ME8692 KSN Notes- By www.EasyEngineering.net 1

Downloaded From: www.EasyEngineering.net ww w.E asy E ngi nee rin g.n et **Note: Other Websites/Blogs Owners Please do not Copy (or) Republish this Materials, Students & Graduates if You Find the Same Materials with EasyEngineering.net Watermarks or Logo, Kindly report us to easyengineeringnet@gmail.com Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net ww w.E a UNIT 1 syE ngi nee rin g.n et Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net UNIT-1 INTRODUCTION Part- A 1. Distinguish one Dimensional bar element and Beam Element (May/June 2011) 1D bar element: Displacement is considered. 1D beam element: Displacement and slope is considered 2. What do you mean by Boundary value problem? The solution of differential equation is obtained for physical problems, which satisfies some specified conditions known as boundary conditions. The differential equation together with these boundary conditions, subjected to a boundary value problem. Examples: Boundary value problem. 2 2 d y/dx - a(x) dy/dx – b(x)y –c(x) = 0 with boundary conditions, y(m) = S and y(n) = T. ww 3. What do you mean by weak formulation? State its advantages. (April/May 2015), (May/June 2013) w.E A weak form is a weighted integral statement of a differential equation in which the differentiation is distributed among the dependent variable and the weight function and also includes the natural boundary conditions of the problem.  A much wider choice of trial functions can be used.  The weak form can be developed for any higher order differential equation.  Natural boundary conditions are directly applied in the differential equation.  The trial solution satisfies the essential boundary conditions. asy En gin eer i 4. Why are polynomial types of interpolation functions preferred over trigonometric functions? ng. (May/June 2013) Polynomial functions are preferred over trigonometric functions due to the following reasons: net 1. It is easy to formulate and computerize the finite element equations 2. It is easy to perform differentiation or integration 3. The accuracy of the results can be improved by increasing the order of the polynomial. 5. What do you mean by elements & Nodes?(May/June 2014) In a continuum, the field variables are infinite. Finite element procedure reduces such unknowns to a finite number by dividing the solution region into small parts called Elements. The common points between two adjacent elements in which the field variables are expressed are called Nodes. 6. What is Ritz method?(May/June 2014) It is integral approach method which is useful for solving complex structural problem, encountered in finite element analysis. This method is possible only if a suitable function is available. In Ritz method approximating functions satisfying the boundary conditions are used to get the solutions 1 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 7. Distinguish Natural & Essential boundary condition (May/June 2009) There are two types of boundary conditions. They are: 1. Primary boundary condition (or) Essential boundary condition The boundary condition, which in terms of field variable, is known as primary boundary condition. 2. Secondary boundary condition or natural boundary conditions The boundary conditions, which are in the differential form of field variables, are known as secondary boundary condition. Example: A bar is subjected to axial load as shown in fig. ww w.E asy In this problem, displacement u at node 1 = 0, that is primary boundary condition. EA du/dx = P, that is secondary boundary condition. En 8. Compare Ritz method with nodal approximation method.(Nov/Dec 2014), (Nov/Dec 2012) gin Similarity: (i) Both methods use approximating functions as trial solution (ii) Both methods take linear combinations of trial functions. (iii) In both methods completeness condition of the function should be satisfied (iv) In both methods solution is sought by making a functional stationary. Difference (i) Rayleigh-Ritz method assumes trial functions over entire structure, while finite element method uses trial functions only over an element. (ii) The assumed functions in Rayleigh-Ritz method have to satisfy boundary conditions over entire structure while in finite element analysis, they have to satisfy continuity conditions at nodes and sometimes along the boundaries of the element. However completeness condition should be satisfied in both methods. eer i ng. net 9. What do you mean by elements & Nodes? In a continuum, the field variables are infinite. Finite element procedure reduces such unknowns to a finite number by dividing the solution region into small parts called Elements. The common points between two adjacent elements in which the field variables are expressed are called Nodes. 10. State the discretization error. How it can be reduced? (April /May 2015) Splitting of continuum in to smallest elements is known as discretization. In some context like structure having boundary layer the exact connectivity can’t be achieved. It means that it may 2 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net not resemble the original structure. Now there is an error developed in calculation. Such type of error is discretization error. To Reduce Error: (i) Discretization error can be minimized by reducing the finite element (or) discretization element. (ii) By introducing finite element it has a curved member. 11. What are the various considerations to be taken in Discretization process? (i) Types of Elements. (ii) Size of Elements. (iii) Location of Nodes. (iv) Number of Elements. 12. State the principleofminimum potential energy. (Nov/Dec 2010) Amongallthedisplacementequationsthatsatisfiedinternalcompatibilityandthe boundaryconditionthosethatalsosatisfytheequationofequilibriummakethe potential minimum is astable system. PART-B ww w.E energya 𝒅𝟐 𝒖 asy The following differential equation is available for a physical phenomenon. 𝑨𝑬 = 𝒅𝒙𝟐 + 1. 𝒅𝒖 𝒂𝒙 = 𝟎, The boundary conditions are u(0) = 0, 𝑨𝑬 = 𝒅𝒙 En 𝒙=𝑳 = 𝟎 By using Galerkin’s technique, find the solution of the above differential equation. Given Data: 𝑑2𝑢 Differential equ. 𝐴𝐸 = 𝑑𝑥 2 + 𝑎𝑥 = 0 Boundary Conditions 𝑢 0 = 0, gin 𝑑2𝑢 eer i 𝐴𝐸 = 𝑑𝑥 2 + 𝑎𝑥 = 0 To Find: u(x) by using galerkin’s technique Formula used 𝐿 ng. net 𝑤𝑖 𝑅 𝑑𝑥 = 0 0 Solution: Assume a trial function Let 𝑢 𝑥 = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + 𝑎3 𝑥 3 …….. (1) Apply first boundary condition i.e) at x=0, u(x) = 0 1 ⟹ 0 = 𝑎0 + 0 + 0 + 0 𝑎0 = 0 3 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝑑𝑢 Apply first boundary condition i.e at x = L, 𝐴𝐸 = 𝑑𝑥 = 0 ⟹ 𝑑𝑢 = 0+𝑎1 + 2𝑎2 𝑥 + 3𝑎3 𝐿2 𝑑𝑥 ⟹ 0 = 𝑎1 + 2𝑎2 𝐿 + 3𝑎3 𝐿2 ⟹ 𝑎1 = −(2𝑎2 𝐿 + 3𝑎3 𝐿2 ) sub 𝑎0 and 𝑎1 in value in equation (1) 𝑢 𝑥 = 0 + − 2𝑎2 𝐿 + 3𝑎3 𝐿2 𝑥 + 𝑎2 𝑥 2 + 𝑎3 𝑥 3 = −2𝑎2 𝐿𝑥 − 3𝑎3 𝐿2 𝑎2 𝑥 + 𝑎2 𝑥 2 + 𝑎3 𝑥 3 = 𝑎2 𝑥 2 − 2𝐿𝑥 + 𝑎3 (𝑥 3 − 3𝐿2 𝑥) ……… (2) We Know That 𝑑2𝑢 Residual, 𝑅 = 𝐴𝐸 𝑑𝑥 2 + 𝑎𝑥 ww ………. (3) 𝑑𝑢 = 𝑎2 2𝑥 − 2𝐿 + 𝑎3 (3𝑥 2 − 3𝐿2 ) 𝑑𝑥 (2) ⟹ w.E 𝑑2 𝑢 = 𝑎2 2 + 𝑎3 (6𝑥) 𝑑𝑥 2 asy 𝑑2 𝑢 = 2𝑎2 + 6𝑎3 𝑥 𝑑𝑥 2 𝑑2𝑢 Sub 𝑑𝑥 2 value in equation (3) 3 ⟹ 𝑅 = 𝐴𝐸 2𝑎2 + 6𝑎3 𝑥 + 𝑎𝑥 Residual, 𝑅 = 𝐴𝐸 2𝑎2 + 6𝑎3 𝑥 + 𝑎𝑥 En gin From Galerkn’s technique eer i ……… (4) ng. 𝐿 𝑤𝑖 𝑅 𝑑𝑥 = 0 . . … … . . . (5) 0 from equation (2) we know that 𝑤1 = 𝑥 2 − 2𝐿𝑥 net 𝑤2 = 𝑥 3 − 3𝐿2 𝑥 sub w1, w2 and R value in equation (5) 𝐿 𝑥 2 − 2𝐿𝑥 𝐴𝐸 2𝑎2 + 6𝑎3 𝑥 + 𝑎𝑥 𝑑𝑥 = 0 5 ⟹ … … … … … (6) 0 𝐿 𝑥 3 − 3𝐿2 𝑥 𝐴𝐸 2𝑎2 + 6𝑎3 𝑥 + 𝑎𝑥 𝑑𝑥 = 0 … … … … … (7) 0 𝐿 𝑥 2 − 2𝐿𝑥 𝐴𝐸 2𝑎2 + 6𝑎3 𝑥 + 𝑎𝑥 𝑑𝑥 = 0 6 ⟹ 0 4 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝐿 𝑥 2 − 2𝐿𝑥 2𝑎2 𝐴𝐸 + 6𝑎3 𝐴𝐸𝑥 + 𝑎𝑥 𝑑𝑥 = 0 0 𝐿 2𝑎2 𝐴𝐸𝑥 2 + 6𝑎3 𝐴𝐸𝑥 3 + 𝑎𝑥 3 − 4𝑎2 𝐴𝐸𝐿𝑥 − 12𝑎3 𝐴𝐸𝐿𝑥 2 − 2𝑎𝐿𝑥 2 = 0 0 ⟹ [2𝑎2 𝐴𝐸 𝑥3 𝑥4 𝑥4 𝑥2 𝑥3 𝑥3 + 6𝑎3 𝐴𝐸 + 𝑎 − 4𝑎2 𝐴𝐸𝐿 − 12𝑎3 𝐴𝐸𝐿 − 2𝑎𝐿 ]𝐿0 = 0 3 4 4 2 3 3 𝐿3 𝐿4 𝐿4 𝐿3 𝐿4 𝐿4 ⟹ 2𝑎2 𝐴𝐸 + 6𝑎3 𝐴𝐸 + 𝑎 − 4𝑎2 𝐴𝐸 − 12𝑎3 𝐴𝐸 − 2𝑎 = 0 3 4 4 2 3 3 2 𝐿4 3 2 ⟹ 3 𝑎2 𝐴𝐸𝐿3 + 2 𝑎3 𝐴𝐸 𝐿4 + 𝑎 4 − 2𝑎2 𝐴𝐸𝐿3 − 4𝑎3 𝐴𝐸𝐿4 − 3 𝑎𝐿4 = 0 2 3 𝐿4 2 − 2 + 𝑎3 𝐴𝐸 𝐿4 − 4 + 𝑎 − 𝑎2 𝐿4 = 0 3 2 4 3 −4 5 2 1 4 5 5 4 ⟹ 𝐴𝐸𝐿3 𝑎2 − 𝐴𝐸𝐿4 𝑎3 = − 𝑎𝐿4 − 𝐴𝐸𝐿3 𝑎2 − 𝐴𝐸𝐿4 𝑎3 = 𝑎𝐿 3 2 3 4 3 2 12 −4 5 5 𝐴𝐸𝐿3 𝑎2 − 𝐴𝐸𝐿4 𝑎3 = − 𝑎𝐿4 ………. 8 3 2 12 ⟹ 𝐴𝐸𝑎2 𝐿3 ww Equation (7) w.E asy En 𝐿 (𝑥 3 − 3𝐿2 𝑥) 𝐴𝐸 2𝑎2 + 6𝑎3 𝑥 + 𝑎𝑥 𝑑𝑥 = 0 ⟹ 0 𝐿 3 ⟹ 2 gin (𝑥 − 3𝐿 𝑥) 2𝑎2 𝐴𝐸 + 6𝑎3 𝐴𝐸𝑥 + 𝑎𝑥 𝑑𝑥 = 0 0 eer i ng. 𝐿 2𝐴𝐸𝑎2 𝑥 3 + 6𝐴𝐸𝑎3 𝑥 4 + 𝑎𝑥 4 − 6𝐴𝐸𝑎2 𝐿2 𝑥 − 18𝐴𝐸𝑎3 𝐿2 𝑥 2 − 3𝑎𝐿2 𝑥 2 𝑑𝑥 = 0 ⟹ 0 𝐿 𝑥4 𝑥5 𝑥5 𝑥2 𝑥3 𝑥3 ⟹ 2𝐴𝐸𝑎2 + 6𝐴𝐸𝑎3 + 𝑎 − 6𝐴𝐸𝑎2 𝐿2 − 18𝐴𝐸𝑎3 𝐿2 − 3𝑎𝐿2 =0 4 5 5 2 3 3 0 net 𝐿 1 6 1 5 4 5 2 2 2 3 2 3 ⟹ 𝐴𝐸𝑎2 𝑥 + 𝐴𝐸𝑎3 𝑥 + 𝑎𝑥 − 3𝐴𝐸𝑎2 𝐿 𝑥 − 6𝐴𝐸𝑎3 𝐿 𝑥 − 𝑎𝐿 𝑥 =0 2 5 5 0 1 6 1 𝐴𝐸𝑎2 𝐿4 + 𝐴𝐸𝑎3 𝐿5 + 𝑎𝐿5 − 3𝐴𝐸𝑎2 𝐿2 (𝐿2 ) − 6𝐴𝐸𝑎3 𝐿2 (𝐿3 ) − 𝑎𝐿2 (𝐿3 ) = 0 2 5 5 1 6 1 ⟹ 𝐴𝐸𝑎2 𝐿4 + 𝐴𝐸𝑎3 𝐿5 + 𝑎𝐿5 − 3𝐴𝐸𝑎2 𝐿4 − 6𝐴𝐸𝑎3 𝐿5 − 𝑎𝐿5 = 0 2 5 5 1 6 1 ⟹ 𝐴𝐸𝑎2 𝐿4 − 3 + 𝐴𝐸𝑎3 𝐿5 − 6 + 𝑎𝐿5 + − 1 = 0 2 5 5 5 24 4 ⟹ 𝐴𝐸𝑎2 𝐿4 − 𝐴𝐸𝑎3 𝐿5 = 𝑎𝐿5 2 5 5 5 24 4 𝐴𝐸𝑎2 𝐿4 + 𝐴𝐸𝑎3 𝐿5 = − 𝑎𝐿5 …………. 9 2 5 5 ⟹ 5 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Solving Equation (8) and (9) 4 5 5 5 24 4 Equation (8) ⟹ 3 𝐴𝐸𝑎2 𝐿3 + 2 𝐴𝐸𝑎3 𝐿4 = − 12 𝑎𝐿4 Equation (9) ⟹ 2 𝐴𝐸𝑎2 𝐿4 + 5 𝐴𝐸𝑎3 𝐿5 = − 5 𝑎𝐿5 5 4 Multiplying Equation (8) 2 𝐿 and Equation (9) by 3 20 25 25 𝐴𝐸𝑎2 𝐿4 + 𝐴𝐸𝑎3 𝐿5 = − 𝑎𝐿5 6 4 24 20 25 16 𝐴𝐸𝑎2 𝐿4 + 𝐴𝐸𝑎3 𝐿5 = − 𝑎𝐿5 6 4 15 Subtracting 25 96 16 25 − 𝐴𝐸𝑎3 𝐿5 = − 𝑎𝐿5 4 15 15 24 375 − 384 384 − 375 𝐴𝐸𝑎3 𝐿5 = 𝑎𝐿5 60 360 −9 9 ⟹ 𝐴𝐸𝑎3 𝐿5 = 𝑎𝐿5 60 360 ww w.E ⟹ −0.15𝐴𝐸𝑎3 = 0.025𝑎 asy 𝑎3 = −0.1666 𝑎 𝐴𝐸 En 𝑎3 = − Substituting a3 value in Equation (8) 4 5 −𝑎 4 −5 4 𝐴𝐸𝑎2 𝐿3 + 𝐴𝐸 𝐿 = 𝑎𝐿 3 2 6𝐴𝐸 12 4 −5 4 5 −𝑎 𝐴𝐸𝑎2 𝐿3 = 𝑎𝐿 − 𝐴𝐸𝐿4 = 3 12 2 6𝐴𝐸 4 −5 4 5 𝐴𝐸𝑎2 𝐿3 = 𝑎𝐿 + 𝐴𝐸𝐿4 3 12 2 4 𝐴𝐸𝑎2 𝐿3 = 0 3 𝑎 6𝐴𝐸 gin … … … . (10) eer i ng. net 𝑎2 = 0 Sub a2 and a3 value in equation (2) −𝑎 ⟹ 𝑢 𝑥 = 0𝑥 𝑥2 − 2𝐿𝑥 + 6𝐴𝐸 𝑎 ⟹𝑢 𝑥 = 3𝐿2 𝑥 − 𝑥 3 6𝐴𝐸 𝑥 3 − 3𝐿2 𝑥 = 0 Result: 𝑢 𝑥 = 𝑎 3𝐿2 𝑥 − 𝑥 3 6𝐴𝐸 6 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 2. Find the deflection at the centre of a simply supported beam of span length “l” subjected to uniformly distributed load throughout its length as shown in figure using (a) point collocation method, (b) sub-domain method, (c) Least squares method, and (d) Galerkin’s method. (Nov/Dec 2014) Given data Length (L) = 𝑙 UDL = 𝜔 𝑁/𝑚 To find ww Deflection Formula used w.E 𝑑4 𝑦 𝐸𝐼 4 − 𝜔 = 0, 𝑑𝑥 0≤𝑥≤𝑙 asy Point Collocation Method R = 0 𝑙 Sub-domain collocation method = 0 𝑅𝑑𝑥 = 0 𝑙 En Least Square Method 𝐼 = 0 𝑅 2 𝑑𝑥 𝑖𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 Solution: gin eer i The differential equation governing the deflection of beam subjected to uniformly distributed load is given by 𝑑4 𝑦 𝐸𝐼 4 − 𝜔 = 0, 𝑑𝑥 0≤𝑥≤𝑙 … … … . (1) ng. The boundary conditions are Y=0 at x=0 and x = l, where y is the deflection. 𝑑4 𝑦 𝐸𝐼 4 = 0, 𝑑𝑥 𝑎𝑡 𝑥 = 0 𝑎𝑛𝑑 𝑥 = 𝑙 net Where 𝑑4𝑦 𝐸𝐼 𝑑𝑥 4 = 𝑀, (Bending moment) E → Young’s Modules I → Moment of Inertia of the Beam. 𝜋𝑥 Let us select the trial function for deflection as 𝑌 = 𝑎𝑠𝑖𝑛 𝑙 ……. (2) Hence it satisfies the boundary conditions ⟹ 𝑑𝑦 𝜋 𝜋𝑥 = 𝑎 . cos 𝑑𝑥 𝑙 𝑙 𝑑2 𝑦 𝜋2 𝜋𝑥 ⟹ 2 = −𝑎 2 . sin 𝑑𝑥 𝑙 𝑙 7 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝑑3 𝑦 𝜋3 𝜋𝑥 ⟹ 3 = −𝑎 3 . cos 𝑑𝑥 𝑙 𝑙 𝑑4 𝑦 𝜋4 𝜋𝑥 ⟹ 4 = 𝑎 4 . sin 𝑑𝑥 𝑙 𝑙 Substituting the Equation (3) in the governing Equation (1) 𝜋4 𝜋𝑥 𝐸𝐼 𝑎 4 . sin −𝜔 = 0 𝑙 𝑙 𝜋4 𝜋𝑥 Take, Residual 𝑅 = 𝐸𝐼𝑎 𝑙 4 . sin 𝑙 − 𝜔 a) Point Collocation Method: In this method, the residuals are set to zero. 𝜋4 𝜋𝑥 ⟹ 𝑅 = 𝐸𝐼𝑎 4 . sin −𝜔 =0 𝑙 𝑙 ww 𝜋4 𝜋𝑥 𝐸𝐼𝑎 4 . sin =𝜔 𝑙 𝑙 w.E 𝑙 To get maximum deflection, take 𝑘 = 2 (𝑖. 𝑒. 𝑎𝑡 𝑐𝑎𝑛 𝑏𝑒 𝑜𝑓 𝑏𝑒𝑎𝑚) 𝜋4 𝜋 𝑙 asy 𝐸𝐼𝑎 𝑙 4 . sin 𝑙 2 = 𝜔 4 𝜋 𝐸𝐼𝑎 4 = 𝜔 𝑙 𝑎= 𝜔𝑙 4 𝜋 4 𝐸𝐼 [∵ sin En Sub “a” value in trial function equation (2) gin 𝜔𝑙 𝜋𝑥 . sin 𝜋 4 𝐸𝐼 𝑙 4 𝐴𝑡 𝑥 = eer i ng. 4 𝑌= 𝜋 = 1] 𝑙 𝑙 𝜔𝑙 𝜋 𝑙 ⟹ 𝑌max = 4 . sin 2 𝜋 𝐸𝐼 2 2 𝜔 𝑙4 𝑌max = 𝜋 4 𝐸𝐼 [∵ sin 𝜔𝑙 4 𝑌max = 97.4𝐸𝐼 net 𝜋 = 1] 2 b) Sub-domain collocation method: In this method, the integral of the residual over the sub-domain is set to zero. 𝑙 𝑅𝑑𝑥 = 0 0 Sub R value 𝜋4 𝜋𝑥 ⟹ 𝑎𝐸𝐼 4 sin − 𝜔 𝑑𝑥 = 0 𝑙 𝑙 8 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝑙 𝜋𝑥 −cos 𝑙 −𝜔 𝑥 = 0 𝜋 𝑙 0 4 𝜋 ⟹ 𝑎𝐸𝐼 4 𝑙 𝑙 𝜋4 𝜋𝑥 𝑙 ⟹ 𝑎𝐸𝐼 4 −cos −𝜔𝑥 = 0 𝑙 𝑙 𝑢 0 ∵ cos 𝜋 = −1 , 𝑐𝑜𝑠0 = 1 𝜋3 ⟹ −𝑎𝐸𝐼 𝑙 3 cos𝜋 − 𝑐𝑜𝑠0 𝜔 𝑙 = 0 𝜋3 −𝑎𝐸𝐼 3 −1 − 1 = 𝜔 𝑙 𝑙 𝜔𝑙 4 𝜔𝑙 4 ⟹ −𝑎 = 3 = 2𝜋 𝐸𝐼 62𝐸𝐼 Sub “a” value in the trial function equation (2) ww 𝜔𝑙 4 𝜋𝑥 𝑌= . sin 62𝐸𝐼 𝑙 w.E 𝑙 𝜔𝑙 4 𝜋 𝑙 𝐴𝑡 𝑥 = , 𝑌𝑚𝑎𝑥 = . sin ( ) 2 62𝐸𝐼 𝑙 2 𝑌𝑚𝑎𝑥 = 𝜔𝑙 4 62𝐸𝐼 c) Least Square Method: asy In this method the functional 𝑙 En gin 𝑅 2 𝑑𝑥 𝑖𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝐼= 0 𝑙 𝜋4 𝜋𝑥 (𝑎𝐸𝐼 4 . 𝑠𝑖𝑛 − 𝜔)2 𝑑𝑥 𝑙 𝑙 𝐼= 0 𝑙 𝜋8 𝜋𝑥 𝜋4 𝜋𝑥 2 2 [𝑎 𝐸 𝐼 8 . 𝑠𝑖𝑛 − 𝜔 − 2𝑎𝐸𝐼𝜔 4 . 𝑠𝑖𝑛 ]𝑑𝑥 𝑙 𝑙 𝑙 𝑙 2 = 0 2 2 𝜋8 1 2𝜋𝑥 = [𝑎 𝐸 𝐼 8 𝑥 𝑠𝑖𝑛 𝑙 2 𝑙 2 eer i 2 2 = 𝑎2 𝐸 2 𝐼 2 ng. net 𝑙 𝜋4 𝜋𝑥 𝑙 𝑙 2 + 𝜔 − 2𝑎𝐸𝐼𝜔 4 . [−𝑐𝑜𝑠 ]] 2𝜋 𝑙 𝑙 𝜋 0 𝜋8 1 𝑙 𝑙 − 𝑠𝑖𝑛2𝜋 − 𝑠𝑖𝑛0 𝑙8 2 2𝜋 𝜋4 𝑙 + 𝜔2 𝑙 + 2𝑎𝐸𝐼𝜔 4 . [−𝑐𝑜𝑠𝜋 − 𝑐𝑜𝑠0] 𝑙 𝜋 ∵ 𝑠𝑖𝑛2𝜋 = 0; 𝑠𝑖𝑛0 = 0; 𝑐𝑜𝑠𝜋 = 0; 𝑐𝑜𝑠0 = 1 𝐼 = 𝑎2 𝐸 2 𝐼 2 𝐼= 𝜋8 𝑙 𝜋3 2 + 𝜔 𝑙 + 2𝑎𝐸𝐼𝜔 . (−1 − 1) 𝑙2 2 𝑙3 𝑎2 𝐸 2 𝐼 2 𝜋 8 𝜋3 2 + 𝜔 𝑙 − 4𝑎𝐸𝐼𝜔 2𝑙 7 𝑙3 𝜕𝜋 Now, 𝜕𝑎 = 0 9 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝑎2 𝐸 2 𝐼 2 𝜋 8 𝜋3 ⟹ = 4𝐸𝐼𝜔 3 2𝑙 7 𝑙 𝑎2 𝐸 2 𝐼 2 𝜋 8 𝜋3 = 4𝐸𝐼𝜔 3 𝑙7 𝑙 4𝐸𝐼𝜔𝑙 5 𝑎= 5 𝜋 𝐸𝐼 Hence the trial Function 4𝜔𝑙 4 𝜋𝑥 𝑌 = 5 . sin 𝜋 𝐸𝐼 𝑙 𝑙 𝜋 𝐴𝑡 𝑥 = 2 , max 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 [∵ 𝑠𝑖𝑛 2 = 1] ww 4𝜔𝑙 4 𝜋 𝑙 𝑌𝑚𝑎𝑥 = 5 𝑠𝑖𝑛 ( ) 𝜋 𝐸𝐼 2 2 𝜔𝑙 4 𝑌𝑚𝑎𝑥 = 76.5 𝐸𝐼 w.E d) Galerkin’s Method: In this method 𝑙 asy 𝑌. 𝑅 𝑑𝑥 = 0 0 𝑙 ⟹ 𝑎𝑠𝑖𝑛 0 𝑙 𝜋𝑥 𝑙 4 𝑎𝐸𝐼 𝜋 𝜋𝑥 𝑠𝑖𝑛 −𝜔 4 𝑙 𝑙 En 𝑑𝑥 = 0 𝜋4 𝜋𝑥 𝜋𝑥 𝑎 𝐸𝐼 4 𝑠𝑖𝑛2 − 𝑎𝜔𝑠𝑖𝑛 𝑑𝑥 = 0 𝑙 𝑙 𝑙 gin 2 ⟹ 0 𝑙 𝑎2 𝐸𝐼 ⟹ 0 eer i ng. 𝜋4 1 2𝜋𝑥 𝜋𝑥 [ (1 − 𝑐𝑜𝑠 ) − 𝑎𝜔𝑠𝑖𝑛 𝑑𝑥 = 0 4 𝑙 2 𝑙 𝑙 𝜋4 1 ⟹ 𝑎 𝐸𝐼 4 [ 1 − 𝑙 2 2 1 2𝜋𝑥 𝑥− 𝑠𝑖𝑛2 2𝜋 𝑙 𝑙 𝑙 𝜋𝑥 + 𝑎𝜔 𝑐𝑜𝑠 =0 𝜋 𝑙 0 net 𝜋4 𝑙 𝑙 𝑎 𝐸𝐼 4 − 2𝑎𝜔 =0 𝑙 2 𝜋 2 2𝜔𝑙 2𝑙 3 ∴𝑎= . 𝜋 𝐸𝐼𝜋 4 4𝜔𝑙 3 𝑎= 5 𝜋 𝐸𝐼 Hence the trial Function 4𝜔𝑙 4 𝜋𝑥 𝑌 = 5 . sin 𝜋 𝐸𝐼 𝑙 10 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝑙 𝜋 𝐴𝑡 𝑥 = 2 , max 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 [∵ 𝑠𝑖𝑛 2 = 1] 4𝜔𝑙 4 𝜋 𝑙 𝑌𝑚𝑎𝑥 = 5 𝑠𝑖𝑛 ( ) 𝜋 𝐸𝐼 2 2 4𝜔𝑙 4 𝑌𝑚𝑎𝑥 = 5 𝜋 𝐸𝐼 𝜔𝑙 4 𝑌𝑚𝑎𝑥 = 76.5 𝐸𝐼 Verification, We know that simply supported beam is subjected to uniformly distributed load, maximum deflection is, ww 𝑌𝑚𝑎𝑥 = 3) i) 5 𝜔𝑙 4 384 𝐸𝐼 = 0.01 w.E 𝜔𝑙 4 𝐸𝐼 What is constitutive relationship? Express the constitutive relations for a linear asy elastic isotropic material including initial stress and strain. [Nov/Dec 2009] En Solution: (4) gin It is the relationship between components of stresses in the members of a structure or in a solid body and components of strains. The structure or solids bodies under consideration are made of elastic material that obeys Hooke’s law. 𝜎 = 𝐷 {𝑒} Where eer i [D] is a stress – strain relationship matrix or constitute matrix. The constitutive relations for a linear elastic isotropic material is 𝜎𝑥 𝜎𝑦 𝜎𝑧 𝛿𝑥𝑦 𝛿𝑦𝑧 𝛿𝑧𝑥 (1 − 𝑣) 𝑣 𝐸 𝑣 = 1 + 𝑣 1 − 2𝑣 0 0 0 0 (1 − 𝑣) 𝑣 0 0 0 0 0 0 0 0 (1 − 𝑣)1 − 2𝑣 0 2 0 0 0 0 0 0 0 0 1 − 2𝑣 2 0 ng. net 0 𝑒𝑥 0 𝑒𝑦 0 𝑒𝑧 0 𝑣𝑥𝑦 0 1 − 2𝑣 𝑣𝑦𝑧 𝑣𝑧𝑥 2 𝒅𝟐 𝒚 ii) Consider the differential equation 𝒅𝒙𝟐 + 𝟒𝟎𝟎𝒙𝟐 = 𝟎 for 𝟎 ≤ 𝒙 ≤ 𝟏 subject to boundary conditions Y(0) = 0, Y(1) = 0. The functions corresponding to this problem, to be eternized 𝒍 𝒅𝒚 𝟐 is given by 𝑰 = 𝟎 −𝟎. 𝟓 𝒅𝒙 + 𝟒𝟎𝟎𝒙𝟐 𝒀 . Find the solution of the problem using Ray Light Ritz method by considering a two term solution as 𝒀 𝒙 = 𝒄𝟏 𝒙 𝟏 − 𝒙 + 𝒄𝟐 𝒙𝟐 (𝟏 − 𝒙) (12) 11 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Given data 𝑑2𝑦 Differential equation = 𝑑𝑥 2 + 400𝑥 2 = 0 for 0 ≤ 𝑥 ≤ 1 Boundary conditions Y(0) = 0, Y(1) = 0 𝑑𝑦 2 𝑙 𝐼 = 0 −0.5 𝑑𝑥 + 400𝑥 2 𝑌 𝑌 𝑥 = 𝑐1 𝑥 1 − 𝑥 + 𝑐2 𝑥 2 (1 − 𝑥) To find: Rayleigh- Ritz method Formula used ww 𝜕𝐼 𝜕𝑐1 =0 w.E 𝜕𝐼 =0 𝜕𝑐2 Solution: asy 𝑌 𝑥 = 𝑐1 𝑥 1 − 𝑥 + 𝑐2 𝑥 2 (1 − 𝑥) En 𝑌 𝑥 = 𝑐1 𝑥 𝑥 − 𝑥 2 + 𝑐2 (𝑥 2 − 𝑥 3 ) 𝑑𝑦 = 𝑐1 1 − 2𝑥 + 𝑐2 (2𝑥 − 3𝑥 2 ) 𝑑𝑥 = 𝑐1 1 − 2𝑥 + 𝑐2 𝑥(2 − 3𝑥) 𝑑𝑦 2 = 𝑐1 1 − 2𝑥 + 𝑐2 𝑥(2 − 3𝑥)2 2 𝑑𝑥 gin eer i ng. = 𝑐12 1 − 4𝑥 + 4𝑥 2 + 𝑐22 𝑥 2 4 − 12𝑥 + 9𝑥 2 + 2𝑐1 𝑐2 𝑥 1 − 2𝑥 (2 − 3𝑥) = 𝑐12 1 − 4𝑥 + 4𝑥 2 + 𝑐22 𝑥 2 4 − 12𝑥 + 9𝑥 2 + 2𝑐1 𝑐2 𝑥(2 − 3𝑥 − 4𝑥 + 6𝑥 2 ) 𝑑𝑦 2 = 𝑐12 1 − 4𝑥 + 4𝑥 2 + 𝑐22 𝑥 2 4 − 12𝑥 + 9𝑥 2 + 2𝑐1 𝑐2 𝑥(2 − 7𝑥 + 6𝑥 2 ) 𝑑𝑥 net We know that 𝑙 𝐼= 0 𝑑𝑦 2 −1 [−0.5 + 400𝑥 2 𝑦] = 𝑑𝑥 2 𝑙 0 𝑑𝑦 2 + 400 𝑑𝑥 𝑙 𝑥2 𝑦 0 𝑙 𝑐12 1 − 4𝑥 + 4𝑥 2 + 𝑐22 𝑥 2 4 − 12𝑥 + 9𝑥 2 + 2𝑐1 𝑐2 𝑥 2 − 7𝑥 + 6𝑥 2 = 0 𝑙 + 400[ 𝑥 2 𝑐1 𝑥 1 − 𝑥 + 𝑐2 𝑥 2 1 − 𝑥 0 By Solving 12 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net −1 𝑐12 2 1 𝑐1 𝑐2 𝐼= + 𝑐22 + 𝑐1 𝑐2 + 400 + 2 3 15 3 20 30 𝐼= −1 2 1 2 1 40 𝑐1 − 𝑐2 − 𝑐1 𝑐2 + 20𝑐1 + 𝑐2 6 15 6 3 𝜕𝐼 =0 𝜕𝑐1 −1 1 × 2𝑐1 − 𝑐2 + 20 = 0 6 6 −1 1 ⟹ × 𝑐1 − 𝑐2 + 20 = 0 3 6 ⟹ … … … . . (1) Similarly, 𝜕𝐼 =0 𝜕𝑐2 ww w ⟹ −2 1 40 𝑐2 − 𝑐1 + =0 15 6 3 … … … . . (2) w ww.E .Eaa ssyyE nggi By Solving (1) and (2) 𝑐1 = 80 200 ; 𝑐1 = 3 3 We know that 𝑌 = 𝑐1 𝑥 1 − 𝑥 + 𝑐2 𝑥 2 (1 − 𝑥) 𝑌= 4) inneee erirni ng.gn. 80 200 2 𝑥 1−𝑥 + 𝑥 1−𝑥 3 3 Consider a 1mm diameter, 50m long aluminum pin-fin as shown in figure used to enhance the heat transfer from a surface wall maintained at 300C. Calculate the ente temperature distribution in a pin-fin by using Rayleigh – Ritz method. Take, 𝒌 = 𝟐𝟎𝟎𝒘 𝒅𝟐 𝑻 𝒎𝐂 for aluminum h= 𝟐𝟎𝟎𝒘 , 𝑻 = 𝟑𝟎𝐂. 𝒎𝟐 𝐂 ∞ 𝑷𝒉 𝒅𝑻 𝒌 𝒅𝒙𝟐 = 𝑨 (𝑻 − 𝑻∞ ) , 𝑻 𝟎 = 𝑻𝒘 = 𝟑𝟎𝟎𝐂, 𝒒𝑳 = 𝑲𝑨 𝒅𝒙 𝑳 = 𝟎 (insulated tip) 13 Downloaded From: www.EasyEngineering.net t Downloaded From: www.EasyEngineering.net Given Data: The governing differential equation 𝑘 𝑑2 𝑇 𝑃𝑕 = (𝑇 − 𝑇∞ ) 𝑑𝑥 2 𝐴 d = 1mm = 1x10-3m Diameter Length L = 50mm = 50x10-3m K = 200𝑤 𝑚C Thermal Conductivity Heat transfer co-efficient h = 200𝑤 𝑚C Fluid Temp 𝑇∞ = 30C. Boundary Conditions 𝑇 0 = 𝑇𝑤 = 300C 𝑑𝑇 ww 𝑞𝐿 = 𝐾𝐴 𝑑𝑥 𝐿 = 0 To Find: w.E Ritz Parameters Formula used asy 𝜋 = 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 − 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 Solution: En The equivalent functional representation is given by, 𝜋 = 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 − 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝜋 =𝑢−𝑣 𝐿 𝜋= 0 𝐿 𝜋= 0 1 𝑑𝑇 2 𝐾 𝑑𝑥 + 2 𝑑𝑥 1 𝑑𝑇 2 𝐾 𝑑𝑥 + 2 𝑑𝑥 𝐿 0 𝐿 0 gin eer i 1 𝑃𝑕 𝑇 − 𝑇∞ 2 𝑑𝑥 − 𝑞𝐿 𝑇𝐿 2 𝐴 … … … … . (1) 1 𝑃𝑕 𝑇 − 𝑇∞ 2 𝑑𝑥 2 𝐴 ………….. 