Solution of Machine Design Problem #M1c: A 1”-12 UNF steel bolt of SAE grade 5 is under direct shear loading. he coefficient coefficient of friction bet!een "ating "ating surfaces is #.$. he bolt is tightened tightened to its full %roof strength. strength. ensile area is #.&&' #.&&' in 2 (roof strength is )5 *%si+ and ,ield , ield strength is 2 *%si a /hat shear force !ould !ould the friction friction carr,0 carr,0 b /hat shear load can the bolt bolt !ithstand !o ,ielding ,ielding if the friction friction bet!een cla"%ed "e"bers "e"bers is co"%letel, lost0 ase the calculation on the thread root area. Data for this bolt3 At4#.&&' in2 S %4)5### %si S,42### %si he initial load in the bolt is3 Fi4AtS %4.&&')5###45&'55 lbs he friction force Ff 4µFi 4 #.$5&'554225$2 lbs he direct shear stress force !o ,ielding is3 Fs 4 2AtS,s42At#.5)S,42#.&&'#.5)2###46#6$# lbs 7f the bolt is sub8ect to shear in the shan* area+ then use the larger shan* area. Problem #M3
Assu"ing a rigid brac*et+ the to% bolts elongate or strain ' ti"es "ore than the lo!er bolts. a*ing a*ing "o"ents about the %i9ot %oint !e get3 F @ L
=
F T D + F B d
5$## $# F T
=
=
1&2##
F T 12 +
F B '
$
lbs
he load %er bolt is half of this a"ount or )1## lbs. 7ncor%orating a factor of safet, of $+ the design load %er bolt is $)1##4'2$## lbs. he re:uired tensile area for each bolt is3 At
=
F S p
=
'2$## 12####
=
#.26
in 2
he a%%ro%riate bolt is a ;rade ) < = >1# UN?
Problem #M4
B
he bolts are ”-1'UN?. he distance bet!een bolts is 1.25”. he load is 26## lbs and B4)”. Find the shear stress on each bolt. he torsional shear force balancing the force cou%le is3 F 1
PL
=
26###) 21.25
=
2d
=
)&$#
lbs
he direct shear stress F 2
=
F
=
'
-##
lbs
he total force is3 2
2
F = F 1 + F 2
=
)&)&
lbs
he shear stress in each bolt using the shan* dia"eter is3 τ =
F A
=
)&)&
π $
.5
2
=
$$2'6
psi
Exercise # M1 :
A pair of spur gears with a pitch of 6 are in mesh. The pinion has 18 teeth and rotates at 1800 rpm transmitting 0.5 Hp. Both gears have 20-degree pressure anges. The num!er of teeth on the gear is "6. #etermine the radia and tangentia forces on the pinion.
F t
=
V =
''###hp V π 1)##'
=
12
and V = 1$1'.6
nd
π
12
and F t
=
''###.5
=
1$1'.6
The radia force is$ F r = F t tan2# = 11.6 @ tan 2# = $.25
11.6
lbs
lbs
Exercise # M2:
A pair of heica gears transmit 15 %& power and the pinion is rotating at 1000 rpm. The hei' ange is 0.50 radians and the norma pressure ange is 0."5 radians. The pitch diameter of the pinion is (0 mm and the pitch diameter of the gear is 210 mm. #etermine the tangentia) radia) and a'ia forces !etween the gear teeth. *Answers$ +0,2) 1(02) 22"6 ewtons
he tangential load is deter"ined fro" the %o!er relationshi%. he relationshi% in S7 units uses the dri9ing tor:ue3 -5$-W = -5$-15 = 1$' .2' N − m T = n 1### !here the %o!er is in C!. he tangential load creating this tor:ue is3
F t =
T d
2
1$'.2'
=
#.#'5
= $#-2 N
he radial and aial forces are3 F r = F t tanφ = $#-2 tan#.'-$ = 16#2 N F a
= F t tanψ = $#-2 tan.5 = 22'5 N
Note that3 tanφ =
tanφ n cosψ
=
tan#.'5 cos#.5
= .$1&
φ = #.'-$
rad
Exercise #M33 A %air of straight-tooth be9el gears as sho!n in the figure abo9e are in "esh trans"itting '5 h% at 1### r%" %inion s%eed. he gear rotates at $## r%". he gear s,ste" has a %itch of & and a 2#-degree %ressure angle. he face !idth is 2 inches
and the %inion has '& teeth. Deter"ine the tangential+ radial+ and aial forces acting on the %inion. Ans!ers )' lbs+ 2)' lbs+ 11' lbs.
d da9g γ
b Fro" the gear geo"etric infor"ation3 d p
=
N p
=
P
'&
=
&
& in
and d g
=
15 in
he %itch cone angle can be obtained fro" the ratio of %itch dia"eters. γ p
= tan −1
d p d g
= tan −1
& 15
= #.$ ⇒ γ p
= 21.) deg .
