Steel Structures I - Connections PDF

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School of Civil Engineering CIVL3206 Steel Structures 1 Tutorial Set 5 – Connections - Solution Properties from OneSteel handbook 150UC30.0 Grade 300, Ag = 3860 mm 2, fy = 320 MPa, fu = 440 MPa. tf =9.4 mm 120×16 plate, A g = 1200 ×16 = 1920 mm2, fy = 300 MPa, fu = 430 MPa. (See Table 2.1 of AS 4100) Bolts in shear (φ = 0.8)(Clause 9.3.2.1) Number of bolts is as yet unknown. M20 bolt, Grade 4.6/S, shear through the threads Ac = 225 mm2, f uf = 400 MPa, Lap length = ??? Correction factor unknown as L j is unknown – so assume kr = 1 Bolt shear, φV f = φ0.62(nnA c+nxAo)k rf uf = 0.8×0.62×(1×225+0)×1.0×400 = 44.64 kN (per bolt) Hence, no. of bolts required = 795/44.64 = 17.8. Therefore use 5 rows of bolts (total of 20 bolts). The lap length should now be checked.

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Lap length, Lj = 4×90 = 360 mm (distance from first to last bolt) Correction factor L j = 1.075 - 360/4000 = 0.985 (Table 9.3.2.1) Bolt shear, φV f = φ0.62(nnA c+nxAo)k rf uf = 0.8×0.62×(1×225+0)×0.985×400 = 43.97 kN (per bolt) Total Bolt shear, φVf = 43.97×20 = 879 kN Other “bolt related” failure modes should be checked as the bolts in shear bear on both the splice plate and UC flange. Ply in bearing on UC (φ= 0.9)(Clause 9.3.2.4) ae can be to edge of ply or to another hole. From diagram it can be seen that the edge of the next hole is closer than the edge of the ply. ae = 90 – 11 x 2 + 20/2 (Distance from edge of hole to edge of hole plus half bolt diameter) ae = 78 mm Tearout, φV b = φaet pf up = 0.9×78 ×9.4×440 = 290 kN (per bolt hole) Pileup, φVb = φ3.2dft p fup = 0.9×3.2×20×9.4×440 = 238 kN (per bolt hole) Pileup controls, hence design capacity is φV b = 20×238 = 4765 kN (20 bolt holes transferring the load) Note that this is much larger than the total bolt shear failure capacity. Ply in bearing on plate (φ = 0.9)(Clause 9.3.2.4) ae = 78 mm (as before) Tearout, φV b = φaet pf up = 0.9×78×16×430 = 483 kN (per bolt) Pileup, φVb = φ3.2dft p fup = 0.9×3.2×20×16×430 = 396 kN (per bolt) Pileup controls, hence design load is φV b = 20×396 = 7925 kN.

Note that this is much larger than the total bolt shear failure capacity. Tension capacity of the UC (φ = 0.9) (Clause 7.2) φNtUC = min[φA gfy ; φ0.85kt A nf u] = 868 kN (given) Tension capacity of the splice plate (φ = 0.9)(Clause 7.2) φNtp = min[φAgf y; φ0.85ktAn fu] = 400 kN (given)

Consider all failure modes Plate tension UC tension Bolt shear UC bearing Splice plate bearing

φNt = 2 × 400 = 800 kN φNt = 868 kN 20 × φVf = 20 × 43.9 = 879 kN 20 × φVb = 20 × 238 = 4765 kN 20 × φV b = 20 × 396 = 7925 kN

Hence the design capacity of the connection is controlled by the splice plate in tension and is 800 kN. This exceeds the design load of 795 kN, and is therefore acceptable. Note that the bearing capacity of the bolts onto the UC or splice plate ply is much larger than the bolt shear capacity. It may be better to use less, but stronger bolts in this connection (eg Grade 8.8). Hence 5 rows of bolts of each flange on each side of the splice are required. The design capacity of the connection is 800 kN, with the critical failure mode being tension capacity of the splice plates.

Welded Connection: Junction of the two UCs Splice Plate

Flange of UC

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As the connection is now welded (ie no holes) the net area for tension section capacity of both the UC and the splice plate must increase. Hence the tension capacities of the UC and plates cannot decrease (they may increase – but not necessarily). Therefore it can be said that the tension capacity of the two splice plates is at least 800 kN. It can be seen that if each weld is L w in length, there is a total of 4L w on each side of the connection (top and bottom flanges) to transfer the load between the splice plates and the UC. Ply in bearing on UC flange and Splice plates - Ply in bearing does not apply as this is a welded connection. Weld in shear (φ = 0.8) (Clause 9.7.3.10) 6 mm fillet weld, SP, E48XX Throat thickness t t = 6/√2 = 4.24 mm, fuw = 480 MPa, weld length L w = ???? mm

Correction factor, k r = ???? as Lw is not known, initially assume k r = 1.0 (Table 9.7.3.10.2) Weld shear (per unit length), φv w = φ0.6ttkrf uf = 0.8×0.6×4.24×1.0×480 = 0.977 kN/mm Due to the symmetry of the connection, and the concentric loading, the distribution of shear in the welds is uniform. Hence v* = N*/(4L w) For strength, φv w ≥ v* = N*/(4Lw). * Lw ≥ N /(4φvw ) = 795 kN / 0.977 kN/mm / 4 = 203.4 mm For simplicity choose 205 mm, and k r = 1.0 since weld length L w < 1700 mm (Table 9.7.3.10.2) so the assumption that kr = 1.0 was correct φvw = 0.977 kN/mm, v * = 795/(4×205) = 0.970 kN/mm, and therefore φvw ≥ v* Hence total weld strength in shear = 4 × 205 × 0.977 = 800 kN based on 205 mm weld. Possible failures: UC tension failure, φN t = 868 kN (perhaps higher) Splice plate tension failure, φN t = 400 × 2 = 800 kN (perhaps higher) Weld shear φv w × 4Lw = 800 kN Critical failure is weld shear, and the capacity of the connection is 800 kN > N* = 795 kN. Note also that the length of the weld (205 mm) is greater than the depth of the 150UC30.0 section (d = 164 mm), so Clause 7.3.2(b) is satisfied, and the assumption that kt = 0.85 for the UC was correct. Also note that weld failure is not necessarily a desirable occurrence, so it may be prudent to choose a slightly longer weld. Tim Wilkinson October 2006 C:\wilko\Docs\teaching\steel_structures_1\2006\CIVL3206_2006_tutorial_5_connections_solution.doc...