16-Civ-B1 Advanced Structural Analysis · December 2013

Question 4 of 9: Least work — beam propped by an elastic strut

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examination, December 2013 — 98-Civ-B1 Advanced Structural Analysis. Closed book, 3 hours, calculator permitted. Nine questions: #1 and #2 are compulsory, two of #3–#5 and two of #6–#9 are selected, and six questions constitute a complete paper (100 marks). Marks are printed in the left margin. All nine questions are solved below, because the set is a study resource rather than an examination script.

Reference texts. R. C. Hibbeler, Structural Analysis, 10th ed. (slope-deflection ch. 11, moment distribution ch. 12, influence lines ch. 6, energy methods ch. 9); A. Kassimali, Structural Analysis, 6th ed. (force/flexibility method ch. 13, matrix stiffness ch. 18); A. Ghali, A. M. Neville & T. G. Brown, Structural Analysis: A Unified Classical and Matrix Approach, 7th ed. (support-movement and lack-of-fit effects, ch. 4 and 6). Canadian design codes (CSA S16, CSA A23.3, NBCC) are not needed here: every question is an elastic-analysis question, not a design question.

Sign convention used throughout (stated once, obeyed everywhere). Global x to the right, y up; moments and rotations are counter-clockwise positive. For a member running from end $i$ to end $j$, let $\mathbf{e}_1$ be the unit vector from $i$ to $j$ and $\mathbf{e}_2$ that vector turned through $+90^\circ$. The chord rotation is $\psi_{ij}=\left[(\boldsymbol{\Delta}_j-\boldsymbol{\Delta}_i)\cdot\mathbf{e}_2\right]/L$, so the slope-deflection equation reads

$$M_{ij}=\frac{2EI}{L}\left(2\theta_i+\theta_j-3\psi_{ij}\right)+\mathrm{FEM}_{ij}$$

with $\mathrm{FEM}=+wL^{2}/12$ and $+PL/8$ at the i end of a member carrying a downward uniform load or a central point load (and the negative of those at the j end). Ordinary sagging-positive bending moments follow from $M_{\text{sag}}(i)=-M_{ij}$ and $M_{\text{sag}}(j)=+M_{ji}$. Every number below was computed twice — once by hand from these equations and once by an independent direct-stiffness solution — and the two agree.

Question 4: Least work — beam propped by an elastic strut (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. A beam continuous from the pin at A to the roller at C over a span of 8 m carries a uniformly distributed load of 6 kN/m over its whole length; at mid-span B a vertical strut 1.76 m long, pinned top and bottom, props the beam off the foundation.

Given data
Quantity	Symbol	Value
Beam span A to C	L	8 m
Uniform load over the whole span	w	6 kN/m
Beam flexural rigidity	EI	10 000 kN·m²
Strut length	L_s	1.76 m
Strut axial rigidity	EA	1650 kN

Find. The bending moment at B, and the maximum bending moment and maximum shear force anywhere in the beam.

Question 4: an 8 m beam under 6 kN/m, propped at mid-span by a pin-ended strut 1.76 m long. The strut force is the single redundant.

Approach. The strut is a two-force member, so the structure is indeterminate to the first degree; taking the strut force $S$ as the redundant and minimising the total strain energy of beam plus strut gives a single compatibility equation in $S$.

