Slope-deflection analysis of a three-member beam with support settlement: 16-Civ-B1 Advanced Structural Analysis, December 2019, Q6

Question 6: Slope-deflection analysis of a three-member beam with support settlement (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.

Quantity	Symbol	Value
Inclined member (1)–(2): run / rise / true length	—	6 m / 2.5 m / 6.5 m
Relative rigidity of (1)–(2)	—	1.3 EI
Member (2)–(3)	$L_{23}$	9 m, 1.6 EI
Overhang (3)–(4)	$L_{34}$	1 m, EI, free tip
Uniformly distributed load	$w$	4.8 kN/m
Reference flexural rigidity	$EI$	$8.1\times10^{4}\ \text{kN}\cdot\text{m}^2$
Settlement of support (3)	$\Delta_3$	8 mm downward

Check: unit typo in the printed data. The paper prints $EI = 8.1\times10^{4}\ \text{kN}\cdot\text{mm}^2$. That value is physically impossible for a 16 m beam — it is roughly the stiffness of a wire. Read as $8.1\times10^{4}\ \text{kN}\cdot\text{m}^2$ the settlement term $3EI_{23}\Delta/L_{23}^{2} = 38.4\ \text{kN}\cdot\text{m}$ lands in the same order as the load term $wL_{23}^{2}/8 = 48.6\ \text{kN}\cdot\text{m}$, which is what makes the question a meaningful comparison of the two effects. All work below uses $\text{kN}\cdot\text{m}^2$.

Find. The end moments of all three members, the reactions, and the shear and bending moment diagrams with their maximum and minimum ordinates.

Question 6: inclined member 1-2 (1.3EI) from the built-in support, member 2-3 (1.6EI) and a 1 m overhang, under 4.8 kN/m.

Approach. Establish first that no joint can translate horizontally, so the problem carries only two rotation unknowns; convert the UDL on the inclined member to a fixed-end moment using the horizontal projection; write the two slope-deflection joint equations including the chord rotation caused by the settlement; then recover the diagrams from the member end forces.

