Question 2 of 4: Rhombic rigid-jointed frame — plane-frame elements, symmetry, and a diagonal tie

Problem 2: Rhombic rigid-jointed frame — plane-frame elements, symmetry, and 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 rhombic frame of four rigidly jointed, inextensible members, each of length $L$ and bending rigidity $EI$, is pulled apart by a self-equilibrated pair of loads $P$ acting outward along its horizontal diagonal; in part 2.2 a pin-ended tie of axial stiffness $EA = 12EI/L^{2}$ spans that diagonal.

Find. The nodal translations and rotations of the unreinforced frame, its shear-force and bending-moment diagrams, and then the deflected shape, shears and moments of the reinforced frame, with a comment on how much the tie changes them.

[Figure not reproduced: Figure 2 (redrawn). Left, Figure 2(a): the unreinforced rhombic frame pulled apart by P along the horizontal diagonal AC. Right, Figure 2(b): the same frame with a pin-ended truss bar spanning AC. See the official exam paper.]

Approach. Exploit the two axes of symmetry to reduce the frame to one unknown, evaluate the sway terms of the printed plane-frame stiffness matrix for that one degree of freedom, obtain the joint displacements from an energy balance, then recover shears and moments element by element; for part 2.2 add the tie as a second spring in parallel with the whole frame.

1. Reduce the kinematics using double symmetry. Label the corners A (left), B (top), C (right) and D (bottom). Loading and geometry are symmetric about both diagonals, so every joint sits on an axis of symmetry and no joint can rotate: $$\theta_A=\theta_B=\theta_C=\theta_D=0.$$ Writing the translations as $\mathbf{u}_A=(-u,0)$, $\mathbf{u}_C=(+u,0)$, $\mathbf{u}_B=(0,v)$ and $\mathbf{u}_D=(0,-v)$, inextensibility of member AB requires the relative displacement of its ends to have no component along the member. With the AB direction $\mathbf{e}=(1,1)/\sqrt2$, $$\left(\mathbf{u}_B-\mathbf{u}_A\right)\cdot\mathbf{e}=\frac{u+v}{\sqrt2}=0\ \Rightarrow\ v=-u.$$ So the rhombus stretches along AC by exactly as much as it closes along BD, and the entire kinematics is governed by the single unknown $u$.
2. Find the sway of each member. With $\mathbf{n}=(-1,1)/\sqrt2$ perpendicular to AB, the relative transverse (sway) displacement of member AB is $$\Delta=\left|\left(\mathbf{u}_B-\mathbf{u}_A\right)\cdot\mathbf{n}\right| = \left|(u,-u)\cdot\frac{(-1,1)}{\sqrt2}\right| = \sqrt2\,u.$$ By symmetry all four members sway by the same amount. Each member is therefore a beam with both end rotations locked at zero and a relative end translation $\Delta$ across it — the pure sway mode of the printed element matrix.
3. Read the sway stiffness off the printed element matrix. In the plane-frame matrix supplied with the question, the transverse-translation and rotation partition contributes, for $\theta_i=\theta_j=0$ and relative transverse displacement $\Delta$, $$V=\frac{12EI}{L^{3}}\Delta,\ \ M_i=M_j=\frac{6EI}{L^{2}}\Delta,$$ and the strain energy stored in one member is $U_1=\tfrac12\left(\dfrac{12EI}{L^{3}}\right)\Delta^{2}=\dfrac{6EI\Delta^{2}}{L^{3}}$. The axial partition $EA/L$ contributes nothing because the members are inextensible.
4. Assemble the energy and solve for the single unknown. Summing four identical members with $\Delta=\sqrt2\,u$, $$U=4\left(\frac{6EI\left(\sqrt2 u\right)^{2}}{L^{3}}\right)=\frac{48EI\,u^{2}}{L^{3}},$$ while the loads do work $W = 2Pu$ (each load $P$ moves outward by $u$). Stationarity of the total potential, $\partial U/\partial u = 2P$, gives $96EIu/L^{3}=2P$, hence $$\boxed{u=\frac{PL^{3}}{48EI}=0.02083\frac{PL^{3}}{EI}}$$ This is the horizontal movement of each loaded corner; the diagonal AC therefore lengthens by $2u = PL^{3}/(24EI)$.
5. State the complete nodal solution. Collecting the results of Steps 1 and 4, and taking $x$ positive to the right and $y$ positive upward, $$u_A=-\frac{PL^{3}}{48EI},\ \ u_C=+\frac{PL^{3}}{48EI},\ \ v_B=-\frac{PL^{3}}{48EI},\ \ v_D=+\frac{PL^{3}}{48EI},$$ with $v_A=v_C=u_B=u_D=0$ and every rotation zero. Corners B and D move inward: the frame narrows across BD by the same amount it widens across AC, which is the inextensibility condition expressing itself.
6. Recover the member end actions. Substituting $\Delta=\sqrt2\,u=\sqrt2\,PL^{3}/(48EI)$ into Step 3, $$V=\frac{12EI}{L^{3}}\cdot\frac{\sqrt2 PL^{3}}{48EI}=\frac{\sqrt2}{4}P=0.3536P,\ \ M=\frac{6EI}{L^{2}}\cdot\frac{\sqrt2 PL^{3}}{48EI}=\boxed{\frac{\sqrt2}{8}PL=0.1768PL}$$ The shear is constant along each member and the end moments are equal in magnitude at the two ends, so the bending moment varies linearly and passes through zero at mid-length. Resolving the member end forces at a joint gives the axial force, $N=\sqrt2 P/4 = 0.3536P$ tension in all four members; the two members meeting at a joint carry equal and opposite end moments, which is how the joint stays in equilibrium with no applied couple.
7. Draw the diagrams. Every member carries the same constant shear $0.3536P$ and the same antisymmetric moment diagram: $0.1768PL$ at one end, zero at mid-length, $0.1768PL$ of opposite sense at the other end. The maximum moment in the frame is therefore $\sqrt2PL/8$ at all four corners, and there is a point of contraflexure at the mid-point of every member.

