as shown. It is used as a compression member to support a vertical load, P. The column is hinged on its top and fixed at the bottom (k = 0.70). Determine the largest allowable P that the column can carry if it has an unsupported length of 4m. Use A36 steel (Fy = 248 MPa) Properties of an L100x100x8: Area = 1551 mm² Moment of Inertia, Ix = ly = 1448 x 10³ mm Centroid, x = y = 27.37 mm Ans. 343.02 KN 8 100 100
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- 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) 2/37 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 437.53 714.48 650.57 512.92A W21 x 57 of A992Gr50 steel column having length of 6 m and is fixed at both ends. (Gr50 steel = 50 ksi = 345 MPa = Fy) is used to carry a load of 900 kN dead load and 400 kN live load. Note: Use NSCP 2015 Properties of W21 x 57 A 10800 mm d = 536 mm bf = 167 mm tr = 16.5 mm tw 10.3 mm Kdes rx = 212 mm Ry = 34.3 mm = 29.2 mm a. Calculate the critical slenderness ratio of the W section if the column is braced at midpoint of the weak axis b. Calculate the compressive capacity of built-up section using ASD and LRFD c. Calculate the demand-capacity factor, Is the section adequate?
- A built-up section shown, 30 feet long, is fixed at both ends and braced in the weak direction at the midheight. A992 steel is used. Determine the following: 11. Ix 12. Ky 13. effective slenderness ratio 14. Fe 15. LRFD design strength 16. ASD allowable strength 18 4 W10 × 49 A 14.4 in² bf 10 in d. 10 in tw 0.340 in tf 0.560 in Ix= 272 in¹ Iy=93 4in¹ W10×54 A 15-3 in² bf 10 in d = 10 lin tw0-370 in tf 0.615 in Ix= 303 in 4 Ty = 103 in 1Estimate 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)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 3m
- A column is built up from four (4)- 125 x 125 x 18 angle shapes as shown. The plates are not continuous but are spaced at intervals along the column length and function to maintain the separation of the angles. They do not contribute to the cross-sectional properties. The effective length is 4 m. Compute the allowable design compressive strength based on flexural buckling. E= 250 MPa. Use ASD. k 375 mm 125mm, HPlate 125mm 4 - 4 125 × 125× l8 section 下好业The tension member shown below is C12 x 20.7 of A36 steel. Will it safely support a service dead load of 160kN and a service live load of 325kN? Use equation 3.1 for U. A. Use LRFD B. Use ASD Note: Use SI units (mm) for the following dimensions; Use CSI Steel for the Steel Properties. 12" 22" 22" 22" ооо ооо оооо 7/8-in.-diameter bolts C12 x 20.7Determine 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 answer
- The beam shown in the figure has continuous lateral support of bothflanges. The uniform load is a service load consisting of 50% deadload and 50% live load. The dead load includes the weight of thebeam. If A992 steel is used, is a W16 × 31 adequate?a. Use LRFD.b. Use ASD.An I-section with two cover plates is used as tension member and is subjected to a load of 120 t over a length of 16 m. If the allowable tensile stress is 15000 kg/cm², is the member design safe? If 9821.6 cm4, so, find the factor of safety. Use ISWB 300@ 48.1 kg/m, A = 61.33 cm², Ixx Ivy 990.1 cm4. There are two rows of fasteners in each flange, thus making atleast two holes in one section. The fastener diameter is 22 mm, cover plate size 260 mm x 8 mm, flange thickness of l-section = 10 mm. The center of fastener is located at 60 mm from center line of the l-section. = =Three plates (14 mm and 16 mm in thickness) are welded to a W10 x 49 to form a built-up shape as shown. K.L= K.L= 7.6 m and F= 345 MPa, compute the design strength for LRFD. Mm x300mm WIOX49 14mm Piate Plate