- #1
why the author taking the moment about B and 2m from A ? (in the first solution , the author taking moment about the point which 2m from A )( in the second solution , the author taking moment about B , am i right ?We are looking for the moment area so the formula would be (area of triangle)(centroidal moment of area) and that would be (0.5)(2)(800) as triangle area and (4/3) as the centroid.
i dont understand , can you explain further?He considered that is why there are two solutions (AreaAB)XA and (AreaAB)XB denoting two moment areas
then, why the author include the moment 400Nm in the second solution ????You should review how to draw moment diagrams. It is already known that when we derive from shear to moment diagrams , any given moment is considered a straight line thus no area and that is why it isn't included in computing the moment area.
what do you mean by separate areas for every reaction???Again , you should read on how to solve moment areas. The 2nd solution is the separate areas for every reaction or simply it is another way to solve the problem. Also you could see that he took the moment area about A to solve the problem and that is why there is no moment diagram for reaction A.