CELE Reinforced & Prestressed Concrete — Reinforced Concrete Fundamentals: WSD and USDExam Answer Templates

Examiner tip

For a 1-mark question, the entire answer fits on one line. Write the formula, substitute, and state the answer with unit. No elaboration needed.

Model answer

Ec = 4700√f'c = 4700√25 = 4700 × 5 = 23 500 MPa

Answer structure

• Line 1: State the formula and substitute the given value, then compute the numerical answer with correct unit [1 mark]

Scoring breakdown

Common mark deductions

• Using Ec = 4700f'c (forgetting the square root) — gives a wildly wrong answer
• Omitting the unit 'MPa' — answer is incomplete
• Using f'c in kPa or kN/m² instead of MPa — wrong numerical result

Key phrases to include

• Ec = 4700√f'c
• 23 500 MPa
• normal-weight concrete

Examiner tip

The NSCP 2015 φ table is a guaranteed board exam item. Memorize all six values: 0.90, 0.75, 0.75 (spiral), 0.65 (tied), 0.65 (bearing), 0.60 (plain).

Model answer

(a) Flexure (tension-controlled): φ = 0.90
(b) Shear and torsion: φ = 0.75

Answer structure

• Line 1: Give φ for flexure with the qualifier 'tension-controlled' [0.5 mark]
• Line 2: Give φ for shear/torsion [0.5 mark]

Scoring breakdown

Common mark deductions

• Writing φ = 0.85 for flexure — this is an older ACI value, not current NSCP 2015
• Omitting the qualifier 'tension-controlled' for the flexure value
• Confusing φ for shear (0.75) with φ for tied columns (0.65)

Key phrases to include

• tension-controlled
• φ = 0.90
• φ = 0.75
• NSCP 2015

Examiner tip

Always anchor the definition to the equation a = β₁c. An abstract definition without the formula earns partial credit at best.

Model answer

β₁ is the ratio that converts the neutral-axis depth c to the equivalent rectangular stress-block depth a, i.e., a = β₁c. Per NSCP 2015: β₁ = 0.85 for f'c ≤ 28 MPa, decreasing by 0.05 per 7 MPa above 28 MPa, with a minimum of 0.65.

Answer structure

• Line 1: Define β₁ through the relation a = β₁c [0.5 mark]
• Line 2: State the NSCP 2015 values/limits [0.5 mark]

Scoring breakdown

Common mark deductions

• Defining β₁ as the ratio of stress-block depth to the beam depth d — this is wrong; it refers to neutral-axis depth c
• Not stating the upper limit (f'c ≤ 28 MPa) or the minimum (0.65)

Key phrases to include

• a = β₁c
• neutral-axis depth
• 0.85
• 28 MPa
• minimum 0.65

Examiner tip

Always write the range check first ('f'c = 42 MPa > 28 MPa') before applying the formula — this earns the first mark and prevents applying the wrong formula.

Model answer

Given: f'c = 42 MPa > 28 MPa → use the reduction formula.

β₁ = 0.85 − 0.05 × [(f'c − 28) / 7]
= 0.85 − 0.05 × [(42 − 28) / 7]
= 0.85 − 0.05 × [14 / 7]
= 0.85 − 0.05 × 2
= 0.85 − 0.10
= 0.75

Check: 0.65 ≤ 0.75 ≤ 0.85 ✓

∴ β₁ = 0.75

Answer structure

• Step 1: Identify which range f'c falls in and state the applicable formula [1 mark]
• Step 2: Substitute and compute to get β₁ = 0.75, and verify against the floor of 0.65 [1 mark]

Scoring breakdown

Common mark deductions

• Computing (42 − 28)/7 as 2.0 but then forgetting to multiply by 0.05 — arithmetic slip
• Not checking whether the result is above the minimum 0.65
• Using f'c = 42 MPa in the formula but mistakenly stating the range as f'c ≤ 28 MPa

Key phrases to include

• β₁ = 0.85 − 0.05[(f'c − 28)/7]
• 42 MPa > 28 MPa
• β₁ = 0.75
• minimum 0.65

Examiner tip

In WSD, n is used to transform steel area to an equivalent concrete area. WSD exam problems typically use n = 8 for f'c = 35 MPa — memorize common pairs.

