07-Str-A4 · December 2015

Question 7 of 9: Frame Analysis by the Flexibility (Force) Method

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

Notes on this paper

Paper format. 07-Str-A4 Advanced Structural Analysis, National Examinations, December 2015 — three hours, closed book (an approved Sharp or Casio calculator is the only aid). Nine questions: Questions 1 and 2 are compulsory (8 and 12 marks); the candidate then answers two of Questions 3, 4, 5 (18 marks each) and two of Questions 6, 7, 8, 9 (22 marks each), so six questions constitute a complete 100-mark paper. Marks are printed in the left margin. All nine questions are solved here, because this set is a study resource rather than an examination script.

Reference texts. R. C. Hibbeler, Structural Analysis, 10th ed. (slope-deflection, moment distribution, influence lines, Castigliano); A. Kassimali, Structural Analysis, 6th ed. (force and displacement methods, degrees of freedom); A. Ghali, A. M. Neville & T. G. Brown, Structural Analysis: A Unified Classical and Matrix Approach, 7th ed. (flexibility and stiffness formulations, lack of fit, support movement); W. Weaver Jr. & J. M. Gere, Matrix Analysis of Framed Structures, 3rd ed. (assembly of [K] and {P}); J. C. McCormac, Structural Analysis: Using Classical and Matrix Methods, 4th ed. Canadian design context: CSA S16, CSA A23.3 and the National Building Code of Canada supply the load and resistance rules that these analyses feed, although this paper is purely an analysis paper.

Check: sign convention used throughout. Member end moments $M_{ij}$ are counter-clockwise positive, which is the convention that matches the standard six-degree-of-freedom stiffness matrix and the vector $\{\theta_2,\theta_3,\delta\}$ the paper itself defines (Question 9 states “positive counter clockwise”). With that choice the fixed-end moment of a downward uniformly distributed load is $\mathrm{FEM}_{ij}=+wL^{2}/12$ at the $i$ end, the chord rotation is $\psi_{ij}=[(\mathbf{d}_j-\mathbf{d}_i)\cdot\mathbf{e}_\perp]/L$ with $\mathbf{e}_\perp$ the member axis turned $+90^{\circ}$, and ordinary sagging moments are recovered as $M_{\text{sag}}(i)=-M_{ij}$, $M_{\text{sag}}(j)=+M_{ji}$. Mixing this with Hibbeler’s clockwise-positive convention produces clean-looking integers that are wrong; every diagram below is plotted in the sagging convention.

Check: printed unit in Question 6. The paper prints “EI = 3.2 × 10⁴ kN.mm²”. Taken literally that is a flexural rigidity some ten orders of magnitude too small for an 18 m beam and would make the 36 mm settlement effect vanish. The intended unit is $\text{kN}\cdot\text{m}^2$: with $EI=3.2\times10^{4}\ \text{kN}\cdot\text{m}^{2}$ the settlement term $3EI\Delta/L^{2}=54\ \text{kN}\cdot\text{m}$ is the same order as the load term $wL^{2}/8=48\ \text{kN}\cdot\text{m}$, which is plainly what the question is testing. The solution below uses $\text{kN}\cdot\text{m}^{2}$ and flags the typographical error.

Question 7: Frame Analysis by the Flexibility (Force) Method (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.

Item	Value
Member 1–2 (horizontal)	4.0 m, carrying $w=51\ \text{kN/m}$ downward
Member 2–3 (inclined)	4.0 m run, 3.0 m drop ⇒ $L=5.0$ m, unloaded
Joint 1	Built in (fixed) at the wall
Joint 3	Pinned member end on a roller bearing on the vertical wall face — horizontal reaction only
Flexural rigidity	Same $EI$ both members, inextensible

Find. The redundant, all internal actions, and the shear and moment diagrams with their extreme ordinates.

Question 7 — the support at joint 3 is a roller running on a vertical plane, so it supplies a horizontal reaction only. Reading it as an ordinary ground roller is the classic wrong start.

