07-Str-A2 · December 2014

Question 5 of 7: B2 — Design of beam AB in a determinate concrete frame

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

Notes on this paper

Paper format. National Examinations, December 2014 — 07-Str-A2 Elementary Structural Design, 3-hour duration, closed book (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, 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 in total, 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, unless otherwise specified”, but the paper never splits the figure loads into dead and live components. Throughout Parts A and B the applied figure loads are therefore treated as live load, factored by 1.5, and self-weight of concrete members as dead load, factored by 1.25 (NBCC 4.1.3.2, combination 1.25D + 1.5L). In Part C the paper does name the components, so 1.25D + 1.5S is used directly. If an examiner intended a different split the method is unchanged — only the numerical factor moves.

Check — figure dimensions. Every figure on this paper is a hand sketch on page 3 whose OCR text is unusable; all geometry below was read from the printed figure of the original PDF. Figure A1 is not drawn to scale horizontally (its printed 600 and 300 dimensions are authoritative, not the drawn proportions). In Figure B1 the dimension “1 m × 1 m” runs from the section centreline to the outer face, so each cell is 1 m square and the box is 2000 mm wide × 1000 mm deep; the “600 × 600” dimension line brackets the void, giving 200 mm walls and a 400 mm central web. That reading is confirmed independently by the “65 typical” cover note: 65 + 65 cover plus two 15M tie legs plus one 30M bar needs exactly 200 mm of wall.

Question 5: B2 — Design of beam AB in a determinate concrete frame (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. An L-shaped frame: horizontal beam AB of 6 m (two 3 m bays) with a roller at A, monolithic with column BC which drops 6 m to a pin at C. Loads as printed are unfactored.

Item	Location	Value
Point load, downward	mid-span of AB (3 m from A)	300 kN
Point load, downward	joint B	400 kN
Point load, horizontal →	3 m below B on column BC	60 kN
Support A	—	roller (vertical only)
Support C	—	pin
f'_c / f_y	—	35 / 400 MPa

Find. A rectangular section for AB plus its flexural and shear reinforcement, with the bar layout drawn.

Figure B2 — the determinate frame ABC. The roller at A supplies only a vertical reaction, so the whole 60 kN horizontal load is carried at the pin C.

Approach. The frame has three reaction components and is statically determinate, so solve it by equilibrium alone; trial a section, add its self-weight, re-solve, then size the flexural steel from the stress block and the stirrups from the A23.3 simplified shear method.

Reactions from the printed (unfactored) loads. Placing the origin at C, with B at (0, 6) and A at (−6, 6), moments about C give $$300(3.0) - 60(3.0) - 6A_y = 0 \;\Rightarrow\; A_y = 120\ \text{kN}\ \uparrow$$ $$C_y = 300+400-120 = 580\ \text{kN}\ \uparrow \qquad C_x = 60\ \text{kN}\ \leftarrow$$
The joint moment at B. Working along the beam, $$M_B = A_y(6.0) - 300(3.0) = 720-900 = -180\ \text{kN}\cdot\text{m}$$ Reading up the column instead gives $-60(6.0)+60(3.0) = -180$ kN·m — the same value, as it must be, and confirming the frame solution. Here the beam moment at B is not zero, because the horizontal load in the column feeds a real moment into the joint.
Trial section and self-weight. Try 500 × 900 mm: $$w_{sw} = 0.5(0.9)(24) = 10.8\ \text{kN/m} \;\Rightarrow\; w_f = 1.25(10.8) = 13.5\ \text{kN/m}$$
Factored analysis. With the point loads at 1.5 and the self-weight at 1.25, moments about C give $$A_{y,f} = \frac{450(3.0)+13.5(6.0)(3.0)-90(3.0)}{6.0} = 220.5\ \text{kN}$$ so that along the beam $$M_f(3.0) = 220.5(3.0)-\tfrac{13.5(3.0)^2}{2} = \boxed{601\ \text{kN}\cdot\text{m}\ \text{(sagging)}}$$ $$M_f(6.0) = 220.5(6.0)-\tfrac{13.5(6.0)^2}{2}-450(3.0) = \boxed{-270\ \text{kN}\cdot\text{m}\ \text{(hogging at B)}}$$ The shear changes sign at mid-span ($+180$ kN to $-270$ kN), confirming that the peak sagging moment sits exactly under the 300 kN load.
Effective depth. With 40 mm cover, 10M stirrups and 30M main bars in a single layer, $$d = 900-40-11.3-\tfrac{29.9}{2}=833.75\ \text{mm}$$
Bottom steel for the sagging moment. Solving $\phi_sA_sf_y(d-a/2)=M_f$ together with $a = \phi_sA_sf_y/(\alpha_1\phi_cf'_cb)$: $$A_s = 2231\ \text{mm}^2 \;\Rightarrow\; \boxed{4\text{-}30\text{M} = 2800\ \text{mm}^2}$$ The four bars fit in one layer with 92.6 mm clear spacing, well above the $\max(1.4d_b, 30\ \text{mm}) = 41.9$ mm minimum. The depth to the neutral axis is $c/d = 0.114$, far inside the ductility limit of 0.5.
Top steel for the hogging moment at B. Strength alone needs only 974 mm², but A23.3 10.5.1.2 requires $$A_{s,\min}=\frac{0.2\sqrt{f'_c}}{f_y}b_th = \frac{0.2\sqrt{35}}{400}(500)(900) = 1331\ \text{mm}^2$$ so the minimum governs: provide 2-30M = 1400 mm² continuous through the joint and anchored into the column.
Shear design. The critical section sits $d_v$ from the face of the joint, where $$d_v = \max(0.9d,\ 0.72h)=750.4\ \text{mm} \qquad V_f = 300\ \text{kN}$$ $$V_c=\phi_c\lambda\beta\sqrt{f'_c}\,b_wd_v = 0.65(1.0)(0.18)\sqrt{35}(500)(750.4)=260\ \text{kN}$$ $$V_s = V_f-V_c = 40.7\ \text{kN}$$ The strength requirement alone would allow a spacing of 1.8 m, so the minimum-reinforcement rule controls: $$s \le \frac{A_vf_y}{0.06\sqrt{f'_c}\,b_w}=\frac{200(400)}{0.06\sqrt{35}(500)}=451\ \text{mm}, \qquad s_{\max}=\min(0.7d_v,600)=525\ \text{mm}$$ $$\Rightarrow\ \boxed{\text{10M closed stirrups at 300 mm throughout}}$$
Reinforcement layout. Bottom: 4-30M running the full length of AB, with at least two bars continuous into the support regions. Top: 2-30M continuous, lapped over the mid-span and fully anchored around the corner into column BC to deliver the 270 kN·m hogging moment into the joint. Stirrups: 10M closed hoops at 300 mm, tightened to 150 mm over the 900 mm adjacent to B where the joint reinforcement is congested.

