strength of materialformulas short notes

7/25/2019 Strength of MaterialFormulas Short Notes

1/19

Strength of

Material(Formula & Short Notes)


2/19

Stress and strain

Stress = Force / Area

( )tL Changeinlength

Tensionstrain eL Initial length


3/19

Brinell Hardness Number (BHN)

2 2( )2

P

DD D d

where, P = Standard load, D = Diameter of steel ball, and d = Diameter of the indent.

Elastic constants:


4/19

STRAIN ENERGY

Energy Methods:

(i) Formula to calculate the strain energy due to axial loads ( tension):

U = P / ( 2AE ) dx limit 0 to L

Where, P = Applied tensile load, L = Length of the member , A = Area of the member, and

E = Youngs modulus.

(ii) Formula to calculate the strain energy due to bending:

U = M / ( 2EI ) dx limit 0 to L

Where, M = Bending moment due to applied loads, E = Youngs modulus, and I = Moment ofinertia.

(iii) Formula to calculate the strain energy due to torsion:

U = T / ( 2GJ ) dx limit 0 to L

Where, T = Applied Torsion , G = Shear modulus or Modulus of rigidity, and J = Polar

moment of inertia

(iv) Formula to calculate the strain energy due to pure shear:

U =K V / ( 2GA ) dx limit 0 to L

Where, V= Shear load

G = Shear modulus or Modulus of rigidity

A = Area of cross section.

K = Constant depends upon shape of cross section.

(v) Formula to calculate the strain energy due to pure shear, if shear stress is given:

U = V/ ( 2G )

Where, = Shear Stress


V = Volume of the material.


5/19

(vi) Formula to calculate the strain energy , if the moment value is given:

U = M L / (2EI)

Where, M = Bending moment

L = Length of the beam

E = Youngs modulus

I = Moment of inertia

(vii) Formula to calculate the strain energy , if the torsion moment value is given:

U = T L / ( 2GJ )

Where, T = Applied Torsion

L = Length of the beam


J = Polar moment of inertia

(viii) Formula to calculate the strain energy, if the applied tension load is given:

U = PL / ( 2AE )

Where,

P = Applied tensile load.

L = Length of the member

A = Area of the member

E = Youngs modulus.

(ix) Castiglianos first theorem:

= U/ P

Where, = Deflection, U= Strain Energy stored, and P = Load

(x) Formula for deflection of a fixed beam with point load at centre:

= - wl3/ 192 EI

This defection is times the deflection of a simply supported beam.


6/19

(xi) Formula for deflection of a fixed beam with uniformly distributed load:

= - wl4/ 384 EI

This defection is 5 times the deflection of a simply supported beam.

(xii) Formula for deflection of a fixed beam with eccentric point load:

= - wa3b3 / 3 EI l3

Fixed end moments for a fixed beam with the given loading conditions:

Type of loading (A--B) MAB MBA

-wl / 8 wl / 8

-wab2

/ l2

wab2

/ l2

-wl2/ 12 wl2/ 12

-wa2 (6l28la + 3a2)/12 l2

-wa2 (4l-3a)/ 12 l2

-wl2/ 30 -wl2/ 30

-5 wl2/ 96 -5 wl2/ 96

M / 4 M / 4


7/19

Eulers formula for different end conditions:

1. Both ends fixed:

PE = 2 EI / ( 0.5L)2

2. Both ends hinged :

PE = 2 EI / (L)2

3. One end fixed ,other end hinged:

PE = 2 EI / ( 0.7L)2

4. One end fixed, other end free:

PE = 2 EI / ( 2L)2 where L = Length of the column

Rakines formula:

PR = f C A / (1+ a (l eff / r)2)

where, PR = Rakines critical load

fC = yield stress

A = cross sectional area

a = Rakines constant

leff = effective length

r = radius of gyration

Eulers formula for maximum stress for ainitially bent column:

max= P /A + ( Mmax/ Z )= P/ A + P a / ( 1- ( P / PE ))Z

Where, P = axial load

A = cross section area

PE= Eulers load

a = constant

Z = section modulus


8/19

Eulers formula for maximum stress for a eccentrically loaded column:

max= P /A+( M max /Z) = P/A + ( P e Sec(leff/2 ) (P/EI) )/((1- (P / PE )) Z )

Where, P = axial load

A = cross section area

PE = Eulers load

e = eccentricity

Z = section modulus

EI = flexural rigidity

General expressions for the maximum bending moment, if the deflection curve

equation is given:

BM = - EI ( d2y / dx2)

Maximum Principal Stress Theory ( Rakines theory):

1 = fy.

where 1 is the maximum Principal Stress, and fyis elastic limit stress.

