15.1 Structural Steel Tension Members and Connections

15.1 Structural Steel Tension Members and Connections

Tension members are structural elements subjected to direct axial tensile forces that tend to elongate the member along its longitudinal axis. Common applications in Philippine civil engineering include truss chords and web diagonals, roof bracing systems, bridge hangers, cable stays, and sag rods for purlins. Because axial tension suppresses global flexural buckling, tension members represent the most efficient structural steel configurations. However, their design is critical because failure frequently initiates at connections where holes, weld heat-affected zones, and stress concentrations reduce cross-sectional capacity.

In the National Structural Code of the Philippines (NSCP 2015 Chapter 5, Section 504 for tension members), which is based on the AISC 360-10 Specification, tension members may be designed using either Load and Resistance Factor Design (LRFD) or Allowable Strength Design (ASD).

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1. Primary Limit States for Tension Members

A structural steel tension member must possess adequate strength to resist factored tensile loads ($P_u$) in LRFD or allowable service loads ($P_a$) in ASD across three fundamental limit states:

1. Tensile Yielding on the Gross Section ($A_g$): Prevents excessive plastic elongation of the member along its main unperforated body under service loads.
2. Tensile Rupture on the Net Effective Section ($A_e$): Prevents sudden catastrophic fracture across the reduced cross-section at bolt holes or connection ends.
3. Block Shear Rupture: Prevents a combined shear and tension tearout mechanism through the connection boundary.

Where:

• $F_y$ = specified minimum yield stress of the structural steel (MPa).
• $F_u$ = specified minimum tensile strength of the structural steel (MPa).
• $A_g$ = gross cross-sectional area of the member (mm²).
• $A_e$ = effective net cross-sectional area (mm²).

Slenderness Recommendation

Although tension members do not suffer from compressive buckling, excessive flexibility causes unsightly sag, wind flutter, and vibrational fatigue during transport and erection. NSCP 2015 Section 504 (AISC D1) recommends a maximum slenderness ratio of:

$\frac{L}{r} \le 300$

where $L$ is the unbraced length and $r$ is the minimum radius of gyration ($r = \sqrt{I/A}$). This limit is not mandatory for rods and cables in tension.

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2. Net Area ($A_n$) and Staggered Bolt Holes

When holes are drilled or punched in a tension member, the cross-sectional area is reduced. The net area ($A_n$) is the gross cross-sectional area minus the projected area of bolt holes, plus any adjustments for staggered alignments.

Design Hole Diameter Calculation

Under AISC 360-10 Section B4.3b (adopted in NSCP 2015), the hole width deducted in tension calculations is the nominal hole plus 2 mm. For standard holes on bolts up to 24 mm, that is 4 mm larger than the nominal bolt diameter ($d_{bolt}$):

$d_h = d_{bolt} + 2\text{ mm (standard fabrication clearance)} + 2\text{ mm (damage allowance)} = d_{bolt} + 4\text{ mm}$

(For U.S. Customary units, $d_h = d_{bolt} + \frac{1}{16}'' + \frac{1}{16}'' = d_{bolt} + \frac{1}{8}''$.)

Staggered Bolt Hole Configurations (Cochrane's Rule)

When bolt holes are staggered along adjacent gage lines, failure may occur along a straight transverse plane or along a zigzag diagonal path traversing multiple holes. To account for combined normal and shear stresses along the diagonal path, V. H. Cochrane formulated the empirical pitch-gage adjustment term $\frac{s^2}{4g}$:

$w_n = w_g - \sum d_h + \sum \frac{s^2}{4g}$

$A_n = A_g - \sum (d_h \cdot t) + \sum \left( \frac{s^2}{4g} \cdot t \right)$

Where:

• $w_g$ = gross width of the plate or unrolled angle (mm).
• $t$ = thickness of the member element (mm).
• $s$ = longitudinal center-to-center pitch between consecutive staggered holes (mm).
• $g$ = transverse center-to-center gage between adjacent hole centerlines (mm).

When evaluating structural angles with holes in both legs, the angle is unfolded into an equivalent flat plate. The gross width equals the sum of the leg lengths minus the angle thickness ($w_g = L_1 + L_2 - t$). The transverse gage distance ($g$) between a hole in one leg and a hole in the other leg equals the sum of their distances from the outer heel minus the angle thickness ($g = g_1 + g_2 - t$).

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3. Effective Net Area and the Shear Lag Factor ($U$)

When an axial tensile force is transferred to a member through only some, but not all, of its cross-sectional elements (for example, a single angle connected only through one leg, or a wide-flange W-beam connected only through its flanges), the connected element carries a disproportionately high stress near the connection. The unconnected elements lag behind in developing stress. This non-uniform tensile stress distribution is called shear lag.

