. 5.23 Mpa
Em!!mmI
PROBLEM 5
A 36 mm shaft uses a sleeve beanng that sustain a load of 4000 N. If the allowable
2
bearing pressure IS 1.3 MN/m . Find the length of the bearing.
A. 75.47 mm
C. 89.23 mm
B. 23.44 mm
O. 85.47 mm
Em!!mmI
F = 4000 N = 4 KN
3
S = 1.3 Mpa = 1.3 x 10 Kpa
Using bearing stress formula
F
Sb = ._..
2:M[J = 0
05 RA = 0.3(2.2)
RA = 1.32 KN = 1320 N
A=LO
But: UO = 1.5
L = 1.5 0
2
A = (150)(0) = 150
F
F
1320
P= - ' = " - -~
1.50"
(1.5 )(22)2
A
' - -
A
2
= 1.818 MN/mm or Mpa
PROBLEM 8
3
4
A sleeve beanng is to have an LO ratio of 1.0 and an allowable bearing pressure of
2
0.5 MN/m . Find the inside diameter and the length of the bearmg If it is to sustain a
load of 2550 N.
A. 23.45 mm
C. 45.34 mm
B.71.41mm
O. 9834 mm
1.3 x 10 = -
A
A = 0.0031 m
2
Solving for bearing length:
A=LO
00031 = L (0.036)
L = 0.08547 m =, 85.47 mm
Em!!mmI
PROBLEM 6
A 20 mm shaft uses sleeve bearings. The total load per bearing is 2000 N. An UO
ratio of 25 is desired. What is the bearing pressure?
A. 1 Mpa
C. 2 Mpa
B. 3 Mpa
O. 4 Mpa
Em!!mmI
LID = 2.5
L = 2.50 = 2.5(20) = 50 mm
Solving for the area:
A = LO = 50(20)
F = 2550 N
F = 000255 MN
UO = 10
L=O
Solving for the area:
S = F/A
2
A = F/S = 000255 I 0.5 = 00051 m
Solving for the diameter'
A = LO
00051 = 0(0)
0=007141 m =71.41 mm
1000 rnrn"
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218
Bearing
Bells
A sleeve bearing has an outside diameter of 1.50 in 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.
A. 100 psi
C. 200 psi
B. 150 psi
D 250 psi
15 BE3I-ts
219
PROBLEM 9
SOLUTION
ro = Do/2 = 1.5/2 = 0.75 in
r, = ro - t = 075 - 3/16
D, = 2 r, = 2(0.5625)
D, = 1125 in
= 0.5625 in
Types of belt transmission
Solving for the bearing pressure:
F
450
.
P = --- = -.----- = 200 PSI
A 2(1.125)
(.
Flat belt - used with flat pulleys and allows long distance
<. V-belt - used with shave or grooved pulleys and provides stronger grip at short
distance between shafts.
PROBLEM 10
A thrust washer has an inside diameter of 12 mm and an outside diameter of 75 mm.
If the allowable bearing pressure is 0.6 Mpa, how much load can it sustain?
A. 2582.9 N
C. 329344 N
D 863445 N
B. 1235.5 N
SOLUTION
Solving for the cross-sectional area:
A=
lJ
J =~
(D/ - D;2)
A = 0.00430476 m
[(0.075)2.- (0.012)2 ]
2
Solving for the load applied:
S = F/A
6)(0.00430476)
F = S A = (10.6 x 10
F = 2582.90 N
<. Toothed belt - paired with toothed pulleys and used as timing belt where speed
ratio must be maintained
Materials for transmission belts:
(.
Oak-tanned leather is the standard material for flat belts.
(.
Chrome leather is used where very pliable material is desired.
(.
Rubber belt is used when exposed to moisture, acids and alkalies.
t,
Fabric and canvas belts are used for light power transmission.
mmmI
For Open Belt Connection:
e
'"
Pulley diameter and speed relation
to,
Belt length:
D,N, = D2 N2
(D -D )
L = 1.57(0 2 + D1) + 2C + - 2- - - 14C
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220
"
Belts
Angle of contact:
e=tc ± 2sid (R 2 ~ R1
1
9 = rt ±
Note:
O 2 -0 1
J
1"1 - Fe
--= eIII
1"2 -Fe
radians
where:
Fe = centrifugal tension
3
w = belt weight, Ib/in
b = belt width, in
t = belt thickness, in
v = belt velocity, fUsec
,radians
Use + sign for larger pulley
Use - sign for smaller pulley
For Crossed Belt Connection
(,
Pulley diameter and speed relation:
0 1 N1 =02 N:z
where:
. D2 + D1
2C +
Net belt pull (tangential force on pulley)
I" = 1"1 - 1"2
(.
Stress in Belt:
S
bt
-
where:
Sw = working stress
= 300 pSI for leather belts
(.. Power Transmitted and Torque Relation by Belt:
T = (1"1 - 1"2) r = I" x r
P = 2nTN, Kw
where:
P = power, Kw
T = torque, KN-m
N = speed, rps
t,.. Angle of contact:
e =n + 2slo,1
(..
e
_ 1"1
W
4C
L = length of belt
0 1 = diameter of smaller pulley
O 2 = diameter of larger pulley
R = radius of larger pulley
r = radius of smaller pulley
e = arc of contact, radians
C = center distance
2
12wbtv /g
(,
e
(., Belt Length:
L = 1.57 ([)2 + D,)
D 2 -D 1
radians
n + -----,
Horsepower transmitted and stress relation
P
C
(1"1 - 1"2 )V, hp
550
bt
Belt Tension
If Centrifugal Tension is Neglected
-1"1 = e fs
1"2
where:
1"1 = tension in tight side
1"2 = tension in slack side
f = coefficient of friction
e = arc of contact, rad
"
221
Be/ts
(.. If Centrifugal Tension is Considered
= [V(Sw_~J2W)]
550
9
~
Formulas:
(..
Belt Length
L = 1.57(0 + d)+2C + (0 - d)2
4C
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[(ets) I(e
ls
- 1)]
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222
Belt s
Bells
223
'- Center Distance
PHOBLEM I
Find the angle of contact on the small pulley for an open belt drive with a 72 In center
distance The pulley diameters are 6 in and 12 in.
C. 185.34°
A 175.22°
D. 165.34°
B. 17034°
c. Arc of contact on small sheave
IE:.1!!.immI
For smaller pulley:
8 = angle of contact
where:
. 1 I D1 - D2
L = pitch length of belt
C = center distance
D = pitch diameter
d = pitch diameter of large sheave
b = 4L -o.28tOtt:l)
8 = 180 - 2sin' [12 -6J = 175.22°
2(72)
'- Standard Pitch Length and Designation of V-Belts:
Table 3.3, p 21(PSME Code)
Example:
B75 IS Section B v-belt with length of 76.8 inches
Horsepower Rating for v-belts:
HP = XS09 1
"Z$3
de
where:
PH = recommended horsepower
X,Y,Z are constants (Table 3.6)
S = belt speed in thousands of feet per minute
de = equivalent diameter of small sheave which is equal to pitch diameter
multiplied by small diameter factor (Table 3.9)
l.
l
l-2C-J
8 = 180 - 2sln
-)is
PROBLEM 2
Find the belt length at a 72 in center distance connected In open belt.
diameters are 6 in and 12 in.
A. 162.34 in
C 123.55 in
B. 173.45 in
D. 172.39 in
ki-I!i';r-UM
L = belt length
IT
L= -(D +D )+2C+
2
1
2
(0
D)2
2- 1
4C
IT
(12-6)2
L = - (12 + 6) + 2(72) + _._-2
4(72)
'" Design Procedure in Determining the number of V-belts required:
Given: Size of belt, sheave diameters, speed, power transmitted
1. Find the length of the belt from Table 3.3.
2. Solve for the center distance and the arc of contact.
3. Find the value of X, Y & Z from Table 3.6.
4. Solve for the speed ratio and find the small diameter factor from Table 3.9, then
solve for de.
5. Compute the HP rating per belt.
6. Find the length correction factor from Table 3.7 and arc of contact correction factor
from Table 3.8, then solve for the corrected HP rating per belt.
7. Find the service factor from Table 3.5, then divide the corrected power transmitted
by the Hp rating per belt.
The pulley
L = 172.39 in
PROBLEM 3
A 6 in diameter pulley turning at 600 rpm is belt connected to a 12 in diameter pulley.
If there is 4% slip, find the speed of the 12 in pulley.
A. 187 rpm
C. 203 rpm
B. 223 rpm
D 288 rpm
__-r1.IlJ~
USing the diameter and speed relation:
D1 N, = D2 N2
6(600) = 12(N z)
Nz = 300 rpm
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225
Belts
224
Belts
F
1
e
IT
)
180
F2
Considenng the 4% slip
N2' = 300 (1 - 0.04)
N 2 ' = 288 rpm
F 1 = 2.38 F2
S = F/A
F=SA
F=S
PROBLEM 4
For a given belt a manufacturer gives a horsepower rating of 2.5 hp per inch of width
based on a belt speed of 2600 ft/min. The drive is to handle 10 hp and the and the
arc of contact correction factor is 0.90. Find the width of belt needed. Assume that
belt widths are available in 1 inch increments.
A. 3 in
C. 4 in
D. 6 in
B. 5 in
(~xd2)
= 400
4
l(~'
XO.25 2 : = 19.631bs
4)
F2 = 19.63/2.38 = 825 Ib
n(1)(100)
.
v = '.-. - . - = 26.179ftl min
12
HP = ~Fl_=
~;0\1
= (196~_=8.25)?7J
33,000
Em!!ImD
= 0.00903 hp
33.000
PROBLEM 6
Solving for the actual hp rating
A 3/8 in flat leather belt is 12 in wide and is used on a 24 in diameter pulley rotating
0
3
600 rpm. The specific weight of the belt is 0.035 Ib/in The angle of contact is 150
If the coefficient of friction is 0.3 and the allowable stress is 300 psi how much hp can
it transmit?
C 78.5 hp
A. 34.5 hp
D 54.5 hp
B 69.5 hp
P = 25 (0.90) = 2.225 hplin
Belt width = 2._5~E
(030)(165.63x
10hp
in
Im!!iit·a.
Belt width = 4.44 in
For flat belts:
Therefore: Use 5 in width of belt
pv 2 e fe_1
F 1 - F 2 = b t (s - - - ) ( - - )
2.68
e f8
v = IT (24/12)(600/60) = 62.83 ft/s
PROBLEM 5
A 1/4 in round endless belt connects a 1 in diameter pulley with a 2 in pulley; spacing
between pulleys is 4 in on centers. The allowable stress is 400 psi. If the 1 in pulley
is the driver and rotates 100 rpm. How much horsepower can be transmitted?
Assume the coefficient of friction to be 0.30 for each pulley.
A.0123hp
C 0.0115hp
B. 1.234 hp
D. 00090 hp
e '8 = e
'o
30)l150)(rr!180)
= 2.193
Pv2
0035(62.83)2
----- = ..
= 51.55
2.68
2.68
12.1932 -1l
F 1 - F2 = (12)(3/8) [300-51.55JI-' - L 2.1932 --'
Em!!ImD
F 1 - F 2 = 608.26 Ibs
Since both have the same coefficient of friction, smaller pulley will be the basis of
computation.
8 = angle of contact
.,
8 = 180 - 2sln
F1
.-.-F2
~
l
l
--2C-J
D2-D1
180 - 2 sin
1
(~1J
l2(4 )
HP =(1"".1..- F2 )v
33,000 .
= 165.63°
e f8
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(60826t(62.83x60) = 69.50 hp
33,000
lOMoARcPSD|3535879
227
l uak c:
Belts
.6.MJ
PROBLEM 7
A 5 mm round belt connects a 20 mm pulley with a 40 mm pulley. The center
distance is 150 mm. The 20 mm pulley rotates at 100 rpm and the coefficient of
friction of the belt is 0.25. Find the horsepower capacity for this arrangement if the
2
allowable belt stress is 2.6 N/mm .
A. 0.12 kw
C 000282 kw
B. 0.23 kw
D. 0.054 kw
~
SOLUTION
F
S = -'.
A
F = SA
F = (2.6) (nI4)(5)2 = 51.05 N
For larger pulley:
8B = 180 + 2
sin'1l-~2
~_f3l
Types of Brakes:
J= 180 + 2 sin' 20.-= ~
PUII:(:R 2 -R J
For smaller
85 = 180·2 sin
Brake· is a device which is used to regulate or stop the motion of a body.
150
'1(20-10J
1= 187.65° = 3.275 rad
)
l'-C'-- = 180·2 sin -1'50- = 172.35
1
°
3 rad
1. Mechanical Brakes
a. band
b. block
c. shoe
d. disk
e. spot
2. Hydrodynamic Brakes:
a utilize fluid friction
For laroer pulley:
3. Electrical Brakes:
a. utilized the strength of electromagnetic fields
e 10 = e(15 25)(3 275) = 2.27
For smaller pulley:
ere = e(O 25)(3) = 2.12
Simple Band Brake
Therefore smaller pulley governs the design.
F1
18
-=e
F2
~05=2.1
Tension in the band
f1. = e f8
F2
where:
F 1 =: force on tight side
F 2 =: force on slack side
F2
F2==24.10N
v = n (0.02)(100/60) = 01047 rn/s
P = (Fl' F2 ) v == (51.05 . 24.1 )(0.1 047) = 2.822 watts =: 0.002822 kw
f =: coefficient of friction
8 = angle of contact. rad
Brake Torque Developed:
T =: (F1 - F2)r
where:
T = brake torque
r =: radius of friction surface on the drum
Actuating Force Required: (by taking moment about the pivot point)
Fa = aF2
L
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228
Brake
229
Brake
Maximum Unit Pressure:
where:
W = weight lowered
h = total distance traveled
where:
w = width of the bond
PROBLEM 1
Stress in band:
The band brake of a band brake has 210 degrees of contact with its drum. By
laboratory tests, it is found that the pull on the tight side is 800 Ibs and the pull on the
slack side is 285 lb. What is the coefficient of friction?
A. 0.281
C. 0 186
B 0.753
D. 0.453
where:
t = thickness of the band
~
Differential Band Brake
By taking moment about the
Using tension ratio of belt:
point:
F,
-=e
F2
fO
800
1(210° x_
Self-Locking Differential Band Brake:
___ = e
A differential band brake is self-locking when Fa is zero or negative.
285
2.807 = e3665f
Block Brake (Vallance p 364)
IT
_)
180
Take In both sides:
In2.807 = In e36651
In2.807 = 3.665f (In e)
f=0.281
PROBLEM 2
where:
T = braking torque
F r = radial force between the drum and each shoe
f = coefficient of friction
h = effective moment arm of the friction force
r = radius of the friction surface of the drum
8 = angle of contact
P max = maximum normal pressure between block and drum
w = axial width of block
Automotive Shoe Brake (Vallance pp 366-370)
Heat Dissipated in Brakes (Vallance p 374)
Icl ""fFrV
where:
H = heat dissipated
f = coefficient of friction
F r = radial force
V = surface velocity
For Brake used in lowering of a weight:
H·.. ;=i•. Wh
2
The inertia load on an electric brake is 32 Ib-ft and the drive shaft is rotating at 2000
rpm. What average torque is required to bring this load to a complete stop in 12 sec?
A. 173.61 ft-Ibs
C. 12345 tt-lbs
B 153.45 ft-lbs
D. 237.23 ft-Ibs
SOLUTION
Using the formula of torque in electric brake:
T=
2
J!v.r )N _
308(1.2)
(32)(2000) = 173.16 ft-Ibs
T = (308)(1.2)
PROBLEM 3
The band of a band brake has a contact angle of 180°. It is found experimentally that
pulls on the bands are 275 kg and 100 kg respectively under certain operating
conditions. Find the coefficient of friction
C 0197
A. 0.973
B. 0.567
0 0322
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230
Brake
elwell
2:31
Ki·)!WDI
Ie
F1
-=e
17 Clutch
F2
275
--- = e
f(180ox no)
180
100
nl
2.75 = e
In2.75 = Itf (In e)
f = 0.322
~
PROBLEM 4
A brake requires 900 in-lb of torque to stop a shaft operating at 840 rpm in a period of
3.5 seconds. What is the inertia load expressed in Ib-ft 2 ?
A. 90.23
C. 34.45
B 74.34
D. 9625
Clutch - is a machine member which is used to connect shafts so that the driven
shaft will rotate with the driving shaft. and to disconnect them at will.
Types of clutches:
Em!!Iir.mI
1. Jaw clutches - jaws or teeth in the two elements interlock
2)N
T= (wr
308 (t)
T = 900/12 = 75 ft-Ib
2. Friction clutches - the drivinq force is transmitted by friction; the major types are'
plate or disk clutch, cone clutch, band clutch, block clutch and expanding-ring
clutch.
2)(840)
75 = lwr
308(3.5)
w ~ = 96.25 ft2 -Ib
3. Hydraulic clutches - the torque is transmitted by a moving fluid
4. Electromagnetic clutches - the torque is transmitted by means of a magnetic lieid.
Disk or Plate Clutch
Power transmitted:
P = 2rrT N
T = n f Fa fl
where:
T = torque transmitted
n = number of pairs of mating friction surfaces
f = coefficient of friction
Fa = axial load
r, = mean friction radius
Mean frictional Radius, rf
For uniform pressure disc clutch
fr
=
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2.'
-
[f
3
O
__ rj3 ]
3 ro 2-'--. rI 2
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233
Clutch
Clutch
(0~) 5
For uniform wear clutch (worm clutch)
(0.200/2)
T =-sln 100
T = 142.53 N-m
PROBLEM 3 (ME Bd. Apr. 97)
Cone clutch:
2
If the inertia is 90 Ib-ft and the speed of the driven shaft is to be increased from 0 to
2500 rpm in 5 seconds, shaft material is of SAE 4140 with yeild strength of 110,000
lb/in", find the clutch starting torque in It-lb.
A. 159
C 146
B. 128
D. 191
T:;: Fa trt
sine
PROBLEM 1
0
A cone clutch has an angle of 10 and a coefficient of friction of 0.42.
Find the axial
force required if the capacity of the clutch is 7 kw at 500 rpm. The mean diameter of
the active conical sections is 300 mm.
A. 234.45 N
C. 368.49 N
B 186.45 N
D. 492.45 N
Em!!immI
Torque =
(0;2 - ~
ex = - - SOLUTION
W2
P=2nTN
7 = 2 n T (500/60)
T = 0.13369 KN-m
T = 133.69 N-m
(L
Using uniform wear method:
T=
l~; J
~
ri
133.69 = (0.42~
= sif;ex
(D2m J
I ex
g
-
t
rev
1min
2nrad
= 2500-- x ----- x --.-min 60 sec
rev
=
(~6Q)
- 0
26180 rad/s
= 52.359 rad/s"
5
(90Ib· ft2 )(52.359rad 1S2)
1S2----
Torque
- 32~ 1t
Torque
146 lb-tt
Alternate Solution:
(0.300/2)
2
T= (~)f'J
= 900(2.?OO) =146.10ft-lb
308 (t)
308(5)
sin 10
F = 368.49 N
PROBLEM 2
PROBLEM 4
How much torque can a cone clutch transmit if the angle of the conical elements is 10
degrees the mean diameter of the conical clutch sections 200 mm and an axial force
of 550 N is applied? Assume the coefficient of friction is 0.45.
A. 123.23 N-m
C 142.53 N-m
B. 34.56 N-m
D. 234.56 N-m
Find the frictional radius for a disc clutch. The disc clutch has an outside diameter of
10 in and an inside diameter of 6 in.
C 5.34 in
A. 454 in
D. 8.44 in
B. 4.08 in
SOLUTION
SOLUTION
ro = 10/2 = 5 in
T=
T=
r, = 6/2 = 3 in
~ ;l tr .J
~-
3
2 ro - r,.3
fF
Sin ex
r'=:3
(rm )
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l
~J a
,
l
2 r 53 3
3
1 = 408 in
3l5" 3"]
lOMoARcPSD|3535879
234
("{llldl
Clutch
PROBLEM 5
In a clutch the outside diameter is 8 in and the inside diameter is 4 in. An axial lorce
of 400 Ib is used to hold the two parts together. If the coefficient of friction of the
mating materials is 0.4, how much torque can the clutch handle?
