RC Design — Chapter 1: Working Stress Design

00 — LEC 1 — FOUNDATION

Why Reinforced Concrete?

One physical truth explains everything: concrete is strong in compression, terrible in tension. Steel handles both. Put them together where each excels — that's RC in one sentence.

🧠 The ruler test: Bend a ruler — the top compresses, the bottom stretches. A concrete ruler would snap at the bottom tensile zone. Fix: embed steel exactly where snapping would happen.

⚡ SIM 1 — MATERIAL TESTER: COMPRESSION vs TENSION — WATCH CONCRETE CRACK & STEEL YIELD

LOAD INTENSITY: 0%

No load applied. Both materials at rest.

🪨

Concrete — Compression ✓

Handles crushing loads like stone. All concrete above the neutral axis in a beam is in compression.

f'c = 3,000–5,000 psi

⚠️

Concrete — Tension ✗

Brittle. Tensile strength ≈ 7.5√f'c ≈ f'c/10. Cracks instantly. Cannot be relied upon for tension.

ft ≈ 7.5√f'c psi

🔩

Steel — Takes the Tension

Rebar placed in the tension zone carries what concrete cannot. High strength, ductile, bonds perfectly.

fy = 40,000–60,000 psi

🔗

Bond — The Secret

Same thermal expansion coefficient. Both strain identically — the fundamental compatibility assumption.

α ≈ 6×10⁻⁶/°F for both

Brief History

ANCIENT ROME

Pozzolana Cement

Volcanic ash + lime. The Pantheon dome still stands 2,000 years later.

1796

Portland Cement (Aspdin, England)

Patents modern Portland cement. Same formulation used worldwide today.

1867 — KEY DATE

Reinforced Concrete (Monier, France)

Gardener Monier patents iron-reinforced concrete flower pots. Idea scales to beams and skyscrapers.

1920s

Working Stress Design (WSD) Codified

Stresses kept below allowable limits at service loads — the method in this chapter.

01 — LEC 2 — MIX DESIGN & UNIT WEIGHTS

The Recipe Defines the Strength

Concrete is a four-component composite. The ratio of water to cement — the w/c ratio — is the single most important variable controlling strength.

🌫️

Air (1–2%)

Voids from mixing. Too much reduces strength.

🟫

Cement

Adhesive binder. Portland cement. Hydrates with water.

💧

Water

Triggers hydration. w/c = most critical ratio.

🏖️

Aggregate

Fine sand + coarse. 70–75% of volume.

⚡ SIM 2 — w/c RATIO EXPLORER: CHANGE THE WATER/CEMENT RATIO — WATCH f'c CHANGE ON THE CURVE

w/c RATIO: 0.50 STRUCTURAL RC

USE	RATIO C:FA:CA	f'c APPROX
Mass concrete	1:2:4	2,500 psi
Structural RC (standard)	1:1.5:3	3,000–4,000 psi
High strength	ACI 211 design	>5,000 psi

🧱

Plain Concrete

No steel rebar.

wc = 145 pcf

🏗️

Reinforced Concrete

Includes rebar weight.

wRC = 150 pcf

02 — LEC 3 — LOADS & MODULAR RATIO

What Forces Act — and Who Carries Them?

WSD uses actual service loads. Safety is built through reduced allowable stresses. First, identify every force. Then, bridge the two-material problem with n.

⬇️

Dead Load (DL)

Permanent: self-weight, finishes. Calculated from unit weights.

🚶

Live Load (LL)

Occupancy: people, furniture. Specified by BNBC/ASCE 7.

40–100 psf typical

💨

Wind / Seismic

Lateral loads. Govern tall structures.

WSD PHILOSOPHY

$$f_{actual} \leq f_{allowable} = \frac{f_{ultimate}}{\text{Factor of Safety}}$$

fc,allow = 0.45 f'c | fs,allow = 20,000 psi (Gr.40) or 24,000 psi (Gr.60)

Modular Ratio n — Bridging Two Materials

🧠 Why n? Steel is ~8–10× stiffer than concrete. At equal strain, the stiffer material carries proportionally more stress. n tells you exactly how much more: 1 in² of steel "behaves like" n in² of concrete.

