# How to Solve Lame's Equation for A and B

• Jud
In summary, Homework Equations state that 140MPa = A + - A(0.12^2)/0.1^2. A(0.12^2) = B, so 140MPa = 120MPa + 8.6MPa
Jud

## Homework Statement

A steel pipe is 200mm internal diameter and 20mm thick. Calculate the safe internal pressure if the allowable stress is not to exceed 140Mpa.

σr = A-B/r^2[/B]
and
σtheta = A+B/r^2

## The Attempt at a Solution

Right so, I've got two equations...

(eq1)140MPa =A+B/0.1^2
and
(eq2)0 = A-B/0.12^2

I need help getting A and B. Answers (A=57.38)and (B=0.826)

I know I need to use simultaneous equations. Could someone please give me a step by step solution as I'm stuck.

I have so far done -(eq2)+(eq1) to cancel out A.
and got
...140MPa = B/0.1^2 + B/0.12^2

I'm not sure what to do next to get B.

Hint: use eq. 2 to get A in terms of B (or vice versa). Substitute back into eq. 1 and solve.

SteamKing said:
Hint: use eq. 2 to get A in terms of B (or vice versa). Substitute back into eq. 1 and solve.
I don't understand, sorry. Could you elaborate with the values please.

Jud said:
I don't understand, sorry. Could you elaborate with the values please.

It's very simple. You've got A - B/0.12^2 = 0. Find A in terms of B, or B in terms of A.

Once you have found an expression for A or B, substitute that expression back into eq. 1, which will then be one equation in one unknown.

You're in algebra 101 here.

SteamKing said:
It's very simple. You've got A - B/0.12^2 = 0. Find A in terms of B, or B in terms of A.

Once you have found an expression for A or B, substitute that expression back into eq. 1, which will then be one equation in one unknown.

You're in algebra 101 here.

I.e. B = A/0.12^2

So,
140MPa = A + (A/0.12^2)/0.1^2?

Jud said:
I.e. B = A/0.12^2

Uh, not quite. Remember, you've got

A - B/0.12^2 = 0

How did you come up with

B = A/0.12^2 ?

Jud, starting with A - B/0.12^2 = 0, solve for B, without skipping steps.

Mark44 said:
Jud, starting with A - B/0.12^2 = 0, solve for B, without skipping steps.

A = -B/0.12^2

A(0.12^2) = -B

Jud said:
A = -B/0.12^2

A(0.12^2) = -B

So you're saying if A - 5 = 0, then A = -5. Check your work again.

SteamKing said:
So you're saying if A - 5 = 0, then A = -5. Check your work again.

-A(0.12^2) = -B

So, 140MPa = A + - A(0.12^2)/0.1^2

Jud said:
-A(0.12^2) = -B

You can multiply both sides of the equation by -1 to obtain

A(0.12^2) = B

So, 140MPa = A + - A(0.12^2)/0.1^2

And how would you solve for A?

Friendly advice: I don't know anything about your academic background, but your apparent lack of understanding of algebra is going to prevent you from solving more complex problems than this. Try to take time and remedy this gap in your mathematical knowledge. ;)

## 1. What is Lame's Equation and why is it important?

Lame's equation is a partial differential equation that describes the steady-state motion of an elastic body. It is important because it allows us to understand how an elastic body responds to external forces and can be used to solve various engineering problems.

## 2. How is Lame's Equation derived?

Lame's Equation is derived from the Navier's equations of motion for an elastic solid. It takes into account the material properties of the body, such as Young's modulus and Poisson's ratio, as well as the external forces acting on it.

## 3. Can Lame's Equation be applied to non-linear materials?

Yes, Lame's Equation can be extended to non-linear materials by incorporating additional terms that account for the non-linear behavior of the material. However, this makes the equation more complex and difficult to solve analytically.

## 4. How is Lame's Equation used in real-world applications?

Lame's Equation is used in various engineering fields, such as structural mechanics, geomechanics, and fluid mechanics. It is used to model the behavior of materials under different loads and boundary conditions, and can be used to design structures that can withstand external forces.

## 5. What are the limitations of Lame's Equation?

Lame's Equation assumes that the material is homogeneous, isotropic, and linearly elastic, which may not always be the case in real-world situations. It also does not take into account the effects of time, temperature, and other environmental factors, which may affect the behavior of the material. Additionally, it can only be applied to small deformations and cannot account for large deformations or failure of the material.

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