A W10x22 used as a column under axial compression. Use A36 steel. For each column length and end conditions given, calculate the design capacity. (a) L = 8 ft, pinned ends (b) L=8 ft, fixed ends (c) L= 16 ft, pinned ends (d) L = 16 ft, fixed ends (e) L = 16 ft, pinned ends, braced at the midspan in the weak axis direction.
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- A W14 x 74 compression member carries a service loads of DL= 2,500 KN and LL= 300 KN. The column is fixed at the top end and pinned at the bottom. The length is 6 m. If A992 steel (E= 350 MPa and F= 450 MPa) is used, check the adequacy of the column. Check also the column stability. Use ASD.2. Determine the lightest W-shape for a column 30' tall in a braced frame. The column is braced at mid-height in the weak direction. The loads are 100k dead load and 200k live load. The bottom of the column is pinned and the top of the column is constrained in rotation.* Use A572 Grade 60 steel. *Not tronslation.Verify the adequacy of column AB, part of a sway frame structure, to carry the loads shown in the figure. The column is made of ASTM A36 steel (Fy = 250 MPa). Use ASD specifications. For determining its effective length factor, neglect inelastic effect. %3D W21X62 C W21X62 L=5 m L=5 m Flanges of Columns and web of girders are in the plane of the frame NOTE: Take note of the moment of inertia to be used (ly) for columns. W18X50 W18X50 L=5 m L=5 m P(DL) = 40 kN NM 001 = (10)d NM 00S = (1)d NY SZI = (1)d W21X62 L-4.2m W21X62 L=4.2 m
- The cantilever beam shown in Figure P5.8-4 is a W10 × 77 of A992 steel. There isno lateral support other than at the fixed end. Use an unbraced length equal to the spanlength and determine whether the beam is adequate. The uniform load is a servicedead load that includes the beam weight, and the concentrated load is a service liveload.a. Use LRFD.b. Use ASD.Determine whether a W24 x 117 of A992 steel is adequate for the beam shown. The uniform load does not include the weight of the beam. Lateral support is provided at A, B, and C. Use LRFD. P= 12 k P = 36 k WD = 1 k/ft WL = 3 k/ft 10 20- 30If the beam in Problem 5.5-9 i5 braced at A, B, and C, compute for the unbr Cb aced length AC (same as Cb for unbraced length CB). Do not include the beam weight in the loading. a. Use the unfactored service loads. b. Use factored loads.
- The frame shown in Figure P4.7-8 is unbraced, and bending is about the x-axis of the members. All beams areW1835, and all columns areW1054. a. Determine the effective length factor Kx for column AB. Do not consider the stiffness reduction factor. b. Determine the effective length factor Kx for column BC. Do not consider the stiffness reduction factor. c. If Fy=50 ksi, is the stiffness reduction factor applicable to these columns?A plate girder must be designed for the conditions shown in Figure P10.7-4. The given loads are factored, and the uniformly distributed load includes a conservative estimate of the girder weight. Lateral support is provided at the ands and at the load points. Use LRFD for that following: a. Select the, flange and web dimensions so that intermediate stiffeners will he required. Use Fy=50 ksi and a total depth of 50 inches. Bearing stiffeners will be used at the ends and at the load points, but do not proportion them. b. Determine the locations of the intermediate stiffeners, but do not proportion them.AW14 x 53 of Grade 70 steel is used as a compression member. It is 30 feet long, pinned at the top and fixed at the bottom, and has additional bracing in the weak direction. The column is braced (pinned) on its weak axis 10 ft form the top and bottom. What is the nominal strength in axial compression for the member? O 659 kips O 761 kips O 856 kips O 733 kips
- Q2) The members of the truss structure shown below is plain concrete. The compressive strength of the concrete is 25 MPa. Compute the maximum load P that can be carried by the structure. (Cross section of each member of the truss is 200 x 200 mm and don't use material factors and do not consider slenderness) Comment on your results briefly. P A& 2m SC 2 m 1380 2m Dsteel Design Civil Engineering Please show solutions: AW 310 x 118 section with a length of 8.0 m is used as a column. Determine the safe axial load the column can carry using AISC specifications with Fy = 345 MPa when: Properties of W 310 x 118 A = 15000 mm² rx = 136 mm ry = 77.6 mm A. column ends are fixed B. one end of the column is fixed; the other free Use LRFD Answers: A.] For Ag = 248.06X15000 100 3720,9kw $en=0.90x3720.9 = 3348.21 kW B.JPn= For Ag =40.72 x 15 000 LOW Pr= 0.90x610,81 = 549.729 KN2. A steel column 10 m long is fabricated from W section and a cover plate. Determine the safe axial load that the column can carry. Fy = 345 MPa, E = 200 GPa. Use AISC/NSCP Specs. 310 mm cover plate AY I 6 mm y2 y d1 10 m d2 W 310 x 107 a) Both ends of column are fixed b) Both ends of column are hinged c) One end fixed, the other end hinged Use design values of k. W 310 x 107 A = 13600 mm? d = 311 mm Ix = 248x10° mm* ly = 81.2x10° mm“ d.