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Lecture 3_Deflection

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Lecture 3_Deflection

  1. 1. www.derby.ac.uk Deflection 5ME517 Structural Analysis and Materials Dr Amal Oraifige a.oraifige@derby.ac.uk
  2. 2. www.derby.ac.uk Deflection is a term that is used to describe the movement of a structure displaced under loading. Deflection is a distance, therefore its measured in “m” OR “mm”. Deflection
  3. 3. www.derby.ac.uk Deflection and Slope Consider a simply supported beam with a point load at the middle.
  4. 4. www.derby.ac.uk Deflection and Slope
  5. 5. www.derby.ac.uk Consider a cantilever beam with a point load at the free end. Deflection and Slope
  6. 6. www.derby.ac.uk Simple Cases
  7. 7. www.derby.ac.uk Deflection Formula Where: F = Load (N) L = Length of the beam (m) E = Young Modulus (GN/m2) I = Second Moment of Area (m4) EI = Flexural Stiffness (Nm2) FL3 EI
  8. 8. www.derby.ac.uk Deflection Formula Halving the load will halve the resultant deflection Halving the length will reduce the deflection by a factor of 8. The higher the E value the lower the resulting deflection. FL3 EI
  9. 9. www.derby.ac.uk Deflection Formula  Increasing the I value will result in greater rigidity, which decrease in deflection without having to increase the weight of the beam. I = 360000mm4 I = 40000mm4 60mm 20mm 20mm 60mm
  10. 10. www.derby.ac.uk Deflection – Case 1 CANTILEVER WITH POINT LOAD AT FREE END Deflection (δmax) = FL3 3EI
  11. 11. www.derby.ac.uk Deflection – Case 2 CANTILEVER WITH A UNIFORMLY DISTRIBUTED LOAD Deflection (δmax) = wL4 8EI
  12. 12. www.derby.ac.uk Deflection – Combined 1 & 2 CANTILEVER WITH A UNIFORMLY DISTRIBUTED LOAD Total Deflection (δmax) = + wL4 8EI FL3 3EI
  13. 13. www.derby.ac.uk Deflection – Case 3 SIMPLY SUPPORTED BEAM WITH POINT LOAD IN MIDDLE Deflection (δmax) = FL3 48EI
  14. 14. www.derby.ac.uk Deflection – Case 4 SIMPLY SUPPORTED BEAM WITH A UNIFORMLY DISTRIBUTED LOAD Deflection (δmax) = 5wL4 384EI
  15. 15. www.derby.ac.uk Deflection – Combined 3 & 4 SIMPLY SUPPORTED BEAM WITH A UNIFORMLY DISTRIBUTED LOAD Deflection (δmax) = + 5wL4 384EI FL3 48EI
  16. 16. www.derby.ac.uk Example 1 A simply supported beam has a point load on the middle of 20kN and a UDL of 24N/m distributed on its entire length. The beam is 2m long and has a solid circular section of 140mm diameter. The beam is made of a material that has a young’s modulus of 70GPa. Calculate: • Flexural Stiffness. • Maximum Deflection.
  17. 17. www.derby.ac.uk Example 2 A cantilever beam is 6m long and has a maximum deflection of 1.5mm at the free end. The beam has a point load of 3kN at the free end and a U.D.L of 300N/m along its entire length. Calculate:  Flexural Stiffness.  The dimensions of the section if the beam has a tube section with an internal diameter 0.8 the outer diameter. Take the Modulus of Elasticity as 200GPa.
  18. 18. www.derby.ac.uk Complex Cases “Macaulay’s Theory”
  19. 19. www.derby.ac.uk When the loads on a beam do not conform to simple cases, the solution for slope and deflection must be found from first principles. Macaulay developed a method for making the integration simpler. Deflection – Macaulay’s Theory
  20. 20. www.derby.ac.uk Macaulay’s Theory The basic equation governing the slope and deflection of beams is: d2y dx2 EI = M where M is the Bending Moment
  21. 21. www.derby.ac.uk Macaulay’s Theory Rules 1.Always start from the left hand side of the beam. 2.The Bending Moment is taken from the last portion of beam after the last load to the right. 3.Terms in the bending moment expressions such as (x-a) must be integrated as a whole bracket and must not be expanded. 4.When values are substituted for x, any resulting terms inside the bracket with (–) sign or 0, the whole bracket must be ignored and given a value of Zero.
  22. 22. www.derby.ac.uk Macaulay’s Theory Rules
  23. 23. www.derby.ac.uk 1. Write down the bending moment equation placing x on the right hand side after the last load. Macaulay’s Theory Steps
  24. 24. www.derby.ac.uk 2. Integrate once treating the square bracket as the variable for the slope. Integrate Macaulay’s Theory Steps
  25. 25. www.derby.ac.uk 3. Integrate again using the same rules for the deflection. Integrate again Macaulay’s Theory Steps
  26. 26. www.derby.ac.uk 4.Use boundary conditions to solve A and B – where deflection (y) is = 0. 5.Solve the slope and deflection by putting in appropriate value of x. 6.Ignore any brackets containing negative values. Macaulay’s Theory Steps
  27. 27. www.derby.ac.uk Macaulay Theory – Example 1 Calculate the deflection at the centre of the beam shown below taking the flexural stiffness as 30MNm2. Answer: 0.14mm
  28. 28. www.derby.ac.uk Macaulay Theory – Example 2 Calculate the maximum deflection at 3.65m of the beam shown below taking the flexural stiffness as 442kNm2. Answer: 1.23mm

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