CIV-331 · Lesson 3 of 12 · 21 min
Tension members
Tension members are the simplest steel members to design, because tension straightens a member rather than buckling it. They are still worth careful study, because the connection at each end removes material and concentrates stress, and most tension members are governed by what happens at their ends. This lesson designs the bottom chord of the maintenance bay roof truss and introduces the three limit states of every tension member: yielding of the gross section, fracture of the net section and block shear.
Where tension members are used
Truss bottom chords and the diagonals of a Pratt truss under gravity load, hangers that carry a floor or a hoist from above, sag rods that support girts and purlins, and the diagonals of tension-only bracing are all tension members. Their cross-sections range from round bars and rods to angles, double angles, tees, channels, W shapes, HSS and plates. The choice usually depends on how the member is connected: a double angle or tee bolts or welds easily to a single gusset plate, and a W shape suits heavy truss chords with connections on both flanges.
Yielding of the gross section
Away from the connections, a tension member yields over its whole cross-section when the stress reaches \(F_y\). Yielding over a long length would stretch the member enough to distort the structure, so it is treated as an ultimate limit state:
Net area
At a bolted connection, the holes remove material and the stress across the remaining net section is higher than in the body of the member. Fracture across the net section is the second limit state. The net area is the gross area minus the area of the holes along the critical path:
Holes are normally 2 mm larger than the bolt. Punching damages a thin ring of material around the hole, so S16 deducts a further 2 mm when holes are punched. For an M20 bolt in a punched hole the allowance is 24 mm; for a drilled hole it is 22 mm. Fabricators decide how to make the holes, so designers usually assume punched holes unless the drawings require drilling.
When holes are staggered, the fracture path can zigzag from hole to hole. A diagonal step is less weakening than a straight cut through the same number of holes, because part of the path is inclined. The empirical correction, proposed by Cochrane in 1922 and still used, adds \(s^2/4g\) for each diagonal link in the path:
where \(s\) is the stagger (the pitch between holes along the member) and \(g\) is the gauge (the transverse spacing between lines of holes). Every reasonable path is checked and the one with the smallest net width governs. For an angle, the gauge between holes in the two legs is measured along the centreline of the thickness: the sum of the two gauges from the heel minus the thickness.
Shear lag and the effective net area
When only part of a cross-section is connected, such as one leg of an angle, the force enters the member through the connected part and spreads into the unconnected part over some length. Near the connection the unconnected part carries less than its share, and the net section fails at a lower load than its full area would suggest. This is shear lag. The longer the connection and the closer the centroid of the member is to the plane of the connection, the smaller the loss.
S16 accounts for shear lag with an effective net area \(A_{ne}\) that is smaller than \(A_n\). For angles connected by one leg, \(A_{ne} = 0.80A_n\) when there are four or more bolts in each line and \(0.60A_n\) when there are fewer. For W shapes and channels connected by the flanges only, and for welded connections, the standard gives other reductions that depend on the shape, the number of bolts and the length of weld. The fracture resistance is
Because the net section uses \(F_u\) with a lower resistance factor, and the gross section uses \(F_y\) with a higher one, a member can be governed by either. For ductile behaviour it is better if the gross section yields first, which needs a net-to-gross ratio of roughly \(F_y/F_u\) or more after the factors.
Block shear
A connection can also tear out as a block. The block separates along a shear plane parallel to the force, through the line of bolts, and a tension plane perpendicular to it. Block shear governs short connections with few bolts and thin material, and it must be checked in the member and in the gusset plate.
S16 combines fracture on the net tension area with yielding on the gross shear area, using an average of \(F_y\) and \(F_u\) for the shear plane:
The efficiency factor \(U_t\) is 1.0 when the block is symmetric and the tension plane is loaded uniformly, as in a gusset with two lines of bolts loaded through its centre. It is smaller when the tension is uneven: 0.6 for an angle connected by one leg and for a coped beam with one line of bolts, and lower still for a coped beam with two lines.
Slenderness
Tension members do not buckle, but a very slender member sags under its own weight, vibrates and is easily damaged in handling. S16 limits the slenderness ratio \(L/r\) of tension members to 300, except for rods and cables, which are usually installed with a little pretension.
Designing the bottom chord
The truss bottom chord L2L3 carries \(T_f = 705\) kN (lesson 1). Owen chooses a double angle, two L102x102x9.5 back to back with a 10 mm gap for the gusset plates. Each angle is connected to the gussets by one line of four M20 ASTM A325 bolts in punched holes, at 75 mm pitch with a 40 mm end distance, on a gauge line 64 mm from the heel.
