STEELWORK DESIGN (CON4334)- SOLUTION
Q1. From Table 9.5 & 9.6, ps = 375 N/mm2, pbb = 1000 N/mm2, and pbs = 550 N/mm2 Bolt capacity: Shear capacity of bolts: Double shear capacity on threads Ps = 2psAs = 2*375*353*10-3= 2*132.4 = 264.8 kN per bolt Comparing main plate thickness 22 mm and TWO cover plate thickness 2 x 12 = 24 mm, tp = 22 mm Bolt:
Bearing capacity of bolts Pbb = pbb*d* tp = 1000*24*22*10-3 = 528 kN per bolt
Plate: Bearing capacity For Grade S355, pbs = 550 N/mm2 and Us = 510 N/mm2, For standard hole, kbs = 1.0, d = 20 mm, tp = 22 mm, e = 40 mm Net distance between the bearing edge and the near edge of adjacent holes in the direction of load transfer Ic = 60 – 26 = 34 mm For grade 8.8, Ub = 800 N/mm2, therefore, Pbs = kbs d tp pbs = 1.0 x 24 x 22 x 550 x 10-3 = 290.4 kN Pbs = 0.5 kbs e tp pbs = 0.5 x 1.0 x 40 x 22 x 550 x 10-3 = 242 kN & Pbs = 1.5 Ic tp Us 2.0 d tp Ub = 1.5 x 34 x 22 x 510 x 10-3 = 572.2 kN 2.0 x 24 x 22 x 800 x 10-3 = 844.8 kN Hence bearing capacity of connecting part = 242 kN per bolt Overall bolt capacity Overall bolt capacity = 242 per bolt (Bearing capacity governs) No. of bolt required = 750 / 242 = 3.1 Use 6 nos. 24 mm dia. Bolts is adequate. Check bolt layout: Min. spacing = 2.5d = 2.5*24 = 60 mm. Max spacing = lesser of 12 t ( = 12 x 12 = 144 mm) or 150 mm Spacing provided = 60 mm O.K. Spacing perpendicular to direction of load = 70 mm < 3 x bolt diameter = 72 mm NOT O.K. Increase the bolt spacing to 75 mm and hence the new edge distance is 75/2 mm.
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STEELWORK DESIGN (CON4334)- SOLUTION
Min. end and edge distance, rolled edge = 30 mm (Table 9.3) Edge distance provided = 37.5 mm O.K. End distance provided = 40 mm, O.K. 275 = 116 mm > 37.5 mm Max. edge distance = 11t1 = 11*12* 355 O.K. 275 = 216 mm > 37.5 mm = 11t2 = 11*22* 345 O.K. Check Strength of Plates
Check Main plate only as it is more critical. (22mm < 12*2 = 24 mm) Gross area = 150*22 = 3300 mm2 Design strength py = 345 N/mm2 for 16 mm < t 40 mm Effective area Ae = Ke*net area = 1.1*22*(150 – 2*26) = 2371 mm2 gross area = 3000 mm2 Capacity Pt = Ae*py = 2371*345*10-3 = 818 kN > 750 kN O.K.
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STEELWORK DESIGN (CON4334)- SOLUTION
Q2. Welding strength: Use 6 mm fillet weld, Strength of weld = 0.7*leg length*pw / 1000 kN/mm = 0.7*6*250/1000 = 1.05 kN/mm
Length required = 850 / 1.05 = 810 mm Balance the weld on 4 sides, length required = 810/4 = 202.5 mm Add 2s to the weld length, hence weld length on each side = 202.5 + 2*6 = 214.5 mm (say 220mm) Min. lap length = 5*12 = 60 mm < 220 mm Min. length of return = 2*6 = 12 mm Provide 15 mm return
O.K. O.K.
Weld length on each side transverse spacing between welds = 130 mm O.K. Provide 220 mm Check Strength of Plates
Main plates: Gross area = 150*22 = 3300 mm2 Design strength py = 345 N/mm2
for 16 mm < t 40 mm
Capacity Pt = Ae*py = 3300*345 = 1138.5 kN > 850 kN O.K. Each cover plate: Gross area = 130*12 = 1560 mm2 Design strength py = 355 N/mm2
for t < 16 mm
Capacity Pt = Ae*py = 1560*355 = 553.8 kN each plate Therefore the capacity of TWO cover plates = 2*553.8 = 1107.6 kN > 850 kN O.K.
