Advanced Strength Of Material
•Download as PPTX, PDF•
7 likes•2,943 views
Elastic Strain Energy due to Gradual Loading. Elastic Strain Energy due to Sudden Loading. Elastic Strain energy due to impact loading. Elastic Strain Energy due to Principal Stresses. Energy of Dilation And Distortion.
Recommended
More Related Content
What's hot
What's hot (20)
Viewers also liked
Viewers also liked (20)
Similar to Advanced Strength Of Material
Similar to Advanced Strength Of Material (20)
More from Akhtar Kamal
More from Akhtar Kamal (10)
Recently uploaded
Recently uploaded (20)
Advanced Strength Of Material
- 1. AKHTAR KAMAL (120450119156) Advanced Strength Of MaterialBranch: Mechanical- 4C(1) Collage: S.V.M.I.T. Akhtar Kamal
- 2. INTRODUCTION 1) Elastic Strain Energy due to Gradual Loading. 2) Elastic Strain Energy due to Sudden Loading. 3) Elastic Strain energy due to impact loading. 4) Elastic Strain Energy due to Principal Stresses. 5) Energy of Dilation And Distortion. Akhtar Kamal
- 3. What is strain energy? When the body is subjected to gradual, sudden or impact load, the body deforms, and work done upon it. The material behave like a perfect spring and oscillates about its mean position. If the elastic limit is not exceeded, this work is stored in the body. This work done or energy stored in the body is called strain energy. Akhtar Kamal
- 4. Some Important Definition And Question (1)Resilience: Total strain energy stored in body is called resilience. It is denoted as ‘’ 𝒖 = 𝝈 𝟐 𝟐𝑬 × 𝑽 Where… 𝝈 = 𝒔𝒕𝒓𝒆𝒔𝒔 𝑽 = 𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒐𝒅𝒚 (2)Proof Resilience: Maximum strain energy which can be stored in a body at elastic limit is called proof resilience. It is denoted as ‘𝒖 𝒑’ 𝒖 𝒑 = 𝝈 𝑬 𝟐 𝟐𝑬 × 𝑽 Where… 𝝈 𝑬 = 𝒔𝒕𝒓𝒆𝒔𝒔 at elastic limit (3)Modulus of resilience: Maximum strain energy which can be stored in a body per unit volume, at elastic limit is called Modulus of resilience. It is denoted as ‘𝒖 𝒎’ 𝒖 𝒎 = 𝝈 𝑬 𝟐 𝟐𝑬 Akhtar Kamal
- 5. Strain Energy Due to Gradual Loading Considr a bar of length is 𝑙 and uniform section area 𝐴, subjected to gradual load 𝑃. Akhtar Kamal
