Two-element beam — reactions and shear and moment diagrams: 16-Civ-B9 The Finite Element Method, May 2015, Q2

Problem 2: Two-element beam — reactions and shear and moment diagrams (one of three equal-value problems)

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.

Problem 2 — data read from Figure 2 and the question statement
Quantity	Symbol	Value
Length of element 1 (node 1 to node 2)	$L_{1}$	6 m
Length of element 2 (node 2 to node 3)	$L_{2}$	4 m
Uniform load on span 1–2 (downward)	$w_{1}$	20 kN/m
Uniform load on span 2–3 (downward)	$w_{2}$	25 kN/m
Modulus of elasticity	$E$	200 GPa $=200\times10^{6}$ kN/m$^{2}$
Second moment of area	$I$	$40\times10^{6}$ mm$^{4}=4\times10^{-5}$ m$^{4}$
Flexural rigidity	$EI$	8000 kN$\cdot$m$^{2}$
Support at node 1	—	fixed: $v_{1}=0$, $\theta_{1}=0$
Support at node 2	—	roller: $v_{2}=0$, $\theta_{2}$ free
Support at node 3	—	fixed: $v_{3}=0$, $\theta_{3}=0$

Find. The three support reactions (two of them with a fixing moment) and the complete shear-force and bending-moment diagrams for the 10 m beam, using a two-element Euler–Bernoulli discretisation with the printed element stiffness matrix.

[Figure not reproduced: Figure 2.1 — The structure of Figure 2, redrawn with the two-element discretisation. Sign convention: $v$ positive upward, $\theta$ positive counter-clockwise. See the official exam paper.]

Approach. Apply the boundary conditions first — they leave a single free degree of freedom, the rotation at node 2 — replace each distributed load by its work-equivalent nodal load vector, solve the one-equation system for $\theta_{2}$, then recover the element end forces as $\{r\}=[k]\{d\}+\{r^{F}\}$ and read the reactions and diagram ordinates from them.

