4. An A992 WF steel section is used for the tension member shown in the figure. The bolts are 20mm in diameter. The WF section is to be connected to a 12mm-thick gusset plate. a. Determine the nominal strength based on gross section yielding. b. Determine the nominal strength based on effective net section fracture. WF Section * ********** + 65 || 85 |4-75 ►| "B" 0 1 2 3 4 5 6 7 8 9 Wide Flange Section W16x67 W16x40 W16x36 W14x74 W14x68 W14x61 W14x48 W12x72 W12x65 W12x53
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- Determine the design tensile strength of plate (200x8 mm) connected to 10-mm thick gusset using 20 mm bolts as shown in the figure, if the yield and the ultimate stress of the steel used are 250 MPa and 410 MPa, respectively. Add 1mm around the bolt for the hole. Use LRFD method. Plate 8-mm thick 2 3 40+ 30 301 T 200 mm Gusset 10-mm thick 3af 30 2_3 *40 40+ 50,54 +40A plate with width of 420 mm and thickness of 13 mm is to be connected to a plate of the same width and thickness by 30 mm diameter bolts, as shown in the Figure 1. The holes are 3mm larger than the bolt diameter. The plate is A36 steel with yield strength Fy = 248MPa. Assume allowable tensile stress on net area is 0.60Fy. It is required to determine the value of b such that the net width along bolts 1-2-3-4 is equl to the net width along bolts 1-2-4. W = 420 mm t = 13 mm a = 64 mm c = 104 mm d = 195 mm Bolt Diameter = 30 mm Holes Diameter = 30 mm + 3 = 33 mm ?? = 248 MPa Allowable Tensile Stress on ?? = 0.60Fy a. Calculate the vaue of b in millimeters. b. Calculate the value of the net area for tension in plates in square millimeters. c. Calculate the value of P so that the allowable tensile stress on net area will not be exceeded.A L 3 x 2 x ¼ is connected to a gusset plate via six bolts. The nominal diameter of the bolt is 0.25 inches, the pitch spacing is 1.75 inches, the gage spacing is 2 inches, and the thickness of the connection is (¼) inch. The yield stress is 50 Ksi and the ultimate stress is 60 Ksi. Consider section line (a-a') for the analysis. . a. What is the effective net area (Ae) of the angle section in inches2 [a-a']?(The shear lag factor is "0.80") b. What is the design tensile yielding strength in Kips for the steel member? c. What is the design tensile rupture strength in Kips for the steel member? d. What is the minimum Factor of Safety if a tensile load of 23 Kips is applied to the angle section? please make sure the answer is correct 100% I only need the final answers
- The given angle bar L125x75x12 with Ag = 2,269 sq.mm. is connected to a gusset plate using 20 mm diameter bolts as shown in the figure. Using A36 steel with Fy = 248 MPa and Fu = 400 MPa, determine the following: 2. Determine the nominal tensile strength of the 12 mm thick, A36 angle bar shown based on: a. Gross yielding b. Tensile rupture Bolts used for the connection are 20 mm in diameter. O O O O O O O Effective net area of the tension member if the shear lag factor is 0.80. Select the correct response: 1,516.1 1,354.4 1,431.2 1,221.63. A plate with width of 300mm and thickness of 20mm is to be connected to two plates of the same width with half the thickness by 24mm diameter bolts, as shown. The rivet holes have a diameter of 2mm larger than the rivet diameter. The plate is A36 steel with yield strength F,-248MPa and ultimate strength F,-400MPa. a. Determine the design strength of the section. b. Determine the allowable strength of the section 24mm 30mmProblem 2 A tension plate shown below is used to support suspended load "T. Gusset Plate F, = 248 MPa Fu = 400 MPa 200 mm a) Determine the allowable tensile capacity of the plate if L= 240 mm, (Assume weld strength is satisfactory).-
- Topic:Bolted Steel Connection - Civil Engineering *Use latest NSCP/NSCP 2015 formula to solve this problem *Please use hand written to solve this problem The bracket shown in the figure is supported by four 22 mm diameter bolts in single shear. The bracket is subject to an eccentric load of 150 kN. Use LRFD. Use A36 steel. Questions a) Determine the critical force on the most stressed bolt. b) Determine the nominal shear stress of the most stressed bolt.The tension member shown is an angle L150 x 90 x 10, of A36 steel, with Fy= 250 MPa and Fu= 400 MPa, and is subjected to a service DL= 150 KN and a service LL= 150 KN The bolts are 20 mm o. Use ASD. 40mm 40mn 40mm 50mm ... Compute the allowable strength based on yielding in KN.0 Compute the nominal strength based on fracture in KN.0 Check the adequacy of the section.0Situation 5. The angular section shown below is welded to a 12 mm gusset plate. Both materials are A36 steel with Fy = 250 MPa. The allowable tensile stress is 0.6Fy. The weld is E80 Electrode and 12 mm thickness. INNOVATIONS Properties of L 150x90x12: y = 50 shear stress of weld = 0.3Fu A = 2750 Allowable REVIEW INNOVATIE a K ➜ www A. 234 KN B. 349 KN b 13. What is the value of P without exceeding the allowable tensile the angle? C. 382 kN p. 413 kN 14. Find required length of the weld based on shear? A. 280 mm C. 300 mm D. 380 mm B. 320 mm 15. Find the required value of a? A. 108 mm B. 97.9 mm D. 185 mm NEW INNOVATIONS REVIEW INNOVATIOf REVIEW NEW INNOVATIONS
- A 15" x 3/8" bar of A572 Gr. 50 steel is used as a tension member. It is connected to a gusset plate with 7/8-in diameter bolts as shown in the figure. Use s = 2.0 and g = 3.0. a) Determine the allowable tensile strength of the section based on tensile rupture of the net area.b) Determine the allowable tensile strength of the section based on yielding of the gross area.1. A tension member is made of 90 mm thick steel plates 225 mm wide by a lap joint made with three (3) rows of 20mm. Use LFRD specifications ASTM A36 (Fy=138 MPa, Fu=220 MPa, compute the following: a. Capacity of the section based on the yielding of the gross area Critical effective net area b. c. Capacity of the section based on the tensile fracture of the critical net area. Dia 20mm DIAMETER OF BOLTS PSituation 2. Two plates each with thicknessF16mm are bolted together with6 -22mm dimater bolts forming a lap conne ction. Bolt spacing are as follows: S1 = 40mm, S2 = 80mm, Sa = 100mm. Bolt hole diameter=25 mm 50 250mm 30 30 60 75 Allowable stress: Tensile stess on gross area of the plate=0.60 Fy Tensile stress on net area of the plate=0.5Fy Shear Stress of the bolt Fv=120MPA Bearing Stress of the bot Fp=1.2 Fu Calculate the permiss ible tensile load P under the following Conditions: 4. Based on shear capacity of bolts 5. Based on bearing capacity of bolts 6. Based on block shear strength (taking into consideration the failure path given in the figure below) 40 80 16 mm 40 180 mm 40 Bearing Failure Path #1 140 209 Bearing Failure Path #2