Mechanics of Materials Question

[brk51]

Homework Statement

Homework Equations

G = Shear Stress/Shear Strain G = E/(2(1+n)) n being poissons ratio

The Attempt at a Solution

So I know the easiest way to solve is simply to manipulate the first equation a little bit >>> Shear Stress = G * Shear Strain >>> Shear Stress = P/A >>>> P = G * Shear Strain * 20mm * 150mm then boom you get 300 kN...

But there is a longer way I want to understand. Correct me in my process because I know I'm wrong and I think I know where..

So...
1) The angle I calculate for shear strain. Where do I measure it from?
2) Is poissons ratio 0? The picture seems to indicate there is no lateral contraction...if not, how do I calculate it?

 

[Chestermiller]

Homework Statement

View attachment 211827

Homework Equations

G = Shear Stress/Shear Strain G = E/(2(1+n)) n being poissons ratio

The Attempt at a Solution

So I know the easiest way to solve is simply to manipulate the first equation a little bit >>> Shear Stress = G * Shear Strain >>> Shear Stress = P/A >>>> P = G * Shear Strain * 20mm * 150mm then boom you get 300 kN...

But there is a longer way I want to understand. Correct me in my process because I know I'm wrong and I think I know where..
View attachment 211829

So...
1) The angle I calculate for shear strain. Where do I measure it from?
2) Is poissons ratio 0? The picture seems to indicate there is no lateral contraction...if not, how do I calculate it?

Poisson's ratio is not equal to zero. Poisson's ratio is a property of the material, not a strain effect in any particular problem. If you want to understand what is happening here (in another way), you need to go back and examine more closely the derivation of the equation relating the shear modulus to Young's modulus and the Poisson ratio.

 

[brk51]

Poisson's ratio is not equal to zero. Poisson's ratio is a property of the material, not a strain effect in any particular problem. If you want to understand what is happening here (in another way), you need to go back and examine more closely the derivation of the equation relating the shear modulus to Young's modulus and the Poisson ratio.

While I understand your advice, that's why I'm here. I don't get it. I wouldn't just go through all this trouble posting a thread if I didn't already try to understand this.

 

[Chestermiller]

While I understand your advice, that's why I'm here. I don't get it. I wouldn't just go through all this trouble posting a thread if I didn't already try to understand this.

OK. I understand now. Let me get back to you in a little while.

 

[brk51]

OK. I understand now. Let me get back to you in a little while.

ok thank you

 

[Chestermiller]

OK. You are interested in the derivation of the shear modulus as a function of Young's modulus and Poisson's ratio. The derivation that I am familiar with starts out with a square object that has a square inscribed within it, oriented at an angle of 45 degrees to the outer square. Is this where your derivation starts out?

 

[brk51]

the book

OK. You are interested in the derivation of the shear modulus as a function of Young's modulus and Poisson's ratio. The derivation that I am familiar with starts out with a square object that has a square inscribed within it, oriented at an angle of 45 degrees to the outer square. Is this where your derivation starts out?

the book just uses variables but yes essentially continue..

 

[Chestermiller]

the book

the book just uses variables but yes essentially continue..

I don't quite follow. Can you please elaborate?

 

[brk51]

ther

I don't quite follow. Can you please elaborate?

Are we talking about the problem I posted above or the literal derivation of the shear modulus

 

[Chestermiller]

ther

Are we talking about the problem I posted above or the literal derivation of the shear modulus

The literal derivation of the shear modulus.

 

[brk51]

The literal derivation of the shear modulus.

Ok yes..

 

[Chestermiller]

The derivation assumes that the outer square is subjected to a tensile stress of $\sigma$ in the x direction and a tensile stress of $-\sigma$ (i.e., a compressive stress) in the y direction. This results in a homogeneous state of stress (and strain) within the object. Are we together so far?

 

[brk51]

yessir

 

[Chestermiller]

The stress loading I described causes the object to stretch in the x direction, and to compress an equal amount in the y direction. It also causes the horizontal diagonal of the inscribed square to stretch in the x direction, and the vertical diagonal to compress in the y direction. The strains in these diagonals are the same as in the x and y directions u for the outer square. Still OK?

 

[brk51]

The stress loading I described causes the object to stretch in the x direction, and to compress an equal amount in the y direction. It also causes the horizontal diagonal of the inscribed square to stretch in the x direction, and the vertical diagonal to compress in the y direction. The strains in these diagonals are the same as in the x and y directions u for the outer square. Still OK?

Apologize wifi in the library is not too hot,

yes I get that.

 

[Chestermiller]

OK. I have something I need to do now. Be back tomorrow.

 

[Chestermiller]

Along the boundary of the inscribed square, the inscribed square exerts a normal stress and a shear stress on each of the four small right triangles that surround it, in order for these four triangles to be held in static equilibrium. From a force balance on the upper-right-hand triangle, what is the normal stress and shear stress that the inscribed square exerts at its boundary with this triangle?