CELE Reinforced & Prestressed Concrete — Reinforced Concrete Beams: Shear and TorsionExam Answer Templates

Examiner tip

One-mark VSA questions reward precision. Write the formula exactly as it appears in NSCP 2015 Section 422.5.5.1 — coefficient 0.17, not 0.166 or 1/6.

Model answer

Vc = 0.17 λ √f'c · bw · d

where λ = 1.0 for normal-weight concrete, f'c in MPa, bw = web width in mm, d = effective depth in mm; result in Newtons.

Answer structure

• Line 1: Write the complete formula with all variable definitions [1 mark]

Scoring breakdown

Common mark deductions

• Writing 0.17√f'c without λ — loses mark if examiner checks for completeness
• Omitting units or unit basis (N, MPa, mm)
• Confusing the coefficient (some students write 0.75 which is φ, not the Vc coefficient)

Key phrases to include

• 0.17
• λ√f'c
• bw·d
• normal-weight concrete λ = 1.0
• NSCP 2015 simplified

Examiner tip

Concept-pair questions (value + reason) are classic 2-mark structures. Always give both the number AND the engineering reason to guarantee both marks.

Model answer

Per NSCP 2015 Section 421.2.1, φ = 0.75 for shear and torsion.

This is lower than the flexural φ = 0.90 because shear failure is brittle and sudden — there is no visible warning (no large deflection or crack widening before collapse), unlike flexural failure which is ductile. The code therefore uses a more conservative φ to provide a larger safety margin against this mode.

Answer structure

• Statement 1: State φ = 0.75 with code reference [1 mark]
• Statement 2: Explain brittleness of shear failure versus ductility of flexural failure [1 mark]

Scoring breakdown

Common mark deductions

• Stating φ = 0.85 (the old ACI 318-99 value — no longer current)
• Giving the value without any justification — earns only 1 out of 2
• Comparing to wrong φ value for flexure

Key phrases to include

• φ = 0.75
• NSCP 2015
• brittle and sudden
• no visible warning
• flexural φ = 0.90
• conservative

Examiner tip

For 2-mark numerical problems, show every substitution on its own line — examiners award marks for process, not just the final number.

Model answer

Given: bw = 300 mm, d = 500 mm, f'c = 28 MPa, λ = 1.0, φ = 0.75

Step 1 — Concrete shear strength:
Vc = 0.17 λ √f'c · bw · d
Vc = 0.17 (1.0)(√28)(300)(500)
Vc = 0.17 (5.292)(150 000)
Vc = 134 928 N ≈ 134.9 kN

Step 2 — Design shear strength:
φVc = 0.75 × 134.9 = 101.2 kN

∴ Vc = 134.9 kN, φVc = 101.2 kN

Answer structure

• Given data block: list all known values [0 marks but essential for traceability]
• Step 1: Write Vc formula and substitute [1 mark]
• Step 2: Multiply by φ = 0.75 to get φVc [1 mark]

Scoring breakdown

Common mark deductions

• Using bw·d in m² instead of mm² — gives answer 1 000 000× too small
• Forgetting to multiply by φ for the second part
• Rounding √28 incorrectly (e.g., using 5.3 without showing calculation)

Key phrases to include

• Vc = 0.17λ√f'c·bw·d
• √28 = 5.292
• λ = 1.0
• φ = 0.75
• 134.9 kN
• 101.2 kN

Examiner tip

Always write both limits as a min() expression — examiners deduct a mark if you give only one of the two values in the minimum.

Model answer

NSCP 2015 provides two maximum-spacing bands based on the stirrup demand Vs:

(1) If Vs ≤ 0.33√f'c · bw · d (low shear demand):
smax = min(d/2, 600 mm)

(2) If Vs > 0.33√f'c · bw · d (high shear demand):
smax = min(d/4, 300 mm)

Rationale: When shear demand is high, closer spacing ensures stirrups intercept any diagonal crack that forms between them.

