WT12×153 05 WT12×153 Determine the design strength of two WT-shapes (WT12×153) connected by plates with 1 in. diameter bolts. The LRFD design strength of a single bolt in a web is 60 kips. Use ASTM A36 steel. 3" 3" 3"3"2" 3"3" 3"3"3" Side View Double Plate WT12×153 E- Top View
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- A channel shape, C8 × 18.75, is used as a tension member. The channels are boltedto a 10mm-thick gusset plate with 22mm- diameter bolts. The tension member is A572 Grade 50 steel (Fy = 345 MPa, Fu = 450 MPa) and the gusset plate is A36 (Fy = 250 MPa, Fu = 410MPa). What is the ult. capacity of the gusset plate in kips, if block shear governs? What is the effective net area of the channel section, in square millimeters?A WT7x21.5 shape made from A36 steel will be bolted through its flange to a gusset plate using two rows of standard A325X 1 inch bolts. Determine the LRFD available strength, φRn, for gross tension yield of the WT shape. How many bolts per row are required for the LRFD available bolt shear strength of the connection to exceed the LRFD available strength for gross tension yeild of the WT (Problem 1). There are two rows of bolts along the direction of loading. Based on the number of bolts, which affects the connection length, compute the LRFD available strength, φRn, for tensile rupture of the WT shape. Assume the bolts in each row are spaced s=3 inch on center. Do not check bolt bearing limit states.3. 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 c) what is the bolt bearing capacity of the beam web of the shear connection in kips? (2 decimal places)
- Calculate the capacity of the shear connection below. A36 Plate: Fyp= 248MPa, Fup= 400MPa. A992 Beam; Fyb= 345MPa, Fub= 450MPa. Fexx= 480MPa. twb= 10.92mm tp= 12, X= 45, nr= 3(no. of bolts) What is the bolt shear capacity in kN? What is the bearing capacity of the shear plate in kN? What us the bearing capacity of the beam in kN? What is the shear yielding capacity of shear plate in kN? What is the shear rupture capacity of the shear plate in kN? What is the block shear rupture capacity of shear plate in kN? What is the weld capacity in kN?A 1 ⁄2 × 5-inch plate of A588 steel is used as a tension member. It is connected to a gusset plate with four 5 ⁄8-inch-diameter bolts as shown in Figure 1. Assume that the effective net area Ae equals the actual net area An. (a) Calculate the design strength for LRFD? (b) Calculate the allowable strength for ASD?Use load and resistance factor design and select a W shape with a nominal depth of 10 inches (a W 10) to resist a dead load of 175 kips and a live load of 175 kips. The connection will be through the flanges with two lines of 11 4 -inch-diameter bolts in each flange, as shown in Figure P3.6-6. Each line contains more than two bolts. The length of the member is 30 feet. Use A588 steal.
- Two plates each with thickness t = 5 mm are bolted together with 6 - 22 mm diameter bolts forming a lap connection. Plate (A36): Fy = 248 Mpa FU = 400 MPa bolts are A325 in standard holes with threads excluded from shear planes. Considering Tension Rupture, compute the the design strength for LFRD. Use NSCP 2015The design shear strength phi rnv of one A325 X-bolt, 1 1/8 inch in diameter (Group A, X type) in double shear is equal to: Group of answer choices 112 kips 125 kips 74.2 kips 101 kipsDesign 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 KN
- the design shear strength phi rnv of one A325-X bolt (Group A, X type), 7/8 inch in diameter in double shear is equal to: 46.8 kips 30.7 kips 27.9 kips or 61.3 kipsAn angle of ASTM A36 steel (with Fy=36 ksi) is connected by two-rows of 7⁄8-in. bolts, as shown. It is exposed to a dead load of 20 k, a live load of 40 k. Design a 4 in. size (4×?) member considering the strength of the angle only.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.