ENGINEERING MECHANICS
PARALLELOGRAM LAW OF FORCESQ P O R a q
The resultant force is given by,
R = P2+Q2+2PQ cosq
Direction of the resultant force can be determined as below: tan a = P+Q sinQ cosqq
where a is the angle between the resultant force of force P. TRIANGLE LAW OF FORCES
R ® = ®P +Q® Q ® R ® P ®
POLYGON LAW OF FORCES
R ® = P1 P2 P3 P4 ® ® ® ® + + + E P4 P3 P2 P1 R A B D C LAMI’S THEOREM R P Q b a g O P sina = Q R sinb =sing PROJECTILES
Important equations used in projectiles are listed below. 1. The time of flight t of a projectile on a horizontal plane is
given by t =2 u sin
g a
2. Horizontal range (R) of a projectile is given by R = 2 u sin 2 g a Maximum range is at 2a =90° Þ a =45° Rmax = 2 u g ANGULAR VELOCITY w = d dt q w = 2 N 60 p rad/s ANGULAR ACCELERATION a = d dt w
FORMULA SHEET ON
RELATIONSHIP BETWEEN CIRCULAR AND LINEAR MOTION
When a body moves in a circular path from point A to B as shown in Figure. O A B S q r S = r q dS dt = d (r ) d r dt dt q q = d v r r dt q = = w dv dt = d (r ) d r r dt dt w = w = a a= ar
These are two components of acceleration tangential and normal acceleration at= r a an= 2 2 v (r ) r r w = = r w2
The total acceleration at0t is the vector sum of the two
components.
atot= a2t +an2
atot= (r )a + w2 (r 2 2)
Angle between the total acceleration and radius is f = 1 t n a tan a - æ ö ç ÷ è ø SIMPLE PENDULUM L m A B O q C
For a simple pendulum Periodic time tP= L 2 g p and
frequency of the oscillation n =
P
1 1
g / L
t = 2p .
Closely Coiled Helical Spring
x A B A B Periodic Time tP= 2 m K p
if d is the deflection of the spring is given by d = mg
K Þ m K = d/g
Þ tP= 2p d/ g
Frequency of the oscillation n =
P
1 1
g / t = 2p d Hz.
If the mass of the spring m1 is also taken into consideration then
n = 1 1 K 2p m+m / 3 COMPOUND PENDULUM w = mg A q h q O G
The periodic time of a compound pendulum is given by tP = 2 2 G K h 2 gh + p
and the frequency of oscillation
h = 2 2
P G
1 1 gh
Where KG = Radius of gyration about an axis
through the centre of gravity G and perpendicular to the plane of motion. and h = Distance of centre of gravity G from the point of suspension O.
TORSIONAL PENDULUM q l A B C f r Body The periodic time is given by
tP =
2 K r g p l and frequency of oscillation
n = P 1 r g / t = 2 Kp l where
r = Distance of each wire from the axis of body K = radius of gyration
l = length of each wire INELASTIC BODIES
The losts of Kinetic energy (EL) during impact of inelastic bodies
is given by ELoss = 1 2 1 2 2 1 2 m m (u u ) 2 (m m ) × -+ where
m1 = mass of the first body
m2 = mass of the second body
u1 and u2 are the velocities of the first and second bodies
respectively.
In case both bodies are elastic bodies the Energy Loss during the impact is given by ELoss = 2 2 1 2 1 2 1 2 m m (u u ) (1 e ) 2 (m m ) × - -+
where e = Coefficient of Restitution.
