Question 2 of 7: Reinforced concrete frame — flexural and shear steel for BC, and its deflection
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 07-Str-A5 Advanced Structural Design, National Exams, December 2013 — 3 hours, closed book (handbooks and textbooks permitted, no notes). Seven questions of equal value; any five constitute a complete paper. All seven are solved here, because the set is intended as a study resource. All loads printed on the figures are unfactored.
Design data given on the paper (SI). Concrete $f'_c = 30$ MPa; structural steel $F_y = 350$ MPa; reinforcing steel $f_y = 400$ MPa. Prestressed concrete: $f'_{ci} = 35$ MPa at transfer, $f'_c = 50$ MPa, $n = 6$, $f_{pu} = 1750$ MPa, $f_{py} = 1450$ MPa, $f_{initial} = 1200$ MPa, losses $= 240$ MPa (hence $f_{pe} = 960$ MPa).
Reference texts. CSA S16 Design of Steel Structures with the CISC Handbook of Steel Construction — plate girders (Cl. 14), lateral–torsional buckling (Cl. 13.6), plastic design (Cl. 8.6, 13.7), beam-columns (Cl. 13.8), composite beams (Cl. 17), connections (Cl. 21); CSA A23.3 Design of Concrete Structures — flexure and shear (Cl. 10, 11), slenderness (Cl. 10.13–10.16), footings (Cl. 15), prestressed concrete (Cl. 18); NBCC for load combinations; C. G. Salmon, J. E. Johnson & F. A. Malhas, Steel Structures: Design and Behavior; J. G. MacGregor & J. K. Wight, Reinforced Concrete: Mechanics and Design; T. Y. Lin & N. H. Burns, Design of Prestressed Concrete Structures; L. S. Beedle, Plastic Design of Steel Frames.
Check: load factors. The paper states only that the printed loads are unfactored, and gives no dead/live split. Every question below therefore treats each printed concentrated load as specified live load and factors it by $\alpha_L = 1.5$ (NBCC principal case $1.25D + 1.5L$); member self-weight, wherever it matters, is estimated from the trial section and factored at $1.25$. A candidate assuming a different split would obtain proportionally different sizes; the method is what is examined.
Check: section properties. All steel sections selected below are quoted by their plate dimensions (flange $b \times t$, web $h \times w$) and every property — $A$, $I_x$, $I_y$, $Z$, $J$, $C_w$, $r_x$, $r_y$ — is computed from those dimensions with root fillets neglected (slightly conservative). This makes every line checkable without a handbook; a rolled shape of equal or greater properties may be substituted directly.
Check: reading of Figures 3 and 4. On the examination drawing the dimension chain under Figure 3 reads 3 m + 4 m + 7 m + 4 m, and the overall dimension reads 15 m. Reading the drawing directly resolves this: the 15 m dimension line spans B to C ($4 + 7 + 4 = 15$ m) and the 3 m is a cantilever overhang projecting beyond support B to the free end A, which carries the 80 kN load. Similarly in Figure 4 the 10.5 m is the overall width of the floor ($3 \times 3.5$ m beam spacing), not a span; the design span is the 15 m stated in the note. Both readings are self-consistent and are used throughout.
Check: support conditions in Figure 2. The frame drawing shows identical triangle-and-roller symbols at A and D, and the frame must resist the 80 kN horizontal load, so both bases are taken as pinned. The frame is then statically indeterminate to the first degree, which is the classical form of this problem for both the elastic (Questions 2, 3) and the plastic (Questions 5, 6) treatments.
The three welded I-sections selected in this paper, drawn to a common scale.
Question 2: Reinforced concrete frame — flexural and shear steel for BC, and its deflection (14 + 6 = 20 marks)
Given. The design starts from a single-bay pinned-base portal frame, columns AB and DC each 10 m high, beam BC spanning 15 m, with $f'_c = 30$ MPa and $f_y = 400$ MPa.
Given data
Quantity
Symbol
Value
Frame height / span
$h$ / $L$
10 000 mm / 15 000 mm
Horizontal load at B
$H$
80 kN (unfactored)
Vertical loads at B and C
$P_B$, $P_C$
600 kN each (unfactored)
Vertical load at mid-span E
$P_E$
500 kN (unfactored)
Concrete / steel strengths
$f'_c$ / $f_y$
30 MPa / 400 MPa
Resistance factors
$\phi_c$ / $\phi_s$
0.65 / 0.85
Find. (a) the area and arrangement of longitudinal and transverse reinforcement in a rectangular beam BC; (b) the mid-span deflection of BC under service load, compared with the CSA A23.3 limits.
Figure 2 — the pinned-base portal frame used in Questions 2, 3, 5 and 6.
Approach. Analyse the frame elastically for the factored loads with prismatic members of equal $EI$, size the beam depth from the peak sagging moment, compute the tension steel from the rectangular stress block, design the stirrups by the A23.3 general method, then check deflection with Branson's effective moment of inertia and the long-term multiplier.
