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CELE Steel & Timber Design — Steel Tension MembersDetailed Explanation

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Detailed explanation of Steel Tension Members for the CELE 2026. Full depth, full reasoning — exactly what you need when Professional Regulation Commission (PRC) — Board of Civil Engineering tests this chapter with applied or scenario-based questions in the CELE Steel & Timber Design subtest.

Exam context

Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Steel & Timber Design section sits under a "Core" weighting, and Steel Tension Members is the 1st chapter in the 5-chapter CELE Steel & Timber Design rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Steel & Timber Design.

Steel Tension Members - Detailed Explanation

Tension members are among the most fundamental structural steel elements encountered in civil engineering practice and in the PRC Civil Engineer Licensure Examination. They appear as truss bottom chords, diagonal braces, hanger rods, sag rods, cable stays, and tie rods — any member whose primary action is to resist a pulling (tensile) force along its longitudinal axis. Unlike compression members, tension members do not buckle; the governing failure modes are (1) gradual yielding across the gross cross-section and (2) sudden fracture through the reduced (net) cross-section at bolt holes or other discontinuities. NSCP 2015, which adopts AISC 360 by reference, codifies both limit states under the Load and Resistance Factor Design (LRFD) and Allowable Strength Design (ASD) frameworks. This chapter systematically develops every concept tested on the board exam — from basic gross- and net-area computations, through staggered-hole geometry, the shear-lag factor U, and block shear — with fully worked numerical examples in SI units and explicit identification of common board-exam traps.

Concepts

Definition and Classification of Tension Members

A tension member is any structural element loaded exclusively (or predominantly) by a direct tensile axial force. Because the stress state is uniform tension, there is no risk of global or local buckling, making tension members the simplest steel elements to analyze. Common tension-member types in Philippine construction: • Truss members — bottom chord, vertical and diagonal ties in roof trusses of industrial buildings. • Bracing members — diagonal bracing in SMRF or CBF lateral systems (NSCP 2015 Section 505). • Hanger rods — suspending floor beams from above in transfer structures or bridges. • Sag rods — intermediate supports for purlins that prevent them from sagging under gravity load components. • Eye bars and cable stays — used in long-span bridges and stadium roofs. Cross-sections used for tension members include: flat plates, round/square solid bars, structural angles (L-shapes), structural tees (WT), double angles (2L), wide-flange sections (W), channels (C), and hollow structural sections (HSS). The choice depends on connection geometry and available rolled shapes. Under NSCP 2015 LRFD, the required strength Pu (factored tensile force from load combinations of Section 203) must satisfy: φt Pn ≥ Pu where Pn is the nominal tensile strength and φt is the appropriate resistance factor.

Examples

All three limit states must be checked for any bolted tension member. Shear lag is critical for angles connected through one leg — a very common board-exam scenario.

Scenario

Identify the governing member type: A roof truss diagonal carries a factored tensile load Pu = 320 kN. The designer proposes a single L75×75×8 angle connected through one leg with 3 bolts. What limit states must be verified?

Solution

Three limit states must be checked: 1. Tensile yielding on the gross area: φPn = 0.90 Fy Ag 2. Tensile rupture on the effective net area: φPn = 0.75 Fu Ae, where Ae = U·An (U < 1.0 because only one leg is connected — shear lag applies). 3. Block shear rupture at the bolt group (AISC 360 Section J4.3). The lowest of the three design strengths is compared with Pu = 320 kN.

Applications

Roof truss design for industrial warehouses and gymnasiums throughout the Philippines.
Lateral bracing design for buildings in Seismic Zone 4 (Metro Manila, Visayas, Mindanao).
Hanger design for pedestrian bridges (foot bridges over irrigation canals).
Sag rod design for long purlin spans in corrugated-metal-roofed structures.
Tie rod design in pre-engineered metal buildings commonly used for commercial and industrial projects.

Misconceptions

Misconception: The same φ = 0.90 applies to both yielding and rupture. FACT: φ = 0.90 for yield, φ = 0.75 for rupture.
Misconception: Tension members never fail suddenly. FACT: Rupture on the net section is sudden and catastrophic; hence the more conservative φ = 0.75.
Misconception: A slenderness limit L/r ≤ 300 is a mandatory code requirement. FACT: It is a recommendation (preferred limit), not a prescriptive code mandate for tension members.

Related Concepts

Gross area (Ag)
Net area (An)
Effective net area (Ae)
Shear-lag factor (U)
Block shear rupture
Bolt hole geometry

Common Exam Questions

Example

A double-angle bottom chord of a Pratt truss — both legs are connected to a gusset plate, so U = 1.0.

Approach

Be able to classify a member as a tension member from the given structural description or free-body diagram. Note whether all cross-sectional elements participate in force transfer (affects U).

Question Type

Identification

Example

Under NSCP 2015, what resistance factor applies to tensile rupture? Answer: φt = 0.75.

Approach

Know which code provisions govern: NSCP 2015 Section 502 (LRFD basics), AISC 360 Chapter D (tension members), AISC 360 Section J4.3 (block shear).

Question Type

Code citation

Key Points To Remember

Tension members resist pulling forces; no buckling occurs, so only strength (yielding and rupture) and serviceability (slenderness) are checked.
NSCP 2015 adopts AISC 360 provisions for steel design; all tension member checks are in AISC 360 Chapter D.
Two LRFD resistance factors apply: φt = 0.90 for yielding, φt = 0.75 for rupture — these are different and must not be interchanged.
Common section types: plates, angles, double angles, W-shapes, HSS. Select based on connection requirements.
In ASD: Ω = 1.67 (yield), Ω = 2.00 (rupture); Pa ≤ Pn/Ω.

