A 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. CO • D A
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- 4.) A 150 x 90 x 12 angular section is welded to a gusset plate as shown in the figure. The angle is A36 steel with FY=248 MPa. The weld is E 80 electrode with Fu = 550 MPa. The allowable tensile stress for the angle is 0.6Fy and the allowable shear stress for the weld is 0.3Fu. The Area of the angular section is 2751 sq.mm. with y=51mm. CS 56 Which of the following most nearly gives the design force P? a.) 400,365.9 N b.) 409,348.8 N c.) 412,793.5 N d.) 420,366.6 N 5.) A W350 x 90 steel is used as a simply supported beam 8m long. The beam carries three equal concentrated loads at every quarter points. It also carries a uniform dead load o 5 kn/m (including its own weight) and a uniform live load of 7.20 kn/m. Properties of W 350 x 90 steel: bf = 250 mm The allowable bending stress is 0.66Fy. The allowable shear stress is 0.40Fy. The allowable deflection is L/360. Use Fx=248 Mpa and E=200 Gpa. tf = 16.40 mm V d = 350 mm Determine the maximum value of each concentrated load based on…The double T-beam is fabricated by welding the three plates together as shown.  Part A Determine the shear stress in the weld necessary to support a shear force of V=78 kN�=78 kN.The aluminum rod AB (Gal = 27 GPa) is bonded to the brass rod BD (Gbr = 39 GPa). The portion CD of the brass rod is hollow and has an inner diameter of 40 mm. If the allowable shear stresses for the aluminum and brass are 90 MPa and 70 MPa, respectively, and the allowable angle of twist of the free end A is 6 degrees, determine the maximum permissible value of T. Please show your complete handwritten solution. Thank you
- The steel shafts are connected together using a fillet weld as shown in the figure below. (Figure 1) Figure ▾ art A Tavg for Part A for Part 12 mm Value T= 60 N-m 50 mm- Determine the average shear stress in the weld along section a-a if the torque applied to the shafts is T = 63 N-m. Note: The critical section where the weld fails is along section a-s. Express your answer to three significant figures and include the appropriate units. Units 12 mm 1 12 mm 12 mm a do for Part 1 of 1 redo for Part A reset for Part A keyboard shortcuts for Part A help for Part A4 A flanged bolt coupling has ten 12-mm di ameter steel bolts on 500 mm diameter b olt circle and six 16 mm diameter aluminu m bolts on 300 mm diameter bolt circle. T he maximum shear stresses of the materi als are 60 MPa in steel and 40 MPa in al uminum. Use G = 80 GPa for steel and 3 0 GPa for aluminum. What is the maximu m required shear strength of each alumin um bolt to determine the maximum torqu e that can be applied to the system? Dra wing not included in this problem.A The member consists of the steel rod AB that is screwed into the end of the bronze rod BC. Find the largest value of P that meets the following design criteria: (i) the overall length of the member is not to change by more than 3 mm; and (ii) the stresses are not to exceed 140 MPa in steel and 120 MPa in bronze. The moduli of Problem 4: Steel A= 480 mm? 1.0 m B elasticity are 200 GPa for steel and 80 GPa for bronze. Problem 5: The compound bar carries the axial forces P and 2P. Find the maximum allowable value of P if the working stresses are 40 ksi for steel and 20 ksi for alumi- num, and the total elongation of the bar is not to exoceed 0.2 in. 3P Bronze A= 650 mm 2 m 3 ft 4 ft 2P Steel A = 0,4 in.2 E = 29 × 10° psi 2P Aluminum A = 0.6 in.2 E= 10 x 10° psi
- A brass sleeve S is fitted over a steel bolt B, and the nut is tightened until it is just snug. The bolt has a diameter of dB = 25 mm, and the sleeve has inside and outside diameters of d1 = 26 mm and d2 = 36 mm, respectively. d2 di dB Sleeve (S) Bolt (B) Calculate the temperature rise AT that is required to produce a compressive stress of 23 MPa in the sleeve. (Use material properties as follows: for the sleeve, as = 21 x 10-6/C and Es = 200 GPa; for the bolt, aB = 10 × 10¬6/°C and EB = 400 GPa.) AT = °C (Round to the nearest whole number.)The following steel member is connected to the gusset plate via two longitudinal welds. The width of the member is 58 mm and the length of the weld is 87 mm. The yield stress is 300 MPa and the ultimate stress is 675 MPa. The gross area of the member is 700 mm^2 . a. What is the tensile design yield strength in KN? b. What is the tensile design rupture strength in KN? c. If a tensile load of 85 KN is requiredfor the member, what is the minimum factor of safety for the steel member?A plate with width of 300mm and thickness of 20mm is to be connected between two plates 12mm thick with same width by 4-25mm diameter rivets a shown. The rivet holes have a diameter 2mm larger than the rivet diameter. The plate is A36 steel with yield strength Fy=248 Mpa and Fu-400MPa. The rivets have nominal shear stress capacity of 165MPa. O 803 O 1490 • Determine maximum load (P1) that can be applied without exceeding design tensile stress in plates (consider yielding and rupture of plates ONLY) in KN O 1180 O 1340 ο ο ο ο ο P3 O 1968 O 590 O 670 O 893 O 707 OO 150 O 1180 P1 P2
- A single unequal angle 100 x 75 x 6 is connected to a 10 mm thick gusset plate at the ends with six 16 mm diameter bolts to the 100 mm of leg to transfer tension as shown in figure. What will be the block shear strength (in kN) of the angle section assuming that the yield and the ultimate stress of steel used are 250 MPa and 410 MPa respectively? ISA (100 × 75 x 6) -T 16 mm o bolt 10 mm 40 mm 40 mm| 40 mm 40 mm 40 mm 40 mm 10 mmSituation 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 INNOVATIONS2- 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