07-Str-A1 · December 2018

Question 4 of 8: Member Forces in Two Trusses

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

Question 4: Member Forces in Two Trusses (18 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.

4(a) — six-panel camelback truss

Given. Six panels at 2.4 m (total 14.4 m) along a horizontal bottom chord L₁…L₇. The top chord joints sit above the interior bottom joints: U₁ above L₂ at 3.2 m, then U₂, U₃, U₄ above L₃, L₄, L₅ on a level top chord 1.0 m higher at 4.2 m, and U₅ above L₆ at 3.2 m. Three downward loads of 84 kN each act at L₂, L₃ and L₄. The truss is pinned at L₁ and on a roller at L₇. Each top joint carries a vertical to the bottom chord below it and a diagonal running down-and-outwards towards mid-span.

Find. The forces in U₁-U₂, U₁-L₃ and U₂-L₃, each with its sense.

Approach. The truss is determinate ($m + r = 21 + 3 = 24 = 2j$). Find the reactions, then take one vertical section between L₂ and L₃ — it cuts exactly three members, two of which are the ones asked for — and finish the third at joint L₃.

1. Reactions, using the symmetry of the loading. The three 84 kN loads have their resultant 252 kN at $x = 4.8$ m, so $$14.4\,R_{L7} = (84)(2.4 + 4.8 + 7.2) = 1209.6 \qquad R_{L7} = 84.0\ \text{kN}\ \uparrow$$ $$R_{L1} = 252 - 84.0 = 168.0\ \text{kN}\ \uparrow, \qquad H_{L1} = 0$$
2. Cut between L₂ and L₃ and take moments about L₃ for the top chord. A vertical line at $x = 3.6$ m severs U₁-U₂, the diagonal U₁-L₃ and the bottom chord L₂-L₃. Moments about L₃ (4.8, 0) eliminate the last two, because both pass through it. The top chord runs from (2.4, 3.2) to (4.8, 4.2), so its length is $\sqrt{2.4^2 + 1.0^2} = 2.6$ m and its unit vector is $(0.9231,\ 0.3846)$; acting at U₁, whose position relative to L₃ is $(-2.4,\ 3.2)$, its moment arm works out at 3.8769 m. Hence $$-(4.8)(168.0) + (2.4)(84) - 3.8769\,F_{U_1U_2} = 0$$ $$F_{U_1U_2} = \frac{-604.8}{3.8769} = \boxed{156.0\ \text{kN compression}}$$
3. Take moments about U₁ for the bottom chord. Both cut members that pass through U₁ drop out, leaving $$-(2.4)(168.0) + 3.2\,F_{L_2L_3} = 0 \qquad F_{L_2L_3} = +126.0\ \text{kN tension}$$ a value we need only as a stepping stone, but one worth recording because it also checks the horizontal equilibrium of the cut.
4. Get the diagonal from vertical equilibrium of the same free body. The diagonal runs from (2.4, 3.2) to (4.8, 0), so its length is exactly 4.0 m and its unit vector is $(0.6,\ -0.8)$. With the top chord known, $$168.0 - 84 + 0.3846(-156.0) - 0.8\,F_{U_1L_3} = 0$$ $$F_{U_1L_3} = \frac{-24.0}{-0.8} = \boxed{30.0\ \text{kN tension}}$$ Horizontal equilibrium is then the independent check: $0.9231(-156.0) + 0.6(30.0) + 126.0 = -144 + 18 + 126 = 0$.
5. Finish at joint L₃ for the vertical. Four members meet there, but two of them are now known. The diagonal U₁-L₃ pulls the joint towards U₁ with components $30.0(-0.6,\ 0.8) = (-18.0,\ 24.0)$ kN, and the vertical U₂-L₃ acts straight up, so $$24.0 + F_{U_2L_3} - 84 = 0 \qquad F_{U_2L_3} = \boxed{60.0\ \text{kN tension}}$$ Horizontal equilibrium at the same joint gives the next bottom-chord force, $-18.0 - 126.0 + F_{L_3L_4} = 0$, i.e. 144.0 kN tension, confirming that the chord force grows towards mid-span as it must.

4(b) — wall-mounted cantilever truss

Given. A bottom chord L₁…L₄ at 6 m centres on a horizontal line, an inclined chord L₁-U₁-U₂-U₃ rising 2.5 m per panel so that U₁ is 2.5 m above L₂, U₂ is 5.0 m above L₃ and U₃ is 7.5 m above L₄, and an intermediate joint M₁ 2.5 m above L₄. Verticals run U₁-L₂, U₂-L₃, U₃-M₁ and M₁-L₄; diagonals run U₁-L₃, L₃-M₁ and U₂-M₁. Three downward loads of 10 kN act at L₁, L₂ and L₃. The truss is anchored to a vertical wall at U₃ and L₄.

Find. The forces in U₁-U₂, U₁-L₃ and U₂-L₃.

Approach. This is a cantilever, so do not begin at the supports at all: walk the joints inwards from the free end L₁, where only two members meet. All three requested forces then fall out before the wall reactions are ever needed — which also means the answers do not depend on which of the two wall supports is the pin.

1. Note the panel geometry, because it repeats. Every inclined chord panel rises 2.5 m over 6 m, so its length is $\sqrt{6^2 + 2.5^2} = 6.5$ m and its direction cosines are $(6/6.5,\ 2.5/6.5) = (0.9231,\ 0.3846)$ — a 12-5-13 triangle again. The truss is determinate: $m + r = 13 + 3 = 16 = 2j$.
2. Start at the free end L₁, which has only two members. With the 10 kN load hanging there, $$0.3846\,F_{L_1U_1} = 10 \qquad F_{L_1U_1} = +26.0\ \text{kN tension}$$ $$F_{L_1L_2} = -0.9231(26.0) = -24.0\ \text{kN, i.e. 24.0 kN compression}$$
3. Move to L₂, where the vertical simply picks up the panel load. The two bottom-chord bars are collinear and the only other member is vertical, so $$F_{U_1L_2} = +10.0\ \text{kN tension} \qquad F_{L_2L_3} = -24.0\ \text{kN compression}$$ The chord force is unchanged across the panel, as it must be when nothing acts horizontally.
4. Resolve at U₁, where the two unknowns separate neatly. Four members meet: the known chord to L₁ (26.0 kN), the known vertical (10.0 kN), and the two unknowns U₁-U₂ and U₁-L₃, whose unit vectors are $(0.9231,\ \pm 0.3846)$. Horizontal equilibrium gives their sum and vertical equilibrium their difference: $$F_{U_1U_2} + F_{U_1L_3} = 26.0 \qquad F_{U_1U_2} - F_{U_1L_3} = \frac{20}{0.3846} = 52.0$$ $$F_{U_1U_2} = \boxed{39.0\ \text{kN tension}} \qquad F_{U_1L_3} = \boxed{13.0\ \text{kN compression}}$$
5. Finish the vertical at joint U₂. Continuing the walk (L₃ then M₁ then the wall, or U₂ directly using the wall reactions) gives U₂-U₃ = 52.0 kN tension and U₂-M₁ = 13.0 kN compression, whence vertical equilibrium at U₂ leaves $$-15.0 - F_{U_2L_3} + 0.3846\bigl[52.0 - (-13.0)\bigr] = 0 \qquad F_{U_2L_3} = \boxed{10.0\ \text{kN tension}}$$ The same value comes out of joint L₃ independently, and it is no accident that it equals the panel load: the vertical's only job is to hang that 10 kN from the inclined chord.