Two Limit States: Tensile Yielding and Tensile Rupture

NSCP 2015 (AISC 360-16, Section D2) specifies that the design tensile strength φtPn is the lesser value from the two limit states: ────────────────────────────────────────────────── LIMIT STATE 1 — TENSILE YIELDING ON THE GROSS SECTION ────────────────────────────────────────────────── Pn = Fy · Ag φt = 0.90 (LRFD) Ω = 1.67 (ASD) This represents gradual, ductile plastic elongation of the entire cross-section remote from the holes. Because the yielding zone is spread over the full member length, the member gives visible warning before failure — the reason for the less-severe penalty (φ = 0.90). ────────────────────────────────────────────────── LIMIT STATE 2 — TENSILE RUPTURE ON THE NET SECTION ────────────────────────────────────────────────── Pn = Fu · Ae φt = 0.75 (LRFD) Ω = 2.00 (ASD) This represents sudden, brittle fracture localised at the bolt holes where the cross-section is reduced. Although Fu > Fy, the effective area Ae < Ag and the resistance factor is only 0.75, making rupture often (though not always) the controlling limit state for heavily connected members. Design Rule: The GOVERNING design strength is: φtPn = min(0.90 Fy Ag, 0.75 Fu Ae) Practical Insight: For most A36 steel (Fy = 248 MPa, Fu = 400 MPa): • The ratio Fu/Fy = 400/248 ≈ 1.61, but the ratio (0.75 Fu)/(0.90 Fy) ≈ 1.35. • So rupture governs only when Ae/Ag < 1/1.35 ≈ 0.74. • If the holes remove less than about 26% of the gross area (and U is near 1.0), yielding controls. For A572 Gr.50 (Fy = 345 MPa, Fu = 450 MPa): • (0.75 Fu)/(0.90 Fy) = 337.5/310.5 ≈ 1.09, meaning rupture governs more easily.

Examples

Even though 3 bolts are used, only ONE hole is deducted for the critical cross-section (the section through the first bolt hole, perpendicular to the load). Yielding governs because the net section loss is relatively small (only 9.6% of Ag removed).

Scenario

BOARD-TYPE PROBLEM. A PL250×16 mm tension plate (A36: Fy = 248 MPa, Fu = 400 MPa) is connected with three 22-mm-diameter bolts in a single row perpendicular to the force. Standard hole diameter = bolt diameter + 2 mm = 24 mm. U = 1.0. Determine the LRFD design tensile strength.

Solution

STEP 1 — Gross area: Ag = 250 × 16 = 4 000 mm² STEP 2 — Net area (one hole deducted at critical section): An = Ag − dh × t = 4 000 − 24 × 16 = 4 000 − 384 = 3 616 mm² (Only ONE hole at the critical cross-section despite 3 bolts; all bolts do not share the same cross-section.) STEP 3 — Effective net area: Ae = U · An = 1.0 × 3 616 = 3 616 mm² STEP 4 — Yielding: φPn(yield) = 0.90 × 248 × 4 000 = 892 800 N = 892.8 kN STEP 5 — Rupture: φPn(rupture) = 0.75 × 400 × 3 616 = 1 084 800 N = 1 084.8 kN STEP 6 — Govern: φPn = min(892.8, 1 084.8) = 892.8 kN ← YIELDING CONTROLS Check: Ae/Ag = 3 616/4 000 = 0.904 > 0.74 → yielding expected to control. ✓

Even with two holes, yielding still controls. With three holes (An = 4 000 − 3×384 = 2 848 mm²), Ae/Ag = 0.712 < 0.74, so rupture would control: φPn = 0.75×400×2 848 = 854.4 kN < 892.8 kN.

Scenario

BOARD-TYPE PROBLEM. Repeat the above but with TWO holes at the critical cross-section (two bolts across the width, say 22-mm bolts, hole = 24 mm). Same plate PL250×16.

Solution

An = 4 000 − 2(24)(16) = 4 000 − 768 = 3 232 mm² Ae = 1.0 × 3 232 = 3 232 mm² φPn(yield) = 0.90 × 248 × 4 000 = 892.8 kN φPn(rupture) = 0.75 × 400 × 3 232 = 969.6 kN φPn = min(892.8, 969.6) = 892.8 kN ← YIELDING STILL CONTROLS Ae/Ag = 3 232/4 000 = 0.808 > 0.74 → yielding still expected. ✓

Applications

Sizing tension plates for gusset-plate connections in truss bridges.
Checking existing bolted hanger capacities during structural assessment/retrofitting.
Determining whether yielding or rupture governs when selecting between plate thicknesses.
ASD approach used for checking temporary bracing during construction where older codes apply.

Misconceptions

Misconception: More bolts always means higher tensile capacity. FACT: More bolts in a transverse row mean more holes deducted from An — this can REDUCE capacity.
Misconception: Use Fu for yielding and Fy for rupture. FACT: It is the exact OPPOSITE — Fy for yielding (gross area), Fu for rupture (net area).
Misconception: The design strength = φPn from rupture only. FACT: Always compute both and take the MINIMUM.
Misconception: ASD and LRFD give the same governing limit state. FACT: Because the Ω ratio ASD (Ω=2.00/1.67=1.20) differs from LRFD (0.90/0.75=1.20), they actually do yield the same controlling limit state — but ASD is applied to service loads, LRFD to factored loads.

Related Concepts

Gross area computation
Net area computation
Effective net area and shear-lag factor
A36 vs A572 steel properties
LRFD load combinations (NSCP 2015 Section 203)

Common Exam Questions

Example

Given: PL200×12, Fy=248, Fu=400, two 20-mm bolts (dh=22 mm), U=1.0. Find φPn. Answer: yield governs at 535.7 kN.

Approach

Set up both limit states, compute each φPn, select the minimum. Show all steps clearly on board exams.

Question Type

Compute design strength

Example

What minimum plate width is needed for a 10-mm-thick plate with Pu=400 kN, one 22-mm hole, A36 steel?

Approach

Set φPn = Pu and solve for Ag or An. Usually involves setting up both inequalities and solving for the unknown dimension.

Question Type

Back-calculate required area

Example

A highly-notched section with Ae/Ag=0.60: rupture controls since 0.60 < 0.74 threshold for A36.

Approach

Compare 0.90FyAg vs 0.75FuAe; state which is smaller and explain why.

Question Type

Identify which limit state controls

Key Points To Remember

ALWAYS compute both limit states and take the MINIMUM design strength.
φ = 0.90 → yielding (gross area, Fy); φ = 0.75 → rupture (net effective area, Fu).
Yielding = ductile/gradual; Rupture = brittle/sudden — hence different φ values.
For A36 steel, yielding often controls when bolt-hole area removed is less than ~26% of Ag.
For higher-strength steel (A572 Gr.50), rupture controls more frequently.
In ASD: check Pa ≤ FyAg/1.67 AND Pa ≤ FuAe/2.00; use the lower allowable.

