Reinforced Concrete Beams: Shear and Torsion - Detailed Explanation
Shear and torsion are two of the most critical—and most frequently tested—topics in the PRC Civil Engineer Licensure Examination under Reinforced and Prestressed Concrete. Unlike flexural failure, which is ductile and provides visible warning signs (excessive deflection, wide cracks), shear failure is sudden, brittle, and catastrophic. This makes shear design a life-safety concern and a cornerstone of NSCP 2015 provisions.
This chapter covers the complete design framework: (1) computing concrete shear strength Vc using the NSCP 2015 simplified formula, (2) designing vertical stirrups for the excess shear Vs, (3) applying spacing limits and minimum reinforcement rules, and (4) determining whether torsional effects must be explicitly designed for or can be neglected. Board exam problems on this topic typically test formula recall, unit consistency, correct identification of the critical section, and application of spacing limit bands. Master each concept here and you will confidently solve any shear-torsion problem in the licensure exam.
Shear Strength of Reinforced Concrete Beams
Examples
A simply supported rectangular beam has bw = 300 mm, d = 500 mm, f'c = 28 MPa, normal-weight concrete. The factored shear at the critical section is Vu = 95 kN. Determine if stirrups are required.
Solution
Step 1: Compute Vc.
Vc = 0.17(1.0)√28 × 300 × 500
√28 = 5.292 MPa^0.5
Vc = 0.17 × 5.292 × 300 × 500 = 134,946 N ≈ 134.9 kN
Step 2: Compute φVc.
φVc = 0.75 × 134.9 = 101.2 kN
Step 3: Compute 0.5φVc.
0.5φVc = 0.5 × 101.2 = 50.6 kN
Step 4: Compare with Vu = 95 kN.
50.6 kN < Vu (95 kN) ≤ φVc (101.2 kN)
Conclusion: Since Vu > 0.5φVc but Vu ≤ φVc, minimum stirrups are required but no excess shear design is needed.
This problem illustrates the three-zone check. Many examinees incorrectly conclude no stirrups are needed because Vu < φVc. The code requires minimum stirrups whenever Vu > 0.5φVc. This is the threshold that triggers transverse reinforcement even when concrete alone could carry the shear.
Board-style quick check: Beam with bw = 250 mm, d = 450 mm, f'c = 21 MPa. Find φVc.
Solution
Vc = 0.17(1.0)√21 × 250 × 450
√21 = 4.583
Vc = 0.17 × 4.583 × 250 × 450 = 87,765 N = 87.8 kN
φVc = 0.75 × 87.8 = 65.8 kN
Quick recall: Vc = 0.17λ√f'c · bw · d. Keep units consistent: bw and d in mm, f'c in MPa → result in N. Divide by 1000 for kN.
The total nominal shear strength of a reinforced concrete beam is the sum of two components: the concrete contribution Vc and the stirrup (transverse reinforcement) contribution Vs.
Vn = Vc + Vs
The design requirement from NSCP 2015 (Section 422) is:
φVn ≥ Vu where φ = 0.75 (shear strength reduction factor)
The factored shear Vu is computed from the structural analysis under factored load combinations (1.2D + 1.6L, etc.). Crucially, NSCP 2015 specifies that the critical section for shear is located at a distance d from the face of the support, not at the centerline or face of the beam. This is because the support reaction introduces compression into the concrete adjacent to the support, which inhibits diagonal cracking in that zone.
CONCRETE SHEAR STRENGTH (NSCP 2015 Simplified, Section 422.5.5.1):
Vc = 0.17 λ √f'c · bw · d
Where:
- λ = 1.0 for normal-weight concrete (λ = 0.75 for lightweight)
- f'c = compressive strength of concrete (MPa)
- bw = web width of the beam (mm)
- d = effective depth from compression face to centroid of tension steel (mm)
- Vc is in Newtons (N)
This simplified formula is the most commonly used in board exams. It is derived from the more detailed formula that accounts for the reinforcement ratio ρw and the ratio Vu·d/Mu, but the simplified version is conservative and code-compliant for most beams.
Key design check:
- If Vu ≤ 0.5φVc: No stirrups required (minimum stirrups may still apply in some cases)
- If 0.5φVc < Vu ≤ φVc: Minimum stirrups required
- If Vu > φVc: Design stirrups for the excess (Vs > 0)
Applications
- Design of shear reinforcement in floor beams of residential and commercial buildings (NSCP 2015 compliance)
- Checking adequacy of existing beams during structural assessment and retrofit
- Girder design for bridges and elevated structures under live load shear
- Preliminary sizing of beam cross-sections to meet shear demand
Common misconceptions
- Using φ = 0.90 (flexure factor) instead of φ = 0.75 for shear — a critical error that underestimates required stirrups
- Checking shear at the support centerline instead of at distance d from the support face — overestimates Vu
- Using gross dimensions instead of effective depth d in the Vc formula
- Forgetting that λ = 0.75 for lightweight concrete — overestimates Vc
- Concluding no stirrups are needed whenever Vu ≤ φVc — ignores the 0.5φVc minimum stirrup threshold
Related concepts
- Stirrup Design (Vs calculation)
- Minimum Shear Reinforcement
- Critical Section Location
- Load Factors and Combinations (NSCP 2015 Section 409)
Common exam questions
Example
A beam with bw = 300 mm, d = 520 mm, f'c = 35 MPa — find φVc.
