Question 2 of 3: Square frame under parting forces, bare and truss-braced

Question 2: Square frame under parting forces, bare and truss-braced (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.

2.1  The bare frame

Given. A closed square frame of side $2L$, each side of uniform flexural stiffness $EI$, with rigid right-angled corners and axially rigid (inextensible) members. Outward horizontal forces $P$ act at the mid-height of the two vertical sides, at points $E$ and $G$; the load set is self-equilibrated.

Find. The deflected shape, the shear force diagram and the bending moment diagram, together with the deflection of the load points.

Approach. Use the double symmetry to reduce the closed frame to a quarter with two symmetry cuts, resolve the equilibrium of that quarter down to one redundant moment, close the problem with the compatibility condition that neither cut face may rotate, then recover the diagrams and the deflections from the moment field by Castigliano's theorem.

1. Establish what the kinematic assumptions leave free. Inextensibility fixes the chord length of every side, and rigid right angles fix the angle between the sides at every corner; together they mean the four corners form a rigid square and therefore cannot translate relative to one another. What is not constrained is the corner rotation and the transverse bowing of each side, and those carry the whole response. This is the observation that makes the problem tractable by hand.
2. Cut on the two axes of symmetry. The structure and the loading are symmetric about both the $x$ and the $y$ axis. At a symmetric section the antisymmetric stress resultant — the transverse shear — must vanish, and the section may not rotate. Cutting at $E$ and at $F$ isolates the first-quadrant quarter $E\!-\!B\!-\!F$ carrying half the load $P$ at $E$.
3. Take moments and forces on the quarter. With only an axial force and a couple crossing each cut, vertical equilibrium gives $V_E=0$ — the side members carry no axial force at all — and horizontal equilibrium gives $H_F=P/2$, so the top and bottom members carry a tension of $P/2$. Moments about $B$ give one equation in the two cut moments: $$M_E+M_F+\frac{PL}{2}=0$$ which leaves a single redundant.
4. Write the moment field in terms of that redundant. Taking $X=M_E$ and measuring $s$ upward from $E$ and $t$ from $B$ toward $F$, $$M(s)=X+\frac{Ps}{2}\quad\text{on }EB,\qquad M(t)=X+\frac{PL}{2}\quad\text{on }BF$$ The moment on the top member is constant, because the only force acting on that free body beyond the cut is the axial $P/2$, whose line of action lies along the member.
5. Impose compatibility. A unit change in $X$ is a self-equilibrating pair of unit couples on the two cut faces, and the conjugate displacement is the relative rotation of those faces, which symmetry forces to zero. Hence $\int M\,(\partial M/\partial X)\,ds/EI=0$ with $\partial M/\partial X=1$ everywhere: $$\int_0^{L}\!\left(X+\frac{Ps}{2}\right)ds+\int_0^{L}\!\left(X+\frac{PL}{2}\right)dt = 2XL+\frac{3PL^{2}}{4}=0$$ $$\boxed{\;X = M_E = -\frac{3PL}{8}\;}$$
6. Read off the complete moment field. Back-substituting, the corner moment is $M_B=X+PL/2=+PL/8$, and since the top member carries a constant moment that same $PL/8$ runs the whole length of the top and bottom members. Along each side member the moment falls linearly from $+PL/8$ at a corner to $-3PL/8$ at the load point, crossing zero $3L/4$ above (and below) the load point. The peak magnitude is therefore $$\left|M\right|_{\max}=\frac{3PL}{8}=0.375\,PL \quad\text{at the two load points.}$$
7. Differentiate for the shears, and collect the axial forces. $V=dM/ds$ gives a constant $P/2$ in each side member, reversing sign as it passes the load point, and zero shear everywhere in the top and bottom members, which is consistent with their uniform moment. The axial forces are zero in the side members and $P/2$ tensile in the top and bottom members — the pair of horizontal members is what actually resists the parting.
8. Obtain the deflections from the strain energy. With $U=4\times\frac{1}{2EI}\left[\int_0^L M_{EB}^{2}\,ds+\int_0^L M_{BF}^{2}\,dt\right]$ and the moment field above, $$U=\frac{5P^{2}L^{3}}{48EI}\quad\Rightarrow\quad \Delta_h=\frac{\partial U}{\partial P}=\frac{5PL^{3}}{24EI}$$ so each load point moves outward by half of that. Adding a dummy inward pair $Q$ at $F$ and $H$ and taking $\partial U/\partial Q$ at $Q=0$ gives the vertical closure: $$\boxed{\;\delta_h=\frac{5PL^{3}}{48EI}\ \text{outward},\qquad \delta_v=\frac{PL^{3}}{16EI}\ \text{inward}\;}$$

2.2  The frame reinforced with four truss bars

Given. The same frame, now braced by four pin-ended truss bars forming a rhombus that joins the four mid-side points $E$, $F$, $G$, $H$. Each bar has length $L\sqrt{2}$ and axial rigidity $EA=12EI\sqrt{2}/L^{2}$.

