Reinforced Concrete Beams: Shear and Torsion - Revision Notes
Shear and torsion are among the most frequently tested topics in the PRC Civil Engineer Licensure Examination under Reinforced and Prestressed Concrete Design. Unlike flexural (bending) failure, shear failure is sudden and brittle — there is virtually no warning before collapse. This is why NSCP 2015 (aligned with ACI 318) imposes conservative limits and requires mandatory minimum stirrups over large portions of a beam. This chapter covers: (1) the concrete contribution to shear strength Vc, (2) stirrup design including spacing limits and minimum reinforcement, and (3) the torsion threshold check. Master these formulas and the step-by-step design procedure, and you will be well-prepared to answer board exam problems on this topic.
1. Fundamentals of Shear in Reinforced Concrete Beams
Formulas
φVn = φ(Vc + Vs) ≥ Vu
Example
If Vu = 250 kN, φVc = 101.2 kN, and φVs = 148.8 kN, then φVn = 250 kN ≥ 250 kN. OK.
Variables
Vu = factored shear force (N); Vc = concrete shear strength (N); Vs = stirrup shear strength (N); φ = 0.75
Application
Fundamental shear design inequality — must be satisfied at every section along the beam.
Exam tips
- Memorize φ = 0.75 for shear. It is different from φ = 0.90 for flexure.
- When the problem says 'at the face of the support,' use that Vu directly. When it says 'at the support centerline,' subtract the shear over distance d to get the critical section value.
- Draw a free-body diagram and mark the critical section before computing Vu.
Key points
- Shear failure produces diagonal tension cracks inclined at approximately 45° to the beam axis — often called 'diagonal tension failure.'
- Shear is most critical near supports where the shear force is maximum.
- The critical section for shear design is located at a distance d (effective depth) from the face of the support — NOT at the support centerline. This is because the inclined compression strut transfers load directly to the support.
- If a concentrated load is applied within distance d from the support face, the critical section moves to the face of the support.
- Total nominal shear strength: Vn = Vc + Vs, where Vc = concrete contribution and Vs = stirrup contribution.
- The strength reduction factor for shear is φ = 0.75 (NSCP 2015 / ACI 318-14 Table 21.2.1).
- Design requirement: φVn ≥ Vu, i.e., φ(Vc + Vs) ≥ Vu.
Definitions
Effective Depth (d)
Distance from the extreme compression fiber to the centroid of the tension reinforcement. For stirrup design, d appears in both Vc and Vs formulas.
Importance
Every shear formula uses d. An error in d propagates through the entire design.
Critical Section for Shear
Located at distance d from the face of the support (not the centerline). Shear at this section governs stirrup design.
Importance
Board exams frequently give Vu at the support and ask you to use it directly — always check whether the problem says 'at the support' or 'at the critical section.'
bw (Web Width)
Width of the beam web. For rectangular beams, bw = b (total width). For T-beams, bw is the stem width.
Importance
Shear formulas use bw, not the flange width — a common confusion for T-beam problems.
λ (Lightweight Concrete Factor)
λ = 1.0 for normal-weight concrete; λ = 0.85 for sand-lightweight; λ = 0.75 for all-lightweight concrete.
Importance
Most board exam problems specify normal-weight concrete (λ = 1.0). Check the problem statement.
Common mistakes
- Using Vu at the support face instead of at the critical section d away from the support face.
- Forgetting that φ = 0.75 for shear (not 0.90 which is for flexure).
- Applying bw = flange width for T-beams instead of the web width.
- Using λ = 1.0 without checking if the concrete is lightweight.
2. Concrete Shear Strength (Vc)
Formulas
Vc = 0.17λ√f'c × bw × d
Example
bw = 300 mm, d = 500 mm, f'c = 28 MPa, λ = 1.0: Vc = 0.17(1.0)(√28)(300)(500) = 0.17 × 5.292 × 150,000 = 134,933 N = 134.9 kN; φVc = 0.75 × 134.9 = 101.2 kN
Variables
λ = 1.0 (NW concrete); f'c = compressive strength (MPa); bw = web width (mm); d = effective depth (mm); Vc in Newtons (N)
Application
Simplified concrete shear strength for non-prestressed members. Used in the vast majority of board exam problems.
