# Question 3.5: Consider again the two-element axial-deformation structure o......

Consider again the two-element axial-deformation structure of Example 2.3 (Section 2.2), shown here in Fig. 1. In Example 2.2 the axial stresses $σ_1$ and $σ_2$ in elements (1) and (2), respectively, were determined. Since the two-element structure is statically determinate, these stresses are independent of the materials from which the elements are made. Now let us determine the displacement $u_C$ of end C if element AB is steel ($E_1$ = 200 GPa) and element BC is aluminum ($E_2$ = 70 GPa).

Step-by-Step
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Plan the Solution The nomenclature for a typical element is shown in Fig. 2. The axial forces were previously obtained in Example 2.3, and Eq. 3.14 can be used to express the elongation of each element in terms of its axial force. Finally, the displacement $u_C$ of end C is simply the sum of the elongations of the two elements.

$e = fF, \text{where} f ≡\frac{L}{AE}$          (3.14)

Equilibrium: From the free-body diagrams and equilibrium equations of Example 2.3 we have

$F_1$ = -20 kN, $F_2$ = 10 kN      Equilibrium      (1)

where $F_i$ is the axial force in element i.

Element Force-Deformation Behavior: For the two uniform elements, the force-deformation equation, Eq. 3.14, gives

$e_i = f_i F_i, \text{where} f_i = \frac{L_i}{A_iE_i}$,    i = 1, 2    Element Force- Deformation        (2)

Inserting numerical values into Eq. (2) gives the two element flexibility coefficients

$f_1 = \frac{L_1} {A_1E_1} = \frac{(300 mm) }{(314.2 mm^2)(200 kN/mm^2)} = 4.77(10^{-3})$ mm/kN

$f_2 = \frac{L_2} {A_2E_2} = \frac{(200 mm) }{(314.2 mm^2)(200 kN/mm^2)} = 1.62(10^{-2})$ mm/kN

Note that element (2) is over three times as flexible as element (1), primarily because of its smaller cross-sectional area and smaller modulus of elasticity. Inserting these flexibility coefficients, together with the element axial forces from Eq. (1), into Eq. (2) gives the two element elongations

$e_1 = f_1F_1$ = -0.0955 mm            (2′)

$e_1 = f_1F_1$ = 0.1617 mm

Thus, element (1) is shortened due to its compressive force while element (2) elongates due to its tensile force.

Geometry of Deformation; Compatibility Equation: Since the two elements are attached end-to-end, the displacement of the right end of element (1) is compatible with (i.e., equal to) the displacement of the left end of element (2).Therefore, the displacement of end C is given by

$u_C = e_1 + e_2$      Geometry of Deformation    (3)

This equation is called a compatibility equation, since it enforces the condition that the respective end displacements of the two elements joined together at B be compatible.
The final solution is

$u_C = e_1 + e_2$ = -0.0955 mm + 0.1617 mm

$u_C = 6.62(10^{-2})$ mm        Ans.

Review the Solution Note how each of the quantities $L_i, A_i$ and $E_i$ enters into the element flexibility coefficients $f_i$, and how these flexibilities and the element forces $F_i$ together determine the elongations of these uniform elements

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