Question

The displacements, in meters, at an arbitrary point (x,y,z) in a given body are:

u(x,y,z)=(x⁴ y+xy³ )×10⁻⁴
v(x,y,z)=x³ y³ ×10⁻⁴
w(x,y,z)=0
​
Determine all the strains in x−y−z coordinates at any point (x,y,z).

Answer 1

Strains at any point (x,y,z) are found by differentiating the displacement field with respect to the coordinates, resulting in strains in the x, y direction, and the shear strains xy, yx, which describe the deformation of the body.

Step-by-step explanation:

The strains in the body at any point (x,y,z) can be determined by differentiating the displacement field with respect to the spatial coordinates. The displacements given for the body are:

• u(x,y,z) = (x⁴ y + xy³) × 10⁻⁴
• v(x,y,z) = x³ y³ × 10⁻⁴
• w(x,y,z) = 0

The strain in the x-direction, εxx, is the partial derivative of u with respect to x, which is 4x³y + y³ scaled by 10⁻⁴. Similarly, the strain in the y-direction, εyy, is the partial derivative of v with respect to y, which is 3x³y² scaled by 10⁻⁴. Since w is zero, the strain in the z-direction, εzz, is zero. The shear strains εxy and εyx are obtained by differentiating u with respect to y and v with respect to x respectively, resulting in x⁴ + 3x²y² scaled by 10⁻⁴ for both since strain is symmetric in its indices. The shear strains εxz and εyz are zero due to w being zero.

The normal strains and shear strains determine how the body deforms under an applied force or displacement. Strains are dimensionless quantities that describe the deformation in terms of relative elongation or shear per unit length.