1. Unit 5 – Kinematics of Machinery
R.Sivakumar,IIYr ‘C’Section
1
1. A loadof 10KN is raisedbymeansof a screw jack,havinga square threadedscrew of 12mm pitchand
of meandiameter500mm.if a force of 100 N isappliedatthe endof a levertoraise the load,what
shouldbe the lengthof the leverused? Coefficientof friction=0.15. What isthe mechanical advantage
obtained?State whetherthe screwisself-lockingornot.
Givendata:
W= 10KN = 10×103
N
P=12mm
P1=100N
D=50mm
µ= tan φ = 0.15
Solution:
Lengthof lever,
tan α = p/πd = 0.0764
p= w (tan α + tan φ / 1-tan α tan φ)
T= P× d/2
T=77.25 N-m
T=P1× l
=100× l N-mm
l = 572.5 mm
Mechanical advantage,
M.A =W/p1 = 10×103
/100 = 100
Self-lockingof screw:
ɳ= tan α / tan (α + φ)
ɳ= tan α (1- tan α tan φ) / tan α + tan φ)
ɳ=0.3335; ɳ=33.35%

2. Unit 5 – Kinematics of Machinery
R.Sivakumar,IIYr ‘C’Section
2
2. The followingdataare relatedtoa screw jack:pitch of the threadscrew = 8 mm; diameterof the
screwthread= 40 mm; co-efficientof frictionbetweenscrew andnut=0.1, load20KN; Assumingthat
the loadrotates withthe screw,determine. i,the ratio of torquesrequiredtoraise andlowerthe load.
ii,the efficiencyof the machine.
Givendata:
P= 18 mm; d= 40 mm;ɳ= tan φ=0.1; w= 20KN = 20×103
N
Solution:
Tan α = pc/πd =18/ π×40 = 0.06367
Case:1 Force requiredtoraise the load:
P1= w (tanα + tan φ/ 1-tan α tan φ)
=20×103
(0.1+0.06367/ 1 - 0.1×0.06367)
P1=3294.21N
Torque,T1= p ×(d/2) = 3294.21 × (40÷2)
T1=65884.22 N mm
Case2:
Force requiredtolowerthe load,
P2=w( tan φ – tan φ / 1+ tan φ tan α)
=20×103
( 0.1- 0.06367/ 1+0.1×0.06367)
P2= 722.16 N
Torque, T2 = p× d/2= 722.16 × 40/2
T1=14443.26 N mm
Ratioof torque, T1/T2 = 65884.22/14443.26
T1/T2=4.56
ii,Efficiencyof the machine
ɳ= tan α / tan(α+φ) = tan α(1-tanα . tan φ)/tan α+tan φ
=0.06367 × (1 - 0.1 × 0.06367)/ 0.1+ 0.06367 = 0.3865 ; ɳ= 38.65 %

3. Unit 5 – Kinematics of Machinery
R.Sivakumar,IIYr ‘C’Section
3
3. A single dryplate clutchtransmits7.5Kw at 900 rpm. The axial pressure islimitedto0.07 N/mm2
.If
coefficientof frictionis0.25, find(i) Meanradiusandface widthof the frictionliningassumingthe ratio
of meanradiusto face widthas4 and 2, (ii) Outerandinnerradii of clutch plate.
Givendata:
P= 7.5 Kw = 7.5 × 103
w; N= 900 rpm ; P =0.03 N/mm2
; µ=0.25
Solution:
Mean Radiusand face width
Ratioof meanradiusto face width,
R/w= 4
A = 2π R . w. P
T = n . µ . wR
= n.µ (2π R. w .P) R
= n.µ(2πR × R/4 ×P) R
=π/2 ×n . µ PR3
T=0.055 R3
N-mm
P= T Ѡ
w- 2πN/60 = 94.26 rad/s
T=7.5×103
/94.26
= 79.56 × 103
N-mm
R3
= 79.56 × 103
/ 0.055
R=113 mm
W= R/4 = 28.25 mm
Outerand InnerRadii
W= r1 - r2 = 28.25 mm (i)
Alsoforuniformwear,the men radiusof clutch plate is
R= r1 + r2/2 → r1 + r2 = 2R→ r1 + r2 = 226 mm (ii)
Solving(i) and(ii) r1=127.125 mm; r2= 98.875 mm