Coefficient of Restitution is the ratio of relative velocities of the bodies after impact and before the impact.
e = 2 1 1 2 v v u u
-value of e lies between 0 and 1 0£ e £ 1
e = 0 for perfectly inelastic bodies e = 1 for perfectly elastic bodies
STRENGTH OF MATERIALS
STRESS TENSOR
Elongation of a bar Subjected to axial load P
P
l P
AE l d =
Elongation of a tapered bar subjected to axial load P
d1 d2 1 2 4 PL E d d d = p P
Elongation of a prismatic bar under its self weight =
2
L 2E g g = self weight per unit volume
L d
Elongation of a conical bar under its self weight =
2 L 6E g
STRA IN
Consider a rod of length Lo subjected to load P
P P Lo Lf long L L L L L f o x o o -D e = e = = lateral ( o f) y z o d d d d d - -D e = e = e = = V V V x y z d e = = e + e + e
Relationship Between Elastic Constants E = 2G(1 + m)
E = 9KG 3K+G G = E 1 2´1+m K = E 1 3´1 2- m Value of any EC ³ 0 Note : mcork = 0 E (Young’s Modulus)
G (Modulus of Rigity/Shear Modulus) K (Bulk Modulus)
m (Poisson’s Ratio)
Young’s Modulus or Modulus of Elasticity s µ elong
s = E = young’s modulus elong E Þ elong ¯ Þ d l ¯
\ A material having higher E value is chosen EMS = 200 GPa
ECI = 100 GPa EAl = 200GPa
3
\ (dl)MS < (dl)CI < (d)Al
Shear Modulus or Modulus of Rigidity
t =Gg \ for a given t, G µ 1. g Bulk Modulus (K) K = v v Normal stress s = e e Poisson’s Ratio m = – lateral strain longitudinal strain
Beams of Uniform Strength
To make beam a beam of uniform strength:— (i) depth is varied.
dx = d x L
\ depth should be varied parabolically. (ii) width ‘b’ is varied
\ bx = b x L é ù ê ú ë û
\ width should be varied linearly.
Euler’s Formulae
Assumptions
® The self weight of column is neglected. ® Crushing effect is neglected.
® Flexural rigidity is uniform. ® Load applied is truly axial.
® Length is very large compared to cross-section. \ Pe µ f [E, Imin, end conditions, L2]
\ Pe = 2 min 2 e E I . L p
Pe : Euler’s buckling load. Imin : min [Ixx and Iyy]. Le : effective length of column. L : actual length of column.
L = Le a
length fixity coefficient
(end fixity coefficient)
2 1 n= a End Conditions ® ¯ Values of a and h Both Ends Hinged (BH) Both Ends Fixed (BF) Fixed and Hinged (F & H) Fixed and Free (FF) a 2 1 h = a 1 1 2 1 2 1 4 1 4 2 2
If remaining all other parameters are same, (Pe)BF > (Pe)FH > (Pe)BH > (Pe)FF Slenderness Ratio S = Le K where K = min I A se : buckling stress se = 2 2 E s p
\ S Þ Pe ¯ Þ buckling tendency is increased \ (S)SC < (SMC) < (SLC) SC : Short Column MC : Medium Column LC : Long Column For steels, if S £ 30 Þ short column S > 100 Þ long column 30 < S £ 100 Þ medium column
STRAIN ENERGY METHODS
Strain energy of bar =
2 2 1 P L AL P AL . 2 2AE 2E 2 s s e d = = ´ =
® Strain energy of solid circular shaft subjected to torsion t = p T , Z where T : twisting moment.
Zp : polar section modulus for circular × section. Zp = d .3 16 p æ ö ç ÷ è ø \ SE = 1T 1 T L2 2
( )
AL . 2 2 GJ 4G t q = =® Strain energy of hollow circular × section shaft. d : Inner diameter. D : Outer diameter. K = d D K = 0 for solid K < 1 Zp = 16 p D3 (1 – K4) \ SE = 2 4G t (AL) (1 + K2), where t = p T . Z
STRAIN ENERGY DUE TO BENDING
U =
(
)
2 b xx xx a M dx 2EIò
U : strain energy Mxx : moment at section x-xTHEORY OF MACHINES
GRUBLER’S CRITERIONIn a mechanism total no. of degrees of freedom is given by F = 3(n – 1) – 2j
where n is no. of links and
j = no. of joints (simple hinges)
most of the mechanism are constrained so F = 1 which produces 1 = 3(n – 1) – 2j
Þ 2j – 3n + 4 = 0 this is called Grubler’s criterion. If there are higher pairs also no. of degrees of freedom is given by
F = 3(n – 1) – 2j – h where h = no. of higher pairs.