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STEELWORK DESIGN (CON4334)- SOLUTION
Q3 Ultimate Load P = 200 kN on the whole bracket Hence ultimate load P on one plate = 200/2 = 100 kN
x 2 8 * 50 2 2 *10 4 y 2 4 * 50 2 1502 10 *10 4 x 2 y 2 2 10*10 4 1.2 *105
Bolt group
r = (502 + 1502)0.5 = 158.1 mm cos = 50 / 158.1 = 0.316
Bolt at the corner is subjected to highest load: Load due to moment Pe r1 100 * 0.3 *10 3 *158.1 FT 39.5 kN ( x 2 y 2 ) 1.2 *10 5 Load due to shear 100 P Fs 12.5 kN 8 No. of bolts Resultant load on bolt FR Fs FT 2 Fs FT cos 2
2
12.5 2 39.5 2 2 *12.5 * 39.5 * 0.316 45.0 kN Try M16 grade 8.8 bolts Ps = ps*As = 375*157*10-3 = 58.9 kN > 45.0 kN
O.K.
As the end distance and size of column are not given, the checking of bearing capacity is NOT required.
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STEELWORK DESIGN (CON4334)- SOLUTION
Q4. Conditions to be satisfied:
Fs Ps ,
FT Pnom ,
Fs F T 1.4 Ps Pnom
Design Load P = 1.4*200 + 1.6*150 = 520 kN Single shear capacity on threads, Ps = ps As = 375*245*10-3 = 91.9 kN Tension capacity, Pnom = 0.8 pt As = 0.8*560*245*10-3 = 109.8 kN
Use approximate method of analysis for conservative results.
y 2 2 * 60 2 1202 1802 2402 3002 2 *198 *103 396 *103
mm 2
The maximum bolt tension is at the top bolt:
FT
Pe y1 520 * 250 * 300 98.5 kN Pnom 109.8 kN 396 *103 y2
The vertical shear per bolt:
Fs
P 520 43.3kN Ps 91.9 kN No. of bolts 12
Fs F 43.3 98.5 T 0.471 0.897 1.368 1.4 Ps Pnom 91.9 109.8 The bolts are satisfactory.
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STEELWORK DESIGN (CON4334)- SOLUTION
Q5. Design Ultimate Load = 350 kN a1 = area of connected leg = 2 x 62 x 6 = 744 mm2 a2 = area of unconnected leg = 2 x 47 x 6 = 564 mm2 Ae = 744 + 564 = 1308 mm2
For double angles connected on both sides of a gusset plate, Tension capacity
Pt =
py (Ae – 0.15 a2) = 355 (1308 – 0.15 x 564) x 10-3 = 434.3 kN > 350 kN The angle is satisfactory.
Try 6 mm fillet weld, strength = 1.05 kN/mm Length required = 350/1.05 = 333 mm Hence weld length for each angle = 333 /2 = 167 mm. From section table, centroid is 20.4 mm from side X, the weld on each side should be balanced according to the centroidal axis. Side X, length = 167 x 44.6 / 65 = 115 mm Add 2s, final length = 115 + 2*6 = 127 mm, say 130 mm Side Y, length = 167 – 115 mm = 52 mm Add 2s, final length = 52 + 2*6 = 64 mm, say 65 mm O.K. Provide 15 mm for the returns.
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STEELWORK DESIGN (CON4334)- SOLUTION
Q6. Design Load on the whole bracket = (1.4*350 + 1.6+300) = 970 kN Design Load P on ONE plate = 970 / 2 = 485 kN Eccentricity e = 200 + 200/2 = 300 mm Weld Length L = 2*(200 +300) = 1000 mm r 100 2 150 2 180.3 mm Moment of inertia: 3003 200 * 300 2 1.35 *107 mm3 Ix 6 2 3 2 200 200 * 300 7.333 *10 6 mm3 Iy 6 2 Ip = (13.5 + 7.333)*106 = 20.833*106 mm3 100 cos 0.555 180.3 Direct shear Fs = 485/1000 = 0.485 kN/mm Shear due to torsion on weld at corner: 485 * 300 *180.3 FT 1.259 kN/mm 20.833 *10 6 Resultant shear: FR 0.485 2 1.259 2 2 * 0.485 *1.259 * 0.555 1.581 kN/mm Required leg length or weld size = required strength of weld*103/(0.7pw) =1.581*1000/(0.7*250) = 9.0 mm Use 10 mm fillet weld
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STEELWORK DESIGN (CON4334)- SOLUTION
Q7. Assume the flanges resist the bending moment and the weld resists the shear force. The flange welds resist the moment F
Pe DB
600 ∗ 250 528.3 ∗ 190 2 ∗ 10
1.67 kN/mm
Required leg length or weld size = (Required strength of weld)*103/(0.7pw) = 1.67*1000/(0.7*250) = 9.5 mm Use 10 mm fillet weld for flanges The web welds resist the shear: F
P 2a
600 2 ∗ 340 2 ∗ 6
0.91 kN/mm
Required leg length or weld size = (Required strength of weld)*103/(0.7pw) = 0.91*1000/(0.7*250) = 5.2 mm Use 6 mm fillet weld for flanges
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