Mechanics of Solids(M2H321546)

Mechanics of Solids (M2H321546)Unit -1

Simple Stresses & Strains

TopicsTypes of stresses & strains, Hookes law, stress-strain diagram, Working stress, Factor of safety, Lateral strain, Poissons ratio, volumetric strain, Elastic moduli, Deformation of simple and compound bars under axial load, Analysis of composite bar with varying cross section.2

2

Types of Stresses & Strains3Direct Stress ()When a force is applied to an elastic body, the body deforms. The way in which the body deforms depends upon the type of force applied to it.

Compressive Stress due to compressive forceA Compression force makes the body shorter.

Tensile Stress due to tensile forceA tensile force makes the body longer

4Resistance offered by the material per unit cross- sectional area is called STRESS

Tensile and compressive forces are called DIRECT FORCES

Stress is the force per unit area upon which it acts.

or Pascal (Pa) Note: Most of engineering fields used kPa, MPa, GPa. ( is called as Sigma)

5Direct Strain () Also called as Longitudinal StrainIn each case, a force F produces a deformation x . In engineering, we usually change this force into stress and the deformation into strain and we define these as follows:

Strain is the deformation per unit of the original length.

Strain, = L/L = Change in length/ Original length( is called as Epsilon)

Strain has no units since it is a ratio of length to length

LDL

6

Hookes law7

Below the yield stressStress Strain (ie) = E Where E is a constant called as Youngs Modulus or Modulus of Elasticity

Stress-Strain diagram8

StrainStressStress- Strain Curve for Mild Steel (Ductile Material)Plastic stateOf materialElastic StateOf materialYield stress Point

E = modulus of elasticity

Ultimate stress point

Breaking stress point

Stress () strain () diagramsOA: Initial region which is linear and proportionalSlope of OA is called modulus of elasticityBC: Considerable elongation occurs with no noticeable increase in stress (yielding)CD: Strain hardening changes in crystalline structure (increased resistance to further deformation)DE: Further stretching leads to reduction in the applied load and fracture OABCE: True stress-strain curve

FIG. 1-10 Stress-strain diagram for a typical structural steel in tension (not to scale)

11

Working stressThe stress to which the material may be safely subjected in the course of ordinary use. Also called as Allowable Load or Allowable stressMax load that a structural member/machine component will be allowed to carry under normal conditions of utilisation is considerably smaller than the ultimate loadThis smaller load = Allowable load / Working load / Design loadOnly a fraction of ultimate load capacity of the member is utilised when allowable load is appliedThe remaining portion of the load-carrying capacity of the member is kept in reserve to assure its safe performance12

Factor of safety13

Elastic moduli14

Modulus of Elasticity: Stress required to produce a strain of unity.Represents slope of stress-strain line OA.

AOstressstrainValue of E is same in Tension & Compression.

=E

E

AO Hookes Law:-Up to elastic limit, Stress is proportional to strain =E ; where E=Youngs modulus=P/A and = / LP/A = E ( / L) =PL /AE

E

Volumetric Strain . . . . 17

Also we know that

Problems18

Deformation of simple and compound bars under axial load19

An elastic rod of 25mm in diameter, 200mm long extends by 0.25mm under a tensile load of 40KN. Find the intensity of stress, the strain and the elastic modulus for the material of the rod.

20

Find the maximum and minimum stresses produced in the stepper bar of 12mm and 25mm diameter shown in Figure due to an axially applied compressive load od 12KN.21

A steel rod of 25mm in diameter and 2M long is subjected to an axial pull of 45KN. Find The intensity of stressThe strain andElongation. Take E= 2x105 N/mm2

22

A load of 4000N has to be raised at the end of a steel wire. If the unit stress in the wire must not exceed 80N/mm2 , what is the minimum diameter required? What will be the extension of 3.50M length of wire? Take E = 2 x 105 N/mm223

A 20mm diameter brass rod was subjected to a tensile load of 40KN. The extension of the rod was found to be 254 divisions in the 200mm extension meter. If each division is equal to 0.001mm, find the elastic modulus of brass.24

A hollow steel column has an external diameter of 250mm and an internal diameter of 200mm. Find the safe axial compressive load for the column, if the safe compressive stress is 120 N/mm225

A hollow steel column of external diameter 250mm has to support an axial load of 2000KN. If the Ultimate stress for the steel column is 480N/mm2 , find the internal diameter of the column allowing a load factor of 4.26

The following data refer to a mild steel specimen tested in a laboratory:Diameter of the specimen = 25mm; Length of the specimen = 300mm; Extension at the load of 15KN = 0.045mm; Load at the yield point = 127.65KN; Max. Load = 208.60KN; Length of the specimen after the failure = 375mm; Neck dia. = 17.75mmFind E, Yield point stress, Ultimate stress, % of elongation, % of reduction in area, Safe stress adopting a FOS of 2.

27

Analysis of composite bar with varying cross section

28

Bars of Varying cross-Section29

A bar ABCD 950mm long is made up of 3 parts AB, BC and CD of lengths 250mm, 450mm, 250mm respectively. The part BC is in square section of 30mm x 30mm. The rod is subjected to a pull of 26000N. Find the stresses in 3 parts, extension of the rod. Take E = 2 x 105 N/mm2 .30

A brass bar having a cross sectional area of 1000 mm2 is subjected to axial forces as shown in figure. Find the total change of length of the bar. Take E= 1.05 x 105 mm2

31

A member ABCD is subjected to point loads P1 ,P2 ,P3 and P4 as shown in Figure. Calculate the force P3 necessary for equilibrium if P1 = 120KN, P2 = 220KN and P4 = 160KN. Also determine the net change in length of the member. Take E = 2 x 105 N/mm2 .32

Composite Bar33

BrassSteel

A compound tube consists of a steel tube of 150mm ID & 10mm thickness and an outer brass tube of 170mm ID & 10mm thickness. The two tubes are of the same length. The compound tubes carries an axial load of 1000KN. Find the stresses and the load carried by each tube and the amount it shortens. Length of each tube is 150mm. Take E for steel= 2 x 105 N/mm2 and E for Brass= 1 x 105 N/mm234

Lateral strain35

Poissons ratio36

Volumetric strainIt is the unit change in volume due to a deformation. It is an important measure of deformation.

37

Consider a rectangular solid of sides x, y and z under the action of principal stresses 1 , 2 , 3 respectively. Then 1 , 2 , and 3 are the corresponding linear strains, than the dimensions of the rectangle and it becomes ( x + 1 . x ); ( y + 2 . y ); ( z + 3 . z )

A steel bar of 50mm wide, 12mm thick and 300mm long is subjected to an axial pull of 84KN. Find the changes in length, width, thickness and volume of the bar. Take E = 2 x 105 and Poissons ratio = 0.3238

A steel rod 4m long and 20mm diameter is subjected to an axial tensile load of 45KN. Find the change in length, diameter and the volume of the rod. Take E = 2 x 105 N/mm2 and Poissons ratio = 0.2539

Thank youfor your attention

40