Determine the available strength of the compression member shown in Figure 15 HSS 8x8x ASTM A500, Grade B steel (Fy= 46 ksi)
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Determine the available strength of the compression member shown in figure below.
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- Problem2. 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) 15'4.3-4 Determine the available strength of the compression member shown in Figure P4.3-4. in each of the following ways: a. Use AISC Equation E3-2 or E3-3. Compute both the design strength for LRFD and the allowable strength for ASD. 15 HSS 10x6x ASTM A500, Grade B steel (Fy=46 ksi) 2/32. A column HP 14 x 102 of A572 Gr. 55 steel has a length of 15 ft is fixed at both ends. Compute the design compressive strength for LRFD and the allowable compressive strength for ASD. (Steel section properties are provided in the next page) ASTM Designation A572 Gr. 42 Gr. 50 Gr. 55 Gr. 60⁰ Gr. 65⁰ Yield Stress (ksi) 42 50 55 60 65 Fu Tensile Stressa (ksi) 60 65 70 75 80
- For the column shown what is the nominal compressive strength (Pn) ? Pn=? W12x50 10 12X50 THT X combilevered WIEstimate the cross-sectional area of a 350S125-27 cold-formed shape. a. If the member is tested in tension, what would be the maximum force thesample could carry before reaching the yield strength if the steel has ayield strength of 225 MPa?b. Would you expect a 2.5 m stud to carry the same load in compression?(explain)7 7a 7b 7c Alaterally supported beam was designed for flexure. The beam is safe for shear & deflection. The most economical section is structural tubing however the said section is not readily available at the time of the construction. If you are the engineer in charge of the construction what alternative section will be the best replacement? Why? The section is 8" x 8" x 7.94 mm thick: Use Fy=248 MPa: E=200,000 MPa AISC wall thickness Ix 106 S x 103 Jx 103 mm4 mm3 mm4 rx =ry Area Ag (mm2) mm Designation Weight/m 8x8 7.94 47.36 6,039 79.25 37.84 371.99 60.35 expla'n briefly your cho'ce. (transform your comparative analys's 'nto a narative form to support your cho'ce) 8x8 14.29 80.61 10,258 76.2 59.52 585.02 99.06 8x8 72.7 9,290 76.96 54.53 539.13 90.32 8x8 9.53 56.09 7,161 78.48 44.12 432.62 70.76 mm 12.7 Zx 103 mm3 437.53 714.48 650.57 512.92
- 4. Calculate the design strength (ocPn) of W24X76 with length of 12 ft. and pinned ends. A572 Grade50 steel is used. E=29x103 ksi. Show your work in detail. ASTM Classification A36 A572 Grade 50 A992 Grade 50 A500 Grade B (HSS rect, sq) A500 Grade B (HSS round) A53 Grade B Yield Strength F, (ksi) 36 50 50 46 42 35 Ultimate Strength F (ksi) 58 65 65 58 58 602) 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 KNUse 2015 NSCP. A compression member shown below with Fy = 50 ksi 30′ W12 x 87 A992 steel 1. Which of the following most nearly gives the value of critical buckling stress in ksi? 2. Which of the following most nearly gives the nominal strength in kips? 3. Considering ASD. Which of the following most nearly gives the allowable strength in kips? 4. Considering LRFD. Which of the following most nearly gives the design strength in kips?
- 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=1.32Determine whether the D = 560 kips L = 68 kips compression member shown is adequate to support the given service loads. Take note Pu = 1.4D. 20' W12 × 79 K = 0.80, r = 3.05 in E = 29000 ksi, Fy = 50 ksi %3D A992 steel Input Yes or No for your final answer. Blank 1 Blank 1 Add your answerDetermine the Design Flexural Strength of W460x52 A992 Steel with the following cases: CIVIL ENGINEERING STEEL DESIGN With Continuous lateral bracing Unbraced length = 4m, Cb = 1.0 Unbraced length= 12m, Cb = 1.0 a.) b.) c.) Fy = 345 MPa Properties: d = 450mm tf = 10.8 mm 1x 212x10mm* Cw 306x10 mm" Sx** 944 x10¹mm¹ bf 142mm MY Ag - 6650mm2 Zx 1090x10'mm' Ho439mm ry = 31mm tw* 7.62mm J-211x10¹mm* ly = 6.37x10 mm rls - 38.4mm