Approach
Apply Vc = 0.17λ√f'c · bw · d. Verify λ value (normal vs. lightweight). Multiply by φ = 0.75 to get φVc. Convert N to kN.
Example
Given Vu = 75 kN and φVc = 85 kN — are stirrups needed?
Approach
Compare Vu to 0.5φVc and φVc to determine the zone: no stirrups, minimum stirrups, or designed stirrups.
Key points to remember
- φ = 0.75 for shear (not 0.90 which is for flexure)
- Critical shear section is at distance d from the face of support, NOT at the support centerline
- Vc = 0.17λ√f'c · bw · d (SI units, result in Newtons)
- λ = 1.0 for normal-weight concrete, 0.75 for all-lightweight
- Vu is factored shear; always use factored loads in design
- φVn = φ(Vc + Vs) ≥ Vu is the design inequality to satisfy
- Minimum stirrups required if Vu > 0.5φVc
Stirrup Design: Required Spacing and Limits
Examples
Complete board-style stirrup design: Beam bw = 300 mm, d = 500 mm, f'c = 28 MPa, fy = 415 MPa. Vu = 250 kN. Use 10 mm U-stirrups (Av = 157 mm²).
Solution
Step 1: From previous example, φVc = 101.2 kN.
Vs = (Vu − φVc)/φ = (250 − 101.2)/0.75 = 148.8/0.75 = 198.4 kN
Step 2: Check crushing limit.
0.66√28 × 300 × 500 = 0.66 × 5.292 × 150,000 = 524,000 N = 524.0 kN
Vs (198.4 kN) < 524.0 kN ✓ Section is adequate.
Step 3: Determine spacing band.
0.33√28 × 300 × 500 = 0.33 × 5.292 × 150,000 = 262,000 N = 262.0 kN
Vs (198.4 kN) < 262.0 kN → Band 1 applies
s_max = min(d/2, 600) = min(250, 600) = 250 mm
Step 4: Compute required spacing.
s = Av·fy·d / Vs = 157 × 415 × 500 / 198,400
s = 32,577,500 / 198,400 = 164.2 mm
Step 5: Choose governing spacing.
s_final = min(164.2, 250) = 164.2 mm → Use s = 150 mm (round down for safety)
Answer: Provide 10 mm U-stirrups at s = 150 mm c/c
This is the full 5-step procedure expected in board exams. Note that the computed spacing (164 mm) already satisfies the spacing limit (250 mm), so the spacing limit is not the governing criterion here. Always report a practical rounded-down value.
High shear case: Same beam but Vu = 380 kN. Check if section is adequate and find spacing.
Solution
Vs = (380 − 101.2)/0.75 = 278.8/0.75 = 371.7 kN
Crushing check: Vs,max = 524.0 kN > 371.7 kN ✓ OK
Band check: 0.33√28 × 150,000 = 262.0 kN
Vs (371.7 kN) > 262.0 kN → Band 2 applies
s_max = min(d/4, 300) = min(125, 300) = 125 mm
s_required = 157 × 415 × 500 / 371,700 = 87.6 mm
s_final = min(87.6, 125) = 87.6 mm → Use s = 80 mm
Note: Very close spacing signals that a larger beam section should be considered for economy.
When Vs exceeds the band threshold (0.33√f'c·bw·d), the maximum allowable spacing halves. This example demonstrates Band 2 where s_max drops to min(d/4, 300 mm). The governing spacing here is the computed value (87.6 mm), not the limit.
When the factored shear Vu exceeds φVc, the excess shear must be carried by transverse reinforcement (stirrups). The design procedure follows these sequential steps:
STEP 1 — COMPUTE REQUIRED Vs:
Vs = (Vu − φVc) / φ
Note carefully: you subtract φVc from Vu, then divide by φ. Board exam pitfall #1 is computing Vs = Vu − φVc without dividing by φ, which underestimates the required stirrup strength.
STEP 2 — CHECK CRUSHING LIMIT:
Vs ≤ Vs,max = 0.66√f'c · bw · d
This is the upper bound on Vs. If Vs exceeds this limit, the concrete between diagonal cracks will crush before stirrups yield. The section must be enlarged — stirrups alone cannot solve this. This is a 'section inadequacy' check.
