Analysis of a Duplex Structure A steel rod of cross-sectional area As and modulus of elasticity Es has been placed inside a copper tube of cross-sectional area Ac and modulus of elasticity Ec (Figure 4.1a). Determine the axial shortening of this system of two members, sometimes called an isotropic

Question

Analysis of a Duplex Structure

A steel rod of cross-sectional area [latex]A_{s}[/latex] and modulus of elasticity [latex]E_{s}[/latex] has been placed inside a copper tube of cross-sectional area [latex]A_{c}[/latex] and modulus of elasticity [latex]E_{c}[/latex] (Figure 4.1a). Determine the axial shortening of this system of two members, sometimes called an isotropic duplex structure, when a force [latex]P[/latex] is exerted on the end plate as shown.

Assumptions: Members have the same length [latex]L[/latex]. The end plate is rigid.

Accepted Answer

The forces produced in the rod and in the tube are designated by [latex]P_{s}[/latex] and [latex]P_{c},[/latex] respectively.

Statics: The equilibrium condition is applied to the free body of the end plate (Figure 4.1b):

[latex]P_{c} + P_{s} = P[/latex] (a)

This is the only equilibrium equation available, and since it contains two unknowns ([latex]P_{c} ext {and} P_{s}[/latex]), the structure is statically indeterminate to the first degree (see Section 1.8).

Deformations: Through the use of Equation 4.1, the shortening of the members are

[latex]\delta = \frac {PL} {AE}[/latex] (4.1)

[latex] \delta_c=\frac{P_c L}{A_c E_c}, \quad \delta_s=\frac{P_s L}{A_s E_s} [/latex]

Geometry: Axial deformation of the copper tube is equal to that of the steel rod:

[latex] \frac{P_c L}{A_c E_c}=\frac{P_s L}{A_s E_s} [/latex] (b)

Solution of Equations (a) and (b) gives

[latex] P_c=\frac{\left(A_c E_c ight) P}{A_c E_c+A_s E_s}, \quad P_s=\frac{\left(A_s E_s ight) P}{A_c E_c+A_s E_s} [/latex] (4.4)

The foregoing equation show that the forces in the members are proportional to the axial rigidities.

Compressive stresses [latex]\sigma _{c}[/latex] in copper and [latex]\sigma _{s}[/latex] in steel are found by dividing [latex]P_{c}[/latex] and [latex]P_{s} [/latex] by [latex]A_{c}[/latex] and [latex]A_{s},[/latex] respectively. Then, applying Hooke’s law together with Equation 4.4, we obtain the compressive strain

[latex] \varepsilon=\frac{P}{A_c E_c+A_s E_s} [/latex] (4.5)

The shortening of the assembly is therefore δ = εL.

Comments: Equation 4.5 indicates thatthe strain equals the applied load divided by the sum of the axialrigidities of the members. Composite duplex structures are treated in Chapter 16.

Question 4.1: Analysis of a Duplex Structure A steel rod of cross-sectiona...

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