Question 3 of 3: Beam element — shape functions, work-equivalent loads and the centre-node solve

Problem 3: Beam element — shape functions, work-equivalent loads and the centre-node solve (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. A plane beam element with the two-node, four-degree-of-freedom Hermite interpolation of Figure 3.1, and the fixed-ended beam of Figure 3.2 carrying a triangular load.

Find. (3.1) the fourth Hermite shape function $N_4(x)$; (3.2) the work-equivalent nodal force and moment vector for each of the two elements, and their assembled value at the centre node; (3.3) the vertical displacement and the rotation of the centre of the beam.

[Figure not reproduced: Figure 3.1 (redrawn). The four degrees of freedom of the plane beam element, and the sign convention: $v$ positive upward, $\phi$ positive counter-clockwise. See the official exam paper.]

[Figure not reproduced: Figure 3.2 (redrawn), with the two-element mesh superimposed. Nodes 1 and 3 are fully restrained; the centre node 2 carries the only free degrees of freedom, $v_2$ and $\phi_2$. See the official exam paper.]

Approach. Build $N_4$ from the four nodal conditions that define it, integrate each shape function against the distributed load to obtain the work-equivalent nodal vector element by element, assemble the two vectors at the shared centre node, assemble the corresponding two rows and columns of the printed stiffness matrix, and solve the resulting two-equation system.

3.1 The fourth shape function

1. State the conditions $N_4$ must satisfy. Each Hermite shape function is the deflected shape of the element when its own degree of freedom is given a unit value and the other three are held at zero. For $N_4$, which is paired with the rotation $\phi_2$ at node 2, that means $$N_4(0)=0,\qquad N_4'(0)=0,\qquad N_4(L)=0,\qquad N_4'(L)=1 .$$ The first pair holds node 1 completely still; the third holds node 2 at zero deflection; the fourth applies the unit rotation.
2. Take the general cubic and impose them. Writing $N_4 = \alpha x^3+\beta x^2+\gamma x+\delta$, the conditions at $x = 0$ give $\delta = 0$ and $\gamma = 0$ immediately. The remaining two conditions become $$\alpha L^3+\beta L^2=0,\qquad 3\alpha L^2+2\beta L=1 .$$ The first gives $\beta = -\alpha L$; substituting into the second, $3\alpha L^2-2\alpha L^2=\alpha L^2=1$, so $\alpha = 1/L^2$ and $\beta = -1/L$.
3. Write the result in the form used for the other three. Substituting the two constants, $$N_4(x)=\frac{x^3}{L^2}-\frac{x^2}{L}=\boxed{\frac{1}{L^3}\left(x^3L-x^2L^2\right)}.$$
4. Check it against the properties the set must have. Three checks are available and all are quick. Substituting $x = L/2$ gives $N_4 = -L/8$, so the mid-span deflection produced by a unit rotation at node 2 alone is $L/8$ downward, matching $N_2(L/2)=+L/8$ by antisymmetry. Differentiating, $N_4' = 3x^2/L^2-2x/L$, which is zero at $x = 0$ and unity at $x = L$ as required. Finally the translation functions alone must reproduce a rigid-body movement, $N_1+N_3 = 1$, which they do, while $N_2$ and $N_4$ contribute nothing to it — consistent with $N_4(0)=N_4(L)=0$.

