A W 5300*101 of Steel Section (fy= 345 MPa & fu= 450 MPa) is used as simply supported beamwith a span of 15 The dead load is the summation of the super dead load (X1=9 kN/m as uniform load) and the beam weight. In addition to the dead load, a uniform live load is applied to the beam. If the lateral support is provided at 3-meter intervals, what is the maximum service live load in kN/m t
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- A W 5300*101 of Steel Section (fy= 345 MPa & fu= 450 MPa) is used as simply supported beamwith a span of 15 The dead load is the summation of the super dead load (X1=9 kN/m as uniform load) and the beam weight. In addition to the dead load, a uniform live load is applied to the beam. If the lateral support is provided at 3-meter intervals, what is the maximum service live load in kN/m that can be supported.
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- The rigid frame shown is unbraced. The members are oriented so that bending is about the strong axis. Support conditions in the direction perpendicular to the plane of the frame are such that Ky = 1.0. The beams are W410 x 85, and the columns are W250 × 89. A992 steel is used, Fy=345 mPa. The axial compressive dead load is 750 kN and the axial compressive live load is 900 kN. Use the AISC alignment chart or Magdy I. Salama Formula for Kx. 4.5 m 5.5 m 6.0 m ly Cw Sx mm^2 mm^4 mm^3 1100 1510 A Ix rx Sy ry Zx Zy SECTION m mm^4 mm^3 378 mm^3 mm^3 mm^3 mm^9 1040 713 716 mm mm W250x89 4500 11400 143 W410x85 5500 10800 315 48.4 65.2 1230 574 1730 310 112 171 18 199 40.8 926 Which of the following best gives the axial strength of column AB? Use the NSCP 2015 LRFD specifications. O 2,340 kN 1,647 kN O 2,745 kN O 2,471 kNThe rigid frame shown is unbraced. The members are oriented so that bending is about the strong axis. Support conditions in the direction perpendicular to the plane of the frame are such that Ky = 1.0. The beams are W410 x 85, and the columns are W250 × 89. A992 steel is used, Fy=345 mPa. The axial compressive dead load is 750 kN and the axial compressive live load is 900 kN. Use the AISC alignment chart or Magdy I. Salama Formula for Kx. B 4.5 m 5.5 m 6.0 m Sx mm^2 mm^4 mm^3 1100 L A Ix Sy Zx Cw ly mm mm^4 mm^3 112 48.4 Zy mm mm^3 mm^3 mm^3 mm19 1040 SECTION ry 65.2 1230 574 713 W250x89 4500 11400 W410x85 5500 10800 315 143 378 1510 171 18 199 40.8 1730 310 926 716 Which of the following best gives the allowable axial strength of column AB? Use the NSCP 2015 LRFD specifications. O 1,647 kN 2,471 kN O 900 kNThe rigid frame shown is unbraced. The members are oriented so that bending is about the strong axis. Support conditions in the direction perpendicular to the plane of the frame are such that Ky = 1.0. The beams are W410 x 85, and the columns are W250 x 89. A992 steel is used, Fy=345 mPa. The axial compressive dead load is 750 kN and the axial compressive live load is 900 kN. Use the AISC alignment chart or Magdy I. Salama Formula for Kx. B 4.5 m 5.5 m 6.0 m Sx ly Sy mm^4 mm^3 A Ix Zx Zy J Cw rx ry mm^3 mm^3 mm^3 mm^g SECTION mm^2 mm^4 mm^3 mm mm W250x89 4500 W410x85 5500 10800 11400 143 1100 112 48.4 378 65.2 1230 574 1040 713 315 1510 171 18 199 40.8 1730 310 926 716 Which of the following best gives the allowable axial strength of column AB? Use the NSCP 2015 LRFD specifications. O 1,647 kN O 2,471 kN O 900 kN
