Reinforced & Prestressed Concrete

9% of exam

ACI flexural and shear strength of beams and one-way slabs, reinforcement ratio limits and development length, short-column axial capacity, and prestressed-concrete service stresses and prestress losses.

6 concepts

Exam areas

A. Horizontal members (beams, slabs)

ACI Flexural Design of RC Beams

The Whitney stress block, nominal moment $M_n$ of a singly-reinforced beam, the tension-controlled $\phi=0.90$ , and the $\rho_{min}$ / $\rho_{max}$ limits that keep failures ductile.

Reinforced-concrete flexure is the first design check you reach for on the PE Civil structural exam, and it is the one place where the NCEES PE Civil Reference Handbook actually carries the method: §4.3.2.2 (Beams—Flexure Strength) gives the equivalent rectangular stress block and the internal-force geometry for a singly-reinforced section. Everything beyond that single section — shear, columns, development, slabs, prestress — lives in ACI 318-14, but the singly-reinforced beam is handbook-grounded, so you can cite §4.3.2.2 for the stress block itself. Points are lost here not on the algebra but on three things: forgetting the strength-reduction factor

\phi

, assuming the section is tension-controlled without checking the strain, and slipping between

\text{kip}\cdot\text{ft}

and

\text{kip}\cdot\text{in}

when you form

M = A_s f_y (d - a/2)

The equivalent rectangular (Whitney) stress block

Concrete's real compressive stress distribution is a curved, parabola-like shape, but ACI replaces it with a uniform block of stress

0.85 f'_c

acting over a depth

a

— chosen so the resultant compression force and its line of action match the real distribution. The block depth

a

relates to the true neutral-axis depth

c

by the factor

\beta_1

, which is

0.85

for

f'_c \le 4000\,\text{psi}

and drops

0.05

for every

1000\,\text{psi}

above that (floor of

0.65

). Setting compression equal to tension,

0.85 f'_c\, b\, a = A_s f_y

, gives the block depth directly. Keep

f'_c

in psi here — it is a pressure, not the

\sqrt{f'_c}

form, so there is no unit trap, but mind that

b

is in inches.

a = \dfrac{A_s f_y}{0.85\, f'_c\, b}, \qquad c = \dfrac{a}{\beta_1}

Nominal and design moment

With the compression resultant

C = 0.85 f'_c\, b\, a

centered at

a/2

below the top and the tension resultant

T = A_s f_y

at the steel centroid, the internal couple has a lever arm of

d - a/2

. The nominal moment is the force times that arm. The design moment is

\phi M_n

, and for a tension-controlled section

\phi = 0.90

. Always carry

M_n

\text{in}\cdot\text{lb}

(or

\text{kip}\cdot\text{in}

) through the arithmetic and convert to

\text{kip}\cdot\text{ft}

only at the end — dividing by

12

at the wrong moment is the most common numeric slip on this problem.

M_n = A_s f_y\!\left(d - \dfrac{a}{2}\right), \qquad \phi M_n \ge M_u

Strain check: is the section tension-controlled?

The

\phi = 0.90

is not automatic — it applies only when the extreme tension steel yields well before the concrete crushes, i.e. the net tensile strain

\varepsilon_t \ge 0.005

(tension-controlled). From similar triangles on the linear strain diagram with

\varepsilon_{cu} = 0.003

at the top,

\varepsilon_t = 0.003(d_t - c)/c

. If

\varepsilon_t

falls between

\varepsilon_y

(

\approx 0.00207

for Grade-60 steel) and

0.005

the section is in the transition zone and

\phi

is interpolated between

0.65

(or

0.75

for spirals) and

0.90

. A properly proportioned beam is designed to land at or above

0.005

so you keep the full

\phi = 0.90

and a ductile, well-warned failure.

