Reinforced Concrete Beams: Shear and Torsion - Summary
Shear and torsion are critical failure modes in reinforced concrete beams that require careful design to ensure structural safety. While bending stress is resisted primarily by longitudinal flexural steel, shear and torsion demand transverse reinforcement (stirrups and closed hoops). This chapter addresses the design methodology for shear according to NSCP 2015, focusing on the contributions of concrete (Vc) and stirrup steel (Vs), spacing rules, and an introduction to torsional design. Unlike bending failure, which is ductile and preceded by observable deflection, shear failure is sudden and brittle—hence the design approach is deliberately conservative. Mastery of this topic is essential for PRC board examinations and professional practice, as inadequate shear reinforcement has caused catastrophic failures in Philippine buildings.
Key concepts
Total Shear Capacity (Vn)
Importance
This is the fundamental design criterion for shear. Without understanding this additive relationship, designers cannot properly size stirrups or verify section adequacy.
The nominal shear capacity is the sum of contributions from concrete (Vc) and stirrup steel (Vs): Vn = Vc + Vs. The design requirement is ΦVn ≥ Vu, where Φ = 0.75 (per NSCP 2015) and Vu is the factored shear force at the critical section (distance d from support face). This two-component model recognizes that concrete carries some shear through diagonal compression, while stirrups intercept diagonal tension cracks.
Concrete Shear Strength (Vc)
Importance
Vc is the baseline; it determines whether stirrups are even needed. If Vu ≤ ΦVc, minimal stirrups suffice. Above this threshold, additional stirrup steel is required. Board exams frequently test this calculation.
NSCP 2015 simplified formula for non-prestressed beams: Vc = 0.17λ√(f'c) bw d, where λ = 1.0 for normal-weight concrete, f'c is in MPa, bw is the web width in mm, and d is the effective depth in mm. The result is in Newtons. This formula is empirical and derived from test data. Higher concrete strength, larger cross-section, and greater depth all increase Vc. This value represents the shear force the concrete section alone can safely resist before diagonal cracking initiates.
Stirrup Shear Strength (Vs)
Importance
This formula governs the selection and spacing of stirrups. A common mistake is forgetting to divide (Vu − ΦVc) by Φ; this error leads to undersized stirrups and unsafe designs.
When Vu > ΦVc, stirrups must carry the excess: Vs = (Vu − ΦVc) / Φ. The stirrups are designed using Vs = (Av fy d) / s, where Av is the total area of steel crossing the diagonal crack (for a U-stirrup, Av = 2 × bar area), fy is the yield strength in MPa, d is the effective depth, and s is the spacing in mm. Solving for spacing: s = (Av fy d) / Vs. The stirrup acts as a mechanical tie across the developing shear crack.
Upper Limit on Vs (Crushing Prevention)
Importance
This is a section-adequacy check. If Vs (calculated from Vu) exceeds the limit, no amount of stirrup detailing will work—the beam must be redesigned structurally. This is a critical verification often overlooked by junior engineers.
To prevent concrete web crushing, NSCP 2015 limits Vs ≤ 0.66√(f'c) bw d. If this limit is exceeded, the section is inadequate and must be enlarged (increase bw or d). This limit reflects the actual bearing capacity of the concrete struts that form the diagonal compression mechanism. Exceeding this limit can cause sudden crushing even if sufficient stirrup steel is present.
Maximum Stirrup Spacing (Spacing Band Rules)
Importance
Failure to apply the correct spacing band is a common board-exam error. Students must compute Vs, check which band applies, then verify that the calculated spacing s does not exceed the appropriate s_max.
NSCP 2015 defines two spacing bands depending on the stirrup demand: (1) If Vs ≤ 0.33√(f'c) bw d, then s_max = min(d/2, 600 mm); (2) If Vs > 0.33√(f'c) bw d, then s_max = min(d/4, 300 mm). The spacing band transitions when shear demand is moderate; high-demand sections require closer stirrups to ensure the diagonal crack is adequately intercepted. This rule is purely empirical but is mandated by the code.