STEP 3 — DETERMINE Av (STIRRUP AREA):
For a standard U-stirrup (2 legs), Av = 2 × Abar
For a 10mm bar: Abar = π/4 × 10² = 78.54 mm² → Av = 157 mm²
For a 12mm bar: Abar = 113.1 mm² → Av = 226 mm²
STEP 4 — COMPUTE REQUIRED SPACING:
From Vs = Av·fy·d/s, rearrange:
s = Av·fy·d / Vs
STEP 5 — APPLY MAXIMUM SPACING LIMITS:
The code imposes spacing limits to ensure no diagonal crack can pass between stirrups:
Band 1 — If Vs ≤ 0.33√f'c · bw · d:
s_max = min(d/2, 600 mm)
Band 2 — If Vs > 0.33√f'c · bw · d:
s_max = min(d/4, 300 mm)
The limit halves when Vs is large because shear demand is high and closer stirrup spacing is needed. The threshold 0.33√f'c·bw·d is exactly half of the crushing limit 0.66√f'c·bw·d.
STEP 6 — USE THE LESSER OF COMPUTED s AND s_max:
Final spacing = min(s_computed, s_max)
Always round down to a practical value (e.g., use 150 mm instead of 164 mm).
MINIMUM STIRRUP REINFORCEMENT (NSCP 2015):
Required when Vu > 0.5φVc:
Av,min = max(0.062√f'c · bw·s/fy, 0.35 · bw·s/fy)
Note the two expressions — take the larger one. The first governs for f'c > 31.9 MPa; the second governs for lower-strength concrete.
Applications
- Stirrup layout drawings and bar schedules in structural design packages
- Determining zones of constant stirrup spacing along a beam span
- Retrofit design for beams with inadequate shear reinforcement
- Optimization of stirrup spacing to reduce labor cost while meeting code requirements
Common misconceptions
- Computing Vs = Vu − φVc without dividing by φ — the most common board exam error
- Using only 1 leg of the stirrup (Av = Abar) for a U-stirrup — doubles the error in spacing
- Applying Band 2 spacing limit when Band 1 is correct — gives an unnecessarily conservative (costly) design
- Not checking the crushing limit — may result in using an inadequate cross-section
- Using Vs in kN in the spacing formula without converting to N — causes unit inconsistency
Related concepts
- Concrete Shear Strength (Vc)
- Shear Strength Reduction Factor φ
- Torsion Design (closed stirrups)
- Minimum Shear Reinforcement
- Beam Cross-Section Sizing
Common exam questions
Example
Design 12 mm stirrups for a beam with bw = 350 mm, d = 600 mm, f'c = 35 MPa, fy = 415 MPa, Vu = 400 kN.
Approach
Follow 5-step procedure: compute Vs, check crushing, identify spacing band, compute s = Av·fy·d/Vs, apply s_max.
Example
Is a 250×450 mm beam with f'c = 21 MPa adequate if Vs demand is 280 kN?
Approach
Compute Vs and compare to 0.66√f'c·bw·d. If Vs exceeds this limit, the section must be enlarged regardless of stirrup spacing.
Example
What is the maximum allowable stirrup spacing if Vs = 150 kN for a beam with bw = 300, d = 500, f'c = 28?
Approach
Compute 0.33√f'c·bw·d and compare with Vs to determine which band applies, then state s_max.
Key points to remember
- Vs = (Vu − φVc)/φ — always divide by φ when computing Vs
- Crushing limit: Vs ≤ 0.66√f'c·bw·d — if exceeded, enlarge the section
- Spacing limit band threshold: 0.33√f'c·bw·d (half of crushing limit)
- s_max = d/2 or 600 mm (when Vs is small); d/4 or 300 mm (when Vs is large)
- U-stirrup has 2 legs: Av = 2 × Abar
- s_required = Av·fy·d / Vs; always use the lesser of s_required and s_max
- Av,min = max(0.062√f'c·bw·s/fy, 0.35·bw·s/fy)
Torsion in Reinforced Concrete Beams
Examples
Torsion threshold check: A 300 × 500 mm beam (f'c = 28 MPa, normal-weight) carries Tu = 4 kN·m. Can torsion be neglected?
Solution
Step 1: Compute Acp and pcp.
Acp = 300 × 500 = 150,000 mm²
pcp = 2(300 + 500) = 2(800) = 1,600 mm
Step 2: Compute threshold torque.
T_threshold = φ(0.083)λ√f'c · (Acp²/pcp)
= 0.75 × 0.083 × 1.0 × √28 × (150,000²/1,600)
= 0.75 × 0.083 × 5.292 × 14,062,500
= 0.75 × 0.083 × 5.292 × 14,062,500
Compute step by step:
0.083 × 5.292 = 0.4392
0.4392 × 14,062,500 = 6,177,750
0.75 × 6,177,750 = 4,633,313 N·mm
T_threshold = 4,633,313 N·mm = 4.63 kN·m
Step 3: Compare.
