Design of Steel Components (FE Civil): Beams, Columns, Tension Members, Connections

NCEES FE Civil spec: Structural Engineering, G. Design of steel components (e.g., codes and design philosophies, beams, columns, tension members, connections). FE Reference Handbook: Civil Engineering chapter, Design of Steel Components section (AISC 360-16).

What the FE Civil Spec Covers

Topic G covers design of steel components under AISC 360: codes and design philosophies (LRFD and ASD), beams, columns, tension members and connections. FE questions typically ask for a design strength, check whether a given W-shape works, or pick the lightest adequate section from a short list.

Key Concepts

LRFD compares factored load effects with design strength: $R_{u} \leq ϕ R_{n}$ . ASD compares service load effects with allowable strength: $R_{a} \leq R_{n} /Ω$ .
Beams fail by yielding (plastic moment), lateral-torsional buckling if the compression flange is not braced, local buckling for noncompact sections, or web shear.
Columns fail by flexural buckling, controlled by the slenderness ratio $K L / r$ about the weaker axis.
Tension members are checked for yielding on the gross area and rupture on the effective net area, which accounts for bolt holes and shear lag.
Connections: bolt shear and bearing, and fillet welds sized by throat area.

Core Formulas (LRFD)

Flexure, compact and fully braced

M_{n} = M_{p} = F_{y} Z_{x}, ϕ_{b} = 0.90

If $L_{p} < L_{b} \leq L_{r}$ :

M_{n} = C_{b} [M_{p} - (M_{p} - 0.7 F_{y} S_{x}) \frac{L _{b} - L _{p}}{L _{r} - L _{p}}] \leq M_{p}

Shear

V_{n} = 0.6 F_{y} A_{w} C_{v}, A_{w} = d t_{w}, ϕ_{v} = 1.00 (most rolled W-shapes)

Compression

F_{e} = \frac{π ^{2} E}{( K L / r ) ^{2}}, F_{cr} = {0.65 8^{F_{y} / F_{e}} F_{y} 0.877 F_{e} K L / r \leq 4.71 E / F_{y} K L / r > 4.71 E / F_{y}

ϕ_{c} P_{n} = 0.90 F_{cr} A_{g}

Tension and welds

ϕ_{t} P_{n} = min (0.90 F_{y} A_{g}, 0.75 F_{u} A_{e}), A_{e} = U A_{n}

ϕ R_{n} = 0.75 (0.60 F_{E X X}) (0.707 w) L_{weld}

Where It Is in the FE Reference Handbook

Search the Civil Engineering chapter, Design of Steel Components section. It has the beam, shear, column and tension member equations, Cb values for common bracing cases, net area rules for bolt holes, and W-shape tables listing $Z_{x}$ , $ϕ_{b} M_{p}$ , $L_{p}$ , $L_{r}$ and $ϕ_{v} V_{n}$ . Use E = 29,000 ksi.

Worked Example: Check a W18x50 Beam

A simply supported floor beam spans 30 ft with continuous lateral bracing of the compression flange. Service loads are D = 0.8 kip/ft (including self-weight) and L = 1.2 kip/ft. Steel is A992, $F_{y} = 50$ ksi. Check a W18x50 ( $Z_{x} = 101 in^{3}$ , d = 18.0 in, $t_{w} = 0.355$ in, compact).

Step 1: Factored load

w_{u} = 1.2 (0.8) + 1.6 (1.2) = 0.96 + 1.92 = 2.88 kip/ft

(1.4D = 1.12 kip/ft does not govern.)

Step 2: Required moment and shear

M_{u} = \frac{w _{u} L ^{2}}{8} = \frac{2.88 ( 30 ) ^{2}}{8} = 324 kip-ft

V_{u} = \frac{w _{u} L}{2} = \frac{2.88 ( 30 )}{2} = 43.2 kips

Step 3: Flexural strength

ϕ_{b} M_{n} = 0.90 (50) (101) /12 = 378.8 kip-ft \geq 324 ✓

Equivalently, the required plastic modulus is $Z_{req} = M_{u} / (ϕ_{b} F_{y}) = 324 (12) / (0.90 \cdot 50) = 86.4 in^{3}$ , and 101 is larger.

Step 4: Shear strength

ϕ_{v} V_{n} = 1.00 (0.6) (50) (18.0) (0.355) = 191.7 kips \geq 43.2 ✓

The W18x50 works, with flexure at about 86 percent of capacity. Shear is rarely critical for long, uniformly loaded beams. In a real design you would also check live load deflection.

Quick Column Check

A W-shape column with no slender elements has KL/r = 100 about its weak axis, $F_{y} = 50$ ksi and E = 29,000 ksi. First find which buckling equation applies:

4.71 E / F_{y} = 4.71 29, 000/50 = 113.4 > 100 \Rightarrow inelastic buckling

F_{e} = \frac{π ^{2} ( 29 , 000 )}{10 0 ^{2}} = 28.6 ksi, F_{cr} = 0.65 8^{50/28.6} (50) = 24.1 ksi

ϕ_{c} F_{cr} = 0.90 (24.1) = 21.7 ksi

Multiply by the gross area to get $ϕ_{c} P_{n}$ . Notice that a slender column reaches less than half of its yield stress, which is why bracing the weak axis is so effective.

