Shear Design & Stirrups in RCC Beams
Diagonal Tension, Nominal Shear Stress, Design Shear Strength, Stirrup Spacing & IS 456:2000 — Fully Solved GATE CE Examples
Last Updated: March 2026
📌 Key Takeaways
- Shear in RCC beams causes diagonal tension — inclined cracks that form at approximately 45° to the beam axis. Shear reinforcement (stirrups or bent-up bars) must cross these cracks to prevent diagonal tension failure.
- Nominal shear stress: τv = Vu / (b × d) — where Vu is the factored shear force, b is beam width, d is effective depth.
- Design shear strength of concrete τc is obtained from IS 456 Table 19 — it depends on the steel percentage pt = 100Ast/(bd) and concrete grade fck.
- If τv < τc: minimum shear reinforcement only. If τv > τc: shear reinforcement must be designed. If τv > τc,max: section must be redesigned (increase dimensions).
- Stirrup spacing formula: Sv = 0.87 fy Asv d / Vus, where Vus = Vu − τcbd is the shear carried by stirrups.
- Maximum stirrup spacing = 0.75d or 300 mm, whichever is smaller (IS 456 Cl. 26.5.1.5).
- Minimum shear reinforcement is always required (IS 456 Cl. 26.5.1.6): Asv/bwSv ≥ 0.4/(0.87fy).
1. Shear in RCC Beams — How It Works
When a beam is loaded transversely, it develops internal shear forces at every cross-section. In a homogeneous elastic beam, shear stress varies parabolically across the depth — zero at the top and bottom fibres, maximum at the neutral axis. In an RCC beam, the situation is different because the tension zone is cracked, fundamentally altering the stress distribution.
In an RCC beam, the primary shear transfer mechanisms are:
- Shear in the uncracked compression zone: The concrete above the neutral axis, being in compression, remains uncracked and can carry shear stress directly.
- Aggregate interlock: The rough surfaces of cracks interlock under shear, transferring force across the crack faces even in the cracked tension zone.
- Dowel action of tension steel: The longitudinal tension bars resist transverse displacement across cracks by acting as dowels — small-scale bending of the bar across the crack.
- Shear reinforcement (stirrups): Once diagonal cracks form, stirrups crossing the cracks carry the additional shear force that the concrete can no longer resist.
IS 456 simplifies this complex behaviour into a practical design approach: the concrete is assumed to carry a certain design shear strength τc, and any shear force above this limit must be carried by shear reinforcement. This is conservative, well-calibrated against test data, and gives safe designs.
2. Diagonal Tension and Cracking
At any point in a beam, the state of stress consists of both normal stress (bending) and shear stress. Using Mohr's circle of stress, the principal stresses can be found. Near the neutral axis where bending stress is zero, the principal stresses are equal and opposite — one is a principal tension and the other is a principal compression, both at 45° to the beam axis.
When the principal tensile stress exceeds the tensile strength of concrete (typically 3–5 N/mm²), diagonal cracks form perpendicular to the principal tension direction — which is at approximately 45° to the beam axis. These are called diagonal tension cracks or shear cracks.
Types of Shear Cracks in RCC Beams
Flexure-shear cracks: Begin as vertical flexural cracks at the bottom and propagate diagonally upward as they approach the support (where shear is high). Most common type.
Web-shear cracks: Inclined cracks that start in the web near the neutral axis (where shear stress is maximum and bending stress is low). More common in prestressed or thin-web beams.
Diagonal tension failure (without shear reinforcement): Sudden, brittle failure when the diagonal crack propagates all the way to the compression zone in one step. This is extremely dangerous — no warning before collapse. Shear reinforcement prevents this.
How stirrups prevent failure: Stirrups (vertical or inclined loops of steel) cross the potential diagonal cracks. When a crack forms, the stirrup is placed in tension — it holds the two faces of the crack together and redistributes the shear force through the truss analogy. In this model, the beam acts like a Pratt truss: the compression zone is the top chord, the tension steel is the bottom chord, the stirrups are the vertical tension members, and the inclined compression struts in the concrete are the diagonal members.
