CE Board Exam Randomizer

Question 1

Problem 1 (Investigation):

A square tied column 450mm on each side is reinforced with 8-25mm bars with fy=415MPa. Determine the safe service axial live load if the axial dead load on the column is 820kN. Use f'c=21MPa.

Answer

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Question 2

Problem 2 (Design):

Design a square tied column that must support an axial dead load of 575kN and an axial live load of 770kN. Assume f'c = 27.6MPa and fy=414MPa. Use 28mm main bars and 10mm ties.

Answer

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Question 3

Problem 3 (Investigation):

A round spiral column having a diameter of 450mm is reinforced with six 25mm bars with fy=345MPa. If the service axial dead load is 900kN, determine the safe axial live load of the column. Use f'c=34MPa.

Answer

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Question 4

Problem 4 (Design):

Design a round spiral column to support an axial dead load of 900kN and an axial live load of 1300kN. Assume that 3% longitudinal steel is desired, f'c=27.6MPa, and fy=414MPa. Use 25-mm main reinforcement. Determine also the minimum spacing of 10-mm spiral (f_yh=275MPa) with 30mm steel covering.

Answer

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Question 5

Problem 5:

The column in the figure shown below has a strain on its compressive edge equal to -0.003 and has a tensile strain of +0.002 on its other edge. It is reinforced with 6-28mm bars. b=350mm and h=600mm. Steel covering to the centroid of reinforcement is 65mm. Determine the nominal values of Pn and Mn that cause this strain distribution if f'c=28MPa and fy=414MPa.

Answer

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Question 6

Problem 6: Shear in Concrete Columns along x and y Axes

Refer to the figure shown below.
Given:
Column dimension, W x L = 400 mm x 600 mm
Main reinforcement, Ast = 10-25 mmØ
Ties of hoop reinforcement = 12 mmØ spaced at 100 mm O.C.
Steel yield strength = 415 MPa
Concrete compressive strength = 28 MPa
Concrete clear cover = 40 mm
Permissible concrete shear stress = 0.82 MPa

a. Which of the following most nearly gives the nominal axial load strength (in kilonewtons) of the column, Pn?
b. Which of the following most nearly gives the nominal shear capacity of the section along the x-direction?
c. Which of the following most nearly gives the nominal shear capacity of the section along the y-direction?

Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 6: – Diagram

Answer

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Question 7

Problem 7: CE Board May 2013 & Nov. 2012 — Bar Substitution

A column is to be reinforced with 8 – 28 mm ∅ bars. If 25 mm ∅ bars are used instead, how many bars are needed if the column reinforcement has to be equal on all four sides?

Answer

Required Steel Area

$$A_s = 8\times\frac{\pi}{4}(28)^2 = 8\times 615.75 = 4926.0 \text{ mm}^2$$

Number of 25 mm Bars

$$n = \frac{A_s}{\pi/4\,(25)^2} = \frac{4926.0}{490.87} = 10.04$$

Since reinforcement must be equal on all four sides, n must be a multiple of 4.

10.04 → round up to next multiple of 4 → n = 12 bars (3 bars per side)

$$A_{s,\text{prov}} = 12\times 490.87 = 5890 \text{ mm}^2 > 4926 \text{ mm}^2 \checkmark$$

Question 8

Problem 8: CE Board May 2016 — Rectangular Tied Column Design

An axially loaded rectangular tied column is to be designed for the following service loads:
Dead load = 1600 kN | Live load = 845 kN
Required strength: U = 1.2DL + 1.6LL | φ = 0.65
Effective cover to centroid of reinforcement = 70 mm
f'_c = 27.5 MPa | f_y = 415 MPa

Using 3% vertical steel ratio, what is the required column width (mm) if architectural considerations limit the width in one direction to 400 mm?
For a column section 400 mm × 500 mm, what is the minimum design moment (kN·m) about the stronger axis?
For a 400 mm × 500 mm column section with 16 bars and ρ = 3%, find the required bar diameter.

Answer

#1 — Required Column Width (ρ = 3%, b = 400 mm)

$$P_u = 1.2(1600) + 1.6(845) = 3272 \text{ kN}$$ $$P_u = \phi(0.80)\left[0.85f'_c(A_g - \rho A_g) + f_y\rho A_g\right]$$ $$3{,}272{,}000 = 0.65(0.80)\,A_g\left[0.85(27.5)(0.97) + 415(0.03)\right]$$ $$3{,}272{,}000 = 0.52\,A_g\left[22.674 + 12.45\right] = 18.267\,A_g$$ $$A_g = 179{,}147 \text{ mm}^2 \qquad h = \frac{179{,}147}{400} = 447.9 \text{ mm} \rightarrow \boxed{\text{use } h = 500 \text{ mm}}$$

