EBK MECHANICS OF MATERIALS
EBK MECHANICS OF MATERIALS
7th Edition
ISBN: 8220102804487
Author: BEER
Publisher: YUZU
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Textbook Question
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Chapter 11.3, Problem 10P

Using E = 200 GPa, determine (a) the strain energy of the steel rod ABC when P = 25 kN, (b) the corresponding strain-energy density in portions AB and BC of the rod.

Chapter 11.3, Problem 10P, Using E = 200 GPa, determine (a) the strain energy of the steel rod ABC when P = 25 kN, (b) the

Fig. P11.10

(a)

Expert Solution
Check Mark
To determine

The strain energy of the steel rod ABC.

Answer to Problem 10P

The strain energy of the steel rod ABC is 12.185J_.

Explanation of Solution

Given information:

The diameter of the steel rod AB is dAB=20mm.

The diameter of the steel rod BC is dBC=16mm.

The length of the rod AB is LAB=1.2m.

The length of the rod BC is LBC=0.8m.

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

The applied load P=25kN.

Calculation:

Calculate the area of the rod (A) as shown below.

A=πd24 (1)

For the steel rod AB.

Substitute 20mm for d in Equation (1).

AAB=π×2024=314.16mm2

For the steel rod BC.

Substitute 16mm for d in Equation (1).

ABC=π×1624=201.06mm2

Calculate the strain energy (U) as shown below.

U=P2L2EA

Calculate the strain energy for rod ABC as shown below.

U=UAB+UBC=P2LAB2EAAB+P2LBC2EABC=P22E(LABAAB+LBCABC)

Substitute 25kN for P, 200GPa for E, 1.2m for LAB, 314.16mm2 for AAB, 0.8m for LBC, and 201.06mm2 for ABC.

U=(25kN×1,000N1kN)22×200GPa×109N/m21GPa(1.2m314.16mm2×(1m1,000mm)2+0.8m201.06mm2×(1m1,000mm)2)=625×106400×109(3,819.7+3,978.9)=1.5625×103×7,798.6=12.185Nm×1J1Nm

=12.185J

Therefore, the strain energy for the steel rod ABC is 12.185J_.

(b)

Expert Solution
Check Mark
To determine

The strain energy density in rod AB and rod BC

Answer to Problem 10P

The strain energy density in rod AB is uAB=15.83kJ/m3_.

The strain energy density in rod BC is uBC=38.65kJ/m3_.

Explanation of Solution

Given information:

The diameter of the steel rod AB is dAB=20mm

The diameter of the steel rod BC is dBC=16mm

The length of the rod AB is LAB=1.2m

The length of the rod BC is LBC=0.8m

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

The applied load P=25kN.

Calculation:

Refer to part (a).

The area of rod AB is AAB=314.16mm2.

The area of the rod BC is ABC=201.06mm2.

Calculate the stress (σ) as shown below.

σ=PA (2)

For the rod AB.

Substitute 25kN for P and 314.16mm2 for A in Equation (2).

σAB=25kN×1,000N1kN314.16mm2×(1m1,000mm)2=79.58×106N/mm2×1Pa106N/mm2=79.58Pa

For the rod BC.

Substitute 25kN for P and 201.06mm2 for A in Equation (2).

σBC=25kN×1,000N1kN201.06mm2×(1m1,000mm)2=124.34×106N/mm2×1Pa106N/mm2=124.34Pa

Calculate the strain energy density (u) as shown below.

u=σ22E (3)

For the rod AB.

Substitute 79.58Pa for σ and 200GPa for E in Equation (3).

uAB=(79.58Pa)22×200GPa×109Pa1GPa=6,332.9764400×109=15.83×109Pa×106N/mm21Pa=15.83×103N/mm2×(1,000mm1m)2

=15.83×103Nm/m3×1kJ103Nm=15.83kJ

Hence, the strain energy density in rod AB is uAB=15.83kJ_.

For the rod BC.

Substitute 124.34Pa for σ and 200GPa for A in Equation (3).

uBC=(124.34Pa)22×200GPa×109Pa1GPa=15,460.4356400×109=38.65×109Pa×106N/mm21Pa=38.65×103N/mm2×(1,000mm1m)2

=38.65×103Nm/m3×1kJ103Nm=38.65kJ

Therefore, the strain energy density in rod BC is uBC=38.65kJ_.

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Chapter 11 Solutions

EBK MECHANICS OF MATERIALS

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