Mechanics of Solids (M2H321546) Unit -1 Simple Stresses & Strains

Topics Types of stresses & strains, Hooke’s law, stress-strain diagram, Working stress, Factor of safety, Lateral strain, Poisson’s ratio, volumetric strain, Elastic moduli, Deformation of simple and compound bars under axial load, Analysis of composite bar with varying cross section.

Types of Stresses & Strains Direct Stress (σ) 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. Tensile Stress due to tensile force A tensile force makes the body longer Compressive Stress due to compressive force A Compression force makes the body shorter.

Tensile and compressive forces are called DIRECT FORCES Resistance 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) ( σ is called as Sigma) Note: Most of engineering fields used kPa, MPa, GPa.

Direct Strain (ε) Also called as Longitudinal Strain In 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, ε = ΔL/L = Change in length/ Original length (ε is called as Epsilon) Strain has no unit’s since it is a ratio of length to length L DL

Hooke’s law σ = E ε Below the yield stress Stress α Strain (ie) σ α ε Where E is a constant called as Youngs Modulus or Modulus of Elasticity

Stress-Strain diagram

E = modulus of elasticity Stress- Strain Curve for Mild Steel (Ductile Material) Yield stress Point Ultimate stress point Breaking stress point Plastic state Of material Stress Elastic State Of material E = modulus of elasticity Strain

Stress (σ) – strain (ε) diagrams FIG. 1-10 Stress-strain diagram for a typical structural steel in tension (not to scale) OA: Initial region which is linear and proportional Slope of OA is called modulus of elasticity BC: 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

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

Factor of safety

Elastic moduli

Modulus of Elasticity: Stress required to produce a strain of unity. Represents slope of stress-strain line OA.  =E   Value of E is same in Tension & Compression. A stress E O  strain

A   O E Hooke’s Law:- Up to elastic limit, Stress is proportional to strain     =E ; where E=Young’s modulus =P/A and  =  / L P/A = E ( / L)  =PL /AE

Volumetric Strain . . . . Also we know that

Problems

Deformation of simple and compound bars under axial load

An elastic rod of 25mm in diameter, 200mm long extends by 0 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.

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.

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

A load of 4000N has to be raised at the end of a steel wire 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/mm2

A 20mm diameter brass rod was subjected to a tensile load of 40KN 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.

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/mm2

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.

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. = 17.75mm Find E, Yield point stress, Ultimate stress, % of elongation, % of reduction in area, Safe stress adopting a FOS of 2.

Analysis of composite bar with varying cross section

Bars of Varying cross-Section

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 .

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

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 .

Composite Bar Steel Brass

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/mm2

Lateral strain "The strain at right angles to the direction of applied load is known as LATERAL STRAIN." (OR) "Change in breadth dimensions to original dimension is also known as LATERAL STRAIN" 𝑳𝒂𝒕𝒆𝒓𝒂𝒍 𝑺𝒕𝒓𝒂𝒊𝒏= 𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝑳𝒂𝒕𝒆𝒓𝒂𝒍 𝒅𝒊𝒎𝒆𝒏𝒔𝒊𝒐𝒏𝒔 𝑶𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒅𝒊𝒎𝒆𝒏𝒔𝒊𝒐𝒏𝒔

Poisson’s ratio

Volumetric strain It is the unit change in volume due to a deformation. It is an important measure of deformation. Consider a rectangular solid of sides x, y and z under the action of principal stresses σ1 , σ2 , σ3 respectively. Then ε1 , ε2 , and ε3 are the corresponding linear strains, than the dimensions of the rectangle and it becomes ( x + ε1 . x ); ( y + ε2 . y ); ( z + ε3 . 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 Poisson’s ratio = 0.32

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 Poisson’s ratio = 0.25

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