A.123.45in-lb
C.213.34in-lb
B. 497.80 in-lb
D. 563.23 in-Ib
4i·"mtmI
Using uniform wear method:
P = 2]"(TN_
33,000
20 = 2]"(T.(1 2~0
33,000
T =105.04 ft-Ib
IIIEm!!IilimI
r, = 4/2 = 2 in
ro = 8/2 = 4 in
T =
Solving for the torque:
T=
2 fP
3
s;;a[~-?I )
IF
T = -.--(rm )
sin c:
[r 3 _ r3 ]
r:2~-
10504 = iO.3Q)~1}
(16 12)
sin 10"
F = 91.202 Ib
2
l(4)3_(2)3]
- _. -~
= 497.8 .in-lb
T = -(0.40)(400)
3
(4)2 -- (2/
slnr~
.
= -
F
r,
sin 10° = 91 202/F"
Fn = 525.21 Ib
PROBLEM 6
A disc clutch has 6 pairs of contracting friction surfaces. The frictional radius is 2 in
and the coefficient of friction is 0.3. An axial force of 100 Ibs acts on the clutch. The
shaft speed is 400 rpm. How much horsepower can the clutch transmit?
A. 2.28 hp
C. 3.23 hp
D. 4.23 hp
B. 1.23 hp
Fe = Fn (sina + I casu)
Fe = 525.21 [(sin10o + 0.3(cos100)] = 246.37 Ib
PROBLEM 8
SOLUTION
In a band clutch, the ratio 01 the pullan the tight side of the band to that 01 the slack
side is 4:1. The band contacts the drum for 250 degrees. What is the coefficient of
friction?
C 0234
A. 0.123
D. 0.462
B. 0.318
Using a uniform pressure method:
T = f P (r,) n
T = (030)(100)(2)(6)
T = 360 in-Ib
P = TC~3
33,000
235
~
2]"((360/12)(400) = 2.28 hp
33,000
F I 1 F2 = 4/1
F,
fO
-- = e
F2
4/1 = e l(250 x ,,/180)
PROBLEM 7
A cone clutch has cone elements at an angle of 10° The clutch transmits 20 hp at a
speed of 100 rpm. The mean diameter of the conical friction sections is 16 in and the
coefficient 01 Iriction ;s 0.3. Find the axial force needed to engage the clutch.
A. 246.37 Ibs
C. 234.56 Ibs
B. 212.561bs
D. 346.781bs
In 4 = In e436331
f = 0318
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236
Clutch
237
('1111 ell
PROBLEM 9
F 1 = F2 e
0
A band clutch has an angle of contact of 270 on a 15 in diameter drum. The
rotational speed of the drum is 250 rpm and the clutch transmits 8 hp. The band is
1/16 in thick and has a design stress of 5000 psi. How wide should the band be?
Assume a coefficient of friction of 0.40.
A.2.123in
C.1014in
B. 3.234 In
D. 4354 in
1o
.'
750 =, F2e
IT
0.4(250 x ----)
180'
F2 = 13094 Ib
T = (F 1 - F2) r
T = (750 - 130.94)(16/2) = 4952.48 in-lb = 412.707 It-lb
SOLUTION
p = 2ITTN
F1 = S b t
T = (F 1 - F2 ) r
33,000
2IT(412.707)(35_0) = 27.50 hp
33,000
-
p = 2n:TN
PROBLEM II
33,000
Find the power capacity under uniform wear of a cone clutch with mean diameter of
250 mm if the conical elements are inclined 8 degrees and the axial force is 450 N.
The rotational speed of the driver is 200 rpm and the coefficient of friction is 0.20.
C. 1.287 kw
A. 1683 kw
D. 3234 kw
B. 2340 kw
8 = _2_ITT_(_25_0_)
33,000
T = 168.067 ft-Ib = 2016.811 in-Ib
F
F2
.-1 = e
to
F 1 = F2 e
EelUi[']~1
fe
"
Using uniform wear method
It
0.4(270 x - - )
180C
F1 = F2e
!o-t:.i] = .lS:?1~450)
L 2
= 6.586 F2
T =_.t£...1
sin (X
(0.25/2) = 80.8345 N-m
sin 8°
Substitute to the equation of torque:
T = 0.0808345 KN-m
2016.8114 = (6.586F 2 - F2)(15/2)
F 2 = 48.14 Ib
F 1 = 6.586 (48.14) = 317.0471b
P = 2 IT TN = 2 IT (0.0808345)(200/60) = 1.683 kw
PROBLEM 12
F1 = S b t
317.047 = (5000)(1/16)(b)
b = 1.0145 in
PROBLEM 10
The angle of contact of a band clutch is 250 degrees. The cross section of the band
is 1/16 in x 1.5 in. The design stress for the band material is 8,000 psi. If the drum is
16 inches in diameter and rotates at 350 rpm, what is the horsepower capacity of the
clutch? The coefficient of friction is 0.4.
A. 20.34 hp
C. 22.34 hp
D. 27.50 hp
B. 1523 hp
"''-e]''!I[eg'
F 1 = S b t = (8000)(1.5)(1/16) = 750 Ib
Determine the power capacity of a cone clutch under uniform pressure and assuming
the following conditions: major diameter = 250 mm; minor diameter = 200 mm; length
of conical elements in contact = 125 mm; rotational speed = 870 rpm; coefficient of
2
friction = 0.30; and allowable pressure = 70,000 N/m .
A. 12.34 kw
C. 14.56 kw
B. 19.09 kw
O. 23.45 kw
SOLUTION
T = P b IT 00f(2/3)
l
r
3
- r3
1
°2. . '2J
ra
- r,
0.2 }
[' (0.125)3 - (0.10)3
I 0.25t-..-'-.
0.3)(2/3)1 - ~T = 70,000(0.125)(IT) I -,2
(0.125)2 - (0.1 0)2
l
T =209.603 N-m = 0209603 KN-m
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J
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23H
239
[I.[u,11 ineries
CLutch
1 B Machineries
P = 2 IT T N = 2 IT (0,209603)(870 160) = 19,096 kw
PROBLEM 13
Find the power capacity under uniform wear of a cone clutch with the following
specification speed = 870 rpm; length of conical elements in contact = 125 mm; major
diameter = 250 mm; minor diameter = 200 mm; coefficient of friction = 0,30; and axial
operating force = 500 N,
A. 7,69 kw
C, 9,34 kw
B, 5,23 kw
D. 3.23 kw
~
~
SOLUTION
.
r0 - r,
Polar Moment of Inertia
.
b
0.125-0.10
Using English units:
sin u. =
sino; =
---
0,125
a = 11,537°
T= ~ -
Jrr:= p.L. J
9
sina
50 ( .3)(g~_135 !
_0.10)
Sin 11 ,53r
2
Where: Jm = polar moments of inertia 01 masses, ft-lb-sec
2
p = Ib/ft
L = length, ft
g = 322 tt/sec"
J = polar moment 01 inertia 01 area with constant cross-section, ft4
= 84375 N-m = 0.084375 KN-m
Using SI units:
P = 2 IT TN = 2 IT (0.084375)(870/60) = 7.69 kw
J m = p LJ
2
PROBLEM 14
Assuming uniform wear, find the power capacity of a single disc clutch with an outside
and inside diameter of 200 mm and 100 mm respectively, a rotational speed 01 1160
rpm, a coefficient of friction of 0,35 and an axial operating force of 800 New1ons.
A. 4,23 kw
C. 6.32 kw
B. 2.55 kw
D. 8.23 kw
Where: J m = kg_m
3
p = kg/m
L = length, m
4
J =m
Radius of Gyration
Using English Units
SOLUTION
Ko = radius of gyration
l
Using uniform wear:
T = fFl-2~
I r + rj
0
T = (0.35)(800{ 0.10; 0,05
Ko
=~Jmg
. •.·•·.
W
Jm= W
k~
g
Where: J m = polar moments of inertia 01 masses, tt-lb-sec"
J = 21 N-m = 0.021 KN-m
P = 2IT TN = 2IT (0.021)(1160/60) = 2,55 kw
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K o = ft
g = 322 It/sec"
W = weight, Ib
lOMoARcPSD|3535879
,\Tn r II i fI ('I WS
240
ModI ineries
!(l~etic
Using 51 Units
Energ:y
Kinetic Energy of Translating body
J m ==
12
KE == ~ mv
2
Where: J m == kg_m2
3
p == kg/m
Ko = m
m == mass, kg
Wv
2
2g
Kinetic Energy of Rotating body
KE== Vz Jrno}
Where: J m == moment of inertia, ft-lb-sec'
(J) == angular velocity, rad/sec
Radius of Oscillation
For isosceles triangle:
For circle:
For parabola:
KE == - ~
3/4 of height
5/8 of diameter
5/7 of the height
Total Kinetic Energy
KE == 172 mv
2
2
+% J m co
Center of Percusion
Q == distance from axis of rotation to center of percussion
Force of a Blow:
WS
Average force of blow == Where:
r == the distance form axis of rotation to center of gravity of body
d
Formulas relating Torque and Angular Acceleration
Where: 5 == total height, It
W == weight of driver in Ibs
o == distance in feet which pile IS driven
Using English units:
Linear Impulse and Momentum
....
...
W
". .
Lmear momentum = rnx v.= ,,­ v
9
Where: To == torque in pounds-feet
J m == moment of inertia, ft-lb-sec 2
K, == radius of gyration, ft
a == angular acceleration, rad per sec"
Using 51 Units:
Where:
To == torque, N-m
J m == moment of inertia, kg_m2
a == angular acceleration, rad per sec 2
Linear impulse: F x t
Where:
F == force
t == time
Angular impulse and Momentum
Angular momentum == Jm ill
Angular impulse == To x t
Angular impulse == Change in angular momentum
To x t == J m(rot - Ulo)
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241
lOMoARcPSD|3535879
242
Mochinerics
ATClchincries
For cast iron having a tensile strength of 19,000 psi the bursting speed would
be:
Centrifugal Force
v == 1~
In English units:
F
=0
243
centrifugal force
0 x 19,000 =; 436 fUsee
Thickness of Cast Iron flywheel:
For solid rim:
y== ~ F R 9
W
Where: W weight, Ibs
v velocity, ftls
R radius, ft
N speed, rpm
=0
For jointed rim:
=0
=0
=0
In 51 units:
F= 0.01097 m R N2
Where:
Where: N speed, rpm
R = radius, m
m == mass, kg
F = force, N
=0
=0
Spokes and Arms of flywheels:
The strength of the arms should equal three­fourths the strength of the shaft In
torsion.
FLYWHEEL
If W equals the width of the arm at the hub and 0 equals the shaft diameter. then
Energy of Flywheel due to change of Velocity:
2
E== W(v 1 - V2 )
2g­-
Where: W = weight, Ibs
v = velocity, ftls
Centrifugal Stress in Flywheel rims:
y2
S == ­'­­­-
Where:
=0
=0
F = 0.01097 h1 R N2
2
t thickness, in
d diameter, in
n = number of arms
v peripheral speed, ftls
10
S = tensile strength of rim material, psi
v = nrn speed, His
W==1';30 for a wheel having 6 arms. For 8 arms, W:;i1:2D
~
For simple pendulum
T periods, sec
L = length of pendulum, ft
2
g = 32.2 ftls
=0
For physical pendulum:
2
T == 2rr Jk o
gr
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L,+,+
/\1u('11 incries
ko = radius of gyration, It
r = radius, It
g = 322 ft/s 2
Section Modulus of square beam, Z
Mue/tillcrics
245
P = internal pressure, psi
Thick-walled cylinders of ductile material; closed ends.
Clavarinc's equation is used:
3
Z =a /6
Where: a = side of square beam
ft = poisson's ratio
~
Strength of Taper Pins:
Thick-walled cylinders of ductile material; open ends.
Clavarinos equation is used:
Using English units:
d=113
JOST
Where: T = torque, Ibs
S = safe unit stress, psi
HP = horsepower
N = speed, rpm
o = shaft diameter, in
d = tapered pin diameter, in
,l1 = poisson's ratio
Collapsing Pressure of Cylinders and Tubes subjected to External Pressures.
p == 50,210,000 (1/0)3
P = collapsing pressure, psi
t = wall thickness, in
o = outside diameter of cylinder or tubes
Using SI units:
~
Where: d = tapered pin diameter, mm
N = speed, rpm
0= shaft diameter, mm
P = power transmitted, watts
S = unit stress, N/mm 2
1. Torsional deflection of circular shafts:
a
where:
For low pressure cylinders of cast iron:
'
,DP
2500
t = t h.ckness, In = --,'~., ,'~
o = inside diameter, in
= angular deflection, deg
T = torque, in-lb
L = length, in
D = shaft diameter, in
G = 11,500 psi (for steel)
(X
2. Shaft diameter for 0.08 degrees per foot of length of shaft
deflection.
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lOMoARcPSD|3535879
24fi
For English units:
D = 0.29
L
it
·iff
247
Mochin eries
!'v[Ucll illcrics
Where:
Where: D = diameter, in
T = torque, in-Ib
P = horsepower
N = speed, rpm
For SI units:
(A + 0) x A x B x K
B = width between drum flanges, in
0= diameter of drum barrel, in
H = diameter of drum flanges, in
K = factor from table
Y = depth no filled on drum
Rope Load due to Bendi ng
Where:
1. St> = bending stress =
D = diameter, mm
P = power, watts
N = speed, rpm
T = torque, N-mm
Pb"'· Sb>A
E = 12,000, 000 psi (average value)
o = sheave diameter
20 times its
3. Shaft deflec tion of 1 degre e for a length of
diame ter.
2. dw = wire diameter
0.10
P = power, hp
o = diameter, in
N = speed, rpm
T = torque, in-lb
3 A=
4. Linea r deflec tion of shafti ng
e)
own
a. Shafting subjected to no bending action of pulleys except its
weight
8:~5,9JE2
b Shafting subjected to bending action of pulleys, etc
Where: L = shaft length, ft
o = shaft diameter, in
mmm
Stren gth of Chain s
~
W20 54,000 0
Lengt h of Wire Rope
2
W = breaking loads, Ibs
o = diameter of bar, in
A = depth of rope space on drum, in
/11' \20(H "-.D .: 2.Y)/2
L = length of wire rope, It
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248
Mach i tier ies
Machi/writ's
Milling machine feed rate:
f m ", It nt N
Milling Cutters
fm = milling machine feed rate, in/min
ft = feed rate, in/tooth
nt = number of teeth of milling cutter
N = speed, rpm
Number of Teeth of Milling cutters
T",,6.3D
Estimating Planer Cutting Speeds
W
T = number of teeth
o = cutter diameter, in
W = width of cut in inches
A = helix angle of cutter
o = depth of cut, in
Vc = cutting speed, fpm
So = number of cutting strokes per minute
L = length of cutting stroke, ft
Planning Time
For high speed milling with sintered carbide:
T= W[LX(~,'
T =.'. ,K H~
FNelW
L"
•
l v: +1)+o,o25]
v,
T = time, min
W = width of surface to be planed, in
L = length of stroke, It
V c = cutting speed, fpm
V r = return speed in fpm
T = number of teeth
H = horsepower
F = feed per tooth, in
N = revolutions per minute of cutter
D = depth of cut, in
W = width of cut, in
~
Cutting Speed
Size and types of rivets:
v= nDN
Rivet diameter falls between:
N = speed, rpm
D = diameter, ft
v = cutting speed, fpm
D= 1:2jt'· to 1.4jt
Where: t = thickness of plate
Cutting time for turning, boring and facing
J=h
IN
T = cutting time, min
f = feed rate, in/rev
N = lathe spindle speed, rpm
Joint Strength
F = safe tensile load
F = n x Ar x 5 s
N = number of rivets
A r = cross-sectional area of rivets
Ss = allowable shearing stress
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Working Strength of Bolts
W ==St(0.55d2 -0.25 d)
~
W == St (A - 0.25d)
Horsepower of friction wheels:
W = working strength of bolt, Ib
St = allowable working stress, psi
d = nominal outside diameter of bolt, in
A = area at the root of the thread, in 2
1. HP = ­ ­ ­ Where:
D = diameter of friction wheel
N = speed, rpm
W = width of face, in
f = coefficient of friction
P = force in Ibs per inch of width
Holding Power of Set Screw:
P =
DNd~·3
,hp
T = 1250 0 d 2 .3 , ltHb
2.
Where: D = shaft diameter. in
d = set screw diameter, in
N = speed, rpm
C
3. HP
Tensile Stress Area of Thread or Screw
S
= factor =
3.1416 Pxf
33,000)(12
DxNxWxC
ur Gear
a. For steels up to 100,000 psi ultimate strength
Gear set Center Distance:
At == U.ll::lo41 u ­ ­ -
b. For steels over 100,000 psi ultimate tensile strength
C = center distance
Dp = pitch diameter of pinion
At =
Additional Gear Formulas from Machineries Handbook
D = basic major diameter, in
n = number of thread per inch
Dm n = minimum pitch diameter of external thread, in
2
At = tensile stress area, in
Sharp V - thread
D = depth of thread
0= P xcos30o
D=
0.866
1. Gear set Center Distance:
C ;"Qp(fP~jHatiO·1)
2
C = center distance
Dp = pitch diameter of pinion
2. Circular Pitch for Given Center Distance and Ratio
no. of thread per inch
p = pitch, in
L
C = center distance
T = number of teeth
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3. Circular Thickness of Tooth when Outside Diameter has been Enlarged
253
~
Wrapped-spring Clutches
t = tooth thickness
Pc = circular pitch
e = amount outside diameter is increased over standard
¢ = pressure angle
1.
HP = horsepower
N = speed, rpm
4. Circular Thickness of Tooth when Outside Diameter has been Enlarged
2. Clutch starting torque
ft-Ibs
t = tooth thickness
Pc = circular pitch
e = amount outside diameter is increased over standard
¢ = pressure angle
2
2
I = W R moment of inertia, Ib-ft
W = weight, Ib
R = radius of gyration, ft
t-N = final rpm - initial rpm
t = time to required speed in seconds
5. Chordal Thickness of Tooth when Outside Diameter is Standard
3. Heat Generated
o = heat generated
tc = chordal thickness
= pitch diameter
N = number of teeth
o
6. Chordal Thickness of Tooth when Outside Diameter is Special
2
WR = total inertia, Ib-ft
N 1 = final rpm
N2 = initial rpm
T, = clutch torque, ft-\b
T1 = torque load, tt-lb
t = circular thickness
7. Chordal addendum
2
PROBLEM 1
A car moving at 60 mph when the brakes are suddenly locked and the car begins to
skid. If it takes 2 seconds to slow the car to 30 mph, at what rate is it being
decelerated, how long the car comes to a halt, and how far will it have traveled?
.i-j'·"[-RI
a = addendum
va = 60 (5280/60) = 88 ft/s
= 30 (5280/60) = 44 ft/s
VI
A.
VI = Va + at
44 = 88 + a (2)
a = -22 ft/s 2
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B.
MGchineries
Machineries
44 ft/s
0 (stop)
VI '= Va + at
o '= 44 + (-22) t
t '" 2 sec
T '= total time
T '= 2 + 2 '= 4 sec
Va
'=
255
2
C. 3.23 m/s
2
D.2.19m/s
A. 4.23 rn/s"
B. 5.23 m/s"
VI '=
c.
S '= Va t + Yz a t
SOLUTION
W,= 50 (9.81)
T '= W +
600
a '=
490.5
-a
g
2
S '= 88 (4) + 112 (-22) (4/
S '= 176 ft
'=
W
490.5 + 490.5 a
9.81
2.19 m/s"
'=
PROBLEM 2
A flywheel on a press rotating at 120 rpm is slowed to 102 rpm during a punching
process that requires 34 sec for the punching portion of the cycle. What angular
deceleration the flywheel experience?
