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Part 3 Beams(1).pdf
- 1. FEA/CFD for Biomedical Engineering Week 4: Axial Members and Beams
- 2. Beams
- 3. Beams โข A structural member whose cross-sectional dimensions are relatively smaller than its length โข A beam is commonly subjected to transverse loading โ Creates bending โข Truss โ loading only applied at nodes โข Beam โ load can be applied anywhere along the beam
- 4. Beams โข Beams can be analytically investigated by considering the neutral axis โข The deflection of the neutral axis at any point x is represented by v โข For small deflections ๐๐ = โ ๐๐๐๐ ๐ผ๐ผ Bending moment at that section Lateral distance from neutral axis Second moment of inertia
- 5. Beams ๐ธ๐ธ๐ธ๐ธ ๐๐2๐ฃ๐ฃ ๐๐๐ฅ๐ฅ2 = ๐๐ ๐ฅ๐ฅ ๐ธ๐ธ๐ธ๐ธ ๐๐3๐ฃ๐ฃ ๐๐๐ฅ๐ฅ3 = ๐๐๐๐ ๐ฅ๐ฅ ๐๐๐๐ = ๐๐ ๐ฅ๐ฅ ๐ธ๐ธ๐ธ๐ธ ๐๐4๐ฃ๐ฃ ๐๐๐ฅ๐ฅ4 = ๐๐๐๐ ๐ฅ๐ฅ ๐๐๐๐ = ๐ค๐ค ๐ฅ๐ฅ The deflection (v) of the neutral axis is related to Internal bending moment Transverse shear The load See Table 4.1 and 4.2 in the course text book for the deflection and slopes of beams under typical loads
- 6. Finite Element formulation of beams โข A simple beam element consists of two nodes At each node there are two degrees of freedom โ Vertical Displacement โ Rotation angle โข The displacement field is represented by a third order polynomial with four unknowns ๐ฃ๐ฃ = ๐๐1 + ๐๐2๐ฅ๐ฅ + ๐๐3๐ฅ๐ฅ2 + ๐๐4๐ฅ๐ฅ3
- 7. Finite Element formulation of beams ๐ฃ๐ฃ = ๐๐1 + ๐๐2๐ฅ๐ฅ + ๐๐3๐ฅ๐ฅ2 + ๐๐4๐ฅ๐ฅ3 โข For Node i: the vertical displacement at x = 0 โข For Node i: The slope at x = 0 โข For Node j: the vertical displacement at x = L โข For Node j: The slope at x = L ๐ฃ๐ฃ = ๐๐1 = ๐๐๐๐๐ ๏ฟฝ ๐๐๐๐ ๐๐๐๐ ๐ฅ๐ฅ=0 = ๐๐2 = ๐๐๐๐๐ ๐ฃ๐ฃ = ๐๐1 + ๐๐2๐ฟ๐ฟ + ๐๐3๐ฟ๐ฟ2 + ๐๐4๐ฟ๐ฟ3 = ๐๐๐๐๐ ๏ฟฝ ๐๐๐๐ ๐๐๐๐ ๐ฅ๐ฅ=๐ฟ๐ฟ = ๐๐2 = ๐๐๐๐๐
