16-Civ-A2 Elementary Structural Design · May 2015

Question 6 of 7: B3 — Design of beam BC

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, May 2015 — 98-Civ-A2 Elementary Structural Design, 3-hour duration, closed book (textbooks and design handbooks permitted; one approved Casio or Sharp calculator). Seven questions in three parts: Part A (A1–A3, structural steel, CSA S16), Part B (B1–B3, reinforced concrete, CSA A23.3) and Part C (C1, timber, CSA O86). A candidate answers two from Part A, two from Part B and the one question in Part C — five solutions, all of equal value. All seven are solved here so the set works as a complete study resource.

Reference texts. CSA S16, Design of Steel Structures, with the CISC Handbook of Steel Construction; CSA A23.3, Design of Concrete Structures, with the CAC Concrete Design Handbook; CSA O86, Engineering Design in Wood, with the CWC Wood Design Manual; National Building Code of Canada (load combinations); Kulak & Grondin, Limit States Design in Structural Steel; MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design.

Check — load factors. Note 6 on page 1 states “All loads shown are unfactored”, but the paper never splits the figure loads into dead and live components. Every applied figure load in Parts A and B is therefore treated as live load and factored by 1.5; concrete self-weight, where it is not expressly excluded, would be dead load at 1.25 (NBCC 4.1.3.2, combination 1.25D + 1.5L). Part C names its components, so 1.25D + 1.5L is applied directly there. If an examiner intended a different split, the method below is unchanged — only the numerical factor moves.

Check — figure dimensions. All four figures are hand sketches on page 3; every dimension below was read from the printed figures. Figure A2 reads 80 – 300 – 80 across the bottom plate (overall 460 mm wide) with the 300 mm dimension spanning the two web centrelines, and 250 mm overall depth. Figure B1 reads 3.0 m top, 2.5 m bottom, 2.0 m deep, walls “300 constant” and cover “70 typical”. Figure B2 reads 5 m + 5 m span, 4 m + 4 m column height, roller at A and hinge at D.

Question 6: B3 — Design of beam BC (12 + 8 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Span BC	10.0 m (5 + 5), rigidly joined to columns AB and CD
Factored loads	525 kN at midspan; end shears delivered from the frame analysis
Factored moments (from B2)	480 kN·m hogging at B, 592.5 kN·m sagging at midspan, 960 kN·m hogging at C
Factored shear	214.5 kN in the left half, 310.5 kN in the right half
Materials	f'_c = 35 MPa, f_y = 400 MPa

Find. A rectangular cross-section, the flexural steel at each critical location, the shear reinforcement, and a drawing of the bar arrangement.

Factored bending-moment diagram for beam BC (kN·m). The end moments equal the moments carried up the adjoining columns: 120 × 4 at B and 120 × 8 at C.

Approach. Recover the beam moment diagram from the frame equilibrium already established in B2, size the section for the largest moment (hogging at C), then compute the steel at each of the three critical sections and design the stirrups from the peak shear.

Beam moments from frame equilibrium. Measuring x from B and taking sagging as positive, the moment at any cut follows from the forces to the left: $$ M(x) = A_y x - 600x - 120(4) - 525\langle x - 5\rangle $$ which gives −480 kN·m at B, +592.5 kN·m at midspan and −960 kN·m at C. The moment at each end is exactly the moment carried up the adjoining column — 120 × 4 at B and 120 × 8 at C — which is the check that the analysis closes. The hogging moment at C is nearly twice that at B because column CD is the one that carries the full 8 m lever of the 120 kN horizontal thrust.
Shear diagram. The end shear at B is $A_y - 600 = 214.5$ kN, and past the midspan load it becomes $214.5 - 525 = -310.5$ kN. The design shear is therefore $$ V_f = 310.5~\text{kN adjacent to C} $$
Select the section. Try 500 mm wide × 900 mm deep. With 40 mm cover, 10M stirrups and 30M bars, $$ d = 900 - 40 - 11 - 15 = 834~\text{mm} $$ This gives a span-to-depth ratio of 11, appropriate for a heavily loaded transfer beam, and a width that will accept six 30M bars in one layer.
Flexural steel at C (the governing section). Iterating $A_s = M_f/[\phi_s f_y(d - a/2)]$ with $a = \phi_s A_s f_y/(\alpha_1\phi_c f'_c b)$ converges to $A_s = 3693~\text{mm}^2$. Provide 6–30M = 4200 mm²: $$ a = \frac{0.85(4200)(400)}{0.7975(0.65)(35)(500)} = 157.4~\text{mm} $$ $$ M_r = 0.85(4200)(400)\left(834 - 78.7\right)\times10^{-6} = \boxed{1079~\text{kN}\cdot\text{m}} \;>\; 960 $$ The neutral axis is at $c = a/\beta_1 = 178$ mm, so $c/d = 0.21$ — well under-reinforced and ductile.
Flexural steel at midspan and at B. The same iteration gives 2199 mm² required at midspan and 1763 mm² at B. Provide 4–30M (2800 mm²) bottom at midspan and 3–30M (2100 mm²) top at B, for which $M_r = 744$ and $567$ kN·m respectively — both comfortably above their demands. Minimum steel is $$ A_{s,\min} = \frac{0.2\sqrt{f'_c}\,b_t h}{f_y} = \frac{0.2(5.92)(500)(900)}{400} = 1331~\text{mm}^2 $$ which every section satisfies.
Check that the top bars fit in one layer. Across a 500 mm width, six 30M bars inside 40 mm cover and 10M stirrups leave $$ \text{clear space} = 500 - 2(40) - 2(11) - 6(29.9) = 218~\text{mm} $$ Divided into five gaps this is 43.6 mm, which exceeds the A23.3 Clause 6.6.5.2 minimum of the greater of $1.4 d_b = 41.9$ mm and 30 mm. A single layer is therefore possible and the full effective depth is preserved.
Shear design. With $d_v = \max(0.9 \times 834,\, 0.72 \times 900) = 751$ mm and at least minimum transverse steel present, the simplified method gives $\beta = 0.18$, $\theta = 35^\circ$: $$ V_c = 0.65(0.18)(5.92)(500)(751)\times10^{-3} = 260~\text{kN} $$ $$ V_s\text{ required} = 310.5 - 260 = 51~\text{kN} $$ The stirrup demand is small, so the spacing is governed by the detailing rules rather than by strength. Since $V_f = 310.5$ kN is well below $0.125\phi_c f'_c b_w d_v = 1067$ kN, the maximum spacing is $$ s_{\max} = \min(0.7 d_v,\; 600) = 525~\text{mm} $$ and the minimum area requirement at 350 mm spacing is $A_{v,\min} = 155~\text{mm}^2$. Adopt 10M closed stirrups at 350 mm ($A_v = 200~\text{mm}^2$), giving $$ V_r = 260 + 208 = 468~\text{kN} \;>\; 310.5~\text{kN} $$
Arrange the steel. Run the 3–30M top bars continuously from B to C and lap in the additional 3–30M over the right half so that six bars are present at C; run 4–30M bottom bars continuously, with at least two carried into each joint for integrity. Curtail the extra top bars no closer to the support than the point of contraflexure plus a development length; from the moment equation the contraflexure points are at 2.24 m and 6.91 m from B.

