Principal stresses 3D, solving for direction cosines n1,m1,n1

[jan2re]

Homework Statement

-4.882L1+M1+2N1=0
L1-2.882M1=0
2L1-0.882N1=0
L1^2+M1^2+N1^2=1
How to solve for L1 ,M1 and N1 ?
0<L1,M1,N1<1

Homework Equations

The Attempt at a Solution

 

[anorlunda]

You must do better than that. We require that you show some effort in the Attempt At Solution part.

 

[jan2re]

I have tried back substition ,gauss jordan,matrices and so on ,but this is only a simple math problem so i don't want to spend 1 hour typing in all this mathematics .

 

[Chestermiller]

I have tried back substition ,gauss jordan,matrices and so on ,but this is only a simple math problem so i don't want to spend 1 hour typing in all this mathematics .

Are you saying you can’t solve this by hand?

 

[Mark44]

I have tried back substition ,gauss jordan,matrices and so on ,but this is only a simple math problem so i don't want to spend 1 hour typing in all this mathematics .

It wouldn't take an hour to type at least some effort.

Are you saying you can’t solve this by hand?

Gauss-Jordan isn't appropriate, since this isn't a linear system. Some ordinary algebra could be used to eliminate a variable or two.

 

[Chestermiller]

It wouldn't take an hour to type at least some effort.

Gauss-Jordan isn't appropriate, since this isn't a linear system. Some ordinary algebra could be used to eliminate a variable or two.

Solve the 2nd equation for M1 in terms of L1 and the 3rd equation for N1 in terms of L1. Then substitute the results in 4th equation to obtain L1. Hopefully when substituting the results into the first equation, it will be satisfied identically.

 

[Ray Vickson]

I have tried back substition ,gauss jordan,matrices and so on ,but this is only a simple math problem so i don't want to spend 1 hour typing in all this mathematics .

So, do I understand that you want US to do the typing instead?

 

[scottdave]

You should post what you have attempted. From my quick look at it, you may not have specific values of L₁ M₁ and N₁ and might just be left with one of them in terms of another. But you should be able to narrow a range of values (possibly better than 0 to 1).

 

[Chestermiller]

You should post what you have attempted. From my quick look at it, you may not have specific values of L₁ M₁ and N₁ and might just be left with one of them in terms of another. But you should be able to narrow a range of values (possibly better than 0 to 1).

The object of this calculation is to get a unit vector in one of the three principal directions (probably of stress). The calculation arrived at the present point by using the Cauchy stress relationship to find the directions in which the traction vector acting on a surface is normal to the surface. This identifies the magnitudes of the three principal stresses, and is designed to determine their directions. The OPs set of equations is supposed to find the principal direction for one of the principal stresses. The three homogeneous linear algebraic equations are supposed to have more than the null solution, so the determinant of the OPs equations should be zero (i.e., the three equations should be linearly dependent). I haven't checked to see if the determinant is zero, but it's easy enough to do, and the OP should already have done this. The purpose of the 4th equation is to establish a unit vector in the principal direction, by setting its magnitude equal to unity.

 

[scottdave]

The object of this calculation is to get a unit vector in one of the three principal directions (probably of stress)...

Thanks. That makes sense.

 

[jan2re]

The object of this calculation is to get a unit vector in one of the three principal directions (probably of stress). The calculation arrived at the present point by using the Cauchy stress relationship to find the directions in which the traction vector acting on a surface is normal to the surface. This identifies the magnitudes of the three principal stresses, and is designed to determine their directions. The OPs set of equations is supposed to find the principal direction for one of the principal stresses. The three homogeneous linear algebraic equations are supposed to have more than the null solution, so the determinant of the OPs equations should be zero (i.e., the three equations should be linearly dependent). I haven't checked to see if the determinant is zero, but it's easy enough to do, and the OP should already have done this. The purpose of the 4th equation is to establish a unit vector in the principal direction, by setting its magnitude equal to unity.

The object of this calculation is to get a unit vector in one of the three principal directions (probably of stress). The calculation arrived at the present point by using the Cauchy stress relationship to find the directions in which the traction vector acting on a surface is normal to the surface. This identifies the magnitudes of the three principal stresses, and is designed to determine their directions. The OPs set of equations is supposed to find the principal direction for one of the principal stresses. The three homogeneous linear algebraic equations are supposed to have more than the null solution, so the determinant of the OPs equations should be zero (i.e., the three equations should be linearly dependent). I haven't checked to see if the determinant is zero, but it's easy enough to do, and the OP should already have done this. The purpose of the 4th equation is to establish a unit vector in the principal direction, by setting its magnitude equal to unity.