- 8. Finite Element formulation of beams โข We now have 4 equations with 4 unknowns. โข Solve for c1 โ c4 Then substitute into ๐ฃ๐ฃ = ๐๐1 + ๐๐2๐ฅ๐ฅ + ๐๐3๐ฅ๐ฅ2 + ๐๐4๐ฅ๐ฅ3 ๐ฃ๐ฃ = ๐๐๐๐1๐๐๐๐๐ + ๐๐๐๐2๐๐๐๐2 + ๐๐๐๐1๐๐๐๐1 + ๐๐๐๐๐๐๐๐๐๐ โข The shape functions are commonly called Hermite Shape functions and are given by ๐๐๐๐๐ = 1 โ 3๐ฅ๐ฅ2 ๐ฟ๐ฟ2 + 2๐ฅ๐ฅ3 ๐ฟ๐ฟ3 ๐๐๐๐2 = ๐ฅ๐ฅ โ 2๐ฅ๐ฅ2 ๐ฟ๐ฟ + ๐ฅ๐ฅ3 ๐ฟ๐ฟ2 ๐๐๐๐1 = 3๐ฅ๐ฅ2 ๐ฟ๐ฟ2 โ 2๐ฅ๐ฅ3 ๐ฟ๐ฟ3 ๐๐๐๐๐ = โ ๐ฅ๐ฅ2 ๐ฟ๐ฟ + ๐ฅ๐ฅ3 ๐ฟ๐ฟ2
- 9. Finite Element formulation of beams โข Stiffness Matrix โ Derivation found in Section 4.3 of Chapter 4 of the course textbook โ I'm not going to go through another one!!!! โข Key components are ฮ ๐๐ = ๐ธ๐ธ๐ธ๐ธ 2 ๏ฟฝ 0 ๐ฟ๐ฟ ๐๐2๐ฃ๐ฃ ๐๐๐ฅ๐ฅ2 2 ๐๐๐๐ ๐๐2๐ฃ๐ฃ ๐๐๐ฅ๐ฅ2 = ๐๐๐๐๐ ๐๐๐๐2 ๐๐๐๐1 ๐๐๐๐๐ ๐๐๐๐๐ ๐๐๐๐2 ๐๐๐๐1 ๐๐๐๐๐ โข Followed by a bunch of matrix manipulationโฆ.The stiffness matrix for a beam element for two degrees of freedom isโฆ.. ๐ฒ๐ฒ (๐๐) = ๐ธ๐ธ๐ธ๐ธ ๐ฟ๐ฟ3 12 6๐ฟ๐ฟ โ12 6๐ฟ๐ฟ 6๐ฟ๐ฟ 4๐ฟ๐ฟ2 โ6๐ฟ๐ฟ 2๐ฟ๐ฟ2 โ12 โ6๐ฟ๐ฟ 12 โ6๐ฟ๐ฟ 6๐ฟ๐ฟ 2๐ฟ๐ฟ2 โ6๐ฟ๐ฟ 4๐ฟ๐ฟ2
- 10. Finite Element formulation of beams โข Load matrix derivation again can be found in section 4.3 chapter 4 โข Table 4.2 also contains the relationships between the actual load and its equivalent nodal loads for some typical loading situations
- 11. Example Analyse a cantilevered balcony beam that is a wide flange and determine the deflection at B and C Properties: โข Cross sectional area = 0.0066 m2 โข Depth = 0.45 m โข Second moment of area = 0.000212 m4 โข Load = 14 600 N/m โข Modulus of Elasticity = 200 GPa 14 600 N/m 3.048 m
- 12. Example โข As we are using a single element the element local stiffness and load matrices are the same as the global matrices ๐ฒ๐ฒ (๐๐) = ๐ฒ๐ฒ (๐บ๐บ) = ๐ธ๐ธ๐ธ๐ธ ๐ฟ๐ฟ3 12 6๐ฟ๐ฟ โ12 6๐ฟ๐ฟ 6๐ฟ๐ฟ 4๐ฟ๐ฟ2 โ6๐ฟ๐ฟ 2๐ฟ๐ฟ2 โ12 โ6๐ฟ๐ฟ 12 โ6๐ฟ๐ฟ 6๐ฟ๐ฟ 2๐ฟ๐ฟ2 โ6๐ฟ๐ฟ 4๐ฟ๐ฟ2 ๐ธ๐ธ๐ธ๐ธ ๐ฟ๐ฟ3 12 6๐ฟ๐ฟ โ12 6๐ฟ๐ฟ 6๐ฟ๐ฟ 4๐ฟ๐ฟ2 โ6๐ฟ๐ฟ 2๐ฟ๐ฟ2 โ12 โ6๐ฟ๐ฟ 12 โ6๐ฟ๐ฟ 6๐ฟ๐ฟ 2๐ฟ๐ฟ2 โ6๐ฟ๐ฟ 4๐ฟ๐ฟ2 ๐๐11 ๐๐12 ๐๐21 ๐๐22 = โ ๐ค๐ค๐ค๐ค 2 โ ๐ค๐ค๐ฟ๐ฟ2 12 โ ๐ค๐ค๐ค๐ค 2 ๐ค๐ค๐ฟ๐ฟ2 12 ๐ญ๐ญ (๐๐) = ๐ญ๐ญ (๐บ๐บ) โ ๐ค๐ค๐ค๐ค 2 โ ๐ค๐ค๐ฟ๐ฟ2 12 โ ๐ค๐ค๐ค๐ค 2 ๐ค๐ค๐ฟ๐ฟ2 12
