- 3 The Diagram Shows A Simply Supported Beam Assume The Beam Has No Mass And Is In Static Equilibrium Udl 18 Kn M 5 K 1 (62.11 KiB) Viewed 18 times
3 The diagram shows a simply supported beam. Assume the beam has no mass and is in static equilibrium. UDL = 18 kN/m 5 K
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3 The diagram shows a simply supported beam. Assume the beam has no mass and is in static equilibrium. UDL = 18 kN/m 5 K
3 The diagram shows a simply supported beam. Assume the beam has no mass and is in static equilibrium. UDL = 18 kN/m 5 KN UDL= W kN/m B 0.5 m 0.5 m 1.5 m 0.5 m Diagram not to scale (a) Calculate the equivalent concentrated load produced by the uniformly distributedload (UDL). (b) Calculate the vertical reaction force at point A. A concentrated load of 20kN is added that acts downwards, at point C. To keep the beam in static equilibrium, an additional concentrated load (D) of unknown magnitude is also added to the beam. UDL = W kN/m 5 kn 20 KN 1 B 10.5 m 0.5 ml 1.5 m 10.5 m Diagram not to scale. (C) Ignore the additional concentrated load D. Calculate the turning moment at point B due to the additional concentrated loadat point C, the UDL and the 5kN concentrated load. (d) Calculate the magnitude of the additional concentrated load (D) to keep the beam in static equilibrium.