2 ng. net ∵ 𝑞𝐿 = 0 Assume a trial function Let 𝑇 𝑥 = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 … … … … … . . (3) Apply boundary condition at x = 0, T(x) = 300 300 = 𝑎0 + 𝑎1 (0) + 𝑎2 (0)2 𝑎0 = 300 Substituting 𝑎0 value in equation (3) 𝑇 𝑥 = 300 + 𝑎1 𝑥 + 𝑎2 𝑥 2 …………….. 4 14 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net ⟹ 𝑑𝑇 = 𝑎1 + 2𝑎2 𝑥 𝑑𝑥 … … … … … … (5) Substitute the equation (4), (5) in (2) 𝑙 𝜋= 0 1 𝑘 (𝑎1 + 2𝑎2 𝑥)2 𝑑𝑥 + 2 𝑙 0 1 𝑃𝑕 270 + 𝑎1 + 𝑎2 𝑥 2 2 𝑑𝑥. 2 𝐴 [∵ 𝑎 + 𝑏 2 = 𝑎2 + 𝑏 2 + 2𝑎𝑏; 𝑎 + 𝑏 + 𝑐 2 = 𝑎2 + 𝑏 2 + 𝑐 2 + 2𝑎𝑏 + 2𝑏𝑐 + 2𝑐𝑎 𝑙 𝑘 𝑃𝑕 𝜋= (𝑎12 + 4𝑎22 𝑥 2 + 4𝑎1 𝑎2 𝑥) + 2 2𝐴 0 𝑙 2702 + 𝑎1 2 𝑥 2 + 𝑎2 2 𝑥 4 + 540𝑎1 𝑥 + 2𝑎1 𝑥 3 + 540𝑎2 𝑥 2 𝑑𝑥 0 50𝑥10 −3 𝑘 4𝑎22 𝑥 3 4𝑎1 𝑎2 𝑥 2 𝜋 = (𝑎12 𝑥 + + 2 3 2 0 ww 𝜋= 50𝑥10 −3 𝑃𝑕 𝑎1 2 𝑥 3 𝑎2 2 𝑥 5 540𝑎1 𝑥 2 2𝑎1 𝑎2 𝑥 4 540𝑎2 𝑥 3 + 72900𝑘 + + + + + 2𝐴 3 5 2 4 3 0 w.E asy 𝑘 4𝑎22 (50 × 10−3 )3 4𝑎1 𝑎2 (50 × 10−3 )2 (50 × 10−3 )𝑎12 + + 2 3 2 + 𝜋= [∵ 𝑙 = 50𝑥10−3 ] En 𝑃𝑕 𝑎1 2 (50 × 10−3 )3 𝑎2 2 (50 × 10−3 )5 72900𝑘 + + 2𝐴 3 5 gin eer i 200 𝜋 × 10−3 × 20 50 × 10−3 𝑎12 + 1.666 × 10−4 𝑎22 + 50 × 10−3 𝑎1 𝑎2 + 𝜋 2 2 × 2 × 10−3 2 ng. = 364.5 + 4.166 × 10−5 𝑎12 + 6.25 × 10−8 𝑎22 + 0.675𝑎1 + 3.125 × 10−6 𝑎1 𝑎2 + 0.0225𝑎2 𝜋= 5𝑎12 + 0.0166𝑎22 + 0.5𝑎1 𝑎2 + 14.58 × 10 −7 net + 1.6691 + 2.5 × 10−3 𝑎22 + 2700 𝑎1 2 + 0.125 𝑎1 𝑎2 + 900𝑎2 ] 𝜋 = 6.66𝑎12 + 0.0191𝑎22 + 0.625𝑎1 𝑎2 + 2700𝑎1 + 900𝑎2 + 14.58 × 107 𝜕𝜋 Apply 𝜕𝑎 = 0 2 ⟹ 13.32𝑎1 + 0.625𝑎2 + 27000 = 0 13.32𝑎1 + 0.625𝑎2 = − + 27000 … … … … … (6) ⟹ 0.625𝑎1 + 0.382𝑎2 + 900 = 0 0.625𝑎1 + 0.382𝑎2 = −900 … … … … . . (7) Solve the equation (6) and (7) 13.32𝑎1 + 0.625𝑎2 = − + 27000 0.625𝑎1 + 0.382𝑎2 = −900 … … … … … (6) ………….. 7 (6) x 0.625 15 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 8.325𝑎1 + 0.3906𝑎2 = −16875 ………….. 8 (7) x -13.32 −8.325𝑎1 − 0.5088𝑎2 = 11988 ………….. 9 −0.1182𝑎2 = −4887 𝑎2 = 41345 Sub 𝑎2 value in equation (6) 13.32𝑎1 + 0.625(41345) = − + 27000 𝑎1 = −3967.01 Sub 𝑎0 , 𝑎1 and 𝑎2 values in equation (3) 𝑇 = 300 − 3697.01𝑥 + 41345𝑥 2 ww 5) Explain briefly about General steps of the finite element analysis. w.E Step: 1 [Nov/Dec 2014] asy Discretization of structure The art of sub dividing a structure into a convenient number of smaller element is known as discretization. En Smaller elements are classified as (i) i) One dimensional element ii) Two dimensional element iii) Three dimensional element iv) Axisymmetric element gin eer i ng. One dimensional element:- net a. A bar and beam elements are considered as one dimensional element has two nodes, one at each end as shown. 1 (ii) 2 Two Dimensional element:Triangular and Rectangular elements are considered as 2D element. These elements are loaded by forces in their own plane. 3 1 4 3 1 2 2 16 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net iii) Three dimensional element:The most common 3D elements are tetrahedral and lexahendral (Brick) elements. These elements are used for three dimensional stress analysis problems. iv) Axisymmetric element:The axisymmetric element is developed by relating a triangle or quadrilateral about a fixed axis located in the plane of the element through 3600. When the geometry and loading of the problems are axisymmetric these elements are used. ww w.E The stress-strain relationship is given by, 𝜎 = 𝐸𝑒 Where, 𝜎 = Stress in 𝑥 direction 𝐸 = Modulus of elasticity Step 2:- Numbering of nodes and Elements:- asy En gin The nodes and elements should be numbered after discretization process. The numbering process is most important since if decide the size of the stiffness matrix and it leads the reduction of eer i memory requirement . While numbering the nodes, the following condition should be satisfied. {Maximum number node} – {Minimum number node} = minimum ng. net 17 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net ww w.E asy En Step 3: gin Selection of a displacement function or a Interpolation function:It involves choosing a displacement function within each element. Polynomial of linear, eer i quadratic and cubic form are frequently used as displacement Function because they are simple to work within finite element formulation. 𝑑 𝑥 . ng. net The polynomial type of interpolation functions are mostly used due to the following reasons. 1. It is easy to formulate and computerize the finite element equations. 2. It is easy to perform differentiation or Intigration. 3. The accuracy of the result can be improved by increasing the order of the polynomial. 18 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Step – 4:Define the material behavior by using strain – Displacemnt and stress. Strain relationship: Strain – displacement and stress – strain relationship and necessary for deriving the equatins for each finite element. In case of the dimensional deformation, the strain – displacement relationship is given by, 𝑑𝑢 𝑒 = 𝑑𝑥 Where, 𝑢 → displacement field variable 𝑥 direction 𝑒 → strain. Step – 5 Deviation of equation is in matrix form as ww 𝑓1 𝑘11 , 𝑘12 , 𝑘13 … . . 𝑘1𝑛 𝑢1 𝑓2 𝑘21 , 𝑘22 , 𝑘23 … . . 𝑘2𝑛 𝑢2 𝑓3 𝑘31 , 𝑘32 , 𝑘33 … . . 𝑘3𝑛 𝑢3 w.E . . . 𝑓4 In compact matrix form as. asy 𝑘𝑛1 , 𝑘42 , Where, 𝑘43 … . . 𝑘4𝑛 . . . En . . . 𝑢𝑛 gin 𝑒 is a element, {𝐹} is the vector of element modal forces, [𝑘] is the element stiffness eer i matrix and the equation can be derived by any one of the following methods. (i) Direct equilibrium method. (ii) Variational method. (iii) Weighted Residual method. ng. Step (6):- net Assemble the element equations to obtain the global or total equations. The individual element equations obtained in step 𝑠 are added together by using a method of super position i.e. direction stiffness method. The final assembled or global equation which is in the form of 𝑓 = 𝑘 {𝑢} 𝐹 → Global Force Vector Where, 𝐾 → Global Stiffness matrix {𝑢} → Global displacement vector. Step (7):Applying boundary conditions: 19 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net The global stiffness matrix [𝑘] is a singular matrix because its determinant is equal to zero. In order to remove the singularity problem certain boundary conditions are applied so that the structure remains in place instead of moving as a rigid body. Step (8):Solution for the unknown displacement formed in step (6) simultaneous algebraic equations matrix form as follows. Deviation of equation is in matrix form as ww 𝑓1 𝑘11 , 𝑘12 , 𝑘13 … . . 𝑘1𝑛 𝑢1 𝑓2 𝑘21 , 𝑘22 , 𝑘23 … . . 𝑘2𝑛 𝑢2 𝑓3 𝑘31 , 𝑘32 , 𝑘33 … . . 𝑘3𝑛 𝑢3 𝑓3 𝑘41 , 𝑘42 , 𝑘43 … . . 𝑘4𝑛 𝑢4 . . . 𝑓4 𝑘𝑛1 , w.E 𝑘42 , asy 𝑘43 … . . 𝑘4𝑛 . . . 𝑢𝑛 . . . These equation can be solved and unknown displacement {𝑢} calculated by using Gauss elimination. Step (9):- En gin Computation of the element strains and stresses from the modal displacements 𝒖 : eer i In structural stress analysis problem. Stress and strain are important factors from the solution of displacement vector {𝑢}, stress and strain value can be calculated. In case of 1D the ng. strain displacement can strain. 𝑒= 𝑑 𝑢 = 𝑢2 − 𝑢1 Where, 𝑢1 and 𝑢2 are displacement at model 1 and 2 net 𝑥1 − 𝑥2 = Actual length of the element from that we can find the strain value, By knowing the strain, stress value can be calculated by using the relation. Stress 𝜎 = 𝐸𝑒 Where, 𝐸 → young’s modulus 𝑒 → strain Step – 10 Interpret the result (Post processing) 20 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Analysis and Evaluation of the solution result is referred to as post-processing. Post processor computer programs help the user to interpret the results by displaying them in graphical form. 6) Explain in detail about Boundary value, Initial Value problems. The objective of most analysis is to determine unknown functions called dependent variables, that are governed by a set of differential equations posed in a given domain. Ω and some conditions on the boundary Γ of the domain. Often, a domin not including its boundary is called an open domain. A domain boundary is called an open domain. A domain Ω with its boundary Γ is called a closed domain. Boundary value problems:- Steady state heat transfer : In a fin and axial deformation of a bar ww shown in fig. Find 𝑢(𝑥) that satisfies the second – order differential equation and boundary conditions. w.E −𝑑 𝑑𝑥 𝑑𝑢 𝑎 𝑑𝑥 + 𝑐𝑢 = 𝑓 for 0 < 𝑥 < 𝐿 asy 𝑑𝑢 𝑢 𝑜 = 𝑢0 , 𝑎 𝑑𝑥 i) 𝑥=𝐿 = 𝑞0 En Bending of elastic beams under Transverse load : find 𝑢 𝑥 that satisfies the fourth order gin differential equation and boundary conditions. 𝑑2 𝑑𝑥 2 𝑑2𝑢 𝑏 𝑑𝑥 2 + 𝑐𝑢 = 𝐹 for 0 < 𝑥 < −𝐿 𝑢 𝑜 = 𝑢0 , 𝑑2𝑢 𝑑 𝑏 𝑑𝑥 2 𝑑𝑥 𝑥=𝐿 = 𝑚0 . 𝑑2𝑢 𝑏 𝑑𝑥 2 0 Ω = (o, L) x=0 eer i 𝑑𝑢 = 𝑑0 𝑑𝑥 𝑥=0 = 𝓋0 x=L ng. net x Initial value problems:i) A general first order equation:Find 𝑢 𝑡 that satisfies the first-order differential equation and initial condition. Equation and initial condition:𝑑𝑢 𝑎 𝑑𝑡 + 𝑐𝑢 = 𝐹 for 0 < 𝑡 ≤ 𝑇 𝑢 0 = 𝑢0 . ii) A general second order equation:Find 𝑢 𝑡 that satisfies the second – order differential equation and initial conditions:21 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝑑2𝑢 𝑑𝑢 𝑎 𝑑𝑡 + 𝑏 𝑑𝑡 2 + 𝑐𝑢 = 𝐹 for 0 < 𝑡 ≤ 𝑇 𝑑𝑢 𝑢 𝑜 = 𝑢0 , 𝑏 𝑑𝑡 𝑡=0 = 𝑣0 Eigen value problems:(i) Axial vibration of a bar: Find 𝑢 𝑥 and 𝑙 that satisfy the differential equation and boundary conditions. −𝑑 𝑑𝑥 𝑑𝑢 𝑎 𝑑𝑥 − 𝜆𝑢 = 0 for 𝑜 < 𝑥 < 𝐿 𝑢 𝑜 = 0, 𝑎 ww (ii) 𝑑𝑢 =0 𝑑𝑥 𝑥=𝐿 Transverse vibration of a membrane:Find 𝑢 (𝑥, 𝑦) and 𝜆 that satisfy the partial differential equation and w.E boundary condition. 𝑑 𝑑𝑢 𝑑 𝑑𝑢 − 𝑑𝑥 𝑎1 𝑑𝑥 + 𝑑𝑦 𝑎2 𝑑𝑦 − 𝜆𝑢 = 0 in Ω asy 𝑢 = 0 on Γq En The values of 𝜆 are called cigen values and the associated functions 𝑢 are called cigen functions. b) gin eer i A simple pendulum consists of a bob of mass 𝒎(𝒌𝒈)attached to one end of a rod of length 𝒍(𝒎) and the other end is pivoted to fixed point 𝟎. ng. Soln:𝑑 𝐹 = 𝑑𝑡 𝑚𝑣 = 𝑚𝑎 𝑑𝑣 𝐹𝑥 = 𝑚. 𝑑𝑡𝑥 −𝑚𝑔 sin 𝜃 = 𝑚𝑙 net 𝑑2 𝑄 𝑑𝑡 2 or 𝑑2 𝑄 𝑔 + sin 𝑄 = 0 𝑑𝑡 2 𝑙 𝑑2 𝑄 𝑠 + 𝑄=0 𝑑𝑡 2 𝑙 𝑑𝑄 + (𝑜) = 𝑈0. 𝑑𝑡 22 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝑄 𝑡 = 𝐴𝑠 𝑖𝑛 𝜆𝑡 + 𝐵 cos 𝜆 𝑡. Where, 𝑠 𝜆= and 𝐴 and 𝐵 are constant to be determined using the initial condition we 𝑙 obtain. 𝜈 𝐴 − 𝜆0 , 𝐵 = 𝜃0 the solution to be linear problem is 𝜈 𝜃 𝑡 = 𝜆0 𝑆𝑖𝑛 ∧ 𝑡 + 0. 𝐶𝑜𝑠 𝜆𝑡 for zero initial velocity and non zero initial position 𝜃0 , we have. ww 𝜃 𝑡 = 𝜃0 cos 𝜆𝑡. 7) w.E A simply supported beam subjected to uniformly distributed load over entire span and it is subject to a point load at the centre of the span. Calculate the bending moment asy and deflection at imdspan by using Rayleish – Ritz method. (Nov/Dec 2008). En Given data:- gin eer i ng. To Find: net 1. Deflection and Bending moment at mid span. 2. Compare with exact solutions. Formula used 𝜋 = 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 − 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 Solution: We know that, πx 3πx Deflection, y = a1 sin l + a2 sin l 1 2 23 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Total potential energy of the beam is given by, π=U−H 2 2 Where, U – Strain Energy. H – Work done by external force. The strain energy, U of the beam due to bending is given by, 2 1 d2y 0 dx 2 EI U= 2 3 dx 2 dy ww dx πx π = a1 cos l × dy a 1 πx = dx πx w.E d2y dx 2 a1 π =− l d2y l2 d2y l πx π sin l × l − asy a1 π2 =− dx 2 a 2 3πx cos l + l + a2 cos l πx l 0 EI U= 2 EI = 2 − a1 l2 l a 2 3π l a π2 πx sin l 3πx sin l × a EI π 4 l 0 l 4 l 2 gin π2 3πx − 9 2l 2 sin l πx 3π 3πx 2 l a1 π2 πx a2 π2 3πx sin + 9 sin l 2 0 l l l2 = 2 l4 l 3πx cos En Substituting dx 2 value in equation (3), 3π × l − 9 2l 2 sin sin l π2 3πx a21 sin2 l + 81a22 sin2 3πx l dx 2 eer i dx ng. πx 3πx + 2 a1 sin l .9 a2 sin l dx [∴ a + b 2 = a2 + b2 + 2ab] EI π 4 U = 2 l4 l 0 πx a21 sin2 l + 81a22 sin2 𝑙 2 πx a sin2 l dx 0 1 1 l1 = a21 0 2 l 1 − cos l 2πx a2 l πx + 18 a1 a2 sin l . sin 2πx = a21 2 0 1 − cos l = 21 3πx 𝑙 dx − 0 dx 3πx net dx l 2 ∴ sin x = 5 1−cos 2x 2 dx 1 2πx cos l dx 0 24 Downloaded From: www.EasyEngineering.net 2 Downloaded From: www.EasyEngineering.net = = = 𝑙 0 a21 sin2 𝑎 12 2 𝑎 12 𝑙 2𝜋𝑥 𝑙 2𝜋 𝑙 sin 𝑥 𝑙0 − 1 0 2𝜋𝑙 𝑙 − 0 − 2𝜋 sin 𝑙 − sin 0 2 𝑎 12 1 𝑙 − 2𝜋 0 − 0 2 𝑎 12 𝑙 = ∴ sin 2𝜋 = 0; sin 0 = 0 2 πx 𝑎12 𝑙 dx = l 2 6 2 Similarly, ww 𝑙 3πx 81 a22 sin2 l dx 0 𝑙1 = 81a22 0 2 1 − cos w.E 1 = = = = 𝑙 0 𝑙 = 81a22 2 0 1 − cos 81a22 sin2 6πx 𝑙 dx − 0 2 81𝑎 22 𝑥 𝑙0 − 2 81𝑎 22 2 81𝑎 22 2 ∴ sin2 x = dx l 1−cos 2x 2 dx l asy 81a 22 6πx 𝑙 6πx cos l dx 0 En 𝑙 6𝜋𝑥 𝑙 6𝜋 𝑙 sin 1 0 gin 6𝜋𝑙 𝑙 − 0 − 6𝜋 sin 𝑙 − sin 0 1 𝑙 − 6𝜋 0 − 0 = 𝑎 12 𝑙 eer i ng. ∴ sin 6𝜋 = 0; sin 0 = 0 2 3πx 81𝑎22 𝑙 dx = l 2 net 7 2 𝑙 πx 3πx 18 a1 a2 sin l . sin l dx 0 𝑙 πx = 18 a1 a2 0 sin l . sin 𝑙 3πx 3πx l dx πx = 18 a1 a2 0 sin l . sin l dx 𝑙1 = 18 a1 a2 0 2 cos 2πx l − cos 4πx l dx ∴ sin 𝐴 sin 𝐵 = = 18 a 1 a 2 2 𝑙 2πx cos l dx − 0 cos 𝐴−𝐵 −cos 𝐴+𝐵 2 𝑙 4πx cos l dx 0 25 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net = 2 𝑙 4𝜋𝑥 𝑙 4𝜋 𝑙 sin − 0 𝑙 0 = 9 a1 a 2 0 − 0 = 0 ∴ sin 2𝜋 = 0; sin 4𝜋 = 0; sin 0 = 0 πx 3πx . sin dx = 0 l l 8 𝑙 0 2𝜋𝑥 𝑙 2𝜋 𝑙 sin 18 a 1 a 2 18 a1 a2 sin 2 Substitute (6), (7) and (8) in equation (5), EI π 4 𝑎 12 𝑙 81𝑎 22 𝑙 + +0 2 2 U = 2 l4 ww EI π 4 𝑙 w.E U = 4 l 4 𝑎12 + 81𝑎22 Strain Energy, U = asy 𝐸𝐼𝜋 4 2 𝑎 + 81𝑎22 4𝑙 3 1 9 En Work done by external forces, gin 𝑙 𝜔 𝑦 𝑑𝑥 + 𝑊 𝑦𝑚𝑎𝑥 0 𝐻= 2 𝑙 2𝜔𝑙 𝑎2 𝜔 𝑦 𝑑𝑥 = 𝑎1 + 𝜋 3 0 eer i ng. 11 𝜋𝑥 3𝜋𝑥 𝑦 = 𝑎1 sin 𝑙 + 𝑎2 sin 𝑙 We know that, 10 net 1 In the span, deflection is maximum at 𝑥 = 2 𝑦𝑚𝑎𝑥 = 𝑎1 sin 𝜋× 1 2 𝑙 𝜋 + 𝑎2 sin 3𝜋× 1 2 𝑙 3𝜋 𝜋 = 𝑎1 sin 2 + 𝑎2 sin 2 3𝜋 ∴ sin 2 = 1; sin 2 = −1 𝑦𝑚𝑎𝑥 = 𝑎1 − 𝑎2 12 Substitute (11) and (12) values in equation (8), 2𝜔𝑙 H= 𝜋 𝑎 𝑎1 + 32 + 𝑊 (𝑎1 − 𝑎2 ) 13 26 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Substituting U and H values in equation (2), we get 𝜋= 𝜋= 𝐸𝐼𝜋 4 2𝜔𝑙 𝑎12 + 81𝑎22 − 4𝑙 3 𝐸𝐼𝜋 4 𝜋 2𝜔𝑙 𝑎12 + 81𝑎22 − 𝜋 4𝑙 3 𝑎 𝑎1 + 32 + 𝑊 (𝑎1 − 𝑎2 ) 𝑎 𝑎1 + 32 − 𝑊 (𝑎1 − 𝑎2 ) 14 For stationary value of 𝜋, the following conditions must be satisfied. 𝜕𝜋 𝜕𝑎 1 𝜕𝜋 𝜕𝑎 1 𝜕𝜋 = 0and𝜕𝑎 = 0 2 𝐸𝐼𝜋 4 2𝜔𝑙 = 4𝑙 3 2𝑎1 − 𝜋 − 𝑊 = 0 ww 𝐸𝐼𝜋 4 2𝑙 3 2𝜔𝑙 𝑎1 − 𝜋 − 𝑊 = 0 𝐸𝐼𝜋 4 2𝜔𝑙 𝑎 = +𝑊 1 2𝑙 3 𝜋 w.E asy 𝑎1 = 2𝑙 3 2𝜔𝑙 +𝑊 𝐸𝐼𝜋 4 𝜋 15 En gin 𝜕𝜋 𝐸𝐼𝜋 4 2𝜔𝑙 1 = 162𝑎2 − +𝑊 =0 3 𝜕𝑎2 4𝑙 𝜋 3 Similarly, 𝐸𝐼𝜋 4 4𝑙 3 eer i ng. 2𝜔𝑙 162𝑎1 − 𝜋 + 𝑊 = 0 𝐸𝐼𝜋 4 2𝑙 3 2𝜔𝑙 162𝑎1 = 𝜋 − 𝑊 𝑎2 = 2𝑙 3 2𝜔𝑙 −𝑊 81𝐸𝐼𝜋 4 3𝜋 net 16 From equation (12), we know that, Maximum deflection, 𝑦𝑚𝑎𝑥 = 𝑎1 − 𝑎2 2𝑙 3 𝑦𝑚𝑎𝑥 = 𝐸𝐼𝜋 4 2𝜔𝑙 2𝑙 3 + 𝑊 − 81𝐸𝐼𝜋 4 𝜋 2𝜔𝑙 3𝜋 −𝑊 27 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 4𝜔 𝑙 4 2𝑊𝑙 3 4𝜔 𝑙 4 2𝑊𝑙 3 𝑦𝑚𝑎𝑥 = 𝐸𝐼𝜋 5 + 𝐸𝐼𝜋 4 − 243𝐸𝐼𝜋 5 + 81𝐸𝐼𝜋 4 𝜔 𝑙4 𝑊𝑙 3 𝑦𝑚𝑎𝑥 = 0.0130 𝐸𝐼 + 0.0207 𝐸𝐼 17 We know that, simply supported beam subjected to uniformly distributed load, maximum deflection 5 𝜔 𝑙4 𝑦𝑚𝑎𝑥 = 384 𝐸𝐼 is, Simply supported beam subjected to point load at centre, maximum deflection is, 𝜔 𝑙3 𝑦𝑚𝑎𝑥 = 48𝐸𝐼 5 𝜔 𝑙4 𝜔 𝑙3 𝑦𝑚𝑎𝑥 = 384 𝐸𝐼 + 48𝐸𝐼 ww So, total deflection, w.E 𝑦𝑚𝑎𝑥 = 0.0130 𝜔𝑙 4 𝑊𝑙 3 + 0.0208 𝐸𝐼 𝐸𝐼 asy 18 From equations (17) and (18), we know that, exact solution and solution obtained by using En Rayleigh-Ritz method are same. Bending Moment at Mid span gin We know that, eer i ng. d2y Bending moment, M = EI dx 2 From equation (9), we know that, d2y dx 2 = − 𝑎1 𝜋2 𝑙2 𝜋𝑥 sin 𝑙 + 𝑎2 9𝜋 2 𝑙2 19 net 3𝜋𝑥 sin 𝑙 Substitute 𝑎1 and 𝑎2 values from equation (15) and (16), d2y dx 2 2𝑙 3 = − 𝐸𝐼𝜋 4 𝜋2 2𝜔𝑙 𝜋𝑥 2𝑙 3 + 𝑊 × 𝑙 2 sin 𝑙 + 81𝐸𝐼𝜋 4 𝜋 2𝜔𝑙 9𝜋 2 3𝜋𝑥 − 𝑊 × 𝑙 2 sin 𝑙 3𝜋 𝑙 Maximum bending occurs at 𝑥 = 2 1 1 𝜋×2 3𝜋 × 2 2𝑙 3 2𝜔𝑙 𝜋2 2𝑙 3 2𝜔𝑙 9𝜋 2 = − + 𝑊 × 2 sin + − 𝑊 × 2 sin 𝐸𝐼𝜋 4 𝜋 𝑙 𝑙 81𝐸𝐼𝜋 4 3𝜋 𝑙 𝑙 28 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 2𝑙 3 2𝜔𝑙 𝜋2 2𝑙 3 2𝜔𝑙 9𝜋 2 = − + 𝑊 × 2 (1) + − 𝑊 × 2 (−1) 𝐸𝐼𝜋 4 𝜋 𝑙 81𝐸𝐼𝜋 4 3𝜋 𝑙 𝜋 3𝜋 ∴ sin 2 = 1; sin 2 = −1 = − 2𝑙 2𝜔𝑙 2𝑙 2𝜔𝑙 + 𝑊 − −𝑊 𝐸𝐼𝜋 2 𝜋 9𝐸𝐼𝜋 2 3𝜋 4𝜔𝑙 2 4𝜔 𝑙 2 2𝑊𝑙 2𝑊𝑙 = − 𝐸𝐼𝜋 3 + 𝐸𝐼𝜋 2 − 27𝐸𝐼𝜋 3 + 9𝐸𝐼𝜋 2 3.8518 𝜔𝑙 2 =− ww 𝐸𝐼𝜋 3 2.222𝑊𝑙 + 𝐸𝐼𝜋 2 d2 y 𝜔𝑙 2 𝑊𝑙 = − 0.124 + 0.225 dx 2 𝐸𝐼 𝐸𝐼 w.E d2y Substitute dx 2 value in bending moment equation, asy d2y 𝜔 𝑙2 𝑊𝑙 Mcentre = EI dx 2 = −𝐸𝐼 0.124 𝐸𝐼 + 0.225 𝐸𝐼 En Mcentre = − 0.124 𝜔𝑙 2 + 0.225 𝑊𝑙 gin 20 eer i (∴Negative sign indicates downward deflection) We know that, simply supported beam subjected to uniformly distributed load, maximum bending moment is, Mcentre = ng. 𝜔 𝑙2 8 net Simply supported beam subjected to point load at centre, maximum bending moment is, Mcentre = 𝑊𝑙 4 Total bending moment, Mcentre = 𝜔 𝑙2 8 𝑊𝑙 + 4 Mcentre = 0.125 𝜔𝑙 2 + 0.25 𝑊𝑙 21 From equation (20) and (21), we know that, exact solution and solution obtained by using Rayleigh-Ritz method are almost same. In order to get accurate results, more terms in Fourier series should be taken. 29 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net ww w.E a UNIT 2 syE ngi nee rin g.n et Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net UNIT – II ONE DIMENSIONAL PROBLEMS PART - A 1. What is truss?(May/June 2014) A truss is an assemblage of bars with pin joints and a frame is an assemblage of beam elements. Truss can able to transmit load and it can deform only along its length. Loads are acting only at the joints. 2. State the assumptions made in the case of truss element. The following assumptions are made in the case of truss element, 1. All the members are pin jointed. 2. The truss is loaded only at the joints 3. The self weight of the members are neglected unless stated. 3. What is natural co-ordinate?