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Strain energy
The document discusses different types of strain energy stored in materials when subjected to loads. It defines strain energy as the work done or energy stored in a body during elastic deformation. The types of strain energy discussed include: elastic strain energy, strain energy due to gradual, sudden, impact, shock and shear loading. Formulas are provided to calculate strain energy due to these different loadings. Examples of calculating strain energy in axially loaded bars and beams subjected to bending and torsional loads are also presented.
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Strain energy
- 1. CREATED BY: RAJESH GOSWAMI
- 2.
- 3. CONTENT: 1. Elastic strainenergy 2. Strain energy due to gradual loading 3. Strain energy due to sudden loading 4. Strain energy due to impact loading 5. Strain energy due to shock loading 6. Strain energy due to shear loading 7. Strain energy due to bending (flexure) 8. Strain energy due to torsion 9. Examples
- 4. When abody 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
- 5. Resilience : Totalstrain energy stored in a body is called resilience. Proof Resilience : Maximum strain energy which can be stored in a body is called proof resilience. ∴ 𝐮 = 𝛔 𝟐 𝟐𝐄 × 𝐕 ∴ 𝐮p = (𝛔 𝐄) 𝟐 𝟐𝐄 × 𝐕 Where, σ = stress V = volume of the body Where, σE = stress at elastic limit
- 6. Modulus ofResilience : Maximum strain energy which can be stored in a body per unit volume, at elastic limit is called modulus of resilience. ∴ 𝐮m = (𝛔 𝐄) 𝟐 𝟐𝐄
- 7. Consider abar of length L placed vertically and one end of it is attached at the ceiling. Let P =Gradually applied load L =length of bar A =Cross-sectional area of the bar δl =Deflection produced in the bar σ =Axial stress induced in the bar. It may be tensile or compressive, depending upon if the bar under consideration is under tensile or compressive load E =Modulus of elasticity of bar material L δl P
- 8. Work done onthe bar = Area of the load – deformation diagram … (1) = 1 2 × 𝑃 × 𝛿𝑙
- 9. Work Stored inthe bar = Area of the resistance – Deformation diagram = 1 2 × R × δl = 1 2 × σ × A × δl… (2) Now, Work done = Work stored ∴ 1 2 P × δl = 1 2 σ × A × δl ∴ P = σ × A ….. stress due to gradual load. ∴ σ = P A
- 10. Strain Energy = 1 2 ×R × δl = 1 2 σ × A × δl R = σ × A = 1 2 σ × A × ε × lε = δl l = 1 2 σ × A × σ E × lE = σ ε = σ2 2E ×A× l u = 𝛔 𝟐 𝟐𝐄 × v … strain energy due to gradual l
- 11. When theload is applied suddenly the value of the load is P throughout the deformation. But, Resistance R increase from O to R Work done on the bar =P× δl ... (1)
- 12. Work stored inthe bar = 1 2 ×R×δl = 1 2 × σ×A× δl...(2) Now, Work done = Work stored ∴ P × δl = 1 2 × σ×A× 𝛿𝑙 P = 1 2 × σ×A∴ σ = 2P A ∴ Hence , the Maximum Stress intensity due to a suddenly applied load is Twice the stress intensity produced by the load of the same magnitude applied gradually.
- 13. L δl P Collar h Load P isdropped through a height h, before it commences to load the bar.
- 14. Work done onthe bar = Force × Deformation =P( h + δl ) =P( h + σ∙l E ) … (1) Work stored in the bar= 1 2 × δl× R =Strain Energy = σ2 2E × V … (2) Now, Work done = Work stored ∴ P( h + δl )= σ2 2E × V 𝛿𝑙 = 𝜎 ∙ 𝑙 𝐸 ∴
- 15. ∴ P( h+ σl E )= σ2 2E × A × l ∴ P × h + p×σl E = σ2 2E × A × l ∴P × h × 2E Al + p × σl E × 2E Al =σ2 ∴ 2EPh Al + 2P × σ A =σ2 ∴ σ2 − 2P × σ A = 2EPh Al ∴ σ2 − 2P × σ A + p2 A2 = 2EPh Al + p2 A2
- 16. ∴(σ − P A )2 = 2EPh Al + p2 A2 ∴(σ− P A ) = 2EPh Al + p2 A2 … Stresses due to impact load ∴ σ = P A + 2EPh Al + p2 A2 If load is applied suddenly, h = 0 ∴ σ = P A + p2 A2 + 0 ∴ σ = 2P A
- 17. When 𝛿𝑙is very small as compered to h , then Work done = P × h σ2 2E × A × l = P ×h 𝜎2 = 2𝐸𝑃ℎ 𝐴𝑙 𝜎 = 2𝐸𝑃ℎ 𝐴𝑙
- 18. Let, Work doneon the bar by shock = u Work stored in the bar = ... max instantaneous stress 1 2 × R × δl = 1 2 × σ × A × σl E = σ2 2E × A × l ∴ U = σ2 2E × A × l ∴ σ2 = 2UE Al ∴ σ = 2UE Al
