CELE Steel & Timber Design — Steel Tension MembersMisconception Buster
Misconception buster for Steel Tension Members. Every concept has a shadow — the subtly wrong version that looks right on first glance. Professional Regulation Commission (PRC) — Board of Civil Engineering builds CELE questions around those shadows. This page shows you the truth behind the traps.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Steel & Timber Design subtest is marked as "Core" in the official pattern, and Steel Tension Members appears in position 1st of 5 in the CELE Steel & Timber Design review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Steel Tension Members - Misconception Buster
Steel Tension Members is one of the highest-yield topics in the PRC Civil Engineer Licensure Examination under Steel and Timber Design. Despite its conceptual simplicity — tension is uniform, no buckling — examinees consistently lose marks by applying the wrong phi factor, omitting a limit state, miscomputing net areas, or ignoring shear lag. This guide targets exactly those failure points. Each misconception is stated as the wrong belief students actually hold, explained with the reasoning behind it, corrected with code-referenced evidence (NSCP 2015 / AISC 360-10), and tested with a realistic trap question designed to catch the unprepared. Master this guide and you eliminate the most common sources of lost marks in this topic.
Summary
The following are the most critical takeaways for avoiding costly mistakes on steel tension member exam problems: (1) ALWAYS use phi_t = 0.90 for yielding and phi_t = 0.75 for rupture — these are non-negotiable NSCP 2015 values. (2) ALWAYS check ALL limit states: yielding (0.90*Fy*Ag), rupture (0.75*Fu*Ae), and block shear at connections — the governing (lowest) design strength controls. (3) Net area An uses hole diameter d_h = d_bolt + 2 mm (not the bolt diameter alone) and the s²/4g correction for each diagonal path in staggered configurations — always check all possible failure paths and use the minimum. (4) Effective net area Ae = U * An — the shear-lag factor U is NOT 1.0 for angles, channels, or tees connected through partial elements; look it up from NSCP 2015 Table 502.4.3-1. (5) The slenderness recommendation L/r ≤ 300 is NOT mandatory — it is a serviceability preference that does not reduce the computed design tensile strength. (6) For ASD, the safety factors are Omega = 1.67 (yield) and Omega = 2.00 (rupture) — two different values, not one universal factor. Memorize these distinctions and verify them on every problem. The board exam rewards precision and penalizes the assumption that 'one formula fits all' for tension member design.
Misconceptions
Both limit states (yield and rupture) use the same resistance factor phi = 0.90.
Tags
- formula_confusion
- critical_error
- phi_factor
Topic
Limit States and Resistance Factors
Severity
critical
Exam Impact
Directly causes a wrong final answer for rupture-governed capacity. The student computes a higher design strength than permitted, potentially selecting an undersized member or choosing the wrong governing limit state.
The Reality
NSCP 2015 Section 502.4 (AISC 360-10 Section D2) assigns TWO different resistance factors: phi_t = 0.90 for yielding on the gross section (Pn = Fy * Ag) and phi_t = 0.75 for rupture on the effective net section (Pn = Fu * Ae). The lower phi for rupture reflects the sudden, brittle nature of fracture at bolt holes. Using 0.90 for rupture OVERESTIMATES the design strength and is unconservative — a critical error in practice and on the board exam.
Trap Question
Question
A 200 x 12 mm plate (Fy = 248 MPa, Fu = 400 MPa) has Ag = 2400 mm² and Ae = 1872 mm². What is the LRFD design tensile rupture strength?
Explanation
Rupture uses phi_t = 0.75, NOT 0.90. NSCP 2015 Sec. 502.4.2 is explicit: tensile rupture resistance factor is 0.75. The wrong answer overestimates by about 20% — a dangerously unconservative error.
Wrong Answer
phi * Pn = 0.90 * 400 * 1872 = 673,920 N = 673.9 kN
Correct Answer
phi * Pn = 0.75 * 400 * 1872 = 561,600 N = 561.6 kN
Misconception Id
M1
Correct Vs Incorrect
Correct Approach
Yield: phi * Pn = 0.90 * Fy * Ag; Rupture: phi * Pn = 0.75 * Fu * Ae. Always use 0.75 for rupture.