(Nov/Dec 2014), (April/May 2011) A natural co-ordinate system is used to define any point inside the element by a set of dimensionless numbers, whose magnitude never exceeds unity, This system is useful inassembling of stiffness matrices. 4. Define shape function. State its characteristics (May/June 2014), (Nov/Dec 2014), (Nov/Dec 2012) In finite element method, field variables within an element are generally expressed by the following approximate relation: u (x,y) = N1(x,y) u1+N2 (x,y) u2+ N3(x,y) u3 Where u,1 u2, u3 are the values of the field variable at the nodes and N1 N2 N3 are interpolation function. N1 N2 N3 is called shape functions because they are used to express the geometry or shape of the element. The characteristics of the shape functions are follows: 1. The shape function has unit value at one nodal point and zero value at the other nodes. 2. The sum of the shape function is equal to one. 5. Why polynomials are generally used as shape function? Polynomials are generally used as shape functions due to the following reasons: 1. Differentiation and integration of polynomials are quite easy. 2. The accuracy of the results can be improved by increasing the order of the Polynomial. 3. It is easy to formulate and computerize the finite element equations. 6. Write the governing equation for 1D Transverse and longitudinal vibration of the bar at one end and give the boundary conditions. (April/May 2015) The governing equation for free vibration of abeam is given by, 𝜕4 𝑣 𝜕2 𝑣 𝐸𝐼 4 + 𝜌𝐴 2 = 0 𝜕𝑥 𝜕𝑡 Where, E – Young’s modulus of the material. I – Moment of inertia Ρ – Density of the material. A – Cross sectional area of the section of beam. ww w.E asy En gin eer ing .ne t The governing equation for 1D longitudinal vibration of the bar at one end is given by d2 U AE + ρAUω2 = 0 dx 2 Where, U – axial deformation of the bar (m) ρ – Density of the material of the bar (kg/m3) 1 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net ω – Natural frequency of vibration of the bar A – Area of cross section of the bar (m2) 7. Express the convections matrix for 1D bar element. (April/May 2015) hPL 6 [ 2 1 ] 1 2 Convection stiffness matrix for 1D bar element: hPTaL 1 1 2 Convection force matrix for 1D bar element: Where, h- Convection heat transfer coefficient (w/m2k) P – Perimeter of the element (m) L – Length of the element (m) Ta – Ambient temperature (k) 8. State the properties of a stiffness matrix.(April/May 2015), (Nov/Dec 2012) The properties of the stiffness matrix [K] are, 1. It is a symmetric matrix 2. The sum of the elements in any column must be equal to zero. 3. It is an unstable element, so the determinant is equal to zero. ww w.E asy En 9. Show the transformation for mapping x-coordinate system into a natural coordinate system for a linear bar element and a quadratic bar element.(Nov/Dec 2012) For example consider mapping of a rectangular parent element into a quadrilateral element gin eer ing .ne t The shape functions of this element are To get this mapping we define the coordinate of point P as, 10. Define dynamic analysis.(May/June 2014) When the inertia effect due to the mass of the components is also considered in addition to the externally applied load, then the analysis is called dynamic analysis. 2 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 11. What are the types of boundary conditions used in one dimensional heat transfer problems? (i) Imposed temperature (ii) Imposed heat flux (iii) Convection through an end node. 12. What are the difference between boundary value problem and initial value problem? (i) The solution of differential equation obtained for physical problems which satisfies some specified conditions known as boundary conditions. (ii) If the solution of differential equation is obtained together with initial conditions then it is known as initial value problem. (iii) If the solution of differential equation is obtained together with boundary conditions then it is known as boundary value problem. PART -B ww w.E 1. For the beam and loading shown in fig. calculate the nodal displacements. Take [E] =210 GPa =210×109 𝑵 𝒎𝟐 , [I] = 6×10-6 m4 NOV / DEC 2013 12 𝐾𝑁 𝑚 6 KN asy En 1m 2m Given data gin eer ing .ne t Young’s modulus [E] =210 GPa =210×109 𝑁 𝑚2 Moment of inertia [I] = 6×10-6 m4 Length [L]1 = 1m Length [L]2 = 1m W=12 𝑘𝑁 𝑚 =12×103 𝑁 𝑚 F = 6KN To find  Deflection Formula used −𝑙 2 −𝑙 2 f(x) 12 −𝑙 2 𝑙2 𝐹1 𝑀 + 1 = 𝐹2 𝑀2 𝐸𝐼 𝑙3 12 6𝑙 6𝑙 4𝑙 2 – 12 6𝑙 – 6𝑙 2𝑙 2 – 12 6𝑙 – 6𝑙 2𝑙 2 12 – 6𝑙 – 6𝑙 12 𝑢1 𝜃1 𝑢2 𝜃2 4𝑙 2 M1,θ1 6 KN M1,θ1 Solution 1 For element 1 𝑣1, F1 3 Downloaded From: www.EasyEngineering.net 2 𝑣2 ,F2 Downloaded From: www.EasyEngineering.net −𝑙 2 −𝑙 2 f(x) 12 −𝑙 2 𝑙2 𝐹1 𝑀1 + = 𝐹2 𝑀2 𝐸𝐼 𝑙3 12 6𝑙 6𝑙 4𝑙 2 – 12 6𝑙 – 6𝑙 2𝑙 2 – 12 6𝑙 – 6𝑙 2𝑙 2 12 – 6𝑙 – 6𝑙 4𝑙 2 𝑢1 𝜃1 𝑢2 𝜃2 12 Applying boundary conditions F1=0N ; F2=-6KN=-6×103 N; M1=M2=0; u1=0; θ1=0; u2≠0; 0 103× 0 = −6 0 – 12 6 –6 2 210×10 9 ×6×10 −6 3 1 – 12 – 6 12 – 6 6 2 –6 4 ww w.E 12 6 −12 6 4 −6 =1.26×106 −12 −6 12 6 2 −6 For element 2 −𝑙 2 −𝑙 2 f(x) 12 −𝑙 2 𝑙2 f(x)=0 θ2≠0 𝐹2 𝑀2 + = 𝐹3 𝑀3 12 6 6 4 6 2 −6 4 0 0 𝑢2 0 asy En 𝐸𝐼 𝑙3 12 6𝑙 6𝑙 4𝑙 2 – 12 6𝑙 – 6𝑙 2𝑙 2 M2,θ2 𝑢2 𝜃2 𝑢3 𝜃3 gin Applying boundary conditions f(x) = -12 𝑘𝑁 𝑚 =12×103 𝑁 𝑚; 12 𝐾𝑁 𝑚 M3,θ3 2 – 12 6𝑙 – 6𝑙 2𝑙 2 12 – 6𝑙 – 6𝑙 4𝑙 2 12 𝑢1 𝜃1 𝑢2 𝜃2 F2=F3=0=M2=M; u2≠0; θ2≠0; u3=θ3=0 0 12 −6 6 103 × −1 + 0 = 1.26×106× −6 0 −12 1 0 6 6 − 12 6 4 −6 2 6 12 − 6 4 −6 4 12 −6 6 103 × −1 = 1.26×106× −6 −12 1 6 6 − 12 6 4 −6 2 6 12 − 6 4 −6 4 𝑣2, F2 3 𝑣3 ,F3 eer ing .ne t 𝑢2 𝜃2 0 0 𝑢2 𝜃2 0 0 4 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Assembling global matrix 12 0 6 0 −12 3 6 −12 10 × = 1.26×10 × 6 −1 −6 0 1 0 6 4 −6 2 0 0 Solving matrix -12×103=1.26×106×24u2=0; -1×103=1.26×106×8θ2=0; Result θ2=-9.92rad u2=-3.96×10-4m ww w.E 2. −12 −6 24 0 −12 6 −6 2 0 8 −6 2 0 0 −12 −6 12 −6 0 0 6 2 −6 4 0 0 𝑢2 𝜃2 0 0 u2=-3.96×10-4m θ2=-9.92rad Determine the axial vibration of a steel bar shown in fig. Take [E] =2.1×105 𝑵 𝒎𝒎𝟐 , [ρ] = 7800 𝒌𝒈 𝒎𝟑 NOV/DEC 2014 1200mm2 900mm2 300mm 400mm asy En Given data A1=1200mm2; A2=900mm2 l1 =300mm; l2=400mm Young’s modulus [E] =2.1×105 𝑁 𝑚𝑚2 Density [ρ] = 7800 𝐾𝑔 𝑚3 =7.8×10-6 𝐾𝑔 𝑚𝑚3 To find  Stiffness matrix  Mass matrix  Natural frequency  Mode shape Formula used General equation for free vibration of bar 𝑘 − 𝑚𝜆 {u}= 0 1 –1 𝐴𝐸 Stiffness matrix [k] = 𝑙 –1 1 𝜌𝐴𝐿 2 1 Consistent mass matrix [m] = 6 1 2 𝜌𝐴𝐿 1 0 Lumped mass matrix [m] = 2 0 1 Mode shape 𝑘 − 𝑚𝜆 U1 = 0 ; Normalization 𝑈1𝑇 M U1 = 1 Solution For element 1 gin u1 eer ing .ne t 1200mm2 u2 300mm 5 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝐴𝐸 Stiffness matrix [k] = [k1] = =8.4×105 1 –1 –1 1 1 –1 ; –1 1 = 1 –1 =105 𝐴1 𝐸1 𝑙1 𝑙 1200 ×2.1×10 5 –1 ; 1 Consistent mass matrix [m] = 𝜌𝐴1 𝐿1 [m1] = 6 = ww w.E 2 1 300 6 1200 ×300×7.8×10 −6 6 8.4 – 8.4 – 8.4 8.4 2 1 ; 1 2 𝜌𝐴𝐿 1 2 1 −1 −1 1 2 1 1 2 2 1 1 2 0.936 0.468 0.468 0.936 = 0.468× [m1] For element 2 Stiffness matrix [k] = = asy En u2 900 mm2 1 –1 –1 1 𝐴𝐸 𝑙 400mm 1 –1 ; 𝑙2 –1 1 900×2.1×10 5 1 −1 = 400 −1 1 1 −1 5 = 4.73×10 −1 1 4.73 – 4.73 [k2] = 105 ; – 4.73 4.73 [k2] = 𝐴2 𝐸2 Consistent mass matrix [m] = [m2] = 𝜌𝐴2 𝐿2 = 6 2 1 𝜌𝐴𝐿 6 1 2 900×400×7.8×10 −6 6 2 1 1 2 0.936 0.468 0.468 0.936 u3 gin 2 1 ; 1 2 2 1 eer ing .ne t 1 2 = 0.468 [m2] = Assembling global matrix 8.4 Stiffness matrix [k] = 105 −8.4 0 0.936 Consistent mass matrix [m] = 0.468 0 −8.4 13.13 −4.73 0.468 1.87 0.468 0 −4.73 4.73 0 0.468 0.936 6 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net General equation for free vibration of bar 𝑘 − 𝑚𝜆 {u} = 0 8.4 −8.4 0 0.936 0.468 0 105 −8.4 13.13 −4.73 - λ 0.468 1.87 0.468 =0 0 −4.73 4.73 0 0.468 0.936 13.13 −4.73 105 −4.73 1.87 –λ 4.73 0.468 13.13 × 105 − 1.87𝜆 −4.73 × 105 − 0.468𝜆 0.468 =0 0.936 −4.73 × 105 − 0.468𝜆 = 0 4.73 × 105 − 0.936𝜆 [(13.13×105 -1.87λ)( 4.73 × 105 − 0.936𝜆) – (−4.73 × 105 − 0.468𝜆)( −4.73 × 105 − 0.468𝜆)] =0 6.2×1011 – 1.23× 106 λ – 8.84×10 5 λ + 1.75×λ2 -2.24×1011 -2.21×105 λ -2.21×105 λ – 0.22 λ2 =0 ww w.E 1.53λ2 -2.55×105 λ+3.96×1011 =0 Solving above equation 𝜆1 = 1.49×106 𝜆2 = 1.73×105 = 0.173×106 To find mode shape asy En 𝑘 − 𝑚𝜆 {𝑢} = 0 ; 𝜆1 = 0.173×106 105 13.13 −4.73 gin −4.73 1.87 – 0.173×106 4.73 0.468 0.99 × 106 −0.55 × 106 −0.55 × 106 0.31 × 106 𝑢2 𝑢3 = 0 0.99×106 u2 – 0.55× 106 u3 =0 - 0.55×106 u2 + 0.31×106 u3 =0 u3 = 1.77u2 𝑘 − 𝑚𝜆 {𝑢} = 0 𝜆2 = 1.49×106 105 13.13 −4.73 −4.73 1.87 – 1.49×106 4.73 0.468 −1.48 × 106 −1.17 × 106 −1.17 × 106 −0.924 × 106 0.468 0.936 eer ing .ne t 0.468 0.936 𝑢2 𝑢3 = 0 𝑢2 𝑢3 = 0 𝑢2 𝑢3 = 0 -1.482×106 u2 – 1.17× 106 u3 =0 - 1.17×106 u2 -0.924×106 u3 =0 𝑢3 =-1.26u2 Normalization 𝑈1𝑇 M U1 = 1 Normalization of 𝜆1 7 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝑢2 𝑢2 1.77𝑢2 =1 1.87 0.468 0.46 0.936 1.77𝑢2 𝑢2 𝑢2 1.77𝑢2 = 1 1.77𝑢2 2.7𝑢22 + 3.79𝑢22 =1 𝑢22 = 6.4 ; 1 𝑢2 = 0.392 𝑢3 =1.78𝑢2 ; 𝑢3 = 0.698 Normalization of 𝜆2 𝑈2𝑇 M U2 = 1 ww w.E 𝑢2 −1.26𝑢2 1.87 0.468 0.46 0.936 1.28𝑢2 −0.707𝑢2 𝑢2 −1.26𝑢2 =1 𝑢2 −1.256𝑢2 = 1 asy En 1.28𝑢22 + 0.88𝑢22 =1 𝑢22 = 0.46; 𝑢3 =-1.268𝑢2 𝑢3 = -0.84 Result Mode shape gin 2 1 3 u2=0.392 Mode 1 eer ing .ne t u1=0 u3=0.698 u2=0.678 u1=0 Mode 2 u3=-0.698 8 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 3. Consider the simply supported beam shown in fig. let the length L=1m, E=2×1011𝑵 𝒎𝟐 , area of cross section A=30cm2, moment of inertia I=100mm4, density[ρ] = 7800𝒌𝒈 𝒎𝟑 . Determine the natural frequency using two types of mass matrix. Lumped mass matrix and consistent mass matrix. APRIL / MAY 2011 L Given data Length = 1m Young’s modulus E=2×1011 𝑁 𝑚2 ww w.E Area A=30cm2 = 3×10-3 m2 Moment of inertia I=100mm4 = 100×10-12 m4 Density[ρ] = 7800 kg/m3=76518 𝑁 𝑚3 To find asy En  Lumped mass matrix  Consistent mass matrix  Natural frequency Formula used gin eer ing .ne t General equation for free vibration of beam 𝑘 − 𝜔2 𝑚 {u} = 0 – 12 6𝑙 – 6𝑙 2𝑙 2 𝐸𝐼 Stiffness matrix[k] = 𝑙 3 – 12 – 6𝑙 12 – 6𝑙 6𝑙 2𝑙 2 – 6𝑙 4𝑙 2 156 22𝑙 22𝑙 4𝑙 2 𝜌𝐴𝐿 Consistent mass matrix [m] = 420 54 13𝑙 −13𝑙 −3𝑙 2 1 0 0 0 𝜌𝐴𝑙 0 0 0 0 Lumped mass matrix [m] = 2 0 0 1 0 0 0 0 0 12 6𝑙 6𝑙 4𝑙 2 54 −13𝑙 13𝑙 −3𝑙 2 156 – 22𝑙 −22𝑙 4𝑙 2 Solution For element 1 9 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 6𝑙1 4𝑙12 −6𝑙1 2𝑙12 12 6𝑙1 𝐸1 𝐼 Stiffness matrix[k]1 = 3 𝑙 1 −12 6𝑙1 −12 −6𝑙1 12 −6𝑙1 6𝑙1 2𝑙12 −6𝑙1 4𝑙12 θ1 θ2 1 2 0.5 m 𝑣1 12 6 × 0.5 −12 6 × 0.5 2×10 11 ×100×−12 = 0.53 12 3 [k]1 =160× −12 3 ww w.E −12 −3 12 −3 3 1 −3 0.5 Lumped mass matrix [m]1 = 6 × 0.5 4 × 0.52 −6 × 0.5 2 × 0.52 𝜌𝐴 𝑙 1 2 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 57.38 0 0 0 0 0 76518 ×3×10 −3 ×0.5 [m]1 2 57.38 = 0 0 0 0 0 0 0 156 22𝑙1 𝜌𝐴 𝑙 Consistent mass matrix [m]1 = 4201 54 −13𝑙1 = 76518 ×3×10 −3 ×0.5 420 156 22 × 0.5 54 −13 × 0.5 42.63 [m]1 = 3 14.74 −1.77 6 × 0.5 2 × 0.52 −6 × 0.5 4 × 0.52 3 0.5 −3 1 asy En = −12 −6 × 0.5 12 −6 × 0.5 𝑣2 3 0.27 1.77 −0.20 0 0 1 0 gin 22𝑙1 4𝑙12 13𝑙1 −3𝑙12 22 × 0.5 4 × 0.52 13 × 0.5 −3 × 0.52 0 0 0 0 eer ing .ne t 54 13𝑙1 156 −22𝑙1 −13𝑙1 −3𝑙12 −22𝑙1 4𝑙12 54 13 × 0.5 156 −22 × 0.5 −13 × 0.5 −3 × 0.52 −22 × 0.5 4 × 0.52 14.74 −1.77 1.77 −0.20 42.63 −3 −3 0.27 10 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net For element 2 6𝑙2 4𝑙22 −6𝑙2 2𝑙22 12 6𝑙2 𝐸𝐼 Stiffness matrix[k]2 = 𝑙 3 −12 2 6𝑙2 6𝑙2 2𝑙22 −6𝑙2 4𝑙22 −12 −6𝑙2 12 −6𝑙2 θ2 θ3 2 3 0.5 m 𝑣2 12 6 × 0.5 −12 6 × 0.5 2×10 11 ×100×−12 = 0.53 12 3 [k]2 = 160× −12 3 ww w.E Lumped mass matrix [m]2 = 𝜌𝐴 𝑙 2 2 3 1 −3 0.5 −12 −3 12 −3 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 asy En = 76518 ×3×10 −3 ×0.5 2 57.38 = 0 0 0 [m]2 0 0 0 0 = 420 42.63 [m]2 = 3 14.74 −1.77 156 22 × 0.5 54 −13 × 0.5 3 0.27 1.77 −0.20 −12 −6 × 0.5 12 −6 × 0.5 6 × 0.5 2 × 0.52 −6 × 0.5 4 × 0.52 3 0.5 −3 1 0 0 1 0 0 0 0 0 57.38 0 0 0 156 22𝑙2 𝜌𝐴 𝑙 Consistent mass matrix [m]2 = 4202 54 −13𝑙2 76518 ×3×10 −3 ×0.5 6 × 0.5 4 × 0.52 −6 × 0.5 2 × 0.52 𝑣3 0 0 0 0 gin 22𝑙2 4𝑙22 13𝑙2 −3𝑙22 22 × 0.5 4 × 0.52 13 × 0.5 −3 × 0.52 eer ing .ne t 54 13𝑙2 156 −22𝑙2 −13𝑙2 −3𝑙22 −22𝑙2 4𝑙22 54 13 × 0.5 156 −22 × 0.5 −13 × 0.5 −3 × 0.52 −22 × 0.5 4 × 0.52 14.74 −1.77 1.77 −0.20 42.63 −3 −3 0.27 Global matrix 11 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 12 3 −12 Stiffness matrix [k] =160× 3 0 0 3 1 −3 0.5 0 0 −12 −3 24 0 −12 3 57.38 0 0 Lumped mass matrix [m]= 0 0 0 0 0 0 0 0 0 0 0 114.77 0 0 0 42.63 3 14.74 Consistent mass matrix[m]= −1.77 0 0 Frequency for lumped mass matrix 𝑘 − 𝜔2 𝑚 {u} = 0 3 0.27 1.77 −0.2 0 0 ww w.E asy En 3 0.5 0 2 −3 0.5 0 0 −12 −3 12 −3 0 0 3 0.5 −3 1 0 0 0 0 0 0 0 0 0 0 57.38 0 0 0 0 0 0 0 14.74 1.77 85.26 0 14.74 −1.77 −1.77 −0.2 0 0.5 1.77 −0.2 0 0 14.74 1.77 42.63 −3 0 0 −1.77 −0.2 −3 0.27 0 0 −12 −3 12 −3 0 57.38 0 0 3 0 2 −𝜔 0.5 0 −3 0 1 0 0 0 0 0 0 0 0 0 114.77 0 0 0 0 0 0 0 0 0 0 0 0 0 57.38 0 0 0 0 0 0 0 𝑣1 𝜃1 𝑣2 𝜃2 =0 𝑣3 𝜃3 Applying boundary conditions 𝑣1 =0=𝜃1 ; 𝑣2 ≠0; 𝜃2 ≠0 12 3 −12 3 0 3 1 −3 0.5 0 −12 −3 24 0 −12 160 × 3 0.5 0 2 −3 0 0 −12 −3 12 0 0 3 0.5 −3 0 57.38 0 0 3 0 2 −𝜔 0.5 0 −3 0 1 0 𝑣3 =0=𝜃3 ; 0 0 0 0 0 114.77 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 57.38 0 0 0 0 0 0 0 0 0 𝑣2 𝜃2 =0 0 0 12 3 −12 160 × 3 0 0 160 × 24 0 3 1 −3 0.5 0 0 −12 −3 24 0 −12 3 3 0.5 0 2 −3 0.5 0 114.7 − 𝜔2 2 0 3840 − 𝜔2 × 114.7 0−0 0 0 gin 𝑣2 𝜃2 = 0 eer ing .ne t 0−0 =0 320 − 0 {(3840 − 𝜔2 × 114.7) × ( 320 − 0)-0-0} =0 1228800-36704𝜔2 = 0 𝜔2 = 33.47 𝜔 = 5.78 𝑟𝑎𝑑 𝑠 Frequency for consistent mass matrix 12 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝑘 − 𝜔2 𝑚 {u} = 0 12 3 −12 160 × 3 0 0 3 1 −3 0.5 0 0 −12 −3 24 0 −12 3 3 0.5 0 2 −3 0.5 0 0 −12 −3 12 −3 0 42.63 0 3 3 14.74 − 𝜔2 0.5 −1.77 0 −3 0 1 Applying boundary conditions 𝑣1 =0=𝜃1 ; 𝑣2 ≠0; 𝜃2 ≠0 12 3 −12 160 × 3 0 0 0 42.63 0 3 3 14.74 2 −𝜔 0.5 −1.77 0 −3 0 1 3 1 −3 0.5 0 0 −12 −3 24 0 −12 3 ww w.E 3 0.5 0 2 −3 0.5 160 × 24 0 0 2 − 𝜔2 3840 − 85.26ω2 0−0 0 0 −12 −3 12 −3 85.26 0 0 0.5 𝑣2 𝜃2 14.74 1.77 85.26 0 14.74 −1.77 3 0.27 1.77 −0.2 0 0 14.74 1.77 85.26 0 14.74 −1.77 gin 2×42.63 29203 .3 ±25359 .28 85.26 85.26 𝑣1 𝜃1 𝑣2 𝜃2 =0 𝑣3 𝜃3 −1.77 −0.2 0 0.5 1.77 −0.2 0 0 14.74 1.77 42.63 −3 0 0 −1.77 −0.2 −3 0.27 0 0 𝑣2 𝜃2 =0 0 0 eer ing .ne t ax2 +bx+c=0; x = 29203 .3 ± 29203 .32 −4×42.63×1.23×10 6 29203 .3+25359 .28 0 0 −1.77 −0.2 −3 0.27 =0 asy En 42.63 λ2 -29203.3 λ+1.23×106 =0 𝜆1 = 0 0 14.74 1.77 42.63 −3 0−0 =0 320 − 0.5ω2 Take λ = 𝜔2 = −1.77 −0.2 0 0.5 1.77 −0.2 𝑣3 =0=𝜃3 ; (3840 − 85.26𝜔2 ) 320 − 0.5𝜔2 = 0 1.23×106-1920𝜔2 -27283.2𝜔2 +42.63𝜔4 =0 𝜆= 3 0.27 1.77 −0.2 0 0 −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 29203 .3−25359 .28 𝜆2 = ; 𝜆1 =639.95; 85.26 𝜆2 =45.08 λ = 𝜔2 𝜔1 = λ1 ; 𝜔 2 = λ2 𝜔1 = 639.95 𝜔2 = 45.08 𝜔1 = 25.3 𝑟𝑎𝑑 𝑠 𝜔2 = 6.7 𝑟𝑎𝑑 𝑠 13 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 4. For a tapered plate of uniform thickness t = 10mm as shown in fig. find the displacements at the nodes by forming in to two element model. The bar has mass density ρ = 7800𝑲𝒈 𝒎𝟑 Young’s modulus E = 2×105𝑴𝑵 𝒎𝟐 . In addition to self weight the plate is subjected to a point load p = 10KN at its centre. Also determine the reaction force at the support. Nov/Dec 2006 80mm 150m m P 300m m ww w.E 40m Given data Mass density ρ = 7800𝑘𝑔 𝑚3 m = 7800 × 9.81=76518 𝑁 𝑚3 = 7.65 × 10-5 𝑁 𝑚𝑚3 Young’s modulus E = 2×105𝑀𝑁 𝑚2 ; = 2×105 × 106 𝑁 𝑚2 = 2×105 𝑁 𝑚𝑚2 asy En Point load P = 10 KN To find gin  Displacement at each node  Reaction force at the support Formula used {F} =[K] {u} 𝐴𝐸 Stiffness matrix [k] = 𝑙 𝜌𝐴𝑙 Force vector 𝐹 = 2 1 – 1 𝑢1 𝐹1 𝐴𝐸 = 𝑙 𝐹2 –1 1 𝑢2 1 1 1 – 1 𝑢1 –1 1 𝑢2 eer ing .ne t {R} =[K] {u} -{F} Solution The given taper bar is considered as stepped bar as shown in fig. W1=80mm W1=80mm P 150mm 150m m 2 300m m 1 10KN 150mm 3 W3=40 mm 14 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net W1 = 80mm W2 = 𝑊1 +𝑊3 2 = 80+40 = 60 mm 2 W3 = 40mm Area at node 1 A1 = Width × thickness =W1 × t1 = 80 × 10 = 800mm2 Area at node 2; A2 = Width × thickness =W2 × t2 = 60 × 10 =600mm2 ww w.E Area at node 1 A1 = Width × thickness = W3 × t3 = 40 × 10 =400mm2 Average area of element 1 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑛𝑜𝑑𝑒 1 +𝐴𝑟𝑒𝑎 𝑜𝑓 𝑛𝑜𝑑𝑒 2 Ā1 = 𝐴1 + 𝐴2 asy En 2 = = 2 800+600 2 = 700mm2 Average area of element 2 Ā2 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑛𝑜𝑑𝑒 2 +𝐴𝑟𝑒𝑎 𝑜𝑓 𝑛𝑜𝑑𝑒 3 2 For element 1 700 ×2×10 5 = 𝐴2 + 𝐴3 150 1 −1 = 2 2 = 500mm2 eer ing .ne t u1,F1 150mm 1 1 7.65×10 −5 ×700×150 2 −4.67 𝑢1 4.67 𝑢2 4.67 = 2× 10 −4.67 𝜌 Ā1 𝑙 1 = −1 𝑢1 1 𝑢2 5 Force vector 𝐹 1 = 600+400 gin 2 – 1 𝑢1 1 𝑢2 1 –1 Ā 𝐸 Stiffness matrix [k]1 = 1𝑙 1 1 = 1 1 = u2,F2 10KN 4.017 4.017 u2,F2 For element 2 Ā 𝐸 Stiffness matrix [k]2 = 2𝑙 2 2 = – 1 𝑢2 1 𝑢3 1 –1 500 ×2×10 5 150 = 2× 105 1 −1 3.33 −3.33 10KN −1 𝑢2 1 𝑢3 150mm u3,F3 −3.33 𝑢2 3.33 𝑢3 15 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Force vector 𝐹 2 = = 𝜌 Ā2 𝑙 2 2 1 1 7.65×10 −5 ×500×150 2 = 2.869 2.869 −4.66 7.99 −3.33 0 −3.33 3.33 1 1 Global matrix 4.66 Stiffness matrix [k] = 2×105 × −4.66 0 4.017 Force vector 𝐹 = 6.88 2.87 Finite element equation ww w.E {F} =[K] {u} 𝐹1 4.66 𝐹2 = 2×105 × −4.66 𝐹3 0 −4.66 7.99 −3.33 0 −3.33 3.33 asy En 𝑢1 𝑢2 𝑢3 Applying boundary conditions 𝑢1 = 0; 𝑢2 ≠ 0; 𝑢3 ≠ 0; 𝐹2 = 10 × 103 N 𝐹1 4.66 𝐹2 = 2×105 × −4.66 𝐹3 0 −4.66 7.99 −3.33 0 −3.33 3.33 gin 4.017 4.66 5 3 = 2×10 × 6.88 + 10 × 10 −4.66 2.87 0 10006.88 7.99 = 2×105 2.86 −3.33 𝑢1 𝑢2 𝑢3 −4.66 7.99 −3.33 −3.33 𝑢2 3.33 𝑢3 2×105 (7.99𝑢2 − 3.33𝑢3 ) = 10006.88 2×105 (-3.33𝑢2 + 3.33𝑢3 ) = 2.86 eer ing .ne t 0 −3.33 3.33 0 𝑢2 𝑢3 Solving above equation 2×105 (4.66 𝑢2 ) = 10009.74 𝑢2 = 0.01074 mm 2×105 (-3.33×0.01074+3.33𝑢3 ) = 2.86 666000𝑢3 = 2.86 + 7152.88 𝑢3 = 0.01074 16 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Reaction force {R} =[K] {u} -{F} 𝑅1 4.66 𝑅2 = 2×105 × −4.66 𝑅3 0 −4.66 7.99 −3.33 𝑢1 𝐹1 0 −3.33 𝑢2 - 𝐹2 𝐹3 3.33 𝑢3 𝑅1 4.66 5 𝑅2 = 2×10 × −4.66 𝑅3 0 −4.66 7.99 −3.33 0 −3.33 3.33 𝑢1 4.017 0.01074 - 10006.88 0.01074 2.87 0 − 0.05 + 0 4.017 =2×105 0 + 0.086 − 0.036 - 10006.88 0 − 0.036 + 0.036 2.87 ww w.E = 2×105 4.017 −0.05 0.05 - 10006.88 2.87 0 −10000 4.017 = 10000 - 10006.88 0 2.87 asy En −10004.017 = −6.88 −2.86 Result 𝑅1 −10004.017 𝑅2 = −6.88 𝑅3 −2.86 5. gin eer ing .ne t A wall of 0.6m thickness having thermal conductivity of 1.2 W/mk. The wall is to be insulated with a material of thickness 0.06m having an average thermal conductivity of 0.3 W/mk. The inner surface temperature in 1000OC and outside of the insulation is exposed to atmospheric air at 30oc with heat transfer coefficient of 35 W/m2k. Calculate the nodal temperature. NOV/DEC 2014 Given Data:Thickness of the wall, l1 = 0.6m Conduction Convection Conduction Thermal conductivity of the wall K1= 1.2W/mk h Thickness of the insulation l2 = 0.06m T1 T3 T2 Thermal Conductivity of the wall K2 = 0.3W/mk Inner surface temperature T1= 1000oC+273 Wall Insulation 𝑙1 𝑙2 = 1273 K 17 Downloaded From: www.EasyEngineering.net 𝑇∞ Downloaded From: www.EasyEngineering.net Atmospheric air temperature T2 = 30 +273 = 303 K Heat transfer co-efficient at outer side h = 35W/m2k To find Nodal temperature T2 and T3 Formula used 1D Heat conduction 𝐴𝑘 1 𝐹1 = 𝐹2 𝑙 –1 – 1 𝑇1 1 𝑇2 1D Heat conduction with free end convection ww w.E 𝐴𝑘 [K]= 𝑙 1 –1 0 + hA 0 –1 1 0 1 Solution For element 1 asy En k1 A1 1 −1 T1 f1 = f2 l1 −1 1 T2 For unit area: A1 = 1m2 1.2 1 −1 T1 = 0.6 −1 1 T2 f1 2 −2 T1 = f2 −2 2 T2 Conduction T1 gin eer ing .ne t For element (2) A2 K 2 1 −1 0 0 T2 0 + hA = h T2 A T 0 1 1 l2 −1 1 3 T 1 X 0.3 1 0 0 −1 0 2 + 35 × 1 =35×303×1× 0.06 −1 T3 0 1 1 1 T1 0 0 5 −5 0 + = 0 35 T2 −5 5 10.605 × 103 5 −5 T1 0 = T −5 5 10.605 × 103 2 T2 L1 Conduction T2 Convection T3 L2 Assembling finite element equation f1 2 −2 0 T1 f2 = −2 7 −5 T2 f3 0 −5 40 T3 Applying boundary conditions f1 = 0 18 Downloaded From: www.EasyEngineering.net h T∞ Downloaded From: www.EasyEngineering.net f2 = 0 f3 = 10.605 x 103 2 −2 0 T1 0 −2 7 −5 T2 = 0 T 10.605 × 103 0 −5 40 3 Step (1) The first row and first column of the stiffness matrix K have been set equal to 0 except for the main diagonal. 