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SUPPORT REACTIONS

SUPPORT REACTIONS. BY GP CAPT NC CHATTOPADHYAY. NEED FOR SUPPORT. THE LOAD CARRYING STRUCTURES NEED SUPPORTS TO AVOID -DEFORMATION -BENDING -INSTABILITY . Types of loads.

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SUPPORT REACTIONS

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  1. SUPPORT REACTIONS BY GP CAPT NC CHATTOPADHYAY

  2. NEED FOR SUPPORT • THE LOAD CARRYING STRUCTURES NEED SUPPORTS TO AVOID -DEFORMATION -BENDING -INSTABILITY

  3. Types of loads • Several types of loads can act on beams, such as concentrated loads, distributed loads, varying loads and couples. • Concentrated loads are idealized from loads applied on a very small area. Magnitudes are known . • Distributed loads are spread along the axes of beams. For example, the weight of the beam can be assumed as a distributed force. • Distributed loads are converted into equivalent forces to make the solution process easier. For vertical loads, the equivalent force is through the centroid of the distributed load. Thus, if the distributed load is a basic shape, the centroid is easy to determine without integration. • Couples are moments applied on the beam. Do not contribute to force . • Uniform varying Load In this case the load varies uniformly along the length (gradual increase, gradual decrease)

  4. POINT LOAD

  5. UDL Length = L Length = l

  6. UNIFORMLY VARYING LOAD

  7. COMBINED UDL AND UVL W= 1500N/m =

  8. TYPES OF SUPPORT • 2 (USUALLY ONE) HINGED SUPPORTS • SUPPORTS TAKE VERTICAL AND HORI…LOAD • USUALLY DESIGNED WITH A ROLLER SUPPORT FOR FREE EXPANSION OF ONE END • VERTICAL AND HORI… LOADS DETERMINE REACTION AND LINE OF ACTION • 2 OR MORE VERTICAL SUPPORTS • JUST PIVOTS • TAKES ONLY VERTICAL LOADS

  9. SIMPLE SUPPORT

  10. ROLLER SUPPORT LOCATION OF ROLLER BEARING TO SUPPORT JET ENGINE ROTOR

  11. HINGED SUPPORT KNEE HINGE

  12. COMBINED SUPPORT DISTRITIBUTED LOAD = w HINGED SUPPORT ROLLER SUPPORT

  13. I THINK…. I CAN THINK….

  14. Calculate the support reactions • Solution: First change UDL in to point load.Resolved all the forces in horizontal and vertical direction. Since roller at B (only one vertical reaction) and hinged at point B (one vertical and one horizontal reaction).Let reaction at hinged i.e., point B is RBH and RBV, and reaction at roller support i.e. point D is RDV Let ΣH & ΣV is the sum of horizontal and vertical component of the forces ,The supported beam is in equilibrium, hence     ΣH = ΣV = 0    RH = RBH = 0    RBH = 0                                       ...(i)    ΣV = RBV –50 –5 – RDV = 0 RBV + RDV = 55                                 ...(ii)Taking moment about point B50 × 0.5 – RBV × 0 – RDV × 5 + 5 × 7 = 0RDV =12 KN                                .......ANSPutting the value of RBV in equation (ii)RBV = 43KN                                 .......ANSHence RBH = 0, RDV = 12KN, RBV = 43KN

  15. NUMERICAL • A 9m long beam ABCD is supported on a hinge at ‘A’ and a roller support at an angle of 45o at ‘D’. The beam has a clockwise moment of 9 kNm at ‘B’ and a point load of 6 kN at ’C’. If the points are equidistant, find the support reactions.

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