CELE Steel & Timber Design — Steel Tension MembersStudy Notes

Steel tension members are designed using the Load and Resistance Factor Design (LRFD) methodology, as specified in NSCP 2015 and AISC 360-16. The design requirement is: φ Pₙ ≥ Pᵤ where φ Pₙ is the design strength (reduced nominal strength) and Pᵤ is the factored (amplified) tensile load. Two limit states govern tension member design: (A) TENSILE YIELDING ON THE GROSS SECTION This represents the onset of inelastic deformation across the entire cross-sectional area. Once yielding initiates, significant strains occur, but the member retains load-carrying capacity and provides visible warning before collapse. This limit state is ductile and is the preferred failure mode. Pₙ = Fy Ag φ = 0.90 (LRFD) Ω = 1.67 (for ASD: allowable = Fy Ag / 1.67) where Fy is the specified yield stress (MPa) and Ag is the gross cross-sectional area (mm²). (B) TENSILE RUPTURE ON THE NET EFFECTIVE SECTION When tension is transmitted through bolt holes or welded connections, the reduced cross-section at these connection points is susceptible to sudden fracture. This is a brittle, catastrophic failure mode with little warning. Pₙ = Fu Ae φ = 0.75 (LRFD) Ω = 2.00 (for ASD: allowable = Fu Ae / 2.00) where Fu is the specified tensile strength (ultimate stress, MPa) and Ae is the effective net cross-sectional area (mm²). The DESIGN STRENGTH is the lower of these two values: φPn(design) = min[0.90 Fy Ag, 0.75 Fu Ae] In practice, rupture typically governs when bolt holes are present because Fu > Fy but Ae << Ag. Yielding may govern only in members with very few or no holes.

The net area is the gross area reduced by the projected area of bolt holes along the critical failure path. Proper computation of net area is essential for rupture checks and is a frequent source of error in the PRC exam. 2.1 BASIC NET AREA (PERPENDICULAR HOLES) When bolt holes are arranged perpendicular to the member axis (straight line failure path): An = Ag – Σ(dh × t) where: - dh = hole diameter for design = bolt diameter + clearance (typically db + 2 to 3 mm, or per specification) - t = thickness of the member - Σ(dh × t) = sum of projected areas of all holes on the failure path Example: A plate 200 mm wide, 10 mm thick, with two 20 mm bolts (design hole 22 mm): An = 200(10) – 2(22)(10) = 2,000 – 440 = 1,560 mm² 2.2 STAGGERED HOLES (GAGE AND PITCH) When bolt holes are arranged in a staggered (zig-zag) pattern, multiple failure paths exist. The designer must examine each path and use the one with the smallest net area (most critical). For a zig-zag path that crosses two or more holes diagonally: Net width along zig-zag = Wg – Σdh + Σ(s²/4g) where: - Wg = gross width of plate - Σdh = sum of hole diameters on the path - s = pitch (longitudinal spacing between holes on the same gage line) - g = gage (transverse distance between gage lines) - s²/4g term = recovered area from each staggered segment (credit for diagonal efficiency) The zig-zag path credit (s²/4g) accounts for the additional length traveled by the failure plane when it steps laterally; this distributes stress over a longer path and partially recovers area. PRACTICAL PROCEDURE: 1. Identify all possible failure paths (usually 1, 2, or 3 segments). 2. For each path, sum the holes crossed and apply the gage-pitch correction. 3. Compute An for each path; use the MINIMUM. 4. An is then width × thickness; if multiple rows of different pitch/gage, sum by row. COMMON MISTAKE: Forgetting to add the zig-zag credit; students subtract holes but forget +s²/4g, underestimating capacity.