Number the degrees of freedom and impose the supports. With three nodes and two degrees of freedom each, the structure has six: $\{d\}=\{v_{1}\;\theta_{1}\;v_{2}\;\theta_{2}\;v_{3}\;\theta_{3}\}^{T}$. Node 1 and node 3 are built in, so $v_{1}=\theta_{1}=v_{3}=\theta_{3}=0$; the roller at node 2 removes $v_{2}$ but leaves the rotation free. Only $\theta_{2}$ survives, and the whole problem collapses to one scalar equation. This is worth doing before any matrix is written out: it turns a $6\times6$ assembly into a single division.
Evaluate the flexural rigidity in consistent units. Working in kN and m, $$EI=(200\times10^{6}\ \text{kN}/\text{m}^{2})(4\times10^{-5}\ \text{m}^{4})=8000\ \text{kN}\cdot\text{m}^{2}.$$ Every stiffness term below is a multiple of this number.
Extract the two stiffness coefficients that matter. From the printed matrix, the diagonal entry associated with a rotation at either end of an element is $4EI/L$. Element 1 contributes through its second node, element 2 through its first, so the assembled stiffness at the single free degree of freedom is $$K_{\theta_{2}\theta_{2}}=\frac{4EI}{L_{1}}+\frac{4EI}{L_{2}} =\frac{4EI}{6}+\frac{4EI}{4}=\frac{5}{3}EI =\frac{5}{3}(8000)=13\,333.3\ \text{kN}\cdot\text{m}.$$ No other term of the $6\times6$ matrix is needed, because every other degree of freedom is restrained.
Replace the distributed loads by work-equivalent nodal loads. For a uniformly distributed load $w$ acting downward on an element of length $L$, the fixed-end force vector, with upward forces and counter-clockwise moments positive, is $$\{r^{F}\}=\left\{\frac{wL}{2},\;\frac{wL^{2}}{12},\;\frac{wL}{2},\;-\frac{wL^{2}}{12}\right\}^{T},$$ and the equivalent nodal load applied to the structure is its negative. Numerically, $$\{r^{F}\}_{1}=\{60,\;60,\;60,\;-60\}^{T},\qquad \{r^{F}\}_{2}=\{50,\;33.333,\;50,\;-33.333\}^{T},$$ in kN and kN$\cdot$m. The fixed-end moments are $w_{1}L_{1}^{2}/12=20(36)/12=60$ and $w_{2}L_{2}^{2}/12=25(16)/12=33.333$ kN$\cdot$m.
Assemble the load term at node 2. Only the moments reach the free degree of freedom, and they oppose one another because the two spans hog in opposite senses at the shared support: $$F_{\theta_{2}}=+\frac{w_{1}L_{1}^{2}}{12}-\frac{w_{2}L_{2}^{2}}{12}=60-33.333=26.667\ \text{kN}\cdot\text{m}.$$ This is the classical unbalanced fixed-end moment of moment distribution, arrived at through the stiffness route.
Solve for the rotation at node 2. Substituting into $K_{\theta_{2}\theta_{2}}\theta_{2}=F_{\theta_{2}}$, $$\theta_{2}=\frac{26.667}{13\,333.3}\quad\Longrightarrow\quad \boxed{\theta_{2}=+2.00\times10^{-3}\ \text{rad}\ \text{(counter-clockwise)}}$$ Equivalently $EI\theta_{2}=16.0$ kN$\cdot$m$^{3}$, which is the form to carry into the next step because it keeps the arithmetic exact. The positive sign says the beam rotates counter-clockwise over the roller, which is what the heavier short span demands.
Recover the end forces of element 1. With $\{d\}_{1}=\{0,\,0,\,0,\,\theta_{2}\}^{T}$, only the fourth column of $[k]_{1}$ is active: $$\{r\}_{1}=[k]_{1}\{d\}_{1}+\{r^{F}\}_{1} =EI\theta_{2}\left\{\frac{6}{L_{1}^{2}},\;\frac{2}{L_{1}},\;-\frac{6}{L_{1}^{2}},\;\frac{4}{L_{1}}\right\}^{T}+\{r^{F}\}_{1}.$$ With $EI\theta_{2}=16$ and $L_{1}=6$ the first term is $\{2.667,\,5.333,\,-2.667,\,10.667\}^{T}$, so $$\{r\}_{1}=\{62.667,\;65.333,\;57.333,\;-49.333\}^{T}.$$
Recover the end forces of element 2. Here $\{d\}_{2}=\{0,\,\theta_{2},\,0,\,0\}^{T}$, so the second column of $[k]_{2}$ is active: $$\{r\}_{2}=EI\theta_{2}\left\{\frac{6}{L_{2}^{2}},\;\frac{4}{L_{2}},\;-\frac{6}{L_{2}^{2}},\;\frac{2}{L_{2}}\right\}^{T}+\{r^{F}\}_{2} =\{6,\;16,\;-6,\;8\}^{T}+\{50,\;33.333,\;50,\;-33.333\}^{T},$$ $$\{r\}_{2}=\{56.000,\;49.333,\;44.000,\;-25.333\}^{T}.$$ The end moments at node 2 come out as $-49.333$ from element 1 and $+49.333$ from element 2, equal and opposite as they must be at a support that applies no external moment — the first arithmetic check.
Assemble the reactions. The vertical reaction at node 2 is the sum of the two element shears meeting there; the fixing moments are the end moments at nodes 1 and 3: $$\boxed{\;R_{1}=62.67\ \text{kN}\uparrow,\quad M_{1}=65.33\ \text{kN}\cdot\text{m},\quad R_{2}=57.333+56.000=113.33\ \text{kN}\uparrow,\quad R_{3}=44.00\ \text{kN}\uparrow,\quad M_{3}=-25.33\ \text{kN}\cdot\text{m}\;}$$ The fixing moments are hogging at both built-in ends, as expected; in the internal bending-moment convention that means $M(0)=-65.33$ and $M(10)=-25.33$ kN$\cdot$m.
Check global equilibrium before drawing anything. Vertically, $$62.667+113.333+44.000=220.0\ \text{kN}=w_{1}L_{1}+w_{2}L_{2}=120+100=220.0\ \text{kN}\ \text{(closes).}$$ Taking moments about node 1, $$113.333(6)+44.000(10)+65.333-25.333-120(3)-100(8)=1120+40-1160=0\ \text{(closes).}$$ Both close exactly, which they should: cubic Hermite interpolation is the exact solution of the governing fourth-order equation for a prismatic beam, so a two-element model with consistent nodal loads returns the exact reactions, not an approximation.
Build the shear diagram. Within each span the shear falls linearly at the rate of the applied load. Measuring $x$ from node 1, $$V(x)=62.667-20x\ \ (0\leq x\leq 6),\qquad V(s)=56.000-25s\ \ (0\leq s\leq 4,\ s=x-6),$$ giving $V=+62.67$ and $-57.33$ kN at the ends of span 1–2 and $+56.00$ and $-44.00$ kN at the ends of span 2–3. The jump at node 2 is $56.000-(-57.333)=113.33$ kN, exactly the reaction there — the second check. Shear vanishes at $x=62.667/20=3.133$ m and at $s=56.000/25=2.240$ m, i.e. $x=8.240$ m.
Build the moment diagram. Integrating the shear from each built-in end, $$M(x)=-65.333+62.667x-10x^{2},\qquad M(s)=-49.333+56.000s-12.5s^{2}.$$ The maximum sagging moments occur where the shear crosses zero: $$M(3.133)=+32.84\ \text{kN}\cdot\text{m},\qquad M(2.240)=+13.39\ \text{kN}\cdot\text{m}.$$ Setting each parabola to zero locates the points of contraflexure at $x=1.321$ and $4.946$ m in span 1–2 and at $x=7.205$ and $9.275$ m in span 2–3. The largest moment anywhere in the beam is the hogging $65.33$ kN$\cdot$m at the built-in end 1, which is the section that governs design.

