PE Exam (Civil) Reinforced Concrete: Flexure, Shear, Axial, Columns (ACI 318)
Last updated: May 2, 2026
Reinforced Concrete: Flexure, Shear, Axial, Columns (ACI 318) questions are one of the highest-leverage areas to study for the PE Exam (Civil). This guide breaks down the rule, the elements you need to recognize, the named traps that catch most students, and a memory aid that scales to test day. Read it once, then practice the same sub-topic adaptively in the app.
The rule
Reinforced concrete design on the PE Civil exam follows ACI 318 strength (LRFD) design: factor the loads to get $U$ (per ACI 318 §5.3 / ASCE 7 §2.3), compute the nominal capacity $M_n$, $V_n$, or $P_n$ from equilibrium and the equivalent rectangular stress block ($a = \beta_1 c$, $\beta_1 = 0.85$ for $f'_c \le 4 \text{ ksi}$), then reduce by the strength reduction factor $\phi$ to require $\phi R_n \ge U$. The four critical $\phi$ values are $\phi = 0.90$ (tension-controlled flexure), $\phi = 0.75$ (shear and torsion), $\phi = 0.65$ (compression-controlled tied member), and $\phi = 0.75$ (compression-controlled spirally reinforced member).
Elements breakdown
Flexure of a singly-reinforced rectangular beam
Tension-controlled rectangular section with bottom steel only; ACI 318 §22.2.
- Verify $A_s$ between $A_{s,min}$ and $A_{s,max}$
- Compute stress block depth: $a = \dfrac{A_s f_y}{0.85 f'_c b}$
- Find neutral axis $c = a/\beta_1$
- Check tensile strain $\varepsilon_t = 0.003(d-c)/c$
- If $\varepsilon_t \ge 0.005$, tension-controlled, use $\phi = 0.90$
- Compute $M_n = A_s f_y (d - a/2)$
- Design strength is $\phi M_n$, must satisfy $\phi M_n \ge M_u$
One-way shear in beams
Concrete plus stirrups resist factored shear $V_u$ at the critical section a distance $d$ from the support face; ACI 318 §22.5.
- Take $V_u$ at distance $d$ from support face
- Concrete contribution: $V_c = 2\lambda\sqrt{f'_c}\, b_w d$ (simplified, psi)
- Required stirrup capacity: $V_s = V_u/\phi - V_c$
- Stirrup spacing: $s = \dfrac{A_v f_{yt} d}{V_s}$
- $A_v$ counts ALL legs crossing the crack (typically 2 for U-stirrup)
- Max spacing $s_{max} = \min(d/2,\ 24\text{ in})$ if $V_s \le 4\sqrt{f'_c}\, b_w d$
- Halve $s_{max}$ to $d/4$ when $V_s$ exceeds that threshold
- If $V_s > 8\sqrt{f'_c}\, b_w d$, section size is inadequate
- Use $\phi = 0.75$ throughout shear design
Short axial column capacity
Concentrically loaded short column; ACI 318 §22.4 limits $P_n$ to account for accidental eccentricity.
- $P_o = 0.85 f'_c (A_g - A_{st}) + f_y A_{st}$
- Tied column max: $P_{n,max} = 0.80\, P_o$
- Spiral column max: $P_{n,max} = 0.85\, P_o$
- Apply $\phi = 0.65$ for tied, $\phi = 0.75$ for spiral
- Design strength: $\phi P_{n,max}$ must $\ge P_u$
- Verify $\rho_g = A_{st}/A_g$ between $0.01$ and $0.08$
- Tie size: #3 for longitudinals up to #10, #4 above
- Tie spacing: $\min(16 d_b,\ 48 d_t,\ \text{least column dim})$
Combined axial + flexure (interaction)
Beam-columns require an interaction diagram or P-M curve check (ACI 318 §22.4).
- Plot factored $(P_u, M_u)$ point
- Compare with $\phi P_n$–$\phi M_n$ envelope
- Balanced point separates tension- and compression-controlled
- $\phi$ varies with $\varepsilon_t$ between $0.65$ and $0.90$
- For slender columns, magnify $M_u$ via moment magnifier (§6.6.4)
Common patterns and traps
The Missing-$\phi$ Distractor
PE writers nearly always include a choice equal to the unreduced nominal capacity ($M_n$, $V_n$, or $P_n$) so candidates who forget the strength reduction factor land on it. The values are typically 11% larger than the correct answer for flexure ($1/0.9$), 33% larger for shear ($1/0.75$), or 54% larger for tied compression ($1/0.65$).
A choice that is exactly $1/\phi$ times the correct answer, expressed in the same units.