Worked problem
Bottom chord of the roof truss
Check two L102x102x9.5 (350W; each angle \(A = 1850\) mm², \(t = 9.5\) mm, \(\bar x = 29.0\) mm) for \(T_f = 705\) kN, with one line of four M20 bolts per angle.
Show solution
- Gross yielding:
- \[T_r = 0.90(3700)(350) = 1166\ \text{kN}\]
- Net fracture. One hole per angle on the critical section, with a 24 mm allowance:
- \[A_n = 2(1850 - 24 \times 9.5) = 3244\ \text{mm}^2 \qquad A_{ne} = 0.80(3244) = 2595\ \text{mm}^2\]
- \[T_r = 0.75(2595)(450) = 876\ \text{kN}\]
- Block shear in each angle, with \(U_t = 0.6\). The shear plane runs 40 + 3(75) = 265 mm along the bolt line, and the tension plane runs from the bolt line to the toe:
- \[A_n = (102 - 64 - 12)(9.5) = 247\ \text{mm}^2 \qquad A_{gv} = 265(9.5) = 2518\ \text{mm}^2\]
- \[T_r = 2(0.75)\Big[0.6(247)(450) + 0.6(2518)\,\frac{350 + 450}{2}\Big] = 1006\ \text{kN}\]
- Slenderness in the plane of the truss: \(L/r = 3000/31.2 = 96 < 300\).
Answer: Net fracture governs: \(T_r = 876\) kN against \(T_f = 705\) kN, a utilization of 0.80. The double angle is adequate.
Note that net fracture governs here, not gross yielding. The holes and the 0.80 shear lag factor remove more than a fifth of the strength. A longer connection does not help in S16 once there are four bolts in a line, but welding the angles to the gusset, as lesson 10 shows, avoids the hole deduction altogether.
Sofia's program makes the same checks and adds a staggered-hole example: a hanger plate for the hoist with four lines of holes, alternate lines staggered.
# Truss bottom chord L2L3: 2L102x102x9.5, 350W, one line of 4 M20 bolts per angle
Fy, Fu = 350.0, 450.0
phi, phiu = 0.90, 0.75
Tf = 705.0 # kN, from lesson 1
A1, t, xbar = 1850.0, 9.5, 29.0 # one angle: area (mm2), thickness, centroid from back
leg, g = 102.0, 64.0 # leg length and gauge from the heel (mm)
ha = 20 + 4 # hole allowance for M20 in punched holes (mm)
n_bolts, pitch, end = 4, 75.0, 40.0
Ag = 2 * A1
Tr_yield = phi * Ag * Fy / 1e3
An = 2 * (A1 - ha * t)
An_eff = (0.80 if n_bolts >= 4 else 0.60) * An
Tr_frac = phiu * An_eff * Fu / 1e3
print(f"Gross yielding: Tr = 0.90 x {Ag:.0f} x {Fy:.0f} = {Tr_yield:.0f} kN")
print(f"Net fracture: An = {An:.0f}, An' = {An_eff:.0f} mm2, Tr = {Tr_frac:.0f} kN")
# Block shear in each angle: shear along the bolt line, tension to the toe of the leg
Ut = 0.6
Agv = (end + (n_bolts - 1) * pitch) * t
Ant = (leg - g - ha / 2) * t
Tr_block = 2 * phiu * (Ut * Ant * Fu + 0.6 * Agv * (Fy + Fu) / 2) / 1e3
print(f"Block shear: Ant = {Ant:.0f}, Agv = {Agv:.0f} mm2 per angle, Tr = {Tr_block:.0f} kN (two angles)")
Tr = min(Tr_yield, Tr_frac, Tr_block)
print(f"Governing Tr = {Tr:.0f} kN, Tf = {Tf:.0f} kN, utilization = {Tf/Tr:.2f}")
r = 31.2 # mm, radius of gyration of the pair in the plane of the truss
print(f"Slenderness L/r = {3000/r:.0f} (limit 300)")
# Staggered holes in a 250 x 12 mm hanger plate: gauge 60 mm, stagger 50 mm
b, tp, gs, s = 250.0, 12.0, 60.0, 50.0
paths = {"straight, 2 holes": b - 2 * ha,
"zigzag, 4 holes": b - 4 * ha + 3 * s**2 / (4 * gs)}
for name, wn in paths.items():
print(f"Plate path {name:18s}: net width = {wn:.1f} mm")
wn = min(paths.values())
print(f"Critical net area = {wn*tp:.0f} mm2")Gross yielding: Tr = 0.90 x 3700 x 350 = 1166 kN Net fracture: An = 3244, An' = 2595 mm2, Tr = 876 kN Block shear: Ant = 247, Agv = 2518 mm2 per angle, Tr = 1006 kN (two angles) Governing Tr = 876 kN, Tf = 705 kN, utilization = 0.80 Slenderness L/r = 96 (limit 300) Plate path straight, 2 holes : net width = 202.0 mm Plate path zigzag, 4 holes : net width = 185.2 mm Critical net area = 2223 mm2
Worked problem
Hanger plate with staggered holes
A 250 × 12 mm hanger plate has four lines of M20 bolt holes at a 60 mm gauge, with alternate lines staggered 50 mm. Find the critical net area.
