16-Civ-A1 Elementary Structural Analysis · Undated paper

Question 6 of 8: Indeterminate frame by moment distribution / slope deflection

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

Notes on this paper

Paper format. Engineers Canada / EGBC national examination 16-Civ-A1 Elementary Structural Analysis, 3 hours, CLOSED BOOK. 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 and stability (Ch. 2), method of joints and sections (Ch. 3), shear and bending-moment diagrams (Ch. 4), influence lines and moving loads (Ch. 6), virtual-work deflections (Ch. 8–9), slope-deflection and moment distribution (Ch. 10–11); A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges, compound (Gerber) beams, three-hinged frames, Maxwell’s law of reciprocal deflections.

Sign convention. Upward reactions positive; sagging bending moment positive (tension on the underside of a beam), hogging negative; member axial force tension positive (T), compression negative (C). Shear is positive when the resultant of the forces on the left of a section acts upward.

Check: figure reading. All geometry, loads and support symbols used below were read directly from the printed figures. Where a support symbol governs the answer it is called out explicitly in the Given.

Question 6: Indeterminate frame by moment distribution / slope deflection (22 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. Take joint 1 (the fixed base) as the origin.

Member	From → to	Length	Rigidity
Left cantilever	5 (−2, 5) → 2 (0, 5)	2 m	EI
Column	1 (0, 0) → 2 (0, 5)	5 m	1.5EI
Sloping member	2 (0, 5) → 3 (12, 0)	13 m (12 m horizontally)	1.3EI
Right cantilever	3 (12, 0) → 4 (14, 0)	2 m	EI

Joint 1 is a fixed support, joint 3 is a pin support, and joints 5 and 4 are free ends. A uniformly distributed load of 5 kN/m acts downward over the full 16 m horizontal projection of the structure, so the total applied load is $5 \times 16 = 80\text{ kN}$.

Find. The shear-force and bending-moment diagrams for every member, labelled with their maximum and minimum ordinates.

Q6 — the frame. The sloping member spans 12 m horizontally and drops 5 m, so its true length is 13 m.

Approach. Show first that there is no sidesway, so the only unknowns are the rotations at joints 2 and 3. Reduce the two cantilevers to known end actions, compute the fixed-end moment of the inclined member from its horizontal projection, then distribute (or write slope-deflection equations) and recover the diagrams from statics.

Degree of indeterminacy and sidesway check. With $m = 4$, $j = 5$ and $r = 3 + 2 = 5$, $$\text{DSI} = 3(4) + 5 - 3(5) = 2$$ The members are inextensible: the vertical column fixes the vertical position of joint 2, and the inclined member, whose far end is a pin at a fixed point, then fixes its horizontal position as well. Joint 2 therefore cannot translate, so there is no sidesway and the only kinematic unknowns are $\theta_2$ and $\theta_3$.
Reduce the cantilevers. Each 2 m overhang carries $5 \times 2 = 10\text{ kN}$ and hands its joint a known shear and moment: $$V = 10\ \text{kN}, \qquad M = \frac{wL^{2}}{2} = \frac{5(2)^{2}}{2} = \boxed{10\ \text{kN}\cdot\text{m (hogging)}}$$ at joint 2 from member 5–2 and at joint 3 from member 3–4. These are applied actions, not unknowns.
Fixed-end moment of the inclined member. A vertical load specified per metre of horizontal projection produces, on an inclined member, exactly the fixed-end moments of the equivalent horizontal beam of the projected span: $$\text{FEM} = \frac{w\,L_{\text{proj}}^{2}}{12} = \frac{5(12)^{2}}{12} = \boxed{60\ \text{kN}\cdot\text{m}}$$ at each end (hogging at 2, sagging sense at 3). Note that the stiffness of the member still uses its true length, $4EI/L = 4(1.3EI)/13 = 0.4EI$.
Solve for the joint rotations. Writing slope-deflection equations for the column ($4(1.5EI)/5 = 1.2EI$ at joint 2) and the inclined member, and enforcing moment balance at joints 2 and 3 with the two cantilever moments as applied loads, gives the member end moments. (Equivalently: distribute the out-of-balance moments between the column and the inclined member at joint 2, with the pin at joint 3 handled by the modified stiffness $3EI/L$ plus its cantilever moment.) The result is: $$M_{21} = 60,\quad M_{23} = 70,\quad M_{12} = 30,\quad M_{32} = 10\ \text{kN}\cdot\text{m}$$
Check joint equilibrium. At joint 2 the cantilever, the column and the inclined member must balance: $$10 + 60 = 70 \quad\checkmark$$ and at joint 3 the inclined member balances its cantilever, $10 = 10$. Both are exact.
Reactions. From the member end forces, the fixed base carries $$H_1 = \boxed{18\ \text{kN}}, \qquad V_1 = \boxed{37.5\ \text{kN} \uparrow}, \qquad M_1 = \boxed{30\ \text{kN}\cdot\text{m}}$$ and the pin at joint 3 carries $H_3 = 18\text{ kN}$ (opposing) and $V_3 = \boxed{42.5\ \text{kN} \uparrow}$. Global checks: $37.5 + 42.5 = 80\text{ kN}$ vertically, the two horizontal reactions cancel, and moments about joint 1 give $-80(6) - 30 + 12(42.5) = 0$.
Shear diagrams. The column carries a constant shear of $18\text{ kN}$ and an axial compression of $37.5\text{ kN}$. Each cantilever runs linearly from 0 at the free tip to 10 kN at the joint. On the inclined member the shear measured normal to the member runs linearly from $\boxed{+32.31\ \text{kN}}$ at joint 2 to $\boxed{-23.08\ \text{kN}}$ at joint 3, crossing zero 7.0 m horizontally from joint 2; its axial compression grows from 6.04 kN at joint 2 to 29.12 kN at joint 3.
Bending-moment diagram of the inclined member. Taking the free body to the left of a section at horizontal distance $x$ from joint 2, with $y = 5 - \tfrac{5}{12}x$, $$M(x) = 30 + 37.5x - 18y - 2.5(x + 2)^{2}$$ This gives $-70\text{ kN}\cdot\text{m}$ at joint 2 and $-10\text{ kN}\cdot\text{m}$ at joint 3, with a sagging maximum where the normal shear vanishes: $$M_{\max} = \boxed{+52.5\ \text{kN}\cdot\text{m at } x = 7.0\ \text{m}}$$ Points of contraflexure occur at $x = 2.42\text{ m}$ and $x = 11.58\text{ m}$.
Column moment diagram. With no load along its length the column moment is linear, from $30\text{ kN}\cdot\text{m}$ at the fixed base to $60\text{ kN}\cdot\text{m}$ of opposite sense at joint 2, passing through zero $30/18 = 1.67\text{ m}$ above the base.

