Statically Indeterminate Structures

Change in Length Due to Axial Loads

Elongation in a prismatic bar (linear elastic region): $\delta = \frac{PL}{EA}$

Bars with Intermediate Axial Loads:

1. Identify each section

2. Determine internal axial loads in each section

3. Determine change of length in each section using: $\delta = \frac{PL}{EA}$

4. Add together all the changes in length to find the total change of length $\Sigma\delta$

Bars with prismatic sections: $\delta = \sum_{i = 1}^{n}\frac{N_{i}L_{i}}{E_{i}A_{i}}$

Bars with varying loads and dimensions: $\delta = \int_{0}^{L}d\delta = \int_{0}^{L}\frac{N(x)dx}{EA(x)}$

Statically Indeterminate Structures

Statically Determinate Structures: Structures where the reactions and internal forces can be found using equilibrium equations

Statically Indeterminate Structures: Structures where the equilibrium equations are not enough to solve for the reactions. The other equations used are compatibility equations.

Compatibility Equations: The change in length of the statically indeterminate bar must match the conditions at the support. If both supports are fixed, the length does not change. $\delta_{A} + \delta_{B} = \delta_{AB}$

Force-Displacement Relations: The relationship between the force of the reactions and the change in length. The force-displacement relationship can be used in the compatibility equations. $\delta_{A} = \frac{R_{A}L_{A}}{EA}$

3 Step Approach:

1. Equilibrium Equations and FBD

2. Constitutive Equations (PLEA equation)

3. Compatibility Equations (relating the PLEA equations)

Degree of indeterminacy = the number of unknowns - the number of equilibrium equations

Example: Problem 2.4-19:

Given: $W = 7200\ lb$, the outer rods are aluminum, $E_{1} = 10 \times 10^{6}psi$, $d_{1} = 0.4\ in.$, $L_{1} = 40\ in.$, the inner rod is magnesium, $E_{2} = 6.5 \times 10^{6}\ psi$, $\sigma_{1} = 24,000\ psi$, and $\sigma_{2} = 13,000\ psi$

Find: $L_{2}$ and $d_{2}$ when all three rods are loaded to their maximum values.

1. FBD and equilibrium equations

$\uparrow \Sigma F_{y} = 0 = 2F_{1} + F_{2} - W$

$7200\ lb = 2F_{1} + F_{2}$

$F_{1} = \sigma_{1}A_{1} = 24,000\ psi \cdot \frac{\pi}{4}{(0.4in)}^{2} = 960\pi\ lb$

$F_{2} = \sigma_{2}A_{1} = 13,000\ psi \cdot \frac{\pi}{4}d_{2}^{2} = 3250\pi d_{2}^{2}\ lb$

$7200\ lb = 2(960\pi\ lb) + 3250\pi d_{2}^{2}\ lb$

$d_{2} = 0.338\ in.$

$F_{2} = 372\pi$

1. Constitutive equations

$\delta_{1} = \frac{F_{1}L_{1}}{E_{1}A_{1}} = \frac{(960\pi\ lb) \cdot (40\ in.)}{(10 \times 10^{6}\ psi) \cdot (\pi/4){(0.4\ in.)}^{2}} = 0.096\ in.$

$\delta_{2} = \frac{F_{2}L_{2}}{E_{2}A_{2}} = \frac{(372\pi\ lb) \cdot L_{2}}{(6.5 \times 10^{6}\ psi) \cdot (\pi/4){(0.338\ in.)}^{2}} = 0.002L_{2}\ in.$

1. Compatibility Equations

Since the change in length of the rods has to be the same, the compatibility equation is:

$\delta_{1} = \delta_{2}$

$0.096\ in.\ = \ 0.002L_{2}\ in. \Rightarrow L_{2} = 48\ in.$