2) (25 points) For a laminate [(902/0)]s, subject to in-plane load Ny only, ?sin

Question

2) (25 points) For a laminate [(902/0)]s, subject to in-plane load Ny only, ?sing the maximum stress criterion, a) compute first ply failure (FPF) load, report the critical ply and stresses at failure, b) last ply failure (degrading all plies). Each layer is 2.5 mm thick. (note that 0 ply fiber direction coincides with x-axis. if you use CLT, consider FPLP1 column as computed stresses in materials coordinate system of the layer). Material properties Ei 54 GPa E2- 18. GPa, V20.25, G12 9 GPa. Xr Xc 1034. MPa Yr 31. MPa Yc 138 MPa S-30 MPa. (be carefull with the units)

Explanation / Answer

Given :

Laminate = [0/904]s

(rest from the question details )

considering end conditions

[Theta] = 0 degree
[Theta] = 90degree

[Theta] = 0 degree

          we will calculate the stresses in the principal material directions:

[sigma] 1 = [sigma] xcos2 [Theta]

[sigma] 2 = [sigma] xsin2 [Theta]

[ au] 12 = - [sigma] xcos [Theta] sin [Theta]

for maximum stress criteria --- considering only tensile stresses i.e. ignoring compressive stresses.

hence,

[sigma] x = [sigma] 1t /cos2 [Theta]

     = 800 / cos2 0deg

     = 800 MPa

for all other formulae [sigma] x is not defined

now ,

thickness of lamina = 0.125 mm

no of ply's = 4

total thickness = 0.125*4 = 0.5mm

area of thickness = 0.5*0.5= 0.25 mm2

hence,

stress = load/ area

axial load = stress * area

               = 800 * 0.25

               = 200 N

    2. [Theta] = 90deg

by the above already stated equations :

[sigma] x= 60 MPa

and load = 60 *0.25 = 15 N

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