Reinforced Concrete Beams: Shear and Torsion - Cheat Sheet
Your 30-minute rapid-fire reference for shear design, stirrup spacing, and torsion thresholds. Every formula, limit, and pitfall you need to pass the exam.
Shear Strength Fundamentals
Formulas
Vₙ = Vₑ + Vₛ → φVₙ ≥ Vᵤ
Meaning
Vₙ = nominal shear capacity (kN); Vₑ = concrete contribution (kN); Vₛ = stirrup contribution (kN); Vᵤ = factored shear (kN); φ = 0.75
Watch out
DO NOT forget φ = 0.75 for shear (different from φ = 0.90 for flexure). The design capacity is φVₙ, not Vₙ alone.
When to use
Always — this is the master shear design equation
Vₑ = 0.17λ√(f'ₑ) × bw × d
Meaning
Vₑ = concrete shear strength (N); λ = 1.0 for normal-weight concrete; f'ₑ = concrete compressive strength (MPa); bw = web width (mm); d = effective depth (mm)
Watch out
√(f'ₑ) is NOT √(f'ₑ in MPa) × 1000 — use the actual MPa value inside the square root. Result is in Newtons, not kN (divide by 1000 for kN).
When to use
Calculate concrete shear capacity as first step in any shear design (NSCP 2015 simplified formula, SI units)
φVₑ = 0.75 × 0.17λ√(f'ₑ) × bw × d
Meaning
Design shear strength provided by concrete alone
Watch out
Students often forget to multiply by 0.75. Your actual available capacity is 0.75 × Vₑ, not Vₑ.
When to use
Immediately after computing Vₑ — needed to determine if stirrups are required and to find Vₛ
Common values
Value
0.75
Symbol
φ
Quantity
φ-factor for shear
Value
0.17
Symbol
k
Quantity
Concrete shear coefficient (NSCP simplified, normal-weight)
Value
1.0
Symbol
λ
Quantity
λ for normal-weight concrete
Important facts
- Shear φ-factor = 0.75 (not 0.90 like flexure) — code is conservative because shear failure is brittle.
- NSCP 2015 uses simplified formula Vₑ = 0.17λ√(f'ₑ) × bw × d (no Vd or M term — assumes conservative estimate).
- λ = 1.0 for normal-weight concrete; λ = 0.85 for lightweight concrete.
- All SI units: f'ₑ in MPa, bw and d in mm → Vₑ in N (divide by 1000 for kN).
- Minimum stirrups required wherever Vᵤ > 0.5φVₑ (halfway mark triggers stirrup demand).
Key definitions
Shear failure
Example
Beam collapses without warning when diagonal tension exceeds Vₙ.
Sudden, brittle failure along diagonal cracks; governed by stirrups and concrete crushing.
Critical section for shear
Example
For a simply supported beam, check shear at distance d = 500 mm from the left support face.
Distance d from the face of support where maximum factored shear Vᵤ is calculated.
Concrete shear strength Vₑ
Example
Vₑ = 134.9 kN for a 300×500 mm beam with f'ₑ = 28 MPa.
Shear capacity provided by concrete matrix alone (no stirrups); controlled by diagonal tension and aggregate interlock.
Diagrams to know
- Diagonal crack pattern in deep and shallow beams
- Shear-moment envelope with critical section at distance d from support
- Free-body diagram of stirrup tension and concrete strut
Stirrup Design & Spacing
Formulas
Vₛ = (Vᵤ − φVₑ) / φ
Meaning
Vₛ = shear force that stirrups must carry (kN); subtraction gives excess demand, then divide by φ to get required nominal strength
Watch out
NEVER forget the division by φ. The numerator is the excess, and you must divide to get the nominal demand. Common error: using (Vᵤ − Vₑ) directly without the φ factors.
