For the tension member shown, determine the total service load (DL + LL) if LL = 4DL. Use A36 steel
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- Determine the capacity of the details shown below.A992: Fy=50ksi, Fu=65ksiA36: Fy=36ksi, Fu=58ksieffective bolt hole diameter = bolt hole + 1/16"beam properties: tw=0.38 in., bf=7.07 in., tf=0.63 in.Fexx=70ksip) what is the tensile yielding capacity of flange plate of the flange connection in kips? (2 decimal places)ASSIGNMENT #5 BIAXIAL BENDING Problem: Check the beam shown for compliance with the NSCP Specification. Lateral support is provided only at the ends, and A992 steel is used, F, = 345 mPa. The 90 kN service loads are 30% dead load and 70% live load. a. Use LRFD b. Use ASD d tw by ty 1x(106) S(10³) mm mm mm mm³ mm mmª W410x100 415 10 260 16.9 398 1920 SECTION Tx mm 177 1m 90 kN 90 KN 1.5m W410x100 I Loaded Section Ly(106) Sy(10³) Ty mmª mm³ mm 49.5 381 62.4 1m Zx(10³) Zy(10³) mm³ mm³ 2130 581A simply supported beam shown below is HEA 400 of 5355 steel. It is laterally supported at points with an "X" The material properties and the section properties are given below. Use ASD load combination DL+LL and determine whether the beam has adequate flexural strength. Material: $355 Fy= 355 MPal Fu= 510 MPa E= 200000 MPa Section properties: A: 15900 mm² d: 390 mm h: 298 mm b: 300mm tr: 19 mm W 2311x10³ mm³ Iy: 8564x10*mm* A 4m t: 11 mm. İy: 73.4 mm W: 2563x10³ mm³ J: 189x104 mm DL: 150 KN LL: 100 KN 17 B 4m Cw: 2942x10⁹ mm C E: 200000 MPa
- 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 the capacity of the details shown below. A992: Fy=50ksi, Fu=65ksiA36: Fy=36ksi, Fu=58ksieffective bolt hole diameter = bolt hole + 1/16"beam properties: tw=0.38 in., bf=7.07 in., tf=0.63 in.Fexx=70ksi g) what is the flexural/local buckling capacity of the shear plate of the shear connection in kips? (2 decimal places)TENSION MEMBERS: THE SINGLE 200 X 10 mm STEEL PLATE IS CONNECTED TO A 12 mm THICK STEEL PLATE BY FOUR 16 mm DIAMETER RIVETS AS SHOWN IN THE FIGURE. THE RIVETS USED ARE A502 GRADE 2, HOT DRIVEN RIVETS. THE STEEL IS ASTM A36 WITH Fy = 248 MPa AND Fu = 400 MPa. DETERMINE THE VALUE OF P. a. P BASED ON TENSION OF GROSS AREA b. P BASED ON TENSION OF NET AREA c. P BASED ON BEARING OF PROJECTED AREA d. P BASED ON SHEAR RUPTURE (BLOCK SHEAR)
- Determine the capacity of the details shown below.A992: Fy=50ksi, Fu=65ksiA36: Fy=36ksi, Fu=58ksieffective bolt hole diameter = bolt hole + 1/16"beam properties: tw=0.38 in., bf=7.07 in., tf=0.63 in.Fexx=70ksi d) what is the shear yielding capacity of the shear plate of the shear connection in kips? (2 decimal places)Find the block shear strength of a tension member as shown in figure below. The steel is of grade Fe410. a d 50 115 18 mm dia hole C b ISA 80 × 50 × 8 45 35Please write the complete solutions and Legibly. Rate will be given! Do not just copy asnwers in CHEGG! Thank you :) A singly reinforced rectangular beam 270 mm x 430 is reinforced with 4-16-mm-dia. rebars with yield strength 420 MPa. If bar centroid is located 70 mm from the bottom edge of the beam and f'c = 21 MPa, calculate the design moment strength in kN-m. Write your answer with 2 decimal places only.
- For the column shown what is the nominal compressive strength (Pn) ? Pn=? W12x50 10 12X50 THT X combilevered WIA 1.5" x1.5" solid steel bar is made of a steel having an Fy = 50 ksi and an Fu = 65 ksi. It is loaded in tension by the following unfactored forces: • DL = 30 kips ● LL = 50 kips • SL 10 kips = Determine if the bar yields in tension. Assume LRFD. Use the ASCE 7-22 Load Combos and Factors.sted ingid 250MM B k steel bar bolt AUR rigid bar 12 mm thick 8 mm Brass 350mm Jointd 16 mene thick 350mm JointB 250 mm Xg = 20x10% Eg = 90 G Pa -6 A, = 12x10/0 Es = 200 G Pa thickness> 16 mm 8 mm bolt %3D 12 mmi thick shear strength f bolf bearing streng th of holt= 100 la 50 MPa %3D Assume no failure will take place in steel or brass. Temperature chauge brass steel on Temperature change on Determine thai can be applied to system - the max