上一篇：有限元方法问答 03：直接刚度法 · 系列总目录 · 下一篇：有限元方法问答 05：二维问题

L4 — Truss Problems（杆/桁架单元）

Q21 ⭐ 杆单元刚度矩阵

EN: For a 1D bar element of length LL, cross-sectional area AA, and Young’s modulus EE:

ke=EAL[1−1−11]\boxed{\mathbf{k}_e = \frac{EA}{L}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}}

中文： 一维杆单元刚度矩阵为 EAL[1−1−11]\frac{EA}{L}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}。EA/L 是杆的轴向刚度。注意杆单元只能承受轴向拉压。

Q22 🔴 杆单元不能承受什么力？

EN: A bar/truss element can ONLY carry axial forces (tension and compression). It CANNOT resist shear forces or bending moments. Each node has only translational DOFs along the bar axis. If the structure is subjected to transverse loads, a truss model alone is insufficient — beam or frame elements must be used.

中文： 杆单元只能承受轴向拉压，不能承受剪力和弯矩。每个节点只有轴向平动自由度。若结构承受横向载荷，必须使用梁单元或框架单元。

Q23 🟡 杆单元形函数和B矩阵

EN: For a 1D bar element, the linear shape functions are:

Ni(x)=1−xL,Nj(x)=xLN_i(x) = 1 - \frac{x}{L}, \quad N_j(x) = \frac{x}{L}

The strain-displacement matrix B\mathbf{B} (1×2) is:

B=[−1L1L]\mathbf{B} = \begin{bmatrix} -\frac{1}{L} & \frac{1}{L} \end{bmatrix}

Since B\mathbf{B} is constant, strain ε=Bd=uj−uiL\varepsilon = \mathbf{Bd} = \frac{u_j - u_i}{L} is constant within the element.

中文： 形函数 Ni=1−x/LN_i=1-x/L, Nj=x/LN_j=x/L（线性插值）；B矩阵 [−1/L,1/L][-1/L, 1/L] 为常数 → 单元内应变为常数。杆单元的应变 = 两端位移差/长度。

Q24 🟡 二维桁架坐标变换

EN: For a 2D truss element oriented at angle θ\theta from the global X-axis, define direction cosines c=cos⁡θc = \cos\theta, s=sin⁡θs = \sin\theta. The element stiffness in global coordinates (4×4) is:

ke=EAL[c2cs−c2−cscss2−cs−s2−c2−csc2cs−cs−s2css2]\boxed{\mathbf{k}_e = \frac{EA}{L}\begin{bmatrix} c^2 & cs & -c^2 & -cs \\ cs & s^2 & -cs & -s^2 \\ -c^2 & -cs & c^2 & cs \\ -cs & -s^2 & cs & s^2 \end{bmatrix}}

where c=(xj−xi)/Lc = (x_j-x_i)/L, s=(yj−yi)/Ls = (y_j-y_i)/L.

中文： 通过方向余弦 c=cos⁡θc=\cos\theta, s=sin⁡θs=\sin\theta 将局部坐标刚度变换到全局坐标。全局刚度矩阵为4×4（每节点2个DOF），含 c2,cs,s2c^2, cs, s^2 项。变换原理：kglobal=TTklocalT\mathbf{k}_{global} = \mathbf{T}^T\mathbf{k}_{local}\mathbf{T}。

Q25 🔴 分布载荷的等效节点力

EN: For a bar element with distributed axial load q(x)q(x), the equivalent nodal forces are calculated using the shape functions:

fse=∫0LNTq(x) dx\mathbf{f}_s^e = \int_0^L \mathbf{N}^T q(x) \, dx

For uniform load qq: fse={qL/2qL/2}\mathbf{f}_s^e = \begin{Bmatrix} qL/2 \\ qL/2 \end{Bmatrix} (equal split).