Net Area Computation and Staggered Holes

The net area An is the gross area Ag reduced by the material removed at holes. NSCP 2015 / AISC 360 requires deducting the DESIGN hole diameter, which is larger than the nominal bolt diameter to account for punching damage: Design hole diameter = bolt diameter + 2 mm (standard practice per AISC; NSCP allows +3 mm for punched holes without sub-drilling) Typically dh = db + 2 mm for drilled/sub-punched-and-reamed holes dh = db + 3 mm for punched holes (to account for material damage at edges) For a SINGLE ROW OF HOLES perpendicular to the load: An = Ag − n·dh·t where n = number of holes at the critical cross-section, t = thickness. ────────────────────────────────────────────────── STAGGERED HOLES — THE s²/4g CORRECTION ────────────────────────────────────────────────── When bolt holes are staggered (offset longitudinally), a diagonal failure path may be more critical than a straight transverse path. For each diagonal segment in the potential failure chain: Add: s²/(4g) where: s = longitudinal center-to-center spacing (pitch) between the two holes g = transverse center-to-center spacing (gage) between the two hole lines For a plate of gross width Wg with holes in the failure path: Net width = Wg − Σdh + Σ(s²/4g) [one term per diagonal pair] An = Net width × t The critical (governing) failure path is the one that gives the MINIMUM net width (most material removed). All possible failure paths — straight across and all possible zigzag routes — must be checked. Notes on the s²/4g term: • If s is large relative to g, the diagonal path is nearly straight → adds very little. • If s is small relative to g, the diagonal path cuts a steep angle → approximation breaks down, but the straight path usually governs anyway. • The formula is a geometric approximation from the Cochrane formula (1922), widely adopted worldwide. For ANGLES, unfold the angle into a flat plate and use the gage measured along the leg centerline (the 'working gage' or 'standard gage' from AISC tables).

Examples

The full zig-zag path through all three holes gives the minimum net width (148 mm) and governs. Always check all possible paths systematically. Many board exam problems specifically test whether students check the diagonal path.

Scenario

BOARD-TYPE PROBLEM. A PL200×10 tension plate (A36) has three 22-mm-diameter bolts (hole dh = 24 mm) arranged with two in one transverse line and one in a second line, staggered. Layout: Line 1 has holes at y = 60 mm and y = 140 mm from one edge; Line 2 has one hole at y = 100 mm (between the two) but offset 40 mm longitudinally (s = 40 mm) from Line 1. Gage g = 40 mm (between line 1-hole at 60 mm and line 2-hole at 100 mm), g = 40 mm (between line 2 and other line 1 hole). Find the governing net area.

Solution

Possible failure paths: PATH A — Straight through Line 1 (two holes): Net width = 200 − 2(24) = 152 mm PATH B — Straight through Line 2 (one hole): Net width = 200 − 24 = 176 mm PATH C — Zigzag: Line1-top → Line2-middle → (diagonal between them): Uses holes at y=60 and y=100: g=40 mm, s=40 mm Net width = 200 − 2(24) + (40²)/(4×40) = 200 − 48 + 10 = 162 mm PATH D — Zigzag: Line2-middle → Line1-bottom: Uses holes at y=100 and y=140: g=40, s=40 (same geometry) Net width = 200 − 2(24) + (40²)/(4×40) = 162 mm (same as C by symmetry) PATH E — Full zigzag through all three holes (Line1-top→Line2-mid→Line1-bottom): Net width = 200 − 3(24) + (40²/4×40) + (40²/4×40) = 200 − 72 + 10 + 10 = 148 mm ← MINIMUM An = 148 × 10 = 1 480 mm² ← GOVERNS

The s²/4g correction (8.33 mm) partially compensates for the hole deductions, making the diagonal path less critical than might initially appear. Always compare this with the straight-line path: Net width (straight, 2 holes) = 200 − 44 = 156 mm → An = 1 560 mm². The straight path governs here (1 560 < 1 643).

Scenario

BOARD-TYPE PROBLEM (reference document Example 2). A 200-mm-wide, 10-mm plate has 20-mm bolts (hole 22 mm) in two lines, staggered with pitch s = 50 mm and gage g = 75 mm. Find the net area for the zig-zag path crossing two holes with one stagger.

Solution

Net width = Wg − 2dh + s²/(4g) = 200 − 2(22) + (50²)/(4×75) = 200 − 44 + 2500/300 = 200 − 44 + 8.33 = 164.33 mm An = 164.33 × 10 = 1 643 mm²

Applications

Gusset plate connections in highway truss bridges (DPWH standard bridges).
Multi-bolt splice plates for moment-resisting frames.
Angle bracing members connected through one leg with multiple staggered bolts.
Tension chord connections in pre-engineered steel buildings (common in Philippine industrial parks).

Misconceptions

Misconception: Deduct only one hole regardless of path length. FACT: Deduct ALL holes in the failure path, then add back s²/4g for each diagonal segment.
Misconception: The hole diameter = bolt diameter. FACT: Add 2 mm (drilled) or 3 mm (punched) to account for hole-making damage.
Misconception: Only check the straight transverse path. FACT: ALWAYS check all zigzag paths — they often govern in exam problems.
Misconception: s²/4g increases the net area by a lot. FACT: It is a partial correction only; the diagonal path still has more holes deducted than the straight path that skips some holes.
Misconception: For angles, use the back-to-back gage directly. FACT: For angles with holes in both legs, unfold the section and use the through-the-bend gage = g1 + g2 − t.

Related Concepts

Gross area
Effective net area and shear-lag factor
Bolt hole sizing (standard, oversize, slotted)
Connection geometry and bolt spacing requirements
Block shear failure path

Common Exam Questions

Example

3-hole staggered arrangement: check straight paths and all zigzag combinations; select path with smallest net width.

Approach

List all possible failure paths. For each path, apply: net width = Wg − Σdh + Σ(s²/4g). Take the minimum net width × thickness = An.

Question Type

Compute net area with staggered holes

Example

For a 22-mm bolt with punched holes: dh = 22 + 3 = 25 mm.

Approach

Add 2 mm (drilled) or 3 mm (punched) to the bolt diameter per AISC 360 Commentary; the exam typically specifies which to use or gives it directly.

Question Type

Find the design hole diameter

Example

A 4-hole staggered arrangement may have 8+ failure paths; systematic tabulation prevents missing the critical one.