4. Unit 5 – Kinematics of Machinery
R.Sivakumar,IIYr ‘C’Section
4
4, A single plate clutch,withbothsideseffective,has outerandinnerdiameters300mm and 200 mm
respectively.The maximumintensityof pressure atanypointincontact surface isnot to exceed0.1
N/mm2
.If coefficientof frictionis0.3,determine the powertransmittedbyaclutch at speed2500 rpm.
Givendata:
d1 =300 mm(or) r1= 150 mm
d2 =200 mm(or) r2 = 100 mm
p= 0.1 N/mm2
; µ= 0.3; N =2500 rpm
Solution:
Ѡ= 2πN/60 =261.8 rad/s
P . r2 = C
C= 0.1 × 100 =10 N/mm
W=2πC (r1 - r2)
=3142 N
R= r1+r2/2 = 125 mm
= 0.125 mm
Torque transmitted,T= n . µ. W R
= 2×0.3×3142×0.125
=235.65 N-mm
Powertransmitted,P=T. Ѡ
= 235.65 × 261.8
=61693 w
=61.693 Kw

5. Unit 5 – Kinematics of Machinery
R.Sivakumar,IIYr ‘C’Section
5
5. Two pulleys,one 450 mmin diameterandthe other200 mm indiameterare on parallel shafts2.1m
apart and are connectedbya belt,asa cross beltdrive.The largerpulleyrotatesat225rpm. The
maximumpermissible tensioninthe beltandthe pulleyis0.25. findthe powerthatcan be transmitted.
Givendata:
d1= 450 mm=0.45 m => r1 = 0.225 m; d2 = 200 mm = 0.2m => r2 = 0.1 m
C= 2.1 m; N= 225 rpm ; T1 =1KN =1000N ; µ=0.25
To find:
Powertransmitted.
Solution:
For a crossedbeltdrive
Sinα = r1 + r2/C = 0.225+0.1 / 2.1
Α = 9.732 ̊
Ɵ= 180 ̊+ 2 α
=180 + 2 × 9.732 ̊
=3.4813 rad
Speedof belt V =πd1N1 /60
=π×0.45×225/60
= 5.3014 m/s
We knowthat,
T1/T2 = eµƟ
T1/T2= e0.25×3.4813
T1/T2=2.3876
T2= T1/2.3876 =1000/2.3876
T2=418.8154 n
Power Transmitted
P = (T1 – T2) V = (1000-418.8154) × 5.3014 = 3081.09 W → P=3.081 kW

6. Unit 5 – Kinematics of Machinery
R.Sivakumar,IIYr ‘C’Section
6
6. A leatherbeltisrequiredtotransmit7.5 KW froma pulley1.2m indiameterrunningat250 rpm.The
angle embracedis165 ̊ and coefficient of friction between the belt and pulley is0.3. if safe working
stressfor leatherbeltis1.5 MPa, densityof leather1Mg/m3
and thicknessof belt10mm, determine the
widthof the belttakingcentrifugal tensionintoaccount.
Givendata:
P 7.5 KW = 7500 W; d= 1.2 mm; N= 250 rpm
Ɵ = 165 ̊= 2.88 rad; µ=0.3; σ =1.5 MPa
Ρ = 1mg/m3
= 1 × 106
g/m3
; t = 10 mm = 0.01 m
Solution:
v=πdN/60
v= 15.71 m/s
P= (T1 – T2) V
T1 – T2=477.4 N
2.3log e(T1/T2) = µƟ
T1/T2 = 2.375
T1= 824.6 N & T2 =347.2 N
m= Area × length× density
= b.t.l.p
m = 10 b kg
Centrifugal tension, Tc= mv2
=10 b (15.71)2
Tc =2468 B n
Maximumtension, T=αbt
T=15000 b N
T= T1 + Tc (or)
15000 b = 824.6 + 2468 b
b = 0.0658 m → b = 65.8 mm