MODULAR RATIO

$$n = \frac{E_s}{E_c} = \frac{29{,}000{,}000}{57{,}000\sqrt{f'_c}} \quad \text{(round to nearest integer)}$$

⚡ SIM 3 — MODULAR RATIO VISUALIZER: SEE HOW n CHANGES WITH f'c AND WHAT IT MEANS PHYSICALLY

f'c (psi): 3000 psi

f'c (psi)	Ec (psi)	n	fc,allow	fs,allow (Gr.40)
2,500	2,850,000	10	1,125	20,000
3,000	3,122,000	9	1,350	20,000
4,000	3,605,000	8	1,800	20,000

03 — LEC 4–5 — BEAM GEOMETRY & CROSS-SECTION

Defining the Beam: d, b, and A_s

Three dimensions define every RC beam cross-section. The most critical is d, the effective depth — the distance from the compression face to the centroid of the steel.

EFFECTIVE DEPTH

$$d = h - \text{cover} - d_{stirrup} - \frac{d_{bar}}{2}$$

Typical cover = 1.5", stirrup ≈ 0.375" (#3), bar radius = d_b/2. Standard assumption: d ≈ h − 2.375"

STEEL RATIO

$$\rho = \frac{A_s}{b \cdot d}$$

ρ typically 0.005–0.018 for structural RC. ρ_min = 200/f_y (psi). ρ_max = 0.75ρ_b (ACI).

⚡ SIM 4 — CROSS-SECTION BUILDER: STEP THROUGH EACH LAYER TO SEE HOW d IS COMPUTED

Click the buttons above to step through how d is calculated layer by layer.

⚙️ CALCULATOR 1 — COMPUTE d AND ρ FROM SECTION DIMENSIONS

h (in) — total height

b (in) — width

Cover (in)

Stirrup dia (in)

Bar: dia,area

# bars

04 — LEC 6 — THREE STAGES OF LOADING

From Zero Load to Collapse

As moment increases, an RC beam passes through three distinct behavioral stages. WSD designs in Stage II — after cracking but well before yielding.

①

Stage I — Uncracked

Concrete handles tension. Full section acts. Neutral axis near center. Linear elastic.

M < Mcr

②

Stage II — Cracked Elastic (WSD)

Concrete cracks in tension. Steel carries all tension. Neutral axis shifts up. Still linear.

WSD operates HERE

③

Stage III — Inelastic

Steel yields. Large deflections. Nonlinear. ACI strength design (USD) targets this stage.

Near failure

CRACKING MOMENT CRITERION

$$f_r = 7.5\sqrt{f'_c} \quad \text{(psi)} \quad\Rightarrow\quad M_{cr} = f_r \cdot \frac{I_g}{y_{bottom}}$$

If computed f_ct > f_r, the section is cracked → use Stage II (k & j) analysis

⚡ SIM 5 — LOADING STAGES: INCREASE LOAD — WATCH CRACKS FORM AND STRESS ZONES EVOLVE

LOAD LEVEL: 0.00 k/ft STAGE I — UNCRACKED ELASTIC

Stage I Check

fct = M·y_bot/I_tr
If fct < fr → Stage I OK

Stage II (WSD)

Use cracked section:
k, j method below

Transition

fct > fr → cracked
→ Use Stage II

05 — LEC 7 — THE TRANSFORMED SECTION METHOD

Two Materials, One Section

The flexure formula σ = My/I assumes homogeneous material. RC has two materials. The Transformed Section converts steel into equivalent concrete so we can use the standard formula.

🧠 The key: Since strain is equal at the steel location (compatibility), and steel is n times stiffer, steel carries n times the stress. Replace each A_s with nA_s of concrete — now we have one material.