To account for shear lag, the net area ($A_n$) is reduced to an effective net area ($A_e$):

$A_e = U A_n$

(For welded connections where tensile force is transmitted directly to all elements without bolt holes, $A_e = U A_g$.)

Determination of the Shear Lag Factor ($U$)

NSCP 2015 Section 504 adopts the shear lag cases of AISC 360-10 Table D3.1:

1. General Case (Bolted or Welded): When tension load is transmitted to some elements by bolts or longitudinal welds: $U = 1 - \frac{\bar{x}}{L}$ Where:

   • $\bar{x}$ = connection eccentricity, defined as the perpendicular distance from the connection shear plane to the centroid of the cross-section resisting the load.
   • $L$ = length of the connection along the line of force (distance between first and last bolt in a line, or length of longitudinal weld).

2. Plates where tension is transmitted solely by bolts: All elements connected $\implies U = 1.0$.

3. Welded Transverse Splices: If welds are purely transverse, $A_e = A_{\text{connected}}$ and $U = 1.0$.

4. Plates with Longitudinal Welds Only:

   • If $L \ge 2w$: $U = 1.00$
   • If $2w > L \ge 1.5w$: $U = 0.87$
   • If $1.5w > L \ge w$: $U = 0.75$ (where $w$ is the plate width between welds, and $L$ must be at least equal to $w$).

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4. Block Shear Rupture Mechanics

Block Shear Rupture is a tearing limit state where a segment or "block" of steel rips away from the member at the connection perimeter. The failure surface consists of two perpendicular planes:

• A shear plane parallel to the applied tensile force.
• A tension plane perpendicular to the applied tensile force.

Governing Formula (AISC 360-10 Eq. J4-5, adopted in NSCP 2015 Section 510)

The nominal block shear rupture strength $R_n$ is given by:

$R_n = 0.60 F_u A_{nv} + U_{bs} F_u A_{nt} \le 0.60 F_y A_{gv} + U_{bs} F_u A_{nt}$

$\phi R_n = 0.75 R_n \quad (\text{LRFD}), \qquad \frac{R_n}{\Omega} = \frac{R_n}{2.00} \quad (\text{ASD})$

Where:

• $A_{gv}$ = gross area subject to shear (total length of shear planes $\times$ thickness).
• $A_{nv}$ = net area subject to shear (gross shear length minus deducted hole diameters $\times$ thickness).
• $A_{nt}$ = net area subject to tension (gross tension width minus deducted hole diameters $\times$ thickness).
• $0.60 F_u$ = ultimate shear rupture stress ($F_{vu} \approx F_u / \sqrt{3} \approx 0.60 F_u$ via von Mises yield criterion).
• $0.60 F_y$ = shear yield stress.
• $U_{bs}$ = shear lag reduction factor for block shear:
  • $U_{bs} = 1.0$ when the tension stress distribution across the tension plane is uniform (typical angles, gusset plates, and coped beams with a single bolt line).
  • $U_{bs} = 0.5$ when tension stress is non-uniform (e.g., coped beams with multiple bolt lines where inner bolts yield before outer bolts).

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5. Bolted Connections in Tension Systems

Bolted steel connections are categorized as either bearing-type connections or slip-critical connections.

Bearing-Type vs. Slip-Critical Connections

• Bearing-Type: Bolts act as pins. Load is transferred through mechanical bearing of the bolt shank against the edge of the hole and shear across the bolt cross-section. Some initial slip of the connection occurs until bolts bear against the holes.
• Slip-Critical: High-strength bolts (ASTM F3125 Grade A325 or A490) are tightened to a specified minimum clamping pretension ($T_b$). Load is transferred purely by friction developed between the mating faying surfaces. Slip-critical design is required in structures subject to fatigue, cyclic load reversal, or where slip would impair structural alignment.

Bearing-Type Limit States

For bearing connections, each bolt must satisfy bolt shear, hole bearing, and tearout:

1. Nominal Bolt Shear Strength ($R_n$): $R_n = F_{nv} A_b$ Where $F_{nv}$ is the nominal shear stress from NSCP Table 510.3.2 (e.g., for Group A / A325 bolts: $F_{nv} = 372\text{ MPa}$ when threads are included in shear planes 'N', and $F_{nv} = 469\text{ MPa}$ when threads are excluded 'X'); $A_b = \frac{\pi d_{bolt}^2}{4}$ is the nominal unthreaded shank area.