Model answer

Given: f'c = 35 MPa, Es = 200 000 MPa

Step 1 — Modulus of elasticity of concrete:
Ec = 4700√f'c = 4700√35 = 4700 × 5.916 = 27 806 MPa

Step 2 — Modular ratio:
n = Es / Ec = 200 000 / 27 806 = 7.19

Round up to nearest integer per practice: n = 8

∴ n ≈ 8 (or 7.19 if exact value is required)

Answer structure

• Step 1: Compute Ec using Ec = 4700√f'c [1 mark]
• Step 2: Divide Es by Ec to obtain n; state rounding convention [1 mark]

Scoring breakdown

Common mark deductions

• Inverting the ratio as n = Ec/Es — a conceptually wrong definition
• Not showing the Ec computation — answer appears to come from nowhere
• Leaving n as a non-integer without stating whether rounding is applied

Key phrases to include

• n = Es/Ec
• Ec = 4700√f'c
• modular ratio
• WSD transformed section

Examiner tip

Always state the code equation number if you know it (NSCP 2015 Eq. 405.3.1). Even if you compute Mu correctly, naming the combination explicitly earns the methodology mark.

Model answer

Given: MD = 80 kN·m, ML = 60 kN·m

NSCP 2015 governing gravity load combination (LRFD):
Mu = 1.2 MD + 1.6 ML
= 1.2(80) + 1.6(60)
= 96 + 96
= 192 kN·m

∴ Mu = 192 kN·m

The section must be designed so that φMn ≥ 192 kN·m.

Answer structure

• Step 1: State the governing NSCP 2015 LRFD load combination 1.2D + 1.6L [1 mark]
• Step 2: Substitute and compute Mu = 192 kN·m; state the design inequality φMn ≥ Mu [1 mark]

Scoring breakdown

Common mark deductions

• Using 1.4D + 1.7L (the old ACI 318-89 / pre-2001 combination) — not current NSCP 2015
• Forgetting to write the design check φMn ≥ Mu — this is the purpose of computing Mu
• Mixing MD and ML subscripts in the formula

Key phrases to include

• 1.2MD + 1.6ML
• Mu = 192 kN·m
• φMn ≥ Mu
• NSCP 2015 LRFD

Examiner tip

Examiners reward answers that explicitly link the name of the method to its core equation. Mention '0.45f'c' for WSD and 'φMn ≥ Mu' for USD — these are the signature expressions.

Model answer

Working Stress Design (WSD) — also called the Alternate Design Method — keeps stresses at service loads within allowable fractions of the material strength. Concrete stress is limited to 0.45f'c and steel stress to allowable fs. The analysis uses the elastic modular ratio n = Es/Ec to transform the reinforced section into an equivalent homogeneous section. WSD is conservative, easy to visualize, but can be inefficient because it does not account for actual load variability or ductility.

Ultimate Strength Design (USD / LRFD) — the method prescribed by NSCP 2015 — factors the service loads upward (e.g., Mu = 1.2MD + 1.6ML) to obtain demand and then reduces the nominal section capacity by the strength-reduction factor φ (e.g., φ = 0.90 for tension-controlled flexure). The design requirement is φMn ≥ Mu. USD uses the equivalent rectangular (Whitney) stress block of intensity 0.85f'c to compute nominal moment capacity, and explicitly accounts for ductility through strain compatibility.

Key distinction: WSD works at service-load stress level; USD works at the factored (ultimate) load level with reduced nominal strength.