3.2 Work-equivalent discrete loads for the triangular load

1. State the work-equivalence principle. The nodal load vector is chosen so that the distributed load and its discrete replacement do the same work through any admissible nodal displacement. With the element displacement field written as $v(x)=\sum N_i(x)d_i$, equating the two work expressions gives, for a downward load of intensity $q(x)$, $$f_i=-\int_0^{L_e}N_i(x)\,q(x)\,dx ,$$ the minus sign carrying the fact that the load acts against the positive (upward) sense of $v$.
2. Carry out the integrals once, for a general linear load. Taking $q(x)=q_1+(q_2-q_1)x/L_e$ and integrating each of the four shape functions against it gives the standard result $$\{f\}=\begin{Bmatrix}F_{v1}\\M_1\\F_{v2}\\M_2\end{Bmatrix}=\begin{Bmatrix}-\dfrac{L_e\left(7q_1+3q_2\right)}{20}\\[6pt]-\dfrac{L_e^{2}\left(3q_1+2q_2\right)}{60}\\[6pt]-\dfrac{L_e\left(3q_1+7q_2\right)}{20}\\[6pt]+\dfrac{L_e^{2}\left(2q_1+3q_2\right)}{60}\end{Bmatrix}.$$ Setting $q_1 = q_2 = q$ recovers the familiar uniform-load vector $\{-qL_e/2,\ -qL_e^2/12,\ -qL_e/2,\ +qL_e^2/12\}$, which is the check that the general expression is right.
3. Read off the load intensities at the three nodes. The load rises linearly from zero at the left support to $w$ at the right over a span of $2L$, so at the centre node it is exactly half its maximum: $$q_1=0,\qquad q_2=\frac{w}{2},\qquad q_3=w .$$
4. Evaluate for element 1 ($q_1 = 0$ to $q_2 = w/2$, length $L$). $$F_{v1}=-\frac{L\left(0+3(w/2)\right)}{20}=-\frac{3wL}{40},\qquad M_1=-\frac{L^{2}\left(0+2(w/2)\right)}{60}=-\frac{wL^{2}}{60},$$ $$F_{v2}=-\frac{L\left(0+7(w/2)\right)}{20}=-\frac{7wL}{40},\qquad M_2=+\frac{L^{2}\left(0+3(w/2)\right)}{60}=+\frac{wL^{2}}{40}.$$
5. Evaluate for element 2 ($q_1 = w/2$ to $q_2 = w$, length $L$). $$F_{v1}=-\frac{L\left(7(w/2)+3w\right)}{20}=-\frac{13wL}{40},\qquad M_1=-\frac{L^{2}\left(3(w/2)+2w\right)}{60}=-\frac{7wL^{2}}{120},$$ $$F_{v2}=-\frac{L\left(3(w/2)+7w\right)}{20}=-\frac{17wL}{40},\qquad M_2=+\frac{L^{2}\left(2(w/2)+3w\right)}{60}=+\frac{wL^{2}}{15}.$$
6. Assemble the two vectors at the shared centre node. Node 2 receives element 1's second node contribution and element 2's first: $$F_2=-\frac{7wL}{40}-\frac{13wL}{40}=\boxed{-\frac{wL}{2}},\qquad M_2=+\frac{wL^{2}}{40}-\frac{7wL^{2}}{120}=\boxed{-\frac{wL^{2}}{30}} .$$ So the centre node carries a downward force of $wL/2$ and a clockwise moment of $wL^2/30$. Summing the four vertical entries gives $-wL$, which is the total area under the triangle, $\tfrac12 w(2L)$ — the equilibrium check that must always be made after forming a work-equivalent vector.

3.3 Displacement and slope at the centre of the beam

1. Identify the free degrees of freedom. Both ends are built in, so $v_1 = \phi_1 = v_3 = \phi_3 = 0$. Of the six degrees of freedom in the two-element model only $v_2$ and $\phi_2$ survive, which is why the whole problem reduces to a $2\times2$ system.
2. Assemble the two surviving rows and columns. The centre node takes element 1's node-2 block, which is the lower-right $2\times2$ partition of the printed matrix, plus element 2's node-1 block, its upper-left partition: $$[K]_{22}=\frac{EI}{L^{3}}\begin{bmatrix}12 & -6L\\ -6L & 4L^{2}\end{bmatrix}+\frac{EI}{L^{3}}\begin{bmatrix}12 & 6L\\ 6L & 4L^{2}\end{bmatrix}=\frac{EI}{L^{3}}\begin{bmatrix}24 & 0\\ 0 & 8L^{2}\end{bmatrix}.$$ The off-diagonal terms cancel identically, because the two elements meet the node from opposite sides. That is a structural fact worth noticing: at an interior node of a uniform mesh, translation and rotation decouple, so the two answers can be written down separately.
3. Solve for the centre deflection. With $F_2 = -wL/2$ from 3.2, $$v_2=\frac{F_2}{24EI/L^{3}}=-\frac{wL}{2}\cdot\frac{L^{3}}{24EI}=\boxed{-\frac{wL^{4}}{48EI}},$$ a downward deflection of $wL^4/(48EI)$. In terms of the full span $\ell = 2L$ this is $w\ell^4/(768EI)$.
4. Solve for the centre rotation. With $M_2 = -wL^2/30$, $$\phi_2=\frac{M_2}{8EI/L}=-\frac{wL^{2}}{30}\cdot\frac{L}{8EI}=\boxed{-\frac{wL^{3}}{240EI}},$$ a clockwise rotation of $wL^3/(240EI)$. Its sign is the one physical sense to check: the load is heavier toward the right support, so the beam sags more to the right of centre and the tangent at mid-span tips down toward the right, which is clockwise.
5. Check the magnitudes against a known case. The same two-element model under a uniform load $w$ over the span $2L$ would give $v_2 = -w(2L)^4/(384EI) = -wL^4/(24EI)$. The triangular load carries half the total, and it puts that half further from mid-span, so a centre deflection of exactly half the uniform value is the right order — and $wL^4/(48EI)$ is exactly half of $wL^4/(24EI)$. The rotation is much smaller than the deflection divided by $L$, as it must be for a doubly built-in beam whose ends cannot rotate at all.