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simply supported beams

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simply supported beams subjected to concentrated and udls

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simply supported beams

  1. 1. ANKIT SOOD CHRISTINA MAYINI DHEER VEER VIKRAM SINGH HITESH SOOD SIDDHARTH MAHAJAN VISHAL MISHRA
  2. 2. The purpose of this calculation is to obtain information about shear, bending moment, and deflection distribution over the length of a beam, which is under various transverse loads: couples, concentrated and linearly distributed loads. The result of calculation is represented by shear force, bending moment and deflection diagrams. The material of the beam is linear-elastic and isotropic with elasticity modulus E. All loads are lateral (forces or moments have their vectors perpendicular to the beam axis) and acting at the same plane. All deflections occur in this plane of bending. Deflections are small compared to the length of the beam. In this case we apply equilibrium equations to the unreformed beam axis (or its parts) and assume that the curvature of the deformed beam axis is equal to the second derivative of the deflection function.
  3. 3. RELATION BETWEEN LOADING, S.F, B.M, SLOPE AND DEFLECTION
  4. 4. Consider a simply supported uniform section beam with a single load F at the centre. The beam will be deflect symmetrically about the centre line with zero slope (dy/dx) at the centre line. It is convenient to select the origin at the centre line.
  5. 5. Consider a simply supported uniform section beam with a Concentrated Load and UDL The B.M Equation is: w W1 b c 3 b c a w b 6 c l b c + Note that Macaulay terms are integrated with respect to, for example, (x -a) and they must be ignored when negative. Substitution of end conditions will then yield the values of the constants A and B in the normal way and hence the required values of slope or deflection.
  6. 6. Simply supported beams when subjected to multiple loadings yields graphs of these nature

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