Determine the maximum allowable eccentricity for the column/axial member. The column is fixed at both ends.External axial load applied → F= 453.589 KNYoungs modulus →E = 200 GPDiameter of the column → d= 0.3048 mCross sectional area of the column → A = 0.073 m²So the area moment of inertia of the column about transverse axis:→/==π64π64X-×0.30484 = 0.00042 m4Calculate and show that the column/axial member is safe and can withstand the load If the applied axial load becomes critical load then the length of the column can be found out by the aboveformulae.So:4π²ElL²→ 453.589 x 10³ =→F=4π²x200 x 109 x 0.00042→ L²=7311 m→ L = 85.50 mSo the column will fail due to buckling if the length of the column will be greater than the above.And in this condition, the critical stress will be:46=A453.589 × 1030.073Because the induced critical stress is less than the yield strength of the material so it is the limit of the stressinduced in the column. Beyond this stress the column will fail due to buckling.= 6.21 MP

Given data in question Load 453.589 kNDiameter 0.3048 mElasticity modulus 200 GPaMOI 0.00042…

Answered: Determine the maximum allowable…

Engineering

Civil Engineering

Determine the maximum allowable eccentricity for the column/axial member

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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Measuring carpentry and miscellaneous items

The modification of planning and drawing of the construction project into the various descriptions and quantities to estimate the value, cost, and price of the project is known as measurement. The quantity supervisor produces the details of the quantities to the contractor or subcontractor to measure the cost of the work and determine the profit of the contractor or subcontractor. The measurement of the construction work depends on its type, usage, and stages in the project. In the early stage of the project, the details of the project depend on the general parameters. As the work starts, more information is searched and the details are dependent on the elemental estimates of the work.

Question

Determine the maximum allowable eccentricity for the column/axial member.

The column is fixed at both ends.
External axial load applied → F= 453.589 KN
Youngs modulus →E = 200 GP
Diameter of the column → d= 0.3048 m
Cross sectional area of the column → A = 0.073 m²
So the area moment of inertia of the column about transverse axis:
→/=
=
π
64
π
64
X
-×0.30484 = 0.00042 m4
Calculate and show that the column/axial member is safe and can withstand the load

If the applied axial load becomes critical load then the length of the column can be found out by the above
formulae.
So:
4π²El
L²
→ 453.589 x 10³ =
→F=
4π²x200 x 109 x 0.00042
→ L²=7311 m
→ L = 85.50 m
So the column will fail due to buckling if the length of the column will be greater than the above.
And in this condition, the critical stress will be:
46=
A
453.589 × 103
0.073
Because the induced critical stress is less than the yield strength of the material so it is the limit of the stress
induced in the column. Beyond this stress the column will fail due to buckling.
= 6.21 MP

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