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- TENSION MEMBERS Given the connection, determine the following: d) The maximum capacity before failure in deformation (yielding) e) The maximum capacity before failure in fracture 10" f) If the member will be subjected to a tensile load of 200 kips, will if fail in yielding? Steel plate, %" thick Fy= 36 ksi Fracture? Fu= 58 ksi 1" diameter boltDetermine the available tensilecapacity of the connection shownin the figure. The yield stress of theHSS 6.000 × 0.280 is Fy = 42 ksiand the tensile strength is Fu = 58ksi. The strength of the 1/2-ingusset plate and the 1/4-in filletweld is adequate.The relevant properties of theHSS 6.000 × 0.280 are obtainedfrom AISC Manual Table 1-13 asAg = 4.69 in²t = 0.260 inD = 6.0 in4 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.
- 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 INNOVATIONS1. Two plates each with thickness 7-160mm are bolted together with 6-22 mm dia. bolts forming Bolt 52= 80mm spacing are as follows S₁ = 40mm Bolt hole dia. 25mm S3 = 100mm L fu- 483 тра ту- 345 тра Solve the allowable istrength and the P a Ultimate strength in: 1- Yielding 2. Rupture 3. Shear 4. Block Shear oo lap connection atat2-A teel ptd is streteky between fwo rigid walls ard carries a tensile load f GO0ON at c f the alowate Stress is not to exceed 130 MPa at -4C, what is the minimum diameter f as ard E= J00 GPa. the nod? Figuro: ST L
- 7 Enercise:- 30 long with is subjected to a Assuming deformation is entirely elastic and manimum allowable elongation tensile 30000y that the is 0.3 mm while manimum. allowable reduction in, digmeter is 0.015mm. choose amory below only. candidate meeting requirent E (GPa) A cylindrical rod 100 mm diameter of 12 mm load of 30000 N. Aluminium alloy Bronze Titanium allow 70 100 110 U(Poiss ons vahi) 0.33 0.34 0.3The assembly consists of two A992 steel bolts AB and EF and an 6061-T6 aluminum rod CD. When the temperature is 30°C, the gap between the rod and rigid member AE is 0.1 mm. Assume BF is also rigid. (Figure 1) Figure 400 mm A B 25 mm 50 mm D 0.1 mm 300 mm E 1 of 1 25 mm Part A Determine the normal stress developed in the bolts if the temperature rises to 180°C. Express your answer to three significant figures and include the appropriate units. Enter negative value in the case of compression and positive value in the case of tension. of = Submit Part B σ₁ = Submit O HA Value Request Answer Provide Feedback ■μÅ Value Determine the normal stress developed in the rod if the temperature rises to 180°C. Express your answer to three significant figures and include the appropriate units. Enter negative value in the case of compression and positive value in the case of tension. Units Request Answer wwwww Units ? wwwww Review ? Next >Which of the following DO NOT match? O Plastic Moment --- FyZ O Long Column -- Fcr = 0.658^(Fy/Fe)[Fy] O Beam Strength --- OMn O Allowable Load --- Rn/2
- 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?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…Tension Member Design Problem 1. A channel shape is under 50 kips dead and 100 kips live tensile axial load as shown in the figure. The member is connected to a gusset plate with 10 inch longitudinal welds. Find the lightest channel shape to carry the loading. Use only vielding and rupture limit states to design. Use 50 ksi steel (Fy=50 ksi, Fu=65 ksi). (a) Assume yielding limit state controls in the design process; (b) After selecting the lightest section, check the rupture limit state. Do not redesing if needed. 1 Pa=50 kips PL=100 kips 10"