The same three quantities now have to be recomputed with the tie in place, and the cleanest route is to treat the frame and the tie as two springs acting in parallel across the loaded diagonal.

1. Express the frame as a diagonal spring. From Step 4, the elongation of the diagonal is $e = 2u = PL^{3}/(24EI)$ under a diagonal pull $P$, so the frame alone offers $$k_f=\frac{P}{e}=\frac{24EI}{L^{3}}.$$
2. Add the truss bar as a second spring. The tie spans corner to corner, a length $L_t=L\sqrt2$, and the question supplies its axial rigidity as $EA = 12EI/L^{2}$. Its axial stiffness is therefore $$k_t=\frac{EA}{L_t}=\frac{12EI/L^{2}}{\sqrt2 L}=\frac{12}{\sqrt2}\frac{EI}{L^{3}}=8.485\frac{EI}{L^{3}}.$$ Being pin-ended it adds nothing to the rotational degrees of freedom, so the frame kinematics of Steps 1 to 3 are unchanged and only the diagonal stiffness grows.
3. Solve the reinforced system. Springs in parallel add, so $$k=k_f+k_t=\left(24+8.485\right)\frac{EI}{L^{3}}=32.485\frac{EI}{L^{3}},\ \ e'=\frac{P}{k}=\boxed{0.03078\frac{PL^{3}}{EI}}$$ against $0.04167\,PL^{3}/EI$ before, and each loaded corner now moves $u'=e'/2=0.01539\,PL^{3}/EI$.
4. Split the load between frame and tie. Each spring carries load in proportion to its stiffness: $$N_t=P\frac{k_t}{k_f+k_t}=0.2612P\ \text{(tension in the bar)},\ \ P_f=P\frac{k_f}{k_f+k_t}=0.7388P.$$ The frame now bends under only 73.88 per cent of the applied load.
5. Scale the member actions. Because the frame remains linear and its deformation pattern is unchanged, every internal action scales with $P_f$: $$V'=0.3536\,(0.7388P)=0.2612P,\ \ M'=0.1768\,(0.7388P)L=\boxed{0.1306\,PL}$$ The shape of both diagrams is exactly as in Figure for the unreinforced frame — constant shear per member, linear moment with contraflexure at mid-length — but every ordinate is multiplied by 0.7388. The deflected shape is likewise identical in form and 26.1 per cent smaller in amplitude.
6. Comment on the reinforcement. The tie removes 26.1 per cent of the corner deflection, the peak moment and the shear, at the cost of one pin-ended bar carrying 0.2612 P in direct tension. The reason the gain is not larger is that the bar is deliberately soft: $EA = 12EI/L^{2}$ makes $k_t$ only about 35 per cent of the frame's own diagonal stiffness. The mechanism is nevertheless the right one — the bar resists the load axially, which is far more efficient than resisting it in bending, so a stiffer bar would keep buying deflection reduction cheaply until the frame moments became negligible and the assembly behaved as a tie with a light bending stiffener around it.