Water Resources HubFE section 11 of 16 · free theory
Structural Design
ASCE 7 load combinations, steel tension members, columns, and beams per AISC, reinforced-concrete flexure and shear per ACI, one-way slabs, development concepts, and timber basics.
Take the free 5-question mini-quiz ↓
Load combinations (ASCE 7)
Design never uses raw service loads directly. LRFD factors the loads up and compares against a reduced nominal strength (φRn); ASD keeps service loads and compares against a reduced allowable strength (Rn/Ω). The exam loves asking which combination governs — run them all, take the largest.
LRFD (factored): 1.4D / 1.2D + 1.6L + 0.5(Lr or S or R) / 1.2D + 1.6(Lr or S or R) + (L or 0.5W) / 1.2D + 1.0W + L + 0.5(Lr or S or R) / 1.2D + 1.0E + L / 0.9D + 1.0W / 0.9D + 1.0E
ASD (service): D / D + L / D + (Lr or S or R) / D + 0.75L + 0.75(Lr or S or R) / D + (0.6W or 0.7E) / D + 0.75L + 0.75(0.6W) + 0.75(Lr or S or R) / 0.6D + 0.6W / 0.6D + 0.7E
| D, L, Lr, S, R, W, E | dead, live, roof live, snow, rain, wind, and earthquake loads |
| 0.9D / 0.6D combos | check uplift and overturning — minimum dead load against maximum wind or seismic |
The 0.5 and 0.75 coefficients are not arbitrary: only one transient load is assumed at its full value at a time. When wind or earthquake appears, the companion live load drops to L (LRFD) or 0.75L (ASD).
Steel tension members (AISC)
A tension member can fail two ways: the whole cross-section yielding, or the net section fracturing at the holes. Check both — the smaller controls.
Gross-section yielding: Pn = Fy Ag φt = 0.90 (LRFD), Ωt = 1.67 (ASD)
Tensile rupture: Pn = Fu Ae φt = 0.75 (LRFD), Ωt = 2.00 (ASD)
Ae = U · An
| Ag | gross cross-sectional area |
| An | net area (gross minus holes) |
| U | shear lag factor ≤ 1.0 — accounts for load entering through only some elements |
Design check: LRFD φtPn ≥ Pu; ASD Pn/Ωt ≥ Pa. Also check block shear rupture around the connection as a separate limit state, and keep the slenderness ratio L/r ≤ 300 (a recommended limit for tension members, not a hard failure check).
Steel columns (AISC)
Column strength is governed by buckling, which is why the slenderness ratio KL/r runs the whole calculation. Short columns crush near Fy; long columns buckle elastically at Fe.
Fe = π²E / (KL/r)²
If KL/r ≤ 4.71√(E/Fy): Fcr = [0.658(Fy/Fe)] · Fy (inelastic buckling)
If KL/r > 4.71√(E/Fy): Fcr = 0.877 Fe (elastic buckling)
Pn = Fcr Ag φc = 0.90 (LRFD), Ωc = 1.67 (ASD)
| KL/r | effective slenderness ratio — use the largest value from all buckling axes |
| K | effective length factor (1.0 for pinned ends; the handbook tabulates the rest) |
Design check: LRFD φcPn ≥ Pu; ASD Pn/Ωc ≥ Pa. Recommended upper limit KL/r ≤ 200 for compression members. The 4.71√(E/Fy) breakpoint is where the two formulas meet — compute it once per steel grade and reuse it.
Steel beams: bending and shear (AISC)
For the compact, laterally supported beams the FE exam favours, flexural strength is simply the plastic moment. Shear strength comes from the web.
Compact section, Lb ≤ Lp: Mn = Mp = Zx Fy φb = 0.90 (LRFD), Ωb = 1.67 (ASD)
Shear: Vn = 0.6 Fy Aw Cv φv = 0.90 (LRFD), Ωv = 1.67 (ASD)
| Zx | plastic section modulus about the strong axis |
| Aw | web area = d · tw |
| Cv | web shear coefficient — 1.0 for most rolled I-shapes (simplified handbook treatment) |
Design checks: φbMn ≥ Mu and φvVn ≥ Vu (LRFD). Longer unbraced lengths (Lb > Lp) bring in lateral-torsional buckling reductions — the handbook gives those formulas, and the exam will point you to them.
Reinforced concrete: flexure (ACI)
The rectangular stress block turns a messy nonlinear problem into two equations. Concrete crushes at a strain of 0.003; steel is assumed elastic-perfectly plastic at fy.
a = As fy / (0.85 f′c b)
Mn = As fy (d − a/2) φ = 0.90 for tension-controlled sections
| a | depth of the equivalent rectangular stress block |
| d | effective depth — from the compression face to the centroid of the tension steel |
| Tension-controlled | steel strain εt ≥ 0.005 when concrete crushes — check with c = a/0.85 and εt = 0.003(d − c)/c |
Design check: φMn ≥ Mu. If the section is not tension-controlled, φ drops (down to 0.65 for compression-controlled) — the worked example below shows the strain check that justifies φ = 0.90.