Answer structure

• Rule 1: State condition (Vs ≤ 0.33√f'c·bw·d) and corresponding smax = min(d/2, 600 mm) [1 mark]
• Rule 2: State condition (Vs > 0.33√f'c·bw·d) and corresponding smax = min(d/4, 300 mm) [1 mark]

Scoring breakdown

Common mark deductions

• Swapping the two limits (applying d/4 rule when Vs is low)
• Stating only d/2 or only d/4 without the companion absolute limit (600 or 300 mm)
• Omitting the threshold condition 0.33√f'c·bw·d

Key phrases to include

• 0.33√f'c·bw·d
• d/2 and 600 mm
• d/4 and 300 mm
• min(d/2, 600)
• min(d/4, 300)

Examiner tip

This is a classic board-exam trap. The word 'face' is the key — memorize: critical section for shear is at d from the FACE, not the centerline.

Model answer

The critical section for shear is located at a distance d (effective depth) from the face of the support. Shear between this section and the support face is carried by the direct compression strut to the support and need not govern stirrup design.

Answer structure

• Line 1: State location (distance d from face of support) and brief reason [1 mark]

Scoring breakdown

Common mark deductions

• Stating 'd from the centerline of the support' — this is incorrect
• Stating '2d' or 'd/2' — wrong distance
• No mention of face of support (just saying 'from the support' is ambiguous)

Key phrases to include

• distance d from the face of the support
• effective depth
• not at the support centerline

Examiner tip

Five-mark shear problems follow a strict five-step sequence. Examiners award one mark per step — if you skip a step (even if your final answer is correct), you lose that step's mark. Write all five steps explicitly.

Model answer

Given: bw = 300 mm, d = 500 mm, f'c = 28 MPa, φ = 0.75
Vu = 250 kN, fy = 415 MPa
Av = 2 × (π/4)(10²) = 2 × 78.54 = 157.1 mm² (two-legged U-stirrup)
φVc = 101.2 kN (from T3)

Step 1 — Required stirrup strength:
Vs = (Vu − φVc) / φ = (250 − 101.2) / 0.75 = 148.8 / 0.75 = 198.4 kN

Step 2 — Check crushing limit (NSCP 2015):
Vs,max = 0.66√f'c · bw · d = 0.66(5.292)(300)(500) = 524,016 N = 524.0 kN
Since Vs = 198.4 kN < 524.0 kN ✓ — section is adequate, no need to enlarge.

Step 3 — Determine spacing band:
Threshold = 0.33√f'c · bw · d = 0.33(5.292)(300)(500) = 262,008 N = 262.0 kN
Since Vs = 198.4 kN < 262.0 kN → LOW shear band
∴ smax = min(d/2, 600) = min(250, 600) = 250 mm

Step 4 — Compute required spacing:
s = Av·fy·d / Vs = 157.1 × 415 × 500 / 198,400 = 32,595,250 / 198,400 = 164.3 mm

Step 5 — Govern by minimum:
s_required = 164 mm < smax = 250 mm ✓

∴ Use s = 150 mm (round down to nearest 25 mm increment for practical detailing)

Final Answer: Provide 10 mm U-stirrups at s = 150 mm c/c.

Answer structure

• Given block: state Vu, Av computation for two-legged stirrup [0 marks, but traceable]
• Step 1: Compute Vs = (Vu − φVc)/φ [1 mark]
• Step 2: Check crushing limit Vs ≤ 0.66√f'c·bw·d [1 mark]
• Step 3: Identify spacing band using 0.33√f'c·bw·d threshold and state smax [1 mark]
• Step 4: Compute s = Av·fy·d / Vs [1 mark]
• Step 5: Compare s with smax, state governing value, round to practical spacing [1 mark]

