Horizontal deflection at C by virtual work: 07-Str-A1, December 2014, Q8

Question 8: Horizontal deflection at C by virtual work (20 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 bent beam ABC pinned to a wall at A and propped by a tie rod DB running down from a second wall pin at D.

Given data
Item	Value
A (pinned support)	$(0,\ 0)$ m
D (pinned support)	$(0,\ 3.75)$ m
B (junction of the two beams and the tie)	$(5,\ 0)$ m
C (free end)	$(9,\ 3)$ m
Tie rod DB	length 6.25 m, $AE = 4.5 \times 10^{4}$ kN
Beams AB and BC	5 m each, inextensible, $EI = 2.4 \times 10^{5}\ \text{kN}\cdot\text{m}^{2}$
Load	60 kN vertically downward at C

Find. The horizontal component of the displacement of point C, in millimetres, stating its direction.

Structure for Question 8. The tie rod is a two-force member, so the reaction at D acts along DB and the structure is determinate.

Approach. The tie is a two-force member, so the structure is determinate; find the real forces, then apply a unit horizontal load at C and evaluate $1 \cdot \Delta = \int M m / EI\ \mathrm{d}x + \sum N n L / AE$, keeping the axial term because only the beams are stated to be inextensible.

Geometry. The tie runs from $(0,\ 3.75)$ to $(5,\ 0)$, so its length is $\sqrt{5^{2}+3.75^{2}} = 6.25$ m and its direction cosines are $0.8$ and $-0.6$ — a 3-4-5 triangle scaled by 1.25. The beam BC runs 4 m across and 3 m up, so it is 5 m long, another 3-4-5.
Force in the tie. Because D connects to nothing but the tie, the reaction there acts along DB and the only unknowns are $T$ and the two components at A. Taking moments about A for the whole structure, the vertical component of the tie force, $0.6T$, acts 5 m from A and the 60 kN load acts 9 m from A: $$0.6T(5) = 60(9) \ \Rightarrow \ 3T = 540 \ \Rightarrow \ T = \boxed{180\ \text{kN (tension)}}$$
Reactions at A. Resolving, $$A_{x} = 0.8(180) = \boxed{144\ \text{kN}}, \qquad A_{y} = 60 - 0.6(180) = \boxed{-48\ \text{kN}}$$ so A pushes 144 kN horizontally and pulls down with 48 kN — the tie lifts the beam harder than the load pushes it down, which is the signature of a well-tensioned prop.
Real bending moments. Along AB, at a distance $x$ from A, the only contribution is the 48 kN vertical component, so $M = 48x$, rising to $240\ \text{kN}\cdot\text{m}$ at B. Along BC, working back from the free end at a distance $s$ from C, the 60 kN load has a lever arm $0.8s$, so $M = 48s$, again $240\ \text{kN}\cdot\text{m}$ at B. The moment is continuous at B, as it must be, and the diagram is two straight lines meeting at a peak of $240\ \text{kN}\cdot\text{m}$.
Virtual system. Remove the real loads and apply a unit horizontal load at C, pointing to the right. Moments about A give a lever arm of 3 m for the unit load and 5 m for the vertical component of the virtual tie force, so $3t(1) = 3$ and $t = 1.0$; the virtual moments are then $m = 0.6x$ on AB and $m = 0.6s$ on BC, reaching $3.0$ m at B. The real and virtual diagrams are similar triangles, which makes the integration trivial.
Evaluate the two contributions. Over each 5 m beam, $$\int_{0}^{5}(48x)(0.6x)\,\mathrm{d}x = 28.8\frac{5^{3}}{3} = 1200\ \text{kN}\cdot\text{m}^{3}$$ so the bending term is $$\Delta_{\text{bending}} = \frac{2(1200)}{2.4 \times 10^{5}} = 0.0100\ \text{m}$$ and the tie contributes $$\Delta_{\text{tie}} = \frac{NnL}{AE} = \frac{180(1.0)(6.25)}{4.5 \times 10^{4}} = 0.0250\ \text{m}$$
Total horizontal movement. Adding, $$\Delta_{Cx} = 0.0100 + 0.0250 = \boxed{0.035\ \text{m} = 35\ \text{mm to the right}}$$ The sign is positive in the sense of the unit load, so C moves away from the wall. Note that the tie supplies 71.4 per cent of the total: stretching the rod by 25 mm is what dominates the movement, and stiffening the beams alone would barely help.

Question 8 — results
Quantity	Value
Force in the tie rod DB	180 kN tension
Horizontal reaction at A	144 kN
Vertical reaction at A	48 kN downward
Maximum bending moment (at B)	$240\ \text{kN}\cdot\text{m}$
Bending contribution to the deflection	10.0 mm
Tie-extension contribution	25.0 mm
Horizontal deflection at C	35 mm to the right

Topic. Virtual work for a structure combining flexural and axial flexibility.

Key relations. $1 \cdot \Delta = \displaystyle\int \frac{Mm}{EI}\,\mathrm{d}x + \sum \frac{NnL}{AE}$, where the first term collects the bending of the beams and the second the extension of every axially loaded bar; a two-force member transmits force only along its own axis, so a support attached to nothing else supplies one unknown rather than two.

Why this works. The virtual-force principle equates the external work of a unit load moving through the real displacement with the internal work of the virtual stress resultants acting through the real strains, and the two kinds of strain simply add. Recognising the tie as a two-force member is what makes the structure determinate: it reduces the unknowns from four to three, and it means the same single quantity $T$ appears in the equilibrium equations and in the axial term of the deflection integral. Because the real and virtual moment diagrams here are similar triangles on each beam, the integrals could equally be evaluated from a product table as $\tfrac{1}{3}M_{\max}m_{\max}L$, giving the same $1200$ per beam.

Common pitfall. Dropping the axial term because "trusses have axial forces and beams do not". Here the tie contributes two and a half times as much as all the bending put together, and omitting it turns a 35 mm answer into a 10 mm one. The complementary error is including axial deformation of the beams as well, which the question explicitly excludes by calling them inextensible.

Source. Hibbeler, Structural Analysis, 10th ed., §9.4–9.7 (virtual work for beams, frames and combined systems); Kassimali, Structural Analysis, 6th ed., §7.4–7.6.

Question 8 of 8: Horizontal deflection at C by virtual work

Question 8: Horizontal deflection at C by virtual work (20 marks)