. For the figure shown, calculate the net area of the 10 x 200 mm plate. The plate is connected at its end with two lines of 20 mm bolts. Using LRFD, determine the load P that the plates can carry against tensile yielding in the gross section and the tensile rupture in the net section. Use A-36 steel. Pu PL 5x200mm PL 5x200mm Pu PL 10x200mm Pu
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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 +40Considering the following steel connection. The plates in Pink are 9mm steel plates. The middle plate (Yellow) is 18mm thick. The width of the plate is 100mm. The maximum allowable tension stresses on any of the plates is 100Mpa in Gross Area Yielding and 150 Mpa for Net Area or Tension Rupture. The bolts used are 8mm in diameter, the holes are 10mm in diameter, no need to add 1.6mm. The bolts allow a maximum of 280 Mpa of shear. Determine the maximum allowable "P" of the connection in kN.The given plate below with width of 200 mm andthickness of 16 mm is to be connected to two plates ofthe same width with half the thickness by 20 mmdiameter rivets as shown. The rivet hole is 2 mm greaterthan the rivet diameter. Allowable tensile stress on netarea is 0.6Fy. Allowable bearing stress is 1.35Fy. Use a501 plate and a502 gr2 rivet a. Determine maximum load P without exceeding allowable tensile stress on plate b. Determine maximum load P without exceeding allowable shear stress on rivets c. Determine maximum load P without exceeding allowable bearing stress between plates and rivets
- 2. For the figure shown, calculate the net area of the 10 x 200 mm plate. The plate is connected at its end with two lines of 20 mm bolts. Using LRFD, determine the load P that the plates can carry against tensile yielding in the gross section and the tensile rupture in the net section. Use A-36 steel. PL 5x200mm PL 5x200mm Pu PL 10x200mm Pu3. A plate with width of 300mm and thickness of 20mm is to be connected to two plates of the same width with half the thickness by 24mm diameter bolts, as shown. The rivet holes have a diameter of 2mm larger than the rivet diameter. The plate is A36 steel with yield strength F,-248MPa and ultimate strength F,-400MPa. a. Determine the design strength of the section. b. Determine the allowable strength of the section 24mm 30mm4. An UPE 300 is to be used as a tension member and it is bolted to a 10mm thick gusset plate with bearing type 8-M20 8.8 bolts as shown in the figure below. a. Check all spacing and edge distance requirements. b. Compute the allowable tensile strength of the channel UPE 300. Don't consider shear strength of the bolts and bearing strength of the bolt holes. [all dimensions are in mm] Material both for gusset plate and UPE 300: S275 F-275N/mm² Fu=430N/mm² UPE 300 Ag = 5660 mm² tw 9.5 mm x = 28.9 mm 70 160 70 + 10mm thick gusset plate +50+ 80 -80-80-50 o +50+80 -80 80-50 UPE 300
- A 15" x 3/8" bar of A572 Gr. 50 steel is used as a tension member. It is connected to a gusset plate with 7/8-in diameter bolts as shown in the figure. Use s = 2.0 and g = 3.0.a) Determine the design tensile strength of the section based on yielding of the gross area.b) Determine the critical net area of the connection shown.Determine 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
- Use ASD to determine the width of the 5/8 inch thick A36 plate subjected to tension as shown below. The dead load effect is PD=85 kip while the live load effect is PL=80 kip. Two rows of 5/8 inch standard bolts will be used for the connection and the plate width can be manufactured in 1 inch increments. Assume that the larger connected plate does not fail. Consider the limit states of gross tensile yield and tensile rupture. Find the required tension strength, Pa, for the controlling ASD load combination. Determine the width necessary for the plate to resist gross tensile yield. Determine the width necessary for the plate to resist tensile rupture. Report the required plate width.The connection shown is subjected to a tensile force of P = 100 kN. The allowable shear stress for the bolts is 100 MPa. Assume each bolt supports an equal portion of the load. Determine the required diameter of the bolts. a. 20.6 mm b. 25.2 mm c.35.7 mm d.17.8 mm choose letter of correct answerSTAGGERED 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.