Three plates (14 mm and 16 mm in thickness) are welded to a W10 x 49 to form a built-up shape as shown. K,Lx= K,L= 7.6 m and F= 345 MPa, compute the design strength for LRFD.
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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 +40An 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.)(1) Given A992 steel Ag = 10 in %3D KLy- 80, ry what is $ Pa=? = 60 the' column streugth design
- 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 INNOVATIONSTENSION MEMBERS: THE SINGLE 200 X 10 mm STEEL PLATE IS CONNECTED TO A 12 mm THICK STEEL PLATE BY FOUR 16 mm DIAMETER RIVETS AS SHOWN IN THE FIGURE. THE RIVETS USED ARE A502 GRADE 2, HOT DRIVEN RIVETS. THE STEEL IS ASTM A36 WITH Fy = 248 MPa AND Fu = 400 MPa. DETERMINE THE VALUE OF P. a. P BASED ON TENSION OF GROSS AREA b. P BASED ON TENSION OF NET AREA c. P BASED ON BEARING OF PROJECTED AREA d. P BASED ON SHEAR RUPTURE (BLOCK SHEAR)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 MPaH.W A Solid Steel bar of dlameter 6omm and length 350 mm is Placed inside an aluminum cylinder öf Inside diameter Femm and out side d lameter Ilomm, a comPressive load of looo lais applied on the assembly, find the stresses in the bar and the cylinder ? take Est = 200GPa and EAl = F0GPan %3D Cover Plate 0.25mm d- do 35omm A lum. cylinder Steel bar10 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 max7 7a 7b 7c Alaterally supported beam was designed for flexure. The beam is safe for shear & deflection. The most economical section is structural tubing however the said section is not readily available at the time of the construction. If you are the engineer in charge of the construction what alternative section will be the best replacement? Why? The section is 8" x 8" x 7.94 mm thick: Use Fy=248 MPa: E=200,000 MPa AISC wall thickness Ix 106 S x 103 Jx 103 mm4 mm3 mm4 rx =ry Area Ag (mm2) mm Designation Weight/m 8x8 7.94 47.36 6,039 79.25 37.84 371.99 60.35 expla'n briefly your cho'ce. (transform your comparative analys's 'nto a narative form to support your cho'ce) 8x8 14.29 80.61 10,258 76.2 59.52 585.02 99.06 8x8 72.7 9,290 76.96 54.53 539.13 90.32 8x8 9.53 56.09 7,161 78.48 44.12 432.62 70.76 mm 12.7 Zx 103 mm3 437.53 714.48 650.57 512.92hexagonnal head 30 60 mm threads bolt diameter @ threads 32 min %3D bolt diameter = 38 mm T=80 kN 1. Calculate the tensile stress in the body of the bolt. 2. Find the tensile stress at the root of the threads. 3. Find the compressive stress at the head as the bolt bears on the surface to resist tensile load.