Figure 2.2 — Shear force and bending moment diagrams. Sagging is plotted upward; the shear jump at node 2 equals the 113.33 kN reaction there.

Check: the support conditions are read from Figure 2, in which node 1 and node 3 are drawn against hatched walls (fixed) and node 2 carries a roller symbol. If node 2 were instead a pin with a rotational release on one side, or if either end were pinned rather than built in, the free degree of freedom count would change and every number above with it. The reading used here is the one that makes the printed two-element instruction sensible: it leaves exactly one unknown rotation.

Problem 2 — results
Quantity	Symbol	Value
Rotation at node 2	$\theta_{2}$	$+2.00\times10^{-3}$ rad (counter-clockwise)
Vertical reaction at node 1	$R_{1}$	62.67 kN (up)
Fixing moment at node 1	$M_{1}$	65.33 kN$\cdot$m (hogging)
Vertical reaction at node 2	$R_{2}$	113.33 kN (up)
Vertical reaction at node 3	$R_{3}$	44.00 kN (up)
Fixing moment at node 3	$M_{3}$	25.33 kN$\cdot$m (hogging)
Shear at node 1 / left of node 2	$V$	$+62.67$ / $-57.33$ kN
Shear right of node 2 / at node 3	$V$	$+56.00$ / $-44.00$ kN
Support moment at node 2	$M_{2}$	$-49.33$ kN$\cdot$m (hogging)
Maximum sagging moment, span 1–2	$M_{\max}$	$+32.84$ kN$\cdot$m at $x=3.133$ m
Maximum sagging moment, span 2–3	$M_{\max}$	$+13.39$ kN$\cdot$m at $x=8.240$ m
Points of contraflexure	$x$	1.321, 4.946, 7.205, 9.275 m
Governing (largest) moment	$M$	65.33 kN$\cdot$m hogging at node 1

Topic. The Euler–Bernoulli beam element and work-equivalent nodal loads — the direct stiffness method applied to a two-span continuous beam with a single active degree of freedom.

Key relations. Element stiffness as printed, $[k]=\dfrac{EI}{L^{3}}\begin{bmatrix}12&6L&-12&6L\\6L&4L^{2}&-6L&2L^{2}\\ -12&-6L&12&-6L\\6L&2L^{2}&-6L&4L^{2}\end{bmatrix}$; fixed-end forces for a uniform load, $\{r^{F}\}=\{wL/2,\;wL^{2}/12,\;wL/2,\;-wL^{2}/12\}^{T}$; solution $[K_{ff}]\{d_{f}\}=\{F_{f}\}$; force recovery $\{r\}=[k]\{d\}+\{r^{F}\}$; and the statics identities $dV/dx=-w$, $dM/dx=V$.

Why this works. The cubic Hermite shape functions of the beam element are the exact solution of $EI\,d^{4}v/dx^{4}=0$, so they reproduce the deflected shape of a prismatic segment loaded only at its ends without any error at all. Distributed load is handled by superposition: first clamp both ends and record the fixed-end forces, then apply their negatives as nodal loads and let the structure rotate. Adding the two states back together in the recovery step $\{r\}=[k]\{d\}+\{r^{F}\}$ returns exact end forces, which is why the equilibrium checks close to machine precision rather than approximately. The interior stress resultants are then obtained by ordinary statics on each span, not by differentiating shape functions — a point candidates often miss. The whole calculation is the matrix restatement of moment distribution: $4EI/L$ is the rotational stiffness, $w L^{2}/12$ the fixed-end moment, and the single equation is the balance of the unbalanced moment at the only joint free to rotate.

Common pitfall. Applying the distributed load as two lumped point loads of $wL/2$ at the nodes and omitting the $wL^{2}/12$ moment terms — that is a different structure and it gets the reactions wrong; the moment terms are what make the fixed-end condition fixed. The second pitfall is a sign slip on the fourth entry of $\{r^{F}\}$: the end moments of the two ends of a span are equal and opposite in the counter-clockwise-positive convention, so the vector is $\{+,+,+,-\}$, not all positive.

Source. Logan, A First Course in the Finite Element Method, 6th ed., Ch. 4 (beam elements, equivalent nodal loads, Table 4-1 of fixed-end reactions); McCormac, Structural Analysis: Using Classical and Matrix Methods, 5th ed., Ch. 22; Hibbeler, Structural Analysis, 10th ed., Ch. 4 for the shear and moment diagram construction.

Question 2 of 3: Two-element beam — reactions and shear and moment diagrams

Problem 2: Two-element beam — reactions and shear and moment diagrams (one of three equal-value problems)