Member forces in two determinate trusses: 07-Str-A1, December 2014, Q4

Question 4: Member forces in two determinate 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.

Given. Two plane trusses with pinned joints, each determinate on inspection ($m + r = 2j$).

Truss geometry and loads read from the drawings
Part	Layout	Supports	Loads
(a)	Span 22 m in bays of 6, 5, 5 and 6 m; the top chord $U_{2}U_{3}U_{4}$ lies 2.5 m above the line of the supports and the bottom chord $L_{1}L_{2}L_{3}$ lies 2.5 m below it	pin at $U_{1}$, roller at $U_{5}$	110 kN down at $U_{2}$, 100 kN down at $U_{3}$
(b)	$L_{1}(0,0)$, $L_{2}(6,0)$, $U_{1}(0,5)$, $M_{1}(6,2.5)$, $U_{2}(6,7.5)$, $U_{3}(12,5)$ in metres	roller at $L_{1}$, pin at $L_{2}$	120 kN down and 36 kN to the right at $U_{1}$; 120 kN down at $U_{2}$; 60 kN down at $U_{3}$

Find. The three listed member forces in each truss, with the sense (tension or compression) stated in each case.

(a) Lens-shaped truss, 22 m span

Truss (a). The three members asked for are drawn heavy. Every diagonal from a support runs on a 6:2.5 slope, so its length is 6.5 m and its direction cosines are 12/13 and 5/13.

Reactions. Moments about the pin at $U_{1}$ give $$R_{U_{5}} = \frac{110(6) + 100(11)}{22} = \frac{660 + 1100}{22} = 80\ \text{kN}, \qquad V_{U_{1}} = 210 - 80 = 130\ \text{kN}$$ and $H_{U_{1}} = 0$ because no horizontal load is applied.
Joint U₁ — the member asked for first. Only two members meet the pin, both on a 6 by 2.5 slope, so each has length $\sqrt{6^{2}+2.5^{2}} = 6.5$ m and direction cosines $12/13$ and $5/13$. Horizontal equilibrium forces the two forces to be equal and opposite, and vertical equilibrium then gives $$2\left(\tfrac{5}{13}\right)F = 130 \ \Rightarrow \ F_{U_{1}L_{1}} = \boxed{169\ \text{kN (tension)}}$$ with $F_{U_{1}U_{2}} = 169$ kN compression in the upper diagonal.
Joint L₁. Three members meet here and no load is applied. Resolving horizontally, $$F_{L_{1}L_{2}} = 169\left(\tfrac{12}{13}\right) = \boxed{156\ \text{kN (tension)}}$$ and resolving vertically gives the post force $F_{U_{2}L_{1}} = -65$ kN, that is 65 kN compression. The bottom chord is in tension, as it must be in a simply supported truss loaded downwards.
Joint U₂ — the diagonal. Four members meet at $U_{2}$, two of them now known, and the 110 kN load acts there. The diagonal $U_{2}L_{2}$ runs 5 m across and 5 m down, so its direction cosines are $1/\sqrt{2}$ each. Vertical equilibrium gives $$169\left(\tfrac{5}{13}\right) + 65 - \tfrac{1}{\sqrt{2}} F_{U_{2}L_{2}} - 110 = 0 \ \Rightarrow \ F_{U_{2}L_{2}} = 20\sqrt{2} = \boxed{28.3\ \text{kN (tension)}}$$ Horizontal equilibrium at the same joint then returns $F_{U_{2}U_{3}} = 176$ kN compression, which is the check that the joint closes.

(b) Kite-shaped truss with a horizontal load

Truss (b). U2 sits 2.5 m above U1 and U3, so U1-U2 and U2-U3 are diagonals on a 6:2.5 slope, not a horizontal top chord.

