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A thin square plate in biaxial stress is subjected to stresses ?? and ??., as shown in part a of the figure. The width of the plate is h = 12.0 in. Measurements show that the normal strains in the x and v directions are s = 427 × 10-6 and s = 113 × l0-6, respectively. With reference to part b of the figure. which shows a Iwo-dimensional view of the plate. determine the following quantities. (a) The increase .d in the length of diagonal Oil. (b) The change . in the angle between diagonal Oti and the x axis, (c) The shear strain y associated with diagonals Oil and cf(that is. find the decrease in angle ced).

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Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347

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Section
BuyFindarrow_forward

Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347
Chapter 7, Problem 7.7.7P
Textbook Problem
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A thin square plate in biaxial stress is subjected to stresses ?? and ??., as shown in part a of the figure. The width of the plate is h = 12.0 in. Measurements show that the normal strains in the x and v directions are s = 427 × 10-6 and s = 113 × l0-6, respectively.

With reference to part b of the figure. which shows a Iwo-dimensional view of the plate. determine the following quantities.

(a) The increase .d in the length of diagonal Oil.

(b) The change . in the angle between diagonal Oti and the x axis,

(c) The shear strain y associated with diagonals Oil and cf(that is. find the decrease in angle ced).

  Chapter 7, Problem 7.7.7P, A thin square plate in biaxial stress is subjected to stresses ?? and ??., as shown in part a of the

(a)

To determine

The increase in the length of the diagonal.

Explanation of Solution

Given information:

The width of the square plate is 12.0 in , the strain along x-direction is 427 × 10 6 , the strain along y-direction is 113 × 10 6 and the angle of inclination is 45 ° .

Explanation:

Write the expression for the length of the diagonal.

   L d = b 2 ...... (I)

Write the expression for strain along x 1 -direction.

   ε x 1 = ε x + ε y 2 + ε x ε y 2 cos ( 2 θ ) + γ x y 2 sin ( 2 θ ) ...... (II)

Write the expression for change in diameter.

   Δ d = ε x 1 L d ...... (III)

Calculation:

Substitute, 12 in for b in Equation (I).

   L d = 12 in 2 = 12 in × 1.41421 = 16.97 in

Substitute, 427 × 10 6 for ε x , 113 × 10 6 for ε y

   0 for γ x y and 45 ° for θ in Equation (II)

(b)

To determine

The change in the angle between diagonal Od and the x-axis.

(c)

To determine

The shear strain associated with the diagonals.

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