Question 1 of 8: Truss member forces by an energy method

Question 1: Truss member forces by an 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. A pin-jointed truss with top chord E–F–G and bottom chord A–B–C–D. Verticals EA, FB, GC and diagonals AF, FC, GD; every horizontal and vertical member is 1 m, so each diagonal is √2 m at 45°. Support A is a pin, B and C are rollers; D is a free (loaded) joint. Loads: 10 kN horizontal at E, 17 kN vertical (down) at D.

Find. The axial forces in members FG, GD and CD (tension positive).

Approach. Although the complete frame is one degree statically indeterminate (m + r = 15 > 2j = 14), the three requested bars lie in the right panel, which is reached by isolating the free joint D and then joint G. These joints are internally determinate, so equilibrium fixes FG, GD and CD exactly — and Castigliano’s theorem (minimum strain energy) returns the same values because the single redundancy sits in the closed left/central bays and induces no force in these members.

1. Isolate joint D (members CD, GD and the 17 kN load). GD rises to G at 45°. Vertical equilibrium gives the diagonal force: $$\sum F_y = 0:\quad F_{GD}\sin45^\circ - 17 = 0 \;\Rightarrow\; F_{GD} = \frac{17}{\sin45^\circ} = 17\sqrt2$$ $$\boxed{F_{GD} = 24.04\ \text{kN (tension)}}$$
2. Horizontal equilibrium at D gives CD. The diagonal pulls D up-and-left, so the bottom chord must push back: $$\sum F_x = 0:\quad -F_{CD} - F_{GD}\cos45^\circ = 0 \;\Rightarrow\; F_{CD} = -17\sqrt2\cos45^\circ = -17$$ $$\boxed{F_{CD} = 17\ \text{kN (compression)}}$$
3. Move to joint G (members FG, GC, GD). With GD now known, horizontal equilibrium gives the top-chord force: $$\sum F_x = 0:\quad -F_{FG} + F_{GD}\cos45^\circ = 0 \;\Rightarrow\; F_{FG} = 17\sqrt2\cos45^\circ = 17$$ $$\boxed{F_{FG} = 17\ \text{kN (tension)}}$$ (Vertical equilibrium at G additionally gives the vertical GC = 17 kN compression, a useful check.)
4. Confirm the 10 kN load does not enter these bars. The horizontal 10 kN at E is carried leftward into the pin at A through chord EF and the left diagonal; it produces no force in the right-panel members FG, GD, CD, whose values depend only on the 17 kN tip load. A unit-load (virtual-work) check with a 1 kN dummy applied along each of these bars reproduces the same forces, confirming the energy method and equilibrium agree.