Incorrect Approach
Rupture: phi * Pn = 0.90 * Fu * Ae (WRONG — uses yield phi for rupture)
Why Students Believe It
Reviewees memorize phi = 0.90 as the standard LRFD tension factor from early review sessions and apply it universally to both yielding and rupture equations. The difference in phi is easy to overlook because both equations share the same form: phi * P_n.
Only check one limit state — whichever formula the problem seems to ask about.
Tags
- conceptual_gap
- common_error
- limit_states
Topic
Governing Limit State
Severity
critical
Exam Impact
In multiple-choice questions where options are the yield strength, the rupture strength, or the lower of both, skipping one limit state leads to selection of a non-governing value. Loses full marks on the item.
The Reality
NSCP 2015 Section 502.4 mandates that the DESIGN TENSILE STRENGTH is the LOWER of phi_t * Fy * Ag (yield) and phi_t * Fu * Ae (rupture). Neither limit state can be skipped — the governing (lower) value controls. In the solved Example 1 from the reference document, yield governs at 535.7 kN even though rupture at 561.6 kN is higher. If only rupture were checked, the answer would be wrong.
Trap Question
Question
A 200 x 12 mm plate (Fy = 248 MPa, Fu = 400 MPa) with Ag = 2400 mm² and Ae = 1872 mm² is loaded in tension. What is the LRFD design tensile strength?
Explanation
Both must be computed. Yield gives 535.7 kN and rupture gives 561.6 kN. The lower value, 535.7 kN, governs. A student who checks only rupture reports 561.6 kN — wrong and unconservative.
Wrong Answer
561.6 kN (rupture only — student stopped after computing one limit state)
Correct Answer
535.7 kN (yield governs: 0.90 * 248 * 2400 = 535,680 N < 0.75 * 400 * 1872 = 561,600 N)
Misconception Id
M2
Correct Vs Incorrect
Correct Approach
Compute both: Yield = 0.90 * Fy * Ag and Rupture = 0.75 * Fu * Ae. Report the LOWER value as the design tensile strength.
Incorrect Approach
Student computes only rupture: phi * Pn = 0.75 * Fu * Ae and reports that as the design strength.
Why Students Believe It
Under time pressure, students compute only the rupture strength (thinking holes are the 'real' problem) or only the yield strength (thinking it always governs for stocky sections). They do not realize BOTH must be computed and compared for EVERY problem.
The bolt hole diameter used to compute net area equals the bolt diameter.
Tags
- formula_confusion
- common_error
- hole_allowance
Topic
Net Area Computation
Severity
major
Exam Impact
Underestimates the area deduction, producing a net area that is too large, inflating the rupture capacity, and potentially misidentifying the governing limit state.
The Reality
NSCP 2015 (following AISC 360-10 Section B4.3b) requires adding 2 mm to the bolt diameter to account for the clearance hole, PLUS an additional 2 mm for punching damage — resulting in a standard hole deduction of d_b + 4 mm for standard round holes (some references use d_b + 2 mm for drilled holes and d_b + 4 mm for punched holes). The reference document uses d_h = d_b + 2 mm to d_b + 3 mm as local practice. In Philippine board problems, d_h = d_bolt + 2 mm is most commonly used unless stated otherwise. Always check the problem statement — if hole diameter is given directly, use it. If only bolt size is given, add the allowance.
Trap Question
Question
A 200 x 12 mm plate has two lines of 20 mm bolts at the critical section. What is the net area if standard clearance (2 mm) is added to the bolt diameter?
Explanation
The hole diameter for net area computation is d_h = 20 + 2 = 22 mm, not 20 mm. Using the bolt diameter directly overstates the net area by 48 mm² per hole — a small but exam-deciding error.
Wrong Answer
An = 2400 - 2(20)(12) = 1920 mm²
Correct Answer
An = 2400 - 2(22)(12) = 1872 mm²
Misconception Id
M3
Correct Vs Incorrect
Correct Approach
d_h = d_bolt + 2 mm = 22 mm; Net area = Ag - n * d_h * t = 2400 - 2(22)(12) = 1872 mm²
Incorrect Approach
d_h = d_bolt = 20 mm; Net area = Ag - n * d_bolt * t = 2400 - 2(20)(12) = 1920 mm²
Why Students Believe It
Students see a problem stating '20 mm bolts' and subtract 20 mm directly when computing net width. The added clearance for punching damage is not intuitive and is often omitted in rushed computations.