\varepsilon_t = 0.003\,\dfrac{d_t - c}{c} \;\ge\; 0.005 \;\Rightarrow\; \phi = 0.90

Minimum reinforcement

Too little steel is dangerous: if the cracking moment exceeds the flexural strength of the reinforced section, the beam fails suddenly the instant it cracks, with no warning. ACI 318-14 §9.6.1 guards against this with a minimum steel area — the larger of

3\sqrt{f'_c}/f_y

and

200/f_y

times

b_w d

(with

f'_c

and

f_y

in psi, and

\sqrt{f'_c}

requiring

f'_c

in psi). The

200/f_y

term governs up to

f'_c \approx 4444\,\text{psi}

; above that the

3\sqrt{f'_c}/f_y

term takes over.

A_{s,min} = \max\!\left(\dfrac{3\sqrt{f'_c}}{f_y},\ \dfrac{200}{f_y}\right) b_w d

Maximum reinforcement via the strain limit

ACI 318-14 does not publish an explicit

\rho_{max}

; instead it caps the steel indirectly by requiring

\varepsilon_t \ge 0.004

for flexural members (and, practically, designing for

\varepsilon_t \ge 0.005

to keep

\phi = 0.90

). That strain limit fixes the largest neutral-axis depth,

c \le \tfrac{0.003}{0.003+\varepsilon_{t,limit}}\,d_t

, and back-substituting through

a = \beta_1 c

and

A_s f_y = 0.85 f'_c b a

gives the corresponding maximum steel ratio. The takeaway: an over-reinforced beam that would crush the concrete before the steel yields is simply not permitted — you add compression steel or a deeper section instead.

\rho_{max} = 0.85\,\beta_1\,\dfrac{f'_c}{f_y}\cdot\dfrac{0.003}{0.003 + \varepsilon_{t,limit}}

Exam strategy

Anchor the analysis problem in three moves: compute

a = A_s f_y/(0.85 f'_c b)

, then

M_n = A_s f_y (d - a/2)

, then

\phi M_n

after confirming

\varepsilon_t \ge 0.005

. For a design problem (given

M_u

, find

A_s

) set

\phi M_n = M_u

and solve the resulting quadratic in

A_s

— or iterate by assuming a lever arm of about

0.9d

, finding a trial

A_s

, recomputing

a

, and repeating once. Always keep

f'_c

in psi inside

\sqrt{f'_c}

for

A_{s,min}

, keep

M_n

\text{kip}\cdot\text{in}

until the last step, and check

A_{s,min} \le A_s

and

\varepsilon_t \ge 0.005

before you trust the answer. Cite handbook §4.3.2.2 for the stress block; cite ACI 318-14 for the

\phi

factors and steel limits.

Key equations

Stress-block depth

a = \dfrac{A_s f_y}{0.85\, f'_c\, b}

Depth of the equivalent rectangular compression block. $A_s$ (in²), $f_y$ , $f'_c$ (psi), $b$ (in); $a$ in inches. From force balance $C=T$ .

Block factor β₁

\beta_1 = \begin{cases} 0.85 & f'_c \le 4000 \\ 0.85 - 0.05\dfrac{f'_c-4000}{1000} & 4000 < f'_c \le 8000 \\ 0.65 & f'_c > 8000 \end{cases}

Ratio of block depth $a$ to neutral-axis depth $c$ (ACI 318-14 Table 22.2.2.4.3). $f'_c$ in psi.

Nominal moment

M_n = A_s f_y\!\left(d - \dfrac{a}{2}\right)

Internal couple of the section. Lever arm $d-a/2$ in inches → $M_n$ in $\text{lb}\cdot\text{in}$ ; divide by 12,000 for $\text{kip}\cdot\text{ft}$ .

Design strength requirement

\phi M_n \ge M_u, \qquad \phi = 0.90\ \text{(tension-controlled)}

Strength condition. $\phi=0.90$ when $\varepsilon_t \ge 0.005$ ; transition between 0.65/0.75 and 0.90 otherwise.

