Engineering science lesson 4

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Engineering science lesson 4

  1. 1. Chapter 1- Static engineering systems 1.1 Simply supported beams 1.1.1 determination of shear force 1.1.2 bending moment and stress due to bending 1.1.3 radius of curvature in simply supported beams subjected to concentrated and uniformly distributed loads 1.1.4 eccentric loading of columns 1.1.5 stress distribution 1.1.6 middle third rule
  2. 2. Types of columnsDepending on the mode of failure, columns canbe categorised in the following ways (a) a short column a column which will fail in true compression (b) a long column a column which buckles before full compressive strength is reached
  3. 3. Types of beams• Beams can be classified according to the manner in which they supported
  4. 4. Loads• Loads can be applied on the beams in one of the following ways
  5. 5. Sheer force and bending moments• When a beam is loaded by forces or couples, internal stresses and strains are created. Consider a cantilever arrangement• It is convenient to reduce the resultant to a shear force, V, and a bending moment, M.
  6. 6. Sign conventions• Positive shear forces always deform right hand face downward with respect to the left hand face. Positive shear stress acts clockwise while negative shear stress acts counter-clockwise• Positive bending moments always elongate the lower section of the beam. Positive moment compresses upper (sagging moments) whereas negative moment compresses lower (hogging moments)
  7. 7. Relationships for continuous loads• Consider the following beam segment with a uniformly distributed load with load intensity q. Note that distributed loads are positive when acting downward and negative when acting upward.• Summing forces vertically• Summing moments and discarding products of differentials because they are negligible compared to other terms
  8. 8. Relationships for concentrated loads • consider the following beam segment with a concentrated load, P. Again, concentrated loads are positive when acting downward and negative when acting upward. • Summing forces vertically • An abrupt change occurs in the shear force at a point where a concentrated load acts. As one moves from left to right through a point of load application, the shear force decreases by an amount equal to the magnitude of the downward load. • Summing the moments
  9. 9. Relationships for couples• Summing the moments
  10. 10. Sheer force and bending momentdiagrams-Concentrated loads
  11. 11. Sheer force and bending momentdiagrams-Uniform loads
  12. 12. Sheer force and bending momentdiagrams-several concentrated loads
  13. 13. Example• Determine the shear force and bending moment at 1m and 4m from the right hand end of the beam shown in the figure. Neglect the weight of the beam.• A cantilever beam that is free at end A and fixed at end B is subjected to a distributed load of linearly varying intensity q. The maximum intensity of the load occurs at the fixed support and is equal to q0. Find the shear force V and bending moment M at distance x from the free end of the beam
  14. 14. Flexural strains

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