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Chapter 17, Problem 1RP

The wood column is 4 m long and is required to support the axial load of 25 kN. If the cross section is square, determine the dimension a of each of its sides using a factor of safety against buckling of F.S. = 2.5. The column is assumed to be pinned at its top and bottom. Use the Euler equation. Ew = 11 GPa, and σY = 10 MPa.Chapter 17, Problem 1RP, The wood column is 4 m long and is required to support the axial load of 25 kN. If the cross section

Expert Solution & Answer
Check Mark
To determine

Find the dimensions a of the square column.

Answer to Problem 1RP

The dimension a of the square column is 103mm_.

Explanation of Solution

Given information:

The length of the column is L=4m.

The axial load applied on the column is P=25kN.

The factor of safety against buckling is F.S.=2.5.

The column is pinned at its top and bottom.

The modulus of elasticity of the wood column is Ew=11GPa.

The allowable yield stress is σY=10MPa.

Calculation:

Find the critical load (Pcr) applied on the column.

Pcr=F.S.×P

Substitute 2.5 for F.S. and 25 kN for P.

Pcr=2.5×25=62.5kN

Calculate the moment of inertia (I) of the square column using the relation:

I=a412

Here, the dimension of the column is a.

The value of effective-length factor for rectangular column, K=1, since both the ends are pinned.

Show the expression for Euler’s formula as shown below:

Pcr=π2EwI(KL)2

Here,Pcr is the maximum load on the material just before it begins to buckle, E is the Young’s modulus of the material, I is the least moment of inertia of the material cross-sectional area, and L is the length of the material.

Substitute 62.5 kN for Pcr, 11GPa for Ew, a412 for I, 1 for K, and 4 m for L.

62.5kN×103N1kN=π2×11GPa×109N/m21GPa×a412(1×4)262.5×103×16×12π2×11×109=a4a=0.103m×1,000mm1m=103mm

Check the validity of using Euler’s formula:

Euler’s formula is valid only when the critical stress (σcr) is less than the yield stress (σY).

Find the cross-sectional area (A) of the column as follows;

A=a2=103×103=10,609mm2

Calculate the critical stress for the column using the relation:

(σcr)=PcrA

Substitute 62.5 kN for Pcr and 10,609mm2 for A.

(σcr)=62.5kN×1,000N1kN10,609=5.89MPa<(σY=10MPa)

Thus, the use of Euler Equation is valid.

Therefore, the dimension a of the square column is 103mm_.

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Chapter 17 Solutions

Statics And Mechanics Of Materials, Student Value Edition Plus Mastering Engineering With Pearson Etext -- Access Card Package (5th Edition)

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    Column buckling; Author: Amber Book;https://www.youtube.com/watch?v=AvvaCi_Nn94;License: Standard Youtube License