- 13. Example โข Applying the boundary condition U11 = U12 = 0 at node 1 โข Simplifying ๐ธ๐ธ๐ธ๐ธ ๐ฟ๐ฟ3 1 0 0 0 0 1 0 0 โ12 โ6๐ฟ๐ฟ 12 โ6๐ฟ๐ฟ 6๐ฟ๐ฟ 2๐ฟ๐ฟ2 โ6๐ฟ๐ฟ 4๐ฟ๐ฟ2 ๐๐11 ๐๐12 ๐๐21 ๐๐22 = 0 0 โ ๐ค๐ค๐ค๐ค 2 ๐ค๐ค๐ฟ๐ฟ2 12 12 โ6๐ฟ๐ฟ โ6๐ฟ๐ฟ 4๐ฟ๐ฟ2 ๐๐21 ๐๐22 = ๐ฟ๐ฟ3 ๐ธ๐ธ๐ธ๐ธ โ ๐ค๐ค๐ค๐ค 2 ๐ค๐ค๐ฟ๐ฟ2 12
- 14. Example 12 โ6๐ฟ๐ฟ โ6๐ฟ๐ฟ 4๐ฟ๐ฟ2 ๐๐21 ๐๐22 = ๐ฟ๐ฟ3 ๐ธ๐ธ๐ธ๐ธ โ ๐ค๐ค๐ค๐ค 2 ๐ค๐ค๐ฟ๐ฟ2 12 14 600 N/m 3.048 m 12 โ6 3.048 โ6 3.048 4 3.048 2 ๐๐21 ๐๐22 = 3.048 3 200 ร 109 0.000212 โ 14600 3.048 2 14600 3.0482 12 Second moment of area (I) = 0.000212 m4 Load (w) = 14 600 N/m Modulus of Elasticity (E) = 200 GPa
- 15. Example 12 โ18.29 โ18.29 37.16 ๐๐21 ๐๐22 = โ1.49 ร 10โ2 7.55 ร 10โ3 12 โ6 3.048 โ6 3.048 4 3.048 2 ๐๐21 ๐๐22 = 3.048 3 200 ร 109 0.000212 โ 14600 3.048 2 14600 3.0482 12 Therefore the deflection at endpoint C is U21 = -0.0037 m and U22 = -0.00163 rad
- 16. Example โข To determine the deflection at point B, we use the deflection equation for the beam element and evaluate the shape functions at x = L/2 ๐ฃ๐ฃ๐ต๐ต = ๐๐11 0 + ๐๐12 0 + ๐๐21 โ0.0037 + ๐๐22 โ0.00163 โข Computing the value of the shape functions at point b ๐๐21 = 3๐ฅ๐ฅ2 ๐ฟ๐ฟ2 โ 2๐ฅ๐ฅ3 ๐ฟ๐ฟ3 ๐๐22 = โ ๐ฅ๐ฅ2 ๐ฟ๐ฟ + ๐ฅ๐ฅ3 ๐ฟ๐ฟ2 = 3 ๐ฟ๐ฟ2 ๐ฟ๐ฟ 2 2 โ 2 ๐ฟ๐ฟ3 ๐ฟ๐ฟ 2 3 = 1 2 = โ ๐ฟ๐ฟ 2 2 ๐ฟ๐ฟ + ๐ฟ๐ฟ 2 3 ๐ฟ๐ฟ2 = โ ๐ฟ๐ฟ 8 ๐ฃ๐ฃ = ๐๐11๐๐11 + ๐๐12๐๐12 + ๐๐21๐๐21 + ๐๐22๐๐22 ๐ฃ๐ฃ๐ต๐ต = 1 2 โ0.0037 + โ 3.048 8 โ0.00163 = โ0.00122 ๐๐
- 17. Conclusions โข You have learned about shape functions that can be used to give approximations of the properties between nodes โข Learned how to deal with beam elements โข Learned how to deal with distributed loads