Answer structure

• Paragraph 1: Define WSD — service-load stresses, allowable fractions (0.45f'c), modular ratio n [1 mark]
• Paragraph 2: Define USD — factored loads (1.2D+1.6L), strength reduction φ, Whitney block, design check φMn ≥ Mu [1 mark]
• Line 3: A concise comparative statement distinguishing the two philosophies [1 mark]

Scoring breakdown

Common mark deductions

• Describing USD without mentioning the φ factor — this is the most defining feature of LRFD
• Writing 'WSD uses ultimate loads reduced by safety factor' — conceptually wrong
• No comparative statement — just two parallel descriptions with no contrast

Key phrases to include

• allowable stress
• 0.45f'c
• modular ratio n = Es/Ec
• factored loads
• 1.2D + 1.6L
• φ = 0.90
• φMn ≥ Mu
• Whitney stress block
• 0.85f'c

Examiner tip

Draw a small sketch showing the beam cross-section, the triangular strain diagram, the actual stress distribution, and the equivalent rectangular block side-by-side. Even a rough sketch earns the conceptual mark.

Model answer

In USD, the actual non-linear (parabolic-rectangular) concrete compressive stress distribution above the neutral axis is replaced by an equivalent rectangular stress block — the Whitney block — for computational convenience without significant loss of accuracy.

The equivalent rectangular block has:

• Uniform stress intensity = 0.85f'c (acts over the full block depth)
• Block depth a = β₁c, where c is the distance from the extreme compression fibre to the neutral axis

The factor β₁ converts c to a:
β₁ = 0.85 for f'c ≤ 28 MPa
β₁ = 0.85 − 0.05[(f'c − 28)/7] for 28 < f'c ≤ 55 MPa (minimum β₁ = 0.65)

The total compression resultant on the section is:
C = 0.85f'c × a × b (for a rectangular beam of width b)

This resultant acts at a/2 from the compression face, giving an internal moment arm of (d − a/2) for computing the nominal moment capacity Mn.

Answer structure

• Sentence 1: State the purpose — replace actual stress distribution with a simpler equivalent [1 mark]
• Sentence 2: Define the two parameters — uniform stress 0.85f'c and block depth a = β₁c [1 mark]
• Sentence 3: Give β₁ values and the resultant force expression C = 0.85f'c·a·b [1 mark]

Scoring breakdown

Common mark deductions

• Stating the stress-block intensity as f'c (not 0.85f'c) — a critical error
• Confusing a (block depth) with c (neutral-axis depth) — they differ by the factor β₁
• Not mentioning what β₁ depends on (f'c value)

Key phrases to include

• 0.85f'c
• a = β₁c
• neutral-axis depth c
• β₁ = 0.85
• 28 MPa
• minimum 0.65
• C = 0.85f'c·a·b

Examiner tip

Always check tension-controlled vs compression-controlled — this determines which φ to use. The threshold is εt = 0.005 for tension-controlled (φ = 0.90) and εt = εty for compression-controlled (φ = 0.65 tied).

Model answer

Given: b = 300 mm, d = 500 mm, As = 1 847 mm², f'c = 28 MPa, fy = 415 MPa, Es = 200 000 MPa

Step 1 — Depth of stress block a:
From equilibrium (T = C):
As·fy = 0.85f'c·a·b
a = As·fy / (0.85f'c·b)
= (1 847 × 415) / (0.85 × 28 × 300)
= 766 505 / 7 140
= 107.4 mm

Step 2 — Neutral-axis depth c (with β₁):
Since f'c = 28 MPa: β₁ = 0.85
c = a / β₁ = 107.4 / 0.85 = 126.4 mm

Step 3 — Net tensile strain εt:
εt = 0.003 × (d − c) / c
= 0.003 × (500 − 126.4) / 126.4
= 0.003 × 2.956
= 0.00887

Check: εt = 0.00887 > 0.005 → Section is TENSION-CONTROLLED ✓
(φ = 0.90 applies for flexure)

∴ a = 107.4 mm; tension-controlled (εt = 0.00887 > 0.005)

Answer structure

• Step 1: Apply equilibrium T = C to get a = As·fy / (0.85f'c·b) [1 mark]
• Step 2: Find c = a/β₁ using β₁ = 0.85 for 28 MPa [0.5 mark]
• Step 3: Compute εt and compare with 0.005 to classify section [1.5 marks]

Scoring breakdown

Common mark deductions

• Using 0.85f'c × a × d instead of 0.85f'c × a × b in the compression resultant
• Forgetting to divide a by β₁ to get c before computing εt
• Using the wrong limit: εt > 0.004 instead of εt > 0.005 for tension-controlled classification

Key phrases to include

• T = C equilibrium
• a = As·fy/(0.85f'c·b)
• β₁ = 0.85
• c = a/β₁
• εt = 0.003(d−c)/c
• tension-controlled
• εt > 0.005

Examiner tip

When explaining concepts like ductility, always link the physical behaviour (how the beam fails) to the code provision (the φ value). Examiners reward mechanistic reasoning.

