Question 8 of 8: Deflection of a Frame 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 shallow trapezoidal frame with node 1 pinned at
(0, 0), node 2 at (1.6 m, 1.2 m), node 3 at (5.6 m, 1.2 m) and node 4 on a
roller at (7.2 m, 0). The dimension string along the base reads
1.6 m + 2 m + 2 m + 1.6 m and the rise to the level of nodes 2 and 3 is
1.2 m.
Given data, Question 8
Quantity
Value
Member 1-2 and member 3-4
rise 1.2 m over 1.6 m, so each is exactly 2.0 m long
Member 2-3
horizontal, 4.0 m long
Applied loads
12 kN vertically downward at node 2 and at node 3
Supports
pin at node 1, roller at node 4
Flexural rigidity
EI = 4000 kN·m2, the same in all members
Find. The vertical deflection at the centre of span 2-3,
that is at x = 3.6 m, considering flexural strain only.
Question 8: the frame, the real bending moment diagram M and the virtual bending moment diagram m for a unit downward load at the centre of span 2-3, both developed along the frame axis.
Approach. Both the structure and the loading are symmetric,
which gives the reactions at once; the real moment diagram then turns out to be
linear on the inclined legs and constant along the horizontal member, so the
virtual work integral is elementary.
Check the geometry. Each leg spans 1.6 m horizontally and
rises 1.2 m, so
$$L_{12} = \sqrt{1.6^{2} + 1.2^{2}} = \sqrt{2.56 + 1.44} = 2.00\ \text{m}$$
exactly — a 3-4-5 triangle scaled by 0.4. The total developed length from
support to support is $2 + 4 + 2 = 8$ m.
Find the reactions from symmetry. The frame, its supports
and the two 12 kN loads are all symmetric about the centre line, so
$$V_1 = V_4 = \boxed{12\ \text{kN} \uparrow}$$
and the roller at node 4 makes $H_1 = 0$.
Write the real bending moment on the legs. Measuring $s$
from node 1 along member 1-2, the section is at horizontal distance $0.8s$ from
the pin and the only force to the left is the 12 kN reaction, so
$$M(s) = 12(0.8s) = 9.6s\ \text{kN}\cdot\text{m}, \qquad 0 \le s \le 2\ \text{m}$$
giving $M_2 = 19.2\ \text{kN}\cdot\text{m}$ at the knee.
Show the real moment is constant along member 2-3. At a
horizontal distance $t$ past node 2 the reaction has a lever arm $1.6 + t$ and
the 12 kN load at node 2 has a lever arm $t$, so
$$M = 12(1.6 + t) - 12t = 19.2\ \text{kN}\cdot\text{m}$$
independent of $t$. The horizontal member carries pure bending with no shear,
which is why the diagram is a flat top.
Set up the virtual system. Apply a downward unit load at
the centre of 2-3, at $x = 3.6$ m. Moments about node 4 give
$R_1 = 3.6/7.2 = 0.5$, and by symmetry $R_4 = 0.5$ as well. On the legs
$$m(s) = 0.5(0.8s) = 0.4s$$
and along 2-3 the virtual moment rises linearly from $m = 0.8$ at node 2 to
$m = 1.8$ at the centre before falling back to $0.8$ at node 3.
Integrate over the two legs. Each contributes
$$\int_{0}^{2} (9.6s)(0.4s)\,\mathrm{d}s = 3.84 \times \frac{2^{3}}{3}
= 10.24\ \text{kN}^{2}\cdot\text{m}^{3}$$
so the pair contributes $20.48$.
Integrate over the horizontal member. Since $M$ is the
constant $19.2$, the integral is $19.2$ times the area under the $m$ diagram.
That area is a trapezoid of base 4 m rising from $0.8$ to $1.8$ and back, equal
to $2 \times \tfrac{1}{2}(0.8 + 1.8)(2) = 5.2\ \text{m}^{2}$, so
$$\int M m\,\mathrm{d}x = (19.2)(5.2) = 99.84\ \text{kN}^{2}\cdot\text{m}^{3}$$
Assemble and divide by EI.
$$\Delta = \frac{1}{EI}\int M m\,\mathrm{d}s
= \frac{20.48 + 99.84}{4000} = \frac{120.32}{4000} = 0.03008\ \text{m}$$
$$\Delta = \boxed{30.1\ \text{mm downward}}$$
The result is positive, so it acts in the direction of the unit load, downward,
as physical sense demands for a frame sagging under two symmetric loads.