1 0 0 T1 0 T 0 7 −5 2 = 0 10.605 × 103 0 −5 40 T3 Step – II The first row of the force matrix is replaced by the known temperature at node 1 ww w.E 1 0 0 T1 1273 0 7 −5 T2 = 0 10.605 × 103 0 −5 40 T3 asy En Step – III The second row first column of stiffness K value is multiplied by known temperature at node 1 -2 × 1273 = -2546. This value positive digit 2546 has been added to the second row of the force matrix. 1 0 0 T1 1273 0 7 −5 T2 = 2546 T 10.605 × 103 0 −5 40 3 ⟹ 7 T2 − 5 T3 = 2546 −5 T2 + 40 T3 = 10.605 × 103 Solving above Eqn ×8 56 T2 − 40T3 = 20.368 × 103 5 T2 − 40T3 = 10.605 × 103 gin 51 T2 = 30973 eer ing .ne t T2 = 607.31 K 7 × 607.31 -5 T3 = 2546 4251.19 - 5 T3 = 2546 -−5 T3 = −1705 T3 = 341.03 K Result Nodal Temp T1 = 1273 K T2 = 607.31K T3 = 341.03 K 19 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 7. Derivation of the displacement function u and shape function N for one dimensional linear bar element. OR Derive the shape function, stiffness matrix and load vector for one dimensional bar element. May / June 2013 Consider a bar with element with nodes 1 and 2 as shown in Fig. 𝜐1 and 𝜐2 are the displacement at the respective nodes. 𝜐1 And 𝜐2 is degree of freedom of this bar element. 𝓍 1 2 𝑢1 ww w.E 𝑢2 𝑙 Fig Two node bar element asy En Since the element has got two degrees of freedom, it will have two generalized co-ordinates. 𝑢 = 𝑎0 + 𝑎1 𝑥 Where, 𝑎0 and 𝑎1 are global or generalized co – ordinates. Writing the equation in matrix form, 𝑎0 𝑢 = 1𝑥 𝑎 1 At node 1, 𝑢 = 𝑢1 , 𝑥 = 0 At node 1, 𝑢 = 𝑢2 , 𝑥 = 1 Substitute the above values ion equation, 𝑢1 = 𝑎0 𝑢2 = 𝑎0 + 𝑎1 𝑙 Arranging the equation in matrix form, 𝑢1 1 0 𝑢2 = 1 𝑙 gin 𝑎0 𝑎1 eer ing .ne t 𝑢∗ 𝐶 𝐴 ∗ Where, 𝑢 ⟶ Degree of freedom. 𝐶 ⟶ Connectivity matrix. 𝐴 ⟶ Generalized or global co-ordinates matrix. 𝑎0 1 0 −1 𝑢1 = 𝑎1 𝑢2 1 𝑙 = 𝑙−0 1 −0 −1 1 𝑎12 −1 𝑎22 = 1 𝑎11 𝑁𝑜𝑡𝑒: 𝑎 21 𝑢1 𝑢2 1 𝑎22 × −𝑎 21 𝑎11 𝑎22 − 𝑎12 𝑎21 −𝑎12 𝑎11 20 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝑎0 1 𝑙 0 𝑢1 𝑎1 = 𝑙 −1 1 𝑢2 𝑎0 Substitute 𝑎 𝑣𝑎𝑙𝑢𝑒𝑠 𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 1 𝑙 0 𝑢1 𝑢 = 1 𝑥 𝑙 −1 1 𝑢2 1 𝑙 0 𝑢1 = 𝑙 1 𝑥 −1 1 𝑢2 𝑢1 1 = 𝑙 1−𝑥 0+𝑥 𝑢 2 ∵ 𝑀𝑎𝑡𝑟𝑖𝑥 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 1 × 2 2 × 2 = 1 × 2 𝑢1 1− 𝑥 𝑥 𝑢 = 𝑙 𝑙 𝑢2 𝑢1 𝑢 = 𝑁1 𝑁2 𝑢 2 Displacement function, 𝑢 = 𝑁1 𝑢1 + 𝑁2 𝑢2 ww w.E 𝑙− 𝑥 Where, Shape function, 𝑁1 = 𝑙 𝑥 ; 𝑠𝑕𝑎𝑝𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 , 𝑁2 = 𝑙 Stiffness matrix for one dimensional linear bar element Consider a bar with element with nodes 1 and 2 as shown in Fig. 𝜐1 and 𝜐2 are the displacement at the respective nodes. 𝜐1 And 𝜐2 is degree of freedom of this bar element. asy En 𝓍 1 𝑢1 gin 𝑙 Stiffness matrix, 𝐾 = 𝑣 B T 𝐷 𝐵 𝑑𝑣 Displacement function, 𝑢 = 𝑁1 𝑢1 + 𝑁2 𝑢2 Shape function, 𝑁1 = 𝑙− 𝑥 𝑙 2 𝑢2 eer ing .ne t 𝑥 ; 𝑠𝑕𝑎𝑝𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 , 𝑁2 = 𝑙 Strain displacement matrix,[B] = 𝑑𝑁1 𝑑𝑁2 𝑑𝑥 −1 1 = 𝑙 𝑑𝑥 𝑙 −1 𝑙 1 [B]T= 𝑙 One dimensional problem [D] = [E] = young’s modulus −1 [K] = 𝒍 𝟎 𝑙 1 ×𝐸× −1 1 𝑙 𝑙 𝑑𝑣 𝑙 21 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 1 𝑙 𝑙2 = 0 −1 𝑙2 −1 1 𝑙 𝑙2 0 −1 𝑙2 −1 = ww w.E 𝑙2 1 × 𝐸 × 𝑑𝑣 𝑙2 𝑙2 1 × 𝐸 × A × dx 𝑙2 1 −1 𝑙2 = AE −1 𝑙2 1 𝑙2 𝑙2 1 −1 𝑙2 = AE −1 𝑙2 1 𝑙2 𝑙2 × 1 −1 𝑙2 1 𝑙2 𝑙2 𝐴𝐸𝑙 1 −1 −1 1 𝐴𝐸 1 –1 –1 1 asy En = 𝑙2 [K] = 𝑙 𝑙 𝑑𝑥 0 1 −1 𝑙2 = AE −1 𝑙2 1 𝑙2 𝑙2 𝑥 𝑙0 (𝑙 − 0) 𝑙2 −1 = AE 𝑙 [dv = A×dx gin Finite element equation for finite element analysis {F} =[K] {u} eer ing .ne t 𝐴𝐸 1 – 1 𝑢1 𝐹1 = 𝐹2 1 𝑢2 𝑙 –1 Load vector [F] Consider a vertically hanging bar of length𝑙, uniform cross section A, density ρ and young’s modulus E. this bar is subjected to self weight Xb The element nodal force vector 𝐹 𝑒= 𝑁 𝑇 Xb Self weight due to loading force Xb = ρAdx x Displacement function, 𝑢 = 𝑁1 𝑢1 + 𝑁2 𝑢2 Where; 𝑁1 = 𝑙− 𝑥 𝑙 𝑥 ; 𝑁2 = 𝑙 ; [N] = 𝑙− 𝑥 𝑥 𝑙 𝑙 xb 22 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝑙− 𝑥 [N]T = 𝑙 𝑥 𝑙 Substitute Xb and [N]T values 𝐹 𝑒= 𝑙 0 𝑙− 𝑥 𝑙 𝑥 𝑙 = ρA ρA dx 𝑥− 𝑥2 2𝑙 = ρA 𝑙 2 𝑙 2 𝜌𝐴𝑙 2 1 1 ww w.E Force vector {F} = 7. 𝑥2 2𝑙 𝑙− 𝑥 𝑙 𝑥 𝑙 𝑙 = ρA 0 𝑙 = ρA 𝑙− 𝑙2 2𝑙 0 dx 𝑙2 2𝑙 = ρA 𝑙− 𝑙 2 𝑙 2 DERIVATION OF SHAPE FUNCTION AN STIFFNESS MATRIX FOR ONEDIMENSIONAL QUADRATIC BAR ELEMENT: May / June 2012 asy En Consider a quadratic bar element with nodes 1,2 and 3 as shown in Fig.(i), 𝑢1 , 𝑢2 𝑎𝑛𝑑 𝑢3 are the displacement at the respective nodes. So, 𝑢1 , 𝑢2 𝑎𝑛𝑑 𝑢3 are considered as degree of freedom of this quadratic bar element. 𝓍 𝜐1 1 3 𝑙 gin eer ing .ne t 2 𝜐2 𝜐3 2 𝑙 Fig. (i). Quadratic bar element Since the element has got three nodal displacements, it will have three generalized coordinates. u = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 Where, 𝑎0 , 𝑎1 𝑎𝑛𝑑 𝑎2 are global or generalized coordinates. Writing the equation is matrix form, 23 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝑈 = 1𝑥 𝑥 𝑎0 𝑎1 𝑎2 2 At node 1, u = u1 , 𝑥 = 0 At node 2, u = u2 , 𝑥 = 1 1 At node 3, u = u3 , 𝑥 = 2 Substitute the above values in equation. u1 = 𝑎0 u2 = 𝑎0 + 𝑎1 𝑙 + 𝑎2 𝑙 2 ww w.E 𝑙 u3 = 𝑎0 + 𝑎1 2 + 𝑎2 𝑙 2 2 Substitute the equation we get u2 = 𝑢1 + 𝑎1 𝑙 + 𝑎2 𝑙 2 asy En 𝑎 𝑙2 𝑎 𝑙 u3 = 𝑢1 + 21 + 24 u2 − u1 = 𝑎1 𝑙 + 𝑎2 𝑙 2 𝑎 𝑙2 𝑎 𝑙 u3 − 𝑢1 = 21 + 24 Arranging the equation in matrix form, gin 𝑙 u2 − u1 = 𝑙 u3 − 𝑢1 𝑙2 2 4 𝑙 𝑙2 ⇒ a1 a2 = a1 a2 𝑙2 𝑙 𝑙2 2 4 −1 u2 − u1 u3 − 𝑢1 𝑙2 = 𝑙3 1 𝑙3 − 4 2 𝑎11 𝑁𝑜𝑡𝑒 ∵ 𝑎 21 ⇒ eer ing .ne t 4 −𝑙 −𝑙 2 2 1 𝑎12 −1 𝑎22 = X 𝑎22 −𝑎21 𝑎11 𝑎22 − 𝑎12 𝑎21 a1 a2 = −4 ⇒ 𝑎1 = 𝑙 3 𝑙2 1 −𝑙 3 4 𝑙2 4 4 −𝑙 2 −𝑙 2 𝑙 𝑙 u2 − u1 u3 − 𝑢1 −𝑎12 𝑎11 u2 − u1 u3 − 𝑢1 u2 − u1 −𝑙 2 u3 − 𝑢1 24 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net −4 −𝑙 ⇒ 𝑎2 = 𝑙 3 Equation u2 − u1 + 𝑙 u3 − 𝑢1 2 𝑙2 𝑢2 −4 𝑎1 = 𝑙 3 𝑙2 𝑢1 − 4 4 −4𝑙 2 𝑢 2 = + 4𝑙 3 − 𝑢2 = 𝑙 𝑙 4𝑙 2 𝑢 1 𝑢 − + 4𝑙 3 4 𝑢3 + 𝑙1 + −3 𝑢 1 𝑎1 = −𝑙 2 𝑢3 + 𝑙 2 𝑢1 𝑢2 − 𝑙3 4𝑙 2 𝑢 1 𝑙3 4 𝑢1 − 𝑙 𝑙 4 𝑢3 + 𝑙 4𝑙 2 𝑢 3 𝑙 Equation −4 −𝑙𝑢 2 𝑙 2 𝑎2 = 3 ww w.E 4𝑙 𝑢 2 = = 𝑙 − 𝑢1 + 𝑙𝑢3 − 𝑙𝑢1 2 4𝑙 4𝑙 4𝑙 2 𝑙3 + 2 𝑙 3 𝑢1 − 𝑙 3 𝑢3 + 𝑙 3 𝑢1 2𝑢 2 2 𝑙2 4 4 − 𝑙 2 𝑢1 − 𝑙 2 𝑢3 + 𝑙 2 𝑢1 asy En 2 2𝑢 4 𝑎2 = 𝑙 2 𝑢1 + 𝑙 2 2 − 𝑙 2 𝑢3 Arranging the equation in matrix form, 1 𝑎0 −3 𝑎1 = 𝑙 2 𝑎2 0 −1 𝑙 2 𝑙2 𝑙2 0 gin 𝑙 −4 𝑙2 Substitution the equation 1 𝑢 = 1 𝑥 𝑢 = 𝑢 = 𝑁1 𝑥 1− 3 𝑙 𝑁2 2 𝑢1 𝑢2 𝑢3 4 −3 −1 0 0 𝑙 2 𝑙 2 𝑙 −4 𝑙2 𝑙2 𝑙2 𝑥+ 4 2 𝑥2 −𝑥 𝑙2 𝑙 + eer ing .ne t 𝑢1 𝑢2 𝑢3 2 𝑥2 4𝑥 𝑙2 𝑙 − 4 𝑥2 𝑙2 𝑢1 𝑢2 𝑢3 𝑢1 𝑢2 𝑢3 𝑁3 𝑢 = 𝑁1 𝑢1 + 𝑁2 𝑢2 + 𝑁3 𝑢3 Where, shape function, 𝑁1 = 1 − 𝑁2 = −𝑥 𝑁3 = 4𝑥 𝑙 3𝑥 + 𝑙 + 2𝑥 2 𝑙2 2𝑥 2 𝑙2 4𝑥 2 − 𝑙2 𝑙 25 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net STIFFNESS MATRIX FOR ONE-DIMENSIONAL QUADRATIC BAR ELEMENT: 𝜐1 1 1 𝑙 𝜐1 2 2 2 3 𝜐2 2 𝑙 Fig. A bar element with three nodes Consider a one dimensional quadratic bar element with nodes 1,2, and 3 as shown in Fig. 2. Let 𝑢1 , 𝑢2 𝑎𝑛𝑑 𝑢3 be the nodal displacement parameters or otherwise known as degree of freedom. ww w.E We know that, asy En Stiffness matrix, 𝑘 = 𝐵 𝑇 𝐷 𝐵 𝑑𝑣 𝑣 In one dimensional quadratic bar element, gin Displacement function, 𝑢 = 𝑁1 𝑢1 + 𝑁2 𝑢2 + 𝑁3 𝑢3 Where, 𝑁1 = 1 − 𝑁3 = + 𝑙2 𝑙 2𝑥 2 −𝑥 𝑁2 = 2𝑥 2 3𝑥 + 𝑙2 𝑙 4𝑥 2 4𝑥 − 𝑙2 𝑙 We know that, 𝑑 𝑁1 𝑑 𝑁2 𝑑 𝑁3 Strain – Displacement matrix, 𝐵 = ⟹ 𝑑 𝑁1 ⟹ 𝑑 𝑁2 ⟹ 𝑑 𝑁3 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 = −3 4𝑥 = −1 𝑙 + 2 4 8𝑥 eer ing .ne t 𝑑𝑥 + 𝑙2 𝑙 4𝑥 𝑙 𝑑𝑥 = 𝑙 + 𝑙2 −3 4𝑥 Substitute the equation 𝐵 = 𝑙 + 𝑙2 −1 𝑙 4𝑥 + 𝑙2 4 𝑙 8𝑥 − 𝑙2 26 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net −3 𝑙 −1 𝐵𝑇= 𝑙 4 4𝑥 + 𝑙2 4𝑥 + 𝑙2 8𝑥 + 𝑙2 𝑙 In one dimensional problems, 𝐷 = 𝐸 = 𝐸 = 𝑌𝑜𝑢𝑛𝑔′ 𝑠𝑀𝑜𝑑𝑢𝑙𝑢𝑠 Substitute 𝐵 𝐵 𝑇 𝑎𝑛𝑑 𝐷 values in stiffness matrix equation 𝐿𝑖𝑚𝑖𝑡 𝑖𝑠 0 𝑡𝑜 𝑙 . −3 4𝑥 + 2 𝑙 𝑙 −1 4𝑥 + 2 𝑙 𝑙 4 8𝑥 − 𝑙 𝑙2 𝑙 ⟹= ww w.E 0 −3 ⟹ 𝑘 = 𝐸𝐴 −3 4𝑥 + 2 𝑙 𝑙 −1 4𝑥 + 2 𝑙 𝑙 4𝑥 −3 4𝑥 −3 4𝑥 𝑙 −1 4𝑥 𝑙 −1 4𝑥 4 8𝑥 −1 + 𝑙2 𝑙 0 + 𝑙2 𝑙 −3 + 𝑙2 𝑙 + 𝑙2 𝑙 ⟹ 𝑘 = 𝐸𝐴 𝑙 0 − 𝑙2 𝑙 −1 4𝑥 𝑙 −1 4𝑥 4 + 𝑙2 𝑙 gin 12𝑥 16𝑥 2 3 12𝑥 4𝑥 16𝑥 2 𝑙 𝑙 12𝑥 𝑙 4𝑥 𝑙4 16𝑥 2 𝑙 𝑙 4𝑥 𝑙 4𝑥 𝑙4 16𝑥 2 𝑙2 −12 𝑙2 − 3 + 3 − 𝑙3 − 𝑙3 + 24𝑥 16𝑥 + 𝑙3 + 𝑙3 − 1 𝑙4 32𝑥 2 𝑙2 −4 𝑙4 𝑙2 9𝑥 12𝑥 2 12𝑥 2 16𝑥 3 − − + 𝑙2 2 𝑙3 2 𝑙3 3 𝑙4 2 2 3𝑥 12𝑥 4𝑥 16𝑥 3 − − + 𝑙2 2 𝑙3 2 𝑙3 3 𝑙4 −12 24𝑥 2 16𝑥 2 32𝑥 2 + + − 𝑙2 2 𝑙3 2 𝑙3 3 𝑙4 ⟹ 𝑘 = 𝐸𝐴 − 2 − 3+ 3 − 𝑙3 − 𝑙3 + 8𝑥 16𝑥 + 𝑙3 + 𝑙3 − 𝑙4 32𝑥 2 𝑙4 −3 4𝑥 𝑙 −1 8𝑥 4 4𝑥 4 4𝑥 𝑙 4 8𝑥 4 + 𝑙2 + 𝑙2 𝑙 − 𝑙2 𝑙 − 𝑙2 𝑙 8𝑥 − 𝑙2 8𝑥 − 𝑙2 𝑙 𝑑𝑥 8𝑥 − 𝑙2 𝑙 eer ing .ne t −12 24𝑥 16𝑥 32𝑥 2 𝑙 −4 𝑙 8𝑥 𝑙 16𝑥 𝑙4 32𝑥 2 32𝑥 𝑙4 64𝑥 2 3𝑥 12𝑥 2 4𝑥 2 16𝑥 3 − − 3+ 𝑙2 2 𝑙3 2𝑙 3 𝑙4 2 2 𝑥 4𝑥 4𝑥 16𝑥 2 − − + 𝑙2 2 𝑙3 2 𝑙3 3 𝑙4 −4 8𝑥 2 16𝑥 2 32𝑥 2 + 3+ − 𝑙2 2𝑙 2 𝑙3 3 𝑙4 9 6 6 16 − − + 𝑙 𝑙 𝑙 3𝑙 3 6 2 16 − − + 𝑙 𝑙 𝑙 3𝑙 −12 12 8 32 + + − 𝑙 𝑙 𝑙 3𝑙 4𝑥 + 𝑙2 + 𝑙2 𝑙 + 𝑙2 𝑙 12𝑥 − 2 × E 𝑑𝑣 4𝑥 + 𝑙2 9 3 = 𝐸𝐴 + 𝑙2 asy En 𝑙 −3 4 8𝑥 − 𝑙 𝑙2 + 2 𝑙2 16 𝑙2 + 3 − 3 + 𝑙3 + 𝑙3 − 32𝑥 − 𝑙3 − 𝑙3 + 𝑑𝑥 𝑙4 −12 24𝑥 2 16𝑥 2 32𝑥 3 + + − 𝑙2 2 𝑙3 2 𝑙3 3 𝑙4 2 2 −4 8𝑥 16𝑥 32𝑥 2 + + − 𝑙2 2 𝑙3 2 𝑙3 3 𝑙4 16 32𝑥 2 32𝑥 64𝑥 2 − − 3+ 𝑙2 2 𝑙3 2𝑙 3 𝑙4 3 6 2 16 − − + 𝑙 𝑙 𝑙 3𝑙 1 2 4 16 − − + 𝑙 𝑙 𝑙 𝑙 −4 4 8 32 + + − 𝑙2 𝑙 𝑙 3𝑙 −12 12 8 32 + + − 𝑙 𝑙 𝑙 3𝑙 −4 4 8 32 + + − 𝑙 𝑙 𝑙 3𝑙 16 16 16 64 − − + 𝑙 𝑙 𝑙 3𝑙 27 Downloaded From: www.EasyEngineering.net 𝑑𝑥 Downloaded From: www.EasyEngineering.net 7 1 −8 3𝑙 3𝑙 3𝑙 1 7 −8 ⟹ 𝑘 = 𝐸𝐴 3𝑙 3𝑙 3𝑙 −8 −8 16 3𝑙 3𝑙 3𝑙 1 −8 𝐸𝐴 7 ⟹ 𝑘 = 1 7 −8 3𝑙 −8 −8 16 LOAD VECTOR FOR ONE DIMENSIONAL QUADRATIC BAR ELEMENT: `We know that, general force vector is, ww w.E 𝐹 = 𝑙 𝑁𝑇 0 Xb 1− 𝑁1 Where, 𝑁 𝑇 = 𝑁2 = 𝑁3 Substitute the equation, asy En 3𝑥 2𝑥 2 −𝑥 𝑙2 2𝑥 2 + 𝑙 𝑙 4𝑥 𝑙 + 𝑙2 4𝑥 2 − 𝑙2 Due to self weight, Xb = ρ A 𝑑𝑥 1− 𝐹 = 𝑙 0 gin 3𝑥 2𝑥 2 −𝑥 𝑙2 2𝑥 2 + 𝑙 𝑙 4𝑥 𝑙 − 𝑙2 eer ing .ne t ρ A 𝑑𝑥 2𝑥 3 −𝑥 2 3 𝑙2 2𝑥 3 + 2𝑙 2𝑙 4𝑥 2 2𝑙 1− = ρA 𝑙− =ρA 𝑙2 4𝑥 2 3𝑥 2 𝑥− 𝐹 =ρA + + − 1 3 𝑙2 4𝑥 3 3 𝑙2 0 3𝑙 2 + 2𝑙 −𝑙 2 + 2𝑙 4 𝑙2 − 2𝑙 2𝑙 3 3 𝑙2 2 𝑙3 3 𝑙2 4𝑙 3 3 𝑙2 3𝑙 2𝑙 2 −𝑙 2 4𝑙 2 + + − 3 2𝑙 3 4𝑙 3 28 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net =ρA 0.166 𝑙 0.166 𝑙 0.166 𝑙 0.166 = ρ A 𝑙 0.166 0.166 ww w.E 𝐹 =ρA𝑙 1 6 1 6 2 3 𝐹1 𝐹2 = ρ A 𝑙 𝐹3 1 6 1 6 2 3 asy En gin eer ing .ne t 29 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net ww w.E a UNIT 3 syE ngi nee rin g.n et Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net UNIT-III TWO DIMENSIONAL SCALAR VARIABLE PROBLEMS PART- A 1. Differentiate CST and LST elements. (Nov/Dec 2014) Three nodded triangular element is known as constant strain triangular element. It has 6unknown degrees of freedom called u1, v1, u2, v2, u3, v3. The element is called CST because it has constant strain throughout it. Six nodded triangular element is known as Linear Strain Triangular element. It has 12unknown displacement degrees of freedom. The displacement function for the element are quadratic instead of linear as in the CST. 2. What do you mean by the terms: C0, C1 and Cn continuity? C0 – Governing differential equation is quasiharmonic, ø has to be continuous. C1 – Governing differential equation is biharmonic, øas well as derivative has to be continuous inside ww w.E a and between the elements. Cn – Governing differential equations is polynomial. 3. How do we specify two dimensional elements? (May/June 2014) syE Two dimensional elements are defined by three or more nodes in two dimensional plane (i.e x and y plane). The basic element useful for two dimensional analysis is a triangular element. 4. What is QST element?(May/June 2014) ngi Ten noded triangular elements are known as Quadratic strain element (QST). nee rin g.n et 5. Write the governing differential equation for two dimensional heat transfer. The governing differential equation for two dimensional heat transfer is given by, 6. Write the governing differential equation for shaft with non-circular cross-section subjected to torsion. The governing differential equation is given by, 1 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 1 𝑑2 ∅ 1 𝑑2 ∅ + + 2𝜃 = 0 𝐺 𝑑𝑥 2 𝐺 𝑑𝑦 2 Where, Ø – Field variable  - Angle of twist per unit length (rad/m) G – Modulus of rigidity or shear modulus (N/m2) 7. What is geometric isotropy?(May/June 2013) An additional consideration in the selection of polynomial shape function for the displacement model is that the pattern should be independent of the orientation of the local coordinate system. This property is known as Geometric Isotropy, Spatial Isotropy or Geometric Invariance. ww w.E a 8.Write the strain displacement matrix of CST element.(Nov/Dec 2012),(April/May 2011) 1 [B]= 2𝐴 𝑞1 0 𝑟1 0 𝑟1 𝑞1 𝑞2 0 𝑞3 0 𝑞 = 𝑦 − 𝑦 𝑞 = 𝑦 − 𝑦 0 𝑟2 0 𝑟3 𝑟1 = 𝑥 2 − 𝑥 3 𝑟2 = 𝑥 3 − 𝑥 1 3 2 2 1 3 𝑟2 𝑞2 𝑟3 𝑞3 1 𝑝1 = 𝑥2 𝑦3 − 𝑥3 𝑦2 𝑝2 = 𝑥3 𝑦1 − 𝑥1 𝑦3 𝑞3 = 𝑦1 − 𝑦2 𝑟3 = 𝑥2 − 𝑥1 𝑝3 = 𝑥1 𝑦2 − 𝑥2 𝑦1 syE 9. Why higher order elements are preferred? Higher order elements are preferred to, (i) Represent the curved boundaries (ii) Reduce the number of elements when compared with straight edge elements to model geometry. ngi nee 10. Evaluate the following area integrals for the three noded triangular element 𝛼 ! 𝛽! 𝛾! 𝑋 2𝐴 𝛼+ 𝛽+ 𝛾+2 𝑁𝑖 𝑁𝑗2 𝑁𝑘3 𝑑𝐴. (May/June 2013), (Nov/Dec 2012) We know that, 1! 2! 3! 𝛽 𝛾 𝐿𝛼𝑖 𝐿2 𝐿𝑘 𝑑𝐴 = 𝑋 2𝐴 (1+ 2+ 3+2)! rin Here, α = 1, β = 2, γ = 3 𝑁𝑖 𝑁𝑗2 𝑁𝑘3 𝑑𝐴 = 1𝑋2𝑋1𝑋3𝑋2𝑋1 𝑋 2𝐴 (8𝑋7𝑋6𝑋5𝑋4𝑋3𝑋2𝑋1) = 1! 2! 3! 𝑋 2𝐴 (8)! g.n et 𝐴 =1680 𝑁𝑖 𝑁𝑗2 𝑁𝑘3 𝑑𝐴 11. Write the strain displacement relation for CST element. 𝑒𝑋 1 𝑞1 𝑒𝑌 = 0 2𝐴 𝑟 𝛾𝑥𝑦 1 0 𝑟1 𝑞1 𝑞2 0 𝑟2 0 𝑟2 𝑞2 𝑞3 0 𝑟3 0 𝑟3 𝑞3 𝑢1 𝑣1 𝑢2 𝑣2 𝑢3 𝑣3 2 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 12. List out the two theories for calculating the shear stress in a solid non circular shaft subjected to torsion. The two theories which helps in evaluating the shear stresses in a solid non circular shaft is proposed by, (i) St. Venant called as St.Venant theory (ii) Prandtl called as Prandtl’s theory. 13. Write down the shape functions associated with three noded linear triangular element (April/May 2015) 1 𝑁1 = 2𝐴 𝑝1 + 𝑞1 𝑥 + 𝑟1 𝑦 1 1 ; 𝑁2 = 2𝐴 𝑝2 + 𝑞2 𝑥 + 𝑟2 𝑦 ; 𝑁3 = 2𝐴 𝑝3 + 𝑞3 𝑥 + 𝑟3 𝑦 ; PART - B 1. For a four Noded rectangular element shown in fig. determine the temperature at the point (7, 4). The nodal values of temperature are T1=420C, T2=540C, T3= 560C, & T4= 460C. Also determine 3 points on the 500C contour line. ww w.E a Given: ϕi= 420C ϕj= 540C ϕk=560C ϕm=460C m (5,5) 460C k(8,5) 560C syE 2b=3 2a=2 b=3/2 a=1 ngi i (5,3) 460C To find: 1. Temperature at point (2,1),ϕ 2. Three points on 500C. Formula used: nee j(8,3) 540C rin g.n et s  t   s  t   Ni= 1  1    1  1    2b  2a   3  2  t   s  t  s  Nj=  1     1    2b  2a   3  2   st  Nk=    4ab     st    =  st   4  3 1   6    2   s   t  s   t  Nm=  1     1    2a  2b   2  3  3 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Solution: The point (7,4) in global coordinate (x,y) is changed in the local coordinate (s,t) S= x-xi  7-5=2 t= y-yi  4-3=1 the temperature at point (2,1) in local coordinate as ϕ = Niϕi + Njϕj + Nkϕk + Nmϕm.  2  1  1 Ni= 1  1   =  3  2  6  2  1  1 Nj=  1   =  3  2  3 ww w.E a  2 1  1 Nk=  =  6  3  1  2  1 Nm =  1   =  2  3  6 syE ngi 1 1 1 1 ϕ =  42   54   56   46 . 