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CECC-483 Reference Formula Sheet — Reinforced Concrete Columns

Complete column formula list (tied, spiral, eccentrically loaded, and slender) per the CECC-483 / NSCP review notes.

Classification of columns:

  1. According to reinforcement: tied or spiral.
  2. According to slenderness: short ($KL_u/r \le 34-12 M_1/M_2$) or slender.
  3. According to load: axially loaded; eccentrically loaded (uniaxial or biaxial bending).
  4. According to section: square/rectangular; round/circular.

Axially loaded resistance (short columns):

$$P_u = 0.80\,\phi\,A_g\!\left[0.85f'_c(1-\rho_g)+\rho_g f_y\right]\quad\text{(tied, }\phi=0.65\text{)}$$ $$P_u = 0.85\,\phi\,A_g\!\left[0.85f'_c(1-\rho_g)+\rho_g f_y\right]\quad\text{(spiral, }\phi=0.70\text{ or 0.75 by NSCP 2015)}$$ $$P_u = \phi\,0.80\,A_g\!\left[0.85f'_c(A_g-A_{st}) + A_{st}f_y\right]\quad\text{(tied, alternative form)}$$

ACI code specs:

Minimum spiral steel ratio:

$$\rho_s = 0.45\!\left[\dfrac{A_g}{A_c}-1\right]\!\dfrac{f'_c}{f_y}$$ $$\rho_s = \dfrac{4A_s(D_c-d_b)}{D_c^2\,s}\quad\text{(geometric)}$$

$A_c$ = core area; $D_c = D-80$ mm = core diameter; $s$ = spiral pitch.

Compressive force of steel (for force balance):

$$C_s = A'_s(f_y - 0.85f'_c)\quad\text{(considering displaced concrete)}$$ $$C_s = A'_s f_y\;(\text{if }f'_s = f_y),\;C_s = A'_s f'_s\;\text{(neglecting displaced concrete)}$$

Eccentrically loaded columns — balanced condition:

$$c_b = \dfrac{600\,d}{600+f_y}\qquad a_b = \beta_1 c_b$$ $$f'_s = 600\,\dfrac{c_b - d'}{c_b}\quad (\text{if }f'_s>f_y\Rightarrow f'_s = f_y)$$ $$T + P_b = C + C'\;\Rightarrow\;A_s f_y + P_b = 0.85f'_c\,a_b\,b + A'_s f'_s$$ $$P_b = 0.85f'_c\,a_b\,b\quad\text{(when steel cancels)}$$ $$M_b = P_b\,e_b$$

Mode-of-failure table:

EccentricityLoadsType of Failure
$e > e_b$$P_u < P_b$Tension failure
$e = e_b$$P_n = P_b$Balanced failure
$e < e_b$$P_n > P_b$Compression failure

Minimum eccentricity (axially loaded check):

$$e_{min} = 0.10\,h\;\text{(rectangular)};\;e_{min} = 0.05\,D\;\text{(circular)}$$ $$e_{min} = 15 + 0.03h\;\text{(mm, tied)};\;e_{min} = 15 + 0.03D\;\text{(mm, spiral)}$$

Analytical method — general eccentric column:

$$T + P_n = C + C'\;\Rightarrow\;A_s f_y + P_n = 0.85f'_c ab + A'_s f'_s$$ $$\Sigma M_T = 0:\;P_n(d_t) = 0.85f'_c\,ab(d-a/2) + A'_s f'_s(d-d')$$ $$a = \beta_1 c$$

