A C9x20 is used as tension member connected to a gusset plate of ½ in thick (Figure 5). The member has to resist service dead load and live load of 40 kips & 80 kips respectively. Steel used is A36. Assume connection is adequate in block shear. Design the bolted connection (check slip critical, shear and bearing strength). Use ASD. t = " C9x 20
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- A rigid bar AC is supported pinned supported at A and bar DB at B. See typical bolt connection detail. A concentrated weight, denoted as W is attached at the end of the rigid bar. L1 = 2.6 m L2 = 2.1 m H= 1.2 m W= 14 KN What is the required diameter of the bolt (in mm) at connection B given the allowable shearing stress to be 40 MPa ? Consider single shear. L1 L2 A B C H D इस W CONNECTION AT BA 16 mm thick tension member is connected by two 6.25 mm spliced plates as shown. The tension member carries a service loads of dead load of 110 kN and a live load of 100 kN. 40, 80 , 40 , 40, 80 40 1625 mm 16 mm 1625 mm Fy 248 MPa Fu = 400 MPa Diam. of bolts = 16 mm Fnv = 300 MPa O Determine the nominal strength for one bolt due to shear. O Determine the nominal strength for one bolt due to bearing strength of the connection. ® Determine the number of bolts required for the connection.Five M20 8.8/ bolts are used in the connection between a beam and a column. Both sides of the cleat plate are welded to column using fillet weld (Manual metal arc AS4855 B-E43XX) with size of 5 mm. The design force transferred from beam to column is 300 kN. The cleat plate has thickness of 12 mm, fy = 300 MPa, and fü = 360 MPa. Determine the unit stress sustained by the weld
- The plate with dimension 18 mm x 400 mm (thickness by width) is to be connected to a gusset plate using 6-Ø24 mm bolts in arrange in 3 row (2 each row) along the x-axis. The plate is to be design for tension. Using A36 steel and assuming A= An. Use NSCP 2015. a. Compute the design strength considering yielding and tensile rupture. (LRFD} b. Compute the allowable strength considering yielding and tensile rupture. (ASD)A bolted connection shown consists of two plates 300mm x12mm connected by 4 - 22 mm diameter bolts. Edge distances = 75mm dhole for tensile and rupture = db + 3 mm dhole for bearing strength for Lc = db + 1.5 mm Fy = 248 Mpa Fu = 400 Mpa Fn = 330 Mpa Use LRFD design method. Determine the design strength due to the gross yielding of plates. (kN)A 15" x 3/8" bar of A572 Gr. 50 steel is used as a tension member. It is connected to a gusset plate with 7/8-in diameter bolts as shown in the figure. Use s = 2.0 and g = 3.0.a) Determine the design tensile strength of the section based on yielding of the gross area.b) Determine the critical net area of the connection shown.
- A 15" x 3/8" bar of A572 Gr. 50 steel is used as a tension member. It is connected to a gusset plate with 7/8-in diameter bolts as shown in the figure. Use s = 2.0 and g = 3.0. Determine the critical net area of the connection shown. Determine the design tensile strength of the section based on yielding of the gross area.The plate with dimension 18 mm x 400 mm (thickness by width) is to be connected to a gusset plate using 6-Ø24 mm bolts in arrange in 3 row (2 each row) along the x-axis. The plate is to be design for tension. Using A36 steel and assuming Ae= An. Use NSCP 2015. a. Compute the design strength considering yielding and tensile rupture. (LRFD) b. Compute the allowable strength considering yielding and tensile rupture. (ASD)Design a welded connection for an MC9x23.9 of A572 Grade 50 steel connected to a 3/8-inch-thick gusset plate (Figure 6). The gusset plate is A36 steel. Show your results on a sketch, complete with dimensions. = 3/8" Figure 6 D = 48 k L = 120 k MC9 x 23.9 a. Use LRFD. b. Use ASD.
- Problem. A bolted connection shown consists of two plates 300 mm x 12 mm connected by 4 - 22 mm diameter bolts. 12 mm bolts t=12 mm t=12 mm P. P- P 300 75 75 75 Edge distances -75 mm dhole for tensile and rupture d, +3 mm dnole for bearing strength for L- de +1.5 mm Fy 248 MPa F.- 400 MPaCompute the maximum acceptable tensile SERVICE LOAD that may act on a single tee section that is connected to a gusset plate using welds 12 inches long, as shown in the figure. The service live load is three times the dead load. Use A992 steel. USE LRFD ONLY, no block shear will occur. WT12 x 38 Longitudinal welds 11.2 in? y = 3.0 in. Given: Properties of WT12 × 38: Ag = Use A992 Steel: F, = 50 ksi Fu = 65 ksi bf = 8.99in. %3D %3D LL = 3 DL %3D %3D tw y = centroidal distance bf C. What is the Governing Ultimate Tensile Capacity based on Net Fracture Round your answer to 3 decimal places.The tension member is a PL 1⁄2 × 6. It is connected to a 3⁄8-inch-thick gusset plate with 7⁄8-inch-diameter bolts. Both components are of A242 steel. Note: A242 Fu = 70ksi dh = db + 1/16’’ Use: Consider deformation at the bolt hole what is the: minimum spacing as per AISC code provisions maximum spacing as per AISC code provisions minimum edge distance as per AISC code provisions maximum edge distance as per AISC code provisions