When to use
Whenever Vᵤ > φVₑ; this is the required stirrup contribution
s = (Av × fy × d) / Vₛ
Meaning
s = stirrup spacing (mm); Av = total leg area of one stirrup (mm²); fy = stirrup yield strength (MPa); d = effective depth (mm); Vₛ = stirrup shear demand (N)
Watch out
Av for a U-stirrup (vertical two-leg) = 2 × (bar area), NOT bar area alone. If using 10 mm bar (78.5 mm²), then Av = 2 × 78.5 = 157 mm². Units: Vₛ in N, not kN.
When to use
After computing Vₛ, to determine required spacing — smaller Vₛ allows larger spacing (more economical)
Vₛ,max = 0.66√(f'ₑ) × bw × d
Meaning
Upper limit on stirrup contribution to prevent concrete crushing along diagonal struts
Watch out
If this limit is exceeded, DO NOT increase stirrups — the concrete itself will fail. Enlarge bw or d, or increase f'ₑ.
When to use
Check after computing Vₛ — if Vₛ > Vₛ,max, the section is too small and must be enlarged
smax = min(d/2, 600 mm) [if Vₛ ≤ 0.33√(f'ₑ) × bw × d]
Meaning
Maximum allowed spacing when shear demand is light; smax is the smaller of d/2 or 600 mm
Watch out
If your computed spacing s exceeds this limit, use smax instead. The code will not allow larger spacing even if calculation permits.
When to use
When Vₛ ≤ 0.33√(f'ₑ) × bw × d (lower shear band)
smax = min(d/4, 300 mm) [if Vₛ > 0.33√(f'ₑ) × bw × d]
Meaning
Maximum spacing when shear demand is high; reduced to d/4 or 300 mm
Watch out
Spacing limit halves (d/2 → d/4) once you cross the 0.33 threshold. Recheck which band applies after computing Vₛ.
When to use
When Vₛ exceeds 0.33√(f'ₑ) × bw × d (upper shear band) — spacing requirement becomes stricter
Av,min = 0.062√(f'ₑ) × (bw × s) / fy ≥ 0.35 × (bw × s) / fy
Meaning
Minimum stirrup area required whenever Vᵤ > 0.5φVₑ; use the larger of the two expressions
Watch out
Two expressions — always compute both and use the larger. First expression is based on concrete strength; second is absolute minimum (0.35 term).
When to use
In regions where stirrups are demanded but calculated Av is small; ensures a minimum steel ratio
Common values
Value
0.33√(f'ₑ) × bw × d
Symbol
V₀.₃₃
Quantity
Concrete contribution factor (lower band)
Value
0.66√(f'ₑ) × bw × d
Symbol
Vₛ,max
Quantity
Crushing limit factor
Value
min(d/2, 600 mm)
Symbol
smax,low
Quantity
Minimum spacing smax (low band)
Value
min(d/4, 300 mm)
Symbol
smax,high
Quantity
Minimum spacing smax (high band)
Important facts
- Stirrups are required only if Vᵤ > 0.5φVₑ (minimum stirrup threshold).
- A U-stirrup has TWO legs → Av = 2 × (area of one bar). Count legs, not bars.
- Spacing rule: compute s from formula, then cap at smax. Use the more restrictive value.
- Once Vₛ > 0.33√(f'ₑ) × bw × d, spacing limit changes from d/2 to d/4 — recheck the band.
- If Vₛ > Vₛ,max, section is undersized; enlarge bw, d, or f'ₑ. More stirrups will NOT help.
- Minimum stirrups in high-shear zones: both expressions must be checked; use the larger Av.
Key definitions
Stirrup (closed stirrup for torsion, open/U-stirrup for shear)
Example
10 mm diameter U-stirrup spaced at 150 mm provides Av = 157 mm² per stirrup.
Transverse reinforcement crossing diagonal cracks; U-stirrup has 2 legs carrying shear, closed stirrup has 4 legs for torsion.
Leg area Av
Example
Two 10 mm bars: Av = 2 × 78.5 = 157 mm².
