Chapter 9: Stability of Columns
The load-carrying capacity of structure members:
- Strength 强度，抵抗破坏
- Rigidity 刚度，抵抗变形
- Stability 稳定性，抵抗失稳
History
- 1729, 穆申布洛克
- Lagrange
- Euler
9.1 Concept of Stability
A pratical example
A wooden bar: \(b \times h = 0.5 \times 3 \, \mathrm{cm}^2, \sigma_c = 40 \, \mathrm{MPa}\)
- 屈服（yielding）：\(P = \sigma_c A = 6 \,\mathrm{kN}\)
- 横向挠度（lateral deflection）：\(P = 30 \,\mathrm{N}\)，就失稳了
横向挠度是由 屈曲（Buckling） 引起的
Mechanical model of stability
- \(P < P_{\text{cr}}\)，杆做振动回到平衡位置，稳定平衡
- \(P > P_{\text{cr}}\)，杆偏离平衡位置，失稳
- \(P = P_{\text{cr}}\)，随遇平衡
Critical force \(P_{\text{cr}}\)：使杆从稳定平衡转变为不稳定平衡的临界载荷
Consider a slender column with pinned ends, which is subjected to a compressive axial load \(P = P_{\text{cr}}\), i.e. the critical buckling load.
Equation of bending moment:
\[ M(x) = -P_{\text{cr}} \cdot y(x) \]
When \(\sigma < \sigma_p\),
\[ \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = \frac{M(x)}{EI} = -\frac{P_{\text{cr}}}{EI} y(x) \]
Let \(k^2 = \frac{P_{\text{cr}}}{EI}\), we get
\[ y = A \sin kx + B \cos kx \]
Boundary conditions: \(\left. y\right|_{x=0} = 0, \, \left. y\right|_{x=l} = 0\), we have \(B = 0, \, \sin kl = 0\).
The solutions are \(kl = n\pi, \, n = 1, 2, \ldots\), so
\[ k^2 = \frac{n^2 \pi^2}{l^2} = \frac{P_{\text{cr}}}{EI} \implies P_{\text{cr}} = \frac{n^2 \pi^2 EI}{l^2} \]
杆要达到弯曲平衡，要求载荷 \(P\) 取一系列离散值！
Select \(n = 1\) (the minimum load), we have
\[ \begin{equation} \tag{9-1} \label{eq:9-1} P_{\text{cr}} = \frac{\pi^2 EI}{l^2} \end{equation} \]
\eqref{eq:9-1} is Euler's formula.
Discussion
- \(P_{\text{cr}} \propto \frac{1}{l^2}\)，\(l\) 越大，失稳载荷越小
- \(P_{\text{cr}} \propto E\)，\(E\) 越大，失稳载荷越大
- \(I\) is the least moment of inertia（可以沿任意方向失稳）
- For \(n = 1\), the deflection curve is
\[ y(x) = A \sin \left(\frac{\pi x}{l}\right) \]
where \(A\) is NOT determined (can be any value).
What if \(P = 1.1 P_{\text{cr}}\) ?
见下面补充内容！
9.3 Columns with Other Supports
\(\kappa = 0\) 处，\(M = 0\)，等效为铰支座！
\(\mu l\) 等效长度
\[ \begin{equation} \tag{9-2} \label{eq:9-2} P_{\text{cr}} = \frac{\pi^2 EI}{(\mu l)^2} \end{equation} \]
\eqref{eq:9-2} is Generalized Euler's formula.
Discussion
- \(\mu\): the effective length factor
- cylindrical pins
Example 9.1
Determine Euler's formula for a column with two built-in ends.
