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Power Exercises 2. 1.0 A double-angle shape is shown. The steel is A36, and the holes are for 12 mm diameter bolts. Assume that Ae = 0.75An. a. Determine the design tensile strength for LRFD. b. Determine the allowable strength for ASD. Ans. a) Pu = 594.0 kN; b) Pa = 396 kN 2.0 The tension member shown contains holes for 19 mm Ø bolts. At what spacing s, will the net area for the section through one hole be the same as a rupture line passing through two holes? Ans. 81.24 mm 3.0 An L200 x 100 x 18.75 is used as a tension member, with 22 mm Ø bolts in each leg at usual gage location. Two rows of bolts are used in the long leg, and one in the short leg. Determine the minimum stagger, s, necessary so that only two holes need be subtracted in determining the net area. What is the net area? Ans. s = 67.8 mm; An = 4,335.94 mm^2
Ag = 1560 mm^2 (Angle) 2L 125 x 75 x 8 LLBB
75 P
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s s^ s s
4.0 Determine the smallest net area of the tension member shown. The holes are for 19 mm Ø bolts at the usual gage locations. The stagger is 37.5 mm. Ans. An = 1,853.57 mm^2 5.0 The tension member shown is a PL 16 × 250, and the steel is A36. The bolts are shown are 22 mm in diameter. a. Determine the design strength for LRFD. b. Determine the allowable strength for ASD. Ans. a) Pu = 892.8 kN; b) Pa = 595.2 kN 6.0 The tension member shown is an L150 × 87.5 × 8. The bolts are 19 mm in diameter. If A36 steel is used, is the member adequate for a service dead load of 138 kN and a service live load of 138 kN? a. Use LRFD. b. Use ASD. Ans. a) Adequate: 400.99 kN > 386.4 kN; b) Not Adequate: 267.33 kN < 276 kN
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2L125 x 87.5 x 6.25 s
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C8 × 11.
PL10 x 137.
10.0 A PL 10 mm × 150 mm tension member is welded to a gusset plate as shown. The steel is A36. Assume that Ae = Ag and compute the following. a. The design strength for LRFD. b. The allowable strength for ASD. Ans. a) Pu = 334.8 kN; b) Pa = 223.2 kN 11.0 A C8 × 11.5 is connected to a gusset plate with 22 mm diameter bolts as shown. The steel is A572 Grade 50. If the member is subjected to dead load and live load only, what is the total service load capacity if the live-to-dead load ratio is 3? Assume that Ae = 0.85An. a. Use LRFD. b. Use ASD. Properties of A572 Grade 50: Fy = 345 MPa Fu = 448 MPa Properties of C8 x 11.5: Ag = 2,181 mm^2 bf = 57.4 mm d = 203.2 mm tw = 5.6 mm tf = 9.9 mm Ans. a) T = 361.95 kN; b) T = 361.95 kN 12.0 The tension member is a PL 10 mm × 137.5 mm of A242 steel. It is connected to a 10 mm thick gusset plate, also of A242 steel, with 19 mm diameter bolts as shown. Determine the nominal block shear strength of the tension member. Properties of A242: Fy = 345 MPa Fu = 483 MPa Ans. a) Tu = 530.7 kN; b) T = 353.8 kN
PL10 x 150
2C8 x 18.
13.0 A double-channel shape, 2C8 × 18.75, is used as a tension member. The channels are bolted to a 10 mm gusset plate with 22 mm diameter bolts. The tension member is A572 Grade 50 steel and the gusset plate is A36. If ASD is used, how much tensile load can be applied? Consider all limit states. Properties of A572 Grade 50: Fy = 345 MPa Fu = 448 MPa Properties of C8 x 18.75: Ag = 3,555 mm^2 bf = 64.2 mm d = 203.2 mm tw = 12.4 mm tf = 9.9 mm 𝑥̅= 14.35 mm Ans. T = 1189.09 kN 14.0 Two plates each of equal thickness are bolted together with 6 -22 mm Ø A325 bolts spaced as follows: S 1 = 40 mm S 3 = 100 mm S 2 = 80 mm t = 16 mm Steel Strength and Shear Stresses are: Yield Strength: Fy = 248 MPa Ultimate Strength: Fu = 400 MPa Allowable bolt shear stress: Fv = 120 MPa Bolt Hole diameter = 25 mm Calculate the allowable and design tensile load T under the following conditions: a) Based on gross area of the plate. b) Based on the net area of the plate. c) Based on block shear strength. Ans. a) Tu 642.82 kN; Ta = 428.54 kN; b) Tu = 624 kN; Ta = 416 kN; c) Tu = 849.12 kN; Ta = 566.08 kN
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