07-Str-B3 · December 2013
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Examinations — December 2013 — 07-Str-B3 Applications of Finite Elements. Three hours, closed book, with two 8.5 in by 11 in pages of handwritten notes and one approved non-communicating calculator. Four pages including the cover; three problems, all to be attempted, all of equal value. Every problem is answered in full below.
Reference texts: Logan, D.L., A First Course in the Finite Element Method (6th ed., Cengage) — the bar element, the Hermite beam element and its work-equivalent load vector, and the bilinear quadrilateral, in the same notation this paper uses; Cook, R.D., Malkus, D.S., Plesha, M.E. & Witt, R.J., Concepts and Applications of Finite Element Analysis (4th ed., Wiley) — element strain fields, plane stress versus plane strain, and joining elements with dissimilar degrees of freedom; Chandrupatla, T.R. & Belegundu, A.D., Introduction to Finite Elements in Engineering (4th ed., Pearson) — the stepped-bar treatment of a tapered member; Bathe, K.-J., Finite Element Procedures (2nd ed., Prentice Hall) — convergence and constraint (multi-point) equations; Zienkiewicz, O.C. & Taylor, R.L., The Finite Element Method: Its Basis and Fundamentals (7th ed., Butterworth-Heinemann) — general theory; Przemieniecki, J.S., Theory of Matrix Structural Analysis (Dover) — the beam stiffness matrix printed on page 4; Hibbeler, R.C., Mechanics of Materials (10th ed., Pearson) — axially loaded members of varying section.
NOTE — sign convention used throughout Problem 1. The paper places the origin at the loaded tip and the built-in end at $x = L$, and states the boundary conditions as $u(L) = 0$ and $EA(x)\,du/dx|_{x=0} = P$. Figure 1 draws $P$ pulling the tip away from the wall, so with $x$ measured toward the wall the bar is in tension and every displacement is negative. Displacements are therefore reported as signed values, with the physical elongation quoted alongside; stresses are reported as positive tensile values. Nothing in the answers depends on this choice, only on its consistency.
Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.
The four-node rectangle carries two translational degrees of freedom at each node, so eight constants are available and each displacement component is interpolated by a complete set of four terms — the bilinear field
$$u(x,y)=a_1+a_2x+a_3y+a_4xy,\qquad v(x,y)=b_1+b_2x+b_3y+b_4xy .$$The $xy$ term is what distinguishes this element from the three-node triangle: it is not part of a complete quadratic, but it is needed so that the four nodal values of each component can be matched independently. Differentiating gives the strains, and through the plane-stress constitutive matrix the stresses follow the same variation:
$$\varepsilon_x=\frac{\partial u}{\partial x}=a_2+a_4y,\qquad \varepsilon_y=\frac{\partial v}{\partial y}=b_3+b_4x,\qquad \gamma_{xy}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=(a_3+b_2)+a_4x+b_4y .$$The answer, then, is that the stresses are not constant — but neither are they fully linear. Specifically:
Two consequences are worth stating because they explain the element's known weaknesses. First, $\sigma_x$ cannot vary along $x$, so a single element cannot represent a bending moment that changes along the span; a mesh of Q4 elements in a beam therefore models a linearly varying moment as a staircase, and needs refinement along the span rather than through the depth. Second, the presence of the $a_4y$ term in $\varepsilon_x$ together with $a_4x$ in $\gamma_{xy}$ means that when the element is bent it develops a spurious shear strain that no real beam in pure bending has; the element is too stiff in bending, the defect called shear locking, and it is the reason incompatible-mode and reduced-integration variants of the Q4 were developed. Both effects vanish as the mesh is refined, since the element does contain the constant-strain and rigid-body states required for convergence.
The difficulty is a mismatch of degrees of freedom, not of geometry. A plane-stress node carries only $u$ and $v$. A plane beam node carries $u$, $v$ and the rotation $\phi$. If the beam is simply attached to a membrane node, the beam's rotational degree of freedom has no counterpart in the membrane mesh, so nothing resists it: the joint behaves as a frictionless pin, the beam sheds its end moment, and the model is not the structure that was intended. Worse, if the rotation is attached to nothing at all the global stiffness matrix acquires a zero on that diagonal and the solution fails outright.
