How to find the length when trying to determine the shear

[sylent33]

Homework Statement

A cuboid made of steel with a shear modulus of G = 79.3 GPa is attached to the dark surface and a force F of 2⋅10^6 N acts as shown. Explain, including formulas and a sketch, what happens in case a) and b). Calculate for a) σ and for b) τ, Δx and the shear γ = Δx/l = τ/G (so use the approximation for small angles).

Relevant Equations

σ =F/A

The sketch that goes with the problem.

My attempt at solving the problem:

a) to get σ (I think this is called tension in english) we need the force,and we need the area.From the sketch we can see that the area is A = 6*2 and we convert those in m and than set in the formula we should get $$ σ = 3,33 Gpa $$ Now the question "explain what happens in this case" is confusing to me.Confusing in that sence,is too broad.What exactly should I be considering what happens with the cuboid what happens with the forces where do they point, all of these ?

b) For b we need to determine the shear stress the formula is the same except the only diffrence will be in the area(see sketch) I get a value of \gamma = 3,33 Gpa. Now the formula for the shear
looks like this $$
\gamma =
\frac{
\tau}{G} $$ The value I get is $\gamma = 0,042$

Now to determine $\Delta x$ is where I am having a few problems. Although the formula is given to me I am not sure what my l should be.As far as I know l should be the length but when looking at the sketch I just don't see it.A hint in the right direction would be great,thanks!

 

[Lnewqban]

I believe that the question is about what type of efforts each shadowed surface resists.
For a, everything is symmetrical, therefore you only have tension stress.
For b, you have a lever against the force, which orientation is parallel to the shadowed surface.
All that induces a moment, therefore you have two types of stresses.

At least for b, I believe that the problem is referring to the linear vertical deviation from the horizontal that the shape suffers under stress (perhaps only for the free end to the right).

 

[Chestermiller]

I believe that the question is about what type of efforts each shadowed surface resists.
For a, everything is symmetrical, therefore you only have tension stress.
For b, you have a lever against the force, which orientation is parallel to the shadowed surface.
All that induces a moment, therefore you have two types of stresses.

At least for b, I believe that the problem is referring to the linear vertical deviation from the horizontal that the shape suffers under stress (perhaps only for the free end to the right).

The problem statement is designed for a new initiate, and they expect the students to apply the simple formulas they have been given. Even though in part b there would more likely be some pure bending present, they expect them to assume shear bending.

In case a, they expect you to assume a homogeneous deformation, even though the sample is constrained at the dark surface. So they expect them to assume that, by whatever mechanism, the sample is able to contract at the dark surface.

In both cases the expect the student to assume that the forces are uniformly distributed on the surfaces upon which they are applied.

 

[Chestermiller]

Homework Statement:: A cuboid made of steel with a shear modulus of G = 79.3 GPa is attached to the dark surface and a force F of 2⋅10^6 N acts as shown. Explain, including formulas and a sketch, what happens in case a) and b). Calculate for a) σ and for b) τ, Δx and the shear γ = Δx/l = τ/G (so use the approximation for small angles).
Relevant Equations:: σ =F/A

The sketch that goes with the problem.

My attempt at solving the problem:

a) to get σ (I think this is called tension in english) we need the force,and we need the area.From the sketch we can see that the area is A = 6*2 and we convert those in m and than set in the formula we should get $$ σ = 3,33 Gpa $$ Now the question "explain what happens in this case" is confusing to me.Confusing in that sence,is too broad.What exactly should I be considering what happens with the cuboid what happens with the forces where do they point, all of these ?

b) For b we need to determine the shear stress the formula is the same except the only diffrence will be in the area(see sketch) I get a value of \gamma = 3,33 Gpa. Now the formula for the shear
looks like this $$
\gamma =
\frac{
\tau}{G} $$ The value I get is $\gamma = 0,042$

Now to determine $\Delta x$ is where I am having a few problems. Although the formula is given to me I am not sure what my l should be.As far as I know l should be the length but when looking at the sketch I just don't see it.A hint in the right direction would be great,thanks!

