Truss Member Forces: 07-Str-A1, May 2016, Q4

Question 4: Truss Member Forces (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.

(a) Diamond truss on a pin and a roller

Given. Joint coordinates read from the drawing, in metres with the origin at $L_1$ and the $L_2$–$L_4$ chord as $y = 0$: $L_1(0,3)$, $L_2(4,0)$, $L_3(8,0)$, $L_4(12,0)$, $L_5(16,3)$, $U_1(4,6)$, $U_2(8,9)$, $U_3(12,6)$. Loads of 24, 48 and 72 kN act downward at $L_2$, $L_3$ and $L_4$. $L_1$ is a pin and $L_5$ a roller on a horizontal plane. Every sloping member lies on a 3:4 slope, so its length is 5 m and its direction cosines are $(0.8,\ \pm 0.6)$.

Find. The forces in $L_1$–$L_2$, $L_4$–$U_3$ and $L_3$–$L_4$, each labelled tension or compression.

Approach. Reactions first; then a joint at which only two unknowns meet ($L_1$) for the first member, a section that cuts exactly three members for the bottom chord, and a joint again ($L_4$) for the vertical.

Check determinacy and find the reactions. The truss has $m = 13$ members and $j = 8$ joints with $r = 3$, so $m + r = 16 = 2j$ and it is determinate. Taking moments about $L_1$, $$R_{L5}(16) = 24(4) + 48(8) + 72(12) = 1344 \;\Longrightarrow\; \boxed{R_{L5} = 84.0\ \text{kN}\ \uparrow}$$ $$\boxed{R_{L1} = 144 - 84 = 60.0\ \text{kN}\ \uparrow}$$ with no horizontal reaction, since no horizontal load acts.
Isolate joint $L_1$ for the first member. Only two members meet there, $L_1$–$U_1$ rising at $(0.8,\ 0.6)$ and $L_1$–$L_2$ falling at $(0.8,\ -0.6)$, together with the 60 kN reaction. Horizontal equilibrium gives $F_{L1U1} = -F_{L1L2}$, and substituting into vertical equilibrium, $$60 + 0.6\,F_{L1U1} - 0.6\,F_{L1L2} = 60 - 1.2\,F_{L1L2} = 0$$ $$\boxed{F_{L_1L_2} = +50.0\ \text{kN}\ \text{(tension)}}$$ and correspondingly $F_{L1U1} = 50.0\ \text{kN}$ compression.
Cut a section between $L_3$ and $L_4$ for the bottom chord. A vertical cut at $8 < x < 12$ severs exactly three members: the bottom chord $L_3$–$L_4$, the diagonal $U_3$–$L_3$ and the top chord $U_2$–$U_3$. Taking the right-hand free body and moments about $U_3(12,6)$ removes both of the other two, and the 72 kN load at $L_4(12,0)$ lies directly below $U_3$ so it too drops out: $$\sum M_{U_3} = 84\,(16-12) - F_{L_3L_4}\,(6) = 0$$ $$\boxed{F_{L_3L_4} = \frac{336}{6} = +56.0\ \text{kN}\ \text{(tension)}}$$ Repeating with the left-hand free body gives $-60(12) + 24(8) + 48(4) + 6F = 0$, the same 56.0 kN.
Return to joint $L_4$ for the vertical member. Four members meet at $L_4$: the two bottom chords, the vertical $L_4$–$U_3$ and the inclined $L_4$–$L_5$ at $(0.8,\ 0.6)$, with the 72 kN load. Horizontal equilibrium first, $$-56.0 + 0.8\,F_{L_4L_5} = 0 \;\Longrightarrow\; F_{L_4L_5} = +70.0\ \text{kN (tension)}$$ and then vertical equilibrium, $$F_{L_4U_3} + 0.6(70.0) - 72 = 0 \;\Longrightarrow\; \boxed{F_{L_4U_3} = +30.0\ \text{kN}\ \text{(tension)}}$$
Check at joint $L_5$. The roller reaction of 84.0 kN must be balanced by $L_4$–$L_5$ at 70.0 kN tension and $U_3$–$L_5$; horizontal equilibrium gives $F_{U_3L_5} = -70.0\ \text{kN}$, and vertical equilibrium then reads $84 - 0.6(70) + 0.6(-70) = 84 - 42 - 42 = 0$, which closes exactly.

