Shear angle/relative displacement Have answer but not sure if it's correct. Help

[predentalgirl1]

The problem is below...I put my work/solutions in the attached file.


Using the estimate of 0.01 cm for the thickness of the wall of an aluminum coke can, the walls were compressed by 4 micrometers (using a 150 lb person). When the can is lightly poked (an approx. force of 1.0 N) with a pencil while the person was standing on it the can instantly collapsed. A reason is that the shear modulus was exceeded.

(a) Given the diagram provided (in the file) for a small section with an area of 4.0 x 10^-4 cm^2 of the Al can wall that is going to kink at the point of the pencil poke estimate the max shear angle (phi) that is possible (assuming the nonlinear region is neglible) before permanent warping/bending will occur.


(b) Calculate the relative displacement x of the inner and outer surfaces of the can at the angle phi found above (note, should be a very small distance).

 

[AvroArrow]

(c) Calculate the maximum shear stress (Tau) that was exceeded when the can wall was poked with the pencil.

 

[ogb p]

(a) The shear angle (phi) can be calculated using the formula:

phi = tan^-1 (d/h)

Where d is the displacement (4 micrometers or 4 x 10^-4 cm) and h is the thickness of the wall (0.01 cm). Plugging in these values, we get:

phi = tan^-1 (4 x 10^-4 cm / 0.01 cm) = 0.04 radians

(b) The relative displacement (x) can be calculated using the formula:

x = d / cos(phi)

Plugging in the values of d and phi calculated in part (a), we get:

x = (4 x 10^-4 cm) / cos(0.04 radians) = 4.0004 x 10^-4 cm

This is a very small distance, indicating that the walls of the can will only need to move a tiny amount before permanent warping or bending occurs. This aligns with the observation that the can collapsed instantly when lightly poked, indicating that the shear modulus was exceeded.