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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



Lame’s Equations are

σh = β/r2 + α

σr = β/r2 – α

Where α and β are Lame’s constants

There is a thick vessel subjected to internal fluid pressure pi at the inner radius ri. It is under external fluid pressure po at the outer radius ro.

We want to find the equations for σh and σr at any radius. It at any radius between ri and ro.

Fig. Thin Cylindrical element  Fig. Longitudinal Failure

Consider a long open ended thick walled cylinder  in Fig (a) and Fig.(b) subjected inside pressure pi and external pressure po. Inside pressure is high.

Considering the failure of the cylinder lengthwise.


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 σh and σr on the element

Applying the equation of equilibrium to the half ring element,

Sum of downward forces= Sum of upward forces

σh dr l+ σh dr l + (σr+ dσr) 2 (r +dr) l= σr 2rl

On simplification, we get

2 σh dr = –2dr σr –2rdσr—2dr dσr

Divide by 2dr and neglect dr dσr

We get

σh  = — σr  –r (dσr/dr)                                 (1)


 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.

ϵlong = σl/E — μσh/E +μ σr/E = constant

σl  is constant and small. σl/E becomes extremely small,

therefore can be neglected.

Therefore — μσh/E +μ σr/E = constant

Taking  out –μ/E, we get

σh — σr = constant= 2α                   (2)

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

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

σr + 2α = — σr  –r (dσr/dr)

2 σr+ r (dσr/dr) =– 2α

Multiply each term by r

2r σr+ r2(dσr/dr) =– 2α r

Write it in the below form

d(r2 σr)/dr =– 2α r

Integrating, we get

r2 σr = —  α r2 + β

σr =  β/r2  — α                                  (3)

Put this value in eq(1), we get

σh =  β/r2  + 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=ri,  σr = pi+              pi = β/ri2  — α

At  r=ro, σr =po                        po= β/ro2  — α

α = (piri2 –poro2) /(ro2 –ri2)

β = (r12 r22[pi—po]) /(ro2 –ri2)

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

we get

σh =  (r12 r22[pi—po])/[r2(ro2 –ri2)]+ (piri2 –poro2) /(ro2 –ri2)

On simplification, we get

σh =  [(piri2 –poro2) + r12 r22/(ro2 –ri2)r2]/ (ro2 –ri2)      (5)

σr =  [(poro2 –piri2) + r12 r22/(ro2 –ri2)r2]/ (ro2 –ri2)      (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 σh will occur at the innermost radius. It is given by

σhmax = [pi(ri2+ro2)—2poro2)]/ (ro2 –ri2)

From eq (6)

It is evident that the maximum value of σr is larger of pi or po. In our analysis it has been assumed that pi > po. Therefore σr will also be maximum at the innermost radius.  Its value is equal to pi.

σr max = pi   with pi >  po   

Longitudinal Stress

Longitudinal stress, σL = [piri2—poro2)]/ (ro2 –ri2).

This contain no variable. Hence it is constant. It is small as compared to the maximum values of σhmax and σr max . Hence σL 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 pi >po. 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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