Question 5 of 7: Moment and shear resistances of a concrete culvert section
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2013 — 07-Str-A2,
Elementary Structural Design. Three hours; a "CLOSED BOOK" examination in which handbooks
and textbooks are permitted. Seven questions in three parts — Part A steel (CAN/CSA-S16),
Part B reinforced concrete (CAN/CSA-A23.3), Part C timber (CAN/CSA-O86). A candidate answers two
questions from Part A, two from Part B and the single question in Part C, five in all, and page 1
records that all questions are of equal value, so each carries 20 marks with the split
A1 (6 + 7 + 7), A2 (5 + 10 + 5), A3 (4 + 4 + 12),
B1 (6 + 8 + 6), B2 (10 + 10), B3 (4 + 4 + 12),
C1 (10 + 10) printed in the marking scheme. All seven questions are solved
below.
Reference texts. CSA S16, Design of Steel Structures, used with the
CISC Handbook of Steel Construction (section tables, Class H HSS, bolt and weld tables);
CSA A23.3, Design of Concrete Structures, used with the CAC Concrete Design Handbook;
CSA O86, Engineering Design in Wood, used with the CWC Wood Design Manual;
National Building Code of Canada (load combinations); MacGregor and Bartlett,
Reinforced Concrete: Mechanics and Design (Canadian edition); Kulak and Grondin,
Limit States Design in Structural Steel.
Check — load factors used throughout. Page 1 note 6 states that all loads
shown are unfactored but the paper nowhere splits them into dead and live. Every applied load in
Figures A2, B1 and B3 is therefore treated as a specified live load and factored by
1.5, and self-weight, where a question asks for it, by
1.25, following NBCC load combination case 2,
1.25D + 1.5L. Question B3 names its 80 kN horizontal load as
wind, so that question is additionally checked under case 4,
1.25D + 1.4W + 0.5L. If a grader intends a different dead/live split
every factored action scales linearly and no design step below changes.
Check — steel grade in Figure A1. The A1 stem attaches "G40.21 300W" to
the plates and is silent on the grade of the hollow section. All three components are taken as
300W, Fy = 300 MPa; if the HSS is
in fact 350W the plates still yield first and the plastic moments quoted rise by less than 8 per cent.
Corner radii of the hollow section are ignored (square corners), which is the usual hand-calculation
idealisation and overstates the HSS area by about 4 per cent relative to the tabulated value.
Question B2: Moment and shear resistances of a concrete culvert section (20 marks: 10 + 10)
Given. A trough-shaped culvert 2 m deep, 4 m wide at the top and
3 m wide at the invert, with walls 300 mm thick, main steel in the base slab and a bar at the
head of each wall.
Given data (Figure B2 and the question stem)
Quantity
Value
Overall depth
2 000 mm
Width, top / invert
4 000 / 3 000 mm
Wall and base thickness
300 mm, constant
Cover to bar centres
70 mm typical
Tension steel (invert)
6 – 30M (As = 4 200 mm2)
Compression steel (wall heads)
2 – 20M (As' = 600 mm2)
Transverse steel
15M stirrups at 200 mm (one leg per wall)
Materials
fc' = 35 MPa, fy = 400 MPa
Find. The factored moment of resistance and the factored shear resistance of the
section.
Figure B2 — the culvert section. Read as a
longitudinally spanning trough, it is a very deep flanged beam: the invert slab is the tension flange
and the two inclined walls are the webs.
Check — how the section is being read. The
reinforcement arrangement settles the interpretation: main steel concentrated in the invert, a single
bar at the head of each wall, and stirrups running up the walls are the signature of a
longitudinally spanning trough, that is, a channel-shaped beam 2 m deep whose tension
flange is the base slab. The alternative reading — a transversely spanning box in which the base
slab is a one-way slab — would put the 6 – 30M bars across the 3 m width and
give, per metre of culvert length, $M_r = 103\ \text{kN}\cdot\text{m/m}$ and
$V_r = 219\ \text{kN/m}$. Those figures are quoted here for completeness; the trough reading is the one
carried through, because it is the only one for which the 2 – 20M wall-head bars and the
wall stirrups have a structural role. The stirrup is taken as a single closed loop following the trough
perimeter, so it crosses each wall once and provides two legs, $A_v = 400\ \text{mm}^2$.
