2. Compute the design strength in tension, P, of the A36 steel plate (1/4 in x 12 in) with a line of 7/8" diameter BOLTS as shown in the sketch. Your calculations must include BOTH yielding and fracture limit states. #2- 2×12 T O SNANS
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- 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 1,221.6Prepare an excel file and to find a) area of the bolt, b) Average Shear Stress Bolt c) Bearing Area Stress and d) allowable stress. Applied force range – 500 to 1000 N (in step of 100 N) Bolt diameter range – 10 to 25 mm (in step of 5 mm) Plate thickness – 10 mm Consider ultimate tensile strength as 460 N/mm2. And consider factor of safety as 1.5.The butt connection shows 8-22 mm diameter bolts spaced as shown below. P- 50 100 50 50 100 50 16 mm +HHHH 40 80 40 12 mm Steel strength and stresses are: Yield strength, Fy = 248 MPa Ultimate strength, Fu = 400 MPa Allowable tensile stress on the gross area = 148 MPa Allowable tensile stress on the net area = 200 MPa Allowable shear stress on the net area = 120 MPa Allowable bolt shear stress, Fv = 120 MPa Based on the gross area of the plate. Based on the net area of the plate. Based on block shear strength. Bolt hole diameter = 25 mm Calculate the allowable tensile load, P, under the following conditions:
- A bearing type connection is shown in Figure 3.19. The diameter of A 325 bolts is 22 mm and the A572 Grade 50 plate material has a width of 150 mm and thickness of 16 mm. Assume diameter hole to be 24 mm. Bolt threads are excluded from the shear plane. Allowable stress of A 325 bolts: Fv = 207 MPa Fp = 1.5 Fu (to prevent excessive hole deformation) Allowable stresses of A572 Grade 50 plate material: Fy = 345 MPa Fu = 450 MPa a. Compute the tensile capacity due to the failure of the plates. b. Compute the tensile capacity due to the failure of the bolts.4. As shown below, the splicing joint of a beam was made with bearing type high-strength bolts. The loads transferred through the joint is V=3000KN, M=90KN.m. The bolts are of grade 10.9, nominal diameter 22mm, effective area Ae=2.45cm2;f=310N/ mm?; f-500N/mm²; End plates made of Q235B steel, thickness 22mm, fe-470N/mm?. Check the strength of this bolt connection. -24mm 35 130 35 80 V 1 og1 081 1 081Topic:Bolted Steel Connection - Civil Engineering *Use latest NSCP/NSCP 2015 formula to solve this problem The butt connection shows 8-22 mm dia. A325 bolts spaced as follows: S1 = 40 mm S3 = 50 mm t1 = 16 mm S2 = 80 mm S4 = 100 mm t2 = 12 mm Steel strength and stresses are: Fy = 248 MPa Fu = 400 MPa Allowable tensile stress on the gross area = 148 MPa Allowable tensile stress on the net area = 200 MPa Allowable shear stress on the net area = 120 MPa Allowable bolt shear stress, Fv = 120 MPa Bolt Hole diameter = 25 mm Questions: Calculate the allowable tensile load T, under the following conditions. a) Based on the gross area of the plate b) Based on the net area of the plate c) Based on block shear strength
- Topic: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.What will be the tensile strength of 160 x 8 mm plate with the holes for 16 mm bolts as shown in figure? Plates are of steel grade Fe410. 1 30 mm 40mm' 40mm 25 mm 2 25 mm 1 3 25 mm 25 mm 30 mm 1 160 mm 2.2. A steel plate is 360 mm wide and 20 mm thick with four bolts hole into the place as shown in the figure. Compute the following: a. Critical net area required by the NSCP specs. b. Max. critical net area required by the NSP specs. Note that, Max. net area is 85% of the critical net area c. Capacity of the joint if the allowable tensile stress is 0.75Fy. Use A36 steel Fy=248 MPa 90m²m 90 mm 90mm 45 45mm In my comin P Scanned with CamScanner CS
- A bolted lap joint is shown in the figure below. The bolts are 19mm in diameter. Effective diameter of the holes is 23mm. The plates are 12 mm thick, x1 = 33 mm, x2 = 76 mm, x3 = 50 mm. Yield strength of plate, Fy = 248 MPa Ultimate tensile strength of plate, Fu = 400 MPa Required: Using allowable strength design: 1. Find the safe load “P” based on gross area yielding. 2. Find the safe load “P” based on net area rupture. 3. Find the safe load “P” based on block shear.4. Calculate the design strength (ocPn) of W24X76 with length of 12 ft. and pinned ends. A572 Grade50 steel is used. E=29x103 ksi. Show your work in detail. ASTM Classification A36 A572 Grade 50 A992 Grade 50 A500 Grade B (HSS rect, sq) A500 Grade B (HSS round) A53 Grade B Yield Strength F, (ksi) 36 50 50 46 42 35 Ultimate Strength F (ksi) 58 65 65 58 58 60A 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