Least-work analysis of a tie-rod-propped beam (Castigliano): 16-Civ-B1 Advanced Structural Analysis, May 2013, Q3

Question 3: Least-work analysis of a tie-rod-propped beam (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.

Given data
Beam span, free tip $A$ to built-in end $C$	$4\text{ m}+6\text{ m}=10\text{ m}$
Uniformly distributed load on $AB$	$w = 2\text{ kN/m}$ over the first 4 m
Tie rod	from $B$ (4 m from the free tip) to the wall, 8 m above the beam
Rod length	$L_r=\sqrt{6^{2}+8^{2}} = 10\text{ m}$ (a 6–8–10 triangle)
Beam flexural rigidity	$EI = 1.44\times10^{4}\ \text{kN}\cdot\text{m}^{2}$, inextensible
Rod axial rigidity	$EA = 1.25\times10^{4}\ \text{kN}$

Find. The force in the tie rod by the least-work theorem, then the maximum shear force and the maximum bending moment in the beam, and the two diagrams.

Q3 — tie-rod-propped cantilever, and the resulting shear force and bending moment diagrams for the beam.

Approach. The structure is once statically indeterminate; take the rod force $T$ as the redundant, write the beam moment and the rod energy in terms of $T$, and impose the least-work condition $\partial U/\partial T = 0$.

Confirm the degree of indeterminacy. The beam is built in at $C$ (three reaction components), the rod is a two-force member pinned at both ends, and the left end of the beam is free. Counting reaction components and members against the equations of equilibrium leaves one redundant, so a single compatibility condition — here supplied by Castigliano's second theorem in its least-work form — closes the problem.
Resolve the rod force at the joint. The rod runs from $B$ at $(4,\,0)$ to the wall at $(10,\,8)$, so its direction cosines are $$c_x=\frac{6}{10}=0.6,\qquad c_y=\frac{8}{10}=0.8 .$$ A tension $T$ therefore lifts the beam at $B$ with a vertical component $0.8T$ and pushes it towards the wall with a horizontal component $0.6T$. The beam is inextensible, so the horizontal component does no work and does not enter the compatibility equation.
Write the bending moment as a function of $T$. Measuring $x$ from the free tip and taking the left-hand free body (sagging positive), $$M(x) = -\frac{wx^{2}}{2}\quad (0\le x\le 4),\qquad M(x) = -wa\!\left(x-\tfrac{a}{2}\right)+0.8T\,(x-4)\quad (4\le x\le 10),$$ with $a=4\text{ m}$ and $wa=8\text{ kN}$. Only the second expression depends on the redundant, and $$\frac{\partial M}{\partial T}=0.8\,(x-4)\quad\text{on }BC,\qquad \frac{\partial M}{\partial T}=0\quad\text{on }AB .$$
Apply the least-work theorem. The total strain energy is the beam bending energy plus the rod axial energy, and the redundant takes the value that minimises it: $$\frac{\partial U}{\partial T} =\frac{1}{EI}\int_{4}^{10} M\,\frac{\partial M}{\partial T}\,dx +\frac{T L_r}{EA}=0 .$$ Substituting $u=x-4$ and integrating over $0\le u\le 6$, $$\int_{0}^{6}\Bigl[-8(u+2)+0.8T\,u\Bigr]\,0.8u\,du = 46.08\,T-691.2 .$$
Solve for the rod force. Inserting the rigidities, $$\frac{46.08\,T-691.2}{1.44\times10^{4}}+\frac{10\,T}{1.25\times10^{4}}=0 \;\Longrightarrow\; 0.0032\,T-0.048+0.0008\,T=0 ,$$ $$\boxed{T = 12.0\ \text{kN (tension)} .}$$ The vertical component is $0.8T=9.60\text{ kN}$ upward and the horizontal component is $0.6T=7.20\text{ kN}$ towards the wall.
Complete the equilibrium of the beam. The total applied load is $wa=8.0\text{ kN}$ downward while the rod lifts $9.60\text{ kN}$, so the built-in end must pull the beam down: $$R_C = 8.0-9.60 = -1.60\ \text{kN}\ (\text{i.e. }1.60\text{ kN downward}) .$$ The shear (taking upward forces to the left of the section as positive) is therefore $V=-2x$ over $AB$, reaching $-8.00\text{ kN}$ just left of $B$, and then jumps by $+9.60\text{ kN}$ to a constant $+1.60\text{ kN}$ over $BC$.
Evaluate the bending moments. On the overhang the moment is the cantilever parabola $M=-x^{2}$, so at the rod joint $$M_B = -\frac{2\times4^{2}}{2}=-16.0\ \text{kN}\cdot\text{m} ,$$ and beyond $B$ the moment rises linearly at the constant shear rate: $$M_C = -16.0 + 1.60\times 6 = -6.40\ \text{kN}\cdot\text{m} .$$ Both are hogging. The largest values are $$\boxed{\left|M\right|_{\max}=16.0\ \text{kN}\cdot\text{m at }B,\qquad \left|V\right|_{\max}=8.00\ \text{kN just left of }B .}$$

Final results — Question 3
Quantity	Value
Force in the steel rod	12.0 kN tension
Vertical / horizontal components delivered to the beam at $B$	9.60 kN up / 7.20 kN towards the wall
Reaction at the built-in end $C$	1.60 kN downward (hold-down)
Maximum shear in the beam	8.00 kN, just left of $B$
Shear in span $BC$	+1.60 kN, constant
Maximum bending moment in the beam	16.0 kN·m hogging, at $B$
Bending moment at the built-in end $C$	6.40 kN·m hogging

Topic. Castigliano's second theorem in its least-work form, applied to a structure whose redundant is an axially loaded member.

Key relations. $U=\displaystyle\int\frac{M^{2}}{2EI}\,dx+\sum\frac{N^{2}L}{2EA}$; the least-work condition $\partial U/\partial X=0$ for a redundant $X$ whose corresponding displacement is zero; equivalently $\displaystyle\int\frac{M}{EI}\frac{\partial M}{\partial X}\,dx+\frac{XL}{EA}=0$.

Why this works. Castigliano's second theorem states that the partial derivative of the total strain energy with respect to a force equals the displacement in the direction of that force. For a redundant restraint the corresponding relative displacement is zero, so the derivative vanishes and the theorem becomes the principle of least work. Here the redundant $T$ is the rod force, and its “displacement” is the gap that would open between the rod and the beam if the rod were cut — which must be zero. Including the rod term $TL_r/EA$ is essential: it contributed $0.0008T$ against the beam's $0.0032T$, so ignoring the rod extension would have overestimated $T$ by 25 %.

Common pitfall. Two traps. First, forgetting that only the component of the rod force normal to the beam does bending work — here $0.8T$, not $T$ — while the axial component is absorbed by the stated inextensibility of the beam. Second, assuming the built-in end must push upward: because the rod lifts more than the load weighs, the fixed end develops a downward reaction, and a solver who forces $R_C\ge0$ will get the wrong sign on the shear in $BC$ and on the fixed-end moment.

Source. Hibbeler, Structural Analysis, 10th ed., §9.7–9.9 (Castigliano's theorem for beams, frames and trusses) and §10.6; Kassimali, Structural Analysis, 6th ed., §7.7; Ghali, Neville and Brown, §7.3 on the least-work principle.

Question 3 of 9: Least-work analysis of a tie-rod-propped beam (Castigliano)

Question 3: Least-work analysis of a tie-rod-propped beam (Castigliano) (16 marks)