The Single-Leg Stirrup Trap
In shear problems, $A_v$ is the total area of ALL stirrup legs crossing the diagonal crack, almost always two legs for a U-stirrup. Candidates who plug in the area of one #3 bar ($0.11 \text{ in}^2$) instead of two legs ($0.22 \text{ in}^2$) compute a stirrup spacing exactly half the correct value.
A spacing answer that is one-half of the correct spacing — e.g., $3 \text{ in}$ when the answer is $6 \text{ in}$.
The Forgotten 0.80 Cap
ACI 318 §22.4.2.1 caps tied column nominal axial capacity at $P_{n,max} = 0.80\,P_o$ to account for unintended eccentricity; spirals get $0.85\,P_o$. Skipping this cap and applying $\phi$ directly to $P_o$ produces an answer roughly 25% too high.
An axial capacity answer about 1.25× the correct value, sometimes paired with a separate distractor that uses the spiral $0.85$ factor on a tied column.
The Wrong-Lever-Arm Slip
The flexural lever arm is $(d - a/2)$, not $d$ and not $(d - a)$. Using $d$ overestimates $M_n$ by roughly $a/(2d)$ (a few percent for typical beams); using $(d-a)$ underestimates it by roughly $a/(2d)$. Both errors regularly appear among the four choices.
A flexural strength answer that brackets the correct value above and below by 5–15%, generated by swapping $(d-a/2)$ for $d$ or $(d-a)$.
The Critical-Section Fumble
For shear, $V_u$ is taken at distance $d$ from the support face (ACI 318 §9.4.3.2), not at the support face itself. Candidates who use the support-face shear get an inflated $V_u$ and a tighter required spacing. The error is small numerically but visible across the choice spread.
A spacing answer slightly tighter than correct (e.g., $5 \text{ in}$ vs. $6 \text{ in}$) traceable to using $V_u$ at the face instead of at $d$.
How it works
Picture a $14 \text{ in} \times 24 \text{ in}$ beam with $d = 21 \text{ in}$, $A_s = 3.00 \text{ in}^2$, $f'_c = 4 \text{ ksi}$, $f_y = 60 \text{ ksi}$. First find the stress block: $a = \dfrac{(3.00)(60)}{(0.85)(4)(14)} = \dfrac{180}{47.6} = 3.78 \text{ in}$. Then $c = 3.78/0.85 = 4.45 \text{ in}$, giving $\varepsilon_t = 0.003(21-4.45)/4.45 = 0.0112 > 0.005$, so the section is tension-controlled and $\phi = 0.90$. Nominal moment is $M_n = (3.00)(60)(21 - 3.78/2) = (180)(19.11) = 3{,}440 \text{ kip-in} = 287 \text{ kip-ft}$. Design strength is $\phi M_n = (0.90)(287) = 258 \text{ kip-ft}$. Notice every term carries units that cancel cleanly to $\text{kip-ft}$: that is the unit-cancellation check the PE rewards. Skip the $\phi$ or the $a/2$ term and you land squarely on a wrong-answer choice the test writers planted.
Worked examples
The Reyes Bridge approach girder is a simply supported reinforced concrete beam with rectangular cross-section $b = 14 \text{ in}$, total depth $h = 25 \text{ in}$, and effective depth $d = 22 \text{ in}$. It is reinforced with four #8 bottom bars (total $A_s = 3.16 \text{ in}^2$). Material properties are $f'_c = 4 \text{ ksi}$ (normal-weight concrete, $\beta_1 = 0.85$) and $f_y = 60 \text{ ksi}$. The section has been verified to be tension-controlled. No compression reinforcement is provided.
Most nearly, what is the design flexural strength $\phi M_n$ of the section?
- A $256 \text{ kip-ft}$
- B $285 \text{ kip-ft}$ ✓ Correct
- C $313 \text{ kip-ft}$
- D $316 \text{ kip-ft}$
Why B is correct: Compute the stress block depth: $a = \dfrac{A_s f_y}{0.85 f'_c b} = \dfrac{(3.16)(60)}{(0.85)(4)(14)} = \dfrac{189.6}{47.6} = 3.98 \text{ in}$. Nominal moment: $M_n = A_s f_y (d - a/2) = (3.16)(60)(22 - 1.99) = (189.6)(20.01) = 3{,}794 \text{ kip-in} = 316.2 \text{ kip-ft}$. Apply $\phi = 0.90$ for tension-controlled flexure: $\phi M_n = (0.90)(316.2) = 285 \text{ kip-ft}$. Units cancel: $(\text{in}^2)(\text{kip}/\text{in}^2)(\text{in}) = \text{kip-in}$, divided by $12$ gives $\text{kip-ft}$.
