Formula Sheet - Applied Mechanics & Design

FORMULA SHEET ON APPLIED MECHANICS & DESIGN ENGINEERING MECHANICS

LAMI’S THEOREM P

PARALLELOGRAM LAW OF FORCES

Q g b

O a

R

Q

q

R

a

O

P

P Q R = sin a = sin b sin g

The resultant force is given by,

PROJECTILES

P 2 + Q 2 + 2PQ cos q

R=

Direction of the resultant force can be determined as below: Q sin q tan a = P + Q cos q

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=

where a is the angle between the resultant force of force P.

TRIANGLE LAW OF FORCES

2.

Horizontal range (R) of a projectile is given by R=

® Q

® R ®

®

®

R = P+Q ® P

®

®

®

®

®

R = P1 + P2 + P3 + P4

u 2 sin 2a g

Maximum range is at 2a =90° Þ a =45° Rmax =

POLYGON LAW OF FORCES

2 u sin a g

u2 g

ANGULAR VELOCITY D

P4

P3

E

C

R

w=

dq dt

w=

2p N rad/s 60

P2 A

P1

B

ANGULAR ACCELERATION a=

dw dt

2

RELATIONSHI P BETWEEN CIR CULAR AND LINEAR MOTION

and frequency of the oscillation

When a body moves in a circular path from point A to B as shown in Figure.

n=

B S

1 1 = t P 2p

g/L .

Closely Coiled Helical Spring

q r

O

A

S = rq

x

dS d (r q) dq =r = dt dt dt

v=r

A

A

B

B

dq = rw dt

Periodic Time tP = 2p

if d is the deflection of the spring is given by

d (r w ) dw dv =r = ra = dt dt dt

d=

a = ra

These are two components of acceleration tangential and normal acceleration at = r a

v 2 (r w)2 = an = = r w2 r r The total acceleration a t0t is the vector sum of the two components. a tot =

a 2t + a n2

a tot =

(r a ) + (r w )

m K

Þ

mg K

m = d/g K

Þ

t P = 2p d / g

Frequency of the oscillation n =

1 1 = t P 2p

g / d Hz.

If the mass of the spring m1 is also taken into consideration then n=

1 2p

K m + m1 / 3

COMPOUND PENDULUM

2

2 2

Angle between the total acceleration and radius is æa ö f = tan -1 ç t ÷ è an ø

O q

h

SIMPLE PENDULUM G q

O

A

q

w = mg

L

The periodic time of a compound pendulum is given by m C

B A

K G2 + h 2 gh

and the frequency of oscillation

For a simple pendulum Periodic time tP = 2p

t P = 2p

L g

1 1 h = t = 2p P

gh K 2G

+ h2

3 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.

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

TORSIONAL PENDULUM f l

Elongation of a bar Subjected to axial load P

A

B

C

q r

d=

l

Pl AE

P Elongation of a tapered bar subjected to axial load P

Body

d1

P

d2

d=

The periodic time is given by tP =

2p K r

l g

Elongation of a prismatic bar under its self weight = g = self weight per unit volume

and frequency of oscillation 1 r n = t = 2p K P

4 PL p d1d 2 E

gL2 2E

L

g/l

d

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 =

m1 × m 2 (u1 - u 2 )2 2 (m1 + m 2 )

Elongation of a conical bar under its self weight =

STRA IN Consider a rod of length Lo subjected to load P

P

P

where

Lo

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 =

m1 × m 2 (u1 - u 2 )2 (1 - e2 ) 2 (m1 + m 2 )

where e = Coefficient of Restitution. Coefficient of Restitution is the ratio of relative velocities of the bodies after impact and before the impact. e =

v 2 - v1 u1 - u 2

Lf

elong = e x =

DL L f - L o = Lo Lo

elateral = e y = e z =

eV =

D d -( d o - d f ) = d do

dV = ex + e y + ez V

Relationship Between Elastic Constants E = 2G(1 + m) E = 3K(1 – 2m)

gL2 6E

4 9KG 3K+G E 1 G= ´ 2 1+ m

E=

K=

E 1 ´ 3 1 - 2m

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 =

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 =

Shear Modulus or Modulus of Rigidity t = Gg

L 2e

.

