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bok:eng:mech:ini [2021/05/15 02:19] anwlur created |
bok:eng:mech:ini [2021/05/16 11:49] (current) anwlur |
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Considerations in design shall, at minimum, include | Considerations in design shall, at minimum, include | ||
- | * Deflections | + | * Deflections |
- | * Deformations | + | * Deformations |
- | * Static Loads vs Strength | + | * Static Loads |
- | * Dynamic Loads vs Strength | + | * Dynamic Loads |
- | Take a beam as an example, with coordinate axis superimposed over it. | + | Fundamental to the above consideration is the deduction of stresses (loads applied over a unit surface area) versus the material strength. If stresses exceed the elastic deformation limit then irreversible plastic deformation will start potentially leading to the eventual catastrophic failure of the material. |
+ | Let's consider a critter of known mass is standing on a fixed beam at a known distance from the beam origin and you want to know at which point in the beam are the stresses highest. | ||
+ | {{ :bok:eng:mech:yoda-on-a-beam.png?400 |}} | ||
+ | |||
+ | This can be abstracted as follows and, with Newton's third law, it is possible to determine reaction forces and moments. | ||
+ | |||
+ | {{ :bok:eng:mech:abstract-beam.png?400 |}} | ||
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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? | ||
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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. | ||
+ | |||
+ | 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}} $ | ||
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+ | </hidden> | ||
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+ | $ V = {{dM} \over {dx}} $ or Shear is equal to the rate of change of Moment along a beam. |