Exam tips
- Quick mental check: For f'c = 28 MPa, √28 ≈ 5.29. For f'c = 21 MPa, √21 ≈ 4.58. Memorize a few common values.
- φVc = 0.75 × 0.17 × √f'c × bw × d = 0.1275√f'c × bw × d. This combined factor saves one calculation step.
- If the problem gives Vu < φVc but says stirrups are required — remember minimum stirrups are required if Vu > 0.5φVc.
Key points
- NSCP 2015 provides a simplified formula and a more detailed formula for Vc. The simplified formula is most commonly used in board exams.
- Simplified formula (NSCP 2015 Section 422.5.5.1, normal-weight concrete): Vc = 0.17λ√f'c × bw × d
- The units must be consistent: f'c in MPa, bw and d in mm → Vc in Newtons (N). Convert to kN by dividing by 1000.
- The φVc value determines three critical thresholds for stirrup design.
- Threshold 1 (no stirrups needed): If Vu ≤ 0.5φVc, stirrups are not required by analysis.
- Threshold 2 (minimum stirrups required): If 0.5φVc < Vu ≤ φVc, minimum stirrups are still required.
- Threshold 3 (stirrups by design): If Vu > φVc, compute Vs and design stirrups.
Definitions
φVc
Design shear strength provided by concrete alone = 0.75 × Vc. This is the benchmark value compared against Vu to determine if stirrups are needed and what type.
Importance
φVc is the single most important intermediate result in shear design. Compute it first, then compare with Vu.
Common mistakes
- Computing √f'c but forgetting to multiply by 0.17 — just taking 0.17 and √f'c as separate factors is fine, but some reviewees skip the 0.17 multiplier.
- Leaving Vc in Newtons and comparing it with Vu in kN — always keep units consistent.
- Using d = total beam height h instead of the effective depth d = h − cover − stirrup diameter − half of main bar diameter.
3. Stirrup Design — Required Spacing and Limits
Formulas
Vs = (Vu − φVc) / φ
Example
Vu = 250 kN, φVc = 101.2 kN: Vs = (250 − 101.2) / 0.75 = 148.8 / 0.75 = 198.4 kN
Variables
Vu = factored shear (N or kN); φVc = design concrete shear strength; φ = 0.75
Application
Required stirrup shear strength. This is Step 1 of stirrup design.
s = Av × fy × d / Vs
Example
Av = 157 mm², fy = 415 MPa, d = 500 mm, Vs = 198,400 N: s = (157 × 415 × 500) / 198,400 = 32,577,500 / 198,400 = 164 mm
Variables
Av = total stirrup bar area for all legs (mm²); fy = stirrup yield strength (MPa); d = effective depth (mm); Vs = required stirrup shear strength (N); s = stirrup spacing (mm)
Application
Compute the required stirrup spacing. Use Vs in Newtons when Av in mm², fy in MPa, d in mm.
Vs,max = 0.66√f'c × bw × d
Example
f'c = 28 MPa, bw = 300 mm, d = 500 mm: Vs,max = 0.66(5.292)(150,000) = 523,914 N = 523.9 kN
Variables
f'c in MPa; bw, d in mm; Vs,max in Newtons
Application
Upper limit on stirrup shear strength to prevent web-crushing failure. If computed Vs exceeds this, the beam cross-section must be made larger.
Vs,band = 0.33√f'c × bw × d
Example
f'c = 28 MPa, bw = 300 mm, d = 500 mm: 0.33(5.292)(150,000) = 261,954 N = 262 kN. Since Vs = 198.4 kN < 262 kN, use smax = min(250, 600) = 250 mm.
Variables
Same as Vs,max but with 0.33 coefficient instead of 0.66
Application
Determines which smax governs. If Vs ≤ this value: smax = min(d/2, 600 mm). If Vs exceeds this: smax = min(d/4, 300 mm).
Exam tips
- Memorize the 5-step stirrup design procedure in order: (1) Compute Vc and φVc; (2) Compute Vs = (Vu−φVc)/φ; (3) Check Vs ≤ 0.66√f'c bw d; (4) Check spacing band (0.33 threshold); (5) Compute s and apply smax.
- A shortcut: 0.33√f'c bw d = (Vs,max)/2. Half of the crushing limit is the band threshold.