In part a, I get 1.67 GPa for the tensile stress. Please describe the nature of the deformation when this stress is applied to the sample. Do you know how to calculate the tensile strain from this data? Do you know how to calculate the upward movement of the upper surface?

In part b, you correctly calculated the shear stress and the shear strain. Please describe the nature of the deformation when this shear stress is applied to the sample. Do you know how to calculate the upward displacement of the right surface when the sample suffers this strain?

 

[sylent33]

Oh you are right,i misstyped. For tensile stress in part a I also got 1,67 Gpa I don't know why I put 3,33 that should be for part b). To clear it up

a) tensile stress 1,67Gpa
b) shear stress 3,33 Gpa shear strain 0,042

Now the deformation according to google should be that the surface extends in axial direction and and contracting in the transverse direction. I do not know how to calculate the tensile strain from that data.And I do not know how to calculate the upward movement of the upper surface ( I don't even really understand this.)

For b and shear deformation should turn the cuboid into a parellogram, and for the other question the same answer.

 

[Chestermiller]

Oh you are right,i misstyped. For tensile stress in part a I also got 1,67 Gpa I don't know why I put 3,33 that should be for part b). To clear it up

a) tensile stress 1,67Gpa
b) shear stress 3,33 Gpa shear strain 0,042

Now the deformation according to google should be that the surface extends in axial direction and and contracting in the transverse direction. I do not know how to calculate the tensile strain from that data.And I do not know how to calculate the upward movement of the upper surface ( I don't even really understand this.)

For b and shear deformation should turn the cuboid into a parellogram, and for the other question the same answer.

First part a. Your assessment of what happens is correct. Do you know how to get Young’s modulus, given the shear modulus G and the Poisson ratio of 0.3 for steel?

 

[sylent33]

Okay I did a quick google search and what comes up is this; $$ E = \frac{G}{e} $$ where e is Poisson ratio of steel.

 

[Chestermiller]

Okay I did a quick google search and what comes up is this; $$ E = \frac{G}{e} $$ where e is Poisson ratio of steel.

This is not correct. The correct relationship is $$G=\frac{E}{2(1+\nu)}$$where $\nu$ is Poisson's ratio.

 

[sylent33]

This is not correct. The correct relationship is $$G=\frac{E}{2(1+\nu)}$$where $\nu$ is Poisson's ratio.

Okay so if we want to get E it should look like this $$ E = 2vG + 2G $$ E = 206,18

 

[Chestermiller]

Okay so if we want to get E it should look like this $$ E = 2vG + 2G $$ E = 206,18

Yes, 206 GPa. So what is the axial strain in part a?

 

[sylent33]

Yes, 206 GPa. So what is the axial strain in part a?

Well the way I see it axial strain is the ratio between the change in length of an object and its original lenght.I can get the original length but how to I get to the change in lenght? Or can I use the fact that we calculated Youngs Modul and that we can express it as tensile stress/tensile strain and we can calculate the tensile strain out of that.But are tensile strain and axial strain the same?

 

[Chestermiller]

Well the way I see it axial strain is the ratio between the change in length of an object and its original lenght.I can get the original length but how to I get to the change in lenght? Or can I use the fact that we calculated Youngs Modul and that we can express it as tensile stress/tensile strain and we can calculate the tensile strain out of that.But are tensile strain and axial strain the same?

Yes, tensile strain and axial strain are the same. So what do you get for that strain?

