Solid Mechanics Reference

    Buckling

    Buckling is the sudden change in shape of a structural component under a compressive load

    Taken from TAM251 Lecture Notes - L13S2

    The beam is still able to withstand normal loads, but buckling causes an instability. Small perturbations make the structure unstable. Failure is elastic (\( \sigma < \sigma_Y \)), but if increased loads are applied, further deformation and plastic failure (yielding) / brittle failure (fracture) can occur (post-buckling failure).

    Single Column

    Taken from TAM251 Lecture Notes - L13S3

    Column \( AB \) is supporting uniaxial compressive load \( P \). To properly design this column, the cross-section must satisfy the following:
    $$ \sigma = \frac{P}{A} \le \sigma_{all}\ $$
    $$ \delta = \frac{PL}{EA} \le \delta_{spec}\ $$
    Increasing the load can cause the column to buckle \( \rightarrow \) instability causing failure.

    Two Rods and a Torsional Spring

    Taken from TAM251 Lecture Notes - L13S4

    Rods \( AC \) and \( CB \) are perfectly aligned and a torsional spring connects them at point \( C \). For small perturbations, point \( C \) moves to the right.
    • If
    • If

    Taken from TAM251 Lecture Notes - L13S5

    The spring restoring moment is
    $$ M_s = K(2\Delta\theta)= \text{restoring moment}\ $$
    The moment resultant from the applied load P tends to move the rod away from the vertical position
    $$ M_{load} = P\frac{L}{2}\sin\Delta\theta = P\frac{L}{2}\Delta\theta = \text{destabilizing moment}\ $$
    • Stable system:
    • Unstable system:
    • Equilibrium position gives:
    The critical load can be found with
    $$ P_{cr} = \frac{4K}{L}\ $$

    **Expandable Derivation**

    $$ M_s = M_{load}\ $$
    $$ K(2\Delta\theta) = P_{cr}\frac{L}{2}\Delta\theta\ $$

    **End Derivation**

    Euler's formula

    Euler's formula can be used to solve for the critical load of a uniaxially loaded column.

    Pinned-end Columns

    Taken from TAM251 Lecture Notes - L13S6

    Rod \( AB \) is pinned on each end. Equilibrium gives
    $$ M = -Py\ $$
    After a small perturbation, the system reaches equilibrium
    $$ M(x) = EIy $$
    $$ EIy $$
    $$ y $$
    Linear, homogeneous differential equation of second order with constant coefficients. The general solution is
    $$ y(x) = A\sin(px) + B\cos(px)\ $$
    With boundary conditions
    $$ y(0) = y(L) = 0\ $$
    Euler's Formula
    $$ P_{cr} = \frac{\pi^2EI}{L^2}\ $$
    Buckling occurs at
    $$ P > P_{cr}\ $$

    **Expandable Derivation**

    $$ y(x) = A\sin(\sqrt{\frac{P}{EI}}x) + B\cos(\sqrt{\frac{P}{EI}}x)\ $$
    $$ y(x=0)=0 \rightarrow A\sin(0)+B\cos(0) = 0\ $$
    $$ B=0\ $$
    $$ y(x=L)=0 \rightarrow A\sin(\sqrt{\frac{P}{EI}}L)+0 = 0\ $$
    $$ A\sin(\sqrt{\frac{P}{EI}}L)=0\ $$
    This has two solutions

    $$ A = 0 \rightarrow \text{not interesting}\ $$

    $$ A = n \rightarrow n \text{any number except where} A\sin(\sqrt{\frac{P}{EI}}L) = n\pi $$
    $$ \frac{P}{EI}L^2 = n^2\pi^2\ $$
    $$ P_{cr} = \frac{n^2\pi^2EI}{L^2}\ $$

    Buckling usually happens at the smallest non-zero value of \( P_{cr} \)

    $$ n=1\ $$

    Higher \( n \) values can be achieved if columns are prevented from buckling at \( n=1 \)

    **End Derivation**

    Other Boundary Conditions

    Taken from TAM251 Lecture Notes - L13S8

    Different boundary conditions change length used in the critical load formula to the effective length (\( L_e \)).

    $$ P_{cr} = \frac{\pi^2EI}{L_e^2}\ $$