7. Unit 5 – Kinematics of Machinery
R.Sivakumar,IIYr ‘C’Section
7
7. A screwjackhas a square threadof meandiameter60 mm and pitch8 mm.coefficientof frictionat
the screwthread is0.09. A loadof 3KN is to be liftedthrough120 mm. Determine the torque required
and the workdone in liftinginliftingthe load120 mm.Findthe efficiencyof jackalso.
Givendata:
d= 60 mm= 0.06 m; p=8 mm = 0.008 m; µ = 0.09
W= 3KN = 3×103
N
Solution:
Helix angle,α =tan-1
(p/πd)
= tan-1
(0.008/π×0.06)
= 2.43 ̊
Frictionangle,φ= tan-1
µ
=tan-1
0.09 =5.14 ̊
(i) Torque requiredtoraise the load:
T1 = W tan (α+φ).d/2
= (3×103
) [tan (2.43 ̊+5.14 ̊)] (0.06/2)
= 11.96 N.m
(ii) Workdone inliftingthe loadthrough12 cm:
In one complete revolutionof the screwedrod,the loadisliftedthroughadistance equal to
pitch.
Numberof turnsrequiredtoliftthe loadthrougha distance of 12cm = 12/0.8 =15
Work done inliftingthe load,W= 2 π N T
= 2 π × 15 × 11.96 =1127 NM
(iii) Efficiencyof the screwjack:
ɳ screw jack = tan α / tan (α+φ) = tan 2.43 ̊ / tan (2.43 ̊+5.14 ̊)
=0.3193 (or) =31.93%

8. Unit 5 – Kinematics of Machinery
R.Sivakumar,IIYr ‘C’Section
8
8, Two pulleys,one 450 mm diameterandthe other200 mm diameterare onparallel shaft2.1 m apart
and are connectedbya crossedbeltis1 KN andthe coefficientof frictionbetweenthe beltandthe
pulleyis0.25. Find:(i) the lengthof the beltrequired,and(ii) the powerthatcanbe transmitted.
Givendata:
Crossbeltdrive;d1= 450 mm = 0.45 m (or) r1= 0.225 m;
d2=200 mm = 0.2 m (or) r2=0.1 m;
N1=225 rpm; x=2.1 m; T1=1 KN= 1000N; µ= 0.25.
Solution:
(i),Lengthof the beltrequired(L):
L =π (r1+r2) +2x + (r1+r2)2
/x
= π (0.225+0.1) +2×2.1 + (0.225+0.1)2
/2.1
= 5.271 m
(ii),Powertransmitted(P):
We knowthatto findpowertransmitted;firstwe needtodetermine the valuesof T1,T2 and v.
T1=1000 N (given)
V= πd1N1 /60 = π ×0.45×200 / 60
=4.71 m/s
To findƟ:
Ɵ= (180 ̊+ 2α) ×π/180 ̊ rad
Sinα = r1+r2/x
=0.225+2.1
=0.1547 (or)
α =sin-1
0.1547 = 8.9 ̊
Ɵ= (180 ̊+ 2α) ×π/180 ̊
Ɵ= (180 ̊+ 2×8.9 ̊) × π/180 ̊ = 3.452 rad