A. -2.52 rad/s"
C. -1.23 rad/s"
D. -8.45 rad/s"
B. -5.45 rao/s"
II'm!!Dm
SOLUTION
+ a. t
Wa '= 120 (2rr/60) '= 12.57 rad/s
w, '= 102 (2rr/60) '= 10.68 rad/s
10.68 '= 12.57 + a. (3/4)
a. '= -2.52 rad/s"
WI
'=
PROBLEM 5
A flywheel has a diameter of 3 ft and weighs 1000 pounds. What torque must be
applied, neglecting bearing friction, to accelerate the flywheel at the rate of 100
revolutions per minute .per second?
A. 265.80 ft-lbs
C. 365.80 ft-Ibs
B. 565.80 tt-Ibs
D. 665.80 ft-Ibs
Wa
a. = 100 rev/sec-min (2rr/60) = 10.47 rad/s"
2
W 2
r ) a.
g
T = (Yz m r ) 0.= (112 -
T = \/2 (1000/32.2)(3/2)2 (10.47)
T = 365.80 ft-lbs
PROBLEM 3
A 100 Ib body is being hoisted by a winch, the tension in the hoisting cable being kept
constant at 110 lb. At what rate the body is accelerated?
2
C. 2.22 ft/s 2
A. 1.22 ftls
2
B. 3.33 ft/s
D. 4.44 ft/s 2
II'm!!Dm
PROBLEM 6
A flywheel has a diameter of 1.5 m, and a mass of 800 kg. What torque is needed to
produce an angular acceleration of 100 revolutions per minute, per second?
A. 5356 N.m
C. 3356 N.m
B. 4356 N.m
D. 2356 N.m
4-NiIDi1m
T '= W + FR
W
T '= W + - a
g
a. = 100 rev/sec-min (2rr/60)
T = (112 m (') a.
T = 112 (800)( 1.5/2/ (10.47)
T = 2356 N.m
100
110 '= 100 + - - a
32.2
2
a '= 3.22 ftls
10.47 rad/s"
PROBLEM 7
PROBLEM 4
A body of mass 50 kg is being hoisted by a winch, and the tension in the cable is 600
N. What is the acceleration?
A 12 inch cube of steel weighing 490 lbs is being moved on a horizontal conveyor belt
at a speed of 6 miles per hour (88 ft/s) What is the kinetic energy of the cube?
A. 323.45 ft-Ibs
C. 534.34 ft-Ibs
0 633.34 ft-Ibs
B. 489.22 ft-Ibs
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Moc h i fl cr ies
Mo ciiin er ie«
SOLUTION
SOLUTION
PE = potential energy
Potential energy = Work produced
PE = W
m (z + d) = F x d
100 (10 + 0.3) = F (0.3)
F = 3433.33 kg (0.00981) = 33.68 KN
W 2
KE = 1'2 - V
g
KE = '12 _490 (8.8)2 = 489.22 tt-lbs
32.2
PROBLEM 8
If a cube of mass 200 kg is being moved on a conveyor belt at a speed of 3 mis, what
is the kinetic energy of the cube?
A. 800 J
C. 850 J
B. 900 J
D. 950 J
Em!!ImD
KE = % mv 2
% 200(3)2
PROBLEM 11
A 1000 Ib block is pulled up a 2-degree incline by a cable exerting a constant force F
of 600 Ibs If the coefficient of friction between the block and the plane is 0.5, how
fast will the block be moving up the plane 10 seconds after the pun is applied?
A. 12.4 mph
C. 17.3 mph
B. 10.2 mph
D. 14.3 mph
w..ie],IIIWl'
KE = 900 J
cosu. + sin o.)
R = F - P = -W(~
R = 600 - 1000(0.5 cos2° + sin 2°)
R = 600 - 535 = 65.41 Ibs
PROBLEM 9
W
W
g
g
A pile driver weighing 200 pounds strikes the top of the pile after being fallen from a
height of 20 ft. If forces a pile into the ground a distance of 1'2 foot. What is the
Rt = --vr--vo
average force of the blow?
A. 8,2001bs
B. 8,6001bs
65.41 x 10 = ~-vf
C. 8,4001bs
D. 8,8001bs
1000
-~(O)
1000
32.2
32.2
v, = 21.06 ftls = 14.3 6 miles/hr
Em!!ImD
PROBLEM 12
A 500 kg block is pulled up a 2 degree incline by a constant force F of 4 KN. The
coefficient of friction between the block and the plane is 0.5. How fast will the block
be moving 10 seconds after the pull is applied?
A. 27.6 rn/s
C 30.4 rn/s
B. 34.2 m/s
D. 38.3 m/s
PE = potential energy
Potential energy = Work produced
PE = W
m (z + d) = F x d
d = Y2 It = 0.5 ft
200 (20 + 0.5) = F (0.5)
F = 8200lbs
Em!!:immI
R = F - P = -mg(~
coso. + sin u.)
R = 4000 -500 x 9.81 (0.5 cos2° + sin 2°)
PROBLEM 10
R t = mv, -mv o
A pile driver of mass 100 kilograms falls 10 meters and moves the pile a distance of
0.3 m. What is the average force of the blow?
A. 23.45 KN
C. 5423 KN
B. 33.68 KN
D. 43.23 KN
Vj
1378 x 10 = 500(vr - 0)
= 27.6 m/s
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1.378 KN
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259
PROBLEM 13
PROBLEM 16
A flywheel having a moment of inertia of 25 lbs-ft-sec" is revolving with an angular
velocity of 10 radians per second when a constant torque of 20 Ibs-ft is applied to
reverse its direction of rotation. For what length of time must this constant torque act
to stop the flywheel and bring it up to reverse
speed of 5 radians per second?
A. 12.34 sec
C. 18.80 sec
B.14.34sec
D.16.34sec
A casting of mass 150 kg is lifted 4 meters in 15 seconds by means of a crane. What
is the power?
A. 392 watts
C. 456 watts
B. 523 watts
D. 634 watts
ID.IDIilmI
'= W x v = W x (Sit)
P '= (150 x 0.00981)(4115)
P '= 0.392 kw = 392 watts
P
SOLUTION
Angular impulse '= Change in angular momentum
To t = J (Wf - wo)
PROBLEM 17
20 t '= 25 ([10 - (-5)]
t '= 18.8 seconds
A cast iron flywheel with a mean rim radius of 9 inches, is rotated at a speed of 800
rpm. If the weight of the nrn is 20 Ibs, what is the centrifugal force?
A. 3169.341bs
C. 3269.341bs
B. 3262.34 Ibs
D. 4269.34 Ibs
PROBLEM 14
2
A flywheel with a moment of inertia of 20 kg_m is revolving with an angular velocity of
10 radians per second when constant torque of 30-N-m is applied to reverse its
direction of rotation. For what length of time must the constant torque act to stop the
flywheel and bring it to reverse speed of 5 radians per second?
A 5sec
C.10sec
B.12sec
D.16sec
ID.IDIilmI
Wv 2
Fe = - . gR
V = 2 It R N = 2 It (9/12)(800/60) = 62.83 ftJs
Fe = 20 (62.83)2
SOLUTION
= 3269.341bs
32.2(9 I 12)
Angular impulse '= Change in angular momentum
PROBLEM 18
To t '= J (Wf - wo)
30t '= 20([10-(-5)]
t = 10 seconds
A steel pulley with a mean rim radius of 120 mm is rotated at a speed of 1100 rpm. If
the mass of the rim is 5 kilograms, what is the centrifugal force?
A. 5344.50 N
C. 6432.30 N
B. 7961.50 N
D. 8734.56 N
PROBLEM 15
A casting weighing 300 Ibs is to be lifted by means of an overhead crane.
casting is lifted 10ft in 12 seconds. What is the horsepower developed?
A. 0.45 hp
C. 6.5 hp
B. 8.5 hp
D. 95 hp
Em!!DrmI
HP '=
HP =
The
SOLUTION
mv 2
Fe = -R
V = 2 It R N = 2 It (0.12)(1100/60)
mv
550
'=
300 (10/12)
-5~
Fe = 5(13.82)2
0.12
m(S/t)
- 550
'=
0.45 hp
13.82 mls
= 796150 N
PROBLEM 19
A round bar made from SAE 1025 low Carbon steel is to support a direct tension load
of 50,000 Ibs.
Using a factor of safety of 4, and assuming that the stress
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concentration factor k = 1. What is the
suitable standard diameter. Yield stress is
40,000 psi.
A. 3 Yz in
C 2 1), in
B. 2 9/16 in
D. 1 Y2 in
Machineries
SOLUTION
Using the formula of strength of tapered pin:
8
SOLUTION
=0
2~?7T
Dd 2
T = F x r = 200 x 800 = 160,000 N.mm
8 a ll
F
A
8 a ll
s, I F8 = 40,000/4
50,~
10 000 =
,
8
10,000 psi
A
A = rr/4 D
2
= _1.27 (160,000) = 406 N/mm or Mpa
50 (1 0)2
PROBLEM 22
A = 5 in
5 = nl4 D
If a shaft of 50 mm diameter is to transmit power of 12 kw at a speed of 500 rpm, find
2
the mean diameter of the pin for a material having a safe unit stress of 40 N/mm .
A10mm
C.12mm
B. 14 mm
D. 16 mm
2
2
&1·l!iiiGD
D = 2.523 in
Therefore, use D = 2 9/16 (standard)
PROBLEM 20
A lever secured to a 2 inches round shaft by a steel tapered pin (d = 3/8") has a pull
of 50 Ibs at a 30 in radius from shaft center. Find the unit working stress of the pin.
A 6770 psi
C. 6790 psi
B. 7433 psi
D. 5234 psi
EmmmD
8
=0
~-S'
mm
112,000
d = 11 0.3 ~ 500 x 50 ~ 40
12.09 mm
Find the thickness of metal required in the hemi-spherical end of a cylindrical vessel,
2 feet in diameter, subjected to an internal pressure of 500 psi. The material is mild
steel and a tensile stress of 10,000 psi is allowable.
A. 0 10 in
C. 0.3 in
B. 0.5 in
D. 0.7 in
Dd 2
T = Fxr
T = 50 x 30 = 1500 in-lb
27 (1500)
8
2(3/8)2
=0
d = 110.3
PROBLEM 23
Using the formula of strength of tapered pin:
8 = 2·7~
261
SOLUTION
2.
Using the formula of spherical shell because of two hem i-spherical ends.
6770 psi
8
PROBLEM 21
A lever secured to a 50 mm round shaft by a steel tapered pin ( d
=0
of 200 N at a radius of 800 mm. Find the working stress on the pin.
A. 32.3 Mpa
C. 45.20 Mpa
B. 40.6 Mpa
D. 56.34 Mpa
10 mm) has a pull
P o,
4 t
P o,
500 x 2 x 12
48
4 x 10,000
0.3 inch
PROBLEM 24
Find the thickness of metal required in the hem i-spherical end of a cylindrical vessel,
2
750 mm in diameter, subjected to an internal pressure of 3 N/mm . The material is
2
mild steel and a tensile stress of 70 N/mm is allowable.
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A.4mm
B.2mm
Machineries
C.6mm
D. 8 mm
SOLUTION
Using the formula of spherical shell because of two hem i-spherical ends.
For main shaft:
3
D N
P =-80
P o,
S
C. 93 mm
D. 85 mm
A. 78 mm
B. 81 mm
Wf':rv.:m:D
4 t
P o,
4S
3 x 750
---
4 x 70
~=
3
(500)
0.746
80
D = 3.18 in = 80.78 mm
8.04 mm
D
PROBLEM 25
PROBLEM 28
What would be the diameter of the line shaft to transmit 10 horsepower if the shaft
makes 150 rpm?
A. 9/16 in
C.29/16in
B. 39/16 in
D. 1 9/16 in
What power would a short shaft, 50 mm in diameter, transmit at 400 rpm?
A. 50 kw
C. 55 kw
B.60kw
D.65kw
SOLUTION
SOLUTION
P =
D3 N
~-
D3 N
P
38
53.5
D
3
(50/25.4)3 (400)
P
(150)
53.5
D = 1.53 in = 1 9/16 in (standard)
10 =
263
38
80.29 hp = 60 kw
PROBLEM 29
PROBLEM 26
What horsepower would be transmitted by a short shaft, 2 inches in diameter,
carrying but two pulleys too close to bearings if the shaft makes 300 rpm?
A. 60 hp
C. 65 hp
B. 75 hp
D. 70 hp
SOLUTION
P =
D
Find the torsional deflection for a solid steel shaft 4 inches in diameter and 48 inches
long, subjected to twistinq moment of 24,000 in-lb.
A. 0.122 deg
C. 0.23 deg
B. 0.052 deg
D. 0.43 deg
IEm!!Im3
Using the derived formula for deflection:
3
N
38
(X
where:
(2)3 (300) = 63 hp
P = 38
PROBlEM 27
What would be the diameter of power-transmitting shaft to transmit 150 kw at 500
rpm?
= -4~' 584 TL
D G
d
eg
ex = angular deflection, deg
T = torque, in-Ib
L = length, in
D = shaft diameter, In
G = 11,500 psi (for steel)
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a =
A = (H - 0 - 2Y)/2
2
584 (24,009)( 48
= 0.23 degree
(4)4 (11,5000,000)
L = length of wire rope, ft
L = (A + 0) x A x B x K
PROBLEM 30
Where:
Find the torsional deflection of a solid steel shaft, 100 mm in diameter and 1300 mm
long, subjected to twisting moment of 3 x 106 N-mm. The torsional modulus of
elasticity is 80,000 N/mm 2 .
A. 0.122deg
c. 0.234 deg
B. 0.285 deg
O. 0.543 deg
B = width between drum flanges, in
o =' diameter of drum barrel, in
H = diameter of drum flanges, in
K = factor from table
Y = depth no filled on drum
Em!!ImD
A = (30 - 18 - 0)/2 = 6 in
L = (6+18)x6x24x0.741 = 2560ft
ex
584 (3 x 10 6 )(1300)
584 TL
[)4G
(100)4 (80,000)
0.285 degree
PROBLEM 31
Find the diameter of steel line shaft to transmit 10 hp at 150 rpm with a torsional
deflection not to exceeding 0.08 degree foot length.
A. 0.834 in
C. 1.23 in
B. 3.234 in
o. 2.35 in
PROBLEM 33
Find the bending stress and equivalent bending load due to the bending of a 6 x 19
(Fiber core) wire rope of % in diameter around a 24 in pitch diameter sheave.
A. 14561bs
C. 19831bs
B. 1590 Ibs
O. 1763 Ibs
E:mimD
Em!!ImD
For 6 x 19 Fiber Core rope:
d., = 0.063 d = 0.063 (0.5~
= 0.0315 in
2
2
A = 0.450 d = 0450 (0.5) = 0.101 in
E = 12,000,000 psi (average value)
Shaft diameter for 0.08 degrees per foot of length of shaft deflection.
For English units:
0=4.6
Where:
265
{f
0 = diameter, in
T = torque, in-Ib
P = horsepower
N = speed, rpm
Pb
m
10
o = 4.6 4 _ _
150
E dw = (12,000,000) (0.0315)
15,750 psi
D
24
Sb A = 15,750 (0.101) = 1590 Ibs
Sb
PROBLEM 34
Find the breaking load in pounds of a wrought iron chain crane if the diameter of bar
from which links are made is 2 inches.
A.216,000Ibs
C.316,000Ibs
B. 416,0001bs
O. 516,0001bs
.
= 2.35 Inches
PROBLEM 32
Find the length in feet of 9/16 inch diameter rope required to fill a drum having the
following dimensions: B = 24 in, 0 = 18 in, H = 30 in. (K = 0.741)
A.1873ft
C.2874ft
B. 2560 ft
O. 2645 ft
4i·)!!ImD
A = depth of rope space on drum, in
&i-)''tmD
W = 54,000 D
Where:
2
W = breaking loads, lbs
o = diameter of bar, in
W
= 54,000 (2)2 = 216,0001bs
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PROBLEM 35
SOLUTION
The specific gravity of cast iron is 7.2. Find the weight of 5 cubic inches of cast iron.
A. 1.1 Ibs
C. 1.2 Ibs
B. 1.31bs
D. 1.4lbs
24(100) = D2 (600)
Weight of any material = 0.0361 (SG), Ib
Weight of any material
02 = 4 in.
= 0.0361 (7.2) = 1.2996 Ib
PROBLEM 39
How many Y2 inch diameter set-screws would be required to transmit 3 horsepower at
PROBLEM 36
The weight of a cubic inch of gold is 0.697 lb. Find the specific gravity.
A. 13.45
C. 17.23
B. 15.34
D. 19.31
&-,e],'''[.];''
a shaft speed of 1000 rpm if the shaft diameter is 1 inch.
A. 1
C.2
B. 1.5
D.2.5
SOLUTION
Weight of any material = 0.0361 (SG)
SG
Using the relation of D and N:
D1 Nl = D2 N2
Em!!ImmI
0.697
267
Holding Power of Set Screw:
= 0.0361 (SG)
P _ ONd2 .3
-
= 19.31
PROBLEM 37
If the diameter of driving pulley is 15 inches and its speed, 180 rpm. The diameter of
driven pulley is 9 inches. Find the speed of driven pulley.
A. 100 rpm
C. 200 rpm
B. 300 rpm
D. 400 rpm
,hp
P '" 1(1000)(1/2)2.3
50
No. of set screw
= 4.1 hp
3
= -4.1
= 0.731
Use 1 set screw
Em!!ImmI
PROBLEM 40
Using the relation of D and N:
D1 N1
How many 3/8 inch diameter set-screws would be required to transmit 3 horsepower
at a shaft speed of 1000 rpm if the shaft diameter is 1 inch.
= D2 N2
A. 1
B.2
C. 1.5
0.3
15(180) = 9 (N2)
SOLUTION
N2 = 300 rpm
Holding Power of Set Screw:
PROBLEM 38
If the diameter of driving pulley is 24 inches and its speed, 100 rpm, and the driven
pulley is to rotate 600 rpm, find the diameter of driven pulley.
A. I in
C. 2 in
[1. 3 in
D. 4 in
23
P = DNd .
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,hp
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Machineries
p = 1(1 000)(3/8)2.3
50
Much incrips
269
= 2.1 hp
26(N 1) = 4(800)
3
No. of set screw = - =1.428
2.1
Use 2 set screws
N1 = 123 rpm
PROBLEM 44
PROBLEM 41
What is the working strength of a 1 inch bolt which is screwed up tightly in a packed
joint when the allowable working stress is 10,000 psi?
A. 3000 Ibs
C. 3500 lbs
B. 4000 Ibs
D. 4000 Ibs
If the dnving gear has 20 teeth and rotates 80 rpm, and the driven gear has 40 teeth,
then the speed of the driven gear is:
A.10rpm
C.20rpm
B. 30 rpm
O. 40 rpm
E'm!!IilmI
Ei1!!mmI
Using the relation of T and N:
W = workinq strength of bolt
T1 N, = T 2 N2
W = S1 (0.55d 2 ~ 0.25d)
20 (80) = 40 (N2)
W = 10,000[0.55(1)2 - 0.25(1)] = 3000 lbs
N2 = 40 rpm
PROBLEM 42
42. If the diameter of driven pulley is 36 inches and its required speed, 150 rpm, and
the speed of driving pUlley is 600 rpm, then the diameter of driving pulley is:
C.7in
A. 5in
B. 9 in
O. 11 in
PROBLEM 45
If the pitch diameter of the driver is 8 inches, its speed, 75 rpm, and the pitch
diameter of the driven gear, 20 rpm is:
A. 20 in
C. 30 in
B.40in
0.50in
. SOLUTION
SOLUTION
Using the relation of 0 and N:
Using the relation of 0 and N:
0 1 N, = O2 N2
0, N, = 02 N2
8 (75) = O2 (20)
0, (600) = 36 (150)
0,
= 9 inches
O2
PROBLEM 43
= 30 inches
PROBLEM 46
If the diameter of the driven pulley is 4 inches, its required speed, 800 rpm, and the
diameter of the driver is 26 inches, then the required speed of the driver is:
A. 112 rpm
C. 123 rpm
B. 134 rpm
O. 145 rpm
2,
If the inertia is 80 Ib-ft and the speed of the driven shaft is to be increased from 0 to
1500 rpm in 3 seconds, find the clutch starting torque in Ibs.
