Determine the LRFD design strength and the ASD allowable strength. Neglect block shear. A36 steel and 7/8-in Ø bolts 224 N N N 00 72 77 24 in 3 in 3/ in 2 in
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- A 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.Plate A, PL 16mm x 375mm, is connected to another plate B with 10- 22 mm o bolts as shown. The steel is A36 steel with Fy= 250 MPa and F,= 400 MPa. Assume that plate B has adequate strength. Use ASD. Plate B 375a 75 m 2 75mm Plate A 75 mm 75mm 150 50150 Compute the nominal strength based on yielding in KN.O Determine the allowable tensile strength of Plate A in KN.O Compute the allowable tensile strength at section a-1-2-2-4-c in KN.0A PL 38 X 6 tension member is welded to a gusset plate as shown. The steel is A36 (Fy = 36ksi, Fu = 58ksi). a. The design strength, Pu based on gross area b. The design strength, Pu based on effective area PL % x6 3/8" 6" Cross Sectional area of PL3/8x6 a) Blank 1 b) Blank 2
- 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)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)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
- A PL 38 X 6 tension member is welded to a gusset plate as shown. The steel is A36 (Fy = 36ksi, Fu = 58ksi). a. The design strength, Pu based on gross area b. The design strength, Pu based on effective area PL % x 6 3/8" 6" Cross Sectional area of PL3/8x6 a) Blank 1 b) Blank 2 Blank 1 Add your answer Blank 2 Add your answerA PL40 mm X 250 mm (smaller member) is connected to a gusset plate (bigger member) as shown. The diameter of the holes are 25 mm. The pitch and gage of the holes are 50 mm and 75 mm, respectively. The yield strength of the steel is 260 MPa while the ultimate tensile strength of the steel is 400 MPa. Determine the design (LRFD) tensile strength of the tension member in kN. Neglect block shear. H FO E. • D A В2- A 2m COPPER BAR SUBJECTEO TO A TENSILE IS SUSPENDED FROM A LOAD IBO KN SUPPORTED BY TWO IDENTICAL PIN THAT S pOSIS AS SHON OF THE LOWER AFPLIED TENSILE POSTS SUPPORT THE STEEL DETERMINE THE TOTAL DEFORMATION END OF THE LOAD HINT: NOTE THAT THE STEEL PIN WHERE THE COFPER BAR IS ATTACHED. COPPER BAR DUE TO STEEL POST L=0.5m A= 4500mm E- 200GPA COPFER BAR L- am A= 4800 Mn E- l20 GPa 180KN