Net tensile strain

\varepsilon_t = 0.003\,\dfrac{d_t - c}{c}

Strain in extreme tension steel from the linear strain diagram with $\varepsilon_{cu}=0.003$ . $\ge 0.005$ → tension-controlled.

Minimum steel

A_{s,min} = \max\!\left(\dfrac{3\sqrt{f'_c}}{f_y},\ \dfrac{200}{f_y}\right) b_w d

ACI 318-14 §9.6.1; prevents brittle failure at cracking. $f'_c$ , $f_y$ in psi; $\sqrt{f'_c}$ needs psi.

Reinforcement ratio

\rho = \dfrac{A_s}{b\,d}

Tension steel ratio; compare against $\rho_{min}=A_{s,min}/(b_w d)$ and the strain-limited $\rho_{max}$ .

Maximum steel ratio (strain limit)

\rho_{max} = 0.85\,\beta_1\dfrac{f'_c}{f_y}\cdot\dfrac{0.003}{0.003+\varepsilon_{t,limit}}

From $c$ at the strain limit ( $\varepsilon_{t,limit}=0.004$ minimum, 0.005 to keep $\phi=0.90$ ). Caps the steel indirectly.

Resistance-factor (design) coefficient form

M_u = \phi A_s f_y\!\left(d-\dfrac{a}{2}\right)

Design equation solved for $A_s$ given $M_u$ ; quadratic in $A_s$ since $a$ depends on $A_s$ .

Concrete modulus

E_c = 57{,}000\sqrt{f'_c}\ \ (\text{psi})

Normalweight-concrete modulus (ACI 318-14 §19.2.2). $f'_c$ in psi; needed for service/deflection checks.

Worked examples

Design moment of a singly-reinforced beam

Problem. A rectangular beam has $b = 14\,\text{in}$ , effective depth $d = 22\,\text{in}$ , and $A_s = 4.00\,\text{in}^2$ (4 #9 bars). Materials are $f'_c = 4000\,\text{psi}$ and $f_y = 60{,}000\,\text{psi}$ . Find the design moment $\phi M_n$ .

Solution. Block depth:

a = \dfrac{A_s f_y}{0.85 f'_c b} = \dfrac{4.00(60{,}000)}{0.85(4000)(14)} = \dfrac{240{,}000}{47{,}600} = 5.04\,\text{in}

. Strain check:

\beta_1 = 0.85

(since

f'_c = 4000

), so

c = a/\beta_1 = 5.04/0.85 = 5.93\,\text{in}

. Then

\varepsilon_t = 0.003\dfrac{22-5.93}{5.93} = 0.0081 \ge 0.005

→ tension-controlled,

\phi = 0.90

. Nominal moment:

M_n = A_s f_y(d - a/2) = 4.00(60)(22 - 2.52) = 240(19.48) = 4{,}675\,\text{kip}\cdot\text{in} = 389.6\,\text{kip}\cdot\text{ft}

. Design moment:

\phi M_n = 0.90(389.6) = 350.6\,\text{kip}\cdot\text{ft}

. **Answer:

\phi M_n \approx 351\,\text{kip}\cdot\text{ft}

.** Sanity:

\rho = 4.00/(14\cdot22) = 0.0130

, between

\rho_{min}=0.0033

and the tension-controlled limit, and

a/d = 0.23

is a typical, well-proportioned beam.

a = \dfrac{4.00(60{,}000)}{0.85(4000)(14)} = 5.04\ \text{in}, \quad \phi M_n = 0.90(4.00)(60)(22-2.52)/12 = 351\ \text{kip}\cdot\text{ft}

Design: required steel for a given moment

Problem. Select $A_s$ for a rectangular beam carrying a factored moment $M_u = 300\,\text{kip}\cdot\text{ft}$ , with $b = 12\,\text{in}$ , $d = 20\,\text{in}$ , $f'_c = 4000\,\text{psi}$ , $f_y = 60{,}000\,\text{psi}$ .

