Structural Steel Design

9% of exam

AISC tension, compression, flexure, and shear member strength by LRFD and ASD, block shear, lateral-torsional buckling, beam-column interaction, and composite-beam behavior.

7 concepts

Exam areas

A. Horizontal members (beams, slabs)

AISC Flexure, Plastic Moment & LTB

The plastic moment Mp = Fy·Zx, the three lateral-torsional-buckling zones set by Lp and Lr, the moment-gradient factor Cb, compactness checks, and φb = 0.90.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

AISC Beam Shear & Web Strength

Web shear strength Vn = 0.6·Fy·Aw·Cv1, the φv = 1.00 break for rolled I-shapes, web local yielding and crippling under concentrated loads, and when stiffeners are needed.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

Using the AISC Manual Tables (Exam Strategy)

Navigate the 15th-ed Manual design tables — Zx (3-2), φcPn (4-1), available moment vs. Lb (3-10), φvVn — to read φ-strengths directly and beat the equations on the clock.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

B. Vertical members (columns, walls)

AISC Compression Member Strength

Compute the flexural-buckling strength of a steel column from KL/r: the elastic Fe, the inelastic/elastic Fcr branches at the 4.71 transition, and the LRFD/ASD strengths.

A steel column rarely fails by crushing — it fails by buckling, and the whole of AISC compression design is an answer to one question: how much does buckling knock the strength below the squash load

F_y A_g

? The PE Civil Structural exam tests this in the vertical-members area as a near-certain question, and the moves are always the same: find the governing effective slenderness

KL/r

, decide which buckling branch you are on, read or compute the critical stress

F_{cr}

, and apply the resistance or safety factor. The NCEES PE Civil Reference Handbook gives you no column-strength table — §4.2 Steel covers only fastener-group geometry and weld symbols — so this is a fully NARRATIVE topic taught from AISC 360-16 Chapter E and the Steel Construction Manual, 15th ed. (Table 4-1).

Slenderness governs everything

The effective slenderness ratio

KL/r

is the master variable.

K

is the effective-length factor that captures end restraint (the ideal pinned-pinned column is

K=1.0

; fixed-fixed is

0.5

; a sway column with one end free can be

2.0

or more),

L

is the unbraced length, and

r=\sqrt{I/A}

is the radius of gyration about the buckling axis. A column can have a different unbraced length and a different

r

about each principal axis, so you must check both axes and let the larger

KL/r

govern — that is where most exam errors hide. For a doubly-symmetric W-shape with equal bracing on both axes, the weak (minor,

y

) axis with the smaller

r_y

always controls.

\left(\dfrac{KL}{r}\right)_{\text{gov}} = \max\left[\dfrac{K_xL_x}{r_x},\ \dfrac{K_yL_y}{r_y}\right]

The elastic buckling stress Fe

The elastic critical stress is the Euler stress written in terms of slenderness — the stress at which a perfectly straight elastic column would buckle. It depends only on the modulus and the slenderness, never on

F_y

, because an elastic buckle happens before any yielding. Everything downstream — the branch decision, the inelastic reduction — is built on

F_e

, so compute it first and carry full precision.

F_e = \dfrac{\pi^2 E}{\left(KL/r\right)^2}

Two branches: inelastic vs. elastic buckling

Real columns have residual stresses and initial crookedness, so AISC splits the strength curve at

KL/r = 4.71\sqrt{E/F_y}

(equivalently

F_y/F_e = 2.25

). Stocky columns below that limit buckle inelastically — partial yielding precedes buckling — and use the transcendental

0.658

form, which smoothly approaches

F_y

as the column gets stubby. Slender columns above the limit buckle elastically, and

F_{cr}

is simply

0.877\,F_e

, the

0.877

accounting for out-of-straightness. Pick the branch by comparing

KL/r

against

4.71\sqrt{E/F_y}

(which is

113

for

F_y=50

ksi,

134

for

F_y=36

ksi).

F_{cr} = \begin{cases} \left(0.658^{\,F_y/F_e}\right)F_y, & \dfrac{KL}{r}\le 4.71\sqrt{\dfrac{E}{F_y}}\\[2ex] 0.877\,F_e, & \dfrac{KL}{r}> 4.71\sqrt{\dfrac{E}{F_y}} \end{cases}

Nominal strength and the design check

Once

F_{cr}

is known, the nominal compressive strength for the flexural-buckling limit state is simply the critical stress acting over the gross area

A_g

(net area never governs compression because the bolts are still there to carry load). Then apply the format you are working in — never mix them. In LRFD the design strength is

\phi_c P_n

with

\phi_c = 0.90

, checked against the factored load

P_u

. In ASD the allowable strength is

P_n/\Omega_c

with

\Omega_c = 1.67

, checked against the service load

P_a

P_n = F_{cr}\,A_g, \qquad \phi_c P_n\ (\phi_c = 0.90)\quad\text{or}\quad \dfrac{P_n}{\Omega_c}\ (\Omega_c = 1.67)

Local buckling and other limit states

The flexural-buckling equations above assume the cross-section's plates do not buckle locally before the member does — i.e., the section is nonslender. For nonslender elements (most rolled W-shapes used as columns, which AISC Table 1-1 confirms), use

A_g

as above. If any element is slender (high

b/t

), Chapter E reduces the strength through an effective-area factor and the section can lose capacity well below

F_y A_g

. Singly-symmetric, unsymmetric, and slender built-up shapes may also be governed by torsional or flexural-torsional buckling rather than pure flexural buckling — but for the doubly-symmetric rolled shapes the PE favors, minor-axis flexural buckling controls.

