Given information:
The normal stress of the nylon element along the x axis is − 3.9 MPa , normal stress along the y axis is − 3.2 MPa , stress along the z axis is − 1.8 MPa , the strain in the x direction is − 640 × 10 − 6 and the strain in the y direction is − 310 × 10 − 6 .
Explanation:
Write the expression for the bulk modulus.
K = σ h Δ V V ...... (I)
Here, the bulk modulus is K , the hydrostatic stress is σ h , the change in volume is Δ V and the original volume of the element is V .
Write the expression for the change in volume in terms of strains.
Δ V V = ε x + ε y + ε z ...... (II)
Here, the strain in the x direction is ε x , strain in the y direction is ε y and the strain in the z direction is ε z .
Write the expression for the hydrostatic stress in terms of stresses.
σ h = σ x + σ y + σ z 3 ...... (III)
Here,the stress along x axis is σ x , stress along y axis is σ y and stress along z axis is σ z .
Write the expression for strain in the x direction using Hooke’s law.
ε x = 1 E ( σ x − ν ( σ y + σ z ) ) ...... (IV)
Here, the modulus of elasticity is E and the Poisson’s ratio is ν .
Write the expression for strain in the y direction using Hooke’s law.
ε y = 1 E ( σ y − ν ( σ x + σ z ) ) ...... (V)
Write the expression for strain in the z direction using Hooke’s law.
ε z = 1 E ( σ z − ν ( σ x + σ y ) ) ...... (VI)
Calculation:
Substitute − 640 × 10 − 6 for ε x , − 3.9 MPa for σ x , − 1.8 MPa for σ z and − 3.2 MPa for σ y in Equation (IV).
− 640 × 10 − 6 = 1 E [ − 3.9 MPa − ν ( − 3.2 MPa − 1.8 MPa ) ] − 640 × 10 − 6 × E = ( ( − 3.9 MPa × 10 6 Pa 1 MPa ) + ( 5 MPa × 10 6 Pa 1 MPa ) × ν ) − 640 × 10 − 6 E − 5 Pa × ν × 10 6 = − 3.9 Pa × 10 6 ...... (VII)
Substitute − 310 × 10 − 6 for ε y , − 3.9 MPa for σ x , − 1.8 MPa for σ z and − 3.2 MPa for σ y in Equation (V).
− 310 × 10 − 6 = 1 E [ − 3.2 MPa × 10 6 Pa 1 MPa − ν ( − 3.9 MPa × ( 10 6 Pa 1 MPa ) − 1.8 MPa × ( 10 6 Pa 1 MPa ) ) ] ( − 310 × 10 − 6 ) E = ( − 3.2 Pa + 5.7 Pa × ν ) × 10 6 − 310 × 10 − 6 × E − 5.7 Pa × ν × 10 6 = − 3.2 Pa × 10 6 ...... (VIII)
Solve the Equations (VII) and (VIII) for the value of E and ν .
2.098 × 10 − 3 E = 6.23 Pa × 10 6 E = 6.23 Pa × 10 6 2.098 × 10 − 3 E = 2.9695 × 10 9 Pa
Substitute 2.9695 × 10 9 Pa for E in Equation (VII).
− 640 × ( 2.9695 × 10 9 ) Pa × 10 − 6 − 5 Pa × ν × 10 6 = ( − 3.9 × 10 6 ) Pa 5 Pa × ν × 10 6 = 1999520 Pa ν = 0.3999
Substitute 2.9695 × 10 9 Pa for E , 0.3999 for ν , − 3.9 MPa for σ x , − 1.8 MPa for σ z and − 3.2 MPa for σ y in Equation (VI).
ε z = 1 2.9695 × 10 9 Pa [ ( − 1.8 Pa − 0.3999 ( − 3.9 Pa − 3.2 Pa ) ) × 10 6 ] = 1.03929 × 10 6 Pa 2.9695 × 10 9 Pa = 350 × 10 − 6
Substitute − 3.9 MPa for σ x , − 1.8 MPa for σ z and − 3.2 MPa for σ y in Equation (III).
σ h = ( − 3.9 MPa ) + ( − 3.2 MPa ) + ( − 1.8 MPa ) 3 = − 8.9 MPa 3 = − 2.9667 MPa
Substitute − 640 × 10 − 6 for ε x , − 310 × 10 − 6 for ε y and 350 × 10 − 6 for ε z in Equation (II).
Δ V V = ( − 640 × 10 − 6 ) + ( − 310 × 10 − 6 ) + ( 350 × 10 − 6 ) = ( − 640 − 310 + 350 ) × 10 − 6 = − 600 × 10 − 6
Substitute − 2.9667 MPa for σ h and − 600 × 10 − 6 for Δ V V in Equation (I).
K = − 2.967 × 10 6 Pa − 600 × 10 − 6 = ( 4.95 × 10 9 Pa ) × ( 1 GPa 10 9 Pa ) = 4 .95 GPa
Conclusion:
The bulk modulus of the nylon is 4.95 GPa .