Approach

Set up a table: Path | Holes crossed | Σdh | Σ(s²/4g) | Net width. Identify the minimum.

Question Type

Compare failure paths

Key Points To Remember

Deduct the DESIGN hole diameter (bolt Ø + 2 mm or +3 mm), NOT the nominal bolt diameter.
For staggered holes: Net width = Wg − Σdh + Σ(s²/4g); check ALL possible failure paths.
The MINIMUM net width (most subtraction) gives the CRITICAL (governing) net area.
For angles with staggered holes in two legs, unfold the angle and use the distance between hole centers measured along the angle's centerline (gage across the bend = g1 + g2 − t, where t = angle thickness).
An can never be greater than Ag; if Σ(s²/4g) terms make the computed An > Ag, cap it at Ag.
In practice, when holes in different bolt lines do NOT stagger, each transverse section is checked independently and the worst governs.

Effective Net Area and the Shear-Lag Factor U

When a tension member is connected to a gusset plate through only SOME of its cross-sectional elements (e.g., an angle bolted through one leg only), the unconnected elements do not immediately pick up their share of the load at the connection. This phenomenon is called SHEAR LAG — the stress lags from the connected element into the unconnected elements over some distance beyond the connection. If not accounted for, the capacity is overestimated. NSCP 2015 / AISC 360 Section D3 accounts for shear lag through the EFFECTIVE NET AREA: Ae = U × An where U is the shear-lag factor (0 < U ≤ 1.0). The general formula for U (AISC 360 Table D3.1, Case 2): U = 1 − x̄/L where: x̄ = distance from the centroid of the connected element to the connection plane (the shear-lag distance) L = length of the connection = (n−1) × s, where n = number of bolts in line, s = bolt pitch Key cases from AISC 360 Table D3.1: CASE 1: All elements connected (plates, W-shapes connected through both flanges, HSS) → U = 1.0 CASE 2: General formula → U = 1 − x̄/L CASE 3: W, M, S, HP shapes connected through the web only (bf/d ≥ 0.67) → U = 0.70 CASE 4: Angles connected through ONE leg: - 4 or more bolts in line: U = 0.80 - 2 or 3 bolts in line: U = 0.60 (Alternatively, use the general formula U = 1 − x̄/L if it gives a higher value.) CASE 5: Single angles with a single bolt: U = 0.60 (special) For WELDED connections, L = weld length (parallel to load). For PLATES bolted through the FULL width: U = 1.0 (all elements connected). Physical Meaning of x̄: For a single angle (L75×75×8), x̄ is the distance from the heel of the angle (the connected leg) to the centroid of the angle cross-section. For equal-leg angles, x̄ ≈ 0.25b (where b = leg width), approximately. IMPORTANT: Per AISC 360, U shall not exceed: - 0.90 for W-shapes with 3 or more bolts per line through the flange - The tabulated values in Table D3.1 when those are given Use the LARGER of (1 − x̄/L) and the tabulated minimum for angles.

Examples

This is a classic board-exam scenario. The shear-lag factor is critical — using U = 1.0 would give a non-conservative (unsafe) overestimate of 48 × 958/827 − 1 ≈ 16% error. Always compute U for single-leg connections.

Scenario

BOARD-TYPE PROBLEM. An L75×75×8 angle (Ag = 1 150 mm², x̄ = 20.5 mm, Fy = 248 MPa, Fu = 400 MPa) is bolted through ONE leg with 3 × 22-mm bolts at 75-mm pitch (dh = 24 mm, t_leg = 8 mm). Find φPn.

Solution

STEP 1 — Gross area: Ag = 1 150 mm² STEP 2 — Net area (one hole in the critical cross-section): An = 1 150 − 24 × 8 = 1 150 − 192 = 958 mm² STEP 3 — Shear-lag factor U: L = (3 − 1) × 75 = 150 mm U = 1 − x̄/L = 1 − 20.5/150 = 1 − 0.137 = 0.863 Table D3.1 minimum for 3 bolts: U_min = 0.60 Use the LARGER: U = 0.863 (general formula governs) STEP 4 — Effective net area: Ae = U × An = 0.863 × 958 = 827 mm² STEP 5 — Yielding: φPn = 0.90 × 248 × 1 150 = 256 680 N = 256.7 kN STEP 6 — Rupture: φPn = 0.75 × 400 × 827 = 248 100 N = 248.1 kN STEP 7 — Govern: φPn = min(256.7, 248.1) = 248.1 kN ← RUPTURE CONTROLS Note: Rupture controls because the combination of shear lag (U = 0.863) and hole deduction made Ae small enough.

U = 0.85 is given directly. With a single bolt, AISC Table D3.1 would give U_min = 0.60 only, but the problem states 0.85 (presumably from 1 − x̄/L calculation). The answer matches the reference document approach.

Scenario

BOARD-TYPE PROBLEM (reference document Exercise 1). An L75×75×8 angle (Ag = 1 150 mm², U = 0.85) is bolted through ONE leg with ONE 20-mm bolt (dh = 22 mm, t = 8 mm). Find φPn (Fy = 248, Fu = 400 MPa).

Solution

An = 1 150 − 22 × 8 = 1 150 − 176 = 974 mm² Ae = 0.85 × 974 = 827.9 mm² φPn(yield) = 0.90 × 248 × 1 150 = 256 680 N = 256.7 kN φPn(rupture) = 0.75 × 400 × 827.9 = 248 370 N = 248.4 kN φPn = min(256.7, 248.4) = 248.4 kN ← RUPTURE CONTROLS

Applications

Single-angle bracing connections in steel-framed commercial buildings.
Double-angle chords of light roof trusses where back-to-back connection makes U = 1.0.
W-shape tension members connected through flanges only vs. web-and-flanges.
Welded connections: U = 1 − x̄/L_weld; longer welds improve efficiency.
HSS round and square sections with gusset-plate connections.

