TAM 2xx References

Navigate

Axial Loading

Notation and Convention

Fig: Notation

Taken from TAM251 Lecture Notes - L4S4

  • Displacement of points in space: the movement of a point relative to its initial position in space (ex; \( \delta_B \) )
  • Change in length of a segment: the relative displacement of a point with respect to another point (ex; \( \delta_{B/A} = \delta_B - \delta_A = \delta_1 \) ). This can be written multiple ways:
    $$ \delta_{AD} \equiv \delta_{DA} \equiv \delta_{A/D} \equiv \delta_{D/A}\ $$
Sign Conventions

BSM: for the forces, it would be wise to note the convention applies specifically to *internal* force in a bar.

  • \( F>0 \) : tension
  • \( F<0 \) : compression
  • \( \delta>0 \) : elongation
  • \( \delta<0 \) : contraction

Saint-Venant's Principle: slender beam case

Stress analysis very near to the point of application of load \( P \). Saint-Venant's principle: the stress and strain produced at points in a body sufficiently removed* from the region of external load application will be the same as the stress and strain produced by any other applied external loading that has the same statically equivalent resultant and is applied to the body within the same region. *farther than the widest dimension of the cross section

BSM: a figure would go a long way to clarifying this point, Hibbeler textbook probably has something to use for inspiration.

Force-Deformation Relation

$$ \delta = \frac{PL}{EA}\ $$
Axial Flexibility:
$$ \delta = fP => f = \frac{L}{EA}\ $$
Axial Stiffness:
$$ P = k\delta => k = \frac{EA}{L}\ $$
*Expandable Derivation*
$$ P = \sigma A\ $$
$$ \sigma = E\varepsilon\ $$
$$ \varepsilon = \delta / L\ $$
$$ => P = E\frac{\delta}{L}A\ $$
**End Derivation**

Axially Varying Properties

For non-uniform load, material property and cross-section area:
$$ \delta = \int_0^L\frac{F(x)}{E(x)A(x)}dx\ $$
Assume variations with \( x \) are "mild" (on length scale longer than cross-sectional length scales)

*Expandable Derivation*

$$ \sigma = E\varepsilon\ $$
$$ \frac{F(x)}{A(x)} = E(x)\varepsilon(x)\ $$
$$ \frac{F(x)}{A(x)} = E(x)\frac{d\delta(x)}{dx}\ $$
$$ d\delta = \frac{F(x)}{E(x)A(x)}dx\ $$

**End Derivation**

Principle of Superposition

Superposition: If the displacements are (1) small and (2) linearly related to the force components acting, the displacements caused by the components can be added up:
$$ \delta = \sum_i \delta_i = \sum_i \frac{F_i L_i}{E_i A_i}\ $$

General Solving Procedure

  1. Draw a FBD
  2. Equilibrium equations: force balance and moment balance
  3. Constitutive equations: stress-strain or force-displacement relations
  4. Compatibility equations: geometric constraints

Statically Determinate Problems

Fig: StaticallyDeterminate

Taken from TAM251 Lecture Notes - L4S9

All internal forces can be obtained from equilibrium analysis only

Statically Indeterminate Problems

Fig: StaticallyIndeterminate

Taken from TAM251 Lecture Notes - L4S10

Equilibrium does not determine all internal forces.

Thermal Effects: Temperature Changes

Notation

  • \( \Delta T > 0, \sigma < 0 \) : Compression
  • \( \Delta T < 0, \sigma > 0 \) : Tension

\( \delta_T \) , \( \varepsilon_T \) present in addition to elastic \( \delta_E \) , \( \varepsilon_E \) (from internal forces). Superposition (small strains):

$$ \varepsilon_{tot} = \varepsilon_{E} + \varepsilon_{T}\ $$
$$ \delta_{tot} = \delta_{E} + \delta_{T}\ $$

Fig: Temperature_NoLoad

Taken from TAM251 Lecture Notes - L4S17

A rod rests freely on a smooth horizontal surface. Temperature of the rod is raised by \( \Delta T \). Rod elongates by an amount.
$$ \delta_{T} = \alpha \Delta T L\ $$
Linear coefficient of thermal expansion \( \alpha \), \( [\alpha] = \frac{1}{K},\frac{1}{°C},... \). This deformation is associated with an average thermal strain:
$$ \varepsilon_{T} = \frac{\delta_T}{L} = \alpha T\ $$
Fig: Temperature_TwoPlates

Taken from TAM251 Lecture Notes - L4S17

Initially, rod of length \( L \) is placed between two supports at a distance \( L \) from each other. With no internal forces, there is no stress or strain.
$$ R_{A} = R_{B} = 0\ $$
$$ R_{A} = F\ $$
After raising the temperature, the total elongation of the rod is still zero. The total elongation is given by:
$$ \delta = \frac{FL}{EA} + \alpha L \Delta T = 0\ $$
The stress in the rod due to change in temperature is given by:
$$ \sigma = -\alpha E \Delta T\ $$

Misfit Problems

A misfit problem is one in which there is difference between a design distance and the manufactured length of a material. Some misfits are created intentionally to pre-strain a member. (e.g. spokes in a bicycle wheel or strings in a tennis racket). This type of problem neither modifies the equilibrium equations (1) nor the force-extension relations, (2) but the compatibility equations, (3) need to be modified.

Stress Concentration

BSM: we do not cover this topic in class/hw/exams. It is covered in ME 330 and ME 371.

Highest at lowest cross-sectional area. Stress concentration factor:
$$ K = \frac{\sigma_{max}}{\sigma_{avg}}\ $$
  • Found experimentally
  • Solely based on geometry

**Reference pages have a broken link image here**