Show solution
- Straight path through two holes in one cross-section:
- \[w_n = 250 - 2(24) = 202\ \text{mm}\]
- Zigzag path through all four holes, with three diagonal links:
- \[w_n = 250 - 4(24) + 3\,\frac{50^2}{4(60)} = 250 - 96 + 31.3 = 185\ \text{mm}\]
Answer: The zigzag path governs: \(A_n = 185(12) = 2223\) mm².
Rods, cables and pin-connected members
Threaded rods are used for hangers, sag rods and light bracing. Their strength is governed by the area at the root of the threads, and S16 treats them like bolts in tension, with a resistance based on a fraction of the gross area times \(F_u\). Cables and strands have their own standards. Pin-connected plates and eyebars, once common in bridges, appear today in exposed structures; their proportions are limited so that the plate does not fracture or dish around the pin.
Practical points
- Size tension members by the connection, not by the body. Start from the bolt or weld pattern and check net fracture and block shear first.
- Keep gussets compact and connections concentric. Eccentric connections add bending that the simple checks do not cover.
- For double angles, add stitch bolts or fillers at intervals so the two angles act together and do not rattle.
- A truss chord also carries load reversal in some combinations: under wind uplift the bottom chord may go into compression and need bracing to the roof.
Common mistakes
MistakeUsing the gross area for fracture.
InsteadFracture uses the effective net area, reduced for holes and shear lag.
MistakeDeducting only the bolt diameter for each hole.
InsteadDeduct the bolt diameter plus 2 mm, plus another 2 mm for punched holes.
MistakeChecking only the straight path when holes are staggered.
InsteadCheck every plausible zigzag path and add \(s^2/4g\) for each diagonal link.
MistakeForgetting block shear because the member area is large.
InsteadBlock shear depends on the connection geometry and often governs short connections.
Summary and key terms
- Gross yielding: \(T_r = \phi A_g F_y\) with \(\phi = 0.90\).
- Net fracture: \(T_r = \phi_u A_{ne} F_u\) with \(\phi_u = 0.75\) and shear lag reductions such as 0.80 or 0.60 for angles.
- Staggered holes: \(w_n = w - \sum d_h + \sum s^2/4g\); the path with the smallest net width governs.
- Block shear combines tension fracture with shear on the gross area at \((F_y + F_u)/2\) and an efficiency factor \(U_t\).
- The bottom chord, two L102x102x9.5, has \(T_r = 876\) kN against 705 kN, governed by net fracture.
- Gross area
- The full cross-sectional area of a member.
- Net area
- The area remaining after deducting the holes along a fracture path.
- Shear lag
- The uneven stress in a member near a connection that engages only part of the cross-section.
- Effective net area
- The net area reduced for shear lag, An′ or Ane in S16.
- Stagger
- The longitudinal spacing between holes in adjacent lines.
- Gauge
- The transverse spacing between lines of holes.
- Block shear
- Tearing out of a block of material along combined shear and tension planes at a connection.
- Hole allowance
- The width deducted for each hole in computing net area.
Practice problems
Work each problem on paper first, then reveal the answer to check.
A single L102x102x9.5 brace is connected by one leg with three M20 bolts. Find \(T_r\) for net fracture.
Show answer
\(A_n = 1850 - 228 = 1622\); \(A_{ne} = 0.60(1622) = 973\) mm²; \(T_r = 0.75(973)(450) = 328\) kN.
Why does a punched hole get a larger deduction than a drilled hole?
Show answer
Punching damages and work-hardens a ring of material around the hole, which may crack.
Find the yield resistance of a 25 mm round rod hanger of 350W steel, ignoring threads.
Show answer
\(T_r = 0.90(491)(350) = 155\) kN on the gross area; the threaded portion will govern.
What happens to block shear strength if the end distance is increased?
Show answer
The shear plane lengthens, so block shear strength rises.
Lesson quiz · 5 questions
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