Tu = 4 kN·m < T_threshold = 4.63 kN·m
Conclusion: Torsion may be NEGLECTED. Design for shear only.
This is a classic board exam torsion problem. The result is close to the threshold (4 vs 4.63 kN·m), which tests careful computation. Note: Acp uses the full gross cross-section (300 × 500), not the net area or stirrup cage dimensions.
A 400 × 600 mm spandrel beam (f'c = 28 MPa) carries Tu = 12 kN·m. Must torsion be designed for?
Solution
Acp = 400 × 600 = 240,000 mm²
pcp = 2(400 + 600) = 2,000 mm
T_threshold = 0.75 × 0.083 × 1.0 × √28 × (240,000²/2,000)
= 0.75 × 0.083 × 5.292 × (57,600,000,000/2,000)
= 0.75 × 0.083 × 5.292 × 28,800,000
= 0.75 × 0.083 × 152,409,600
= 0.75 × 12,649,997
= 9,487,498 N·mm ≈ 9.49 kN·m
Tu = 12 kN·m > T_threshold = 9.49 kN·m
Conclusion: Torsion MUST be designed. Closed stirrups and longitudinal torsion bars are required in addition to shear stirrups.
When Tu exceeds the threshold, torsion design is mandatory. This spandrel beam example is typical of situations where edge beams receive slab loads along their span, generating significant twisting moments.
Torsion (twisting moment) occurs when a beam carries loads that are applied eccentrically or when spandrel beams receive loads from adjacent slabs. Torsional stress adds to diagonal tension in the web, and unlike flexural shear, torsion requires CLOSED stirrups (a full rectangular loop with 135° hooks) and additional longitudinal bars to resist the resulting spiral tension.
NSCP 2015 TORSION THRESHOLD CONCEPT:
The code recognizes two types of torsion:
- Compatibility torsion — Occurs in statically indeterminate structures where torsion can redistribute to adjacent members (e.g., spandrel beams). Code allows reduction to the threshold value.
- Equilibrium torsion — Occurs when torsion is essential for stability (e.g., a curved beam). Full torsion must be resisted.
CAN TORSION BE NEGLECTED? (NSCP 2015 Section 422.7.1):
For solid sections (non-prestressed), torsion may be neglected if:
Tu < φ · 0.083 · λ · √f'c · (Acp² / pcp)
Where:
- φ = 0.75
- λ = 1.0 (normal-weight concrete)
- Acp = area enclosed by the outside perimeter of the cross-section (mm²)
For a solid rectangle: Acp = b × h (full gross section, not just bw × d) - pcp = outside perimeter of the cross-section (mm)
For a solid rectangle: pcp = 2(b + h) - Tu = factored torsional moment (N·mm); convert kN·m to N·mm by ×10⁶
If Tu < threshold: Design for shear only (torsion neglected)
If Tu ≥ threshold: Design for combined shear and torsion using closed stirrups + longitudinal bars
WHEN TORSION MUST BE DESIGNED:
The combined shear-torsion design involves:
- Closed stirrups sized for both Vs and torsional stirrup demand At
- Additional longitudinal steel Al distributed around the perimeter
- The torsional reinforcement is additive to the shear stirrups
Note: The torsion threshold uses the OUTSIDE dimensions (Acp, pcp), not the stirrup centerline dimensions (Aoh, ph). The stirrup centerline dimensions are used later in the detailed torsion design (beyond the scope of most board exam threshold questions).
IMPORTANT UNIT NOTE: Tu in the threshold formula should be in N·mm to match the units of the right-hand side (N·mm). Convert from kN·m: 1 kN·m = 10⁶ N·mm.
Applications
- Spandrel beams at building perimeters receiving one-sided slab loads
- Curved beams in spiral ramps, balconies, and curved floor plans
- Canopy beams and overhanging structures with eccentric loading
- Transfer beams supporting offset columns
Common misconceptions
- Using Aoh (area inside stirrup centerline) instead of Acp (full gross area) in the threshold formula
- Using ph (stirrup perimeter) instead of pcp (outer perimeter) in the threshold formula
- Forgetting to convert Tu from kN·m to N·mm (factor of 10⁶) — causes threshold comparison errors
- Thinking U-stirrups can resist torsion — only CLOSED stirrups with 135° hooks are effective
- Applying the torsion threshold check at the support instead of the critical section
Related concepts
- Closed Stirrups vs. Open Stirrups
- Longitudinal Torsion Reinforcement (Al)
- Combined Shear and Torsion Design
- Spandrel Beam Design
- Equilibrium vs. Compatibility Torsion
Common exam questions
Example
A 350×550 mm beam, f'c=21 MPa, Tu=5 kN·m. Is torsion design required?
