Simple Stress and Strain relationship stress and strain in two dimensions, principle stresses, stress transformation, mohr circle
Normal stress σ = Eε
Shear stress ζ = G γ
Where, ε, γ are normal and shear strain respectively
E, G are Young’s modulus and shear modulus respectively A – Proportional limit point
B – Elastic limit point C – Upper yield point D – Lower yield point
E – True breaking stress point F – Nominal breaking stress point
Stress-Strain Relationship For mid steel
Proof stress is equivalent to 0.1% to 0.2% stress
Proof stress = Yield stress
Factor of safety =
Let k = bulk modulus of elasticity, μ = Poisson’s ratio a) E = 2G (1+ μ)
b) E = 3 K (1- 2μ) Relations Between E,K,G,μ c) E =
d) μ =
Theoretically poisson’s ratio μ varies from -1 to +0.5, Practically poisson’s ratio μ varies from 0 to 0.5.
Elongation of bars of different section 1. For uniformly tapered circular bars
Stress A B C D
E
F
Strain
2. Uniformly tapered rectangular bars
3. Elongation of a uniform section bar due to self weight =
Where 2
W = Weight of bar A = Area of cross section
4. Elongation of a bar uniformly tapered due to self weight
= 2 =
Where A = Average area of cross section = Unit weight of material of a bar 6 5. Elongation of a stepped bar
=
( )
= 4PL
Ed d
= + +
= [ + + ]
Analysis of a compound bar 1. Condition of equilibrium
P + P = P
2. Condition of compatibility P= L
E = P L E
3. Equivalent modulus of elasticity PL
For composite material consisting of steel and copper rod,
For thermal expansion, as >
Copper will be in compression and steel will be in
tension.
Governing equation will be ( )∆t = + Temperature Stresses
1. In uniform section fixed from both sides change in length = tL
where
= coefficient of thermal expansion t = change in temperature
σ = (
L) E σ = t E It support yields by ‘a’
then σ =( a)
L E = ( tL a)E L
2. In a tapered circular section fixed from both sides Force ‘P’ induced due to temperature change ‘t’
= Where t is temperature change
Strain energy = ∪ =
Module of resilience is the strain energy per unit volume = u = ∪
= = Eε
Proof resistance is applicable when σ = yield stress = σ up = Ζ, ζ’ are complementary shear stress σ + σ = σ + σ
Normal stress, σ = . / + . / cos2θ - Sin 2θ Shear stress, = sin 2θ + = cos 2θ
Principle stresses are direct normal stresses acting on mutually perpendicular planes on which shear stresses are zero.
If σ and σ are the principle stresses (major and minor respectively) σ , σ = . /±√. / +
The plane of maximum shear stress lies at 45 to the plane of principal stress
= and on that plane of ,σ =
If the plane on which σ is maximum makes θ angle with vertical the plane on which is maximum, makes (45+ θ) angle with vertical,
Tan 2 θ = σ maximum Cot 2 θ =
ζ maximum
In case of pure shear element, the principle stresses act at 45 to the plane of pure shear stress
Stress invariants
I = (σ + σ + σ ) = σ + σ + σ = σ + σ + σ i.e. sum of stresses on any plane is constant I = (σ σ + σ σ + σ σ ) = σ σ + σ σ + σ σ Mohr’s Circle
Transformation equation for strains
=
=
=
,
( +
)
= ( +
) =
ε =ε + ε
2 + .ε ε
2 / cos 2θ +γ
2 = sin 2θ
=ε + ε
2 .ε ε
2 / cos 2θ γ
2 = sin 2θ
2 = .ε ε
2 / cos 2θ +γ
2 = sin 2θ Maximum normal strain
= ε + ε
2 + √.ε ε
2 / + .γ 2 /
= ε + ε
2 √.ε ε
2 / + .γ 2 / To find plane of principal strain use tan 2 =
Maximum shear strain
To find plane of maximum shear strain
tan 2 = (ε ε γ )
= √.ε ε
2 / + .γ 2 / Sign convention for strain
+
+ , +
Properties of Mohr’s Circle
a) Center of Mohr’s Circle, C = = b) Radius of Mohr’s Circle r = =
= √. / + xy c) Major principle stress, σ = C + r
d) Minor principle stress, σ = C – r
e) In case of pure shear elements σ = + and σ = - and centre of the circle coinsides with the origin
Octahedral stresses
Stress on a octahedral plane (a plane which is equally inclined it all three principal axis of reference
For such plane
= = = 1
√3
Where , , are direction cosines
= + +
3 =
3
=( ) + ( ) + ( ) 9
=√2
3 √ 3
Simple bending theory, flexural and shears tresses, unsymmetrical bending
Let, y = distance of any particular section from neutral axis stress due to bending, σ = y
σ
σ σ
σ = +
σ = -
Where M = Bending moment
I = Moments of inertia about neutral axis Also, =
Where E = Young’s modulus R = Radius of curvature Flexure formula
= =