Total cross-sectional area of all stirrup legs crossing a diagonal crack (2 for U-stirrup, 4 for closed).
Spacing s
Example
s = 150 mm means one stirrup every 150 mm of beam length.
Vertical distance between successive stirrups along beam length.
Diagrams to know
- Stirrup spacing diagram showing s, d/2, d/4, and 600 mm / 300 mm limits
- Shear-demand band (low vs. high) with corresponding spacing rules
- U-stirrup cross-section showing 2-leg configuration and Av calculation
Torsion Design & Threshold
Formulas
Tᵤ < φ × 0.083λ√(f'ₑ) × (Acp² / pcp) → NEGLECT torsion
Meaning
Tᵤ = factored torque (N·mm); Acp = area enclosed by outside perimeter (mm²); pcp = outside perimeter (mm); φ = 0.75; if inequality holds, design for shear only
Watch out
Acp and pcp are based on OUTSIDE dimensions (bw + 2 cover), NOT stirrup centerline. Many students use wrong perimeter. Neglect threshold is usually small — torque must be very light to qualify.
When to use
First step in any torsion problem: check if torque is below neglect threshold before committing to torsion design
Acp = bw × h
Meaning
Acp = area enclosed by the outside perimeter of the cross-section (mm²); bw and h are the outside dimensions
Watch out
NOT the area of concrete; it is the bounding rectangular area (bw × h). Do not subtract reinforcement or voids.
When to use
Calculate Acp for torsion threshold check; use full section dimensions, not reduced by cover
pcp = 2(bw + h)
Meaning
pcp = outside perimeter (mm); the perimeter around the full section
Watch out
Perimeter is 2(bw + h), not just bw + h. Include all four sides.
When to use
Use with Acp in the torsion threshold formula
Common values
Value
0.083
Symbol
kt
Quantity
Torsion coefficient (NSCP simplified)
Value
0.75
Symbol
φ
Quantity
φ-factor for torsion
Value
≈ 0.85
Symbol
α
Quantity
Reduction factor for Acp → Ao
Important facts
- Torsion is RARELY neglected — torque threshold is very small (typically 3–5 kN·m for common sections).
- If Tᵤ exceeds threshold, design requires CLOSED stirrups (not open U-stirrups) and extra longitudinal steel.
- Torsion + shear are combined: the stirrup carries both Vₛ and torsional shear; interaction is complex.
- Acp and pcp use outside dimensions (full bw × h section, including cover).
- If torsion must be designed: closed stirrup Av,t = (At × fy × (2Ao)) / (Tᵤ × s), where Ao ≈ 0.85 Acp.
Key definitions
Torsion
Example
A curved beam or cantilever with an eccentric load experiences torsion.
Twisting moment applied to a beam; resisted by closed stirrups and longitudinal bars forming a truss.
Torsion threshold (neglect limit)
Example
For a 300×500 mm section with f'ₑ = 28 MPa, neglect threshold ≈ 4.6 kN·m.
Maximum torque below which code allows torsion to be ignored and beam designed for shear only.
Closed stirrup
Example
A 10 mm closed rectangular stirrup has 4 legs, Av = 4 × 78.5 = 314 mm².
Rectangular or polygon-shaped stirrup with four legs all crossing the diagonal; required for torsion resistance.
Longitudinal torsion steel
Example
Extra bars at corners of beam perimeter, spaced per ACI 318-19 / NSCP 2015.
Extra steel placed around the perimeter (in addition to main flexural steel) to resist torsional tension.
Diagrams to know
- Torsion threshold flowchart: does Tᵤ < threshold?
- Closed stirrup vs. U-stirrup (4 legs vs. 2 legs)
- Diagonal tension pattern under torsion (circular pattern, not just linear cracks)
Critical Sections & Spacing Rules
Important facts
- Stirrups extend from d above the support face to 2d beyond the inflection point (for continuous beams).