Thanks for your answer, maybe i should have posted this in another topic.This is from a lecture in Advanced Mechanics and not homework, i have the complete solution M=0.019, L=0.055 N=0.1256. So the case is how to manage to solve for this values the easiest way. i have written 6-7 pages i attempt to solve this so i didnt see why i should bother to write it here. I have tried to get rid of variables by isolate one of the variables ex. L and put it into the other equations and so on but it got messy and i also was left with terms with 2 unknowns like M=2.345N .When i then used this terms in my fourth equation L^2+M^2... i didnt manage to get it right.Also tried to add equations to get rid of variables but because of the nulls i didnt manage to do so. So my question is how to solve this sets of equations ? You don't have to solve it ,just which method is the best. Sorry to have used the homwork topic,and not explaining it properly.

 

[Chestermiller]

Step 1: Check the determinant of the 3 linear equations to make sure that they are linearly dependent (such that the equations have more than just the null solution). I calculated the determinant, and it did indeed come out to 0. So you can discard one of the equations. In this problem, the easiest one to discard is the first equation.

Step 2: Solve for two of the variables in terms of the third:

M = 0.347 L
N = 2.27 L

Step 3: Substitute into the direction cosines equation:
$$L^2+(0.347L)^2+(2.27L)^2=1$$This yields $$L=0.40$$$$M=0.139$$$$N=0.907$$
Sorry, but these don't agree with your results. The sum of the squares of your results don't add up to 1.

 

[Ray Vickson]

Thanks for your answer, maybe i should have posted this in another topic.This is from a lecture in Advanced Mechanics and not homework, i have the complete solution M=0.019, L=0.055 N=0.1256. So the case is how to manage to solve for this values the easiest way. i have written 6-7 pages i attempt to solve this so i didnt see why i should bother to write it here. I have tried to get rid of variables by isolate one of the variables ex. L and put it into the other equations and so on but it got messy and i also was left with terms with 2 unknowns like M=2.345N .When i then used this terms in my fourth equation L^2+M^2... i didnt manage to get it right.Also tried to add equations to get rid of variables but because of the nulls i didnt manage to do so. So my question is how to solve this sets of equations ? You don't have to solve it ,just which method is the best. Sorry to have used the homwork topic,and not explaining it properly.

The equations you start with are inconsistent, so have no solution. If you substitute in your alleged "solution" into those equations, you can see that it does not work.

How do I know they are inconsistent? Well, just try solving one variable at a time (using notation $L,M,N$ instead of $L1,M1,N1$):
$$ \begin{array}{ccc} L-2.882 M = 0 & \Rightarrow & M = .3469812630\: L \\
2 L-.882 N = 0 & \Rightarrow & N = 2.267573696\: L
\end{array}
$$
Thus
$$-4.882 L+M+2 N = 0.1286550\times 10^{-3} \: L$$
In order for this last quantity to be zero we would need $L = 0$, but that would give $M=N=L=0$, making the fourth equation invalidl

 

[Chestermiller]

The equations you start with are inconsistent, so have no solution. If you substitute in your alleged "solution" into those equations, you can see that it does not work.

How do I know they are inconsistent? Well, just try solving one variable at a time (using notation $L,M,N$ instead of $L1,M1,N1$):
$$ \begin{array}{ccc} L-2.882 M = 0 & \Rightarrow & M = .3469812630\: L \\
2 L-.882 N = 0 & \Rightarrow & N = 2.267573696\: L
\end{array}
$$
Thus
$$-4.882 L+M+2 N = 0.1286550\times 10^{-3} \: L$$
In order for this last quantity to be zero we would need $L = 0$, but that would give $M=N=L=0$, making the fourth equation invalidl

The coefficient of L you calculated is just roundoff. For all intents and purposes, this coefficient is zero. So, as I said in my response, anyone of the linear equations is redundant, and can be discarded. The solution I gave in my response is the correct one.

 

[Ray Vickson]

The coefficient of L you calculated is just roundoff. For all intents and purposes, this coefficient is zero. So, as I said in my response, anyone of the linear equations is redundant, and can be discarded. The solution I gave in my response is the correct one.

When I apply your method (but keeping more accuracy during the computations and just rounding at the end) I get
$$\begin{array}{c}
L = 0.397\\
M = 0.174\\
N = 0.901
\end{array}
$$
I just let Maple do all the calculations.