Fro" geo"etr,+ the a9erage dia"eter of the %inion is3
= d p − b sinγ p = & − 2 sin 21.) = 5.2& in
d p + avg
he %inions a9erage %itch-line 9elocit, is3 V avg =
dn
π
5.2&1###
π
=
12
=
12
1'66
ft "in
No! !e can find the tangential force 3 F t
=
''###W V avg
=
''###'5. 1'66
=
)'- lbs
he aial and radial forces are3 F r = F t tanφ tanγ = )'- tan2# cos21.) = 2)' lbs F a
= F t tanφ sinγ = )'- tan2# sin21.) = 11'
lbs
Exercise #M4: A !or" gear reducer is dri9en b, a 12## r%" "otor. he !or" has ' threads and the gear has $5 teeth. he circular %itch of the gear is %4#.5”+ the center distance is $.5 inches+ the nor"al %ressure angle is 2# degrees+ and face-!idth of the gear is b41 inch. Use a coefficient of sliding friction of #.#2. Deter"ine3 a ;ear and !or" dia"eters+ and !or" Bead. 6.1&+ 1.)$+ 1.5 in b he !or" gear efficienc, )).& c 7s the unit self-loc*ing No
he gear %itch dia"eter can be found fro" the nu"ber of teeth and its circular %itch3
N g p
$5.5
d g
=
d w
= 2c − d g = 2 $.5 − 6.1& = 1.)$
=
=
6.1&
π π he %itch dia"eter of the !or" can be obtained fro" the center distance and the %itch dia"eter of the gear3 in
he lead is the nu"ber of threads in the !or" ti"es the !or"s aial %itch !hich is the sa"e as the gears circular %itch3 L
= pa N W = #.5' = 1.5
in
he gear efficienc, can be obtained *no!ing the sliding friction as !ell as the n or"al %ressure angle and the lead angle of the !or". he lead angle of the !or" is3 tanλ =
L
1.5
=
π d w
=
π 1.)$
λ = 1$.5
#.2& ⇒
deg
he efficienc,3 e=
cosφ n − f tanλ
=
cosφ n + f cotλ
cos2# − #.#2- tan1$.5
=
cos2# + #.#2- cot1$.5
#.))&
e = )).&F
he gear set is o9erhauling re9ersible because the friction coefficient f4#.#2 is less than that re:uired for loc*ing !hich is3 cosφ n tanλ = cos 2# cos1$ .5 = #.2$ Problem M #73 A 1#”-!ide flat belt is used !ith a dri9ing %u lle, of dia"eter 1&” and a dri9en %ulle, of ri" dia"eter '&” in an o%en configuration. he center distance bet!een the t!o %ulle,s is 15 feet. he friction coefficient bet!een the belt and the %ulle, is #.)#. he belt s%eed is re:uired to be '&## ft"in. he belts are initiall, tensioned to 5$$ lbs. Deter"ine the follo!ing. ans!ers are in %arentheses3 elt engage"ent angle on the s"aller %ulle, '.#' radians. Force in belt in the tight side 8ust before sli%%age. 1### lbs. Mai"u" trans"itted G%. .$ h% he engage"ent angle is3 θ d
D − d
= π − 2 sin −1
2C
−
= π − 2 sin 1
'& − 1&
21512
= '.#'
rad
At the 9erge of sli%%age3 F 1 F 2
=
e
µθ d
=
e
#.) '.#'
⇒
F 1 = 11.'F 2
Also3 F 1 + F 2
=
2 F i = 25$$
he trans"itted G% is3 F − F 2 V HP = 1 ''###
Problem #M8
=
⇒
F 1 = 1###
lbs
1### − ))'&## ''###
=
and F 2
--.5 hp
=
)) lbs
;i9en3 A "ulti-%late dis* clutch d4#.5” D4&” ("a41## %si ?oefficient of friction4#.1 (o!er trans"itted4 15 h% at 15## r%" Find3 Nu"ber of friction surfaces He:uired or:ue ?a%acit, for unifor" !ear T =
&'###hp
&'###15
=
n
=
15##
&'# lb − in
or:ue ca%acit, %er surface T =
fpa d
π
)
D
2
−
2
d =
.11##.5
π
)
&
2
−
2
#.5 = 6#.2 in − lb
he re:uired nu"ber of surfaces is &'#6#.2 !hich is a%%roi"atel, surfaces. For unifor" %ressure bra*e %ads !hen the, are ne!3 T =
fpa
π
12
D
'
−
'
d =
.11##
π
12
'
&
−
'
.5 = 5&5 in − lb
Inl, t!o surfaces are needed for constant %ressure %ads. Problem #M9: A bra*e !ith bra*ing tor:ue ca%acit, of 2'# ft-lb brings a rotational inertia 71 to rest fro" 1)## r%" in ) seconds. Deter"ine the rotational inertia. Also+ deter"ine the energ, dissi%ated b, the bra*e.