Minimum Stirrup Reinforcement
Importance
This requirement prevents brittle failure in low-shear zones and ensures crack control. Many inexperienced designers omit stirrups where Vu is small, only to find the section violated code on minimum ties.
Even where Vu ≤ ΦVc, minimum stirrups are required wherever Vu > 0.5ΦVc (NSCP 2015). The minimum area is Av,min = 0.062√(f'c) (bw s / fy) ≥ 0.35(bw s / fy). This ensures a minimum transverse tie exists across the section to control flexural and torsional cracking. The two expressions allow engineers to choose the more stringent condition.
Critical Section for Shear
Importance
Using Vu at the wrong location is a common error. Board problems often specify shear at mid-span or at various points; students must recognize that shear design applies only at and beyond distance d from the support.
The critical section for shear design is located at a distance d (the effective depth) from the face of the support, not at the support itself. This is because of the redistribution of stress over the effective depth. At the support face, concentrated reaction stresses are high; by distance d, these stresses have dispersed into the cross-section and the shear design becomes applicable. For cantilevers, the critical section is at the face of the fixed end.
Torsion Threshold (Torque Neglect Limit)
Importance
This threshold allows engineers to avoid the complexity of combined shear-torsion design when torsion is minor. However, for corner beams, spandrels, and edge beams in Philippine buildings (common in seismic zones), torsion is often significant and must be designed for.
For beams carrying both shear and torsion, NSCP 2015 allows torsion to be neglected if the factored torque Tu is below a threshold: Tu < Φ × 0.083λ√(f'c) × (Acp² / pcp), where Acp is the area enclosed by the outside perimeter of the section and pcp is that perimeter in mm. The Φ factor is 0.75. If Tu falls below this limit, design for shear alone; if Tu exceeds the limit, closed stirrups and longitudinal torsion bars must be added to the design.
Closed Stirrups and Longitudinal Torsion Steel
Importance
Torsion design is more involved than shear-only design. For board-exam purposes, understanding the neglect threshold is often sufficient; full torsion-design calculations are less commonly tested but may appear in advanced problems.
When torsion design is required (Tu above the threshold), both closed stirrups and longitudinal bars along the corners are necessary. Closed stirrups form a complete loop and resist the diagonal shear stresses induced by torque; longitudinal bars resist longitudinal cracking along the face. The combined system must carry both the shear Vu and the torque Tu, with combined strength formulas (outside the scope of this introductory treatment but noted for completeness).
Reduction Factor Φ = 0.75 for Shear
Importance
This factor must be applied consistently. A frequent algebraic error is forgetting to multiply Vc by Φ when checking if stirrups are needed, or forgetting to divide by Φ when calculating Vs. Board exams test this rigorously.
NSCP 2015 specifies Φ = 0.75 for shear (and torsion) design, reflecting the greater uncertainty and brittleness of these failure modes compared to flexure (Φ = 0.90). The design requirement ΦVn ≥ Vu incorporates this safety factor to account for variability in material properties, construction quality, and analytical assumptions.
Effective Depth (d) in Shear Design
Importance
Accuracy in determining d is essential. Errors here propagate through both Vc and Vs calculations. For composite beams or beams with multiple layers of steel, d must reflect the centroid of the main tension steel.
The effective depth d is the distance from the extreme compression fiber to the centroid of the tension reinforcement, typically measured perpendicular to the neutral axis. In shear calculations, d represents the lever arm over which stirrups act and is the key geometric parameter controlling Vc. A larger d increases both Vc and the effective interception length of stirrups, but deeper beams may have different bending reinforcement patterns.
Important points
- Shear failure is sudden and brittle, unlike bending failure; therefore, design is conservative (Φ = 0.75, not 0.90).
- The design criterion is ΦVn = Φ(Vc + Vs) ≥ Vu. Both concrete and stirrup contributions must be verified.
- Vc is calculated using the NSCP 2015 simplified formula: Vc = 0.17λ√(f'c) bw d (SI units: MPa, mm, N).
- When Vu > ΦVc, additional stirrups are required. The stirrup demand is Vs = (Vu − ΦVc) / Φ; do not forget the Φ divisor.