3. Nominal Shear Stress τv
Nominal Shear Stress (IS 456 Cl. 40.1)
τv = Vu / (b × d)
Where:
Vu = factored shear force at the section [N]
b = breadth of the beam web [mm] (for T-beams, use bw not bf)
d = effective depth [mm]
τv = nominal shear stress [N/mm²]
Note for T-beams: Always use the web width bw in the shear stress calculation, not the effective flange width bf. The flange is not effective in shear transmission near the web.
Note for varying depth beams: IS 456 Cl. 40.1.1 provides a modification for beams of varying cross-section. For beams with haunches or tapers, the effective shear force is modified by the vertical component of the prestress or the inclined tension/compression force.
4. Design Shear Strength of Concrete τc
The design shear strength τc is the maximum shear stress that concrete alone (without shear reinforcement) can resist at the ultimate limit state. It is obtained from IS 456 Table 19, which gives τc as a function of the percentage of longitudinal tension steel pt and the concrete grade.
Steel Percentage for Shear
pt = 100 Ast / (b × d)
Where Ast = area of tension steel that is fully anchored beyond the section considered (not all bars may be considered if some are curtailed before the section).
For T-beams: use b = bw (web width) — same as for τv.
τc Values — IS 456 Table 19 (Extract) ⭐ GATE
| pt (%) | M20 (fck=20) | M25 (fck=25) | M30 (fck=30) |
|---|---|---|---|
| ≤ 0.15 | 0.28 | 0.29 | 0.29 |
| 0.25 | 0.36 | 0.36 | 0.37 |
| 0.50 | 0.48 | 0.49 | 0.50 |
| 0.75 | 0.56 | 0.57 | 0.59 |
| 1.00 | 0.62 | 0.64 | 0.66 |
| 1.25 | 0.67 | 0.70 | 0.71 |
| 1.50 | 0.72 | 0.74 | 0.76 |
| 1.75 | 0.75 | 0.78 | 0.80 |
| 2.00 | 0.79 | 0.82 | 0.84 |
| 2.25 | 0.81 | 0.85 | 0.88 |
| 2.50 | 0.82 | 0.88 | 0.91 |
| 3.00 and above | 0.82 | 0.90 | 0.92 |
Units: N/mm²
Key observations: τc increases with pt (more tension steel = more dowel action and better aggregate interlock). τc increases slightly with fck. τc has a maximum value (plateaus at high pt) — adding more tension steel beyond ~3% gives no further benefit in shear capacity.
Enhancement of τc for Slabs (IS 456 Cl. 40.2.1)
For solid slabs, the design shear strength of concrete is taken as k × τc, where k is an enhancement factor:
| Overall depth D (mm) | k |
|---|---|
| 300 or more | 1.00 |
| 275 | 1.05 |
| 250 | 1.10 |
| 225 | 1.15 |
| 200 | 1.20 |
| 175 | 1.25 |
| 150 or less | 1.30 |
Thinner slabs have higher τc because the size effect in shear is less pronounced for shallow members.
5. Maximum Shear Stress τc,max
Maximum Shear Stress — IS 456 Table 20
τc,max is the maximum shear stress that an RCC section can carry under any circumstances — even with maximum shear reinforcement. If τv > τc,max, the section dimensions must be increased (beam is too small for the applied shear).
| Concrete Grade | τc,max (N/mm²) |
|---|---|
| M15 | 2.5 |
| M20 | 2.8 |
| M25 | 3.1 |
| M30 | 3.5 |
| M35 | 3.7 |
| M40 and above | 4.0 |
τc,max represents the crushing strength of the inclined compression struts in the truss model. Beyond this limit, the concrete itself fails in compression — no amount of shear steel can help.
6. IS 456 Shear Design Logic — Three Cases
Three Cases — IS 456 Cl. 40.2, 40.3, 40.4
Case 1: τv < τc
Nominal shear stress is less than concrete's design shear strength.
The concrete alone can carry the shear.
→ Provide minimum shear reinforcement only (IS 456 Cl. 26.5.1.6)
Case 2: τc ≤ τv ≤ τc,max
Shear exceeds concrete's capacity but is within the section's ultimate limit.
→ Design shear reinforcement to carry the excess shear Vus = Vu − τcbd
Case 3: τv > τc,max
Shear is so high that the section itself is inadequate — compression struts would crush.
