CELE Reinforced & Prestressed Concrete — Reinforced Concrete Beams: Shear and TorsionStudy Notes

Shear stress in a beam arises from the vertical component of load and induces diagonal tension in the concrete. Unlike the uniform bending stress distribution, shear stresses are highest near the supports and zero at midspan. The shear force diagram (SFD) shows this variation; critical sections for design occur where the shear magnitude is largest.

A concrete beam without stirrups fails suddenly when diagonal cracks propagate through the web. The crack typically initiates at 45° and runs from the neutral axis toward the compression zone. Once this crack forms, the uncracked concrete above can no longer carry shear; failure is imminent. This brittle behavior is why building codes require designer conservatism: all shear strength contributions (concrete and steel) are multiplied by a reduction factor φ = 0.75, lower than the 0.85 used for bending.

The total shear capacity is the sum of two components:

V_n = V_c + V_s

where:

• V_c = shear strength contributed by concrete (intrinsic to the beam geometry and strength)
• V_s = shear strength contributed by stirrups (reinforcement crossing the diagonal crack)

The design criterion is:

φV_n ≥ V_u (with φ = 0.75)

where V_u is the factored shear force at the critical section. The critical section is located at a distance d from the support face (not at the support itself). This distance recognizes that compressive arch action near the support can carry some shear without stirrups.

Practical insight for the exam: A beam with inadequate stirrups does not yield gradually; it can collapse during a single load application. Therefore, the design code is deliberately conservative, and you must never skip minimum stirrup requirements or exceed spacing limits.

NSCP 2015 (Chapter 11, Shear and Torsion) provides a simplified formula for concrete shear strength in non-prestressed beams, analogous to ACI 318:

V_c = 0.17 × λ × √(f'_c) × b_w × d

where:

• λ = modification factor for concrete type (λ = 1.0 for normal-weight concrete; λ = 0.85 for sand-lightweight; λ = 0.75 for all-lightweight). For most Philippine design, use λ = 1.0.
• f'_c = concrete compressive strength in MPa (typical: 28 or 35 MPa)
• b_w = web width in mm (for rectangular beams, b_w = beam width b)
• d = effective depth in mm (distance from extreme compression fiber to centroid of tensile steel)
• Result V_c is in Newtons (N); convert to kN by dividing by 1000.

This formula applies when:

• The beam is not prestressed
• No special loading conditions (e.g., no concentrated load very close to support)
• Shear is not combined with significant axial force

The physical meaning: V_c represents the intrinsic ability of concrete (through aggregate interlock and dowel action) to resist diagonal tension. It depends on the square root of strength (not linearly), the cross-sectional area (b_w × d), and a calibrated coefficient (0.17).

Once V_c is calculated, the design shear capacity from concrete is:

φV_c = 0.75 × V_c

If the factored shear force V_u exceeds this, stirrups must carry the excess. If V_u is less than 0.5φV_c, no stirrups are required (minimum reinforcement rule). If V_u is between 0.5φV_c and φV_c, minimum stirrups are still provided.

Practical note for licensure candidates: This formula is an empirical fit to test data. Do not try to derive it from first principles—memorize it, use it correctly, and move to the next part of the design. The constant 0.17 is calibrated specifically for SI units (MPa, mm, N); if you switch to English or mixed units, you will need to recalibrate.

When the factored shear V_u exceeds the concrete's contribution φV_c, vertical stirrups (also called web reinforcement or links) must carry the excess shear. The stirrup design follows a straightforward logic:

Step 1: Calculate the required stirrup strength.

The shear not carried by concrete must be supplied by stirrups:

V_s = (V_u − φV_c) / φ

Note the division by φ: this converts the shear deficit into the required nominal strength from stirrups. Algebraically,
φV_n = φ(V_c + V_s) ≥ V_u
φV_c + φV_s ≥ V_u
φV_s ≥ V_u − φV_c
V_s ≥ (V_u − φV_c) / φ

Many candidates forget this division by φ and produce incorrect stirrup spacing. Remember: the formula is V_s = (V_u − φV_c) / φ, not just V_u − φV_c.

Step 2: Relate V_s to stirrup area and spacing.

The vertical component of stirrup force is determined by the amount of steel crossing a diagonal crack per unit distance:

V_s = (A_v × f_y × d) / s

where:

• A_v = total area of stirrup legs crossing the shear crack (mm²)

  • For a U-stirrup (closed on three sides): A_v = 2 × A_bar
  • For a full closed stirrup: A_v = 4 × A_bar
  • Example: Two 10 mm bars in a U-stirrup: A_v = 2 × 78.5 = 157 mm²
• f_y = yield strength of stirrup steel (typically 415 MPa for deformed bars in the Philippines)
• d = effective depth (mm), same as for concrete contribution
• s = stirrup spacing along beam axis (mm)

Rearranging for spacing:

s = (A_v × f_y × d) / V_s

This is the fundamental stirrup spacing formula. Once you know the bar size (A_v), yield strength (f_y), and effective depth (d), you can compute the spacing required to provide the necessary V_s.