Scoring breakdown

Common mark deductions

• Forgetting to divide by φ when computing Vs from Vu — one of the most common board-exam errors
• Using Av = 78.5 mm² (one leg only instead of two) — gives spacing twice as large
• Skipping the crushing check — loses Step 2 mark
• Not identifying the correct spacing band — leads to wrong smax
• Using d in meters instead of millimeters in the spacing formula

Key phrases to include

• Vs = (Vu − φVc)/φ
• two-legged U-stirrup, Av = 157 mm²
• crushing limit 0.66√f'c·bw·d
• 0.33√f'c·bw·d threshold
• smax = min(d/2, 600) = 250 mm
• s = Av·fy·d / Vs
• 164 mm < 250 mm

Examiner tip

Three-mark concept problems expect three distinct, testable statements. Structure your answer in three numbered parts — this signals to the examiner exactly where your three marks are.

Model answer

Trigger Condition (NSCP 2015 Section 409.6.3.3):
Minimum shear reinforcement is required wherever:
Vu > 0.5φVc
(i.e., factored shear exceeds half the design concrete shear capacity)

Minimum Reinforcement Formula:
Av,min = max[
0.062√f'c · (bw · s / fy),
0.35 · (bw · s / fy)
]

Both limits must be evaluated; the larger value governs. The 0.35 limit controls for lower-strength concrete (approximately f'c < 32 MPa). Spacing s is the stirrup spacing being checked.

Answer structure

• Part 1: State trigger condition Vu > 0.5φVc [1 mark]
• Part 2: Write first limit 0.062√f'c·bw·s/fy [1 mark]
• Part 3: Write second limit 0.35·bw·s/fy and state that the larger governs [1 mark]

Scoring breakdown

Common mark deductions

• Writing only one of the two Av,min limits — loses third mark
• Stating the trigger as Vu > φVc (instead of 0.5φVc)
• Not clarifying which limit governs (just writing both formulas without comparison)

Key phrases to include

• Vu > 0.5φVc
• 0.062√f'c · bw·s/fy
• 0.35 · bw·s/fy
• larger value governs
• NSCP 2015

Examiner tip

The 0.66 coefficient is twice 0.33 — both come from the same code provision. Think of 0.33 as the 'halfway warning' and 0.66 as the 'hard ceiling.'

Model answer

Upper limit on Vs (NSCP 2015 Section 422.5.10.1):
Vs,max = 0.66√f'c · bw · d

If the required Vs exceeds this limit, the concrete section is too small to develop adequate diagonal compression strut capacity — increasing stirrup area will NOT help. The designer must ENLARGE THE BEAM SECTION (increase bw, d, or both) and re-compute Vc and Vs.

Answer structure

• Statement 1: State the formula Vs,max = 0.66√f'c·bw·d [1 mark]
• Statement 2: Consequence if exceeded — must enlarge the beam section [1 mark]

Scoring breakdown

Common mark deductions

• Confusing Vs,max = 0.66√f'c·bw·d with the band threshold 0.33√f'c·bw·d
• Stating 'add more stirrups' as the solution — this is incorrect and loses the second mark

Key phrases to include

• 0.66√f'c·bw·d
• concrete crushing
• enlarge the beam section
• adding stirrups will not help

Examiner tip

Always show √f'c numerically (e.g., √35 = 5.916) — it demonstrates you did not blindly punch numbers, and avoids arithmetic errors that cannot be traced.

Model answer

Given: bw = 300 mm, d = 550 mm, f'c = 35 MPa, fy = 415 MPa, Vu = 180 kN, s = 200 mm

Step 1 — Compute φVc:
Vc = 0.17(1.0)√35 × 300 × 550 = 0.17(5.916)(165 000) = 165 844 N = 165.8 kN
φVc = 0.75 × 165.8 = 124.4 kN

Step 2 — Check trigger:
0.5φVc = 0.5 × 124.4 = 62.2 kN
Vu = 180 kN > φVc = 124.4 kN → Stirrups required (shear demand exceeds concrete capacity)