- 19. If tis the uniform shear stress produce in the material by external forces applied within elastic limit, the energy Stored due to shear Loading is given by, 𝐮 = 𝛕 𝟐 𝟐𝐆 × 𝐕 Where, t = shear stress G = Modulus of rigidity
- 20. Consider asquare block ABCD of length l , Faces BC and AD are subjected to shear stress τ , Let face AD is fixed. The section ABCD will deform to AB1C1D through the ang ∅ = Shear strain tan ∅ = BB1 l ∅ is very small ∴ tan ∅ = ∅ ∴ ∅ = BB1 l ….. Shear Strain
- 21. Force P onface BC P = τ × BC × l When P in applied gradually In case of gradual load. u = P 2 × BB1 average force = 1 2 × (τ × BC × l ) × BB1 = 0+P 2 = P 2 = 1 2 × (τ × BC × l ) × ∅ ∙ lBB1 = ∅ ∙ l = 1 2 × (τ × A) ∙ τ G ∙ l G = τ ∅ = 1 2G × τ2 × A × l BC × l= Au = τ2 2G × V The elastic energy stored due to shear loading is known as shear resilience
- 22. Consider twotransverse section 1-1 and 2-2 of a beam distant dx apart as shown in fig. Consider a small strip of area da at distant y from the neutral axis. B.M. in small portion dx will be constant. M I = σ y ∴ σ = M × y … (1)
- 23. ∴ Strain energystored in small strip of area da. u = σ2 2E × v = σ2 2E × (da ∙ dx) = 1 2E × ( M I ∙ y)2× (da ∙ dx) = 1 2E ∙ M2∙y2 I2 × da ∙ dx …(2) ∴ Starin energy stored in entire section of a beam. utotal = y=yt y=yc 1 2E ∙ M2 ∙ y2 I2 ∙ da ∙ dx
- 24. = 1 2E ∙ M2∙dx I2 yt yc y2 ∙da y2 ∙ da = I = second moment of = 1 2E ∙ M2∙dx I2 I area. utotal = M2 2EI dx …(3) Now, for strain energy in entire beam, integrate between limits 0 to l. ... Strain energy due to bending. ∴ u = 0 l M2 2EI ∙ dx
- 25. We haveseen that, when a member is subjected to a uniform shear stress 𝛕, the strain energy stored in the member is τ2 2G × V. Consider a small elemental ring of thickness dr, at radius r.
- 26. ur = τr 2 2G × V ur= ( r R ∙ τ )2 2G × V = r2 ∙ τ2 R2 ∙ 2G × V = r2 ∙ τ2 R2 ∙ 2G × (2πr × dr) × l ur = τ2 ∙ r3 R2 ∙ G ∙ π ∙ l ∙ dr … strain energy for one
- 27. Total strainenergy for whole section, is obtained by integrating over a range from r = 0 to r = D/2 for a solid shaft. u = 0 D/2 τ2 ∙ r3 R2 ∙ G ∙ π ∙ l ∙ dr = τ2 ∙ π ∙ l R2 ∙ G 0 D/2 r3 ∙ dr = τ2 ∙ π ∙ l R2 ∙ G [ r4 4 ]0 D/2 = τ2 ∙ π ∙ l R2 ∙ G ( D 2 )4 4
- 28. = τ2 ∙ π∙ l R2 ∙ G ∙ D4 64 = τ2 ∙ π ∙ l D 4 2 ∙ G ∙ D4 64 = τ2 ∙ π ∙ l ∙ D2 16G = τ2 ∙ π ∙ l ∙ D2 4 × 4 × G = τ2 ∙ l ∙ A 4G ∴ R = D 2 ∴ A = π 4 × D2 u = τ2 4G × V … Strain energy due to tors
- 29. Ex-1 : Anaxial pull of 50 kN is suddenly applied to a steel bar 2m long and 1000 mm2 in cross section. If modulus of elasticity of steel is 200 kN/ mm2. Find, (i) maximum instantaneous stress (ii) maximum instantaneous extension (iii) Strain energy (iv) modulus of resilience. Solution : here, P = 50 kN (Sudden load) A = 1000 mm2 l = 2m = 2000 mm E = 200 kN/mm2 = 200× 103 N/mm2
- 30. (i) Maximum instantaneousstress : σ = 2P A = 2 × 50 × 103 1000 (ii) Maximum instantaneous extension : E = σ ε ε = σ E = 100 200 × 103 = 5 × 10−4 ε = δl l δl = ε ∙ l = 5 × 10−4 × 2000 = 100 N/mm2 = 1 mm
- 31. (iii) Strain energy(u) : u = σ2 2E × V = (100)2 2 × 200 × 103 × 1000 × 2000 (iv) Modulus of resilience (um) : um = σ2 2E = (100)2 2 × 200 × 103 = 50,000 N ∙ mm = 0.025 N ∙ mm/mm3
- 32. EX –2: A 1500 mm long wire of 25 mm2 cross sectional area is hanged vertically. It receives a sliding collar of 100 N weight and stopper at bottom end. The collar is allowed to full on stopper through 200 mm height. Determine the instantaneous stress induced in the wire and corresponding elongation. Also determine the strain energy stored in the wire. Take modulus of elasticity of wire as 200 GPa. Solution : here, P = 100 N A = 25 mm2 l = 1500 mm
- 33. σ = P A + P2 A2 + 2EPh A ∙l 100 25 1002 252 + 2 × 200 × 103 × 100 × 200 25 × 1500 = 4 + 461.89 = 465.89 N/mm2
- 34. δl = σ ∙l E = 465.89 × 1500 200 × 103 Strain energy, u = σ2 2E × V u = (465.89)2 2 × 200 × 103 × (25 × 1500) = 3.49 mm = 20,348.76 N.mm
- 35.