Approach. With four reaction components and three equilibrium equations the frame is redundant to one degree; release the horizontal restraint at joint 3, compute the horizontal displacement there under the load and under a unit redundant, and solve the single compatibility equation.

Count the redundancy and choose the release. $m=2$, $j=3$, $r_{\text{reac}} = 3\ (\text{fixed}) + 1\ (\text{roller}) = 4$: $$r = 3(2)+4-3(3) = \boxed{1}$$ Take the horizontal reaction at joint 3 as the redundant $X$. Removing it leaves a determinate cantilever tree: the beam built in at joint 1, with the inclined member hanging free from joint 2.
Write the moments in the primary structure. Under the load alone, a section of the beam at distance $x$ from joint 1 sees only the UDL to its right: $$M_{0}(x) = -\frac{w(4-x)^{2}}{2}$$ and the hanging inclined member is entirely unstressed. Under a unit horizontal force at joint 3 (acting in the $+x$ sense) the beam sees a constant moment equal to the 3.0 m lever arm, and the inclined member sees a moment proportional to the height $y$ of the section above joint 3: $$m_{1}(x) = 3.0\ \text{(beam)},\qquad m_{1}(s) = y(s) = 0.6s\ \text{(inclined member)}$$
Evaluate the two flexibility coefficients. $$f_{10} = \frac{1}{EI}\int M_{0}m_{1}\,\mathrm{d}s = -\frac{3.0\,w\,(4.0)^{3}}{6EI} = -\frac{1632}{EI}$$ $$f_{11} = \frac{1}{EI}\left[(3.0)^{2}(4.0) + \frac{(3.0)^{2}(5.0)}{3}\right] = \frac{36+15}{EI} = \frac{51}{EI}$$ The $1/EI$ cancels in the ratio, which is why the question can decline to give a numerical $EI$.
Solve the compatibility equation. The support is rigid, so the total horizontal displacement at joint 3 must vanish: $$f_{10}+X f_{11} = 0 \;\Longrightarrow\; X = -\frac{f_{10}}{f_{11}} = \frac{1632}{51} = \boxed{32.00\ \text{kN}}$$ Because $X$ is positive in the assumed direction, the wall pushes joint 3 away from itself with 32.0 kN.
Recover the reactions at joint 1. Vertical equilibrium takes the whole load, and moment equilibrium about joint 1 closes the frame: $$V_{1} = w(4.0) = 204.0\ \text{kN},\qquad H_{1} = -32.0\ \text{kN}$$ $$M_{1} = -\frac{w(4.0)^{2}}{2} + X(3.0) = -408.0 + 96.0 = \boxed{-312.0\ \text{kN}\cdot\text{m}\ \text{(hogging)}}$$
Build the moment field. Along the beam, with sagging positive, $$M(x) = -312.0 + 204.0x - 25.5x^{2}$$ This rises monotonically (the shear $V(x)=204.0-51x$ does not vanish before $x=4.0$), crosses zero at $$x = \frac{204.0-\sqrt{204.0^{2}-4(25.5)(312.0)}}{2(25.5)} = 2.060\ \text{m}$$ and reaches $M_{2} = \boxed{+96.0\ \text{kN}\cdot\text{m}}$ at joint 2. The inclined member carries no load, so its moment falls linearly from $+96.0\ \text{kN}\cdot\text{m}$ at joint 2 to zero at the pin at joint 3; its transverse shear is therefore constant at $96.0/5.0 = 19.20\ \text{kN}$ and its axial force is $25.60\ \text{kN}$ compression.
Check the frame globally. Taking moments about joint 1: the redundant contributes $32.0(3.0)=+96.0$, the load contributes $-204.0(2.0)=-408.0$, and the fixed-end moment reaction is $+312.0$; the sum is exactly zero. Horizontal equilibrium gives $-32.0+32.0=0$, and vertical equilibrium $204.0-204.0=0$. Note that the beam shear at joint 2 is zero: the entire vertical load is carried straight back into the wall, and the inclined member acts as a pure strut delivering the 32.0 kN thrust.

