CE8395-STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS
DEPARTMENT OF MECHANICAL ENGINEERING
R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS
Strength of Material For Mechanical Engineers CE 8395
(Formula & Short Notes) repared by,
VIJAYAN R Assistant Professor, VEL TECH, Avadi, Chennai.
2
R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS
Stress and strain
Stress = Force / Area
L
Tension strain(e ) =
=
Change in length
L
Initial Initial length length
3
R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS
Brinell Hardness Number (BHN)
Elastic constants:
where, P = Standard load, D = Diameter of steel ball, and d = Diameter of the indent.
4 R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS
Axial Elongation of Bar Prismatic Prismatic Bar Due Due to External Load Load
∆=
Elongation of Prismatic Bar Due to Self Weight
Where is specific weight
∆= =
Elongation of Tapered Bar •
Circular Tapered
∆=
•
Rectangular Tapered
∆= .( − ) Stress Induced by Axial Stress and Simple Shear •
Normal stress
•
Tangential stress
Principal Stresses and Principal Planes •
Major principal stress
•
Major principal stress
5 R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS
Principal Strain
Mohr’s CircleCircle-
STRAIN ENERGY Energy Methods: (i) Formula to calculate the strain energy due to axial loads (tension):
U = ∫ P ² / ( 2AE)dx
limit 0 toL
Where, P = Applied tensile load, L = Length of the member member , A = Area of the member, and
E = Young’smodulus. (ii) Formula to calculate the strain energy due tobending:
U = ∫ M ² / ( 2EI) dx
limit 0 toL
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 totorsion:
U = ∫ T ² / ( 2GJ) dx
limit 0 toL 6
R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS
Where, T = Applied Torsion , G = Shear modulus or Modulus of rigidity, and J = Polar moment ofinertia (iv) Formula to calculate the strain energy due to pureshear:
U =K ∫ V ² / ( 2GA) dx Where,
limit 0 to L
V= Shearload G = Shear modulus or Modulus M odulus 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 isgiven:
U = τ ² V / ( 2G ) Where,
τ = ShearStress G = Shear modulus or Modulus M odulus of rigidity V = Volume of the material.
(vi) Formula to calculate the strain energy , if the moment value isgiven:
U = M ² L / (2EI) (2EI) Where,
M = Bending moment L = Length of the beam
E = Young’smodulus I = Moment ofinertia (vii) Formula to calculate the strain energy , if the torsion moment value isgiven:
U= Where,
T ²L / ( 2GJ)
T = AppliedTorsion L = Length of the beam G = Shear modulus or Modulus M odulus of rigidity J = Polar moment of inertia 7
R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS
(viii) Formula to calculate the strain energy, if the applied tension load isgiven:
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: 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 / 192EI
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 / 384EI
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 EIl3
Stresses due to •
Gradual Loading:-
•
Sudden Loading:-
8 R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS
•
Impact Loading:-
Deflection,
Thermal Stresses:-
∆ = ∆ = ∆ When bar is not totally free to expand and can be expand free by “a”
= ∆ ∆ − Temperature Stresses in Taper Bars:-
= ∆ =
Tempertaure Stresses in Composite Bars
Hooke's Law (Linear elasticity):
Hooke's Law stated that within elastic ela stic limit, the linear relationship between simple stress and strain for a bar is expressed by equations.
9 R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS
,
= E
= ∆
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:
10
R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS
Relationship between E, G, K and µ: •
Modulus of rigidity:-
•
Bulk modulus:-
•
= 2( 2 (1 + ) = 3( 3(11 − 2) = 9 + 3 − 2 = 3 + 3
Shear Stress in Rectang ular Beam
Compound Stresses •
Equation of Pure Bending
•
Section Modulus
•
Shearing Stress
Where, V = Shearing force
̅ =First moment of area 11 R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH CE8395-STREN GTH OF MATERIALS FOR MECHANICAL ENGINEERS
•
Shear Stress Circular Beam
Moment of Inertia and Section Modulus
12 R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH CE8395-STREN GTH OF MATERIALS FOR MECHANICAL ENGINEERS
•
Direct Stress
=
where P = axial thrust, A = area of cross-section •
=
Bending Stress
where M = bending moment, y- distance of fibre from neutral axis, I = moment of inertia. •
=
Torsional Shear Stress
where T = torque, r = radius of shaft, J = polar moment of inertia.
√ + + √ +
Equivalent Torsional Moment Equivalent Bending Moment
13 R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH CE8395-STREN GTH OF MATERIALS FOR MECHANICAL ENGINEERS
Shear force and Bending Moment Relation
= −
14 R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH CE8395-STREN GTH OF MATERIALS FOR MECHANICAL ENGINEERS
Euler’s Buckling Load
=
For both end hinged = l For one end fixed and other free = 2l For both end fixed = l/2 For one end fixed and other hinged = l/ √ √ Slenderness Ratio ( λ )
15 R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH CE8395-STREN GTH OF MATERIALS FOR MECHANICAL ENGINEERS
Rankine’s Formula for Columns
•
P R =
•
P cs cs
Crippling load by Rankine’s formula
= σcs A = Ultimate crushing load for column
Crippling load obtained by Euler’s formula
•
Deflection in different Beams
Torsion
Where, T = Torque, • • • •
J = Polar moment of inertia G = Modulus of rigidity, θ = Angle of twist L = Length of shaft,
16 R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH CE8395-STREN GTH OF MATERIALS FOR MECHANICAL ENGINEERS
Total angle of twist
• • •
• •
GJ = Torsional rigidity
= Torsional stiffness = Torsional flexibility = Axial stiffness = Axial flexibility
Moment of Inertia About polar Axis •
Moment of Inertia About polar Axis
•
For hollow circular shaft
Compound Shaft •
Series connection
Where, θ1 = Angular deformation of 1 st shaft θ2 = Angular deformation of 2 nd shaft •
Parallel Connection
17 R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH CE8395-STREN GTH OF MATERIALS FOR MECHANICAL ENGINEERS
Strain Energy in Torsion
For solid shaft,
For hollow shaft,
Thin Cylinder •
Circumferential Circumferential Stress /Hoop / Hoop Stress
η = Efficiency of joint •
Longitudinal Stress
•
Hoop Strain
•
Longitudinal Strain
•
Ratio of Hoop Strain to Longitudinal Strain
18 R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH CE8395-STREN GTH OF MATERIALS FOR MECHANICAL ENGINEERS
Stresses in Thin Spherical Shell •
Hoop stress/longitudinal stress/longitudinal stress
•
Hoop stress/longitudinal strain
•
Volumetric strain of sphere
Thickness ratio of Cylindrical Shell with Hemisphere Ends
Where v=Poisson Ratio
19 R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING
CE8395-STRENGTH CE8395-STREN GTH OF MATERIALS FOR MECHANICAL ENGINEERS
20 R VIJAYAN
DEPARTMENT OF MECHANICAL ENGINEERING