Approach
Compute Acp = b×h and pcp = 2(b+h). Apply threshold formula with φ=0.75. Convert Tu to N·mm. Compare Tu to threshold.
Example
Find the maximum Tu that can be neglected for a 300×500 mm beam with f'c=35 MPa.
Approach
Direct substitution into φ(0.083)λ√f'c·Acp²/pcp. Keep units consistent.
Key points to remember
- Torsion threshold: Tu < φ(0.083)λ√f'c · Acp²/pcp (with φ = 0.75)
- Acp = full gross area (b × h), pcp = full outer perimeter 2(b+h) — not net or stirrup centerline dimensions
- Units: Acp in mm², pcp in mm → threshold in N·mm; convert Tu from kN·m by multiplying ×10⁶
- Below threshold → design for shear only (torsion neglected)
- Above threshold → use closed stirrups + longitudinal bars for torsion
- Closed stirrups have 135° hooks at corners — open U-stirrups are NOT acceptable for torsion
- Compatibility torsion may be reduced to the threshold value in indeterminate structures
Minimum Shear Reinforcement and Practical Layout
Examples
For a beam bw = 300 mm, d = 500 mm, f'c = 28 MPa, fy = 415 MPa using s = 250 mm stirrups, verify the minimum area requirement.
Solution
Expression 1: 0.062√28 × 300 × 250 / 415
= 0.062 × 5.292 × 300 × 250 / 415
= 0.062 × 5.292 × 75,000 / 415
= 24,607 / 415 = 59.3 mm²
Expression 2: 0.35 × 300 × 250 / 415
= 26,250 / 415 = 63.3 mm²
Av,min = max(59.3, 63.3) = 63.3 mm²
For 10 mm U-stirrups: Av = 157 mm² > 63.3 mm² ✓
Minimum requirement satisfied.
Note that Expression 2 (0.35·bw·s/fy) governs because f'c = 28 MPa < 31.9 MPa. This is the typical case in Philippine construction where f'c ranges from 21 to 35 MPa. The 10 mm U-stirrup easily exceeds minimum requirements.
NSCP 2015 requires minimum transverse reinforcement in beams to prevent sudden shear failure when diagonal cracks form. Without minimum stirrups, concrete alone must resist all shear after cracking — which can lead to brittle collapse with little warning.
MINIMUM STIRRUP REQUIREMENT TRIGGER (NSCP 2015 Section 409.6.3):
Minimum stirrups are required when: Vu > 0.5 · φ · Vc
Minimum area of shear reinforcement:
Av,min/s = max[ 0.062√f'c · bw/fy , 0.35 · bw/fy ]
The two expressions give the minimum Av per unit length of beam. The larger of the two governs:
- 0.062√f'c · bw/fy: governs when f'c > 31.9 MPa (high-strength concrete)
- 0.35 · bw/fy: governs when f'c ≤ 31.9 MPa (most common in PH practice)
For standard spacing s, rearrange:
Av,min = max[ 0.062√f'c · bw·s/fy , 0.35 · bw·s/fy ]
EXCEPTIONS (Minimum stirrups NOT required, NSCP 2015 Section 409.6.3.1):
- Slabs and footings
- Concrete joist construction
- Beams with total depth ≤ 250 mm
- Beams integral with slabs where total depth ≤ greater of (2.5tf or 0.5bw)
- Beams constructed with steel fiber-reinforced concrete meeting specific criteria
PRACTICAL STIRRUP LAYOUT ALONG A BEAM SPAN:
Shear demand varies along the span. Stirrups are typically arranged in zones:
- Zone 1 (near supports): Close spacing for high Vu
- Zone 2 (mid-span): Wider spacing as Vu decreases
- Zone 3 (near mid-span): Minimum spacing or minimum stirrups only
The point where stirrups can stop (transition to minimum or none) is where the shear envelope intersects 0.5φVc. In practice, stirrups are continued at least to the point where Vu = 0.5φVc.
STIRRUP HOOK REQUIREMENTS:
- 135° hooks for torsion (closed stirrups)
- 90° hooks are permitted for shear-only stirrups in most cases
- Hooks must extend into the compression zone
FIRST STIRRUP PLACEMENT:
The first stirrup from the face of the support is typically placed at s/2 from the support face (not full spacing), to provide reinforcement close to the support without exceeding the required spacing.