Quick Tension Member Check

A 6 in by 1/2 in A36 flat bar ( $F_{y} = 36$ ksi, $F_{u} = 58$ ksi) is connected by a single line of 3/4 in bolts, so one hole is deducted at any cross section. For a flat bar, U = 1.0.

A_{g} = 6 (0.5) = 3.00 in^{2}, A_{n} = [6 - (0.75 + 0.125)] (0.5) = 2.56 in^{2}

ϕ P_{n, yield} = 0.90 (36) (3.00) = 97.2 kips, ϕ P_{n, rupture} = 0.75 (58) (2.56) = 111.5 kips

The smaller value, 97.2 kips from gross section yielding, is the design tensile strength.

LRFD or ASD?

Both methods use the same nominal strength $R_{n}$ . LRFD multiplies it by $ϕ$ and compares with factored loads; ASD divides it by $Ω$ (1.67 for flexure and compression) and compares with service load combinations. Unless an FE question says ASD, assume LRFD. When choosing the lightest adequate W-shape from a list, sort by weight per foot and take the first one whose design strength meets the demand, rather than the one with the largest capacity.

Common Mistakes

Using the elastic modulus $S_{x}$ instead of $Z_{x}$ for the plastic moment.
Forgetting to convert kip-in to kip-ft (divide by 12).
Using $F_{y}$ with net area or $F_{u}$ with gross area in tension. Yielding goes with gross area, rupture with effective net area.
Taking KL/r about the strong axis when the weak axis is less braced and gives a larger slenderness.
Using hole diameter equal to bolt diameter. Add 1/16 in for the standard hole and another 1/16 in for damage when computing net area.

Practice Problems

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

A compact, doubly symmetric steel beam section is sufficiently laterally braced to develop its full plastic moment. Fy = 250 MPa, plastic section modulus Zx = 800,000 mm³, elastic section modulus Sx = 700,000 mm³, and φb = 0.90. Use Mn = Fy·Zx and ignore other limit states. Find the design flexural strength φbMn. See the worked solution or practice the Steel Design deck
For a specified compact beam use Mn = Cb[Mp − (Mp − Mr)(Lb − Lp)/(Lr − Lp)] ≤ Mp, valid for Lp < Lb < Lr. Given Mp = 300 kN·m, Mr = 180 kN·m, Lp = 1.5 m, Lb = 4.0 m, Lr = 5.0 m, Cb = 1.0 and φb = 0.90, find the design flexural strength φbMn. See the worked solution or practice the Steel Design deck
A steel column has no slender elements, KL/r = 70, E = 200,000 MPa, Fy = 250 MPa, Ag = 4,000 mm² and φc = 0.90. Use Fe = π²E/(KL/r)². If Fy/Fe ≤ 2.25, Fcr = 0.658^(Fy/Fe)·Fy; otherwise Fcr = 0.877Fe. Only flexural buckling governs. Find φcPn. See the worked solution or practice the Steel Design deck
A steel tension member has net area An = 1,500 mm² (hole deductions already included), shear-lag factor U = 0.80 and Fu = 450 MPa. For net-section rupture use Ae = U·An and φ = 0.75. Gross-section yielding has a higher strength. Find the design rupture strength. See the worked solution or practice the Steel Design deck
An A36 steel plate 8 in wide × 1/2 in thick (F_y = 36 ksi, F_u = 58 ksi) is a tension member. It is connected with three 3/4-in diameter bolts arranged in a single line across the width, with no stagger, and all of the cross-section is connected so U = 1.0. Per AISC LRFD, the design tensile strength φP_n is most nearly: See the worked solution or practice the Steel Design deck
A rolled steel beam web has shear area Aw = 2,000 mm² (total section area is 5,000 mm²), Fy = 345 MPa, Cv = 1.0, and for this specified case φv = 1.0. Use Vn = 0.6·Fy·Aw·Cv. Find the design web shear strength φvVn. See the worked solution or practice the Steel Design deck

Related Calculators and Examples

Frequently Asked Questions

What resistance factors should I know for FE steel design?

Flexure 0.90, compression 0.90, tension yielding on the gross section 0.90, tension rupture on the net section 0.75, and fillet welds 0.75. Shear in the web of most rolled W-shapes uses 1.00; other webs use 0.90.

Do I need the AISC Steel Construction Manual for the FE exam?

No. The FE Reference Handbook includes the AISC equations you need plus tables of common W-shapes with properties and design strengths, so you can look up values such as the design moment of a W18x50 directly.

When does lateral-torsional buckling matter?

When the unbraced length of the compression flange, Lb, is longer than Lp. If Lb is at or below Lp the beam can reach its full plastic moment. Between Lp and Lr the strength drops linearly, and beyond Lr elastic buckling controls.

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