- In multiple-choice problems, if one answer option has an error in the φ factor for Vs, it will give an answer roughly 33% higher than the correct one — be alert to this trap.
- If the problem asks for the maximum spacing only, go directly to steps 3 and 4 without computing the exact spacing.
Key points
- When Vu > φVc, stirrups must carry the excess shear. Compute required Vs then find spacing s.
- Step 1: Vs = (Vu − φVc) / φ. This is the most commonly missed step — dividing by φ, NOT just subtracting.
- Step 2: Check crushing limit — Vs must not exceed 0.66√f'c × bw × d. If exceeded, the section must be enlarged.
- Step 3: Determine spacing band — compare Vs with 0.33√f'c × bw × d to select the correct smax.
- Step 4: Compute required spacing — s = Av × fy × d / Vs.
- Step 5: Apply smax and use the smaller of computed s and smax.
- For a standard U-stirrup (two legs), Av = 2 × (π/4 × db²). For 10 mm bars: Av = 2 × 78.54 = 157.1 mm². For 12 mm bars: Av = 2 × 113.1 = 226.2 mm².
- Spacing limits (NSCP 2015 Section 409.7.6.2.2):
- — If Vs ≤ 0.33√f'c × bw × d: smax = min(d/2, 600 mm)
- — If Vs > 0.33√f'c × bw × d: smax = min(d/4, 300 mm)
- The 0.33 threshold is exactly half of 0.66 — the spacing limit halves when shear demand is high.
Definitions
Av (Stirrup Area)
Total cross-sectional area of all stirrup legs crossing the inclined crack plane. For a single U-stirrup with bar diameter db: Av = 2 × (π db² / 4).
Importance
The leg count is the most commonly missed point. A single U-stirrup always has 2 legs crossing the shear crack. Never use the area of just one leg.
Stirrup Spacing (s)
Center-to-center distance between stirrups along the beam axis (mm). Must satisfy both the computed requirement (s = Av fy d / Vs) and the code maximum spacing (smax).
Importance
Always round DOWN to a practical value (e.g., 164 mm → use 150 mm or 160 mm, not 170 mm) for conservatism.
Maximum Stirrup Spacing (smax)
Code-imposed upper limit on spacing. Governed by two criteria: (1) the shear demand level (Vs vs. 0.33√f'c bw d), and (2) absolute limits of 600 mm or 300 mm.
Importance
The spacing band check is frequently the deciding factor in board exam MCQs. Always perform this check even if the computed s is already small.
Common mistakes
- NOT dividing by φ when computing Vs — writing Vs = Vu − φVc instead of Vs = (Vu − φVc)/φ. This underestimates Vs and gives a non-conservative (unsafe) spacing.
- Using only one leg area for a U-stirrup (Av = 78.54 mm² for 10 mm bar instead of 157.1 mm²).
- Forgetting to check the crushing limit (Vs ≤ 0.66√f'c bw d) before proceeding with spacing calculations.
- Not checking which spacing band applies — defaulting to smax = d/2 = 600 mm without checking if the high-shear band (d/4 = 300 mm) governs.
- Rounding spacing UP instead of DOWN — rounding up gives less reinforcement than required (unsafe).
- Using fy of the main bars instead of the stirrup fy in the spacing formula.
4. Minimum Stirrup Reinforcement
Formulas
Av,min = max(0.062√f'c × bw × s / fy , 0.35 × bw × s / fy)
Example
bw = 300 mm, s = 250 mm, f'c = 28 MPa, fy = 415 MPa: (1) 0.062(5.292)(300×250/415) = 59.4 mm²; (2) 0.35(300×250/415) = 63.3 mm². Av,min = 63.3 mm². A U-stirrup of 10 mm bars gives Av = 157 mm² > 63.3 mm². OK.
Variables
f'c in MPa; bw in mm; s = stirrup spacing (mm); fy = stirrup yield strength (MPa); Av,min in mm²
Application
Minimum stirrup area requirement when Vu > 0.5φVc but ≤ φVc. This ensures a minimum level of shear ductility.
Exam tips
- Boundary condition: No stirrups required → Vu ≤ 0.5φVc. Minimum stirrups → 0.5φVc < Vu ≤ φVc. Designed stirrups → Vu > φVc.