For linearly varying load q=Cxq = Cx: fse={CL2/6CL2/3}\mathbf{f}_s^e = \begin{Bmatrix} CL^2/6 \\ CL^2/3 \end{Bmatrix} (1/3 to left, 2/3 to right).

The total equivalent force equals the total applied load (∫0Lq(x) dx\int_0^L q(x)\,dx).

中文： 等效节点力通过形函数积分得到：fse=∫NTq dx\mathbf{f}_s^e = \int\mathbf{N}^T q\,dx。均布：各得一半；线性分布：1/3给近端，2/3给远端。等效节点力的总和等于实际总载荷（静力等效）。

Q26 ⭐ 收敛性要求

EN: For convergence of FEA, two requirements must be satisfied:

  1. Completeness: The shape functions must be able to represent rigid-body displacements (constant terms in the polynomial) and constant strain states (linear terms). This is verified by the patch test.

  2. Compatibility: The displacement field must be continuous across element boundaries. For bar elements, this means C0C^0 continuity (displacement continuous, but its derivative — strain — may be discontinuous).

Together: Completeness + Compatibility → Convergence (as mesh is refined).

中文： 收敛的两个必要条件——(1)完备性：形函数能表示刚体位移（常数项）和常应变（线性项），通过分片试验验证；(2)相容性：位移在单元边界连续。杆单元满足C⁰连续（位移连续，导数/应变可不连续）。完备+相容+稳定→收敛。

Q27 🔴 C⁰ vs C¹ 连续性

EN: Continuity classification for FEM elements:

  • C⁰ continuity: The field variable itself is continuous across element boundaries, but its first derivative may be discontinuous. Examples: bar elements, plane stress/strain elements, 3D solid elements. (Displacement is continuous, strain/stress may jump.)

  • C¹ continuity: Both the field variable AND its first derivative are continuous across boundaries. Example: beam elements (both deflection vv AND slope θ=dv/dx\theta = dv/dx are continuous).

中文： C⁰连续——场变量本身连续，但一阶导数可间断（如杆单元：位移连续，应变不连续）；C¹连续——场变量及其一阶导数都连续（如梁单元：挠度和转角都连续）。板的Kirchhoff理论要求C¹连续。

Q28 🟡 梁挠度插值多项式的阶次（易错）

EN: The beam element uses a cubic (3rd-order) polynomial for deflection:

v(x)=a0+a1x+a2x2+a3x3v(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3

This is exact when no distributed load acts on the element (EI d4v/dx4=0EI\,d^4v/dx^4 = 0). However, under a uniform distributed load qq, the exact solution is 4th-order (v∝x4v \propto x^4 term from integrating q/EIq/EI four times). The cubic FEM approximation is therefore NOT exact for distributed loads — the solution matches exactly only at the nodes, while interior points have some error.

中文： ⚠️ 易错点——梁单元位移插值是三次多项式。无分布载荷时三次是精确解；有均布载荷时，精确解是四次（因为 EI v′′′′=qEI\,v''''=q，积分四次产生 x4x^4 项），FEM的三次解仅是近似，只在节点处匹配。考试常设陷阱。

Q29 ⭐ 梁单元等效节点力

EN: For a beam element under uniform transverse load qq:

fe={qL/2qL2/12qL/2−qL2/12}←vi←θi←vj←θj\boxed{\mathbf{f}_e = \begin{Bmatrix} qL/2 \\ qL^2/12 \\ qL/2 \\ -qL^2/12 \end{Bmatrix} \begin{matrix} \leftarrow v_i \\ \leftarrow \theta_i \\ \leftarrow v_j \\ \leftarrow \theta_j \end{matrix}}

These are the fixed-end forces: shear qL/2qL/2 at each end, and moments qL2/12qL^2/12 (counter-clockwise at ii, clockwise at jj). Sign conventions depend on the local coordinate system.

