The beam supports the distributed load with $w_{0} = 6.0 \text{ kN/m}$ as shown. The reactions at the supports A and B are vertical. Part A Determine the resultant internal loadings acting on the cross section at point D. Express your answers, separated by commas, to three significant figures. $N_{D}, V_{D}, M_{D}$ kN, kN, kN\cdot m
Added by Chad W.
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Step 1
First, let's determine the reaction at support A. Since the beam is in equilibrium, the sum of the vertical forces must be zero. The distributed load of 8.0 kN/m acts over a length of 4 m, so the total load on the beam is 8.0 kN/m * 4 m = 32 kN. Therefore, the Show moreā¦
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