Question 2 of 4: Rhombus plane frame with inextensible members, and the effect of a diagonal tie

Problem 2: Rhombus plane frame with inextensible members, and the effect of a diagonal tie (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 closed rhombus frame, four rigidly jointed members.

Find. Part (a) the translations and rotations of all four nodes; part (b) the shear force and bending moment diagrams; part 2.2 the same three diagrams for the tied frame, with a comment on what the reinforcement achieves.

Approach. Because the members cannot extend, the four axial constraints reduce the twelve nodal degrees of freedom to a single independent unknown, and the double symmetry of the loading kills all four nodal rotations. The plane frame element stiffness matrix printed on the paper then supplies the member end actions directly from the sway (chord rotation) of each member, and one equilibrium — equivalently one energy — equation fixes the amplitude.

1. Part (a) — Reduce the frame to one degree of freedom. Place the origin at the centre of the rhombus, so the nodes sit at $1(-a,0)$, $2(0,a)$, $3(a,0)$, $4(0,-a)$ with $a=L/\sqrt{2}$. The axial term of the printed stiffness matrix, $EA/L$, becomes infinite, so each member enforces zero relative displacement along its own axis. For member 1–2, whose unit axis is $\mathbf{n}=(1,1)/\sqrt{2}$, that constraint reads $$(\mathbf{u}_2-\mathbf{u}_1)\cdot\mathbf{n}=0$$ Writing $\mathbf{u}_1=(-\delta,0)$ and $\mathbf{u}_2=(0,\eta)$ by symmetry gives $(\delta+\eta)/\sqrt{2}=0$, hence $\eta=-\delta$. The whole kinematics is therefore $$\boxed{u_1=-\delta,\quad u_3=+\delta,\quad v_2=-\delta,\quad v_4=+\delta}$$ in words: the horizontal diagonal lengthens by $2\delta$ while the vertical diagonal shortens by the same amount, and $\delta$ is the only unknown translation.
2. All four nodal rotations vanish. The structure and its loading are symmetric about both diagonals. Reflecting in the horizontal axis maps member 1–2 onto member 1–4 and leaves node 1 fixed, so the rotation at node 1 must equal its own negative; the same argument at nodes 2, 3 and 4 (using the vertical axis) gives $$\boxed{\theta_1=\theta_2=\theta_3=\theta_4=0}$$ This is the payoff of the symmetry: the rotational degrees of freedom never enter the solution, and a four-element frame model returns exactly zero for them.
3. Chord rotation of a typical member. With the joints unable to rotate, each member deforms purely by sway. For member 1–2 the relative end displacement is $\mathbf{u}_2-\mathbf{u}_1=(\delta,-\delta)$, and the component transverse to the member, along $\mathbf{p}=(-1,1)/\sqrt{2}$, is $$\Delta=(\mathbf{u}_2-\mathbf{u}_1)\cdot\mathbf{p}=-\sqrt{2}\,\delta$$ so every member suffers a transverse relative displacement of magnitude $\sqrt{2}\,\delta$, i.e. a chord rotation $\psi=\sqrt{2}\,\delta/L$.
4. Member end actions from the printed stiffness matrix. Taking the local element equation $\{f\}=[k]\{d\}$ with $\theta_i=\theta_j=0$, $v_i=0$ and $v_j=\Delta$, the rows of the given $6\times6$ matrix give $$m_i=-\frac{6EI}{L^{2}}\Delta,\quad m_j=-\frac{6EI}{L^{2}}\Delta,\quad f_{yi}=-\frac{12EI}{L^{3}}\Delta,\quad f_{yj}=+\frac{12EI}{L^{3}}\Delta$$ Both end moments have the same sign, which is the signature of double curvature: the bending moment is linear along the member and passes through zero at midspan.
5. Solve for the amplitude by total potential energy. A fixed-ended member swayed by $\Delta$ stores $U=\tfrac12(12EI/L^{3})\Delta^{2}=6EI\Delta^{2}/L^{3}$. With $\Delta^{2}=2\delta^{2}$ and four identical members, $$\begin{aligned}U&=4\left(\frac{6EI(2\delta^{2})}{L^{3}}\right)=\frac{48EI\delta^{2}}{L^{3}} \\ W&=2P\delta\end{aligned}$$ Stationarity of $\Pi=U-W$ gives $96EI\delta/L^{3}=2P$, hence $$\boxed{\delta=\frac{PL^{3}}{48EI}=0.020833\,\frac{PL^{3}}{EI}}$$ A four-element plane frame model, assembled from the printed matrix with a very large $EA$ and only the three rigid-body restraints, returns the identical value and confirms that all four rotations are zero.
6. Displacements and rotations at all nodal points. Collecting the kinematics of steps 1 and 2, with $\delta=PL^{3}/48EI$:

Signs follow the global axes of Figure 2.1: positive $u$ to the right, positive $v$ upward. The loaded corners move apart and the transverse corners draw in by the same amount, which is the inextensibility constraint expressing itself.

