Lack of Fit — a Link Fabricated 0.06 m Too Long: 16-Civ-B1 Advanced Structural Analysis, May 2016, Q5

Question 5: Lack of Fit — a Link Fabricated 0.06 m Too Long (21 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
Beam spans	1–2 = 6.0 m; 2–3 = 8.0 m
Columns 2–5 and 3–4	6.0 m each
Joint 1	pin support at the beam end
Member 2–5	hinged at joint 2, pinned at joint 5 → two-force member
Member 3–4	rigid at joint 3, pinned at joint 4
Lack of fit	member 2–5 made 0.060 m too long, forced into place
Flexural rigidity	EI = 6.0 × 10⁴ kN·m²; all members inextensible

Find. The complete set of member end moments, the force locked into the mis-fitted link, and the shear force and bending moment diagrams with their maximum and minimum ordinates.

Question 5 — no applied load. Member 2–5 is hinged at joint 2 and pinned at joint 5, so it is a two-force member; it was fabricated 0.06 m too long and forced into place after the rest of the frame was erected. All members inextensible.

Approach. Because member 2–5 is pinned at one end and hinged at the other it is a two-force member, and because every member is inextensible it cannot shorten: the whole 0.06 m misfit must appear as a prescribed upward displacement of joint 2. Delete the link, impose $v_2 = +0.060$ m, solve the resulting beam-and-column frame by slope-deflection, and read the link force off as the reaction at that prescribed degree of freedom.

Establish the kinematics before any stiffness is written. The beam is inextensible and joint 1 is a pin, so $u_1 = u_2 = u_3 = 0$; column 3–4 is inextensible with a pin at 4, so $v_3 = 0$. There is therefore no sway, and the only imposed movement is$$v_2 = +0.060\ \text{m (upward)}$$A member made too long and forced in must push its neighbour away, which is why the sign is positive; if the arithmetic later returned tension in the link, the sign of $\psi$ would be inverted.
Chord rotations. With the clockwise-positive convention,$$\psi_{12} = -\frac{0.060}{6} = -0.01000,\qquad \psi_{23} = +\frac{0.060}{8} = +0.00750,\qquad \psi_{34} = 0$$The beam is jacked up in the middle, so its left half tilts counter-clockwise and its right half clockwise — opposite signs, as the geometry demands.
Slope-deflection equations. Joint 1 is a pin and joint 4 is a pin, so use the modified stiffness $3EI/L$ for members 2–1 and 3–4 and eliminate $\theta_1$ and $\theta_4$ at once. With no applied load every fixed-end moment is zero:$$M_{21} = \tfrac{3EI}{6}\left(\theta_2 - \psi_{12}\right) = 0.5EI\left(\theta_2 + 0.01\right)$$$$M_{23} = \tfrac{2EI}{8}\left(2\theta_2 + \theta_3 - 3\psi_{23}\right),\qquad M_{32} = \tfrac{2EI}{8}\left(2\theta_3 + \theta_2 - 3\psi_{23}\right)$$$$M_{34} = \tfrac{3EI}{6}\,\theta_3 = 0.5EI\,\theta_3$$
Joint equilibrium. The link delivers axial force only, so it puts no moment into joint 2:$$\text{joint 2:}\quad M_{21}+M_{23}=0 \;\Rightarrow\; \theta_2 + 0.25\theta_3 = 6.250\times10^{-4}$$$$\text{joint 3:}\quad M_{32}+M_{34}=0 \;\Rightarrow\; 0.25\theta_2 + \theta_3 = 5.625\times10^{-3}$$Solving the pair,$$\theta_2 = \boxed{-0.8333\ \text{mrad}},\qquad \theta_3 = \boxed{+5.8333\ \text{mrad}}$$(clockwise positive).
Member end moments. Substituting back, with $EI = 6.0\times10^{4}$ kN·m²,$$M_{12}=0,\qquad M_{21}=+275.0,\qquad M_{23}=-275.0$$$$M_{32}=-175.0,\qquad M_{34}=+175.0,\qquad M_{43}=0 \qquad (\text{kN}\cdot\text{m})$$Both joints balance exactly, and the two pinned ends are moment-free as they must be. In the sagging convention the beam therefore runs from zero at joint 1 to$$M_{\text{sag}}(2) = \boxed{-275.0\ \text{kN}\cdot\text{m (hogging)}},\qquad M_{\text{sag}}(3) = \boxed{+175.0\ \text{kN}\cdot\text{m (sagging)}}$$and column 3–4 carries 175.0 kN·m at its head falling linearly to zero at the pin.
Shears and the link force. With no span load the shear is constant in each member:$$V_{1\text{-}2} = \frac{275.0}{6} = -45.83\ \text{kN},\qquad V_{2\text{-}3} = \frac{175.0-(-275.0)}{8} = +56.25\ \text{kN}$$$$V_{3\text{-}4} = \frac{175.0}{6} = 29.17\ \text{kN}$$The step in beam shear at joint 2 is the force the link hands to the beam:$$S = 56.25 - (-45.83) = \boxed{+102.08\ \text{kN, compression in member 2--5}}$$Compression is the physically correct answer for a member that was too long and had to be squeezed into place.