Maximum Principal Strain Theory ( St. Venants theory):

e1 = fy / E

In 3D, e1= 1/E[ 1 (1/m)( 2 + 3) ] = fy / E [ 1 (1/m)( 2 + 3) ] = fy

In 2D, 3= 0 e1= 1/E[ 1 (1/m)( 2 ) ] = fy / E [ 1 (1/m)( 2 ) ] = fy

Maximum Shear Stress Theory (Trescas theory):

In 3D, ( 1 - 3) / 2 = fy/2 ( 1 - 3) = fy

In 2D, ( 1 - 2) / 2 = fy/2 1 = fy

Maximum Shear Strain Theory (Von Mises- Hencky theory or Distortion energy

theory):

In 3D, shear strain energy due to distortion:

U = (1/ 12G)[ ( 1 - 2)2 + ( 2 - 3) 2+ ( 3 - 1) 2 ]


9/19

Shear strain energy due to simple tension:

U = fy2 / 6G

(1/ 12G)[ ( 1 - 2)2 + ( 2 - 3) 2+ ( 3 - 1) 2 ] = fy2 / 6G

[ ( 1 - 2)2 + ( 2 - 3) 2+ ( 3 - 1) 2 ] = 2 fy2

In 2D, [ ( 1 - 2)2 + ( 2 - 0) 2+ ( 0- 1) 2 ] = 2 fy2

Maximum Strain Energy Theory (Beltrami Theory):

In 3D, strain energy due to deformation:

U = (1/ 2E)[ 12+ 22+ 32 -(1/m)( 12 + 22 + 22 )]

Strain energy due to simple tension:

U = fy2 / 2E

(1/ 2E)[12+ 22+ 32 -(2/m)( 12 + 22 + 22 )] = fy2 / 2E

[12+ 22+ 32 -(2/m)( 12 + 22 + 22 )] = fy2

In 2D, [ 12+ 22 - (2/m)( 12 )] = fy2

Failure theories and its relationship between tension and shear:

1. Maximum Principal Stress Theory ( Rakines theory):

y= fy2. Maximum Principal Strain Theory( St. Venants theory):

y= 0.8 fy3. Maximum Shear Stress Theory ( Trescas theory):

y=0.5 fy

4. Maximum Shear Strain Theory ( Von Mises Hencky theory or Distortion energy

theory):

y= 0.577 fy

4. Maximum Strain Energy Theory ( Beltrami Theory):

y= 0.817fy.

Volumetric strain per unit volume:

fy2 / 2E


10/19

Torque, Power, and Torsion of Circular Bars:

Relation between torque, power and speed of a rotating shaft:

63000

TnH

Where His power in Hp, Tis torque in lb-in, andnis shaft speed in rpm.

In SI units:

TH

Where His power in Watts, Tis torque in N-m, and is shaft speed in rad/s.

The shear stress in a solid or tubular round shaft under a torque:

The shear stress:

J

Tr

Jis the area polar moment of inertia and for a solid (di=0) or hollow section,

)(32

44

io ddJ

The angle of rotation of a shaft under torque:

GJ

TL

Axial deflection of a bar due to axial loading

The spring constant is:

L

EAK

Lateral deflection of a beam under bending load:


11/19

3

48

L

EIK

For cantilevered beams of length L:

3

3

L

EIK

Torsional stiffness of a solid or tubular bar is:

L

GJKt

The units are pounds per radians.

Load Distribution between parallel members:

If a load (a force or force couple) is applied to two members in parallel, each member takes

a load that is proportional to its stiffness.

K2K1

F

TKt1

Kt2


12/19

The force F is divided between the two members as:

FKK

KFF

KK

KF

21

22

21

11

The torque T is divided between the two bars as:

TKK

KTT

KK

KT

tt

t

tt

t

21

22

21

11

Direct shear stress in pins:

A

F

2

The clevis is also under tear-out shear stress as shown in the following figure (top view):

Tear-out shear stress is:

A

F

4

t

FF


13/19

In this formula A= (Ro-Ri) is approximately and conservatively the area of the dotted

cross-section. Roand Riare the outer and inner radii of the clevis hole. Note that there are

4 such areas.

