Unit 2 simple stresses and strain

Simple stress and strain for various material

1. MECHANICS OF STRUCTURE Simple Stresses and Strains Presented by Mr. P. D. Akarte

2. Mechanical properties of material Introduction : The Various type of structure and its various member subjected to various type of load are made from various compound and composite material. Fundamental concept likes various types of stress , strain, elastic constant and property of material are basically needed to understand the subject in proper way. Mechanical Property of material 1. Strength 2. Stiffness 3. Elasticity 4. Plasticity 5. Malleability 6. Ductility 7. Brittleness 8. Hardness 9. Toughness 10. Fatigue 11. Creep

3. Type of Load

4. Axial or Direct Load : When a load whose line of action is passing through axis of body Eccentric Load : When load whose line of action is not passing through axial of body

5. Shear Force or Load : A Force or load which act tangentially to the plane of body Stress : It is the internal resisting force per unit area Unit ; N/mm2 Strain : It is the ratios of change length to the original length. Unit : No Unit

6. Shear stress : ( q) When two equal and opposite forces acts tangentially on the body so the body tends to slide across the section due to this stress is developed is called as shear stress. Unit : N/mm2

7. Types of Strain Longitudinal Strain : It is define as the ratio of change in length to the original length. Lateral Strain: It the ratio of change in lateral dimension to the original dimension Volumetric strain: Volumetric strain is defined as the ratio of change in volume to original volume.

8. Elasticity : It is define as the property of the material due to which the material regain its original shape and size after removal of the external load is called as elasticity. Hooke’s law : It is that when a elastic material is loaded within its elastic limit, then stress produce is directly proportional to strain. Mathematically,

9. Modulus of Elasticity Or Young’s Modulus ( E ) Within elastic limit, the ratio of stress to strain is a constant and this constant is called as constant of proportionality.

10. Change in length or deformation of body Suppose a bar of length L is subjected to force of P KN so that deformation produce shown in below fig. Where, P = Force applied. A = Cross section area. E = Modulus of elasticity. L = length of bar.

11. Some Important formulae

12. Q.1 Calculate longitudinal stress developed in 2 cm diameter bar undergo tensile force of 120 kN. Solution: Diameter d = 2 cm = 2 x 10 = 20 mm. Tensile force P = 120 kN = 120 x 10³ N we know, Stress ( σ ) = P / A A = л /4 x 20² = 120 x 10³/ 314.15 = -----------

13. Q.2 Determine the tensile force on a steel bar of circular cross section 25 mm diameter, if strain equal to 0.75 x 10 -3 . Consider E for steel is 200 Gpa . Given Data : Diameter of steel bar : 25 mm Strain ( e ) : 0.75 x 10 -3 E= 200 Gpa = 200 x 10 3 To find : P = ?

14. Step 1: To find stress in the bar We know σ = E x e = 200 x 10 3 x 0.75 x 10 -3 = ------ N/ mm 2 Step 2: To find Tensile force in the bar σ = P/A P = σ x A =

15. Que: A steel rod of cross sectional area 20 x 20 mm and length of 1.5m long subjected to a axial pull of 50 kN. Calculate the Stress developed in rod. Also calculate change in length. Take E= 200 Gpa. Solution: - 1. Stress developed in rod = axial pull /area = 50 x 1000 / 20 x 20 = 12.5 N/ mm 2 2.Change in length (∂ l ) = Pl/AE = 50 x 1000 x 1500 / 20 x 20 x 200 x 1000 =0.937 mm.