→ Redesign the section — increase b or d. No amount of shear reinforcement can help.
7. Shear Reinforcement — Vertical Stirrups
Vertical stirrups are the most common form of shear reinforcement. They are closed rectangular or U-shaped loops of bars placed perpendicular to the beam axis at regular intervals. They cross the potential diagonal tension cracks and carry the excess shear force by acting in tension.
Shear Carried by Stirrups Vus
Vus = Vu − τc × b × d
Vus is the shear force that must be carried by the stirrups (shear reinforcement) — the excess above what concrete can carry.
Stirrup Spacing Formula (IS 456 Cl. 40.4)
For vertical stirrups:
Sv = 0.87 fy Asv d / Vus
Where:
Sv = spacing of stirrups [mm]
fy = yield strength of stirrup steel [N/mm²]
Asv = total cross-sectional area of stirrup legs at the section [mm²]
For a 2-legged stirrup of diameter φs: Asv = 2 × (π/4) × φs²
For a 4-legged stirrup: Asv = 4 × (π/4) × φs²
d = effective depth [mm]
Vus = shear carried by stirrups [N]
Alternative form: Vus = 0.87 fy Asv d / Sv
(Useful when spacing is known and shear capacity is being checked)
Maximum Stirrup Spacing (IS 456 Cl. 26.5.1.5)
Sv,max = minimum of:
• 0.75 d (most commonly governing)
• 300 mm
The spacing must not exceed this limit even if the calculated spacing from the Vus formula is larger.
For inclined stirrups (at angle α): Sv,max = d (slightly more relaxed)
For Inclined Stirrups (angle α to beam axis)
Sv = 0.87 fy Asv d (sinα + cosα) / Vus
For vertical stirrups: α = 90°, sinα = 1, cosα = 0 → reduces to Sv = 0.87fyAsvd/Vus ✓
For 45° inclined stirrups: sinα = cosα = 1/√2 → Sv = 0.87fyAsvd×(√2)/Vus
Inclined stirrups are more efficient than vertical stirrups but less practical to construct.
8. Bent-Up Bars as Shear Reinforcement
In addition to stirrups, some of the main longitudinal tension bars can be bent up (at 45° typically) near the supports where shear is highest and the full area of tension steel is no longer needed for bending. These bent-up bars act as inclined tension members crossing the diagonal cracks.
Shear Capacity of Bent-Up Bars (IS 456 Cl. 40.4(b))
Vus = 0.87 fy Asb sinα
Where:
Asb = cross-sectional area of bent-up bars at the bend [mm²]
α = angle of bending (typically 45°) → sinα = sin45° = 0.707
Vus = 0.87 × fy × Asb × 0.707
IS 456 Restriction (Cl. 40.4): Bent-up bars alone shall not be used as shear reinforcement in a critical zone of length 2d from the support. They must be used in combination with stirrups, and their contribution shall not exceed 50% of the total shear reinforcement required.
9. Minimum Shear Reinforcement
Minimum Shear Reinforcement (IS 456 Cl. 26.5.1.6) ⭐ GATE
Minimum shear reinforcement is always required, even when τv < τc, except:
- Members of minor structural importance such as lintels
- Where the overall depth does not exceed 150 mm
- Where the overall depth does not exceed 2.5 times the flange width (for T and L beams)
Minimum reinforcement condition:
Asv / (bw × Sv) ≥ 0.4 / (0.87 fy)
Rearranging for spacing:
Sv ≤ 0.87 fy Asv / (0.4 × bw)
For 2-legged 8mm stirrups (Asv = 100.5 mm²) with Fe 415 in a 250mm wide beam:
Sv ≤ 0.87 × 415 × 100.5 / (0.4 × 250) = 36,285 / 100 = 362.85 mm
But also Sv ≤ 0.75d — say d = 450 mm: 0.75×450 = 337.5 mm (governs)
So Sv = 300 mm (adopt practical value ≤ 337.5 mm and ≤ 300 mm max spacing)
Purpose of minimum shear reinforcement: Even when concrete can carry the shear, minimum stirrups prevent sudden brittle failure if unexpected overloading occurs. They also hold the longitudinal bars in position, prevent the buckling of compression bars, and improve ductility and post-cracking behaviour.
