07-Str-B3 · May 2015

Question 2 of 3: Two-Element Beam Model — Reactions and Internal Force Diagrams

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. Professional Engineers Ontario / Engineers Canada national examination 07-Str-B3 — Applications of the Finite Element Method, May 2015. Three hours; four pages. Closed book, with two 8½ × 11 in pages of handwritten notes permitted and one approved non-communicating calculator. Three problems, all of equal value; candidates are instructed to attempt all three. Problem 1 is a ten-part concept paper, Problems 2 and 3 are calculations.

07-Str-B3 is a finite element methods paper rather than a member-design paper: bar and beam elements, isoparametric quadrilaterals, the constant-strain triangle, numerical integration and structural dynamics. The reference list below is therefore the finite-element literature.

Reference texts.

Logan, D. L., A First Course in the Finite Element Method, 6th ed. — Ch. 3 (bar element), Ch. 4 and 5 (beam element, work-equivalent loads), Ch. 6 (constant-strain triangle, plane stress), Ch. 10 (isoparametric elements, Gauss quadrature), Ch. 16 (structural dynamics, mass matrices).
Cook, R. D., Malkus, D. S., Plesha, M. E. and Witt, R. J., Concepts and Applications of Finite Element Analysis, 4th ed. — Ch. 3–4 (formulation and convergence), Ch. 6 (isoparametric elements and numerical integration), Ch. 7 (element quality, the patch test, mesh transitions), Ch. 11 (dynamics), Ch. 15 (mixed and hybrid formulations).
Bathe, K.-J., Finite Element Procedures, 2nd ed. — Ch. 4 (variational basis and convergence), Ch. 5 (numerical integration, reduced and selective integration), Ch. 9 (time integration).
Chandrupatla, T. R. and Belegundu, A. D., Introduction to Finite Elements in Engineering, 4th ed. — Ch. 5 (CST, plane stress and plane strain), Ch. 6 (beams and frames).
Zienkiewicz, O. C., Taylor, R. L. and Zhu, J. Z., The Finite Element Method: Its Basis and Fundamentals, 7th ed. — Ch. 6 (the patch test), Ch. 11 (mixed and hybrid forms).
Hughes, T. J. R., The Finite Element Method: Linear Static and Dynamic Finite Element Analysis — Ch. 4 (mixed methods and reduced integration), Ch. 5 (beam and plate elements), Ch. 7 (consistent and lumped mass matrices).
CSA A23.3:19 Design of Concrete Structures and CSA S806:12 Design and Construction of Building Structures with Fibre-Reinforced Polymers — the Canadian code frame behind the strengthened-beam modelling question (1.9).

Question 2: Two-Element Beam Model — Reactions and Internal Force Diagrams (33.3 marks — one of three problems of equal value)

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.

Given.

Quantity	Symbol	Value
Modulus of elasticity	$E$	200 GPa = $200\times10^{6}$ kPa
Second moment of area	$I$	$40\times10^{6}\ \text{mm}^{4}=4.0\times10^{-5}\ \text{m}^{4}$
Flexural rigidity	$EI$	$8000\ \text{kN}\,\text{m}^{2}$
Span 1–2 (element 1)	$L_{1}$	6.0 m, uniformly loaded at $w_{1}=20$ kN/m
Span 2–3 (element 2)	$L_{2}$	4.0 m, uniformly loaded at $w_{2}=25$ kN/m
Supports	—	node 1 fixed, node 2 roller, node 3 fixed

Find. The three support reactions (two vertical forces plus a fixing moment at each built-in end, and the vertical reaction at the roller), and the complete shear-force and bending-moment diagrams, obtained from a two-element finite element model.

[Figure not reproduced: Figure 2 (redrawn) — propped fixed–fixed beam: node 1 fixed, node 2 on a roller 6.0 m to the right, node 3 fixed a further 4.0 m along. Two Hermite beam elements are used, one per span. See the official exam paper.]

Approach. Number the degrees of freedom, discover that the only unconstrained one is the rotation $\theta_{2}$ at the roller, replace each distributed load by its work-equivalent nodal load vector, solve the resulting single scalar equation for $\theta_{2}$, and recover the reactions as element end forces from $\mathbf{f}=\mathbf{k}\mathbf{d}-\mathbf{F}^{\text{eq}}$.