6 3 3 6 0 ϕ = 51.4 C The x,y coordinates of 500C contour line are m (5,5) 460C 𝜙 𝑗 −𝜙 𝑖 nee = 𝑥 𝑗 −𝑥 𝑥 𝑗 −𝑥 𝑖 = rin 𝑦 𝑗 −𝑦 𝑦 𝑗 −𝑦 𝑖 k(8,5) 560C g.n et j(8,3) 540C i 460C (5,3) i,j 𝜙 𝑗 −𝜙 500C 54  50 8  x 3  y   54  42 8  5 3  3 (1) (2) (3) Equating(1),(2) equating (1),(3) 4 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 4 8 x  12 3 4 3 y  12 0 x  7cm m,k 𝜙 𝑘 −𝜙 𝜙 𝑘 −𝜙 𝑚 y  3cm = 𝑥 𝑘 −𝑥 𝑥 𝑘 −𝑥 𝑚 = 𝑦 𝑘 −𝑦 𝑦 𝑘 −𝑦𝑚 56  50 8  x 5  y   56  46 8  5 5  5 (1) (2) (3) Equating (1),(2) equating (1),(3) 6 8 x  10 3 ; 6 5 y  10 0 x  6.2cm ; y  5cm ww w.E a Third point y=4 syE [lower point yi=3, upper point ym=5] Centre line between the sides i,j&k,m Local coordinates t = y-yi= 4-3 = 1 ngi nee ϕ = Niϕi + Njϕj + Nkϕk + Nmϕm 50= s 1  s  1  1  1  42  1  54 3 2  3  2  rin 1 s  s 1     56  1  46 2  3  6  g.n et  s  s 1  21  93  9.33s  231    3  3 50= 21  73  9s  9.33s  23  7.66s s  1.63cm (6.2,5) s  x  xj 1.63  5  x (6.7,4) x  6.7cm y  4cm 500C (7,3) 5 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 2. For the plane stress element shown in Fig, the nodal displacements are: [Anna University, May 2002] U1=2.0mm; v1=1.0mm; U2=0.5mm; v2=0.0mm; ww w.E a U3=3.0mm; v3=1.0mm. Determine the element stresses σx, σy, σ1, and σ2 and the principal angle θp, let E=210 GPA, ν= 0.25 and t=10 mm. All coordinates are in millimetre. Given: Nodal Displacements: syE U1=2.0mm; v1=1.0mm; ngi U2=0.5mm; v2=0.0mm; nee U3=3.0mm; X1= 20mm y1=30mm X2= 80mm y2=30mm X3=50mm y3=120mm v3=1.0mm. rin g.n et Young’s modulus, E= 210 GPa =210x109 Pa = 210x109N/m2 = 210x103 N/mm2 =2.1x 105 N/mm2 Poisson’s ratio, Thickness, ν=0.25 t= 10mm 6 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net To find: 1. Element stress a) Normal stress, σx b) Normal stress, σy c) Shear stress,  xy d) Maximum normal stress, σ1 e) Minimum normal stress, σ2 2. Principle angle,θp Formula used:  Stress {σ} = [D] [B] {u}  Maximum normal stress, σmax = σ1 = ww w.E a  Minimum normal stress, σmin = σ2 =  principle angle, tan 2θp= Solution: we know that 2  x  y 2  x  y    2 xy   2   2  x  y    2 xy   2   2 2 xy  x  y syE 1 x1 1 Area of the element, A= 1 x 2 2 1 x3 =  x  y ngi y1 1 20 30  1  y 2  1 80 30  2 1 50 120 y3 nee rin 1 x[ 1x(80x120-50x30)-20(120-30)+30(50-80)] 2 1 = x [8100-1800-900] 2 A=2700 mm2 ….. (1) g.n et We know that, Strain Displacement matrix,  q1 1  0 [B]= 2A   r1 Where, 0 q 2 0 q3 0  r1 0 r 2 0 r 3  q1 r 2 q 2 r 3 q3 …… (2) q1 = y2 – y3 = 30-120 = -90 q2= y3 – y1 = 120- 30 = 90 q3= y1- y2 = 30 – 30 = 0 7 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net ww w.E asy E ngi nee rin g.n et **Note: Other Websites/Blogs Owners Please do not Copy (or) Republish this Materials, Students & Graduates if You Find the Same Materials with EasyEngineering.net Watermarks or Logo, Kindly report us to easyengineeringnet@gmail.com Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net r1= x3- x2 = 50-80 = -30 r2= x1- x3 = 20-50 = -30 r3= x2- x1 = 80-20 = 60 Substitute the above values in equation no. (2), 0 90 0  90 1  0  30 0  30  [B] = 2A   30  90  30 90 0  60  0  0 0 60 Substitute Area, A value, 0 90 0  90 1  0  30 0  30 [B] = 2  2700   30  90  30 90 60 3 0  3 0 30  0 1 0 1 = 2  2700    1  3  1 3 0 2  0  ww w.E a 0 0 syE 2 0  60  0  0 0 3 0  3 0 -3   [B] = 5.555 x 10  0  1 0  1   1  3  1 3 0 2  0  ngi We know that 0 0 2 nee Stress strain relationship matrix [D] for plane stress problem is, [D]=   1 v 0   E v 1 0  1 v2  1 v  0 0  2   ………(3) rin g.n et   1 0.25 0   2.1 10 0.25 1 0  = 1  (0.25) 2  1  0.25  0  0  2   5 0.25 0   1 2.1 10 5  = 0.25 1 0   0.9375  0 0 0.375 8 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 4 1 0  2.1x10 5  0.25  = 1 4 0   0.9375 0 0 1.5 4 1 0  = 56  10 1 4 0    0 0 1.5 …. (4) 3 3 0  3 0 4 1 0  -3    [D] [B] = 56  10 1 4 0 x 5.555 x 10  0  1 0  1     1  3  1 3 0 0 1.5 3 3 0  4 1 0   3 0    = 311.08 x 1 4 0  0  1 0  1   0 0 1.5   1  3  1 3 ww w.E a  12  0  0  = 311.08  3  0  0   0  0  1.5 0 0 2 0 2  0  0 0 2 0 1  0 040 12  0  0 3 0 0 0 1 0 040 000 000 0  0  4.5 0  0  1.5 0  0  4.5 003 1 12   12  4 3 =311.08 x  3    1.5  4.5  1.5 syE 1 4 0 0 4.5 3 ngi 2  8  0  We know that Stress { σ} = [D] [B] {u} nee  u1  v   1 u 2  = [D] [B]   v2  u 3     v3  1 12   12  4 3 = 311.08   3   1.5  4.5  1.5 1 4 0 0 4.5 3 0 2  0  rin 020  0  8  0  0  0  0  g.n et 2 1  2    0.5 8  X   0 0    3    1  9 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net  (12  2)  (1 1)  (12  0.5)  (1 0)  (0  3)  (2  1)    = 311.08  (3  2)  (4  1)  (3  0.5)  (4  0)  (0  3)  (8  1)   (1.5  2)  (4.5  1)  (1.5  0.5)  (4.5  0)  (3  3)  (0  1)     17    {σ} =311.08  0.5  0.75     x   5288.36      y  =   155.54     233.31   z    Normal stress, σx =  5288.36 N/mm2 Normal stress, σy =  155.54 N/mm2 ww w.E a Shear stress,  xy = 233.31 N/mm2 We know that,  x  y syE Maximum normal stress, σmax = σ1 = 2  x  y    2 xy ….. (7)   2   2 ngi =  5288.36  155.54    5288.36  155.54   (233.31) 2 2 2   σ1 = -144.956 N/mm2 Minimum normal stress, σmin = σ2 = 2  x  y 2 nee 2  5288.36  155.54   5288.36  155.54  2     (233.31) 2 2   2 = rin g.n  x  y    2 xy …… (8)   2   et σ2 = -5298.9N/mm2 We know that principle angle, tan 2θp= 2 xy  x  y  2 xy   tan 2θp = tan-1    x   y  2  233.31   = tan-1    5288.36  155.54  2θp=-5.1940 10 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net  θp = -2.590 Result: 1. Element stress a) Normal stress, σx=  5288.36 N/mm2 b) Normal stress, σy=  155.54 N/mm2 c) Shear stress,  xy = 233.31 N/mm2 d) Maximum normal stress, σ1= -144.956 N/mm2 e) Minimum normal stress, σ2= -5298.9N/mm2 2. Principle angle,θp= -2.590 3. Calculate the element stiffness matrix and the temperature force vector for the plane stress element as shown in figure. The element experiences a 20°C increase in temperature, Assume coefficient of thermal expansion is 6 x 10-6/°C. Take Young’s modulus E = 2 X 105N/mm2,possion ratio v=0.25,Thickness t= 5mm. ww w.E a Given data: X1 = 0; Y1 = 0 X2 = 2; Y2 = 0 X3 = 1; Y3 = 3 syE ngi nee E = 2 X 105N/mm2 V = 0.25 t= 5mm ΔT = 10 C ° rin g.n et α = 6 x 10-6/°C To find: 1. Element stiffness matrix [K] 2. The temperature force vector [F] Formula used:  Stiffness matrix [K] = [B] T [D] A t  Temperature force vector, {F} = [B] T [D] {eo} A t Solution: We know that, stiffness matrix [K] = [B] T [D] A t Where A = Area of the element 11 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 1 X 1 Y1 1 1 X 2 Y2 = 2 1 X 3 Y3 1 = 2 = 1 0 0 1 2 0 1 1 3 1 [1(6-0)-0+0]; A= 3 mm2. 2  q1 1  0 Strain –Displacement matrix [B] = 2A   r1 0 r1 q2 0 0 r2 q3 0 q1 r2 q2 r3 Where, q1 = y2 – y3 = 0-3 = -3; r1 = x3 – x2 = 1-2 = -1 q2 = y3 – y1 = 3-0 = 3; r2 = x1 – x3 = 0-1 = -1 q3 = y1 – y2 = 0-0 = 0; r3 = x2 – x1 = 2-0 = 2 ww w.E a 0 r3  q3  Substitute the above values in [B] matrix equation 1 [B] = 2A 3 0  3 0  0 1 0 1    1  3  1 3 2 syE 1 Substitute “A” value, [B] = 23 [B] = 0.1667 0 2  0  0 0 3 0  3 0  0 1 0 1    1  3  1 3 3 0  3 0  0 1 0 1    1  3  1 3 ngi 0 2  0  0 0 2 0 0 2 nee 0 2  0  rin We know that, stress-strain relationship matrix [D] for plane stress problem is E [D] = 1V 2 2 x105 0.25 = 0.9375 We know [B] = 0.1667 1 v 0   v 1 0   1 v 0 0 2   4 1 0  1 4 0    0 0 1.5 2105 = 10.252  1 0.25 0   0.25 1 0   1  0.25  0  0 2   ; [D] = 53.33 x 10 3 0  3 0  0 1 0 1    1  3  1 3 0 0 2 3 g.n et 4 1 0  1 4 0    0 0 1.5 0 2  0  12 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net [B]T = 0.1667 T [B] [D] = 0.1667   3 0  1  0  1  3   3 0  1    0 1 3  0 0 2   2 0   0   3 0  1  0  1  3   3 0  1    0 1 3  0 0 2   2 0   0 ww w.E a = 0.1667 X 53.33 X 103 [B]T [D] = 8.890 X 103   12  1   12   1  0   2 x 53.33 x 10   12  1   12   1  0   2 3 4 3 4 3 4 3 4  1.5   4.5  1.5   4.5  3   0  syE 0 8 3 4 1 0  1 4 0    0 0 1.5  1.5   4.5  1.5   4.5  3   0  ngi 0 8 nee   12  1   12   1  0   2 3 4 3 4  37.5  7.5   34.5 [B]T [D] [B] = 1.482 X 103    1.5  3    6 7.5 17.5 1.5  9.5  34.5 1.5 37.5  7.5  1.5  9.5  7.5 17.5 3 9 3 9 9 8 3 6 9 8 6 0 [B]T [D] [B] == 8.890 X 103 0 8 rin g.n  1.5   4.5 3 0  3 0  1.5    x 0.1667  0  1 0  1 4.5    1  3  1 3 3   0  0 0 2 et 0 2  0  6   8  6   8  0   16  Substitute [B]T [D] [B] and A, t values in stiffness matrix 13 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Stiffness matrix [K] = [B] T [D] A t  37.5  7.5   34.5    1.5  3    6 7.5 17.5 1.5  9.5  34.5 1.5 37.5  7.5  1.5  9.5  7.5 17.5 3 9 3 9 9 8 3 6 9 8 6 0  37.5  7.5   34.5 [K] = 22.23 X 103    1.5  3    6 7.5 17.5 1.5  9.5  34.5 1.5 37.5  7.5  1.5  9.5  7.5 17.5 3 9 3 9 9 8 3 6 9 8 6 0 [K] = 1.482 X 103 ww w.E a 6   8  6   x3x5 8  0   16  6   8  6   8  0   16      We know that, for plane stress problem, Initial strain {eo} =       syE 6 x10 6 x10 60    -6   6 {eo} = 6 x10 x10 = 1 x 10 60 60   0     ngi nee We know that, Temperature force vector, {F} = [B] T [D] {eo} A t 3 {F} = 8.890 x 10   12  1   12   1  0   2 3 4 3 4 0 8  1.5   4.5  1.5   4.5  3   0  x 1 x 10 -6 60   60 60   xAxt Substitute “A” and “t” values 3 = 8.890 x 10 x 1 x 10 -6 x3x5   12  1   12   1  0   2 3 4 3 4 0 8  1.5   4.5  1.5   x 4.5  3   0  rin g.n et 60   60 60   14 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net (12 x60)  (3 x60)  0  (1x60)  (4 x60)  0     (12 x60)  (3 x60)  0    = 0.1335  (1x60)  (4 x60)  0   000     (2 x60)  (8 x60)  0     900  300   900     300 0    600  = 0.1335  120.15  40.05    120.15  {F} =    40.05  0    80.10  ww w.E a Result: syE Stiffness matrix [K] = 22.23 X 103 Temperature force vector, {F} = 4.  37.5  7.5   34.5    1.5  3    6 7.5 17.5 1.5  9.5  34.5 1.5 37.5  7.5  1.5  9.5  7.5 17.5 3 9 3 9 9 8 3 6 9 8 6 0 ngi  120.15  40.05    120.15     40.05  0    80.10  nee rin 6   8  6   8  0   16  g.n et A thin plate is subjected to surface traction as shown in figure. Calculate the global stiffness matrix. fig (i) Take Young’s modulus E = 2 X 105N/mm2, possion ratio v=0.30, Thickness t=25mm.Assume plane stress condition. 15 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Given data: E = 2 X 105N/mm2; Breath b =250mm V = 0.25; length l =500mm t= 25mm; tensile surface traction T= 0.4 N/mm2 1 “T” is converted into nodal force F = 2 T A = ½ x T x (b x t) 1 = 2 x 0.4 x 250 x 25 ww w.E a F = 1250 N Fig (ii) Discretized plate To find: Global stiffness matrix [K]. Formula used: syE  Global Stiffness matrix [K]1 = [B] T [D][B] A t Solution: ngi nee rin g.n et Fig (iii) For element (1) - Nodal displacements are u1, v1, u3, v3 and u4 v4 Fig (iv) Take node 1 as origin; 16 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net For node 1: X1= 0, Y1=0; For node 3: X2=500, Y2=250; For node 4: X3= 0, Y3=250; We know that, stiffness matrix [K]1 = [B] T [D][B] A t 1 X 1 Y1 1 X 2 Y2 1 Where A =Area of the triangular element = 2 = 1 X3 1 0 0 1 500 250 1 = 2 Y3 1 0 250 1 x 1 (500x250 -0) = 62500mm2 2 A = 62.5 X 103 mm2 1 Strain –Displacement matrix [B] = 2A ww w.E a q1 0 0 r1 q2 0 0 r2 q3 0 0 r3 r1 q1 r2 q2 r3 q3 Where, q1 = y2 – y3 = 250-250 = 0 r1 = x3 – x2 = 0-500 = -500 q2 = y3 – y1 = 250-0 = 250 r2 = x1 – x3 = 0-0 = 0 q3 = y1 – y2 = 0-250 =-250 r3 = x2 – x1 = 500-0 = 500 syE Substitute the above values in [B] matrix equation 1 [B] = 2A 0 0 0  500 250 0  500 0 0 1 Substitute “A” value, [B] = 2  62.5  103 [B] = 250 2  62.5  103 0 0  0 2   2 0 0 0  250 0 250 500 ngi 0  0  0  500    500 0 1 0 0 0 1 0 0 1 2 0 500 nee  250 250 0 0 0  250 0 0 250 500 0 2   1 rin 0  500   250 g.n et We know that, stress-strain relationship matrix [D] for plane stress problem is E [D] = 1V 2 2105 = 0.91 1 v 0   v 1 0   1 v 0 0 2   21 0 5 = 1 ( 0.3) 2  1 0.3 0   0.3 1 0   1  0.3  0 0 2   0   1 0.3 0.3 1 0    0 0 0.35 17 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net [D][B] = = 439.56 0   1 0.3 0.3 1 0    0 0 0.35 2 x105 0.91  0  0    0.7  0.6 2 1 0.3 0 0 1  0.3 0 0 0.35 0.7 250 We know that, [B] = 2  62.5  103 0 0  0 2   2 0 0 0 1 0 0 1 2 0 2   1 0.6  2   0.35 1 0 0 0 1 0 0 1 2 syE 0  2 0  0 2 0     0    1 0 0 [B]T [D] [B] = 2 x 10-3 x 439.56  x 0 0 1   0   0.7 1 0 2    2  1   0 0 2   1  0.6 2 1 0.3 0 0 1  0.3 nee 0.35 0.7 ngi = 0.8791 1 0 0  2 0  0 2 0    1 0 0   0 1  0 1 0 2   2  1   0 ww w.E a [B]T = 2 x 10-3 0 0  0 2   2 0 250 x 2 x62.5 x10 3  1.4  0   0    0.7   1.4   0.7 0 4  0.6 0 0  0.6 1 0  0.7 0 0 0.35  1.4 0.6 1 0.7 0.6 4 1 0.6 0.7  0.35 2.4  1.3 0 0 0.7   4  0.6    0.35  1.3   4.35  rin 0.6  2   0.35 g.n et Substitute [B]T [D] [B] and A, t values in stiffness matrix Stiffness matrix [K]1 = [B] T [D] A t  1.4  0   0 Stiffness matrix [K]1 =0.8791    0.7   1.4   0.7 0 4  0.6 0 0  0.6 1 0  0.7 0 0 0.35  1.4 0.6 1 0.7 0.6 4 1 0.6 0.7  0.35 2.4  1.3 0.7   4  0.6  3  x 6.25x 10 x25  0.35  1.3   4.35  18 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net  1.4  0   0    0.7   1.4   0.7 =1373.59 x 103 U1  1923.026  0   0 [K]1 =1x103    961.513   1923.026   961.513 0 4  0.6 0 0  0.6 1 0  0.7 0 0 0.35  1.4 0.6 1 0.7 0.6 4 1 0.6 0.7  0.35 2.4  1.3 v1 v3 0 5494.36  824.154 0 0  824.154 1373.59 0 824.154  5494.36  1373.59 824.154 ww w.E a For element (2): u3 0.7   4  0.6    0.35  1.3   4.35  syE u4 v4  961.513 0 0 480.7565  1923.026 961.513  u1 824.154  5494.36  v1  1373.59 824.154  u3  961.513  480.7565 v3 961.513 3296.616  1785.667  u 4   480.7565  1785.667 5975.1165  v4 ngi nee rin fig(v) Nodal displacements are u1, v1, u3, v3 and u4 v4 g.n et Take node 1 as origin; For node 1: X1= 0, Y1=0; For node2: X2=500, Y2=0; For node 3: X3= 500, Y3=250; We know that, stiffness matrix [K]2 = [B] T [D][B] A t 1 Where A =Area of the triangular element = 2 = 1 X 1 Y1 1 X 2 Y2 1 X3 Y3 1 = 2 1 0 1 500 0 0 1 500 250 1 x 1 (500x250 -0) = 62500mm2 2 A = 62.5 x 103 mm2 19 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net  q1 1  0 Strain –Displacement matrix [B] = 2A   r1 0 r1 q2 0 0 r2 q3 0 q1 r2 q2 r3 0 r3  q3  Where, q1 = y2 – y3 = 0-250 = -250; r1 = x3 – x2 = 500-500 = 0 q2 = y3 – y1 = 250-0 = 250; r2 = x1 – x3 = 0-500 = -500 q3 = y1 – y2 = 0-0 =0; r3 = x2 – x1 = 500-0 = 500 Substitute the above values in [B] matrix equation 1 [B] = 2A 0 250 0  250  0 0 0  500   0  250  500 250 ww w.E a 0 0 500  1 250 0 Substitute “A” value, [B] = 2  62.5  103   0 0 0 0  500  0  1 0 0 2 0 0 1  2 1 2 0 2  0  We know that, stress-strain relationship matrix [D] for plane stress problem is E [D] = 1V 2 1 2 x105 = 0.91 0.3   0 2 x105 [D][B] = 0.91 = 439.56  1   0.3   0 syE 1 v 0   v 1 0   1 v 0 0 2   0.3 1 0 0  0  0.35 ngi 2 x10 = 1 ( 0.3) 2 nee 0   1 0.3 250 0.3 1  x 0   2 x62.5 x10 3  0 0 0.35 0 0 1 0.3  0.6 2 0 0  0.35  0.7 0.35 0.7 250 We know that, [B] = 2  62.5  103  1 0.3 0   0.3 1 0   1  0.3  0 0 2   5 rin  1 0 1 0  0 0 0 2   0  1  2 1 0 0 2 g.n 0 2  0  et 0.6  2  0   1 0 1 0  0 0 0 2   0  1  2 1 0 0 2 0 2  0  20 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net [B]T = 2 x 10-3 0 1 0 0 0  1   1 0  2    0 2 1  0 0 2   2 0   0 0 1 0 0 0  1    1 1 0  2  T -3 [B] [D] [B] = 2 x 10 x 439.56     0.3  0 2 1   0 0 0 2    2 0   0 ww w.E a = 0.8791  1  0   1   0.6  0    0.6 0 0.35 0.7  0.35 1 0.7 2.4  1.3 0.6  0.35  1.3 4.35 0  0.7  1.4 0.7  0.7 0  1.4 0.6 0.7 4 1.4 0 syE 0 0 1 0.3  0.6 2 0 0  0.35  0.7 0.35 0.7 0.6  2  0  ngi  0.6  0  0.6   4  0   4  Substitute [B]T [D] [B] and A, t values in stiffness matrix Stiffness matrix [K]1 = [B] T [D] A t Stiffness matrix [K]1 =0.8791  1  0   1   0.6  0    0.6 =1373.59 x 103 u1  1  0   1   0.6  0    0.6 0 0.35 0.7  0.35 1 0.7 2.4  1.3  0.7 0 0 0.35 0.7  0.35  0.7 0 v1 nee rin 0.6  0.35  1.3 4.35 0  0.7  1.4 0.7  1.4 0.6 0.7 4 1.4 0 1 0.7 2.4  1.3 0.6  0.35  1.3 4.35 0  0.7  1.4 0.7  1.4 0.6 0.7 4 1.4 0  0.6  0  0.6   4  0   4  u3 v3  0.6  0  0.6  3  x 6.25x 10 x25 4  0   4  u4 g.n et v4 21 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 0  1373.59 824.154 0 480.7565 961.513  480.7565  961.513 961.513 3296.616  1785.667  1923.026  480.7565  1785.667 5975.1165 961.513  1373.59  0    1373.59 [K]2 =1x103   824.154  0    824.154  961.513 0  1923.026 824.154 961.513  5494.36  824.154  u1 v 0  1 824.154  u3   5494.36  v3  u4 0  5494.36  v4 1923.026 0 Global stiffness matrix [K]: Assemble the stiffness matrix equations [K]1 & [K]2 = 1 x 103 x u1 v1 1923.026+ 0+0 u2 v2 -1373.59 824.154 u3 v3 u4 v4 -961.513+ -1923.026 961.513 u1 0+0 824.154 -5494.36 v1 -1923.026 824.154 0 0 u2 0+0 1373.59 -824.154 ww w.E a 0+0 5494.36+ 961.513 -480.7565 480.7565 -824.154+ -961.513 -1373.59 961.513 3296.616 824.154 -480.7565 -1785.667 5975.116 961.513 -5494.36 0 0 v2 0+0 -824.154+ 0+ 1373.59+ 0+0 -1373.59+ 824.154+ u3 -961.513 -1923.026 961.513 1923.026 0 0 0+0 0+ 0+0 961.513+ -480.7565 0 +0 961.513 3296.616 -1785.667 u4 -480.7565 -1785.667 5975.116 v4 v3 u4 -961.513+ -824.154 -1785.667 0+ syE 824.154 0+ -5494.36 -1923.026 824.154 0 0 961.513 -5494.36 0 0 480.7565 + ngi -1373.59 824.154 5494.36 nee Global stiffness matrix [K] = 1 x 103 x u1 v1 u2 v2 v3 u3 rin g.n v4 et 3296.616 0 -1373.59 824.154 0 -1785.667 -1923.026 961.513 u1 0 5975.1165 961.513 -480.7565 -1785.667 0 824.154 -5494.36 v1 -1373.59 961.513 9296.616 -1785.667 -1923.026 824.154 0 0 u2 824.154 -480.7565 -1785.667 5975.116 5 961.513 -5494.36 0 0 v2 0 -1785.667 -1923.026 961.513 9296.616 0 -1373.59 824.154 u3 -1785.667 0 824.154 -5494.36 0 5975.116 5 961.513 -480.7565 v3 -1923.026 824.154 0 0 -1373.59 961.513 3296.616 -1785.667 u4 961.513 -5494.36 0 0 824.154 -480.7565 -1785.667 5975.116 v4 22 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 5. Derive the Shape function for the six noded triangular elements. Fig. A six noded triangular element Consider a six-noded triangular element is shown in figure. It belongs to the serendipity family of elements. It consists of six nodes, which are located on the boundary. ww w.E a We know that, shape function N1=1 at node 1 and 0 at all other nodes. The natural coordinates of the nodes are indicated in the figure. By following our earlier procedure, the shape functions can be obtained as, At node 1: (Coordinates L1 =1, L2 =0, L3 =0) Shape function N1=1 at node 1 N1=0 at all other nodes, N1 has to be in the form of Substitute L1= 1 and N1 =1 syE N1 = C L1 (L1 - 1 ); where C is constant. 