Solution. Required nominal moment:

M_n = M_u/\phi = 300/0.90 = 333.3\,\text{kip}\cdot\text{ft} = 4000\,\text{kip}\cdot\text{in}

(assume tension-controlled, verify later). Set

M_n = A_s f_y(d - a/2)

with

a = A_s f_y/(0.85 f'_c b)

. Substituting

a

gives a quadratic in

A_s

M_n = A_s f_y d - \dfrac{f_y^2}{2(0.85 f'_c b)}A_s^2

, i.e.

4000 = 60(20)A_s - \dfrac{60^2}{2(0.85)(4)(12)}A_s^2 = 1200 A_s - 44.12 A_s^2

. Solving:

A_s = 3.89\,\text{in}^2

. Then

a = 3.89(60)/(0.85\cdot4\cdot12) = 5.72\,\text{in}

c = 6.73\,\text{in}

\varepsilon_t = 0.003(20-6.73)/6.73 = 0.0059 \ge 0.005

→

\phi = 0.90

confirmed. **Answer:

A_s = 3.89\,\text{in}^2

; provide 4 #9 (

4.00\,\text{in}^2

).** Sanity:

A_{s,min} = (200/60{,}000)(12)(20) = 0.80\,\text{in}^2 < 3.89

, OK; back-check the provided

A_s = 4.00\,\text{in}^2

a = 4.00(60)/(0.85\cdot4\cdot12) = 5.88\,\text{in}

, so

\phi M_n = 0.90(4.00)(60)(20-5.88/2)/12 = 307\,\text{kip}\cdot\text{ft} \ge 300

, satisfied.

1200\,A_s - 44.12\,A_s^2 = 4000 \;\Rightarrow\; A_s = 3.89\ \text{in}^2

Common pitfalls

•Dropping the strength-reduction factor: the deliverable is $\phi M_n$ , not $M_n$ . For a tension-controlled beam $\phi = 0.90$ ; forgetting it overstates capacity by 11%.
•Assuming $\phi = 0.90$ without checking $\varepsilon_t \ge 0.005$ . A heavily-reinforced section can fall in the transition zone where $\phi < 0.90$ , or be illegal if $\varepsilon_t < 0.004$ .
•Unit mixing in $M_n = A_s f_y(d - a/2)$ : $f_y$ in ksi and $d$ in inches gives $\text{kip}\cdot\text{in}$ — divide by 12 for $\text{kip}\cdot\text{ft}$ , not 1000. Converting at the wrong step is the classic numeric error.
•Putting $f'_c$ in ksi inside $\sqrt{f'_c}$ for $A_{s,min}$ (or $E_c$ ). The $3\sqrt{f'_c}$ and $57{,}000\sqrt{f'_c}$ forms require $f'_c$ in psi.
•Using $\beta_1 = 0.85$ for high-strength concrete. Above $4000\,\text{psi}$ , $\beta_1$ decreases $0.05$ per $1000\,\text{psi}$ ; this changes $c$ and the strain check.
•Citing ACI 318-19 or a different edition. NCEES supplies ACI 318-14 — match it; the $\phi$ -factor and $A_{s,min}$ provisions are referenced by their 318-14 section numbers.
•Forgetting $A_{s,min}$ entirely on a lightly-loaded or shallow member — the minimum can govern and its omission yields a brittle, code-illegal section.

References

NCEES PE Civil Reference Handbook — §4.3.2.2 Beams—Flexure Strength (equivalent rectangular stress block, β₁)
ACI 318-14, Building Code Requirements for Structural Concrete — §22.2 (flexural strength), §21.2 (φ factors), §9.6.1 (minimum reinforcement)
ACI 318-14 Table 22.2.2.4.3 — β₁ as a function of f′c

ACI Shear Design & Development Length

Concrete shear $V_c = 2\lambda\sqrt{f'_c}\,b_w d$ , stirrup design $V_s = A_v f_{yt} d/s$ , the $\phi V_n \ge V_u$ check with $\phi = 0.75$ , and tension development length $\ell_d$ .