Question 3 of 8: Vertical Deflection of a Beam by Virtual Work
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2016 —
07-Str-A1 Elementary Structural Analysis, 3 hours, CLOSED BOOK (an
approved Sharp or Casio calculator is permitted). Six questions constitute a
complete paper: Questions 1–5 are compulsory and one only of
Questions 6, 7 or 8 is attempted, for 100 marks
(6 + 18 + 18 + 20 + 18 + 20). Marks are shown in the left margin.
All eight questions are worked below, because the three
alternatives exercise three quite different methods — three-hinged-frame
statics, influence lines and virtual work — and the complete set is the
more useful study resource.
Every structure on this paper is defined by a hand-drawn figure and nothing
else: the printed text states the task, the drawing carries the geometry, the
loads and the support types.
Sign conventions used throughout. Shear force is positive
when the resultant of the forces to the left of a section acts upwards along
the positive local normal of the member; bending moment is positive when it
sags the member (tension on the underside), and sagging ordinates are plotted
above the axis in every diagram. Truss member forces are quoted as
T for tension and C for compression. In the slope-deflection
work of Question 5 the member-end moments follow the usual convention,
clockwise on the member end taken as positive, so that the sagging moment is
$+M_{ij}$ at the left end of a member and $-M_{ji}$ at its right end.
Given. A beam with the five points marked A, B, C, D and E
on the figure, dimensioned 2 m + 4 m + 4 m + 2 m from the left-hand end, so that
A is at x = 0, B at x = 2 m, C at x = 6 m, D at x = 10 m and E at x = 12 m. The
supports are a pin under B and a roller under D; the vertical broken line at C
carries the centre-line symbol and is the axis of symmetry, not a support.
Given data, Question 3
Quantity
Value
Overhang A-B and D-E
2 m each, loaded at 12 kN/m
Interior span B-D
8 m, loaded at 6 kN/m
Supports
pin at B (x = 2 m), roller at D (x = 10 m)
Flexural rigidity
EI = 2000 kN·m2, uniform
Total applied load
12(2) + 6(8) + 12(2) = 96 kN
Find. The vertical deflection of the overhang tip A, with
its direction stated.
Question 3: the loaded beam and its real bending moment diagram. The virtual system is a single downward unit load applied at A.
Approach. Use the unit-load form of virtual work,
$\Delta = \int M m\,\mathrm{d}x / EI$, with $M$ the real bending moment and $m$
the moment produced by a unit downward load at A; symmetry supplies both
reactions immediately and reduces the real moment diagram to two expressions.
Take advantage of symmetry for the reactions. The beam,
the supports and the loading are all symmetric about C, so the two reactions
are equal and each carries half the total:
$$R_B = R_D = \frac{96}{2} = \boxed{48\ \text{kN} \uparrow}$$
The beam is determinate, so no compatibility condition is needed to get
there.
Write the real bending moment. Measuring $x$ from A, the
left overhang gives
$$M(x) = -\frac{12x^{2}}{2} = -6x^{2}, \qquad 0 \le x \le 2\ \text{m}$$
so $M_B = -24\ \text{kN}\cdot\text{m}$. Inside the span, adding the reaction and
the 6 kN/m ruling,
$$M(x) = -24(x - 1) - 3(x - 2)^{2} + 48(x - 2), \qquad 2 \le x \le 10\ \text{m}$$
which returns $-24\ \text{kN}\cdot\text{m}$ at D, as symmetry demands, and
$+24\ \text{kN}\cdot\text{m}$ at mid-span.
Set up the virtual system. Apply a single downward unit
load at A and re-solve the same determinate beam. Moments about B give
$R_D = -0.25$ (downward) and hence $R_B = 1.25$ upward, so
$$m(x) = -x \quad (0 \le x \le 2), \qquad m(x) = 0.25x - 2.5 \quad (2 \le x \le 10)$$
with $m = 0$ over the right-hand overhang, because nothing beyond D is loaded
in the virtual system. Note that the virtual system is not symmetric
even though the real one is — symmetry is being used for the real
moments, not for the unit load.
Integrate over the left overhang. Here $Mm = (-6x^{2})(-x)
= 6x^{3}$, so
$$\int_{0}^{2} 6x^{3}\,\mathrm{d}x = \frac{6(2)^{4}}{4} = +24\ \text{kN}^{2}\cdot\text{m}^{3}$$
a positive contribution: hogging real moment multiplied by hogging virtual
moment.
Integrate over the interior span. Carrying out
$\int_{2}^{10} \left[ -24(x-1) - 3(x-2)^{2} + 48(x-2) \right]\left( 0.25x - 2.5
\right)\mathrm{d}x$ gives $-64\ \text{kN}^{2}\cdot\text{m}^{3}$. The sagging
real moment in the middle of the span works against the hogging virtual moment
there, which is why this term is the larger and reverses the sign of the
total.
Assemble and divide by the flexural rigidity.
$$\int \frac{M m}{EI}\,\mathrm{d}x = \frac{24 - 64}{2000} = \frac{-40}{2000}
= -0.0200\ \text{m}$$
$$\Delta_A = \boxed{20.0\ \text{mm upward}}$$
The negative sign is measured against the downward unit load, so A rises.
Sanity-check the sign physically. The interior span sags
under 6 kN/m and rotates the beam anticlockwise at B, which lifts the tip of
the left overhang; the 24 kN of ruling on the overhang itself pushes the tip
down. The 8 m span wins, and the tip finishes 20 mm above its unloaded
position.