Sol. According to symmetry, we have
\[ \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = \frac{M(x)}{EI} = \frac{m - Py}{EI} \]
Let \(k^2 = \frac{P}{EI}\),
\[ \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} + k^2 y = \frac{m}{EI} \]
The general solution is
\[ y = A \sin kx + B \cos kx + \frac{m}{P} \]
The boundary conditions are
\[ \left \{ \begin{aligned} \left. y\right|_{x=0} &= 0 \\ \left. y\right|_{x=l} &= 0 \end{aligned} \right. \quad \left \{ \begin{aligned} \left. y'\right|_{x=0} &= 0 \\ \left. y'\right|_{x=l} &= 0 \end{aligned} \right. \]
then we get
\[ \left \{ \begin{aligned} A &= 0 \\ B &= -\frac{m}{P} \\ \cos kl &= 1 \\ \sin kl &= 0 \end{aligned} \right. \implies kl = n\pi, \, n = 0, 2, 4, \ldots \]
长细比（slenderness ratio）\(\lambda = \frac{\mu l}{i}\) dimensionless
应力不能超过比例极限
\[ \sigma_{\text{cr}} = \frac{\pi^2 E}{\lambda^2} \leq \sigma_{\text{p}} \implies \lambda \geq \pi \sqrt{\frac{E}{\sigma_{\text{p}}}} = \lambda_{\text{p}} \]
Euler's formula 不适用时（\(\lambda < \lambda_{\text{p}}\)），用经验公式
\[ \sigma_{\text{cr}} = a - b \lambda \leq \sigma_{\text{s}} \implies \lambda \geq \frac{a - \sigma_{\text{s}}}{b} = \lambda_{\text{s}} \]
- 若 \(\lambda < \lambda_{\text{s}}\)，则材料屈服，属于强度问题
| Type | Definition | \(P_{\text{cr}}\) | \(\sigma_{\text{cr}}\) | Property |
| Long | \(\lambda > \lambda_{\text{p}}\) | \(P_{\text{cr}} = \frac{\pi^2 EI}{(\mu l)^2}\) | \(\sigma_{\text{cr}} = \frac{\pi^2 E}{\lambda^2}\) | Stability |
| Intermediate length | \(\lambda_{\text{p}} < \lambda < \lambda_{\text{s}}\) | $$ | \(\sigma_{\text{cr}} = a - b \lambda\) | Stability |
| Short | \(\lambda < \lambda_{\text{s}}\) | \(P_{\text{cr}} = \sigma_{\text{s}} A\) | \(\sigma_{\text{cr}} = \sigma_{\text{s}}\) | Strength |
长细比由几何尺寸和边界条件决定
9.6 Improving the Stability of Columns
\[ \left \{ \begin{aligned} &\text{large flexibility:} \, \sigma_{\text{cr}} = \frac{\pi^2 E}{\lambda^2} \\ &\text{medium flexibility:} \,\sigma_{\text{cr}} = a - b \lambda \\ &\lambda = \frac{\mu l}{i} \end{aligned} \right. \]
- Reduce the length
- \(l \downarrow \implies \lambda \downarrow \implies \sigma_{\text{cr}} \uparrow\)
- Reasonable cross-section
- \(i \uparrow \implies \lambda \downarrow \implies \sigma_{\text{cr}} \uparrow\)
- local buckling
- Let \(I\) be the same in all directions
- Improve end supports
- \(\mu \downarrow \implies \lambda \downarrow \implies \sigma_{\text{cr}} \uparrow\)
- Rigidly fixed \(>\) Pinned end \(>\) Free end
- Reasonable materials
- \(E \uparrow \implies \sigma_{\text{cr}} \uparrow\)
补充：平衡与稳定
能量（势能面）：\(E(x)\)
- 平衡位置：\(E'(x) = 0\)
- 稳定性
- 稳定平衡：\(E''(x) > 0\)
- 不稳定平衡：\(E''(x) < 0\)
- 临界：\(E''(x) = 0\)
假想一个小球分别在山谷、平地、山顶上
材料力学中的“能量”是什么？
总势能
\[ \Pi = \Pi_1 + \Pi_2 \]
- \(\Pi_1\)：应变能
- \(\Pi_2\)：外力势能 \(\left(-\int \mathbf{F} \cdot \mathrm{d} \mathbf{u}\right)\)
一根杆沿轴向施加力，为什么出现了拉伸，而不是弯曲/扭转？\(\implies\) 实际的变形已经是能量最低的结果。
- 对于同一杆件：不同外载对应着不同的势能面 \(\Pi[\small{\text{变形}}]\)
- 对于给定外载，最小势能原理给出，在所有几何可能变形中，精确解（真实变形情况）是使得 \(\Pi\) 取最小值的变形
- 凹凸性
单轴拉伸
悬臂梁，施加轴向拉力 \(F\).
