Deflection at an internal hinge by Castigliano: 07-Str-A4, May 2014, Q4

Question 4: Deflection at an internal hinge by Castigliano (16 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. A five-point beam, encastré at ① and roller-supported at ⑤, with an internal hinge at ③ and equal point loads at ② and ④.

Given data
Quantity	Value
Segment lengths ①–②–③–④–⑤	3 m each (12 m overall)
Flexural rigidity ①–②	$2EI$
Flexural rigidity ②–⑤	$EI$
Point loads at ② and ④	12 kN downward each
Internal hinge	at ③
$EI$	$1.8 × 10^{4}$ kN·m$^{2}$

Find. The vertical deflection of point ③, the hinge.

Question 4: hinged beam, encastré at ①, roller at ⑤, stepped EI.

Approach. Check the determinacy first: four reaction components, three equilibrium equations and one hinge condition make the beam statically determinate, so no redundant has to be found. Split it at the hinge, replace the right-hand simple span by the shear it delivers, and integrate $\int M m\,\mathrm{d}s/EI$ over the resulting cantilever with a unit load at ③.

Confirm the structure is determinate. With $r_{\text{sup}}=3+1=4$ and one hinge, $4=3+1$: the beam is determinate, so $M$ can be written from statics alone and no least-work step is needed.
Analyse the suspended span. Segment ③–⑤ is a simple span of 6 m carrying 12 kN at its midpoint ④, hinged at ③ and rollered at ⑤, so $R_5=12(3)/6=6.0$ kN and the hinge hands 6.0 kN down to the cantilever.
Reduce to a cantilever. Segment ①–③ is now a 6 m cantilever built in at ① carrying 12 kN at ② and the 6.0 kN hinge shear at its tip ③. Its tip deflection is the answer, because the hinge is a common point of both pieces.
Apply the dummy load at the hinge. Measure $s$ from ③ towards ①. A unit downward load applied exactly at ③ goes straight into the cantilever and leaves the suspended span untouched, so $m=-s$ over the whole cantilever while $M=-6s$ for $0\le s\le 3$ and $M=-(18s-36)$ for $3\le s\le 6$.
Integrate segment by segment, respecting the step in $EI$. Over ③–② the rigidity is $EI$ and $\int_0^{3}6s^{2}\,\mathrm{d}s=54$; over ②–① the rigidity is $2EI$ and $\int_3^{6}(18s^{2}-36s)\,\mathrm{d}s=648$, which contributes $648/2=324$.
Divide by the given rigidity. Adding the two contributions and dividing by $EI=1.8\times10^{4}$ kN·m$^{2}$ gives the deflection in metres; multiply by 1000 to quote it in millimetres.

Castigliano’s theorem for the deflection under a dummy load $Q$ at ③ reads

$$\Delta_3=\left.\frac{\partial U}{\partial Q}\right|_{Q=0}=\int_0^{6}\frac{M}{EI(s)}\,\frac{\partial M}{\partial Q}\,\mathrm{d}s=\frac{54}{EI}+\frac{648}{2EI}=\frac{378}{EI}$$

and substituting the given rigidity,

$$\Delta_3=\frac{378}{1.8\times10^{4}}=0.0210\text{ m}=\boxed{21.0\text{ mm}\ \downarrow}$$

A direct stiffness solution of the same beam, with the hinge modelled as a single-end moment release, reproduces 21.0 mm exactly, together with the built-in reactions $R_1=18.0$ kN and $M_1=-72.0$ kN·m.

Result	Value
Shear delivered through the hinge	6.00 kN
Reaction at ⑤	6.00 kN upward
Reaction at ①	18.0 kN upward
Fixing moment at ①	−72.0 kN·m (hogging)
Flexibility integral $\int Mm\,\mathrm{d}s$	$378/EI$ kN·m$^{3}$
Vertical deflection at ③	21.0 mm downward

Topic. Deflection of a determinate Gerber beam by the unit-load form of Castigliano's theorem.

Key relations. $\Delta=\int M\,(\partial M/\partial Q)\,\mathrm{d}s/EI$ evaluated at $Q=0$; for a determinate structure $\partial M/\partial Q$ is the bending moment $m$ of a unit load applied at, and in the direction of, the required displacement. A hinge in a beam transfers shear but no moment.

Why this works. Recognising that an internal hinge plus a roller makes the whole beam determinate is what turns a forbidding-looking five-point problem into two small determinate pieces. The suspended span hands a known shear to the cantilever, and the deflection of the cantilever tip is the deflection of the hinge, because the two pieces share that point. Applying the dummy load exactly at the hinge is what keeps the suspended span out of the integral: a load there is carried entirely by the cantilever, so $m=0$ to the right of ③. The stepped rigidity then enters only as a divisor on the corresponding integration range — do not average it.

Common pitfall. Averaging $2EI$ and $EI$ into a single rigidity, or integrating the dummy moment over the suspended span as well. The first shifts the answer by about 30 per cent; the second adds a term that should be exactly zero.

Source. Hibbeler, Structural Analysis, 10th ed., §9.8 (Castigliano's theorem applied to beams); Kassimali, Structural Analysis, 6th ed., §7.7 and §3.4 (Gerber beams).

Question 4 of 9: Deflection at an internal hinge by Castigliano

Question 4: Deflection at an internal hinge by Castigliano (16 marks)