Model answer

In USD, a cross-section is classified by the net tensile strain εt in the extreme tension steel at the nominal strength condition (when the extreme concrete compression fibre reaches εcu = 0.003).

Tension-Controlled Section:

• εt ≥ 0.005 (steel is well into the yield plateau)
• The section fails by steel yielding — ductile failure mode
• φ = 0.90 (NSCP 2015) for flexure
• Preferred design condition for beams

Compression-Controlled Section:

• εt ≤ εty = fy/Es (steel has not yet yielded)
• e.g., for fy = 415 MPa: εty = 415/200 000 = 0.00208
• The section fails by concrete crushing — brittle failure mode
• φ = 0.65 (tied columns) or φ = 0.75 (spiral columns) per NSCP 2015

Transition Zone (εty < εt < 0.005):

• φ is interpolated linearly between the compression-controlled and tension-controlled values.

Beams must be designed to be tension-controlled to ensure a ductile, forewarned failure.

Answer structure

• Part A: Define tension-controlled with εt ≥ 0.005, ductile mode, φ = 0.90 [1 mark]
• Part B: Define compression-controlled with εt ≤ εty, brittle mode, φ = 0.65/0.75 [1 mark]
• Part C: Describe the transition zone and state why beams must be tension-controlled [1 mark]

Scoring breakdown

Common mark deductions

• Defining limits in terms of c/d ratios without relating back to εt — the code defines these in terms of strain
• Saying φ = 0.85 for tension-controlled — outdated ACI value
• Not computing εty for the given fy — the threshold is material-specific

Key phrases to include

• εt ≥ 0.005
• tension-controlled
• φ = 0.90
• εt ≤ εty = fy/Es
• compression-controlled
• φ = 0.65
• φ = 0.75
• ductile failure
• εcu = 0.003

Examiner tip

The transformed area concept is: steel is n times stiffer than concrete, so one mm² of steel 'acts like' n mm² of concrete in elastic analysis. This physical meaning helps you avoid the n vs 1/n confusion.

Model answer

(a) Allowable concrete compressive stress in flexure (WSD):
fc,allow = 0.45 f'c
(e.g., for f'c = 28 MPa: fc,allow = 0.45 × 28 = 12.6 MPa)

(b) Modular ratio:
n = Es / Ec = Es / (4700√f'c)
where Es = 200 000 MPa for deformed steel bars

(c) Transformed area of one φ25 mm bar (n = 9):
Steel bar area: As = π(25)²/4 = 490.9 mm²
Transformed area = n × As = 9 × 490.9 = 4 418 mm²
(The steel bar is replaced by an equivalent concrete area of 4 418 mm² for elastic analysis.)

Answer structure

• Part (a): State fc,allow = 0.45f'c with a numerical example [1 mark]
• Part (b): Write n = Es/Ec with the Ec formula [1 mark]
• Part (c): Compute As for φ25 mm then multiply by n to get transformed area [1 mark]

Scoring breakdown

Common mark deductions

• Using 0.40f'c or 0.33f'c for the allowable compressive stress — wrong fraction
• Computing the transformed area as As/n instead of nAs — the steel is stiffer, so its transformed area is larger
• Using diameter instead of radius in the area formula: A = π(25)²/4, not π(25)²

Key phrases to include

• fc,allow = 0.45f'c
• n = Es/Ec
• transformed area = nAs
• 4700√f'c
• 490.9 mm²

Examiner tip

For 5-mark numerical questions, the marks are distributed across the steps, not just the final answer. A wrong final answer due to arithmetic but with correct methodology can still earn 3–4 marks. Always show every step clearly.

