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Question 3.1: A bimetallic bar is made of two linearly elastic materials, ......

A bimetallic bar is made of two linearly elastic materials, material 1 and material 2, that are bonded together at their interface, as shown in Fig. 1.
Assume that E_2 > E_1. Determine the distribution of normal stress that must be applied at each end if the bar is to undergo axial deformation, and determine the location of the point in the cross section where the resultant force P must act. Express your answers in terms of P, E_1, E_2, and the dimensions of the bar.

لقطة الشاشة 2023-02-13 142110
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Plan the Solution The bar is said to undergo axial deformation, so we know from Eq. 3.1 that the strain, ε(x), is constant on every cross section.
And since the bar is prismatic (constant cross section) and is loaded only at its ends, we can assume that ε(x) = ε = const for the entire bar. Since there are two values of E, Eq. 3.18 will lead to different values of stress in the two materials. We can use the stress-resultant integrals of Eqs. 3.5 to relate these two stresses to the resultant force.

ε(x) = \frac{du(x)}{dx}      Strain-Displacement Equation    (3.1)

F(x) = \int_A σ  dA

M_y(x) = \int_A zσ  dA         Stress Resultants        (3.5)

M_z(x) = -\int_A yσ  dA

σ(x, y, z) = E(x, y, z)ε (x)          (3.18)

Strain Distribution: For the reasons noted above,

ε(x) = ε = const               (1)

Stress Distribution: From Eq. (1) and Eq. 3.4, the stresses in the two parts of the bar will be

σ ≡ σ_x= Eε(x)          Hooke’s Law     (3.4)

σ_1 = E_1ε,  σ_2 = E_2ε         (2)

as illustrated in Fig. 2.

Resultant Force: The resultant force and moments on the cross section are given by Eqs. 3.5.

F(x) = \int_A {σ  dA}

M_y(x) = \int_A zσ  dA              (3)

M_z(x) = -\int_A yσ  dA

However, we know that the resultant of a constant normal stress distribution is a force acting through the centroid of the area on which the constant stress acts. Therefore, we can replace the stress distribution of Fig. 2 and Fig. 3a by two axial forces, P_1 and P_2, acting at the centroids of their respective areas of the cross section, as shown in Fig. 3b.
From Eqs. (2) and the F(x) equation in Eqs. (3),

P_1 = E_1εA_1 = E_1ε \left(\frac{bh}{3}\right),          P_2 = E_2εA_2 = E_2ε \left(\frac{2bh}{3}\right),       (4)

Taking the summation of forces in the x direction, we get

\underrightarrow{+} \sum{F_x}:           P_1 = P_2 = P                   (5)

Thus, the resultant force acting on the ends of the bar (and at every cross section, for that matter) is related to the extensional strain ε by the equation

P =\frac{εbh} {3} \left)(E_1 + 2E_2\right)              (6)

Solving Eq. (6) for ε and inserting the result into the Eqs. (2), we get the stresses

σ_1 = \frac{3PE_1} {bh(E_1 + 2E_2)} ,    σ_2 = \frac{3PE_2} {bh(E_1 + 2E_2)}                (7)

In Fig. 3b the forces P_1 and P_2 are shown acting at the centroids of the areas on which they act. The location of the single resultant force P in Fig. 3c can be determined by taking moments about a z axis passing through point A.

+\circlearrowright \left(\sum{M} \right)_A:       P_1 \left(\frac{5h} {6}\right) +  P \left(\frac{h}{3}\right) = Py_P                     (8)

Thus, the resultant force acting on each end of the bar (and at every cross section) is a force P located at

y_P = h \left( \frac{5E_1 + 4E_2}{6E_1 + 12E_2}\right),       z_P = 0     (9)

Review the Solution A good check on the results above is to let E_1 = E_2 = E. Then, we get

σ_1 = σ_2 = \frac{P}{bh},   y_P =\frac{h}{2}

which is consistent with Eq. 2.4.

σ(x) = \frac{F(x)} {A(x)}      Axial Stress Equation        (2.4)

لقطة الشاشة 2023-02-13 171359
لقطة الشاشة 2023-02-13 171425

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