- 6. Stress Due to Gradual Load Since the load is applied gradually,(i.e. it increases from 0 to P), average load is considered. Work done on the bar = Area of the load – Deformation diagram. = 𝟏 𝟐 × 𝑷 × 𝜹𝒍 … . . (𝟏) Work stored in the bar = Area of the resistance – Deformation diagram. = 𝟏 𝟐 × 𝑹 × 𝜹𝒍 = 𝟏 𝟐 × (𝝈 ∙ 𝑨) × 𝜹𝒍 … . . (𝟐) Work done = Work stored 𝟏 𝟐 × 𝑷 × 𝜹𝒍 = 𝟏 𝟐 × (𝝈 ∙ 𝑨) × 𝜹𝒍 𝝈 = 𝑷 𝑨 … . . 𝐒𝐭𝐫𝐞𝐬𝐬 𝐝𝐮𝐞 𝐭𝐨 𝐠𝐫𝐚𝐝𝐮𝐚𝐥 𝐥𝐨𝐚𝐝.Akhtar Kamal
- 7. Strain Energy Due to Gradual Loading Strain energy = 1 2 × 𝑅 × 𝛿𝑙 = 1 2 × (𝜎 ∙ 𝐴) × 𝛿𝑙 = 1 2 × 𝜎 ∙ 𝐴 × 𝜀 ∙ 𝑙 (∵ 𝜀 = 𝛿𝑙 𝑙 ) = 1 2 × 𝜎 ∙ 𝐴 × 𝜎 𝐸 𝑙 ∵ 𝐸 = 𝜎 𝜀 = 𝜎2 2𝐸 ∙ 𝐴𝑙 𝑢 = 𝜎2 2𝐸 × 𝑉 Akhtar Kamal
- 8. Elastic Strain Energy due to Sudden Loading When the load is applied suddenly the value of the load is P throughout the deformation. But, Resistance R increases from 0 to R. Work done on the bar= 𝑃 × 𝛿𝑙 … . . 1 Work store in the bar= 1 2 × 𝑅 × 𝛿𝑙 = 1 2 × 𝜎 × 𝐴 × 𝛿𝑙 … . . (2) Akhtar Kamal
- 9. Work done = work store 𝑷 × 𝜹𝒍 = 𝟏 𝟐 × 𝝈 × 𝑨 × 𝜹𝒍 𝝈 = 𝟐𝑷 𝑨 Hence, the maximum stress intensity due to a suddenly applied load twice the stress intensity produced by the load of the same magnitude applied gradually. Akhtar Kamal
- 11. STARIN ENERGY DUE TO IMPACT LOADING Work done on the bar= 𝑓𝑜𝑟𝑐𝑒 × 𝑑𝑒𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 = 𝑷(𝒉 + 𝜹𝒍) = 𝑷[𝒉 + 𝝈×𝒍 𝑬 ] … … . (𝟏) Akhtar Kamal
- 12. Work stored in the bar= 𝟏 𝟐 × 𝜹𝒍 × 𝑹 = strain energy = 𝝈² 𝟐𝑬 × 𝑽 … … … … (𝟐) Work done = Work stored 𝑷(𝒉 + 𝜹𝒍) = 𝝈² 𝟐𝑬 × 𝑽 𝑷 𝒉 + 𝝈 × 𝒍 𝑬 = 𝝈 𝟐 𝟐𝑬 × 𝑨 × 𝒍 𝑷 × 𝒉 + 𝑷 × 𝝈 × 𝒍 𝑬 = 𝝈 𝟐 𝟐𝑬 × 𝑨 × 𝒍 𝟐𝑬𝑷𝒉 𝑨𝒍 + 𝟐𝑷 × 𝝈 𝑨 = 𝝈² 𝝈² − 𝟐 × 𝑷 × 𝝈 𝑨 = 𝟐 × 𝑬 × 𝑷 × 𝒉 𝑨 × 𝒍 𝝈 − 𝑷 𝑨 𝟐 = 𝑷 𝟐 𝑨 𝟐 + 𝟐 × 𝑬 × 𝑷 × 𝒉 𝑨 × 𝒍 Akhtar Kamal
- 13. 𝝈 = 𝑷 𝑨 + 𝑷 𝟐 𝑨 𝟐 + 𝟐 × 𝑬 × 𝑷 × 𝒉 𝑨 × 𝒍 … . . 𝑺𝒕𝒓𝒆𝒔𝒔 𝒅𝒖𝒆 𝒕𝒐 𝒊𝒎𝒑𝒂𝒄𝒕 𝒍𝒐𝒂𝒅 𝝈 = 𝑷 𝑨 + [ 𝑷 𝟐 𝑨 𝟐 + 𝟎] 𝝈 = 𝑷 𝑨 + 𝑷 𝑨 = 𝟐𝑷 𝑨 When 𝜹𝒍 is very small as compared to 𝒉,then work done = 𝑷 × 𝒉 𝝈 𝟐 𝟐𝑬 × 𝑨 × 𝒍 = 𝑷 × 𝒉 𝝉 = 𝟐𝑬×𝑷×𝒉 𝑨×𝒍 Impact factor : the ratio of maximum dynamic deformation to the static deformation is called the impact factor. 𝒊 = [𝟏 + (𝟏 + 𝟐𝒉 𝜹 ] But 𝑷 𝑾 = 𝜹𝒍 𝜹 The ratio 𝑷 𝑾 is sometimes known as load factor. Akhtar Kamal
