hoop stress or circumferential stress and radial stress Lame's equation, t > d/20

1. Thick Cylinders
by
Venkata Sushma chinta

2. Thick cylinders
The thickness of the cylinder is large compared to that of thincylinder. i. e., in case of
thick cylinders, the metal thickness ‘t’ is more than ‘d/20’, where ‘d’ is the internal
diameter of the cylinder.
Magnitude of radial stress (pr) is large and hence it cannot be neglected. The
circumferential stress is also not uniform across the cylinder wall. The radial stress is
compressive in nature and circumferential and longitudinal stresses are tensile in
nature. Radial stress and circumferential stresses are computed by using ‘Lame’s
equations’.
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3. LAME’S EQUATION Assumptions:
 Plane sections of the cylinder normal to its axis remain plane and normal even under
pressure.
 Longitudinal stress (σL) and longitudinal strain (εL) remain constant throughout the
thickness of the wall.
 Since longitudinal stress (σL) and longitudinal strain (εL) are constant, it follows that
the difference in the magnitude of hoop stress and radial stress (pr) at any point on
the cylinder wall is a constant.
 The material is homogeneous, isotropic and obeys Hooke’s law. (The stresses are
within proportionality limit).

4. Consider a thick cylinder of external radius r1 and
internal radius r2, containing a fluid under
pressure ‘p’ as shown in the fig. Let ‘L’ be the
length of the cylinder.

5. Consider an elemental ring of radius ‘r’ and thickness ‘δr’ as shown in the figures.
Let pr and (pr+ δpr) be the intensities of radial pressures at inner and outer faces of
the ring.
pr+δpr
pr
r
pr+δpr
pr
r
σc
σc
r δr
Pr
pr+δpr
External
pressure
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6. Consider the longitudinal section XX of the ring as shown
in the fig.
The bursting force is evaluated by
considering the projected area:
‘2rL’ for the inner face and
‘2(r+δr)L’ for the outer face .
The net bursting force, P = pr.2rL - (pr+δpr)2.(r+ δr).L
=( -pr. δr– r.δpr - δpr . δr) 2L
Bursting force is resisted by the hoop tensile force developing at the level
of the strip i.e., Fr=σc 2 δr.L

7. Thus, for equilibrium, P = Fr
(-pr. δr– r.δpr- δpr . δr) 2L = σ c.2. δr.L
-pr.δr – r.δpr- δpr . δr = σc.δr
Neglecting products of small quantities, (i.e., δpr . δr)
σc = - pr –(r . δpr )/ δr ...…………….(1)
Longitudinal strain is constant. Hence we have,
𝜀 𝐿 =
1
𝐸
𝜎𝐿 − 𝜈 𝜎𝑐 − pr
𝜎𝑐 − pr= 2a;
σc = pr + 2a …………(2)

8. pr + 2a = - pr –(r . δpr )/
δr
2(pr +a)= –(r . δpr )/ δr
𝛿pr
pr+a
= −2.
δr
r
...........(3)
Integrating (3),
(-2 ×loge r) + c = loge (pr +
a)
loge (pr+a) = -2 ×loge r +
loge b
=> log (p +a) = log (
𝑏
)
i.e., pr  a 
b
r2
............(5)
Substituting it in equation 2, we get
radial stress, pr  -a +
b
r2
Hoopstress, σc  a +
b
r2
...........(4)
σc = pr + 2a
σc = -a +
b
r2 + 2a

9. Points to remember:
1. Variations of Hoop stress and Radial stress are parabolic
across the cylinder wall.
2. At the inner edge, the stresses are maximum.
3. The value of ‘Permissible or Maximum Hoop Stress’ is to be
considered on the inner edge.

10. Q1.A steel cylinder of 160mm internal diameter and 320mm outer
diameter. If it is subjected to an internal pressure of 150MPa, find the
maximum radial and tangential stress and the maximum shear stress.
Assume the ends are closed.
ri= 80mm
ro=160mm
Pi=150MPa
radial stress, pr  -a +
b
r2
The boundary conditions to solve constants are :
At radius r= ri , internal pressure is applied so pr =150
At radius r= ro , no external pressure is applied acting , so pr =0

11. pr  -a +
b
r2
150=-a+
b
802 ….(1)
0 =-a+
b
1602 …..(2)
Solving equation (1) and (2)
a= 50 ,b= 1280000
pr  -50 +
1280000
r2
σc = a +
b
r2
σc =50 +
1280000
r2
Hoop stress at inner radius is ri= 80mm
σ c(max) = 50 +
1280000
802 = 50 + 200 = 250𝑀𝑃𝑎
Hoop stress at outer radius is ro=160mm
σ c(min) = 50 +
1280000
1602 = 50 + 50 = 100𝑀𝑃𝑎

12. Maximum shear stresses τmax =
=
𝜎 𝑥−𝜎 𝑦
2
=
𝜎 𝑐−(−pr)
2
=
250+150
2
= 200𝑀𝑃𝑎

13. Q2. A cylinder is 150 mm ID and 450 mm OD. The internal pressure is 160 MPa and the
external pressure is 80 MPa. Find the maximum radial and tangential stresses and the
maximum shear stress. The ends are closed
ri= 75mm
ro=225mm
Pi=160MPa
Po=80MPa
radial stress, pr  -a +
b
r2
The boundary conditions to solve constants are :
At radius r= ri =75mm , internal pressure is applied so pr =160 MPa
At radius r= ro =225mm ,external pressure is applied , so pr =80 MPa
160=-a+
b
752 ….(1)
80 =-a+
b
2252 …..(2)
Solving equation (1) and (2)
a= -70 ,b= 506250
pr  70 +
506250
r2

14. σc = a +
b
r2
σc =-70 +
506250
r2
Hoop stress at inner radius is ri= 75mm
σ c(max) = -70 +
506250
752 = −70 + 90 = 20𝑀𝑃𝑎
Hoop stress at outer radius is ro=225mm
σ c(min) = -70 +
506250
2252 = −70 + 10 = −60𝑀𝑃𝑎
a= -70 ,b= 506250
Maximum shear stresses
τmax =
𝜎 𝑥−𝜎 𝑦
2
=
𝜎 𝑐+P
2
=
20+160
2
= 90𝑀𝑃𝑎

15. Q3. A cylinder has an ID of 100 mm and an internal pressure of 50 MPa. Find the needed
wall thickness if the factor of safety n is 2.0 and the yield stress is 250 MPa. Use the
maximum shear stress theory, i.e. maximum shear stress = yield strength/2n.
ri= 50mm
ro=?
ro - ri = t
Pi=50MPa
Po=0MPa
Maximum shear stresses
τmax =
𝜎 𝑐+Pr
2
= yield strength/2n
𝜎 𝑐+Pr
2
=
250
2∗2
𝜎𝑐 + Pr
2
= 62.5𝑀𝑃𝑎
𝜎𝑐 + Pr= 125
𝜎𝑐 + 50= 125
𝜎𝑐= 75

16. radial stress, pr  -a +
b
r2 ; hoop stress is σc = a +
b
r2
The boundary conditions to solve constants are :
At radius r= ri =50mm , internal pressure is applied so pr =50 MPa
At radius r= ri =50mm , hoop stress is σc =62.5 MPa
50=-a+
b
502 ….(1)
75 =a+
b
502 …..(2)
Solving equation (1) and (2)
a= 12.5,b= 156250
pr  -12.5 +
156250
r2
At radius r= ro, No external pressure is applied , so pr =0 MPa
0  -12.5 +
156250
ro
2 On solving ro = 111.8 mm => t= 111.8 -50=61.8mm