Misconceptions

Misconception: U = 1.0 whenever holes are in both legs of an angle. FACT: U = 1.0 only if the load is introduced uniformly through ALL elements — having bolts in both legs via separate connections does not automatically mean U = 1.0 unless the geometry satisfies AISC criteria.
Misconception: Always use the tabulated U from Table D3.1 (e.g., 0.60 or 0.80 for angles). FACT: Use the LARGER of the tabulated minimum and U = 1 − x̄/L.
Misconception: Longer connections always mean lower U. FACT: Longer L → larger denominator in x̄/L → SMALLER x̄/L → LARGER U. Longer connections IMPROVE shear-lag efficiency.
Misconception: x̄ is the same as the eccentricity at the connection. FACT: x̄ is the distance from the centroid of the connected elements to the shear-transfer plane, a section property, not a connection eccentricity.

Related Concepts

Net area computation
Block shear rupture
Bolt patterns and spacing requirements (AISC 360 Section J3)
Weld length and shear lag in welded connections
Section properties of standard angles (AISC Manual Table 1-7)

Common Exam Questions

Example

L100×75×8 angle with 4 bolts at 65 mm pitch, connected through the 100-mm leg: compute U and Ae.

Approach

Determine x̄ (given or from tables), compute L = (n−1)×s, compute U = 1−x̄/L, then Ae = U×An.

Question Type

Compute Ae given section and connection data

Example

A W200×52 connected through both flanges by bolts: U = 1.0 → Ae = An.

Approach

If all elements are connected → U = 1.0, no shear lag. If only one leg or flange → U < 1.0, check Table D3.1.

Question Type

Determine if shear lag is significant

Example

What minimum number of bolts (at 75 mm pitch) for L75×75×8 (x̄ = 20.5 mm) to achieve U ≥ 0.85?

Approach

Use U = 1 − x̄/L = 1 − x̄/[(n−1)s]; solve for n given target U.

Question Type

Find minimum number of bolts to reach a target U

Key Points To Remember

Shear lag REDUCES the effective net area when not all cross-section elements are connected.
Ae = U × An; for plates with all elements connected, U = 1.0.
For angles with one bolt: U = 0.60; 2–3 bolts: U = 0.60; 4+ bolts: U = 0.80. These are MINIMUM values — use 1 − x̄/L if it gives HIGHER U.
For the general formula: U = 1 − x̄/L; longer connections → smaller x̄/L → larger U → better efficiency.
A longer connection (more bolts, larger L) reduces the shear-lag penalty.
x̄ for a section is tabulated in AISC Steel Construction Manual (or computed from section geometry).

Block Shear Rupture

Block shear is a combined failure mode where a block of material tears away from the member or gusset at the bolt group — part of the failure surface is in TENSION (perpendicular to the load) and part is in SHEAR (parallel to the load). This is governed by AISC 360 Section J4.3 and must be checked whenever a tension member has a bolted or welded connection. The nominal block shear strength is: Pn = 0.60 Fu Anv + Ubs Fu Ant ≤ 0.60 Fy Agv + Ubs Fu Ant where: Agv = gross area subject to shear (parallel to load) Anv = net area subject to shear (parallel to load, holes deducted) Ant = net area subject to tension (perpendicular to load) Ubs = 1.0 when tension stress is uniform (most plates, angles with 1 line of bolts) Ubs = 0.5 when tension stress is non-uniform (e.g., gussets, some connections) The LRFD design block shear strength: φRn = 0.75 × Pn The limit in the formula prevents the shear-yielding component from exceeding 0.60Fy Agv — this is the upper bound on shear resistance. Physical interpretation: • Shear fracture along the bolt line + tensile fracture across the back of the bolt group. • In angles bolted through one leg, this is the classic L-shaped tear-out. Design check: φRn (block shear) ≥ Pu Compare with φPn from yielding and rupture; the governing strength is the MINIMUM of all three.

Examples

Block shear capacity = 677.7 kN. This must be compared with the tensile yielding and rupture capacities. The governing limit state is the smallest of all three. Block shear is checked with φ = 0.75, same as rupture.

Scenario

BOARD-TYPE PROBLEM. A PL150×12 tension plate (A36: Fy = 248, Fu = 400 MPa) is connected with 3 × 20-mm bolts in a single row, 70 mm pitch, edge distance = 35 mm. dh = 22 mm. Check block shear.

Solution

GEOMETRY: Shear path length = 35 + 2(70) = 175 mm Tension path = 150/2 − 70/2 = ... (For a centered bolt line, tension width = half the plate = 75 mm from bolt line to plate edge, but here the bolts are along the centerline.) Since bolts are in the center of the 150-mm plate, the tear-out block forms along TWO shear planes (both sides of the bolt row): SIMPLIFIED for a centered row (Ubs=1.0): Agv = 2 × (175 × 12) = 4 200 mm² [both shear planes] Anv = 2 × [(175 − 2.5 × 22) × 12] = 2 × [(175 − 55) × 12] = 2 × 1 440 = 2 880 mm² [2.5 holes: 1 full + 2 half-holes at ends for each shear plane; conservative = 2.5 holes per plane] Ant = [150 − 2(35) − 22] × 12 = [150 − 70 − 22] × 12 = 58 × 12 = 696 mm² Pn = min(0.60×400×2 880 + 1.0×400×696, 0.60×248×4 200 + 1.0×400×696) = min(691 200 + 278 400, 625 248 + 278 400) = min(969 600, 903 648) = 903 648 N φRn = 0.75 × 903 648 = 677 736 N ≈ 677.7 kN

Applications

Gusset plate connection design in truss bridges.
Beam end connections (shear tabs, clip angles) — block shear of the beam web.
Angle-to-gusset connections for bracing members.
Coped beam connections where the top flange is removed.

Misconceptions

Misconception: Block shear only applies to gusset plates, not to tension members. FACT: Block shear must be checked for ALL bolted tension member connections.
Misconception: Use φ = 0.90 for the yielding component of block shear. FACT: The overall φ for block shear is 0.75, applied to the entire expression.
Misconception: Ubs is always 0.5. FACT: Ubs = 1.0 for most standard connections; 0.5 is for non-uniform tension cases.

Related Concepts

Net area computation
Bolt pattern and spacing
Connection design (AISC 360 Chapter J)
Tensile rupture limit state
Shear yielding of plates

Common Exam Questions

Example

2-bolt connection in a single angle — identify the L-shaped failure block and compute φRn.

Approach

Identify the shear and tension failure planes from the bolt pattern. Compute Agv, Anv (deduct holes), Ant (deduct holes). Apply both expressions in the formula and take the minimum × 0.75.