ACKERMAN STEERING GEAR MECHANISM
Equation for the correct steering is
cot f – cot q = c/b
where c = Distance between the pivots of the front axles b = Wheel base
f and q are angle through which the axis of the outer wheel and inner whel turns respectively.
DAVIS STEERING GEAR MECHANISM
tan a = c/2b
where c = Distance between the pivots of the front axles b = Wheel base
a = Angle of inclination of the links to the vertical
HOOK’S JOINT 2 2 Driving Driven N 1 cos sin = N cos - q a a
where NDriving= Speed of the driving shaft in r.p.m.
NDriven= Speed of the driven shaft in r.p.m.
q = Angle through which the arms of the cross turn a = Angle of intersection of two shafts
FRACTIONAL TORQUE IN PIVOT AND COLLAR BEARING
(i) Frictional torque transmitted in a flat bearing is given by T = 2 W R
3´ m while considering uniform pressure
And in case of uniform wear T = 1 WR
2´ m
where m = Coefficient of friction
W = Load transmitted to the bearing R = Radius of the shaft
(ii) Frictional torque transmitted in a Conical Pivot bearing is given by
T = 2 W R cosec
3´ m a
while considering uniform pressure And in case of uniform wear
T = 1 WR cosec 2´ m a where
a = semi angle of the cone
(iii) Frictional torque transmitted in a trapezoidal or truncated conical pivot bearing is given by
T = 3 3 1 2 2 2 1 2 r r 2 W cosec 3 r r é - ù ´ m ê ú a -ê ú ë û
while considering uniform pressure. And in case of uniform wear T = 1 W (r + r ) cosec1 2
where r1 and r2 are the external and internal radii of the
conical bearing respectively R = r1 r2
2 +
is the mean radius of the bearing.
(iv) Frictional torque transmitted in a flat collar bearing is given b y T = 3 3 1 2 2 2 1 2 r r 2 W 3 r r é - ù ´ m ê ú -ê ú ë û
while considering uniform pressure And in case of uniform wear
T = 1 W (r1 r )2 2´ m + BELT DRIVE
Centrifugal Tension Tc= mV2
where m = Mass per unit length of the belt V = Linear velocity of the belt Velocity Ratio
The velocity ratio of speeds of driver and driven pulleys is given b y 2 2 1 1 1 2 N d t S = = 1 N d t 100 w + æ - ö ç ÷ w + è ø
where d1, d2= diameters of driver and driven pulleys
w1, w2= angular velocities of driver and driven pulleys
N1, N2= rotational speeds of driver and driven pulleys
expressed in revoluations per minute (r.p.m.) S = S1 + S2 + 0.01S1S2 is percentage of total
effective slip
S1= Percentage slip between driver and the belt
S2= Percentage slip between belt and the follower
(driven pulleys)
GEARS AND GEAR DRIVE
Dedendum circle:
Root circle diameter = Pitch circle diameter ´ cos f where f is the pressure angle.
