Mechanics of Materials, 7th Edition
Mechanics of Materials, 7th Edition
7th Edition
ISBN: 9780073398235
Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek
Publisher: McGraw-Hill Education
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Chapter 10.4, Problem 106P
To determine

Find the thickness of the lightest tube.

Expert Solution & Answer
Check Mark

Answer to Problem 106P

The thickness of the lightest steel tube is 15mm_.

Explanation of Solution

Given information:

The length of the steel tube is L=2.2m.

The outer diameter of the steel tube is do=80mm.

The magnitude of the axial load is P=165kN.

The eccentricity of the load in steel tube is e=15mm.

The allowable yield stress of the steel tube is σY=250MPa.

The modulus of elasticity of the steel tube is E=200GPa.

The allowable stress in bending is (σall)bending=150MPa

Calculation:

The effective length of the column (Le) is equal to the length of the column (L).

Le=L=2.2m

Find the inner diameter of the steel tube (di) using the relation.

di=do2t

Here, the thickness of the steel tube is t.

Substitute 80 mm for do.

di=802t

Find the cross sectional area of the steel tube (A) using the equation.

A=π(do2di2)4

Substitute 80 mm for do and (802t) for di.

A=π×(802(802t)2)4

Find the moment of inertia of the steel tube (I) using the equation.

I=π(do4di4)64

Substitute 80 mm for do and (802t) for di.

I=π(804(802t)4)64

Find the minimum radius of gyration (r) using the relation.

r=IA

Substitute π(804(802t)4)64 for I and π×(802(802t)2)4 for A.

r=(π(804(802t)4)64)(π×(802(802t)2)4)=(804(802t)4)16(802(802t)2)

Find the distance between the neutral axis to the extreme fibre (c) using the relation.

c=do2

Substitute 80 mm for do.

c=802=40mm

Find the slenderness ratio Lr using the equation.

Lr=4.71EσY

Here, the modulus of elasticity of the material is E and the allowable yield strength is σY

Substitute 200 GPa for E and 250 MPa for σY:

Lr=4.71200GPa×1,000MPa1GPa250=133.22

Find the ratio of the effective length to the minimum radius of gyration.

Ler=2.2m×1,000mm1m(804(802t)4)16(802(802t)2)=8,800(804(802t)4)(802(802t)2) (1)

Consider 8,800(804(802t)4)(802(802t)2)<Lr=133.22.

Find the effective stress (σe) using the equation.

σe=π2E(Le/r)2

Substitute 200 GPa for E and 8,800(804(802t)4)(802(802t)2) for Le/r.

σe=π2×200GPa×1,000 MPa1GPa(8,800(804(802t)4)(802(802t)2))2=π2×200GPa×1,000 MPa1GPa×(804(802t)4)(802(802t)2)(8,800)2=0.02549(804(802t)4)(802(802t)2)

Find the critical stress (σcr) using the relation.

σcr=[0.658σYσe]σY

Substitute 250 MPa for σY and 0.02549(804(802t)4)(802(802t)2) for σe.

σcr=[0.658250[0.02549(804(802t)4)(802(802t)2)]]250=[0.6589,807.891(802(802t)2)(804(802t)4)]250

Find the allowable stress (σall)bending due to centric load using the relation.

(σall)centric=σcr1.67

Substitute [0.6589,807.891(802(802t)2)(804(802t)4)]250 for σcr.

(σall)centric=[0.6589,807.891(802(802t)2)(804(802t)4)]2501.67=149.701[0.6589,807.891(802(802t)2)(804(802t)4)]

Find the maximum moment (M) using the relation.

M=Pe

Here, the allowable load is P and the eccentricity of the load is e.

Substitute 165 kN for P and 15 mm for e.

M=165kN×1,000N1kN×15=2,475,000N-mm

Find the thickness of the lightest tube (t) using the centric and bending equation.

P/A(σall)centric+Mc/I(σall)bending1

Substitute 165 kN for P, π×(802(802t)2)4 for A, 149.701[0.6589,807.891(802(802t)2)(804(802t)4)] for (σall)centric, 2,475,000 N-mm for M, 40 mm for c, π(804(802t)4)64 for I, and 150 MPa for (σall),bending.

(165kN×1,000N1kNπ×(802(802t)2)4)149.701[0.6589,807.891(802(802t)2)(804(802t)4)]+(2,475,000×40π(804(802t)4)64)150=1

Solve the equation;

The thickness is t=13.901mm.

The nearest 3 mm increment of the thickness is 15 mm.

Check:

Substitute 15 mm for t in Equation (1).

Ler=8,800(804(802(15))4)(802(802(15))2)=8,80034,710,0003,900=93.28<Lr=133.22

Therefore, the thickness of the lightest steel tube is 15mm_.

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Chapter 10 Solutions

Mechanics of Materials, 7th Edition

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