2. Bearing and Tearout at Bolt Holes: Under NSCP 2015 Section 510.3.10, the nominal strength per hole is:

   • When deformation around the bolt hole at service load is a design consideration: $R_n = 1.2 l_c t F_u \le 2.4 d_{bolt} t F_u$
   • When deformation around the bolt hole at service load is not a design consideration: $R_n = 1.5 l_c t F_u \le 3.0 d_{bolt} t F_u$ Where:
   • $l_c$ = clear distance in the direction of force between the edge of the hole and the edge of the adjacent hole or edge of the material (mm).
   • $t$ = thickness of the connected material (mm).
   • $d_{bolt}$ = nominal diameter of the bolt (mm).
   • The term $1.2 l_c t F_u$ represents the tearout limit state, while $2.4 d_{bolt} t F_u$ represents the hole bearing deformation limit state.

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6. Welded Connections: Fillet Welds

Welded connections join steel components through metallurgical fusion using electric arc processes (SMAW, GMAW, FCAW, SAW). Fillet welds are the most common structural weld type due to minimal edge preparation requirements.

Fillet Weld Geometry and Throat

A standard fillet weld has a triangular cross-section with leg size $w$. The critical failure plane is the effective throat thickness ($t_e$), defined as the shortest distance from the root of the joint to the face of the diagrammatic weld:

$t_e = w \cdot \cos 45^\circ = \frac{w}{\sqrt{2}} \approx 0.707 w$

Nominal Strength of Fillet Welds

The nominal shear strength of the weld metal per unit area is $F_{nw} = 0.60 F_{EXX}$, where $F_{EXX}$ is the classification strength of the welding electrode (e.g., for E70xx electrodes, $F_{EXX} = 70\text{ ksi} \approx 485\text{ MPa}$):

$R_n = F_{nw} A_{we} = (0.60 F_{EXX}) \cdot (0.707 w L)$

$\phi R_n = 0.75 R_n \quad (\text{LRFD}), \qquad \frac{R_n}{\Omega} = \frac{R_n}{2.00} \quad (\text{ASD})$

Directional Strength Enhancement

NSCP 2015 Section 510.2.4 permits an increase in nominal fillet weld strength when the angle of loading ($\theta$) deviates from the longitudinal axis of the weld:

$R_n = 0.60 F_{EXX} (1.0 + 0.50 \sin^{1.5} \theta) A_{we}$

• For a purely longitudinal weld ($\theta = 0^\circ$): $1.0 + 0.50 \sin^{1.5}(0^\circ) = 1.00$.
• For a purely transverse weld ($\theta = 90^\circ$): $1.0 + 0.50 \sin^{1.5}(90^\circ) = 1.50$ (a 50% increase in nominal shear capacity).

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7. Eccentrically Loaded Bolt Groups (Elastic Method)

When a bracket load $P$ acts at eccentricity $e$ from the centroid of a bolt group, each bolt resists a direct shear and a torsional shear. The plates are assumed rigid, and the group rotates about its centroid.

• Direct shear on each of $n$ bolts: $R_{d} = P/n$, parallel to $P$.
• Torsional moment: $M = P e$.
• Torsional shear on a bolt at distance $r$ from the centroid: $R_t = M r / \sum r^2$, perpendicular to $r$.
• Components, with $\sum r^2 = \sum (x^2 + y^2)$: $R_{tx} = M y / \sum r^2$ and $R_{ty} = M x / \sum r^2$.
• Combine vectorially for the most remote bolt: $R = \sqrt{(R_{tx})^2 + (R_{ty} + R_d)^2}$ for a vertical load $P$.

Example. A vertical load $P = 90\text{ kN}$ acts at $e = 250\text{ mm}$ from the centroid of 6 bolts in two vertical lines $150\text{ mm}$ apart, with rows at $y = -100, 0, +100\text{ mm}$.

• $\sum r^2 = 6(75)^2 + 4(100)^2 = 33{,}750 + 40{,}000 = 73{,}750\text{ mm}^2$.
• $M = 90(250) = 22{,}500\text{ kN}\cdot\text{mm}$.
• For a corner bolt: $R_{tx} = 22{,}500(100)/73{,}750 = 30.5\text{ kN}$ and $R_{ty} = 22{,}500(75)/73{,}750 = 22.9\text{ kN}$.
• Direct shear $R_d = 90/6 = 15.0\text{ kN}$.
• $R = \sqrt{30.5^2 + (22.9 + 15.0)^2} = \sqrt{930 + 1{,}436} = 48.6\text{ kN}$.

Each bolt must have a design shear strength $\phi R_n \ge 48.6\text{ kN}$. The elastic method is conservative compared with the AISC instantaneous-center method.

