DESIGN OF R.C. SHORT COLUMN - ACI 318-25

1.0 INPUT DATA

Geometry & Material

b = mm Width of column

h = mm Depth of column

f'_c = MPa Specified compressive strength of concrete [ACI 19.2.2]

f_y = MPa Specified yield strength of reinforcement [ACI 20.2.2]

E_s = MPa Modulus of elasticity of steel

Figure 1: Geometry

RC Column Dimensions

Reinforcement

n_bars = Total (equal on all sides)

Bar Ø = mm Main bar diameter

A_b = mm² Main single bar area

Cover = mm Concrete cover

Shear Ties

Tie Ø = mm Secondary bar diameter

s = mm Secondary bar vertical spacing

n_legs = Number of legs

Design Loads (Factored)

P_u = kN M_u = kN-m V_u = kN

2.0 CALCULATIONS

2.1 Section Properties

d = h - cover - tie - bar/2 Effective depth

d = 400 - 40 - 10 - 12.5 = 337.5 mm

A_g = 400 × 400 = 160000 mm² Concrete gross area

A_s = 8 × 500 = 4000 mm² Total steel area

2.2 Interaction Diagram Key Points

A. Pure Compression (Point A)

ACI 318-19 Eq. 22.4.2.2

P_o = 0.85f'_c(A_g - A_s) + f_yA_s

P_o = 0.85(30)(156000) + 420(4000) = 5658 kN

φP_n,max = 0.65 × 0.80 × P_o = 2942 kN

B. Balanced Condition (Point B)

ε_y = f_y / E_s = 420/200000 = 0.0021

c_b = [0.003 / (0.003 + ε_y)] × d = 198.5 mm

a_b = β₁ × c_b = 0.85 × 198.5 = 168.7 mm

Balanced: φP_n = 1500 kN, φM_n = 300 kN-m

C. Pure Tension

φP_nt = 0.90 × f_y × A_s = 1512 kN

3.0 INTERACTION DIAGRAM

Checking...

2.3 Shear Check (ACI 318-19)

Step 1: Concrete Capacity (V_c)

Table 22.5.5.1 (Criterion: Compression member)

N_u/A_g = 2000000/160000 = 12.5 MPa

V_c = 0.17λ√f'_c bd [1 + N_u/(14A_g)]

V_c = 0.17(1.0)√30(400×337.5)[1 + 12.5/14]

V_c = 238.5 kN

Step 2: Steel Tie Capacity (V_s)

Eq. 22.5.8.5.3

A_v = n_legs × A_bar = 2 × 78.5 = 157 mm²

V_s = (A_v f_yt d) / s

V_s = (157 × 420 × 337.5) / 150 = 148.4 kN

Max V_s Check (0.66√f'_cbd): 488 kN (OK)

Step 3: Total Shear Strength (φV_n)

φ = 0.75 (Shear)

φV_n = 0.75 (V_c + V_s)

φV_n = 0.75 (238.5 + 148.4) = 290.2 kN