Steel Tension Members - Cheat Sheet
Your 30-minute exam survival guide for tension member design. Covers the two limit states, net-area calculation (staggered holes), shear lag, and block shear. Every formula, definition, and pitfall you need for the PRC licensure exam.
Two Limit States for Tension Design
Formulas
φₜ Pₙ = 0.90 Fᵧ Aₘ (yielding)
Meaning
φₜ = 0.90 resistance factor; Fᵧ = specified yield strength (MPa); Aₘ = gross cross-sectional area (mm²); Pₙ = nominal strength
Watch out
DO NOT use net area here — use the full gross area Ag. This represents plastic deformation over the entire member.
When to use
Ductile failure — applies to the full, unpierced section; typically governs when Fy·Ag is low or holes are small.
φₜ Pₙ = 0.75 Fᵤ Aₑ (rupture)
Meaning
φₜ = 0.75 resistance factor (lower!); Fᵤ = ultimate (tensile) strength (MPa); Aₑ = effective net area (mm²); accounts for sudden brittle failure at holes.
Watch out
The φ factor is 0.75 here (not 0.90) because rupture is sudden. Must use Aₑ, not Ag. Many students flip these φ values.
When to use
Sudden, brittle fracture through the weakest cross-section (at bolt holes). Always calculate both and take the LOWER result.
Common values
Value
0.90
Symbol
φₜ (tension yield)
Quantity
Resistance factor (yield)
Value
0.75
Symbol
φₜ (tension rupture)
Quantity
Resistance factor (rupture)
Value
1.67
Symbol
Ωₜ
Quantity
Safety factor — yield (ASD)
Value
2.00
Symbol
Ωₜ
Quantity
Safety factor — rupture (ASD)
Important facts
- Design strength = 0.90 Fy Ag (yield) or 0.75 Fu Ae (rupture) — CHOOSE THE LOWER ONE.
- Yielding uses φ = 0.90 (higher) on Fy (lower); rupture uses φ = 0.75 (lower) on Fu (higher). This balance is why both must be checked.
- LRFD required capacity: φ Pₙ ≥ Pu (factored load). ASD required capacity: Pₙ / Ω ≥ P (service load); Ωt = 1.67 (yield), 2.00 (rupture).
- Ductility: yielding is gradual (allows redistribution); rupture is sudden (brittle). The lower φ for rupture reflects this severity.
- If Fy Ag > Fu Ae, then rupture controls (lower design strength). If Fu Ae > Fy Ag, then yield controls.
Key definitions
Gross Area (Aₘ)
Example
A 200 × 12 mm plate has Aₘ = 200 × 12 = 2400 mm²
Full cross-sectional area of the member before subtracting holes.
Net Area (Aₙ)
Example
Aₘ = 2400 mm²; two 22 mm holes → Aₙ = 2400 − 2(22)(12) = 1872 mm²
Gross area minus the area of all bolt holes along the failure path.
Effective Net Area (Aₑ)
Example
For an angle bolted through one leg, U ≈ 0.85; if Aₙ = 1200 mm², then Aₑ = 0.85 × 1200 = 1020 mm²
Net area multiplied by shear-lag factor U to account for non-uniform stress transfer when only part of the section is connected.
Shear-Lag Factor (U)
Example
Plate bolted through full width: U = 1.0; L-shape bolted one leg: U ≈ 0.85
Reduction factor (0 < U ≤ 1.0) accounting for stress concentration near the bolted connection; U = 1.0 only when all elements are connected.
Hole Diameter (dₕ)
Example
20 mm bolt → dₕ = 22 mm (use 22 mm in net-area calculations, NOT 20 mm)
Nominal bolt diameter plus clearance allowance; typically dₕ = dᵦ + 2 mm (or + 3 mm per Philippine practice).
Diagrams to know
- Stress-strain diagram: flat yield plateau (ductile region) then upturn to failure (brittle rupture region).
- Net section diagram: show all holes on one 'slice' of the member and shade the pierced area being subtracted.
Net Area Calculation: Straight & Staggered Holes
Formulas
Aₙ = Aₘ − Σ(dₕ · t)
Meaning
Aₙ = net area; dₕ = hole diameter (bolt size + 2–3 mm allowance); t = material thickness; Σ = sum for all holes in the failure path.
