Lack of fit in an unloaded frame: 16-Civ-B1 Advanced Structural Analysis, May 2014, Q6

Question 6: Lack of fit in an unloaded frame (24 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. An unloaded frame whose only action is a fabrication error of 60 mm in the left-hand link.

Given data
Item	Value
Member ①–②	6.06 m made, 6.00 m intended (pin at ①, hinge at ②)
Member ②–③	2 m, $EI$
Member ③–④	8 m, $EI$
Column ③–⑤	8 m, $EI$, encastré at ⑤
Column ④–⑥	12 m, $1.5EI$, encastré at ⑥
$EI$	$4.0 × 10^{4}$ kN·m$^{2}$
Applied loads	none
Fabrication error	$\delta = 0.060$ m

Find. The member end moments caused by forcing the over-length member into place, hence the shear and bending moment diagrams and the force locked into the link.

Question 6: unloaded frame with a 60 mm lack of fit in member ①-②.

Approach. Recognise the over-length member as a two-force member, convert the lack of fit into a prescribed horizontal displacement of the frame, and run slope-deflection with that displacement as data rather than as an unknown. The link force then falls out of horizontal equilibrium.

Classify member ①–②. It has a pin at ①, a hinge at ② and no transverse load, so it is a two-force member: it can carry only an axial force, directed along its own (horizontal) axis.
Turn the misfit into a displacement. The member is inextensible, so it cannot be squeezed by 60 mm. Forcing it home therefore pushes joint ② outward by exactly the error, $u_2=+0.060$ m, and the link ends up in compression — the sign check that catches an inverted $\psi$.
Propagate the displacement through the frame. Beam ②–③ and beam ③–④ are horizontal and inextensible, so $u_3=u_4=0.060$ m; the two columns are vertical and inextensible, so $v_3=v_4=0$. The whole beam line simply slides sideways by 60 mm.
Show that beam ②–③ is unstressed. The hinge gives $M_{23}=0$, and because the link can deliver no vertical force, the shear in ②–③ is zero as well; a member with zero end moment and zero shear carries zero moment everywhere, so $M_{32}=0$ too. The bending problem reduces to the portal ③–④ on its two columns.
Write the chord rotations. For column ③–⑤, $\psi=-\delta/8=-0.00750$; for column ④⑥, $\psi=-\delta/12=-0.00500$; for the beam, both ends translate equally, so $\psi=0$. Every member happens to share the same stiffness coefficient, $2EI/8=2(1.5EI)/12=1.0\times10^{4}$ kN·m.
Solve the two joint equations. Joint ③ gives $4\theta_3+\theta_4=-3\delta/8$ and joint ④ gives $\theta_3+4\theta_4=-3\delta/12$, whose solution is $\theta_3=-0.00500$ rad and $\theta_4=-0.00250$ rad.
Recover the moments and the link force. Back substitution gives the six end moments; the column shears then follow as $(M_{ij}+M_{ji})/h$ and their sum is the compression carried by the over-length link.

Solving the joint equations,

$$\theta_3=-5.00\times10^{-3}\ \text{rad},\qquad \theta_4=-2.50\times10^{-3}\ \text{rad}$$

and back-substituting into the slope-deflection equations,

$$\boxed{M_{34}=-125,\ M_{43}=-100,\ M_{35}=+125,\ M_{53}=+175,\ M_{46}=+100,\ M_{64}=+125\ \text{kN}\cdot\text{m}}$$

Both joints balance exactly ($-125+125=0$ and $-100+100=0$), which is the arithmetic check worth doing before drawing anything.

Question 6: bending moment along the beam; members ①-② and ②-③ carry none.

The shears follow from the end moments. Column ③–⑤ carries $(125+175)/8=37.5$ kN, column ④–⑥ carries $(100+125)/12=18.75$ kN, and horizontal equilibrium of the whole frame therefore requires

$$N_{12}=37.5+18.75=\boxed{56.25\text{ kN compression}}$$

Beam ③–④ carries a constant shear of 28.125 kN, which passes into the columns as equal and opposite axial forces — a self-equilibrating couple, exactly as it must be on a structure carrying no external load. It is worth noticing how severe a 60 mm error is: a 1 per cent length error in one member locks 175 kN·m into a column base of a completely unloaded frame.

Quantity	Value
Joint rotations	$\theta_3=-5.00×10^{-3}$ rad, $\theta_4=-2.50×10^{-3}$ rad
Beam ③–④	$M_{34}=-125$, $M_{43}=-100$ kN·m; shear 28.125 kN
Column ③–⑤	$M_{35}=+125$, $M_{53}=+175$ kN·m; shear 37.5 kN
Column ④–⑥	$M_{46}=+100$, $M_{64}=+125$ kN·m; shear 18.75 kN
Members ①–② and ②–③	no moment, no shear
Force locked into the link	56.25 kN compression
Largest ordinate anywhere	175 kN·m at the base ⑤

Topic. Lack of fit (fabrication error) analysed as a prescribed displacement.

Key relations. For an inextensible two-force member made $\delta$ too long, the far joint is displaced by $\delta$ along the member axis. Slope-deflection then runs with $\psi_{ij}=[(\mathbf{d}_j-\mathbf{d}_i)\cdot\mathbf{e}_2]/L$ known and only the joint rotations unknown; the member force is recovered afterwards from equilibrium.

Why this works. Nothing about a lack-of-fit problem needs new theory — it needs the right question asked first: which quantity is prescribed? Here the inextensible link cannot shorten, so the displacement is data and the force is the unknown, the exact reverse of a load case. That reversal removes the storey-shear equation entirely: instead of solving for the sway, we impose it and use horizontal equilibrium at the end to read off the link force. The two-force member argument is what makes the reduction legitimate, and it also proves that everything to the left of joint ③ carries no bending at all.

Common pitfall. Modelling the link as an ordinary axially stiff member and then prescribing $u_2$, which invents an enormous spurious force; and getting the sign of $\psi$ backwards, which puts the over-length member in tension. If the arithmetic says tension, the chord rotation is inverted.

Source. Ghali, Neville and Brown, Structural Analysis: A Unified Classical and Matrix Approach, 7th ed., Ch. 5 (lack of fit, temperature and support movement); Kassimali, Structural Analysis, 6th ed., §16.4.

Question 6 of 9: Lack of fit in an unloaded frame

Question 6: Lack of fit in an unloaded frame (24 marks)