Shear and moment diagrams for a 4°-indeterminate rigid frame (18 marks): 24-Bld-A1 Elementary Structural Analysis, December 2016, Q5

Question 5: Shear and moment diagrams for a 4°-indeterminate rigid frame (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. Node 1 (0,4) fixed into a wall; node 2 (8,4); node 3 (14,4) roller; node 4 (15,4) free tip; node 5 (8,0) fixed, joined to node 2 by a vertical column. UDL 12 kN/m on 1–2 (8 m), UDL 8 kN/m on 2–3–4 (7 m total). Uniform $EI$ throughout — a rigid frame indeterminate to the 4th degree (three reaction components at each fixed support plus one at the roller, minus three equilibrium equations).

Find. All reactions and the shear/moment diagram (with extremes) for every member.

Fig. Q5 — two fixed supports (1 and 5) and one roller (3); member 2–5 is the vertical leg.

Approach. With uniform $EI$, distribute the fixed-end moments at the rigid joint 2 (where members 1–2, 2–3 and 2–5 meet) by moment distribution using relative stiffnesses $4EI/L$; because member 2–3–4 carries an overhang past the roller at 3, its fixed-end moments must include the overhang's transferred moment before distribution begins. The moment-distribution result is cross-checked here against a full stiffness-method (matrix) solve of the same frame, which is exact for a linear frame and used as the authoritative number set.

Fixed-end moments (member 1–2, $L=8$, $w=12$). $\text{FEM}_{12}=-\dfrac{wL^2}{12}=-64\text{ kN}\cdot\text{m}$, $\text{FEM}_{21}=+64\text{ kN}\cdot\text{m}$.
Overhang 3–4 (1 m, $w=8$) transferred onto joint 3, then onto span 2–3 as an end moment/shear, combined with span 2–3's own UDL (6 m, $w=8$) FEM $=\mp24\text{ kN}\cdot\text{m}$, before the joint-2 distribution.
Distribute at joint 2 (members 1–2, 2–3, 2–5 meeting rigidly; relative stiffness $\propto 1/L$ since $EI$ is common: $1/8:1/6:1/4=3:4:6$, distribution factors $3/13:4/13:6/13$) and carry over to the far, fixed ends 1 and 5; iterating to convergence (or, equivalently, solving the joint-2 rotation directly) gives the final end moments below.
Reactions (final, cross-checked by the matrix stiffness solve). Node 1: $H=\boxed{5.63\text{ kN}}$, $V=49.41\text{ kN}$, $M=67.75\text{ kN}\cdot\text{m}$. Node 3 (roller): $V=25.75\text{ kN}$. Node 5: $H=-5.63\text{ kN}$, $V=76.84\text{ kN}$, $M=7.50\text{ kN}\cdot\text{m}$. Global check: $\sum V = 49.41+25.75+76.84=152.0\text{ kN} = 12(8)+8(7)$ ✓.
Member 1–2. $V$: $+49.41\text{ kN}$ at 1, falling to $-46.59\text{ kN}$ at 2, crossing zero at $x=4.12\text{ m}$ where $M=\boxed{+33.96\text{ kN}\cdot\text{m}}$ (local sag peak). $M_1=-67.75$, $M_2=-56.5\text{ kN}\cdot\text{m}$ (both hogging, i.e. tension on the top fibre).
Member 2–3. $V$: $+30.25\to-17.75\text{ kN}$, crossing zero at $x=3.78\text{ m}$ where $M=\boxed{+15.69\text{ kN}\cdot\text{m}}$. $M_2=-41.5$, $M_3=-4.0\text{ kN}\cdot\text{m}$.
Member 3–4 (overhang). $V$: $+8\to0\text{ kN}$; $M$: $-4.0\to0\text{ kN}\cdot\text{m}$, monotonic.
Member 2–5 (column). No transverse load, so $V=5.63\text{ kN}$ constant; $M$ varies linearly from $-15.0\text{ kN}\cdot\text{m}$ at node 2 to $+7.5\text{ kN}\cdot\text{m}$ at node 5.

Q5 – reactions and member diagram extremes
Location / member	Quantity	Value
Node 1 (fixed)	$H,V,M$	5.63 kN, 49.41 kN, 67.75 kN·m
Node 3 (roller)	$V$	25.75 kN
Node 5 (fixed)	$H,V,M$	−5.63 kN, 76.84 kN, 7.50 kN·m
1–2	$V$ range / $M$ range	+49.41 to −46.59 kN / −67.75 to +33.96 kN·m
2–3	$V$ range / $M$ range	+30.25 to −17.75 kN / −41.5 to +15.69 kN·m
3–4	$V$ range / $M$ range	+8 to 0 kN / −4.0 to 0 kN·m
2–5	$V$ / $M$ range	5.63 kN (const.) / −15.0 to +7.5 kN·m

Check: numbers above come from a full plane-frame stiffness (matrix) analysis using the given uniform EI, which is mathematically identical to a converged moment-distribution/slope-deflection solution — equivalent to carrying the classical hand method to full convergence rather than 2–3 distribution cycles.

Topic: Moment distribution / slope-deflection for an indeterminate rigid frame with a rigid three-member joint.

Key relations: Distribution factor at a joint $= \dfrac{(4EI/L)_i}{\sum(4EI/L)}$; carry-over factor $=1/2$ to the far fixed end; fixed-end moment for a UDL span, $-wL^2/12$ and $+wL^2/12$.

Why this works: Locking every joint against rotation first (fixed-end moments), then releasing one joint at a time and distributing its out-of-balance moment in proportion to each member's rotational stiffness, converges to the same joint rotations — and hence the same final member-end moments — as solving the full compatibility equations directly.

Common pitfall: forgetting that an overhang beyond a roller (member 3–4 here) contributes a known end moment to joint 3 before any distribution starts, and omitting the carry-over to the FAR fixed ends 1 and 5 after each distribution cycle.

Source: Hibbeler §12.1–12.6 (moment distribution); Kassimali §16–17.

Question 5 of 8: Shear and moment diagrams for a 4°-indeterminate rigid frame (18 marks)

Question 5: Shear and moment diagrams for a 4°-indeterminate rigid frame (18 marks)