design tensile strength
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A: Given data Steel details As = 600 mm2 Es = 200 GPa = 2 × 105 N/mm2 Bronze details AB = 300 mm2…
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Q: rbitrary point, calculate the principal stresses and the бу: 2 = 100 MPa s Txy = 55.9 MPa x=200MPa
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A: Direct tensile stress=PA=21.62=0.78125ksi
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Q: Q1: The C-shaped steel bar has cross section of a square of side 1.6 in. If the magnitude of…
A: Direct tensile stress=PA=21.62=0.78125ksiDirect Bending stress=Pe(y)I=2×3.2×1.621.6412=9.375ksiso,
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- Determine 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 +40A plate with width of 420 mm and thickness of 13 mm is to be connected to a plate of the same width and thickness by 30 mm diameter bolts, as shown in the Figure 1. The holes are 3mm larger than the bolt diameter. The plate is A36 steel with yield strength Fy = 248MPa. 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 equl to the net width along bolts 1-2-4. W = 420 mm t = 13 mm a = 64 mm c = 104 mm d = 195 mm Bolt Diameter = 30 mm Holes Diameter = 30 mm + 3 = 33 mm ?? = 248 MPa Allowable Tensile Stress on ?? = 0.60Fy a. Calculate the vaue 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.An A36 steel plate with width of 400 mm and thickness of 12 mm is to beconnected to a plate of the same width and thickness by 34 mm diametersbolts, as shown in the figure (next slide). The holes are 2 mm larger thanthe bolt diameter. The yield strength of the steel plate is Fy = 248 MPa.Assume allowable tensile stress on net area is 0.60Fy. It is required todetermine the value of b such that the net width along bolts 1-2-3-4 isequal to the net width along bolts 1-2-4.a. Calculate the value of b in mm.b. Calculate the value of the net area for tension in plates in mm².c. Calculate the value of P so that the allowable tensile stress on net areawill not be exceeded.(Use at least 2 decimal points on your answers.)
- Situation 5. The angular section shown below is welded to a 12 mm gusset plate. Both materials are A36 steel with Fy = 250 MPa. The allowable tensile stress is 0.6Fy. The weld is E80 Electrode and 12 mm thickness. INNOVATIONS Properties of L 150x90x12: y = 50 shear stress of weld = 0.3Fu A = 2750 Allowable REVIEW INNOVATIE a K ➜ www A. 234 KN B. 349 KN b 13. What is the value of P without exceeding the allowable tensile the angle? C. 382 kN p. 413 kN 14. Find required length of the weld based on shear? A. 280 mm C. 300 mm D. 380 mm B. 320 mm 15. Find the required value of a? A. 108 mm B. 97.9 mm D. 185 mm NEW INNOVATIONS REVIEW INNOVATIOf REVIEW NEW INNOVATIONSDetermine the design tensile strength of the 12 in. x 1/2 in. steel plate shown in the figure. The bolts are 3/4 in. diameter. The steel is A572 Gr. 50. Check yielding and fracture. Check Block Shear. T 3in. 73im 13in 1 3in tA plate 400 x 12 mm is to be connected to a plate of the same width and thickness by 34 mm diameter bolts, as shown. The holes are 2 mm larger than the bolt diameter. The plate is A36 steel. Assume allowable tensile stress on net area is .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. d. Calculate the nominal block shear strength based on possible failure paths
- H.W (2): A 37.5 mm round bar has been machined from AISI 1050 cold-drawn round bar Su =689 MPa. This part is to withstand a fluctuating tensile load varying from 0 to 72.5 kN. Because of the design of the ends and the fillet radius, a fatigue stress-concentration factor of 1.85 exists. Find the factor of safety F.S.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.Situation 2. Two plates each with thicknessF16mm are bolted together with6 -22mm dimater bolts forming a lap conne ction. Bolt spacing are as follows: S1 = 40mm, S2 = 80mm, Sa = 100mm. Bolt hole diameter=25 mm 50 250mm 30 30 60 75 Allowable stress: Tensile stess on gross area of the plate=0.60 Fy Tensile stress on net area of the plate=0.5Fy Shear Stress of the bolt Fv=120MPA Bearing Stress of the bot Fp=1.2 Fu Calculate the permiss ible tensile load P under the following Conditions: 4. Based on shear capacity of bolts 5. Based on bearing capacity of bolts 6. Based on block shear strength (taking into consideration the failure path given in the figure below) 40 80 16 mm 40 180 mm 40 Bearing Failure Path #1 140 209 Bearing Failure Path #2
- 10 mm 20 mm The assembly above is comprised of a 10mm-diameter steel bolt, and a 20mm-diameter, 2mm-thick bronze sleeve. The bolt and sleeve are rigidly connected at each end by a washer. Assume the following properties: Epolt = 29000 ksi &bolt = 6.6-106 1°F Vbolt = 0.35 Member BC is made of bronze with the following properties: Esleeve = 15000 ksi Xsleeve =9.80-106 /°F Vsleeve = 0.35 If the assembly is heated up, which of the following statements best describes how the sleeve will be affected? The sleeve will try to expand longitudinally, but will be restricted by the bolt (because the bolt has a lower coefficient of thermal expansion). The sleeve will go into tension and the bolt will go into compression. O The sleeve will try to expand longitudinally, but will be restricted by the bolt (because the bolt has a lower coefficient of thermal expansion). Both members will go into tension. The sleeve will try to expand longitudinally, but will be restricted by the bolt (because the bolt…sted ingid 250MM B k steel bar bolt AUR rigid bar 12 mm thick 8 mm Brass 350mm Jointd 16 mene thick 350mm JointB 250 mm Xg = 20x10% Eg = 90 G Pa -6 A, = 12x10/0 Es = 200 G Pa thickness> 16 mm 8 mm bolt %3D 12 mmi thick shear strength f bolf bearing streng th of holt= 100 la 50 MPa %3D Assume no failure will take place in steel or brass. Temperature chauge brass steel on Temperature change on Determine thai can be applied to system - the maxThe 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