Establish the kinematics before counting unknowns. Support (1) is built in and support (2) is a roller, so $v_2 = 0$; member (1)–(2) is inextensible and inclined, and the inextensibility condition $u_2 c + v_2 s = 0$ with $c = 6/6.5 \ne 0$ therefore forces $u_2 = 0$ as well. Member (2)–(3) is horizontal and inextensible, so $u_3 = u_2 = 0$, and the same argument carries to joint (4). There is no sway. The overhang (3)–(4) has a free tip, so it offers no rotational restraint at all — it merely delivers a known moment and shear to joint (3). The unknowns are $\theta_2$ and $\theta_3$ only.
Fixed-end moments, remembering that the UDL is per metre of horizontal projection. The load band is drawn horizontally across the whole structure, so on the inclined member the intensity is $w$ per metre of horizontal projection. Resolving, the intensity normal to the member is $$q_n = w\cos^{2}\alpha = 4.8\left(\frac{6}{6.5}\right)^{2} = 4.0899\ \text{kN/m},$$ and the fixed-end moment is $$\mathrm{FEM}_{12} = \frac{q_n L_{12}^{2}}{12} = \frac{w\,a^{2}}{12} = \frac{4.8(6)^{2}}{12} = 14.40\ \text{kN}\cdot\text{m},$$ i.e. the projection $a = 6\ \text{m}$ governs the fixed-end moment while the true length 6.5 m still governs the stiffness. For the horizontal member, $\mathrm{FEM}_{23} = wL_{23}^{2}/12 = 4.8(81)/12 = 32.40\ \text{kN}\cdot\text{m}$, and the determinate overhang supplies a known $M_3 = -wL_{34}^{2}/2 = -2.40\ \text{kN}\cdot\text{m}$ together with a 4.80 kN shear.
Chord rotations from the settlement. Joint (2) does not move, and joint (1) is fixed, so $\psi_{12} = 0$. Joint (3) drops 8 mm relative to joint (2), so $$\psi_{23} = \frac{v_3 - v_2}{L_{23}} = \frac{-0.008}{9} = -8.889\times10^{-4}\ \text{rad}$$ (clockwise, hence negative in the counter-clockwise convention). The overhang is determinate and is therefore completely unaffected by the settlement.
Write and solve the two joint equations. With $k_{12} = 1.3EI/6.5 = 0.2EI$ and $k_{23} = 1.6EI/9$, the slope-deflection expressions are $$M_{12} = 2k_{12}\theta_2 + 14.40, \qquad M_{21} = 4k_{12}\theta_2 - 14.40,$$ $$M_{23} = 2k_{23}\left(2\theta_2 + \theta_3 - 3\psi_{23}\right) + 32.40,\qquad M_{32} = 2k_{23}\left(2\theta_3 + \theta_2 - 3\psi_{23}\right) - 32.40 .$$ Joint (2) requires $M_{21} + M_{23} = 0$; joint (3) requires $M_{32} + M_{34} = 0$ with $M_{34} = +2.40\ \text{kN}\cdot\text{m}$ supplied by the overhang. Solving the pair gives $\theta_2 = -6.611\times10^{-4}\ \text{rad}$ and $\theta_3 = -4.819\times10^{-4}\ \text{rad}$.
Back-substitute for the moments. Converting to the ordinary sagging convention, the moment diagram ordinates at the joints are $$M_1 = \boxed{+7.02\ \text{kN}\cdot\text{m}},\qquad M_2 = \boxed{-57.24\ \text{kN}\cdot\text{m}},\qquad M_3 = \boxed{-2.40\ \text{kN}\cdot\text{m}},\qquad M_4 = 0 .$$ The moment at the built-in end has come out sagging, which is surprising until the effects are separated (next step). Joint (3) reproduces the overhang cantilever moment exactly, as it must.
Separate the load effect from the settlement effect. Running the analysis twice, once with the load and no settlement and once with the settlement and no load, $$M_1 = \underbrace{-4.50}_{\text{load}} + \underbrace{+11.52}_{\text{settlement}} = +7.02, \qquad M_2 = \underbrace{-34.20}_{\text{load}} + \underbrace{-23.04}_{\text{settlement}} = -57.24 .$$ So the 8 mm settlement more than reverses the fixed-end moment at (1) — its own contribution, $+11.52\ \text{kN}\cdot\text{m}$, is two and a half times the load moment there — while at joint (2) the two effects add and drive the hogging up by two thirds. This is exactly the design point the question is making: a differential settlement of 8 mm on a 9 m span is a serviceability matter, yet it dominates the strength check on the inclined member. The settlement moment at (1) is one half of that at (2) because member (1)–(2) has a fixed far end and the carry-over factor is one half.
Reactions and the load-path check. Vertical equilibrium of each member gives $$R_1 = 3.69\ \text{kN},\qquad R_2 = 52.80\ \text{kN},\qquad R_3 = 20.31\ \text{kN},$$ and $3.69 + 52.80 + 20.31 = 76.80 = 4.8 \times 16$, the whole applied load. Because both (2) and (3) are rollers and no horizontal load is applied, the horizontal reaction at the built-in support is exactly zero — the inclined member's axial force and its shear resolve to a purely vertical force at each end.
Shear force diagram. Working along the member axes and plotting the shear normal to each member, the ordinates are $+3.41\ \text{kN}$ at (1) falling linearly to $-23.18\ \text{kN}$ at (2), a jump to $+27.69\ \text{kN}$ on the far side of (2) falling to $-15.51\ \text{kN}$ at (3), and $+4.80\ \text{kN}$ at (3) on the overhang falling to zero at the free tip. The zero crossings sit at $3.41/4.0899 = 0.833\ \text{m}$ along member (1)–(2) and $27.69/4.8 = 5.769\ \text{m}$ along member (2)–(3).

