**Column Design Capacity Analysis***Objective:* Determine the design capacity of a W10x22 column made from A36 steel under axial compression. Assess various lengths and end conditions to see how they influence capacity.**Column Scenarios:**1. **Case (a):** - Length \( L = 8 \) ft - End Condition: Pinned ends2. **Case (b):** - Length \( L = 8 \) ft - End Condition: Fixed ends3. **Case (c):** - Length \( L = 16 \) ft - End Condition: Pinned ends4. **Case (d):** - Length \( L = 16 \) ft - End Condition: Fixed ends5. **Case (e):** - Length \( L = 16 \) ft - End Condition: Pinned ends, with bracing at the midspan in the weak axis direction*Notes:*- Analyze each condition considering the differences between pinned and fixed ends, which affect the effective length factor and column capacity.- Midspan bracing affects the buckling resistance in the weak axis, altering the effective length used in calculations.

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.

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.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Use A36 steel and compute the nominal strength of the column shown below. The member ends are fixed in all directions (x, y, and z). Use the AISC specs, equations. Do not use column tables. Check flexural buckling and flexural-torsional buckling. Assume local buckling does not govern. Hint: Consider which axis needs to only be checked for flexural buckling and which axis only needs to be checked for flexural-torsional buckling.
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 (Ex= 350 MPa and F₁- 450 MPa) is used, check the adequacy of the column. Check also the column stability. Use ASD.
A W18x60 is used as a column to carry a dead load of 850 kN and a live load of 1450 kN. Design a base plate to support the column such that the depth of the base plate is twice its width. The concrete compression block is 700 mm x 700 mm with fc' = 24 MPa. Use A50 steel base plate. USE LRFD.
Select a WT section for the compression member shown in Figure . The load is the total service load, with a live-to-dead load ratio of 2.5:1. Use Fy 5 50 ksi. a. Use LRFD. b. Use ASD
A W12x79 section is used as partially restrained column for a floor 8 meters high. The steel used is Grade 60 (Fy = 415 MPa). The W12x79 section has a gross cross-sectional area of 14,968 sq.mm, radius of gyration along strong axis is 133.65 mm, and radius of gyration along weak axis is 77 47 mm. Using k = 0.70, Calculate the design axial strength (KN) of the column by LRFD. The nominal compressive strength, Pn, shall be determined based on the limit state of flexural buckling. Pn= FerAg (505-3-1) The flexural buckling stress, Fr, is determined as follows: 1. when 2. when KL r ≤ 4.71 (or F. ≥ 0.44F,) For = [0.65872|F₂ KL * >4.71 F (or F, <0.44F,) For = 0.877F. (505.3-2) (505.3-3)
A 20-foot long column is pinned at the bottom and fixed against rotation but free to translate at the top. It must support a service dead load of 110 kips and a service live load of 110 kips. a. Select a W12 of A992 steel. Use the column load tables. 1. Use LRFD. 2. Use ASD. b. Select a W18 of A529 Grade 55 steel. Use the trial-and-error approach covered in Section 4.6. 1. Use LRFD. 2. Use ASD.
Using LRFD select the lightest W 14 section to carry a service load P of 100 kips dead load and 400 kips live load. The compression load acts with an eccentricity of 12-inches with respect to the strong axis. Use A992 steel (Fy = 50 ksi). Take the unbraced length to be L. This beam-column is part of a braced frame (BF) system. Please determine Cb and use it. Take Cm = 1.0. Since this can involve many iterations, start the problem by selecting a column meant to resist an ‘equivalent’ axial load of Pu plus 140% of the Mu. (use kips as the unit for Mu). I.E.: Pu,eq = Pu + 1.4*Mu (gives result in kips). Also, note that regardless of the results of this first attempt, the beam-column must meet the AISC criteria of H1-1a or H1-1b as usual. Note: This ‘equivalent’ method was first published in the AISC Engineering Journal by Uang, Watter, and Leet.
You are requested to design a column for PD = 225 k and PL = 400 k, using A992 steel with KL = 16 ft. A W14 x 68 is available but may not provide sufficient capacity. If it does not, cover plates may be added to the section to increase the W14’s capacity. Design the cover plates to be 12 inch wide and welded to the flanges of the section to enable the column to support the required load (see Fig. P6-18). Determine the minimum plate thickness required, assuming the plates are available in 1/16th-in increments.
A W18 × 97 with Fy = 60 ksi is used as a compression member. The length is 13 feet.Compute the nominal strength for Kx = 2.2 and Ky = 1.0.
why do we need to know the Specific Gravity and Absorption of Fine Aggregate, Surface Moisture of Fine Aggregate, and how it is important in concrete
Select a WT section for the compression member shown in Figure. The load is the total service load, with a live-to-dead load ratio of 2.5:1. Use Fy 5 50 ksi. a. Use LRFD. b. Use ASD 21' 175 k 1
A Wx35 steel column has an unsupported height of 6m. Use A36 steel with Fy = 248 MPa and E = 200,000 MPa. A = 6504 rx=88.9 mm ry = 51.56 mm Determine the safe axial load if both ends are hinged.