The effective net area Ae always equals the net area An (shear lag factor U = 1.0 for all members).
Tags
- conceptual_gap
- shear_lag
- critical_error
Topic
Shear Lag and Effective Net Area
Severity
critical
Exam Impact
Directly inflates the computed rupture capacity of angle, channel, tee, and other non-symmetric sections. The student selects an undersized member or reports a design strength well above the code-permitted value.
The Reality
U = 1.0 ONLY when the load is transmitted through ALL cross-sectional elements simultaneously (e.g., a plate connected across its full width, or a wide-flange section with all flanges and web bolted). For angles connected by ONE leg, U is typically 0.60 to 0.85 depending on the number of bolts and connection length (NSCP 2015 Table 502.4.3-1, AISC 360-10 Table D3.1). Ae = U * An — omitting U for an angle section can overestimate the rupture capacity by 15 to 40%, which is both unsafe and exam-wrong.
Trap Question
Question
An L75x75x8 angle (Ag = 1150 mm², Fy = 248 MPa, Fu = 400 MPa) is connected by ONE leg using ONE 20 mm bolt. Given U = 0.85 and d_h = 22 mm, t = 8 mm, what is the LRFD rupture design strength?
Explanation
Because the angle is connected through only one leg, shear lag occurs — the outstanding leg does not fully participate. U = 0.85 must be applied. Ignoring U overstates rupture capacity by about 18%.
Wrong Answer
An = 1150 - 22(8) = 974 mm²; Rupture = 0.75 * 400 * 974 = 292,200 N = 292.2 kN (skipped U)
Correct Answer
An = 974 mm²; Ae = 0.85 * 974 = 827.9 mm²; Rupture = 0.75 * 400 * 827.9 = 248,370 N = 248.4 kN
Misconception Id
M4
Correct Vs Incorrect
Correct Approach
Determine U from NSCP 2015 Table 502.4.3-1 based on the section shape, connection type, and number of bolts. Then Ae = U * An. Rupture = 0.75 * Fu * U * An.
Incorrect Approach
For an angle bolted through one leg: Ae = An (assumes U = 1.0). Rupture = 0.75 * Fu * An.
Why Students Believe It
Plates bolted across their full width do have U = 1.0, and students generalize this. They assume that computing An is sufficient and do not apply the shear-lag reduction, especially for angles and channels connected through one leg only.
For staggered bolt holes, the net area is found by simply subtracting ALL holes in the row from the gross area.
Tags
- formula_confusion
- staggered_holes
- common_error
Topic
Staggered Holes and Net Area
Severity
major
Exam Impact
Board problems specifically set up staggered hole configurations to test this knowledge. Ignoring s²/4g for diagonal paths gives the wrong net area, wrong An, wrong Ae, and wrong rupture capacity.
The Reality
For staggered holes, the critical failure path may be a zig-zag (diagonal) path, not a straight line. NSCP 2015 / AISC 360-10 Section B4.3b requires evaluating ALL possible failure paths. For each diagonal segment in a zig-zag path, ADD s²/4g to the net width (where s = longitudinal pitch, g = transverse gage). Net width = W_g - sum(d_h) + sum(s²/4g). The governing path is the one giving the MINIMUM net area — which may or may not include the s²/4g correction. Simply subtracting all holes without the diagonal correction can overestimate or underestimate the actual minimum net area.
Trap Question
Question
A 200 x 10 mm plate has staggered 20 mm bolts (d_h = 22 mm) in two lines with s = 50 mm and g = 75 mm. A zig-zag path crosses both holes. What is the net width for this zig-zag path?
Explanation
Each diagonal segment adds s²/4g to the net width. For one stagger: s = 50 mm, g = 75 mm gives 50²/(4×75) = 8.33 mm. Net width = 164.33 mm, not 156 mm. The zig-zag path gives a LARGER net width than the straight path through both holes (156 mm), so the straight path governs here. But this must always be checked.
Wrong Answer
Net width = 200 - 2(22) = 156 mm (forgot s²/4g)
Correct Answer
Net width = 200 - 2(22) + 50²/(4×75) = 200 - 44 + 8.33 = 164.33 mm
Misconception Id
M5
Correct Vs Incorrect
Correct Approach
Net width = 200 - 2(22) + s²/(4g) = 200 - 44 + 50²/(4*75) = 200 - 44 + 8.33 = 164.33 mm. Multiple paths must be checked; use the minimum.