2 ngi N1 = C x 1 (1 C=2 1 ) 2 Substitute C value in the above equation N1 = 2 L1 (L1 - 1 ) 2 nee N1 = L1 (2L1 -1) At node 2: (Coordinates L1 =0, L2 =1, L3 =0) Shape function N2=1 at node 2 rin g.n et N2=0 at all other nodes, N2 has to be in the form of N2 = C L2 (L2 - Substitute L2= 1 and N2 =1 1 ); where C is constant. 2 N2 = C x 1 (1 - 1 ) 2 C=2 Substitute C value in the above equation N2= 2 L2 (L2 - 1 ) 2 N2 = L2 (2L2 -1) 23 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net At node 3: (Coordinates L1 =0, L2 =0, L3 =1) Shape function N3=1 at node 3 N3=0 at all other nodes, N3 has to be in the form of N3 = C L3 (L3 - Substitute L3= 1 and N3 =1 1 ); where C is constant. 2 N3 = C x 1 (1 - 1 ) 2 C=2 Substitute C value in the above equation N3= 2 L3 (L3 - ww w.E a 1 ) 2 N3 = L3 (2L3 -1) Now, we define N4, N5 and N6 at the mid-points. At node 4: (Coordinates L1 = 1 1 , L2 = , L3 =0) 2 2 Shape function N4=1 at node 4 syE N4=0 at all other nodes, N4 has to be in the form of Substitute L4= 1 1 and L2 = 2 2 N4 = C L1L2; ngi where C is constant. N4 = C x C=4 1 1 x 2 2 Substitute C value in the above equation nee N4 = 4L1 L2 At node 5: (Coordinates L1 =0, L2 = 1 1 , L3 = ) 2 2 Shape function N5=1 at node 5 N5=0 at all other nodes, N5 has to be in the form of Substitute L2= N5 = C L2L3; 1 1 and L3 = 2 2 N5= C x rin g.n et where C is constant. 1 1 x 2 2 C=4 Substitute C value in the above equation N5 = 4L2 L3 At node 6: (Coordinates L1 = 1 1 , L2 =0, L3 = ) 2 2 Shape function N6 =1 at node 6 N6=0 at all other nodes, 24 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net N6 has to be in the form of Substitute L1= N6 = C L1L3; 1 1 and L3 = 2 2 where C is constant. 1 1 x 2 2 N6= C x C=4 Substitute C value in the above equation N6 = 4L1 L3 Shape functions are, N1 = L1 (2L1 -1) N2 = L2 (2L2 -1) N3 = L3 (2L3 -1) N4 = 4L1 L2 ww w.E a N5 = 4L2 L3 N6 = 4L1 L3 6. Derive the Shape function for the Constant Strain Triangular element (CST). syE We begin this section with the development of the shape function for a basic two dimensional finite element, called constant stain triangular element (CST). ngi We consider the CST element because its derivation is the simplest among the available two dimensional elements. nee rin g.n et Fig. Three noded CST elements. Consider a typical CST element with nodes 1, 2 and 3 as shown in fig. let the nodal displacements to be u1, u2, u3, v1, v2, v3.  u1  u   2 u3  Displacement u     v1   v2     v3  25 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Since the CST element has gat two degrees of freedom at each node (u, v), the total degree of freedom are 6. Hence it has 6 generalised coordinates. Let, u  a1  a2 x  a3 y … (3.1) v  a 4  a5 x  a 6 y … (3.2) Where a1, a2, a3, a4, a5, and a6 are globalised coordinates  u1  a1  a2 x1  a3 y1 u 2  a1  a2 x2  a3 y 2 u3  a1  a2 x3  a3 y3 Write the above equations in matrix form, ww w.E a u1  1 x1    u 2   1 x 2 u  1 x 3  3  y1   a1    y 2  a 2  y 3  a3   a1  1 x1    a 2   1 x 2 a  1 x 3  3  y1  y 2  y 3  1 x1  Let D = 1 x 2  1 x3 y1  y 2  y 3  We know, D-1 = CT D 1 u1    u 2  u   3 syE … (3.3) ngi nee Find the co-factor of matrix D. C11 =  x2 y2 x3 y3 C12 =  1 y2 C13 =  1 x2 C21 =  x1 y1 x3 y3 1 y3 1 x3  ( x 2 y 3  x3 y 2 ) rin g.n … (3.4) et  ( y 3  y 2 )  y 2  y 3  ( x3  x 2 )  ( x1 y3  x3 y1 )  x3 y1  x1 y3 26 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net C22 =  1 y1 C23 =  1 x1 1 y3 1 x3  ( y3  y1 )  ( x3  x1 )  x1  x3 x1 x2 y1  x1 y 2  x2 y1 y2 C32 =  1 y1  ( y 2  y1 )  y1  y 2 C33 =  1 x1  ( x2  x1 ) 1 x2 C31 =  1 y2 ww w.E a  x 2 y 3  x3 y 2   y 2  y 3   x3  x 2  x3 y1  x1 y3   y3  y1  x1  x3  C = x1 y 2  x 2 y1   y1  y 2  x 2  x1  syE x2 y3  x3 y 2  x3 y1  x1 y3  x1 y 2  x2 y1   y 2  y3   y3  y1   y1  y 2   T C =  x3  x 2  x1  x3  x2  x1  1 x1 y1 We know that, D= 1 x 2 y2 1 x3 y3 ngi D = 1 ( x2 y3  x3 y 2 )  x1  y3  y 2   y1 x3  x2  Substitute CT and D value in equation (3.4) D-1 = 1  ( x2 y3  x3 y 2 )  x1  y3  y 2   y1 x3  x2  nee …(3.5) rin g.n …(3.6) et x2 y3  x3 y 2  x3 y1  x1 y3  x1 y2  x2 y1   y 2  y3   y3  y1   y1  y2   x3  x 2  x1  x3  x2  x1  Substitute D-1 value in equation (3.3)  a1  1 x1    a 2   1 x 2 a  1 x 3  3  y1  y 2  y 3  1 u1    u 2  u   3 27 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net  a1  x2 y3  x3 y2    1 a   y2  y3   2 =  a  ( x2 y3  x3 y2 )  x1  y3  y 2   y1 x3  x2  x3  x2   3 x3 y1  x1 y3  x1 y2  x2 y1  u1   y3  y1   y1  y2   u2  x1  x3  x2  x1  u3  ..(3.7) The area of the triangle can be expressed as a function of the x,y coordinate of the nodes 1,2 and 3. 1 x1 1 1 x2 2 A= 1 x3 A y1  y 2  y 3  1 ( x2 y3  x3 y 2 )  x1  y3  y 2   y1 x3  x2  2 2 A  ( x2 y3  x3 y 2 )  x1  y3  y 2   y1 x3  x2  ww w.E a ... (3.8) Substitute 2A value in equation (3.7),  a1  x2 y3  x3 y2    1  y 2  y3  a 2  = 2A     a3  x3  x2   a1   p1   1 q1 a 2  = 2A     a3  r1 x3 y1  x1 y3  x1 y2  x2 y1  u1   y3  y1   y1  y2   u2  u  x1  x3  x2  x1   3 syE ngi  p 2  p3 u1  q 2 q3  u 2  r 2 r 3 u3  nee ...(3.9) p1  x2 y3  x3 y2  p 2  x3 y1  x1 y3  p3  x1 y2  x2 y1  q1   y2  y3  Where, r1  x3  x2  q 2   y3  y1  q3   y1  y2  r 2  x1  x3  r 3  x2  x1  From eq (3.1) we know that u= 1 x  a1   y a 2  a3   rin g.n et  a1    Sub a 2 values from Eq (3.10)  a3   u= 1 x  p1 1 q1 y  2A r1  p 2  p3 u1  q 2 q3  u 2  r 2 r 3 u3  28 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 1  1 x 2A  p1  p2  p3 u1    y  q1 q 2 q3  u 2  r1 r 2 r 3 u3  1  p1  q1 x  r1 y 2A u=  u!    p3  q3 x  r3 y  u2  u   3 p2  q2 x  r2 y  p1  q1 x  r1 y  2A  ww w.E a  u!  p3  q3 x  r3 y      u2  2A    u3  p2  q2 x  r2 y 2A The above equation is in the form of u = N1 u= V = N2  u1    N 3 u 2  u   3 syE N1 N2 Similarly, Where shape function , N1= N2= N3= … (3.11) ngi  v1    N 3 v2  v   3 p1  q1 x  r1 y 2A p1  q1 x  r1 y 2A p3  q3 x  r3 y 2A nee … (3.12) rin g.n Assembling the equations (3.11) and (3.12) in matrix form Displacement matrix u = u ( x, y )   N1    v ( x, y )   0 0 N1 N2 0 0 N2 N3 N3 u1  v   1 0   u    2  0   v2  u3      v3   et … (313) 29 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net ww w.E a UNIT 4 syE ngi nee rin g.n et Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net UNIT IV – TWO DIMENSIONAL VECTOR VARIABLE PROBLEMS PART - A 1. What is meant by axisymmetric field problem? Give example.(April/May 2010) In some of the three dimensional solids like flywheel, turbine, discs etc, the material is symmetric with respect to their axes. Hence the stress developed is also symmetric. Such solids are known as axisymmetric solids. Due to this condition, three dimensional solids can be treated as two dimensional elements. 2. List the required conditions for a problem assumed to be axisymmetric. (April/May 2011) The condition to be axisymmetric is as follows:  Problem domain must be symmetric about the axis of revolution.  All boundary conditions must be symmetric about the axis of revolution.  All loading conditions must be symmetric about the axis of revolution. ww 3. What is Plane stress and Plane strain condition? (April/May 2015), (May/June 2013) Plane stress - A state of plane stress is said to exist when the elastic body is very thin and there is no load applied in the coordinate direction parallel to the thickness. w.E Example: A ring press-fitted on a shaft in a plane stress problem. In plane stress problem σz, τyz, τzx are zero. asy Plane strain – A state of plane strain is said to exist when the strain at the plane perpendicular to the plane of application of load is constant. En 4. What are the forces acting on shell elements? Give its applications gin The two forces in which the shell element is subjected to are: Bending moments Membrane forces eer i Shell elements can be employed in the analysis of the following structures, Example:  Sea shell, egg shell (the wonder of the nature);  Containers, pipes, tanks;  Car bodies;  Roofs, buildings (the Superdome), etc. ng. net 5. Write the constitutive relations for axisymmetric problems. 6. Define body force. A body force is distributed force acting on every elemental volume of the body. Unit: force per unit volume. 7. Write the governing equation for 2D bending of plates. 1 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 8. Write the stress strain relationship for plane stress problems. ww w.E 9. Differentiate material non linearity and geometric non linearity. (Nov/Dec 2012) Material Non linearity Geometric non linearity (i) The stress – strain relation for the (i) The Strain – Displacement relations material may not be linear. are not linear. (ii) Basic non-linear relations are (ii) Need consideration of actual strain time dependent complex constitutive displacement relations rather than linear relations strain displacement. asy En gin 10. Write the equilibrium equations for two dimensional elements. (Nov/Dec 2012) In elasticity theory, the stresses in the structure must satisfythe following equilibrium equations, eer i ng. wherefx and fy are body forces (such as gravity forces) per unit volume. PART - B net 1. For the axe symmetric element shown in fig .Determine the element stresses. Take E= 2.1 x 105 N/mm2 𝝂 = 0.25. The co-ordinates shown in fig are in mm. The nodal displacements are u1=0.05 mm, u2=0.02 mm, u3=0.0 mm, 𝝎𝟏 = 𝟎. 𝟎𝟑 𝒎𝒎, 𝝎𝟐 = 𝟎. 𝟎𝟐 𝒎𝒎, 𝝎𝟑 = 𝟎. 𝟎 𝒎𝒎. Z (0,0) 1 3 (30,50) 2 (60,0) 2 r Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Given data: r1 = 0 mm z1=0 mm u1=0.05 mm 𝜔1 = 0.03 𝑚𝑚 r2 = 60 mm z2=0 mm u2=0.02 mm 𝜔2 = 0.02 𝑚𝑚 r3 = 30 mm z3=50 mm u3=0.0 mm 𝜔3 = 0.0 𝑚𝑚 E= 2.1 x 105 N/mm2, 𝜈 = 0.25 To find ww i. Radial stress 𝜎𝑟 ii. Circumferential stress 𝜎𝜃 iii. Longitudinal stress 𝜎𝑧 iv. Shear stress 𝜏𝑟𝑧 Formula used w.E {σ} 𝜎𝑟 𝜎𝜃 𝜎𝑧 𝜏𝑟𝑧 = [D] [B] {u} 𝑢1 𝑤1 𝑢2 = [D] [B] 𝑤 2 𝑢3 𝑤3 asy En Solution: gin {σ} = [D] [B] {u} D = Stress - Strain relationship matrix 𝐸 D= 1+𝜈 1−2𝜈 = 1−𝜈 𝜈 𝜈 0 𝜈 1−𝜈 𝜈 0 2.1 x 105 1 + 0.25 1 − 2 × 0.25 3 1 [D] = 336 × 103 × 0.25 1 0 𝜈 𝜈 1−𝜈 0 0 0 0 eer i ng. 1− 2𝜈 2 1 − 0.25 0.25 0.25 0.25 1 − 0.25 0.25 0.25 0.25 1 − 0.25 0 1 3 1 0 1 1 3 0 0 0 0 0 1 0 3 1 = 84 × 103 1 0 1 3 1 0 1 1 3 0 net 0 0 0 1 − 2 × 0.25 2 0 0 0 1 [B] =Strain displacement relationship matrix or gradient matrix [B] = 1 2𝐴 𝛽1 𝛾₁𝑧 + 𝛽₁ + 𝑟 𝑟 0 𝛾1 𝛼₁ 0 0 𝛾1 𝛽1 𝛽2 𝛾2 𝑧 + 𝛽 + 2 𝑟 𝑟 0 𝛾2 𝛼2 0 0 𝛾2 𝛽2 𝛽3 𝛾3 𝑧 + 𝛽 + 3 𝑟 𝑟 0 𝛾3 𝛼3 0 0 𝛾3 𝛽3 3 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝛼1 = 𝑟2 𝑧3 − 𝑟3 𝑧2 = 60 × 50 − 30 × 0 = 3000𝑚𝑚2 𝛼2 = 𝑟3 𝑧1 − 𝑟1 𝑧3 = 30 × 0 − 0 × 50 = 0 𝛼3 = 𝑟1 𝑧2 − 𝑟2 𝑧1 = 0 × 0 − 60 × 0 = 0 𝛽1 = 𝑧2 − 𝑧3 = 0 − 50 = −50 ; 𝛾1 = 𝑟3 − 𝑟2 = 30 − 60 = −30 ; 𝛽2 = 𝑦3 − 𝑦1 = 50 − 0 = 50 ; 𝛾2 = 𝑟1 − 𝑟3 = 0 − 30 = −30 ; 𝑟= 𝑟1 + 𝑟2 + 𝑟3 0 + 60 + 30 = = 30 𝑚𝑚 3 3 𝑧= 𝑧1 + 𝑧2 + 𝑧3 0 + 0 + 50 = = 16.67 𝑚𝑚 3 3 𝛽3 = 𝑦1 − 𝑦2 = 0 − 0 = 0 𝛾3 = 𝑟2 − 𝑟1 = 60 − 0 = 60 𝛼₁ 𝛾1 𝑧 3000 (−30 × 16.67) + 𝛽1 + = + −50 + = 33.33 𝑚𝑚 𝑟 𝑟 30 30 ww 𝛼2 𝛾2 𝑧 (−30 × 16.67) + 𝛽2 + = 0 + 50 + = 33.33 𝑚𝑚 𝑟 𝑟 30 w.E 𝛼3 𝛾3 𝑧 60 × 16.67 + 𝛽3 + = 0+0+ = 33.33 𝑚𝑚 𝑟 𝑟 30 1 𝑟1 𝐴 = 2 1 𝑟2 1 𝑟3 1 asy 𝑧1 1 1 𝑧2 = 1 2 𝑧3 1 1 0 0 60 0 30 50 En gin = 2 [1 3000 − 0 − 0 50 − 0 + 0 30 − 60 ]=1500 𝑚𝑚2 [B] = −50 0 33.33 0 2 × 1500 0 −30 −30 −50 1 eer i 0 0 0 50 0 33.33 0 33.33 −30 0 60 0 50 60 0 −30 3 1 1 0 1 3 1 0 [D] [B] = 84 × 103 × 3.34 ×10-4 1 1 3 0 0 0 0 1 0 0 0 −50 0 50 0 33.33 0 33.33 0 33.33 −30 0 60 0 −30 0 50 60 0 −30 −50 −30 −116.67 49.99 = 28 −16.67 −30 −30 −30 −90 −50 𝜎𝑟 −116.67 −30 𝜎𝜃 49.99 −30 = 28 𝜎𝑧 −16.67 −90 𝜏𝑟𝑧 −30 −50 183.33 −30 149.99 −30 83.33 −90 −30 50 183.33 149.99 83.33 −30 ng. net 33.33 60 99.99 60 33.33 180 60 0 −30 33.33 60 −30 99.99 60 −90 33.33 180 50 60 0 0.05 0.03 0.02 0.02 0 0 4 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝜎𝑟 −3.66 −102.65 𝜎𝜃 4 112 = 28 = 𝜎𝑧 −3.66 −102.65 𝜏𝑟𝑧 −2.6 −72.8 Results Radial stress 𝜎𝑟 = −102.65 N/mm2 Circumferential stress 𝜎𝜃 = 112 N/mm2 Longitudinal stress 𝜎𝑧 = −102.65 N/mm2 Shear stress 𝜏𝑟𝑧 = −72.8 N/mm2 2. Calculate the element stiffness matrix and the thermal force vector for the axisymmetric triangular element shown in figure. The element experiences a 15 0c increase in temperature. The co-ordinates are in mm. Take α= 10 x 10-6/0c ; E= 2x 105 N/mm2 , 𝝂 = 0.25 ww Z w.E 3 (9,10) asy (6,7) 1 2 (8,7) r En Given data: r1 = 6 mm z1=7 mm r2 = 8 mm z2=7 mm r3 = 9 mm z3=10 mm 5 2 E= 2 × 10 N/mm , 𝜈 = 0.25, α= 10 × 10-6/0c To find Thermal force vector {F}t Formula used [K]=[𝐵]T D B 2πr A gin eer i ng. {F}= 𝐵 T D 𝑒𝑡 2πr A Solution: [B] =Strain displacement relationship matrix or gradient matrix 0 𝛽3 𝛽1 0 𝛽2 𝛼₁ 𝛾₁𝑧 𝛼2 𝛾2 𝑧 𝛼3 𝛾 𝑧 + 𝛽₁ + 0 + 𝛽2 + 0 + 𝛽3 + 3 1 𝑟 𝑟 𝑟 𝑟 𝑟 [B] = 2𝐴 𝑟 0 𝛾1 0 𝛾2 0 𝛾1 𝛽1 𝛾2 𝛽2 𝛾3 0 0 net 𝛾3 𝛽3 𝛼1 = 𝑟2 𝑧3 − 𝑟3 𝑧2 = 8 × 10 − 9 × 7 = 17𝑚𝑚2 𝛼2 = 𝑟3 𝑧1 − 𝑟1 𝑧3 = 9 × 7 − 6 × 10 = 3𝑚𝑚2 𝛼3 = 𝑟1 𝑧2 − 𝑟2 𝑧1 = 6 × 7 − 8 × 7 = 13𝑚𝑚2 𝛽1 = 𝑧2 − 𝑧3 = 7 − 10 = −3𝑚𝑚 ; 𝛾1 = 𝑟3 − 𝑟2 = 9 − 8 = 1𝑚𝑚 ; 𝛽2 = 𝑦3 − 𝑦1 = 10 − 7 = 3 ; 𝛾2 = 𝑟1 − 𝑟3 = 6 − 9 = −3 ; 𝛽3 = 𝑦1 − 𝑦2 = 7 − 7 = 0 𝛾3 = 𝑟2 − 𝑟1 = 8 − 6 = 2 5 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝑟= 𝑟1 + 𝑟2 + 𝑟3 6 + 8 + 9 = = 7.67 𝑚𝑚 3 3 𝑧= 𝑧1 + 𝑧2 + 𝑧3 7 + 7 + 10 = = 8 𝑚𝑚 3 3 𝛼₁ 𝛾1 𝑧 17 (1 × 8) + 𝛽1 + = + −3 + = 0.26 𝑚𝑚 𝑟 𝑟 7.67 7.67 𝛼2 𝛾2 𝑧 3 (−3 × 8) + 𝛽2 + = +3+ = 0.26 𝑚𝑚 𝑟 𝑟 7.67 7.67 𝛼3 𝛾3 𝑧 −14 2 × 8 + 𝛽3 + = +0+ = 0.26 𝑚𝑚 𝑟 𝑟 7.67 7.67 ww 1 𝑟1 1 𝐴 = 2 1 𝑟2 1 𝑟3 𝑧1 1 1 𝑧2 = 1 2 𝑧3 1 −3 1 0.26 2×3 0 1 0 0 1 −3 3 0 0.26 0 0 −3 −3 3 1−𝜈 𝜈 𝜈 0 𝜈 1−𝜈 𝜈 0 w.E [B] = [D] 𝐸 = 1+𝜈 1−2𝜈 6 7 1 8 7 = 2 [1 80 × 63 − 6 10 − 7 + 7(9 − 8)=3 𝑚𝑚2 9 10 asy 2 × 10 5 = 1+ 0.25 1−2× 0.25 En 𝜈 𝜈 1−𝜈 0 1 − 0.25 0.25 0.25 0 3 1 = 320 × 105 × 0.25 1 0 −3 0.26 0 0 [B]T[D] = 0.167 3 0.26 0 0 0 0.26 0 0 0 0.26 0 2 1 3 1 0 1 1 3 0 0 0 0 0 1 1 −3 0 −3 −3 3 0 2 0 2 gin 1− 2𝜈 2 eer i 0.25 0.25 1 − 0.25 0.25 0.25 1 − 0.25 0 0 1 3 1 0 1 1 3 0 0 0 0 ng. 1− 2× 0.25 2 0 0 0 1 0 1 3 1 −3 0 −3 × 8×104 1 −3 3 1 0 2 0 0 2 −8.7 −2.2 1 1 = 13.36×103 9.26 3.78 −3 −3 0.26 0.78 2 2 −3 0.26 0 0 [B]T = 0.167 3 0.26 0 0 0 0.26 0 0 0 0 ; 2 0 net 0 0 0 1 −2.7 1 3 −3 3.26 −3 −9 3 0.26 2 6 0 6 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net −8.7 −2.2 1 1 T 3 9.26 3.78 [B] [D][B] = 13.36×10 −3 −3 0.26 0.78 2 2 −2.7 1 −3 0.26 3 −3 0 0 3.26 −3 × 0.167 3 0.26 −9 3 0 0 0.26 2 0 0.26 6 0 0 0 0 1 1 −3 0 −3 −3 3 0 2 0 2 1.42 −5.4 26.63 −5.7 −29.79 11.21 −5.7 −18 −5.7 12 12.26 6 −5.01 3 −29.79 12.26 −18.78 37.76 6.5 [K]= 321.27 × 10 36 11.21 −18 −18.78 5.2 −18 5.2 1.42 −5.7 −5.01 4.2 0.52 −18 −5.4 6 6.5 0.52 12 Thermal force vector {F}= 𝐵 T D 𝑒𝑡 2πr A ww 𝑒𝑡 = 𝛼∆𝑡 10 × 10−6 × 15 150 −6 𝛼∆𝑡 10 × 10 × 15 -6 150 = =10 0 0 0 −6 𝛼∆𝑡 150 10 × 10 × 15 w.E −8.7 −2.2 1 1 {F}= [B]T[D] 𝑒𝑡 2πr A = 13.36×103 9.26 3.78 −3 −3 0.26 0.78 2 2 asy En −1493.46 −150 1506.54 = 1.927 −450 456.54 600 gin 𝐹1𝑢 −2878.25 𝐹1𝑤 −289.08 𝐹2𝑢 2903.45 Thermal force vector {F} = = 𝐹2𝑤 −867.25 879.86 𝐹3𝑢 1156.34 𝐹3𝑤 3. −2.7 1 3 −3 150 3.26 −3 × 10-6 150 × 2π × 7.67 × 3 −9 3 0 0.26 2 150 6 0 eer i ng. net DERIVE THE EXPRESSION FOR STRESS – STRAIN RELATIONSHIP FOR A 2D- ELEMENT? EQUATION OF ELASTICITY 1. Stress – strain relationship matrix for a two dimensional element Consider a three dimensional body as shown in fig. which is subjected to a stress σx σy and σz 7 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Y σy σz σx σ x x ww Hook’sZlaw σz w.E σy σ = Ee 𝜎 asy e=𝐸 The stress in the x direction produces a positive strain in x direction as shown in fig. En 𝜎 ex = 𝐸𝑥 gin The positive stress in the y direction produces a negative strain in the x direction ey = −𝜈𝜎 𝑦 𝐸 eer i ng. The positive stress in the z direction produces a negative strain in the x direction ez = ex = −𝜈𝜎 𝑧 𝐸 𝜈𝜎 𝑦 𝜎𝑥 𝜈𝜎 − 𝐸 − 𝐸𝑧 𝐸 𝜎𝑦 𝜈𝜎 𝜈𝜎 ey = − 𝐸 𝑥 + 𝐸 − 𝐸 𝑧 ez = − 𝜈𝜎 𝑥 𝐸 − 𝜈𝜎 𝑦 𝐸 net 𝜎 + 𝑧 𝐸 Solving 3 equations 𝐸 e𝑥 1−𝑣 +𝑣 𝑒𝑦 +𝑉 𝑒2 𝐸 v e𝑥 1−𝑣 − 𝑒𝑦 +𝑉 𝑒2 𝐸 v e𝑥+𝑣 𝑒𝑦 + 1−𝑣 𝑒2 𝜎𝑥 = 1+𝑣 1−2𝑣 𝜎𝐽 = 1+𝑣 1−2𝑣 𝜎2 = 1+𝑣 1−2𝑣 The shear stress and shear strain relationship 𝜏 = 𝐺𝛾 where, 𝜏 - Shear Stress 8 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝛾 – Shear Strain G – Modular of rigidity 𝜏 𝑥𝑦 = G𝛾𝑥𝑦 𝜏 𝑦𝑧 = G𝛾𝑦𝑧 𝜏 𝑧𝑥 = G𝛾𝑧𝑥 𝐸 G Modular of rigidity = 2 1+𝑣 𝐸 𝐸 1−2𝑣 𝜏 𝑥𝑦 = 2 1+𝑣 𝛾𝑥𝑦 ; 𝜏 𝑥𝑦 = 2 1+𝑣 1−2𝑣 𝛾 𝑦 2 𝜏 𝑦𝑧 = ww 𝜎𝑥 𝜎𝑦 𝜎𝑧 𝜎𝑥𝑦 𝜎𝑦𝑧 𝜎𝑧𝑥 𝐸 1−2𝑣 1+𝑣 1−2𝑣 2 𝐸 𝜏 𝑧𝑥 1+𝑣 1−2𝑣 w.E 𝐸 = 𝜎 1−2𝑣 = = 𝐷 𝐸 1−2𝑣 1+𝑣 1−2𝑣 2 1−v v v 1− v v v 0 0 v v 1− v 0 0 0 0 0 0 𝑒 En 0 gin D- in a stress strain relation ship matrix 𝐸 𝐷 = 1+𝑣 1−2𝑣 1−v v v 1− v v v 0 0 v v 1− v 0 0 0 0 0 0 0 𝛾𝑦𝑧 𝛾𝑧𝑥 2 asy 1+𝑣 1−2𝑣 𝛾𝑥𝑧 ;𝜏 𝑦𝑧 = 2 0 0 0 0 0 0 0 1−2𝑣 2 0 0 0 0 0 0 0 0 0 1−2𝑣 0 0 0 1−2𝑣 2 2 1−2𝑣 1−2𝑣 0 0 0 0 0 0 𝑒𝑥 𝑒𝑦 𝑒𝑧 𝛾𝑥𝑦 𝛾𝑦𝑧 𝛾𝑧𝑥 2 eer i 0 2 0 0 0 ng. 0 net 1−2𝑣 2 Where E – Yours Modules V – Poisson Ratio (i) PLANE STRESS CONDITION:Plane stress is defined to be a state of stress in which the normal stress 𝜎 and shear stress 𝜏 cleared perpendicular to the plane are assumed to be zero. Normal stress 𝜎𝑧 = 0; Shear Stress 𝜏𝑥𝑧 + 𝜏𝑦𝑧 = 0 𝜎𝑧 = 𝜏 𝑥𝑧 = 𝜏 𝑦𝑧 = 0 9 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝜎 𝑒𝑥 = 𝑥 - v 𝜎𝑦 𝐸 𝑒𝑥 = 𝜎𝑥 -v 𝐸 𝜎 𝜎𝑦 𝐸 𝐸 ; 𝑒𝑦 = -v 𝑥 + 𝐸 𝜎𝑦 𝐸 𝜎𝑦 2 𝜎𝑥 v𝑒𝑦 = −𝑣 +𝑣 𝐸 𝐸 𝜎 𝑒𝑥 + v𝑒𝑦 = 𝐸𝑥 - 𝑣 2 𝜎𝑥 𝐸 𝜎𝑥 𝑒𝑥 + v𝑒𝑦 = 𝐸 - 1 − 𝑣 2 𝐸 𝜎𝑥 = 1−𝑣 2 𝑒𝑥 + 𝑣 𝑒𝑦 𝜎𝑦 𝜎 v 𝑒𝑥 = v 𝐸𝑥 -V 2 𝐸 𝜎𝑦 𝜎 𝑒𝑦 = -v 𝐸𝑥 + 𝐸 ww 𝜎𝑦 𝜎𝑦 w.E v 𝑒𝑥 + 𝑒𝑦 = -V 2 𝐸 + 𝐸 𝜎𝑦 v 𝑒𝑥 + 𝑒𝑦 = 𝐸 1 − 𝑣 2 𝐸 asy 𝑣𝑒𝑥 + 𝑒𝑦 Where G Modular of rigidity = 𝜎𝑦 = 1−𝑣 2 En Share Stress 𝜏 𝑥𝑧 = G 𝛾𝑥𝑧 𝛾𝑥𝑦 gin Share Strain V – Poisson ratio 𝐸 𝜏𝑥𝑦 = 1+𝑣 1−𝑣 × 𝐸 𝜏𝑥𝑦 = 1−𝑣 2 × 1−𝑣 1−𝑣 2 2 1+𝑣 eer i 𝜏𝑥𝑦 = 2 1+𝑣 𝛾𝑥𝑦 𝐸 𝐸 𝛾𝑥𝑦 ng. × 𝛾𝑥𝑦 2 net Above equation matrix form 𝜎𝑥 𝜎𝑦 𝜏𝑥𝑦 𝐸 = 1−𝑣 1 𝑣 0 𝑣 1 0 0 0 1−𝑣 2 𝑒𝑥 𝑒𝑦 𝜏𝑥𝑦 Two dimensional stress strain relationship matrix for phase stress location. 