- Check shear at d from support face; DO NOT check at support face itself (accounts for load distribution).
- Spacing is NOT uniform: it may be tighter near supports (higher shear) and looser toward midspan (lower shear).
- Maximum spacing limits are ABSOLUTE — calculated spacing s must not exceed smax even if analysis suggests it.
- In transition zones (e.g., from smax = d/2 to d/4), use the more restrictive limit.
Key definitions
Critical section for shear
Example
Simply supported beam, left support: critical section at x = d from left face.
Located at distance d from the face of support; shear force Vᵤ is maximum here for cantilevers, minimum for interior spans of continuous beams.
Spacing band (low vs. high shear)
Example
If Vₛ = 150 kN and 0.33 threshold is 262 kN, use smax = d/2 = 250 mm (low band).
Two regimes: Vₛ ≤ 0.33√(f'ₑ) bw d → smax = min(d/2, 600); Vₛ > 0.33√(f'ₑ) bw d → smax = min(d/4, 300).
Diagrams to know
- Shear diagram showing critical section at distance d from support
- Stirrup layout along beam length with variable spacing
- Spacing band diagram showing transition at Vₛ = 0.33√(f'ₑ) threshold
Minimum Stirrup Requirements
Formulas
Av,min = 0.062√(f'ₑ) × (bw × s) / fy or 0.35 × (bw × s) / fy [use larger]
Meaning
Minimum stirrup area required when Vᵤ > 0.5φVₑ; compute both and use the larger value
Watch out
Do NOT forget the 'or' condition — always check both expressions. Second expression (0.35) is strength-independent; it ensures a minimum percentage.
When to use
Determine minimum Av if calculated Av is smaller than Av,min
Common values
Value
0.5φVₑ
Symbol
V_threshold
Quantity
Minimum stirrup threshold fraction
Important facts
- Minimum stirrups apply to regions where Vᵤ > 0.5φVₑ (halfway point of Vₑ).
- Two expressions must both be computed; use the larger Av — this is MANDATORY.
- Even in low-shear regions, a minimum steel ratio is enforced to control cracking.
- Minimum stirrups usually result in spacing = smax (economical design).
Key definitions
Minimum stirrup threshold
Example
If φVₑ = 101 kN, minimum stirrups required wherever Vᵤ > 50.5 kN.
Vᵤ > 0.5φVₑ triggers requirement for minimum stirrups (typically 10 mm dia. spaced at smax).
Diagrams to know
- Minimum stirrup trigger diagram (threshold line at 0.5φVₑ on shear-demand graph)
Must remember
- φ for shear = 0.75 (NOT 0.90). The design inequality is φVₙ ≥ Vᵤ, where φ = 0.75.
- Vₑ = 0.17λ√(f'ₑ) bw d (NSCP 2015 simplified, SI units). Result is in Newtons; divide by 1000 for kN.
- After finding Vₛ = (Vᵤ − φVₑ) / φ, check if Vₛ ≤ Vₛ,max = 0.66√(f'ₑ) bw d. If exceeded, enlarge section.
- Stirrup spacing s = (Av fy d) / Vₛ, but cap at smax. Spacing limit depends on shear band: d/2 or 600 mm (low) vs. d/4 or 300 mm (high).
- Av for U-stirrup = 2 × (bar area). Two 10 mm bars = 2 × 78.5 = 157 mm². DO NOT use bar area alone.
- Minimum stirrups triggered at Vᵤ > 0.5φVₑ. Compute Av,min using BOTH expressions; use the larger.
- Torsion neglect threshold: Tᵤ < φ × 0.083λ√(f'ₑ) Acp²/pcp. Use OUTSIDE dimensions (full bw × h); if exceeded, design closed stirrups + long. steel.
- Critical section for shear is at distance d from support face (NOT at support itself). This is where Vᵤ is checked.
