🎓 Lesson 12
D4
Combined Shear-Torsion Design per ACI 22.7
It's the process of designing concrete beams and columns to safely resist both twisting (torsion) and sideways pushing (shear) forces at the same time.
🎯 Learning Objectives
- ✓ Calculate the required area of transverse reinforcement for combined shear-torsion using ACI 22.7.2 interaction criteria
- ✓ Design longitudinal torsional reinforcement based on the computed torsional moment and section geometry
- ✓ Analyze whether a given RC member satisfies ACI 22.7 strength limits for Vu, Tu, and their interaction
- ✓ Explain the physical rationale behind the √(Vu/Vn)² + (Tu/Tn)² ≤ 1.0 interaction equation
- ✓ Apply ACI 22.7 provisions to select appropriate closed stirrup spacing and longitudinal bar distribution
📖 Why This Matters
In mining infrastructure—such as ore pass linings, crusher foundations, conveyor trestles, and blast-resistant silos—members often experience simultaneous twisting (e.g., from eccentric equipment loads or seismic torsion) and lateral forces (e.g., earth pressure or blast-induced ground motion). Ignoring the shear–torsion interaction leads to under-reinforced sections, premature diagonal cracking, and catastrophic brittle failure. ACI 22.7 ensures safety and serviceability where real-world loading is never purely one type—and it’s explicitly required for structural elements in critical mine support systems per MSHA and ACI 318 compliance.
📘 Core Principles
Torsion in reinforced concrete generates a 'space truss' action: longitudinal bars resist tensile chord forces while closed stirrups act as tension ties resisting the diagonal compression field. When shear and torsion coexist, their diagonal crack patterns overlap, reducing effective concrete contribution and demanding synergistic reinforcement. ACI 22.7 models this via interaction: the normalized demand (Vu/Vn + Tu/Tn) must not exceed unity—but uses a more conservative elliptical interaction limit √(Vu/Vn)² + (Tu/Tn)² ≤ 1.0. Crucially, torsion triggers minimum longitudinal steel requirements—even if calculated demand is low—ensuring crack control and ductility. The theory also distinguishes 'equilibrium torsion' (required for static equilibrium, e.g., cantilevered chutes) from 'compatibility torsion' (induced by rotational restraint), with different design implications per ACI 22.7.1.
📐 Key Interaction Check & Reinforcement Formulas
ACI 22.7 requires two primary checks: (1) the combined strength interaction, and (2) minimum transverse and longitudinal reinforcement. The interaction equation ensures combined demand does not exceed capacity; transverse steel (Av + At) is sized for both shear and torsion contributions; longitudinal steel (Al) is derived from torsional equilibrium and detailing rules.
ACI 22.7.2 Interaction Limit
√(Vu/Vn)² + (Tu/Tn)² ≤ 1.0Checks combined demand-to-capacity ratio for shear and torsion simultaneously.
Variables:
| Symbol | Name | Unit | Description |
|---|---|---|---|
| Vu | Factored shear force | kN | Applied design shear at section |
| Vn | Nominal shear strength | kN | Sum of concrete (Vc) and steel (Vs) contributions |
| Tu | Factored torsional moment | kN·m | Applied design torsion at section |
| Tn | Nominal torsional strength | kN·m | Concrete-only torsional capacity per ACI 22.7.4.1 |
Typical Ranges:
Mine infrastructure beams: 0.6 – 0.95 (target design range)
High-seismic or blast-loaded members: ≤ 0.85 (conservative target)
💡 Worked Example
Problem: A 600 mm × 400 mm rectangular beam has f'c = 35 MPa, fy = 420 MPa. Factored shear Vu = 180 kN, factored torsional moment Tu = 32 kN·m. Effective depth d = 540 mm, core dimensions x1 = 540 mm, y1 = 340 mm, Acp = 240,000 mm², Pcp = 2000 mm. Compute Vn and Tn, then verify interaction.
1.
Step 1: Compute nominal shear capacity Vn = 0.17λ√f'c Acp + (ϕVs) — but first check if torsion threshold exceeded: Tu > 0.25√f'c (Acp)²/Pcp = 0.25×√35×(240000)²/2000 ≈ 15.2 kN·m → Tu = 32 kN·m > threshold → torsion reinforcement required.
2.
Step 2: Compute nominal torsional capacity Tn = 0.17λ√f'c (Acp)² / Pcp = 0.17×1.0×√35×(240000)²/2000 ≈ 15.2 kN·m (per ACI 22.7.4.1). For design, use Tn = 15.2 kN·m.
3.
Step 3: Compute Vn per ACI 22.5.5.1: Vn = Vc + Vs, where Vc = 0.17λ√f'c b w d = 0.17×1×√35×400×540 ≈ 217 kN. Assume no shear reinforcement initially → Vn ≈ 217 kN.
4.
Step 4: Apply interaction: √(Vu/Vn)² + (Tu/Tn)² = √(180/217)² + (32/15.2)² = √(0.685)² + (2.105)² = √0.470 + 4.432 = √4.902 ≈ 2.21 > 1.0 → FAILS. Therefore, need both increased transverse reinforcement (to raise Vn and Tn) and longitudinal steel.
Answer:
The interaction ratio is 2.21 > 1.0 → design fails. Transverse and longitudinal reinforcement must be added per ACI 22.7.6 and 22.7.7 to increase Vn and Tn, and satisfy Al ≥ 1.7×(Tu×x1×y1)/(ϕfy×(x1+y1)).
🏗️ Real-World Application
At the Bingham Canyon Mine (Utah), a reinforced concrete ore pass liner experienced significant eccentric flow-induced torque and dynamic blast-load shear. Initial design considered shear only, but post-construction monitoring revealed spiral cracking near the top transition zone. Forensic analysis confirmed combined shear–torsion overstress. The retrofit applied ACI 22.7-compliant closed No. 4 (13M) stirrups at 100 mm spacing plus four additional longitudinal bars (No. 14, 43M) — matching the minimum Al requirement and satisfying √(Vu/Vn)² + (Tu/Tn)² ≤ 0.92. Post-retrofit instrumentation showed no further cracking over 5+ years of operation.