位移 \(U|_{x=0} = 0, \, U|_{x=l} = kl\)
应变 \(\varepsilon = \dfrac{\partial U}{\partial x} = k\)
\[ \left \{ \begin{aligned} \Pi_1 &= \int_V \frac{1}{2} E \varepsilon^2 \,\mathrm{d} V = \frac{1}{2} E k^2 Al \\ \Pi_2 &= -F \cdot U|_{x=l} = -F \cdot kl \end{aligned} \right. \]
最小势能原理
\[ \frac{\partial \Pi}{\partial k} = 0 \implies k = \frac{F}{EA} \]
也是对的选取，只不过精确解不在此类解中，只能求得近似解。
不限定 \(U(x)\) 的形式，严格求解，实际上就是变分法
- 工程应用中，数值求解：基函数展开法（Ritz 法）
以多项式展开为例，\(U(x) = a_0 + a_1 x + a_2 x^2 + \ldots\)
泛函极值问题转化为函数极值问题：\(\Pi[U(x)] = \Pi(a_0, a_1, a_2, \ldots)\)
\(\Pi_1(k)\) 形状固定，\(\Pi_2(k)\) 的直线斜率随着 \(F\) 的变化而变化，从而调节势能面
外力看成 \(k \to 0, l \to \infty\) 的弹簧，\(\Pi_2(k)\) 即为其弹性势能，\(\Pi\) 为体系的焓
弯曲
悬臂梁，端部施加外力 \(F\).
猜测挠度形式为 \(w(x) = ax + bx^2 + cx^3 + dx^4\)
边界条件：\(w|_{x=0} = 0, \, w'|_{x=0} = 0 \implies a = 0\)
\[ M = EI w'' = EI (2b + 6cx + 12dx^2) \]
应变能
\[ \begin{aligned} \Pi_1 &= \int_0^l \frac{M^2(x)}{2EI} \,\mathrm{d} x \\ &= \frac{EI}{2} \int_0^l (2b + 6cx + 12dx^2)^2 \,\mathrm{d} x \\ &= \frac{EI}{2} \left[ 4b^2 l + 36c^2 \frac{l^3}{3} + 144d^2 \frac{l^5}{5} + 24bc \frac{l^2}{2} + 48bd \frac{l^3}{3} + 144cd \frac{l^4}{4} \right] \\ &= EIl \left[ 2b^2 + 6c^2 l^2 + \frac{72}{5} d^2 l^4 + 6bcl + 18 cd l^3 + 8bd l^2 \right] \end{aligned} \]
外力势能
\[ \Pi_2 = -(-F) \cdot w|_{x=l} = F \cdot (bl^2 + cl^3 + dl^4) \]
能量极值条件：
\[ \begin{aligned} \frac{\partial \Pi}{\partial b} &= EIl [4b + 6cl + 8d l^2] + F l^2 = 0 \\ \frac{\partial \Pi}{\partial c} &= EIl [12c l^2 + 6bl + 18 d l^3] + F l^3 = 0 \\ \frac{\partial \Pi}{\partial d} &= EIl \left[\frac{144}{5} d l^4 + 18 c l^3 + 8 b l^2 \right] + F l^4 = 0 \end{aligned} \]
写成矩阵形式
\[ \begin{bmatrix} 4 & 6 & 8 \\ 6 & 12 & 18 \\ 8 & 18 & \frac{144}{5} \end{bmatrix} \begin{bmatrix} b \\ cl \\ dl^2 \end{bmatrix} + \frac{Fl}{EI} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 0 \]
解得
\[ \begin{bmatrix} b \\ cl \\ dl^2 \end{bmatrix} = \frac{Fl}{EI} \begin{bmatrix} -\frac{1}{2} \\ \frac{1}{6} \\ 0 \end{bmatrix} \]
若改成均匀载荷 \(q\)，则应变能 \(\Pi_1\) 不变，外力势能改为
\[ \begin{aligned} \Pi_2 &= -\int_0^l (-q) \cdot w(x) \,\mathrm{d} x = q \int_0^l w(x) \,\mathrm{d} x \\ &= q \left[ b \frac{l^3}{3} + c \frac{l^4}{4} + d \frac{l^5}{5} \right] \end{aligned} \]
若改成外力偶 \(M_{\text{e}}\)，外力势能改为 \(\Pi_2 = - (-M_{\text{e}}) \cdot w'|_{x=l}\)
单轴压缩 Reprise
假设梁的总长度 \(L\) 恒定