Question 7 — shear in the loaded beam 1–2, falling linearly from 204 kN at the wall to zero at joint 2.

Question 7 — bending moment in beam 1–2, sagging positive. The inclined member (not plotted) is a straight line from $+96$ kN·m at joint 2 to zero at joint 3.

Quantity	Value
Flexibility coefficients	$f_{10}=-1632/EI$, $f_{11}=+51/EI$
Redundant (horizontal reaction at joint 3)	32.00 kN
Reactions at joint 1	$V=204.0$ kN, $H=32.0$ kN, $M=312.0$ kN·m hogging
Beam 1–2 shear	$+204.0$ kN at joint 1 to $0$ at joint 2
Beam 1–2 moment	min $-312.0$ kN·m at joint 1; max $+96.0$ kN·m at joint 2; zero at $x=2.06$ m
Member 2–3 moment	$+96.0$ kN·m at joint 2 falling linearly to $0$ at joint 3
Member 2–3 shear and axial	19.20 kN shear (constant); 25.60 kN compression

Topic. The flexibility (force) method with a single redundant.

Key relations. $f_{i0} + \sum_j X_j f_{ij} = \Delta_i$ with $f_{ij}=\int m_i m_j\,\mathrm{d}s/EI$ and $f_{i0}=\int M_0 m_i\,\mathrm{d}s/EI$; for a rigid support $\Delta_i = 0$.

Why this works. Releasing one restraint converts the frame into a determinate primary structure whose displacements can be written down directly. Superposing the released displacement with $X$ times the unit-load displacement and forcing the sum back to the true (zero) value recovers the redundant. The method is at its most economical when, as here, the released structure is a plain cantilever tree and the unit-load moment diagram is trivial — a constant on one member and a straight line on the other. Because both flexibility terms carry $1/EI$, the redundant is independent of the absolute rigidity, so the question needs to state only that both members share the same $EI$.

Common pitfall. Mis-identifying the support at joint 3. The rollers are drawn between the hatched wall and the support triangle, so they roll vertically and the reaction is horizontal; treating it as an ordinary ground roller (vertical reaction) produces a frame that cannot equilibrate the 204 kN vertical load through joint 1 alone and gives a completely different moment field. A second trap is forgetting that the released inclined member is unstressed in the primary structure, which makes $f_{10}$ a beam-only integral.

Source. Kassimali, Structural Analysis, 6th ed., §13.1–13.5; Ghali, Neville & Brown, 7th ed., §4.1–4.4; Hibbeler, Structural Analysis, 10th ed., §10.1–10.4.

← Question 6 Next: Question 8 →

Item	Value
Member 1–2 (horizontal)	4.0 m, carrying \(w=51\ \text{kN/m}\) downward
Member 2–3 (inclined)	4.0 m run, 3.0 m drop ⇒ \(L=5.0\) m, unloaded
Joint 1	Built in (fixed) at the wall
Joint 3	Pinned member end on a roller bearing on the vertical wall face — horizontal reaction only
Flexural rigidity	Same \(EI\) both members, inextensible

Quantity	Value
Flexibility coefficients	\(f_{10}=-1632/EI\), \(f_{11}=+51/EI\)
Redundant (horizontal reaction at joint 3)	32.00 kN
Reactions at joint 1	\(V=204.0\) kN, \(H=32.0\) kN, \(M=312.0\) kN·m hogging
Beam 1–2 shear	\(+204.0\) kN at joint 1 to \(0\) at joint 2
Beam 1–2 moment	min \(-312.0\) kN·m at joint 1; max \(+96.0\) kN·m at joint 2; zero at \(x=2.06\) m
Member 2–3 moment	\(+96.0\) kN·m at joint 2 falling linearly to \(0\) at joint 3
Member 2–3 shear and axial	19.20 kN shear (constant); 25.60 kN compression