Bar arrangement for beam BC at the section adjacent to C. The six top bars fit in one layer with 43.6 mm clear spacing, preserving the full 834 mm effective depth.

Check — self-weight. Question B2 instructs that self-weight be ignored, and the same instruction is carried into B3 because both questions refer to the one frame. For information, a 500 × 900 beam weighs 10.8 kN/m, which at 1.25 would add roughly 110 kN·m to the support moments — about 11 percent at C. The 6–30M selected at C carries 1079 kN·m against a demand that would rise to about 1070 kN·m, so the section as chosen still works, but with no reserve. A designer not bound by the exam instruction should either deepen the beam to 950 mm or add a seventh bar at C.

Location	M_f (kN·m)	A_s required (mm²)	Bars provided	M_r (kN·m)
B (hogging, top steel)	480	1763	3–30M	567
Midspan (sagging, bottom)	592.5	2199	4–30M	744
C (hogging, top steel)	960	3693	6–30M	1079
Section 500 mm × 900 mm, d = 834 mm; A_s,min = 1331 mm²
Shear (adjacent to C)	V_f = 310.5 kN		10M at 350 mm	V_r = 468 kN
Contraflexure points	2.24 m and 6.91 m from B — curtail the added top bars beyond these plus l_d

Topic. Flexural and shear design of a continuous-behaviour rectangular beam in a rigid frame.

Key relations. $A_s = M_f/[\phi_s f_y (d - a/2)]$ solved iteratively with $a = \phi_s A_s f_y/(\alpha_1\phi_c f'_c b)$; $A_{s,\min} = 0.2\sqrt{f'_c}b_t h/f_y$; $V_c = \phi_c\lambda\beta\sqrt{f'_c}b_w d_v$; $V_s = \phi_s A_v f_y d_v\cot\theta/s$; $s_{\max} = \min(0.7d_v, 600)$.

Why this works. Although the frame is statically determinate, the beam still behaves as a continuous member because the rigid joints at B and C force the column moments into it. Those end moments are not something extra to guess at — they follow directly from the column statics, and computing them twice (once up the column, once along the beam) is the standard check that the analysis is consistent. The asymmetry in the answer, with C carrying twice the moment of B, traces entirely to the horizontal load being applied at 4 m rather than at the top: column AB has only a 4 m lever above the load while column CD has the full 8 m. Once the diagram is right, the flexural design is the ordinary singly-reinforced iteration, and the shear design turns out to be governed by detailing rules rather than strength — a common outcome in deep beams where d_v is large.

Common pitfall. Treating BC as a simply supported beam with a midspan point load, which would give a single sagging moment of 1312 kN·m and no top steel at all — catastrophically wrong at the supports where the real member hogs. A second pitfall is checking only the largest moment and detailing that steel everywhere, which wastes reinforcement and misses the requirement to show the arrangement. A third is forgetting to confirm the bars fit in one layer; two layers at C would drop d by about 60 mm and cost roughly 7 percent of the capacity.

Source. CSA A23.3, Design of Concrete Structures, Clauses 6.6.5, 10.1, 10.5, 11.2 and 11.3; CAC Concrete Design Handbook, Chapters 2 and 4; MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design, Chapters 5 and 6.

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