Diaphragm Design Calculation Sheet – Collector and Shear Checks As ACI318-19
- Adisorn O.
- May 10
- 3 min read
Updated: 5 days ago
by alpsconsult
1. Collector (Drag Strut) Design (with Ω₀)
Given:
- Axial force from ETABS (Fx) = 220 kN (tension)
- Overstrength factor (Ω₀) = 2.5 (special shear wall system)
- φ = 0.9 (ACI tension strength reduction)
- Steel yield strength (fy) = 500 MPa
Step 1: Amplify collector force
Pu = Fx × Ω₀ = 220 × 2.5 = 550 kN
Step 2: Required steel area
As = Pu / (φ × fy)
= 550 × 10^3 / (0.9 × 500 × 10^6) = 1.22 × 10⁻³ m² = 1222 mm²
Step 3: Rebar selection
Use 4-#20 bars (4 × 314 mm² = 1256 mm²) – OK
Remark:
To compute the axial force in collector (strut) we can do in the follwoing step
1. compute shear flow q = Vu/L ; L is total length of load path
2. At support (shear wall) the shear flow to support is qw = Vu/lw in the opposite direction to the load
3. Compute the axial force diagram
Pu = q x l1 (compression)
Pu = q x l3 (tension)
2.1 Shear at Interface with Shear Wall (without Ω₀)
Given:
- Total shear at wall = 320 kN
- Effective contact length = 1.2 m
- Slab thickness = 0.2 m
-Shear force per m of shear wall qw = V/lw = 320/12 = 26.7 kN/m
Depends on the interface shear provision (roughened, dowels, or friction).
Check against μ × φ × Avf/s × fy > qw = V/lw or dowel capacity per ACI 318 §22.9.4
Proof: For a given transfer shear Vu along the interface length L μ × φ × Asf/s*L × fy > Vu (Sum of transfer shear along the interface) hence, μ × φ × Asf/s × fy > Vu/L
2.2 Shear at Interface with Collectors (without Ω₀)
The same concept of shear-friction strength of the dowel bars is also checked at interface along the collectors. The collectors are used at the load path where shear wall is misssing (see figure 8.4).
Depends on the interface shear provision (roughened, dowels, or friction).
Check against μ × φ × Avf/s × fy > q = V/L (It's not L-lw, it's just a shear flow) or dowel capacity per ACI 318 §22.9.4
Since the slab and collector are usually cast at the same time, μ = 1.4 is normally used.
3. Diaphragm Shear at Distance from Support (without Ω₀)
Given:
- Diaphragm shear = 250 kN at 3 m from wall
- Effective width = 10 m
- Slab thickness = 0.2 m
- f'c = 30 MPa
Step 1: Shear stress
v = V / (b × t) = 250 / (10 × 0.2) = 125 kN/m²
Step 2: Concrete shear capacity
Vc = 0.17 × √f'c × b × d
= 0.17 × √30 × 1000 × 160 = 149 kN
φVc = 0.75 × 149 = 112 kN/m (use b = 1 m)
→ 125 > 112 → Provide reinforcement to increase shear strength in the direction of lateral load.
4. Chord Reinforcement Design (without Ω₀)
Given:
- Diaphragm span = 12 m - Diaphragm shear = 300 kN - Diaphragm depth = 10 m
Step 1: Compute moment in diaphragm (treated like a beam)
M = V × L / 2 = 300 × 12 / 2 = 1800 kNm (diaphragm force is treated as concentrate load at CM of diaphragm, normally at mid span)
Step 2: Compute chord tension/compression
T = M / c = 1800 / 10 = 180 kN (c is about 0.9-0.95 L)
Step 3: Required steel area
As = T / (φ × fy) = 180 × 10³ / (0.9 × 500 × 10⁶) = 0.0004 m² = 400 mm²
Step 4: Select rebar
Use 2-#20 bars (2 × 314 mm² = 628 mm²) – OK
Further Study
A presentation by Dr. David Fanella at CRSI provides a brief information how to design the diaphragm.