#2 — Minimum Design Moment (400 mm × 500 mm)

Per §22.4.2.1, the design accounts for accidental eccentricity using e = 0.10h (tied column):

$$e = 0.10\,h = 0.10(500) = 50 \text{ mm}$$ $$M_u = P_u \times e = \frac{3{,}272{,}000 \times 50}{10^6} = \boxed{163.6 \text{ kN·m}}$$

#3 — Bar Diameter (16 bars, ρ = 3%, 400 × 500 mm)

$$\rho = \frac{n\cdot\frac{\pi}{4}d_b^2}{A_g} \implies 0.03 = \frac{16\cdot\frac{\pi}{4}d_b^2}{400\times 500}$$ $$\frac{\pi}{4}d_b^2 = \frac{0.03\times 200{,}000}{16} = 375 \text{ mm}^2$$ $$d_b = \sqrt{\frac{375\times 4}{\pi}} = \sqrt{477.5} = 21.85 \text{ mm} \rightarrow \boxed{d_b = 22 \text{ mm}}$$

Question 9

Problem 9: CE Board Nov. 2015 — Rectangular Tied Column (P_n & Shear)

Given:
Column dimension W × L = 400 mm × 600 mm | Main reinforcement: A_st = 10 – 25 mm ∅ bars
Ties = 12 mm ∅ at 100 mm o.c. | f'_c = 28 MPa | f_y = 415 MPa
Clear concrete cover = 40 mm | Permissible concrete shear stress v_c = 0.88 MPa

What is the nominal axial load strength P_n of the column (kN)?
Find the nominal shear capacity V_nx along the x-direction (bending about y-axis).
Find the nominal shear capacity V_ny along the y-direction (bending about x-axis).

Answer

#1 — Nominal Axial Strength P_n

$$A_{st} = 10\times\frac{\pi}{4}(25)^2 = 4908.7 \text{ mm}^2 \qquad A_g = 400\times 600 = 240{,}000 \text{ mm}^2$$ $$P_n = 0.80\left[0.85f'_c(A_g - A_{st}) + f_y A_{st}\right]$$ $$= \frac{0.80\left[0.85(28)(235{,}091) + 415(4908.7)\right]}{1000} = \boxed{6105.8 \text{ kN}}$$

#2 — V_nx (x-direction: b_w = 400 mm)

d = 600 − 40 − 12 − 25/2 = 535.5 mm | A_v = 3 legs × π/4(12)² = 339.3 mm²

$$V_{nx} = \frac{v_c\,b_w\,d + \dfrac{A_v\,f_{yt}\,d}{s}}{1000} = \frac{0.88(400)(535.5) + \dfrac{339.3(415)(535.5)}{100}}{1000}$$ $$= \frac{188{,}496 + 754{,}013}{1000} = \boxed{942.5 \text{ kN}}$$

#3 — V_ny (y-direction: b_w = 600 mm)

d = 400 − 40 − 12 − 25/2 = 335.5 mm | A_v = 4 legs × π/4(12)² = 452.4 mm²

$$V_{ny} = \frac{0.17\lambda\sqrt{f'_c}\,b_w\,d + \dfrac{A_v\,f_{yt}\,d}{s}}{1000} = \frac{0.17(1)\sqrt{28}(600)(335.5) + \dfrac{452.4(415)(335.5)}{100}}{1000}$$ $$= \frac{181{,}102 + 629{,}884}{1000} = \boxed{811.0 \text{ kN}}$$

Question 10

Problem 10: CE Board Nov. 2015 — Spiral Column (Slenderness & Bar Count)

A spiral column 600 mm in diameter has an unsupported height of 2.4 m. The column is bent in single curvature and is braced against sidesway.
Service loads: DL = 3200 kN, LL = 1420 kN | Required strength: U = 1.2D + 1.6L
f'_c = 27.5 MPa | f_y = 413 MPa

What is the slenderness ratio of the column? (Assume pinned ends, k = 1.0)
If the required steel ratio is 1.7%, find the number of 32 mm ∅ bars.
Find the number of 32 mm ∅ bars required at the ultimate design load (U = 1.2D + 1.6L).