Q4(a). All sloping members lie on a 3:4 slope and are 5 m long. The section used for $L_3$–$L_4$ is any vertical cut between $L_3$ and $L_4$; it severs three members, and moments about $U_3$ eliminate two of them.

Member	Force	Sense
$L_1$–$L_2$	$50.0\ \text{kN}$	Tension
$L_4$–$U_3$	$30.0\ \text{kN}$	Tension
$L_3$–$L_4$	$56.0\ \text{kN}$	Tension
Reaction at $L_1$	$60.0\ \text{kN}$	Upward
Reaction at $L_5$	$84.0\ \text{kN}$	Upward

(b) Parallel-chord truss with an interrupted bottom chord

Given. Joint coordinates in metres with the origin at $L_1$: $L_1(0,0)$, $U_1(0,6)$, $U_2(4,6)$, $U_3(8,6)$, $U_4(12,6)$, $U_5(16,6)$, $U_6(20,6)$, $L_2(4,3)$, $L_3(12,3)$, $L_4(16,3)$, $L_5(20,3)$. Loads of 30 kN act downward at $U_3$, $U_4$ and $U_5$. Both $L_1$ and $L_5$ are pinned. The bottom chord is deliberately interrupted between $L_2$ and $L_3$: no member is drawn there, and $L_1$, $L_2$ and $U_3$ are collinear on a 3:4 line.

Find. The forces in $U_3$–$L_3$, $L_4$–$U_6$ and $L_3$–$L_4$, each labelled tension or compression.

Approach. Two pins give four reaction components, one more than statics allows — but the gap in the bottom chord supplies the missing equation, because a vertical cut through the gap severs only two members and both of them radiate from $U_3$. That makes the whole left-hand assembly a two-force body and fixes the direction of the reaction at $L_1$.

Confirm the count and locate the condition equation. Counting off the drawing, $m = 18$, $j = 11$ and $r = 4$, so $m + r = 22 = 2j$: determinate, in spite of the two pins. A vertical cut anywhere in $4 < x < 8$ crosses only the top chord $U_2$–$U_3$ and the inclined $L_2$–$U_3$, because the bottom chord is missing there. Both pass through $U_3(8,6)$, so moments about $U_3$ for the left-hand free body contain nothing but the reaction at $L_1$: $$\sum M_{U_3} = 6H_1 - 8V_1 = 0 \;\Longrightarrow\; H_1 = \tfrac{4}{3}V_1$$ i.e. the reaction at $L_1$ must act along the line $L_1U_3$.
Close the global equations. With $H_5 = -H_1$ and $V_5 = 90 - V_1$, moments about $L_1$ give $$-30(8) - 30(12) - 30(16) + 20V_5 - 3H_5 = 0$$ $$-1080 + 20(90 - V_1) + 3\left(\tfrac{4}{3}V_1\right) = 0 \;\Longrightarrow\; 720 = 16 V_1$$ $$\boxed{V_1 = V_5 = 45.0\ \text{kN}\ \uparrow,\qquad H_1 = 60.0\ \text{kN}\rightarrow,\quad H_5 = 60.0\ \text{kN}\leftarrow}$$ The resultant at each pin is $75.0\ \text{kN}$ — a genuine horizontal thrust of 60 kN, exactly as in a three-hinged arch, because the missing chord cannot carry the tie force across the gap.
Cut at $x = 10$ for the diagonal $U_3$–$L_3$. That cut severs only $U_3$–$U_4$ (horizontal) and $U_3$–$L_3$, so vertical equilibrium of the left-hand free body — which carries the reaction at $L_1$ and the 30 kN load at $U_3$ — involves one unknown only: $$45 - 30 - 0.6\,F_{U_3L_3} = 0 \;\Longrightarrow\; \boxed{F_{U_3L_3} = +25.0\ \text{kN}\ \text{(tension)}}$$
Work in from the right-hand pin for $L_4$–$U_6$. At joint $L_5$ only the horizontal $L_4$–$L_5$ and the vertical $U_6$–$L_5$ meet the reaction $(-60,\ 45)$, giving $F_{L_4L_5} = -60.0\ \text{kN}$ and $F_{U_6L_5} = -45.0\ \text{kN}$, both compression. Moving up to joint $U_6$, where the top chord, that vertical and the diagonal $U_6$–$L_4$ at $(-0.8,\ -0.6)$ meet, vertical equilibrium reads $$45.0 - 0.6\,F_{L_4U_6} = 0 \;\Longrightarrow\; \boxed{F_{L_4U_6} = +75.0\ \text{kN}\ \text{(tension)}}$$
Finish at joint $L_4$ for the bottom chord. The joint carries $L_3$–$L_4$, $L_4$–$L_5$ at $-60.0$ kN, the vertical $L_4$–$U_5$ and the diagonal $L_4$–$U_6$ at $+75.0$ kN, with no applied load. Horizontal equilibrium alone settles it: $$-F_{L_3L_4} + (-60.0) + 0.8(75.0) = 0$$ $$F_{L_3L_4} = 0$$ so $L_3$–$L_4$ is a zero-force member. The check comes from a section between $L_3$ and $L_4$: taking moments about $U_5(16,6)$, the reaction at $L_5$ has zero moment arm about $U_5$ because it acts along $U_5L_5$ produced, and the 30 kN load at $U_5$ passes through the centre, leaving $3F_{L_3L_4} = 0$.
Sanity-check the left-hand assembly. Because it is a two-force body loaded only at $L_1$ and $U_3$, every member in it except the collinear strut $L_1$–$L_2$–$U_3$ must be unstressed. Working joint by joint confirms exactly that: $F_{L_1L_2} = F_{L_2U_3} = -75.0\ \text{kN}$ (compression, equal to the pin resultant) and $U_1$–$U_2$, $U_2$–$U_3$, $U_1$–$L_1$, $U_1$–$L_2$ and $U_2$–$L_2$ are all zero.