Step 3 — Minimum Av (at s = 200 mm):
Limit 1: Av,min = 0.062√35 × (300 × 200) / 415 = 0.062(5.916)(60 000/415) = 0.062 × 5.916 × 144.6 = 53.0 mm²
Limit 2: Av,min = 0.35 × (300 × 200) / 415 = 0.35 × 144.6 = 50.6 mm²
Governing: Av,min = 53.0 mm² (Limit 1 controls since f'c = 35 MPa > 32 MPa)

∴ Minimum Av = 53.0 mm² for s = 200 mm. (Two 6 mm bars = 2 × 28.3 = 56.5 mm² > 53 mm² ✓)

Answer structure

• Step 1: Compute Vc and φVc [1 mark]
• Step 2: Check trigger condition (compare Vu to φVc and to 0.5φVc) [1 mark]
• Step 3: Evaluate both Av,min limits and state which governs [1 mark]

Scoring breakdown

Common mark deductions

• Not checking both Av,min formulas — selecting only one loses the third mark
• Using the wrong trigger (comparing to φVc/2 only instead of also checking Vu vs φVc)
• Substituting s = 200 m instead of 200 mm in the Av,min formula

Key phrases to include

• Vc = 0.17λ√f'c·bw·d
• √35 = 5.916
• φVc = 124.4 kN
• Vu > φVc → stirrups required
• 0.062√f'c·bw·s/fy
• 0.35·bw·s/fy
• limit 1 controls

Examiner tip

Acp uses the OUTSIDE perimeter (the gross section), while Aoh uses the stirrup centerline. Examiners specifically test this distinction — always label which you are using.

Model answer

Torsion Threshold (NSCP 2015 Section 409.5.4.1):

Tu,threshold = φ · 0.083 · λ · √f'c · (Acp² / pcp)

where:
φ = 0.75 (strength-reduction factor for torsion)
λ = 1.0 for normal-weight concrete
f'c = specified compressive strength of concrete (MPa)
Acp = area enclosed by the outside perimeter of the cross-section (mm²)
pcp = outside perimeter of the cross-section (mm)

If the factored torque Tu < Tu,threshold, the torsional effect is negligible and the member is designed for shear only.

Answer structure

• Line 1: Write complete formula with φ = 0.75 and coefficient 0.083 [1 mark]
• Line 2–6: Define Acp and pcp clearly as OUTSIDE perimeter quantities [1 mark]

Scoring breakdown

Common mark deductions

• Using Aoh (stirrup centerline area) instead of Acp (outside perimeter area) — common confusion
• Omitting φ from the threshold formula
• Writing coefficient as 0.083 × 10⁻³ or misplacing the decimal

Key phrases to include

• φ = 0.75
• 0.083
• λ√f'c
• Acp²/pcp
• outside perimeter
• torsion may be neglected
• NSCP 2015

Examiner tip

Always end torsion-threshold problems with a one-sentence decision: 'Since Tu [</>] T_th, torsion [may be neglected / must be designed for].' This conclusion sentence is worth a dedicated mark.

Model answer

Given: bw = 300 mm, h = 500 mm, f'c = 28 MPa, λ = 1.0, φ = 0.75, Tu = 4.0 kN·m

Step 1 — Section properties (outside perimeter):
Acp = bw × h = 300 × 500 = 150 000 mm²
pcp = 2(bw + h) = 2(300 + 500) = 1 600 mm

Step 2 — Compute torsion threshold:
T_th = φ (0.083)(λ)(√f'c)(Acp² / pcp)
T_th = 0.75 (0.083)(1.0)(5.292)[(150 000)² / 1 600]
T_th = 0.75 × 0.083 × 5.292 × [22 500 000 000 / 1 600]
T_th = 0.75 × 0.083 × 5.292 × 14 062 500
T_th = 4 630 313 N·mm = 4.63 kN·m

Step 3 — Comparison:
Tu = 4.0 kN·m < T_th = 4.63 kN·m

∴ Torsion may be NEGLECTED. Design the beam for shear only per NSCP 2015 Section 409.5.4.1.

