CELE Steel & Timber Design — Steel Tension MembersRevision Notes

Formulas

Exam Tips

• Memorize the φ-pair: 0.90/0.75 for yield/rupture. Write it at the top of your scratch paper before solving.
• Always compute BOTH limit states even if one seems obviously larger — board exams test this awareness.
• For A36 steel: Fy = 248 MPa, Fu = 400 MPa. For A572 Gr.50: Fy = 345 MPa, Fu = 448 MPa. Memorize these.
• If the problem says 'find the design tensile strength,' it means min(0.90FyAg, 0.75FuAe).

Key Points

• LRFD design requirement: φPn ≥ Pu. The LOWER of the two limit-state strengths governs the member capacity.
• Limit State 1 — Tensile Yielding on the Gross Area: Pn = Fy × Ag, with φt = 0.90 (LRFD) or Ωt = 1.67 (ASD). This limit state is gradual and ductile — the entire gross cross-section yields before failure.
• Limit State 2 — Tensile Rupture on the Effective Net Area: Pn = Fu × Ae, with φt = 0.75 (LRFD) or Ωt = 2.00 (ASD). This limit state is sudden and brittle — fracture occurs at the net section through the holes.
• Note that Fy < Fu but Ag > Ae, and φ for yielding (0.90) > φ for rupture (0.75), so either limit state can govern depending on section geometry and connection details.
• In practice, yielding often governs for lightly bolted members with few holes; rupture can govern when many holes significantly reduce the net section.
• For LRFD: φtPn (yield) = 0.90 FyAg and φtPn (rupture) = 0.75 FuAe. Compare both; the smaller value is the design tensile strength.

Definitions

Section Title

1. Two Governing Limit States (NSCP 2015 / AISC 360-16 Section D2)

Common Mistakes

• Using φ = 0.90 for BOTH limit states — rupture always uses φ = 0.75.
• Reporting only one limit state result and not comparing both to find the governing value.
• Using net area An in the yielding formula instead of gross area Ag.
• Forgetting to convert N to kN at the final step, leading to unit errors in comparison.

Formulas

Exam Tips

• Sketch the bolt layout. Label all holes, pitch s, and gage g. Enumerate each failure path systematically.
• For 3 holes in two lines (one staggered): typically 3 possible paths — straight through line 1, straight through line 2, zig-zag through both.
• The stagger correction s²/4g is always POSITIVE — it represents area 'saved' by the diagonal path vs. cutting straight across two hole rows.
• If s is very small and g is very large, s²/4g ≈ 0, meaning the staggered path behaves like a straight-line path through two holes.
• Watch for the phrase 'design hole diameter' in the problem — if given explicitly, use it directly without adding clearance again.

Key Points

• The net area An is the gross area minus the material removed by bolt holes along the critical failure path.
• Hole diameter for design: dh = db + 2 mm (punching clearance) + 2 mm (damage allowance) = db + 4 mm per AISC 360-16 Section B4.3b. Some Philippine references and NSCP 2015 use dh = db + 3 mm total. Always read the problem statement for the hole allowance given.
• For a straight (non-staggered) failure path: An = Ag − Σ(dh × t), where t is the plate thickness.
• For STAGGERED holes, each zig-zag diagonal path is also checked. For each inclined segment between holes, ADD s²/(4g) to the net width, where s = longitudinal pitch and g = transverse gage.
• The net width for each potential failure path = Wg − Σdh + Σ(s²/4g). Multiply by t to get An for that path.
• Check ALL possible failure paths (straight and staggered) and take the MINIMUM net area as the governing An.
• For structural shapes (angles, channels, W-sections), compute An by deducting hole areas from the gross area: An = Ag − Σ(dh × t_element).
• For angles with one hole in each leg, the critical failure path may be a diagonal across the heel — check all paths.

Definitions

Section Title

2. Net Area Computation — Standard and Staggered Holes

Common Mistakes

• Using bolt diameter db instead of design hole diameter dh = db + 2 to 4 mm.
• Failing to check the straight-line path in addition to the staggered path — both must be evaluated.
• Applying the stagger correction s²/4g per hole rather than per diagonal SEGMENT between two holes.
• Confusing s (pitch along the load direction) and g (gage transverse to load). Switching them gives a wrong answer.
• For L-sections or C-sections: forgetting that the leg outstanding (not connected) still has its area in Ag but may not have a hole deduction.

Formulas

Exam Tips

• Memorize these U values for the board exam: Plates (full width) = 1.0; Angles ≥3 bolts = 0.85; Angles 2 bolts = 0.70; W-shapes flanges connected = 0.90.
• When the problem gives U explicitly, use it directly — don't recompute from the formula unless asked.
• The three-step area sequence: Ag (from tables) → An (subtract holes) → Ae (multiply by U). Never skip a step.
• For a welded connection of an angle to a gusset through one leg, AISC also provides U values based on weld lengths — the concept is the same.

Key Points

• When a tension member is connected through only some of its cross-sectional elements (e.g., an angle bolted through one leg only), the stress is NOT uniformly distributed across An at the connection. This phenomenon is called SHEAR LAG.
• Shear lag reduces the effectiveness of the unconnected elements at the connection, so the effective net area Ae = U × An, where U ≤ 1.0.
• U = 1.0 only when ALL cross-sectional elements are connected (e.g., a plate bolted through its full width, or a W-section bolted through both flanges and the web).
• For angles, tees, channels, and W-sections connected by only one leg or flange, U < 1.0. The exact value depends on the number of fasteners and the eccentricity of the connection.
• AISC 360-16 / NSCP 2015 Table D3.1 provides U values. For angles with 3 or more bolts in the connected leg: U = 0.80. For angles with 2 bolts: U = 0.60. For plates: U = 1.0.
• The general AISC formula for U: U = 1 − x̄/L, where x̄ = distance from the centroid of the connected element to the connection plane, and L = length of the connection (end-to-end bolt spacing along load direction).
• For board exam problems, U is usually given explicitly or the applicable category is specified — memorize the Table D3.1 common values.

