ME 555 Stress Analysis
Ing. Prof. P. Y. Andoh, PhD, MGhIE
Phone: 050 797 0658
Email: andohp_2@yahoo.com
Jan 2014
II
Part
STRESS AND STRAIN IN
ROTATING DISCS, RINGS
AND CYLINDERS
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Rotating Solid Shafts
• Consider an element of a disc at radius r as shown in Fig.4-1.
• Assuming unit thickness:
• volume of element = 𝑟𝛿𝜃𝛿𝑟
• mass of element = 𝜌𝑟𝛿𝜃𝛿𝑟
• Therefore centrifugal force acting on the element
= 𝑚𝜔2 𝑟 = 𝜌𝑟𝛿𝜃𝛿𝑟𝜔2 𝑟 = 𝑝𝑟 2 𝜔2 𝛿𝜃𝛿𝑟
• Now for equilibrium of the element radially
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𝛿𝜃
2 2
2𝜎𝐻 𝛿𝑟 sin
+ 𝜎𝑟 𝑟𝛿𝜃 − 𝜎𝑟 + 𝛿𝜎𝑟 𝑟 + 𝛿𝑟 𝛿𝜃 = 𝑝𝑟 𝜔 𝛿𝜃𝛿𝑟
2
Rotating Solid Shafts: General Equation
𝛿𝜃
𝛿𝜃
• If δθ is small, sin 2 = 2 𝑟𝑎𝑑𝑖𝑎𝑛
• Therefore in the limit, as δr → 0, the above equation reduces to
𝑑𝜎
𝜎𝐻 − 𝜎𝑟 − 𝑟 𝑑𝑟𝑟 = 𝜌𝑟 2 𝜔2
(4.1)
If there is a radial movement or “shift” of the element by an amount s as the disc rotates, the
radial strain is given by
𝜀𝑟 =
𝑑𝑠
𝑙
=
𝑑𝑟
𝐸
𝜎𝑟 − 𝑣𝜎𝐻
(4.2)
𝑠
Now it has been shown in (a)’ that the diametral strain is equal to the circumferential strain. 𝑟
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𝑟𝑙
𝑙
= 𝐸 𝜎𝐻 − 𝑣𝜎𝑟
; 𝑠 = 𝐸 𝜎𝐻 − 𝑣𝜎𝑟
(4.3)
Rotating Solid Shafts: General Equation
𝑑𝑠
𝑙
𝑟 𝜎
𝜎
• Differentiating, 𝑑𝑟 = 𝐸 𝜎𝐻 − 𝑣𝜎𝑟 + 𝐸 𝑑𝑟𝐻 − 𝑑𝑟𝑟
(4.4)
• Equating (4.2) and (4.4) and simplifying,
𝜎𝐻 − 𝜎𝑟 1 + 𝑣
𝑑𝜎𝐻
𝑑𝜎𝑟
+ 𝑟 𝑑𝑟 − 𝑣𝑟 𝑑𝑟 = 0
(4.5)
• Substituting for 𝜎𝐻 − 𝜎𝑟 from eqn. (4.1),
𝑑𝜎𝑟
𝑟
+ 𝜌𝑟 2 𝜔2
𝑑𝑟
∴
𝑑𝜎𝐻
𝑑𝜎𝑟
1+𝑣 +𝑟
− 𝑣𝑟
=0
𝑑𝑟
𝑑𝑟
𝑑𝜎𝐻 𝑑𝜎𝑟
+
= −𝜌𝑟𝜔2 (1 + 𝑣)
𝑑𝑟
𝑑𝑟
• Integrating,
𝜌𝑟 2 𝜔2
𝜎𝐻 +𝜎𝑟 = −
2
1 + 𝑣 + 2𝐴
(4.6)
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Rotating Solid Shafts: General Equation
𝑑𝜎𝑟
𝜌𝑟 2 𝜔2
• Subtracting eqn. (4.1), 2𝜎𝑟 + 𝑟 𝑑𝑟 = − 2
𝑑𝜎𝑟
𝑑
𝑙
2
• But
2𝜎𝑟 + 𝑟 𝑑𝑟 = 𝑑𝑟 𝑟 𝜎𝑟 × 𝑟
3 + 𝑣 + 2𝐴
𝑑 2
𝜌𝑟 2 𝜔2
𝑟 𝜎𝑟 = 𝑟 −
3 + 𝑣 + 2𝐴
𝑑𝑟
2
4 2
2
𝜌𝑟
𝜔
2𝐴𝑟
𝑟 2 𝜎𝑟 = −
3+𝑣 +
−𝐵
8
2
• where -B is a second convenient constant of integration
𝐵
𝜎𝑟 = 𝐴 − 𝑟 2 −
3+𝑣
𝜌𝑟 2 𝜔2
8
(4.7)
• From eqn. (4.5),
𝐵
𝜎𝐻 = 𝐴 + 𝑟 2 −
1 + 3𝑣
𝜌𝑟 2 𝜔2
8
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(4.8)
Rotating Solid Shafts: General Equation
• For a solid disc the stress at the centre is given when r = 0.