Question Type

Compute block shear strength

Example

Often block shear governs for short connections (few bolts); yielding governs for long connections with few holes.

Approach

Compute all three: φPn(yield), φPn(rupture), φRn(block shear). The smallest is the design strength.

Question Type

Determine governing failure mode

Key Points To Remember

Block shear is a THIRD limit state for tension members with bolted connections; it must ALWAYS be checked.
φ = 0.75 for block shear (same as tensile rupture).
Ubs = 1.0 for most practical cases (single line of bolts, plates, angles).
Block shear formula: Pn = min(0.60FuAnv + UbsFuAnt, 0.60FyAgv + UbsFuAnt).
The tensile component uses Fu (fracture); the shear component uses both Fu (fracture) and Fy (yield) — take the smaller.
Block shear often governs for short connection lengths (few bolts, small L).

Slenderness Recommendation and Serviceability

Unlike compression members, tension members have NO slenderness limit tied to strength — there is no buckling. However, a very slender tension member can sag excessively under its own weight, vibrate under dynamic loads (wind-induced flutter, foot traffic), or be easily bent during fabrication and erection. NSCP 2015 / AISC 360 Section D1 RECOMMENDS (but does not mandate): L/r ≤ 300 for tension members other than rods (Rods are exempt — they are often used as long, slender tie rods.) where: L = member length (mm) r = least radius of gyration of the cross-section (mm) This is a serviceability and good-practice criterion, NOT a strength limit. Its violation does not automatically make the member unsafe in strength, but the engineer should justify exceeding it. For a single angle L75×75×8: r_min = r_z (the radius of gyration about the weakest axis, the diagonal z-axis) ≈ 14.7 mm (from AISC tables) For a 4-m member: L/r = 4000/14.7 = 272 < 300 ✓ For a round rod of diameter d: r = d/4 For d = 20 mm: r = 5 mm A 1.5-m rod: L/r = 1500/5 = 300 — just at the limit (rods exempted anyway). Practical design note: When a tension member also carries incidental compression (e.g., a diagonal brace in a seismic frame that reverses), the compression check governs and usually controls slenderness more strictly (L/r ≤ 200 for most compression members per AISC 360 Section E2 commentary).

Examples

The check is straightforward. Note that 160 is well within the limit, leaving ample margin. For context: a member with r = 13 mm would give L/r = 4000/13 = 308 > 300, which is marginally non-conforming — the engineer should reconsider the section.

Scenario

BOARD-TYPE PROBLEM (reference document Example 3). A tension member is 4 m long with a least radius of gyration r = 25 mm. Check the recommended slenderness.

Solution

L/r = 4 000 / 25 = 160 < 300 ✓ The member satisfies the NSCP 2015 / AISC 360 recommended slenderness limit for tension members.

This illustrates both the formula and the exemption for rods. The board exam may present this as a calculation trap — knowing the exemption is key.

Scenario

BOARD-TYPE PROBLEM. A sag rod spans 6 m between purlins. What is the minimum round-bar diameter such that L/r ≤ 300?

Solution

For a round bar: r = d/4 L/r = 6 000/(d/4) ≤ 300 6 000/(d/4) ≤ 300 d/4 ≥ 6 000/300 = 20 mm d ≥ 80 mm However, note: rods are EXEMPT from the L/r ≤ 300 recommendation. In practice, sag rods are sized by their tensile capacity (force from purlin lateral loads), not by slenderness. A 12–16 mm rod is typical for light roof loads.

Applications

Checking long diagonal braces in industrial buildings against excessive sag.
Evaluating sag rod slenderness in roof structures with long purlin spans.
Comparing single-angle vs. double-angle sections for a given span (double angle has larger r_min).
Deciding whether to add intermediate supports to a very long tension diagonal.

Misconceptions

Misconception: L/r > 300 means the member fails and cannot be used. FACT: It is a recommendation — the member can still be used with proper engineering justification.
Misconception: Use the radius of gyration about the strong axis for the slenderness check. FACT: Use r_MIN (the least radius of gyration) for the most conservative slenderness check.
Misconception: The L/r limit for tension and compression members is the same. FACT: Tension: 300 (recommended); Compression: 200 (recommended per AISC 360 Commentary).
Misconception: Rods must satisfy L/r ≤ 300. FACT: Rods are explicitly exempted from this recommendation.

Related Concepts

Radius of gyration (r = √(I/A))
Section properties of standard shapes
Compression member slenderness (L/r ≤ 200)
Serviceability criteria in NSCP 2015
Vibration and sag of long structural members

Common Exam Questions

Example

L=5 m, single angle L75×75×8 (r_z = 14.7 mm): L/r = 5000/14.7 = 340 > 300 — marginally exceeds recommendation.

Approach

Compute L/r_min. Compare with 300. State pass/fail and note that it is a recommendation.

Question Type

Check slenderness adequacy

Example

For a 3-m tension rod: r_min = 3000/300 = 10 mm → if rod: d = 4r = 40 mm (but rods are exempt).

Approach

Set L/r = 300, solve r_min = L/300. Select section with r ≥ r_min from tables.

Question Type

Find minimum r or section size for a given length

Key Points To Remember

L/r ≤ 300 is RECOMMENDED, not a mandatory strength requirement for tension members.
Rods (circular solid bars) are exempt from the 300 limit.
Use the LEAST radius of gyration (minimum r) for the most conservative check.
Exceeding L/r = 300 requires engineering judgment and justification (excessive sag, vibration concerns).
When a brace reverses under seismic load, compression governs and L/r ≤ 200 (typical) is more restrictive.
For a single angle, r_min = r_z (about the z-axis); obtain from AISC Manual tables or compute from Iz and A.

Practice Problems

Standard two-bolt, full-plate connection. With dh = 23 mm (punched hole, +3 mm), the net area is slightly less than with drilled holes (+2 mm). Yielding controls because the hole removal is less than 23% of Ag. Note: the reference document used dh = 22 mm (+2 mm); using +3 mm is conservative and also acceptable.

Problem

PROBLEM 1 — Design Tensile Strength of a Plate. A PL200×12 mm tension plate (A36: Fy = 248 MPa, Fu = 400 MPa) is connected with two 20-mm-diameter bolts across the critical section (standard punched hole: dh = bolt Ø + 3 mm = 23 mm, U = 1.0). Determine: (a) The gross area Ag. (b) The net area An. (c) The effective net area Ae. (d) The LRFD design tensile strength φPn.

