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Good morning!

Good morning!. New words. Beam 梁 Tension 拉伸 Compression 压 Shear 剪 Torsion 扭转 Bending 弯曲

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Good morning!

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  1. Good morning!

  2. New words Beam 梁 Tension 拉伸 Compression 压 Shear 剪 Torsion 扭转 Bending 弯曲 Concave side 凹面 Convex side 凸面

  3. Exersize • measure the tendency of one part of a beam to be slided with respect to the other part • -------(shear force) • the tendency for one part to be rotated with respect to the other part • -------(bending moment)

  4. Review: In last week, we have learned columns made of : timber steel concrete

  5. Question: in considering the beam what are we concerned with? Answer: the effects of the forces New member –beam

  6. Five forms of deformation Tension Compression Shear Bending Torsion

  7. Internal forces For example: the internal forces that result from bending deformations are compressive on the concave side and tensile on the convex side

  8. SF&BM measure the tendency of one part of a beam to be slided with respect to the other part (SF effect),and the tendency for one part to be rotated with respect to the other part(BM effect) • The question we need resolve is to calculate the SF&BM

  9. Shear Force

  10. Calculation The SF at any section A in a straight beam is the algebraic sum of all vertical forces lefts of that point SFA=∑VLA

  11. Eg1: 3KN 5KN A B C D 2M 2M 1M 8KN

  12. STEPS • Cut the beam at point B ,so we will get part of the shear force of the beam 8KN 3KN A B

  13. If we cut the beam at point C ,what will happen to the SF?TRYING… What about point D? TRYING…

  14. SFD • If we plot the value of the SF at all point along the axis of a beam, we’ll obtain the Shear Force Diagram 8KN 5KN

  15. Principle for solving the SFD • Calculate the shear forces at the following significant positions: • At the start and end of the beam • At every support • At every point load • At the start and end of every distributed load

  16. Conclusion: • We find that at a point load the SF changes instantaneously and the SF diagram shows a step.

  17. Eg2: A B C D 2M 2M 1M

  18. Steps: • First calculate the reaction at support A. • Then cut the beam at point D

  19. Finally plot the SFD 15 A B C D 2M 2M 1M

  20. Conclusion: • When there is an UDL,we see the SF change by equal intervals ,and the SFD will have a constant slope.

  21. Have a Try 4KN 6KN 3KN A ① ② B ③

  22. Bending Moment BML=∑ALL • The BM at any section A in a straight beam is the algebraic sum of areas of the SFD left of that point

  23. Eg3 A B C D 2M 2M 1M 8KN 5KN A B C D

  24. NOTE • Locate the positions of zero SF, as these will also be positions of maximum BM.

  25. The result is: 0 16KN 26KN

  26. HAVE A TRY • 4KN 6KN 3KN • A ① ② B ③

  27. CONSIDER • All we learned today is the simply supported beam, then what about continuous beam? • Let’s talk about it on next lesson!

  28. Homework: Exersize1 a/b/c/f

  29. Bye bye See you next class

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