- Stirrup spacing is computed as s = (Av fy d) / Vs. For a U-stirrup, Av equals the total area of both legs.
- The crushing limit Vs ≤ 0.66√(f'c) bw d must be checked; if exceeded, the section must be enlarged.
- Spacing limits are band-dependent: s_max = min(d/2, 600) when Vs ≤ 0.33√(f'c) bw d; s_max = min(d/4, 300) when Vs exceeds this value.
- Minimum stirrups (Av,min = 0.062√(f'c) bw s / fy or 0.35 bw s / fy, whichever is larger) are required wherever Vu > 0.5ΦVc.
- The critical section for shear is at distance d from the support face, not at the support itself.
- Torsion may be neglected if Tu < Φ × 0.083λ√(f'c) × (Acp² / pcp); otherwise, closed stirrups and longitudinal bars are required.
- For combined shear-torsion sections, the design becomes significantly more complex and requires interaction equations (advanced topic).
- All formulas in NSCP 2015 assume non-prestressed concrete unless otherwise noted; prestressed beams have modified Vc expressions.
- Board exams often test stirrup spacing calculation, spacing-band selection, and identification of the critical section.
Chapter objectives
- Understand the mechanics of shear failure in reinforced concrete beams and the roles of concrete and steel in resisting shear
- Apply NSCP 2015 simplified formulas to calculate concrete shear strength (Vc) and design stirrups (Vs) for non-prestressed beams
- Design vertical stirrups, determine appropriate spacing, and verify compliance with maximum-spacing and minimum-reinforcement requirements
- Recognize the critical section for shear design (distance d from the support face) and properly apply the Φ = 0.75 reduction factor
- Evaluate whether torsion must be designed for or may be neglected based on the torque threshold, and understand the role of closed stirrups and longitudinal torsion bars
- Solve board-style numerical problems involving shear and torsion design with practical column and beam geometry
- Identify common pitfalls in shear design (forgetting Φ divisor, leg-count errors, spacing-band confusion) and avoid them in practice
Concept relationships
Concrete shear strength Vc is the baseline. If Vu ≤ ΦVc, stirrups are not strictly required for shear (only minimum ties for control). If Vu > ΦVc, stirrup demand Vs is calculated as (Vu − ΦVc) / Φ, and stirrup spacing and area are sized accordingly.
Relationship
Vc → Stirrup Requirement
Larger web width bw and greater effective depth d both increase Vc, reducing stirrup demand. However, a larger d also permits greater spacing (s_max depends on d), so there is a beneficial geometric effect. Conversely, increasing bw does not change the spacing limit directly but increases the load Vc can carry.
Relationship
Section Size → Vc & Spacing
Higher fy yields closer spacing (s decreases as Vs = Av fy d / s). Thus, high-strength steel allows wider spacing, reducing labor and congestion—a practical design consideration.
Relationship
Stirrup Yield Strength → Spacing
When Vs reaches 0.33√(f'c) bw d, the spacing band transitions from d/2 to d/4. This reflects the code's philosophy: moderate shear can be handled with looser stirrups, but high shear demands tight control to intercept cracks.
Relationship
Shear Demand & Spacing Band Transition
The check Vs ≤ 0.66√(f'c) bw d is independent of stirrup design; it is a statement about the concrete's ability to form the diagonal compression strut. If this limit is violated, no stirrup detailing will prevent crushing—the section must be enlarged. This is a hard constraint.
Relationship
Crushing Check → Section Adequacy
The critical section (distance d from support) is where the diagonal cracks typically initiate and shear design applies. Near the support, stress redistribution is complex; beyond distance d, simpler design assumptions hold. For cantilevers, the critical section is at the fixed end, not at the free end.
Relationship
Critical Section & Support Condition
The torsion neglect threshold depends on Acp (outside cross-sectional area) and pcp (outside perimeter). Wider, deeper sections have larger Acp and pcp, raising the threshold—thus larger beams can carry more torque before design is required. This reflects the better torsional rigidity of larger sections.
