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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).

لقطة الشاشة 2023-02-13 231559
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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

لقطة الشاشة 2023-02-13 232014

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