Establish the flexural rigidity in consistent units. Working in kilonewtons and metres, $$EI=\bigl(200\times10^{6}\ \text{kPa}\bigr)\bigl(4.0\times10^{-5}\ \text{m}^{4}\bigr)=8000\ \text{kN}\,\text{m}^{2}$$ The same rigidity applies to both elements, as the question states.
Set out the degrees of freedom and apply the boundary conditions. With three nodes and two degrees of freedom per node the model has six: $v_{1},\theta_{1},v_{2},\theta_{2},v_{3},\theta_{3}$, with $v$ positive upward and $\theta$ positive counter-clockwise. The fixed end at node 1 removes $v_{1}$ and $\theta_{1}$; the fixed end at node 3 removes $v_{3}$ and $\theta_{3}$; the roller at node 2 removes $v_{2}$ but leaves the rotation free. The structure therefore has exactly one unknown, $$\text{free DOF}=\{\theta_{2}\}$$ and only the single diagonal entry of the stiffness matrix associated with $\theta_{2}$ is needed. This is the whole economy of the finite element method here: an apparently indeterminate two-span beam collapses to one scalar equation.
Extract the rotational stiffness at node 2. The rotational diagonal term of the beam element supplied on the question paper is $4EI/L$. Node 2 receives one such contribution from each element: $$K_{\theta_{2}\theta_{2}}=\frac{4EI}{L_{1}}+\frac{4EI}{L_{2}}=\frac{4(8000)}{6}+\frac{4(8000)}{4}=5333.33+8000=13\,333.33\ \text{kN}\,\text{m/rad}$$
Replace each distributed load by its work-equivalent nodal load vector. For a downward uniformly distributed load $w$ on a Hermite beam element the consistent load vector, in the same degree-of- freedom order as the stiffness matrix, is $$\mathbf{F}^{\text{eq}}=\Bigl[-\frac{wL}{2},\ -\frac{wL^{2}}{12},\ -\frac{wL}{2},\ +\frac{wL^{2}}{12}\Bigr]^{T}$$ Applying it to each element: $$\mathbf{F}^{\text{eq}}_{1}=[-60,\ -60,\ -60,\ +60]^{T},\qquad\mathbf{F}^{\text{eq}}_{2}=[-50,\ -33.333,\ -50,\ +33.333]^{T}$$ in kN and kN·m.
Assemble the load acting on the free degree of freedom. The rotation $\theta_{2}$ collects the fourth entry of element 1 and the second entry of element 2: $$P_{\theta_{2}}=+\frac{w_{1}L_{1}^{2}}{12}-\frac{w_{2}L_{2}^{2}}{12}=60.000-33.333=26.667\ \text{kN}\,\text{m}$$ The two spans fight each other, as they must: the longer, lighter span tries to rotate node 2 one way and the shorter, heavier span the other.
Solve the reduced system. One equation in one unknown gives $$\theta_{2}=\frac{P_{\theta_{2}}}{K_{\theta_{2}\theta_{2}}}=\frac{26.667}{13\,333.33}$$ $$\boxed{\theta_{2}=+2.000\times10^{-3}\ \text{rad}\ \ (\text{counter-clockwise})}$$
Recover the end forces of element 1. The element end forces are the stiffness product less the work-equivalent loads, $\mathbf{f}=\mathbf{k}\mathbf{d}-\mathbf{F}^{\text{eq}}$, with $\mathbf{d}_{1}=[0,\ 0,\ 0,\ \theta_{2}]^{T}$. Only the fourth column of $\mathbf{k}_{1}$ contributes: $$\mathbf{k}_{1}\mathbf{d}_{1}=\frac{EI}{L_{1}}\Bigl[\frac{6}{L_{1}},\ 2,\ -\frac{6}{L_{1}},\ 4\Bigr]^{T}\theta_{2}=[2.667,\ 5.333,\ -2.667,\ 10.667]^{T}$$ Subtracting $\mathbf{F}^{\text{eq}}_{1}$, $$\mathbf{f}_{1}=[\,62.667\ \text{kN},\ \ 65.333\ \text{kN}\,\text{m},\ \ 57.333\ \text{kN},\ \ -49.333\ \text{kN}\,\text{m}\,]^{T}$$ Check: the two end shears sum to $62.667+57.333=120$ kN, exactly $w_{1}L_{1}$.
