2-PL10x1/2 A-36 Bolts are 7/8 in A-325, threads are not excluded. All edge spacings are minimum | 3" | 3" | о PL 10x3/4 A-572Gr50 a) Determine the nominal bolt capacity. b) Determine the nominal bearing capacity for each plate. c) Determine the nominal rupture capacity of each plate. d) What is the nominal capacity of the connection? e) Using LRFD, is this connection capable of supporting a dead load of 150 kip and a live load of 100 kip?
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- Two steel plate tension members have been connected using 0.72 inch diameter bolts arranged in an equally-spaced 9 by 3 rectangular formation. Both plates have a thickness equal to 1/3 in. Take shear fracture stress as 51.2 ksi, clear distance for each edge bolt as 0.12 in, and the clear distance for the other bolts as 1.072 in. a)Sketch the connection showing all the details and measurements with units. b)Find the maximum allowable service dead load and live load assuming live load is half as much as dead load. Consider ASD.STAGGERED CONNECTIONS: A PLATE WITH WIDTH OF 400 mm AND THICKNESS OF 12 mm IS TO BE CONNECTED TO A PLATE OF THE SAME WIDTH AND THICKNESS BY 34 mm DIAMETER BOLTS, AS SHOWN IN THE FIGURE. THE HOLES ARE 2 mm LARGER THAN THE BOLT DIAMETER. THE PLATE IS A36 STEEL WITH YIELD STRENGTH Fy = 248 MPa. 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 EQUAL TO THE NET WIDTH ALONG BOLTS 1-2-4. a. CALCULATE THE VALUE 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.Select an American Standard Channel shape for the following tensile loads: dead load = 54 kips, live load = 80 kips, and wind load = 75 kips. The connection will be with longitudinal welds. Use an estimated shear lag factor of U = 0.85. (In a practical design, once the member was selected and the connection designed, the value of U would be computed and the member design could be revised if necessary.) The length is 17.5 ft. Use Fy=50 ksi and Fu=65 ksi. a. Use LRFD. b. Use ASD.
- The truss member shown in the accompanying illustration consists of two C12 x 25s(A36 steel) connected to a 25 mm gusset plate. How many 22 mm A325 bolts (threads excludedfrom shear plane) are required to develop the full design tensile capacity of the member if it isused as a bearing-type connection? Assume U = 0.85. Use ASD method.Design a weld connection for an UNP 280×80×9 of S 355 (fy= 355 MPa and fu= 470MPa) steel connected to a 10 mm gusset plate. The guesst plate is S275. Please draw all calculated values on a sketch with full dimensions. D= 335 KNL= 670 KNConsider the tension member shown below. The channel is made of ASTM A572 (Fy = 50 ksi; Fu = 65 ksi), and is connected to a 3/8-inch thick gusset plate made of ASTM A36 steel (Fy = 36 ksi; Fu = 58 ksi). The bolts are 7/8 inch in diameter, made of ASTM A307 bolts. The bolted connection is bearing-type. (a) Check if the given connection follows the code provisions on spacing and edge distance limits. (b) Determine the maximum allowable service tensile load that the member is allowed to carry. Consider all applicable limit states for tension member and connection. Consider only block shear and bearing limit state for gusset plate. Use ASD specifications. use AISC book (3decimals) PLEASEEE.
- Consider the tension member shown below. The channel is made of ASTM A572 (Fy = 50 ksi; Fu = 65 ksi), and is connected to a 3/8-inch thick gusset plate made of ASTM A36 steel (Fy = 36 ksi; Fu = 58 ksi). The bolts are 7/8 inch in diameter, made of ASTM A307 bolts. The bolted connection is bearing-type. (a) Check if the given connection follows the code provisions on spacing and edge distance limits. (b) Determine the maximum allowable service tensile load that the member is allowed to carry. Consider all applicable limit states for tension member and connection. Consider only block shear and bearing limit state for gusset plate. Use LRFD specifications. use AISC bookTopic: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 channel shown is attached to a 12 mm gusset plate with 9-22 mm diameter A 325 bolts as shown. Use LRFD. Fnv = 300 MPa Fu = 400 MPa Fy = 248 MPa Ag = 3354 mm2 Question: Determine the capacity of the channel based on the bearing strength of the connection.Determine the maximum service load P (ASD) that the connection can resist. Use 27 mm Ø A490-N bolts, and 10 mm A-36 plate and a load P with a slope 1H: 2V. Assume that edge distance and bolt spacing are not critical. ***Note: Lod P is acting at a slope of 1H: 2V
- The connection shown is subjected to a tensile force of P = 100 kN. The allowable shear stress for the bolts is 100 MPa. Assume each bolt supports an equal portion of the load. Determine the required diameter of the bolts. a. 20.6 mm b. 25.2 mm c.35.7 mm d.17.8 mm choose letter of correct answer4. Determine the design strength of the connection shown in the Figure below. Thebolts are 25 mm diameter A490 bolts with the threads not in the plane of shear.A36 steel is used (Fy = 250 MPa, Fu = 400 MPa).a. Compute the shear strength for all bolts.b. Compute the bearing strength for the tension member on all bolts.c. Compute the bearing strength for the gusset plate on all bolts.d. Compute the tensile strength of the tension member.e. Compute the design strength of the connection.Determine the capacity of the details shown below. A992: Fy=50ksi, Fu=65ksiA36: Fy=36ksi, Fu=58ksieffective bolt hole diameter = bolt hole + 1/16"beam properties: tw=0.38 in., bf=7.07 in., tf=0.63 in.Fexx=70ksi i) what is the weld capacity of the shear connection in kips? (2 decimal places)