- For standard f'c values (21–35 MPa), criterion 2 (0.35 bw s / fy) often governs for minimum stirrups — verify by calculating both.
- On the board exam, if the question asks 'what is the maximum spacing when only minimum stirrups are required,' use: smax = Av fy / (0.35 bw) and compare with d/2 and 600 mm.
Key points
- Even when the concrete alone can theoretically carry the shear, minimum stirrups are required to prevent sudden brittle failure and to maintain ductility.
- Minimum stirrups are required wherever Vu > 0.5φVc.
- Exceptions where minimum stirrups may be waived (NSCP 2015 Section 409.6.3): hollow core slabs, footings, certain slabs, and shallow members with h ≤ 250 mm or h ≤ max(2.5tf, 0.5bw) for flanged sections.
- The minimum stirrup area requirement ensures two criteria are satisfied simultaneously:
- (1) Av,min = 0.062√f'c × (bw × s / fy)
- (2) Av,min ≥ 0.35 × (bw × s / fy)
- The governing (larger) value of Av controls. In practice, criterion (1) governs for f'c > 32 MPa; criterion (2) governs for f'c < 32 MPa.
- Rearranged to find maximum spacing for minimum steel: smax = Av × fy / (0.062√f'c × bw) and smax = Av × fy / (0.35 × bw) — use the smaller.
Definitions
Minimum Shear Reinforcement Threshold
The shear force level (Vu = 0.5φVc) below which no stirrups are required by the code. Above this level, at least minimum stirrups must be provided regardless of whether concrete alone is sufficient.
Importance
This is a very common board exam question: 'Where can stirrups be discontinued?' Answer: where Vu ≤ 0.5φVc.
Common mistakes
- Thinking stirrups can be eliminated once Vu drops below φVc — minimum stirrups are still required down to 0.5φVc.
- Applying only one of the two Av,min criteria instead of taking the larger of both.
- Confusing Av,min (for minimum stirrups) with the computed Av from the s = Av fy d / Vs formula.
5. Introduction to Torsion Design
Formulas
Tth = φ × 0.083λ√f'c × (Acp² / pcp)
Example
300×500 mm beam, f'c = 28 MPa: Acp = 300×500 = 150,000 mm²; pcp = 2(300+500) = 1,600 mm. Tth = 0.75(0.083)(5.292)(150,000²/1,600) = 0.75(0.083)(5.292)(14,062,500) = 4,633,000 N·mm = 4.63 kN·m. If Tu = 4 kN·m < 4.63 kN·m, torsion may be neglected.
Variables
φ = 0.75; λ = 1.0 (NW concrete); f'c in MPa; Acp = area enclosed by outside perimeter of concrete cross-section (mm²); pcp = outside perimeter of cross-section (mm); Tth in N·mm
Application
Torsion threshold — if Tu < Tth, torsion effects are negligible and may be ignored per NSCP 2015 Section 422.7.1.
Exam tips
- The most common torsion question in board exams is: 'Can torsion be neglected?' — compute Tth and compare with Tu. If Tu < Tth, say yes and move on.
- Acp² / pcp has units of mm⁴/mm = mm³. Multiplying by √f'c (MPa^0.5 = N^0.5/mm) and the constants gives N·mm (torque). Check units to verify your calculation.
- Remember: φ = 0.75 is used for both shear and torsion (same strength reduction factor).
Key points
- Torsion (twisting moment) induces diagonal tension that spirals around the beam — resisted by CLOSED stirrups (not open U-stirrups) and additional longitudinal bars.
- NSCP 2015 allows torsion to be neglected (designed for shear only) if the factored torque Tu is below the threshold torque Tth.
- The threshold is based on the gross concrete section properties Acp and pcp (OUTSIDE perimeter dimensions, not stirrup centerline).
- If Tu ≥ threshold: design closed stirrups for combined shear + torsion, add longitudinal torsion steel Al, and check combined stress interaction.
- For equilibrium torsion (statically determinate torsion — cannot be reduced by redistribution): always design for the full Tu.
- For compatibility torsion (statically indeterminate — torsion from member deformation compatibility): may reduce Tu to the threshold value in design.
