GENG2004 - Lecture 7

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GENG2004 - Lecture 6
Axially Loaded Members - Slides

Deformation Under Axial Loading

• Force-displacement relation $-$ relationships between forces acting on a bar and its changes in lengths

• Axial means that a force is applied directly along the longitudinal or central axis of an object

  
  i.e. force acts parallel to length of object

• Displacement is change in length → has units of length

  • Do not confuse with strain which is dimensionless
  • Your answers for displacement should usually be in $\text{mm}$ scale

Displacement formula:
$\delta =\frac{PL}{AE}$

• $P$ $-$ pressure
• $L$ $-$ length

When parameters vary, we use integration:
$\delta ={\int }_0^L\frac{P(x)}{EA(x)}dx$

• ? Cantilever $-$ structural element that is firmly attached at one end and unsupported at the other

Example: for cantilever beams, start at free end

We get values for the internal forces:
$P_1=240\text{kN},P_2=-60\text{kN},P_3=120\text{kN}$
We can then solve our displacement equation for each segment of the beam and sum the results together to get total displacement:
$\delta ={\sum }_i\frac{P_i{L}_i}{A_i{E}_i}=\frac{1}{E}(\frac{P_1{L}_1}{A_1}+\frac{P_2{L}_2}{A_2}+\frac{P_3{L}_3}{A_3})$

Statically Indeterminate Structures

Recap: ENSC2004 - Statically Indeterminate Structures

• Structures that have more unknown forces than equilibrium equations are known as statically indeterminate
  • Not possible to determine reaction forces using equilibrium alone when unknowns > equations
• To analyse such structures → supplement equilibrium eqns with additional eqns pertaining to the displacements of the structure

Compatibility Equations

• Compatibility equation expresses the fact that the change in length of the bar must be compatible with the conditions at the supports

Example of Using a Compatibility Equation

• Imagine a concrete column reinforced with steel rods
• When you press down on it, the steel and concrete must shorten by the exact same amount because they are bonded together
• We represent this using a compatibility equation:
  ${\delta }_{steel}={\delta }_{concrete}$
• Without it, you wouldn’t be able to figure out how much of the load each material is carrying

Full example in slides $-$ determine reactions at $A$ and $B$:

• More supports than required for equilibrium → statically indeterminate

  

• Redundant reactions replaced with unknown loads

  

  • This is known as superposition method

• Deformations due to actual loads and redundant reactions are determined separately and then added
  $\delta ={\delta }_L+{\delta }_R=0$
  ​This is our complexity equation!

  • We know that total deformation must be zero as the object is fixed at both ends

• Our equilibrium equation:
  $R_A+R_B=300+600=900\text{kN}$

Temperature-Displacement Relation

• Temperature change results in change in length or thermal strain
  ${ϵ}_T=\alpha \Delta T$
  • $\alpha$ $-$ coefficient of thermal expansion
• No stress associated with thermal strain unless elongation is restrained by the supports; unrestrained displacement:
  ${\delta }_T=\alpha \Delta T\cdot L$
• If restrained: treat additional support as redundant, apply superposition
  $\delta ={\delta }_T+{\delta }_P=0$
  • Thermal strain is prevented as object is constrained
  • Mechanical displacement: ${\delta }_P=\frac{PL}{AE}$
• These equations rearranged to:
  $P=-AE\alpha (\Delta T),\sigma =-E\alpha (\Delta T)$

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Stress Concentrations - Slides

Stress Concentrations

• Our equations for stress in axially loaded bars assume that stress distribution is uniform

• In reality, bars may have holes, groves, notches, etc.

  • These disruptions create high concentrations of stress in the bar
    • Known as stress concentrations
  • The discontinuities themselves are known as stress raisers

• $K$ $-$ stress concentration factor
  $K=\frac{{\sigma }_{max}}{{\sigma }_{avg}}$

Stress Concentration: Circular Hole

For a circular hole in an object (example uses a flat bar under axial loading):
$\begin{array}{cc}K=3& (x_1=\alpha ,x_2=0)\\ K=-1& (x_1=0,x_2=\alpha )\end{array}$
At these specific points on the hole, $K$ has this value

• Stress is forces to curve sharply around the hole above:
  • The top and bottom of the hole are under tension
  • The left and right sides of the hole are under compression
• ! This logic seems reversed to what is intuitive, but it is due to the way stress flows around the hole
• We also get a ${\sigma }_{max}=3{\sigma }_{avg}$ as seen by our first $K$ equation

Assumptions:

• The bar/plate is infinite or “very large”
• Linear elasticity & small deformation
• Perfectly circular hole (in reality we would have defects)

Stress Concentration: Fillet

• Fillets are very common in engineering
  • Found in shafts, shoulders, brackets, etc.
• As $r\to 0,K\to \infty$; fillets are useful for reducing stress concentration

• No single closed-form equation for fillets, unlike circular holes
  • This is because fillets have more relevant dimensions ($r,d,D$)
    $K=f(\frac{D}{d},\frac{r}{d})$
    
• You will need to use a graph like the one above to find $K$
  • Using values for $r,d$ and $D$, find corresponding $K$ on graph

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GENG2004 - Lecture 8