Q4(b). The gap between $L_2$ and $L_3$ is the whole point of the question: it is what makes a truss on two pins determinate, and any vertical cut inside it severs only two members.

Check: the absence of a member between $L_2$ and $L_3$ is read from the printed figure, where the bottom chord line stops at $L_2$ and restarts at $L_3$. Inventing the missing chord would make the truss indeterminate to the first degree and the question unanswerable by statics alone, which is the strongest confirmation that the gap is deliberate.

Member	Force	Sense
$U_3$–$L_3$	$25.0\ \text{kN}$	Tension
$L_4$–$U_6$	$75.0\ \text{kN}$	Tension
$L_3$–$L_4$	$0$	Zero-force member
Reaction at $L_1$ (pin)	$60.0\ \text{kN}\rightarrow,\ 45.0\ \text{kN}\uparrow$	Resultant 75.0 kN
Reaction at $L_5$ (pin)	$60.0\ \text{kN}\leftarrow,\ 45.0\ \text{kN}\uparrow$	Resultant 75.0 kN

Topic. Method of joints, method of sections, and condition equations in trusses.

Key relations. Determinacy: $m + r = 2j$. At a joint, $\sum F_x = \sum F_y = 0$; across a section that severs no more than three members whose lines of action are neither all parallel nor all concurrent, $\sum F_x = \sum F_y = \sum M = 0$. A gap in a chord, or any arrangement that lets a cut sever only two members, supplies a condition equation that makes an apparently over-restrained truss determinate.

Why this works. The method of sections is the method of joints applied to a group of joints at once, and its power comes entirely from the choice of moment centre: put the centre at the intersection of the two members you do not want, and one equation delivers the third force. In part (a) that point is $U_3$, and the 72 kN load conveniently sits directly below it. In part (b) the interrupted chord means a cut severs just two members, and since both radiate from a single joint, moments about that joint involve nothing but the support reactions. Recognising that the left-hand assembly is consequently a two-force body — loaded only at $L_1$ and at $U_3$ — both fixes the reaction direction and predicts, before any arithmetic, that all but one of its members carry zero force.

Common pitfall. Assuming that two pinned supports automatically mean external indeterminacy. Here $r = 4$ but the chord gap gives a fourth equation, and a candidate who “repairs” the truss by drawing in the missing bottom chord makes the problem unsolvable by statics. The second pitfall is calling $L_3$–$L_4$ zero by pattern-matching a standard zero-force rule; here it is zero only because the horizontal components of the neighbouring chord and diagonal happen to cancel, and it must be demonstrated.

Source. Hibbeler, Structural Analysis, 10th ed., §3.4–3.6; Kassimali, Structural Analysis, 6th ed., §4.5–4.7 and §4.9.

Question 4 of 8: Truss Member Forces

Question 4: Truss Member Forces (18 marks)

(a) Diamond truss on a pin and a roller

(b) Parallel-chord truss with an interrupted bottom chord