- 14. Strain energy due to bending : consider two transverse section 1-1 and 2-2 of a beam distant 𝒅𝒙 apart as showing. 𝑴 𝑰 = 𝝈 𝒚 𝜺 = 𝑴 𝑰 × 𝒚 … … . (𝟏) strain energy stored in small strip pf area da. 𝝁 = 𝝈 𝟐 𝟐𝑬 × 𝒗 = 𝟏 𝟐𝑬 × (𝒅𝒂 × 𝒅𝒙) = 𝟏 𝟐𝑬 𝑴 𝟐 × 𝒚 𝟐 𝑰 𝟐 × 𝒅𝒂 × 𝒅𝒙 … … . (𝟐) strain energy stored in entire section of a beam 𝐮 = 𝟏 𝟐𝑬 𝑴 𝟐 × 𝒚 𝟐 𝑰 𝟐 × 𝒅𝒂 × 𝒅𝒙 = 𝟏 𝟐𝑬 × 𝑴 𝟐 × 𝒅𝒙 𝑰 𝟐 × 𝑰 𝒖 = 𝑴 𝟐 𝟐𝑬𝑰 × 𝒅𝒙 … … . . (𝟑) Akhtar Kamal
- 15. Strain energy due to torsion : consider a small elements ring of thickness 𝒅𝒓 , at radius 𝒓. 𝒖 = 𝝉 𝟐 𝟐𝑮 × 𝑽 = 𝒓 𝑹 ×𝝉 𝟐 𝟐𝑮 × 𝑽 = 𝝅 × 𝒍 × 𝒅𝒓 × 𝝉 𝟐 × 𝒓 𝟑 𝑹 𝟐×𝑮 𝑹𝒂𝒏𝒈𝒆 𝒇𝒓𝒐𝒎 𝒓 = 𝟎 𝒕𝒐 𝒓 = 𝑫 𝟐 𝒇𝒐𝒓 𝒂 𝒔𝒐𝒍𝒊𝒅 𝒔𝒉𝒂𝒇𝒕 𝒖 = 𝟎 𝑫 𝟐 𝝅 × 𝒍 × 𝒅𝒓 × 𝝉 𝟐 × 𝒓 𝟑 𝑹 𝟐 × 𝑮 = 𝝅 × 𝒍 × 𝝉 𝟐 × 𝟏 𝑹 𝟐×𝑮 𝟎 𝑫 𝟐 𝒓 𝟑 𝒅𝒓 = 𝝉 𝟐 𝝅𝒍 𝑮𝑹 𝟐 × 𝑫 𝟑 𝟔𝟒 , 𝑹 = 𝑫/𝟐 = 𝝉 𝟐 𝝅𝒍 × 𝑫 𝟐 𝟏𝟔𝑮 = 𝝉 𝟐 𝟒𝑮 × 𝑽 Akhtar Kamal
- 16. For hollow shaft : For a hollow shaft the range of integrating will be 𝒇𝒓𝒐𝒎 𝒅 𝟐 𝒕𝒐 𝑫 𝟐 . 𝒖 = 𝝉 𝟐 𝟒𝑮 × 𝑫 𝟐 + 𝒅 𝟐 𝑫 𝟐 × 𝑽 Akhtar Kamal
- 17. Akhtar Kamal
- 18. Energy Of Dilation and Distortion Total strain energy given by equation (1) of article 1.5 can be separated into the following two strain energies . a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension). b) Strain Energy of distortion(shear strain energy) To accomplish this , Let the principal strains be 𝜺 𝟏, 𝜺 𝟐 𝒂𝒏𝒅 𝜺 𝟑,in the deration of principal stresses 𝝈 𝟏, 𝝈 𝟐 and 𝝈 𝟑 respectively. Akhtar Kamal
- 19. Energy Of Dilation and Distortion 𝜺 𝟏 = 𝟏 𝑬 [𝝈 𝟏 − 𝝁 𝝈 𝟐 + 𝝈 𝟑 ] 𝜺 𝟐 = 𝟏 𝑬 [𝝈 𝟐 − 𝝁 𝝈 𝟑 + 𝝈 𝟏 ] 𝜺 𝟑 = 𝟏 𝑬 [𝝈 𝟑 − 𝝁 𝝈 𝟏 + 𝝈 𝟐 ] 𝜺 𝟏 + 𝜺 𝟐+𝜺 𝟑= 𝟏 𝑬 [(𝝈 𝟏 + 𝝈 𝟐 + 𝝈 𝟑) − 𝟐𝝁(𝝈 𝟏 + 𝝈 𝟐 + 𝝈 𝟑)] = (𝝈 𝟏+𝝈 𝟐+𝝈 𝟑) 𝑬 [𝟏 − 𝟐𝝁] But, 𝜺 𝟏 + 𝜺 𝟐+𝜺 𝟑= 𝜺 𝒗 = 𝒗𝒐𝒍𝒖𝒎𝒆𝒕𝒓𝒊𝒄 𝒔𝒕𝒓𝒂𝒊𝒏 𝜺 𝒗 = 𝟏−𝟐𝝁 𝑬 (𝝈 𝟏 + 𝝈 𝟐 + 𝝈 𝟑) Akhtar Kamal
- 20. Energy Of Dilation and Distortion From the above discussion, following conclusions can be made. (a) If 𝝈 𝟏 = 𝝈 𝟐 = 𝝈 𝟑 𝜺 𝟏= 𝜺 𝟐 = 𝜺 𝟑, This means that there is no distortion (so that no shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs. (b) If 𝝈 𝟏 + 𝝈 𝟐 + 𝝈 𝟑 = 𝟎, 𝜺 𝒗 = 𝟎 This means that if the sum of three principal stress is zero, there is no volumetric change(dilation), but only the distortion occurs. The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only, while the other produces distortion (shear stresses) only. Akhtar Kamal