- Determine the safe load of the column section shown, if it has a yield strength of 25 MPa. E = 200000 MPa. Use NSCP Specifications. Fyz248 mpa Properties of Channel Section d = 305 mm t₂ = 7.2 mm A = 3929 mm² t₁ = 12.7 mm Ix=53.7 x 10mm¹ x = 117 mm Properties of W 460 x 74 A = 9450 mm² b = 190 mm ly= 1.61 x 10 mm x = 17.7 mm tw = 9.0 mm rx = 188 mm ry = 41.9 mm d = 457 mm tr = 14.5 mm Ix = 333 x 10 mm Iy = 16.6 x 10mm* 7.21 When the height of column is 6 m. When the height of column is 10 m. Assume K= 1.0 457 CIVIL ENGINEERING- STEEL DESIGNDetermine whether the column shown in Figure is adequate to support 850 kN dead load 2400 kN live load. The steel material is $275. The column is HE450B. P min 4500 mm T 4500 mm Homework 4 - Compression Members - Flexural Buckling Strength QUESTION 7 According to the calculation above, we can say that the flange and web elements are slender elements nonslender elements QUESTION 8 Calculate the buckling length according to the x axis (Lcx) (unit will be mm and no need to write the unit). QUESTION 9 Calculate the buckling length according to the y axis (Lcy) (unit will be mm and no need to write the unit). ПThe AD beam, which is fixed at left end, is under various loads as shown below. a) Draw internal load diagrams (Nx, Vy, M2) using the method of section. b) Determine the state of stress at points G, H and I for section B. c) For the corresponding points, indicate the results (i.e. stress components) on a differential element. b2 A B G М F2 2l h bi GIVEN DATA: Fo = 10000 N, F1 1200 N, F2 = 1000 N, qo = 50 N/cm, Mo = 50000 Nem, l = 25 cm bi = 10 cm, b2 = 20 cm, h 12 cm, t 1 cm
- QI/ A beam carries the loads shown in figure, if the tensile stress must not exceed 20 MPa and the compression stress 70MPa. Find the maximum value of load P Som, r W=2-5p W/m - P Lflj Ko ; - 3,‘, ) o Jo 1 er -1 /i e,PROBLEM 2: The rigid frame shown is unbraced. The members are oriented so that bending is about the strong axis. Support conditions in the direction perpendicular to the plane of the frame are such that Ky = 1.0. The beams are W410 x 85, and the columns are B W250 x 149. A992 steel is used. The axial 4.5 m compressive dead load is 400 kN, and the axial compressive live load is 490 kN. a. Determine the axial compressive design strength of column AB. 5.5 m 6.0 m b. Determine the allowable axial compressive strength of column AB.NO. 1. What is the maximum bending stress in the uniformly loaded cantilever beam as shown if the depth of the beam is h = 250mm, the maximum deflection is 8max = 5mm, the span is L = 3m, and the modulus of elasticity is E = 70 GPa. Use any method. Wo L max A B 4-T
- The beam shown is simply supported and has lateral support only at its ends. The only service dead load is the weight of the beam. Determine whether it is satisfactory for the load shown. A992 steel (E= 345 MPa and F= 450 MPa) is used, and the 30 KN/m is a service live load. Use LRFD 30 KN/m WiL = 30 KN/m W16x 40 -Centroid W16 x 40 3mFor the column shown below, create an interaction diagram on Excel for bending about the strong axis and email it to the instructor. Show the nominal strength as well as the ACI strength including the strength reduction factor, . Utilize the full capacity of the ACI 318. For the graph, the y-axis shall be the axial load and the x-axis shall be the moment. Plot the factored loads, as points, for the various combinations and indicate them in a key. Additionally, plot the same load on the appropriate interaction diagram from the textbook and the provided a copy of that interaction diagram along with the spreadsheet. All work must be your own. Do not collaborate with classmates. Given: f= 4,000 psi fy = 60,000 psi Comb. 120 Design Loads Pu K 780 525 404 570 160 3 4 5 Muy FT-K 62 180 275 145 320 18" 2.0"-> ..... V 20" 10-#7 Total 2.0" XA column 3 m long and pinned at both ends is carries an axial load of 190 kN. The column is made up of 2 angles of unequal legs with long legs back to back and separated by a gusset plate whose thickness is 11 mm. Use A36 steel with Fy = 248 MPa and E = 200 GPa. Three sections are being considered, as follows, with their respective properties relevant of this problem: Section Area ( m? ) rx ( m ) ry ( m ) 2L 125 x 75 x 12 0.00454 0.0390 0.0160 2L 150 x 90 x 10 0.00463 0.0480 0.0195 2L 150 x 90 x 12 0.00550 0.0500 0.0251 a. Which of the sections gives the largest allowable compressive stress? b. Which of the sections gives the most economical (lightest) section for the given load?