Determine the design strength of the figure as shown. The steel grade is A992. *
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- A PL ½ x 8 tension member is connected with six 1-inch diameter bolts as shown in the figure below. The steel is ASTM A36. Shear lag factor, u = 1.0. 1. What is the allowable tensile strength (ASD) for yielding? 2. What is the diameter of the hole? 3. What is the ΦPn based on rupture?The angle L 8 x 8 x ½ in tension shown in the figure below must resist aservice dead load of 35 kips, live load of 50 kips, and a roof live load of 2kips. The steel used is A50 (Fy = 50 ksi, Fu = 65 ksi) and its welding length is 4”.Determine if the member has sufficient fracture and yield strength.Determine the design tensile strength of the plate (200 × 100) mm connected to a 12 mm thick gussets, using 20 mm bolts shown below, if the yield and ultimate stress of the steel used are 250 MPa and 420 MPa respectively, fy= 250 MPa fu= 450 MPa
- The properties of a given steel section are as follows; A = 6000 mm2, Ix = 50 x 106 mm4, Iy = 120 x 106mm4, Ixy = 75 x 106 mm4. Compute the minimum radius of gyration. a. 19.32 mm c. 21.36 mmb. 22.43 mm d. 15.48 mmDetermine the design tensile strength of plate (200x8 mm) connected to 10-mm thick gusset using 20 mm bolts as shown in the figure, if the yield and the ultimate stress of the steel used are 250 MPa and 410 MPa, respectively. Add 1mm around the bolt for the hole. Use LRFD method. Plate 8-mm thick 2 3 40+ 30 301 T 200 mm Gusset 10-mm thick 3af 30 2_3 *40 40+ 50,54 +40The 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 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 allowable tensile strength of the section based on tensile rupture of the net area.b) Determine the allowable tensile strength of the section based on yielding of the gross area.Given that the Gerber beam is drawn. Calculate and draw SFD,BMD,NFD Note : 1 ton = 1000kg Solve max in 25-30 mintes thank uThe Aluminum plates below are connected by two 3/4" diameter bolts. What is the shear stress in the bolts given the load applied?
- complete solution and drawing. if u can please add some explanation thankyou FN = 49 MN = 101 SN = 94 The figure below is a bracket which fastened by a bolts with a diameter of 0.5 inch. Compute the corresponding shear stresses. Assume single shear. Note: Round off your answer in two decimal places.The figure shows a plate having a width of 400 mm and thickness of 12 mm is to be connected to another plate by 34 mm diameter bolts as shown in the figure. Assume diameter of holes to be 2 mm larger than the diameter of the bolts. Use A 36 steel plate with yield strength Fy = 248 MPa and minimum tensile strength Fu = 400 MPa. For a chain holes extending across a part in any diagonal or zigzag line, the net width of the part shall be obtained by deducting from the gross width the sum of the diameters or slot dimensions of all holes in the chain and adding, for each gage space in the chain, the quantity S24g where S is the longitudinal center to center (pitch) of any consecutive holes, in millimetres and g is the transverse center to center spacing (gage) between fastener gage lines in millimetres. If a = 60 mm, c = 150 mm and d = 100 mm. Determine the nearest value of “b” so that the net width along bolts 1-2-3-4 is equal to the net width along bolts 1-2-4. Determine the nearest value…5. a bolted lap joint shown in the figure below. The bolts are 20m in diameter in 23 mm holes. The plates are 12m thick X350 Allowable stress of Plates: Tenaion in gross area0.60Fy Tension in net area0.50 Fu Shear on Net Area 0.30Fu Yield Strength of plate, Fy 248 Npa Ultinate tensile strength of Plate. Fu400 Mpa x, Find the safe load P based on a- Gross area yielding b. net area rapture c. block shear 2 of 3