A. 100 rt-Ib
C. 120 tt-lb
B.110ft-lb
0.130ft-lb
E'm!!IilmI
SOLUTION
Using the relation of 0 and N:
0, N,
= 02 N2
Te = clutch starting torque, ft-Ibs
T = Ix,A.~
c
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Mac iiin eti es
Machineries
2
C = 3.1416 (150) (0.2) = 0.00024
33,000 x 12
HP = 10 x 200 x 2 x 0.00024 = 0.96 hp
= 130 tt-Ib
r, = 80308x 1500
(3)
PROBLEM 49
PROBLEM 47
If the inertia is 80 lb-ft", and the speed of the driven shaft is to be increased from 0 to
1500 rpm in 3 seconds. Calculate the heat generated for each engagement if clutch
starting torque is 10 in-lb.
A. 20.50 Btu
C. 30.50 Btu
B. 41.50 Btu
D. 51.50 Btu
A body weighing 28 Ibs rests on a horizontal surface. The force required to keep it in
motion along the surface is 7 Ibs. Find the coefficient of friction.
A. 0.12
C. 0.25
B. 0.45
D. 0.85
4-t.l'iimD
Fr = frictional resistance
Fr = f N = f W
7 = f (28)
f = 0.25.
E·] "imB
Q = heat generated
Q = Tc x WR
2
2
2
X (N1 -
2
N2 )
(Tc - T1 ) x 4.7 x 10 6
WR = total inertia, Ib-tt
N1 = final rpm
N2 = initial rpm
Te = clutch torque, tt-Ib
T 1 = torque load, tt-Ib
Q
271
f = coefficient of friction
P = force in Ibs per inch of width
I = W R = moment of inertia. Ib-tt 2
W = weight, Ib
R = radius of gyration, ft
tlN = final rpm - initial rpm
t = time to required speed in seconds
PROBLEM 50
Btu
Measurement M = 3.495 inches at the gaging notch of a 3 in pipe thread and the wire
diameter is 0.07217 in. Find the pitch diameter. (P = 0.125 in)
A. 1.23 in
C. 2.34 in
B. 0.34 in
D. 3.39 in
2
= 130X80x[(1500)2_ 02]
(130-10)x4.7x10 6
SOLUTION
41.50 Btu
M = E -(0.86603x P)+3 x W
1.00049
E = effective pitch diameter, in
PROBLEM 48
Find the horsepower transmitted by a pair of friction wheels; the diameter of driving
wheel is 10 inches, and it revolves at 200 rpm. The width of the wheel is 2 inches.
The force per inch width of face is 150 pound and the coefficient of friction is 0.20.
A. 0.96 hp
C. 1.2 hp
B. 023 hp
D. 1.6 hp
SOLUTION
C = 3.1416 P x f
33,000 x 12
HP = 0 x N x W x C
Where:
3.495 = E - (0.86603 x 0.125) + 3 x 0.07217
1.00049
E = 3.3885 in
PROBLEM 51
Find the tooth thickness on the tooth circle of a 14 Y2 degree full depth tooth of 12
diametral pitch.
C. 0.455 in
A. 0.131 in
D. 0.864 in
B. 0.234 in
FM'rmmII
o = diameter of friction wheel
N = speed, rpm
W = width of face, in
Tooth Thickness = 1.5708/P
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Machineries
Mach in eties
Tooth thickness = 0.1309 in
t = Pc - e tan <j>
2
PROBLEM 52
Find the tooth thickness on the tooth circle of a 20 degree full depth involute tooth
having a diametral pitch of 5.
A. 0.863 in
C. 0.108 in
B. 0.314 in
D. 0563 in
t =Q.:6283 -(0.2746xtan14.5 0 )
2
Tooth Thickness = 1.5708/P
1.57
Tooth thickness
5
Tooth thickness = 0.3142 in
0.2432 in
PROBLEM 55
A pinion has 15 teeth of 3 diametral pitch. Find the chordal thickness at the standard
pitch diameter.
A. 0.653 in
B. 0.523 in
SOLUTION
C. 0.863 in
D. 0.234 in
Em!!ImD
Chordal Thickness of Tooth when Outside Diameter is Standard
PROBLEM 53
The outside diameter of a pinion having 10 teeth of 5 diametral pitch and pressure
angle of 14 V2 degrees is to be increased by 0.2746 in. The circular pitch equivalent
to 5 diametral pitch is 0.6283. Find the arc tooth thickness at the standard pitch
diameter.
A. 0.385 in
C. 0.543 in
B. 0.863 In
D. 0.534 in
tc = D sin (90° 1 N) = 5 sin [9
10:
SOLUTION
Circular Thickness of Tooth when Outside Diameter has been Enlarged
t = tooth thickness
t= ~
P
2
273
+ e tan o
t = Pc + e tan ¢ = (0.6283/2) + 0.2746 tan14.5° = 0.3852 in
2
PROBLEM 54
The outside diameter of a gear having a pressure angle of 14 Y2 degrees is the be
reduced by 0.2746 in or an amount equal to the increase in diameter of its mating
pinion. The circular pitch is 0.6283 in. Determine the circular tooth thickness at the
standard pitch diameter.
A. 0.434 in
C. 0.243 in
B. 0.843 in
D. 0.672 in
SOLUTION
Circular Thickness of Tooth when Outside Diameter has been Enlarged
t = tooth thickness
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J = 0.5226 in
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Sit WU ioua I Problems
P ~/ -l.
R··'
'
T""
~ 2'}>i
\J ct . . .. t .<'~
r ~,
i.
"t,
C. 1211.47
D. 1311.47
A.1011.47Ibs
B 1111.47
6. the load at slack side
A.5591bs
B. 5691bs
C 5791bs
D. 589 Ibs
SOLUTION:
S "ltY'"'Ur::A"'T'"}'1""0/''N'"A'''
J
>, L;
., ~i
s ., ,
,,'co;'
, .'.
;1 .,t,
.",,?
,,'..
275
Situat ion ul Problems
{{<}
t ,,'
Driven
c = 4m
SiM~EL,BOR'P
1.
p
p
SITUATIONAL PROBLEM 1 (ME Board April 1981)
An open belt drive connects a 450 mrn driving pulley to another driven pulley 1000
mm in diameter. The belt is 300 mm wide and 10 mm thick. The coefficient of friction
of the belt drive is 0.30 and the mass of the belt is 2.8 kg per meter of belt length.
Other data are as follows:
Center distance between shafts = 4 meters
Maximum allowable tensile stress of the belt = 1500 kpa
Speed of driving pulley = 900 rpm
Determine:
1. the belt density in lb/in"
A. 0.014
B, 0.024
C, 0.034
D. 0.044
2. the belt speed in ftlsec
A. 63.34
B. 65.03
C. 67.37
D. 69.57
3. the angle of contact, rad.
A. 3 rad
B. 4 rad
C. 5 rad
D. 6 rad
4. the power transmitted, kw
A. 37.45
B. 40.80
5. the load at tight side
2.
v ==
3.
e
180 0 - 2 sin-t D22- 1
CD
e
180 -
4.
2sin-{1045t7°X_~=3
l
L 2(4000)
J
rad
180 0
Sw = working stress
15 00 k pa x
14.7 psi
21
.
= 7.52 PSI
101.325kpa
[e J
f8
550 hp
b t - -------'-___=_- v(Sw -12pv 2 /g) ef8 -1
12 p v
g
2
= 12(0.034)(69.57)2 = 60.836 psi
32.2
300 1 10
[ 25.4 25.4
C. 43.24
D. 48.34
'I I
r
hp = 54.7 = 40.8 KW
5.
F 1 = b t Sw
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550 hp
IL 69.573(217.52 - 60.836)
II eO 3(3)
Jl eO
3(3) -1
l
F2
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Sit [lot irmal Problems
SittlO{iOfWI
300
10
F1 = - - X - - x 217.52 ~ 1011.47 Ibs
25.4
25.4
6.
J
For hollow shaft: mH = (I -71: (Do2 ... Dj2) L w
,4
.
.
(F, - F2) V = 550 hp
(1011.47 - F2) (69.573) = 550 (54.70)
F2 = 5791bs
Do =
C. 3.42 in
D. 4.03 in
3. the volume of the hollow shaft per foot length.
3
A 53.23 in
C. 60.30 in 3
3
B. 57.74 in
D. 63.48 in 3
4. the shearing stress of hollow shaft if it is used to transmit 150 kw at 600 rpm, Mpa.
A 12.84
C. 16.23
B. 14.56
D. 17.63
For solid shaft: S = 16 T
71: d3
S=
-~4
Solving for the inner diameter:
2
Do - D,2 = 6.125
2
(4.225)2 - D, = 6.125
D, = 3.424 in = 87 mm
3.
Solving for hollow shaft volume
L = 1 ft = 12 in
71:
2
2 '\
VH = - (Do - D, ) I L
[ 4
)
.rr. ((4.225)2 - (3.424)2) (12)
4
57.74 in
3
P=2TTTN
150 = 2 TT (T)(600/60)
T = 2.387 KN-m
S=
16 TD o
Equation 2
- (·-3.5) ± A-3.5)2 - 4(1)(-3.0625) = 4.225 in = 107 mm
2(1)
2.
4.
SOLUTION:
For hollow shaft:
1
2 ) L w = -1[71:
-71: (D 2
- D)
- d2\)I L w
2 4
[ 4 0,
2
2
D,4 = (D 0 - 6.125l
D0 - D,2 = 6.125 or
Equate equation 1 and 2.
4
2
D 0 - (D 0 - 6.125)2 = 42.87 Do
2
Simplify: D 0 - 3.5 Do - 3.0625 = 0
SITUATIONAL PROBLEM 2 (ME Board April 1981)
1.
.
Equating the two equations: mH = - m,
2
A solid transmission shaft is 3.5 inches in diameter. It is desired to replace it with a
hollow shaft of the same material and same torsional strength but its weight should
only be half as much as the solid shaft. Find:
1. the outside diameter of the shaft
A. 4.23 in
C. 7.38 in
B. 5.34 in
D. 9.30 in
2. the inside diameter of the shaft
A 1.92in
B. 2.56 in
277
Problems
16 T Do
= 16 ~387)
(0.107) = 1762827 kpa = 17.63 Mpa
-t::4
~
'
.
4
4
7I:l(0.107) - (0.087) J
71: (Do - D j )
71: (Do - Dj )
16 T
Equating the stress:
--3
71: d
SITUATIONAL PROBLEM 3 (ME Board April 1981)
16 T Do
= ----4- 4
Do 4 _ DI4
A double threaded right handed worm gear transmits 15 hp at 1150 rpm. The pitch of
71: (Do - Dj )
the worm is 0.75 inches and pitch diameter of 3 inches. The pressure angle is 14 Yz 0
and the coefficient of friction is 0 12. Find:
1. the lead angle
C. 7.23°
A. 1.63°
D. 9.043°
B. 3.83°
= d3 D-o
Do4 - D,4 = (3.5) 3 D0
D0 4 _ D;4 = 42.87 Do
Equation
For solid shaft: w = ~-
or
m = Vw
V
w = density of shaft material
L = length of shaft
rn,
=
l~d2 J
L w
~
.-
~_:>'-,
;~"-
- - - ,; -.-
I
I
2. the normal pressure angle
A 10.327°
B. 12.327°
C. 14.327°
D. 16327°
3 the worm gear efficiency
A. 55.13%
B. 50.23%
C 58.34%
D 60.34%
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Sit liCIt inned Problems
F v
Powe r = -""'---"'550
Fw (15.053)
15
550
Fw = 5481bs
4. the power transmitted to gear (output)
A. 6.39 hp
C. 1027 hp
B. 8.27 hp
D.13.47hp
5. the tangential force on gear
A. 1,798 Ibs
B. 1,898 Ibs
C. 1,998 Ibs
D. 2,3981bs
7.
6. the tangential load on worm
A. 5481bs
B. 5581bs
C. 5681bs
D. 5781bs
7. the separating force
A. 4031bs
B. 453/bs
C. 501 Ibs
D. 5671bs
S
= 9.043°
2.
tan <!>n = tan <!> cos x = (tan 14.5°) (cos 9.043°) = 0.255
10.255
<!>n = tan·
= 14.33°
3.
e = tanx
l
cos<bn-t..t an x
cos o., tanx + f
j
=(0.159)l' cos14.327-0.12(0.159)
(cos14.327)(0.159) + 0.12
e =0.5513 = 55.13%
4.
5.
~-
e
0.5513
Pi
£!Q.
Po = 8.27 hp
15
1150)
. a f gear = l(1.51
v g = ve Iocrty
-~
= 237 ft / sec
12
60
Ft v g
J
Vw
cos <l>n sin x + f cos x
cOS14.37°)(sin9.043°) + (0.12)(cos9.043°)
SOLUTION:
1.
2.
= ~t
(2.37)
550
Ft = 1,898 Ibs
6.
548(sin14.327°)
A band brake is installed on a drum rotating at 250 rpm, and a diameter of 900 mm.
The angle of contact is 1.5IT radians and one end of the brake band is fastened to a
fixed pin while the other end to the brake arm 150 mm from the fixed pin. The
coefficient of friction is 0.25 and the straight brake arm is 1000 mm long and js placed
perpendicular to the diameter bisecting the angle of contact. Determine:
1. the torque transmitted.
A. 2.73 KN-m
C. 4.16 KN-m
B. 3.92 KN-m
D. 1.91 KN-m
2. the tension at the slack side of the brake drum.
A. 1.89 KN
C. 3.28 KN
B. 2.63 KN
D. 4.93 KN
3. the tension at the tight side of the brake drum.
A. 5.33 KN
C. 7.83 KN
B. 6.14 KN
D. 8.38 KN
4. the minimum force in Newtons applied at the end of the brake arm necessary to
stop the drum if 50 kw is being absorbed.
A.125N
C.175N
B. 150 N
D. 200 N
5. A steel band with a maximum tensile stress of 55 Mpa and 3.0 mm thick will be
used. What should be its width in millimeters?
A. 37 mm
C. 41 mm
B. 39 mm
D. 43 mm
Power = - 550
8.27
Fw sin o,
SITUATIONAL PROBLEM 4 (ME Board April 1981)
For double thread,
L = 2 P = 2 (0.75) = 1.5 in
Lead
1.5
tanx = ~=-0.159
IT Ow
IT (3)
x = tan' 0 159
Fs
Fs = 501 Ibs
SOLUTION:
1.
279
Sit ua t ional Problems
Power = 2 IT T N
50 = 2 IT (250/60)
T = 1.91 KN-m
T = (F 1 - Fz) r
F
0.90)
1.91 = (F 1 - F)
Z (--
2
= IT 0 N = velocity of worm
IT
(~r2.150
l12
. 60 )1=15.053 fUsec
F1 - Fz = 4.24 KN
5. = efO = e0 2 5 (1 5 1T )
F2
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r-:
3.25
1000 mm
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Si[IU~1
i()llGI
SitUG [lOTI cd Pr ob! ems
Problems
281
Om == 0, + d == 2.5 + d
F 1 == 3.25 F2
Substitute:
3.25F 2 - F2 == 4.24
F2 == 1.89 KN
l
63,000 == 8 (1.25) (825 (2.5 + dl
It d
Try: d == V2 in
63,000 == 8 (1.?5) (825) (2.5 !.Q.501
It (0.50)3
3
F1 == 3.25F2 == 3.25 (1.89) == 6.14 KN
4.
Summation of moments about the fixed end equal to zero.
F2 sin45° (150) - F (1000) == 0
F == 0.106F2 == 0.106(1.89) == 0.20 KN == 200N
63,000 == 63,000
Therefore: d == V2 inch
Om == 2.5 + 0.5 == 3 in
5.
S ==
5_ == -'i
A
5 0 = -6~
C == Om =~6v
d
0.5
tb
,
(0.003)(b)
3.
b == 0.037 m == 37 mm
400 == 825_
Y
Y == 2.0625 in
SITUATIONAL PROBLEM 5 (ME Board October 1981)
A helical compression spring has a scale of 400 Ibs/inch, an inside diameter of 2.5
inches, a free length of 8 inches and with squared and ground ends. Material is to be
chrome vanadium steel with working stress of 63,000 psi and G == 10,800,00 psi. For
a load of 825 lbs, and for average service, Whaal factor, k == 1.25, determine:
3
8FC n
Y==GT
2.0625 == )52~.3!-
(6)3 (n) _
(10,800,000)(0.5)
n == 7.8125 coils
1. the standard size wire diameter
c. 3,4 in
A. % in
D. 1 % in
B. V2 inch
2. the spring index.
A. 6
B. 7
C. 8
O. 9
3. number of active coils
A. 8.23
B. 1123
C. 534
O. 7.81
4. the solid height
A. 2.83 in
B. 3.26 in
C. 4.91 in
O. 6.34 in
5. the stress at solid height
A. 94,546 pSI
B. 95,735 psi
C. 96,735 psi
O. 97,735 psi
SOLUTION:
1.
8 K F Om
S ­­­­­­­s­
It d3
F
k == .Y
4.
For square and ground end,
Solid length == (n + 2) d == (7.8125 + 2)(0.5) == 491 in
5.
At solid height,
ys == FL ­ SL == 8 ­ 4.91 == 3.09 in
Fs == force at solid == k Ys == 400 (3.09) == 1237.6 Ibs
Ss == 8 K Fs Om = ~ (1.25) (1237.6)(~
It d3
It (0.5)3
== 94 546 psi
'
SITUATIONAL PROBLEM 6 (ME Board October 1981)
A 2 ]/2 inches double square thread with two threads/inch is to be used. The frictional
radius of the collar is 2 inches and the coefficient of friction are 0.10 for the threads
and 0.15 for the collar. The velocity of the nut is 10ft/min. For a lifting load of 4000
lbs, find:
1. the lead angle
A. 5.23°
C. 7.94°
B. 6.27°
O. 8.05°
2. the torque required to turn the screw
A. 1102.35 in­Ibs
C. 1493.65 in­Ibs
B. 1293.65 in­Ibs
O. 1693.65 in­Ibs
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282
Sitllnl iO/lClI Problems
3. the torque required to overcome collar friction
A. 1000 in-Ibs
C 1400 in-Ibs
B. 1200 in-Ibs
O. 1600 in-Ibs
4. the efficiency of the screw
A. 25.58%
B. 26.58%
C. 2765%
O. 28.58%
5. the power input of the screw
A. 1.37 hp
B. 4.38 hp
C. 7.37 hp
O. 9.37 hp
SITUATIONAL PROBLEM 7 (ME Board April 1982)
A 16 mm plate is lapped over and secured by fillet weld on the inside and outside to
form a penstock 1.5 m diameter. An allowable stress of 140 Mpa on the plate and an
allowable shearing stress of 90 Mpa on the throat side 01 the 14 mm fillet weld.
Determine
1. the internal pressure that can be carried by the plate
A. 3 Mpa
C. 7 Mpa
B. 5 Mpa
O. 9 Mpa
2. the internal pressure that can be carried by the welded joint.
C. 3.38 Mpa
A. 1.38 Mpa
B. 2.38 Mpa
O. 4.38 Mpa
SOLUTION:
1.
Lead '= 2p
Lead '= 2 (1/2) '= 1 in
For square thread:
3. the thickness of the plate if the internal pressure is 2 Mpa (neglecting welded
joint).
A. 16.67 mm
C. 14.67 mm
B. 18.67 mm
O. 12.67 mm
Om = Do - 2 p = 2.5 - 2 (1 / 2) = 2.25 in
2
2
4. the thickness of the plate if the internal pressure is 20 Mpa (neglecting welded
joint).