Applications
- Bar scheduling and detailing on structural plans
- Determining extent of stirrup zones along beam span
- Quality control checking during concrete construction inspection
- Retrofit assessment of existing beams for code compliance
Common misconceptions
- Thinking slabs require minimum stirrups — they are explicitly exempt
- Using the smaller of the two Av,min expressions instead of the larger
- Placing the first stirrup at full spacing s from the support instead of s/2
- Stopping stirrups where Vu = φVc instead of where Vu = 0.5φVc
Related concepts
- Stirrup Design (Vs)
- Shear Envelope
- Beam Detailing (ACI 315 / NSCP 2015)
- Beam Exemptions from Stirrup Requirements
Common exam questions
Example
Is Av = 100 mm² sufficient for minimum requirements if bw = 250 mm, s = 200 mm, f'c = 21 MPa, fy = 415 MPa?
Approach
Apply both expressions, take the maximum. Check if provided Av meets or exceeds Av,min.
Example
If the shear varies linearly from 200 kN at the support to 0 at mid-span, and 0.5φVc = 50 kN, at what distance from the support can stirrup spacing be relaxed?
Approach
Find the location along the span where Vu = 0.5φVc. Stirrups may stop at this point (or minimum stirrups continue to end of beam).
Key points to remember
- Minimum stirrups required when Vu > 0.5φVc
- Av,min = max(0.062√f'c·bw·s/fy, 0.35·bw·s/fy); take the larger value
- 0.35·bw·s/fy governs for f'c ≤ 31.9 MPa (most Philippine practice)
- Stirrups may be stopped where Vu ≤ 0.5φVc along the span
- First stirrup is placed at s/2 from the support face
- Slabs, footings, joists, and shallow beams are exempt from minimum stirrup requirements
- Critical section is at d from support face — shear here, not at the support itself
Practice problems
PROBLEM 1 (Shear Design — Complete Solution)
A simply supported beam has bw = 350 mm, d = 600 mm, f'c = 35 MPa (normal-weight), fy = 415 MPa. The factored shear at the critical section is Vu = 320 kN. Design 12 mm vertical U-stirrups. Report the required spacing and verify all code limits.
Solution
STEP 1: Compute Vc.
Vc = 0.17(1.0)√35 × 350 × 600
√35 = 5.916 MPa^0.5
Vc = 0.17 × 5.916 × 350 × 600
= 0.17 × 5.916 × 210,000
= 211,420 N = 211.4 kN
STEP 2: Compute φVc.
φVc = 0.75 × 211.4 = 158.6 kN
Check: Vu (320 kN) > φVc (158.6 kN) → Stirrups required for excess shear.
STEP 3: Compute Vs.
Vs = (Vu − φVc)/φ = (320 − 158.6)/0.75 = 161.4/0.75 = 215.2 kN
STEP 4: Check crushing limit.
Vs,max = 0.66√35 × 350 × 600
= 0.66 × 5.916 × 210,000
= 820,742 N = 820.7 kN
Vs (215.2 kN) < Vs,max (820.7 kN) ✓ Section is adequate.
STEP 5: Determine spacing band.
Band threshold = 0.33√35 × 350 × 600
= 0.33 × 5.916 × 210,000
= 410,371 N = 410.4 kN
Vs (215.2 kN) < 410.4 kN → Band 1
s_max = min(d/2, 600) = min(300, 600) = 300 mm
STEP 6: Compute Av for 12 mm U-stirrup.
Av = 2 × π/4 × 12² = 2 × 113.1 = 226.2 mm²
STEP 7: Compute required spacing.
s = Av·fy·d / Vs
= 226.2 × 415 × 600 / 215,200
= 56,313,800 / 215,200 = 261.7 mm
STEP 8: Govern spacing.
s_final = min(261.7, 300) = 261.7 mm → Use s = 250 mm
FINAL ANSWER: Provide 12 mm U-stirrups at 250 mm c/c.
All six code checks are demonstrated: Vc, φVc, Vs, crushing limit, band threshold, and s_max. The computed spacing (261.7 mm) is very close to Band 1's limit (300 mm), so the limit does not govern. Rounding down to 250 mm is practical and conservative. This type of comprehensive problem is the most common shear design question in the board exam.
PROBLEM 2 (Torsion Threshold)
A rectangular spandrel beam 400 mm wide × 600 mm total depth is made of normal-weight concrete with f'c = 28 MPa. The factored torque is Tu = 12 kN·m. Must torsion be explicitly designed?
Solution
STEP 1: Compute Acp and pcp using gross outside dimensions.
Acp = 400 × 600 = 240,000 mm²
pcp = 2(400 + 600) = 2,000 mm
STEP 2: Compute torsion threshold.
T_threshold = φ(0.083)λ√f'c · (Acp²/pcp)
= 0.75 × 0.083 × 1.0 × √28 × (240,000²/2,000)
Acp²/pcp = (240,000)²/2,000 = 57,600,000,000/2,000 = 28,800,000 mm³
√28 = 5.292
0.083 × 5.292 = 0.4392
0.4392 × 28,800,000 = 12,648,960
T_threshold = 0.75 × 12,648,960 = 9,486,720 N·mm
= 9.49 kN·m
STEP 3: Compare Tu to threshold.