Recover the end forces of element 2. Here $\mathbf{d}_{2}=[0,\ \theta_{2},\ 0,\ 0]^{T}$, so the second column of $\mathbf{k}_{2}$ is the active one: $$\mathbf{k}_{2}\mathbf{d}_{2}=\frac{EI}{L_{2}}\Bigl[\frac{6}{L_{2}},\ 4,\ -\frac{6}{L_{2}},\ 2\Bigr]^{T}\theta_{2}=[6.000,\ 16.000,\ -6.000,\ 8.000]^{T}$$ and therefore $$\mathbf{f}_{2}=[\,56.000\ \text{kN},\ \ 49.333\ \text{kN}\,\text{m},\ \ 44.000\ \text{kN},\ \ -25.333\ \text{kN}\,\text{m}\,]^{T}$$ with $56.000+44.000=100$ kN $=w_{2}L_{2}$, as required. Note also that the two element moments at node 2, $-49.333$ and $+49.333$ kN·m, cancel exactly: no external couple is applied there, so the internal moment must be continuous through the roller.
Assemble the reactions. Each support reaction is the sum of the element end forces that meet at that node: $$R_{1}=62.667\ \text{kN}\uparrow,\qquad M_{1}=65.333\ \text{kN}\,\text{m}\ (\text{counter-clockwise})$$ $$R_{2}=57.333+56.000=113.333\ \text{kN}\uparrow$$ $$R_{3}=44.000\ \text{kN}\uparrow,\qquad M_{3}=-25.333\ \text{kN}\,\text{m}\ (\text{clockwise})$$ $$\boxed{R_{1}=62.67\ \text{kN},\quad R_{2}=113.33\ \text{kN},\quad R_{3}=44.00\ \text{kN},\quad M_{1}=65.33\ \text{kN}\,\text{m},\quad M_{3}=-25.33\ \text{kN}\,\text{m}}$$ Global check: $62.667+113.333+44.000=220.0$ kN, which is exactly $w_{1}L_{1}+w_{2}L_{2}=120+100=220$ kN.
Build the shear-force diagram. Within each span the shear falls off linearly at the rate of the applied load. In span 1, measuring $x$ from node 1, $$V(x)=R_{1}-w_{1}x=62.667-20x\ \ \Rightarrow\ \ V(0)=+62.67\ \text{kN},\quad V(6^{-})=-57.33\ \text{kN}$$ crossing zero at $x=62.667/20=3.133$ m. At the roller the diagram jumps upward by the full reaction $R_{2}=113.33$ kN, from $-57.33$ to $+56.00$ kN. In span 2, with $x'$ measured from node 2, $$V(x')=56.000-25x'\ \ \Rightarrow\ \ V(4^{-})=-44.00\ \text{kN},$$ crossing zero at $x'=56.000/25=2.240$ m.
Build the bending-moment diagram. Integrating the shear, and taking sagging as positive, span 1 gives $$M(x)=R_{1}x-M_{1}-\frac{w_{1}x^{2}}{2}=62.667x-65.333-10x^{2}$$ so $M(0)=-65.33$ kN·m (hogging at the fixed end), $M(6)=-49.33$ kN·m (hogging over the roller) and the maximum sagging moment occurs where the shear vanishes: $$M(3.133)=62.667(3.133)-65.333-10(3.133)^{2}=+32.84\ \text{kN}\,\text{m}$$ Span 2 gives, with $x'$ from node 2, $$M(x')=56.000x'-49.333-12.5x'^{2}$$ so $M(4)=-25.33$ kN·m at the right-hand fixed end and $$M(2.240)=56.000(2.240)-49.333-12.5(2.240)^{2}=+13.39\ \text{kN}\,\text{m}$$ $$\boxed{M_{\text{hog,max}}=-65.33\ \text{kN}\,\text{m}\ \text{at node 1};\qquad M_{\text{sag,max}}=+32.84\ \text{kN}\,\text{m}\ \text{at}\ x=3.133\ \text{m}}$$