Bresler's reciprocal equation (biaxial bending):

$$\dfrac{1}{P_{ni}} = \dfrac{1}{P_{nx}} + \dfrac{1}{P_{ny}} - \dfrac{1}{P_{o}}$$

Slenderness — ACI moment magnifier method:

$$M_c = \delta_b M_{2b} + \delta_s M_{2s}$$ $$\delta_b = \dfrac{C_m}{1-P_u/(\phi P_c)}\ge 1.0\qquad \delta_s = \dfrac{C_m}{1-\Sigma P_u/(\phi\,\Sigma P_c)}\ge 1.0$$ $$C_m = 0.60 + 0.40\,M_1/M_2 \ge 0.40\quad\text{(braced, no transverse loads)}$$ $$C_m = 1.0\quad\text{(all other cases)}$$ $$P_c = \dfrac{\pi^2 EI}{(KL_u)^2}\qquad EI = \dfrac{E_c I_g / 2.5}{1+\beta_d}$$ $$\beta_d = \dfrac{\text{factored axial dead load}}{\text{factored axial total load}}$$ $$r = 0.30h\;\text{(rectangular)},\;r = 0.25D\;\text{(round)}$$

Slenderness can be ignored if $KL_u/r \le 34 - 12 M_1/M_2$ for braced (non-sway) frames; $KL_u/r \le 22$ for unbraced columns. $M_1/M_2$ is negative for single-curvature and positive for reverse-curvature bending.

Steps in solving slender columns:

  1. Check slenderness: $KL_u/r > 34-12 M_1/M_2$ ?
  2. Compute $E_c = 4700\sqrt{f'_c}$.
  3. Compute $I_g = bh^3/12$.
  4. Compute $\beta_d$.
  5. Compute $EI$.
  6. Compute $P_c$.
  7. Compute $C_m$.
  8. Compute $\delta_b$.
  9. Compute the magnified moment $M_c = \delta\,M_u$.
  10. Compute the magnified eccentricity $e = M_u / P_u$ and compare with $e_{min}$.

Reinforced Concrete Columns

Concept Concept Concept Concept Concept Concept Concept Concept
    ACI Code Specifications:
    Tied Column:
  • \(\rho_g=0.01-0.08\)
  • Minimum Reinforcement:
    1. 4-16mm\(\phi\) for square or rectangular column
    2. 6-16mm\(\phi\) for round tied column
  • Minimum Gross Area Ag=60,000mm2 or 250mmx250mm
  • Lateral Ties:
    1. 10mm\(\phi\) for 32mm\(\phi\) or smaller vertical bars
    2. 12mm\(\phi\) for vertical bars greater than 32mm\(\phi\)
  • Spacing of Lateral Ties (choose least):
    1. 16db (main bars)
    2. 48dt (tie bar)
    3. least column dimension
  • When there are more than four vertical bars, additional ties shall be provided so that no longitudinal bar shall be spaced more than 150mm on each side.
    • Concept
      Spiral Column:
  • \(\rho_g=0.01-0.08\)
  • Minimum Reinforcement: - 6-16mm\(\phi\)
  • Spacing of spirals: 25mm\(\le\)S\(\le\)75mm
  • Minimum spiral diameter: 10mm\(\phi\)
      • \(\rho_s=\) ratio of the volume of spiral reinforcement to the volume of concrete core

    \(\rho_s=0.45(\frac{A_g}{A_{ch}}-1)\frac{f'c}{fy_t}\)

    Ag=gross area of column
    Ach=cross-sectional area of concrete core

    Problem (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.

    Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 1 (Investigation): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 1 (Investigation): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 1 (Investigation): – Diagram

    See images:

    Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 1 (Investigation): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 1 (Investigation): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 1 (Investigation): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 1 (Investigation): – Diagram

    Problem (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.

    Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 2 (Design): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 2 (Design): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 2 (Design): – Diagram

    See images:

    Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 2 (Design): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 2 (Design): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 2 (Design): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 2 (Design): – Diagram

    Problem (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.

    Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 3 (Investigation): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 3 (Investigation): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 3 (Investigation): – Diagram

    See images:

    Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 3 (Investigation): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 3 (Investigation): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 3 (Investigation): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 3 (Investigation): – Diagram

    Problem (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 (fyh=275MPa) with 30mm steel covering.

    Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 4 (Design): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 4 (Design): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 4 (Design): – Diagram

    See images:

    Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 4 (Design): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 4 (Design): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 4 (Design): – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 4 (Design): – Diagram

    Problem:

    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.

    Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 5: – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 5: – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 5: – Diagram

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    Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 5: – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 5: – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 5: – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 5: – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 5: – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 5: – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 5: – Diagram Reinforced Concrete Columns | Principles of Reinforced Concrete – Problem 5: – Diagram

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

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

    Problem: 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?

    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)

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

    Problem: 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  |  fy = 415 MPa

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

    #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)

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

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

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

    1. What is the nominal axial load strength Pn of the column (kN)?
    2. Find the nominal shear capacity Vnx along the x-direction (bending about y-axis).
    3. Find the nominal shear capacity Vny along the y-direction (bending about x-axis).

    #1 — Nominal Axial Strength Pn

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

    #2 — Vnx (x-direction: bw = 400 mm)

    d = 600 − 40 − 12 − 25/2 = 535.5 mm  |  Av = 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 — Vny (y-direction: bw = 600 mm)

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

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

    Problem: 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  |  fy = 413 MPa

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

    #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 Pu

    For spiral column: φ = 0.75, Pn,max = 0.85Po

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

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

    Given:
    Column: b × h = 600 mm × 450 mm  |  Main reinforcement: 8 – 28 mm ∅ bars
    Lateral ties = 10 mm ∅ at 100 mm o.c.  |  fy = 415 MPa  |  fyv = 275 MPa
    f'c = 28 MPa  |  Ec = 25,000 MPa  |  Clear cover to centroid of main bars = 70 mm
    Unsupported height Lu = 2.5 m  |  Effective length factor K = 1.0

    1. What is the nominal axial strength Pn of the column (kN)?
    2. Find the critical buckling load Pc (kN).
    3. Calculate the nominal shear strength Vn for bending about the y-axis.

    #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 Pc

    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)

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

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

    Additional board-style practice items for this topic.

    Question Bank: q19

    PSAD - Reinforced Concrete / Eccentrically Loaded Columns / Engr. Janclyde Espinosa (Clidez)

    A tied column which is subjected to an eccentric load has dimensions of 300mmx500mm which is reinforced with 4-28mm⌀ bars arranged as shown in the figure. The concrete cylinder strength f'c=27.6MPa and the stee yield strength fy=414.7MPa. The column carries a nominal load Pn at an eccentricity e from the y-axis of the column section. If the neutral axis is 125mm from the right edge of the column section and neglecting the area of concrete displaced by the compression steel bar,

    q19

    Determine the value of the nominal load Pn assuming that the compression steel will not yield.

    1. 607
    2. 706
    3. 504
    4. 405

    Compute the nominal moment capacity of the column section.

    1. 312.25
    2. 506.22
    3. 447.74
    4. 392.01

    Determine the eccentricity where the nominal load Pn is acting.

    1. 514
    2. 555
    3. 888
    4. 717

    Solution pending in psadquestions/q19.json.

    Question Bank: q35

    PSAD - Reinforced Concrete / Eccentrically Loaded Columns / Engr. Janclyde Espinosa (Clidez)

    The section of a column is shown. d1=320mm, d2=510mm,b1=400mm, b2=260mm. f'c=20.7MPa and fy=415MPa.

    q35

    Determine the location of the centroid of the gross concrete area measured from the y-axis in mm.

    1. 371.2
    2. 354.9
    3. 320.6
    4. 402.3

    Determine the location of the plastic section (in mm) of the column measured from the y-axis. Neglect the area of concrete displaced by the steel.

    1. 403.75
    2. 395.7
    3. 326.8
    4. 311.1

    Determine the factored moment Mux (in kN-m) due to a factored load Pu=5200kN applied 425mm from the y-axis and 190mm from the x-axis. Assume that the column is reinforced such that the plastic centroid is 400mm from the y-axis and 160mm from the x-axis.

    1. 156
    2. 130
    3. 148
    4. 123

    Solution pending in psadquestions/q35.json.