10. Critical Section for Shear
Critical Section Location (IS 456 Cl. 22.6.2)
For beams supported on columns, walls, or other beams, the critical section for shear is located at a distance d (effective depth) from the face of the support.
At this location, the concrete in the support zone provides additional compression that enhances the shear capacity, allowing the design shear force to be taken at d from the support face rather than at the support face itself.
Shear force at critical section:
Vu,critical = Vu,support face − (wu × d) for UDL
Where wu = factored UDL intensity.
This reduction is significant — for a beam with heavy UDL, the shear at d from face can be considerably less than at the support face, leading to a more economical stirrup design.
Exception: When the reaction force introduces compression in the end region (concentrated loads near support, corbels), the critical section is at the face of the support — the d reduction does not apply.
11. Design Procedure — Step by Step
- Step 1 — Find Vu: Calculate the factored shear force at the critical section (at d from support face for UDL loading). Vu = 1.5 × (dead load shear + live load shear) at the critical section.
- Step 2 — Calculate τv: τv = Vu / (b × d). For T-beams use bw.
- Step 3 — Find pt and τc: pt = 100Ast/(bd). Read τc from IS 456 Table 19 by interpolation if needed.
- Step 4 — Check τc,max: If τv > τc,max → increase section size. Stop here if this applies.
- Step 5 — Determine case:
- τv < τc → minimum stirrups only (go to Step 7)
- τv ≥ τc → design stirrups (go to Step 6)
- Step 6 — Design stirrups:
Vus = Vu − τcbd
Choose stirrup diameter φs and number of legs → Asv
Sv = 0.87 fy Asv d / Vus
- Step 7 — Check maximum spacing: Sv ≤ 0.75d and ≤ 300 mm. Adopt the smaller of calculated and maximum spacing.
- Step 8 — Check minimum reinforcement: Verify Asv/(bwSv) ≥ 0.4/(0.87fy).
- Step 9 — Vary spacing along span: Repeat for sections at multiple distances from the support. Closer to midspan, Vu decreases and spacing can be increased.
12. Worked Example 1 — Stirrup Design for Simply Supported Beam
Problem: A simply supported RCC beam has span = 6 m, b = 300 mm, d = 500 mm. Factored load (including self-weight) = 60 kN/m. Ast = 1884 mm² (3–28 mm dia). fck = 20 N/mm², fy = 415 N/mm² for both main steel and stirrups. Design 2-legged vertical stirrups.
Step 1 — Shear Force at Critical Section
Vu,support = wu × L/2 = 60 × 6/2 = 180 kN
Critical section at d = 500 mm = 0.5 m from support face
Vu,critical = 180 − 60 × 0.5 = 180 − 30 = 150 kN
Step 2 — Nominal Shear Stress τv
τv = Vu/(bd) = 150,000/(300 × 500) = 150,000/150,000 = 1.0 N/mm²
Step 3 — Steel Percentage and τc
pt = 100 × 1884/(300 × 500) = 188,400/150,000 = 1.256%
From IS 456 Table 19 (M20), interpolating between pt = 1.25 (τc = 0.67) and pt = 1.50 (τc = 0.72):
τc = 0.67 + (1.256−1.25)/(1.50−1.25) × (0.72−0.67)
= 0.67 + (0.006/0.25) × 0.05 = 0.67 + 0.0012 ≈ 0.671 N/mm²
Approx: τc ≈ 0.67 N/mm²
Step 4 — Check τc,max
τc,max for M20 = 2.8 N/mm²
τv = 1.0 < τc,max = 2.8 ✓ Section size is adequate.
Step 5 — Determine Case
τv = 1.0 N/mm² > τc = 0.67 N/mm²
→ Case 2: Shear reinforcement required.
