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- ISA 100 x 100 x 10mm (Cross sectional area = 1908mm2) serves as tensile member. This angle is welded to a gusset plate along A and B appropriately as shown. Assuming the yield strength of the steel to be 260 N/mm2 the tensile strength of the member can be taken to be approximately A ISA 100x100×10 Gusset Plate BQuestion 4 Complete the table below for the steel sections shown. All sections are bending about their horizontal axis. Section Location hep Ney Slenderness (show calculations) of element bf Flange te ld ++ tw be = 800 mm Web tf = 20 mm d = 1440 mm tw = 16 mm Grade 300, heavily 入。= Whole Asp= Asy= welded plates section br Flange tr d Web bf = 800 mm te = 20 mm d = 1420 mm tw = 16 mm Grade 300, heavily welded plates 入。= Whole Asp= Asy= sectionTopic:Bolted Steel Connection - Civil Engineering -Steel Design *Use latest NSCP/NSCP 2015 formula to solve this problem *Please use hand written to solve this problem A 12.5 mm x150 mm plate is connected to a gusset plate having a thickness of 9.5 mm.Diameter of bolt = 20 mm Both the tension member and the gusset plate are of A 36 steel. Fy = 248 MPa, FU = 400 MPa. Shear stress of bolt Fnv = 300MPa. Questions: a.Determine the shear strength of the connection. b.Determine the bearing strength of the connection. c. Determine the block shear strength of the connection.
- A single unequal angle 100 x 75 x 6 is connected to a 10 mm thick gusset plate at the ends with six 16 mm diameter bolts to the 100 mm of leg to transfer tension as shown in figure. What will be the block shear strength (in kN) of the angle section assuming that the yield and the ultimate stress of steel used are 250 MPa and 410 MPa respectively? ISA (100 × 75 x 6) -T 16 mm o bolt 10 mm 40 mm 40 mm| 40 mm 40 mm 40 mm 40 mm 10 mmThe 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 1,221.6Determine 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 +40
- A PL 38 x 6 tension member is welded to a gusset plate as shown in figure. The steel is A36. PL ½ x 6 The design strength based on yielding is nearest to: The design strength based on rupture is nearest to: The design strength for LRFD is nearest to: The allowable strength based on yielding is nearest to: The allowable strenath based on rupture is nearest to: The allowable strength for ASD.TENSION MEMBERS: THE SINGLE 200 X 10 mm STEEL PLATE IS CONNECTED TO A 12 mm THICK STEEL PLATE BY FOUR 16 mm DIAMETER RIVETS AS SHOWN IN THE FIGURE. THE RIVETS USED ARE A502 GRADE 2, HOT DRIVEN RIVETS. THE STEEL IS ASTM A36 WITH Fy = 248 MPa AND Fu = 400 MPa. DETERMINE THE VALUE OF P. a. P BASED ON TENSION OF GROSS AREA b. P BASED ON TENSION OF NET AREA c. P BASED ON BEARING OF PROJECTED AREA d. P BASED ON SHEAR RUPTURE (BLOCK SHEAR)A steel plate is to be attached to a support with three bolts. The cross-sectional area of the plate is 800 mm² and the yield strength of the steel is 260 MPa. The ultimate shear strength of the bolts is 570 MPa. A factor of safety of 1.67 with respect to yield is required for the plate. A factor of safety of 4.0 with respect to the ultimate shear strength is required for the bolts. Determine the minimum bolt diameter required to develop the full strength of the plate. Note: consider only the gross cross-sectional area of the plate-not the net area. Support Plate ↓ P
- Determine the tensile strength of the welded A992 WT4×20. WT4× 20 PL – Transverse weldA 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.Determine the design tensile strength of plate 150 mm x 12 mm size with holes or 16 mm diameter bolts as shown in figure below. Use Fe410 T 35 14 4 42.5 65 42.5 65 2 3 (All dimensions in mm) 150