Circular pitch:
Pc=
D T p
where D = Pitch circle diameter
T = Number of teeth on the wheel. Diametral pitch: Diameter pitch Pd = c T = D P p Þ Pc ´ Pd = p Module: m = d D 1 = T P Þ m ´ Pd = 1 Arc of recess:
Contact Ratio = Length of arc of contact
Circular Pitch
Contact ratio is the number pairs of teeth in contact. Length of Arc of contact:
Length of Arc of contact = Length of path of contact
cosf
where f is pressure angle Gear Trains
1. Simple gear train:
Velocity ratio = Speed of the driving wheel
Speed of the driven wheel
= no. of teeth on the driven wheel
no. of teeth on the driving wheel
1 4 N N = 3 3 1 2 2 4 4 2 3 4 1 2 3 1 N T N N T T T = = N ´ N ´ N T ´T ´ T T Train value = 4 1 1 4 N T = N T
2. Compound gear train:
Velocity ratio = 1 1 3 5 2 4 6 6 2 4 6 1 3 5 N N T T T N N = = N N N N T T T ´ ´ ´ ´ ´ ´
3. Reverted gear train: If D1, D2, D3, D4 be the pitch circle
diameters of the respective gears and corresponding speeds are N1, N2,N3, N4 then 3 4 1 2 D D D D = 2 2 + + Þ D1 + D2 = D3 + D4 Velocity ratio = 1 2 4 4 1 3 N T T = N T T ´ ´
4. Epicyclic gear train: Velocity Ratio B A C B N T 1 N = + T FLY WHEEL
Coefficient of fluctution of energy: DEmax= Emax – Emin
Cenergy= max per cycle E W D where
DEmax= maximum fluctuation of energy
Cenergy= coefficient of fluctuation of energy
Coefficient of fluctuation of speed: Dwmax=wmax – wmin
Cs= max mean Dw
w
POWER OF GOVERNOR
Power =Mean effort ´ Lift of sleeve Types of Governors
(1) Simple governor-Watt type: Height of the governor is given b y
h =
2
895 N metres
where N = speed of the arm and ball about the spindle axis. (2) Porter governor:
If hp is the height of porter governor (when length of arms
and links are equal).
and hw is height of watt’s governor then p
w
h m + M =
h m
where m = mass of the ball M = mass of the sleeve (3) Hartnell governor:
lsleeve = Xcompression = (r2 – r1)
y x where r1 = Minimum radius of rotation
r2 = Maximum radius of rotation
x = Lenth of ball arm of lever y = Length of sleeve arm of lever Stiffness of the spring is given by
S = S2 S1 h
-where S1 = Spring force at minimum radius of rotation
S2 = Spring force at maximum radius of rotation
BALANCING OF RECIPROCATING MASSES Inertia force due to reciprocating parts is given by FI =FR = m × w2 × r cos 2 cos n q æ q + ö ç ÷ è ø
where q = angle made by the crank. Tractive Force:
FT= (1 – C) mw2 × r (cos q – sin q)
where m = mass of the reciprocating parts w = angular velocity of crank
r = radius of crank
C = fraction of balanced reciprocating mass FTmax or FTmin occurs when q = 135° or 315°
FTmax= 2 (1 – C) mw2 r
FTmin= - 2 (1 – C) mw2 r
Swaying Couple:
Swaying couple = (1 – C) mw2 r ´ a
2 (cos q + sin q) where a = distance between the central line of two cylinders
Swaying couple is maximum or minimum when q = 45° or 225°
Value of minimum and maximum swaying couple = a 2 ± (1 – C) mw2 r Hammer Blow: P = B w2 × b
VIBRATIONS
Mass Moment of Inertia about z-axis and passing through centre of mass Rod IM= 2 M 12 l Iyy = 2 M 3 l l : length of rod M : mass of rod Circular disc IM= 2 MR 2 M : mass of disc R : radius of disc Sphere (Hollow) IM= 2 3 MR 2
M : mass of hollow sphere R : radius of hollow sphere Sphere (Solid)
IM= 2 5 MR2
M : mass of solid sphere R : radius of sphere Ring IM= 2 MR 2 M : mass of ring R : radius of ring Solid Cylinder IM= 2 MR 2
M : mass of solid cylinder R : radius of cylinder Linear Springs
F = kx where F : force applied
x : deflection
k : spring constant / stiffness For a linear spring,
Potential Energy = 1 2 kx
2
Linear Torsion Spring t (q) = Kt q
t : moment
q : angular deformation of spring PE (Potential Energy) of a torsion spring is
PE = 1 2 Ktq
2
Combination of Springs (i) Springs in Series
eq 1 k = 1 2 1 1 k + k Þ 1 2 eq 1 2 k k k k k = + w = keq m
For ‘n’ number of springs in series having stiffness k1, k2, ... kn eq 1 k = 1 2 n 1 1 1 ... k + k k
(ii) SPRINGS IN PARALLEL keq= k1 + k2, w = k / meq
For ‘n’ number of springs in parallel having stiffness k1, k2, ... kn
keq= k1 + k2 + k3 ... + kn ENERGY METHOD E = 1mV2 1sx2 2 +2 dE 0 dt = Þ 1m (2V)dv+ 1ks (2n)dx 0 2 dt 2 dt = Þ x + k x = 0 m && VIBRATIONS IN BEAMS Longitudinal Vibrations in Beams
· Longitudinal vibrations mean along the axis of the rod. Consider a rod the end of which contains a mass ‘m’.