    Question Bank: q46

    PSAD - Reinforced Concrete / Spiral Columns / Engr. Janclyde Espinosa (Clidez)

    A spiral column carries a gravity factored axial dead load of 1200kN and factored axial lived load of 2100kN. The column is of average height and it will be assumed that there is no reduction in strength due to the effects of slenderness. Use approximately 1.5% reinforcement. Using f'c=27.6MPa and fy=415MPa,

    Determine the required diameter of the spiral column.

    1. 450
    2. 400
    3. 420
    4. 430

    Determine the number of 28-mm diameter bar longitudinal bars.

    1. 6
    2. 7
    3. 8
    4. 5

    Determine the practical center to center spacing of 10-mm diameter spiral reinforcement.

    1. 50
    2. 40
    3. 60
    4. 70

    Solution pending in psadquestions/q46.json.

    Question Bank: q47

    PSAD - Reinforced Concrete / Tied Columns / Engr. Janclyde Espinosa (Clidez)

    CE Board May 2022
    A tied column 450mmx450mm is reinforced with 8-28 mm bars equally distributed on its sides. The unsupported length of the column is 2.6m and is prevented to sidesway due to shear walls. K=1.0, f'c=20.7MPa, and fy=415MPa. Use 40mm covering measured from the center of reinfocement. The diameter is 12mm. Es=200GPa.

    Determine the nominal load that the column could carry.

    1. 4416.1kN
    2. 4428.3kN
    3. 4372.86kN
    4. 4226.73kN

    Determine the balanced load.

    1. 1684.07kN
    2. 1573.60kN
    3. 1783.34kN
    4. 1593.52kN

    Determine the balanced moment.

    1. 482.64kN-m
    2. 428.64kN-m
    3. 436.78kN-m
    4. 463.87kN-m

    Solution pending in psadquestions/q47.json.

    Question Bank: q148

    PSAD - Reinforced Concrete / Tied Columns / Engr. Janclyde Espinosa (Clidez)

    Refer to the column section shown.
    W x L = 600 mm x 800 mm
    Main reinforcement Ast = 16 – 25 mm diameter bars
    Lateral ties = 10 mm diameter bars
    Reinforcing steel yield strength:
    Main reinforcement, fy = 415 MPa
    Lateral ties, fyv = 275 MPa
    Concrete, fc’ = 28 MPa
    Clear concrete cover = 40 mm
    Spacing of ties = 150 mm

    q148

    Which of the following most nearly gives the nominal shear strength (in kN) of the column for the shear parallel to the x-axis?

    1. 717
    2. 833
    3. 912
    4. 878

    Which of the following most nearly gives the nominal shear strength (in kN) of the column for the shear parallel to the y-axis?

    1. 833
    2. 912
    3. 717
    4. 878

    Which of the following nearly gives the the nominal compressive load (in kN) at zero eccentricity?

    1. 11600
    2. 11100
    3. 10800
    4. 12300

    Solution pending in psadquestions/q148.json.

    Question Bank: q533

    PSAD - Reinforced Concrete / Tied Columns / Engr. Deguma

    Refer to the figure shown:
    Given:
    Column dimension, WxL = 400mmx600mm
    Main reinforcement, Ast = 10-25mm∅
    Ties of hoop reinforcement = 12mm∅ spaced at 100mm O.C.
    Steel Yield Strength = 415MPa
    Concrete Compressive Strength = 28MPa
    Concrete Clear Cover = 40mm
    Permissible concrete shear stress = 0.82MPa

    q533

    Which of the following most nearly gives the nominal axial load strength (in kN) of the column, Pn?

    1. 6106
    2. 1606
    3. 6016
    4. 6610

    Which of the following most nearly gives the nominal shear capacity (in kN) of the section along the x-direction?

    1. 930
    2. 943
    3. 807
    4. 795

    Which of the following most nearly gives the nominal shear capacity (in kN) of the section along the x-direction?

    1. 795
    2. 851
    3. 943
    4. 807

    Solution pending in psadquestions/q533.json.