Some Special Cases
For the same square beam, and for a given stress, the ratio of moments of resistance, = √2
To achieve same strength under same stresses, the relation regardings weight among rectangular, circular and square section is
W <W <W
The ratio of depth to width of the strongest beam that can be cut from a circular log is 1.414
Moment of resistance m = σZ M Z where Z section modulus
As at the top of beam, y = maximum and at the neutral axis y = 0, bending stress at top is maximum and at neutral axis is zero
Shear stress for Different Section
i) For rectangular section with shear force V,
Shear stress =
. y /, d = depth
Average shear stress =
, b = width
Maximum shear stress = ii) For circular section with shear force v,
Shear stress = (r y ), r = radius of circle
Average shear stress =
y
X X
y
b
b
b b y
y x x
Position (ii) Position (i)
Maximum shear stress =
At y = local shear stress is equal to the average shear stress
iii) For triangular section, with shear force v,
Shear stress =
(h-y)y, b = base width h = height
Average shear stress =
Maximum shear stress = iv) For tubular section, with shear force V
Average shear stress =
( )
Same typical shear distribution diagram
Shear centre, thin walled pressure vessels, uniform torsion, bulking of column, combined and direct bending stress.
Shear centre
Lateral load acting on a beam through shear centre will produce bending without torsion
Shear centre is the centre of flexure
T-section I-section
For channel sections,
Shear centre distance e =
Where, I = moments of inertia about axis of symmetry For semi circular sections,
e =
Thin walled Pressure vessels
Let p = internal pressure, t = thickness d= diameter μ = = Poisson’s ratio E = Young’s modulus
For Cylinders
1. Hoop stress or circumferential stress, F = 2. Longitudinal stress, F =
3. Maximum shear stress, = =
4. Hoop strain, E =
.2 /
5. Longitudinal strain, ε = .1 / 6. Volumetric strain ε =
.5 /
For Spherical shells
1. Hoop stress (f ) = longitudinal stress (f ) = 2. Maximum shear stress = 0
3. Strain in any direction, ϵ = .1 / 4. Volumetric strain, ϵ =
.1 / Uniform Torsion
Let T = Total torque acting on the section e
tw tf h/2
h/2
b c
v
e r
0
V
J = Polar moment of inertia
= Shear stress at any radial distance R = Radius of shaft
G = Rigidity modulus
= Angular momentum due to strain in length of shaft L = Length of shaft
θ = Angular strain in cross-section T
J = = Gθ L
For solid shaft of diameter d, T =
For hollow shaft of outer diameter do and inner diameter di, T =
(d d )
Power transmitted by shaft, p =
KW Where N = Rotation per minute (rpm)
T = means Torque (KN.m)
Strain energy in torsion U =1
2 T θ
For shaft in series,
Torque will be same for both the shafts
1. Twists will be different for both the shafts Let twist are θ for AB and θ for BC
= . /
T + T = T = + +
⁄at = ⁄at
A Dia = d B Dia = d C
l l
A B
T d
d
C
l l l l l
T
= + + =
Shaft of varying diameter 1. θ = ( )0 1
2. Indeterminate system of shafts
=
Column stability
Short column – fails by crushing Long column – fails by bucking
Intermediate columns – fails by combinations of crushing & buckling
Buckling Of Columns A. Euler’s Theory
Assumptions
1. Flexural rigidity EI is uniform
2. After unloading axis the column should be perfectly straight 3. Material is isotropic and homogeneous
4. The line of thrush will coincide exactly with the unrestrained axis of the column
Buckling load, P =
( ) ( )
=
. / Where L
r = called stenderness ratio Where, L = effective length of the column
End conditions Effective length (of original length L )
Both end hanged L =L
B. Variation of bucking load/Euler load with slenderness ratio If is less than a certain limit , . /
We cannot use Euler’s theorem
Rankine – Gordon formula (for intermediate columns) 1
P = 1 P + 1
P Where
P = Rankine buckling load P = Direct compressive load P = Euler’s load
Combined bending and direct stress
Let normal stress σ is due to bending M and is shear stress due to tourism T for shaft of diameter d,
Then, σ =
= =
If the major and minor principal stress are σ = σ respectively,
σ , = ±√. / +
Maximum shearing stress = = ±√. / +
σ = (M + √M + T ) and
= √M + T
d = √ (M + √M + 2)
Where, σ = allowable working stress intension Again d = √ √M + T
Where, = allowable working stress in shear
σ σ