Watch out
Use dₕ (full hole size with clearance), NOT the bolt diameter. Example: 20 mm bolt has dₕ = 22 mm. Subtract this for each hole.
When to use
Straight-line hole arrangement (all holes in one transverse line perpendicular to member axis).
Net width = Wₘ − Σ dₕ + Σ(s²/4g)
Meaning
Wₘ = gross width; dₕ = hole diameter; s = longitudinal pitch (spacing along member axis); g = transverse gage (perpendicular spacing); s²/4g term added once per staggered diagonal segment.
Watch out
The s²/4g term ADDS back (reduces the subtraction). It represents the 'benefit' of stagger. Easy to forget or apply incorrectly. Check every possible zig-zag path.
When to use
Staggered bolt holes. Multiple failure paths exist; the minimum net area governs. Calculate all possible paths and use the smallest.
Aₙ = (Net width) × t
Meaning
Net area equals the minimum net width multiplied by thickness.
Watch out
Always check ALL possible failure paths through staggered holes — don't assume the straight path is minimum.
When to use
After computing the minimum net width (especially with staggered holes), convert to area by multiplying by thickness.
Important facts
- For staggered holes, draw and check EVERY possible path (not just the obvious straight line). The minimum net area from all paths governs.
- The s²/4g term is added only for diagonal segments; it partially restores width lost to holes in the zig-zag path.
- Common practice: dₕ = dᵦ + 2 mm for Standard holes (Philippine codes often use +3 mm); clarify with local standards.
- If multiple holes exist in one transverse line, subtract all of them in that line.
- Stagger is beneficial: it allows load to bypass some holes, so net area improves vs. all holes in one line.
Key definitions
Pitch (s)
Example
Holes at 50 mm spacing along a truss member: s = 50 mm
Distance between consecutive hole centers measured along the member axis (longitudinal direction).
Gage (g)
Example
Two bolt lines 75 mm apart: g = 75 mm
Distance between hole centers measured perpendicular to the member axis (transverse direction), between two lines of holes.
Failure Path
Example
A staggered pattern may have paths that drop diagonally (zig-zag) to avoid some holes, reducing loss vs. a straight path.
The imaginary line connecting holes that represents the weakest cross-section through the member; can be straight or zig-zag (staggered).
Diagrams to know
- Plan view of staggered bolt pattern: show grid with hole positions, draw zig-zag failure paths, label s and g.
- Cross-section showing net width reduction: shade the bolt-hole area subtracted and label the stagger 'recovery'.
Shear-Lag Factor (U) & Effective Net Area
Formulas
Aₑ = U · Aₙ
Meaning
Aₑ = effective net area used in rupture check; U = shear-lag reduction factor (0 < U ≤ 1.0); Aₙ = net area.
Watch out
Do NOT use Aₑ in the yielding formula — use gross area Ag. Shear lag only affects rupture. Confusing these is a major exam mistake.
When to use
Always, when computing rupture strength (0.75 Fu Aₑ). U accounts for non-uniform stress distribution in the connected element.
Common values
Value
1.0
Symbol
U
Quantity
Shear-lag factor — plate (full width)
Value
0.85 (typical, per AISC)
Symbol
U
Quantity
Shear-lag factor — angle (one leg bolted)
Value
0.90 (typical, per AISC)
Symbol
U
Quantity
Shear-lag factor — tee (one flange bolted)
Important facts
- U accounts for the fact that stress at the bolted edge is higher than at the free edge due to shear lag.
- NSCP 2015 / AISC 360 provides U-value tables for standard shapes (angles, tees, channels). Typical: angles ≈ 0.85, tees bolted one flange ≈ 0.90.
- For a plate bolted through the full width with holes going through both edges, U = 1.0 (no lag).
- If the distance from the bolts to the far edge is large relative to the connected length, U decreases further.
- Aₑ is ONLY used in the rupture formula (0.75 Fu Aₑ), never in the yield formula.
Key definitions
Shear Lag
Example
An L-section bolted through only one leg; the unbolted leg stretches slightly more (lower stress), reducing average stress.
Non-uniform stress distribution near a bolted connection where some cross-sectional elements (e.g., one leg of an angle) are connected, leaving other parts (e.g., the other leg) to lag behind in load transfer.
