Vertical Deflection at the Overhang Tip by Virtual Work: 07-Str-A1, May 2017, Q3

Question 3: Vertical Deflection at the Overhang Tip by Virtual Work (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.

Quantity	Value
Overhang A–B (free end to pin)	3 m, carrying 8 kN/m downward
Span B–C (pin to roller)	9 m
Point load	60 kN downward, 3 m from B (6 m from C)
Flexural rigidity	$EI = 2.7\times10^{4}\ \text{kN}\,\text{m}^{2}$, constant

Find. The vertical deflection of point A, in magnitude and direction.

Question 3: the beam with its reactions, the real bending moment diagram $M$, and the virtual moment diagram $m$ produced by a unit downward load at A.

Approach. Unit-load (virtual work) method: find the real moment diagram $M$, apply a 1 kN downward dummy load at A to get $m$, and evaluate $\delta_A = \int M m \,\mathrm{d}x / EI$ segment by segment.

Real reactions. Measure $x$ from B, positive toward C. The overhang load is $8(3) = 24$ kN acting 1.5 m to the left of B: $$\sum M_B = 0:\qquad 9R_C = 60(3) - 24(1.5) = 180 - 36 = 144 \;\Rightarrow\; R_C = \boxed{16.0\ \text{kN}\ \uparrow}$$ $$R_B = 24 + 60 - 16 = \boxed{68.0\ \text{kN}\ \uparrow}$$ Checking about C: $9R_B = 60(6) + 24(10.5) = 360 + 252 = 612$, giving $R_B = 68.0$ kN as required.
Real moment diagram. On the overhang, with $d$ measured from A, $M = -4d^2$, which reaches $-36$ kN·m at B. In the span, $$M(x) = 68x - 24(x + 1.5) = 44x - 36 \quad (0 \le x \le 3), \qquad M(x) = -16x + 144 \quad (3 \le x \le 9)$$ so $M$ runs $-36 \to +96$ kN·m under the load $\to 0$ at C.
Virtual system. Remove the real loads and apply 1 kN downward at A. Moments about B give $R_C = -1/3$ (downward) and $R_B = 4/3$ upward, so $$m = -d \quad \text{on the overhang}, \qquad m(x) = \tfrac{x}{3} - 3 \quad \text{in the span}$$ with $m = -3$ kN·m at B and zero at C, as it must be.
Overhang contribution. With $M = -4d^2$ and $m = -d$, $$\int_0^{3} (-4d^2)(-d)\,\mathrm{d}d = \int_0^3 4d^3\,\mathrm{d}d = \left[d^4\right]_0^3 = +81\ \text{kN}^2\text{m}^3$$
Span, load-side segment ($0 \le x \le 3$). Here $Mm = (44x-36)\left(\tfrac{x}{3}-3\right) = \tfrac{44}{3}x^2 - 144x + 108$, so $$\int_0^3 Mm\,\mathrm{d}x = \tfrac{44}{3}(9) - 144(4.5) + 108(3) = 132 - 648 + 324 = -192$$
Span, far segment ($3 \le x \le 9$). Here $Mm = (-16x+144)\left(\tfrac{x}{3}-3\right) = -\tfrac{16}{3}x^2 + 96x - 432$, and $$\int_3^9 Mm\,\mathrm{d}x = -1248 + 3456 - 2592 = -384$$
Assemble. Adding the three contributions, $$\int Mm\,\mathrm{d}x = 81 - 192 - 384 = -495\ \text{kN}^2\text{m}^3$$ $$\delta_A = \frac{-495}{2.7\times10^{4}} = -0.018333\ \text{m} = \boxed{18.33\ \text{mm}\ \text{UPWARD}}$$
Check the sense physically. The negative sign means the deflection opposes the assumed downward unit load. That is right: the 60 kN in the main span makes B rotate so that the tangent rises to the left, and the resulting lift at the 3 m overhang tip more than cancels the sag the overhang's own 24 kN produces. Quoting the magnitude without the direction throws away the point of the question.

Quantity	Value
Reaction at the pin B	68.0 kN ↑
Reaction at the roller C	16.0 kN ↑
$\int Mm\,\mathrm{d}x$	−495 kN²m³
Vertical deflection at A	18.33 mm upward (0.01833 m)

Topic. Deflection of a determinate beam by the unit-load (virtual work) method.

Key relations. $1 \cdot \delta = \displaystyle\int_0^L \frac{M m}{EI}\,\mathrm{d}x$, where $M$ is the bending moment under the real loads and $m$ the bending moment under a unit load applied at the point and in the direction of the required deflection. A positive result means the deflection is in the direction of the unit load. For an overhang, both diagrams must be written in the same coordinate and both must be continuous through the support.

Why this works. The virtual-work statement equates the external work of the unit load moving through the real displacement with the internal work of the virtual moments acting through the real curvatures $M/EI$. Because both diagrams here are piecewise linear or quadratic, the integral is exact segment by segment; the same result follows from the diagram-multiplication (Vereshchagin) rule, area of one diagram times the ordinate of the straight one at the first diagram's centroid, whenever one of the two is straight over the segment.

Common pitfall. Assuming that a loaded overhang must deflect downward. The rotation the main span imposes at the support is usually the larger effect, and it lifts the tip. The other frequent slip is writing $m$ for the overhang as $+d$ instead of $-d$ — the unit load hogs the overhang exactly as the real load does, so the two diagrams have the same sign there and that segment contributes positively.

Source. Hibbeler, Structural Analysis 10th ed., §8.4 and §9.4 (method of virtual work for beams); Kassimali 6th ed., §7.4–7.6.

Question 3 of 8: Vertical Deflection at the Overhang Tip by Virtual Work

Question 3: Vertical Deflection at the Overhang Tip by Virtual Work (16 marks)