Definitions
Acp (Area Enclosed by Outside Perimeter)
For a rectangular section bw × h: Acp = bw × h. This uses the FULL outside dimensions, not the centerline of stirrups.
Importance
Board exam trap: some problems give Aoh (stirrup centerline area) and Acp (outside area) separately. Use Acp for the threshold check.
pcp (Outside Perimeter)
Perimeter of the gross cross-section: pcp = 2(bw + h) for rectangular sections. Again, full outside dimensions.
Importance
pcp vs. ph (stirrup centerline perimeter) — both appear in torsion formulas. Tth uses pcp; torsion spiral formulas use ph.
Closed Stirrups
Stirrups with a full closed loop — hooks bent 135° back into the core on both ends. Required for torsion because torsional shear flows around the entire perimeter.
Importance
Open U-stirrups work for shear only. Torsion REQUIRES closed stirrups. A common board exam distractor is specifying U-stirrups in a torsion-critical member.
Equilibrium vs. Compatibility Torsion
Equilibrium torsion: torsion required for static equilibrium (e.g., a spandrel beam carrying a slab edge); cannot be reduced. Compatibility torsion: torsion from rotation compatibility (e.g., a frame beam); may be reduced to Tth.
Importance
Knowing the type of torsion determines whether Tu can be reduced in design — a key conceptual point in licensure exams.
Common mistakes
- Using the stirrup centerline area Aoh instead of the outside gross area Acp in the threshold formula.
- Forgetting that torsion requires CLOSED stirrups — specifying standard U-stirrups is incorrect for torsion design.
- Using φ = 0.90 (flexure factor) instead of φ = 0.75 in the torsion threshold formula.
- Neglecting to add the required longitudinal torsion bars Al when Tu exceeds the threshold.
6. Board Exam Strategy and Common Pitfalls Summary
Exam tips
- Approach each problem with the design inequality: φ(Vc + Vs) ≥ Vu. Every sub-question traces back to this inequality.
- Write the formula first, then substitute values — this earns partial credit and reduces algebraic errors.
- For time management in multiple-choice: eliminate answers that violate obvious code limits (e.g., spacing > 600 mm is always wrong for standard beams).
- Commit to memory: φ = 0.75 for shear/torsion; Vc = 0.17λ√f'c bw d; Vs,max = 0.66√f'c bw d; band = 0.33√f'c bw d; smax = d/2 or d/4.
Key points
- The PRC board exam typically tests: (1) computing φVc, (2) finding stirrup spacing for a given Vu, (3) determining where stirrups can stop, and (4) torsion threshold checks.
- Always show the φ factor explicitly — examiners award marks for the design inequality φVn ≥ Vu.
- Critical section location (distance d from support face) is a frequently tested concept — read the problem carefully.
- The 5-step shear design procedure is the exam roadmap: Vc → φVc → Vs → s(required) → s(adopted).
- Double-check units before submitting: Vc in N, Vu in N (or both in kN consistently), spacing in mm.
- Torsion threshold problems are straightforward once you memorize Acp = bw × h and pcp = 2(bw + h).
Common mistakes
- Not dividing by φ when computing Vs from Vu and φVc — the most dangerous and most common error.
- Using the flexure φ = 0.90 instead of shear/torsion φ = 0.75.
- Counting stirrup legs incorrectly for unusual stirrup configurations (4-leg stirrups have Av = 4 × single bar area).
- Failing to verify the crushing limit (0.66√f'c bw d) — if the section is inadequate, the correct answer is 'enlarge the section,' not a computed spacing.
- In torsion: using Aoh (stirrup-enclosed area) in place of Acp (gross section area) in the threshold formula.
Connections
- Shear and torsion design builds directly on flexural beam design (Chapter on Singly/Doubly Reinforced Beams) — you need the effective depth d and cross-section dimensions from flexural design before computing Vc.
- The strength reduction factor concept (φ = 0.75 for shear/torsion, φ = 0.90 for flexure) connects to NSCP 2015 Section 421.2 and the general LRFD/USD philosophy used throughout all RC design chapters.
- Torsion design connects to the study of T-beams and flanged sections — the Acp and pcp for an L-shaped or T-shaped section are computed using the full gross section including overhanging flanges.