9. Unit 5 – Kinematics of Machinery
R.Sivakumar,IIYr ‘C’Section
9
To findT2:
T1/T2 =eµƟ
1000/T2 =e0.25×3.452
=2.37 (or) T2 =1000/2.37=421.94 N
To findP:
P= (T1-T2) v
= (1000-421.94) × 4.71 =2722.66 W (or) 2.723 kW
9. In a thrustbearing,the external andinternaldiametersof the contactingsurfacesare 320 mmand
200 mmrespectively.The total axial loadis80 KN and the intensityof pressure is350 KN/m2
.The shaft
rotatesat 400 rpm. Takingthe coefficientof frictionas,calculate the powerlostinovercomingthe
frictionandthe numberof collarsrequired.
Givendata:
d1 = 320 mm (or) r1 = 160 mm = 0.16 m;
d2 = 200 mm (or) r2 = 100 mm= 0.1 m;
W= 80 KN; p= 350 KN/m2
= 350×103
N/m2
;
N= 400 rpm ; µ= 0.06.
Solution:
Ѡ=2πN / 60 = 2π (400)/60 = 41.89 rad/s
Powerlostinovercomingfriction(P)
T = 2/3 µ W [r1
3
-r2
3
/r1
2
-r2
2
]
=2/3×0.06×80×103
[0.163
-0.13
/0.162
-0.12
]
PowerlostfrictionP= T.Ѡ
=635×41.89 = 26600 W = 26.6 kW
We knowthatthe intensityof pressure,
P = W/nπ (r1
2
–r2
2
)
Or = 350 ×103
= 80×103
/n×π[0.162
-0.12
] Or n=4.66, say 5

10. Unit 5 – Kinematics of Machinery
R.Sivakumar,IIYr ‘C’Section
10
10. A multiplediscclutchtransmits55 kW of powerat 1800 rpm.Coefficientof frictionforthe friction
Surface is0.1. Axial intensityof pressureisnottoexceed160 kN/m2
.The internal radiusis80mm
And0.7 timesthe external radius.Findthe numberof platesneededtotransmitthe requiredtorque.
Givendata: P= 55 kW = 55 × 103
W; N= 1800 rpm; π=0.1; pmax = 160kN/m2
= 160×103
N/m2
;r2 = 80mm =
0.08 m ; r2=0.7 r1
Solution:Give thatr2 = 0.7 r1 = 0.7 r1. r1 = r2/0.7 = 0.08/0.7= 0.1143 m
Assuminguniform wear,the axialforce exertedisgivenby
W = 2 πc(r1-r2) = 2π p max × r2 (r1- r2)
= 2π × 160 × 103
× 0.08 (0.1143 – 0.08) = 2758.57 N
Torque transmittedbya multi plate discclutch,foruniformwear,isgivenby
T =1/2 n µ W (r1 + r2)
= ½ × n ×0.1 × 2758.57 (0.1143 + 0.08) = 26.8 n
55 × 103
2π ×1800 × T/ 60 Or T = 291.78 N.m
Nowequations(i) and(ii),we get
26.8 n = 291.78 or n = 10.887 = 11+
Numberof frictionsurfacesrequired=11
Then,Total numberof plates= Numberof pairs of contact surface + 1
= 11 + 1 = 12 plates
Important Points
Law of belting:The centre lineof the of the belt,asit approachesthe pulley,mustlie inaplane
perpendiculartothe axisof that pulleyormustlie inthe plane of the pulley.Otherwisethe beltwill run
off the pulley.
The centrifugal tensionhasnoeffectonthe powertransmitted.
[Proof:P= (Tt1 – Tt2) v = [(T1+Tc) – (T2 + Tc)] v= = (T1 – T2) v → hence proved]
Maximumtension,T= Maximumstressx Crosssectional areaof belt= 𝑇 = 𝜎 𝑏 𝑡
Where T is maximumtensioninN, σis maximumsafe stressinN/m2
,bisbreadthof beltinm
and t isthicknessof beltinm.
Initial Tensionof belt, 𝑇0 =
𝑇1+𝑇2+2𝑇𝑐
2