16. Q.4 In a tension test a metallic rod of diameter 16 mm. produces an elongation of 48 mm when subjected to 90 kN load. The length of the bar is 150 mm. Find modulus of elasticity. Given Data D = 16 mm δl = 48 mm P = 90 kN. = 90 x 10³ N L = 150 mm

17. Q.5 A steel rod of 25mm in diameter and 2 m long is subjected to an axial pull of 45 kN. Find i ) the stress ii) Strain iii) Elongation. Take E = 200 Gpa. Given Data: D = 25 mm L = 2 m P = 45 kN. = 45 x 10³ N. E = 200 Gpa = 200 x 10 ³ Mpa

18. Q.6 A steel Flat 100 mm wide, 12 mm thick and 5 m long carries an axial load of 20 kN Find the stress, strain and change in length. E = 2.1 x 10 5 Mpa

19. Q. 7 A wire 4 mm in diameter, 4 m long is subjected to an axial pull of 1890 N is stretched by 3mm under the axial pull. Find stress, strain induced. Also find modulus of elasticity. Given data: 1. d = 4 mm 2. L = 4 x 10³ mm. 3. P = 1890 N 4. δ l = 3 mm .

20. Step 1; To find the stress σ = P/A A = л /4 x 4² = 1890 / 12.56 Step 2 : To find the strain e = δ l / l Step 3 : To find the Modulus of elasticity E= σ / e

21. Q.8 : A load of 6 kN is to be raised with help of steel cable. Find the minimum diameter of steel cable if stress is not exceed 110 N/mm2. Given Data : 1. load P = 6 kN. 2. stress σ = 110 N/mm2. 3. To find d =? We know σ = P/A

22. We know σ = P/A A = P / σ = 6 x 10³ / 110 A = л / 4 x d² 54.54 = л / 4 x d² d =

23. Q.9 A steel rod is subjected to an axial pull of 25 kN Find the maximum diameter if the stress is not exceed to 100 N /mm2. The length of the rod is 2000 mm and take E = 2.1 x 10 5 Mpa.

24. Deformation of body or bars of cross section subjected to axial load Total Elongation : δ l = δl 1 + δl 2 + δl 3

25. Q. 1 A mild steel stepped bar 6 m length is 25 mm in diameter of 2 m and 15 mm diameter for remaining length. It is subjected to a tensile load of 70 kN Calculate the change in length if its modulus if elasticity is 200 kN/mm2.

26. Change in length δ l = δl 1 + δl 2 + δl 3 = (Pl/AE) + (PL/ AE ) + (PL/AE) =

27. Q.2 A Bar as shown in fig. is axially loaded. If the maximum stress induced in bar is 100 Mpa, calculate the value of P. Take E = 102 GPa for both part.

28. Q.3 A brass bar having cross section area 1000 mm 2 is subjected to axial forces of 50 kN as shown in fig. Find the total change in length. Take E = 1.05 x 10 5 N/ mm 2

29. Q.4 A bar having cross-section as given in fig. No. 3 is subjected to a tensile load of 150 kN. Calculate the change in length of each part along with the total change in length if E= 2 x 10 5 N/m. ( W-19)

30. Q.5 A bar ABCD of varying cross section is subjected to an axial pull of 75 kN Part AB is 300 mm long, hollow circular with external diameter 30 mm and internal diameter 26 mm. Part BC is square in section of side 15 mm and is 150 mm long. Part CD is 200 mm long and solid circular section of diameter 20 mm. determine deformation of each part and net deformation. Take Eab = 200 Gpa, Ebc = 100 Gpa and Ecd = 75 Gpa.

31. Q.6 A circular bar of 1000 mm length has cross – section as given below : First 200 mm has a diameter 10 mm, second 500 mm has a diameter 25 mm and the last 300 mm has diameter of 15 mm. Determine the maximum axial pull which the bar may be subjected if the maximum stress is limited to 150 MPa . Find total elongation of the bar. Take E = 200 GPa. (S-18)  

32. Numerical on forced applied in intermediate section Q.1 Find the total elongation of the bar shown in fig. Take E = 200 kN/mm2.