Answer structure

• Step 1: Compute Acp and pcp using the outside perimeter [1 mark]
• Step 2: Substitute into threshold formula and compute T_th [1 mark]
• Step 3: Compare Tu to T_th and state conclusion [1 mark]

Scoring breakdown

Common mark deductions

• Using Acp in m² — gives threshold value 10⁶ times smaller, leading to wrong conclusion
• Forgetting φ = 0.75 in the threshold formula — gives T_th too large by 33%
• Not writing an explicit conclusion sentence — examiners want to see a clear decision

Key phrases to include

• Acp = 150 000 mm²
• pcp = 1 600 mm
• T_th = φ(0.083)λ√f'c(Acp²/pcp)
• 4.63 kN·m
• Tu < T_th
• torsion may be neglected
• NSCP 2015 Section 409.5.4.1

Examiner tip

On 5-mark problems, examiners often give partial credit even with an arithmetic error in Step 1 — if your subsequent steps follow logically from your (wrong) φVc. Show all steps so carry-forward credit applies.

Model answer

Given: bw = 400 mm, d = 640 mm, f'c = 28 MPa, fy = 415 MPa, Vu = 400 kN, φ = 0.75
Av = 2 × (π/4)(12²) = 2 × 113.1 = 226.2 mm²

Step 1 — Concrete shear capacity:
Vc = 0.17(1.0)(√28)(400)(640) = 0.17(5.292)(256 000) = 230 180 N = 230.2 kN
φVc = 0.75 × 230.2 = 172.6 kN

Step 2 — Required stirrup strength:
Vs = (Vu − φVc) / φ = (400 − 172.6) / 0.75 = 227.4 / 0.75 = 303.2 kN

Step 3 — Check crushing limit:
Vs,max = 0.66√28 × 400 × 640 = 0.66(5.292)(256 000) = 894,100 N = 894.1 kN
Vs = 303.2 kN < 894.1 kN ✓ Section is adequate.

Step 4 — Spacing band:
Band threshold = 0.33√28 × 400 × 640 = 0.33(5.292)(256 000) = 447,050 N = 447.1 kN
Vs = 303.2 kN < 447.1 kN → LOW demand band
∴ smax = min(d/2, 600) = min(320, 600) = 320 mm

Step 5 — Required spacing:
s = Av·fy·d / Vs = 226.2 × 415 × 640 / 303 200
= 60,089,280 / 303,200 = 198.2 mm

Step 6 — Govern:
s = 198 mm < smax = 320 mm ✓

∴ Use 12 mm U-stirrups at s = 175 mm c/c (round down to nearest 25 mm for practical detailing).

Answer structure

• Av calculation for two-legged 12 mm stirrup [shown in given block]
• Step 1: Compute Vc and φVc [1 mark]
• Step 2: Compute Vs = (Vu − φVc)/φ [1 mark]
• Step 3: Check crushing limit Vs ≤ 0.66√f'c·bw·d [1 mark]
• Step 4: Identify spacing band and determine smax [1 mark]
• Step 5–6: Compute s = Av·fy·d/Vs, compare to smax, state final spacing [1 mark]

Scoring breakdown

Common mark deductions

• Using single-leg area Av = 113 mm² — halves the correct answer
• Computing Vs = Vu/φ − Vc instead of (Vu − φVc)/φ — numerically different and wrong
• Skipping crushing check — automatic loss of Step 3 mark
• Using d = 700 mm (total depth) instead of d = 640 mm (effective depth)

Key phrases to include

• Av = 226.2 mm² (2 legs)
• φVc = 172.6 kN
• Vs = (Vu − φVc)/φ = 303.2 kN
• crushing limit 894 kN > Vs
• band threshold 447 kN
• smax = 320 mm
• s = 198 mm < smax

Examiner tip

Concept-explanation questions reward engineering understanding over formula recitation. Include at least one physical mechanism (aggregate interlock, dowel action, truss analogy) to show depth of knowledge.