Answer

#1 — Slenderness Ratio

For circular sections: r = 0.25D

$$SR = \frac{kL}{r} = \frac{1.0(2400)}{0.25(600)} = \frac{2400}{150} = \boxed{16}$$

#2 — Number of Bars for ρ = 1.7%

$$A_g = \frac{\pi}{4}(600)^2 = 282{,}743 \text{ mm}^2$$ $$A_s = \rho\,A_g = 0.017(282{,}743) = 4806.6 \text{ mm}^2$$ $$n = \frac{4806.6}{\pi/4\,(32)^2} = \frac{4806.6}{804.25} = 5.98 \rightarrow \boxed{n = 6 \text{ bars}}$$

#3 — Number of Bars from P_u

For spiral column: φ = 0.75, P_n,max = 0.85P_o

$$P_u = 1.2(3200) + 1.6(1420) = 6112 \text{ kN}$$ $$P_u = \phi(0.85)\left[0.85f'_c(A_g - A_{st}) + f_y A_{st}\right]$$ $$6{,}112{,}000 = 0.75(0.85)\left[0.85(27.5)(282{,}743 - A_{st}) + 413\,A_{st}\right]$$ $$6{,}112{,}000 = 0.6375\left[6{,}609{,}875 + (413 - 23.375)\,A_{st}\right]$$ $$9{,}587{,}255 = 6{,}609{,}875 + 389.625\,A_{st}$$ $$A_{st} = \frac{2{,}977{,}380}{389.625} = 7642 \text{ mm}^2$$ $$n = \frac{7642}{804.25} = 9.50 \rightarrow \boxed{n = 10 \text{ bars}}$$

Question 11

Problem 11: CE Board Dec. 2014 — Rectangular Column (P_n, Euler Buckling, Shear)

What is the nominal axial strength P_n of the column (kN)?
Find the critical buckling load P_c (kN).
Calculate the nominal shear strength V_n for bending about the y-axis.

Answer

#1 — Nominal Axial Strength

$$A_{st} = 8\times\frac{\pi}{4}(28)^2 = 4926.0 \text{ mm}^2 \qquad A_g = 600\times 450 = 270{,}000 \text{ mm}^2$$ $$P_n = \frac{0.80\left[0.85(28)(270{,}000-4926) + 415(4926)\right]}{1000} = \boxed{6682.4 \text{ kN}}$$

#2 — Critical Buckling Load P_c

I is about the weak axis (bending about axis parallel to the 600 mm side, so h = 450 mm governs):

$$I = \frac{b\,h^3}{12} = \frac{600(450)^3}{12} = 4.556\times10^9 \text{ mm}^4$$ $$P_c = \frac{\pi^2 E_c I}{(KL_u)^2} = \frac{9.8696(25{,}000)(4.556\times10^9)}{(1.0\times 2500)^2} = \boxed{179{,}873 \text{ kN}}$$

#3 — Nominal Shear Strength (bending about y-axis)

b_w = 450 mm | d = 600 − 70 = 530 mm | A_v = 3 × π/4(10)² = 235.6 mm²

$$V_n = \frac{0.17\lambda\sqrt{f'_c}\,b_w\,d + \dfrac{A_v\,f_{yt}\,d}{s}}{1000}$$ $$= \frac{0.17(1)\sqrt{28}(450)(530) + \dfrac{235.6(275)(530)}{100}}{1000}$$ $$= \frac{214{,}544 + 343{,}418}{1000} = \boxed{557.9 \text{ kN}}$$

Reinforced Concrete Columns

Problem 1 (Investigation):

Problem 2 (Design):

Problem 3 (Investigation):

Problem 4 (Design):

Problem 5:

Problem 6: Shear in Concrete Columns along x and y Axes

Problem 7: CE Board May 2013 & Nov. 2012 — Bar Substitution

Required Steel Area

Number of 25 mm Bars

Problem 8: CE Board May 2016 — Rectangular Tied Column Design

#1 — Required Column Width (ρ = 3%, b = 400 mm)

#2 — Minimum Design Moment (400 mm × 500 mm)

#3 — Bar Diameter (16 bars, ρ = 3%, 400 × 500 mm)

Problem 9: CE Board Nov. 2015 — Rectangular Tied Column (Pn & Shear)

#1 — Nominal Axial Strength Pn

#2 — Vnx (x-direction: bw = 400 mm)

#3 — Vny (y-direction: bw = 600 mm)

Problem 10: CE Board Nov. 2015 — Spiral Column (Slenderness & Bar Count)

#1 — Slenderness Ratio

#2 — Number of Bars for ρ = 1.7%

#3 — Number of Bars from Pu

Problem 11: CE Board Dec. 2014 — Rectangular Column (Pn, Euler Buckling, Shear)

#1 — Nominal Axial Strength

#2 — Critical Buckling Load Pc

#3 — Nominal Shear Strength (bending about y-axis)

Problem 9: CE Board Nov. 2015 — Rectangular Tied Column (P_n & Shear)

#1 — Nominal Axial Strength P_n

#2 — V_nx (x-direction: b_w = 400 mm)

#3 — V_ny (y-direction: b_w = 600 mm)

#3 — Number of Bars from P_u

Problem 11: CE Board Dec. 2014 — Rectangular Column (P_n, Euler Buckling, Shear)

#2 — Critical Buckling Load P_c