ENSC2004 - Lecture 9

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ENSC2004 - Lecture 8
Bending Moment & Shear Force Diagrams - Slides

Goal: Construct shear force and bending moment diagrams for beams for help in analysing critical sections in engineering design

Generalising Internal Forces

• We want a diagram that shows shear force at every point along beam
• We can do this through using functions:
  • Make imaginary cut at $x$ distance away from $O$
  • $x$ is a variable, meaning we solve for $\text{V}$ & $\text{M}$ in terms of $x$
  • We end up with a function of $\text{V}$ and $\text{M}$:
    $\text{V}=\text{V}(x)$
    $\text{M}=\text{M}(x)$

Shear Force and Bending Moment Diagrams

• We can plot this data on a SFD or BMD

  • SFD $-$ shear force diagram
  • BMD $-$ bending moment diagram

• On your diagram, you must include key points!

  • Label function
  • Label max/min
  • Label endpoints

• A shear force diagram is simply a plot of the shear force function

  • e.g. for $V=15-5x$:
    

• Bending moment diagrams sometimes plotted positive moments down

  • i.e. the vertical axis is flipped (not always though!)
    e.g. for $M=15x-2.5x^2$:
    

Sometimes we may have different functions for $\text{V}$ and $\text{M}$ over different ranges for $x$

e.g. if a load is near the middle of the beam, both sides of the load may have different functions

• In this case, we ‘split’ the beam

For a function to be valid for a region of the beam, the distributed loading must be continuous over that region

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Beam Bending Stresses - Slides

Choosing a Beam

• Bending moments stress > shear stress
  $\therefore$ beams always break under bending moments $(\text{M})$
  • ${\text{M}}_{max}$ on our BMD is the critical section
• We need to choose a beam that can hold up under the given moment
  • Recap: Deformation Behaviour
  • Need to limit stress within the linear area of a stress-strain graph

Internal Stresses

• When a beam is bent, top under compression, bottom under tension
  
  • At some point, there is neither tension nor compression
  • This point is known as the neutral axis (N.A.)
    • Located at centroid of cross-section
• Therefore, the bending strain ${ϵ}_b$ is proportional to coordinate $y$
  • $y$ is the distance from the neutral axis

Bending stress (${\sigma }_b$) related to bending strain through Young’s modulus
${\sigma }_b=E{ϵ}_b$

• This implies that the bending stress is also proportional to the distance from the N.A.
  ${\sigma }_{b(max)}=\frac{M_{max}{y}_{max}}{I}\le {\sigma }_{all}$

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ENSC2004 - Lecture 10