\[ \int_0^{L - \Delta} \sqrt{1 + w'^2} \,\mathrm{d} x = L \]
若 \(w' \ll 1\)，有以下近似
\[ \int_0^{L - \Delta} \left(1 + \frac{1}{2} w'^2\right) \,\mathrm{d} x = L \]
移项消去 \(L\)，并把积分上限近似为 \(L\)，
\[ \Delta = \frac{1}{2} \int_0^L w'^2 \,\mathrm{d} x. \]
从直杆压到弯杆，
\[ \begin{equation} \tag{*9-1} \begin{aligned} \Delta \Pi_1 &= \int_0^L \frac{M^2}{2EI} \,\mathrm{d} x = \int_0^L \frac{1}{2} EI w''^2 \,\mathrm{d} x \\ \Delta \Pi_2 &= -F \cdot \Delta = -F \cdot \frac{1}{2} \int_0^L w'^2 \,\mathrm{d} x \end{aligned} \end{equation} \]
Ritz 法
取一组完备函数系 \(\psi_i(x), \, i = 1, 2, \ldots, n\)，并且满足边界条件。将 \(w(x)\) 展开为
\[ w(x) = \sum_{i=1}^n a_i \psi_i(x) \]
总势能为
\[ \begin{equation} \tag{*9-2} \begin{aligned} \Delta \Pi &= \int_0^L \frac{1}{2} EI \left(\sum_{i=1}^n a_i \psi_i''(x)\right)^2 \,\mathrm{d} x - F \cdot \frac{1}{2} \int_0^L \left(\sum_{i=1}^n a_i \psi_i'(x)\right)^2 \,\mathrm{d} x \\[1em] &= \frac{1}{2} A(a_i) - \frac{F}{2} B(a_i) \end{aligned} \end{equation} \]
平衡状态时，\(\Delta \Pi \to \min\)，即
\[ \begin{equation} \tag{*9-3} \begin{aligned} \frac{\partial \Delta \Pi}{\partial a_i} &= 0, \, i = 1, 2, \ldots, n \\ \implies \frac{\partial A(a_i)}{\partial a_i} &- F \frac{\partial B(a_i)}{\partial a_i} = 0 \end{aligned} \end{equation} \]
下面计算 \(\frac{\partial A(a_i)}{\partial a_i}\) 和 \(\frac{\partial B(a_i)}{\partial a_i}\).
\[ \begin{aligned} \frac{\partial A(a_i)}{\partial a_i} &= EI \int_0^L \left(\sum_j a_j \psi_j''\right) \cdot \psi_i'' \,\mathrm{d} x = EI \sum_j 2 a_j \underset{K_{ij}}{\underbrace{\int_0^L \psi_j'' \psi_i'' \,\mathrm{d} x}} \\ \frac{\partial B(a_i)}{\partial a_i} &= \int_0^L 2 \left(\sum_j a_j \psi_j'\right) \cdot \psi_i' \,\mathrm{d} x = 2 \sum_j a_j \underset{S_{ij}}{\underbrace{\int_0^L \psi_j' \psi_i' \,\mathrm{d} x}} \end{aligned} \]
代回上式，得
\[ \begin{equation} \tag{*9-4} \sum_j a_j \left[ EI K_{ij} - F S_{ij} \right] = 0 \end{equation} \]
展开来写就是
\[ \left[EI \begin{bmatrix} K_{11} & K_{12} & \cdots & K_{1n} \\ K_{21} & K_{22} & \cdots & K_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ K_{n1} & K_{n2} & \cdots & K_{nn} \end{bmatrix} - F \begin{bmatrix} S_{11} & S_{12} & \cdots & S_{1n} \\ S_{21} & S_{22} & \cdots & S_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ S_{n1} & S_{n2} & \cdots & S_{nn} \end{bmatrix} \right] \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix} = 0 \]
i.e.