P_n = F_{cr}A_g \quad\text{(nonslender)}; \qquad P_n = F_{cr}A_e \quad\text{(slender elements)}

Reading it straight from Manual Table 4-1

On the exam the fast path is almost always AISC Manual Table 4-1, which tabulates

\phi_c P_n

(and

P_n/\Omega_c

) directly versus effective length

KL

taken about the

y

-axis for every W-shape at

F_y = 50

ksi. If your column is braced more tightly about the weak axis, you convert the strong-axis length to an equivalent

y

-axis length by dividing by

r_x/r_y

before entering the table. Knowing the closed-form Chapter E equations is still essential — they let you handle

F_y = 36

ksi, odd

K

values, and confirm a table read — but reach for the table first when the shape and

F_y

match.

(KL)_{\text{eff},y} = \dfrac{K_x L_x}{r_x/r_y}

Exam strategy

Work in this fixed order: (1) get both

K_xL_x/r_x

and

K_yL_y/r_y

and circle the larger; (2) compute

4.71\sqrt{E/F_y}

and compare to decide the branch; (3) compute

F_e

, then

F_{cr}

from the correct formula; (4)

P_n = F_{cr}A_g

; (5) apply

\phi_c = 0.90

(LRFD) or

\Omega_c = 1.67

(ASD) — and confirm which the problem wants before the last line. When the shape is a standard W at

F_y = 50

ksi, Table 4-1 gives

\phi_c P_n

in one read; use it to check your hand calc. The classic traps: using the wrong axis, comparing

F_y/F_e

against

2.25

but then using the wrong branch, and forgetting that

\phi

multiplies while

\Omega

divides.

Key equations

Governing slenderness

\dfrac{KL}{r} = \max\left[\dfrac{K_xL_x}{r_x},\ \dfrac{K_yL_y}{r_y}\right]

Larger of the two principal-axis values governs. $K$ = effective-length factor, $L$ = unbraced length, $r=\sqrt{I/A}$ (in). Keep $KL/r$ preferably below 200.

Elastic (Euler) buckling stress

F_e = \dfrac{\pi^2 E}{\left(KL/r\right)^2}

$E = 29{,}000$ ksi for steel. $F_e$ in ksi. Independent of $F_y$ . AISC 360-16 Eq. E3-4.

Branch transition (inelastic/elastic)

\dfrac{KL}{r}\le 4.71\sqrt{\dfrac{E}{F_y}} \iff \dfrac{F_y}{F_e}\le 2.25

Limit = 113 for $F_y=50$ ksi, 134 for $F_y=36$ ksi. At or below: inelastic branch; above: elastic branch.

Inelastic critical stress

F_{cr} = \left(0.658^{\,F_y/F_e}\right)F_y

AISC 360-16 Eq. E3-2, for $KL/r \le 4.71\sqrt{E/F_y}$ . Approaches $F_y$ as the column gets stocky.

Elastic critical stress

F_{cr} = 0.877\,F_e

AISC 360-16 Eq. E3-3, for slender columns above the transition. The 0.877 covers initial out-of-straightness.

Nominal compressive strength

P_n = F_{cr}\,A_g

$A_g$ = gross area (in²); $P_n$ in kips. Use gross area — bolt holes do not reduce compression. AISC 360-16 Eq. E3-1.

LRFD design strength

\phi_c P_n,\qquad \phi_c = 0.90

Compare to factored load $P_u$ from ASCE 7-16 LRFD combinations. Require $\phi_c P_n \ge P_u$ .

ASD allowable strength

\dfrac{P_n}{\Omega_c},\qquad \Omega_c = 1.67

Compare to service load $P_a$ from ASD combinations. Require $P_n/\Omega_c \ge P_a$ . Never mix with LRFD.

Equivalent weak-axis length for Table 4-1

(KL)_{\text{eff},y} = \dfrac{K_x L_x}{r_x/r_y}

Converts a strong-axis effective length to the $y$ -axis basis the table uses, so you enter Table 4-1 with the larger of $K_yL_y$ and this value.