Shear-force and bending-moment diagrams built from the element end forces. Because the Hermite element carries the exact solution of $EIv'''' = w$ at its nodes, these ordinates are the exact ones for this structure.

One point deserves emphasis before the results are tabulated. These are not approximate diagrams. The Hermite cubic shape functions of the beam element are the exact solution of $EI\,v''''=0$, and once the distributed loads have been converted to work-equivalent nodal loads the finite element model reproduces the exact nodal rotation and hence the exact reactions. Two elements are enough because the structure has exactly two spans; adding more elements would refine the deflected shape within each span but would not change a single number in the table below.

Quantity	Symbol	Result
Rotation at the roller (only free DOF)	$\theta_{2}$	$+2.000\times10^{-3}$ rad (counter-clockwise)
Vertical reaction, fixed end 1	$R_{1}$	62.67 kN (upward)
Fixing moment, end 1	$M_{1}$	65.33 kN·m (counter-clockwise; hogging)
Vertical reaction, roller	$R_{2}$	113.33 kN (upward)
Vertical reaction, fixed end 3	$R_{3}$	44.00 kN (upward)
Fixing moment, end 3	$M_{3}$	25.33 kN·m (clockwise; hogging)
Shear at node 1 / just left of node 2	$V$	$+62.67$ kN / $-57.33$ kN
Shear just right of node 2 / at node 3	$V$	$+56.00$ kN / $-44.00$ kN
Hogging moment over the roller	$M_{2}$	$-49.33$ kN·m
Maximum sagging moment, span 1	$M_{\max}^{(1)}$	$+32.84$ kN·m at $x=3.133$ m
Maximum sagging moment, span 2	$M_{\max}^{(2)}$	$+13.39$ kN·m at $x=8.240$ m
Equilibrium check	$\sum R$	220.0 kN $=w_{1}L_{1}+w_{2}L_{2}$ ✓

Topic: The Hermite beam element, work-equivalent nodal loads, and the recovery of reactions and internal forces from element end forces.

Key relations: element stiffness $\mathbf{k}=\dfrac{EI}{L}\bigl[\,12/L^{2},\ 6/L,\ -12/L^{2},\ 6/L\ ;\ 6/L,\ 4,\ -6/L,\ 2\ ;\ \dots\bigr]$ as printed on the paper; consistent load vector for a uniform load $\mathbf{F}^{\text{eq}}=[-wL/2,\ -wL^{2}/12,\ -wL/2,\ +wL^{2}/12]^{T}$; reduced system $\mathbf{K}_{ff}\mathbf{d}_{f}=\mathbf{P}_{f}$; element end forces $\mathbf{f}=\mathbf{k}\mathbf{d}-\mathbf{F}^{\text{eq}}$; shear and moment recovery $V(x)=V_{0}-wx$, $M(x)=V_{0}x-M_{0}-wx^{2}/2$.

Why this works: A beam element's shape functions are cubic in $x$, and the homogeneous solution of the Euler–Bernoulli equation $EI\,v''''=0$ is also cubic. The element can therefore reproduce the exact deflected shape of an unloaded segment. When a distributed load acts, the exact shape is quartic and the element cannot follow it inside the span, but the work-equivalent load vector is derived from the very same shape functions, $\mathbf{F}^{\text{eq}}=\int\mathbf{N}^{T}w\,dx$, and this makes the nodal displacements and rotations exact anyway. Because the nodal values are exact, so are the end forces, so are the reactions, and so are the two internal-force diagrams once the span load is superimposed. The same numbers would come out of a slope-deflection or moment-distribution analysis; the finite element route simply organises the bookkeeping. Notice that the fixed-end moment terms in $\mathbf{f}=\mathbf{k}\mathbf{d}-\mathbf{F}^{\text{eq}}$ are the classical fixed-end moments $wL^{2}/12$: the two methods are the same method in different notation.

Common pitfall: The two errors that cost the most marks here are (i) forgetting the $-\mathbf{F}^{\text{eq}}$ term when recovering element end forces, which produces the tiny nodal-stiffness contribution alone and reactions that do not sum to the applied load; and (ii) taking the shear and moment diagram from the element's own cubic $v(x)$ by differentiating three and two times. That gives a linear moment and a constant shear inside each element, which contradicts the applied uniform load. The correct route is always to superimpose the span loading on the end forces, as done in steps 10 and 11.

Source: Logan, A First Course in the Finite Element Method, 6th ed., §4.1–4.5 (beam element, boundary conditions, distributed loading) and §5.5 (work-equivalent nodal loads); Chandrupatla & Belegundu, Introduction to Finite Elements in Engineering, 4th ed., §6.1–6.4; Cook, Malkus, Plesha & Witt, 4th ed., §2.7–2.9 (consistent nodal loads, element force recovery).

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