Q6 bending-moment diagram, plotted member by member on a developed axis. Sagging positive.

Member	Shear (max / min)	Moment (max / min)
Cantilever 5–2 (2 m)	0 to 10 kN	0 at the tip, −10 kN·m at joint 2
Column 1–2 (5 m, 1.5EI)	18 kN, constant	30 kN·m at the base, 60 kN·m at joint 2 (opposite sense)
Sloping 2–3 (13 m, 1.3EI)	+32.31 kN at 2, −23.08 kN at 3	−70 at 2, +52.5 at x = 7 m, −10 at 3 (kN·m)
Cantilever 3–4 (2 m)	10 kN to 0	−10 kN·m at joint 3, 0 at the tip
Reactions	Joint 1 (fixed): H = 18 kN, V = 37.5 kN ↑, M = 30 kN·m — Joint 3 (pin): H = 18 kN, V = 42.5 kN ↑

Topic. Analysis of a two-degree indeterminate frame with an inclined member by moment distribution or slope deflection.

Key relations. Slope-deflection $M_{ij} = \dfrac{2EI}{L}\left(2\theta_i + \theta_j - 3\psi\right) + \text{FEM}_{ij}$; member stiffness $4EI/L$ (or $3EI/L$ to a pinned far end) and carry-over factor $\tfrac12$; for a vertical UDL specified per unit horizontal projection on an inclined member, $\text{FEM} = w L_{\text{proj}}^{2}/12$ with the stiffness still based on the true length $L$.

Why this works. Moment distribution and slope deflection both take the joint rotations as the unknowns, so the size of the problem is set by the kinematic indeterminacy, not the static one. That makes the sidesway check the first and most important step: here inextensibility of a vertical column pins the vertical movement of joint 2 and inextensibility of the inclined member then pins the horizontal movement, so the sway degree of freedom disappears and only two rotations remain. The projection rule for the inclined member follows because a vertical load per horizontal metre produces exactly the same vertical shear and moment distribution as the equivalent horizontal beam — only the axial component differs.

Common pitfall. Using the sloping length of 13 m in the fixed-end moment ($5(13)^2/12 = 70.4$ instead of 60), or conversely using the projected 12 m in the stiffness. The two lengths play different roles. The second trap is forgetting the cantilevers: each contributes a known 10 kN·m at its joint and must be entered as an applied moment before any distribution, otherwise the joint balance at 2 and 3 will never close. Finally, the answers here are deliberately not round — 37.5 and 42.5 kN, 52.5 kN·m — which is normal for an indeterminate frame and not a sign of error.

Source. Hibbeler, Structural Analysis, 10th ed., §10.1–10.4 (slope deflection) and §11.1–11.4 (moment distribution, including frames without sidesway); Kassimali, Structural Analysis, 6th ed., Ch. 15–16.

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