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A cube with a side length of 10cm is made up of aluminum metal (G= 25 x 109 N/m2) A horizontal shear force of 50000 N is applied to this cube. (a) How far will the top of the cube move in the horizontal directin relative to the bottom of the cube? (b) What is the shear stress appied to the cube? (c) Calculate the shear strain on the cube.​

Respuesta :

We know that shearing stress=  

F/A=10^

4

Nm

^−2

Length of side of cube=10cm=10/100

m=0.1m

Shearing displacement=â–³n=0.05cm

=0.05

/100m=0.0005m

We know that modulus of rigidity is η=  

F/Aθ

= FL/

A△x  [tanθ≈θ=  △x/L

] = =  

10^4x0.1/5x10^-4

⇒10^7/5^1⇒2x10^6Nm^-2

Hence, modulus of rigidity is equal to 2x10^6Nm^-2

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