 

[sylent33]

Okay so than its like this $$ E = \frac{G}{ \epsilon} $$ Out of that we get $\epsilon$ and it looks like this

$$ \epsilon = \frac{G}{E} = 0,38 $$

And now we can calculalte $\Delta l$ and I think I know how. $$ \epsilon = \frac{\Delta l}{l} $$ we know the original length was 6 so we can get $\Delta l$ and it should look like this. $$ \Delta l = \epsilon * l $$ so $\Delta l$ = 0,0228 m

 

[Chestermiller]

Okay so than its like this $$ E = \frac{G}{ \epsilon} $$ Out of that we get $\epsilon$ and it looks like this

$$ \epsilon = \frac{G}{E} = 0,38 $$

And now we can calculalte $\Delta l$ and I think I know how. $$ \epsilon = \frac{\Delta l}{l} $$ we know the original length was 6 so we can get $\Delta l$ and it should look like this. $$ \Delta l = \epsilon * l $$ so $\Delta l$ = 0,0228 m

This is incorrect. $$\frac{\sigma}{\epsilon}=E$$So $$\epsilon=\frac{\sigma}{E}$$

 

[sylent33]

Oh right its tensile stress and not shear modulus,silly mistake.Okay to calculate it again;

$$\epsilon = 0,0080 $$ and to get $\Delta l$ it should be like this $$\Delta l = 0,00048 m$$

 

[Chestermiller]

This is incorrect. $$\frac{\sigma}{\epsilon}=E$$So $$\epsilon=\frac{\sigma}{E}$$

I get 3 x 0.008 = 0.024 cm

 

[sylent33]

Okay now I am a tad bit confused.If we were looking for the deformation in a) and we can see that the dark surface(the one where the stress is applied) its length is 6cm (at least the way I see it) and that should be 0,048cm, in b I get the same,since the length of the surface is 3cm.

 

[Chestermiller]

Okay now I am a tad bit confused.If we were looking for the deformation in a) and we can see that the dark surface(the one where the stress is applied) its length is 6cm (at least the way I see it) and that should be 0,048cm, in b I get the same,since the length of the surface is 3cm.

For normal loading, the length change and strain take place in the same direction as the applied force, and the length in the applied force direction in part a is 3 cm.

 

[sylent33]

For normal loading, the length change and strain take place in the same direction as the applied force, and the length in the applied force direction in part a is 3 cm.

Oh okay than it makes sence, and for part b it would be the same or the opposite direction.

 

[Chestermiller]

Oh okay than it makes sence, and for part b it would be the same or the opposite direction.

No. In part b, the left hand side does not move, the right hand side moves upward a distance u, and the lower and upper sides tilt slightly to make an angle with the horizontal whose tangent is the shear strain $\gamma$. This being the case, how much does the right hand side move up?

 

[sylent33]

No. In part b, the left hand side does not move, the right hand side moves upward a distance u, and the lower and upper sides tilt slightly to make an angle with the horizontal whose tangent is the shear strain $\gamma$. This being the case, how much does the right hand side move up?

Well the distance it moves is whatever distance $\Delta l$ moves. So I would need to calculate the $\Delta l$ in the same fasion as I did for a) ??

 

[Chestermiller]

Well the distance it moves is whatever distance $\Delta l$ moves. So I would need to calculate the $\Delta l$ in the same fasion as I did for a) ??

No. I said that the tangent of the angle is $\frac{\Delta u}{6}$, where u is the upward displacement of the entire right side of the object, relative to its original position. Can you draw this angle on a diagram? This is equal to the shear strain $\gamma$.

I'm afraid your teachers have not explained this to you well at all, and you need to get a book on "strength of materials" to learn this properly. Physics Forums is not a proper venue for presenting a comprehensive tutorial on this subject.

 

[sylent33]

No. I said that the tangent of the angle is $\frac{\Delta u}{6}$, where u is the upward displacement of the entire right side of the object, relative to its original position. Can you draw this angle on a diagram? This is equal to the shear strain $\gamma$.

I'm afraid your teachers have not explained this to you well at all, and you need to get a book on "strength of materials" to learn this properly. Physics Forums is not a proper venue for presenting a comprehensive tutorial on this subject.

Maybe I just didnt pay attention in class :D :D.Jokes aside,I skipped quite a few lectures this year because of COVID,its fine I'll look at this on my own.

Thank you for your help!