\[ \left[ EI [K] - F [S] \right] [a] = 0 \]
要使该一元 \(n\) 次方程有非平凡解，必须有
\[ \begin{equation} \tag{*9-5} \det \left( EI [K] - F [S] \right) = 0 \end{equation} \]
得到一组解 \(\left \{ \begin{array}{l} F_1^* & F_2^* & \ldots & F_n^* \\ [a]_1^* & [a]_2^* & \ldots & [a]_n^* \end{array} \right.\). \(F\) 只有取离散值，\([a]\) 才有非零解。得到的特征向量 \([a]^*\) 等比例缩放后仍为解，在参数空间 \(\{a_i\}\) 中每一个特征向量对应着一条过原点的直线。总势能 \(\Delta \Pi\) 在这些特征方向上取极值。
确定临界压力
稳定性
上面得到了压杆平衡的条件。那么，平衡态是否稳定？
临界条件：\(\Delta \Pi = 0\). 结合式 \eqref{*9-2} 得到
\[ \begin{equation} \tag{*9-6} F_w = \frac{\displaystyle\int_0^L EI w''^2 \,\mathrm{d} x}{\displaystyle\int_0^L w'^2 \,\mathrm{d} x} = \frac{A(a_i)}{B(a_i)} \end{equation} \]
- \(F_w\) 是与 \(w(x)\) 对应的临界力。每个可能的 \(w(x)\) 都有一个对应的 \(F_w\)，当外力 \(F > F_w\) 时，杆件就会失稳。
- 找到 \(w(x)\) 使得 \(F_w\) 最小，\(F_w\) 就是临界载荷 \(F_{\text{cr}}\)。
Example
假设几何变形为
\[ w(x) = a_1 \sin \frac{\pi x}{L} \]
此时基函数个数 \(n = 1\)，\([K], [S]\) 都是 \(1 \times 1\) 矩阵（标量）
\[ \left \{ \begin{aligned} \psi'_1 &= \frac{\pi}{L} \cos \frac{\pi x}{L} \\ \psi''_1 &= -\frac{\pi^2}{L^2} \sin \frac{\pi x}{L} \end{aligned} \right. \implies \left \{ \begin{aligned} K_{11} &= \int_0^L \psi''_1 \psi''_1 \,\mathrm{d} x = \left(\frac{\pi}{L}\right)^4 \frac{L}{2} \\ S_{11} &= \int_0^L \psi'_1 \psi'_1 \,\mathrm{d} x = \left(\frac{\pi}{L}\right)^2 \frac{L}{2} \end{aligned} \right. \]
代回式 \eqref{*9-5}，得到
\[ EI \left(\frac{\pi}{L}\right)^4 \frac{L}{2} - F_w \left(\frac{\pi}{L}\right)^2 \frac{L}{2} = 0 \implies F_w = EI \frac{\pi^2}{L^2} \approx 9.87 \frac{EI}{L^2} \]
取
\[ w(x) = a_1 x(x-L) \]
可求得 \(F_w = 12\frac{EI}{L^2}\)。可见在这个 \(w(x)\) 的选取下，得到的解没有上面好，但也是正确的。
再加一项，参数空间维数为 \(2\)
\[ w(x) = a_1 \sin \frac{\pi x}{L} + a_2 \sin \frac{2\pi x}{L} \]