Step 6 — Design Stirrups
Vus = Vu − τcbd = 150,000 − 0.67 × 300 × 500 = 150,000 − 100,500 = 49,500 N
Try 2-legged 8 mm stirrups: Asv = 2 × (π/4) × 8² = 2 × 50.27 = 100.53 mm²
Sv = 0.87 × 415 × 100.53 × 500 / 49,500
= 0.87 × 415 × 100.53 × 500 / 49,500
= 18,160,260 / 49,500 = 366.9 mm
Step 7 — Check Maximum Spacing
Sv,max = min(0.75d, 300) = min(0.75×500, 300) = min(375, 300) = 300 mm
Calculated Sv = 366.9 mm > Sv,max = 300 mm
→ Adopt Sv = 300 mm (maximum spacing governs)
Step 8 — Check Minimum Reinforcement
Asv/(b × Sv) = 100.53/(300 × 300) = 100.53/90,000 = 0.001117
0.4/(0.87 × 415) = 0.4/361.05 = 0.001107
0.001117 > 0.001107 ✓ Minimum reinforcement satisfied.
Final: 2-legged 8 mm stirrups @ 300 mm c/c throughout
(In practice, the spacing would be varied — closer near supports, wider near midspan — as shown in Example 2)
13. Worked Example 2 — Variable Stirrup Spacing Along Span
Problem: For the same beam as Example 1 (b=300, d=500, wu=60 kN/m, span=6m), design stirrups with variable spacing. Use 2-legged 8 mm stirrups (Asv = 100.53 mm²) with Fe 415. τc = 0.67 N/mm² (as found above).
Shear Capacity of Concrete Alone
Shear carried by concrete: Vc = τc × b × d = 0.67 × 300 × 500 = 100,500 N = 100.5 kN
Location where Vu = Vc: 100.5 = 180 − 60x → 60x = 79.5 → x = 1.325 m from support
Beyond 1.325 m from support, Vu < τcbd → minimum stirrups only (Sv = 300 mm).
Stirrup Spacing at Various Sections
| Distance from support (m) | Vu (kN) | Vus = Vu−Vc (kN) | Sv = 0.87×415×100.53×500/Vus (mm) | Adopted Sv (mm) |
|---|---|---|---|---|
| 0.5 (critical section) | 150 | 49.5 | 367 | 300 (max governs) |
| 0.75 | 135 | 34.5 | 527 | 300 (max governs) |
| 1.0 | 120 | 19.5 | 932 | 300 (max governs) |
| 1.325 onwards | ≤100.5 | ≤0 | Min. stirrups only | 300 |
In this example, the maximum spacing of 300 mm governs throughout the shear reinforcement zone. The beam requires 2-legged 8 mm stirrups @ 300 mm c/c for the full span.
Practical note: For beams with very high shear near supports, the stirrup spacing near the support is much tighter — perhaps 100–150 mm — opening up to 300 mm near midspan. This variable spacing is efficient and economical.
14. Worked Example 3 — GATE-Style Shear Problem
Problem (GATE-style): A rectangular RCC beam (b = 250 mm, d = 450 mm) has a factored shear force of 200 kN at the critical section. The beam has Ast = 1256 mm² (4–20 mm dia). fck = 25 N/mm², fy = 415 N/mm². Stirrups are 2-legged 10 mm dia Fe 415 bars. Find the required stirrup spacing.
Solution
τv = 200,000/(250 × 450) = 200,000/112,500 = 1.778 N/mm²
pt = 100 × 1256/(250 × 450) = 125,600/112,500 = 1.116%
τc from IS 456 Table 19 (M25): interpolating between pt=1.00 (τc=0.64) and pt=1.25 (τc=0.70):
τc = 0.64 + (1.116−1.00)/(1.25−1.00) × (0.70−0.64)
= 0.64 + (0.116/0.25) × 0.06 = 0.64 + 0.0278 = 0.668 N/mm²
Check τc,max: M25 → τc,max = 3.1 N/mm². τv = 1.778 < 3.1 ✓
Vus = Vu − τcbd = 200,000 − 0.668×250×450 = 200,000 − 75,150 = 124,850 N
Asv = 2 × (π/4) × 10² = 2 × 78.54 = 157.08 mm²
Sv = 0.87 × 415 × 157.08 × 450 / 124,850
= 0.87 × 415 × 157.08 × 450 / 124,850
= 25,486,986 / 124,850
= 204.1 mm
Check max spacing: 0.75 × 450 = 337.5 mm and 300 mm → max = 300 mm
Sv = 204.1 mm < 300 mm ✓
Adopt 2-legged 10 mm stirrups @ 200 mm c/c (rounded down to nearest 25 mm for practicality)
15. Common Mistakes Students Make
- Using the shear force at the support face instead of at d from the support: IS 456 allows the design shear force to be taken at the critical section located at d from the face of the support (not from the centreline). For a beam with heavy UDL, this can reduce Vu significantly. Students who take Vu at the centreline of support over-design the stirrups near the end. Always locate the critical section at d from the support face for beams with standard support conditions.