33. Solution: Given Data: d 1 = 30 mm. P 1 = 50 kN = 50 x 10 3 N d 2 = 50 mm. P 2 = 30 kN = 30 x 10 3 N d 2 = 15 mm. P 3 = 10 kN = 10 x 10 3 N L 1 = 2 m = 2 x 10 3 mm L 1 = 2 m = 2 x 10 3 mm L 1 = 2 m = 2 x 10 3 mm

34. Total Change in length δ l = δl 1 + δl 2 + δl 3 = (Pl/AE) + (PL/ AE ) + (PL/AE ) =

35. Q.2 A Compound bar having steel rod of dia. 35 mm and solid copper rod of dia. 20mm and aluminum square rod of 10 mm is as shown in following figure. Find change in length of bar. Take modulus of elasticity Es = 210 kN/mm 2 , Ec = 110 GPa and Eal = 70GPa. (S-19)

36. To find unknown force P ∑ Fx =0  -30+P-5+10=0 P-25=0 P=25kN

37.   Total Change in length δ l = δl 1 + δl 2 + δl 3 = (Pl/AE) + (PL/ AE ) + (PL/AE ) =

38. Q.3 A bar of uniform cross section area 100 mm 2 is subjected to axial forces as shown in fig . Calculate the net change in length of the bar. Take E = 25 x 10 5 N/mm 2 .

39. To calculate, P ∑ Fx =0  1-P + 4 -2 =0 P = 5 - 2 P = 3 kN Total Change in length δ l = δl 1 + δl 2 + δl 3 = (Pl/AE) + (PL/ AE ) + (PL/AE ) =

40. Q.4 A Composite bar comprising of aluminum and steel as show in fig. Find the value of P if net deformation produced in the bar is 2 mm. Take Es = 2 x 10 5 N/mm2 and Eal = 7 x 10 4 N/mm2.

41. Q.5 Determine the magnitude of P for equilibrium and total elongation of the bar shown in fig. Take E = 210 Gpa. Also calculate minimum stress induced.

42. Q. 6 A compound bar having steel rod of dia of 35 mm and copper rod of 20 mm and aluminum square rod of 10 mm is as shown in fig. Find the change in length of bar take modulus of elasticity Es = 210 Gpa, Ec = 110 Gpa and Eal = 70 Gpa. ( S-19)

43. Stress Strain Curve for Mild Steel

44. A : Proportionality limit B : Elastic limit. C & D : Yield point. E : Ultimate stress F : Breaking or failure point.

45. Stress Strain Curve for HYSD Bar 2 : Proportionality Limit 4: Breaking point.

46. Factor of safety It is the ratio of maximum stress to the working stress. FOS = Max stress / working stress It is based on the yield point stress. For ductile material FOS = Yield point stress/ working Stress

47. Composite Section A structural member consist of two of more dissimilar material joined together to act as a unit is called as composite section

48. Modular Ratio It is the ratio of young’s modulus of elasticity of two different material in construction by composite material. m = (E 1 / E 2 )

49. Stresses in Composite material Procedure to Solve Numerical Total Load P = P1 + P2 P = σ 1 A 1 + σ 2 A 2 2) Compare Elongation for both material δ l1 = δ l2 Make relation between stresses of both material using modular ration m = (E 1 / E 2 )

50. Q.1. A steel cube 40 mm inside diameter and 4 mm metal thickness is filled with concrete. Determine the stress in the each material due to an axial thrust of 100 kN. Take Es = 2.1 x 10 5 N/mm2 and Ec = 0.14 x 10 5 N/mm2. d D

51. 1 ) Total Load P = P1 + P2 P = σ 1 A 1 + σ 2 A 2 A s = л /4 ( D 2 - d 2 ) A c = л /4 x d 2 2) Compare Elongation of both material δ ls = δ lc (Pl/AE) s = (Pl/AE)c σ s / Es = σ c / Ec ------------ σ = P /A σ s / 2.1 x 10 5 = σ c / 0.14 x 10 5 σ s = ----------- σ c

52. Total Load P = P1 + P2 100 x 10 3 = σ s As + σ c Ac

53. Q.2 A composite Bar consisting of steel rod 30 mm in diameter enclosed in a copper tube of external diameter 60 mm and internal diameter 30 mm. It is subjected to compressive axial load of 110 kN Calculate stresses developed in steel and copper. Take Es = 210 kN /mm2. Ec = 105 kN / mm2. Given: Es = 210 kN /mm2. = 210 x 10 3 N / mm2 Ec = 105 kN / mm2. = 105 x 10 3 N / mm2