Pe : Euler’s buckling load. Imin : min [Ixx and Iyy]. Le : effective length of column. L : actual length of column. Le = a L length fixity coefficient n=

1 a2 (end fixity coefficient)

200 GPa 3

\ (dl)MS < (dl)CI < (d)Al

p 2 E Imin

End Conditions ® Both Ends Both Ends Fixed and Fixed and Free Hinged Fixed Hinged ¯ Values of (FF) (F & H) (BF) (BH) a and h a

1 \ for a given t, G µ . g

h=

1 a

2

1

1 2

1

1

4

2

2

Bulk Modulus (K) Normal stress s = K= ev ev

Poisson’s Ratio – lateral strain m= longitudinal strain

Beams of Uniform Strength To make beam a beam of uniform strength:— (i) depth is varied. x L depth should be varied parabolically.

dx = d \ (ii)

width ‘b’ is varied \ \

éxù bx = b ê ú ëLû width should be varied linearly.

If remaining all other parameters are same, (Pe)BF > (Pe)FH > (Pe)BH > (Pe)FF

Slenderness Ratio Le where K = K se : buckling stress

S=

se =

Imin A

p2E

s2 \ 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

2 1 4

5

STRAIN ENERGY METHODS 1 P2 L s 2 s e AL Pd = . = ´ AL = 2 2AE 2E 2 ® Strain energy of solid circular shaft subjected to torsion

Strain energy of bar =

T , Zp

t=

cot f – cot q = c/b 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. where

DAVIS STEERING GEAR MECHANISM tan a = c/2b c = Distance between the pivots of the front axles b = Wheel base a = Angle of inclination of the links to the vertical

where T : twisting moment. Zp : polar section modulus for circular × section.

where

æ pö Zp = ç ÷ d3 . è 16 ø

HOOK’S JOINT

1 1 T 2 L t2 \ SE = Tq = = ( AL) . 2 2 GJ 4G ® Strain energy of hollow circular × section shaft. d : Inner diameter. D : Outer diameter.

d D K = 0 for solid K FD Diametral clearance ratio: Diametral clearance ratio =

C1 d

Eccentricity ratio (Attitude): e Î= C2 Sommerfield number: 2

æCö Rating Life L = ç ÷ èPø where P = load C = dynamic basic load rating 1/3

æ1ö Þ P= Cç ÷ èLø If N is r.p.m. the Life in hours is given by 3

10 6 æCö L= ç ÷ ´ hours è P ø 60 N

2

æ d ö æ l ö 2 ÷ ç ÷ N/ mm 6 ç 4.75 ´ 10 è C1 ø è l + d ø where N = Journal speed in r.p.m. Z = Absolute viscosity of the lubricant Coefficient of friction: ZN

ZN is known as bearing characteristic number and it P 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 W , W = Load on the journal P= l ×d Dynamic load rating:

The factor

3

æ ZN ö æ d ö Sommerfield number = ç ÷ç ÷ è P ø è C1 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=

Bearing Characteristic Number

33 é ZN ù é d ù m= 8 ê úê ú+K 10 ë P û ë C1 û where K is a factor for end leakages l for 0.75 < < 2.8, K = 0.002 d Short and long bearings: l < 1 then bearing is said to be short If d l = 1 bearing is called square bearing d l > 1 then bearing is said to be long d 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

1/3

or

é 106 ù P = C´ê ú êë 60 NL úû

Cone clutch: é r 3 - r 32 ù 2 m W ê 12 ú cosec a 2 3 ëê r 1 - r 2 úû 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 3 w1 = w 4 n = number of shoes 9 7 C – S = m r w2 – m r w2 = m r w2 16 16 Tcone =

Single Block or Shoe Brake : Braking torque is given by Tb = m Rn r Þ

Tb =

mPl r a

11

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