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16-Civ-A1 Elementary Structural Analysis · December 2018

Question 7 of 8: Truss deflection by virtual work (answer one of Q6/Q7/Q8)

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

Notes on this paper

Paper: National Exams — December 2018, 16-Civ-A1 Elementary Structural Analysis (closed book, 3 h; approved Casio/Sharp calculator permitted). Six questions constitute a complete paper: answer Q1–Q5 and one of Q6/Q7/Q8. For completeness all eight questions are worked here.

Reference texts: R. C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy (Ch. 2), method of joints & sections (Ch. 3), shear & moment diagrams (Ch. 4), influence lines (Ch. 6), virtual-work deflections (Ch. 8–9), moment distribution (Ch. 11); A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges, compound (Gerber) beams, three-hinged & indeterminate frames.

Sign convention. Upward reactions positive; sagging bending moment positive (tension on the underside), hogging negative; member axial force tension positive (T), compression negative (C).

These A1 papers are defined entirely by their figures, so every structure is redrawn to scale below.

Question 7: Truss deflection by virtual work (22 marks) (answer one of Q6/Q7/Q8)

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. Cantilever truss off a vertical wall: $L_1(0,0),L_2(3,0),L_3(6,0)$ and the collinear top chord $L_1\!-\!U_1(3,1.25)\!-\!U_2(6,2.5)$, with vertical $L_2U_1$ and diagonal $L_2U_2$; supported at $U_2$ and $L_3$ on the wall. Loads $10\text{ kN}\downarrow$ at $L_1$ and $30\text{ kN}\downarrow$ at $U_1$; all members prismatic, $AE=36.0\times10^{3}\text{ kN}$. Find. The vertical deflection of joint $L_1$.

10 kN30 kNL1L2L3U1U23 m3 m2.5 m
Question 7 — wall-mounted truss; deflection sought at the free tip $L_1$.
  1. Support model. Taking $U_2$ as the pin (carrying the $40\text{ kN}$ vertical) and $L_3$ as a horizontal roller on the wall makes the truss determinate; the member forces then follow from statics.
  2. Real forces $N$ (tension +). By joints/sections: $L_1L_2=-24,\;L_2L_3=-60,\;L_1U_1=+26,\;U_1U_2=+26,\;L_2U_1=-30,\;L_2U_2=+46.9,\;L_3U_2=0\text{ kN}$.
  3. Virtual forces $n$. Apply a unit downward load at $L_1$ on the same truss: only the top chord and its first bottom panel pick it up — $L_1L_2=-2.4,\;L_2L_3=-2.4,\;L_1U_1=+2.6,\;U_1U_2=+2.6$; the remaining bars carry $n=0$.
  4. Unit-load summation. $\delta_{L_1}=\displaystyle\sum \frac{N\,n\,L}{AE}=\frac{172.8+432+219.7+219.7}{36\,000}=\frac{1044.2}{36\,000}=\boxed{0.0290\text{ m}=29.0\text{ mm}\;(\downarrow)}$.

Taking $U_2$ as the pin (as above) gives $\delta_{L_1}=29.0\text{ mm}$; an independent direct-stiffness solution of the truss treated as fully pin-supported at both $U_2$ and $L_3$ (the literal, mildly indeterminate reading) returns the identical $29.0\text{ mm}$, so the reported deflection is robust to that support ambiguity.

Member$N$ (kN)$n$$L$ (m)$NnL$
$L_1L_2$$-24$$-2.4$$3.0$$172.8$
$L_2L_3$$-60$$-2.4$$3.0$$432.0$
$L_1U_1$$+26$$+2.6$$3.25$$219.7$
$U_1U_2$$+26$$+2.6$$3.25$$219.7$
others$n=0$$0$
$\sum NnL$$1044.2$