Count the redundants. Unknowns are $A_x$, $A_y$, $C_y$ and the strut force $S$, against three equations of statics for the beam: the structure is indeterminate to the first degree, and $S$ is the obvious redundant.
State least work as a compatibility statement. Minimising $U=U_{\text{beam}}+U_{\text{strut}}$ with respect to $S$, $\partial U/\partial S=0$, is identical to saying that the beam and the top of the strut deflect by the same amount at B: $$\delta_{B0}-S\,f_{bb}=S\,\frac{L_s}{EA},$$ where $\delta_{B0}$ is the free mid-span deflection of the beam and $f_{bb}$ its mid-span flexibility.
Evaluate the three coefficients. $$\delta_{B0}=\frac{5wL^{4}}{384EI}=\frac{5(6)(8)^{4}}{384(10^{4})}=0.03200\ \text{m},$$ $$f_{bb}=\frac{L^{3}}{48EI}=\frac{512}{48(10^{4})}=1.0667\times10^{-3}\ \text{m/kN},\qquad \frac{L_s}{EA}=\frac{1.76}{1650}=1.0667\times10^{-3}\ \text{m/kN}.$$ The examiner has chosen the strut so that its flexibility is exactly equal to the beam’s — a useful check that the data has been read correctly.
Solve for the redundant. $$S=\frac{0.03200}{2\times1.0667\times10^{-3}}=\boxed{15.0\ \text{kN (compression)}}$$ Because the two flexibilities are equal, the strut takes exactly half of the load it would take if it were rigid; a rigid prop would attract 30 kN.
Return to statics for the beam. With the 15 kN prop force acting upward at mid-span, $$R_A=R_C=\frac{wL}{2}-\frac{S}{2}=24-7.5=\boxed{16.5\ \text{kN}}$$ and the bending moment at B follows from the left free body: $$M_B=16.5(4)-\frac{6(4)^{2}}{2}=66-48=\boxed{+18.0\ \text{kN}\cdot\text{m}}$$ sagging.
Locate the true maximum moment. The greatest sagging moment occurs where the shear vanishes, which is not at the prop: $$x=\frac{R_A}{w}=\frac{16.5}{6}=2.75\ \text{m from A},$$ $$M_{\max}=16.5(2.75)-\frac{6(2.75)^{2}}{2}=45.375-22.688=\boxed{22.69\ \text{kN}\cdot\text{m}}$$ with an identical value 2.75 m from C by symmetry. The maximum shear is at the end supports, $$V_{\max}=\boxed{16.5\ \text{kN}},$$ the shear immediately either side of the prop being only $\mp7.5$ kN.

Question 4: bending-moment diagram. The peak is 22.69 kN·m at the quarter points, not the 18.0 kN·m at the prop.

Question 4: shear-force diagram, showing the 15 kN step at the strut.

Question 4 — results
Quantity	Value
Strut force S	15.0 kN compression
End reactions R_A = R_C	16.5 kN
Bending moment at B	+18.0 kN·m sagging
Maximum bending moment	22.69 kN·m at 2.75 m from each support
Maximum shear	16.5 kN at A and C
Shear each side of the prop	−7.5 kN / +7.5 kN

Topic. The least-work (minimum strain energy) form of Castigliano’s theorem for a beam on an elastic support.

Key relations. $\partial U/\partial R=0$ for a redundant $R$ that produces no external work; superposition form $\delta_0=R(f_{bb}+L_s/EA)$; $\delta_{B0}=5wL^{4}/384EI$; $f_{bb}=L^{3}/48EI$; strut shortening $SL_s/EA$.

Why this works. Least work is compatibility in disguise. The strut does no external work on the structure, so the total strain energy is stationary with respect to its force, and setting the derivative to zero simply says that the two bodies meeting at B move together. Writing it as a flexibility equation makes the physics visible: the redundant is shared in inverse proportion to the two flexibilities, so a stiff strut attracts load and a slender one sheds it. That is why $EA$ appears at all — treat the prop as rigid and the answer doubles.

Common pitfall. Reporting $M_B$ as the maximum moment. Propping a uniformly loaded span at mid-length drives the peak away from the prop, so the design moment is 22.69 kN·m at the quarter points, 26 % larger than the value at B. The second pitfall is omitting the strut’s own flexibility, which here would give $S=30$ kN and a completely different diagram.

Source. Hibbeler, Structural Analysis, 10th ed., §9.9 and §10.4 (force method with elastic supports); Kassimali, 6th ed., §13.4.

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