A. 130.17 mm
C. 170.17 mm
B. 160.17 mm
0 190.17 mm
Lead
1
tanx = - - = - - - =0.1415
IT Om
IT (2.28)
X c~
tan- 1 0.1415 = 8.05 0
WO m ~ax+f
1= (4000)(2.2_5
0.1415+0.10 = 110235 in-Ibs
2 l1 - f tan x
2
l1- 0.1 O( 0.141 5)
I
2.
T =
3.
Dc = 2 rc = 2 (2) = 4 in
Tc =
4.
e
J
fcW(r o Hi) = ~
5. Factor of safety 01 the plate il ultimate stress is 450 Mpa.
A. 3
C. 5
B. 4
O. 6
WO,,-= 0.15(4000)(4) =1200 in-Ibs
2
I ~.15li4)[_
l
2.25
-
1.
I
I
I
S =~_Oi
I
2t
140=~21­
I(
I
=0.2765
27.65%
01 0(0.1415)J
2 (0.016)
I
From Faires, Eq. b, p.507, for welded joint:
F = 2 S L b cos45°
F=~OL
But
=P(1.5)L
2
~
10x12=(1)N
N = 120 rpm
=4.38 h
P
3.
2
(1.5) L = 2 (90)(L)(0 .014) cos 45 0
2
P = 2.376 Mpa
~
I
I
F
2.
V,= L N
3~
PDL
I
P '= 2.99 Mpa
TT = T + Te = 1102.35+1200 = 2302.35in-lbs = 191.86ft-lbs
=~IT(1.86)20
P, = IT~2
33,000
F
SOLUTION:
(0.1415)[1-0.10(0.1415)J
0.1415 + 0.10 +
5.
J
11
tan x(1 - f tan x)
f 0
tan x + 1 + (__c£ )(1 - f tan x)
Om
e=
283
Sit uatioTwl Problems
S= PO,
2t
90 000 = 2.0(")i~_
.
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284
Sit liUt iOllal Problems
t = 0.01667 m
SOLUTION:
4.
S= POi
2t
1.
90000 = 20,000 ~
,
2t
1.5
Since 0.111 is greater than 0.07, then the vessel is thick wall.
lVs-
P
~ 1j=~r
2
7t-~32)
32mm
F
p
)
(0.
4
F = 1.11 KN
.Q: 1667 = 0.111
1 rs+~
Force
Pr essure = Area
1380 C~
t = 0.1667 m
Checking the ratio of tiD,.
t=.l2.
2
285
Sit lIul iorwl Problems
16.67 mm
~JjO+20,
l ~ 90,000 - 20,000
2.
.~
C = Om = Do ~d
d
d
C = 115 -13 = 7.85
13
-1j=0.19017 m
t=190.17mm
5.
FS = Su _ 450,000
S - 90,000 = 5
SITUATIONAL PROBLEM 8 (ME Board April 1982)
A safety valve spring having 9 - Y2 coils has the ends squared and ground. The
outside diameter is 115 mm and wire diameter is 13 mm. It has a free length of 203
mm. The spring must be initially compressed to hold the boiler pressure of 1.38 Mpa
on the valve seat of 32 mm diameter. Modulus of rigidity is taken as G = 80 Gpa.
Determine:
1. the force acting on the spring.
A. 0.811 KN
C. 1.45 KN
B. 1.11 KN
D.1.87KN
3.
3n
8FC
8(1.)753·~=0
y=
G d - = 80,000,000(0.013)
4.
Compressed Length = Free length - y = 203 - 31
5.
For square and ground end:
Actual no. of coil = n + 2 = 9.5
n = 7.5
SL = (n + 2)d = (95)(13) = 1235 mm
Ys = FL - SL
Ys = 203 - 123.5 = 795 mm
m=31 mm
172 mm
I
3
2. the spring index
A. 6.85
B. 7.85
3. the coil deflection.
A. 31 mm
B. 33 mm
C. 8.85
D. 9.85
C. 35 mm
D. 45 mm
4. the compressed length of spring.
A. 142 mm
C. 162 mm
B.152mm
D.172mm
5. the spring stress if it is compressed to its solid length.
A. 360 Mpa
C. 400 Mpa
B. 380 Mpa
D. 420 Mpa
8 Fs C n
Ys =-G(j8(Fs)(7.85)3 (7.5)
0.0795 = 80,000,000(0.013)
(
c
Dm = 102 rom
Do = 115 mm
)/1
1
)
Fs = 2.85 KN
4C-1 0.615 _ 4(7.85)-1 +2.: 615 =1.188
K= 4C-4"+C-- 4(7.85)-4
7.85
8 K Fs Om _ 8 (1.1_88) (2.85)(0.1 02) = 400,287 kpa = 400.29 Mpa
11:(0.013)3
s, =--;T- -
SITUATIONAL PROBLEM 9 (ME Board April 1982)
The large diameter and face of the disk of a multiple disk clutch are 255 mm and 25
mm respectively. The helical compression spring used to engage the clutch has 91/2 effective coils of 10 mm steel wire. The outer coil diameter is 80 mm. The free
length of the spring is 185 rnrn. When in place with clutch engaged, its length is 130
mm. Assuming that there are 10 pairs of friction surface in contact, that the motor
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Sitllal iOl1nl Problems
Sit uat iOT1Cl1 Problems
286
~ l' r()~_.3j\
runs at 1200 rpm, with coefficient of friction on contact surface f = 0.15, (G = 80 Gpa)
,determine:
1. the axial force available for the clutch.
A. 1.69 KN
C. 3.69 KN
D. 469 KN
B. 269 KN
2. the mean frictional radius.
A. 125.45 mm
B. 145.45 mm
C. 115.45 mm
D. 155.45 mm
3. the torque transmitted.
A. 0.192 KN-m
B. 0.292 KN-m
C. 0.392 KN-m
D. 0.492 KN-m
4. the power the clutch can transmit
A. 30.69 kw
C. 34.69 kw
B. 32.69 kw
D. 36.69 kw
SOLUTION:
251mm
_-J1 =J~'20
3.
T = n f Fa rj = 10 (0.15) (1.688) (0.11545)
4.
. ower
115.45 mm
= 0.292 KN-m
= 2 TT TN = 2 TT (0.292) (1200/60) = 3669 kw
SITUATIONAL PROBLEM 10 (ME Board April 1982)
An electric motor running at 1200 rpm drives a punch press shaft at 200 rpm by
means of a 130 mm wide and 8 mm thick belt. When the clutch is engaged the belt
slips. To correct this condition, an idler pulley was installed to increase the angle of
contact but the same belt and pulley were used. The original contact angle on the
200 mm motor pulley is 160 degrees. The original tension ratio is 2.4 and the net
tension is 12 N/mm of the belt width. If an increase in transmission capacity of 20%
will prevent slippage, determine:
1. the coefficient of friction.
C 0.513
A. 0.313
D. 0.613
B. 0.413
4. the tension at the slack side when idler is installed.
C. 0.802 KN
A. 0.452 KN
B. 0.653 KN
D. 1.290 KN
Dm = Do - d = 80 - 10 = 70 mm
y = 185 - 130 = 55 mm
C = Dm = 70 = 7
d
5. the new angle of contact.
A. 210°
B. 220
0
0
C. 230
0
D. 240
SOLUTION:
10
3
8FC n
y=---Gd
1.
3
F2
8Fa(7) (9.5)
(80,000,000) (0.01)
For the clutch:
D, = Do - 2d = 255 - 2(25)
Io = 255/2 = 127.5 mm
r, = 205/2 = 102.5 mm
Under the original installation:
5. = et8 = 2.4
.
e t8 = e l (160 )(1[/ 180 ) = 2.4
Fa = 1.688 KN
2.
r
(127.5)3
3l(127.5)2-(102.5)2
3. the tension at the slack side in the original installation.
A. 2.22 KN
C. 4.44 KN
B.3.33KN
D.1.11KN
251mm
0.055=
=.~
2. the tension at the tight side in the original installation
A. 2.37 KN
C. 2.97 KN
B. 2.67 KN
D. 3.67 KN
255mm
1.
rl' 3 ro 2 __ rI2
287
f = 0.313
2.
205mm
F I - F 2 = F = 12(130) = 1560 N = 1.56 KN
T = (F , - F2 ) r = 156 (0.200/2) = 0 156 KN-m
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288
~
F
F2
= 2.4
or
5. the axle diameter for an allowable torsional stress at 12,000 psi, the efficiency of
the bevel gear differential is 80%.
A. 4 V2 in
C. 2 V2 in
D. 3 V2 in
B. 1 V2 in
F1
F2 = 2.4
F1
- = 1.56
2.4
F1 = 2.674 KN
F1 -
3.
289
Sit uational Problems
Sill/ut ional Problems
SOLUTION:
Wheel
F1 _2.674=1.114KN
F2 = 2.4 - 2.4
4. Under new Installation, the idler should be placed on slack side so that tight side
tension remains the same, that is, F 1 = 2.674 KN
T' = 1.20T = 1.20(0.156) = 0.1872KN-m
(F 1 - F2) r = T'
(2.674 - F 2)(0.100) = 0.1872
F2 = 0.802 KN
5.
F1
f 0
-=e
F2
Wheel
F2
1.
.2.674
- - - =e 031350
0.802
400 (33,000) = 2 TT I 42~,0
0
8 = 3.841 rad 180 = 220
T = torque at wheels
T = F x r = 17,500 (48/2) = 420,000 in-Ibs
P = power at wheels = 500 (0.80) = 400 hp
P=2TTTN
0
TI
IN
12
N = 60 rpm speed of wheels
SITUATIONAL PROBLEM 11 (ME Board October 1982)
A four-wheel vehicle must developed a draw-bar pull of 17,500 Ibs. The engine,
2.
T s N s = TN
34 Ns = 51 (60)
Ns = 90 rpm speed of shaft
3.
Transmission gear ratio
which develops 500 hp and drives through a gear transmission a 34-tooth spiral bevel
pinion gear which meshes with a spiral bevel gear having 51 teeth. This gear is
keyed to the drive shaft of the 48 in diameter rear wheels of the vehicle. Determine:
1. the speed of the wheels.
A. 55 rpm
B. 60 rpm
2. the speed of shaft
A. 80 rpm
B. 85 rpm
-
Transmission gear ratio
C. 65 rpm
D. 70 rpm
I
4.
T w = torque required per wheel
Tw =
C. 90 rpm
D. 95 rpm
3. the transmission gear ratio should be used if the engine develops maximum torque
at 1500 rpm.
A. 12.67:1
C. 14.67:1
B. 13.67:1
D. 16.67:1
5.
420,000 = 210,000 in -Ibs
2
S =~6T
3
TId
12,000 = 16(2~0)
3"
nd
4. the torque required per wheel, in-Ibs.
A. 210,000
C. 230,000
B. 220,000
'J. 240,000
1500 =16.67
90
16.67:1
d = 4.47
Therefore: use 4 1/2 in
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290
Sitwtli(Jllul Problems
The smaller of two concentric helical springs is made of 22 mm diameter steel spring
and has an outside diameter of 100 mm with 10 active coils. The outer spring is
made of 32 mm steel spring and has an outer diameter of 250 mm with 8 active coils.
Before the load is applied, the outer spring is 25 mm longer than the inner spring. If
the load of 90 KN is applied to the nest of the springs. (G = 80 Gpa)
1. Find the spring rate of the inner spring.
A. 493.80 KN/m
C. 693.80 KN/m
B. 593.80 KN/m
D. 793.80 KN/m
3.
3. Find the load carried by the Inner spring
A. 49.13 KN
C. 69.13 KN
B. 59.13 KN
D. 79.13 KN
yl
1.
::Jl
Y2
90KN
~
C = Dm = Do-d = 100-22 =3.545
1
d
d
22
3
8FC n
y=--Gd
2.
£"1 = ~
Y1
= 80,000,000 (0.:-02?l = 493.80 KN / m
8 C3 n
8 (3.545)3 (10)
For outer spring:
3
8FC n
Gd
80,(.3~=1265
KN/m
8 (6.81)3(8)
F2 = 90-F 1
or
F, + F2 = 90
Y2 - Yi = 0.025
F1
F
--.l=493.8
or
Y1 = 493.8
Y1
F2
F
~
= 126.50
or
Y2 = 126.50
Y2
Substitute equation 3 and 4 to 2:
Eqn.1
Eqn.2
Eqn. 3
Eqn. 4
Eqn. 5
1 _ = 0025
90 - F1 + _F_
126.5 493.8
F1 = 69.13KN
For inner spring:
K1 =
Y=
= 6.81
_ = 0.025
126.5 4938
Substitute 1 in 5:
5. Find the percent load carried by the inner spring.
A. 70.81 %
C. 74.81 %
B. 72.81 %
D. 76.81 %
c: ==:;: .
_ Dm _ 250 - 3~
d- 32
~_+F1
4. Find the load carried by the outer spring.
A. 10.87 KN
C. 30.87 KN
B. 20.87 KN
D. 40.87 KN
25mm
C2 ~
F2 -~_=
2
K = Y; - 8 C 3 n
2. Find the spring rate of outer spring.
A. 126.50 KN/m
C. 146.50 KN/m
B. 136.50 KN/m
D. 156.50 KN/m
SOLUTION:
291
Sit u a i i(mal Problems
SITUATIONAL PROBLEM 12 (ME Board October 1982)
4.
F2 = 90- F1 = 90 - 69.13 = 20.87KN
5.
F
69.13
. d
P ercent carne = - 1 =- - = 76.81 %
F
90
SITUATIONAL PROBLEM 13 (ME Board April 1983)
A flange coupling connects two 2" diameter shafts. The flanges are fitted with 6 bolt
of SAE 1040 steel on a 7" bolt circle. The shafts runs at 300 rpm and transmits 4
hp. Assume a factor of safety of 5, ultimate tension of 70,000 psi, and ultimate shes
of 55,000 psi. Determine:
1. the torque transmitted.
A. 8,453.80 in-los
B. 9,453.80 in-lbs
C. 10,453.80 in-lbs
D. 11,453.80 in-los
2. the force applied per bolt.
A. 451.171bs
B. 430.171bs
C. 450.17 Ibs
D. 470.17 Ibs
3. the diameter of the bolt required.
C. 0.428 in
A. 0.228 in
B. 0.328 in
D. 0.528 in
4. the thickness of flange to be used.
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SitUQ t iUTI (II Problems
A. 0141 in
B. 0.241 in
Sit uat iOTiul Problems
C. 0.341 in
D. 0.541 in
293
C. 777.80 N
D. 779.80 N
A. 773.80 N
B. 775.80 N
SOLUTION:
1.
Power = 2 IT T N
45 (33,000) = 2 IT T (300)
T = 787.82 ft-Ibs = 9.453.80 in-Ibs
2.
T = F (r)
4. the width of a steel band for a tensile stress of 50 Mpa.
A. 40.40 mm
C. 46.40 mm
B. 44.40 mm
D. 48.40 mm
SOLUTION:
= F (D, 12)
9,453.80 = F (7/2)
F = 2701 Ibs total force
Fb = force applied per bolt
Fb = Fin = 2701/6 = 450.171bs
F
t~
3.
For shearing of key:
" : : " 1~
I~
s, f1, f1,
S =­­=_._=­FS
A n d2
70,000
4
450.17
5
2'. d2
--------
1.5 m
Dc = 7"
4
1.
d = 0.228 in
4
Torque = F x r
Torque = 820 (60/2) = 24,600 kg­cm = 2.413.29 N­m
For compresslvs of key:
S = f1,
2.
edt
70,000
450.17
5
0.228 t
3.
with a brake lining having a coefficient of friction of 0.25. The arc of contact is 245°.
The drum is attached to a 60 cm hoisting drum, that sustains a rope load of 820 kg.
The operating force has a moment arm of 1.50 m and the band is attached 12 cm
from the pivot point. Find:
1. the torque just required to support the load (clockwise).
A. 2,013.26 N­m
C. 2,413.26 N­m
B. 2,213.26 N­m
D. 2,613.26 N­m
2. the tension at the slack side of the brake.
A. 3321.50 N
C. 3521.50 N
B. 3421.50 N
D. 3621.50 N
0.25(2450 X
11:_)
180° = 2.912
F2
F 1 = 2.912 F2
Torque = (F, ­ F2) r
2413.26 = (F, ­ F2 ) (0.76/2)
F 1 ­ F2 = 6350.7 N
2.912F 2 ­ F2 = 6350.70
F2 = 3321.50 N
t = 0.141 in
SITUATIONAL PROBLEM 14 (ME Board November 1983)
A simple band brake has a 76 cm drum and fitted with a steel band 2/5 cm thick lined
£L = e f 8 = e
4.
F 1 = 2.912 F, = 2.912 (33215) = 9672.20 N
Summation of moments about pivot point = 0
F(150) = 9672.2 (12)
F = 773.8CJ N
S=~!1
A
bt
t = 2/5 0.4 cm
50,000,000
=
= 0.004 m
9672.20
----
b (0.004)
b = 0.0484 m = 48.40 mm
3. the operating force at the end of brake arm.
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SitIW( iOTlo[ Problems
295
Siluu.lioTWl Problems
SITUATIONAL PROBLEM 15 (ME Board November 1983)
A mechanical press is used to punch 6 holes per minute on a 25 mm thick plate. The
hole is 25 mm in diameter and the plate has an ultimate strength in shear of 420 Mpa.
The normal operating speed is 200 rpm and it slows down to 180 rpm during the
process of punching. The flywheel has a mean diameter of one meter and the rim
width is 3 times the thickness. Assume that the hub and arm account for 5% of the
rim weight concentrated at the mean diameter and density of cast iron is 7,200 kg per
cubic meter. Find:
10.31 == ---'!"- [(10472)2 - (9425)2 ]
2(9.81)
4.
W == Wr + Wah
9.71 = W r + 0.05W r
W r = 9.248 KN = 942.70 kg
Wr = volume x density
= (IT D b t) w
b = 3t
942.70 = IT (1) (3t) (t) (7200)
t == 0.118 m == 118 mm
1. the maximum force required to punch a hole.
C. 844.67 KN
A. 804.67 KN
B. 824.67 KN
D. 864.67 KN
2. the power required to drive the press.
A. 0.833 kw
C. 1.56 kw
B. 1.031 kw
D.2.13kw
b == 3t
W == 9.71 KN
5.
D=1
b = 3 t == 3 (0.118) = 0.354 m
b = 354 mm
Rim
SITUATIONAL PROBLEM 16 (ME Board November 1983)
3. the flywheel weight.
A. 3.23 KN
B. 5.39 KN
C. 7.38 KN
D. 9.71 KN
4. the thickness of the rim.
A 118 mm
B. 130 mm
C. 162 mm
D. 184 mm
A flange coupling is to connect two 57 mm shafts. The hubs of the coupling are eacl
111 mm in diameter and 92 mm thick and the flange web are 19 mm thick. Six 1(
mm bolts in a 165 mm diameter circle connect the flanges. The keyway is 6 mn
shorter than the hub's thickness and key is 14 mm x 14 mm Coupling is to transmi
45 kw at 160 rpm. For all parts, yield point value in shear is one-half the yield valu:
which is 448 Mpa. Find the following:
5. the width of the rim.
A. 324 mm
B. 334 mm
C. 344 mm
D. 354 mm
1. the force on shaft.
A. 9425 KN
B. 90.23 KN
SOLUTION:
1.
2.
Maximum Force per hole
Maximum Force per hole
Stress x Shear area == S (IT X d x t p )
420,000 (IT x 0.025 x 0.025) == 824.67 KN
·
·
1 (60)
T rrne required per hole == - - == 10 sec
6
Power == (824.67/2)(0025) == 1.031 kw
10
3. the factor of safety in bearing of key.
C. 3.20
A. 5.34
B. 4.23
'J. 2.86
5. the factor of safety in shear of bolts.
A. 6.39
C. 8.30
B. 7.39
D. 9.30
v, == IT D N1 == IT (1) (200/60) == 1047 rn/s
V2
2. the factor of safety in shear of key.
A. 1.83
C. 348
B. 2.86
D. 5.34
4. the force applied per bolt on coupling.
