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- 1. INDUS INSTITUTE OFTECHNOLOGY & ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING Name : Keval Patel Subject Code : Subject Name : Mechanics of solid B.E Mechanical Engineering , 1st Semester 1
- 2. LOAD• LOAD:- any external force acting upon a machine part Dead Live Force Suddenly applied Impact 2
- 3. Stress (σ) = F / AF= 1 MPa = 1 × 106 N/m2 = 1 N/ mm2R= 3
- 4. Strain (ε) = δl / L hδl= L - l l 4
- 5. Tensile Stress and Strain• When a body is subjected to two equal and opposite axial pulls P (also called tensile load) , then the stress induced at any section of the body is known as tensile stress• Tensile load, there will be a decrease in cross-sectional area and an increase in length of the body. The ratio of the increase in length to the original length is known as tensile strain. 5
- 6. 6
- 7. Compressive Stress and Strain• When a body is subjected to two equal and opposite axial pushes P (also called compressive load) , then the stress induced at any section of the body is known as compressive stress• Compressive load, there will be an increase in cross-sectional area and a decrease in length of the body. The ratio of the decrease in length to the original length Is known as compressive strain 7
- 8. 8
- 9. Youngs Modulus or Modulus of Elasticity Hookes law:- states that when a material is loaded within elastic limit, the stress is directly proportional to strain, σ ∝ ε or σ= E×ε 9
- 10. Example1. A circular rod of diameter 16 mm and 500 mm long is subjected to a tensile force 40 kN. The modulus of elasticity for steel may be taken as 200 kN/mm2. Find stress, strain and elongation of the bar due to applied load. 10
- 11. Shear Stress and Strain• When a body is subjected to two equal and opposite forces acting tangentially across the resisting section, as a result of which the body tends to shear off the section, then the stress induced is called shear stress (τ), The corresponding strain is known as shear strain (φ) 11
- 12. Shear Modulus or Modulus of Rigidity• It has been found experimentally that within the elastic limit, the shear stress is directly proportional to shear strain.Mathematically τ∝φ or τ=G.φ or τ/φ=G• where, τ = Shear stress,• φ = Shear strain,• G = Constant of proportionality, known as shear modulus or modulus of rigidity. It is also denoted by N or C. 12
- 13. Stress-strain Diagram 13
- 14. 14
- 15. 15
- 16. 16
- 17. Working Stress• When designing machine parts, it is desirable to keep the stress lower than the maximum or ultimate stress at which failure of the material takes place. This stress is known as the working stress or design stress.• It is also known as safe or allowable stress. working stress = design stress = safe stress = allowable stress 17
- 18. Factor of Safety The ratio of the maximum stress to the working stress. 18
- 19. 1: A hollow steel tube is to be used to carry an axial compressive load of 160 kN. The yield stress for steel is 250 N/mm2. A factor of safety of 1.75 is to be used in the design. The following three class of tubes of external diameter 101.6 mm are available.Class Thickness• Light 3.65 mm Medium 4.05 mm Heavy 4.85 mmWhich section do you recommend?2: A specimen of steel 20 mm diameter with a gauge length of 200 mm is tested to destruction. It has an extension of 0.25 mm under a load of 80 KN and the load at elastic limit is 102 KN. The maximum load is 130 KN. The total extension at fracture is 56 mm and diameter at neck is 15 mm. Find (i) The stress at elastic limit. (ii) Young’s modulus.(iii) Percentage elongation. (iv) Percentage reduction in area.(v) Ultimate tensile stress. 19
- 20. Stresses in Composite Bars• A composite bar may be defined as a bar made up of two or more different materials, joined together, in such a manner that the system extends or contracts as one unit, equally, when subjected to tension or compression.1. The extension or contraction ofthe bar is being equal2. The total external load on the bar isequal to the sum of the loads carriedby different materials. 20
- 21. • P1 = Load carried by bar 1,• A1 = Cross-sectional area of bar 1,• σ1 = Stress produced in bar 1,• E1 = Youngs modulus of bar 1,• P2, A2, σ2, E2 = Corresponding values of bar 2,• P = Total load on the composite bar,• l = Length of the composite bar, and• δl = Elongation of the composite bar.• We know that P = P1 + P2• Stress in bar 1,• strain in bar 1, 21
- 22. • Elongation in bar -1:• Elongation in bar -2: There fore, δl1 = δl2• The ratio E1 / E2 is known as modular ratio of the two materials 22
- 23. P = P1 + P2 = = 23
- 24. δl1 = δl2P = P1 + P2 = σ1.A1 + σ2.A2 24
- 25. EX:3 A bar 3 m long is made of two bars, one of copper havingE = 105 GN/m2 and the other of steel having E = 210 GN/m2.Each bar is 25 mm broad and 12.5 mm thick. This compoundbar is stretched by a load of 50 kN. Find the increase in lengthof the compound bar and the stress produced in the steel andcopper. The length of copper as well as of steel bar is 3 meach. Ans: Pc=16.67N Ps=33.33N δl=1.52 mm 25
- 26. EX:4• A central steel rod 18 mm diameter passes through a copper tube 24 mm inside and 40 mm outside diameter, as shown in Fig. It is provided with nuts and washers at each end. The nuts are tightened until a stress of 10 MPa is set up in the steel. Find out stress generated in copper tube. Hint: Ps=Pc σs × As = σc × Ac
