07-Str-A4 · December 2015

Question 8 of 9: Sway Frame with an Inclined Member by Slope-Deflection

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 8: Sway Frame with an Inclined Member by 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.

Item	Value
Member 1–2	4.0 m horizontal, rigidity $1.2EI$, UDL $w=27\ \text{kN/m}$
Member 2–3	4.0 m run, 3.0 m drop ⇒ $L=5.0$ m, rigidity $3EI$, unloaded
Joint 1	Built in (fixed)
Joint 3	Pinned member end on a ground roller — vertical reaction only
Members	Inextensible; only relative $EI$ is given, and only relative $EI$ is needed

Find. All member end moments, the reactions, and the shear and moment diagrams with extreme ordinates.

Question 8 — a two-member sway frame. Joint 2 can translate vertically, and the inclined member converts that vertical movement into a horizontal movement of joint 3.

Approach. Establish the sway pattern from inextensibility, write the two slope-deflection equations with the modified stiffness at the pinned roller, add the sway (translation) equation from virtual work, and solve the two-unknown system.

Count and set up the unknowns. $m=2$, $j=3$, $r_{\text{reac}}=3+1=4$, so $r = 6+4-9 = 1$; the frame is redundant to one degree. Kinematically, the horizontal beam is inextensible and built in at joint 1, so $u_2=0$; the vertical movement $v_2$ of joint 2 is free. The inclined member cannot change length, and joint 3 has $v_3=0$, so $$0.8\,u_3 + 0.6\,v_2 = 0 \;\Longrightarrow\; u_3 = -0.75\,v_2$$ The roller carries no moment, so with the modified stiffness the unknowns reduce to $\theta_2$ and the single sway parameter $v_2$.
Write the fixed-end moments. Only member 1–2 is loaded: $$\mathrm{FEM}_{12} = +\frac{wL^{2}}{12} = +\frac{27(4.0)^{2}}{12} = +36.0\ \text{kN}\cdot\text{m},\qquad \mathrm{FEM}_{21} = -36.0\ \text{kN}\cdot\text{m}$$
Write the slope-deflection equations. With $\psi_{12}=v_2/4.0$ for the beam and the corresponding chord rotation for the inclined member, $$M_{12}=\frac{2(1.2EI)}{4.0}\left(\theta_2-3\psi_{12}\right)+36.0,\qquad M_{21}=\frac{2(1.2EI)}{4.0}\left(2\theta_2-3\psi_{12}\right)-36.0$$ $$M_{23}=\frac{3(3EI)}{5.0}\left(\theta_2-\psi_{23}\right)$$ the last using the modified stiffness because the far end is a pinned roller.
Add the equilibrium equations and solve. Moment equilibrium at joint 2 gives $M_{21}+M_{23}=0$; the second equation is the translation equation obtained by virtual work on the sway pattern, using the equivalent nodal loads rather than the total applied load. Solving the pair yields the end moments $$M_{12}=+115.2,\qquad M_{21}=+50.4,\qquad M_{23}=-50.4,\qquad M_{32}=0\ \ (\text{kN}\cdot\text{m})$$ so in sagging terms the wall hogs $-115.2\ \text{kN}\cdot\text{m}$ and joint 2 sags $+50.4\ \text{kN}\cdot\text{m}$. The joint balances exactly: $50.4-50.4 = 0$.
Recover the shears and reactions. For the beam, $$V_{1} = \frac{w L}{2} + \frac{M_{2}-M_{1}}{L} = 54.0 + \frac{50.4-(-115.2)}{4.0} = 54.0+41.4 = \boxed{95.40\ \text{kN}}$$ $$V_{2} = 95.40 - 27(4.0) = -12.60\ \text{kN}$$ and $R_{3} = 12.60\ \text{kN}$. Global vertical equilibrium: $95.40+12.60 = 108.0 = 27(4.0)\ \checkmark$. Because the roller supplies no horizontal reaction and no horizontal load acts, $H_{1}=0$ exactly.
Complete the beam moment diagram. $$M(x) = -115.2 + 95.4x - 13.5x^{2}$$ Zero shear at $x = 95.4/27 = 3.533\ \text{m}$ gives the peak $$M_{\max}= -115.2 + 95.4(3.533) - 13.5(3.533)^{2} = \boxed{+53.34\ \text{kN}\cdot\text{m}}$$ with the point of contraflexure at $x = 1.546\ \text{m}$ and $M = +50.4\ \text{kN}\cdot\text{m}$ at joint 2.
Complete the inclined member. Carrying no span load, member 2–3 has a straight moment diagram from $+50.4\ \text{kN}\cdot\text{m}$ at joint 2 to zero at the roller, so its transverse shear is constant at $50.4/5.0 = 10.08\ \text{kN}$ and its axial force is $7.56\ \text{kN}$ compression — the vertical roller reaction of 12.60 kN resolved onto the member axes.

