# LAME’S EQUATIONS THICK CYLINDER CLASS NOTES FOR MECHANICAL ENGINEERING

**LAME’S EQUATIONS**

** THICK CYLINDER CLASS NOTES**

**FOR MECHANICAL ENGINEERING**

**These equations deal with the hoop and**

**radial stresses variation in a thick cylinder.**

**Various cases considered are thick cylinder**

**subjected to fluid pressure only**

**internal/external or internal and external.**

**Lame’s equation determine the maximum**

**stresses and their location in the thick**

**cylinder.**

### Lame’s Equations are

#### σ_{h} = β/r^{2} + α

#### σ_{r} = β/r^{2} – α

#### Where α and β are Lame’s constants

#### There is a thick vessel subjected to internal fluid pressure p_{i} at the inner radius r_{i}. It is under external fluid pressure p_{o} at the outer radius r_{o}.

#### We want to find the equations for σ_{h} and σ_{r} at any radius. It at any radius between r_{i} and r_{o}.

**Fig. Thin Cylindrical element Fig. Longitudinal Failure**

#### Consider a long open ended thick walled cylinder in Fig (a) and Fig.(b) subjected inside pressure p_{i }and external pressure p_{o}. Inside pressure is high.

#### Considering the failure of the cylinder lengthwise.

**DERIVATION LAME’S EQUATIONS**

#### Fig. a Consider two concentric sections at radius r and r + dr of length ‘l’ .

#### Cut this thin ring into two halves.

#### Fig. b

#### Consider its equilibrium under the forces acting due to σ_{h} and σ_{r} on the element

#### Applying the equation of equilibrium to the half ring element,

#### Sum of downward forces= Sum of upward forces

#### σ_{h} dr l+ σ_{h} dr l + (σ_{r}+ dσ_{r}) 2 (r +dr) l= σ_{r }2rl

#### On simplification, we get

#### 2 σ_{h} dr = –2dr σ_{r} –2rdσ_{r}—2dr dσ_{r}

#### Divide by 2dr and neglect dr dσ_{r}

#### We get

#### σ_{h} = — σ_{r} –r (dσ_{r}/dr) (1)

#### ASSUMPTIONS IN LAME’S EQUATIONS

#### That a transverse plane remains a transverse plane before and after the fluid pressures are applied.

#### As per this assumption, longitudinal strain has to be constant.

#### ϵ_{long} = σ_{l}/E — μσ_{h}/E +μ σ_{r}/E = constant

#### σ_{l } is constant and small. σ_{l}/E becomes extremely small,

#### therefore can be neglected.

#### Therefore — μσ_{h}/E +μ σ_{r}/E = constant

#### Taking out –μ/E, we get

#### σ_{h} — σ_{r} = constant= 2α (2)

#### (2α has been assumed to be constant because of convenience)

#### Substitute from eq (1) in eq (2), we get

#### σ_{r} + 2α = — σ_{r} –r (dσ_{r}/dr)

#### 2 σ_{r}+ r (dσ_{r}/dr) =– 2α

#### Multiply each term by r

#### 2r σ_{r}+ r^{2}(dσ_{r}/dr) =– 2α r

#### Write it in the below form

#### d(r^{2} σ_{r})/dr =– 2α r

#### Integrating, we get

#### r^{2} σ_{r} = — α r^{2} + β

#### σ_{r} = β/r^{2} — α (3)

#### Put this value in eq(1), we get

#### σ_{h} = β/r^{2} + α (4)

#### Eqs.(3) and (4) are Lames equations.

**Lame’s Constants in Lame’s equations**

#### The values of the Lame’s constants α and β can be found from the physical boundary conditions. These physical boundary conditions are applied to eq (3 ), we get

#### At r=r_{i}, σ_{r} = p_{i+ } p_{i} = β/r_{i}^{2} — α

#### At r=r_{o}, σ_{r} =p_{o } p_{o}= β/r_{o}^{2} — α

#### α = (p_{i}r_{i}^{2} –p_{o}r_{o}^{2}) /(r_{o}^{2} –r_{i}^{2})

#### β = (r_{1}^{2} r_{2}^{2}[p_{i}—p_{o}]) /(r_{o}^{2} –r_{i}^{2})

#### Substitute the values of α and β in lame’s equations (3) and (4),

#### we get

#### σ_{h} = (r_{1}^{2} r_{2}^{2}[p_{i}—p_{o}])/[r^{2}(r_{o}^{2} –r_{i}^{2})]+ (p_{i}r_{i}^{2} –p_{o}r_{o}^{2}) /(r_{o}^{2} –r_{i}^{2})

#### On simplification, we get

#### σ_{h} = [(p_{i}r_{i}^{2} –p_{o}r_{o}^{2}) + r_{1}^{2} r_{2}^{2}/(r_{o}^{2} –r_{i}^{2})r^{2}]/ (r_{o}^{2} –r_{i}^{2}) (5)

#### σ_{r} = [(p_{o}r_{o}^{2} –p_{i}r_{i}^{2}) + r_{1}^{2} r_{2}^{2}/(r_{o}^{2} –r_{i}^{2})r^{2}]/ (r_{o}^{2} –r_{i}^{2}) (6)

#### From equations (5) and(6)

#### it is evident that the stresses are inversely proportional to the square of the radius. These stresses vary parabolically.

**Maximum Stresses**

#### From eq (5) it is evident that the maximum value of σ_{h} will occur at the **innermost radius. It** is given by

#### σ_{hmax} = [p_{i}(r_{i}^{2}+r_{o}^{2})—2p_{o}r_{o}^{2})]/ (r_{o}^{2} –r_{i}^{2})

#### From eq (6)

#### It is evident that the maximum value of σ_{r} is larger of p_{i} or p_{o}. In our analysis it has been assumed that p_{i} > p_{o.} Therefore σ_{r} will also be maximum at the **innermost radius. ** Its value is equal to p_{i}.

#### σ_{r max} = p_{i} with p_{i} > p_{o }

**Longitudinal Stress**

#### Longitudinal stress, σ_{L} = [p_{i}r_{i}^{2}—p_{o}r_{o}^{2})]/ (r_{o}^{2} –r_{i}^{2}).

#### This contain no variable. Hence it is constant. It is small as compared to the maximum values of σ_{hmax} and σ_{r max} . Hence σ_{L} is neglected.

**Variations of hoop and radial stresses in a thick shell when plotted.**

**Fig. Variation of hoop and radial stresses in a thick under internal & external pressures**

Hoop stress is tensile. It is maximum at the inner radius. Radial stress is compressive. It is also maximum at the inner radius with p_{i} >p_{o}. Maximum hoop stress is much greater than maximum radial stress. Hence it is designed on the basis of hoop stress.

https://www.mesubjects.net/wp-admin/post.php?post=4047&action=edit MCQ Thick shells

https://www.mesubjects.net/wp-admin/post.php?post=7718&action=edit Stresses Thick Cylinder

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