8. Comprehensive Worked Examples

Worked Example 1: Staggered Tension Plate Analysis

Problem: An A36 steel plate ($F_y = 248\text{ MPa}$, $F_u = 400\text{ MPa}$) with dimensions $250\text{ mm} \times 12\text{ mm}$ carries an axial tensile load. The plate is connected using $20\text{ mm}$ diameter bolts arranged in two gage lines spaced $75\text{ mm}$ apart, with an edge distance of $87.5\text{ mm}$ on each side. The bolts have a staggered pitch of $s = 50\text{ mm}$. Assuming all elements are connected ($U = 1.0$), compute: (a) the net width and governing net area $A_n$, (b) the design tensile rupture strength $\phi P_n$ (LRFD), and (c) the design tensile yielding strength $\phi P_n$ (LRFD).

Solution:

• Step 1: Bolt Hole Diameter Deduction: $d_h = d_{bolt} + 4\text{ mm} = 20 + 4 = 24\text{ mm}$

• Step 2: Net Width Evaluation along Potential Paths:

  • Path A (Straight path through 1 hole): $w_{n1} = w_g - 1 \cdot d_h = 250 - 24 = 226\text{ mm}$
  • Path B (Zigzag path through 2 staggered holes): $w_{n2} = w_g - 2 \cdot d_h + \frac{s^2}{4g} = 250 - 2(24) + \frac{50^2}{4(75)} = 202 + \frac{2500}{300} = 202 + 8.33 = 210.33\text{ mm}$ The critical failure path is Path B because $w_{n2} < w_{n1}$. Governing net width $w_n = 210.33\text{ mm}$.

• Step 3: Net Area: $A_n = w_n \cdot t = 210.33\text{ mm} \times 12\text{ mm} = 2,524\text{ mm}^2$ With $U = 1.0$, $A_e = U A_n = 2,524\text{ mm}^2$.

• Step 4: Design Tensile Rupture Strength (LRFD): $\phi_t P_n = 0.75 \cdot F_u A_e = 0.75 \times 400\text{ MPa} \times 2,524\text{ mm}^2 = 757,200\text{ N} = 757.2\text{ kN}$

• Step 5: Design Tensile Yielding Strength (LRFD): $A_g = 250 \times 12 = 3,000\text{ mm}^2$ $\phi_t P_n = 0.90 \cdot F_y A_g = 0.90 \times 248\text{ MPa} \times 3,000\text{ mm}^2 = 669,600\text{ N} = 669.6\text{ kN}$ Conclusion: The design strength of the member in LRFD is governed by gross section yielding: $\phi P_n = 669.6\text{ kN}$.

Worked Example 2: Block Shear Rupture of a Gusset Plate

Problem: A $10\text{ mm}$ thick gusset plate of A36 steel ($F_y = 248\text{ MPa}$, $F_u = 400\text{ MPa}$) connects a tension truss diagonal using a single line of four $22\text{ mm}$ diameter bolts ($d_h = 26\text{ mm}$). The bolt pitch is $75\text{ mm}$, and the end distance along the line of load is $40\text{ mm}$. The edge distance perpendicular to load is $40\text{ mm}$. Tensile stress distribution is uniform ($U_{bs} = 1.0$). Calculate the nominal and design block shear rupture strength (LRFD).

Solution:

• Step 1: Calculate Geometric Areas:

  • Total length of shear line = $3 \times 75\text{ mm} + 40\text{ mm} = 265\text{ mm}$.
  • $A_{gv} = 265\text{ mm} \times 10\text{ mm} = 2,650\text{ mm}^2$.
  • Net shear length = $265 - 3.5(d_h) = 265 - 3.5(26) = 265 - 91 = 174\text{ mm}$.
  • $A_{nv} = 174\text{ mm} \times 10\text{ mm} = 1,740\text{ mm}^2$.
  • Gross tension length = $40\text{ mm}$.
  • Net tension length = $40 - 0.5(d_h) = 40 - 0.5(26) = 27\text{ mm}$.
  • $A_{nt} = 27\text{ mm} \times 10\text{ mm} = 270\text{ mm}^2$.

• Step 2: Evaluate Block Shear Equation Terms:

  • Rupture Term: $0.60 F_u A_{nv} + U_{bs} F_u A_{nt} = 0.60(400)(1,740) + (1.0)(400)(270) = 417,600 + 108,000 = 525,600\text{ N} = 525.6\text{ kN}$.
  • Yield Cap: $0.60 F_y A_{gv} + U_{bs} F_u A_{nt} = 0.60(248)(2,650) + (1.0)(400)(270) = 394,320 + 108,000 = 502,320\text{ N} = 502.32\text{ kN}$.
  • Governing Nominal Strength: Since $525.6\text{ kN} > 502.32\text{ kN}$, the upper yield cap governs: $R_n = 502.32\text{ kN}$.

• Step 3: Design Strength (LRFD): $\phi R_n = 0.75 \times 502.32\text{ kN} = 376.74\text{ kN}$

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9. Licensure Exam Pitfalls & Review Notes