Question 6: shear force plotted normal to each member along the developed axis 1-2-3-4.

Bending moment diagram. Each member carries a parabola between the joint ordinates already found, with its stationary value at the zero-shear point: $$M_{\max,12} = \boxed{+8.44\ \text{kN}\cdot\text{m}}\ \text{at } 0.833\ \text{m from (1)},\qquad M_{\max,23} = \boxed{+22.65\ \text{kN}\cdot\text{m}}\ \text{at } 5.769\ \text{m from (2)} .$$ The governing ordinate on the whole structure is the hogging $-57.24\ \text{kN}\cdot\text{m}$ over support (2).

Question 6: bending moment along the developed axis, combining the UDL with the 8 mm settlement of support (3).

Quantity	Location	Value
Bending moment	built-in support (1)	+7.02 kN·m (sagging)
Bending moment	support (2)	−57.24 kN·m (hogging, governs)
Bending moment	support (3)	−2.40 kN·m
Maximum sagging moment, member (1)–(2)	0.833 m from (1)	+8.44 kN·m
Maximum sagging moment, member (2)–(3)	5.769 m from (2)	+22.65 kN·m
Shear, member (1)–(2)	at (1) / at (2)	+3.41 / −23.18 kN
Shear, member (2)–(3)	at (2) / at (3)	+27.69 / −15.51 kN
Shear, overhang (3)–(4)	at (3) / at (4)	+4.80 / 0 kN
Reactions	(1) / (2) / (3)	3.69 / 52.80 / 20.31 kN (sum 76.80 kN)
Settlement contribution alone	at (1) / at (2)	+11.52 / −23.04 kN·m

Topic. Slope-deflection with a prescribed support movement, on a structure containing an inclined member and a determinate overhang.

Key relations. $M_{ij} = 2k\left(2\theta_i + \theta_j - 3\psi_{ij}\right) + \mathrm{FEM}_{ij}$ with $k = EI/L$; $\psi_{ij} = (v_j - v_i)/L$ for a horizontal member; for a vertical load of $w$ per metre of horizontal projection on an inclined member, $\mathrm{FEM} = wa^{2}/12$ with $a$ the projection while the stiffness uses the true length; a determinate overhang contributes a known joint moment $-wc^{2}/2$ and no stiffness; carry-over factor one half to a fixed far end.

Why this works. Proving that no joint translates is the step that turns a frame-looking problem into a two-unknown beam problem: an inclined inextensible member whose far end is on a vertical-only support cannot let that support move horizontally, so the whole beam line is locked laterally. Once that is settled, the settlement enters only as a chord rotation, and because slope-deflection is linear the load case and the settlement case superpose exactly — which is why quoting them separately is worth the extra run. Support settlement produces a self-equilibrating moment set: it changes no reaction sum and no global equilibrium, only the distribution, and it would vanish entirely if the structure were determinate (which is precisely why the overhang is unaffected).

Common pitfall. Using the true member length 6.5 m in the fixed-end moment, which gives $4.8(6.5)^2/12 = 16.9$ instead of 14.4 — the projection governs the load, the true length governs the stiffness, and the two are not interchangeable. Second, adding a sway degree of freedom for the inclined member: the system then becomes singular or, worse, returns a plausible wrong answer. Third, getting the sign of $\psi$ backwards, which makes the settlement relieve the hogging at (2) instead of increasing it; the cheap check is that a propped cantilever whose prop settles must change its fixed-end moment by exactly $3EI\Delta/L^{2}$ in the relieving sense, and that number, 38.4 kN·m here, brackets the answer.

Source. Hibbeler, Structural Analysis, §11.3–11.4 (slope-deflection for beams, including support displacements) and §12.5; Kassimali §13.2–13.3. Canadian design context: NBC 2020 Part 4 and CSA S6:19 on imposed deformations.

Question 6 of 9: Slope-deflection analysis of a three-member beam with support settlement

Question 6: Slope-deflection analysis of a three-member beam with support settlement (22 marks)