Question 8 — shear in beam 1–2.

Question 8 — bending moment in beam 1–2, sagging positive.

Quantity	Value
End moments (CCW positive)	$M_{12}=+115.2$, $M_{21}=+50.4$, $M_{23}=-50.4$, $M_{32}=0$ kN·m
Moment at the wall (sagging convention)	$-115.2$ kN·m (hogging)
Moment at joint 2	$+50.4$ kN·m (sagging)
Beam 1–2 shear	$+95.40$ kN at joint 1; $-12.60$ kN at joint 2
Beam 1–2 maximum sagging moment	$+53.34$ kN·m at $x=3.533$ m
Point of contraflexure	$x=1.546$ m from the wall
Member 2–3	Moment $+50.4$ to 0 kN·m; shear 10.08 kN; axial 7.56 kN compression
Reactions	$V_1=95.40$ kN, $H_1=0$, $M_1=115.2$ kN·m; $R_3=12.60$ kN

Topic. Slope-deflection analysis of a sway frame containing an inclined member.

Key relations. $M_{ij}=\frac{2EI}{L}(2\theta_i+\theta_j-3\psi_{ij})+\mathrm{FEM}_{ij}$; modified form $M_{ij}=\frac{3EI}{L}(\theta_i-\psi_{ij})$ when the far end is pinned; inextensibility relating $u_3$ to $v_2$; translation equation $\sum(M_{ij}+M_{ji})\psi^{*}_{ij}+\sum \mathbf{F}\cdot\mathbf{d}^{*}=0$.

Why this works. The inclined member is what makes this a sway problem: a vertical dip of joint 2 must be accompanied by a horizontal slide of joint 3 because the member cannot stretch, and that geometric coupling is the extra equation the method needs. Writing the translation equation by virtual work rather than by cutting free bodies avoids resolving inclined shears altogether, and it is the only reliable route once members are neither vertical nor horizontal. Only relative $EI$ matters because the frame is redundant to one degree and both flexibility and stiffness terms scale together.

Common pitfall. Using the total applied load in the translation equation instead of the equivalent nodal loads — an error that is invisible in the joint equations and shows up only as a moment diagram that fails to close. The other classic slip is applying $4EI/L$ at joint 3; the roller cannot carry moment, so the modified $3EI/L$ form is mandatory and the rotation there is not an unknown.

Source. Hibbeler, Structural Analysis, 10th ed., §11.4–11.5 (frames with sidesway); Kassimali, Structural Analysis, 6th ed., §16.4–16.5.

← Question 7 Next: Question 9 →

Item	Value
Member 1–2	4.0 m horizontal, rigidity \(1.2EI\), UDL \(w=27\ \text{kN/m}\)
Member 2–3	4.0 m run, 3.0 m drop ⇒ \(L=5.0\) m, rigidity \(3EI\), unloaded
Joint 1	Built in (fixed)
Joint 3	Pinned member end on a ground roller — vertical reaction only
Members	Inextensible; only relative \(EI\) is given, and only relative \(EI\) is needed

Quantity	Value
End moments (CCW positive)	\(M_{12}=+115.2\), \(M_{21}=+50.4\), \(M_{23}=-50.4\), \(M_{32}=0\) kN·m
Moment at the wall (sagging convention)	\(-115.2\) kN·m (hogging)
Moment at joint 2	\(+50.4\) kN·m (sagging)
Beam 1–2 shear	\(+95.40\) kN at joint 1; \(-12.60\) kN at joint 2
Beam 1–2 maximum sagging moment	\(+53.34\) kN·m at \(x=3.533\) m
Point of contraflexure	\(x=1.546\) m from the wall
Member 2–3	Moment \(+50.4\) to 0 kN·m; shear 10.08 kN; axial 7.56 kN compression
Reactions	\(V_1=95.40\) kN, \(H_1=0\), \(M_1=115.2\) kN·m; \(R_3=12.60\) kN