Find. The deflected shape, the shear force diagram and the bending moment diagram, with the extreme values on the shear and moment diagrams.

Approach. Replace the bars by their nodal force resultants, so the frame is again the structure of part 2.1 but under two symmetric load pairs; write its flexibility coefficients from 2.1; then close the problem with one compatibility equation stating that the bar elongation equals the relative movement of the two points it joins.

1. Reduce the bar rigidity to an axial spring stiffness. Each bar spans a diagonal of the rhombus, $L_b=L\sqrt{2}$, so $$k_b=\frac{EA}{L_b}=\frac{12EI\sqrt{2}}{L^{2}}\cdot\frac{1}{L\sqrt{2}}=\frac{12EI}{L^{3}}$$ The $\sqrt{2}$ in the printed data is there exactly to make this come out as the sway stiffness $12EI/L^{3}$ of a beam element — a deliberate piece of examiner's arithmetic.
2. Replace the bars by the forces they apply to the frame. By symmetry all four bars carry the same tension $T$. At $E$ the two bars meeting there pull along $(-1,+1)/\sqrt{2}$ and $(-1,-1)/\sqrt{2}$, whose resultant is purely horizontal and inward, of magnitude $R=\sqrt{2}\,T$; at $F$ the resultant is purely vertical and inward, also $\sqrt{2}\,T$. The frame therefore sees $$P'=P-\sqrt{2}\,T \ \text{outward at } E,G,\qquad Q'=\sqrt{2}\,T \ \text{inward at } F,H$$ and both pairs are doubly symmetric, so every result of 2.1 can be reused.
3. Write the frame flexibilities. From part 2.1, an outward pair $P$ opens the horizontal span by $5PL^{3}/24EI$ and closes the vertical span by $PL^{3}/8EI$. By Maxwell reciprocity the cross terms are equal, and the four-fold symmetry of a square makes the two direct terms equal, so $$f_{11}=f_{22}=\frac{5L^{3}}{24EI},\qquad f_{12}=f_{21}=\frac{L^{3}}{8EI}=\frac{3L^{3}}{24EI}$$ Hence $\Delta_h=f_{11}P'+f_{12}Q'$ and $\Delta_v=f_{12}P'+f_{22}Q'$, both measured as relative movements of the opposite pairs of points.
4. Express the bar elongation. The bar $EF$ has unit vector $(-1,1)/\sqrt{2}$, so with $E$ moving out by $\Delta_h/2$ and $F$ moving in by $\Delta_v/2$ its elongation is $$e=\frac{\Delta_h-\Delta_v}{2\sqrt{2}}=\frac{L^{3}}{24\sqrt{2}\,EI}\left(P-2R\right)$$ after substituting the flexibilities, where $R=\sqrt{2}T$ as before.
5. Close the problem with the bar constitutive law. Setting $T=k_b e$ and multiplying by $\sqrt{2}$, $$R=\frac{12EI}{L^{3}}\cdot\frac{L^{3}\left(P-2R\right)}{24\,EI}=\frac{P-2R}{2}\quad\Rightarrow\quad 4R=P$$ $$\boxed{\;R=\sqrt{2}\,T=\frac{P}{4},\qquad T=\frac{\sqrt{2}P}{8}=0.1768\,P\ \text{(tension)}\;}$$ The bars are in tension, as they must be: the frame stretches more horizontally than it draws in vertically, so the rhombus diagonals lengthen.
6. Restate the frame loading and superpose the moment fields. With $P'=3P/4$ and $Q'=P/4$, the general result of 2.1 — $X=-3P'L/8-Q'L/8$, $M(s)=X+P's/2$ on the side member and $M(t)=X+P'L/2+Q't/2$ on the top member — gives $$X=M_E=-\frac{5PL}{16},\qquad M_{\text{corner}}=+\frac{PL}{16},\qquad M_F=+\frac{3PL}{16}$$ so the moment now varies linearly along the top member too, from $PL/16$ at each corner to $3PL/16$ at its mid-point.
7. Differentiate for the shears and collect the axial forces. $$V_{\text{side}}=\frac{P'}{2}=\frac{3P}{8},\qquad V_{\text{top}}=\frac{Q'}{2}=\frac{P}{8}$$ each reversing sign at its own mid-point, while the axial forces become $P/8$ compressive in the side members and $3P/8$ tensile in the top and bottom members. The zero-moment point on the side member has migrated to $5L/6$ from the load point.
8. Recover the deflections. Substituting $P'$ and $Q'$ into the flexibility relations, $$\Delta_h=\frac{3PL^{3}}{16EI},\qquad \Delta_v=\frac{7PL^{3}}{48EI}$$ so each load point moves out by $3PL^{3}/32EI$ and each mid-side point draws in by $7PL^{3}/96EI$. The bracing cuts the horizontal spread by 10 % but increases the vertical draw-in by 16.7 %, because the bars pull $F$ and $H$ inward as they restrain $E$ and $G$.