Question 8 of 8: Deflection of a Strut-Propped Beam by Virtual Work
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 07-Str-A1 Elementary Structural Analysis,
National Examinations, May 2017. Three hours, CLOSED BOOK (an approved Sharp or
Casio calculator is permitted). Page 1 states that six questions
constitute a complete paper: Questions 1–5 are compulsory and the candidate
answers one only of Questions 6, 7 or 8. All eight questions are worked
below.
Sign conventions used throughout. Sagging bending moment is
positive and is plotted above the axis; shear is the sum of the upward forces on
the portion to the left of the section. Truss member forces are quoted as
T for tension and C for compression. In the slope-deflection
work of Question 7 the member-end moments follow the usual convention —
clockwise on the member end taken as positive — so the sagging moment is
$+M_{ij}$ at the left end of a member and $-M_{ji}$ at its right end. For the
bent members of Questions 2(c) the diagrams are developed: the abscissa
is distance measured along the member axis, not the horizontal projection.
A(0,0) pin support → B(4,0) → C(6,0) free end, all on one line
Flexural rigidity
A–B: $2EI$; B–C: $EI$, with $EI = 3500\ \text{kN}\,\text{m}^{2}$
Strut
from the wall pin D(0,3) down to B(4,0); length $\sqrt{4^2+3^2} = 5$ m, $AE = 25000$ kN
Loading
uniformly distributed load over B–C; the magnitude is not printed on the examination figure
Find. The vertical deflection at C, including the strut's
axial flexibility.
Question 8: the beam propped at B by the strut DB. The load intensity on B–C is not given on the paper, so the answer is carried symbolically in $w$.
Check — missing datum. The
examination figure shows the load block over B–C but carries no intensity
label, and none appears in the question text. The deflection is therefore derived
as an exact multiple of the intensity $w$ (kN/m), and a worked numerical case at
$w = 10$ kN/m is given at the end. Substitute the printed value if a legible copy
of the paper is available; the coefficient is unaffected. Note also that D lies
directly above A, so the strut geometry is the 3–4–5 triangle
$4$ m by $3$ m.
Approach. The structure is determinate (pin at A gives two
components, the two-force strut one, total three). Find the strut force and the
real moment diagram; repeat with a 1 kN dummy load at C; then add the bending
term $\int Mm/EI$ and the axial term $NnL/AE$ for the strut.
Strut force in the real system. The strut is a two-force
member along DB, so its unit vector at B pointing toward D is $(-0.8,\,0.6)$.
The UDL resultant is $2w$ acting at $x = 5$ m. Taking moments about A,
$$\sum M_A = 0:\qquad 0.6S(4) = 2w(5)
\;\Rightarrow\; S = \frac{10w}{2.4} = \frac{25w}{6} = 4.1667w$$
positive, i.e. tension: although the member is labelled STRUT, the load
sits beyond the prop, so DB has to hold B up.
Reactions at A.
$$A_y = 2w - 0.6(4.1667w) = -0.75w \ \ (\text{i.e. } 0.75w \ \text{downward}),
\qquad A_x = 0.8(4.1667w) = 3.3333w$$
The downward pin reaction is the expected hold-down for a beam levered up at
its interior prop.
Real bending moments. Nothing but $A_y$ acts on
A–B, so the moment there is linear; on the overhang B–C only the UDL
acts. With $x$ measured from A and $s$ from C,
$$M(x) = -0.75w\,x \quad (0 \le x \le 4), \qquad
M(s) = -\tfrac{1}{2}w s^{2} \quad (0 \le s \le 2)$$
Both give $-2w$ at B, so the diagram is continuous; the beam hogs throughout.
Virtual system. Remove the UDL and apply 1 kN downward at
C, 6 m from A. The same moment equation gives
$$0.6\,s_{\text{strut}}(4) = 1(6) \;\Rightarrow\; s_{\text{strut}} = 2.5,
\qquad a_y = 1 - 0.6(2.5) = -0.5$$
$$m(x) = -0.5x \quad (0 \le x \le 4), \qquad m(s) = -s \quad (0 \le s \le 2)$$
Bending contribution, B–C (rigidity $EI$).
$$\int_0^2 \frac{Mm}{EI}\,\mathrm{d}s
= \frac{1}{EI}\int_0^2 \tfrac{1}{2}w s^{3}\,\mathrm{d}s
= \frac{1}{EI}\left(\tfrac{1}{2}w\cdot 4\right) = \frac{2w}{EI}$$
Together the bending terms give
$$\frac{1}{EI}\left(\frac{8}{3}+2\right)w = \frac{14w}{3EI}
= \frac{4.6667w}{3500} = 1.3333\times10^{-3}\,w \ \text{m}$$
Strut contribution. The strut is the one member whose axial
flexibility is asked for:
$$\frac{NnL}{AE} = \frac{(4.1667w)(2.5)(5)}{25000}
= \frac{52.083w}{25000} = 2.0833\times10^{-3}\,w \ \text{m}$$
It is larger than the whole bending term — with $AE$ only
25 000 kN the tie stretches about 0.83 mm per kN/m of load, and the lever ratio
$6/4$ turns that into 1.25 mm of tip movement per kN/m.
Total. Adding the two,
$$\delta_C = \left(\frac{14}{3EI} + \frac{25\cdot2.5\cdot5}{6\,AE}\right)w
= \left(\frac{1}{750} + \frac{1}{480}\right)w
= \frac{41w}{12000}$$
$$\boxed{\delta_C = 3.417\,w\ \text{mm downward}\quad (w \text{ in kN/m})}$$
For an illustrative $w = 10$ kN/m this is $\boxed{34.2\ \text{mm downward}}$;
for $w = 20$ kN/m, 68.3 mm.
Check the strut term kinematically. The strut elongation is
$e = SL/AE = 4.1667w(5)/25000 = 8.333\times10^{-4}w$ m. Dividing by the vertical
direction cosine $0.6$ gives the drop of B, $1.3889\times10^{-3}w$ m, and scaling
by the lever ratio $6/4$ about the pin at A gives
$2.0833\times10^{-3}w$ m at C — identical to the $NnL/AE$ term, which
confirms both the strut force and the virtual force.
Quantity
Value
Strut force (real system)
$S = 4.167\,w$ kN, tension
Reaction at A
$A_y = 0.75\,w$ kN downward, $A_x = 3.333\,w$ kN
Bending contribution to $\delta_C$
$1.333\times10^{-3}\,w$ m (28.6% of the total)
Strut contribution to $\delta_C$
$2.083\times10^{-3}\,w$ m (50.8% — the larger share)
Vertical deflection at C
$\delta_C = \dfrac{41w}{12000}$ m $= \mathbf{3.417\,w}$ mm downward