U = 1.0 Condition
Example
Flat plate bolted through its full width (bolt holes span the entire width); U = 1.0
Full effective net area applies when the member geometry and connection transfer load uniformly to all elements (entire section is actively connected).
U < 1.0 Condition
Example
Angle L75×75 bolted through the short leg only; U ≈ 0.85 (some typical values: U = 0.85 for angles, 0.90 for tees)
Reduction applies when only part of the cross-section is bolted (e.g., one leg of an angle or one flange of a tee).
Diagrams to know
- Shear-lag diagram: angle bolted through one leg, showing stress concentration at the bolted leg vs. lower stress at the free leg.
- Comparison of U-values for different member types: plate (U=1.0), angle (U≈0.85), tee (U≈0.90).
Block Shear (Tension-Shear Rupture)
Formulas
Rₙ = 0.60 Fᵤ Aₘᵥ + 0.50 Fᵧ Aₙₜ ≤ 0.60 Fᵤ Aₙᵥ + 0.50 Fᵧ Aₙₜ (LRFD)
Meaning
Rₙ = nominal block shear strength; Aₘᵥ = gross shear area; Aₙᵥ = net shear area; Aₙₜ = net tension area; Fᵤ = ultimate strength; Fᵧ = yield strength.
Watch out
Block shear is often OMITTED in undergraduate courses but appears on the PRC exam. Always check it for bolted connections. Use the LOWER of the two terms.
When to use
When bolt group or welded connection could fail by combined shear along one plane and tension across another (e.g., at the perimeter of a bolt group attached to a plate or gusset).
Common values
Value
0.60 Fu
Symbol
block shear
Quantity
Shear rupture stress coefficient
Value
0.50 Fy
Symbol
block shear
Quantity
Tension rupture stress coefficient
Value
0.75
Symbol
φ
Quantity
Resistance factor — block shear
Important facts
- Block shear check is MANDATORY for connections (often forgotten by students).
- Design strength for block shear = φ Rₙ, where φ = 0.75 (same as rupture).
- The formula has two terms: shear rupture (0.60 Fu Aₙᵥ) and combined shear-tension. Use the LOWER result.
- Common in bolted angles and tees where a corner 'block' of material can tear free.
- Calculate using the lesser of: (1) all gross shear + net tension, or (2) net shear + net tension.
Key definitions
Block Shear
Example
Angle connection where bolt group tears along the edge (shear) and tears through the net section (tension); material 'blocks' out.
A combined failure mode where a block of material tears out due to shear along one direction and tension along a perpendicular direction simultaneously.
Shear Area (Aᵥ)
Example
For an angle bolted at two bolt lines, Aᵥ = (number of bolt rows) × (edge distance) × (thickness)
Gross or net cross-sectional area along which shear rupture can occur (parallel to the applied load or connection edge).
Tension Area (Aₜ)
Example
Net width across the bolt holes times thickness.
Net cross-sectional area perpendicular to the shear area where tension failure (rupture through holes) occurs.
Diagrams to know
- Block shear tear pattern: show a rectangular block of material tearing along two perpendicular faces (one shear, one tension).
- Bolt pattern with block shear boundary highlighted: corner distance and edge spacing labeled.
Slenderness & Practical Design
Formulas
L/r ≤ 300 (recommended)
Meaning
L = member length; r = least radius of gyration; recommendation to limit vibration/sag in tension members.
Watch out
This is a RECOMMENDATION, not a hard code requirement. However, exam questions may ask you to verify it. For rods, the check is often waived.
When to use
Check after designing the member size. This is NOT a strength limit (tension members do not buckle) but a serviceability recommendation.
Common values
Value
≤ 300
Symbol
L/r
Quantity
Recommended slenderness ratio limit
Value
≤ 400
Symbol
L/r
Quantity
Typical rod (relaxed limit)
Important facts
- Tension members do NOT buckle (no compression); slenderness limits are for sag, vibration, and handling only.
- L/r ≤ 300 is good practice; sometimes relaxed to 400 for rods in remote locations.
- Radius of gyration r = √(I/A), where I = second moment of inertia, A = cross-sectional area.
- Always use the SMALLEST r value (about the the weakest axis, usually the minor axis).
