Mechanics of Composite Materials: Orthotropic Lamina
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Question 4: An orthotropic lamina with a fiber orientation of 45 degrees is subjected to stresses sigma_x = 0, sigma_y > 0, and tau_xy.
Determine the failure envelope and the non-failure region in the sigma_y - tau_xy plane using Tsai-Hill theory for the following cases:
i) tau_xy > 0
ii) tau_xy < 0
Take y axis as sigma_y and x axis as tau_xy.
DATA
sigma_1T = 250 MPa
sigma_1C = 200 MPa
sigma_2T = 20 MPa
sigma_2C = 40 MPa
tau_12 = 10 MPa
Question 5: A unidirectional lamina with a fibre orientation of 0 degrees is subjected to stresses sigma_1 and sigma_2 in the 1 (fibre) and 2 (transverse) directions as well as a shear stress of tau_12 = 25 MPa. Determine the non-failure region in the sigma_1 - sigma_2 plane using Tsai-Hill criterion.
Note that sigma_1 and sigma_2 can be tensile or compressive.
DATA
sigma_1T = 250 MPA
sigma_1C = 150 MPa
sigma_2T = 30 MPa
sigma_2C = 90 MPa
tau_12 = 50 MPa
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Problem 1: (Question 4)
An orthotropic lamina with a fiber orientation of 45 is subjected to stresses
x = 0, y > 0 and xy.
Determine the failure envelope and the non-failure region in the y - xy plane using Tsai-Hill theory for following cases:
i) xy > 0
ii) xy < 0
Take (y) axis as y and (x) axis as xy .
Data:
Solution:
Since the ultimate properties of material are given with respect to the main directions (along the fibers and transversally), we need to find out the actual stresses in a frame reference oriented accordingly:
By rotating a reference frame in a 2D stress state by angle (), the new values of stresses are given by:
( 1)
( 2)
( 3)
In our problem, the x stresses are missing, so that the above formulae become simpler:
( 4)
( 5)
( 6)
For = 45 we will have:
( 7)
( 8)
( 9)
where subscript [ ]1 indicates properties along fibers and [ ]2 indicates properties transverse to fibers.
The Tsai-Hill criterion tells that:
( 10)
Now, we need to replace the terms in above equation with (7), (8) and (9) and ultimate strength data, upon each ...
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