Question 4 of 7: B1. Design of member ABC in the concrete frame (6 + 8 + 6 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2014 — 98-Civ-A2
Elementary Structural Design, 3 hours, closed book (handbooks and textbooks permitted).
Seven questions in three parts: Part A (A1–A3, steel), Part B (B1–B3, reinforced concrete)
and Part C (C1, timber). The candidate answers two from Part A, two from Part B and the single
Part C question — five solutions in all, of equal value. All seven are solved here,
because the set is intended as a study resource. Page 1 states that
all loads shown are unfactored, so the load factors are applied below rather than assumed
to be built in.
CISC, Handbook of Steel Construction — section property tables and the standard
beam load/moment diagrams referenced by Question A1.
CSA A23.3, Design of Concrete Structures — Clauses 10 (flexure and axial load) and
11 (shear, simplified method).
CSA O86, Engineering Design in Wood — Clause 6 (sawn lumber), Tables 6.3.1B
(Beam and Stringer specified strengths) and 6.4.2 (service-condition factors).
National Building Code of Canada — load combinations
1.25D + 1.5L.
Check: the figures on page 3 of the paper are hand-drawn; every dimension, support type and load below was read from the printed figures. The 300 kN, 250 kN, 300 kN and 80 kN forces are treated as
live load (factor 1.5) because the paper labels them only as unfactored applied loads;
concrete self-weight is the sole dead load and carries 1.25.
Question 4 — B1. Design of member ABC in the concrete frame (6 + 8 + 6 marks)
Find. Cross-sectional dimensions for ABC together with the flexural and shear
reinforcement, and their layout.
Approach. The internal hinge at B makes the frame determinate: take moments about
the hinge on the free body A–B to get the roller reaction, carry the hinge shear across into
B–C to get the hogging moment at C, size a section for that moment, then check shear with the
A23.3 simplified method. Self-weight is added once the section is chosen and the analysis repeated.
Factor the loads and choose a trial section. The applied loads are live, so
$$P_{f,1} = 1.5(250) = 375\ \text{kN}, \qquad P_{f,2} = 1.5(300) = 450\ \text{kN}, \qquad
H_f = 1.5(80) = 120\ \text{kN}$$
Trying $b = 450$ mm, $h = 750$ mm, the factored self-weight is
$w_f = 1.25(24)(0.45)(0.75) = 10.13$ kN/m.
Use the hinge to find the roller reaction. Taking moments about B for the free
body A–B (the hinge transmits no moment):
$$A_y(3.0) = 375(1.5) + 10.13(3.0)(1.5) \;\Longrightarrow\; A_y = \frac{562.5 + 45.6}{3.0} = 202.7\ \text{kN}$$
The shear delivered across the hinge into segment B–C is then
$V_B = 375 + 10.13(3.0) - 202.7 = 202.7$ kN acting downwards.
Build the moment diagram for ABC. Between A and the point load the moment rises
to
$$M_{sag} = 202.7(1.5) - \tfrac{10.13(1.5)^2}{2} = 292.6\ \text{kN}\cdot\text{m}$$
falls to zero at the hinge B, then hogs over B–C under the transferred shear plus self-weight:
$$M_C = 202.7(3.0) + 10.13(3.0)(1.5) = 653.6\ \text{kN}\cdot\text{m} \quad \text{(hogging, tension on top)}$$
$$\boxed{M_{f,hog} = 653.6\ \text{kN}\cdot\text{m at C}, \quad M_{f,sag} = 292.6\ \text{kN}\cdot\text{m},
\quad V_{f} = 233.1\ \text{kN at C}}$$
The 450 kN at C is applied at the joint and passes straight into the column, so it does not appear in
the beam shear.
Size the top steel for the hogging moment. With 40 mm cover, 10M stirrups and
30M bars, $d = 750 - 40 - 11.3 - 14.95 = 683.8$ mm. Equating resistance to demand through
$$M_r = \phi_s A_s f_y\left(d - \frac{a}{2}\right), \qquad a = \frac{\phi_s A_s f_y}{\alpha_1\phi_c f_c' b}$$
gives $A_s = 3\,083$ mm$^2$. Providing 5–30M ($A_s = 3\,500$ mm$^2$) in one
layer, $a = 165.3$ mm and
$$\boxed{M_r = 726.9\ \text{kN}\cdot\text{m} > 653.6\ \text{kN}\cdot\text{m}}$$
with $c/d = 0.242$, comfortably below the 0.636 limit, so the section is ductile and tension-
controlled. Five 30M bars need $5(29.9) + 4(1.4)(29.9) = 317$ mm of clear width against the 347 mm
available inside the stirrups — they fit in a single layer, which is why $b$ was raised from 400
to 450 mm.
Size the bottom steel for the sagging moment. Repeating with 25M bars
($d = 686.1$ mm) requires 1 307 mm$^2$; provide 3–25M ($A_s = 1\,500$ mm$^2$),
giving $M_r = 334.0$ kN·m against 292.6 kN·m. Minimum steel,
$A_{s,min} = 0.2\sqrt{f_c'}\,b h/f_y = 998$ mm$^2$, is satisfied by both faces.
Check shear by the simplified method. With
$d_v = \max(0.9d,\,0.72h) = 615.4$ mm and minimum stirrups present, $\beta = 0.18$ and
$$V_c = \phi_c \lambda \beta \sqrt{f_c'}\, b_w d_v = 0.65(1.0)(0.18)(5.916)(450)(615.4) = 191.7\ \text{kN}$$
which falls short of the 233.1 kN demand, so stirrups must carry the balance. Using 10M closed
stirrups ($A_v = 200$ mm$^2$) at 300 mm with $\theta = 35^\circ$,
$$V_s = \frac{\phi_s A_v f_y d_v \cot\theta}{s} = \frac{0.85(200)(400)(615.4)(1.428)}{300} = 199.2\ \text{kN}$$
$$\boxed{V_r = 191.7 + 199.2 = 390.9\ \text{kN} > 233.1\ \text{kN}}$$
The spacing also satisfies $s \le \min(0.7d_v, 600) = 431$ mm, and
$A_{v,min} = 0.06\sqrt{f_c'}\,b_w s/f_y = 120$ mm$^2 < 200$ mm$^2$.
The governing action is plainly the hogging moment at C, which is more than twice the sagging
moment, because the cantilever action of segment B–C carries the whole hinge shear over a 3 m
lever arm. The reinforcement layout follows that diagram directly: top steel concentrated over the
joint and lapped well past the point of contraflexure at B, bottom steel through the A–B region.
