07-Str-A2 · May 2017

Question 4 of 7: B1 — Design of the column of a determinate concrete frame

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

Question 4: B1 — Design of the column of 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.

Find. Rectangular dimensions for column BC and its longitudinal and transverse reinforcement, satisfying flexure, axial load and shear, with the bar layout shown.

Approach. Solve the determinate frame for the column's axial force, moment and shear, factor them, and — because the eccentricity turns out to be enormous — design the section by strain compatibility at the actual P_f rather than with an axial-capacity formula. Then check shear and slenderness.

1. Solve the frame. Place C at the origin, B at (0, 10) and A at (7, 10). The pin at C supplies two reactions and the roller at A one, so r = 3 on a single rigid body and the frame is determinate. Taking moments about C, $$7A_y = 350(3.5) + 80(6.0) = 1225 + 480 = 1705 \;\Rightarrow\; A_y = 243.6\ \text{kN}$$ $$C_y = 350 - 243.6 = 106.4\ \text{kN}\ \text{(up)}, \qquad C_x = 80\ \text{kN}\ \text{(to the left)}$$
2. Internal actions in column BC. Working up from the pin, the only horizontal force below the 80 kN load is C_x, so the column carries a constant shear of 80 kN over the lower 6 m and a moment growing linearly to $$M = 80 \times 6.0 = 480\ \text{kN}\cdot\text{m}$$ Above the load the two horizontal forces cancel, so the shear is zero and the moment stays at 480 kN·m all the way to B. Reading the same moment along the beam confirms the joint: M_B = 243.6(7) − 350(3.5) = 1705 − 1225 = 480 kN·m. The axial force in the column is C_y = 106.4 kN throughout.
3. Factor the actions. All applied loads are occupancy loads: $$P_f = 1.5(106.4) = \boxed{159.6\ \text{kN}}, \quad M_f = 1.5(480) = \boxed{720\ \text{kN}\cdot\text{m}}, \quad V_f = 1.5(80) = \boxed{120\ \text{kN}}$$ The eccentricity is e = M_f/P_f = 4510 mm — five times the section depth. This member is a beam that happens to stand upright, and the axial-capacity formula P_r,max = 0.80[…] would be meaningless here.
4. Choose trial dimensions. Bending is in the plane of the frame, so the depth must lie in that plane. Try 400 mm wide × 900 mm deep with 4–30M in each of the two faces perpendicular to the bending axis (eight bars, 5600 mm², ρ = 1.56 %, between the A23.3 Clause 10.9.1 limits of 1 % and 8 %). With 40 mm cover and 10M ties, $$d = 900 - 40 - 11.3 - 14.95 = 833.8\ \text{mm}, \qquad d' = 66.3\ \text{mm}$$
5. Solve the section by strain compatibility. Set the extreme-fibre strain at 0.0035, guess the neutral-axis depth c, take the concrete compression as α₁φ_cf_c'b(β₁c) and each bar force as φ_sE_sε_sA_s capped at φ_sf_y, and iterate until the net axial force equals P_f. That converges at $$c = 94.9\ \text{mm}, \qquad a = \beta_1 c = 83.8\ \text{mm}$$ At this depth the near-face bars sit 66.3 mm from the compression face, just inside c, so they are in mild compression (211 MPa) while the far-face bars are far past yield in tension.
6. Moment resistance at that axial load. Taking moments of the three forces about mid-depth, $$M_r = \boxed{807\ \text{kN}\cdot\text{m}} \;>\; M_f = 720\ \text{kN}\cdot\text{m}$$ a utilisation of 0.89. The section is strongly tension-controlled — c/d = 0.11 — so it will warn before it fails, which is exactly what one wants in a member whose demand is almost pure flexure.
7. Shear. With d_v = max(0.9d, 0.72h) = 750.4 mm and the simplified method (β = 0.18, minimum ties provided), $$V_c = \phi_c \lambda \beta \sqrt{f_c'}\, b_w d_v = 0.65(1.0)(0.18)\sqrt{35}(400)(750.4) = 208\ \text{kN} \;>\; 120\ \text{kN}$$ The concrete alone carries the shear — and this ignores the favourable axial compression — so minimum ties govern. Their maximum spacing is the least of 16d_b = 478 mm, 48 tie diameters = 542 mm and the least column dimension = 400 mm. Use 10M ties at 300 mm, closed, with every corner bar and every alternate bar restrained by a tie corner.
8. Slenderness. The roller at A cannot restrain the frame horizontally on its own, but a sway of the column would rotate the rigid joint at B and therefore lift or drop A, which the roller prevents — so the frame is non-sway. Taking k = 1.0 conservatively and r = 0.3h = 270 mm, $$\frac{k l_u}{r} = \frac{10\,000}{270} = 37.0 \;>\; 34 - 12\left(\frac{M_1}{M_2}\right) = 34$$ so Clause 10.15 requires a magnifier to be evaluated. With E_c = 4500√35 = 26 622 MPa, EI = 0.4E_cI_g and P_c = π²EI/(kl_u)² = 25 540 kN, and with C_m = 0.6 + 0.4(0) = 0.6, $$\delta = \frac{C_m}{1 - P_f/(0.75P_c)} = \frac{0.6}{1 - 159.6/19\,155} = 0.605 \;\Rightarrow\; \delta = 1.0$$ The axial load is so small beside P_c that no magnification arises; the design moment stays at 720 kN·m and the section stands.

Reinforcement layout. A 400 × 900 mm column, the 900 mm dimension in the plane of the frame. Four 30M bars in each 400 mm face, one in each corner and two evenly spaced between, with 40 mm clear cover. Closed 10M ties at 300 mm throughout, reduced to 150 mm over the 900 mm above C and below B to confine the bar cages at the pin and at the rigid joint. The eight bars must be lapped into the beam at B and anchored into the pin detail at C.