Buckling and failure criteria
導入済み| English | 意味 |
|---|---|
| buckling | Loss of structural stability, often causing a compressed slender member to deflect sideways. |
| eccentric loading | Loading whose line of action does not pass through the section's centroid, producing bending as well as direct stress. |
Sideways failure
- A slender push member fails by buckling 屈曲 at a fraction of its yield load. Euler for a pinned column: $P_{cr} = \pi^2EI/(KL)^2$ - a steel rod of $d = 20\ \mathrm{mm}$, $L = 1.0\ \mathrm{m}$ carries only $P_{cr} = \mathbf{15.5}$ kN, at $\sigma_{cr} = \mathbf{49.3}$ MPa.
Steel rod d = 20 mm, L = 1.0 m pinned: Euler load (one decimal, kN)?
難易度 2/5 · 140 コイン + 今日初回達成時の小ボーナス
pi^2 E I over L^2.
Its critical stress (one decimal, MPa)?
難易度 2/5 · 140 コイン + 今日初回達成時の小ボーナス
P_cr over A.
Slenderness KL/r with r = 5 mm (integer)?
難易度 2/5 · 140 コイン + 今日初回達成時の小ボーナス
1000/5.
Two curves, one factor
- Slenderness $\lambda = KL/r$: the rod runs $1000/5 = \mathbf{200}$, well past the transition $\lambda_t = \sqrt{2\pi^2E/\sigma_y} = \mathbf{125.7}$ where $\sigma_{cr} = \sigma_y/2 = 125$ MPa exactly; below it Johnson gives $250 - 28.5 = \mathbf{221.5}$ MPa at $\lambda = 60$.
- Design at a factor: $P_{allow} = P_{cr}/n = \mathbf{6.2}$ kN at $n = 2.5$ - the factor is part of the answer, not a afterthought.
Euler, Johnson, or superpose?
Route each stability question to its regime.
Transition slenderness sqrt(2 pi^2 E / sigma_y) (one decimal)?
難易度 2/5 · 140 コイン + 今日初回達成時の小ボーナス
with E = 200 GPa, sigma_y = 250 MPa.
Johnson stress at lambda = 60 (one decimal, MPa)?
難易度 2/5 · 140 コイン + 今日初回達成時の小ボーナス
250 - 28.5.
Below the transition slenderness, the column stress comes from:
難易度 1/5 · 100 コイン + 今日初回達成時の小ボーナス
Euler would exceed yield there; Johnson meets it at sigma_y/2.
Allowable load for P_cr = 15.5 kN at n = 2.5 (one decimal, kN)?
難易度 2/5 · 140 コイン + 今日初回達成時の小ボーナス
P_cr / n.
Judging the state
- Eccentric loading 偏心加载 superposes: a $30\ \mathrm{mm}$ rod with $P = 5\ \mathrm{kN}$ at $e = 60\ \mathrm{mm}$ carries axial $7.07$ plus bending $113.2$ MPa, peaking at $\mathbf{120.3}$ MPa.
- Failure criteria read the principals: von Mises on $(109.1, 30.9)$ gives $\sigma' = \sqrt{\sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2} = \mathbf{97.4}$ MPa, under the 250 MPa yield; max-shear reads $\mathbf{39.1}$ against its $125$ MPa limit - both agree, no yield.
Rod d = 30 mm, P = 5 kN at e = 60 mm: far-fibre stress (one decimal, MPa)?
難易度 2/5 · 140 コイン + 今日初回達成時の小ボーナス
P/A + Pec/I.
A column load applied off the axis adds bending stress to the axial stress.
難易度 1/5 · 100 コイン + 今日初回達成時の小ボーナス
Superpose P/A and Pec/I on the extreme fibre.
Von Mises stress for principals (109.1, 30.9) MPa (one decimal)?
難易度 2/5 · 140 コイン + 今日初回達成時の小ボーナス
sqrt(s1^2 - s1 s2 + s2^2).
Three stability slips
- Euler below the transition. It would exceed yield - impossible.
- $K$ by habit. Fixity comes from the end conditions.
- Criteria on the x-y stresses. Transform first, then judge.
Recap
- $P_{cr} = 15.5$ kN at $\lambda = 200$; $\lambda_t = 125.7$ splits Euler from Johnson (221.5 MPa at $\lambda = 60$).
- $P_{allow} = P_{cr}/n = 6.2$ kN at $n = 2.5$.
- Eccentric peak 120.3 MPa; von Mises $\sigma' = 97.4$ MPa on (109.1, 30.9) - no yield.