Approach. Establish the geometry of the compression zone at the top of the two
inclined webs, solve horizontal equilibrium by strain compatibility including the compression steel,
take moments about the tension steel for Mr, and then apply the CSA A23.3 simplified shear
method to the two webs.
(a) Moment of resistance (10 marks)
Geometry of the compression zone. Each wall leans out by
$(4\,000 - 3\,000)/2 = 500$ mm over the 2 000 mm height, so its inclination from the vertical
is $\theta = \arctan(500/2\,000) = 14.04^\circ$. A wall 300 mm thick measured perpendicular to its
own face therefore has a horizontal width of $300/\cos\theta = 309.2$ mm, and the two walls
together present
$$b = 2(309.2) = 618.5\ \text{mm}$$
of concrete to a horizontal compression block near the top. The effective depth to the invert steel is
$d = 2\,000 - 70 = 1\,930\ \text{mm}$, and the wall-head bars sit at $d' = 70\ \text{mm}$.
Horizontal equilibrium with compression steel. The effective concrete stress is
$\alpha_1\phi_cf_c' = 0.7975(0.65)(35) = 18.14\ \text{MPa}$. Writing
$$T = \alpha_1\phi_cf_c'\,b\,a + \phi_sA_s'f_s' - \alpha_1\phi_cf_c'A_s'$$
with $f_s' = 0.0035\,E_s(c-d')/c$ limited to $f_y$, and $c = a/\beta_1$, and iterating gives
$$a = 113.7\ \text{mm}, \qquad c = 128.8\ \text{mm}, \qquad f_s' = 320\ \text{MPa}$$
so the wall-head bars are stressed but do not quite yield. The two compressive forces are
$C_c = 1\,275.8$ kN and $C_s = 152.2$ kN, which sum to 1 428 kN, closing the equilibrium
check. The strain in the invert steel is
$\varepsilon_s = 0.0035(1\,930 - 128.8)/128.8 = 0.049$, some twenty-four times the yield strain, so the
yielding assumption is amply justified and $c/d = 0.067$ confirms a very ductile, tension-controlled
section.
Take moments about the invert steel.
$$M_r = C_c\left(d - \frac{a}{2}\right) + C_s\left(d - d'\right)
= 1\,275.8(1\,873.1) + 152.2(1\,860)$$
$$\boxed{M_r = 2\,673\ \text{kN}\cdot\text{m}}$$
Neglecting the compression steel altogether would give
$2\,665\ \text{kN}\cdot\text{m}$, a difference of only 0.3 per cent — a useful reminder
that compression reinforcement adds ductility far more than it adds strength.
(b) Shear resistance (10 marks)
Effective shear depth and web width.
$$d_v = \max(0.9d,\ 0.72h) = \max(1\,737,\ 1\,440) = 1\,737\ \text{mm}$$
For shear the web width is taken as the two wall thicknesses measured perpendicular to the walls,
$b_v = 2(300) = 600\ \text{mm}$, which is 3 per cent conservative against the horizontal
projection used for flexure.
Concrete contribution. The walls carry 15M stirrups at 200 mm, well above the
minimum required by Clause 11.2.8.2, so the simplified method applies with $\beta = 0.18$ and
$\theta = 35^\circ$:
$$V_c = \phi_c\lambda\beta\sqrt{f_c'}\,b_vd_v = 0.65(1.0)(0.18)\sqrt{35}(600)(1\,737) = 721.4\ \text{kN}$$
Stirrup contribution. With two 15M legs, $A_v = 400\ \text{mm}^2$, at
$s = 200\ \text{mm}$,
$$V_s = \frac{\phi_sA_vf_yd_v\cot\theta}{s}
= \frac{0.85(400)(400)(1\,737)(1.4281)}{200} = 1\,686.9\ \text{kN}$$
The spacing is within the $0.7d_v = 1\,216$ mm and 600 mm limits of Clause 11.3.8.
Total, and the crushing check.
$$\boxed{V_r = V_c + V_s = 721.4 + 1\,686.9 = 2\,408\ \text{kN}}$$
The upper limit set by crushing of the diagonal struts,
$0.25\phi_cf_c'b_vd_v = 5\,928\ \text{kN}$, is not approached, so the computed value stands. Note that
the stirrups supply seventy per cent of the shear resistance: at this depth the concrete term, which
scales with $\sqrt{f_c'}$ and not with the reinforcement, is the smaller share.