Compatibility is restored by supplying the missing kinematic link explicitly, and there are four standard ways to do it, in rough order of preference:
Constraint (multi-point) equations, or a rigid link. The cleanest device. A rotation of the beam end about the joint corresponds, in the membrane, to equal and opposite horizontal movements of two nodes lying above and below the joint. Writing that relationship as a constraint,
$$\phi_{\text{beam}}=\frac{u_t-u_b}{d},$$where $u_t$ and $u_b$ are the in-plane displacements of two membrane nodes separated by the distance $d$, ties the beam rotation into the membrane's translational degrees of freedom. The constraint is condensed into the global equations before solution, so the beam's end moment is delivered to the membrane as a couple of horizontal forces, exactly as it would be in the real structure. This is what commercial codes call a rigid link, a beam-to-surface offset, or an MPC.
Elements with a drilling degree of freedom. Membrane elements have been formulated with an in-plane rotation at each node — the Allman triangle and its quadrilateral relatives. When such an element is used, its rotational degree of freedom is the same quantity as the beam's $\phi$, and the two simply share it at the node. This is the simplest option when the element library offers it, and it is why most modern surface (plate and membrane) elements carry a drilling degree of freedom whether or not the analyst asks for one.
Model the beam region with membrane elements too. If the member is stocky enough to be meshed through its depth with two or more plane-stress elements, the beam element disappears from the model and with it the mismatch: only translations meet. This is the honest option when the "beam" is really a deep spandrel or a coupling beam, but it is expensive and it forfeits the direct read-out of bending moment that a beam element gives.
A short rigid stub of membrane elements, or a penalty stiffness. Embedding the end of the beam in the mesh over a short length, or attaching an artificially stiff rotational spring, will also transmit the moment. Both are approximations: the penalty stiffness must be large enough to enforce the constraint but small enough not to wreck the conditioning of the matrix, so the constraint-equation route is preferred wherever it is available.
Whichever device is used, the same check applies afterwards: apply a unit moment at the joint and confirm that the membrane picks it up as a self-equilibrating pair of in-plane forces. If the joint rotates freely, the compatibility has not actually been enforced.
The two idealisations are distinguished by which out-of-plane quantity is set to zero. Plane stress assumes $\sigma_z = \tau_{xz} = \tau_{yz} = 0$ and suits a thin body loaded in its own plane, free to contract through the thickness. Plane strain assumes $\varepsilon_z = \gamma_{xz} = \gamma_{yz} = 0$ and suits a long prismatic body, loaded identically on every cross-section, whose ends are restrained against axial movement. Anything loaded out of its plane, or twisted, is neither: it needs a plate-bending or three-dimensional model.
| Column 1 — structure | Column 2 | Reason |
|---|---|---|
| A flat slab floor of a building with vertical loading perpendicular to the slab | Neither | The load is out of the plane of the slab; the action is plate bending, needing a plate-bending or thick-plate element. |
| A wall subjected to wind loading, acting as a shear wall with loads in the plane of the wall | Plane stress | Thin body, loaded in its own plane, faces free, so $\sigma_z \approx 0$ through the thickness. |
| A tensile plate with a hole drilled transversally through it | Plane stress | The classic thin plate with a hole in in-plane tension; the free faces make $\sigma_z = 0$. (A very thick plate would tend toward plane strain at the hole.) |
| A concrete dam subjected to the hydrostatic pressure of the reservoir | Plane strain | A long prismatic body of constant section with the same loading on every section; the adjacent material prevents axial strain, so $\varepsilon_z = 0$ on a unit-thickness slice. |
| A soil mass subjected to a strip footing load | Plane strain | A strip footing is long compared with its width and loads every cross-section alike; the surrounding soil restrains longitudinal strain. |
| A wrench subjected to a force in its plane | Plane stress | Thin flat body loaded in its own plane with free faces. |
| A wrench subjected to twisting forces acting out of the plane of the wrench | Neither | Out-of-plane loading producing torsion and bending; requires a plate-bending or three-dimensional solid model. |
| A triangular plate connection with loads in the plane of the triangle | Plane stress | A thin gusset plate loaded in its own plane — the textbook plane-stress problem. |
The two "plane strain" entries share a tell worth remembering for the exam: both are structures whose third dimension is long and whose loading does not change along it, so a unit slice is representative and the neighbouring slices restrain it. The two "Neither" entries share the opposite tell: in both, the load has a component out of the plane being drawn.