UNCRACKED TRANSFORMED SECTION — NEUTRAL AXIS FROM TOP

$$\bar{y} = \frac{b h^2/2 + (n-1)A_s \cdot d}{b h + (n-1)A_s}$$

MOMENT OF INERTIA — TRANSFORMED UNCRACKED SECTION

$$I_{tr} = \frac{b h^3}{12} + b h\!\left(\bar{y} - \frac{h}{2}\right)^{\!2} + (n-1)A_s(d - \bar{y})^2$$

⚡ SIM 6 — STRESS PROBE: HOVER/CLICK ANYWHERE ON THE CROSS-SECTION TO READ THE STRESS AT THAT DEPTH

M (k·in): 200

Hover over the cross-section to probe the stress at any depth.

NEUTRAL AXIS — CRACKED TRANSFORMED SECTION (from top)

$$\frac{b(kd)^2}{2} = n A_s (d - kd) \quad\Longrightarrow\quad k = \sqrt{(\rho n)^2 + 2\rho n} - \rho n$$

06 — LEC 8 — WSD ANALYSIS: THE k & j METHOD

Cracked Section Analysis — Two Numbers Do Everything

Two dimensionless numbers — k (neutral axis depth ratio) and j (moment arm ratio) — completely define the cracked RC stress state. Here's the derivation and two simulations to make them intuitive.

🧠 Start here: After cracking, only two internal forces exist: C (compression in concrete above N.A.) and T (tension in steel). They must balance: C = T. Their separation (jd) times either force = resisting moment.

The compression resultant C acts at kd/3 from the top (centroid of triangle). The tension resultant T acts at the steel centroid. Their separation = jd = d − kd/3 = d(1 − k/3).

M = C · jd = (½ f_c · k · b · d) · jd = ½ f_c · k · j · b · d²

M = T · jd = A_s · f_s · jd = A_s · f_s · j · d

Compatibility — equal strain at steel

Since concrete and steel are bonded, they strain together at the steel level.

$$\frac{\varepsilon_s}{\varepsilon_c} = \frac{d - kd}{kd} \quad\Rightarrow\quad \frac{f_s/E_s}{f_c/E_c} = \frac{1-k}{k} \quad\Rightarrow\quad f_s = n \cdot f_c \cdot \frac{1-k}{k}$$

Equilibrium — C = T

Internal compression equals internal tension.

$$C = T \quad\Rightarrow\quad \frac{1}{2}f_c \cdot b \cdot kd = A_s \cdot f_s$$

Substitute and simplify

Replace f_s from compatibility into C = T:

$$\frac{1}{2}f_c \cdot b \cdot kd = A_s \cdot n f_c \cdot \frac{1-k}{k}$$

$$\frac{bk^2}{2} = \rho n (1-k) \quad\Rightarrow\quad k^2 + 2\rho n \cdot k - 2\rho n = 0$$

Solve the quadratic

$$k = \sqrt{(\rho n)^2 + 2\rho n} - \rho n$$

$$j = 1 - \frac{k}{3}$$

COMPLETE WSD ANALYSIS FORMULAS

$$k = \sqrt{(\rho n)^2 + 2\rho n} - \rho n \qquad j = 1 - \frac{k}{3}$$ $$f_c = \frac{2M}{k \cdot j \cdot b \cdot d^2} \leq 0.45 f'_c \qquad f_s = \frac{M}{A_s \cdot j \cdot d} \leq f_{s,allow}$$

Where: ρ = A_s/(bd), n = E_s/E_c (integer), M in lb·in, b & d in inches.