- Verify L/r after you've determined the member size; if too large, increase the cross-section.
Key definitions
Slenderness Ratio (L/r)
Example
A rod 5 m long with r = 20 mm: L/r = 5000/20 = 250 (acceptable, < 300)
The ratio of member length to the least radius of gyration; a measure of how 'thin' or 'stretched' the member is.
Diagrams to know
- Sag and vibration diagram: show a long, thin member deflecting sideways under its own weight or dynamic loading.
Connection & Bolt Details
Common values
Value
+2 to +3 mm over bolt diameter
Symbol
dₕ
Quantity
Hole size allowance (standard)
Value
1.5 × dₕ
Symbol
edge distance
Quantity
Minimum edge distance (standard hole)
Value
3 × dᵦ (approximately)
Symbol
s
Quantity
Minimum pitch
Important facts
- All tension member connections must be checked for bolt shear, bearing, and block shear (in addition to member strength).
- Staggered holes are beneficial: they increase net area vs. straight-line holes.
- Welds (fused connections) do NOT have net area reduction, so Aₑ = Aₘ for welded tension members.
- Use dₕ = dᵦ + 2 or 3 mm for punched/drilled holes; clarify the standard for your exam jurisdiction.
- Edge distance must meet minimum code requirements (typically 1.5 × hole diameter for standard holes).
Key definitions
Standard Hole
Example
Ø20 bolt → standard hole diameter = 22 mm (not 20 mm)
Bolt hole with nominal diameter = bolt diameter + 2 mm (or +3 mm per local Philippine practice). Use for net-area calculations.
Edge Distance
Example
Bolt 50 mm from the edge; edge distance = 50 mm
Distance from the center of a bolt hole to the nearest edge of the member, measured perpendicular to the edge.
Minimum Pitch
Example
Ø20 bolts: minimum pitch ≈ 60 mm
Minimum spacing between bolt centers along the axis of the member; typically 3 × bolt diameter.
Diagrams to know
- Bolt hole layout: plan view showing pitch, gage, and edge distances; staggered pattern with failure paths.
- Cross-section through bolted joint: show member thickness, hole diameter, and washer bearing area.
Must remember
- ALWAYS CHECK BOTH LIMIT STATES: φ Pₙ = 0.90 Fy Aₘ (yield) AND φ Pₙ = 0.75 Fu Aₑ (rupture). The LOWER value governs the design strength.
- Use dₕ = dᵦ + 2 or 3 mm for net area calculations, NOT the bare bolt diameter. Subtract this full hole size from gross area.
- For staggered holes, evaluate ALL possible zig-zag failure paths. The path with the MINIMUM net area governs; s²/4g term adds (recovers) width on diagonal segments.
- Aₑ = U·Aₙ: apply the shear-lag factor U ONLY in the rupture formula (0.75 Fu Aₑ), NEVER in the yield formula. U = 1.0 for plates bolted full width; U < 1.0 for angles, tees (typical: 0.85, 0.90).
- DO NOT forget block shear (combined shear + tension rupture). It is a common exam item and often the governing failure mode at connections. φ Rₙ = 0.75(shear + tension).
- For LRFD design: φ Pₙ ≥ Pu (required). For ASD: Pₙ/Ω ≥ P. Use φ = 0.90, Ω = 1.67 (yield); φ = 0.75, Ω = 2.00 (rupture).
- Tension members do NOT buckle, so there is no buckling limit state. L/r ≤ 300 is a RECOMMENDATION for sag/vibration, not a code requirement.
- Shear-lag factor accounts for non-uniform stress when only part of the section is bolted. U < 1.0 reduces the effective net area, making rupture more likely.
- For welds (fused connections), there is no net area reduction: Aₑ = Aₘ and U = 1.0 (no holes, full section transfers load uniformly).
- Always verify that your final member size satisfies L/r ≤ 300 (serviceability) and passes both yield and rupture strength checks under factored loads.
Last minute tips
- Rig your sketches: always draw the member with bolt holes marked, label dₕ, s, g, and show which limit state controls. A clear sketch catches mistakes (e.g., forgetting a hole, wrong failure path).
- Plug φ-factor and area into BOTH formulas first, then compare: if 0.90(248)(2400) = 535.7 kN and 0.75(400)(1872) = 561.6 kN, the first (535.7 kN) governs, so take that one. Order matters: yield first, rupture second, then compare.