𝐸 𝐷 = 1−𝑣 1 𝑣 𝑣 1 0 0 0 0 1−𝑣 2 10 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net (ii) PLANE STRAIN CONDITION Plane strain is defined to be a state of strain in which the strain normal to the xy plane and the shear strain are assumed to be zero. Normal strain 𝑒𝑧 =0 Shear Stress 𝛾𝑥𝑧 = 0 =𝛾𝑦𝑧 𝜎𝑥 𝜎𝑦 𝜎𝑧 𝜎𝑥𝑦 𝜎𝑦𝑧 𝜎𝑧𝑥 𝐸 = 1+𝑣 1−2𝑣 1−v v v 1− v v v 0 0 v v 1− v 0 0 0 0 0 0 0 𝑒𝑧 =0 ; 𝛾𝑥0 =𝛾𝑦𝑧 =0 Sub in above matrix. 𝜎𝑥 𝐸 𝜎𝑦 = 1+𝑣 1−2𝑣 𝛾𝑥𝑦 1−𝑣 𝑣 0 ww w.E 𝑣 1−𝑣 0 0 0 0 0 0 0 0 0 0 0 0 1−2𝑣 0 0 0 1−2𝑣 2 2 1−2𝑣 𝑒𝑥 𝑒𝑦 𝑒𝑧 𝛾𝑥𝑦 𝛾𝑦𝑧 𝛾𝑧𝑥 2 𝑒𝑥 𝑒𝑦 𝛾𝑥𝑦 1−2𝑣 2 asy 0 0 0 Stress Strain relationship matrix for phase strain condition. 𝐷 = 𝐸 1+𝑣 1−2𝑣 1−𝑣 𝑣 0 𝑣 1−𝑣 0 0 0 En 1−2𝑣 2 gin 4. A long hollow cylinder of inside diameter 100 mm and outside diameter 140 mm is subjected to an internal pressure of 4 N/mm2 as shown in figure.(i) By using two elements on the 15 mm length shown in figure. (ii) Calculate the displacements at the F1 inner radiusTake E=2×105 N/mm2. V=0.3. 4 eer i 1 Z 1 ng. net 15 mm Element Cylinder Axis of the hallow Cylinder Element 2 2 3 F2 50 mm 100 mm 70 mm 140 mm 11 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Given data: Inner diameter, de= 100mm Inner radius re= 50 mm Outer diameter De=140 mm Outer radius Re=70mm Internal pressure P=4N/mm2 Length le=15mm Young’s modulus E=2×105 N/mm2 Poison’s ratio v= 0.3 ww To Find W1 u1, w1, u2, w2, u3, w3, u4, w4 w.E Formula used Solution 𝑈 asy En For element (1) U1 Element (r1 Z1) Axis of the hallow cylinder 𝐹 =𝐾 Co ordinates At node 1 Z r1=50mm 15 mm eer i W2 2 (r2 Z2 ) 70 mm At node 2 r r1=50mm U2 ng. 50mm z1=15mm U4 (r3 Z3 ) 1 gin (Nodal displacements u1, w1, u2, w2, u4, w4) W4 net z1=0mm At node 3 r1=70mm z1=15mm We know that, 𝑤ℎ𝑒𝑟𝑒 𝑟 = 𝑟1 +𝑟2 +𝑟3 3 = 50+50+70 3 r = 56.6667mm 𝑧 +𝑧2 +𝑧3 𝑧= 1 3 = 15+0+15 3 ; z= 10 mm 1 Area of the triangle element = × 𝐵𝑟𝑒𝑎𝑑𝑡ℎ × 𝐻𝑒𝑖𝑔ℎ𝑡 2 12 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 1 = 2 × 20 × 15 ; A = 150 mm We know that, Stiffness matrix for axisymmetric triangular element (1), 𝐾 1 =2 𝜋 rA 𝐵 T 𝐷 B 1−𝜈 𝜈 𝜈 0 𝐸 Stress strain relationship matrix 𝐷 = 1+𝜈 1−2𝜈 2𝑋10 5 Stress strain relationship matrix 𝐷 = 1+0.3 1−(2×0.3) ww w.E = 2×10 5 0.5 asy 𝜈 1−𝜈 𝜈 0 1 B = 2𝐴 0 0 𝛾1 𝛽1 𝛽2 𝛾2 𝑧 + 𝛽 + 2 𝑟 𝑟 0 𝛾2 𝛼2 gin 0 0 𝛾2 𝛽2 eer i 𝛽3 𝛾3 𝑧 + 𝛽 + 3 𝑟 𝑟 0 𝛾3 𝛼3 𝛼2 = 𝑟3 𝑧1 − 𝑟1 𝑧3 0 0 𝛾3 𝛽3 𝛼2 = 70 × 15 − 50 × 15 𝛼1 = 750 𝑚𝑚2 𝛼2 = 300𝑚𝑚2 𝛽3 = 𝑦1 − 𝑦2 𝛾3 = 𝑟2 − 𝑟1 𝛽1 = 0 − 15 𝛾1 = 70 − 50 𝛽2 = 15 − 15 𝛾2 = 50 − 70 𝛽3 = 15 − 0 𝛾3 = 50 − 50 𝛽1 = −15𝑚𝑚 𝛾1 = 20𝑚𝑚 𝛽2 = 0 𝛾2 = −20𝑚𝑚 𝛽3 = 15𝑚𝑚 𝛾3 = 0 𝛼₁ 𝛾₁𝑧 𝛼2 + 𝛽2 + 2𝑟 = + 𝛽3 + 3𝑟 = 56.6667 + 15 + 0 𝛾 𝑧 2 = 750 20×10 + (−15) + 56.6667 56.6667 300 56.6667 −750 (−20×10) + 0 + 56.6667 net 𝛼3 = −750𝑚𝑚2 𝛽2 = 𝑦3 − 𝑦1 𝛾2 = 𝑟1 − 𝑟3 𝛾 𝑧 1−(2×0.3) 𝛼3 = 50 × 0 − 50 × 15 𝛽1 = 𝑧2 − 𝑧3 𝛾1 = 𝑟3 − 𝑟2 + 𝛽₁ + 𝑟 𝑟 0 0 0 ng. 𝛼3 = 𝑟1 𝑧2 − 𝑟2 𝑧1 𝛼1 = 50 × 15 − 70 × 0 𝑟 2 0.7 0.3 0.3 0 0.3 0.7 0.3 0 = 384.6153×103 𝜈 𝜈 0.7 0 0 0 0 0.2 𝛼1 = 𝑟2 𝑧3 − 𝑟3 𝑧2 𝑟 𝛼3 1− 2𝜈 0.7 0.3 0.3 0 0.3 0.7 0.3 0 𝜈 𝜈 0.7 0 0 0 0 0.2 We know that , strain-Displacement matrix 𝛽1 𝛾₁𝑧 + 𝛽₁ + 𝑟 𝑟 0 𝛾1 0 0 0 1 − 0.3 0.3 0.3 0.3 1 − 0.3 0.3 𝜈 𝜈 1 − 0.3 0 0 0 En 𝛼₁ 𝜈 𝜈 1−𝜈 0 =1.7647 mm = 1.7647 mm =1.7647mm 13 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝛼₁ 𝛾₁𝑧 𝛼 𝛾 𝑧 𝛼3 Substitute𝛽1, 𝛽2 , 𝛽3, 𝑟 + 𝛽₁ + 𝑟 , 𝑟2 + 𝛽2 + 2𝑟 , 𝑟 𝛾 𝑧 + 𝛽3 + 3𝑟 , 𝛾1 , 𝛾2, 𝛾3 and A values in equations no 5, we get, −15 1 1.7647 B = 2×150 0 20 0 0 20 −15 −15 1.7647 B =3.333 × 10−3 0 20 0 1.7647 0 −20 0 0 −20 0 15 1.7647 0 0 0 0 0 15 0 0 20 −15 0 1.7647 0 −20 0 0 −20 0 15 1.7647 0 0 0 20 20 −15 0 −20 −20 0 0 0 0 15 −15 1.7647 0 0 0 1.7647 B T=3.333 × 10−3 0 0 15 1.7647 0 0 ww w.E 0 0 0 15 0.7 0.3 0.3 0 0.7 0.3 0 D B = 384.6153×10 × 0.3 0.3 0.7 0 0 0 0 0.2 −15 0 1.7647 0 3.33310−3 0 20 20 −15 asy 3 0.3 En −9.9706 0.3 D B = 1.282×103 0.3 0 6 0.5294 0.7 0.3 0.3 0.7 0 0 0 1.7647 0 −20 gin 0 0 −20 0 15 1.7647 0 0 eer i −6 11.0294 0 −6 5.7353 0 −14 5.0294 0 0 0 0 0 0 0 15 ng. net −9.9706 6 0.5294 −6 11.0294 0 0.3 0.7 0.3 −6 5.7353 0 X3.33 10-3 D B B T =1.282×103 𝜈 𝜈 0.7 −14 5.0294 0 0 0 0 0 0 0 0 20 −15 1.7647 20 −15 0 0 0 1.7647 0 −20 0 0 −20 0 15 1.7647 0 0 0 0 0 15 223.798 −139.4118 −85.7611 −139.412 325 70.588 −85.7612 70.588 82.18 D B B 𝑇 = 4.2733 79.412 −280 −10.588 −155.3202 100.5882 10.1210 60 −45 −60 79.4118 −155.32 60 −280 100.588 −45 −10.588 10.1211 −60 280 −100.588 0 −100.588 175.5621 0 0 0 45 14 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Substitute D B B 𝑇 value in equ no 4 𝐾 1 = 2 𝜋 ×56.6667×150×4.2733 223.798 −139.4118 −85.7611 −139.412 325 70.588 −85.7612 70.588 82.18 × 79.412 −280 −10.588 −155.3202 100.5882 10.1210 60 −45 −60 79.4118 −155.32 60 −280 100.588 −45 −10.588 10.1211 −60 280 −100.588 0 −100.588 175.5621 0 0 0 45 223.798 −139.4118 −85.7611 −139.412 325 70.588 70.588 82.18 𝐾 1 =228224.6× −85.7612 79.412 −280 −10.588 −155.3202 100.5882 10.1210 60 −45 −60 79.4118 −155.32 60 −280 100.588 −45 −10.588 10.1211 −60 280 −100.588 0 −100.588 175.5621 0 0 0 45 ww u1 w1 u2 51.076 −31.817 −31.817 74.173 𝐾 1= −19.573 16.110 18.124 −63.903 −35.448 22.597 13.693 10.270 w2 −19.573 16.110 18.755 −2.416 2.310 −13.693 w.E asy u4 w4 18.124 −35.448 13.693 −63.903 22.597 −10.270 −2.416 2.310 −13.693 63.903 −22.597 0 −22.597 40.068 0 0 0 10.270 En gin For element (2) (Nodal displacements, u2, w2, u3, w3, u4, w4) Co ordinates At node 2 Z eer i W4 (r3,z3) U4 ng. r1=50mm z1=0mm At node 3 4 net 15 mm r1=70mm z1=0mm Element At node 4 r1=70mm W2 z1=15mm U2 𝑟1 +𝑟2 +𝑟3 We know that, 𝑤ℎ𝑒𝑟𝑒 𝑟 = 2 3 = 2 3 (r,z1) W3 U3 (r2,z2) 50+70+70 3 50mm r = 63.3333mm, 𝑧 +𝑧2 +𝑧3 𝑧= 1 3 = 0+0+15 3 ;z= 5 mm 70 mm 15 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 1 Area of the triangle element = 2 × 𝐵𝑟𝑒𝑎𝑑𝑡ℎ × 𝐻𝑒𝑖𝑔ℎ𝑡 1 = 2 × 20 × 15 A = 150 mm We know that, Stiffness matrix for axisymmetric triangular element (2), 𝐾 2 =2 𝜋 rA 𝐵 T 𝐷 B 1−𝜈 𝜈 𝜈 0 𝐸 Stress strain relationship matrix 𝐷 = 1+𝜈 1−2𝜈 ww 2𝑋10 5 Stress strain relationship matrix 𝐷 = 1+0.3 1−(2×0.3) w.E 2×10 5 = 0.5 asy 0.7 0.3 0.3 0 0.3 0.7 0.3 0 𝜈 𝜈 0.7 0 0 0 0 0.2 𝜈 1−𝜈 𝜈 0 𝜈 𝜈 1−𝜈 0 En eer i B= 1 2𝐴 0 𝛾1 𝛾1 𝛽1 𝛽2 𝛾 𝑧 + 𝛽2 + 2𝑟 𝑟 𝛼2 0 𝛾2 𝛼1 = 𝑟2 𝑧3 − 𝑟3 𝑧2 𝛼1 = 70 × 15 − 70 × 0 𝛽1 = 𝑧2 − 𝑧3 𝛾1 = 𝑟3 − 𝑟2 0 0 𝛽3 𝛾 𝑧 + 𝛽3 + 3𝑟 𝑟 𝛼3 𝛾2 𝛽2 0 𝛾3 𝛼2 = 𝑟3 𝑧1 − 𝑟1 𝑧3 1−(2×0.3) 2 𝛼2 = −750𝑚𝑚2 𝛽2 = 𝑦3 − 𝑦1 𝛾2 = 𝑟1 − 𝑟3 0 0 net 𝛾3 𝛽3 𝛼3 = 𝑟1 𝑧2 − 𝑟2 𝑧1 𝛼2 = 70 × 0 − 50 × 15 𝛼1 = 1050 𝑚𝑚2 0 0 0 ng. We know that, strain-Displacement matrix 0 0 2 gin 0.7 0.3 0.3 0 0.3 0.7 0.3 0 =384.6153×103 𝜈 𝜈 0.7 0 0 0 0 0.2 𝛽1 𝛾₁𝑧 + 𝛽₁ + 𝑟 𝑟 1− 2𝜈 1 − 0.3 0.3 0.3 0.3 1 − 0.3 0.3 𝜈 𝜈 1 − 0.3 0 0 0 0.7 0.3 0.3 0 0.3 0.7 0.3 0 =384.6153×103 𝜈 𝜈 0.7 0 0 0 0 0.2 𝛼₁ 0 0 0 𝛼3 = 50 × 0 − 70 × 0 𝛼3 = 0 𝛽3 = 𝑦1 − 𝑦2 𝛾3 = 𝑟2 − 𝑟1 16 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝛼₁ 𝑟 𝛾₁𝑧 + 𝛽₁ + 𝑟 𝛼2 𝛾 𝑧 = 𝛽1 = 0 − 15 𝛾1 = 70 − 70 𝛽2 = 15 − 0 𝛾2 = 50 − 70 𝛽3 = 0 − 0 𝛾3 = 70 − 50 𝛽1 = −15𝑚𝑚 𝛾1 = 0 𝛽2 = 15𝑚𝑚 𝛾2 = −20𝑚𝑚 𝛽3 = 0 𝛾3 = −20𝑚𝑚 1050 63.333 −750 + (−15) + 0 (−20×5) 𝑟 + 𝛽2 + 2𝑟 = 𝑟 + 𝛽3 + 3𝑟 = 0 + 0 + 63.333 𝛼3 𝛾 𝑧 63.333 𝛼₁ =1.579 mm + 15 + 63.333 (20×5) = 1.579 mm =1.579mm 𝛾₁𝑧 𝛼 𝛾 𝑧 Substitute 𝛽1, 𝛽2 , 𝛽3, 𝑟 + 𝛽₁ + 𝑟 , 𝑟2 + 𝛽2 + 2𝑟 , 𝛼3 𝑟 𝛾 𝑧 + 𝛽3 + 3𝑟 , 𝛾1 , 𝛾2, 𝛾3 and A values in equations no 10, we get, −15 1 1.579 B = 2×150 0 0 ww 0 0 0 −15 w.E 0 15 0 1.579 −20 0 15 −20 −15 0 1.579 0 B =3.333 × 10−3 0 0 0 −15 asy 15 1.579 0 −20 0 1.579 0 20 0 0 20 0 0 0 0 1.579 −20 0 15 20 En gin D B = 384.6153×103 0.7 0.3 0.3 0 −15 0.3 0.7 0.3 0 1.579 × 3.333 × 10−3 0.3 0.3 0.7 0 0 0 0 0 0.2 0 −10.0263 3 −3.3947 D B = 1.282×10 −4.0263 0 0 10.9737 0 5.6053 0 4.9737 −3 −4 0 0 20 0 eer i 0 0 0 −15 0 15 0 1.579 −20 0 15 −20 −6 0.4737 6 −6 1.1053 6 −14 0.4737 14 3 4 0 0 1.579 0 20 ng. 0 0 20 0 net We know that −15 0 1.579 0 B =3.333 × 10−3 0 0 0 −15 15 1.579 0 −20 −15 1.579 0 0 T −3 15 1.579 B =3.333 × 10 0 0 0 1.579 0 0 0 0 0 1.579 −20 0 15 20 0 0 20 0 0 0 0 −15 0 −20 −20 15 0 20 20 0 17 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net −10.0263 −3.3947 D B B T =1.282×103 −4.0263 0 0 0 0 −3 10.9737 5.6053 4.9737 −4 −15 1.579 0 0 15 1.579 0 0 0 1.579 0 0 −6 0.4737 6 −6 1.1053 6 3.333 × 10−3 −14 0.4737 14 3 4 0 0 0 0 −15 0 −20 −20 15 0 20 20 0 145.034 0 −155.755 0 45 60 −155.755 60 253.456 D B B T =4.2733 80.526 −45 −159.474 −5.360 −60 −71.149 −80.526 0 99.474 ww w w ww.E .Eaa ssyyE nggi 80.526 −5.360 −80.526 −45 −60 0 −159.474 −71.149 99.474 325 50.256 −280 50.526 81.745 9.474 −280 9.474 280 Substitute D B B 𝑇 value in equ no 8 145.034 0 −155.755 0 45 60 −155.755 60 253.456 𝐾 2 =2 𝜋 ×63.333×150×4.2733× 80.526 −45 −159.474 −5.360 −60 −71.149 −80.526 0 99.474 80.526 −5.360 −80.526 −45 −60 0 −159.474 −71.149 99.474 325 50.256 −280 50.526 81.745 9.474 −280 9.474 280 inneee erirni ng.gn. 145.034 0 −155.755 0 45 60 −155.755 60 253.456 𝐾 2 =255.074X103 80.526 −45 −159.474 −5.360 −60 −71.149 −80.526 0 99.474 36.994 0 −39.729 0 11.478 15.304 −39.729 15.304 64.650 𝐾 2 =106 20.540 −11.478 −40.678 −1.367 −15.304 −18.148 −20.540 0 25.373 80.526 −5.360 −80.526 −45 −60 0 −159.474 −71.149 99.474 325 50.256 −280 50.526 81.745 9.474 −280 9.474 280 ente 20.540 −1.367 −20.540 −11.478 −15.304 0 −40.678 −18.148 25.373 82.899 12.877 −71.421 12.877 20.851 2.417 −71.421 2.417 71.421 18 Downloaded From: www.EasyEngineering.net t Downloaded From: www.EasyEngineering.net Assemble the equations. Global stiffness matrix, [ K ] = 51.076 +0 -31.817 +0 -19.573 +0 -31.817 +0 74.173 +0 16.110 +0 18.124 +0 0 -63.903 +0 0 0 w.E ww -35.448 +0 13.693+ 0 0 22.597+0 -10.270+0 -19.573 +0 16.110 +0 18.755+ 36.994 18.124+0 0 63.903+0 -2.416 + 0 -2.416 63.903 +0 +11.478 0+ 0+ (-39.729) 15.304 20.540+0 11.478+0 2.310-22.597 1.367 -15.304 -13.693 0+0 -20.540 asy 0 -35.448+0 13.693+0 0 0 22.957+0 -10.270+0 -39.729 +0 20.540 +0 2.3101.367 -13.693 -20.540 -22.957 -15.304 0+ (-18.148) 12.887+0 0+0 40.068+20 .851 0+2.417 0+2.417 Global stiffness matrix, [ K ] = En 51.076 -31.817 -19.573 -31.817 74.173 16.110 -19.573 16.110 55.749 18.124 -63.903 -2.416 -35.448 22.957 0.943 13.693 -10.270 -34.233 18.124 0 0 -35.448 13.693 -63.903 0 0 22.597 -10.270 -2.416 (-39.729) 20.540 0.943 -34.233 75.381 15.304 -11.478 -38.261 0 0+15.304 0 -11.478 0+64.650 0+ (-40.678) 82.899+0 40.678+0 0 0+12.887 -18.148 0+25.373 0-71.421 gin eer i 0 0 -39.729 0 0 20.540 15.304 64.650 -40.678 18.148 25.373 -11.478 -40.678 82.899 12.887 71.421 ng. 0+25.373 -71.421+0 10.270 +71.421 net -38.261 -18.148 12.887 60.919 2.417 0 25.373 -71.421 2.417 81.691 We know that 𝐹 =𝐾 𝑈 51.076 −31.817 −19.573 𝐹1𝑢 −31.817 74.173 16.110 𝐹2𝑢 −19.5573 16.110 55.759 𝐹3𝑢 18.124 −63.903 −2.416 6 =10 𝐹4𝑢 0 0 −39.729 0 0 20.540 𝐹5𝑢 −35.448 22.957 0.943 𝐹6𝑢 13.693 −10.270 −34.233 18.124 0 0 −63.903 0 0 −2.416 −39.729 20.540 75.381 15.304 −11.478 15.304 64.650 −40.678 −11.478 −40.678 82.899 −38.261 −18.148 12.887 0 25.373 −71.421 −35.448 13.693 22.957 −10.270 0.943 −34.233 −38.261 0 −18.148 25.373 12.887 −71.421 60.919 2.417 2.417 81.691 19 Downloaded From: www.EasyEngineering.net 𝑢1 𝑤1 𝑢2 𝑤2 𝑢3 𝑤3 𝑢4 𝑤4 Downloaded From: www.EasyEngineering.net Forces we know that F1u = F2u = 2 𝑃𝜋 𝑟 𝑒𝑙 𝑒 2 = 2× 𝜋×50×15×4 2 = 9424.77 N The remaining forces are zero F1w, F2w, F3u, F3w, F4w, are zero. Displacements 1. Node 1 is moving in r direction. u1 0 but w1 =0 2. Node 2 is moving in r direction. u2 0 but w2 =0 3. Node 3 & 4 are fixed. So u3, w3 u4 and w4 are zero. Substitute nodal force and nodal displacements values in eqn 12 ww 18.124 0 0 51.076 −31.817 −19.573 −35.448 13.693 −63.903 0 0 −31.817 74.173 16.110 22.957 −10.270 9424.77 −2.416 −39.729 20.540 −19.5573 16.110 55.759 0.943 −34.233 0 9424.77 18.124 −63.903 −2.416 −38.261 0 75.381 15.304 −11.478 =106 × 0 0 0 −39.729 15.304 64.650 −40.678 −18.148 25.373 0 12.887 −71.421 0 0 20.540 −11.478 −40.678 82.899 0 −38.261 −18.148 12.887 60.919 2.417 −35.448 22.957 0.943 0 25.373 −71.421 2.417 81.691 13.693 −10.270 −34.233 w.E asy En 𝑢1 0 𝑢2 0 0 0 0 0 Delete second row, second column, fourth row, fourth column, fifth row, fifth column, sixth row, sixth column, seventh row, seventh column, and eighth row and eight column of the above matrix. Hence the Equation reduces to gin 9424.77 51.706 =106 9424.77 −19.5573 6 −19.5573 55.759 𝑢1 𝑢2 eer i X 9424.77 = 10 (51.706u1-19.573u2) 9424.77 = 106 (-19.573u1-55.749u2) Above equations we solving and we get u1 =2.88×10-4mm ng. u2 =2.70×10-4mm net RESULTS DISPLACEMENTS u1 =2.88×10-4mm w1=0 u2 =2.70×10-4mm w2=0 u3 =0 w3=0 u4 =0 w4=0 20 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 5. DERIVE THE EXPRESSION FOR STRAIN-DISPLACEMENT RELATIONSHIP FOR AXISYMMETRIC ELEMENT. Shape function are given below U = N1u1+N2u2+N3u3 --------------------------- 1 W = N1w1+N2w2 +N3w3 --------------------- 2 𝜕𝑢 Radial strain er = 𝜕𝑟 Eqn 1 d.w.r to “r “ 𝜕𝑢 𝜕𝑁 𝜕𝑁 𝜕𝑁 ww er = 𝜕𝑟 = 𝜕𝑟1 𝑢1 + 𝜕𝑟2 𝑢2 + 𝜕𝑟3 𝑢3 ------------------- 3 𝑢 Circumferential strain e Ɵ = 𝑟 w.E 𝑁 𝑁 𝑁 e Ɵ = 𝑟1 𝑢1 + 𝑟2 𝑢2 + 𝑟3 𝑢3 --------- 4 asy 𝜕𝑤 Longitudinal strain ez = En 𝜕𝑁1 ez = 𝜕𝑁 gin 𝜕𝑤 Shear strain ϒ rz = 𝜕𝑧 + 𝜕𝑟 𝜕𝑁 𝜕𝑁 𝑤1 + 𝜕𝑧2 𝑤2 + 𝜕𝑧3 𝑤3 ---------- 5 𝜕𝑧 𝜕𝑢 𝜕𝑧 𝜕𝑁 𝜕𝑁 eer i 𝜕𝑁 𝜕𝑁 𝜕𝑁 ng. ϒ rz = 𝜕𝑧1 𝑢1 + 𝜕𝑧2 𝑢2 + 𝜕𝑧3 𝑢3 + 𝜕𝑟1 𝑤1 + 𝜕𝑟2 𝑤2 + 𝜕𝑟3 𝑤3 ------ 6 Arranging equation 3, 4, 5 & 6 in matrix form 𝜕𝑁1 𝑒𝑟 𝑒𝜃 = 𝑒𝑧 𝛾𝑟𝑧 0 𝜕𝑟 𝑁1 0 𝑟 𝜕𝑁1 0 𝜕𝑁1 𝜕𝑧 𝜕𝑁1 𝜕𝑧 𝜕𝑟 𝜕𝑁2 𝜕𝑟 𝑁2 𝑟 0 𝜕𝑁3 0 𝜕𝑟 𝑁3 0 0 𝑟 𝜕𝑁2 0 𝜕𝑁3 0 𝜕𝑁2 𝜕𝑧 𝜕𝑁2 𝜕𝑁3 𝜕𝑧 𝜕𝑁3 𝜕𝑧 𝜕𝑟 𝜕𝑧 𝜕𝑟 net 𝑢1 𝑤1 𝑢2 𝑤2 ------------- 7 𝑢3 𝑤3 Shape function 1 𝑁1 = 2𝐴 𝛼1 + 𝛽1 𝑟 + 𝛾1 𝑧 ; 1 𝑁2 = 2𝐴 𝛼2 + 𝛽2 𝑟 + 𝛾2 𝑧 ; 1 𝑁3 = 2𝐴 𝛼3 + 𝛽3 𝑟 + 𝛾3 𝑧 ; 21 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝜕𝑁1 𝛽1 = 𝜕𝑟 2𝐴 𝑁1 1 ∝1 𝛾1 𝑧 = + 𝛽1 + 𝑟 2𝐴 𝑟 𝑟 𝜕𝑁1 𝛾1 = 𝜕𝑧 2𝐴 𝜕𝑁2 𝛽2 = 𝜕𝑟 2𝐴 𝑁2 1 ∝2 𝛾2 𝑧 = + 𝛽2 + 𝑟 2𝐴 𝑟 𝑟 𝜕𝑁2 𝛾2 = 𝜕𝑧 2𝐴 ww 𝜕𝑁3 𝛽3 = 𝜕𝑟 2𝐴 w.E 𝑁3 1 ∝3 𝛾3 𝑧 = + 𝛽3 + 𝑟 2𝐴 𝑟 𝑟 asy 𝜕𝑁3 𝛾3 = 𝜕𝑧 2𝐴 En gin Above values substitute in eqn 7 𝛽1 𝑒𝑟 𝑒𝜃 = 𝑒𝑧 𝛾𝑟𝑧 𝛼₁ 𝛾₁𝑧 1 + 𝛽₁ + 𝑟 𝑟 2𝐴 0 {e} = [B]{u} 𝛾1 𝛽1 = 𝑧2 − 𝑧3 𝛾1 = 𝑟3 − 𝑟2 𝛼1 = 𝑟2 𝑧3 − 𝑟3 𝑧2 1 [B] = 2𝐴 𝛽1 𝛾₁𝑧 + 𝛽₁ + 𝑟 𝑟 0 𝛾1 𝛼₁ 0 0 𝛾1 𝛽1 0 0 𝛾1 𝛽1 𝛽2 𝛼2 𝛾2 𝑧 + 𝛽2 + 𝑟 𝑟 0 𝛾2 𝛽2 = 𝑧3 − 𝑧1 𝛾2 = 𝑟1 − 𝑟3 𝛼2 = 𝑟3 𝑧1 − 𝑟1 𝑧3 𝛽2 𝛾 𝑧 + 𝛽2 + 2𝑟 𝑟 0 𝛾2 𝛼2 eer i 0 0 𝛾2 𝛽2 0 0 𝛾2 𝛽2 𝛼3 𝛾3 𝑧 + 𝛽3 + 𝑟 𝑟 0 𝛾3 0 0 𝑢1 𝑤1 𝑢2 𝑤2 𝑢3 𝑤3 net 𝛾3 𝛽3 𝛽3 = 𝑧1 − 𝑧2 𝛾3 = 𝑟2 − 𝑟1 𝛼3 = 𝑟1 𝑧2 − 𝑟2 𝑧1 𝑏3 𝛾 𝑧 + 𝛽3 + 3𝑟 𝑟 0 𝛾3 𝛼3 ng. 𝑏3 0 0 𝛾3 𝛽3 22 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net ww w.E a UNIT 5 syE ngi nee rin g.n et Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net UNIT V ISOPARAMETRIC FORMULATION PART A 1. What do you mean by uniqueness of mapping? It is absolutely necessary that a point in parent element represents only one point in the isoperimetric element. Some times, due to violent distortion it is possible to obtain undesirable situation of nonuniqueness. Some of such situations are shown in Fig. If this requirement is violated determinant of Jacobiam matrix (to be explained latter) becomes negative. If this happens coordinate transformation fails and hence the program is to be terminated and mapping is corrected. ww w.E Non Uniqueness of Mapping 2. What do you mean by iso parametric element?(April/May 2011) asy If the shape functions defining the boundary and displacements are the same, the element is called as isoparametric element and all the eight nodes are used in defining the geometry and displacement. En gi nee rin g.n e 3. What do you mean by super parametric element? The element in which more number of nodes are used to define geometry compared to the number of nodes used to define displacement are known as superparametric element. t 4. What do you mean by sub parametric element? The fig shows subparametric element in which less number of nodes are used to define geometry compared to the number of nodes used for defining the displacements. Such elements can be used advantageously in case of geometry being simple but stress gradient high. 1 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 5. What do you mean by iso parametric formulation?(April/May 2011) The principal concept of isoparametric finite element formulation is to express the element coordinates and element displacements in the form of interpolations using the natural coordinate system of the element. These isoparametric elements of simple shapes expressed in natural coordinate system, known as master elements, are the transformed shapes of some arbitrary curves sided actual elements expressed in Cartesian coordinate system. ww 6. What is a Jacobian matrix of transformation?(April/May 2011) It‟s the transformation between two different co-ordinate system. This transformation is used to evaluate the integral expression involving „x‟ interms of expressions involving ε. w.E asy XB 1 xA 1  f ( x)dx   f ( )d En gi The differential element dx in the global co-ordinate system x is related to differential element dε in natural co-ordinate system ε by dx = dx/ dε . dε nee dx = J . dε 𝐽 Jacobian matrix of transformation J =dx/ dε = 11 𝐽21 𝐽12 𝐽22 7. Differentiate the serendipity and langrangian elements Serendipity elements langrangian elements rin g.n e In discretized element In discretized element, if nodes If nodes lies on corner, then the are present in both centre of element element are known as serendipity and corner are known as langrangian elements. elements. t 8. Explain Gauss quadrature rule.(Nov/Dec 2012), (April/May 2011) The idea of Gauss Quadrature is to select “n” Gauss points and “n” weight functions such that the integral provides an exact answer for the polynomial f(x) as far as possible, Suppose if it is necessary to evaluate the following integral using end point approximation then 1 I=  f ( x)dx 1 2 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net The solution will be 1  f ( x)dx  w f ( x )  w f ( x )  .........  