The effective net area incorporates a reduction factor U to account for shear lag: Ae = U × An Shear lag occurs when the load is not transmitted uniformly to all elements of a cross-section. In bolted or welded connections, the applied force enters through the connection (e.g., one bolt or one row of bolts) and must spread across the full section. During this spreading, some elements experience less stress than others, creating a non-uniform stress distribution. 3.1 WHEN TO APPLY SHEAR-LAG REDUCTION U < 1.0 when: - An angle is bolted through only one leg (the load-bearing leg transfers load through the bolts; the other leg lags behind in stress). - A tee section is bolted through its stem only (the flange lags). - A channel is bolted through its web only (the flanges lag). - A bolted connection covers only part of the member's cross-section. U = 1.0 when: - A flat plate is bolted entirely across its width (no lagging element). - A solid round bar or solid rectangular section is bolted through its entire cross-section. - A welded connection extends across the full section (or is designed to eliminate lag). - Multiple bolts are arranged such that load spreads uniformly. 3.2 COMPUTATION OF U For bolted connections, AISC 360 and NSCP 2015 provide tables or formulas. Common cases: (a) BOLTED ANGLE (one leg connected): U = 1 – (distance from centroid to connected leg) / (length of connection) = 1 – x̄/Lc where x̄ is the distance from centroid to the unconnected leg and Lc is the length of the bolted connection (distance along the member from first to last bolt). Typical: U ≈ 0.80–0.85 for angles (varies with bolt pattern). (b) WELDED ANGLE (diagonal weld on one leg): U ≈ 1.0 – 0.3 × (leg ratio) for partial weld length; varies. For full-length weld of both legs: U = 1.0. (c) FOR DESIGN, if angle bolted through one hole: Common shortcut: U ≈ 0.85 (unless otherwise specified). (d) FOR BOLTED PLATE: U = 1.0 (load spreads uniformly across the full width immediately at the bolts). 3.3 EFFECTIVE NET AREA IN PRACTICE Once An is found and U is established, the effective net area is: Ae = U × An For rupture check: φPn(rupture) = 0.75 × Fu × Ae = 0.75 × Fu × U × An Forgetting or mis-applying U is a frequent exam error. Always ask: "Is the load transmitted uniformly to all parts of the cross-section?" If not, reduce the net area by U.

Unlike compression members, tension members have no theoretical buckling limit under pure tension. However, practical considerations lead to a recommended slenderness limit. 4.1 SLENDERNESS RATIO The slenderness ratio is defined as: L/r where L is the effective length (usually the span or member length for tension) and r is the least radius of gyration. For tension members in NSCP 2015 (AISC 360), slenderness is not a mandatory check for strength, but it is RECOMMENDED to limit: L/r ≤ 300 (recommendation, not requirement) 4.2 RATIONALE FOR THE RECOMMENDATION - Sag and Vibration: A long, slender tension member (e.g., a tie rod or cable) can sag under its own weight and vibrate excessively in wind or traffic. - Constructability: Excessively slender members are difficult to position and connect accurately; they bend and sway. - Durability: Cyclic motion (wind-induced, pedestrian traffic) accelerates fatigue and connection loosening. - Aesthetics and public confidence: Visible sag reduces perceived structural integrity. 4.3 PRACTICAL COMPUTATION For common shapes: (a) SOLID ROUND ROD (diameter d): r = d/4 L/r = 4L/d For L/r ≤ 300: d ≥ 4L/300 = L/75 (b) FLAT PLATE (width w, thickness t, direction parallel to width): r = t/√12 ≈ 0.289t L/r ≈ 3.46L/t (c) ANGLE OR STRUCTURAL SHAPE: r = (least Ixx)^0.5 / A (from tables) Check against L/r ≤ 300 4.4 WHEN THE LIMIT IS CRITICAL Slenderness becomes critical (difficult to satisfy) in: - Long-span tie rods (e.g., bridge bracing, mining conveyor lines) - Rods with small diameter for a given span - Members in areas with high wind or vibration exposure If L/r > 300 is unavoidable, the member must still satisfy the strength checks (yield and rupture). However, the code commentary notes that serviceability and fatigue concerns should be addressed (stiffening ribs, cable ties, etc.).

Block shear is a compound failure mode that occurs at bolted connections when a block of material tears out along both a tensile and shear plane simultaneously. It is a limit state that must be checked in addition to yielding and rupture. 5.1 WHAT IS BLOCK SHEAR? When bolts are arranged in a pattern (e.g., a row or multiple rows), the connection must transfer load to the bolts. If the bolts are too close to the edge or clustered too tightly, a block of material may fail by: - Tearing (rupture) along a transverse plane perpendicular to the member axis (at the bolt row) - Shearing (yielding or rupture) along a longitudinal plane adjacent to the bolts The block shear failure path forms an L-shaped or rectangular prism that tears out of the connection region. 5.2 BLOCK SHEAR STRENGTH AISC 360 (and NSCP 2015) compute block shear strength as: Pn(block) = 0.6 Fu Agv + Fy Ant (rupture-dominated) or Pn(block) = 0.6 Fy Agv + Fu Ant (yield-dominated, less common) where: - Agv = gross shear area (sum of shear planes in tension direction) - Ant = net tension area (sum of tensile rupture planes) - Fu and Fy are stress values as before - φ = 0.75 (same as rupture, LRFD) The code prescribes: φPn(block shear) = 0.75 × min of the two equations above Simplified: typically φPn(block) = 0.75[0.6 Fu Agv + Fy Ant] (Note: Some older codes use the lesser of yield or rupture separately; modern AISC uses the combined formula above.) 5.3 WHEN BLOCK SHEAR MATTERS Block shear is most critical when: - Bolts are very close to the edge of the member (small edge distance) - The member is short and wide relative to bolt spacing (high Fy term dominates) - Connection angles have bolts near the outer edges - The tension member is an angle or tee (shapes with limited dimensions) For well-designed connections with standard edge distances (e.g., 1.5db to 2db), block shear often does not govern. However, it must always be checked in final design. 5.4 GEOMETRY DEFINITIONS - Shear area Agv = length of shear plane × thickness - For a tension member with bolts in a row, it is typically (distance along the member from first to last bolt + edge distance) × thickness - Tension area Ant = (net width perpendicular to shear plane) × thickness - After subtracting one hole at the failure plane Careful sketching of the failure plane is essential to identify which areas are in shear and which are in tension.

6.1 STEP-BY-STEP TENSION MEMBER DESIGN CHECKLIST When designing or checking a steel tension member, follow this systematic procedure: 1. DETERMINE FACTORED LOAD Pu - Apply LRFD load factors: Pu = 1.2D + 1.6L (typical combination) - Document the load case and factors 2. COMPUTE GROSS CROSS-SECTIONAL AREA Ag - For standard shapes, obtain from tables (e.g., AISC Manual of Steel Construction) - For custom shapes, calculate directly from dimensions 3. CHECK YIELD STRENGTH φPn(yield) = 0.90 × Fy × Ag Verify: φPn(yield) ≥ Pu 4. DETERMINE NET AREA An - Subtract gross hole areas: An = Ag – Σ(dh × t) - For staggered holes, examine all failure paths and use the minimum - Include gage-pitch credits: +Σ(s²/4g) for each zig-zag segment 5. APPLY SHEAR-LAG FACTOR U - U = 1.0 for bolted plates or full-section connections - U < 1.0 for bolted angles/tees/channels (connected through one element) - Compute: Ae = U × An 6. CHECK RUPTURE STRENGTH φPn(rupture) = 0.75 × Fu × Ae Verify: φPn(rupture) ≥ Pu 7. CHECK BLOCK SHEAR (if bolted) φPn(block) = 0.75 × [0.6 Fu Agv + Fy Ant] Verify: φPn(block) ≥ Pu 8. DETERMINE DESIGN STRENGTH φPn(design) = min(φPn(yield), φPn(rupture), φPn(block)) The lowest governs. 9. CHECK SLENDERNESS L/r ≤ 300 (recommended) If exceeded, document mitigating measures or accept with justification 10. VERIFY: φPn(design) ≥ Pu If not, increase member size (increase Ag or improve An/U) and repeat 6.2 TYPICAL GOVERNING MODES (a) YIELD GOVERNS: - Member has very few or no bolt holes - Ratio Fu/Fy is moderate (~1.6) - Gross area is controlling Result: Large capacity; member is strength-capable (b) RUPTURE GOVERNS: - Member has significant bolt holes and/or small effective net area - Shear lag reduces Ae substantially Result: Capacity drops due to An and U Common in angles, tees, and single-leg bolted shapes (c) BLOCK SHEAR GOVERNS (rare): - Bolts are very close to edges - Connection is compact; short span Result: Requires larger edge distances or more bolts 6.3 SIZING A TENSION MEMBER Given Pu (factored load), select a member (shape, size) such that φPn ≥ Pu. Iterative approach: 1. Assume a trial section from a handbook or design table 2. Compute Ag, An, and Ae using the assumed section 3. Compute φPn(yield) and φPn(rupture) 4. If φPn(design) < Pu, try a larger section and repeat 5. Once φPn(design) ≥ Pu, check slenderness and block shear 6. Finalize the selection

7.1 PITFALL 1: USING BOLT DIAMETER INSTEAD OF DESIGN HOLE DIAMETER Error: Subtracting db (bolt diameter) instead of dh (design hole diameter) from the gross area. Why: NSCP 2015 and AISC 360 specify that the hole diameter for design equals the bolt diameter plus an allowance for fabrication clearance (typically 2–3 mm, depending on bolt type and local practice). Fix: Always use dh = db + 2 or 3 mm (as specified in the problem or code table). If not stated, assume +2 mm for standard clearance. Example: - Bolt size: 20 mm - Hole diameter for design: dh = 20 + 2 = 22 mm (NOT 20 mm) - When subtracting: use 22 mm, not 20 mm Impact: Using db underestimates the area lost and overestimates capacity (unconservative). 7.2 PITFALL 2: FORGETTING TO CHECK BOTH LIMIT STATES Error: Computing only yielding OR rupture, but not both; selecting the lower without checking the higher. Why: LRFD design requires the lower of the two limit states. Neither can be omitted. Forgetting rupture is especially common in exams. Fix: Always compute: φPn(yield) = 0.90 × Fy × Ag φPn(rupture) = 0.75 × Fu × Ae Compare and select the lower (minimum) as the design strength. Example: A plate with few holes may have φPn(yield) = 500 kN and φPn(rupture) = 520 kN; yield (lower) governs. If rupture alone had been checked, a dangerous overestimate of 520 kN would result. 7.3 PITFALL 3: NOT APPLYING THE SHEAR-LAG FACTOR U Error: Using An directly for rupture without reducing by U; Or using U = 1.0 for an angle bolted through one leg. Why: Shear lag is a real phenomenon that reduces effective stress transfer. Angles bolted through one leg have U ≈ 0.80–0.85, not 1.0. Fix: Always ask: "Is the load transmitted uniformly to all cross-sectional elements?" If no (bolted angle, tee, channel through one part), reduce by U < 1.0. For flat plates bolted across their full width, U = 1.0. Impact: Omitting U overestimates rupture capacity (unconservative). For an angle, the error can be 10–20% on the high side. 7.4 PITFALL 4: INCORRECT GAGE-PITCH CREDIT FOR STAGGERED HOLES Error: Subtracting all hole areas without adding back the zig-zag credit s²/4g; or misidentifying which path is critical. Why: Staggered holes allow multiple failure paths. The designer must examine each path, apply the gage-pitch credit correctly, and use the minimum (most critical) net area. Fix: 1. Sketch all plausible failure paths (straight line, zig-zag to one side, zig-zag to the other). 2. For each path: Net width = Wg – (number of holes on path) × dh + (number of zig-zag segments) × (s²/4g) 3. Calculate An = net width × t for each path. 4. Use An(min) for rupture. Example: 3 holes staggered; path 1 (straight): An = 1,340 mm²; path 2 (zig-zag): An = 1,080 mm². Use 1,080 mm². Common mistake: Forgetting to add s²/4g for even one segment → underestimates An (too conservative, but wrong method). 7.5 PITFALL 5: WRONG PHI VALUES Error: Using φ = 0.75 for yield (should be 0.90) or φ = 0.90 for rupture (should be 0.75). Why: Yield is ductile (large margin before failure) → higher φ. Rupture is sudden (less margin) → lower φ. Fix: Memorize the values: Yield: φ = 0.90 (⇒ Ω = 1.67 for ASD) Rupture: φ = 0.75 (⇒ Ω = 2.00 for ASD) Block shear: φ = 0.75 Impact: Swapping the factors can reverse the governing limit state or produce an unconservative answer. 7.6 PITFALL 6: IGNORING SLENDERNESS OR MISAPPLYING THE LIMIT Error: Computing L/r and claiming it "fails" when L/r > 300; or not computing it at all. Why: L/r ≤ 300 is a RECOMMENDATION, not a mandatory code requirement. Tension members do not buckle. However, slenderness affects serviceability. Fix: Always compute L/r. If L/r ≤ 300, the member is acceptable (both strength and serviceability). If L/r > 300, document that slenderness is addressed by other means (e.g., intermediate tie, turnbuckle, acceptable sag under service load) or justify the exception. Example: A long tie rod (L/r = 450) may still be acceptable if the designer verifies that sag and vibration are controlled and acceptable to the owner. 7.7 PITFALL 7: FORGETTING BLOCK SHEAR Error: Checking yield and rupture but not block shear, particularly for bolted angles or short connections. Why: Block shear is a third limit state that must be checked for bolted members. It is less common than rupture but can govern in specific geometries (e.g., angle bolted close to edges). Fix: For any bolted tension member, compute: φPn(block shear) = 0.75 × [0.6 Fu Agv + Fy Ant] And compare to the other two limit states. Impact: Omitting the check risks an incomplete design; if block shear governs, the member is undersized. 7.8 PITFALL 8: MIXING UNITS OR USING INCONSISTENT CONVERSION Error: Entering dimensions in mm but stresses in ksi, or vice versa; not converting consistently to SI units. Why: NSCP 2015 specifies SI units (MPa, mm, kN). Mixing units introduces order-of-magnitude errors. Fix: Always work in SI throughout: Stress: MPa (not ksi) Length: mm (not inches) Force: N or kN (not lb or kip) Area: mm² (not in²) Example: Fy = 248 MPa (not 36 ksi) Ag = 1,500 mm² (not 2.33 in²) Load: 200 kN (not 45 kips) 7.9 PITFALL 9: ASSUMING ALL ELEMENTS OF A CROSS-SECTION ARE CONNECTED Error: Using U = 1.0 for a tee section bolted through its stem only; or for an angle bolted through one leg with bolts in only one hole. Why: If not all elements transmit load directly (e.g., flange of a tee only connected via the stem), shear lag reduces U. Fix: Carefully assess the geometry. Use U = 1.0 only if: - A flat plate is bolted across its full width - A solid round or rectangular section is bolted entirely - All legs/flanges are connected with bolts or welds For any other case (angle-one-leg bolted, tee-stem-only bolted, etc.), reduce U. 7.10 PITFALL 10: CHOOSING THE WRONG NET AREA PATH FOR STAGGERED HOLES Error: Computing multiple paths but selecting the maximum instead of the minimum; or missing a valid path. Why: The critical failure path is the one with the smallest (minimum) net area. It is often the zig-zag path, not the straight path. Fix: Always compute all likely paths: 1. Straight across (if applicable) 2. Zig-zag to the left 3. Zig-zag to the right 4. Other multi-segment paths if complex geometry Select the minimum An and use it for rupture. Example: Straight path = 1,600 mm², Zig-zag = 1,400 mm² → Use 1,400 mm² (smaller, more critical).

8.1 STANDARD BOLT AND HOLE SIZES (per NSCP 2015 and AISC 360) Bolt Diameter (mm) | Design Hole Diameter dh (mm) | Typical Clearance 12 | 14 | +2 16 | 18 | +2 20 | 22 | +2 24 | 26 | +2 27 | 30 | +3 (for larger bolts) 30 | 33 | +3 Note: Always verify the specific hole size per the project specification or drawing; values above are typical but not universal. 8.2 REDUCTION FACTORS U FOR COMMON SHAPES (Bolted Connections) Shape / Configuration | U Value | Notes Flat plate (bolted across width) | 1.00 | Load distributed uniformly Solid round or rectangular bar | 1.00 | Full cross-section engaged Angle (bolted one leg, 1 hole) | 0.80 | Conservative estimate; exact depends on geometry Angle (bolted one leg, 2+ holes) | 0.85 | Multi-hole reduces lag effect Angle (bolted both legs) | 1.00 | Full connectivity Welded angle (1 leg, length = leg width)| 0.85 | Reduced for partial weld Welded angle (both legs, full length) | 1.00 | Full connectivity Tee (bolted through stem) | 0.75 | Flanges lag; conservative Channel (bolted through web) | 0.80 | Flanges lag moderately For precise values, refer to AISC 360 Table D3.1 or NSCP 2015 equivalent. 8.3 YIELD AND ULTIMATE STRESSES FOR COMMON PHILIPPINE STEEL GRADES Steel Grade | Fy (MPa) | Fu (MPa) | Fu/Fy Ratio | Common Use Grade 248 (A36 equiv.) | 248 | 400 | 1.61 | General structural, bolted/welded Grade 350 (A992 equiv.) | 350 | 450 | 1.29 | Higher strength, often welded Grade 275 (A572 Gr. 50) | 275 | 450 | 1.64 | Intermediate grade Note: Exact values depend on thickness and product form; consult mill certs or ASEAN steel standards. NSCP 2015 often cites ASTM or Japanese standards adapted to SI units. 8.4 QUICK STRENGTH CALCULATION FORMULAS For a simple plate bolted with n holes (straight-line failure): Yield: φPn = 0.90 Fy (W t) Rupture: φPn = 0.75 Fu [(W – n dh) t] (where W = plate width, t = thickness) For a round rod (no holes): φPn = 0.90 Fy (π d²/4) ≈ 0.71 Fy d² (d in mm, result in N) For an angle (bolted one leg): φPn(rupture) = 0.75 Fu [(width – holes) × t × U] U ≈ 0.80–0.85 (use appropriate value) 8.5 SLENDERNESS QUICK CHECK Member Type | L/r Recommendation | Typical Diameter/Thickness Ratio for L/r = 300 Round rod, L = 3 m| ≤ 300 | d ≥ 3000/300 ÷ 4 = 25 mm Round rod, L = 5 m| ≤ 300 | d ≥ 5000/300 ÷ 4 = 41.7 mm ≈ 45 mm Flat plate, L = 3 m| ≤ 300 | t ≥ 3000 × √12 / 300 ≈ 34.6 mm Angle, L = 4 m | ≤ 300 | Use r from handbook; check directly (Note: These are approximate; always verify with actual section properties.) 8.6 DESIGN SUMMARY TABLE: WHEN EACH LIMIT STATE GOVERNS Limit State | Typical Condition | Governing φ | Design Strength Yielding | Few/no bolt holes; large gross area | 0.90 | 0.90 Fy Ag Rupture | Multiple bolt holes; reduced net area; U<1 | 0.75 | 0.75 Fu Ae = 0.75 Fu U An Block Shear | Bolts very close to edges; compact shape | 0.75 | 0.75(0.6 Fu Agv + Fy Ant) Slenderness | Long span; small cross-section; L/r > 300 | N/A | Serviceability concern; redesign or justify