Flexibility (force) method — symmetric frame under a distributed load: 07-Str-A4, December 2018, Q6

Question 6: Flexibility (force) method — symmetric frame under a distributed load (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. The same trapezoidal frame as Question 5 — roller at ①, pin at ②, roller at ③, roller at ④, legs 6.0 m (rise 3.6 m, run 4.8 m), top member 8.0 m — now carrying a uniformly distributed load of 10.8 kN/m downward on member ②–③ and no support movement.

Given data
Quantity	Symbol	Value
Top member span	L₂₃	8.0 m
Leg length	L_leg	6.0 m (rise 3.6, run 4.8)
Distributed load on ②–③	w	10.8 kN/m
Flexural rigidity	EI	3.6 × 10⁵ kN·m² (cancels)
Total applied load	wL	86.4 kN

Find. The redundant reactions, then the shear-force and bending-moment diagrams with their extreme ordinates.

Question 6 — same geometry as Question 5, now loaded on the top member. Symmetry means the reactions at ① and ④ are equal, so the two redundants collapse to one.

Approach. The frame is two degrees indeterminate. Release the vertical reactions at ① and ④; the primary structure is then a simple span ②–③ on a pin and a roller with two unloaded cantilever legs hanging from it, which is determinate. Because both structure and load are symmetric, the two redundants are equal and one compatibility equation suffices: the vertical displacement of ① must vanish.

The legs are unstressed in the primary structure. With their outer ends released, nothing loads them, so they stay straight and simply carry the rotation of the joints they hang from. That is what makes the primary structure worth choosing.
Compute Δ₁₀. Under the distributed load the simple span rotates its end ② by $\theta_{2} = -wL_{23}^{3}/24EI = -0.640 \times 10^{-3}$ rad, and the rigid leg swings the released support up through the horizontal offset: $$\Delta_{10} = -r\,\theta_{2} = 4.8 \times 0.640 \times 10^{-3} = \boxed{+3.072\ \text{mm (upward)}}$$
Compute f₁₁. A unit upward pair at ① and ④ hands each joint a couple $1 \times 4.8 = 4.8$ kN·m, which puts a constant moment in the top member (the symmetric pair cancels the shear). The resulting end rotation is $mL/2EI = 53.33 \times 10^{-6}$ rad, and the leg also bends as a cantilever under the 0.8 kN transverse component of the unit load: $$f_{11} = \underbrace{4.8 \times 53.33\times10^{-6}}_{0.256\ \text{mm}} + \underbrace{\dfrac{0.8^{2}\,L_{\text{leg}}^{3}}{3EI}}_{0.128\ \text{mm}} = \boxed{0.384\ \text{mm/kN}}$$ Dropping the second term — the leg’s own flexibility — would overstate the redundant by 50 %.
Solve the compatibility equation. $$\Delta_{10} + X f_{11} = 0 \;\Rightarrow\; X = -\dfrac{3.072}{0.384} = \boxed{-8.00\ \text{kN}}$$ The negative sign is the answer’s most interesting feature: the outer supports must hold the frame down with 8.00 kN each, not prop it up.
Back-substitute for the reactions. $$R_{1} = R_{4} = -8.00\ \text{kN}, \qquad R_{2} = R_{3} = +51.20\ \text{kN}$$ and $2(-8.00) + 2(51.20) = 86.40$ kN, the whole applied load.
End moments and the span moment. Each leg carries the hold-down through its 4.8 m offset, so $$M_{2} = M_{3} = 8.00 \times 4.8 = \boxed{38.40\ \text{kN}\cdot\text{m}} \ \text{(hogging)}$$ and the top member’s mid-span value is the free moment less the end moment: $$M_{\text{mid}} = \dfrac{wL^{2}}{8} - 38.40 = 86.40 - 38.40 = \boxed{+48.00\ \text{kN}\cdot\text{m}}$$
Shears. The top member is symmetric, so its shear runs from +43.20 kN at ② through zero at mid-span to −43.20 kN at ③. Each leg carries a constant transverse shear of $38.40/6 = 6.40$ kN together with a 4.80 kN axial force, and those two combine to exactly the 8.00 kN vertical hold-down — a free check on the whole solution.

Question 6 bending moment. The parabolic span sits on a 38.40 kN·m hog at each top joint, and each leg tapers to zero at its released support.

Question 6 — results
Quantity	Value
Δ₁₀ (primary displacement at ①)	+3.072 mm up
f₁₁ (flexibility coefficient)	0.384 mm/kN
Redundant X = R₁ = R₄	−8.00 kN (hold-down)
Reactions at ② and ③	+51.20 kN each
Moment at joints ② and ③	−38.40 kN·m (minimum)
Mid-span moment, top member	+48.00 kN·m (maximum)
Shear, top member	±43.20 kN at the joints, 0 at mid-span
Shear / axial in each leg	6.40 kN / 4.80 kN

Topic. The flexibility (force) method with a symmetric pair of redundants.

Key relations. $\Delta_{10} + Xf_{11} = 0$; simple-span end rotation $wL^{3}/24EI$; end rotation under a constant span moment $mL/2EI$; cantilever tip deflection $PL^{3}/3EI$; mid-span moment $wL^{2}/8 - M_{\text{end}}$.

Why this works. Choosing the released reactions the question suggests turns a two-unknown problem into a one-unknown one, because symmetry guarantees the two redundants are equal and the compatibility equations are identical. The primary structure is then unusually clean: the legs carry nothing at all, so $\Delta_{10}$ reduces to a rigid-body swing driven by one standard beam rotation. The hold-down result is the physically instructive part. The legs splay outward and downward from the loaded span; when the span sags, the joints rotate in the sense that swings the feet up, so the supports there must pull down to keep them on the ground. Real frames of this shape need genuine anchorage, not just a bearing pad.

Common pitfall. Treating the legs as rigid links when computing $f_{11}$. Their own cantilever flexibility contributes a third of the total here, and omitting it gives $X = -12.0$ kN and a 57.6 kN·m joint moment — wrong by half. The same error class appears in every tied or braced frame in this subject: the member you think is only a link is usually carrying enough flexibility to matter.

Source. Hibbeler, Structural Analysis, 10th ed., §10.1–10.5; Kassimali, Structural Analysis, 6th ed., Ch. 12; Ghali, Neville and Brown, 7th ed., Ch. 4.

Question 6 of 9: Flexibility (force) method — symmetric frame under a distributed load

Question 6: Flexibility (force) method — symmetric frame under a distributed load (22 marks)