C. 7.39 KN
A. 543 KN
B. 6.94 KN
D. 8.39 KN
Energy
Fave t p
(Fmax /2) t p
' - == ~ - Power == - - _ . == time
time
time
3.
== IT D N2 == IT (1)(180/60) == 9425 rn/s
824.67 )
Energy = Fave t p == ( --2-. 0.025) C~ 10.31 KN - m
W
2
2
KE == -(V1 - V2 )
2g
C. 86.34 KN
D. 80.34 KN
SOLUTION:
1.
Power = 2 IT T N
45 == 2 IT T (160/60)
T = 2.69 KN-m
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296
Sit Ilalional Problems
Sit UCI t ionul Problems
r = 57/2 = 28.5 mm
r = 0.0285 m
T = F. r
F
Ss = ---- =
~ d2 ~
19 mm---71
~r
4
2.69 = F (0.0285)
F = 94.25 KN
2.
Length of key
Length of key
Length of key
De = 165 mm
For shearing of key:
£..
_
I~
---7 [
92 mm
= 78,277 Kpa = 78.28 Mpa
448
Ssu = = 224 Mpa
2
FS = Ssu = 224 = 2.86
78.28
s,
3.
14 m.m.
14 m.m.
For bearing or compressive stress in key:
F
94.25
= 156,555 kpa = 156.55 Mpa
Sc = 'h- co 0.014
- L
- ..- . (0.086)
2
Siu = 448 Mpa
FS = ~tu
4.
2
_ 448
Sc - 156.55 = 2.86
For the coupling:
T = F (Dc / 2)
2.69 = F( 0.
\
165
2
1
)
SITUATIONAL PROBLEM 17 (ME Board April 1984)
A hydraulic control for a straight motion utilizes a spherical pressure tank "A" that is
connected to a work cylinder "B" as shown in the sketch. "A" maintains pressure in
the tank at 400 psi.
1. The tank is 32 inches in diameter, welded with 100% joint efficiency and made of
steel plates with an allowable tensile strength of 7,500 psi. Calculate the required
thickness of the tank.
C. 0.54 in
A. 0.23 In
D. 0.65 in
B. 0.43 in
2. The pressure drop between tank and the cylinder is 30 psi. Assume allowance for
friction in the cylinder and packing to be 10 percent of F, the operating force.
Calculate the diameter of the piston if value of F is 5000 Ibs.
A. 1.27 in
C. 3.29 in
B. 2.39 in
D. 4.35 in
3. Determine the wall thickness of work cylinder B if it is made of cast iron having an
allowable tensile stress of 2,000 psi.
A. 0.448 in
C. 0.653 in
B. 0.560 in
D. 0.837 in
4. The piston stroke of the work cylinder is 18 inches; the time of a work stroke is 5
seconds. Calculate the horsepower output of the cylinder.
A. 1.29 hp
C. 2.73 hp
B. 4.23 hp
D. 5.34 hp
5. Assume that the work cycle of the piston rod occurs every 30 seconds, that the
overall efficiency of the hydraulic control is 80 percent, and the pump efficiency is
60 percent, determine the horsepower of the motor continuously operating the
pump.
C. 0.95 hp
A. 0.25 hp
D. 1.27 hp
B. 5.34 hp
SOLUTION:
F = 32.557 KN
F 32.557
Force per bolt =- = ' - - = 5.43 KN
n
6
5.
4
= 26,987 Kpa = 26.99 Mpa
(0.016)2
FS= 224 =8.3
26.99
92 - 6
86mm
0.086 m
~ ~
94.25
S _
s - A - wL - 0.014(0.086)
5.43
297
For shear in bolts:
448
Ssu = - - = 224 Mpa
2
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298
Si ; IIUt ;ollal Problems
Sir uat ;OT!GI Problems
S = PO,
4t
1.
~ITLJAON
7500 = 400 (32)
4t
t = 0.427 in
2
For cylinder B:
W = Force = Pressure x Area
5000 + 0.10(5000) = (400 - 30)
(~0
2
J
o = 4.35 in
3.
For cylinder B:
S=fJEL
2t
P = 400 - 30 = 370 psi
2000 = ~70(435)
2t
t = 0.402 in
t
0.402
- = - - - = 0.09 (greater than 0.07)
o 4.35
Therefore, use Lame's Equation for thick cylinder.
t=~(
!S+P
21 YS - P
-1]=~·35l"
2
\
4.
v = Sit = 18/5
Consider a 304.8 mm section of a single-riveted lap joint made up with plates of 6.35
mm thickness and 6 rivets, 15.87 mm in diameter. Assume that rivet holes and rivets,
15.87 mm in diameter. Assume that rivet holes are 1.58 mm larger in diameter than
the rivets. In this joint, the entire load is transmitted from one plate to the other by
means of the rivets. Each plate and the six rivets carries the entire load. Design
2
2.
stress for shear is 598 kq/crn'', for bearing, 1406 kg/cm and for tension, 703 kg/cm
Assume that the rivets will not tear or shear through the plate to the edge of the joints.
Find:
1. the force that can be carried by unpunched plate.
A. 13,606.40 kg
C. 17,606.40 kg
B. 15,606.40 kg
O. 19,606.40 kg
2. the force to cause tensile stress on plate between rivets.
A. 6,394.56 kg
C. 8,932.6 kg
B. 7,083.09 kg
O. 9,099.45 kg
3. the force to cause shearing stress on rivets.
A. 7,580.90 kg
C. 9,580.90 kg
B. 8,580.90 kg
O. 10,580.90 kg
4. the force to cause bearing stress on rivets.
A. 9,347.70 kg
C. 11,347.70 kg
B. 10,347.70 kg
O. 12,347.70 kg
5. the efficiency
C. 60.45%
A. 563.4%
O. 76.34%
B. 63.06%
SOLUTION:
12000+370 -1']=0.448 in
V2000 - 370
F ..
~
,)
I
.... ,
,
0
0
0
0
0
0
= 3.6 in/sec = 0.3 ft/sec
HP= Fv =?OOO (~=2.73
550
550
h
P
5.
..
Work output
Efficiency = Work input
2.73 (5)
0.80(0.60) = (Motor Hp)(30)
299
PROBLEM 18 (ME Board October 1984)
F
I
~.' .
1304.s m ... (
• F
'~
•.
65.35
mm
.....
~
....' .t:
.
I
L
t
F
•
1.
Fu = the force that can be carried by unpunched plate = S A
Fu = 703 (30.48)(0.635) = 13,606.40 kg
2.
Ft = the force to cause tensile stress on plate between rivets
F t = (703) [30.48 - 6(1.745)] (0.635) = 8,932.6 kg
3.
F s = the force to cause shearing stress on rivets.
F s = 6 (IT/4)(1745)2 (598) = 8,580.90 kg
4.
Fb = the force to cause bearing stress on rivets.
Fb = 6 (1745)(0.635)(1406) = 9,347.70 kg
5.
F = safe load = smaller force = 8,580.90 kg
Motor Hp = 0.95
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300
SU'lo ( i Oil (II Problems
For solid shaft: S = _16 T
3
TI: d
1. Torque to be delivered by the clutch in N-m.
A. 224 N-m
C. 245 N-m
B. 239 N-m
O. 265 N-m
2. Axial force on the clutch in KN. Outside and inside diameter of the clutch are 300
mm and 240 mm respectively. There are two pairs of mating surface with a
coefficient of friction of 0.30. (Use uniform wear method)
A. 2.34 KN
C. 3.25 KN
B. 2.95 KN
O. 4.39 KN
3. Torque on wheels
c. 3.56 KN-m
A. 1.234 KN
O. 4.476 KN-m
B. 256 KN
4. Draw bar pult developed in KN.
C. 17.30 KN
A. 12.57 KN
B. 15.23 KN
O. 19.39 KN
5. Speed of travel of the vehicle in km/hr.
C. 10.74 km/hr
A. 5.34 km/hr
B. 7.49 krn/hr
O. 12.03 km/hr
For hollow shaft:
SOLUTION:
SITUATIONAL PROBLEM 19 (ME Board April 1985)
A 76 mm solid shaft is to be replaced with a hollow shaft of equal torsional strength.
The outside diameter of the hollow shaft is 100 mm.
Find:
1. the inside diameter of hollow shaft
C. 90.28 mm
A. 86.55 mm
B. 88.34 mm
O. 92.34 mm
2. the percentage of weight saved
A. 50.34%
C. 56.56%
B. 53.34%
O. 65.34%
SOLUTION:
1.
301
Situa( iorwl Problems
8,580.90 ~ 6306%
Efficiency = 13,606.40
S=
16 TOo
11:(0 0 4-0,4)
Wheel
Equating the stress:
16T _
16TO o
11: d3 - 11:(0 4 -0;4)
0
16 T
_
16 T (1 00)
11: (76)3
-
~ [(100)4 - 0;4]
(100)4 - 0,4 = (76)3 (100)
0, = 86.55 mm
2.
For solid shaft: w
~
m
V
or
m = Vw
1.
Power = 2 rr T N
50 = 2 rr T (2000/60)
T = 0.239 KN-m = 239 N-m
2.
ro = 300/2 = 150 mm = 0.15 m
r, = 240/2 = 120mm = 0.12m
w = density of shaft material
L = length of shaft
rn, = l ~ (76)2 J L w = 4536.46 w L
For hollow shaft:
mH =
l~(0/-12)L
W=l~(10)2_86.5L
T= nfFa(ro
2
w=1970.64 wL
239 = 2(0.30)(Fa)(0.15+0.12)
2
Fa = 2,951 N = 2.95 KN
.
m -mH
4536.46wL-1970.64wL
Percentage of weight saved =
s
= -= 56.56%
ms
4536.46 w L
SITUATIONAL PROBLEM 20 (ME Board April 1985)
The engine of a motor vehicle with a wheel diameter of 712 mm develops 50 kw at
2,000 rpm. The combined efficiency of the differential and transmission is 75% with
an overall speed reduction of 25 is to 1. Determine:
H;2
3.
Power on wheels = 50 (0.75) = 37.50 kw
Speed of wheels = 2000/25 = 80 rpm
P = 2nTwN
375 = 2nTw(80/60)
T w = 4.476 KN-m torque on wheels
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lO.74kph
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J02
SUllO! ionul Problems
Tw,=Fxr
4.476 '= F x (0.712/2)
F = 12.57 KN draw bar pull
5.
Speed = TT 0 N = TT (0 712)(80) = 178.945 m/min
Speed = 178.945 (60/1000) = 10.74 km/hr
Lead
4
tan x = --- = - - = 0.0707
n(18)
nOm
T
3.
SITUATIONAL PROBLEM 21 (ME Board April 1985)
A single threaded trapezoidal metric thread has a pitch of 4 mm and a mean diameter
of 18 mm. It is used as a translation screw in conjunction with a collar having an
outside diameter of 37 mm and an inside diameter of 27 mm. The load is 400 kg and
coefficient of friction is 0.3 for both threads and collar. Find:
1. the collar torque
A. 1920 kg-mm
C. 1890 kg-mm
B. 1763 kg-mm
O. 2003 kg-mm
2. the torque required for thread.
A. 1293 kg-mm
B. 1192 kg-mm
C.1598kg-mm
O. 1401 kg-mm
3. the total torque
A. 1083 kg-mm
B. 2089 kg-mm
C. 3321 kg-mm
O. 4873 kg-mm
303
Situu t lonul Problems
4.
SOLUTION:
5
= 5.4~0)
(18)
2
lr(COS )7~0.(°541
l
+ 0.30 =- 1401 kg - mm
cos 14S - (0.3)(0.0707) J
TT = total torque = Te + T 5
TT = 1920 + 1401 = 3321 kg-mm
SITUATIONAL PROBLEM 22 (ME Board April 1985)
It is found that a shearing machine requires 205 joules of energy to shear a specific
gauge of sheet metal. The mean diameter of the flywheel is to be 76.20 cm. The
normal operating speed is 200 rpm, and slows down to 180 rpm during shearin9
process. The rim width IS 30.48 ern, and the weight of cast iron is 7,196.60 kg/m .
The arms and hub account 10% of the rim weight concentrated at mean diameter.
Determine:
1. the rirn weight
C. 37.23 kg
A. 33.88 kg
O. 39.45 kg
B. 35.23 kg
2. the thickness of the rim
C. 0764 cm
A. 0.145 cm
O. 0587 cm
B. 0.345 cm
SOLUTION:
1.
v, = TT 0 N, = TT (0762) (200/60) = 7.98 m/s
V2 = TT 0 N2 = TT (0762)(180/60) = 7.182 m/s
KE = 205 J = 205 N-m
W
2
2
KE = -_. (V, - V2 )
2g
205 =
30.48 em
»»>.
---'!!...._- [(798)2 - (7.182)2 ]
2(9.81)
W = 332.4 N = 33.88 kg
2.
37 27
0.3(400)(-- + -)
2 =1920kg-mm
22
1.
,)_
Tc = -fW(ro+r
-2 --- -
2.
For ACME of trapezoidal thread
I
l
T5 = W Om cos <jltan x + f]
2
cos <jl - f tan x
<tJ
D =76.20 e
W = W r + Wah
33.88 = W r + 0.10W r
W r = 30.804 kg
W r = volume x density = (TT 0 b t) w
30.804 = TT (0 762)(03048)(t)(7196.6)
t = 0.00587 m = 0587 cm
Rim
= 14.5°
SITUATIONAL PROBLEM 23 (ME Board April 1985)
A double reduction gear assembly is mounted on three parallel shafts located on the
same horizontal plane. Shaft A is driven from a 2 kw source at 3,500 rpm. The
pinion on shaft A has 24 teeth and meshes with a gear on shaft B. Another pinion on
shaft B is in mesh with a gear on shaft C with 160 teeth. The centerline distance
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:304
C = !JA +OSl+ °B2+ 0C
between shaft A and shaft C is 154 times the module. The overall gear train ratio is
14: 1. Assume module for all gears are the same, determine:
A
"--
..,I
~NA"'.I
~
24 M + TB1 M + ~ B2M + 160 M
2---
2
TB2=124-TB1 ____. __ ._ _..
Eqn. 3
Substitute Equations simultaneously:
(124 - T B1) NB = 160 (250)
124 NB -
B~O
Ns == 40,000
NB = 1000 rpm speed of shaft B
I
c
B
2-
2
154 M =
1. the speed of shaft B.
A. 1000 rpm
C. 1200 rpm
B. 1100 rpm
O. 1300 rpm
2. the number of teeth on gears at shaft B.
A. 84 & 40 teeth
C. 84 & 60 teeth
B. 84 & 50 teeth
O. 84 & 70 teeth
3. the torque on shaft A.
A. 3.28 N-m
C. 5.46 N-m
B. 4.34 N-m
O. 6.94 N-m
4. the torque on shaft B
A. 10.23 N-m
C. 16.39 N-m
B. 13.45 N-m
0.18.14N-m
5. the torque on shaft C.
A. 56.34 N-m
C. 73.23 N-m
B. 68.95 N-m
O. 7830 N-m
SOLUTION:
305
Sit llU/ iOTwl Problems
Sit u a t i{JTwl Problems
2.
TB1
24( 3500 == 84 teeth
-1000
T B2
160(250) == 40 teeth
1000
2
3.
Power = 2 TT T N
PA = 2 TT TA NA
2 = 2 TT TA (3500/60)
T A == 0.00546 KN-m = 5.46 N-m
4.
PB = 2 TT TB NB
Ps
c::::::: .'1 =Nc
e = --
PA
0.95 == PB
2
P B = 1.9 kw
1.9 = 2TTTB(1000/60)
T B == 0.01814 KN-m == 18.14 N-m
5.
e = -Pc
PB
Tb2
0.95 == £'52-
c = 154 M
1.
Nc = speed of shaft C
T A NA = T B1 NB
24 (3500) = TB1 NB = 84,000
or T
S1
M = - or 0 = MT
T
C = center distance between shaft A and C
\
=
Eqn 1
= 84,000
Ns
TB2 N B = r, N c = 160 (250) = 40,000
o
19
Pc = 1.805
2 TT TN
Pc
1.805 = 2 TT r. (250/60)
T, = 0.06895 KN-m == 68.95 N-m
Eqn.2
SITUATIONAL PROBLEM 24 (ME Board October 1985)
A key is to be desired for a 12 7 cm, shaft which will transmit power .ot 150 kw at 360
2
rpm. If the allowable shear stress for the key is 920 kgICm and the allowable
2.
compressive stress is 1200 kg/cm determine the following
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Sitl r~(
306
iorlU{ Problems
Sit u at iorw! Problems
1. Cross-sectional dimensions of the flat key to be used.
A. 1.11 em
C. 3.33 em
B. 2.22 em
D. 4.44 em
2. Force acting on a key
A. 62.66 KN
C. 66.04 KN
B. 64.34 KN
D. 68.34 KN
3. Length of key under shearing stress.
A. 7.34 em
C 4.23 em
B.6.31cm
D.2.19cm
4. Length of key under compressive stress.
A. 2.03 em
C. 4.79 em
B. 3.49 cm
D. 5.34 em
5. Safe length of key to be used
A. 2.03 em
C. 6.34 em
B. 4.79 em
D. 8.03 cm
6. Axial force to remove the hub from the shaft if the coefficient of friction is 0.45.
A. 45.23 KN
C. 58.34 KN
0 65.45 KN
B. 56.40 KN
SOLUTION:
1.
From Doughtie and Vallance, Table 5-1, p. 100
For shaft diameter of 12.7 em or 5 in,
w = 1 14 in = 3.175 em
h = 7/8 in = 2.22 em
2.
P = 2 IT TN
150 = 2 IT T (360/60)
T = 3.99 KN-m = 397.89 KN-cm
397.89 = F (12.7/2)
F = 62.66 KN
3.
S=~
•
wL
920 (0.00981) =
62.~
3.175 L
L
4.
F
5.
The safe length of key to be used should be the longer length.
Therefore choose L = 4.79 em
6.
Fa = axial force to remove hub
Fa = 2 f F
Fa = 2 (0.45)(62.66) = 56.40 KN
SITUATIONAL PROBLEM 25 (ME Board October 1985)
A pinion rotating at 1800 rpm and supported on each side by a ball bearing transmits
18 kw to a mating spur gear. If the pressure angle is 20 degrees and the pitch
diameter is 102 mm, determine the following:
Note: Add 3 degrees to the pressure angle to compensate for the friction.
1. Pitch line velocity
C. 11.02 m/sec
A. 8.10 rn/sec
D. 13.03 m/sec
B. 9.61 m/sec
2. Torque transmitted by the pinion.
A. 65.34 N-m
C. 87.23 N-m
D. 95.49 N-m
B. 7523 N-m
3. Tangential load on gears.
C. 1,872 N
A. 1,563 N
B. 1,754 N
D. 1,908 N
4. Separation load on gears.
C. 982.3 N
A. 794.6 N
D. 1,735.8 N
B. 872.3 N
5. Total load on gears.
A. 2,034 N
C. 4,034 N
D. 5,034 N
B. 3,034 N
1.
V
V
2.
= 2.19 cm
For compressive of key:
3.
TTDN
TT (0.102) (1800/60)
9.61 rn/sec
P = 2TTTN
18 = 2 TT (T) (1800/60)
T = 0.09549 :<:N-m = 95.49 N-m
Ft = tangential load
T = Ft • r
95.49 = Ft (0.102/2)
F t = 1.872 KN = 1,872 N
F
Sc='h
- L
4.
2
1200(0.00981) = E·66
2.22
--- L
2
307
4.79 em
SOLUTION:
Key
For shearing of key:
L ~
Fs = load of separation between gears
tan e =
~
Ft
e = 20 + 3 = 23°
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308
:-iiI lIul iOllul Problems
r,
5.
Situational I'rohl ern s:
F
o
= ---s
1872
= 794.6 N
tan 23
F
W (h + y) = - Y
2
h = 3 m = 300 cm
F n = total load on gears
F
cos e = ~
Fe
cos 23
o
309
~
100 (300 + 30.48)
2
(30.38)
F = 2168.50 kg
1872
= --Fn
3.
