   # An element of a material is subjected to plane stresses as shown in the figure. The stresses o, cry., and are 10 MPa, —15 MPa, and 5 MPa. respectively. Assume E = 200 GPa and v = 0.3. (a) Calculate the normal strain in the x, v. and z directions and the shear strain. (b) Calculate the strain-energy density of the element. ### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347
Chapter 7, Problem 7.5.4P
Textbook Problem
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## An element of a material is subjected to plane stresses as shown in the figure. The stresses o, cry., and are 10 MPa, —15 MPa, and 5 MPa. respectively. Assume E = 200 GPa and v = 0.3. (a) Calculate the normal strain in the x, v. and z directions and the shear strain. (b) Calculate the strain-energy density of the element. (a)

To determine

The normal strain in the x direction.

The normal strain in the y direction.

The normal strain in the z direction.

### Explanation of Solution

Given information:

The normal stress in the x direction is 10 MPa , the normal stress in y direction is 15 MPa , modulus of elasticity is 200 GPa , shear in the x y plane is 10 MPa, and the poison’s ratio is 0.3 .

Explanation:

Write the expression for the modulus of rigidity.

G = E 2 ( 1 + v ) ...... (I)

Here, the modulus of Rigidity is G , the shear modulus is E and Poisson’s ratio is v .

Write the expression for the strain in x direction.

ε x = 1 E ( σ x v σ y ) ...... (II)

Here, the normal stress in x direction is σ x , the normal stress in y direction is σ y , modulus of Elasticity is E and the Poisson’s ratio is v .

Write the expression for the strain in y direction.

ε y = 1 E ( σ y v σ x ) ...... (III)

Write the expression for strain in the z direction.

ε z = v E ( σ x + σ y ) ...... (IV)

Write the expression for the shear strain in x y plane.

γ x y = τ x y G ...... (V)

Here, the shear stress in the x y plane is τ x y and the shear modulus is G .

Calculation:

Substitute 10 MPa for σ x , 15 MPa for σ y , 200 GPa for E and 0.3 for v in Equation (I).

G = 200 GPa 2 ( 1 + 0.3 ) = 76.923 GPa

Substitute 10 MPa for σ x , 15 MPa for σ y , 200 GPa for E and 0.3 for v in Equation (II).

ε x = 1 200 GPa ( 10 MPa 0.3 ( 15 MPa ) ) = 1 200 GPa ( 10 9 Pa 1 Gpa ) ( 10 MPa ( 10 6 Pa 1 MPa ) 0.3 ( 15 MPa ( 10 6 Pa 1 MPa ) ) ) = 1 200 × 10 9 Pa ( 14.5 × 10 6 Pa ) = 7.25 × 10 5

Substitute 10 MPa for σ x , 15 MPa for σ y , 200 GPa for E and 0

(b)

To determine

The strain energy density of the element.

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