- For staggered holes, the trick is the s²/4g term: draw a grid, mark every hole, then trace each zig-zag path from left to right. The path that avoids the most holes (or has the largest stagger) wins. Use the SMALLEST net width you find.
- Block shear sneaks onto many exams: the formula is complex, but the concept is simple—material tears in an L-shaped pattern at the connection corner. If a question gives you bolt spacing and edge distance, block shear is coming.
- Don't panic if U is not given explicitly: AISC 360 Appendix D has standard U-values by shape (angle = 0.85, etc.). In exam mode, if you see 'angle bolted one leg,' assume U = 0.85 unless told otherwise.
Comparison tables
Yield vs. Rupture Limit States
| Aspect | Tensile Yield | Tensile Rupture |
|---|---|---|
| Failure Mode | Gradual plastic deformation (ductile) | Sudden brittle fracture (brittle) |
| Area Used | Gross area Aₘ (full section) | Effective net area Aₑ = U·Aₙ (at holes) |
| Stress Used | Fy (yield strength, ~248–400 MPa) | Fu (ultimate strength, ~400–500 MPa) |
| Resistance Factor φ | 0.90 (higher, safer) | 0.75 (lower, reflects suddenness) |
| Safety Factor Ω (ASD) | 1.67 | 2.00 |
| Design Strength | 0.90 Fy Aₘ | 0.75 Fu Aₑ |
| Which Governs? | The LOWER of the two results | The LOWER of the two results |
| Typical When | Few/small holes or low Fu/Fy ratio | Multiple large holes or high Fu/Fy ratio |
Shear-Lag Factor (U) Values by Member Type
| Member Type | Connection Type | Typical U Value | Notes |
|---|---|---|---|
| Flat Plate | Full width bolted | 1.0 | Entire section connected uniformly |
| Angle | One leg bolted | 0.85 | Other leg lags; AISC specifies 0.85 standard |
| Tee | One flange bolted | 0.90 | Web lags slightly |
| Channel | Bolted web | 0.75–0.85 | Both flanges lag; varies with connection pattern |
| I-Beam | Bolted flanges only | 0.75 | Web is unconnected; significant lag |
Common Mistakes & Corrections
| Mistake | Correct Approach | Exam Consequence |
|---|---|---|
| Using bolt diameter instead of hole diameter in net area | Use dₕ = dᵦ + 2 or 3 mm, not dᵦ | Net area too large → underestimated strength → structural unsafe → FAIL |
| Using net area in yield formula | Use gross area Aₘ for yield; Aₑ only for rupture | Yield strength overstated → safe, but inefficient design |
| Forgetting shear-lag factor U | Always apply U for rupture: Aₑ = U·Aₙ | Rupture strength overstated → UNSAFE |
| Flipping φ = 0.90 and 0.75 | Yield: φ = 0.90; Rupture: φ = 0.75 | Design strength wrong → either unsafe or overly conservative |
| Not checking block shear | Always include block shear check for bolted connections | Hidden failure mode → unsafe design → FAIL on exam |
| Not considering all zig-zag failure paths | Draw and evaluate every possible staggered path; use minimum net area | Missed true failure path → overestimated strength → UNSAFE |
| Confusing L/r as a STRENGTH limit | L/r ≤ 300 is a SERVICEABILITY (sag/vibration) recommendation, not strength | May incorrectly reject a valid design or vice versa |
Diagrams
Steel Tension Member Design Decision Tree
Complete decision tree for tension member analysis: from area calculation through limit-state checks to final verification. This is the order to follow in every exam problem.
Net Area Calculation for Staggered Holes
Step-by-step flowchart for finding net area with staggered holes. Key insight: evaluate all failure paths and use the minimum area. The s²/4g term adds back width on diagonal segments.
Tension Member Failure Progression
State diagram showing the progression from elastic loading through yield to rupture. Illustrates why yielding (ductile, gradual) has higher φ than rupture (brittle, sudden).
Typical Failure Mode Distribution in Exam Problems
Pie chart showing approximate frequency of each failure mode in PRC licensure exams. Block shear is often overlooked but appears regularly. Rupture slightly more common than yield, reflecting real-world design trends.