C = Dm_ = 8
d
or
Om = 8 d
F n = 2,034 N
S=8KF~r: l
A dumb waiter designed to travel a height of 3 m when loaded will have a maximum
gross weight of 100 kg. A coil spring is provided below to absorb shock in the event
that the dumb waiter should fall freely because of sudden breakage of the wire rope
carrying it. If the coil spring will be deflected to 30.48 em, if the dumb waiter should
accidentally fall and it hit the spring from its maximum height of travel, determine the
following:
1. Whaal factor
A. 1.045
C. 1.564
B. 1.863
D. 1.184
2. Maximum force acting on spring.
A. 1872.34 kg
C. 2235.40 kg
B. 2168.50 kg
D. 2634.66 kg
3. Wire diameter
A. 1.20 cm
C. 3.68 cm
B. 2.34 cm
D. 4.73 cm
4. Mean coil diameter.
A. 29.42 cm
C. 37.45 cm
B. 34.23 cm
D.41.23cm
5. Number of active coils.
A. 10.20
C. 16.34
B. 14.23
D. 18.34
Note: Use spring index of 8 and assume maximum induced stress and shear
modulus of elasticity to be 3868 kq/cm" and 808,720 kg/cm 2 , respectively.
SOLUTION:
1.
K = 4C =-1 + 0.615
4C-4
C
K = 4(8)~+
. 4(8)-4
2.
0.615 = 1.184
8
F = maximum force on spring
For impact load on spring:
3
TC d
SITUATIONAL PROBLEM 26 (ME Board October 1985)
--1
3m
3868 = 8(1.184)(2168.50)(8d)
TC d
3
d = 3.68 cm
= 8 (3.68)
4.
Om = 8 d
5.
8FC 3 n
y=---Gd
29.42 cm
3
30.48= 8(2168.50)(8) n
808,720 (3.68)
n = 10.2 active coils
SITUATIONAL PROBLEM 27 (ME Board April 1986)
The maintenance shop of PAL has a motor operated hoisting winch which consist of
the following mechanical parts: Motor shaft "A" is fitted with a double threaded left
handed worm in mesh with a 56 tooth worm wheel at shaft "B" directly below.
Compounded on shaft "8" is a 20 tooth 5 module pinion in mesh with a spur gear
keyed on shaft "C" with a horizontal center line distance of 150 mm. Mounted on the
same shaft "C" is a 200 mm diameter hoisting drum. The cable wrap around the
drum sustains a load of 500 kg at motor speed of 1750 rpm. Determine the following:
1. Pitch diameter of pinion
A. 100 mm
B.102mm
2. Pitch diameter of spur gear.
A. 180 mm
B.190mm
3. Speed of shaft C .
A. 27.34 rpm
B. 31.25 rpm
4. Power at shaft C.
A. 1.45 kw
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C.104mm
D. 108 mm
C 200 mm
D. 210 mm
C. 34.24 rpm
D. 39.45 rpm
C. 1.894 kw
lOMoARcPSD|3535879
310
Situnl iOflul t'ronl ems
B. 1.605 kw
5. Torque of shaft A.
A. 7.234 N-m
B. 9.563 N-m
6. Velocity of the load
A. 0.213 m/sec
B. 0.189 m/sec
Sit untiorwl Probl ems
D. 2083 kw
Power at shaft C
C. 11.23 N-m
D. 13.47 N-m
5.
C. 0.327 m/sec
D. 0.485 m/sec
Pc
e=PA
0.65 = 1.605
PA
PA = 2.47 kw
PA = 2 IT TA NA
2.47 = 2 IT TA (1750/60)
TA = 0.01347 KN-m = 13.47 N-m
SOLUTION:
Shaft
"A': 1750 rpm
6.
V = IT Dc Nc
V
200 mm
hoisting drum
,....A
1.
Module = ~
!
Shaft "en
•
o
T
Dp = Module x T = 5 (20) = 100 mm
2.
C = center distance
150 = 102P~
Dp +D s
2
2
Os = 200 mm
3.
Number of teeth of spur gear = 200/5 = 40 teeth
S pee d ra ti10 = No. of teeth of worm =-56 = 28
No. of thread of worm
2
1750
N B = speed of shaft B = - -- = 62.5 rpm
28
Tc Nc = TBNB
(40) Nc = 20 (62.50)
Nc = 31.25 rpm
4.
l __
311
1.605 kw
Torque at shaft C = F x r = (500 x 9.81) (0.2/2) = 490.50 N-m
Power at shaft C = 2 IT TN = 2 IT (490.50/1000)(3125/60)
= IT (0.20) (31.25) = 19.63 m/min = 0.327 m/sec
SITUATIONAL PROBLEM 28 (ME Board April 1986)
A machine shop somewhere in Quezon City fabricated a pair of spur gear 2.5 module
and to be mounted on shafts with a center line distance of 90 mm. The speed ratio
required is 3: 1. Determine the following:
1. Pitch diameter of the gear.
C. 145 mm
A. 135 mm
D. 150 mm
B. 140 mm
2. Number of teeth of the gear.
C. 54 teeth
A. 50 teeth
D. 56 teeth
B. 52 teeth
3. Circular pitch
A. 6.23 mm
B. 7.85 mm
4. Addendum distance
A. 2.50 mm
B. 3.0 mm
5. Clearance
A.0.125mm
B. 0.250 mm
6. Dedendum distance
A. 2.33 mm
B.3.125mm
7. Whole depth
A. 2.38 mm
B. 3.02 mm
8. Working depth
A. 3 mm
B. 4 mm
9. Tooth thickness
A. 3.93 mm
B. 4.34 mm
10. Space width
A. 3 mm
B. 4 mm
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C. 8.37 mm
D. 9.02 mm
C. 3.5 mm
D. 4.0 mm
C. 0.425 mm
D. 0.625 mm
C. 4.66 mm
D. 5.99 mm
C. 4.36 mm
D. 5.625 mm
C.5mm
0.6 mm
C. 5.38 mm
D. 6.34 mm
C. 5 mm
0.6 mm
lOMoARcPSD|3535879
312
Situational Problems
SitllUl iorwl Problems
11. Outside diameter of the gear
A. 120 mm
B.130mm
12. Root diameter of pinion.
A. 28.75 mm
B. 32.34 mm
0.123 in = 3.125 mm
Dedendum
7.
Whole Depth
8
Working Depth
DP = 10.16 = 0.197 in = 5 mm
2
o, +D g
9.
Tooth thickness
1.5708 _ 1.5708 == 0.1546 in = 3.93 mm
10.16
2
10.
Space Width = Backlash + Tooth thickness
0035
0.035
03
.
Bac klash = - - - . =: - - = 0.0 445 rn = 00875 mm
DP
10.16
Space width =: 0.0875 + 3.93 =: 4.02 mm
11.
Outside diameter =: Pitch diameter + 2 (addendum)
OD p =: 45 + 2(2.5) =: 50 mm
OD g =: 135 + 2(2.5) =: 140 mm
12.
Root diameter =: Pitch diameter - 2 (dedendum)
RD p =: 45 - 2(3.125) =: 38.75 mm
RD g =: 135 - 2(3.125) =: 128.75 mm
C. 17.45 mm
D. 38.75 mm
SOLUTION:
1.
1.25 _ ~
-DP-10.16
6
C 140 mm
D. 150 mm
Dp+D g
Gear
C=----
313
2.25 =0.2214in= 5.625 mm
10.16
2
2
j
-Dr -
90,= .-.-..-
180 = Dp + Dg
o, N p = o, Ng
Np
Dg
-=-=3
Ng
90mm
o,
o, = 3 o,
Dp + 3D p = 180
Dp = 45 mm
Dg = 3 (45) '= 135 mm
2.
D
M==N
SITUATlONAL PROBLEM 29 (ME Board April 1986)
A pulley is keyed to a 2 Y2 inches diameter shaft by a 5/8 x 7/16 in x 3 in flat key. The
shaft rotates at 50 rpm. The allowable shearing stress for the key is 22 ksi. The
allowable compressive stress for the key, hub and shaft are 66 ksi, 59 ksi and 72 ksi,
respectively. Determine:
2.5 = 45
Tp
T p =: 18 teeth
2.5 = 135
Tg
T g =: 54 teeth
3.
4.
5.
TC D
TC (45)
Pc = - - =--=7.85 mm
T
18
From Faires, p. 362
' h = 25.4
25.4
-=- - = 10.16
DP = diiarnetraIpitc
M
2.5
Addendum
. = 2.5 mm
1 = 00
- 1 =.-~
. 984 In
DP
10.16
Clearance
0.25 _ 0.25 = 0.0246 in = 0.625 mm
DP' 10 16
1. the torque that can be carried due to shearing stress of key.
C. 71,562.50 in-Ibs
A. 51,562.50 in-Ibs
D. 81,562.50 in-Ibs
B. 61,562.50 in-Ibs
2. the torque that can be carried due to compressive stress of key.
C. 54,11060 in-Ib
A. 50,110.60 in-Ib
B.52,110.60in-lb
D.56,110.60rn-lb
3. the torque that can be carried by the shaft.
A 57,495.10 in-Ibs
C. 77,495.10 in-Ibs
B. 67,495.10 in-Ibs
D. 87,49510 in-Ibs
4. the torque that can be carried by the hub.
C. 46,398.44 in-Ibs
A 42,398.44 in-Ibs
D. 48,398.44 in-Ibs
B. 44,398.44 in-Ibs
5. the maximum torque the pulley can safely deliver.
C. 46,398.44 in-Ibs
A. 42,398.44 in-lbs
B. 44,398.44 in-Ibs
D. 48,398.44 in-Ibs
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Situational Problems
Situa tiona[ Problems
SOLUTION:
1.
36 mm. The coefficient of friction on threads is 0.15. The friction torque on the thrust
bearing of the motor is taken as 20% of the total torque input. Determine:
For shearing of key:
F
Ss=wL
F
22000 = - - ,
(5/8)(3)
Hub
F = 41,250 Ibs
T = F (0/2) = 41,250 (2.5/2)
T = 51,562.50 in-Ibs
2.
For compressive of key:
F
=-h-- L
s,
2
66,000 = 7
1~
(3)
2
F = 43312.50 Ibs
T, = F. r = (43312.50)(2.5/2)
3.
SOLUTION:
= 54,110.60 in-Ib
1.
s, = 16 T
~
3.
It (2.5)3
Lead
25
tan x = - - = - . - - = 0.221
It Om
It (36)
x = 12.46°
Torque required to turn the screw, T,
I
l
T = 67,495.10 in-los
T = W Om cos <jJ tan x + f]
s
2
cos <jJ - f tan x
Torque carried by the hub:
F
For ACME thread: <jJ = 14.5°
I
Sc=-
~L
T = 900\0.036) cos 14.5° (0.221) + 0.15] = 6.31 kg _ m
s
cos 14,50 - 0.15(0.221)
2
l
2
F
59,000 = 7 I 16
- (3)
2
F = 38,718.75 Ibs
T = F . r = (38,718.75)(2.5/2)
5.
v = NL
10 = 400 L
L = 0.025 m = 25 mm
2.
3
22.000 =
4.
I
1. the lead
A. 25 mm
C. 35 mm
B 30 mm
D. 40 mm
2. the lead angle
A. 10.23°
C. 14.34°
B. 12.46°
D. 16.23°
3. the torque required to turn the screw.
C. 8.45 kg-m
A. 4.23 kg-m
B. 6.31 kg-m
D. 10.23 kg-m
4. the total torque input
A. 0.017327 KN-m
C. 0.057327 KN-m
B. 0.037327 KN-m
D. 0.u7428 KN-m
5. the motor power required to operate the screw.
A. 1.239 kw
C. 3.110 kw
B. 2.239 kw
D. 4.239 kw
Assuming the shaft and key are of the same material.
For the shaft:
It d
315
4.
= 48,398.44 in-Ibs
The maximum value of torque that can be safely deliver is the smaller value.
Therefore: Safe torque = 48,398.4 in-Ibs
SlTUATIONAL PROBLEM 30 (ME Board October 1986)
A double thread ACME thread screw driven by a motor at 400 rpm raises the
attached load of 900 kg at a speed of 10 m/min. The screw has a pitch diameter of
5.
T =
T =
T =
T =
torque input = Ts + Te
Ts + 0.20T
6.31 + 0.20 (6.31)
7.572 kg-m x (0.00981) = 0.07428 KN-m
Power = 2 rr TN = 2 rr (0.07428)(400/60)
Power = 3.11 kw
SITUATIONAL PROBLEM 31 (ME Board October 1987)
A gear having 60 teeth is being driven by a 12-tooth gear running at 800 rpm.
Determine the following:
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SitzuLl iOTlll1 Problems
1. the speed of driven gear.
A.150rpm
C.170rpm
B.160rpm
D.180rpm
2. The speed of driven gear if a 24-tooth idler IS placed between the driving and
driven gear.
A.150rpm
C.170rpm
B. 160 rpm
D. 180 rpm
3. The speed of driven gear if two 24-tooth idler gears were placed between the
driving and driven gears.
A. 150 rpm
C. 170 rpm
D. 180 rpm
B. 160 rpm
4. The direction of rotation of the driven gear when one idlers are used.
A. the same
C. ahead
B. opposite
D. none of these
5. The direction of rotation of the driven gear when two idlers are used.
C. reverse
A. the same
B. opposite
D. none of these
317
SitlLal iOTlul Problems
SITUATIONAL PROBLEM 32 (ME Board October 1987)
A disc clutch having an outside diameter of 32 cm and an inside diameter of 12.7 cm
is connected to an engine that turns at 750 rpm. The coefficient of friction is 0.60
while the pressure between the friction suriaces IS 2 kq/crn",
1. Force on the clutch pedal necessary to disengage the clutch disc from the engine
using uniform pressure.
A. 1,155.14 kg
C 1,355.14kg
B. 1,255.14 kg
D 1,455.14 kg
2. Mean friction radius
C. 13.87 cm
A. 11.87 cm
B. 1287 cm
D.14.87cm
3. Clutch torque
A 0.6467 KN-m B. 0.7467 KN-m C. 0.8467 KN-m D. 0.9467 KN-m
4. Power transmitted by the engine
A. 71.35 kw
C. 73.35 kw
B 72.35 kw
D. 74.35 kw
SOLUTION:
SOLUTION:
1.
1.
T1 N 1 = T2N2
12 (800) = 60 (N 2)
N2 = 160 rpm
2.
2.
3.
T, N, = T2N2 = T 3N3
12 (800) = 24 N 2 = 60 (N3)
N3 = 160 rpm
Driven
60 teeth
3.
N:J
T1 N, = T2N2 = T3N3 = T4 N4
12 (800) = 24 N2 = 24 N 3 = 60 N4
N4 = 160 rpm
12 teeth
800 rpm
4.
Refer to question 2 with one idler used:
The direction of driver is the same as driven gear.
5.
Refer to question 3 with two idler used:
The direction of driver is opposite as driven gear
Driven
60 teeth
N:J
1,355.14 kg
3
21 R - r3l
rt
3l R _r
rt
3-1 (16)3 - (6.35):l = 11.87 cm
3l (16)2 - (6.35)
T
T
torque
n f F rt = 1 (0.6) (1355.142) (11.87)
965132 kq-crn = 0.9467 KN-m
T
4.
~"icD
For uniform pressure:
R = 32/2 = 16 cm
r = 12.7/2 = 6.35 cm
2
2)
F = P TT (R - r = 2 (TT) [(16)2 - (6.35)2]
2
Power
Power
2
2TTTN
2 TT (0.9467)(750/60)
7435 kw
SITUATIONAL PROBLEM 33 (ME Board October 1987)
Two short shafts having identical diameters of 38.1 mm and rotating at 400 rpm are
connected by a flange coupling havtnq 4 bolts with a 100 mm bolt circle. The design
shearing stress of the bolt is 12 Mpa and design compressive stress of the flange is
15 Mpa.
1. What is the power transmitted by the short shaft?
A.20.50kw
C 26.50kw
D.28.50kw
B 24.50kw
2. What is the torque transmitted by the shaft?
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Siluat iOTiu/ Problems
SITUATIONAL PROBLEM 34 (ME Board April 1988)
Two shafts are connected by spur gears. The pitch radii of the gears A and Bare 207
mm and 842 mm respectively, If shaft A makes 350 rpm and is subjected to twisting
moment of 236 N-m. Pressure angle is 14.5° What is:
1. Rpm of gear B.
A. 86 rpm
B. 88 rpm
2. Torque in shaft B.
A 920 N-m
B. 940 N-m
3. Separation load of two gears.
A. 294.82 N
B. 296.82 N
4. Total load on gears.
A 1166 N
B.1177N
SOLUTION:
1.
For short shaft, from PSME Code
D3 N
P=-~
38
P = (38.1/254)3 (400) = 35.53 hp = 26.50 kw
38
p = 2rrTN
2650 = 2 IT T (400/60)
T = 0632 KN-m
2.
3.
T = F. r = F (DJ2)
0.632 = F (0.10/2)
F = 12.654 KN
F = 12,654 N
12000 =
,
4.
For compressive of bolt:
F
Sc =~
dt
150=~63
,
C. 960 N-m
D. 980 N-m
C. 298.82 N
D. 300.82 N
C.1188N
D.1199N
1.
DA N A = DBNB
(842 x 2) (350) = (207 x 2) NB
NB = 86 rpm
2.
T A = F1 . rA
0.236 = F1 (0.207)
F1 = 1.140 KN = 1140 N
TB = F1 . rB
T B = 1140 (0.842) = 960 N-m
Dc = 100 mm
3.
F
tan 8 = .2..
Ft
tan 14.5° =
Fs
~
1140
Fs = 294.82 N
3.163
It d 2
4
d = 0.01832 m = 18.32 mm = 0.72 in
Use standard d = 3,4 in (19.05 mm)
C. 90 rpm
D. 92 rpm
SOLUTION:
Fb = ~
= 12.654 = 3.163 KN
4
n
For shearing of bolt:
F
Ss =-~
It d 2
4
319
Situal ional Problems
A 0.032 KN-m
C. 0432 KN-m
B 0.232 KN-m
D. 0.632 KN-m
3. What diameter of bolt should be used?
A. \4 in
C. 3,4 in
B. Y2 in
D. 1 in
4. How thick should the flange be in mrn?
A 7.07 mm
C. 9.07 mm
B. 5.07 mm
D. 11.07 mm
4.
cos 8 =
.5..
Fn
cos 14.5° =
~O
IE,f\ )1< ~
207 mm .:»
)1
' - 842 mm
Fn
F n = 1177 N
(0.01905) t
t = 0.01107 m = 11.07 mm
SITUATIONAL PROBLEM 35 (ME Board April 1988)
A rectangular key was used in a pulley connected to a line shaft with a power of 746
KW at a speed of 1200 rpm. If the shearing stress of the shaft and key are 30 Mpa
and 240 Mpa, respectively.
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Si t u at i o n a! Problems
1. What is the diameter of the shaft?
A. 1860 mm
C. 26.60 mm
B 21.60 mm
D 2960 mm
2. What is the length of the rectangular key if the width is one fourth that of the shaft
diameter?
A. 324 mm
C. 524 mm
B. 4.24 mm
D 624 mm
SOLUTION:
1.
Fl
p = 2rrTN
746 = 2 rr T (1200/60)
T = 00593 KN-m
16 T
S = ---nd 3
30,000 = ~ 6 (0. 05931
nd 3
d = 0.0216 m = 21.60 mm
2.
T = F. r
0.059365 = F (0.0216/2)
F = 5497 KN
02 16
= 0.0054 m
w = -~ = .2:
4
S
S
2.
F 1 - F2 = F = 2.505 KN = 2,505 N
3.
0.35(144 a x-"- )
F
_J..=e f 8=e
180' =2410
F2
r
F
F2 = - -1
241
F
F1 - - 1- - = 2.505
241
F1 = 4.282 KN
T
S = F1
A
4
2100 =_ 4 2 8 ~
,
b(0.006)
=_£...
w L
b = 0.3398 m = 340 mm
5.497
240,000 = - - - -..