⚙️ CALCULATOR 2 — WSD CRACKED SECTION ANALYSIS

b (in)

d (in)

As (in²)

f'c (psi)

fy (psi)

M (k·ft)

WP-1WSD Analysis — Find fs and fc in cracked RC beam

GIVEN

b=12", d=21.5"	As=3.0 in² (3-#9)
f'c=3,000 psi, n=9	M=200 k·in, fy=40ksi

ρ and ρn

ρ = 3.0/(12×21.5) = 0.01163 | ρn = 0.1047

k and j

k = √(0.1047²+2×0.1047)−0.1047 = 0.365 | j = 1−0.365/3 = 0.878

Stresses

fs = 200,000/(3.0×0.878×21.5) = 3,536 psi ✓ < 20,000 psi

fc = 2×200,000/(0.365×0.878×12×21.5²) = 254 psi ✓ < 1,350 psi

✓ Section OK. Both stresses well below allowables.

⚡ SIM 7 — NEUTRAL AXIS VISUALIZER: CHANGE ρ AND n — WATCH kd MOVE AND SEE HOW STRESS PROFILE CHANGES

ρ: 0.010

n: 9

Key insight: More steel (↑ρ) or stiffer steel (↑n) pulls the neutral axis deeper. Deeper N.A. = larger compression zone = smaller jd = more force needed for same moment.

⚡ SIM 8 — C = T FORCE BALANCE: WATCH COMPRESSION AND TENSION FORCES GROW TOGETHER IN EQUILIBRIUM AS MOMENT INCREASES

M (k·in): 0 Apply moment to see C and T forces

07 — LEC 9+ — WSD DESIGN PROCEDURE

Design from Scratch

Analysis asks "What are the stresses?" Design asks "What beam do I need?" We work backward from required moment M to find b, d, and A_s.

🧠 Balanced section: The most efficient beam is one where concrete and steel reach their allowable stresses at exactly the same time — the balanced design target.

BALANCED DESIGN CONSTANTS

$$k_b = \frac{n \cdot f_{ca}}{n \cdot f_{ca} + f_{sa}} \qquad j_b = 1 - \frac{k_b}{3} \qquad R = \frac{1}{2} f_{ca} \cdot k_b \cdot j_b$$

R is the "resistance coefficient" in psi. Then: $bd^2 = M/R$ and $A_s = M/(f_{sa} \cdot j_b \cdot d)$

Design Procedure — Step by Step

Find design moment M from structural analysis

M = wL²/8 (simply supported UDL)

Compute n, fc,allow, fs,allow from material properties

n = 29,000,000 / (57,000√f'c) | fc = 0.45f'c | fs = 20,000 (Gr.40)

Balanced kb, jb, R

kb = n·fc/(n·fc + fs) | jb = 1−kb/3 | R = ½·fc·kb·jb

Required bd²

bd² = M/R → choose b (≥12", even) → solve for d = √(bd²/b)

Round up d → compute h

h = d + cover + dstirrup + dbar/2 → round to nearest ½"

Required As

As = M/(fs·j·d) → select bars from table

Check bar spacing, cover, actual stresses

Verify bars fit in width b, spacing ≥ 1", recompute fs and fc with final d.

⚙️ CALCULATOR 3 — WSD BEAM DESIGN (FIND b, d, As FROM M)

M (k·ft)

f'c (psi)

fy

b (in) ≥12, even

WP-2WSD Design — Design beam for M=80 k·ft (f'c=3ksi, Gr.40, b=12")

GIVEN

M=80 k·ft=960 k·in	f'c=3,000 psi, n=9
fc,allow=1,350 psi	fs,allow=20,000 psi, b=12"

Balanced kb and jb

kb = (9×1350)/(9×1350+20000) = 0.378 | jb = 1−0.378/3 = 0.874

R factor

R = ½×1350×0.378×0.874 = 223 psi

Required bd²

bd² = 960,000/223 = 4,305 in³ → b=12" → d = √(358.8) = 18.94" → use d = 19"

h and As

h = 19+1.5+0.375+0.5 = 21.4" → use h = 22" → actual d = 22−2.375 = 19.625"

As = 960,000/(20,000×0.874×19.625) = 2.80 in² → Use 4-#8 (As=3.16 in²)

✓ Result: b=12", h=22", 4-#8 bars (As=3.16 in²)

📋 QUICK REFERENCE

Formula Sheet — All RC WSD Formulas

Every formula from this chapter, organized by topic.