Select a trial section and factor the loads. Take BC as $500 \times 1400$ mm ($L/h \approx 10.7$, appropriate for a heavily loaded frame beam). Its self-weight is $0.5(1.4)(24) = 16.8$ kN/m, factored to $w_f = 21.0$ kN/m; the concentrated loads become $900$, $750$ and $900$ kN and the horizontal load $120$ kN.
Analyse the frame. The stiffness solution of the pinned-base portal gives the factored beam moments
$$M_{f,\text{B}} = 646.2, \qquad M_{f,\text{E}} = 2157.0\ \text{(sagging)}, \qquad M_{f,\text{C}} = 1846.2\ \text{kN}\cdot\text{m},$$
with beam shears of 452.5 kN at B and 612.5 kN at C. The 600 kN loads sit directly over the columns and so contribute only column axial force; the asymmetry between the two corners is entirely the sway effect of the horizontal load, which adds 600 kN·m of moment at one corner and removes it at the other.
Size the bottom (sagging) steel. With $\alpha_1 = 0.85 - 0.0015f'_c = 0.805$, $\beta_1 = 0.895$ and two layers of 30M bars giving $d = 1310$ mm, equilibrium of the rectangular stress block $M_r = T(d - a/2)$ with $a = T/(\alpha_1\phi_c f'_c b)$ gives $T = 1805$ kN, hence
$$A_s = \frac{T}{\phi_s f_y} = \frac{1.805 \times 10^6}{0.85 \times 400} = 5309\ \text{mm}^2 \; \Rightarrow \; \boxed{8 - 30\text{M} \;(5600\ \text{mm}^2)}$$
in two layers of four. The depth to the neutral axis is $c = a/\beta_1 = 257$ mm, so $c/d = 0.196$ — comfortably tension-controlled and well inside the A23.3 ductility limit.
Size the top (hogging) steel. Repeating the calculation at the corners gives $A_s = 4476$ mm² at C and 1487 mm² at B. The minimum area, $A_{s,min} = 0.2\sqrt{f'_c}\,b_t h/f_y = 1917$ mm², governs at B. Provide $7 - 30\text{M}$ (4900 mm²) over C, reducing to $3 - 30\text{M}$ (2100 mm²) over B, with three bars carried continuously through the span as compression steel and as hangers for the stirrups.
Design the shear reinforcement. With $d_v = \max(0.9d,\,0.72h) = 1179$ mm, minimum stirrups present ($\beta = 0.18$) and $\theta = 35^\circ$,
$$V_c = \phi_c \lambda \beta \sqrt{f'_c}\,b_w d_v = 0.65(0.18)(5.477)(500)(1179) = 377.8\ \text{kN},$$
so at the worst end $V_s = 612.5 - 377.8 = 234.7$ kN and the required spacing of 10M double-leg stirrups is
$$s = \frac{\phi_s A_v f_y d_v \cot\theta}{V_s} = \frac{0.85(200)(400)(1179)(1.428)}{234.7 \times 10^3} = 488\ \text{mm}.$$
Minimum-steel and detailing rules cap the spacing at about 490 mm, so adopt $\boxed{10\text{M stirrups at 450 mm}}$ throughout the span, tightened to 250 mm over the first 1.5 m from each column face for confinement at the joints.
(b) Establish the cracked stiffness. $E_c = 4500\sqrt{30} = 24\,648$ MPa and $n = E_s/E_c = 8.11$. The cracking moment is $M_{cr} = f_r I_g / y_t$ with $f_r = 0.6\lambda\sqrt{f'_c} = 3.29$ MPa, giving $M_{cr} = 536.8$ kN·m. The transformed neutral axis is at $x = 405.5$ mm and $I_{cr} = 48.29 \times 10^9$ mm⁴. Branson's expression with the maximum service moment $M_a = 1480.4$ kN·m gives
$$I_e = \left(\frac{M_{cr}}{M_a}\right)^{3}I_g + \left[1 - \left(\frac{M_{cr}}{M_a}\right)^{3}\right]I_{cr} = 51.44 \times 10^9\ \text{mm}^4.$$
Compute the service deflection. Superposing the uniform load, the mid-span point load and the restraining end moments (service values $M_B = 467.1$, $M_C = 1267.1$ kN·m),
$$\Delta = \frac{5wL^4}{384E_cI_e} + \frac{PL^3}{48E_cI_e} - \frac{(M_B + M_C)L^2}{16E_cI_e} = 8.74 + 27.73 - 19.24 = \boxed{17.2\ \text{mm}}.$$
Splitting this into its sustained and transient parts gives $\Delta_D = 3.90$ mm and $\Delta_L = 13.33$ mm. With $\rho' = 2100/(500 \times 1310) = 0.0032$ and $\zeta = 2.0$, the additional long-term deflection is $\zeta\Delta_D/(1 + 50\rho') = 6.72$ mm.
Compare with the code limits. The immediate live-load deflection is 13.3 mm against $L/360 = 41.7$ mm, and the long-term-plus-live deflection is $6.72 + 13.33 = 20.1$ mm against $L/480 = 31.3$ mm for members supporting non-structural elements liable to damage. Both are satisfied, so the $500 \times 1400$ mm section needs no increase and the serviceability requirement is met.
