Design of Reinforced Concrete Components (FE Civil): Beams and Columns

NCEES FE Civil spec: Structural Engineering, H. Design of reinforced concrete components (e.g., codes and design philosophies, beams, columns). FE Reference Handbook: Civil Engineering chapter, Structural Design section (Reinforced Concrete, ACI 318).

What the FE Civil Spec Covers

Topic H is the design of reinforced concrete components under ACI 318: codes and design philosophy (strength design), beams and columns. FE questions focus on singly reinforced rectangular beams in flexure, checking strain and the strength reduction factor, shear and stirrup spacing, and the axial capacity of short tied columns. Slab and footing questions also appear in the practice banks.

Key Concepts

Strength design: $ϕ M_{n} \geq M_{u}$ , $ϕ V_{n} \geq V_{u}$ , $ϕ P_{n} \geq P_{u}$ , with factored loads from the ASCE 7 combinations.
Whitney stress block: at nominal strength, concrete compression is modeled as a uniform stress $0.85 f_{c}^{'}$ over depth $a = β_{1} c$ . Concrete tensile strength is ignored.
Strain compatibility: concrete crushes at a strain of 0.003. The tension steel strain sets the section class and the value of $ϕ$ .
Under-reinforced is good: tension-controlled beams yield and deflect visibly before failing, which is why the code rewards them with the higher $ϕ = 0.90$ .

Core Formulas

Flexure, singly reinforced rectangular beam

a = \frac{A _{s} f _{y}}{0.85 f _{c}^{'} b}, c = \frac{a}{β _{1}}, M_{n} = A_{s} f_{y} (d - \frac{a}{2})

ε_{t} = 0.003 \frac{d _{t} - c}{c}

Net tensile strain	Section	Tied members
$ε_{t} \geq 0.005$	Tension-controlled	$ϕ = 0.90$
$0.002 < ε_{t} < 0.005$	Transition	$ϕ = 0.48 + 83 ε_{t}$
$ε_{t} \leq 0.002$	Compression-controlled	$ϕ = 0.65$

Minimum flexural steel (psi units):

A_{s, m i n} = max (\frac{3 f _{c}^{'}}{f _{y}}, \frac{200}{f _{y}}) b_{w} d

Shear (psi units, normal-weight concrete)

V_{c} = 2 λ f_{c}^{'} b_{w} d, V_{s} = \frac{A _{v} f _{y} d}{s}, ϕ = 0.75

Short tied column

ϕ P_{n, m a x} = 0.80 ϕ [0.85 f_{c}^{'} (A_{g} - A_{s t}) + f_{y} A_{s t}], ϕ = 0.65

Where It Is in the FE Reference Handbook

Search the Civil Engineering chapter, Structural Design section, for the reinforced concrete pages: resistance factors, net tensile strain, singly reinforced beams, the shear and stirrup spacing table, short columns and a reinforcing bar table with areas. Units there are psi, in. and lb, so convert ksi carefully.

Worked Example: Design Moment of a Beam

A rectangular beam has b = 12 in, d = 20 in and three #8 bars ( $A_{s} = 3 (0.79) = 2.37 in^{2}$ ). Use $f_{c}^{'} = 4, 000$ psi and $f_{y} = 60, 000$ psi. Find the design moment strength.

Step 1: Stress block depth

a = \frac{2.37 ( 60 )}{0.85 ( 4 ) ( 12 )} = \frac{142.2}{40.8} = 3.49 in

Step 2: Neutral axis and steel strain

β_{1} = 0.85 \Rightarrow c = \frac{3.49}{0.85} = 4.10 in

ε_{t} = 0.003 \frac{20 - 4.10}{4.10} = 0.0116 \geq 0.005 \Rightarrow ϕ = 0.90

Step 3: Nominal and design moment

M_{n} = 2.37 (60) (20 - \frac{3.49}{2}) = 142.2 (18.26) = 2596 kip-in = 216.3 kip-ft

ϕ M_{n} = 0.90 (216.3) = 194.7 kip-ft

Step 4: Minimum steel check

\frac{200}{60 , 000} = 0.00333 > \frac{3 4000}{60 , 000} = 0.00316 \Rightarrow A_{s, m i n} = 0.00333 (12) (20) = 0.80 in^{2}

2.37 in² exceeds 0.80 in², so the beam has a design moment strength of about 195 kip-ft and is tension-controlled.

Beam Design Flow on the Exam

Find the factored moment and shear from the governing load combination (usually 1.2D + 1.6L).
For a design question, assume a lever arm of about 0.9d to estimate $A_{s} \approx M_{u} / (ϕ f_{y} \cdot 0.9 d)$ , then pick bars from the bar table.
Compute a, then c, then the steel strain, and confirm the value of phi you assumed.
Check that $ϕ M_{n} \geq M_{u}$ and that the steel area is at least the minimum.
Check shear and choose stirrup spacing from the spacing rules.