(0.0054) L
SITUATIONAL PROBLEM 37 (ME Board April 1989)
A steel shaft transmits 40 hp at 1400 rpm. Considering allowable shearing stress
based on pure torsion to be 5000 psi, find:
L = 0.00424 m = 4.24 mm
SITUATIONAL PROBLEM 36 (ME Board April 1(88)
A pulley 610 mm in diameter transmits 40 kw at 500 rpm. The arc of contact between
the belt and pulley is 144 degrees, the coefficient of friction between belt and pulley is
0.35 and the safe working stress of the bel' is 2.1 Mpa. It is required to find:
1. The tangential force at the rim of the pulley in Newtons.
A. 2,405 N
C. 2,605 N
B. 2,505 N
D. 2,705 N
2. The effective belt pull in Newton.
A. 2,305 N
C. 2,505 N
B. 2,405 N
D. 2,605 N
3. The width of the belt used if its thickness is 6 mm.
A. 340 mm
C. 360 mm
B. 250 mm
D 370 mm
P = 2rrTN
40 = 2 rr T (500/60)
T = 0.763 KN-m
T = F . r
0.763 = F (0.610/2)
F = 2.505 KN = 2.505 N
SOLUTION:
1.
321
Situational Problems
1. the shaft diameter of nearest commercial size.
A.11/4in
C.27/16in
B. 1 7/16 in
D. 1 3/4 in
2. the torsional deflection of the shaft in degrees per foot
A. 0246 deg
C. 0266 deg
B. 0.256 deg
D. 0276 deg
SOLUTION:
1.
P = 2rrTN
40 (33,000) = 2 rr T (1400)
T = 150 ft-Ibs = 1800 in-Ibs
_ ~6T_
S
s
-
TC d3
5000 = ] 6 (1800)
7( D"
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2.
323
Sit u a; ional Problems
Sit u at iorlUl Problems
D = 1.224 in
Therefore use D = 1 7/16 in = 1A375 in (standard)
D + d == OAO
D = OAO - d
e=~
Equate 1 and 2:
2]
5.305 = 314[d(OA - d) - d
2
628 d - 125.6 d + 5.305 = 0
J G
L = 1 ft = 12 in
4
4
J =~
= 1t (1A375) . = 0.4192 in4
32
32
6
G = 12 x 10 for steel
d =
_ _ _ _ _ _ _ _ _ _ Eqn.2
125.6 ± J(=.125.6)2 - 4(628)(5:305)
- - = 0.139 m = 139 mm
2(628)
D = OA - 0.139 == 0.261 m = 261 mm
.
D- d
261 - 139
Face Width = - - - =
= 61 mm
2
8 = .. __1800 (12)
OA192 (12,000,000) = 0.00429 rad = 0246 deg
2
SITUATIONAL PROBLEM 39 (ME Board October 1989)
SITUATIONAL PROBLEM 38 (ME Board October 1989)
A Pajero jeep wagon's engine develops 40 kw at 1200 rpm, with a single plate clutch
with two pairs of friction surfaces transmitting the power. Consider the coefficient of
friction to be 0.30 and mean diameter of disc to be 200 mm, determine
1. Axial force required to engage the clutch and transmit the power.
A. 5.305 KN
C. 5.505 KN
B. 5A05 KN
D. 5.605 KN
2. Inside and outside diameter of the discs and the face width if the maximum
pressure is limited to 200 Kpa.
A. 61 mm
C. 65 mm
B. 63 mm
D. 67 mm
SOLUTION:
1.
Fa
S = PD j
2 t
200mm ID
2.
5305
D-d ,'\
2 IT (200) ( -d I -
5.305
314 (Dd - d
Dj = !?+d
2
0.20 ==
d:+-~
2
2)
2)
-
.
Eqn. 1
r
0 m
i
3
Volume = IT h = TT (6/2)2 (6) == 169.646 m
3/s
Q == 169.646/2 == 84.646 m%r = 0.0235 m
P = w Q h == 9.81 (0.0235) (10 + 6 + 5) == 4.854 kw
4.854
Input power == - - - - - = 8.16 kw
(0.70)(0.85)
SITUATIONAL PROBLEM 40 (ME Board June 1990)
A 48 in diameter diamond saw blade is mounted on a pulley driven steel shaft,
requiring a blade perimeter linear speed of 150 ft/s. Motor drive is 125 hp at 1200
rpm, with 6 inches diameter pulley, Determine:
1. The shaft rpm to attain blade peripheral speed required .
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...........".._-,,.
1
40000 = 58.86 (6)
,
2t
t = 0.0044 m = 4AO mm
2 IT Pmax r, (ro - r,)
\2
1~
SOLUTION:
1.
P == w h
P = 9,81 (6) == 58.86 kpa
P = 2ITTN
40 = 2 IT T (1200/60)
T = 0.31831 KN-m
rj = 200/2 = 100 mm = 0.10 m
T = n f Fa rf
0.31831 = 2 (0.30) (F a)(0.1 0)
Fa = 5.305 KN
2.
With the present water interruptions prevailing in Metro Manila, you have been asked
to design an upright cylindrical water tank 6 meter in diameter and 6 m high, vented,
and to be filled completely with water.
1. Determine the minimum thickness of the tank plate if the stress is limited to
40 Mpa.
C. 4AO mm
A. 2.20 mm
B. 3.30 mm
D. 5.50 mm
2. If the water tank had to be elevated 10 meters from its bottom and has to be filled
up in 2 hours, determine the pump capacity needed to fill the tank and motor
power to drive the pump, if velocity and head loss is 5 meters. Consider pump
efficiency of 70% and motor efficiency of 85%,
A. 5. 16 kw
C. 7. 16 kw
B. 6.16 kw
D. 8.16 kw
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Sit uat u nu i] ['rob/ems
A. 71220 rpm
B. 71420 rpm
2. The shaft diameter.
A. 2.10 in
B. 2.20 in
3. The shaft pulley diameter.
A. 5.053 in
B. 10.053 in
2.
SITlJATIONAL PROBLEM 42 (ME Board June 1990)
A flange coupling has an outside diameter of 200 mm and connects two 40 mm
shafts. There are four 16 mm bolts on a 140 mm bolt circle. The radial flange
thickness is 20 mm. If the torsional stress in the shaft is not to exceed 26 Mpa,
determine'
1. The power that can be transmitted at 600 rpm.
A. 17.12 kw
C. 24.23 kw
B. 20.53 kw
D. 32.23 kw
C. 230 in
D. 2.40 in
C 15.053 in
D 20.053 in
SOLUTION:
1.
V=nDN
150 (60) = rr (48/12) Ns
Ns = 716.20 rpm
2. The shearing stress in the bolts if uniformly distributed.
A. 1.29 Mpa
C. 3.04 Mpa
B. 2.34 Mpa
D. 5.804 Mpa
1200 rpm
125bp~
For line shaft:
3. The maximum shearing stress induced in the bolts.
A. 111.65 Mpa
C. 115.65 Mpa
B. 113.65 Mpa
D. 117.65 Mpa
3
D N
p=-53.5
125 = Q~716.20
3.
11 I'
53.5
D = 2.10 in
o, Nm = o, Ns
6 (1200) = D s (716.20)
D, = 10.053 in
325
Situa! iOrl a I I'robl ems
C. 716.20 rpm
D 71820 rpm
Is
t10
in
4. The bearing pressure in the bolts.
A. 2.34 Mpa
C. 3.65 Mpa
B. 1.03 Mpa
D. 5.34 Mpa
,
SOLUTION:
1~20m
.-:="'~
SITUATIONAL PROOLEM 41 (ME Board June 1990)
In the LRT II project, steel railroad rails of 10m long are to be installed. If lowest
temperature considered is 16°C, and a maximum temperature of 36°C is designed
for, assuming coefficient of thermal expansion of steel to be 11.6 x 10.6 rn/rn- °c and
modulus of elasticity of steel to be 207,000 Mpa,
1. Determine the clearance between rails such that the adjoining rail will just touched
at maximum design temperature
A. 1.23 mm
C. 2.32 mm
B. 1.56 mm
D. 3.23 mm
2. Should the excessive high summer temperature of 40°C occur, determine the
induced stress in the rails at that temperature.
A. 8.34 Mpa
C. 10.56 Mpa
B. 9.605 Mpa
D. 12.34 Mpa
200m
1.
S = 16 T
n D3
26.000 =
16 T
n (0040)3
T = 0.327 KN-m
p = 2 rr TN = 2 rr (0.327) (600/60) = 2053 KW
SOLUTION:
y/2
}
y'~
.·Tl~ p
y/2
1rr
1.
Clearance
Clearance
Clearance
Clearance
2
Stress = k E (b - t1)
6
Stress = 11.6 x 10. (207,000) (40 - 36) = 9605 Mpa
y/2 +
lei mm
k L (t2 - t1)
I c I
6
11.6 X 10. (10) (36 - 16)
0.00232 m = 2.32 mm
10 mm
2.
10 mm
I
Torque = F. r = F (D, /2)
0.327 = F (0.140/2)
F = 4.67 KN
F 4.67
Fb = force per bolt = - = ---- = 1.167 KN
n
4
Fb
1.167
Ss = . - - = - - - - - = 5804 kpa = 5.804 Mpa
rt (0.016)2
n d2
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4
De = 140 mm
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3.
327
Sit uat ional Probl ems
Situational Problems
C. 18.23%
D. 16.29%
A. 12.39%
B. 14.39%
The maximum induced shearing stress in the bolt is caused by the direct
shear and initial tension due to tightening. From Vallance p. 134.
SOLUTION:
F, = initial tension due to tightening = 16,000 d
Where d = diameter of bolt, in
F, = 16,000 (16/25.4) = 10,078.74Ibs = 44.84 KN
SI = tensile stress due to initial tension
44.84
SI =
= 233,000 Kpa = 233 Mpa
n (0.016)2
Sft
S=~_F
1.
~D
A
6000 = ~20l
Ssmax =-~St
4.
2
~2-
1
I
2
+4S s =-.y(223) +4(5.804)
2
2
4
=111.65 Mpa
2
r
2' 0 2
4
1
4
r
Dr = 2.53 in
Therefore use Dr = 2.75 in (standard)
Bearing pressure or compressive stress in bolts:
S
=~
edt
1167
=3.65 M a
(0.016) (0.020)
P
2.
SITUATIONAL PROBLEM 42 (ME Board June 1990)
A square single thread jackscrew has 2 threads per inch. It is to lift 15 tons. The
friction radius of the collar is 1 inch. The coefficient of friction between the treads of
the screw and base is 0.15, that between the screw and collar is 0.13. Determine:
1. The diameter of the root of the screw if the allowable stress in compression is
6,000 psi.
C. 2.75 in
A. 0.45 in
D. 3.85 in
B. 1.50 in
6. The power input of the screw.
A. 10.28 hp
B. 12.46 hp
Lead
P = 050 in (for single thread)
no. of thread per inch
2
Ts
2
W Om tan_~
lJ = ~O.il
2
l1-ftanx
r_~?53
I
2
l
~
l1- 0.15(0.053)J
T s = 9,210.60 in-Ibs
= fe W rc = 0.13 (30,000)(1) = 3,900 in-Ibs
T = total torque = Ts + T, = 9,210.60 + 3,900
T = 13,110.60 in-lbs
T = F x Length of arm
13,110.60 = P (5x12)
P = 218.51 Ibs
3.
Te
4.
Twisting moment or torque at the root of the threads
T, = 9,210.60 in-lbs
5.
V = L N = (0.5/12) (60) = 2.5 ft/min
Po = _W V = ~?.90)(25
33,000
C. 14.93 hp
D. 16.93 hp
6.
7. The efficiency of the screw and collar.
2
tan x = Lead. = 0 )0 = 0.053
n Om n (3)
4. The twisting moment exerted at the root of the threads.
A. 9,210.60 m-lbs
C. 9304.56 in-lbs
B. 9563.40 in-Ibs
D. 9873.45 in-Ibs
5. The power output of the screw if it is turning at 60 rpm.
C. 3.04 hp
A. 1.87 hp
B. 4.20 hp
D. 2.27 hp
___ ._
Om = Dr + £'Itch = 2.75,. 0.50 = 30 in
2. The torque required to turn the screw.
A. 6826.45 in-lbs
C. 8734.57 in-Ibs
B. 7834.40 in-Ibs
D. 9,210.60 in-Ibs
3. The pull required at the end of a 5 ft bar raises the load.
A. 218.51 Ibs
C. 265.301bs
B. 245.231bs
D. 287.451bs
1
= - = 0
.50,In
Pitch
P,
~.TN
33,000
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= 2.2727 h
33,000
P
= 2n(218.15x5)(60) _ 12.46 h
33,000
P
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7.
e=
PPo -__~27 12.46
=c
329
Situutiona[ Problems
SituatioTlul Problems
663,595.37 =
18.23%
F
-n ~ 2n
(0.175)
2
I
F = 1633.36 KN
SITUATIONAL PROBLEM 44 (ME Board October 1991)
A 1200 mm cast iron pulley is fastened to a 112.50 mm shaft by means of a 28.13
mm square key 175 mm long. The key and shaft are SAE 1030 steel annealed with
allowable shearing of key of 29.400 psi and compressive of 6765.3 kg/cm 2 .
1. Force required to shear the key
A. 997.59 KN
C. 756.03 KN
B. 873.45 KN
D. 652.34 KN
2. Force needed to compress the key.
C. 1633.36 KN
A. 1431.93 KN
B. 1596.24 KN
D. 1763.48 KN
3.
Therefore, the safe force to operate the key is 997.59 KN (minimum force)
T = F. r = 997.59 (0.1125/2) = 56.1144 KN-m
P = 2rrTN = 2rr(56.114)(600/60) = 3525.77KW
4.
T=T'
F.r = F.R
997.59 (112.5/2)
F = 93.52 KN
5.
3. Safe power of the shaft at 600 rpm.
A. 2083.54 'r<:VV
C. 4823.49 'r<:VV
B. 3525.77 'r<:VV
D. 5623.74 KW
5. What force acting at the pulley rim will crush the key?
A. 143.37 KN
C. 165.39 KN
B.153.13KN
D.173.29KN
1.
--
---..
For shearing of key:
F
Ss=' w L
T=T'
F.r = F.R
1633.36(112.5/2) = F(1200/2)
F = 153.13 KN
SITUATIONAL PROBLEM 45 (ME Board October 1992)
A 76 mm bearing using oil with an absolute Viscosity of 0.70 poise running at 500 rpm
2
The bearing
gives satisfactory operation with a bearing pressure of 14 kg/cm
clearance is 0.127 mm.
1. Determine the unit pressure at which the bearing should operate if the speed is
changed to 600 rpm.
2
2
A. 168 kg/cm
C 20.30 kg/cm
2
2
B. 18.30 kg/cm
D. 22.04 kg/cm
2. If the bearing is given a total clearance of 0076 mm, what change should be made
in the oil?
C 0.364 poise
A. 0.176 poise
D. 0.472 poise
B. 0.251 poise
4. What force acting at the pulley rim will shear this key?
A. 93.52 KN
C. 75.23 KN
B. 85.27 KN
D. 64.23 KN
SOLUTION:
= F (1200/2)
r = 112.5 mm
SOLUTION:
1.
The bearing characteristic number (Sommerfield number) will be equated for
both conditions.
2
S
S, = 29,400 (101.325/14.7) = 202,650 kpa
F
202,650=(0.02813)(0.175)
F = 997.59 KN
2.
For compressive stress of key:
F
S C =-h
L
2
Sc = 67653 (101.325/1.033) = 663,595.37 Kpa
_ u n, ( 01
- p'lcdj
When speed is changed to 600 rpm:
S1 = S2
~0.7)(591
2
1 _ (0.7)6~
14
0.127 )
P2 = 16.8 kg/cm
2
-
2
P2
l.
76. 1
0.127
2
When the total clearance is changed to 0.076:
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j
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Sit u a t i onri! Problems
Sit uci! ional Problems
S1 = S2
iQ2~l(
(' - 76_ J2 = ~2)°l
14
0,127
.
16,8
331
1. How much energy does the wheel loss?
A. 120,763 N-m
C. 130,458 N-m
B.125,159N-m
D.145,609N-m
'2
r _?6 - j'
0,076
1J2 = 0251 poise
SITUATIONAL PROBLEM 46 (ME Board October 1992)
A Mitsubishi car transmission has two spur gears in mesh having a velocity ratio of
1.4. The number of teeth in the driven gear is 36 and the diametral pitch of 3.
1. Determine the number of teeth in the driver.
A. 22 teeth
C. 30 teeth
B. 26 teeth
D. 34 teeth
2. What moment opposes the motion of the flywheel?
C. 75.45 N-m
A. 65.34 N-m
D. 81.63 N-m
B. 7034 N-m
SOLUTION:
2. Determine the distance between centers.
A. 6.23 in
C. 10.34 in
B. 8.30 in
D. 12.09 In
3. What is the circular pitch?
A. 0.456 in
B. 0.847 in
Journal
Bearing
C. 128 in
D. 1.047 in
SOLUTION:
1.
1.
T 1 N 1 = T2 N2
T2
T2 = 36 teeth
j
=T{~
36 = T 1 (1.4)
T 1 = 25.7 say 26 teeth
2.
T
DP =--
or
o
0 = ~
Driven
Driver
T
DP
c
6.KE = 91 0 (1~524
26
.
O 1 = - = 8.67 In
3
36
.
02 = - = 12 In
3
C = center distance
C =
3.
~
From Faires, p. 535
m k2
2
2
,'I,KE = - - (W1
w2)
2
where: m = 910 kg
k radius of gyration = 1524 mm = 1.524 m
120 (2rr)
W1 = ------.-- = 4 rr rad
60
60 (2rr)
w1 =---=2rr rad
60
2.
)2 [( 4rr)2 _ (2rr)2]= 125,159 N _ m
Opposing moment = Tangential force x Radius = (f W) r
910 (0.06) (0.3048/2)
= 8.321 kg-m = 81.63 N-m
+ O2 = 8.67 + 12 = 1034 in
2
2
rr 0 _ rr (1 2) = 1.047 in
Pc = T - 36
SITUATIONAL PROBLEM 47 (ME Board October 1992)
A flywheel weighing 910 kg has a radius of gyration of 1524 mm. The shaft journals
are 304.8 mm in diameter and have a coefficient of friction of 006 After the wheel
reaches 120 rpm, the driving force is withdrawn and the wheel slows to 60 rpm.
SITUATIONAL PROBLEM 48 (ME Board April 1993)
A 2 Y2 in (8.9 cm) diameter shaft is driven at 3600 rpm by a 400 Hp (298.3 KW)
motor. The shaft drives a 48 in. (121.9 ern) diameter chain sprocket having an output
efficiency of 85%. Determine:
1. Torque in the shaft.
A. 7,003 in-Ibs
C. 7,563 in-Ibs
B 7,197 in-Ibs
D. 8,408In-lbs
2. The output force on the sprocket.
A, 291.79 Ibs
C. 327.451bs
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lOMoARcPSD|3535879
332
SUuol iOT!ul Problems
B. 302.301bs
D. 367.451bs
3. The power delivered by the sprocket.
A. 320 hp
C. 340 hp
B. 330 hp
D. 350 hp
SOLUTION:
1.
P=2nTN
400 (33,000) = 2 rr T (3600)
T = 583.57 ft-Ibs = 7,003 in-Ibs
2.
Torque = F. r
7003 = F. (48/2)
F = 291.79 Ibs
3.
Power delivered = 400 x 0.85 = 340 hp
ALCORCO~
ENGINEERING REVIEW CENTER
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TO OUR BOARD TOPNOTCHERS AND
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National Passing: 48%
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National Passing: 47%
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NATIONAL PASSING: 45%
October, 2001 ME Board Examination
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NATIONAL PASSING = 43%
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ALCORCON PASSING = 83%