Strength of

Material (Formula & Short Notes)

Stress and strain

Stress = Force / Area

( )t

L ChangeinlengthTensionstrain e

L Initial length

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:

STRAIN ENERGY Energy Methods:

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

U = ∫ P ² / ( 2AE ) dx limit 0 to L

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

E = Young’s modulus.

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

U = ∫ M ² / ( 2EI ) dx limit 0 to L

Where, M = Bending moment due to applied loads, E = Young’s modulus, and I = Moment of

inertia.

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

U = ∫ T ² / ( 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 ² / ( 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 = τ ² V / ( 2G )

Where, τ = Shear Stress

G = Shear modulus or Modulus of rigidity

V = Volume of the material.

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

U = M ² L / (2EI)

Where, M = Bending moment

L = Length of the beam

E = Young’s modulus

I = Moment of inertia

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

U = T ²L / ( 2GJ )

Where, T = Applied Torsion

L = Length of the beam

G = Shear modulus or Modulus of rigidity

J = Polar moment of inertia

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

U = P²L / ( 2AE )

Where,

P = Applied tensile load.

L = Length of the member

A = Area of the member

E = Young’s modulus.

(ix) Castigliano’s first theorem:

δ = Ә U/ Ә P

Where, δ = 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 ¼ times the deflection of a simply supported beam.

(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 (6l2 – 8la + 3a2)/

12 l2

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

-wl2 / 30

-wl2 / 30

-5 wl2/ 96

-5 wl2/ 96

M / 4

M / 4

Euler’s formula for different end conditions:

1. Both ends fixed:

PE = л 2 EI / ( 0.5L)2

2. Both ends hinged :

PE = л 2 EI / (L)2

3. One end fixed ,other end hinged:

PE = л 2 EI / ( 0.7L)2

4. One end fixed, other end free:

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

Rakine’s formula:

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

where, PR = Rakine’s critical load

fC = yield stress

A = cross sectional area

a = Rakine’s constant

leff = effective length

r = radius of gyration

Euler’s formula for maximum stress for ‘a’ initially bent column:

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

Where, P = axial load

A = cross section area

PE = Euler’s load

a = constant

Z = section modulus

Euler’s formula for maximum stress for a eccentrically loaded column:

σmax = 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 = Euler’s 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 ( d 2y / dx 2 )

Maximum Principal Stress Theory ( Rakine’s theory):

σ 1 = f y.

where σ 1 is the maximum Principal Stress, and f y is elastic limit stress.

Maximum Principal Strain Theory ( St. Venant’s theory):

e 1 = f y / E

In 3D, e 1 = 1/E[ σ 1 – (1/m)( σ 2 + σ 3) ] = f y / E → [ σ 1 – (1/m)( σ 2 + σ 3) ] = f y

In 2D, σ 3 = 0 → e 1 = 1/E[ σ 1 – (1/m)( σ 2 ) ] = f y / E → [ σ 1 – (1/m)( σ 2 ) ] = f y

Maximum Shear Stress Theory (Tresca’s theory) :

In 3D, ( σ 1 - σ 3) / 2 = f y /2 → ( σ 1 - σ 3) = f y

In 2D, ( σ 1 - σ 2) / 2 = f y /2 → σ 1 = f y

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

theory):

In 3D, shear strain energy due to distortion:

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

Shear strain energy due to simple tension:

U = f y 2 / 6G

(1/ 12G)[ ( σ 1 - σ 2)2 + ( σ 2 - σ 3) 2 + ( σ 3 - σ 1) 2 ] = f y 2 / 6G

[ ( σ 1 - σ 2)2 + ( σ 2 - σ 3) 2 + ( σ 3 - σ 1) 2 ] = 2 f y 2

In 2D, [ ( σ 1 - σ 2)2 + ( σ 2 - 0) 2 + ( 0 - σ 1) 2 ] = 2 f y 2

Maximum Strain Energy Theory (Beltrami Theory):

In 3D, strain energy due to deformation:

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

Strain energy due to simple tension:

U = f y 2 / 2E

(1/ 2E)[σ 12 + σ 22 + σ 32 -(2/m)( σ 1σ 2 + σ 2σ 2 + σ 2σ 2 )] = f y 2 / 2E

[σ 12 + σ 22 + σ 32 -(2/m)( σ 1σ 2 + σ 2σ 2 + σ 2σ 2 )] = f y 2

In 2D, [ σ 12 + σ 22 - (2/m)( σ 1σ 2 )] = f y 2

Failure theories and its relationship between tension and shear:

1. Maximum Principal Stress Theory ( Rakine’s theory):

ζ y = f y

2. Maximum Principal Strain Theory( St. Venant’s theory):

ζ y = 0.8 f y

3. Maximum Shear Stress Theory ( Tresca’s theory):

ζ y =0.5 f y

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

theory):

ζ y= 0.577 f y

4. Maximum Strain Energy Theory ( Beltrami Theory):

ζ y= 0.817f y .

Volumetric strain per unit volume:

f y 2 / 2E

Torque, Power, and Torsion of Circular Bars:

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

63000

TnH

Where H is power in Hp, T is torque in lb-in, and n is shaft speed in rpm.

In SI units:

TH

Where H is power in Watts, T is 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

J is 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:

3

48

L

EIK

For cantilevered beams of length L:

3

3

L

EIK

Torsional stiffness of a solid or tubular bar is:

L

GJK t

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.

K2 K1

F

T Kt1

Kt2

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

F F

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

cross-section. Ro and Ri are 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

y1

b

A1

y1

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:

t

Pdit

2

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

iooiooii

rrr

r

PPrrrPrP

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

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 Poisson’s 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 ri and 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:

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 flex under 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 k is 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:

Where,

M: Bending Moment [N*m]

σ: 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

Macaulay’s Method (Singularity functions):

<x-a>ndx=1

n+1<x-a>

n+1

x>a

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

brackets will be equal to ZERO.

0<x<a<x-a>n= 0

x>a<x-a>n= (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=σ

y

E Id y

2

dx2 = M

,

E

P lE

A l

Where, E = Young's modulus of elasticity

P = Applied load across a cross-sectional area

l = Change in length

l = Original length

Poisson’s 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

Where, p = Intensity of internal pressure

d = Diameter of the shell

t = Thickness of shell

η = Efficiency of joint

Longitudinal stress

4 4

l l

pd pd

t t