54. 1 ) Total Load P = P1 + P2 P = σ 1 A 1 + σ 2 A 2 A c = л /4 ( D 2 - d 2 ) A s = л /4 x d 2 2) Compare Elongation of both material δ ls = δ lc (Pl/AE) s = (Pl/AE)c σ s / Es = σ c / Ec ------------ σ = P /A σ s / 210x 10 3 = σ c / 105 x 10 3 σ s = ----------- σ c

55. Total Load P = P1 + P2 110 x 10 3 = σ s As + σ c Ac

56. Q.3 A RCC column 400 x 400 mm is reinforced with 4 bars of 20mm diameter. Determine the stresses induced in steel and concrete. It is subjected to an axial load of 500 kN and modular ratio is 13. Given : P = 500 kN = 500 x 10 3 N Modular ratio m = 13 Es / Ec = 13 Es = 13 Ec

57. 1 ) Total Load P = Ps + Pc P = σ s A s + σ c A c A s = n л /4 x d 2 A s = 4 x л /4 x 20 2 A g = 400 x400 Ag = As + Ac 400 x 400 = As + Ac 2) Compare Elongation of both material δ ls = δ lc (Pl/AE) s = (Pl/AE)c σ s / Es = σ c / Ec ------------ σ = P /A

58. Q.4 RCC column is 300 x 300 mm in section. It is provided with 8 bars of 20 mm diameter. Determine the stresses induced in concrete and steel bars, if it carries a load of 180 kN. Take Es = 210 Gpa and Ec = 14 Gpa . Given : P = 180 kN = 180 x 10 3 N Es = 210 Gpa = 210 x Mpa Ec = 14 Gpa = 14 x 10 3 Mpa

59. 1 ) Total Load P = Ps + Pc P = σ s A s + σ c A c A s = n x л /4 x d 2 = 8 x л /4 x 20 2 A g = 300 x300 Ag = As + Ac 300 x 300 = As + Ac 2) Compare Elongation of both material δ ls = δ lc (Pl/AE) s = (Pl/AE)c σ s / Es = σ c / Ec ------------ σ = P /A

60. Q. 5 A square R.C.C. column of 300mm X 300 mm in section with 8 steel bars of 20 mm diameter carries a load of 360 kN. Find the stresses induced in steel and concrete. Take modular ratio = 15. Given A=300×300 mm 2 , d =20 mm No. of steel bar = 8, P =360kN, m =15 Find: σc , σs ,

61. 1 ) Total Load P = Ps + Pc P = σ s A s + σ c A c A s = n x л /4 x d 2 = 8 x л /4 x 20 2 A g = 300 x300 Ag = As + Ac 300 x 300 = As + Ac 2) Compare Elongation of both material δ ls = δ lc (Pl/AE) s = (Pl/AE)c σ s / Es = σ c / Ec ------------ σ = P /A

62. Q.6 A RCC Column 450 mm in diameter is reinforced with 6 bar of 16 mm diameter. Find safe load that column carry. If permissible stresses in concrete and steel are 5 N/mm2 and 125 N/mm2 respectively. Take Ec = 0.14 x 10 5 N / mm2 Es = 2.1 x 10 5 N / mm2

63. Q.7 A steel tube with 40 mm inside diameter and 4 mm thickness is filled with concrete. Determine load shared by each material due to axial thrust of 60 kN. Take E steel = 210 N/mm2 E concrete = 14 x 103 N/mm2

64. 1 ) Total Load P = Ps + Pc P = σ s A s + σ c A c A s = n x л /4 x d 2 = 8 x л /4 x 20 2 A g = 300 x300 Ag = As + Ac 300 x 300 = As + Ac 2) Compare Elongation of both material δ ls = δ lc (Pl/AE) s = (Pl/AE)c σ s / Es = σ c / Ec ------------ σ = P /A

65. Q. 8 A brass bar of 250 mm length and 20 mm diameter is fixed inside a steel tube of 40 mm external and 20 mm internal diameter and of same length. The composite bar is subjected to an axial force pull of 140 kN. Find the stress in each metal. Take Es = 200 Gpa and Eb = 110Gpa.