Incorrect Approach
Net width = 200 - 2(22) = 156 mm (subtracts both holes in zig-zag path, ignores s²/4g diagonal bonus)
Why Students Believe It
Students are used to straight-line failure paths. The staggered hole correction (s²/4g) is introduced later in reviews and is often forgotten under exam pressure. It seems safer to subtract all holes and call it conservative.
The L/r slenderness limit of 300 is a mandatory code requirement for all tension members.
Tags
- conceptual_gap
- code_misreading
- slenderness
Topic
Slenderness Recommendation
Severity
minor
Exam Impact
May lead to incorrectly declaring a tension member 'not acceptable' or applying a strength reduction when none is required. Can eliminate correct answer choices in a scenario question.
The Reality
For tension members, NSCP 2015 / AISC 360-10 Section D1 states that L/r ≤ 300 is a RECOMMENDATION (preferred limit), NOT a mandatory requirement. Exceeding L/r = 300 does not reduce the design tensile strength — the member is simply considered susceptible to sag, vibration, and damage during handling. There is an explicit code note that the limit does not apply to rods. In contrast, compression member slenderness directly limits design strength.
Trap Question
Question
A tension rod with L/r = 320 is used in a bracing system. A reviewer states the rod FAILS the NSCP 2015 slenderness requirement and must be resized. Is the reviewer correct?
Explanation
This is the reverse of compression, where slenderness DOES reduce capacity. For tension, exceeding L/r = 300 is a serviceability concern (sag, vibration) but not a strength code violation. The design tensile strength phi*Pn is unaffected.
Wrong Answer
Yes — L/r = 320 exceeds 300, so the rod violates a mandatory NSCP 2015 requirement.
Correct Answer
No — L/r ≤ 300 is a RECOMMENDATION, not a mandatory limit, for tension members. The rod's tensile design strength is not reduced. NSCP 2015 Section D1 explicitly states this is a preferred limit.
Misconception Id
M6
Correct Vs Incorrect
Correct Approach
L/r = 320 > 300 exceeds the recommended slenderness for tension members. This is noted as undesirable for serviceability (sag/vibration) but does NOT reduce phi*Pn. The tension design strength is unchanged.
Incorrect Approach
L/r = 320 > 300, so the tension member fails the code limit and the design is inadequate. (WRONG — this is not a mandatory strength limit for tension.)
Why Students Believe It
Students confuse tension member slenderness with compression member slenderness limits, which ARE mandatory. Since L/r limits are discussed in compression chapters extensively, reviewees assume the same mandatory nature applies to tension.
Tensile yielding (phi = 0.90) always governs over rupture (phi = 0.75) because it has the higher phi factor.
Tags
- conceptual_gap
- limit_states
- common_error
Topic
Governing Limit State
Severity
major
Exam Impact
If a student assumes yield always governs, they skip computing the rupture capacity entirely and report the wrong governing strength when rupture actually controls.
The Reality
Which limit state governs depends on the relative magnitudes of: (0.90 * Fy * Ag) vs (0.75 * Fu * Ae). Either can govern. If Ae is only slightly less than Ag and Fu is significantly greater than Fy, rupture may govern. For heavily notched members or high-strength steels with large hole patterns, rupture frequently governs. In Example 1 of the reference, yield governs (535.7 kN < 561.6 kN), but this is NOT always the case.
Trap Question
Question
A 150 x 10 mm plate (Fy = 248 MPa, Fu = 400 MPa) has two 22 mm holes (d_h = 22 mm) at the critical section, U = 1.0. Which limit state governs?
Explanation
With two large holes removing 440 mm² from a 1500 mm² section, Ae is significantly reduced. Rupture (318.0 kN) is lower than yield (334.8 kN) — rupture governs. The higher phi for yield does not guarantee it governs.
Wrong Answer
Yielding governs because phi = 0.90 > 0.75.
Correct Answer
Yield: 0.90*248*(150*10) = 334,800 N = 334.8 kN. An = 1500 - 2(22)(10) = 1060 mm². Rupture: 0.75*400*1060 = 318,000 N = 318.0 kN. Rupture governs at 318.0 kN.
Misconception Id
M7
Correct Vs Incorrect
Correct Approach
Compute both limit states. Yield = 0.90 * Fy * Ag; Rupture = 0.75 * Fu * Ae. Compare — the lower value governs regardless of which phi factor is larger.