Most FE questions stop after one or two of these steps, so read what is being asked and skip the rest.

Follow-Up: Shear in the Same Beam

Using the same beam (normal-weight concrete, so $λ = 1.0$ ), the concrete shear strength is

V_{c} = 24000 (12) (20) = 30, 360 lb = 30.4 kips, ϕ V_{c} = 0.75 (30.4) = 22.8 kips

With two-legged #3 stirrups ( $A_{v} = 0.22 in^{2}$ ) at s = 10 in:

V_{s} = \frac{0.22 ( 60 ) ( 20 )}{10} = 26.4 kips, ϕ V_{n} = 0.75 (30.4 + 26.4) = 42.6 kips

Because $V_{s}$ is below $4 f_{c}^{'} b_{w} d = 60.7$ kips, the maximum spacing is the smaller of d/2 = 10 in and 24 in, so s = 10 in is acceptable.

Follow-Up: Short Tied Column

A 16 in square tied column has eight #9 bars ( $A_{s t} = 8.00 in^{2}$ , about 3.1 percent of $A_{g} = 256 in^{2}$ ), with the same materials:

ϕ P_{n, m a x} = 0.80 (0.65) [0.85 (4) (256 - 8) + 60 (8)] = 0.52 (843.2 + 480) = 688 kips

Common Mistakes

Using total height h instead of effective depth d (to the centroid of the tension steel).
Forgetting to divide a by beta1 before computing strain. The strain check uses c, not a.
Plugging f'c in ksi into the square root terms. The shear and minimum steel formulas use psi.
Assuming phi = 0.90 without checking strain. Heavily reinforced beams can fall into the transition zone.
Leaving out the 0.80 factor (tied) or 0.85 factor (spiral) on the maximum column axial strength.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

A singly reinforced rectangular concrete beam has width b = 300 mm, effective depth d = 500 mm and tension steel A_s = 1,500 mm². Material strengths are f′_c = 28 MPa and f_y = 420 MPa. Assuming the steel yields, the nominal flexural strength M_n is most nearly: See the worked solution or practice the Reinforced Concrete deck
A singly reinforced rectangular concrete beam has b = 300 mm, effective depth d = 450 mm, f′c = 30 MPa, fy = 400 MPa, and factored moment Mu = 160 kN·m. Use a = Asfy/(0.85f′cb), Mn = Asfy(d − a/2), and φ = 0.90; there is no compression steel, and the steel is assumed to yield. Find the theoretical required tension-steel area As (the small physical root, not a bar layout). See the worked solution or practice the Reinforced Concrete deck
At nominal strength, a reinforced concrete section has neutral-axis depth c = 150 mm and extreme tension-steel depth dt = 500 mm, both measured from the extreme compression face. Plane sections remain plane and the extreme concrete compressive strain is 0.003. Using the supplied ACI 318-14 rule (εt ≥ 0.005 is tension controlled), what is the tensile strain εt and the classification? See the worked solution or practice the Reinforced Concrete deck
A beam has vertical two-legged stirrups at s = 200 mm spacing. Each effective leg has area 100 mm², fy = 400 MPa, and d = 450 mm. Use Vs = Avfyd/s with both legs crossing the shear crack and anchored. Find the nominal stirrup shear contribution Vs (no concrete contribution, no φ). See the worked solution or practice the Reinforced Concrete deck
A short tied column has Ag = 160,000 mm², f′c = 30 MPa, fy = 420 MPa, and factored axial load Pu = 3,000 kN. Use Pu ≤ 0.80φ[0.85f′c(Ag − Ast) + fyAst] with φ = 0.65, ignoring slenderness and eccentricity. Find the theoretical minimum total steel area Ast (it should fall within a 1%-8% ratio). See the worked solution or practice the Concrete: Columns, Slabs & Footings deck

Related Calculators and Examples

Frequently Asked Questions

What is beta1 in reinforced concrete design?

Beta1 relates the depth of the equivalent rectangular stress block to the neutral axis depth, a = beta1 times c. It is 0.85 for concrete strengths up to 4,000 psi, drops by 0.05 for each 1,000 psi above that, and is never less than 0.65.

How do I know if phi is 0.90 for a concrete beam?

Compute the net tensile strain in the extreme tension steel. If it is at least 0.005 the section is tension-controlled and phi is 0.90. If it is 0.002 or less it is compression-controlled (0.65 for tied members), and in between phi is interpolated.

When are stirrups required in a concrete beam?

Under the FE handbook rules, no stirrups are needed if the factored shear is no more than half of phi times Vc. Above that, at least minimum stirrups are required, and where Vu exceeds phi Vc the stirrups must be designed to carry the excess shear.

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