## Euler's Formula

For a specific material, the modulus of elasticity and proportional limit are 35,000 ksi and 50 ksi respectively. Consider a square column made from this same material that has a 10 in^2 area, and a 7 in^4 moment of inertia. What is the max column length based on the Euler formula? Assume a pin-pin connection for the column so the effective length factor (K) = 1.0.

##
__
__**Hint**

**Hint**

Critical buckling stress for long columns:

$$$\sigma _{cr}=\frac{\pi ^{2}E}{(KL/r)^{2}}$$$

where
$$E$$
is the modulus of elasticity,
$$L$$
is the unbraced column length,
$$K$$
is the effective-length factor to account for end supports, and
$$r$$
is the radius of gyration.

##
__
__**Hint 2**

**Hint 2**

Radius of gyration:

$$$r=\sqrt{\frac{I}{A}}$$$

where
$$I$$
is the moment of inertia, and
$$A$$
is the area.

Critical buckling stress for long columns:

$$$\sigma _{cr}=\frac{\pi ^{2}E}{(KL/r)^{2}}$$$

where
$$E$$
is the modulus of elasticity,
$$L$$
is the unbraced column length,
$$K$$
is the effective-length factor to account for end supports, and
$$r$$
is the radius of gyration, which is:

$$$r=\sqrt{\frac{I}{A}}=\sqrt{\frac{7\:in^4}{10\:in^2}}=0.836\:in$$$

where
$$I$$
is the moment of inertia, and
$$A$$
is the area.

A basis assumption for the buckling Euler formula is that the critical stress is less than the elastic buckling, which is the proportional limit (the point on a stress-strain curve where elastic deformation transitions into plastic deformation which is basically where the material yields). To find the max length, we can set the proportional limit and critical stress equal.

$$$\sigma _{cr}=\frac{\pi ^{2}E}{(KL/r)^{2}}\leq 50$$$

Solving for L:

$$$\frac{\pi \sqrt{E}\cdot r}{KL}=\sqrt{50ksi}$$$

$$$L=\frac{\pi \sqrt{E}\cdot r}{K\sqrt{50ksi}}=\frac{\pi\sqrt{35,000ksi}\cdot0.836in}{(1.0)\sqrt{50ksi}}=\frac{491.1in}{7.071}=69.5\:in$$$

69.5 in