Why each wrong choice fails:
- A: Uses lever arm $(d - a) = 18.02 \text{ in}$ instead of $(d - a/2) = 20.01 \text{ in}$, giving $\phi M_n = 0.9(3.16)(60)(18.02)/12 = 256 \text{ kip-ft}$. The full $a$ is the depth of the stress block; the resultant compression force acts at its centroid, $a/2$ below the top. (The Wrong-Lever-Arm Slip)
- C: Uses lever arm $d = 22 \text{ in}$ instead of $(d - a/2)$: $\phi M_n = 0.9(3.16)(60)(22)/12 = 313 \text{ kip-ft}$. Ignores the depth of the rectangular stress block entirely. (The Wrong-Lever-Arm Slip)
- D: Computes $M_n = 316 \text{ kip-ft}$ correctly but forgets to apply $\phi = 0.90$, reporting nominal capacity as if it were design capacity. (The Missing-$\phi$ Distractor)
A simply supported RC beam at the Liu Civic Center has $b_w = 14 \text{ in}$, $d = 22 \text{ in}$, $f'_c = 4{,}000 \text{ psi}$ (normal-weight, $\lambda = 1.0$), and $f_{yt} = 60 \text{ ksi}$. At the critical section (distance $d$ from support face), the factored shear is $V_u = 65 \text{ kips}$. Vertical #3 U-stirrups (two legs, $A_v = 0.22 \text{ in}^2$) are to be used. Use the simplified concrete contribution $V_c = 2 \lambda \sqrt{f'_c}\, b_w d$ (psi units). Assume the section is adequate in size and the shear-resisting region requires $V_s$ less than $4\sqrt{f'_c}\, b_w d$.
Most nearly, what is the required maximum stirrup spacing at the critical section?
- A $3 \text{ in}$
- B $6 \text{ in}$ ✓ Correct
- C $8 \text{ in}$
- D $11 \text{ in}$
Why B is correct: Concrete contribution: $V_c = 2(1.0)\sqrt{4000}(14)(22) = (126.5)(308) = 38{,}960 \text{ lb} = 38.9 \text{ kips}$. Required steel contribution: $V_s = V_u/\phi - V_c = 65/0.75 - 38.9 = 86.7 - 38.9 = 47.8 \text{ kips}$. Spacing: $s = \dfrac{A_v f_{yt} d}{V_s} = \dfrac{(0.22)(60)(22)}{47.8} = \dfrac{290.4}{47.8} = 6.07 \text{ in}$. Round down to $6 \text{ in}$. Maximum permitted spacing $d/2 = 11 \text{ in}$ does not govern.
Why each wrong choice fails:
- A: Uses $A_v = 0.11 \text{ in}^2$ (one leg of a #3 bar) instead of the two legs that actually cross the diagonal crack, giving $s = (0.11)(60)(22)/47.8 = 3.04 \text{ in}$. A U-stirrup wraps the bottom and crosses the inclined crack twice. (The Single-Leg Stirrup Trap)
- C: Forgets to divide $V_u$ by $\phi$ when finding $V_s$: uses $V_s = V_u - \phi V_c = 65 - 29.2 = 35.8 \text{ kips}$, which yields $s = (0.22)(60)(22)/35.8 = 8.11 \text{ in}$. The $\phi$ must apply to the full nominal capacity $V_n = V_c + V_s$, not to $V_c$ alone. (The Missing-$\phi$ Distractor)
- D: Reports the code maximum spacing $s_{max} = d/2 = 11 \text{ in}$ without first computing the spacing required to resist $V_s$. The required spacing always governs when $V_u > \phi V_c$.
A short tied reinforced concrete column on the Patel Warehouse project has a square cross-section measuring $16 \text{ in} \times 16 \text{ in}$. Longitudinal reinforcement consists of eight #8 bars ($A_{st} = 8(0.79) = 6.32 \text{ in}^2$) tied with #3 lateral ties at code-permitted spacing. Material properties: $f'_c = 4 \text{ ksi}$ and $f_y = 60 \text{ ksi}$. The column is short (no slenderness effects) and is loaded concentrically with negligible bending moment.
Most nearly, what is the maximum design axial compressive strength $\phi P_{n,max}$ of the column per ACI 318?