Model answer

Why Shear Failure is More Dangerous:
Shear failure is brittle and sudden — there is no significant ductile deformation before collapse. Unlike flexural failure (where extensive cracking and deflection provide visible warning), a shear-critical beam can fail without any prior warning signs. The Philippine code (NSCP 2015) reflects this by assigning φ = 0.75 for shear versus φ = 0.90 for flexure.

Two Components of Shear Resistance:
(1) Concrete contribution Vc: Shear is carried by the uncracked compression zone, aggregate interlock along diagonal cracks, and dowel action of the longitudinal bars. In design, Vc = 0.17λ√f'c·bw·d (NSCP 2015 simplified).

(2) Stirrup contribution Vs: Stirrups crossing the diagonal crack act as tension ties in a truss model. Each stirrup carries Av·fy, and collectively Vs = Av·fy·d/s.

Total nominal shear strength: Vn = Vc + Vs, with design requirement φVn ≥ Vu.

Answer structure

• Part 1: Explain brittleness/suddenness of shear failure and tie to lower φ [1 mark]
• Part 2: Define Vc and give its physical meaning or formula [1 mark]
• Part 3: Define Vs, explain stirrup mechanism, and state Vn = Vc + Vs [1 mark]

Scoring breakdown

Common mark deductions

• Describing shear failure as 'gradual' or 'ductile' — factually wrong
• Listing Vc and Vs without explaining the physical mechanism of either
• Omitting the design inequality φVn ≥ Vu

Key phrases to include

• brittle and sudden
• no visible warning
• φ = 0.75 vs φ = 0.90
• aggregate interlock
• dowel action
• Vc = 0.17λ√f'c·bw·d
• Vs = Av·fy·d/s
• Vn = Vc + Vs

Examiner tip

This is a classic trap: Vu < φVc does NOT automatically mean no stirrups. The minimum-stirrup rule triggers at 0.5φVc. Boards frequently test this exact scenario.

Model answer

Given: bw = 250 mm, d = 450 mm, f'c = 21 MPa, λ = 1.0, φ = 0.75, Vu = 45 kN

Step 1 — Compute φVc:
Vc = 0.17(1.0)√21 × 250 × 450 = 0.17(4.583)(112 500) = 87 782 N = 87.8 kN
φVc = 0.75 × 87.8 = 65.8 kN

Step 2 — Minimum stirrup trigger:
0.5φVc = 0.5 × 65.8 = 32.9 kN
Vu = 45 kN > 0.5φVc = 32.9 kN → Minimum stirrups are REQUIRED

Note: Since Vu = 45 kN < φVc = 65.8 kN, stirrups are not required for strength, but minimum stirrups must still be provided because Vu > 0.5φVc.

∴ Provide minimum shear reinforcement per NSCP 2015 Section 409.6.3.3.

Answer structure

• Step 1: Compute φVc [1 mark]
• Step 2: Compare Vu to 0.5φVc and φVc; state correct conclusion [1 mark]

Scoring breakdown

Common mark deductions

• Comparing Vu only to φVc (not to 0.5φVc) — leads to wrong conclusion that 'no stirrups needed'
• Stating 'no stirrups required' because Vu < φVc — missing the minimum-stirrup trigger

Key phrases to include

• φVc = 65.8 kN
• 0.5φVc = 32.9 kN
• Vu > 0.5φVc
• minimum stirrups required
• Vu < φVc (no strength demand)
• NSCP 2015 Section 409.6.3.3

Examiner tip

The decision for torsion always has two parts: (1) the comparison result and (2) what to do next. Always name 'closed stirrups' and 'longitudinal torsion bars' — these terms signal code knowledge.