Worked examples

Stocky column — inelastic branch (LRFD and ASD)

Problem. A W10×49 ( $A_g = 14.4\ \text{in}^2$ , $r_y = 2.54\ \text{in}$ , $r_x = 4.35\ \text{in}$ ) of $F_y = 50$ ksi steel is pinned top and bottom and braced about both axes only at the ends. The unbraced length is $14\ \text{ft}$ each way ( $K = 1.0$ ). Find $\phi_c P_n$ and $P_n/\Omega_c$ .

Solution. Both lengths equal, so the smaller

r

governs: minor axis.

KL/r_y = (1.0)(14\times12)/2.54 = 168/2.54 = 66.1

. Branch limit:

4.71\sqrt{E/F_y} = 4.71\sqrt{29000/50} = 113

. Since

66.1 < 113

, use the inelastic branch.

F_e = \pi^2(29000)/66.1^2 = 286200/4373 = 65.4

ksi.

F_{cr} = 0.658^{(50/65.4)}\,(50) = 0.658^{0.764}(50) = 0.726(50) = 36.3

ksi.

P_n = F_{cr}A_g = 36.3(14.4) = 523

kips. LRFD:

\phi_c P_n = 0.90(523) = 471

kips. ASD:

P_n/\Omega_c = 523/1.67 = 313

kips. Sanity:

F_{cr} = 36.3

ksi is well below

F_y = 50

ksi (buckling reduced it ~27%), as expected for a moderately slender column; Manual Table 4-1 lists

\phi_c P_n \approx 471

kips at

KL = 14\ \text{ft}

— a match.

F_{cr} = 0.658^{\,50/65.4}\,(50) = 36.3\ \text{ksi}

Slender column — elastic branch

Problem. A W8×31 ( $A_g = 9.13\ \text{in}^2$ , $r_y = 2.02\ \text{in}$ ) of $F_y = 50$ ksi steel acts as a pinned strut with an unbraced length of $22\ \text{ft}$ about its weak axis ( $K = 1.0$ ). Find the LRFD design compressive strength.

Solution.

KL/r_y = (1.0)(22\times12)/2.02 = 264/2.02 = 130.7

. Limit

4.71\sqrt{29000/50} = 113

. Since

130.7 > 113

, use the elastic branch.

F_e = \pi^2(29000)/130.7^2 = 286200/17082 = 16.76

ksi.

F_{cr} = 0.877(16.76) = 14.70

ksi.

P_n = 14.70(9.13) = 134

kips, so

\phi_c P_n = 0.90(134) = 121

kips. Sanity: at

KL/r = 131

the column is firmly in the slender range, so

F_{cr}

should sit far below

F_y

— here

14.7

ksi, under a third of

50

ksi — confirming elastic buckling controls.

F_{cr} = 0.877\,F_e = 0.877(16.76) = 14.7\ \text{ksi}

Common pitfalls

•Using the strong-axis $r_x$ when the weak axis controls. For a W-shape with equal bracing both ways, the smaller $r_y$ gives the larger $KL/r$ and governs — always check both axes.
•Citing a handbook page for column strength. The PE Civil Handbook §4.2 has no member-strength tables; compression design is AISC 360-16 Ch E. Cite the code, not a handbook page.
•Putting the strength on the wrong branch: $KL/r \le 4.71\sqrt{E/F_y}$ is inelastic ( $0.658$ form); above it is elastic ( $0.877F_e$ ). Memorize the limit values 113 (Fy=50) and 134 (Fy=36).
•Mixing LRFD and ASD: $\phi_c = 0.90$ MULTIPLIES $P_n$ ; $\Omega_c = 1.67$ DIVIDES it. Decide the format up front and never compare a factored load against an allowable strength.
•Reducing compression area for bolt holes. Compression uses gross area $A_g$ — the fasteners fill the holes and carry bearing. Net area is a tension concept.
•Wrong code edition: NCEES supplies AISC 360-16 / Manual 15th ed. Do not use AISC 360-22 constants or $K$ tables from another edition.

References

AISC 360-16 — Specification for Structural Steel Buildings, Chapter E (Design of Members for Compression) — Eqs. E3-1 through E3-4 and the 4.71√(E/Fy) transition.
AISC Steel Construction Manual, 15th ed. — Table 4-1 (Available Strength in Axial Compression) — Tabulated φcPn and Pn/Ωc vs. KL about the y-axis for Fy = 50 ksi.
NCEES PE Civil Reference Handbook v2.2 — §4.2 Steel — Confirms the handbook supplies no compression-strength method; topic is taught from AISC 360-16.

AISC Beam-Column Interaction

The H1-1a/H1-1b axial-flexure interaction equations, biaxial bending, in-plane vs. out-of-plane checks, and second-order amplification of moments via B1 and B2.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

C. Systems (trusses, frames, composite)

AISC Tension Members & Block Shear

Gross-section yielding vs. net-section rupture, the shear-lag factor U and effective net area, and block-shear rupture at bolted/welded end connections.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

Composite Steel Beams

Steel beam acting compositely with a concrete slab via shear studs: effective width, the plastic neutral axis from force balance, Mn, full vs. partial composite action, and stud strength.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders