This discussion focuses on calculating shear forces and bending moments in a beam subjected to various loads, specifically analyzing the behavior at points along the beam. The key point is that the shear force at x = 6 remains 49 kN due to the cumulative nature of shear forces and bending moments, which are integrals of the load curve. The sudden jump in shear force from -40 kN to 89 kN at x = 2 is attributed to the reaction force RA of 129 kN being applied. The shear force diagram illustrates these discontinuities effectively, showing how concentrated loads affect shear force values.
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one more thing , why at x = 6 , the shearing force is still 49kN as in the calculation ? why shouldn't it be = 0 ?David Lewis said:
Because the reaction RA is located at x = 2. As you move along the beam from left to right, the shear force has a sudden jump from V = -40 kN to V = 89 kN, as the reaction RA = 129 kN is added.foo9008 said:
Why should the shear be zero at x = 6?foo9008 said:
sudden jump means there are 2 values of RA at x =2 ? why ?SteamKing said:
Because they are each integrals. The shear force is the integral of the load curve w.r.t. length, and the bending moment is the integral of the shear curve, also w.r.t. length.foo9008 said:
No, there is only one value of RA, namely RA = 129 kN.foo9008 said:
