Steel Tension Members - Study Notes
Steel tension members are structural elements that carry axial tensile forces without bending or buckling effects. Common applications include truss ties, bracing members, sag rods, and hangers in buildings and bridges. Unlike compression members, tension members do not face instability concerns, making their design more straightforward. However, the presence of bolt holes at connection points introduces a critical failure mode: rupture on the net section. According to NSCP 2015 (which adopts AISC 360 Load and Resistance Factor Design principles), tension member design must satisfy two distinct limit states: yielding on the gross cross-sectional area and rupture on the net effective area. The lower of these two strength predictions governs the member's design capacity. This chapter equips PRC Civil Engineer Licensure Examination reviewees with the essential knowledge to accurately compute net areas (including staggered bolt patterns), apply the shear-lag reduction factor, and select members that meet both strength and serviceability requirements.
1. Fundamental Principles and Limit States
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.
Examples
A steel rod 20 mm diameter, Grade 248 (Fy = 248 MPa, Fu = 400 MPa), with no holes. Compute the design strength using LRFD.
Solution
Step 1: Compute gross area.
Ag = π(20)²/4 = 314.16 mm²
Step 2: Check yielding.
Pₙ(yield) = Fy Ag = 248 × 314.16 = 77,911 N
φPₙ(yield) = 0.90 × 77,911 = 70,120 N ≈ 70.1 kN
Step 3: Check rupture (An = Ag for rod with no holes).
Ae = 1.0 × 314.16 = 314.16 mm² (U = 1.0 for solid round bar)
Pₙ(rupture) = Fu Ae = 400 × 314.16 = 125,664 N
φPₙ(rupture) = 0.75 × 125,664 = 94,248 N ≈ 94.2 kN
Step 4: Select the lower (governing) strength.
Design Strength = min(70.1, 94.2) = 70.1 kN → YIELDING GOVERNS
Conclusion: The 20 mm rod can safely carry 70.1 kN in tension (solid section with no holes, yielding is the controlling limit state).
A flat plate 150 mm × 10 mm, Fy = 248 MPa, Fu = 400 MPa, bolted with two 16 mm bolts. Hole diameter dh = 18 mm. Compute design strength.
Solution
Step 1: Gross area.
Ag = 150 × 10 = 1,500 mm²
Step 2: Yielding.
φPₙ(yield) = 0.90 × 248 × 1,500 = 334,800 N = 334.8 kN
Step 3: Net area and rupture.
An = Ag – Σ(dh × t) = 1,500 – 2(18)(10) = 1,500 – 360 = 1,140 mm²
(For bolted connection through full width, U = 1.0)
Ae = 1.0 × 1,140 = 1,140 mm²
φPₙ(rupture) = 0.75 × 400 × 1,140 = 342,000 N = 342.0 kN
Step 4: Compare.
Design Strength = min(334.8, 342.0) = 334.8 kN → YIELDING GOVERNS (narrow margin!)
Remark: With only two bolt holes, the gross area is so large that yielding barely wins. If holes were smaller or fewer, yielding would dominate more clearly. The rupture capacity is close because Fu is much higher than Fy, partially offsetting the loss of area.
Key points
- LRFD combines safety factors into the resistance side: φ Pn ≥ Pu (load factors already applied to loads)
- Yielding (φ = 0.90) is ductile with warning; rupture (φ = 0.75) is brittle and sudden
- The lower design strength (yield or rupture) governs—both must be checked every time
- NSCP 2015 adopted AISC 360 directly; Philippine practice uses SI units but identical methodology
- Typical steel grades: Fy = 248 MPa (Grade 248) or 350 MPa (Grade 350); Fu ≈ 1.6–1.7 × Fy
2. Net Area Computation
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:
- Identify all possible failure paths (usually 1, 2, or 3 segments).
- For each path, sum the holes crossed and apply the gage-pitch correction.
- Compute An for each path; use the MINIMUM.
- 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.
Examples
A plate 250 mm wide, 12 mm thick, has four 20 mm bolts (dh = 22 mm) in a single line across the width. Find the net area.
Solution
Gross area: Ag = 250 × 12 = 3,000 mm²
Hole area: Σ(dh × t) = 4 × 22 × 12 = 1,056 mm²
Net area: An = 3,000 – 1,056 = 1,944 mm²
Verification: Remaining width = 250 – 4(22) = 250 – 88 = 162 mm
An = 162 × 12 = 1,944 mm² ✓
A plate 200 mm × 10 mm has six 20 mm bolts arranged in two rows (gage g = 100 mm apart) with staggered pitch s = 60 mm. Holes in both rows. Find the critical net area.
[Diagram: Row 1 at top with holes at 0, 60, 120 mm; Row 2 at bottom offset by 30 mm]
Solution
Gross area: Ag = 200 × 10 = 2,000 mm²
Path 1 (straight across Row 1): Crosses 3 holes
Net width = 200 – 3(22) = 200 – 66 = 134 mm
An1 = 134 × 10 = 1,340 mm²
Path 2 (zig-zag: Row 1 hole at 0 mm → diagonal to Row 2 hole at 30 mm → diagonal to Row 1 hole at 60 mm → diagonal to Row 2 hole at 90 mm → diagonal to Row 1 hole at 120 mm):
Crosses 5 holes total with 2 staggered segments
Net width = 200 – 5(22) + 2 × (60²)/(4 × 100)
= 200 – 110 + 2 × (3,600/400)
= 200 – 110 + 2 × 9
= 200 – 110 + 18
= 108 mm
An2 = 108 × 10 = 1,080 mm²
Critical net area: An(min) = min(1,340, 1,080) = 1,080 mm²
Design uses An = 1,080 mm² (zig-zag path governs due to more holes crossed despite the gage-pitch credit).
Plate 180 mm × 8 mm with two 18 mm bolts staggered: one at 0 mm, the other at s = 50 mm along a gage g = 80 mm. Find net area.
Solution
Gross area: Ag = 180 × 8 = 1,440 mm²
Path 1 (straight): If both holes aligned vertically, An = (180 – 2×18) × 8 = 144 × 8 = 1,152 mm²
Path 2 (zig-zag through both holes diagonally):
Net width = 180 – 2(18) + (50²)/(4×80)
= 180 – 36 + 2,500/320
= 180 – 36 + 7.8125
= 151.81 mm
An2 = 151.81 × 8 = 1,214.5 mm²
Critical net area: An = min(1,152, 1,214.5) = 1,152 mm² (straight path is slightly tighter)
Note: The zig-zag actually improved capacity slightly here, but the straight path is still critical because it is the minimum.
Key points
- Net area = Ag – hole areas; use design hole diameter dh = bolt diameter + 2–3 mm
- For staggered holes, examine multiple failure paths and select the minimum net area
- The zig-zag correction +s²/4g per segment accounts for diagonal path efficiency
- If multiple paths exist (common in riveted/bolted connections), tabulate each and take the minimum
- If no holes along a section, An = Ag (solid member)
- Net area always ≤ gross area; the difference is the total projected hole area minus zig-zag credits
3. Shear-Lag Reduction Factor (U)
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.
Examples
An L75×75×8 angle (Ag = 1,150 mm², x̄ = 20.6 mm from the centroid to the unconnected leg) is bolted through one leg with two 16 mm bolts spaced 80 mm apart (bolt pattern length Lc = 80 mm). Fy = 248 MPa, Fu = 400 MPa. Find the design strength.
Solution
Step 1: Gross area and yielding.
Ag = 1,150 mm²
φPn(yield) = 0.90 × 248 × 1,150 = 256,620 N ≈ 256.6 kN
Step 2: Net area (one hole per bolt, dh = 18 mm, at the connection edge).
Assuming the bolts are in the connected leg, the failure path crosses both holes:
An = 1,150 – 2(18)(8) = 1,150 – 288 = 862 mm²
(Thickness t = 8 mm for angle)
Step 3: Shear-lag factor U.
U = 1 – x̄/Lc = 1 – 20.6/80 = 1 – 0.2575 = 0.7425 ≈ 0.74
Ae = 0.74 × 862 = 638 mm²
Step 4: Rupture check.
φPn(rupture) = 0.75 × 400 × 638 = 191,400 N ≈ 191.4 kN
Step 5: Compare limit states.
Design Strength = min(256.6, 191.4) = 191.4 kN → RUPTURE GOVERNS (shear lag and bolt holes drastically reduce capacity)
Conclusion: The angle can safely carry 191.4 kN. The shear-lag factor U = 0.74 significantly reduced the effective net area from 862 to 638 mm², making rupture the governing mode.
A plate 200 × 12 mm (Grade 248, Fu = 400 MPa) bolted entirely across its width with two 20 mm bolts (dh = 22 mm). Find design strength; compare to the angle in Example 3.1.
Solution
Step 1: Gross area and yielding.
Ag = 200 × 12 = 2,400 mm²
φPn(yield) = 0.90 × 248 × 2,400 = 535,680 N ≈ 535.7 kN
Step 2: Net area.
An = 2,400 – 2(22)(12) = 2,400 – 528 = 1,872 mm²
Step 3: Shear-lag factor.
U = 1.0 (plate bolted across full width, no lagging element)
Ae = 1.0 × 1,872 = 1,872 mm²
Step 4: Rupture check.
φPn(rupture) = 0.75 × 400 × 1,872 = 561,600 N ≈ 561.6 kN
Step 5: Compare.
Design Strength = min(535.7, 561.6) = 535.7 kN → YIELDING GOVERNS
Conclusion: The plate carries 535.7 kN (nearly 3× the angle from Example 3.1!). The full-width bolting and higher gross area (2,400 vs. 1,150 mm²) dominate. No shear lag (U = 1.0) keeps the effective net area high. For a plate, yielding governs when holes are modest relative to width.
Key points
- U reduces net area to account for shear lag (unequal stress distribution in the connection region)
- U < 1.0 for angles/tees/channels bolted through one element; U = 1.0 for plates and full-section connections
- U depends on member geometry (distance to centroid) and bolt pattern length
- Common angle-one-leg bolted: U = 0.80–0.85; exact value from code tables or formula 1 – x̄/Lc
- Ae = U × An is essential for rupture strength; omitting U underestimates rupture capacity and is unconservative
4. Slenderness and Serviceability
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.).
Examples
A 4 m long tie rod (diameter d = 25 mm) is used in a building bracing system. Check whether the recommended slenderness limit is satisfied.
Solution
Step 1: Radius of gyration for solid round rod.
r = d/4 = 25/4 = 6.25 mm
Step 2: Slenderness ratio.
L/r = 4,000 / 6.25 = 640
Step 3: Compare to recommendation.
640 > 300 → Does NOT satisfy the slenderness recommendation
Step 4: Assessment.
The tie rod is excessively slender. It will sag noticeably under its own weight and may vibrate in wind.
Remedies:
- Increase diameter: For L/r = 300, d = 4(4,000)/300 = 53.3 mm (requires change to design)
- Install mid-span stiffening (e.g., cable tie, turnbuckle) to reduce effective length
- Accept the slenderness if aesthetics and serviceability concerns are addressed (design stiffener)
- Document in design notes that serviceability is secondary to strength here
Conclusion: If strength alone governs, the 25 mm rod may be acceptable if rupture/yield checks pass. However, a 50+ mm rod is strongly recommended for this 4 m span.
Size a round tie rod for a 6 m span, P = 150 kN, Fy = 248 MPa. Ensure L/r ≤ 300.
Solution
Step 1: Strength requirement.
φPn ≥ Pu
0.90 × 248 × Ag ≥ 150,000 N
Ag ≥ 150,000 / (0.90 × 248) = 671.1 mm²
For a solid round rod:
Ag = π d²/4 ≥ 671.1
d² ≥ 2,684.4 / π = 854.6
d ≥ 29.2 mm
Try d = 30 mm: Ag = π(30)²/4 = 706.9 mm² ✓
Step 2: Slenderness check.
r = 30/4 = 7.5 mm
L/r = 6,000 / 7.5 = 800
800 > 300 → Not acceptable.
For L/r = 300:
d ≥ 4(6,000) / 300 = 80 mm
Try d = 80 mm: Ag = π(80)²/4 = 5,027 mm²
Strength: φPn = 0.90 × 248 × 5,027 = 1,124 kN >> 150 kN ✓
Slenderness: L/r = 6,000 / 20 = 300 ✓
Conclusion: A 80 mm diameter rod satisfies both strength and slenderness. The large diameter (driven by slenderness, not strength) is necessary to keep the ratio below 300 over a 6 m span.
Key points
- Slenderness L/r ≤ 300 is recommended for tension members to avoid sag and vibration, not a strength requirement
- Tension members do not fail by buckling; the limit addresses serviceability and durability
- For round rods: d ≥ L/75 to satisfy L/r ≤ 300
- Always compute L/r and verify it satisfies the recommendation; if not, document the reason (e.g., unavoidable geometry) in the design
- Even if L/r > 300, the member must still pass strength checks (yield and rupture); serviceability is secondary to safety
5. Block Shear Failure
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.
Examples
An L75×75×8 angle (from Example 3.1) is bolted through one leg with two 16 mm bolts as follows:
- Bolt spacing along the leg: 100 mm (distance between centerlines)
- Edge distance (from end of angle to first bolt centerline): 40 mm
- Edge distance from outer edge to last bolt: 40 mm
- Thickness t = 8 mm
- Fy = 248 MPa, Fu = 400 MPa, dh = 18 mm
Check whether block shear is critical.
Solution
Step 1: Identify the block shear failure path.
The bolts are in one line along the length of the leg. The block shear failure plane is along the outer edge of the leg (longitudinal shear) and across the width at the bolt row (transverse tension).
Shear plane (along leg edge):
Length = edge distance + bolt spacing + edge distance = 40 + 100 + 40 = 180 mm
Area Agv = 180 × 8 = 1,440 mm²
Tension plane (across perpendicular direction, net of one hole):
Gross width perpendicular to shear ≈ 75 mm (leg width)
Width after subtracting one hole: 75 – 18 = 57 mm
Area Ant = 57 × 8 = 456 mm²
Step 2: Compute block shear strength (NSCP 2015 / AISC).
Pn(block) = 0.6 Fu Agv + Fy Ant
= 0.6(400)(1,440) + 248(456)
= 345,600 + 113,088
= 458,688 N
Design strength:
φPn(block) = 0.75 × 458,688 = 344,016 N ≈ 344.0 kN
Step 3: Compare to rupture from Example 3.1.
Rupture check gave φPn(rupture) = 191.4 kN
Block shear (344.0 kN) >> Rupture (191.4 kN)
Conclusion: Block shear does NOT govern; rupture (governed by shear lag and net area loss at the bolts) remains the limiting failure mode. For this compact angle connection, the bolts remove enough material in direct tension that rupture dominates before block shear can develop.
Key points
- Block shear combines tensile rupture on one plane and shear yield/rupture on perpendicular planes
- φPn(block) = 0.75[0.6 Fu Agv + Fy Ant] (NSCP 2015 / AISC 360)
- Critical for angles, tees, and short-span connections; less common for long, well-proportioned plates
- Always check block shear in addition to yield, rupture, and slenderness
- Edge distance and bolt spacing must be adequate to avoid block shear governing
6. Design Procedure and Examples
6.1 STEP-BY-STEP TENSION MEMBER DESIGN CHECKLIST
When designing or checking a steel tension member, follow this systematic procedure:
DETERMINE FACTORED LOAD Pu
- Apply LRFD load factors: Pu = 1.2D + 1.6L (typical combination)
- Document the load case and factors
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
- CHECK YIELD STRENGTH
φPn(yield) = 0.90 × Fy × Ag
Verify: φPn(yield) ≥ Pu 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
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
- CHECK RUPTURE STRENGTH
φPn(rupture) = 0.75 × Fu × Ae
Verify: φPn(rupture) ≥ Pu - CHECK BLOCK SHEAR (if bolted)
φPn(block) = 0.75 × [0.6 Fu Agv + Fy Ant]
Verify: φPn(block) ≥ Pu - DETERMINE DESIGN STRENGTH
φPn(design) = min(φPn(yield), φPn(rupture), φPn(block))
The lowest governs. - CHECK SLENDERNESS
L/r ≤ 300 (recommended)
If exceeded, document mitigating measures or accept with justification - 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:
- Assume a trial section from a handbook or design table
- Compute Ag, An, and Ae using the assumed section
- Compute φPn(yield) and φPn(rupture)
- If φPn(design) < Pu, try a larger section and repeat
- Once φPn(design) ≥ Pu, check slenderness and block shear
- Finalize the selection
Examples
Design a solid round steel tie rod for a 5 m span to carry a factored tensile load Pu = 200 kN. Fy = 248 MPa, Fu = 400 MPa, no bolts (solid section). Check all limit states and slenderness.
Solution
Step 1: Yield strength estimate.
For yielding alone: Ag ≥ Pu / (0.90 × Fy) = 200,000 / (0.90 × 248) ≈ 895 mm²
For round rod: d ≈ √(4 × 895 / π) ≈ 33.8 mm → Try d = 35 mm
Step 2: Check yield with d = 35 mm.
Ag = π(35)²/4 = 962.1 mm²
φPn(yield) = 0.90 × 248 × 962.1 = 214,628 N ≈ 214.6 kN ✓ (> 200 kN)
Step 3: Rupture (no holes, An = Ag; U = 1.0).
Ae = 962.1 mm²
φPn(rupture) = 0.75 × 400 × 962.1 = 288,630 N ≈ 288.6 kN ✓ (>> 200 kN)
Step 4: Block shear (no bolts, not applicable).
Step 5: Design strength.
φPn(design) = min(214.6, 288.6) = 214.6 kN (yield governs) ✓
Step 6: Slenderness.
r = 35/4 = 8.75 mm
L/r = 5,000 / 8.75 = 571 >> 300
Slenderness is EXCESSIVE. Design mitigations: Install intermediate cable tie or turnbuckle to reduce effective span; or increase diameter.
To satisfy L/r ≤ 300:
d ≥ 4(5,000) / 300 = 66.7 mm → Try d = 70 mm
Check: L/r = 5,000 / (70/4) = 5,000 / 17.5 ≈ 286 ✓
Final selection: d = 70 mm (satisfies slenderness)
Capacity: φPn(yield) = 0.90 × 248 × π(70)²/4 = 0.90 × 248 × 3,848 ≈ 858 kN >> 200 kN ✓
Conclusion: Use a 70 mm diameter rod; sized by slenderness, not by strength. The 200 kN load is easily resisted by much smaller diameters, but the 5 m span requires the larger diameter to prevent excessive sag and vibration.
Design a steel plate (Fy = 248 MPa, Fu = 400 MPa) for a factored load Pu = 400 kN. Bolted connection through the full width with two 20 mm bolts (dh = 22 mm). Assume U = 1.0 and no block shear concern (edge distances adequate). Find the required plate dimensions.
Solution
Step 1: Trial size for yield.
Ag ≥ 400,000 / (0.90 × 248) ≈ 1,792 mm²
Try a plate 200 × 10 mm: Ag = 2,000 mm² (> 1,792 ✓)
Step 2: Check yield.
φPn(yield) = 0.90 × 248 × 2,000 = 446,400 N ≈ 446.4 kN ✓ (> 400 kN)
Step 3: Check rupture.
An = 2,000 – 2(22)(10) = 2,000 – 440 = 1,560 mm²
Ae = 1.0 × 1,560 = 1,560 mm²
φPn(rupture) = 0.75 × 400 × 1,560 = 468,000 N ≈ 468.0 kN ✓ (> 400 kN)
Step 4: Design strength.
φPn(design) = min(446.4, 468.0) = 446.4 kN ✓ >> 400 kN
Step 5: Slenderness (assume L = 3 m span, least r = t/√12 = 10/√12 ≈ 2.89 mm).
L/r = 3,000 / 2.89 ≈ 1,038 >> 300
Slenderness is unacceptable for typical spans. Increase thickness:
For L = 3 m and L/r = 300: t ≥ 3,000 × √12 / 300 ≈ 34.6 mm
Try plate 200 × 35 mm:
Ag = 200 × 35 = 7,000 mm²
φPn(yield) = 0.90 × 248 × 7,000 ≈ 1,560.0 kN >> 400 kN (very conservative on strength)
L/r = 3,000 / (35/√12) = 3,000 / 10.1 ≈ 297 ✓
Alternatively, reduce span or use a different shape (e.g., angle, tube) if plate is too heavy.
Final selection: Plate 200 × 35 mm (satisfies strength, rupture, and slenderness for 3 m span)
Or: Document that for shorter spans (< 2 m), a smaller plate like 200 × 10 or 200 × 12 mm is acceptable.
An L 100 × 100 × 10 angle (Ag = 1,900 mm², assume x̄ = 28.5 mm) is bolted through one leg with four 18 mm bolts (dh = 20 mm) spaced 120 mm apart (Lc = 360 mm). Fy = 248 MPa, Fu = 400 MPa. Find the design strength and compare to a solid (unwelded) angle.
Solution
Step 1: Yielding.
φPn(yield) = 0.90 × 248 × 1,900 = 424,680 N ≈ 424.7 kN
Step 2: Net area and shear lag.
An = 1,900 – 4(20)(10) = 1,900 – 800 = 1,100 mm²
(Thickness t = 10 mm for the angle leg)
U = 1 – 28.5/360 = 1 – 0.0792 ≈ 0.921
Ae = 0.921 × 1,100 = 1,013 mm²
Step 3: Rupture.
φPn(rupture) = 0.75 × 400 × 1,013 = 303,900 N ≈ 303.9 kN
Step 4: Design strength.
φPn(design) = min(424.7, 303.9) = 303.9 kN (rupture governs)
Comparison to solid angle (Pu = 303.9 kN):
Bolted: 303.9 kN (limited by rupture due to holes and shear lag)
If fully welded across both legs (no holes, U = 1.0): φPn ≈ 0.75 × 400 × 1,900 = 570 kN (1.9× higher)
Conclusion: The bolted connection reduces capacity by ~50% due to net area loss (4 holes) and shear lag (U < 1.0). If full strength is needed, consider full-penetration welds or a different connection detail.
Key points
- Always check all three limit states: yield, rupture, and block shear (if bolted)
- The lowest of the three limit states governs; design strength is the minimum
- Common sequence: Yield >> Rupture (governs) >> Block Shear (seldom governs if bolts are well-placed)
- Factored load Pu is established from load combinations (LRFD factors); design must provide φPn ≥ Pu
- Slenderness check is recommended, not mandatory; serviceability concern more than safety
7. Common Exam Pitfalls and Solutions
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:
- Sketch all plausible failure paths (straight line, zig-zag to one side, zig-zag to the other).
- For each path:
Net width = Wg – (number of holes on path) × dh + (number of zig-zag segments) × (s²/4g) - Calculate An = net width × t for each path.
- 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:
- Straight across (if applicable)
- Zig-zag to the left
- Zig-zag to the right
- 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).
Key points
- Use dh (bolt diameter + 2–3 mm clearance) for hole diameter, not bolt diameter alone
- Always check yield AND rupture; take the lower design strength
- Apply shear-lag factor U for bolted angles/tees/channels; U < 1.0 unless all elements are connected
- For staggered holes, examine all failure paths, apply s²/4g credit, and use the minimum net area
- Yield: φ = 0.90; Rupture: φ = 0.75; Block shear: φ = 0.75 (memorize these)
- Slenderness L/r ≤ 300 is recommended, not mandatory; tension members do not buckle
- Always check block shear for bolted connections, especially angles and tees
- Work consistently in SI units (MPa, mm, kN) throughout the design
- U = 1.0 only for fully-connected sections (plates across width, solid bars, full-area connections); U < 1.0 for partial connections
- For staggered holes, the minimum (most critical) net area governs, often the zig-zag path
8. Practical Design Tables and Quick Reference
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 |
Key points
- Always cross-check bolt and hole sizes against the NSCP 2015 or AISC 360 tables; typical clearance is +2 to +3 mm
- Shear-lag factor U varies with shape and connection geometry; 0.75–1.0 is typical
- Philippine steel grades (248, 350 MPa) have Fu/Fy ratios in the range 1.3–1.6; higher-strength steel has lower ratio
- Use the quick-reference formulas only for verification; always show full calculations in exam work
- Slenderness limits are serviceability-driven; always compute L/r even if not mandatory for strength
- The governing limit state depends on member and connection geometry; always check all three (yield, rupture, block shear)
Diagrams
Steel Tension Member Design Decision Tree
A comprehensive flowchart showing the step-by-step decision process for designing a steel tension member. It guides reviewees through load determination, trial section selection, checks for yield, rupture, and block shear limit states, and finally slenderness verification. The chart shows iteration loops when any limit state fails, emphasizing that all checks must pass before design completion.
Steel Tension Member - Core Concepts Mind Map
A comprehensive mind map organizing the key concepts of steel tension member design. The central hub connects five major branches: Limit States (yielding, rupture, block shear), Net Area Calculation methods, Strength Checks, Serviceability concerns, and Common Design Errors. Each branch further subdivides into specific details, allowing reviewees to see relationships between concepts and understand the hierarchical structure of tension member design.
Failure Modes in Tension Members - State Transitions
A state diagram illustrating the progression of tension member behavior from applied load through to failure. The diagram shows three distinct failure pathways: gross yielding (ductile, with warning), rupture at net section (sudden, brittle), and block shear (sudden, at bolted regions). Color coding differentiates between undesirable sudden fractures (red) and acceptable ductile yielding (yellow/green), reinforcing the preference for yield-governed designs.
Typical Distribution of Failure Modes in PRC Exam Questions
A pie chart showing the approximate frequency of different failure modes tested in PRC Civil Engineer Licensure Examination questions on tension members. Rupture (45%) is the most common, reflecting the prevalence of bolted connections and the need to carefully compute net areas and shear-lag factors. Yielding (30%) follows, while block shear and slenderness/serviceability are tested less frequently but remain important for comprehensive preparation.
In summary
Steel tension members are fundamental structural elements that transmit axial tensile forces. Their design, governed by NSCP 2015 and AISC 360, requires verification of two primary limit states—yielding on the gross cross-section and rupture on the effective net section—plus a check on block shear for bolted connections. The design strength is the lower of these three, ensuring both ductility and safety. Key challenges include accurate net-area computation (especially for staggered holes with gage-pitch corrections), proper application of the shear-lag reduction factor U for non-uniform load distribution, and awareness of common pitfalls such as using incorrect bolt-to-hole conversions or omitting required checks. Slenderness (L/r ≤ 300) is recommended to minimize sag and vibration, though not mandatory for strength. A systematic, step-by-step design procedure—establishing factored loads, computing gross and net areas, applying reduction factors, checking all limit states, and verifying slenderness—ensures reliable and code-compliant designs. Exam success in this topic depends on disciplined application of formulas, careful attention to unit consistency (SI units throughout), and thorough examination of all failure modes before arriving at a final member selection.