The beam is governed by the mid-span sagging moment, not by the joint, and the self-weight matters more than it looks: it adds 45 kN·m (about 8 %) to the peak sagging moment and pushes the bar requirement from three 30M bars to four. Note too that the hogging steel is set entirely by the code minimum — a common outcome in stocky beams, and a reminder that the 500 × 900 section is generous for the hogging demand even while it is nearly fully worked in sagging.

Quantity	Value
Reactions (unfactored)	A_y = 120 kN, C_y = 580 kN, C_x = 60 kN
Joint moment at B (unfactored)	180 kN·m hogging
Section	500 × 900 mm, d = 833.75 mm
Factored sagging moment	601 kN·m at mid-span
Factored hogging moment	270 kN·m at B
Bottom steel	4-30M (2800 mm² > 2231 required)
Top steel	2-30M (1400 mm² > 1331 minimum)
Factored shear at d_v	300 kN (V_c = 260 kN)
Stirrups	10M closed at 300 mm (150 mm near B)

Topic: Analysis and flexure/shear design of a beam in a statically determinate concrete frame.

Key relations: determinacy check $r = 3$; $M_f$ from equilibrium; $a=\phi_sA_sf_y/(\alpha_1\phi_cf'_cb)$ and $M_r=\phi_sA_sf_y(d-a/2)$; $A_{s,\min}=0.2\sqrt{f'_c}b_th/f_y$; $c/d \le 0.5$ for ductility; $V_c$ and $s_{\max}$ as above.

Why this works: The roller at A supplies one reaction and the pin at C supplies two, giving exactly the three equations of planar statics — so the frame can be solved without any reference to stiffness, section size or reinforcement. That is what makes the problem tractable by hand, and it is also why the section can be chosen freely: changing it alters only the self-weight, not the distribution of the applied actions. Within the beam, the rectangular stress block converts a nonlinear concrete stress distribution into a single resultant, so flexural design reduces to a quadratic in the steel force. The 1.25 factor on self-weight and 1.5 on the applied loads follow the NBCC principal-load combination.

Common pitfall: Assuming the beam moment at B is zero. With a roller at A and a pin at C that is true for purely vertical loading, but the 60 kN horizontal load on the column breaks the symmetry and delivers a genuine 180 kN·m into the joint. Skipping it leaves the top of the beam unreinforced where it is most needed. A second common slip is omitting self-weight from a beam this size, which here would silently cost a bar.

Source: CSA A23.3, Design of Concrete Structures, Clauses 10.1, 10.5, 11.2 and 11.3; National Building Code of Canada, Clause 4.1.3.2.

← Question 4 Next: Question 6 →