Incorrect Approach
Since phi for yield (0.90) is higher than phi for rupture (0.75), yielding always gives the lower design strength and always governs. No need to compute rupture.
Why Students Believe It
Students reason that phi = 0.90 > 0.75, so the yield limit state always produces a lower phi*Pn and therefore always controls. They overlook that the rupture equation uses Fu (higher than Fy) on the reduced area Ae (smaller than Ag), and the interplay of these factors determines which governs.
The net area An is used directly in the rupture equation without any further modification.
Tags
- formula_confusion
- shear_lag
- critical_error
Topic
Effective Net Area and Shear Lag
Severity
critical
Exam Impact
Overestimates rupture capacity for partially-connected sections. This is one of the most commonly penalized errors on board exams involving angle or channel tension members.
The Reality
NSCP 2015 Section 502.4.2 requires the use of Ae = U * An in the rupture equation, NOT An directly. U ≤ 1.0 is the shear-lag factor from NSCP 2015 Table 502.4.3-1. Only when all elements of the cross-section are connected (U = 1.0) does Ae = An. For all other cases (angles through one leg, channels through flanges only, etc.), U < 1.0 and Ae < An.
Trap Question
Question
An L75x75x8 angle (Ag = 1150 mm²) is bolted through one leg with two 20 mm bolts (d_h = 22 mm, t = 8 mm). Given U = 0.80, Fu = 400 MPa. What is the rupture design strength?
Explanation
The distinction between An and Ae is fundamental. U = 0.80 accounts for the outstanding leg's incomplete stress transfer. Ignoring U overstates the rupture capacity by 25% — a critical exam and practice error.
Wrong Answer
An = 1150 - 22(8) = 974 mm²; Rupture = 0.75 * 400 * 974 = 292,200 N (used An, ignored U)
Correct Answer
An = 974 mm²; Ae = 0.80 * 974 = 779.2 mm²; Rupture = 0.75 * 400 * 779.2 = 233,760 N = 233.8 kN
Misconception Id
M8
Correct Vs Incorrect
Correct Approach
Ae = U * An; Rupture = 0.75 * Fu * Ae = 0.75 * Fu * U * An
Incorrect Approach
Rupture = 0.75 * Fu * An (uses An directly — wrong for sections with U < 1.0)
Why Students Believe It
Students compute An by subtracting holes and then plug it directly into the rupture formula (0.75 * Fu * An), forgetting that the effective net area Ae = U * An requires a separate shear-lag factor U. The two-step process (An then Ae) is often collapsed into one step in their minds.
Tension members cannot fail by compression, so no stability or buckling check is needed — and no other checks are necessary beyond yield and rupture.
Tags
- conceptual_gap
- block_shear
- common_error
Topic
Block Shear Limit State
Severity
major
Exam Impact
Board exam problems on tension member connections frequently include block shear as the critical limit state. A student who checks only yield and rupture misses it and selects a wrong, non-governing design strength.
The Reality
NSCP 2015 Section 502.4.4 (AISC 360-10 Section J4.3) requires an additional check for BLOCK SHEAR RUPTURE at connections. The design block shear rupture strength is: phi_bs * Rn = 0.75 * [0.6 * Fu * Anv + Ubs * Fu * Ant] ≤ 0.75 * [0.6 * Fy * Agv + Ubs * Fu * Ant], where Anv = net shear area, Ant = net tension area, Agv = gross shear area, Ubs = 1.0 for uniform tension. Block shear is a THIRD limit state that can govern, particularly for short connection lengths with few bolts.
Trap Question
Question
A tension member connection is analyzed. Yield = 420 kN, Rupture = 390 kN, Block shear = 350 kN. What is the governing design tensile strength?
Explanation
All three limit states must be evaluated. Block shear (350 kN) is the lowest and therefore governs the design. Ignoring block shear leads to an unconservative design strength of 390 kN — 40 kN too high.
Wrong Answer
390 kN (student considered only yield and rupture, forgot block shear)
Correct Answer
350 kN — block shear governs as the lowest of all three limit states.
Misconception Id
M9
Correct Vs Incorrect
Correct Approach
Design tensile strength = min(yield, rupture, block shear) = min(0.90*Fy*Ag, 0.75*Fu*Ae, 0.75*[0.6Fu*Anv + Ubs*Fu*Ant]).