- Minimum reinforcement rules for stirrups (Av,min) parallel the minimum flexural steel rules (As,min = 0.25√f'c bw d / fy) — both exist to ensure ductile behavior and prevent sudden failure.
- The concept of critical section (distance d from support face) connects to the truss analogy model of shear — the inclined compression strut transfers load directly to the support within this distance.
- Shear design is directly relevant to foundation design (wide beams in footings use a similar Vc formula) and to slab design (two-way punching shear uses a different Vc formula but the same φ = 0.75).
- The torsion threshold formula introduces Acp and pcp — these reappear in the full torsion design equations for closed stirrups (At/s) and longitudinal steel (Al), covered in the advanced torsion chapter.
- RA 544 (Philippine Civil Engineering Law) governs professional responsibility — a civil engineer who designs a beam without adequate shear reinforcement may face legal liability for structural failure, underscoring why code compliance in shear design is both a technical and professional obligation.
Quick review questions
A beam has bw = 300 mm, d = 500 mm, f'c = 28 MPa, and λ = 1.0. What is φVc in kN?
Answer
φVc = 101.2 kN
Vc = 0.17(1.0)(√28)(300)(500) = 0.17 × 5.292 × 150,000 = 134,934 N = 134.9 kN. φVc = 0.75 × 134.9 = 101.2 kN.
For the beam above, Vu = 250 kN. What is the required Vs in kN?
Answer
Vs = 198.4 kN
Vs = (Vu − φVc) / φ = (250 − 101.2) / 0.75 = 148.8 / 0.75 = 198.4 kN. Note: divide by φ = 0.75, do NOT just subtract.
Using 10 mm U-stirrups (Av = 157 mm²) and fy = 415 MPa with Vs = 198,400 N, what is the required stirrup spacing?
Answer
s = 164 mm → use 150 mm
s = Av fy d / Vs = (157 × 415 × 500) / 198,400 = 32,577,500 / 198,400 = 164 mm. Round down to 150 mm for practicality.
For the same beam (f'c = 28 MPa, bw = 300 mm, d = 500 mm), what is the maximum allowable Vs before the beam section must be enlarged?
Answer
Vs,max = 0.66√28 × 300 × 500 = 523,914 N ≈ 523.9 kN
Crushing limit = 0.66√f'c bw d = 0.66(5.292)(150,000) = 523,914 N = 523.9 kN. Since Vs = 198.4 kN < 523.9 kN, the section is adequate.
What is the smax spacing band threshold for the same beam? Which smax applies?
Answer
Band threshold = 262 kN. Since Vs = 198.4 kN < 262 kN, smax = min(d/2, 600) = min(250, 600) = 250 mm.
0.33√f'c bw d = 0.33(5.292)(150,000) = 261,954 N = 262 kN. Because Vs < band threshold, the lower-shear spacing limit governs: smax = d/2 = 500/2 = 250 mm. Final adopted spacing = 150 mm < 250 mm. OK.
At what Vu level can stirrups be completely discontinued along a beam?
Answer
Stirrups may be discontinued where Vu ≤ 0.5φVc.
NSCP 2015 requires minimum stirrups wherever Vu > 0.5φVc. Below this threshold, stirrups are not required by the code. For the example beam, 0.5φVc = 0.5 × 101.2 = 50.6 kN — stirrups can stop where Vu drops to 50.6 kN or less.
A 300×500 mm beam (f'c = 28 MPa, λ = 1.0) carries Tu = 4 kN·m. Can torsion be neglected?
Answer
Yes. Tth = 4.63 kN·m > Tu = 4 kN·m, so torsion may be neglected.
Acp = 300 × 500 = 150,000 mm²; pcp = 2(300+500) = 1,600 mm. Tth = 0.75(0.083)(√28)(150,000²/1,600) = 0.75(0.083)(5.292)(14,062,500) = 4,633 kN·mm = 4.63 kN·m. Since Tu = 4 kN·m < 4.63 kN·m, torsion is negligible per NSCP 2015 Section 422.7.1.
A 4-legged stirrup uses 12 mm bars. What is the total Av?
Answer
Av = 4 × (π/4 × 12²) = 4 × 113.1 = 452.4 mm²
Each 12 mm bar has an area of π(12²)/4 = 113.1 mm². With 4 legs, Av = 4 × 113.1 = 452.4 mm². This higher Av allows a larger stirrup spacing s for the same Vs.