Definitions

Section Title

3. Effective Net Area and the Shear-Lag Factor U

Common Mistakes

• Assuming U = 1.0 for angles or channels bolted through one leg — U < 1.0 for partially connected shapes.
• Applying U to the gross area instead of the net area: Ae = U × An, NOT U × Ag.
• Using the formula U = 1 − x̄/L without knowing which distance x̄ refers to (centroid of the connected element to the connection plane, not the full section centroid).
• Confusing the tabulated U values — U = 0.85 is for ≥3 bolts in an angle, not 2 bolts (which gives U = 0.70).

Formulas

Exam Tips

• For block shear: identify the block, then label one face as 'shear' (parallel to load) and one face as 'tension' (perpendicular to load).
• On shear planes: count full holes interior and half holes at ends. On the tension plane: count half holes at each side.
• When the problem only asks for plate or member capacity away from the connection, block shear does NOT apply — it is a connection-zone check.

Key Points

• Block shear is the third limit state for tension members at connections. It involves a block of material tearing out of the connection, with shear failure on one plane and tension failure on a perpendicular plane simultaneously.
• Block shear strength per NSCP 2015 / AISC 360-16 Section J4.3: Rn = 0.60 Fu Anv + Ubs Fu Ant ≤ 0.60 Fy Agv + Ubs Fu Ant
• Where: Anv = net area subject to shear; Ant = net area subject to tension; Agv = gross area subject to shear; Ubs = 1.0 for uniform tension stress, 0.50 for non-uniform.
• Resistance factor: φ = 0.75 (same as rupture).
• For a plate or angle bolted in a single line, block shear is usually straightforward. For multiple bolt lines, all possible block shear paths must be checked.
• Board exam tip: Block shear is frequently tested as an additional limit state after yielding and rupture. Always check if the problem asks for 'the design tensile strength' — it may require comparing all three limit states.

Definitions

Section Title

4. Block Shear Limit State

Common Mistakes

• Forgetting to check block shear entirely when the problem asks for the overall design tensile strength.
• Computing Anv and Ant without properly accounting for the number of hole diameters on each plane (fractional holes at the end bolts).
• Using φ = 0.90 for block shear — the correct φ is 0.75 (same as rupture).

Formulas

Exam Tips

• Quick memory aid: Tension = 300 (T comes after C, 300 > 200). Compression = 200.
• If the problem asks 'does the member satisfy the slenderness recommendation?', compute L/r and compare to 300. State clearly that it is a recommendation, not a mandatory limit.
• For round rods: r = d/4 (where d = rod diameter). This is useful for sizing rod members.

Key Points

• Tension members have no elastic buckling limit state (unlike compression members), so there is no mandatory maximum L/r.
• However, NSCP 2015 / AISC 360-16 Section D1 RECOMMENDS L/r ≤ 300 to prevent excessive sag, vibration, and handling problems during erection.
• This is a serviceability guideline, NOT a strength requirement. Violation does not make the member structurally inadequate for strength, but it is poor practice.
• Exception: Rods and hangers in tension are exempted from the L/r ≤ 300 recommendation.
• For board exam purposes, treat L/r ≤ 300 as a check that must be verified if slenderness data is given.
• r = least radius of gyration of the cross-section = √(I/A), where I is the minimum moment of inertia.

Definitions

Section Title

5. Slenderness Recommendation

Common Mistakes

• Using the compression member slenderness limit (L/r ≤ 200) for tension members. For tension members, the limit is L/r ≤ 300.
• Using the radius of gyration about the strong axis instead of the LEAST (minimum) radius of gyration.
• Treating L/r > 300 as a strength failure — it is only a recommendation violation for tension members.

Formulas

Exam Tips

• For sizing problems: start with yielding to get minimum Ag, then select a standard size, then check rupture with the actual An and Ae.
• For a round rod of diameter d: Ag = πd²/4. For the threaded section, the root area is smaller — use body area unless threads are at the critical section.
• Board exam sizing problems often include: (a) find minimum diameter, (b) check slenderness. Practice this two-step sequence.

Key Points

• Example 1 (Plate with two bolts): Establishes baseline computation of both limit states — yielding governs here.
• Example 2 (Staggered holes): Net width calculation with the s²/4g correction — rupture path through staggered holes is critical.
• Example 3 (Slenderness check): Simple ratio check — verify L/r ≤ 300.
• Example 4 (Angle with shear lag): Demonstrates the U-factor reduction in Ae and its impact on the rupture limit state.
• Example 5 (Sizing a rod): Back-calculation from Pu to required Ag for a threaded rod.

Definitions

Section Title

6. Worked Board-Style Problems

Common Mistakes

• When sizing, finding Ag from the yielding formula without checking that the net area at holes also satisfies the rupture limit state.
• Using service loads (unfactored D and L) directly instead of factored loads for LRFD.
• For rods: using the root area (at threads) as Ag — the problem will specify whether to use the full body area or root area.