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--- bok:eng:mech:ini [2021/05/15 12:38]
anwlur
+++ bok:eng:mech:ini [2021/05/16 11:49] (current)
anwlur
@@ Line 17: / Line 17: @@
 Let's determine the forces and moments on the beam on an arbitrary distance from the beam origin. To so we make examine the forces and moments at x that keep the beam in equilibrium. This force is parallel with the beam's cross section is called **Shear** and V is its symbol. But how are Moment at x related to Shear at x?
-We know that $ R = F_1 + F_2 $ and $ M = F_1 . x_1 + F_2 . x_2 $
+<hidden Click here to see how the below formula is derived>
-We also know that $ M_x = R.x $ but what about $ R(x + \delta x) $?
+For a body in equilibrium we know $ \Sigma F = 0 $ and $ \Sigma M = 0 $ Taking a positive force in direction of axis and a positive moment follows the Right Hand about axis.
+Therefore $ R = F_1 + F_2 $ and $ M = F_1 . x_1 + F_2 . x_2 $
+If the moment at distance x is $ M_x $ then $ M_x = M - R.x $ then $ M_x + \delta M = M - R(x + \delta x) $
+This rearranges to $ \delta M = M - R.x - R. \delta x - M_x $
+Which simplifies to $ \delta M = -R. \delta x $
+Which rearranges to $ R = - {{dM}  \over {dx}} $ but as the force acting on the cross section is opposite to the reaction force at the origin of the beam then we can generalize that $ V = {{dM}  \over {dx}} $
+</hidden>
+$ V = {{dM}  \over {dx}} $ or Shear is equal to the rate of change of Moment along a beam.

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