Example 6.8: Plate with Staggered Holes and Single and Double Angle Sections Determine the tension capacity of the plate and the angle sections shown in Figures (a), (b) and (c), considering the strength of the sections only (i.e. ignoring block shear and bolt strength).

(a) Plate: Clause 3.4.3 Ae = ae = Ke an ≤ ag ≤ 1.2 (an1 + an2 ) Clause 3.4.4.3 A series of staggered failure patterns must be considered to determine the critical case.

Tension Member Example Monday, October 26, 2015 8:27 AM

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Path A: (two holes) Area to be deducted = 2 (t × D) = (2 × 8 × 22) = 352 mm2

Path B: (two holes and one diagonal path) Area to be deducted = [2 (t × D) − 0.25s2t/g] = [352 − (0.25 × 502 × 10)/50] = 227 mm2

Path C: (three holes and two diagonal paths) Area to be deducted = [3 (t × D) − 2 (0.25s2t/g)] = [(3 × 8 × 22) − (2 × 0.25 × 502 × 10)/50] = 278 mm2

Path A is the most critical and hence: An = [(160 × 8) − 352)] = 928 mm2 Ag = (160 × 8) = 1280 mm2

Clause 3.4.3

Ke = 1.2 for S 275 steel Ae = Ke an = (1.2 × 928) = 1113.6 mm2

Clause 4.6.1

Pt = (py× Ae ) = (275 × 1113.6)/103 = 306.2 kN

(b) Single angle: (i) Consider the connection to be bolted Clause 4.6.3.1 Pt = py× (Ae − 0.5a2) Clause 3.4.3 Ae = (ae1 + ae2) ≤ 1.2 (an1 + an2 ) ae1 = effective area of the connected leg = Ke an1 ≤ a1

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Clause 3.4.3 Ae = (ae1 + ae2) ≤ 1.2 (an1 + an2 ) ae1 = effective area of the connected leg = Ke an1 ≤ a1 ae2 = effective area of the unconnected leg = Ke an2 ≤ a2 an1 = (a1 − area of bolt holes) an2 = (a2 − area of bolt holes) Section Tables Ag = 9.41cm2

Clause 4.6.3.1 a1 = (75 × 8) = 600 mm2 a2 = (Ag − a1) = (941 − 600) = 341 mm2 an1 = [600 − (8 × 18)] = 456 mm2 an2 = 341 mm2 Clause 3.4.3 Ke = 1.2 ae1 = (1.2 × 456) = 547.2 mm2 ≤ a1 ae2 = 341 mm2 ≤ a2 Ae = (ae1 + ae2) = (547.2 + 341) = 888.2 mm2 1.2 (an1 + an2 ) = 1.2 (456 + 341) = 956.4 mm2 ∴Ae ≤ 1.2 (an1 + an2 ) Clause 4.6.3.1 Pt = py× (Ae − 0.5a2) = {275 × [888.2 − (0.5 × 341)]}/103 = 197.4 kN (ii) Consider the connection to be welded Clause 4.6.3.1 Pt = py× (Ag − 0.3a2) = {275 × [941.0 − (0.3 × 341)]}/103 = 230.6 kN

Bolted Single Angle Capacity: 197.4 kN

Welded Single Angle Capacity: 230.6 kN

(c) Double angle: (i) Consider the connection to be bolted Clause 4.6.3.2 Pt = py× (Ae − 0.25a2) Section Tables Ag = 18.8 cm2 (= 9.41 cm2/angle) Clause 4.6.3.2 week 8 Page 3

Section Tables Ag = 18.8 cm2 (= 9.41 cm2/angle) Clause 4.6.3.2 Evaluate area/angle:

a1 = (75 × 8) = 600 mm2 a2 = (Ag − a1) = (941 − 600) = 341 mm2 an1 = [600 − (8 × 14)] = 488 mm2 an2 = 341 mm2 Clause 3.4.3 Ke = 1.2 ae1 = (1.2 × 488) = 585.6 mm2 ≤ a1 ae2 = 341 mm2 ≤ a2 Ae = (ae1 + ae2) = (585.6 + 341) = 926.6 mm2 1.2 (an1 + an2 ) = 1.2 (488 + 341) = 994.8 mm2 ∴ Ae ≤ 1.2 (an1 + an2 )

Tension capacity for both angles: Clause 4.6.3.2

Pt = 2 [py× (Ae − 0.25a2)] = 2 {275 × [926.6 − (0.25 × 341)]}/103 = 462.7 kN (ii) Consider the connection to be welded Clause 4.6.3.2 Pt = 2 [py× (Ag − 0.15a2)] = 2 {275 × [941.0 − (0.15 × 341)]}/103 = 489.0 k

Bolted Double Angle Capacity: 462.7 kN Welded Double Angle Capacity: 489.0 kN

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Comp. member example Monday, October 26, 2015 8:55 AM

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Example Com.Monday, October 26, 2015 9:05 AM

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