EBK MECHANICS OF MATERIALS
EBK MECHANICS OF MATERIALS
7th Edition
ISBN: 8220100257063
Author: BEER
Publisher: YUZU
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Chapter 10.3, Problem 80P

A centric load P must be supported by the steel bar AB. Using allowable stress design, determine the smallest dimension d of the cross section that can be used when (a) P = 108 kN, (b) P = 166 kN. Use σY = 250 MPa and E = 200 GPa.

Fig. P10.80

Chapter 10.3, Problem 80P, A centric load P must be supported by the steel bar AB. Using allowable stress design, determine the

(a)

Expert Solution
Check Mark
To determine

Find the smallest dimension d of the cross section.

Answer to Problem 80P

The smallest dimension d of the cross section is 30.1mm_.

Explanation of Solution

Given information:

The length of the column is L=1.4m.

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

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

The centric load acting in the column is P=108kN.

Calculation:

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

Le=L=1.4m

Find the cross sectional area (A) using the equation.

A=bd

Here, the width of the column is b and the depth of the column is d.

Substitute 3d for b.

A=(3d)d=3d2

Find the moment of inertia (I) of the cross section using the equation.

I=bd312

Substitute 3d for b.

I=(3d)d312=d44

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

r=IA

Substitute d44 for I and 3d2 for A.

r=(d44)3d2=d12

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 effective length (Le) and the minimum radius of gyration (r) as follows;

Ler=1.4m×1,000mm1md12=1,40012d (1)

Consider 1,40012dLr=133.22

Find the effective stress (σe) using the equation.

σe=π2E(Le/ry)2

Substitute 200 GPa for E and 1,40012d for Le/ry.

σe=π2×200GPa×1,000MPa1GPa(1,40012d)2=1,973,920.88d223,520,000

Find the critical stress (σcr) using the relation.

σcr=0.877σe

Substitute 1,973,920.88d223,520,000 for σe.

σcr=0.877×1,973,920.88d223,520,000=1,731,128.612d223,520,000

Calculate the allowable stress (σall) using the relation.

σall=σcr1.67

Substitute 1,731,128.612d223,520,000 for σcr.

σall=(1,731,128.612d223,520,000)1.67=1,731,128.612d239,278,400

Calculate the allowable load (Pall) using the equation.

Pall=σallA

Substitute 1,731,128.612d239,278,400 for σall and 3d2 for A.

Pall=1,731,128.612d239,278,400×3d2=5,193,385.836d439,278,400

Consider the allowable load is equal to the centric load.

Substitute 108 kN for Pall.

108kN×1,000N1kN=5,193,385.836d439,278,400d=30.1mm

Check:

Substitute 30.1 mm for d in Equation (1).

Ler=1,4001230.1=161.12>Lr=133.22

The assumed condition is correct.

Therefore, the smallest dimension d of the cross section is 30.1mm_.

(b)

Expert Solution
Check Mark
To determine

Find the smallest dimension d of the cross section.

Answer to Problem 80P

The smallest dimension d of the cross section is 33.5mm_.

Explanation of Solution

Given information:

The length of the column is L=1.4m.

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

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

The centric load acting in the column is P=166kN.

Calculation:

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

Le=L=1.4m

Find the cross sectional area (A) using the equation.

A=bd

Substitute 3d for b.

A=(3d)d=3d2

Find the moment of inertia (I) of the cross section using the equation.

I=bd312

Substitute 3d for b.

I=(3d)d312=d44

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

r=IA

Substitute d44 for I and 3d2 for A.

r=(d44)3d2=d12

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 effective length (Le) and the minimum radius of gyration (r) as follows;

Ler=1.4m×1,000mm1md12=1,40012d (2)

Consider 1,40012dLr=133.22

Find the effective stress (σe) using the equation.

σe=π2E(Le/ry)2

Substitute 200 GPa for E and 1,40012d for Le/ry.

σe=π2×200GPa×1,000MPa1GPa(1,40012d)2=1,973,920.88d223,520,000

Find the critical stress (σcr) using the relation.

σcr=0.877σe

Substitute 1,973,920.88d223,520,000 for σe.

σcr=0.877×1,973,920.88d223,520,000=1,731,128.612d223,520,000

Calculate the allowable stress (σall) using the relation.

σall=σcr1.67

Substitute 1,731,128.612d223,520,000 for σcr.

σall=(1,731,128.612d223,520,000)1.67=1,731,128.612d239,278,400

Calculate the allowable load (Pall) using the equation.

Pall=σallA

Substitute 1,731,128.612d239,278,400 for σall and 3d2 for A.

Pall=1,731,128.612d239,278,400×3d2=5,193,385.836d439,278,400

Consider the allowable load is equal to the centric load.

Substitute 166 kN for Pall.

166kN×1,000N1kN=5,193,385.836d439,278,400d=33.5mm

Check:

Substitute 33.5 mm for d in Equation (2).

Ler=1,4001233.5=144.77>Lr=133.22

Therefore, the smallest dimension d of the cross section is 33.5mm_.

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Chapter 10 Solutions

EBK MECHANICS OF MATERIALS

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