Question 5 — bending moment diagram for the beam 1–2–3, sagging positive. The point of contraflexure lies 4.889 m to the right of joint 2.

Question 5 — shear force diagram for the beam; the 102.08 kN step at joint 2 is the thrust delivered by the over-length link.

Question 5 — bending moment in column 3–4, plotted from joint 3 (left) to the pin at joint 4 (right); the column shear is a constant 29.17 kN.

Reactions and the global check. The pin at joint 1 must pull down and so must the pin at joint 4, because the link is pushing the beam up in the middle:$$V_1 = -45.83\ \text{kN},\qquad V_5 = +102.08\ \text{kN},\qquad V_4 = -56.25\ \text{kN}$$$$H_4 = +29.17\ \text{kN},\qquad H_1 = -29.17\ \text{kN}$$Vertical equilibrium closes to zero, and moments about joint 1 give $6(102.08) - 14(56.25) + 6(29.17) = 0$. The hold-downs are the practical message of the question: a fabrication error of 60 mm on a 6 m member generates uplift at two supports of an otherwise unloaded frame.
Where the diagram crosses zero. The beam moment runs straight from $-275.0$ at joint 2 to $+175.0$ at joint 3, so the point of contraflexure lies$$x = \frac{275.0}{275.0+175.0}\times 8 = 4.889\ \text{m to the right of joint 2}$$There is no other zero: the whole of span 1–2 hogs.

Check: the misfit is taken entirely by the frame. The question states that all members are inextensible, so none of the 0.06 m can be absorbed by axial shortening of the link itself. If the link were given a finite $EA$, part of the misfit would be relieved and every moment above would fall in proportion — with $EA = 4\times10^{5}$ kN, for example, the link force drops by roughly a sixth. The inextensible reading is the one the question intends and is the conservative one for the frame.

Quantity	Value
Imposed displacement of joint 2	+0.060 m upward (no sway)
Joint rotations (clockwise positive)	$\theta_2 = -0.8333$ mrad; $\theta_3 = +5.8333$ mrad
Beam moment over joint 2	275.0 kN·m hogging (maximum ordinate)
Beam moment at joint 3	175.0 kN·m sagging
Column 3–4	175.0 kN·m at the head, 0 at the pin; shear 29.17 kN
Beam shears	−45.83 kN (span 1–2); +56.25 kN (span 2–3)
Force in the mis-fitted link 2–5	102.08 kN compression
Support reactions	$V_1 = 45.83$ kN down, $V_4 = 56.25$ kN down, $V_5 = 102.08$ kN up, $H = 29.17$ kN
Point of contraflexure	4.889 m right of joint 2

Topic. Lack-of-fit (fabrication error) analysis by slope-deflection, with the misfitted member treated as a prescribed displacement.

Key relations. $M_{ij} = 2Ek(2\theta_i+\theta_j-3\psi_{ij})$; modified stiffness $3EI/L$ at a pinned far end; $\psi = \Delta/L$; shear $V = -(M_{ij}+M_{ji})/L$; link force = reaction at the prescribed degree of freedom.

Why this works. A member that is pinned at one end and hinged at the other can transmit only axial force, so it is kinematically equivalent to a rigid strut of the wrong length. If every member is inextensible, forcing it into place cannot compress it; the discrepancy has to be swallowed by the rest of the structure as an imposed joint displacement, and the force in the strut is then simply the support reaction at that imposed displacement. Modelling it this way avoids the trap of leaving a very stiff axial element in the model and prescribing the displacement, which makes the reported bar force meaningless. Because there is no applied load, every internal action is proportional to $EI\delta$: a stiffer frame is punished harder by the same fabrication error.

Common pitfall. Getting the sign of $\psi$ backwards is the classic error and it is self-diagnosing — an over-length member forced into place must end up in compression, so tension means the imposed displacement was applied the wrong way. The second trap is forgetting that the hinge at joint 2 lets the beam run continuously through it, so the beam moment there is not zero even though the link carries no moment.

Source. Hibbeler, Structural Analysis, 10th ed., §11.3–11.4 and §12.4 (settlement and fabrication error); Kassimali, 6th ed., §16.4; Ghali, Neville & Brown, 7th ed., §4.5.

Question 5 of 8: Lack of Fit — a Link Fabricated 0.06 m Too Long

Question 5: Lack of Fit — a Link Fabricated 0.06 m Too Long (21 marks)