Solution hints3 ?on9ert r%" to radsec3 ω1 4 1)) radsec Note that ω24# Find the ratio 7 17271J72 using ti"e and tor:ue4K.6 Note that 72 is infinitel, large 4K 714.6 slugs-ft Find energ, fro" e:uation4K16'### ft-lb
ω = 1)##
2π &#
= 1)).5 rad s
he ti"e to sto% is3 t =
1 2 ω 1 − ω 2
⇒
T 1 + 2
)=
1 2
1)).5 − #
1 + 2
1 2'#
= -.6&
Since 72KK71+ !e conclude3 7 14.6& slugs-ft 2 he energ, is3 " =
1 2 ω 1 − ω 2 2 2 1 + 2
=
-.6& 2
1)).5 − # 2
= 16''-6 ft − lb
" = 1).& BT!
Example #M 10: ?onsider a helical co"%ression s%ring !ith the follo!ing infor"ation not all are necessaril, needed3
Ends3 S:uared and ground S%ring is not %reset Material3 Music !ire steel !ith S ut42)' *si d4.#55 inches and D4#.$) inches Bf 41.'& inches and N t41# Find the follo!ing. Ans!ers are gi9en in %arentheses. S%ring constant+ C 1$.)6 lbin Bength at "ini"u" !or*ing load of 5 lbs 1.#2” Bength at "ai"u" load of 1# lbs #.&” Solid length #.55” Boad corres%onding to solid length 12.#$ lbs ?lash allo!ance #.1'6” Shear stress at solid length 66&6& %si Surge fre:uenc, of the s%ring $15 GL he s%ring constant and 9arious lengths3 $
d
=
$
) D
L W + "in L W + "aD
=
=
#
'
L f L f
=
N t d
L s%lid
=
L f
$
) #.$)
N
L s%lid
?lash allo!ance3
.#55
=
−
−
−
δ F δ F
=
=
"in
"aD
'
1
=
1# #.#5
δ Fs%lid
⇒
CA Lw+"aD =
−
L s%lid
=
#.&)6 .55 #.1'6M −
=
Stresses τ = $ s
) FD π d '
= 1+
$ s
= 1.#56 )12.#$#.'$) = -'$-& psi π .#55
1
= 1.#56
2C
he surge fre:uenc, is3 f n
=
f n
=
1'-##d ND2 1'-##d ND
2
H' =
1'-###.#55 )#.$) 2
=
$15 H&
Problem #M12
A%%lication Bife BD 4 5###&#$)# 4 1$$# "illion c,cles 7f the reliabilit, !as different than #+ !e di9ide the a%%lication life b, the reliabilit, factor. ut this %roble" onl, as*s for # reliabilit,. A%%lication Boad FD 4 1.$&1#$.$5 4 ')## N Note that one %ound is $.$5 Ne!tons. Using the Bife-Boad relationshi% !ith a%%lication condition on one side and catalog conditions on the other !e get3 #.''' F D L D = C 1# 1######.''' he catalog life rating is based on 1"illion c,cles. herefore the e:ui9alent ca%acit, for increased life is3 C 1#
=
')##1$$# .'''
=
$2-## N
Problem # M13
Ma8or dia"eter d = d m
+
p 2
=
1+
#.25 2
=
1.125
in
or:ue re:uired to lift the load Fd m L + π fd m Ff c d c + T ( = 2 π d m − fL 2 /ith T (
=
125##1 .#25 + π #.11 125####.11.5 + 2 2 π 1 − #.1.25
= 2#&-
in − lb
Mini"u" friction for self-loc*ing f =
L
π d m
=
#.25
π 1
=
#.#)
he scre! is self-loc*ing. he efficienc,3 e
=
FL 2π T (
=
125###.25 2π 2#6#
=
#.2$
