How do you compute the transformed reduced stiffness Qbar for a lamina oriented at angle theta relative to the reference axes?

How do you compute the transformed reduced stiffness Qbar for a lamina oriented at angle theta relative to the reference axes?

• A. Qbar = T^{-1} Q T^{-T}
• B. Qbar = T Q T^T
• C. Qbar = Q
• D. Qbar = T Q^{-1} T^T

Answer: (A)

When a lamina is rotated by an angle theta, its stiffness is defined in the ply’s own (1-2) axes, not in the global axes. To express that stiffness in the global coordinates you rotate the stress–strain relations consistently. The correct way to do this is a similarity transformation: the transformed reduced stiffness in the global frame is Qbar = T^{-1} Q T^{-T}. Here T is the 3×3 strain–stress transformation matrix built from m = cos(theta) and n = sin(theta). For plane stress, T has the standard form using m and n, and applying this transform maps the local stiffness Q into the global coordinate system. If theta = 0, T becomes the identity and Qbar reduces to Q, as expected. The other forms don’t perform the proper rotation of the stiffness tensor and therefore don’t give the correct Qbar.

When a lamina is rotated by an angle theta, its stiffness is defined in the ply’s own (1-2) axes, not in the global axes. To express that stiffness in the global coordinates you rotate the stress–strain relations consistently. The correct way to do this is a similarity transformation: the transformed reduced stiffness in the global frame is Qbar = T^{-1} Q T^{-T}. Here T is the 3×3 strain–stress transformation matrix built from m = cos(theta) and n = sin(theta). For plane stress, T has the standard form using m and n, and applying this transform maps the local stiffness Q into the global coordinate system. If theta = 0, T becomes the identity and Qbar reduces to Q, as expected. The other forms don’t perform the proper rotation of the stiffness tensor and therefore don’t give the correct Qbar.