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bok:eng:mech:ini [2021/05/15 14:47] anwlur |
bok:eng:mech:ini [2021/05/16 11:49] (current) anwlur |
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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? | 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? | ||
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+ | <hidden Click here to see how the below formula is derived> | ||
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. | 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. | ||
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Therefore $ R = F_1 + F_2 $ and $ M = F_1 . x_1 + F_2 . x_2 $ | Therefore $ R = F_1 + F_2 $ and $ M = F_1 . x_1 + F_2 . x_2 $ | ||
- | If $ M_x = R.x + M $ then $ R(x + \delta x) = M + \delta M $ | + | 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) $ |
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+ | This rearranges to $ \delta M = M - R.x - R. \delta x - M_x $ | ||
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+ | Which simplifies to $ \delta M = -R. \delta x $ | ||
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+ | 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}} $ | ||
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+ | </hidden> | ||
- | Therefore $ (R.x - M) + R.\delta x = \delta M $ | + | $ V = {{dM} \over {dx}} $ or Shear is equal to the rate of change of Moment along a beam. |