How many M24 bolts Grade 8.8 will be needed in a tension splice connected to a 20 mm cover plates on the longer leg of a 200 x 100 x 15mm angle in S355 steel, if the full strength of the angle is to be developed?
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How many M24 bolts Grade 8.8 will be needed in a tension splice connected to a 20 mm cover plates on the longer leg of a 200 x 100 x 15mm angle in S355 steel, if the full strength of the angle is to be developed?
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- Two steel plate tension members have been connected using 0.72” diameter bolts arranged in an equally-spaced four by four square formation.Total plate self-weight is specified as 912 pounds.Design of the elements adhered to the set of values that are twice as much as the minimum requirements. Both plates have a thickness equal to 1/3 in. Take shear fracture stress as 54ksi and the min required edge distance as 0.125ft. a)Sketch the connection showing all the details and measurements with units. b)Find the maximum allowable service dead load(excluding the self-weight)and live load assuming live load is half as much as dead load including the self-weight of the plates.Calculate the allowable (ASD) block shear rupture strength (kN) of a C230x22, attached through the web with six M20 bolts as shown. The yield strength of the steel is 345 MPa and the ultimate tensile strength is 483 MPa. C230x22 64.5 Ο Ο Ο OO 64.5 50 75 75 +The butt connection shows 8-22 mm diameter bolts spaced as shown below.Use NSCP2001 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 Bolt hole diameter = 25 mm Calculate the allowable tensile load, P, under the following conditions: 1Based on the gross area of the plate. 2Based on the net area of the plate. 3Based on block shear strength. 50 100 50 50 100 50 40 80 40 12 mm 16 mm
- QUESTION 5 3. check the capacity of the shear connection. A992: Fy=50ksi, Fu=65ksi A36: Fy=36ksi, Fu=58ksi effective bolt hole diameter = bolt hole + 1/16" beam properties: tw=0.295 in. Fexx=70ksi what is the bolts shear capacity in kips? (2 decimal places) 0.5"| 1.5" 1.5" 5/16 _ 4-3/40 A325 SSLT Bolts 5/16What is the working strength of a 2" bolt which is screwed up tightly in a packed joint when the allowable working stress is 12000 psi? (A) 20,120 lbs. B 20,200 lbs. C) 25,347 lbs. D) 20,443 lbs.Consider the following bolted connection. Slippage is permitted. Member: L6x6x3/4 3/16 “ gusset plate Pu Pu A36 steel 3.5* 1"diameter A325-X bolts 1 1/4" 3 " 3* 1 1/4" Question 3.1: Please, indicate if the combined bolt shear strength is closer to: Question 3.2: Please, indicate if the combined bolt bearing strength is closer to: Question 3.3: Please, indicate if the angle yielding strength is closer to: Question 3.4: Please, indicate if the angle fracture strength is closer to: Question 3.5: Please, indicate if the angle block shear strength is closer to: Question 3.6: Please, indicate if the overall connection strength is closer to:
- 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:Two steel plate tension members have been connected using 0.72” diameter bolts arranged in an equally-spaced four by four square formation.Total plate self-weight is specified as 912 pounds.Design of the elements adhered to the set of values that are twice as much as the minimum requirements. Both plates have a thickness equal to 1/3 in. Take shear fracture stress as 54ksi and the min required edge distance as 0.125ft. Find the maximum allowable service dead load(excluding the self-weight)and live load assuming live load is half as much as dead load including the self-weight of the plates.For the cross section shown, calculate the tensile capacity of the steel plate. Assume that the bolt hole is 1/16 inch larger than bolt diameter. Plate 1/2" x 11" .T 7/8" dia. bolt, typ. ASTM A325N
- The plate shown is 8 in wide and 1/2 in thick. The bolts are 7/8 in. The smallest net area (An) of this bolted plate is equal to:a] Compute LRFD 6.) Compute ASD CIVIL STEEL DESIGN ENGINEERING- BOLTED CONNECTION Sample Problem 5: The 175 x 100 x 16 mm angle shown connected to a gusset plate with three 25 mm bolts (A490) when threads are excluded from shear planes. The angle consists of A36 steel. Properties of 4 175 x 100 x 16 A36 Steel Properties of A490 bolts A = 4181 mm² t = 16 mm Fy = 248 MPa F₁ = 400 MPa Fy = 276 MPa Nominal hole dia. = 27 mm 75 T+ 100 Gusset Plate 100 F 100Determine the maximum factored LRFD tensile force capacity in tension only (no block shear). The angle is ASTM A36 steel. X = 0.7 Y = 3.8 Z=1/2 Round your answer to 2 decimal places. Your Answer: Incorrect The answer is 77.15 ± 1%. 1/2" X" bolts 1½" Gusset plate -L3 X 3 X Z Tu I