- Spacing band transition: at Vₛ = 0.33√(f'ₑ) bw d, smax changes from d/2 to d/4. Recheck after computing Vₛ.
- If calculated spacing s > smax, use smax. If calculated Av < Av,min, use Av,min. Code limits are absolute — do not round down.
Last minute tips
- Unit trap: Vₑ formula gives result in Newtons (N), not kN. Divide by 1000 immediately. Double-check units: f'ₑ in MPa, bw and d in mm → Vₛ in N.
- Spacing band rule: compute 0.33√(f'ₑ) bw d and compare Vₛ to this threshold. If Vₛ is above it, smax drops from d/2 to d/4 — spacing gets stricter, not looser. Always recheck the band.
- Stirrup count: U-stirrup = 2 legs (shear), closed = 4 legs (torsion). Read the diagram carefully. If two separate U-stirrups are used, total Av = 4 bar areas.
- Torsion threshold uses OUTSIDE perimeter (bw + cover, full h + cover), NOT stirrup centerline. Acp = bw × h, pcp = 2(bw + h). Memorize this — it is a common mistake.
- Minimum stirrups: always check BOTH Av,min expressions (0.062√(f'ₑ) term AND 0.35 term). Do not skip the second one — it catches low-f'ₑ cases where the first formula gives a smaller value.
Comparison tables
Shear vs. Torsion Design Parameters
| Parameter | Shear (Vᵤ) | Torsion (Tᵤ) |
|---|---|---|
| φ-factor | 0.75 | 0.75 |
| Capacity formula | φVₙ = φ(Vₑ + Vₛ) | φTₙ = φ(Tₑ + Tₛ) |
| Concrete contribution | Vₑ = 0.17λ√(f'ₑ) bw d | Tₑ = 0.083λ√(f'ₑ) Acp²/pcp |
| Stirrup type | U-stirrup (2 legs) | Closed stirrup (4 legs) |
| Neglect condition | Not typically neglected | Tᵤ < φ × 0.083λ√(f'ₑ) Acp²/pcp |
| Spacing limit (light) | min(d/2, 600 mm) | min(pₛ/4, 300 mm) |
| Crushing limit | Vₛ ≤ 0.66√(f'ₑ) bw d | Tₛ ≤ 0.66√(f'ₑ) Acp²/pcp |
Spacing Limits by Shear Demand Band
| Shear Band | Condition | smax (smaller of two) | When to Apply |
|---|---|---|---|
| Low shear | Vₛ ≤ 0.33√(f'ₑ) bw d | min(d/2, 600 mm) | Lower demand zone near midspan or supports with light shear |
| High shear | Vₛ > 0.33√(f'ₑ) bw d | min(d/4, 300 mm) | High demand zone near supports; spacing halves |
| Minimum (always) | Vᵤ > 0.5φVₑ | smax per band + Av,min check | Ensures minimum steel ratio even if calculated Av < Av,min |
Stirrup Leg Count & Av Calculation
| Stirrup Type | Number of Legs | Example (10 mm bar, 78.5 mm² per bar) | Total Av |
|---|---|---|---|
| U-stirrup (shear only) | 2 | 2 × 78.5 mm² | 157 mm² |
| Closed stirrup (torsion) | 4 | 4 × 78.5 mm² | 314 mm² |
| Double-leg U-stirrup | 4 (two U-stirrups) | 2 × (2 × 78.5) mm² | 314 mm² |
Diagrams
Shear Design Decision Tree
Step-by-step flowchart for shear design. Start with Vᵤ, compute Vₑ and φVₑ, check if stirrups are needed, verify section size, determine spacing band, compute actual spacing, and cap at limits. This is the exact sequence to follow in an exam.
Shear & Torsion Formula Map
Mind map showing hierarchy of shear and torsion formulas. Core branches split into concrete contribution, design equations, stirrup spacing, and torsion. Emphasizes key parameters (φ = 0.75, λ = 1.0, unit conversions) and spacing band logic.