Question 4.1: Analysis of a Duplex Structure A steel rod of cross-sectiona...
Analysis of a Duplex Structure
A steel rod of cross-sectional area A_{s} and modulus of elasticity E_{s} has been placed inside a copper tube of cross-sectional area A_{c} and modulus of elasticity E_{c} (Figure 4.1a). Determine the axial shortening of this system of two members, sometimes called an isotropic duplex structure, when a force P is exerted on the end plate as shown.
Assumptions: Members have the same length L. The end plate is rigid.
The forces produced in the rod and in the tube are designated by P_{s} and P_{c}, respectively.
Statics: The equilibrium condition is applied to the free body of the end plate (Figure 4.1b):
P_{c} + P_{s} = P (a)
This is the only equilibrium equation available, and since it contains two unknowns (P_{c} \text {and} P_{s}), 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
\delta = \frac {PL} {AE} (4.1)
\delta_c=\frac{P_c L}{A_c E_c}, \quad \delta_s=\frac{P_s L}{A_s E_s}
Geometry: Axial deformation of the copper tube is equal to that of the steel rod:
\frac{P_c L}{A_c E_c}=\frac{P_s L}{A_s E_s} (b)
Solution of Equations (a) and (b) gives
P_c=\frac{\left(A_c E_c\right) P}{A_c E_c+A_s E_s}, \quad P_s=\frac{\left(A_s E_s\right) P}{A_c E_c+A_s E_s} (4.4)
The foregoing equation show that the forces in the members are proportional to the axial rigidities.
Compressive stresses \sigma _{c} in copper and \sigma _{s} in steel are found by dividing P_{c} and P_{s} by A_{c} and A_{s}, respectively. Then, applying Hooke’s law together with Equation 4.4, we obtain the compressive strain
\varepsilon=\frac{P}{A_c E_c+A_s E_s} (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.