Question 3 of 8: Member forces in a Pratt truss

Question 3: Member forces in a Pratt truss (16 marks)

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 parallel-chord truss of five 4 m panels, span 20 m, depth 3 m. The bottom chord runs L₁…L₆ at 4 m centres; the top chord carries four joints U₁…U₄ sitting directly over L₂, L₃, L₄ and L₅. The web has a vertical under every top joint, rising diagonals in the first three panels (L₁U₁, L₂U₂, L₃U₃) and falling diagonals in the last two (U₃L₅, U₄L₆). Three 30 kN downward loads hang at L₂, L₃ and L₄. L₁ is a pin and L₆ a roller.

Find. The axial forces in L₁U₁, L₂L₃, L₂U₂ and L₂U₁, each labelled tension or compression.

Approach. Confirm determinacy, take the reactions from global statics, then walk joints L₁, U₁ and L₂ in that order, because each of them adds only two new unknowns; finally re-derive the bottom chord by a section cut as an independent check.

1. Check determinacy and note the geometry. The truss has $m = 17$ members, $j = 10$ joints and $r = 3$ restraint components, so $$m + r = 20 = 2j$$ and it is statically determinate. Every diagonal spans one 4 m panel over the 3 m depth, so its length is $\sqrt{4^{2}+3^{2}} = 5$ m exactly and its direction cosines are $4/5 = 0.8$ horizontally and $3/5 = 0.6$ vertically. That exactness is why the answers come out as integers.
2. Reactions from global statics. Taking moments about the pin, $$\sum M_{L_1} = 0:\qquad 30(4) + 30(8) + 30(12) = 20\,R_{L_6}$$ $$R_{L_6} = \frac{720}{20} = 36\ \text{kN},\qquad R_{L_1} = 90 - 36 = 54\ \text{kN}$$ both upward, and there is no horizontal reaction because there is no horizontal load.
3. Joint L₁ gives the end diagonal. Only two members meet the pin: the horizontal chord L₁L₂ and the diagonal L₁U₁. Vertical equilibrium involves the diagonal alone, $$\sum F_y = 0:\qquad 54 + 0.6\,F_{L_1U_1} = 0 \quad\Rightarrow\quad F_{L_1U_1} = -\frac{54}{0.6} = -90\ \text{kN}$$ $$\boxed{F_{L_1U_1} = 90\ \text{kN compression}}$$ The negative sign means the bar pushes on the joint, which is what an end diagonal rising away from a support must do. Horizontal equilibrium then gives the first bottom chord, $F_{L_1L_2} = -0.8(-90) = +72$ kN tension.
4. Joint U₁ gives the vertical. Three members meet there: the diagonal just found, the vertical down to L₂ and the top chord to U₂. The top chord is horizontal, so the vertical equation contains only two terms, $$\sum F_y = 0:\qquad 0.6(90) - F_{U_1L_2} = 0$$ $$\boxed{F_{L_2U_1} = 54\ \text{kN tension}}$$ — the vertical simply hangs the whole pin reaction down onto the loaded bottom joint. Horizontal equilibrium at the same joint returns $F_{U_1U_2} = -72$ kN, i.e. 72 kN compression in the first top chord.
5. Joint L₂ gives the second diagonal. Four members meet L₂ — the two bottom chords, the vertical (now known) and the rising diagonal to U₂ — and a 30 kN load hangs there. Vertically, $$\sum F_y = 0:\qquad 54 + 0.6\,F_{L_2U_2} - 30 = 0 \quad\Rightarrow\quad F_{L_2U_2} = -\frac{24}{0.6} = -40\ \text{kN}$$ $$\boxed{F_{L_2U_2} = 40\ \text{kN compression}}$$ Notice the pattern: a rising diagonal carries the net shear left of its panel divided by 0.6, and the shear has dropped from 54 to 24 kN across the first loaded joint.
6. Horizontal equilibrium at L₂ gives the requested chord. With the diagonal settled, $$\sum F_x = 0:\qquad -72 + F_{L_2L_3} + 0.8(-40) = 0$$ $$\boxed{F_{L_2L_3} = 104\ \text{kN tension}}$$ as a bottom chord in a simply supported truss must be.
7. Check the chord independently by a section cut. Cut vertically through the second panel, severing U₁U₂, L₂U₂ and L₂L₃, and take moments about U₂ for the left-hand piece. The two cut members that pass through U₂ drop out, leaving one equation in one unknown: $$\sum M_{U_2} = 0:\qquad -54(8) + 30(4) + 3\,F_{L_2L_3} = 0 \quad\Rightarrow\quad F_{L_2L_3} = \frac{432 - 120}{3} = 104\ \text{kN}$$ which reproduces the joint-walk value exactly. Summing vertical forces on the same free body returns $0.6\,F_{L_2U_2} = -(54 - 30)$, i.e. $-40$ kN, confirming the diagonal too.