Model answer

Given:
MD = 120 kN·m, ML = 90 kN·m, MLr = 30 kN·m

NSCP 2015 gravity load combinations for flexure (per Section 405.3):

Combination 1 (D + L only):
Mu₁ = 1.2MD + 1.6ML + 0.5MLr
= 1.2(120) + 1.6(90) + 0.5(30)
= 144 + 144 + 15
= 303 kN·m

Combination 2 (D + Lr dominant):
Mu₂ = 1.2MD + 1.6MLr + 1.0ML
= 1.2(120) + 1.6(30) + 1.0(90)
= 144 + 48 + 90
= 282 kN·m

Combination 3 (Dead only — check if 1.4D governs):
Mu₃ = 1.4MD = 1.4(120) = 168 kN·m

Comparison:
Mu₁ = 303 kN·m ← GOVERNS
Mu₂ = 282 kN·m
Mu₃ = 168 kN·m

∴ The governing factored moment is Mu = 303 kN·m

The beam must be designed to satisfy: φMn ≥ 303 kN·m (with φ = 0.90 for tension-controlled flexure).

Answer structure

• Step 1: Identify and list all applicable NSCP 2015 load combinations for gravity loads [1 mark]
• Step 2: Compute Mu for Combination 1 (1.2D + 1.6L + 0.5Lr) [1 mark]
• Step 3: Compute Mu for Combination 2 (1.2D + 1.6Lr + 1.0L) [1 mark]
• Step 4: Compute Mu for Combination 3 (1.4D) [1 mark]
• Step 5: Compare all three, identify the governing value, and state the design requirement φMn ≥ Mu [1 mark]

Scoring breakdown

Common mark deductions

• Evaluating only Combination 1 and assuming it always governs — Combination 2 or 3 can govern in different load ratios
• Missing the 0.5Lr companion action in Combination 1 or using coefficient 0.5L instead of 0.5Lr
• Not stating the final design check φMn ≥ Mu — the question asks you to determine Mu for design, not just as a number
• Adding all three load effects without grouping into separate combinations — loads are NOT all applied simultaneously at their full factored value

Key phrases to include

• 1.2D + 1.6L + 0.5Lr
• 1.2D + 1.6Lr + 1.0L
• 1.4D
• Mu = 303 kN·m
• φMn ≥ Mu
• NSCP 2015

Examiner tip

For 5-mark problems, write each step as a numbered sub-step with a label (Step 1 — β₁, Step 2 — depth a, etc.). This structure shows the examiner you know the procedure and earns partial credit even if you make one arithmetic slip.

Model answer

Given: b = 250 mm, d = 450 mm, f'c = 35 MPa, fy = 415 MPa, As = 1 500 mm²

--- PART (a): Stress block depth a and nominal moment Mn ---

Step 1 — β₁ for f'c = 35 MPa:
β₁ = 0.85 − 0.05 × (35 − 28)/7 = 0.85 − 0.05(1) = 0.80

Step 2 — Depth of stress block a (from T = C equilibrium):
As·fy = 0.85f'c·a·b
a = (1 500 × 415) / (0.85 × 35 × 250)
= 622 500 / 7 437.5
= 83.7 mm

Step 3 — Nominal moment capacity Mn:
Mn = As·fy·(d − a/2)
= 1 500 × 415 × (450 − 83.7/2)
= 622 500 × (450 − 41.85)
= 622 500 × 408.15
= 254 073 375 N·mm
= 254.1 kN·m

--- PART (b): φMn and tension-controlled check ---

Step 4 — Neutral-axis depth c:
c = a / β₁ = 83.7 / 0.80 = 104.6 mm

Step 5 — Net tensile strain εt:
εt = 0.003 × (d − c) / c
= 0.003 × (450 − 104.6) / 104.6
= 0.003 × 3.301
= 0.00990

εt = 0.00990 > 0.005 → Section is TENSION-CONTROLLED ✓
→ φ = 0.90

Step 6 — Design flexural strength:
φMn = 0.90 × 254.1 = 228.7 kN·m

∴ a = 83.7 mm, Mn = 254.1 kN·m, φMn = 228.7 kN·m (tension-controlled, φ = 0.90)

