Question 5 The A572 Grade 50 (fy = 50 ksi, fu = 65 ksi) tension member shown in Figure 3 is connected with three - in. bolts. Determine the LRFD design strength (OP, or ORn) and the ASD allowable strength (Pn/N or Rn/N) of the member. Include Block Shear strength. The uniform tensile stress distribution is assumed. 3.0 in - - 3.0 in = l %3D 3 in = 1, L6 x 4 x (A = 4.75 in?, x in unconnected leg = 0.981 in) 5 in 13 in 5 in Shear plane Tension plane Figure 3.
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- 1. A tension member is made of 90 mm thick steel plates 225 mm wide by a lap joint made with three (3) rows of 20mm. Use LFRD specifications ASTM A36 (Fy=138 MPa, Fu=220 MPa, compute the following: a. Capacity of the section based on the yielding of the gross area Critical effective net area b. c. Capacity of the section based on the tensile fracture of the critical net area. Dia 20mm DIAMETER OF BOLTS PA 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.Topic:Bolted Steel Connection - Civil Engineering *Use latest NSCP/NSCP 2015 formula to solve this problem *Please use hand written to solve this problem The bracket shown in the figure is supported by four 22 mm diameter bolts in single shear. The bracket is subject to an eccentric load of 150 kN. Use LRFD. Use A36 steel. Questions a) Determine the critical force on the most stressed bolt. b) Determine the nominal shear stress of the most stressed bolt.
- 2) Find the axial stresses of menbers FD, GD, GE State if it is tensile or Compressive. 4M 3M A LE 3m G 20 RN Go KNDetermine 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 tProblem2. The compression member is shown in figure. Find the following: a. The Euler stress Fe. b. The buckling stress Fcr c. The design strength d. The allowable strength e. Does the member satisfactorily meet the design requirements? Why? HSS 8x 8x4 ASTM AS00, Grade B steel (Fy = 46 ksi) 15'
- Topic: COMBINED STRESS-AXIAL TENSION AND FLEXURE BENDING: STEEL DESIGN Please solve your Solution in a handwritten Note: It should be handwritten pleaseeee Questions A Tension member with no holes is subjected to axial loads of PD=68kN & PL=64kN.It is also subjected with bending moments of MDy=40kN-m & MLy=55kN-m. Is themember adequate? Steel is of A992 Gr 50 Specs. Use LRFD. Neglect the weight of the beam. Section properties:Lp = 1.863 md=459.99mmtw=9.02mmzx= 1835x10^3 mm^4Lr = 5.305 mLb = 4.8 mbf=191.26mmtf=16mmSx=1611x10^3 mm^4 Sy=195x10^3 mm^4Zy=303x 10^3 mm^4 Cb=1.32A 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 G P P D B F C A ESituation 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 INNOVATIONS
- 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. • D A ВDetermine 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 tension member shown is an angle L150 x 90 x 10, of A36 steel, with Fy= 250 MPa and Fu= 400 MPa, and is subjected to a service DL= 150 KN and a service LL= 150 KN The bolts are 20 mm o. Use ASD. 40mm 40mn 40mm 50mm ... Compute the allowable strength based on yielding in KN.0 Compute the nominal strength based on fracture in KN.0 Check the adequacy of the section.0