w f ( x ) 1 1 2 2 n n 1 w1,w2,…………..…., wnare weighted function, x1,x2……………….., xnare Gauss points 9. What are the differences between implicit and explicit direct integration methods? Implicit direct integration methods: (i) Implicit methods attempt to satisfy the differential equation at time „t‟ after the solution at time “t∆t”is found (ii) These methods require the solution of a set of linear equations at each time step. (iii) Normally larger time steps may be used. (iv) Implicit methods can be conditionally or unconditionally stable. ww w.E Explicit direct integration methods: (i) asy These methods do not involve the solution of a set of linear equations at each step. (ii) Basically these methods use the differential equations at time „t‟ to predict a solution at time “t+∆t” En gi (iii) Normally smaller time steps may be used (iv) All explicit methods are conditionally stable with respect to size of time step. nee (v) Explicit methods initially proposed for parabolic PDES and for stiff ODES with widely separated time constants. 10. State the three phases of finite element method. The three phases of FEM is given by, (i) Preprocessing (ii) Analysis (iii) Post Processing rin g.n e t 11. List any three FEA software.(Nov/Dec 2014) The following list represents FEA software as, (i) ANSYS (ii) NASTRAN (iii) COSMOS 3 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net PART-B 1. A four noded rectangular element is shown in Fig. Determine the following 1. jacobian matrix 2. Strain – Displacement matrix 3. Element Stresses. ww w.E asy En gi T Take E = 2 10 N/mm ; v = 0.25 ; u = 0, 0, 0, 0.003, 0.004, 0.006, 0.004, 0, 0 5 2 Assume the plane Stress condition. Given Data nee Cartesian co – ordinates of the points 1,2,3 and 4 𝑥1 = 0; 𝑦1 = 0 𝑥2 = 2; 𝑥3 = 2; 𝑥4 = 0; 𝑦2 = 0 𝑦3 = 1 𝑦4 = 1 Young‟s modulus, E = 2 105 N/mm2 Poisson‟s ratio v = 0.25 0 0 0.003 0.004 Displacements, u = 0.006 0.004 0 0 Natural co-ordinates , ε = 0 ,  = 0 ε=0;=0 rin g.n e t To find: 1. Jacobian matrix, J 2. Strain – Displacement matrix [B] 3. Element Stress σ. 4 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Formulae used 𝐽11 𝐽12 𝐽21 𝐽22 J = J22 −J12 0 0 1 1 0 −J21 J11  𝐵 = 𝐉 0 4 −J21 J11 J22 −J12 0 0 0 −(1 − ) (1 − ) (1 + ) 0 −(1 + ) 0 0 0 0 −(1 − 𝜀) −(1 + 𝜀) (1 + 𝜀) (1 − 𝜀) −(1 − ) (1 − ) −(1 + ) 0 0 0 (1 + ) 0 −(1 − 𝜀) −(1 + 𝜀) 0 (1 + 𝜀) (1 − 𝜀) 0 0 0 ww Solution :Jacobian matrix for quadrilateral element is given by, w.E 𝐽11 𝐽12 𝐽21 𝐽22 J = asy Where , 1 J11 = 4 −(1 − )𝑥1 + (1 − )𝑥2 +(1 + )𝑥3 −(1 + )𝑥4 J12 = 1 4 En gi (1) −(1 − )𝑦1 + (1 − )𝑦2 +(1 + )𝑦3 −(1 + )𝑦4 nee 1 (2) J21 = 4 −(1 − 𝜀)𝑥1 − (1 + 𝜀)𝑥2 +(1 + 𝜀)𝑥3 +(1 − 𝜀)𝑥4 1 J22 = 4 −(1 − 𝜀)𝑦1 − (1 + 𝜀)𝑦2 +(1 + 𝜀)𝑦3 +(1 − 𝜀)𝑦4 (3) (4) rin g.n e Substitute 𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑦1, 𝑦2, 𝑦3, 𝑦14, ε and  values in equation (1), (2),(3) and (4) 1 (1) J11 = 4 0 + 2 + 2 − 0 𝐉𝟏𝟏 = 1 (2) t 1 J12 = 4 0 + 0 + 1 − 1 J12 = 0 (3) 1 J21 = 4 0 − 2 + 2 − 0 J21 = 0 (4) 1 J22 = 4 −0 − 0 + 1 + 1 J22 = 0.5 5 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net J = 𝐽11 𝐽12 𝐽21 𝐽22 J = Jacobian matrix 1 0 0 0.5 (5) J = 10.5- 0 J = 0.5 We Know that, Strain – Displacement matrix for quadrilateral element is, J22 −J12 0 0 1 0 0 −J21 J11  4 −J21 J11 J22 −J12 1 𝐵 = 𝐉 ww 0 0 0 −(1 − ) (1 − ) (1 + ) 0 −(1 + ) 0 0 0 0 −(1 − 𝜀) −(1 + 𝜀) (1 + 𝜀) (1 − 𝜀) −(1 − ) (1 − ) −(1 + ) 0 0 0 (1 + ) 0 −(1 − 𝜀) −(1 + 𝜀) 0 (1 + 𝜀) (1 − 𝜀) 0 0 0 w.E asy Substitute 𝐉𝟏𝟏 , 𝐉𝟏𝟐, 𝐉𝟐𝟏, 𝐉𝟐𝟐 𝐉 , 𝜺 𝐚𝐧𝐝  𝐯𝐚𝐥𝐮𝐞𝐬 1 𝐵 = 0.5 0.5 0 0 0 0 0 0 1 0.5 −1 0 1 0 10−1 0 0 1 −1 0 −1 0 10 1 0 1 4 0 −1 0 1 01 0 −1 1 0 −1 0 −101 0 1 En gi nee −0.5 0 0.5 0 0.5 0 −0.5 0 1 𝐵 = 0.54 0 −1 0 −1 0 1 0 1 −1 −0.5−10.5 1 0.5 1 −0.5 −1 0 1 0 10−1 0 0.5 = 0.54 0 −2 0 −202 0 2 −2−1−2 1 21 2 −1 rin g.n e t −1 0 1 0 10−1 0 𝐵 = 0.25 0 −2 0 −202 0 2 −2−1−2 1 21 2 −1 We know that, Element stress, σ = 𝐃 𝑩 𝒖 For plane stress condition, 6 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Stress- strain relationship matrix, D = 1𝑣 0 𝑣 1 0 1−𝑣 2 0 0 1−𝑣 𝐸 2 = 2 10 5 1− (0.25)2 1 0.25 0 0 0.25 1 1−0.25 0 0 2 1 0.25 0 = 213.33 103 0.25 1 0 0 0.375 0 41 0 = 213.33103 0.25 1 4 0 0 0 1.5 ww 41 0 = 53.333103 1 4 0 0 0 1.5 w.E Substitute 𝐷 , 𝐵 and 𝑢 asy En gi −1 0 1 0 10−1 0 41 0 σ = 53.333103 1 4 0 0.25 0 −2 0 −202 0 2 0 0 1.5 −2−1−2 1 21 2 −1 0 0 0.003 0.004 0.006 0.004 0 0 nee −4 2 4 −24 2 −4 2 = 53.333103 0.25 −1 −8 1 −81 8 −1 8 −3−1.5−31.531.5 3 −1.5 =13.33310 3 0 0 0.003 0.004 0.006 0.004 0 0 rin g.n e t 0 + 0 + 4 × 0.003 + −2 × 0.004 + 4 × 0.006 + 2 × 0.004 + 0 + 0 0 + 0 + 1 + 0.003 + −8 × 0.004 + 1 × 0.006 + 8 × 0.004 + 0 + 0 0 + 0 + −3 × 0.003 + 1.5 × 0.004 + 3 × 0.006 + 1.5 × 0.004 + 0 + 0 0.036 𝜎 = 13.333103 0.009 0.021 7 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 480 𝜎 = 120 N/m2 280 Result : J = 0.5 480 𝜎 = 120 N/m2 280 2. For the isoparametric quadrilateral element shown in Fig. the Cartesian co-ordinate of point P are (6,4). The loads 10KN and 12KN are acting in x and y direction on the point P. Evaluate the nodal equivalent forces. ww w.E asy En gi nee Givendata : Cartesian coordinates of point P, X = 6; y=4 rin g.n e t The Cartesian co-ordinates of point 1,2,3 and 4 are 𝑥1 = 2; 𝑦1 = 1 𝑥2 = 8; 𝑦2 = 4 𝑥3 = 6; 𝑦3 = 6 𝑥4 = 3; 𝑦4 = 5 Loads ,F𝑥 = 10𝐾𝑁F𝑦 = 12𝐾𝑁 To find : Nodal equivalent forces for x and y directions, 8 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net i,e., F1𝑥 , F2𝑥 , F3𝑥 , F4𝑥 , F1𝑦 , F2𝑦 , F3𝑦 , F4𝑦 Formulae Used 1 N1 = 4 (1-ε) (1-) 1 N2 = 4 (1+ ε) (1-  ) 1 N3 = 4 (1+ ε) (1+) 1 N4 = 4 (1-ε) (1+) ww Fx Element force vector, F e = N T F y w.E solution: Shape functions for quadrilateral elements are, 1 asy N1 = 4 (1-ε)(1-)(1) 1 N2 = 4 (1+ ε) (1-  ) 1 N3 = (1+ ε) (1+) 4 1 N4 = 4 (1-ε) (1+) En gi (2) nee Cartesian co-ordinates of the point,P(x,y) (3) (4) 𝑥 = N1 𝑥1 +N2 𝑥2 + N3 𝑥3 + N4 𝑥4 𝑦 = N1 𝑦1 +N2 𝑦2 + N3 𝑦3 + N4 𝑦4 rin g.n e (5) (6) Substitute 𝑥,𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 , 𝑁1 , 𝑁2 , 𝑁3 , 𝑎𝑛𝑑 𝑁4 values in equation. t 1 6 = 4 [(1-ε) (1-) 2 +(1+ε) (1- )8 + (1+ ε) (1+)6 +(1 - ε) (1+)3] 24= [(1--ε+ε)2+(1-+ε-ε)8+(1++ε+ε)6+(1+-ε-ε)3] 24 = 19-+9ε-3ε 5 = -+9ε - 3ε 9ε -  - 3ε = 5 (7) Substitute 𝑦,𝑦1 , 𝑦2 , 𝑦3 , 𝑦4 , 𝑁1 , 𝑁2 , 𝑁3 , 𝑎𝑛𝑑 𝑁4 values in equation. 9 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 1 4 = 4 [(1-ε) (1-) 1 +(1+ε) (1- )4 + (1+ ε) (1+)6 +(1 - ε) (1+)5] 16 = [1--ε+ε+4-4+4ε-4ε+6+6+6ε+6ε+5+5-5ε-5ε] 16= [16+6+4ε-2ε] 4ε + 6 - 2ε = 0 (8) Equation (7) multiplied by 2 and equation (8) multiplied by (-3). 18ε - 2 - 6ε = 10 (9) -12ε - 18 + 6ε = 0 ww (10) 6ε – 20  = 10 w.E -20  = 10 - 6ε asy 20 = 6ε -10 = 6𝜀−10 En gi 20  = 0.3ε – 0.5 nee Substituting  value in equation (7), 9ε – (0.3ε – 0.5) - 3ε (0.3ε – 0.5) = 5 10.2ε – 0.9ε2 – 4.5 = 0 0.9ε2 - 10.2ε + 4.5 = 0 ε= = 10.2± (−10.2)2 −4 0.9 (4.5) 2(0.9) 10.2−9.372 (11) rin g.n e t 1.8 ε = 0.46 Substitute ε and  values in equation (1),(2),(3) and (4) (1) N1 = 1 4 (1 - 0.46) (1+ 0.362) N1 = 0.18387 (2) N2 = 1 4 (1 + 0.46) (1+ 0.362) 10 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net N2 = 0.49713 N3 = (3) 1 4 (1 + 0.46) (1 - 0.362) N3 = 0.23287 N4 = (4) 1 4 (1 - 0.46) (1 - 0.362) N3 = 0.08613 We know that, Fx Element force vector, F e = N T F ww (12) y w.E F1𝑥 F2𝑥 F3𝑥 F4x 𝑁1 𝑁2 = 𝑁3 𝑁4 F1𝑥 F2𝑥 F3𝑥 F4x 0.18387 0.49713 = 0.23287 0.08613 F1𝑥 F2𝑥 F3𝑥 F4x 1.8387 4.9713 = KN 2.3287 0.8613 F𝑥 asy En gi 10 nee rin g.n e t Similarly, F1𝑦 F2𝑦 F3𝑦 F4y 𝑁1 𝑁2 = 𝑁3 𝑁4 F𝑦 11 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net F1𝑦 F2𝑦 F3𝑦 F4y 0.18387 0.49713 = 0.23287 0.08613 F1𝑦 F2𝑦 F3𝑦 F4y 2.20644 5.96556 = KN 2.79444 1.03356 12 Result: Nodal forces for x directions, ww F1𝑥 F2𝑥 F3𝑥 F4x w.E 1.8387 4.9713 = KN 2.3287 0.8613 Nodal forces for y directions, asy F1𝑦 F2𝑦 F3𝑦 F4y 4. 2.20644 5.96556 = KN 2.79444 1.03356 En gi nee Derive the shape function for the Eight Noded Rectangular Element Consider a eight noded rectangular element is shown in fig. It belongs to the serendipity family of elements. It consists of eight nodes, which are located on the boundary. rin g.n e We know that, shape function N1 = 1 at node 1 and 0 at all other nodes. t 12 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net N1=0 at all other nodes N1 has to be in the form of N1 =C(1- ε)(1-)(1+ε+) (1) Where C is constant Substitute ε = -1 and  = -1 in equation (1) N1 = C (1+1)(1+1)(-1) 1 = -4C ww 1 C =-4 Substitute C value in equation w.E 1 N1= -4 (1+ ε) (1 +) (1+ε+) asy (2) At node 2 :(Coordinates ε =1,= -1) Shape Function N2 = 1 at node 2 En gi N2 = 0 at all other nodes N2has to be in the form of N2 =C(1 +ε)(1-)(1-ε+) Substitute ε = 1 and  = -1 in equation (3) N2 = C (1+1) (1+1) (-1) nee 1 = -4C 1 C =-4 (3) rin g.n e Substitute C value in equation (3) 1 N2= -4 (1+ ε) (1 - ) (1- ε +) t (4) At node 3 :(Coordinates ε =1,= 1) Shape Function N3 = 1 at node 3 N3 = 0 at all other nodes N3has to be in the form of N3 =C(1+ε)(1+)(1- ε - ) (5) Substitute ε = 1 and  = 1 in equation (5) 13 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net N3 = C (1+1) (1+1) (-1) 1 = -4C 1 C =-4 Substitute C value in equation (5) 1 N3= − 4 (1+ ε) (1+ ) (1- ε - ) (6) At node 4 :(Coordinates ε =- 1,= 1) Shape Function N4 = 1 at node 4 ww N4 = 0 at all other nodes N4 has to be in the form of N4 =C(1- ε)(1 + )(1+ε - ) w.E (7) Substitute ε = -1 and  = 1 in equation (7) asy N4 = C (1+1) (1+1) (-1) 1 = -4C 1 C = −4 Substitute C value in equation (3) 1 En gi N4= - 4 (1- ε) (1 + ) (1+ ε -) nee Now , we define N5,N6,N7 and N8 at the mid points. At node 5 :(Coordinates ε = - 1,= - 1) Shape Function N5 = 1 at node 5 rin g.n e (8) N5 = 0 at all other nodes t N5has to be in the form of N5 =C(1- ε)(1 -)(1+ε ) N5 = C (1- ε2)(1 - ) (9) Substitute ε = 0 and  = -1 in equation (9) N5 = C (1-0)(1+1) 1 = 2C 1 C=2 14 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Substitute C value in equation (9) 1 N5= 2 (1- ε2)(1 - ) (10) At node 6 :(Coordinates ε = 1,= - 1) Shape Function N6 = 1 at node 6 N6 = 0 at all other nodes N6 has to be in the form of N6 =C (1+ε)(1 - )(1+ ) N6 = C (1 + ε)(1 - 2) (11) Substitute ε = 1 and  = 0 in equation (11) ww N6 = C (1+1) (1 - 0) 1 = 2C w.E 1 C=2 Substitute C value in equation (11) N6 = asy 1 2 (1+ ε)(1 - 2) (12) En gi At node 7 :(Coordinates ε = 1,= 1) Shape Function N7 = 1 at node 7 nee N7 = 0 at all other nodes N7 has to be in the form of N7 =C (1+ε)(1 + )(1- ε ) N7 = C (1 – ε2)(1 + ) rin g.n e (13) Substitute ε = 0 and  = 1 in equation (12) N7 = C (1-0) (1 + 1) 1 = 2C t 1 C=2 Substitute C value in equation (13) N7 = 1 2 (1 – ε2)(1 + ) (14) At node 8 :(Coordinates ε = -1,= 1) Shape Function N8 = 1 at node 8 N8 = 0 at all other nodes N8 has to be in the form of N8 =C (1-ε)(1 + )(1-  ) 15 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net N8 = C (1 – ε)(1 -2) (15) Substitute ε = -1 and  = 0 in equation (15) N8 = C (1+1) (1 - 0) 1 = 2C 1 C=2 Substitute C value in equation (15) 1 N8 = 2 (1 – ε)(1 -2) (16) Shape Functions are, ww 1 N1= - 4 (1+ ε) (1 +) (1+ε+) w.E 1 N2 = - 4 (1+ ε) (1 - ) (1- ε + ) 1 N3= − 4 (1+ ε) (1 + ) (1- ε - ) asy 1 N4= - 4 (1- ε) (1 + ) (1+ ε -) N5 = N6 = N7 = N8 = 5. 1 2 1 En gi (1- ε2)(1 - ) (1+ ε)(1 - 2) 2 1 2 1 2 nee (1 – ε2)(1 + ) (1 – ε)(1 - 2) rin g.n e Derive the shape function for 4 noded rectangular parent element by using natural coordinate system and co-ordinate transformation η 4 (-1,1) η (+1) 3 (1,1) t ε ε (+1) ε (-1) 1(-1,-1) η (-1) 2 (1,-1) Consider a four noded rectangular element as shown in FIG. The parent element is defined in ε and η co-ordinates i.e., natural co-ordinates ε is varying from -1 to 1 and η is also varying -1 to 1. 16 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net We know that, Shape function value is unity at its own node and its value is zero at other nodes. At node 1: (co-ordinate ε = -1, η = -1) Shape function N1 = 1 at node 1. N1 = 0 at nodes 2, 3 and 4 N1has to be in the form of N1 = C (1 - ε) (1 -η) (1) Where, C is constant. Substitute ε = -1 and η = -1 in equation (1) ww N1 = C (1+1)(1+1) w.E N1= 4C 1 asy C=4 Substitute C value in equation (1) 1 En gi (2) N1 = 4(1 - ε) (1 -η) nee At node 2: (co-ordinate ε = 1, η = -1) Shape function N2 = 1 at node 2. N2 = 0 at nodes 1, 3 and 4 N1has to be in the form of N2 = C (1 + ε) (1 -η) rin g.n e (3) Where, C is constant. Substitute ε = 1 and η = -1 in equation (3) t N2 = C (1+1) (1+1) N2 = 4C 1 C=4 Substitute C value in equation (1) 1 (4) N2 = 4(1 + ε) (1 -η) 17 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net At node 3: (co-ordinate ε = 1, η = 1) Shape function N3 = 1 at node 3. N3 = 0 at nodes 1, 2 and 4 N1has to be in the form of N3 = C (1 + ε) (1 +η) (5) Where, C is constant. Substitute ε = 1 and η = 1 in equation (5) N3 = C (1+1)(1+1) ww N3 = 4C 1 w.E C=4 Substitute C value in equation (1) asy 1 (6) N3 = 4(1 +ε) (1 + η) En gi At node 4: (co-ordinate ε = -1, η = 1) Shape function N4 = 1 at node 4. N4 = 0 at nodes 1, 2 and 3 N1has to be in the form of N4 = C (1 - ε) (1 +η) nee Where, C is constant. Substitute ε = -1 and η = 1 in equation (1) N4 = C (1+1) (1+1) rin g.n e (7) t N4 = 4C 1 C=4 Substitute C value in equation (1) 1 (8) N4 = 4(1 - ε) (1 +η) Consider a point p with co-ordinate (ε ,η). If the displacement function u = 𝑢 represents the 𝑣 displacements components of a point located at (ε ,η) then, 18 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net u = N1 𝑢1 +N2 𝑢2 +N3 𝑢3 +N4 𝑢4 v = N1 𝑣1 +N2 𝑣2 +N3 𝑣3 +N4 𝑣4 It can be written in matrix form as, u= 𝑢 = 𝑣 𝑁1 0 𝑁2 0 0 𝑁1 0 𝑁2 𝑁3 0 𝑁4 0 0 𝑁3 0 𝑁4 ww w.E 𝑢1 𝑣1 𝑢2 𝑣2 𝑢3 𝑣3 𝑢4 𝑣4 (9) In the isoparametric formulation i,e., for global system, the co-ordinates of the nodal points are 𝑥1 , 𝑦1 , 𝑥2 , 𝑦2 , 𝑥3 , 𝑦3 , and 𝑥4 , 𝑦4 . In order to get mapping the co-ordinate of point p is defined as asy En gi 𝑥 = N1 𝑥1 +N2 𝑥2 +N3 𝑥3 +N4 𝑥4 nee 𝑦 = N1 𝑦1 +N2 𝑦2 +N3 𝑦3 +N4 𝑦4 rin g.n e t 19 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net The above equation can be written in matrix form as, 𝑥 u= 𝑦 = ww 𝑁1 0 𝑁2 0 0 𝑁1 0 𝑁2 𝑁3 0 𝑁4 0 0 𝑁3 0 𝑁4 w.E 6. asy 𝑥1 𝑦1 𝑥2 𝑦2 𝑥3 𝑦3 𝑥4 𝑦4 (10) For the isoparametric four noded quadrilateral element shown in fig. Determine the Cartesian co-ordinates of point P which has local co-ordinatesε= 0.5 , η =0.5 En gi nee rin g.n e t Given data Natural co-ordinates of point P ε= 0.5 η =0.5 20 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net Cartesian co-ordinates of the point 1,2,3 and 4 P 𝑥, 𝑦 𝑥1 = 1; 𝑦1 = 1 𝑥2 = 5; 𝑦2 = 1 𝑥3 = 6; 𝑦3 = 6 𝑥4 = 1; 𝑦4 = 4 To find : Cartesian co-ordinates of the point P(x,y) Formulae used: ww Co -ordinate, 𝑥 = N1 𝑥1 +N2 𝑥2 +N3 𝑥3 +N4 𝑥4 Co-ordinate, 𝑦 = N1 𝑦1 +N2 𝑦2 +N3 𝑦3 +N4 𝑦4 w.E Solution asy En gi Shape function for quadrilateral elements are, 1 N1 = 4(1 - ε) (1 -η) 1 N2 = (1 + ε) (1 -η) 4 1 N3 = 4(1 +ε) (1 + η) 1 N4 = 4(1 - ε) (1 +η) nee Substitute ε and η values in the above equations, 1 N1 = 4(1 – 0.5) (1 –0.5) = 0.0625 1 N2 = 4(1 + 0.5) (1 –0.5) = 0.1875 rin g.n e t 1 N3 = 4(1 +0.5) (1 + 0.5) =0.5625 1 N4 = 4(1 – 0.5) (1 +0.5) = 0.1875 We know that, Co-ordinate, 𝑥 = N1 𝑥1 +N2 𝑥2 +N3 𝑥3 +N4 𝑥4 = 0.0625×1+0.1875×5+0.5625×6+0.1875×1 21 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝑥 = 4.5625 Similarly, Co-ordinate, 𝑦 = N1 𝑦1 +N2 𝑦2 +N3 𝑦3 +N4 𝑦4 = 0.0625×1+0.1875×1+0.5625×6+0.1875×4 y = 4.375 7. 𝟏 𝟏 𝒙 𝟐 𝒆 + 𝒙 + dx using Gaussian integration with one, −𝟏 𝒙+𝟕 ,two , three integration points and compare with exact solution Evaluate the integral I = ww Given: w.E 1 −1 I= 𝑒 𝑥 + 𝑥2 + asy To Find: 1 𝑥+7 En gi dx Evaluate the integral by using Gaussian. Formulae used: 1 −1 I= 𝑒 𝑥 + 𝑥2 + 1 f 𝑥1 ,w1 f 𝑥1 , w1 f 𝑥1 + w2 f 𝑥2 + w3 f 𝑥3 Solution 1. point Gauss quadrature 𝑥1 = 0; w1 = 2 f 𝑥 = f 𝑥1 = 𝑒 𝑥 + 𝑥2 + 𝑒0 + 0 + nee 𝑥+7 dx rin g.n e t 1 𝑥+7 1 0+7 f 𝑥1 = 1.1428 w1 f 𝑥1 = 2 ⨯1.1428 = 2.29 22 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 2. point Gauss quadrature 𝑥1 = 1 3 =0.5773; 1 𝑥2 = − 3= -0.5773; w1 = w2 = 1 f 𝑥 = 𝑒 𝑥 + 𝑥2 + 1 𝑥+7 ww f 𝑥1 = 𝑒 0.5773 + 0.57732 + 1 0.5773 +7 f 𝑥1 = 1.7812 + 0.33327 + 0.13197 w.E f 𝑥1 = 2.246 asy w1 f 𝑥1 = 1 ⨯2.246 = 2.246 En gi f 𝑥2 = 𝑒 −0.5773 + (−0.5773)2 + 1 −0.5773 +7 = 0.5614 + 0.3332+0.15569 f 𝑥2 = 1.050 w2 f 𝑥2 = 1 ⨯1.050 = 1.050 nee rin g.n e t w1 f 𝑥1 + w2 f 𝑥2 = 2.246 + 1.050 = 3.29 3. point Gauss quadrature 𝑥1 = 3 5 =0.7745; 𝑥2 = 0: 23 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net 𝑥1 = − 3 5 = - 0.7745; 5 w1 = 9 = 0.5555; 8 w2 = 9 = 0.8888 5 w2 = 9 = 0.5555 1 f 𝑥 = 𝑒 𝑥 + 𝑥 2 + 𝑥+7 1 f 𝑥1 = 𝑒 0.7745 + 0.77452 + 0.7745 +7 ww f 𝑥1 = 2.1697 + 0.6 + 0.1286 w.E f 𝑥1 = 2.898 w1 f 𝑥1 = 0.55555⨯2.898 asy = 1.610 1 f 𝑥2 = 1+ 7 f 𝑥2 = 1.050 En gi nee w2 f 𝑥2 = 0.888⨯1.143 = 1.0159 w1 f 𝑥1 + w2 f 𝑥2 + w3 f 𝑥3 = 1.160 + 1.0159 +0.6786 = 2.8545 1 1 I = −1 𝑒 𝑥 + 𝑥 2 + 𝑥+7 dx Exact Solution = 𝑒 𝑥 1−1 + 1 𝑥3 1 3 −1 rin g.n e t + ln(𝑥 + 7) 1−1 −1 = 𝑒 +1 − 𝑒 −1 + 3 − 3 + ln(1 + 7) − ln(−1 + 7) 2 = 2.7183 − 0.3678 + 3 + ln(8) − ln(6) = 2.3505 +0.6666 + 2.0794 − 1.7917 = 3.0171 + 0.2877 = 3.3048 24 Downloaded From: www.EasyEngineering.net Downloaded From: www.EasyEngineering.net ww w.E asy E ngi nee rin g.n et **Note: Other Websites/Blogs Owners Please do not Copy (or) Republish this Materials, Students & Graduates if You Find the Same Materials with EasyEngineering.net Watermarks or Logo, Kindly report us to easyengineeringnet@gmail.com Downloaded From: www.EasyEngineering.net

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