16-Civ-A1 Elementary Structural Analysis · December 2018
Question 6 of 8: Frame by moment distribution (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 6: Frame by moment distribution (22 marks)(answer one of Q6/Q7/Q8)
Given. A beam $1(0)\!-\!2(8)\!-\!3(14)\!-\!4(16\text{ m})$ on a roller at $1$ and a pin at $3$, rigidly joined at $2$ to an $8\text{ m}$ column built in (fixed) at its base $5$; $48\text{ kN}\downarrow$ at $x{=}4$ and UDL $12\text{ kN/m}$ over $2\!-\!4$ (nodes $x{=}8$ to $16$). All members equal $EI$, inextensible. Find. Reactions and the SFD/BMD (max/min ordinates).
Question 6 — propped beam rigidly tied to a fixed column; solved by moment distribution.
Fixed-end moments. Span $1\!-\!2$ (point load mid): $PL/8=\pm48\text{kN}\cdot\text{m}$ (with the roller end $1$ released, so a modified $3EI/L$ stiffness carries the $2$-end); span $2\!-\!3$ (UDL): $wL^2/12=\pm36\text{kN}\cdot\text{m}$; the overhang $3\!-\!4$ applies a known cantilever moment $-wL^2/2=-24\text{kN}\cdot\text{m}$ at joint $3$; the unloaded column has zero FEM.
Distribute at joints 2 and 3 (no sidesway — the pin at $3$ and fixed base restrain horizontal translation). Balancing and carrying over to convergence gives the member-end moments below.
Balanced moments. At the rigid joint $2$ the beam brings $-63.8$, the far beam $+52.9$ and the column top $+10.9\text{kN}\cdot\text{m}$ (they sum to zero). Under the $48\text{ kN}$ load the beam sags to $\boxed{+64.1\text{kN}\cdot\text{m}}$; over the propped pin $3$ it hogs the overhang value $\boxed{-24\text{kN}\cdot\text{m}}$; the fixed base carries $\boxed{M_5=+5.5\text{kN}\cdot\text{m}}$.
Reactions from the balanced shears. $\boxed{V_1=16.0},\;\boxed{V_3=55.2},\;\boxed{V_5=72.8\text{ kN}}$ (sum $=144\text{ kN}$, the total load), with small horizontal reactions $H_3=+2.1,\;H_5=-2.1\text{ kN}$ from the frame action.