Relationship
Torsion Threshold & Cross-Section Geometry
Minimum stirrups are required wherever Vu > 0.5ΦVc, a lower threshold than the point where calculated stirrups become necessary (Vu > ΦVc). This ensures crack control even in low-shear zones.
Relationship
Minimum Stirrup Trigger & Vu Threshold
Practical applications
A three-story office building in Metro Manila uses 300 mm wide, 600 mm deep rectangular beams (f'c = 28 MPa, fy = 415 MPa) spanning 8 m. Shear analysis shows Vu = 280 kN at the critical section (distance d from support). Design appropriate stirrups.
Key steps
- Calculate Vc = 0.17(1.0)√28 × 300 × 600 = 181.9 kN. 2. Calculate ΦVc = 0.75 × 181.9 = 136.4 kN. 3. Since Vu = 280 > ΦVc, stirrups are needed. 4. Calculate Vs = (280 − 136.4) / 0.75 = 191.5 kN. 5. Check crushing: 0.66√28 × 300 × 600 = 591.6 kN > 191.5 ✓. 6. Determine spacing band: 0.33√28 × 300 × 600 = 296 kN > Vs, so s_max = min(600/2, 600) = 300 mm. 7. Select 10 mm U-stirrups (Av = 2 × 78.5 = 157 mm²): s = (157 × 415 × 600) / 191,500 = 203 mm. 8. Use s = 200 mm (practical). 9. At distance 2d from support (where Vu drops to ~150 kN), recalculate and widen spacing to 300 mm.
Application
Design of Rectangular Beam Stirrups for Office Building
A corner spandrel beam in a 20-story residential building (400 mm × 600 mm, f'c = 30 MPa) experiences Tu = 15 kN·m from the eccentric loading of exterior curtain wall and floor. Must torsion be designed for?
Key steps
- Calculate Acp = 400 × 600 = 240,000 mm². 2. Calculate pcp = 2(400 + 600) = 2000 mm. 3. Calculate threshold: Φ × 0.083 × √30 × (240,000)² / 2000 = 0.75 × 0.083 × 5.477 × (240,000)² / 2000 = 12.36 kN·m. 4. Since Tu = 15 > 12.36 kN·m, torsion design is required. 5. Closed stirrups (instead of U-stirrups) and longitudinal bars at the corners must be added to handle both shear and torque.
Application
Torsion Threshold Check for Corner Beam
In the same office-building beam, at mid-span (4 m from support), shear analysis yields Vu = 50 kN. Calculate whether stirrups are required and, if so, what minimum area applies.
Key steps
- Vc = 181.9 kN (same as before). 2. ΦVc = 136.4 kN. 3. Check: Vu = 50 < ΦVc = 136.4, so calculated stirrups are not needed. 4. Check minimum requirement: 0.5ΦVc = 0.5 × 136.4 = 68.2 kN. 5. Since Vu = 50 < 68.2, even minimum stirrups are not strictly required (but may be placed for crack control). 6. If placed, use Av,min = 0.062√28 × 300 × s / 415 = 0.0247s (mm²), or 0.35 × 300 × s / 415 = 0.252s. The latter governs: Av,min ≥ 0.252 × 300 = 75.6 mm² per stirrup. Use 10 mm bars (78.5 mm² each).
Application
Minimum Stirrup Requirement in Low-Shear Zone
An engineer proposes a 250 mm wide beam with d = 500 mm and f'c = 25 MPa to carry Vu = 400 kN. Check if the section is adequate.
Key steps
- Vc = 0.17 × √25 × 250 × 500 = 106.3 kN. 2. ΦVc = 0.75 × 106.3 = 79.7 kN. 3. Vs = (400 − 79.7) / 0.75 = 426.4 kN required. 4. Crushing limit: 0.66√25 × 250 × 500 = 412.5 kN. 5. Since Vs = 426.4 > 412.5, the section is inadequate—crushing will occur before stirrups can be fully effective. Recommendation: Increase bw to 300 mm or d to 550 mm, or increase f'c to 30 MPa.