中文： 均布载荷q下，梁单元等效节点力 = {qL/2,qL2/12,qL/2,−qL2/12}\{qL/2, qL^2/12, qL/2, -qL^2/12\}。物理含义：两端各承担一半剪力，固端弯矩 qL2/12qL^2/12。注意符号取决于坐标方向约定。

Q30 🔴 梁单元等效节点力与支座反力的区别

EN: A critical distinction in beam problems: when computing reaction forces at supports after solving for displacements, the effect of equivalent nodal loads must be properly accounted for. The total reaction = stiffness contribution + equivalent load contribution.

Specifically, for a beam with distributed load qq, the element equation is kede=fext+feq\mathbf{k}_e\mathbf{d}_e = \mathbf{f}_{ext} + \mathbf{f}_{eq}, where feq\mathbf{f}_{eq} is the fixed-end force vector. When computing reactions at constrained nodes, the fixed-end force contribution at that node must be included.

中文： ⚠️ 常错——计算支座反力时，不仅要考虑刚度×位移的贡献，还要加上等效节点力在支座处的分量。因为分布载荷被转化为等效节点力后，这些"等效节点力"也参与了力的平衡。如均布载荷下的固支梁，即使位移为零，固端反力也不为零。

Q31 ⭐ 梁弯曲正应力公式

EN: The bending normal stress in a beam is:

σ=−MyI\sigma = -\frac{My}{I}

where MM is the bending moment, yy is the distance from the neutral axis, and II is the second moment of area. The negative sign indicates compressive stress on the positive-yy side when MM is positive (conventional beam sign convention).

中文： σ=−My/I\sigma=-My/I，M是弯矩，y是距中性轴的距离，I是截面惯性矩。负号表示正弯矩时上压下拉（或取决于坐标约定）。

Q32 🔴 平面框架单元

EN: A 2D frame element combines axial and bending behavior. Each node has 3 DOFs: {ui,vi,θi}\{u_i, v_i, \theta_i\}. The element has 6 DOFs total. The 6×6 stiffness matrix is assembled by combining:

  • Axial stiffness (EA/LEA/L terms at DOF positions 1 and 4)
  • Bending stiffness (12EI/L312EI/L^3, 6EI/L26EI/L^2, 4EI/L4EI/L, etc. at DOF positions 2,3,5,6)

The full matrix is given in L5 course notes. For 3D frames, each node has 6 DOFs (3 translations + 3 rotations), requiring a 12×12 stiffness matrix including axial, torsional (GJ/LGJ/L), and bending in two planes.

中文： 平面框架单元 = 杆单元（轴向）+ 梁单元（弯曲）的组合，每节点3个DOF（u,v,θu,v,\theta），共6个。三维框架单元每节点6个DOF（3个平动+3个转动），共12×12刚度矩阵（含轴向+扭转GJ/LGJ/L+两个方向的弯曲）。

Q33 🟡 梁铰接点处理

EN: When a hinge (pin connection) exists within a beam, the bending moment at the hinge is zero and the slope is discontinuous. To model this:

  1. Static condensation: Eliminate the rotational DOF at the hinge from the stiffness matrix. For a hinge at the right end, the condensed stiffness matrix reduces from 4×4 to 3×3.
  2. Hinge element: Use a specially derived element with the zero-moment condition pre-applied.

The condensed stiffness for a beam with hinge at right end (DOFs: vi,vj,θjv_i, v_j, \theta_j):

k=3EIL3[1L−1LL2−L−1−L1]\mathbf{k} = \frac{3EI}{L^3}\begin{bmatrix} 1 & L & -1 \\ L & L^2 & -L \\ -1 & -L & 1 \end{bmatrix}

中文： 铰接点弯矩为零，转角不连续。处理方法：用静力凝聚消去铰接处的转动自由度。右端铰接的梁单元刚度矩阵缩聚为3×3。

上一篇：有限元方法问答 03：直接刚度法 · 系列总目录 · 下一篇：有限元方法问答 05：二维问题