Answer structure

• Step 1: Compute β₁ for f'c = 35 MPa → β₁ = 0.80 [0.5 mark]
• Step 2: Apply T = C to get a = 83.7 mm [1 mark]
• Step 3: Compute Mn = As·fy·(d − a/2) = 254.1 kN·m [1 mark]
• Step 4: Find c = a/β₁ = 104.6 mm [0.5 mark]
• Step 5: Compute εt and verify tension-controlled (εt > 0.005) [1 mark]
• Step 6: Apply φ = 0.90 to get φMn = 228.7 kN·m [1 mark]

Scoring breakdown

Common mark deductions

• Not applying β₁ reduction for f'c = 35 MPa (using 0.85 instead of 0.80) — loses the first mark
• Computing Mn = 0.85f'c·a·b·(d − a/2) — wrong formula; should be As·fy·(d − a/2)
• Forgetting to convert N·mm to kN·m (dividing by 10⁶)
• Skipping the tension-controlled verification — this is explicitly required by the question

Key phrases to include

• β₁ = 0.80
• a = As·fy/(0.85f'c·b)
• Mn = As·fy·(d − a/2)
• c = a/β₁
• εt = 0.003(d−c)/c
• εt > 0.005
• tension-controlled
• φ = 0.90
• φMn

Examiner tip

The three-range piecewise formula for β₁ is frequently tested. Memorize all three conditions and always check whether f'c is below 28, between 28 and 55, or at/above 55 MPa before substituting.

Model answer

Complete β₁ formula (NSCP 2015):

β₁ = 0.85 for f'c ≤ 28 MPa
β₁ = 0.85 − 0.05[(f'c − 28)/7] for 28 < f'c ≤ 55 MPa
β₁ = 0.65 (minimum) for f'c ≥ 55 MPa

For f'c = 55 MPa:
β₁ = 0.85 − 0.05[(55 − 28)/7]
= 0.85 − 0.05[27/7]
= 0.85 − 0.05(3.857)
= 0.85 − 0.193
= 0.657 → but β₁ ≥ 0.65, so β₁ = 0.657 ≈ 0.66
(note: at exactly 55 MPa the formula gives ~0.657; floor applies at > 55)

For f'c = 60 MPa:
f'c > 55 MPa → β₁ = 0.65 (minimum governs)

Answer structure

• Part 1: Write all three ranges of the β₁ piecewise formula [1 mark]
• Part 2: Compute β₁ for 55 MPa and state β₁ = 0.65 for 60 MPa [1 mark]

Scoring breakdown

Common mark deductions

• Applying the decreasing formula for f'c > 55 MPa instead of using the floor of 0.65
• Using 56 MPa or 54 MPa as the cutoff for the minimum β₁ — the floor kicks in at 55 MPa

Key phrases to include

• β₁ = 0.85
• f'c ≤ 28 MPa
• 0.85 − 0.05[(f'c − 28)/7]
• minimum β₁ = 0.65
• f'c ≥ 55 MPa

Examiner tip

For long-answer questions asking to 'enumerate and explain', a numbered list followed by a brief summary table is the gold standard format. It shows thoroughness and earns the organization mark. Practice writing this table from memory in under 3 minutes.

Model answer

Per NSCP 2015 (adopting ACI 318 provisions), the strength-reduction factors φ for reinforced concrete are:

1. Tension-Controlled Flexure (εt ≥ 0.005)
   φ = 0.90
   Applied to beams and one-way slabs where steel yields well before concrete crushes. The high φ reflects predictable, ductile behavior.
2. Shear and Torsion
   φ = 0.75
   Shear failure is less ductile and more sudden than flexural failure; the lower φ provides an additional safety margin.
3. Compression-Controlled Members with Spiral Reinforcement (εt ≤ εty)
   φ = 0.75
   Spiral columns have better ductility and confinement than tied columns; the higher φ (vs tied) rewards this.
4. Compression-Controlled Members with Tied Reinforcement (εt ≤ εty)
   φ = 0.65
   Tied columns fail more suddenly under overload; the lowest φ for structural members is assigned here.
5. Bearing on Concrete
   φ = 0.65
   Applies to bearing areas (e.g., under column bases); brittle failure mode similar to tied columns.
6. Plain (Unreinforced) Concrete
   φ = 0.60
   No steel ductility reserve; concrete alone is brittle, so the most conservative φ is used.