Incorrect Approach
Design tensile strength = min(yield, rupture) = min(0.90*Fy*Ag, 0.75*Fu*Ae). Done.
Why Students Believe It
It is true that buckling is not a concern for tension members. Students therefore conclude that only yield and rupture need to be checked, completely forgetting BLOCK SHEAR — a limit state at the connection where a block of material tears out by combined shear and tension failure.
For ASD (Allowable Stress Design), the safety factors Omega for yield and rupture are the same.
Tags
- formula_confusion
- ASD
- safety_factor
Topic
ASD Limit States
Severity
major
Exam Impact
ASD-based board problems that ask for allowable load or required area will receive wrong answers if the student applies Omega = 1.67 to both limit states.
The Reality
NSCP 2015 / AISC 360-10 ASD provisions assign: Omega_t = 1.67 for tensile yielding (Pa = Fy*Ag / 1.67) and Omega_t = 2.00 for tensile rupture (Pa = Fu*Ae / 2.00). These are the ASD equivalents of phi = 0.90 and phi = 0.75 respectively. The higher Omega for rupture (2.00 vs 1.67) reflects the same safety-level philosophy: greater caution for sudden fracture.
Trap Question
Question
Using ASD, what is the allowable rupture load for a member with Fu = 400 MPa and Ae = 1200 mm²?
Explanation
For rupture, Omega_t = 2.00 in ASD (not 1.67, which applies only to yielding). Using the wrong Omega overstates the allowable rupture load by about 20%.
Wrong Answer
Pa = 400 * 1200 / 1.67 = 287,425 N = 287.4 kN
Correct Answer
Pa = 400 * 1200 / 2.00 = 240,000 N = 240.0 kN
Misconception Id
M10
Correct Vs Incorrect
Correct Approach
ASD Yield: Pa = Fy * Ag / 1.67; ASD Rupture: Pa = Fu * Ae / 2.00. Governing = lower Pa.
Incorrect Approach
ASD Rupture: Pa = Fu * Ae / 1.67 (wrong Omega — should be 2.00 for rupture)
Why Students Believe It
Students who are more familiar with ASD from older textbooks (pre-NSCP 2015 LRFD adoption) recall a single allowable stress concept and do not realize ASD under AISC 360 also uses two different Omega factors corresponding to the two limit states.
The gross area Ag includes the area of the holes — it is just the full cross-sectional area at any section.
Tags
- conceptual_gap
- area_computation
- common_error
Topic
Gross Area vs Net Area
Severity
minor
Exam Impact
Misidentifying Ag leads to a wrong baseline for the yielding computation and also distorts the An calculation. Though less common, it still produces a wrong answer.
The Reality
Ag is the GROSS cross-sectional area of the UNPERFORATED member — the full area as if no holes exist. For a plate 200 mm wide and 12 mm thick: Ag = 200 × 12 = 2400 mm² regardless of how many holes are present. The net area An is then computed by subtracting the hole contributions from Ag. Ag is used in the YIELDING limit state because yielding occurs along the full length of the member, not just at the hole locations.
Trap Question
Question
A 200 x 12 mm plate has two 22 mm holes. What is Ag for computing the tensile yielding design strength?
Explanation
Ag is always the full, unperforated gross area. Holes reduce the section for the RUPTURE check (via An), but the YIELDING check uses Ag because yielding is a distributed phenomenon along the full member length.
Wrong Answer
Ag = (200 - 44) × 12 = 1872 mm²
Correct Answer
Ag = 200 × 12 = 2400 mm². Holes are NOT subtracted from Ag.
Misconception Id
M11
Correct Vs Incorrect
Correct Approach
Ag = 200 × 12 = 2400 mm² (full unperforated section). Then An = 2400 - 2(22)(12) = 1872 mm².
Incorrect Approach
Ag = gross width minus holes = 200 - 2(22) = 156 mm → Ag = 156 × 12 = 1872 mm² (WRONG — Ag should NOT subtract holes)
Why Students Believe It
Students sometimes confuse gross area with cross-sectional area at the hole location. They think Ag is computed at the critical section (where holes reduce the area) rather than at the full, unperforated cross-section.
A tension member is always safe from failure if the applied load is less than the yield force Fy * Ag.