E : Young modulus of rod A : Cross-section area of rod
L : Length of rod w = k
m where k =
EA L
Transverse Vibrations in Beams
D = mgL3 Cantilever subjected to concentrated load at free end.
3 EI
ì ü
í ý
î þ
DESIGN
Bearing stress or crushing stress:sb or sc= P d t n× ×
where d = diameter of the rivet t = thickness of the plate d× t = projected area of the rivet
n = no. of rivets per pitch length in bearing or crushing. Bearing Pressure: Pb= P P = A ld
where P is load along the radius of the journal l = length of journal in contact d = diameter of the journal l × d = projected area is contact Stress concentration factor:
Kt= max 0 s
s
where Kt= stress concentration factor
smax= maximum stress at the discontinuity
s0= nominal stress at the same point
Notch sensitivity:
q = f
t
Increase in actul stress
K 1 over nominal value
=
K 1 Increase in theoretical stress over nominal value
-Theories of Failure under the Static Load
Maximum principal or normal stress theory (Rankin’s Theory): Failure occurs when
smax=slimiting
Maximum shear stress theory (Guest’s theory): tmax=tyield at that point
Maximum principal strain theory (Saint Venant’s theory): emax= elimiting or yield at that point.
where e = Strain
Maximum strain energy theory (Haigh’s theory):
S.E.
V =
yield S.E.
V
where S.E. = Strain energy V = Volume
Maximum distortion energy theory (Hencky and Von Mises theory):
Shear Stress Energy Shear Strain Energy at Yield Point =
V V
where V = Volume Factor of Safety
F.S. = Maximum strength of the material
Design or working stress of the material
F.S.ductile materials=
Yield point strength
Working or design stress For static loading
F.S.brittle materials=
Ultimate strength
Design or working stress For static loading
F.S.fatigue loading=
Endurance limit Design or working stress
Fatigue Failure Criteria for Fluctuating Stress Table
Method Name Mathematical Relation
Gerber Method 2 v m u e 1 = F.S. F.S. æs ö s ´ + ç ÷ s s è ø
valid for ductile material Goodman Method m v u e 1 = F.S. s s + s s Soderberg Method m y y e 1 = F.S. s s + s s Elliptic Method 2 2 y m e y F.S. s æF.S. ö 1 æ ö s ´ +ç ´ ÷ = ç ÷ ç s ÷ ç s ÷ è ø è ø
where sm= mean stress = max min
2 s + s su= ultimate stress
se= endurable limit for reverse loading stress
sv= variable stress = max min
2 s - s sy= yield point stress
F.S. = Factor of safety
Torsional shear stress caused by the frictional resistance of the threads during its tightening: Value of torsional shear stress is
given by t = 3 c 16 T (d ) p
where T = applied torque dc= minor diameter
Compression or crushing stress on threads: The value of crushing stress on threads is given by
sc= 2 2 c 4 P d (d ) n é ù pë - û where d = major diameter
dc= minor diameter
n = number of threads employed during engage-ment Design of Keys Condition : c w = t 2 s t
where w = width of the key t = thickness of the key sc= permissible crushing stress