Materials

Modular Ratio

$n = E_s/E_c = 29{,}000{,}000\,/\,(57{,}000\sqrt{f'_c})$

Round to nearest integer. f'c in psi.

Materials

Allowable Stresses

$f_{ca} = 0.45\,f'_c \quad f_{sa} = 20{,}000$ (Gr.40) or $24{,}000$ psi (Gr.60)

Geometry

Effective Depth

$d = h - \text{cover} - d_s - d_b/2$

Typical: d ≈ h − 2.375"

Geometry

Steel Ratio

$\rho = A_s\,/\,(b\cdot d)$

Typical range: 0.005 – 0.018

Cracking

Modulus of Rupture

$f_r = 7.5\sqrt{f'_c}$ (psi)

If f_ct > f_r → section is cracked → use Stage II

Stage II — Cracked

Neutral Axis Ratio k

$k = \sqrt{(\rho n)^2 + 2\rho n} - \rho n$

k = kd/d where kd = neutral axis depth from top

Stage II — Cracked

Moment Arm Ratio j

$j = 1 - k/3$

jd = internal moment arm (C to T)

Analysis

Concrete Stress

$f_c = 2M\,/\,(k\,j\,b\,d^2)$

Check: f_c ≤ 0.45f'_c

Analysis

Steel Stress

$f_s = M\,/\,(A_s\,j\,d)$

Check: f_s ≤ f_sa

Design

Balanced k_b

$k_b = n\,f_{ca}\,/\,(n\,f_{ca} + f_{sa})$

Design

Resistance R

$R = \tfrac{1}{2}\,f_{ca}\,k_b\,j_b$ (psi)

j_b = 1 − k_b/3

Design

Required bd²

$b\,d^2 = M\,/\,R$

Choose b → solve for d → round up → find h

Design

Required A_s

$A_s = M\,/\,(f_{sa}\,j_b\,d)$

Use actual d after rounding h

Bar Sizes (US)

Common Rebars

#3: 0.375" / 0.11 in²   #4: 0.5" / 0.20
#5: 0.625" / 0.31   #6: 0.75" / 0.44
#7: 0.875" / 0.60   #8: 1.0" / 0.79
#9: 1.128" / 1.00   #10: 1.27" / 1.27

Reinforced Concrete:Math First. Build Second.

Why Reinforced Concrete?

Concrete — Compression ✓

Concrete — Tension ✗

Steel — Takes the Tension

Bond — The Secret

Brief History

Pozzolana Cement

Portland Cement (Aspdin, England)

Reinforced Concrete (Monier, France)

Working Stress Design (WSD) Codified

The Recipe Defines the Strength

Air (1–2%)

Cement

Water

Aggregate

Plain Concrete

Reinforced Concrete

What Forces Act — and Who Carries Them?

Dead Load (DL)

Live Load (LL)

Wind / Seismic

Modular Ratio n — Bridging Two Materials

Defining the Beam: d, b, and As

From Zero Load to Collapse

Stage I — Uncracked

Stage II — Cracked Elastic (WSD)

Stage III — Inelastic

Stage I Check

Stage II (WSD)

Transition

Two Materials, One Section

Cracked Section Analysis — Two Numbers Do Everything

Compatibility — equal strain at steel

Equilibrium — C = T

Substitute and simplify

Solve the quadratic

GIVEN

ρ and ρn

k and j

Stresses

Design from Scratch

Design Procedure — Step by Step

Find design moment M from structural analysis

Compute n, fc,allow, fs,allow from material properties

Balanced kb, jb, R

Required bd²

Round up d → compute h

Required As

Check bar spacing, cover, actual stresses

GIVEN

Balanced kb and jb

R factor

Required bd²

h and As

Formula Sheet — All RC WSD Formulas

Reinforced Concrete:
Math First. Build Second.

Defining the Beam: d, b, and A_s