How Is the Shear Modulus of a Sponge Calculated?

[mikefitz]

A large sponge has forces of magnitude 17 N applied in opposite directions to two opposite faces of area 43 cm2 (see the figure below ). The thickness of the sponge (L is ")3 cm. The deformation angle (g is ")8.4°. (a) What is delta x? (b) What is the shear modulus of the sponge?

delta x = 4.43 mm

stress = 17N/430mm = .03953N/mm^2
strain = delta L/L, 4.43/30= .1476

shear modulus = stress/strain
=.03953/.1476=.28 Pa?

what did I do wrong?

 

[PhanthomJay]

A large sponge has forces of magnitude 17 N applied in opposite directions to two opposite faces of area 43 cm2 (see the figure below ). The thickness of the sponge (L is ")3 cm. The deformation angle (g is ")8.4°. (a) What is delta x? (b) What is the shear modulus of the sponge?

delta x = 4.43 mm

stress = 17N/430mm = .03953N/mm^2
strain = delta L/L, 4.43/30= .1476

shear modulus = stress/strain
=.03953/.1476=.28 Pa?

what did I do wrong?

I see for one thing that you have an incorrect conversion factor: 43 cm^2 = 4300mm^2. Beyond that, with my limited understanding of shear deformation (I used to understand it once 40 years ago), and my disdain for Pascals (we use psi in the States!), the Shear Modulus seems very low, even for a sponge.. Are you sure of the magnitude of the deformation angle? You should probably repost in the Engineering forum, if no further help is forthcoming here.