Transition Zone Note: For sections with εty < εt < 0.005, φ is interpolated linearly between the compression-controlled value (0.65 or 0.75) and 0.90.

Summary Table:
Action | φ
Tension-controlled flexure | 0.90
Shear and torsion | 0.75
Spiral columns (CC) | 0.75
Tied columns (CC) | 0.65
Bearing | 0.65
Plain concrete | 0.60

Answer structure

• Item 1: φ = 0.90 — tension-controlled flexure with reason (ductile) [1 mark]
• Items 2–3: φ = 0.75 — shear/torsion and spiral columns with brief reasons [1 mark]
• Items 4–5: φ = 0.65 — tied columns and bearing with brief reasons [1 mark]
• Item 6: φ = 0.60 — plain concrete with reason (no ductility) [0.5 mark]
• Transition zone: linear interpolation between CC and TC values explained [0.5 mark]
• Summary table with all six values — demonstrates systematic organization [1 mark]

Scoring breakdown

Common mark deductions

• Listing only 3–4 of the six φ values — missing items lose marks proportionally
• Assigning φ = 0.75 to tied columns and φ = 0.65 to spiral — these are swapped
• Not providing any reasoning — the question says 'briefly explain', so a one-word explanation is expected for each
• Forgetting φ = 0.60 for plain concrete — often overlooked

Key phrases to include

• φ = 0.90
• tension-controlled
• φ = 0.75
• shear and torsion
• spiral
• φ = 0.65
• tied
• bearing
• φ = 0.60
• plain concrete
• transition zone
• linear interpolation

Do

• Write the key formula on the first line (e.g., Ec = 4700√f'c)
• Substitute the given value immediately on the same or next line
• State the numerical answer with correct unit (MPa, kN·m, mm)
• Use standard notation that matches NSCP 2015

Don't

• Do not write a lengthy introduction or restate the question
• Do not leave answers without units
• Do not use approximation without justification (e.g., n ≈ 8 without showing the exact value)

Do

• Label each step clearly (Step 1, Step 2) to guide the examiner to your marks
• State the formula before substituting — the formula line often earns one mark by itself
• Check your answer (e.g., β₁ ≥ 0.65) and state the check explicitly
• Use 'Given:' and 'Solution:' headers for numerical problems

Don't

• Do not skip the formula and write only the final numerical answer
• Do not write the two steps as one long continuous sentence — use line breaks
• Do not use non-SI units (kips, psi, ft) without conversion

Do

• Use GIVEN / REQUIRED / SOLUTION / ANSWER headings for numerical problems
• Show intermediate results (e.g., a = ... mm, then c = ... mm, then εt = ...)
• End with a concluding statement that directly answers the question
• Include a labeled sketch for stress-block or strain problems

Don't

• Do not combine three separate ideas into one run-on paragraph
• Do not omit the final answer statement — even if all steps are correct, no clear answer loses the last mark
• Do not assume β₁ = 0.85 without checking f'c

Do

• Write 'GIVEN' and 'REQUIRED' sections to show you understand the problem
• Number every step and state what you are computing at the start of each step
• Show the governing equation, substitute, and box the intermediate result before moving to the next step
• Apply and state all code-specified limits (e.g., β₁ ≥ 0.65, εt vs 0.005)
• Conclude with a boxed or underlined summary of all required answers
• State which NSCP 2015 load combination governs if the question involves load combinations

Don't

• Do not write a wall of equations without explanatory labels
• Do not skip the tension-controlled verification — it is almost always explicitly worth 1 mark in flexure problems
• Do not use rounded-off intermediate values that propagate large errors to the final answer
• Do not forget to multiply by φ at the end — the question asks for φMn, not Mn alone