Model answer

Given: bw = 400 mm, h = 600 mm, f'c = 28 MPa, φ = 0.75, Tu = 12 kN·m

Step 1 — Outside perimeter section properties:
Acp = 400 × 600 = 240 000 mm²
pcp = 2(400 + 600) = 2 000 mm

Step 2 — Torsion threshold:
T_th = φ(0.083)(λ)(√f'c)(Acp²/pcp)
T_th = 0.75(0.083)(1.0)(5.292)[(240 000)²/2 000]
T_th = 0.75 × 0.083 × 5.292 × [57 600 000 000 / 2 000]
T_th = 0.75 × 0.083 × 5.292 × 28 800 000
T_th = 9 476 784 N·mm = 9.48 kN·m

Step 3 — Decision:
Tu = 12 kN·m > T_th = 9.48 kN·m

∴ Torsion MUST be designed for. Provide closed stirrups and additional longitudinal torsion reinforcement per NSCP 2015 Section 409.5.4.

Answer structure

• Step 1: Compute Acp and pcp (outside perimeter) [1 mark]
• Step 2: Compute T_th using threshold formula [1 mark]
• Step 3: Compare Tu to T_th and state design decision [1 mark]

Scoring breakdown

Common mark deductions

• Using bw only (ignoring h) in Acp — uses 400 × 400 or similar wrong value
• Using stirrup centerline dimensions instead of outside perimeter
• Not mentioning closed stirrups in the design decision — loses conclusion mark

Key phrases to include

• Acp = 240 000 mm²
• pcp = 2 000 mm
• T_th = φ(0.083)λ√f'c(Acp²/pcp)
• 9.48 kN·m
• Tu > T_th
• torsion must be designed
• closed stirrups
• longitudinal torsion bars

Do

• Write the formula or value exactly as it appears in NSCP 2015
• Include units in formula definitions (N, MPa, mm)
• State λ = 1.0 for normal-weight concrete even if not asked — it shows completeness
• Circle or underline the final answer

Don't

• Do not spend more than 2 minutes on a 1-mark question
• Do not write a paragraph — examiners skim 1-mark answers
• Do not leave blank; a partially correct formula still earns the mark in most rubrics

Do

• Number or label each scorable point (1) and (2)
• Write φ = 0.75 explicitly in every shear calculation
• Show √f'c as a decimal (e.g., √28 = 5.292) to enable error tracing
• State the NSCP 2015 section reference for definition questions

Don't

• Do not merge both points into one run-on sentence — examiners need to see two separate items
• Do not skip units on the final answer
• Do not use f'c = 28 kPa (wrong unit — always MPa)

Do

• Write 'Step 1 —', 'Step 2 —', 'Step 3 —' headers to guide the examiner
• Include a brief engineering reason with each calculation (e.g., 'Check crushing limit to ensure concrete does not fail before stirrups yield')
• Write conclusion: '∴ [your answer]' at the end — the ∴ symbol signals to the examiner you are done
• Verify your answer makes physical sense (e.g., spacing < maximum spacing)

Don't

• Do not skip checking whether your computed value satisfies the code limit
• Do not write steps out of order — always: given → formula → substitution → result → check
• Do not mix kN and N in the same equation without conversion

Do

• Begin with a complete 'Given:' block — lists all data and derived quantities (Av, √f'c) in one place
• Write crushing check even when obviously satisfied — examiners mark for its presence
• State both limits in the min() expression for smax (d/2 AND 600 mm, or d/4 AND 300 mm)
• Round final spacing DOWN to practical increment (25 mm or 50 mm) and state why
• End with a boxed or bolded final answer: 'Use [bar size] stirrups at s = [value] mm c/c'

Don't

• Do not combine steps — examiners need to see each step on its own line to award marks
• Do not use total depth h in place of effective depth d in any shear formula
• Do not forget to double the single-bar area for Av in a U-stirrup
• Do not round intermediate values — carry at least 4 significant figures until the final step