t = permissible shearing stress Torsional load:
t = 16T3
d
p N/m2 for solid shaft
where d = shaft diameter in m T = torsional moment in N-m Bending load:
sb= 3 32 M
d
p for solid shaft
where sb= bending stress
and for a hollow shaft sb= 4 3 i o o 32 M d d 1 d é æ ö ù ê ú p - ç ÷ ê è ø ú ë û Lewis Equation
Maximum value of bending stress = sw =
My I where M is maximum bending moment (i.e. at BC)
M = Ft ´ h Ft = M h Barth Formula: sw=so ´ Cv
Cv= 4.5
4.5+v for carefully cut gears operating at
velocities upto 12.5 m/s = 3
3+v for ordinary cut gears operating at
velocities upto 12.5 m/s Static Tooth Load
FS=se b Pc y = se b p my
where se= Flexural endurance limit
For safety against breakage FS > FD
Diametral clearance ratio: Diametral clearance ratio = C1
d
Eccentricity ratio (Attitude): Î = 2 e C Sommerfield number: Sommerfield number = 2 1 ZN d P C æ ö æ ö ç ÷ ç ÷ è ø è ø
where N = Journal speed in r.p.m., Z = lubricant viscosity, P = bearing pressure normally we take its value as 14.3 ´ 106
Critical pressure in journal bearing:
P = 2 2 6 1 Z N d N/ mm C d 4.75 10 l l æ ö æ ö ç ÷ ç + ÷ ´ è ø è ø
where N = Journal speed in r.p.m.
Z = Absolute viscosity of the lubricant Coefficient of friction: m = 8 1 33 ZN d K P C 10 é ù é ù + ê ú ê ú ë û ë û
where K is a factor for end leakages for 0.75 <
d l
< 2.8, K = 0.002 Short and long bearings:
If
d l
< 1 then bearing is said to be short
d l
= 1 bearing is called square bearing
d l
> 1 then bearing is said to be long Heat generation and rejection in bearing:
Qgen=mWV N-m/s
where W = load on the bearing V = rubbing velocity in m/s Heat rejection is given by
Qrejection= Kh A (tb – ta) J/S
where Kh= heat dissipation coefficient in W/m2/C
A = prejected area of the bearing tb= bearing surface temperature
ta= ambient temerature
Bearing Characteristic Number
The factor ZN
P is known as bearing characteristic number and it is a dimensionless number.
where Z = Absolute viscosity of the lubricant in kg/m-s N = Speed of journal in r.p.m.
P = Bearing pressure on the projected bearing area in N/mm2
P = W
d
l× , W = Load on the journal
Dynamic load rating:
Rating Life L = 3 C P æ ö ç ÷ è ø where P = load
C = dynamic basic load rating
Þ P = 1/3 1 C L æ ö ç ÷ è ø
If N is r.p.m. the Life in hours is given by L = 3 6 C 10 P 60 N æ ö ´ ç ÷ è ø hours or P = 1/ 3 6 10 C 60 NL é ù ´ ê ú ê ú ë û Cone clutch: Tcone= 3 3 1 2 2 2 1 2 r r 2 W 3 r r é - ù m ê ú -ê ú ë ûcosec a
where a = semi-angle of frictional surfaces with the clutch axis.
Centrifugal clutch:
T =m (C – S) ri ´ n
where C = Spring force acting on shoe = m r w2
m = mass of shoe
r = distance of centre of gravity of shoe from centre
w = angular velocity of rotating pulley in rad/s ri= inside radius of pulley rim
S = Inward force due to spring-m (w12) r
w1= 3 w 4 n = number of shoes C – S = m r w2 – 9 16 m r w 2 = 7 16 m r w 2
Single Block or Shoe Brake : Braking torque is given by Tb=m Rn r Þ Tb= P r a l m