- Using the flange width bf instead of web width bw for T-beams: The nominal shear stress τv = Vu/(bd) uses the web width bw for T-beams and L-beams. The flange is not effective in direct shear transmission. Using bf dramatically underestimates τv, giving a false impression that the shear is safe when it is not.
- Forgetting to check both conditions for stirrup spacing: The stirrup spacing must satisfy two conditions simultaneously: (1) the spacing from the Vus formula Sv = 0.87fyAsvd/Vus, AND (2) the maximum spacing of 0.75d or 300 mm. Students often calculate Sv from the formula and report it without checking against the maximum spacing limit. The adopted spacing must be the smaller of the two.
- Not providing minimum shear reinforcement when τv < τc: Even when the concrete alone can carry the shear, IS 456 Cl. 26.5.1.6 requires minimum shear reinforcement throughout the beam. Omitting stirrups entirely when τv < τc violates IS 456 and creates a brittle failure risk.
- Using the total number of tension bars for pt even when some bars are curtailed: When calculating pt for the shear design at a specific section, only bars that are fully anchored beyond that section should be counted. Bars that have been curtailed before the section are not effective and should be excluded from Ast in the pt calculation. Using all bars gives an artificially high pt and τc, which is unconservative.
16. Frequently Asked Questions
Why does τc depend on the tension steel percentage?
The design shear strength τc of concrete increases with pt because the tension steel contributes to shear resistance through two mechanisms. First, dowel action — the bars act as dowels crossing the diagonal cracks, resisting vertical displacement between the crack faces. More steel means more dowel action. Second, aggregate interlock is enhanced when more tension steel is present because the wider compression zone (with more steel holding the tension zone together) provides better crack bridging. Both effects increase with steel percentage, which is why IS 456 Table 19 shows τc increasing with pt.
What is the difference between shear span and shear failure modes?
The shear span av is the distance from a concentrated load to the nearest support. The ratio av/d (shear span to effective depth ratio) determines the failure mode. For av/d > 6: flexural failure dominates — the beam bends and fails in tension before shear failure occurs. For 2.5 < av/d < 6: diagonal tension failure — inclined cracks form and propagate. This is the normal shear failure mode addressed by IS 456. For 1 < av/d < 2.5: shear compression failure — the diagonal crack forms but is arrested by the compression zone. For av/d < 1: direct strut action (deep beam behaviour) — load is transferred directly to the support through an inclined compression strut. This regime is not covered by the standard IS 456 shear formula and requires special deep beam design.
Why is there a maximum shear stress τc,max beyond which the section must be redesigned?
The maximum shear stress τc,max represents the crushing strength of the inclined concrete compression struts that form between the diagonal cracks in the truss analogy. These struts carry compressive force at approximately 45° to the beam axis. When the shear force is very large, the compressive stress in these struts approaches the concrete's compressive strength and the struts crush — regardless of how much shear reinforcement is provided. No stirrup can prevent this crushing mode of failure. The only solution is to increase the beam dimensions (b or d) to reduce the shear stress below τc,max. This is why τc,max is an absolute upper limit that cannot be overcome by adding more steel.
Can we use the same steel grade for stirrups and main bars?
Yes — IS 456 permits Fe 415 or Fe 500 for both main bars and stirrups. Using high-strength steel for stirrups (Fe 415 or Fe 500 instead of Fe 250) allows larger stirrup spacing for the same shear capacity, which reduces congestion during construction. However, very large stirrup spacing (approaching 0.75d) can be impractical in heavily congested beam cages. In practice, Fe 415 is the standard choice for stirrups. For closed stirrups in columns and seismic detailing, Fe 415 is universally used. The 0.87fy design stress applies to stirrup steel just as it does to main bars.