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Modern Engineering Thermodynamics

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114 CHAPTER 4: The First Law of <strong>Thermodynamics</strong> and Energy Transport Mechanisms<br />

4.6.3 Elastic Work<br />

Whenever we compress or extend an elastic solid (like a spring), we perform elastic work. Consider a force ± F !<br />

applied on the end of an elastic rod (see Figure 4.5c). The normal stress σ in the rod is<br />

σ =± j F! j<br />

A<br />

(4.35)<br />

where j F ! j is the magnitude of the force and A is the cross-sectional area of the rod. Since the force F ! and its<br />

corresponding displacement d x ! are always in the same direction, the vector dot product F ! .d x ! always reduces<br />

to Fdx, whereF = j F ! j and dx = jd x ! j, and when the force is applied on the system from the surroundings<br />

rather than being produced by the system, the work is negative and its increment is<br />

The strain ε in the rod is defined as<br />

dW = − F ! .d x ! = − Fdx = − σAdx (4.36)<br />

dε = dx<br />

L = Adx<br />

AL = Adx<br />

V<br />

= dV V<br />

(4.37)<br />

where L is the length of the rod and AL is its volume V: Then,<br />

and Eq. (4.36) becomes<br />

Adx = dV = Vdε (4.38)<br />

Therefore, for elastic work,<br />

dW = −σAdx = −σVdε (4.39)<br />

Elastic work<br />

ð1W 2 Þ elastic<br />

= −<br />

Z 2<br />

1<br />

σVdε<br />

(4.40)<br />

EXAMPLE 4.6<br />

Determine an expression for the work involved in deforming a constant volume elastic solid that obeys Hooke’s law of elasticity<br />

(see Figure 4.13).<br />

L<br />

State 1<br />

L +ΔL<br />

State 2<br />

1 W 2 = ?<br />

F = EA(ΔL/L)<br />

FIGURE 4.13<br />

Example 4.6.<br />

Solution<br />

Here we have V = constant. Also, from strength of materials we can write Hooke’s law as σ = Eε, where E is Young’s modulus<br />

of elasticity. Then, Eq. (4.40) becomes<br />

Z 2<br />

Z 2<br />

ð 1<br />

W 2 Þ elastic<br />

= − σV dε = − EVε dε = − EV<br />

1 <br />

1<br />

= − EV ε2 2 − <br />

ε2 1<br />

= − V 2 2E ðσ2 2 − σ2 1 Þ<br />

Thus, if ε 2 2 > ε2 1 , then ð 1 W 2Þ elastic<br />

is negative and work is being put into the system; and if ε 2 2 < ε2 1 , then ð 1 W 2Þ elastic<br />

is positive<br />

and work is being produced by the system. Note that both tensile strains (ε > 0) and compressive strains (ε < 0) are possible<br />

here. But, the resulting work formula deals only with ε 2 and consequently gives the correct result regardless of the strain<br />

direction.<br />

Z 2<br />

1<br />

ε dε

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