- A $640 \text{ kips}$ ✓ Correct
- B $740 \text{ kips}$
- C $800 \text{ kips}$
- D $980 \text{ kips}$
Why A is correct: Gross area: $A_g = (16)(16) = 256 \text{ in}^2$; net concrete area $A_g - A_{st} = 249.7 \text{ in}^2$. Nominal axial capacity at zero eccentricity: $P_o = 0.85 f'_c (A_g - A_{st}) + f_y A_{st} = (0.85)(4)(249.7) + (60)(6.32) = 848.9 + 379.2 = 1{,}228 \text{ kips}$. For tied columns, $P_{n,max} = 0.80\, P_o = 982 \text{ kips}$, and $\phi = 0.65$ governs. $\phi P_{n,max} = (0.65)(982) = 638 \text{ kips} \approx 640 \text{ kips}$.
Why each wrong choice fails:
- B: Uses the spiral-column $\phi = 0.75$ on a tied column: $(0.75)(982) = 737 \text{ kips}$. ACI 318 assigns $\phi = 0.65$ to compression-controlled tied members because the failure mode is brittle without the confining effect of a spiral.
- C: Skips the $0.80$ accidental-eccentricity cap and applies $\phi$ directly to $P_o$: $(0.65)(1{,}228) = 798 \text{ kips}$. ACI 318 §22.4.2.1 requires the cap to account for unintended eccentricity even on "concentrically" loaded columns. (The Forgotten 0.80 Cap)
- D: Computes $P_{n,max} = 0.80\,P_o = 982 \text{ kips}$ but reports it as the design capacity, omitting $\phi = 0.65$. This is the unreduced nominal value, not the design strength to compare against $P_u$. (The Missing-$\phi$ Distractor)
Memory aid
"Find a, find c, check strain, then $\phi M_n$." For columns: "$0.85$-$0.80$-$0.65$" (tied) or "$0.85$-$0.85$-$0.75$" (spiral) — the three multipliers stacked from concrete to cap to $\phi$.
Key distinction
Tension-controlled vs. compression-controlled is the master switch. It sets $\phi$ (0.90 vs. 0.65/0.75), determines whether you can ignore strain compatibility (you cannot in compression-controlled), and changes which equations apply. Always compute $\varepsilon_t$ before locking in $\phi$.
Summary
On every ACI 318 problem: factor loads, compute nominal capacity from equilibrium with the rectangular stress block, classify the section by tensile strain, apply the matching $\phi$, and verify $\phi R_n \ge U$ with units that cancel.
Practice reinforced concrete: flexure, shear, axial, columns (aci 318) adaptively
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Start your free 7-day trialFrequently asked questions
What is reinforced concrete: flexure, shear, axial, columns (aci 318) on the PE Exam (Civil)?
Reinforced concrete design on the PE Civil exam follows ACI 318 strength (LRFD) design: factor the loads to get $U$ (per ACI 318 §5.3 / ASCE 7 §2.3), compute the nominal capacity $M_n$, $V_n$, or $P_n$ from equilibrium and the equivalent rectangular stress block ($a = \beta_1 c$, $\beta_1 = 0.85$ for $f'_c \le 4 \text{ ksi}$), then reduce by the strength reduction factor $\phi$ to require $\phi R_n \ge U$. The four critical $\phi$ values are $\phi = 0.90$ (tension-controlled flexure), $\phi = 0.75$ (shear and torsion), $\phi = 0.65$ (compression-controlled tied member), and $\phi = 0.75$ (compression-controlled spirally reinforced member).
How do I practice reinforced concrete: flexure, shear, axial, columns (aci 318) questions?
The fastest way to improve on reinforced concrete: flexure, shear, axial, columns (aci 318) is targeted, adaptive practice — working questions that focus on your specific weak spots within this sub-topic, getting immediate feedback, and revisiting items you missed on a spaced-repetition schedule. Neureto's adaptive engine does this automatically across the PE Exam (Civil); start a free 7-day trial to see your sub-topic mastery climb in real time.
What's the most important distinction to remember for reinforced concrete: flexure, shear, axial, columns (aci 318)?
Tension-controlled vs. compression-controlled is the master switch. It sets $\phi$ (0.90 vs. 0.65/0.75), determines whether you can ignore strain compatibility (you cannot in compression-controlled), and changes which equations apply. Always compute $\varepsilon_t$ before locking in $\phi$.
Is there a memory aid for reinforced concrete: flexure, shear, axial, columns (aci 318) questions?
"Find a, find c, check strain, then $\phi M_n$." For columns: "$0.85$-$0.80$-$0.65$" (tied) or "$0.85$-$0.85$-$0.75$" (spiral) — the three multipliers stacked from concrete to cap to $\phi$.
What's a common trap on reinforced concrete: flexure, shear, axial, columns (aci 318) questions?
Forgetting the $\phi$ factor or using the wrong $\phi$ for the limit state
What's a common trap on reinforced concrete: flexure, shear, axial, columns (aci 318) questions?
Mixing units (psi vs ksi, in vs ft) inside the same formula
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