Midspan Deflection by Castigliano’s Second Theorem: 07-Str-A4, December 2016, Q3

Question 3: Midspan Deflection by Castigliano’s Second Theorem (18 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. The frame consists of a horizontal beam 2–3 hung from two pin-ended tension members, 1–2 (vertical) and 1–3 (diagonal), which are anchored at the pinned joint 1.

Given data
Quantity	Symbol	Value
Beam span 2–3 (horizontal)	$L$	4.0 m
Height of joint 1 above joint 2	$h$	3.0 m
Length of the diagonal tie 1–3	$L_{13}$	$\sqrt{3^{2}+4^{2}}=5.0$ m
Uniformly distributed load on 2–3	$w$	3.6 kN/m
Flexural rigidity of the beam	$EI$	$6000\ \text{kN}\cdot\text{m}^{2}$
Axial rigidity of both ties	$EA$	1900 kN

Find. The vertical deflection at the midspan of beam 2–3, counting both the flexural strain energy of the beam and the axial strain energy of the two ties.

Question 3: the beam 2-3 is held up by the vertical tie 1-2 and the diagonal tie 1-3; joint 2 is a roller on the vertical wall, so it carries horizontal force only.

Approach. Show the structure is statically determinate, add a dummy load $Q$ at midspan, express the two tie forces and the beam moment in terms of $Q$, and apply Castigliano's second theorem $\Delta=\partial U/\partial Q$ evaluated at $Q=0$.

Confirm determinacy. The external restraints are the pin at joint 1 (two components) and the roller on the vertical wall at joint 2 (horizontal only), giving three reaction components against three equilibrium equations. Taking the beam plus the two pin-ended ties as one free body, the three unknowns $T_{13}$, $F_{12}$ and $R_{x2}$ follow from statics alone, so Castigliano's second theorem can be applied directly without a compatibility equation.
Statics with the dummy load in place. Place a downward dummy load $Q$ at midspan. The diagonal pulls joint 3 towards joint 1 along $(-4,3)/5$, so its vertical component is $0.6\,T_{13}$. Moments about joint 2 give $$0.6\,T_{13}(4)=w(4)(2)+Q(2)\;\Longrightarrow\; T_{13}=12+0.8333\,Q,$$ and vertical equilibrium of the beam gives $$F_{12}=4w+Q-0.6\,T_{13}=7.2+0.5\,Q.$$ At $Q=0$ the bar forces are $T_{13}=12.0$ kN and $F_{12}=7.2$ kN, both tensile as the drawing's label promises, and the roller reaction is $R_{x2}=0.8(12.0)=9.6$ kN.
Bending moment in the beam. Measuring $x$ from joint 2, $$M(x)=F_{12}\,x-\frac{wx^{2}}{2}-Q\langle x-2\rangle,\qquad \frac{\partial M}{\partial Q}\Big|_{Q=0}= \begin{cases}0.5x, & 0\le x\le 2\\[2pt] 2-0.5x, & 2\le x\le 4.\end{cases}$$ With $M(x)=7.2x-1.8x^{2}$ at $Q=0$, each half of the span contributes $6.0\ \text{kN}\cdot\text{m}^{3}$, so $$\int_0^{4}M\frac{\partial M}{\partial Q}\,\mathrm{d}x =12.0\ \text{kN}\cdot\text{m}^{3} \;\Longrightarrow\;\Delta_{\text{bending}} =\frac{12.0}{6000}=0.00200\ \text{m}.$$
Axial contribution of the two ties. With $\partial F_{12}/\partial Q=0.5$ and $\partial T_{13}/\partial Q=0.8333$, $$\Delta_{\text{axial}}=\frac{F_{12}}{EA}\frac{\partial F_{12}} {\partial Q}L_{12}+\frac{T_{13}}{EA}\frac{\partial T_{13}}{\partial Q}L_{13} =\frac{7.2(0.5)(3)}{1900}+\frac{12.0(0.8333)(5)}{1900} =0.005684+0.026316=0.03200\ \text{m}.$$
Combine. Castigliano's second theorem gives the total midspan deflection $$\boxed{\Delta_{\text{mid}}=0.00200+0.03200=0.0340\ \text{m} =34.0\ \text{mm}\ \text{downward}.}$$ The ties supply $94\,\%$ of it. That is the engineering point of the question: a slender tie of $EA=1900$ kN is an extremely soft support, and the beam is carried down almost bodily rather than bent.

Question 3 — results
Quantity	Value
Force in the diagonal tie 1–3	12.0 kN tension
Force in the vertical tie 1–2	7.2 kN tension
Horizontal reaction at the roller, joint 2	9.6 kN
Flexural contribution to the midspan deflection	2.0 mm
Axial (tie) contribution	32.0 mm
Vertical deflection at midspan of 2–3	34.0 mm downward

Topic. Castigliano's second theorem applied to a determinate assembly whose flexibility is dominated by axial members rather than by bending.

Key relations. $\Delta_i=\dfrac{\partial U}{\partial P_i}$ with $U=\displaystyle\int\frac{M^{2}}{2EI}\mathrm{d}x+\sum\frac{N^{2}L}{2EA}$, so $\Delta=\displaystyle\int\frac{M}{EI}\frac{\partial M}{\partial Q}\mathrm{d}x+\sum\frac{NL}{EA}\frac{\partial N}{\partial Q}$, evaluated at $Q=0$ when the dummy load is fictitious.

Why this works. Castigliano's second theorem is the partial derivative of the complementary strain energy with respect to a force, and it returns the displacement conjugate to that force. Because the structure here is determinate, the internal forces are linear in the dummy load $Q$ and both the derivatives $\partial M/\partial Q$ and $\partial N/\partial Q$ can be read off before any integration. The dummy load must be applied where and in the direction the deflection is wanted, which is why $Q$ sits at midspan pointing down; the Macaulay bracket in $M(x)$ then makes $\partial M/\partial Q$ a tent function that peaks at midspan, exactly the unit-load moment diagram that the virtual-work method would have used.

Common pitfall. The commonest failure is to drop the axial term altogether, on the reflex that axial strain energy is negligible in a frame. Here it is sixteen times the bending term, and omitting it would report 2.0 mm instead of 34.0 mm. The second slip is to forget that $Q$ also changes the tie forces, not just the beam moment: if $\partial T_{13}/\partial Q$ and $\partial F_{12}/\partial Q$ are set to zero the answer collapses again.

Source. Hibbeler, Structural Analysis, 10th ed., §9.7–9.9 (Castigliano's theorem for trusses, beams and frames); Ghali, Neville & Brown, Structural Analysis: A Unified Classical and Matrix Approach, 7th ed., Ch. 7.

Question 3 of 9: Midspan Deflection by Castigliano’s Second Theorem

Question 3: Midspan Deflection by Castigliano’s Second Theorem (18 marks)