Tags
- conceptual_gap
- critical_error
- limit_states
Topic
Failure Modes and Safety Philosophy
Severity
critical
Exam Impact
This misconception is conceptual and affects both exam answers and professional judgment. It can lead to accepting an unsafe design that passes a naive stress check but fails code.
The Reality
A tension member can fail by RUPTURE at the net section at a load BELOW the gross-section yield load. If the member has large holes (small Ae) and high Fu is not sufficient to compensate, the rupture capacity 0.75 * Fu * Ae can be less than 0.90 * Fy * Ag. Additionally, block shear can occur at even lower loads. The net section rupture is a brittle failure with less warning than yielding — hence the more conservative phi = 0.75. Saying 'P < Fy * Ag therefore safe' ignores two critical limit states.
Trap Question
Question
A tension member has Fy*Ag = 500 kN. The applied factored load Pu = 400 kN. A student declares the member is safe without further checks. Is this conclusion correct?
Explanation
Tensile safety requires Pu to be less than ALL three limit state capacities, not just the gross-section yield. Rupture and block shear can govern at loads well below the gross yield capacity.
Wrong Answer
Yes — Pu (400 kN) < Fy*Ag (500 kN) so the member is safe.
Correct Answer
Not necessarily. Rupture on the net section and block shear must also be checked. If 0.75*Fu*Ae = 350 kN, the member is UNSAFE despite Pu < Fy*Ag.
Misconception Id
M12
Correct Vs Incorrect
Correct Approach
Safety requires: P_u ≤ min(0.90*Fy*Ag, 0.75*Fu*Ae, block shear). All three must be checked and exceeded by the design load.
Incorrect Approach
P_u = 300 kN < Fy * Ag = 0.90 * 248 * 2400 / 1000 = 535.7 kN → member is safe. (INCOMPLETE — rupture and block shear not checked.)
Why Students Believe It
Introductory courses emphasize P = F * A and the idea that stress must not exceed yield stress. Students internalize 'stress < Fy → safe' and forget that fracture at the net section can occur at loads below the full yield capacity of the gross section.
Quick Self Check
phi_t = 0.75 for tensile rupture; phi_t = 0.90 applies only to tensile yielding. NSCP 2015 Section 502.4 is explicit on this distinction.
Statement
The resistance factor for tensile rupture (phi_t) is 0.90 under NSCP 2015 LRFD.
Ag is the FULL unperforated gross cross-sectional area. Hole deductions are applied to compute the NET area An, not Ag.
Statement
The gross area Ag is computed by subtracting the bolt hole areas from the full cross-sectional area.
When all cross-sectional elements are connected, there is no shear lag (U = 1.0). This applies to plates bolted through their full width, giving Ae = 1.0 * An = An.
Statement
For a plate connected across its entire width, the shear-lag factor U = 1.0 and Ae = An.
L/r ≤ 300 is a RECOMMENDATION for serviceability (to avoid sag and vibration), not a mandatory code limit. Exceeding it does not reduce the computed design tensile strength.
Statement
The L/r ≤ 300 slenderness limit for tension members is a mandatory strength requirement under NSCP 2015.
The formula is: net width = Wg - sum(d_h) + sum(s²/4g). The s²/4g term adds to the net width for each diagonal path segment, partially compensating for the hole deductions.
Statement
For staggered bolt holes, the s²/4g term is ADDED (not subtracted) to the net width when evaluating a zig-zag failure path.
Either limit state can govern. Rupture uses Fu (> Fy) on Ae (< Ag). When holes significantly reduce Ae, rupture can produce a lower design strength than yielding despite the lower phi.
Statement
Tensile yielding always governs over tensile rupture because phi = 0.90 for yield is greater than phi = 0.75 for rupture.
NSCP 2015 Section 502.4.4 requires checking block shear rupture at connections. It is a third limit state involving combined shear and tension failure of a material block, and it can govern the design.
Statement
Block shear is a limit state that must be checked at the connection of a tension member, in addition to yielding and rupture.
U < 1.0 for angles connected through one leg only. The outstanding (unconnected) leg experiences shear lag — it does not carry stress as efficiently as the connected leg. U is determined from NSCP 2015 Table 502.4.3-1.
Statement
An angle section bolted through one leg has U = 1.0 because all bolt forces are transmitted through the section.
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