Solution

(a) Ag = 200 × 12 = 2 400 mm² (b) An = Ag − n × dh × t = 2 400 − 2(23)(12) = 2 400 − 552 = 1 848 mm² (c) Ae = U × An = 1.0 × 1 848 = 1 848 mm² (d) Yielding: φPn = 0.90 × 248 × 2 400 = 535 680 N = 535.7 kN Rupture: φPn = 0.75 × 400 × 1 848 = 554 400 N = 554.4 kN Governing: φPn = min(535.7, 554.4) = 535.7 kN ← YIELDING CONTROLS Ae/Ag = 1 848/2 400 = 0.770 > 0.74 → yielding expected to control. ✓

Path 1 (straight through the two outer holes) gives the smallest net width (198 mm) and governs. The full zigzag (Path 5) gives 208 mm — less critical than Path 1 in this case because the s²/4g correction (36 mm total) does not fully compensate for the third hole deduction (26 mm). Always check all paths and tabulate systematically.

Problem

PROBLEM 2 — Net Area with Staggered Holes. A tension plate PL250×12 mm (A36) has three 24-mm bolt holes arranged as follows: Holes A and C are in Line 1 (left and right), each 75 mm from the plate edges (so g_AC = 250−2×75 = 100 mm between line centers). Hole B is in Line 2, centered (y = 125 mm), staggered 60 mm longitudinally (s = 60 mm). All holes: dh = 26 mm. Find the governing net width and net area.

Solution

Possible failure paths: PATH 1 — Straight through A and C (both holes in Line 1): Net width = 250 − 2(26) = 198 mm PATH 2 — Straight through B only: Net width = 250 − 26 = 224 mm PATH 3 — Zigzag A→B: g = 125 − 75 = 50 mm (A at y=75, B at y=125) s = 60 mm Net width = 250 − 2(26) + (60²)/(4×50) = 250 − 52 + 3600/200 = 198 + 18 = 216 mm But this path cuts only 2 holes. PATH 4 — Zigzag B→C: Same geometry by symmetry → 216 mm PATH 5 — Full zigzag A→B→C: Two diagonal segments, each with s=60 mm, g=50 mm: Net width = 250 − 3(26) + 2×(60²)/(4×50) = 250 − 78 + 2×18 = 250 − 78 + 36 = 208 mm Governing: PATH 1 → Net width = 198 mm (minimum) An = 198 × 12 = 2 376 mm²

With 4 bolts and a long connection (L=195 mm), U=0.909 is quite high — close to 1.0. Yielding controls because the effective net area is relatively large. If only 2 bolts were used (L=65 mm, U=1−17.8/65=0.726), rupture would likely control.

Problem

PROBLEM 3 — Angle with Shear Lag. An L100×75×8 angle (Ag = 1 370 mm², x̄_100-leg = 17.8 mm for the 100-mm leg connected) is used as a tension brace. It is connected through the 100-mm leg with FOUR 20-mm bolts at 65-mm pitch. dh = 22 mm, Fy = 248 MPa, Fu = 400 MPa. Determine the LRFD design tensile strength (exclude block shear).

Solution

STEP 1 — Ag = 1 370 mm² STEP 2 — Net area (one hole in critical section on the 100-mm leg): An = 1 370 − 22 × 8 = 1 370 − 176 = 1 194 mm² STEP 3 — Shear-lag factor: L = (4 − 1) × 65 = 195 mm U = 1 − x̄/L = 1 − 17.8/195 = 1 − 0.0913 = 0.909 Table D3.1: for 4+ bolts in one leg, U_min = 0.80 Use LARGER: U = 0.909 STEP 4 — Ae = 0.909 × 1 194 = 1 085 mm² STEP 5 — Yielding: φPn = 0.90 × 248 × 1 370 = 305 784 N = 305.8 kN STEP 6 — Rupture: φPn = 0.75 × 400 × 1 085 = 325 500 N = 325.5 kN STEP 7 — Govern: φPn = min(305.8, 325.5) = 305.8 kN ← YIELDING CONTROLS

Tie rod design illustrates both limit states for round bars. The 0.75Ag approximation for threaded area is commonly used; actual thread root areas are tabulated in AISC Manual Table 7-18. The slenderness exemption for rods is an important exam fact.

Problem

PROBLEM 4 — Design a Round Tie Rod. A round solid bar tie rod (A36: Fy = 248 MPa, Fu = 400 MPa, threaded ends) must carry Pu = 200 kN. The threaded section is the critical area (Ae ≈ 0.75Ag for threaded rods, approximate). Find the minimum rod diameter. (a) Using yielding on gross area, (b) Using rupture on the effective (threaded) area.

Solution

(a) Yielding governs: φPn = 0.90 Fy Ag ≥ Pu = 200 000 N Ag ≥ 200 000 / (0.90 × 248) = 200 000 / 223.2 = 896.1 mm² Ag = π d²/4 → d² = 4 × 896.1/π = 1 141.2 → d ≥ 33.8 mm Use d = 36 mm (next standard size) (b) Rupture at threads (Ae ≈ 0.75 Ag): φPn = 0.75 Fu (0.75 Ag) = 0.75 × 400 × 0.75 Ag = 225 Ag ≥ 200 000 Ag ≥ 200 000/225 = 888.9 mm² d ≥ √(4 × 888.9/π) = 33.7 mm → Use d = 36 mm Both checks give essentially the same answer: use d = 36 mm. Slenderness check (if L = 3 m): r = d/4 = 36/4 = 9 mm; L/r = 3000/9 = 333 > 300, but rods are EXEMPT from the L/r ≤ 300 recommendation.

Close result — rupture and yielding are nearly equal (differ by only 2%). This is the transition zone where Ae/Ag ≈ 0.74. Accurate computation of dh is critical; an error of 1 mm in dh can shift the governing limit state. The problem highlights the importance of computing BOTH limit states.

Problem

PROBLEM 5 — Compare Plate Capacities (reference document Exercise 3). A PL150×10 mm plate (A36: Fy = 248 MPa, Fu = 400 MPa) has two 16-mm bolts (dh = 18 mm) at the critical section, U = 1.0. (a) Compute φPn for yielding. (b) Compute φPn for rupture. (c) Which limit state governs? (d) What maximum factored tensile load Pu can the plate carry?