Tu = 12 kN·m > T_threshold = 9.49 kN·m
CONCLUSION: Torsion MUST be designed. Closed stirrups with 135° hooks and additional longitudinal steel Al are required.
The spandrel beam example is classic because edge beams along a building perimeter always receive one-sided slab loads that produce torsion. The 12 kN·m torque (26% above threshold) is substantial and cannot be neglected. Note that Acp uses the FULL 400×600 gross section, not a net or reduced area.
PROBLEM 3 (Minimum Stirrups + Section Check)
A beam with bw = 280 mm, d = 450 mm, f'c = 21 MPa, fy = 275 MPa has Vu = 45 kN at a certain section. Determine: (a) whether stirrups are required, (b) if required, the minimum Av for s = 200 mm stirrups, and (c) whether 10 mm U-stirrups satisfy the minimum.
Solution
(a) Compute φVc.
Vc = 0.17(1.0)√21 × 280 × 450
√21 = 4.583
Vc = 0.17 × 4.583 × 280 × 450 = 98,475 N = 98.5 kN
φVc = 0.75 × 98.5 = 73.9 kN
0.5φVc = 0.5 × 73.9 = 36.9 kN
Since Vu (45 kN) > 0.5φVc (36.9 kN) but Vu (45 kN) < φVc (73.9 kN):
→ Minimum stirrups ARE required.
(b) Minimum Av for s = 200 mm.
Expression 1: 0.062√21 × 280 × 200 / 275
= 0.062 × 4.583 × 56,000 / 275
= 15,900 / 275 = 57.8 mm²
Expression 2: 0.35 × 280 × 200 / 275
= 19,600 / 275 = 71.3 mm²
Av,min = max(57.8, 71.3) = 71.3 mm²
(c) Check 10 mm U-stirrups.
Av = 2 × 78.54 = 157.1 mm² > 71.3 mm² ✓
CONCLUSION: 10 mm U-stirrups at 200 mm c/c satisfy the minimum shear reinforcement requirement.
This problem tests the 0.5φVc threshold for minimum stirrups — a frequently missed check. Since Vu falls between 0.5φVc and φVc, concrete alone is sufficient but minimum transverse reinforcement is still mandated by NSCP 2015. Expression 2 (0.35·bw·s/fy) governs because f'c = 21 MPa < 31.9 MPa.
PROBLEM 4 (Section Adequacy — Crushing Limit)
A beam bw = 250 mm, d = 400 mm, f'c = 21 MPa has Vu = 280 kN. Can this section carry the load with stirrups alone, or must the section be enlarged?
Solution
Step 1: Compute Vc.
Vc = 0.17(1.0)√21 × 250 × 400
= 0.17 × 4.583 × 100,000 = 77,911 N = 77.9 kN
Step 2: Compute φVc.
φVc = 0.75 × 77.9 = 58.4 kN
Step 3: Compute Vs demand.
Vs = (Vu − φVc)/φ = (280 − 58.4)/0.75 = 221.6/0.75 = 295.5 kN
Step 4: Compute crushing limit.
Vs,max = 0.66√21 × 250 × 400
= 0.66 × 4.583 × 100,000
= 302,478 N = 302.5 kN
Vs demand (295.5 kN) < Vs,max (302.5 kN) ✓
Section is MARGINALLY adequate.
Band check: 0.33√21 × 250 × 400 = 151,239 N = 151.2 kN
Vs (295.5 kN) > 151.2 kN → Band 2 applies
s_max = min(d/4, 300) = min(100, 300) = 100 mm
Note: Very tight margin — in practice, a larger section would be recommended.
This problem is at the boundary of section adequacy. While technically the section does not need to be enlarged (Vs < Vs,max), the required stirrup spacing will be very tight (likely <50 mm with standard bars), making this uneconomical. Board exams may present borderline cases to test if examinees correctly apply the crushing check before proceeding to spacing calculations.
Exam preparation tips
- FORMULA RECALL FIRST: Memorize Vc = 0.17λ√f'c·bw·d, Vs = (Vu−φVc)/φ, and the threshold torque formula before attempting any shear-torsion problem. These are non-negotiable.
- UNIT DISCIPLINE: Keep bw, d in mm and f'c in MPa. Results come out in Newtons (N). Convert to kN by dividing by 1,000. For torsion, convert kN·m to N·mm by multiplying by 10⁶.
- THE φ DIVIDE TRAP: Always divide by φ = 0.75 when solving for Vs from Vu − φVc. This is the #1 computational error in board exams on this topic.
- TWO-LEG RULE: A standard U-stirrup has 2 legs. Av = 2 × Abar. Bar areas to memorize: 10mm → 78.54 mm², 12mm → 113.1 mm², 16mm → 201.1 mm².