66. 1 ) Total Load P = Ps + Pc P = σ s A s + σ c A c A s = n x л /4 x d 2 = 8 x л /4 x 20 2 A g = 300 x300 Ag = As + Ac 300 x 300 = As + Ac 2) Compare Elongation of both material δ ls = δ lc (Pl/AE) s = (Pl/AE)c σ s / Es = σ c / Ec ------------ σ = P /A

67. Q.9 A composite bar of length 500 mm consist of a mild steel circular rod of 20 mm diameter enclosed in a brass tube of 30 mm external diameter and 22 mm internal diameter. The composite bar is subjected to an axial pull of 60 kN. Find Stresses in mild steel rod and brass tube. Es = 210 Gpa and Ebr = 100 Gpa

68. Temperature stresses and strain Changes in temperature produce expansion or contraction of materials and result in thermal strains and thermal stresses For most structural materials, thermal strain εT is proportional to the temperature change ΔT : ε T = α (Δ T) α : Coefficient of thermal expansion.

69. Temperature Stress : σ = E α (Δ T) Temperature Strain : e = α t Free deformation or expansion δ l = L α T

70. Q.1 Square Rod 10 mm x10mm in cross section and 100 mm long is fixed at both ends. Determine end reaction due to rise in temperature of 50ºc. Take E = 2 x 10 5 N / mm2 and α = 12 x 10 6 /ºC

71. Q.2 A rod of 10 m long at 10ºc is heated to 70ºc. If the free expansion is prevented, find the magnitude and nature of stress induced. Take E = 2.1 x 10 5 N /mm2. and α = 12 x 10 6 /ºC

72. Q.3 A rod is 2 m long at 10ºc. Find the expansion of the rod when temperature is raised to 80ºc. If this expansion is prevented the stress in the material. Take E = 2 x 10 5 N /mm2 and α = 0.000012 Per ºc.

73. Q.4 A steel rod 15m long is at temperature of 15ºc. Find the free expansion of the length when the temperature is raised to 65ºc. Find the temperature stresses when the expansion of the rod is fully prevent. Take E = 2 x 10 5 N /mm2. and α = 12 x 10 6 /ºC

74. Q.5 An aluminum rod of 22 mm diameter is fixed at both end at the temperature of 150ºC. Find the stress and force induced along with nature in rod when temperature fall to 100ºC and 30ºC. Take Ea = 70 Gpa and α = 23 x 10 -6 /ºC

75. Q.6 A metal rod 16 mm diameter, 1500 mm long is loosely held. If support slip by 0.5 mm due to rise of temperature of 60ºC, Find stress developed in the rod and its nature. Take E = 110 Gpa and Coefficient of thermal expansion α = 10 x 10 -6 / ºC.

76. Strain Energy When a body is subjected to gradual, sudden or impact load, the body deforms and work is done upon it. If the elastic limit is not exceed, this work is stored in the body. This work done or energy stored in the body is called strain energy Energy is stored in the body during deformation process and this energy is called “Strain Energy ”. Strain energy = Work done

77. Resilience Total strain energy stored in a body is called resilience. ∴ 𝐮 = σ² / 2E x V Where, σ = stress V = volume of the body Proof Resilience : Maximum strain energy which can be stored in a body is called proof resilience. 𝐮p = (𝛔𝐄 )² x V/ 𝟐𝐄

78. Modulus of Resilience : Maximum strain energy which can be stored in a body per unit volume, at elastic limit is called modulus of resilience . 𝐮m = (𝛔𝐄 )² / 𝟐

79. Application of load 1.Gradually applied Load It is the type of loading in which load is applied on the bar from zero and slowly increasing uniformly is called as gradually applied load. ∴ σ = P / A 2. Suddenly applied Load 3. Impact Load

80. Thank you!