Shear stresses in beams under bending forces:

bI

VQ

Z

11yAQ

Torsion of Thin-walled Tubes:

F

Y

y1b

A1

y1


14/19

Shear stress:

At

T

2

GtA

TSL24

Where S is the perimeter of the midline, L is the length of the beam, and G is shear modulus.

Stress in Thin-Walled Cylinders

The tangential or hoop stress is:

tPdi

t2

The axial stress is:

t

Pdia

4

Stresses in Thick-walled Cylinders

The tangential stress:

22

2

2222

io

iooiooii

trr

r

PPrrrPrP

The radial stress is:

22

2

2222

io

io

oiooii

rrr

r

PP

rrrPrP

When the ends are closed, the external pressure is often zero and the axial stress is:


15/19

22

2

io

iia

rr

rP

Stresses in rotating rings

)3

31)(

8

3( 2

2

22222 r

r

rrrr oioit

))(8

3( 2

2

22222 r

r

rrrr oioir

where is the mass density and is the Poissons ratio.

Interface pressure as a result of shrink or press fits

The interface pressure for same material cylinders with interface nominal radius of R and

inner and outer radii of riand ro:

)(2

))((222

2222

io

ior

rrR

rRRr

R

EP

Impact Forces

For the falling weight:

Wh

F

WW

hkF

st

e

e

211

211

IF h=0, the equivalent load is 2W. For a moving body with a velocity of V before impact, the

equivalent force is:

mkVFe

Failure of columns under compressive load (Buckling)

The critical Euler load for a beam that is long enough is:


16/19

2

2

L

EICPcr

C is the end-condition number.

The following end-condition numbers should be used for given cases:

When both end are free to pivot use C=1. When one end is fixed (prevented from rotation and lateral movement) and the

other is free, use C= 1/4 . When one end is fixed and the other end can pivot, use C=2 when the fixed end is

truly fixed in concrete. If the fixed end is attached to structures that might flexunder load, use C=1.2 (recommended).

When both ends are fixed (prevented from rotation and lateral movement), use C=4.Again, a value of C=1.2 is recommended when there is any chance for pivoting.

Slenderness ratio:

An alternate but common form of the Euler formula uses the slenderness ratio which is

defined as follows:

A

Ikwhere

k

LRatiosSlendernes

Where kis the area radius of gyration of the cross-sections.

Range of validity of the Euler formula

Euler formula is a good predictor of column failure when:

yS

EC

k

L 22

If the slenderness ratio is less than the value in the RHS of the formula, then the better

predictor of failure is the Johnson formula:

CEk

LSSAP y

ycr

1

2

2

Determinate Beams

Equations of pure bending:


17/19

Where,

M: Bending Moment [N*m]

: normal stress [N/m2]

E: Modulus of elasticity [N/m2]

R: Radius of Curvature [m]

y: Distance from neutral surface [m]

I: Moment of inertia [m4]

Indeterminate Beams

Macaulays Method (Singularity functions):

n

dx=1

n+1n+1

If positive then the brackets (< >) can be replaced by parentheses. Otherwise the

brackets will be equal to ZERO.

0n=(x-a)

n

Hooke's Law (Linear elasticity):

Hooke's Law stated that within elastic limit, the linear relationship between simple

stress and strain for a bar is expressed by equations.

M

I =

E

R =

y

E I d y2

dx2= M


18/19

,

E

P lEA l

Where, E = Young's modulus of elasticity

P = Applied load across a cross-sectional area

l= Change in length

l= Original length

Poissons Ratio:

Volumetric Strain:

V

Changeinvolume Ve

Initial volume V

Relationship between E, G, K and :

Modulus of rigidity:

2(1 )

EG

Bulk modulus:

9

3(1 2 ) 3

E KGK or E

K G

3 2

6 2

K G

K G

Stresses in Thin Cylindrical Shell

Circumferential stress (hoop stress)

2 2

c c

pd pd

t t


19/19

Where, p = Intensity of internal pressure

d = Diameter of the shell

t = Thickness of shell

= Efficiency of joint

Longitudinal stress

4 4

l l

pd pd

t t

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