Kindly also answer the column base plate for steel 1. Thank you. PROBLEM: Consider the cross section shown. All structural steel is A36. Use LRFD and ASDmethod and NSCP(AISC) specifications.DETERMINE THE FOLLOWING:1. Determine the governing slenderness ratio.2. Determine the design strength and the allowable strength of the column.3. Determine the maximum service live load that the column can support if the liveload is twice the dead load.10 02-C8x18.75C8x 18.75A = 3,555 mmbf = 64.2 mmtf = 9.9 mmtw = 12.4 mmd = 203.2 mm* = 14.35 mmIx = 18.31 x 10 mmly = 0.82 x 10 mmMINOR AXISMA JOR AXISCONDITIONS OF LATERAL BRACINGSAND END SUPPORTS000600050007 1. Enumerate possible failures of an insufficient column baseplates subjected to axial loadand moment.2. Why stiffeners are provided for column baseplates.Problem: Redesign the column baseplate in sample problem 03 using the same data forsteel properties and sections using LRFD method and for the loading use the following:P dead load = 500 kN.ive load = 800 kN.

1) Introduction of slenderness ratio : Slenderness is a term used in structural engineering to…

PROBLEM: Consider the cross section shown. All structural steel is A36. Use LRFD and ASD method and NSCP(AISC) specifications. DETERMINE THE FOLLOWING: 1. Determine the governing slenderness ratio. 2. Determine the design strength and the allowable strength of the column. ICA 3. Determine the maximum service live load that the column can support if the live load is twice the dead load. 1O CL 2-CBx18.75 Свх 18.75 A= 3,555 mm? bf = 64.2 mm If = 9.9 mm tw = 12.4 mm d = 203.2 mm = 14.35 mm Ix = 18.31 x 10 mm ly = 0.82 x10 mm MINOR AXIS MA JOR AXIS CONDITIONS OF LATERAL BRACINGS AND END SUPPORTS 0006 0005 0007

PROBLEM: Consider the cross section shown. All structural steel is A36. Use LRFD and ASD method and NSCP(AISC) specifications. DETERMINE THE FOLLOWING: 1. Determine the governing slenderness ratio. 2. Determine the design strength and the allowable strength of the column. ICA 3. Determine the maximum service live load that the column can support if the live load is twice the dead load. 1O CL 2-CBx18.75 Свх 18.75 A= 3,555 mm? bf = 64.2 mm If = 9.9 mm tw = 12.4 mm d = 203.2 mm = 14.35 mm Ix = 18.31 x 10 mm ly = 0.82 x10 mm MINOR AXIS MA JOR AXIS CONDITIONS OF LATERAL BRACINGS AND END SUPPORTS 0006 0005 0007

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Bars

In civil engineering, bars refer to steel reinforcement bars or rebars. These are structural members primarily used for reinforcement purposes in masonry works. A bar is characterized by its cross-sectional area, which is negligible compared to its length. Bars are usually manufactured by steel which is uniform throughout the length, or by meshing multiple steel wires. Other materials of the bar include carbon steel, glass fiber, carbon fiber, and basalt fiber. The reinforcing bars are coated with epoxy resins and other materials to enhance some of their properties including corrosion resistance.

Question

Kindly also answer the column base plate for steel 1. Thank you.

PROBLEM: Consider the cross section shown. All structural steel is A36. Use LRFD and ASD
method and NSCP(AISC) specifications.
DETERMINE THE FOLLOWING:
1. Determine the governing slenderness ratio.
2. Determine the design strength and the allowable strength of the column.
3. Determine the maximum service live load that the column can support if the live
load is twice the dead load.
10 0
2-C8x18.75
C8x 18.75
A = 3,555 mm
bf = 64.2 mm
tf = 9.9 mm
tw = 12.4 mm
d = 203.2 mm
* = 14.35 mm
Ix = 18.31 x 10 mm
ly = 0.82 x 10 mm
MINOR AXIS
MA JOR AXIS
CONDITIONS OF LATERAL BRACINGS
AND END SUPPORTS
0006
0005
0007

1. Enumerate possible failures of an insufficient column baseplates subjected to axial load
and moment.
2. Why stiffeners are provided for column baseplates.
Problem: Redesign the column baseplate in sample problem 03 using the same data for
steel properties and sections using LRFD method and for the loading use the following:
P dead load = 500 kN.
ive load = 800 kN.

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