ENSC2004 - Lecture 4

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ENSC2004 - Lecture 3
Distributed Loads - Slides

Distributed Loads

• A distributed load is a force spread over an area, rather than being concentrated at a single point
  • Forces in real life are distributed loads
• Normally, when doing physics problems, we only consider forces that are concentrated at a point
  • We need a way to translate a distributed load to a point force
    i.e. we need to find the resultant force $F_R$

Magnitude of Resultant Force

• Typically, a distributed load is represented using a function
  • $w$ is a function of $x$ $[w=w(x)]$ and has units of force per length
  • The force magnitude $dF$ acting on it is given as:
    $dF=w(x)dx$
  • Which can be written as:$F_R={\int }_0^{L}w(x)dx=A$
    $A$ is the area under the loading curve $w(x)$

Location of Resultant Force

• The force $dF$ will produce a moment of $(x)(dF)$ about point $O$
• The total moment about point $O$ is given as: $\circlearrowleft +(M_o{)}_w=-\sum dF\times x=-{\int }_0^{L}x\times w(x)dx$
• Assuming that $F_R$ acts at $\bar{x}$, it will produce the moment about point $O$ as: $\circlearrowleft +(M_o{)}_R=-\bar{x}{F}_R$
  
• Comparing and simplifying the last two equations we get: $\bar{x}=\frac{{\int }_0^{L}x\times w(x)dx}{{\int }_0^{L}w(x)dx}=\frac{{\int }_0^{L}xdA}{{\int }_0^{L}dA}$

$F_R$ acts through a point “$C$,” which is called the geometric centre or centroid of the area under the loading curve $w(x)$

Centroid Table

• The location of the force is always at the centroid of the load distribution (diagram)
  • We can use a centroid table to find the centroid (geometric centre) of a composite shape

Note: for triangles, the values for centroid are going from the wide end. To go from the narrow end, the formula is $\frac{2L}{3}$.

• Using the centroid table may make the process easier, however, the method is mathematically the same

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Equilibrium of Rigid Bodies - Slides
Recap: Basic Force Vector Techniques & Resultant for Point Loads

Conditions for Rigid-Body Equilibrium

• For a rigid body to be in equilibrium:
  1. Net force = 0
  2. Net moment around any point = 0
• All physical bodies are 3D, however, we can treat many of them in statics as 2D when forces are applied or projected in the same plane

Conditions for 2D Equilibrium:

• $\sum F_x=0$
• $\sum F_y=0$
• $\sum M=0$

Support Reaction in 2D

• If a support prevents translation of a body in a given direction, then a force is developed on the body in the opposite direction
• Similarly, if rotation is prevented, a couple moment is exerted on the body in the opposite direction

Solving Rigid-Body Equilibrium Problems

1. Create idealised model
   • Model that approximately represents the real situation as closely as possible
     
     More examples on slides
2. Draw free-body diagram showing all external (active and reactive) forces
   • Make sure to include dimensions between all key points
3. Apply equations of equilibrium to solve for any unknowns

As an engineering in the real world, you need an idealised model to represent your scenario! Creating idealised models is a key skill to become an engineer.

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ENSC2004 - Lecture 5