Reinforced concrete: shear (ACI)
Concrete carries some shear on its own; stirrups carry the rest. The exam typically asks for the concrete contribution, the required stirrup spacing, or the total design strength.
Vc = 0.17 λ √(f′c) · b · d (f′c in MPa; SI)
Vs = Av fyt d / s φ = 0.75
φ(Vc + Vs) ≥ Vu
| λ | lightweight-concrete factor (1.0 for normalweight) |
| Av | area of stirrup steel within one spacing s |
| s | stirrup spacing along the beam |
The 0.17 coefficient is the SI form (f′c in MPa); the US-units form is 2λ√(f′c) with f′c in psi. If Vu ≤ φVc/2, stirrups may be omitted in some cases — otherwise minimum stirrups apply.
One-way slabs (ACI)
A one-way slab is a wide, shallow beam: design a 1 m strip as a singly reinforced beam with the flexure equations above, spanning in the short direction. Deflection control usually governs the thickness.
Minimum thickness h (members not supporting damage-sensitive partitions):
| Simply supported | h ≥ l / 20 |
| One end continuous | h ≥ l / 24 |
| Both ends continuous | h ≥ l / 28 |
| Cantilever | h ≥ l / 10 |
l is the clear span in the short direction. Temperature and shrinkage steel goes perpendicular to the main steel, at a minimum ratio the handbook tabulates — look it up rather than memorising it.
Development and splices
A bar is only as strong as its anchorage. Development length — the embedment needed to develop the bar’s yield strength — grows with bar diameter and steel stress and shrinks with concrete strength:
ld increases with fy and db, decreases with √(f′c)
| ld | development length — use the handbook formula with its modification factors (epoxy coating, bar position, lightweight concrete) |
| Hooks | a standard 90° or 180° hook shortens the required development length |
| Splices | lap splices must develop the full bar strength; lap length is a multiple of ld |
Concept-level on the FE exam: know which direction each variable pushes ld, and that hooks and confinement reduce it. The exact formula and factors come from the handbook on exam day.
Timber beams (NDS concepts)
Timber design starts from reference values and multiplies by adjustment factors — the exam tests whether you know the framework, not the table values (never memorise those).
F′b = Fb · CD · CM · Ct · CL · CF · … check: fb = M / S ≤ F′b
Horizontal shear (rectangular section): fv = 1.5 V / A ≤ F′v
| CD | load duration factor (shorter load → higher allowed stress) |
| CM | wet service factor |
| CL | beam stability (lateral-torsional) factor |
| CF | size factor |
Deflection is commonly checked against span/360 for live load (a widely used criterion — confirm against the governing code for real design). Bearing perpendicular to grain is a separate check the exam occasionally includes.
Worked example Governing LRFD load combination
Given:
- Dead load D = 40 kN, live load L = 60 kN, wind load W = 50 kN on a beam.
- Find the governing LRFD factored load.
Solution:
- 1.4D = 1.4(40) = 56 kN.
- 1.2D + 1.6L = 1.2(40) + 1.6(60) = 48 + 96 = 144 kN.
- 1.2D + 1.0W + L + 0.5(Lr or S or R) = 48 + 50 + 60 + 0 = 158 kN.
- 0.9D + 1.0W = 36 + 50 = 86 kN (uplift check — not governing here).
- The largest is combination 3.
Answer: 1.2D + 1.0W + L governs at 158 kN factored.
Worked example Steel tension member — which limit state governs?
Given:
- A36 steel: Fy = 250 MPa, Fu = 400 MPa.
- Gross area Ag = 2000 mm²; effective net area Ae = 1600 mm².
- LRFD. Find the design strength.
Solution:
- Gross-section yielding: φtPn = 0.90 × 250 × 2000 / 1000 = 450 kN.
- Tensile rupture: φtPn = 0.75 × 400 × 1600 / 1000 = 480 kN.
- The smaller value controls — yielding governs even though rupture uses the higher φ-penalised stress, because the net area here is not reduced enough to take over.
Answer: LRFD design strength = 450 kN (gross-section yielding governs).
Worked example Singly reinforced beam — design moment strength
Given:
- Rectangular beam: b = 300 mm, effective depth d = 550 mm.
- f′c = 28 MPa, fy = 420 MPa, As = 1500 mm² (tension steel only).
Solution:
- Stress-block depth: a = Asfy / (0.85 f′c b) = 1500(420) / (0.85 × 28 × 300) = 630,000 / 7,140 = 88.2 mm.
- Nominal moment: Mn = Asfy(d − a/2) = 630,000 × (550 − 44.1) / 106 = 630,000 × 505.9 / 106 = 318.7 kN·m.
- Check tension-controlled: c = a/0.85 = 103.8 mm; εt = 0.003(d − c)/c = 0.003(446.2)/103.8 = 0.0129 ≥ 0.005, so φ = 0.90.
- Design strength: φMn = 0.90 × 318.7 = 286.8 kN·m.
Answer: φMn = 287 kN·m (tension-controlled, φ = 0.90).