Step 3: Check the upper limit on V_s.

The stirrups cannot carry unlimited shear; if too much shear is demanded, the concrete struts between diagonal cracks crush. NSCP 2015 limits:

V_s ≤ 0.66 × √(f'_c) × b_w × d

If your computed V_s exceeds this limit, the beam section is inadequate—you must enlarge b_w or d, increase f'_c, or reduce the applied load. Increasing stirrup area or tightening spacing will not solve an oversized shear demand; it is a section problem, not a reinforcement problem.

Step 4: Apply maximum spacing limits.

Even if the stirrup spacing formula gives a value less than this limit, code imposes maximum spacing to ensure adequate distribution:

• If V_s ≤ 0.33 × √(f'_c) × b_w × d (relatively light shear):
  s_max = min(d/2, 600 mm)
• If V_s > 0.33 × √(f'_c) × b_w × d (heavier shear):
  s_max = min(d/4, 300 mm)

The design spacing is the minimum of the calculated s and s_max:
s_design = min(s, s_max)

The spacing threshold of 0.33√(f'_c) b_w d marks a transition: below it, stirrups can be more widely spaced; above it, they must be tighter. This reflects code philosophy: under heavy shear, cracks propagate more aggressively, and closer stirrups are needed to control them.

Step 5: Minimum stirrup requirement.

Even where V_u is low, some stirrups are required for constructability and durability. Specifically, stirrups must be provided whenever:

V_u > 0.5 × φ × V_c

When stirrups are required, the minimum cross-sectional area is:

A_v,min = 0.062 × √(f'_c) × (b_w × s) / f_y (but not less than 0.35 × (b_w × s) / f_y)

For a U-stirrup (two legs of diameter d_b):
A_v = 2 × (π × d_b² / 4) = π × d_b² / 2

The minimum rule ensures that even lightly sheared sections have some transverse reinforcement. If minimum A_v is larger than required by the strength formula, use minimum.

Practical exam tips:

1. Always check if stirrups are needed: Is V_u > 0.5φV_c?
2. If yes, compute required V_s using V_s = (V_u − φV_c) / φ.
3. Select a stirrup bar size and geometry (U-stirrup, full stirrup, multiple legs).
4. Calculate s from the V_s formula.
5. Check against V_s,max (if exceeded, enlarge section).
6. Compare with s_max (use the smaller value).
7. Round down to a practical spacing (typically 50 mm, 75 mm, 100 mm, 150 mm, 200 mm, etc.).

Torsion (twisting) is a third shear-related phenomenon that occurs when loads are applied off the center of shear or when the beam is part of a frame resisting lateral forces. Unlike bending (which creates compression and tension over the depth) or vertical shear (which creates diagonal tension), torsion induces diagonal tension on all four faces of the beam section, requiring closed (fully enclosed) stirrups and longitudinal bars throughout the perimeter.

Torsional cracks spiral around the section, typically at 45° to the axis. A torsional failure is sudden and can be catastrophic. However, for many practical beams, torsion is secondary or can be safely neglected under code allowances.

Neglect Criteria (NSCP 2015, Section 11.6.1):

You may neglect torsion design if the factored torque is less than the threshold:

T_u < φ × 0.083 × λ × √(f'_c) × (A_cp² / p_cp)

where:

• T_u = factored torque (kN⋅m)
• φ = 0.75 (reduction factor)
• λ = 1.0 for normal-weight concrete
• f'_c = concrete compressive strength (MPa)

• A_cp = area enclosed by the outer perimeter of the section (mm²)

  • For a rectangle: A_cp = b × h
  • Do not deduct the hollow area; this is the gross enclosed area

• p_cp = outer perimeter of the section (mm)

  • For a rectangle: p_cp = 2(b + h)
• Result is in N⋅mm; convert to kN⋅m by dividing by 10⁶

This formula comes from the square of the section's longest diagonal: torsional resistance is proportional to the cross-sectional properties (A_cp, p_cp) and concrete strength.

Physical interpretation: If the applied torque is below this threshold, the concrete's tensile capacity (even without reinforcement) is sufficient to prevent major cracking. For practical beams in buildings, this threshold is often greater than the actual torque, allowing you to design for shear alone.

Conditions where torsion must be considered:

• Curved or skewed beams (e.g., in ramps or circular structures)
• Edge beams of floor systems (spandrel beams) that resist lateral sway
• Beams that are part of closed frames (e.g., perimeter beams in a rigid frame)
• Beams with eccentric or offset loading
• Cantilevered sections with lateral loads

For an introductory licensure exam, torsion design details are often abbreviated. The key skill is recognizing when torsion can be neglected and when it must be considered. Once torsion must be designed, the method combines closed stirrups (to resist both shear and torsion) with longitudinal bars at the corners and faces.

Important note: If torsion must be designed, the stirrup design becomes more complex—you cannot simply use the vertical shear formula. Torsion and shear interact, and combined reinforcement must satisfy both demands simultaneously. This is beyond the scope of an introductory study, but be aware that checking the neglect criterion is your first step in any shear-dominant problem.

A complete shear design follows a systematic sequence. Mastering this workflow is essential for the licensure exam, where time management and accuracy are both critical.

Design Sequence:

1. Determine loads and find shear force diagram (SFD).

   • Calculate factored loads (dead + live with appropriate factors per NSCP).
   • Sketch the SFD; identify the maximum shear and its location.
   • Remember that the critical section for design is at distance d from the support face, not at the support itself.

2. Calculate concrete shear capacity.

   • Use V_c = 0.17√(f'_c) b_w d.
   • Multiply by φ = 0.75 to get φV_c.

3. Check if stirrups are required.

   • Compare V_u at critical section with 0.5φV_c.
   • If V_u < 0.5φV_c, stirrups may be omitted (though code may require minimum for ductility).
   • If V_u ≥ 0.5φV_c, stirrups are mandatory.

4. Torsion screening (if applicable).

   • Check if T_u < threshold; if yes, neglect torsion and proceed with shear only.
   • If T_u ≥ threshold, design for combined shear + torsion (beyond this chapter's scope).

5. Design stirrups in regions where V_u > φV_c.

   • Compute V_s = (V_u − φV_c) / φ.
   • Select stirrup bar size (typically 8, 10, or 12 mm).
   • Choose geometry (U-stirrup for simple beams, closed stirrup if torsion is present).
   • Calculate required spacing: s = (A_v f_y d) / V_s.
   • Apply maximum spacing limit based on V_s relative to 0.33√(f'_c) b_w d.
   • Use s_design = min(calculated s, s_max).

6. Check stirrup limits and minimize required area.

   • Verify V_s ≤ 0.66√(f'_c) b_w d (crushing limit); if exceeded, enlarge section.
   • Check minimum A_v where required.
   • If feasible, round spacing to practical increments (50, 75, 100, 150, 200, 250, 300, 600 mm).

7. Determine stirrup extension.

   • Stirrups are required from distance d from one support to distance d from the opposite support (approximately from x = d to x = L − d).
   • Beyond x = L − d, stirrups can be stopped or spaced more widely.
   • Always provide at least two stirrups at each end, regardless of shear magnitude (constructability).

8. Document the design.

   • Show the stirrup layout on drawings: identify bar size, number of legs (e.g., "2 legs 10 mm ø"), spacing, and extent.
   • Example notation: "2ø10 U-stirrups @ 150 mm c/c from x = 500 mm to x = 7,500 mm" (for an 8 m beam).

Practical considerations for the exam:

• Unit consistency: SI units (N, MPa, mm) throughout. Convert final answers to kN and m for clarity.
• Spacing selection: Always round calculated spacing down to a practical value. Tighter spacing is safer and easier to construct.
• Effective depth (d): This is from the extreme compression fiber to the centroid of the tension reinforcement. For a beam with 40 mm cover and 20 mm diameter main bars, d = h − cover − half-bar diameter. Do not confuse d with overall depth h.
• Web width (b_w): For a solid rectangular beam, b_w = b (width). For flanged sections (T-beam, I-beam), b_w is the width of the web (the narrow part), not the flange.
• Factor of safety: The φ = 0.75 factor is inherent in the code formulas. Do not apply an additional safety factor by hand.
• Critical section: Shear is critical at x = d from the support. This is where you compute V_u for stirrup design. Do not use shear at the support itself.

Common errors to avoid:

1. Forgetting the division by φ when computing V_s. It is V_s = (V_u − φV_c) / φ, not (V_u − φV_c).
2. Confusing A_v (total stirrup area) with individual bar area. For two 10 mm bars, A_v = 2 × 78.5 = 157 mm², not 78.5 mm².
3. Applying the wrong s_max band. Check whether V_s ≤ 0.33√(f'_c) b_w d to determine if s_max = d/2 (600) or d/4 (300).
4. Using overall depth h instead of effective depth d. These differ by the cover and half the main bar diameter.
5. Assuming stirrups extend the full beam length. They actually run from x = d to x = L − d (approximately).
6. Neglecting minimum stirrup check. Even if strength demand is low, minimum A_v may govern, and spacing must be tightened accordingly.
7. Not rounding spacing to a practical value. Spacing should be 50 mm, 100 mm, 150 mm, etc.—not 163.9 mm, even if that is what the formula gives.

As you prepare for the PRC licensure examination, be aware of several nuanced topics and common traps that appear in past papers.

Critical Section Location:

The most frequently missed concept is the location of the critical section for shear design. The code specifies that shear is checked at a distance d from the face of the support, not at the support itself. This is because near the support, a compressive strut can form (like an arch action), allowing the beam to transfer load without stretching the main steel. Beyond distance d, the full shear must be carried by stirrups.

For a cantilever, the critical section is at d from the fixed end (not at the fixed end).
For a simply supported beam with a concentrated load at midspan, the maximum shear is at the supports, but design is based on V_u at x = d.
For a distributed load, the shear diagram is continuous, and you check at x = d from each support.

Effective Depth (d) vs. Overall Depth (h):

Many candidates use the overall beam height h when the formula requires effective depth d. Remember:

d = h − concrete cover − half the main bar diameter

Example: For a 600 mm deep beam with 40 mm cover and 25 mm main bars:
d = 600 − 40 − 12.5 = 547.5 mm ≈ 550 mm

The difference between d and h is typically 50–80 mm. Using h instead of d can reduce your computed shear capacity by 10–15%, risking unsafe design.

Web Width (b_w) for Flanged Sections:

For T-beams, inverted T-beams, and I-beams, the shear formula uses b_w (the web width), not the flange width. This is the narrow part connecting the compression and tension zones.

Example: A T-beam has b_f = 1000 mm (flange) and b_w = 250 mm (web). Use b_w = 250 mm in the V_c formula, not 1000 mm.

Two-Way Shear (Punching):

While this chapter focuses on one-way shear in beams, be aware that flat slabs and two-way systems experience punching shear around columns. The formula and design procedure are different (NSCP 11.12). If the exam includes a punching question, apply the appropriate criterion (not the one-way formula).

Shear Reinforcement in the Flexural Region:

Where bending moment is high, longitudinal steel ratio is high, and the beam may be over-reinforced. NSCP permits reduced concrete shear strength in these cases using a more refined formula involving ρ_w (the web reinforcement ratio). The simplified formula (used here) is conservative and safe but may overestimate stirrup requirements slightly. For licensing exam purposes, the simplified formula (V_c = 0.17√(f'_c) b_w d) is sufficient and is the one most commonly tested.

Torsion Combined with Shear:

Once torsion becomes significant (T_u exceeds the neglect threshold), the stirrup design becomes coupled. The stirrups must resist both shear and torsional forces simultaneously. The stirrup area and spacing are determined by a combined formula, and you must add longitudinal torsion reinforcement at the corners and sides. This topic is typically abbreviated in introductory exams but may appear as a more detailed question. If it does, reference ACI 318 Section 11.6.3 (or NSCP equivalent) for the combined formulation.

Minimum Reinforcement in Flanges:

For wide-flange beams (T or I shape), stirrups must be distributed across the width of the flange if the flange width exceeds a certain ratio relative to the web width. The code requires that stirrups (or additional transverse reinforcement) be placed to prevent web-flange separation. For narrow flanges, this is not critical; for very wide flanges, additional stirrups parallel to the main stirrups may be needed. This is a more advanced topic often tested in specialized exams.

Precast Connections and Shear Transfer:

In precast construction, shear must be transferred across the joint between the precast member and the cast-in-place topping or connection. Shear friction (a separate design check) may govern. Ensure that you identify whether a problem involves precast elements and apply the appropriate shear transfer criterion.

Exam Strategy:

1. Always start by calculating V_c and φV_c. These are the foundation for all subsequent decisions.
2. Sketch the shear force diagram if load distributions are complex. This prevents errors in identifying the critical section.
3. Check the units at every step. A common error is mixing MPa and Pa, or mm and m.
4. Round down spacing to a practical value (always safer). Spacing of 150 mm is better than 163.9 mm.
5. Document your work in a logical sequence (loads → shear → concrete capacity → stirrup design → spacing → extent). This makes it easy to spot errors if your answer seems wrong.
6. Compare your answer with the magnitude of the shear. If V_u is 200 kN and your stirrup area works out to 5 mm², something is wrong. Conversely, if stirrup spacing is 5 mm, something is also wrong.
7. Check reasonableness. For a typical 300×500 mm beam with f'_c = 28 and V_u = 150 kN, expect stirrup spacing between 100–250 mm. If you get 50 mm or 500 mm, re-examine your calculation.