Solution

(a) Ag = 150 × 10 = 1 500 mm² φPn(yield) = 0.90 × 248 × 1 500 = 334 800 N = 334.8 kN (b) An = 1 500 − 2(18)(10) = 1 500 − 360 = 1 140 mm² Ae = 1.0 × 1 140 = 1 140 mm² φPn(rupture) = 0.75 × 400 × 1 140 = 342 000 N = 342.0 kN (c) Yielding governs: φPn = 334.8 kN < 342.0 kN Check: Ae/Ag = 1 140/1 500 = 0.76 > 0.74 → yielding expected to control ✓ (d) Maximum Pu = φPn = 334.8 kN

Exam Preparation Tips

KNOW YOUR φ VALUES COLD: φ = 0.90 for yielding (Fy, gross area), φ = 0.75 for rupture (Fu, net effective area), φ = 0.75 for block shear. These are constants that MUST be memorized — mixing them up is the single most common and costly error on the board exam.
ALWAYS CHECK BOTH LIMIT STATES: Never declare yielding or rupture as governing without computing BOTH values. A quick ratio test — if Ae/Ag > 0.74 for A36 steel, yielding likely controls — is a useful mental check, but always verify numerically.
DESIGN HOLE DIAMETER: The problem will usually give you the hole diameter directly, or state 'standard hole.' If not specified, use bolt Ø + 2 mm (drilled) or bolt Ø + 3 mm (punched). Never use the bolt diameter itself for An calculation.
STAGGERED HOLES — TABULATE ALL PATHS: Draw the hole layout. List every possible failure path (straight + all zigzags). For each path: Net width = Wg − Σdh + Σ(s²/4g). The minimum net width governs. Forgetting the full-zigzag path through all holes is a classic exam trap.
SHEAR LAG — KNOW WHEN U < 1.0: If ALL elements are connected (plate through full width, angle through both legs per AISC criteria) → U = 1.0. If only ONE leg of an angle is connected → U < 1.0 per Table D3.1 or U = 1−x̄/L, whichever is LARGER. Never use U = 1.0 for an angle bolted through one leg without checking.
BLOCK SHEAR IS THE THIRD LIMIT STATE: For every bolted connection, identify the shear and tension failure planes. The formula Pn = 0.60FuAnv + UbsFuAnt (≤ 0.60FyAgv + UbsFuAnt) must be evaluated and compared with the other limit states. Omitting block shear is a common source of non-conservative answers.
L/r ≤ 300 IS RECOMMENDED, NOT MANDATORY: The board exam may test whether you know this is a 'preferred limit,' not a code mandate. State it clearly as a recommendation. Rods are EXEMPT.
UNITS DISCIPLINE: All inputs in N and mm → results in N; convert to kN at the end. Mixing kN and mm (using F in kN and A in mm²) is a common arithmetic error. Set up: φPn = φ × F(MPa=N/mm²) × A(mm²) = result in N.
RATIO TEST FOR GOVERNING LIMIT STATE: For A36 (Fy=248, Fu=400): rupture controls when Ae/Ag < (0.90×248)/(0.75×400) = 223.2/300 = 0.744. For A572 Gr.50 (Fy=345, Fu=450): rupture controls when Ae/Ag < (0.90×345)/(0.75×450) = 310.5/337.5 = 0.920 — rupture controls much more easily for higher-strength steel.
PRACTICE WITH AISC SHAPES TABLES: The board exam gives section properties (Ag, r, x̄) in the problem stem, but familiarity with typical values helps you sense-check: an L75×75×8 has Ag ≈ 1 150 mm², r_z ≈ 14.7 mm, x̄ ≈ 20.5 mm. These serve as quick sanity checks.
RA 544 AWARENESS: Republic Act 544 (Civil Engineering Law of the Philippines) defines the scope of civil engineering practice. While structural steel design specifics are in NSCP 2015, knowing that NSCP 2015 is the mandated design code for structures in the Philippines (not the US-IBC directly) is important for any code-reference question on the board exam.
SHOW ALL STEPS IN THE BOARD EXAM: Partial credit is awarded in most board exam problems. Always write the formula, substitute values, and box the answer with units. A correct setup with arithmetic error gets more credit than a numerical answer without working.

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In summary

Steel tension members, despite being the structurally simplest steel elements, demand rigorous application of multiple interacting code provisions — a characteristic that makes them a recurring and high-yield topic in the PRC Civil Engineer Licensure Examination. The key takeaways from this chapter are: 1. TWO MANDATORY LIMIT STATES: Every tension member design must satisfy both φtPn(yield) = 0.90FyAg and φtPn(rupture) = 0.75FuAe. The governing (lower) design strength is compared with Pu. There is no shortcut — both must be computed. 2. NET AREA DISCIPLINE: The design hole diameter (bolt Ø + 2 or 3 mm) must be used, not the bolt diameter. For staggered holes, ALL failure paths must be checked using the net width formula (Wg − Σdh + Σs²/4g), and the minimum net width governs. 3. SHEAR LAG IS CRITICAL: Ae = U·An, where U < 1.0 whenever not all cross-sectional elements are connected. For single-leg angle connections — the most common board-exam scenario — U is obtained from AISC 360 Table D3.1 or computed as 1 − x̄/L, using the LARGER value. Assuming U = 1.0 for single-leg connections is non-conservative and incorrect. 4. BLOCK SHEAR IS THE THIRD LIMIT STATE: At every bolted connection, a combined shear-tension fracture path must be verified using φRn = 0.75 × (0.60FuAnv + UbsFuAnt), capped by the yielding expression. Neglecting block shear is among the most common sources of error in connection design. 5. SLENDERNESS IS RECOMMENDED, NOT MANDATORY: L/r ≤ 300 is a practical guideline for avoiding sag and vibration in tension members; exceeding it requires engineering judgment but does not constitute a code violation. Rods are explicitly exempt. For the board examination, mastery of these concepts requires not just formula memorization but sound procedural discipline: always draw the section, label hole positions, enumerate all failure paths, select the correct φ and U, and explicitly state which limit state governs. Candidates who internalize these five principles and practice with board-level numerical problems will be well-prepared for this topic on examination day.

Next chapter

Steel Compression Members

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