Question Paper: Finite Element Analysis Question Paper - December 2016 - Mechanical Engineering (Semester 6) - Mumbai University (MU)
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Finite Element Analysis - December 2016

MU Mechanical Engineering (Semester 6)

Total marks: --
Total time: --
INSTRUCTIONS
(1) Assume appropriate data and state your reasons
(2) Marks are given to the right of every question
(3) Draw neat diagrams wherever necessary
1(a) Principal Minimum Potential Energy 5 marks

1(b) Explain the significance of Jacobin matrix. 5 marks

1(c) Types of Boundary Conditions 5 marks

1(d) Sources of error in FEM 5 marks

2(a) Solve the following differential Equation using Galerkin Method.<mfrac><mrow><msup><mi>d</mi><mn>2</mn></msup><mi>ϕ</mi></mrow><mrow><mi>d</mi><msup><mi>x</mi><mn>2</mn></msup></mrow></mfrac><mo>+</mo><mi>cos</mi><mo>⁡</mo><mi>π</mi><mi>x</mi><mo>=</mo><mn>0</mn><mn>0</mn><mo><</mo><mi>x</mi><mo><</mo><mn>1.</mn></math>" role="presentation" style="font-size: 125%; text-align: center; position: relative;">d2ϕdx2+cosπx=00<x<1.<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><mfrac><mrow><msup><mi>d</mi><mn>2</mn></msup><mi>ϕ</mi></mrow><mrow><mi>d</mi><msup><mi>x</mi><mn>2</mn></msup></mrow></mfrac><mo>+</mo><mi>cos</mi><mo>⁡</mo><mi>π</mi><mi>x</mi><mo>=</mo><mn>0</mn><mn>0</mn><mo><</mo><mi>x</mi><mo><</mo><mn>1.</mn></math><script type="math/tex; mode=display" id="MathJax-Element-1">\frac{d^2\phi }{dx^2}+\cos \pi x=0 0< x< 1.</script> Boundary Conditions are: φ(0)=0, φ(1)=0 Find φ(0.25), φ(0.5), φ(0.75). Compare your answer with exact solution. 5 marks

2(b) Figure shows the beam of uniform rectangular cross section 10cm×12cm, sunjected to point load and uniformly distributed load. Young's modulus is 2 Mpa and Poisson's ratio is 0.3, Determine the deflections and slopes.
!mage.
5 marks

3(a) Consider the steady laminar flow of a viscous fluid through a long circular cylindrical tube. The govering equation is −1rddr(rμdwdr)=P0−PLLf0−1rddr(rμdwdr)=P0−PLLf0 -\frac{1}{r}\frac{d}{dr}\left ( r\mu \frac{dw}{dr} \right )=\frac{P_0-P_L}{L}f_0 / where w is the axial (i.e.;z) component of velocity , μ is the viscosity, and f0 is the gradient of pressure ( which includes the combined effect of static pressure in gravitational force). The boundary conditions are (rdwdr)∣∣∣t=0=0,  w(R0)=0(rdwdr)|t=0=0,  w(R0)=0 \left ( r\frac{dw}{dr} \right )\bigg|_{t=0}=0,\ \ w(R_0)=0 / Using the symmetry and two linear elements, determine th e velocity field and compare wtih the exact solutions at the nodes: <mi>W</mi><mo stretchy="false">(</mo><mi>r</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac><mrow><msub><mi>f</mi><mn>0</mn></msub><msubsup><mi>R</mi><mn>0</mn><mn>2</mn></msubsup></mrow><mrow><mn>4</mn><mi>μ</mi></mrow></mfrac><mrow><mo>[</mo><mrow><mn>1</mn><mo>−</mo><msup><mrow><mo>(</mo><mfrac><mi>r</mi><msub><mi>R</mi><mn>0</mn></msub></mfrac><mo>)</mo></mrow><mn>2</mn></msup></mrow><mo>]</mo></mrow></math>" role="presentation" style="font-size: 125%; text-align: center; position: relative;">W(r)=f0R204μ[1(rR0)2]<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><mi>W</mi><mo stretchy="false">(</mo><mi>r</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac><mrow><msub><mi>f</mi><mn>0</mn></msub><msubsup><mi>R</mi><mn>0</mn><mn>2</mn></msubsup></mrow><mrow><mn>4</mn><mi>μ</mi></mrow></mfrac><mrow><mo>[</mo><mrow><mn>1</mn><mo>−</mo><msup><mrow><mo>(</mo><mfrac><mi>r</mi><msub><mi>R</mi><mn>0</mn></msub></mfrac><mo>)</mo></mrow><mn>2</mn></msup></mrow><mo>]</mo></mrow></math><script type="math/tex; mode=display" id="MathJax-Element-4">W(r)=\frac{f_0R_0^2}{4\mu }\left [ 1-\left ( \frac{r}{R_0} \right )^2 \right ]</script> 5 marks

3(b) Derive shape function for ID linear element in natural co-ordinates. 5 marks

4(a) The circular rod depicted in Figure has an outside diameter of 60 mm, length of 1m, and is perfectly insulated on its circumference. The left half of the cylinder is aluminum, for which kx=200 W/m-°C and the right half is copper having kx=389 W/m-°C. The extreme right end of the cylinder is maintained at a temperature of 80°C, while the left end is subjected to a heat input rate 4000 W/m2 !mage. 5 marks

4(b) Analyze the truss completely for displacement, stress and strain as shown in figure.
!mage
5 marks

5(a) Find nodal displacement and element stress for the bar as shown in figure using FEM. Take E=200GPa.
!mage
5 marks

5(b) A CST element has nodal coordinates (10,10), (70,35) and (75,25) for nodes 1,2 and 3 respectively. The element is 2mm thick and is of materiaol with propertises E-70 GPA. Poission's ratio is 0.3. After applying the load to element the nodal deformation were found to be u1=0.01mm, v1=-0.04mm, u2=0.03mm, v2=0.02mm, u3=0.02mm, u1=-0.04mm. Determine the strains ex,ey,exy and corresponding element stresses. 5 marks

6(a) Consider a uniform cross section bar of length L made up of a material whose Yong's modulus and density are given by E and ρ. Estimate the natural frequences of axial vibration of the bar using both consistent and lumped mass matrices.
!mage.
5 marks

6(b) Coordinates of the nodes of finite element are given by P(4,0) and Q(s,0).Find the expression of x in terms of ξ when:
1) Third node R is taken at (6,0)
2) Third node R is taken at (5,0) Comment on the result.
5 marks

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