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The beam shown in Figure (2) is subjected to a uniform live load of 2.4 kN/m, a dead load of 1.0 kN/m, and a single live load

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sol given that, uniform Live load = 2.4 kN/m dead load = 1.0KNlm Single live load = 80 KN Hinge B * am 2m am am given loadingf=80KN 3.4 kN/m f forgetting max om put under e axd=1 am B am (+) am E am (8.mhat E = ( LD Area) x Loading ode + fx Cordinate

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