- 21. Consider a small block of length δℓ, width δb and height δh subjected to three principal stresses σ1, σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb × δh) σ2 = Principal stress on face of area (δℓ × δh) σ3 = Principal stress on face of area (δℓ × δb) μ = Poisson’s ratio for the material. Akhtar Kamal
- 22. Akhtar Kamal
- 23. .˙., Extention of the block in the direction of σ1 δℓı =εı · δı δℓı = 1 𝐸 [σ1 – μ (σ2+σ3 )] δℓ Akhtar Kamal
- 24. .˙. Strain energy due to σ1 = 1 2 (Load due to σ1 in the direction of σ1) × δℓ1 = 1 2 [σ1.δb.δh] x 1 𝐸 [σ1‒ μ (σ2+ σ3)] δℓ = 1 2𝐸 [σ1² ‒ μ (σ1 σ2 +σ1 σ3)] (δb.δh.δℓ)] = 1 2𝐸 [σ1² ‒ μ (σ1 σ2 +σ1 σ3)] δV Where, δV= volume of block = δb.δh.δl Akhtar Kamal
- 25. Similarly, Strain energy due to σ2 = 1 2𝐸 [σ2² ‒ μ(σ2σ1 + σ2 σ3)]δV Strain Energy due to σ3 = 1 2𝐸 [σ3² ‒ μ σ3 σ1 + σ3 σ2)]δV Akhtar Kamal
- 26. .˙. δu = Total Strain energy for volume δV = Sum of strain energies due to σ1,σ2 and σ3 = 1 2𝐸 [σ1² ‒ μ(σ1 σ2 +σ1 σ3)]δV + 1 2𝐸 [σ2 ² ‒ μ(σ2 σ1 +σ2 σ3)]δV + 1 2𝐸 [σ3 ² ‒ μ(σ3 σ1 + σ3 σ2)]δV .˙. δu = 1 2𝐸 [σ1²+ σ2 ²+ σ3 ²- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV Akhtar Kamal
- 27. Thus for a body of Volume V Subjected to the principal Stresses σ1,σ2 and σ3, total strain energy is given by, u= 𝟏 𝟐𝑬 [σ1²+ σ2 ²+ σ3 ² ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V Sign for Principal Stresses, Tension = + ve Compression = -ve Akhtar Kamal
- 28. The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy. Akhtar Kamal
- 29. .˙. u= 𝟏 𝟐𝑬 [σ1²+ σ2 ² ‒ 2μ(σ1 σ2)] V Akhtar Kamal
- 30. σ1=σ , σ2 =0 , σ3 = 0 .˙. u= 1 2𝐸 (σ² × V) .˙. u= σ² 𝟐𝑬 V Akhtar Kamal
- 31. Let τ be simple shear in volume V Then the principle stress will be, σ1= τ , σ2 = ‒ τ , σ3 =0 Substituting these values in (1) we get, u= 1 2𝐸 [τ ²(‒ τ )²+0 ‒ 2μ(τ)(‒ τ)] V = 1 2𝐸 [2 τ ²+ 2 μ τ ²]V = 𝜏 ² 𝐸 (1+ μ) V But,E =2G (1+ μ) G = Modulus of rigidity Therefore 1+ μ 𝐸 = 1 2𝐺 u= τ 𝟐𝑮 V Strain energy per unit volume = u = τ ² 𝟐𝑮 Akhtar Kamal
- 32. Let p= hydrostatic tension or hydrostatic pressure .˙. Either σ1 =p , σ2 =p and σ3=p Or σ1= -p , σ2 = -p and σ3 = -p Substituting any one in equation (1) we get, u= 1 2𝐸 [ p² + p² + p² ‒ 2μ( p.p + p.p + p.p)] V = 1 2𝐸 [3p² ‒ 2μ(3p²)]V = 3p² 2𝐸 (1- 2μ)V but E = 3k(1- 2μ) .˙. (1− 2μ) 𝐸 = 1 3𝑘 k = Bulk modulus .˙. u = p² 𝟐𝒌 V Akhtar Kamal
- 33. Akhtar Kamal