Reactions. With the roller at $L_{1}$ carrying vertical load only, horizontal equilibrium gives $H_{L_{2}} = 36$ kN acting to the left. Moments about the pin at $L_{2}$, remembering that the 36 kN horizontal load acts 5 m above the base line, give $$6 V_{L_{1}} = 120(6) - 36(5) - 60(6) = 720 - 180 - 360 = 180 \ \Rightarrow \ V_{L_{1}} = 30\ \text{kN}$$ and $V_{L_{2}} = 300 - 30 = 270$ kN.
Start where only two members meet: joint U₃. $U_{3}$ carries the 60 kN load on just two members, $U_{3}U_{2}$ and $U_{3}M_{1}$, both on the same 6 by 2.5 slope but mirrored. Horizontal equilibrium makes them equal and opposite, and vertical equilibrium gives $$2\left(\tfrac{5}{13}\right)F = 60 \ \Rightarrow \ F_{U_{2}U_{3}} = 78\ \text{kN tension}, \quad F_{M_{1}U_{3}} = 78\ \text{kN compression}$$
Joint U₂. The two sloping members $U_{1}U_{2}$ and $U_{2}U_{3}$ are mirror images about the vertical through $U_{2}$, so horizontal equilibrium gives them equal forces at once: $$F_{U_{1}U_{2}} = F_{U_{2}U_{3}} = \boxed{78\ \text{kN (tension)}}$$ Vertical equilibrium then gives the post force $F_{U_{2}M_{1}} = -180$ kN, that is 180 kN compression.
Joint U₁ — the second member asked for. The 36 kN horizontal load acts here, together with 120 kN down. Resolving horizontally, with $U_{1}U_{2}$ and $U_{1}M_{1}$ both on the 12/13 slope, $$\tfrac{12}{13}\left(78 + F_{U_{1}M_{1}}\right) + 36 = 0 \ \Rightarrow \ F_{U_{1}M_{1}} = -117 = \boxed{117\ \text{kN (compression)}}$$ Vertical equilibrium at $U_{1}$ then gives the column force $F_{U_{1}L_{1}} = -45$ kN, that is 45 kN compression.
Joint L₁ — the last member. The roller reaction of 30 kN acts up, the column delivers 45 kN of compression from above, and the diagonal $L_{1}M_{1}$ rises on the 5/13 slope. Vertical equilibrium gives $$-45 + \tfrac{5}{13}F_{L_{1}M_{1}} + 30 = 0 \ \Rightarrow \ F_{L_{1}M_{1}} = \boxed{39\ \text{kN (tension)}}$$ and horizontal equilibrium returns $F_{L_{1}L_{2}} = -36$ kN, 36 kN compression in the bottom chord — exactly the applied horizontal load, which is the final check on the whole solution.

Question 4 — member forces requested
Truss	Member	Force	Sense
(a)	$U_{1}\!-\!L_{1}$	169 kN	Tension
(a)	$L_{1}\!-\!L_{2}$	156 kN	Tension
(a)	$U_{2}\!-\!L_{2}$	28.3 kN	Tension
(b)	$U_{1}\!-\!U_{2}$	78 kN	Tension
(b)	$U_{1}\!-\!M_{1}$	117 kN	Compression
(b)	$L_{1}\!-\!M_{1}$	39 kN	Tension

Topic. Analysis of determinate plane trusses by the method of joints, with the method of sections as a cross-check.

Key relations. At every pin, $\sum F_{x} = 0$ and $\sum F_{y} = 0$; a member of run $a$ and rise $b$ has length $\sqrt{a^{2}+b^{2}}$ and contributes components in the ratio $a : b$; a joint where only two non-collinear members meet, with no load, gives two zero-force members; and for the whole truss $m + r = 2j$ is the determinacy condition.

Why this works. Both trusses in this question are built on scaled 5-12-13 triangles, so every diagonal has exact direction cosines of $12/13$ and $5/13$ and none of the arithmetic needs a protractor. The order of attack is what makes the work short: always begin at a joint where the number of unknowns is two, which here means the pin at $U_{1}$ in part (a) and the two-member joint $U_{3}$ in part (b), and then walk outwards. Where two members at a joint are mirror images about the line of the resultant load, horizontal equilibrium equates them immediately and the vertical equation alone finishes the joint — that shortcut solves two of the six answers here in one line each.

Common pitfall. Assuming that a top chord drawn near the top of the figure is horizontal. In part (b) $U_{2}$ sits 2.5 m above $U_{1}$ and $U_{3}$, so $U_{1}U_{2}$ is a diagonal; treating it as horizontal makes the joint at $U_{2}$ insoluble and the whole truss appear indeterminate. The other recurring error is mixing up the supports — here the roller is at $L_{1}$ and the pin at $L_{2}$, so the horizontal reaction appears at $L_{2}$, and swapping them puts the 36 kN into the wrong bottom-chord force.

Source. Hibbeler, Structural Analysis, 10th ed., §3.4–3.6 (method of joints, zero-force members, method of sections); Kassimali, Structural Analysis, 6th ed., §4.5–4.7.

Question 4 of 8: Member forces in two determinate trusses

Question 4: Member forces in two determinate trusses (18 marks)

(a) Lens-shaped truss, 22 m span

(b) Kite-shaped truss with a horizontal load