Application
Comparison of Section Adequacy: When Vs Exceeds Crushing Limit
In the office-building beam (Vc = 181.9 kN, ΦVc = 136.4 kN), at what distance from the support can 12 mm U-stirrups (Av = 2 × 113 = 226 mm², s = 150 mm) be discontinued?
Key steps
- The stirrups can be stopped where Vu drops to 0.5ΦVc = 0.5 × 136.4 = 68.2 kN. 2. From shear diagram (assumed linear over 8 m span from 280 kN at support), Vu = 68.2 kN occurs at distance x from support where: (280 − 68.2) / 8 × 2 = 280 − 53.05x, solving x = 4.0 m. 3. Stirrups can be stopped at approximately 4.0 m from the support. 4. However, continue at minimum spacing for crack control from 4.0 m to mid-span (best practice).
Application
Determination of Stirrup Stop-off Point
Diagrams
Shear Design Decision Tree and Calculation Flow (NSCP 2015)
This flowchart guides the designer through the complete shear-design process: identification of critical shear force, calculation of concrete and stirrup contributions, verification of crushing, determination of spacing band, and final stirrup sizing. It shows decision points and pathways when stirrups are not needed, when minimum stirrups apply, and when full design is required.
Reinforced Concrete Beam Shear and Torsion: Concept Map
A comprehensive mind map showing the hierarchical structure of shear and torsion design concepts. It organizes the key topics (shear framework, stirrup rules, critical section, torsion, code citations, and common errors) to show how they relate and form the complete design methodology.
Shear Crack Development and Stirrup Interception (Conceptual Sequence)
This sequence diagram illustrates the mechanism of shear failure and how stirrups intervene. As shear load increases, concrete first carries part of it (Vc). When this is exceeded, diagonal tension cracks initiate. Stirrups crossing these cracks develop tension and provide the additional capacity (Vs) needed to carry the excess shear.
Beam Section State Transitions Based on Shear Demand
This state diagram maps the progression of design decisions as shear demand increases. Each state represents a different design condition (low shear, minimum stirrups needed, full design required) and shows the transitions and checks required to move from one state to another.
Torsion Design Decision: Neglect vs. Design (NSCP 2015)
This flowchart shows the decision process for torsion design. It calculates the neglect threshold based on section geometry and concrete strength, then determines whether full torsion design is required or if shear-only design suffices. For practical board-exam purposes, mastering the neglect criterion is often the key learning objective.
Typical Load Distribution in Shear-Critical Beam Section
This pie chart illustrates a typical shear load-sharing scenario in a reinforced concrete beam where stirrups are designed. In many practical designs, stirrup steel carries 60–70% of the shear demand, while concrete provides the baseline capacity. The exact split depends on the ratio Vu/ΦVc. (This is illustrative; the actual ratio varies with each design.)
In summary
Shear and torsion design in reinforced concrete beams is governed by NSCP 2015 and requires a systematic approach balancing empirical formulas with practical construction constraints. The core concept—that nominal shear capacity Vn = Vc + Vs must satisfy ΦVn ≥ Vu with Φ = 0.75—is straightforward, but execution demands careful attention to the critical section (distance d from support), the spacing-band transition rule, and the crushing limit check. Concrete provides the baseline resistance Vc = 0.17λ√(f'c) bw d; when shear demand exceeds ΦVc, stirrups are sized to carry Vs = (Vu − ΦVc) / Φ, with spacing determined by s = Av fy d / Vs and constrained by code-mandated maximum values. A common pitfall is forgetting to divide by Φ when calculating Vs—this error leads to undersized stirrups and unsafe designs. Equally important is recognizing the crushing limit Vs ≤ 0.66√(f'c) bw d; if this is exceeded, no stirrup detailing will suffice and the section must be enlarged. Torsion, often significant in corner beams and spandrels in Philippine multi-story buildings, may be neglected if Tu falls below a threshold depending on outside section dimensions; otherwise, closed stirrups and corner longitudinal bars are required. Mastery of these concepts, reinforced by solving board-style numerical problems, is essential for passing the PRC Civil Engineer Licensure Examination and for designing safe, code-compliant structures in professional practice.