Why must closed stirrups be used for torsion design while open U-stirrups suffice for shear only?
Answer
Torsional shear stress flows continuously around the entire cross-section perimeter (shear flow), requiring a closed loop to resist it. Open stirrups cannot resist the tension on the open face.
Shear causes diagonal tension primarily on the beam web faces, which open U-stirrups can intercept. Torsion causes a spiral pattern of diagonal tension that wraps around all four faces — only a closed stirrup loop provides a continuous tension path. If open stirrups are used in torsion, the unstirruped top leg of the U is unresisted, and failure occurs at that face.
Compute Vs for Vu = 180 kN and φVc = 120 kN. A common wrong answer is Vs = 60 kN. What is the correct value and why?
Answer
Correct Vs = 80 kN. Wrong answer of 60 kN results from forgetting to divide by φ.
Correct: Vs = (Vu − φVc) / φ = (180 − 120) / 0.75 = 60 / 0.75 = 80 kN. The wrong answer (60 kN) is Vu − φVc before dividing by φ. Using 60 kN instead of 80 kN gives a stirrup spacing that is too large (unconservative, unsafe). This is the single most important formula to get right.
Diagrams
Chapter Overview: Shear and Torsion in RC Beams
Mind map summarizing all major topics in the shear and torsion chapter — organized around the central design requirement φVn ≥ Vu, branching into concrete shear strength, stirrup design steps, minimum reinforcement, and torsion threshold checks. Use this for a quick visual orientation before diving into each section.
Step-by-Step Stirrup Design Procedure
Complete decision flowchart for stirrup design, from computing Vc through all code checks to the final adopted spacing. Follow this flowchart for every board exam shear problem to avoid missing steps. The critical branch is whether Vs exceeds the crushing limit (0.66 factor) — if it does, the only correct answer is to enlarge the section.
Torsion Threshold Decision Flowchart
Decision flowchart for the NSCP 2015 torsion threshold check — the most common torsion question in board exams. Shows the complete path from computing Acp and pcp through comparing Tu to Tth, distinguishing equilibrium vs. compatibility torsion, and proceeding to full torsion design when required.
Distribution of Board Exam Questions on Shear and Torsion
Approximate distribution of shear and torsion question types in PRC Civil Engineer Licensure Examinations based on review patterns. Stirrup spacing problems are most frequent, followed by Vc computation and torsion threshold checks. Focus your review time accordingly — spend 40% of shear practice time on the full stirrup spacing design procedure.
Shear Zone Classification Along a Beam
State diagram showing the three shear zones along a simply supported beam from support to midspan. As the shear force decreases away from the support, the required reinforcement transitions from designed stirrups to minimum stirrups to no stirrups. This spatial understanding is critical for answering 'where can stirrups be discontinued?' problems on the board exam.
Exam strategy
For PRC board exam shear and torsion problems, follow this disciplined attack sequence: (1) Identify the beam dimensions (bw, h, d) and material strengths (f'c, fy). (2) Compute Vc = 0.17λ√f'c bw d and φVc = 0.75Vc. (3) Compare Vu at the critical section (d from support face) with φVc to classify the stirrup requirement zone. (4) If Vu > φVc, compute Vs = (Vu − φVc)/0.75 — write this step explicitly to avoid the most common error. (5) Check Vs ≤ 0.66√f'c bw d (crushing limit); if violated, state 'enlarge section' and stop. (6) Check Vs vs. 0.33√f'c bw d to select the correct smax (d/2 & 600 mm vs. d/4 & 300 mm). (7) Compute s = Av fy d / Vs and round DOWN to a practical increment. (8) For torsion problems, compute Tth = 0.75(0.083)√f'c (Acp²/pcp) and compare with Tu — if Tu < Tth, state torsion is negligible. Allocate about 3–4 minutes per shear problem and 2–3 minutes for a torsion threshold check. Memorize four key coefficients: 0.17 (Vc), 0.66 (Vs max), 0.33 (spacing band), and 0.083 (torsion threshold). In multiple-choice problems, always eliminate answers with φ = 0.90 for shear or spacings exceeding 600 mm — these are always wrong for standard beams.