Corbel Design - HK CoP 2013

DESIGN OF R.C. CORBEL - HK CoP 2013

1.0 INPUT DATA

Material Properties

f_cu = MPa Characteristic concrete cube strength

f_y = MPa Characteristic steel yield strength

Section Dimensions

b = mm Breadth of corbel

h = mm Overall depth of corbel

c_c = mm Cover to tension reinforcement

a_v = mm Load position from column face

Figure 1: Dimension Requirement for a Corbel

Loading

V_u = kN Design ultimate vertical shear force

N_uc = kN Design ultimate horizontal force (if any)

Figure 2: Strut-and-Tie Action of a Corbel

Assumed Reinforcement Sizes

φ_main = mm Main bar diameter

n_main = Number of main bars

φ_stirrup = mm Stirrup bar diameter

n_legs = Number of stirrup legs

2.0 CALCULATIONS

2.1 Geometry Check

Effective depth

d = h − c_c

d = 750 - 40 = 710.00 mm

Short corbel check

a_v/d = 500/710 = 0.704

SHORT CORBEL (a_v/d < 1.0) - OK

2.2 Shear Check (Clause 6.1.2.5)

Maximum shear stress

v_max = min(0.8·√f_cu, 5 MPa)

v_max = min(0.8·√45, 5) = 5.000 MPa

Maximum shear capacity

V_max,cap = v_max·b·d

V_max,cap = 5.000×250×710/10³ = 887.50 kN

Shear check

check_V_u = PASS (V_u = 620.00 kN ≤ V_max,cap = 887.50 kN)

2.3 Strut-and-Tie Analysis

Moment at column face

M_u = V_u·a_v + N_uc·(h − d)

M_u = 620×500 + 0×(750-710)/10³ = 310.00 kNm

Solving for neutral axis depth (x) from equilibrium

x = 152.40 mm (Solved from Vu = 0.405·fcu·b·x·av / [av² + (d-0.45x)²])

Tension force from strut-and-tie model

T = V_u·a_v/(d − 0.45·x)

T = 620×500/(710-0.45×152.40) = 483.40 kN

Design tension force

T_u = max(T, 0.5·V_u)

T_u = max(483.40, 310.00) = 483.40 kN

2.4 Main Tension Reinforcement

Required area of tension reinforcement

A_s1 = M_u·10⁶/(0.87·f_y·0.8·d) + N_uc·10³/(0.87·f_y)

A_s1 = 310×10⁶/(0.87×500×0.8×710) + 0×10³/(0.87×500) = 1273.38 mm²

A_s2 = T_u·10³/(0.87·f_y)

A_s2 = 483.40×10³/(0.87×500) = 1111.26 mm²

Minimum reinforcement

A_s,min = 0.0013·b·h

A_s,min = 0.0013×250×750 = 243.75 mm²

Design reinforcement

A_s,req = max(A_s1, A_s2, A_s,min)

A_s,req = 1273.38 mm²

Provided reinforcement

A_s,prov = n_main·(φ_main/2)²·π

A_s,prov = 2×(20/2)²×π = 628.32 mm²

check_A_s = FAIL (A_s,prov = 628.32 mm² < A_s,req = 1273.38 mm²)

2.5 Horizontal Stirrups (Shear Reinforcement)

Applied shear stress

v = V_u·10³/(b·d)

v = 620×10³/(250×710) = 3.49 MPa

Enhanced concrete shear stress

v_c = 0.556·(2d/a_v)·(f_cu/25)^1/3

v_c = 0.556×(2×710/500)×(45/25)^1/3 = 0.80 MPa

Shear reinforcement required check

Shear reinforcement IS required (v = 3.49 > v_c = 0.80 MPa)

Required horizontal stirrups

A_h,req = (v − v_c)·b·d/(0.87·f_y)

A_h,req = (3.49-0.80)×250×710/(0.87×500) = 1094.00 mm²

A_h,min = 0.5·A_s,prov = 0.5×628.32 = 314.16 mm²

A_h,design = max(A_h,req, A_h,min) = 1094.00 mm²

Provided stirrups

A_stirrup = n_legs·(φ_stirrup/2)²·π

A_stirrup = 2×(10/2)²×π = 157.08 mm²

Number of sets over 2/3·d = (2/3)×d = 473.33 mm

Sets required = ceil(A_h,design/A_stirrup) = ceil(1094.00/157.08) = 7 set/s

Spacing = floor((2/3·d)/(sets−1)) = floor(473.33/6) = 78 mm c/c

Figure 3: Typical Detailing of a Corbel

* OUTPUT SUMMARY *

Design Summary

Effective depth, d = 710.00 mm

Short corbel check: PASS

Shear check: PASS

Applied shear stress, v = 3.49 MPa

Concrete shear stress, v_c = 0.80 MPa

Required tension steel, A_s,req = 1273.38 mm²

Provided tension steel, A_s,prov = 628.32 mm²

Tension reinforcement check: FAIL - Increase reinforcement

Required horizontal stirrups, A_h,req = 1094.00 mm²

Design stirrup provision: 7 sets T10 @ 78mm c/c

RESULTS

DESIGN CHECK - INCREASE TENSION REINFORCEMENT AND PROVIDE SHEAR REINFORCEMENT