Question 4 of 8: Truss Member Forces by the Method of Sections

Question 4: Truss Member Forces by the Method of Sections (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.

Given. Part (a): a polygonal-chord truss of six 6 m panels spanning 36 m, pinned at $L_1$ and on a roller at $L_7$. The top-chord joints sit above $L_2$ to $L_6$ at heights 4.0, 6.5, 7.5, 6.5 and 4.0 m. Vertical loads of 26 kN act at $L_4$, $L_5$ and $L_6$. Part (b): an inclined truss with $U_1(0,\,0)$, $U_2(6.4,\,4.8)$, $U_3(12.8,\,9.6)$, $L_1(5,\,0)$ and $L_2(11.4,\,4.8)$ metres, so both chords lie on a 3:4 slope and the perpendicular depth of the truss is the 3 m marked on the drawing. $U_1$ is a pin and $U_3$ a roller bearing on a vertical wall, so its reaction is horizontal. Each of $L_1$ and $L_2$ carries 24 kN downward and 24 kN to the right.

Find. The six named member forces, each labelled tension or compression.

Approach. Take the reactions from global equilibrium, then pass one section through each group of three unknown members and choose the moment centre as the intersection of the two members not wanted, so that each force falls out of a single equation.

1. Part (a) — reactions. The three 26 kN loads sit at 18, 24 and 30 m from $L_1$. Moments about $L_1$ give $$R_{L7} = \frac{26(18 + 24 + 30)}{36} = \frac{1872}{36} = 52\ \text{kN}$$ and vertical equilibrium gives $R_{L1} = 78 - 52 = 26$ kN, both upward.
2. Cut section a-a between $L_5$ and $L_6$. A vertical cut at $x = 27$ m severs exactly three members: the top chord $U_4U_5$, the diagonal $U_5L_5$ and the bottom chord $L_5L_6$. Work with the portion to the right, which carries the 26 kN load at $L_6$ and the 52 kN reaction at $L_7$.
3. Moments about $L_5$ give the top chord. Both $L_5L_6$ and $U_5L_5$ pass through $L_5$, so only $U_4U_5$ survives. Its line runs from $(24,\,6.5)$ to $(30,\,4)$, a length of 6.5 m, and its moment arm about $L_5$ works out to exactly 6 m. Hence $$6F_{U_4U_5} + \bigl[52(36-24) - 26(30-24)\bigr] = 0 \;\Rightarrow\; F_{U_4U_5} = -\frac{468}{6} = \boxed{78\ \text{kN compression}}$$ The negative sign in the tension-positive convention is what marks it as compression, as one expects of a top chord.
4. Moments about the chord intersection give the diagonal. The lines of $U_4U_5$ and $L_5L_6$ meet at $(39.6,\,0)$. Taking moments there for the same right-hand portion leaves only $U_5L_5$, whose length is $\sqrt{6^2 + 4^2} = 7.211$ m, and the equation reduces to $$8.653F_{U_5L_5} + 62.4 = 0 \;\Rightarrow\; F_{U_5L_5} = \boxed{7.21\ \text{kN compression}}$$ a small force, because the panel loads either side of $U_5$ very nearly balance.
5. The bottom chord from horizontal equilibrium, then the panel chord. Resolving horizontally on the right-hand portion, $0.9231(78) + 0.8321(7.211) = F_{L_5L_6}$, so $F_{L_5L_6} = 78$ kN tension. Moving to joint $L_5$ and resolving horizontally, $$F_{L_4L_5} = F_{L_5L_6} + F_{U_5L_5}\left(\frac{6}{7.211}\right) = 78 - 6 = \boxed{72\ \text{kN tension}}$$ The same value follows from a cut between $L_4$ and $L_5$ with moments taken about $U_4$: $468 = 6.5F_{L_4L_5}$, and $468/6.5 = 72$ kN. Vertical equilibrium at $L_5$ then gives the post $U_4L_5$ as 30 kN tension, which closes the joint.

1. Part (b) — reactions. Vertical equilibrium gives $U_{1y} = 48$ kN up at once, since the wall roller at $U_3$ is horizontal. Moments about $U_1$ collect the two 24 kN vertical loads at $x = 5$ and $x = 11.4$, the horizontal 24 kN at $L_2$ acting 4.8 m above $U_1$, and the unknown $U_{3x}$ acting 9.6 m above it: $$9.6\,|U_{3x}| = 24(5) + 24(11.4) + 24(4.8) = 508.8 \;\Rightarrow\; U_{3x} = 53\ \text{kN to the left}$$ Horizontal equilibrium then gives $U_{1x} = 53 - 48 = 5$ kN to the right.
2. Cut section b-b through $U_2U_3$, $U_2L_2$ and $L_1L_2$. That cut isolates a right-hand portion containing only $U_3$ and $L_2$ and the member between them, which makes every subsequent moment equation short. Its external forces are the 53 kN wall reaction and the pair of 24 kN loads at $L_2$.
3. Moments about $U_2$ give the bottom chord. Both $U_2U_3$ and $U_2L_2$ pass through $U_2$, and the perpendicular distance from $U_2$ to the line $L_1L_2$ is the 3 m printed on the drawing, so $$3F_{L_1L_2} = 53(4.8) - 24(5) = 254.4 - 120 = 134.4 \;\Rightarrow\; F_{L_1L_2} = \boxed{44.8\ \text{kN tension}}$$ It is worth pausing on that 3 m: the right-angle mark beside it says it is the perpendicular offset of $U_2$ from the bottom chord, which is exactly the lever arm this equation needs, so the examiner has handed the candidate the hardest piece of geometry in the question.
4. Moments about $L_2$ give the top chord. Now $U_2L_2$ and $L_1L_2$ both pass through the moment centre and the two 24 kN loads act there as well, so only the 53 kN reaction and $F_{U_2U_3}$ remain. The perpendicular offset of $L_2$ from the top chord is again 3 m: $$3F_{U_2U_3} + 53(4.8) = 0 \;\Rightarrow\; F_{U_2U_3} = \boxed{84.8\ \text{kN compression}}$$
5. Horizontal equilibrium of the cut portion gives the web member. With the chords known and both lying on the 3:4 slope, $$F_{U_2L_2} = 0.8(84.8) - 0.8(44.8) - 53 + 24 = \boxed{3.0\ \text{kN tension}}$$ Vertical equilibrium of the same portion is then a pure check: $0.6(84.8) - 0.6(44.8) - 24 = 0$. As a second check, joint $L_2$ gives $F_{U_3L_2} = 53$ kN tension and closes horizontally to zero.