- 27. BARS WITH CROSS-SECTIONS VARYING IN STEPS• A typical bar with cross-sections varying in steps and subjected to axial load• length of three portions L1, L2 and L3 and the respective cross- sectional areas are A1, A2, A3• E = Young’s modulus of the material• P = applied axial load. 27
- 28. • forces acting on the cross-sections of the three portions. It is obvious that to maintain equilibrium the load acting on each portion is P only. 28
- 29. Stress, strain and extension of each of these portions are:Portion Stress Strain Extension 1 σ1 = P/ A1 e1 = σ1 / E δ1 = P L1 / A1 E 2 σ2 = P/ A2 e2 = σ2 / E δ2 = P L2 / A2 E 3 σ3 = P/ A3 e3 = σ3 / E δ3 = P L3 / A3 E 29
- 30. Total Elongation: δδ1 + δ2 + δ3 = [P L1 / A1 E] + [P L2 / A2 E] + [P L3 / A3 E] 30
- 31. EX:5The bar shown in Fig. 8.16 is tested in universal testingmachine. It is observed that at a load of 40 kN the totalextension of the bar is 0.280 mm. Determine the Young’smodulus of the material. 31
- 32. Thermal Stresses Stresses due to Change in Temperature Whenever there is some increase or decrease in the temperature of a body, it causes the body to expand or contract. If the body is allowed to expand or contract freely, with the rise or fall of the temperature, no stresses are induced in the body. But, if the deformation of the body is prevented, some stresses are induced in the body. Such stresses are known as thermal stresses. 32
- 33. • l = Original length of the body,• t = Rise or fall of temperature,• α = Coefficient of thermal expansion,∴ Increase or decrease in length, δl = l × α × t• If the ends of the body are fixed to rigid supports, so that its expansion is prevented, then compressive strain induced in the body,• ∴ Thermal stress, 33
- 34. If the free expansion is prevented fully• Since support is not permitting it, the support force P develops to keep it at the original position.• Magnitude of this force is such that contraction is equal to free expansion δl = l × α × t 34
- 35. If free expansion is prevented partiallyExpansion prevented Δ = α tL – δ 35
- 36. EX:6: A steel rail is 12 m long and is laid at a temperature of 18°C. The maximum temperature expected is 40°C.(i) Estimate the minimum gap between two rails to be left so that the temperature stresses do not develop.(ii) Calculate the temperature stresses developed in the rails, if:(a) No expansion joint is provided.(b) If a 1.5 mm gap is provided for expansion.(iii) If the stress developed is 20 N/mm2, what is the gap provided between the rails?Take E = 2 × 105 N/mm2 and α = 12 × 10–6/°C. 36
- 37. Linear and Lateral Strain• Consider a circular bar of diameter d and length l, subjected to a tensile force P• Due to tensile force, the lengthof the bar increases by an amount δl and the diameter decreases byan amount δd• Similarly, if the bar is subjected to a compressive force,• Every direct stress is accompanied by a strain in its own direction is known as linear strain and an opposite kind of strain in every direction, at right angles to it, is known as lateral strain. 37
- 38. Poissons Ratio• When a body is stressed within elastic limit, the lateral strain bears a constant ratio to the linear strain.• This constant is known as Poissons ratio and is denoted by (1/m) or μ. 38
- 39. Volumetric Strain• When a body is subjected to a system of forces, it undergoes some changes in its dimensions. The volume of the body is changed.• The ratio of the change in volume to the original volume is known as volumetric strain.• Volumetric strain, εv = δV / V ; δV = Change in volume, ; V = Original volume.• Volumetric strain of a rectangular body subjected to an axial force is given as• Volumetric strain of a rectangular body subjected to three mutually perpendicular forces is given by 39
- 40. Bulk Modulus• When a body is subjected to three mutually perpendicular stresses, of equal intensity, then the ratio of the direct stress to the corresponding volumetric strain is known as BULK MODULUS.• It is usually denoted by K.• Bulk modulus, 40
- 41. Relation Between Bulk Modulus and Young’s Modulus Relation Between Young’s Modulus and Modulus of Rigidity 41
- 42. EX:7A bar of 25 mm diameter is tested in tension. It is observed thatwhen a load of 60kN is applied, the extension measured over agauge length of 200 mm is 0.12 mm and contraction in diameter is0.0045 mm. Find Poisson’s ratio and elastic constants E, G, K.EX:8A circular rod of 25 mm diameter and 500 mm long issubjected to a tensile force of 60 KN. Determine modulus ofrigidity, bulk modulus and change in volume if Poisson’s ratio= 0.3 and Young’s modulus E = 2 × 105 N/mm2. 42
- 43. EX:9 A 400 mm long bar has rectangular cross-section 10 mm × 30 mm. This bar is subjected to(i) 15 kN tensile force on 10 mm × 30 mm faces,(ii) 80 kN compressive force on 10 mm × 400 mm faces, and(iii) 180 kN tensile force on 30 mm × 400 mm faces.Find the change in volume if E = 2 × 105 N/mm2 and μ = 0.3. 43
- 44. Stress concentration• Whenever a machine component changes the shape of its cross- section, the simple stress distribution no longer holds good.• This irregularity in the stress distribution caused by abrupt changes of form is called stress concentration.• It occurs for all kinds of stresses in the presence of fillets, notches, holes, keyways, splines, surface roughness or scratches etc. 44
- 45. Thank You… 45

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