- SPACING BAND CHECK: After computing Vs, always compare to 0.33√f'c·bw·d. This determines whether s_max = d/2 (or 600 mm) or d/4 (or 300 mm).
- CRUSHING LIMIT ALWAYS FIRST: Before computing spacing, check Vs ≤ 0.66√f'c·bw·d. If this fails, no stirrup spacing solves the problem — the section must be enlarged.
- CRITICAL SECTION HABIT: In every shear problem, mentally note 'shear is checked at d from the face of the support' before reading off Vu from the shear diagram.
- TORSION OUTSIDE DIMENSIONS: For the threshold formula, use Acp = b×h (total gross area) and pcp = 2(b+h) with OUTSIDE dimensions. Never use stirrup centerline dimensions here.
- MINIMUM STIRRUP THRESHOLD: Draw a quick mental flowchart: Vu vs 0.5φVc → minimum stirrups? Vu vs φVc → design stirrups? This prevents missing the minimum stirrup requirement.
- PRACTICE THE 5-STEP PROCEDURE: Most board exam shear problems can be solved with the same 5 steps: (1) Compute Vc, (2) Compute Vs, (3) Check crushing, (4) Determine band and s_max, (5) Compute s = Av·fy·d/Vs and govern. Internalize this sequence.
- TORSION TYPES: Know that compatibility torsion (indeterminate structures) can be redistributed and reduced to the threshold; equilibrium torsion (statically determinate) cannot be reduced.
- LAMBDA FOR LIGHTWEIGHT CONCRETE: If the problem specifies lightweight aggregate concrete, use λ = 0.75 (all-lightweight) or λ = 0.85 (sand-lightweight). Normal-weight → λ = 1.0. Mixing these up reduces Vc unnecessarily or dangerously overestimates it.
Diagrams
Shear and Torsion Design: Complete Concept Map
A comprehensive mind map showing all major concepts in shear and torsion design: the Vc and Vs formulas, the stirrup design procedure, spacing band limits, minimum reinforcement rules, and the torsion threshold check. Use this as a quick review reference before the exam.
Step-by-Step Stirrup Design Procedure
A complete decision-tree flowchart for the stirrup design procedure from NSCP 2015. Follow each decision node in sequence: check if stirrups are needed, check crushing adequacy of the section, identify which spacing band applies, then compute and govern the final spacing.
Torsion Design Decision Flowchart
A decision flowchart for torsion design starting from checking the NSCP 2015 threshold. Key decision points: (1) Is Tu below the threshold? If yes, neglect torsion. If no, identify torsion type and design closed stirrups plus longitudinal bars.
Shear Zones Along a Beam Span and Stirrup Layout
Illustrates how shear demand varies along a beam span and how stirrups are zoned accordingly: dense spacing near supports, transitioning to minimum stirrups, then no stirrups near mid-span. Also highlights the critical section location and first stirrup placement rule.
Common Board Exam Errors in Shear-Torsion Problems
A pie chart showing the relative frequency of common mistakes in board exam shear and torsion problems, based on common review patterns. The largest source of errors is forgetting to divide by φ when computing Vs, followed by incorrect stirrup leg counting and wrong spacing band selection.
Evolution of Shear Design Provisions
A timeline showing the historical evolution of shear design provisions in Philippine practice, from early ACI 318 adoption through successive NSCP editions to the current NSCP 2015 provisions used in board exams.
In summary
Shear and torsion design is one of the most formula-intensive and procedure-driven topics in the PRC Civil Engineer Licensure Examination. Mastery requires both formula memorization and procedural discipline — knowing not just the equations but the correct sequence of checks.
The five critical actions to perform in every shear problem are: (1) locate the critical section at d from the support face, (2) compute Vc = 0.17λ√f'c·bw·d with φ = 0.75, (3) compute Vs = (Vu − φVc)/φ (never forget the φ in the denominator), (4) check the crushing limit Vs ≤ 0.66√f'c·bw·d, and (5) determine the spacing band to find s_max, then govern with the computed spacing.
For torsion, the key is the threshold check using full gross dimensions: Tu vs. φ(0.083)λ√f'c·(Acp²/pcp). Below the threshold, neglect torsion and design for shear only. Above it, closed stirrups with 135° hooks and longitudinal bars are mandatory.
Philippine construction predominantly uses f'c = 21 to 35 MPa normal-weight concrete with fy = 275 or 415 MPa steel — these ranges define your benchmark values for intuitive checking. A φVc in the range of 50–200 kN for typical building beams is a good sanity check.
With systematic practice of the worked examples and the four practice problems in this chapter, you will develop the speed and accuracy needed to solve any board exam shear-torsion problem within the allotted time. Good luck, and approach each problem with the disciplined five-step procedure — it never fails.