7. Part (b) — Member end actions. Substituting $\Delta=\sqrt{2}\,\delta$ into step 4 with $\delta=PL^{3}/48EI$, $$M=\frac{6EI}{L^{2}}\left(\sqrt{2}\,\frac{PL^{3}}{48EI}\right)=\frac{\sqrt{2}}{8}PL =0.17678\,PL$$ $$V=\frac{12EI}{L^{3}}\left(\sqrt{2}\,\frac{PL^{3}}{48EI}\right)=\frac{\sqrt{2}}{4}P =0.35355\,P$$ and the axial force follows from joint equilibrium at node 2, where the two members must between them balance nothing vertically. Each member meets the joint at 45°, so the two axial forces contribute $2\,(N/\sqrt{2})=\sqrt{2}N$ vertically and the two end shears contribute $2\,(V/\sqrt{2})=\sqrt{2}V$ in the opposite sense; hence $\sqrt{2}N=\sqrt{2}V$, i.e. $N=V$, giving $$\boxed{N=\frac{\sqrt{2}}{4}P=0.35355\,P\ \text{(tension in every member)}}$$ Note the internal consistency check $V=2M/L$, which must hold for a member whose moment runs linearly from $+M$ to $-M$ over its length.
8. Bending moment diagram. Each member carries a linear moment from $0.1768\,PL$ at one end, through zero at midspan, to $0.1768\,PL$ at the other end, with the tension face switching sides at midspan. Reading the sign from the element end actions, the outer fibre is in tension at the two loaded corners (nodes 1 and 3) and the inner fibre is in tension at nodes 2 and 4.
9. Shear force diagram. With no load between the joints the shear is constant within each member, at $0.3536\,P$, and it changes sign from one member to the next around the ring — a direct consequence of the moment diagram sloping one way in member 1–2 and the other way in member 2–3.
10. Part 2.2 — Add the pin-ended truss member. The tie spans the horizontal diagonal, so its length is $L_t=2a=\sqrt{2}L$ and its axial stiffness is $$k_t=\frac{EA_t}{L_t}=\frac{12EI/L^{2}}{\sqrt{2}L}=\frac{6\sqrt{2}\,EI}{L^{3}} =8.4853\,\frac{EI}{L^{3}}$$ Being pinned at both ends it attracts no moment; it simply adds an axial spring across the two loaded corners, which move apart by $2\delta$.
11. Re-solve the energy equation with the tie present. The tie contributes $\tfrac12 k_t(2\delta)^{2}=2k_t\delta^{2}$ to the strain energy, so $$\frac{96EI}{L^{3}}\delta+4k_t\delta=2P \;\Longrightarrow\; \delta'=\frac{PL^{3}}{\left(48+12\sqrt{2}\right)EI}$$ $$\boxed{\delta'=0.015392\,\frac{PL^{3}}{EI} =0.7388\,\delta}$$
12. Force in the tie, and the actions left for the frame. The tie force is its stiffness times its extension, $$F_t=k_t(2\delta')=\left(\frac{6\sqrt{2}EI}{L^{3}}\right) \left(\frac{2PL^{3}}{(48+12\sqrt{2})EI}\right)=\frac{12\sqrt{2}}{48+12\sqrt{2}}P =0.2612\,P\ \text{(tension)}$$ so the bending frame now sees only $P-F_t=0.7388\,P$ at each loaded corner. Every frame action therefore scales by that same factor: $$\begin{aligned}M'&=0.7388(0.17678\,PL)=0.13060\,PL \\ V'&=0.7388(0.35355\,P)=0.26120\,P\end{aligned}$$
13. Diagrams for the reinforced frame, and the comment asked for. The shapes of all three diagrams are unchanged — the same flattening mechanism, the same double-curvature moment linear to zero at midspan, the same constant alternating shear — because the tie does not alter the kinematics, only the stiffness that resists it. The magnitudes all fall to 73.9 per cent of their bare-frame values. Physically the tie short-circuits the load path: 26.1 per cent of each applied load now travels straight across the diagonal as pure axial tension instead of being carried around the ring in bending. That is a very cheap 26 per cent, because a bar in tension uses its whole cross-section while a member in bending uses only the fibres near its faces — the printed $EA_t = 12EI/L^{2}$ is a deliberately modest tie, and a stiffer one would take a correspondingly larger share.