Vertical deflection of joint U₂ of a symmetric truss: 16-Civ-A1 Elementary Structural Analysis, May 2013, Q3

Question 3: Vertical deflection of joint U₂ of a symmetric 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 symmetric roof (Fink) truss, span $2\times7.8=15.6$ m, apex $U_2$ at height 3.25 m; pin at $L_1$, roller at $L_3$. Two 26 kN loads act at $U_1$ and $U_3$ perpendicular to the top chords. Axial rigidity $EA=6.5\times10^{5}$ kN for every member.

Find. The vertical deflection of the apex joint $U_2$.

Approach. Use the unit-load (virtual-work) method: $\delta=\sum \dfrac{N\,n\,L}{EA}$, where $N$ are the member forces under the real 26 kN loads and $n$ the member forces under a unit downward load at $U_2$. Both force sets come from the method of joints on this determinate truss.

Q3 truss: apex U₂ at 3.25 m; 26 kN loads normal to the rafters at U₁ and U₃.

Geometry. The rafter $L_1U_2$ has length $\sqrt{7.8^2+3.25^2}=8.45$ m (a scaled 5–12–13 triangle: $\cos\theta=12/13$, $\sin\theta=5/13$). The sub-strut $U_1L_2$ is the perpendicular from $L_2$ to that rafter, so $U_1$ lies 7.2 m along the rafter at $(6.65,\,2.77)$; by symmetry $U_3=(8.95,\,2.77)$.
Resolve the applied loads. Each 26 kN load is normal to the rafter: components $26\sin\theta=10$ kN horizontal (inward) and $26\cos\theta=24$ kN vertical (down). Total vertical load $=48$ kN; by symmetry the horizontal components cancel.
Reactions. $L_{1y}=L_{3y}=\tfrac{48}{2}=24.0$ kN; horizontal reaction at the pin $=0$.
Member forces $N$ (real). By joints/symmetry the four rafter segments carry $-62.4$ kN (compression) and the two bottom-chord panels $+57.6$ kN (tension); the sub-struts carry $U_1L_2=U_3L_2=-26$ kN and the king post $U_2L_2=+48$ kN.
Virtual system ($n$). A unit downward load at $U_2$ gives rafter forces $n=-1.30$ and bottom-chord $n=+1.20$; the web members carry essentially zero. Only the chords contribute to the sum.
Assemble. $\delta_{U_2}=\dfrac{1}{EA}\Big[4(-62.4)(-1.30)(8.45)+2(57.6)(1.20)(7.8)\Big]$. With $EA=6.5\times10^{5}$ kN, $\delta_{U_2}=\dfrac{2449}{6.5\times10^{5}}$ m $\Rightarrow$ ==**$\delta_{U_2}\approx 3.77$ mm (downward)**==.

Quantity	Value
Rafter force (each)	62.4 kN C
Bottom-chord force (each)	57.6 kN T
Vertical deflection of U₂	3.77 mm ↓

Check: the "3.25 m" dimension is read as the apex height, making the rafter a clean 8.45 m (5-12-13) member for which the drawn 90° sub-struts are exactly perpendicular; this is the self-consistent reading of the figure.

Topic. Virtual work (unit-load method) for truss joint deflection.

Key relations. $\delta=\displaystyle\sum_i \frac{N_i\,n_i\,L_i}{(EA)_i}$, with $N$ from the real loads and $n$ from a unit load applied at, and in the direction of, the desired deflection.

Why this works. The dummy unit load creates a statically-admissible virtual force field $n$; multiplying by the real elongations $NL/EA$ and summing is the internal virtual work, which by the work-energy theorem equals the external virtual work $1\times\delta$. Equal-$EA$ members let $EA$ factor out.

Common pitfall. Sign slips: keep tension positive throughout both the $N$ and $n$ sets so that $Nn$ has the correct sign. Also, use the true member lengths (the inclined rafters are 8.45 m, not 7.8 m).

Source. Hibbeler §8.4–8.5 (virtual work for trusses); Kassimali §7.3.

Question 3 of 8: Vertical deflection of joint U₂ of a symmetric truss

Question 3: Vertical deflection of joint U₂ of a symmetric truss (16 marks)