• With r equal to zero the above equations will yield infinite stresses whatever the speed of
rotation unless B is also zero, i.e. B = 0 and hence B/r2 = 0 gives the only finite solution.
• Now at the outside radius R the radial stress must be zero since there are no external forces
to provide the necessary balance of equilibrium if a,. were not zero.
• Therefore from eqn. (4.7), 𝜎𝑟 = 0 = 𝐴 − 3 + 𝑣
• 𝐴 = 3+𝑣
𝜌𝑟 2 𝜔2
8
𝜌𝑟 2 𝜔2
8
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Rotating Solid Shafts: General Equation
• Substituting in eqns. (4.7) and (4.8) the hoop and radial stresses at any
radius r in a solid disc are given by
𝜌𝜔2
𝜎𝐻 = 8
3 + 𝑣 𝑅 2 − 1 + 3𝑣 𝑟 2
𝜎𝑟 = 3 + 𝑣
𝜌𝜔2
8
𝑅2 − 𝑟 2
(4.9)
(4.10)
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Rotating Solid Shafts: Maximum Stresses
• At the centre of the disc, where r = 0, the above equations yield equal values of hoop and
radial stress which may also be seen to be the maximum stresses in the disc, i.e. maximum
hoop and radial stress (at the centre) 𝜎𝑟 = 𝜎𝐻 = 3 + 𝑣
𝜌𝜔2 𝑅2
8
(4.11)
• At the outside of the disc, at r = R, the equations give
𝜎𝑟 = 0 𝑎𝑛𝑑 𝜎𝐻 = 1 − 𝑣
𝜌𝜔2 𝑅2
4
(4.12)
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Rotating Disc with Central Hole: General Equations
• The general equations for the stresses in a rotating hollow disc may be obtained in
precisely the same way as those for the solid disc of the previous section,
𝐵
𝜎𝑟 = 𝐴 − 𝑟 2 −
3+𝑣
𝜌𝑟 2 𝜔2
;
8
𝐵
𝜎𝐻 = 𝐴 + 𝑟 2 −
1 + 3𝑣
𝜌𝑟 2 𝜔2
8
• The only difference to the previous treatment is the conditions which are required to
evaluate the constants A and B since, in this case, B is not zero.
• However, returning to the rotation only case, the required boundary conditions are zero
radial stress at both the inside and outside radius,
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Rotating Disc with Central Hole: General Equations
i.e. at r = R1, 𝜎𝑟 = 0; ∴
and at r = R2, 𝜎𝑟 = 0; ∴
𝐵
0 = 𝐴 − 𝑅2 −
1
𝜌𝑅12 𝜔2
8
3+𝑣
𝐵
0=𝐴− 2−
𝑅2
3+𝑣
𝜌𝑅22 𝜔2
8
(a)
(b)
• Subtracting (a) from (b) and simplifying,
𝜌𝑅12 𝑅22 𝜔2
;
8
𝐵 = 3+𝑣
𝐴= 3+𝑣
𝜌𝜔2 (𝑅12 +𝑅22 )
8
• Substituting in eqns. (4.7) and (4.8) yields the final equation for the stresses
𝑅 𝑅
𝑅12 + 𝑅22 − 1𝑟 2 2 − 𝑟 2
3+𝑣
𝑅1 𝑅2
2
2
𝑅1 + 𝑅2 − 𝑟 2
𝜎𝑟 = 3 + 𝑣
𝜌𝜔2
𝜎𝐻 = 8
2 2
𝜌𝜔2
8
2 2
− (1 + 3𝑣)𝑟
(4.13)
2
(4.14)
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Rotating Disc with Central Hole: Maximum
Stresses
• The maximum hoop stress occurs at the inside radius where r = R1,
• i.e.
𝜌𝜔2
𝜎𝐻𝑚𝑎𝑥 = 8
3 + 𝑣 𝑅12 + 𝑅22 + 𝑅22 − (1 + 3𝑣)𝑅12
𝜌𝜔2
=
4
•
3 + 𝑣 𝑅22 + (1 − 𝑣)𝑅12
(4.15)
• As the value of the inside radius approaches zero the maximum hoop stress value
approaches
𝜌𝜔2
4
3 + 𝑣 𝑅22
• This is twice the value obtained at the centre of a solid disc rotating at the same speed. Thus
the drilling of even a very small hole at the centre of a solid disc will double
the maximum
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hoop stress set up owing to rotation.
Rotating Disc with Central Hole: Maximum
• At the outside of the disc when r = R2 Stresses
𝜌𝜔2
• 𝜎𝐻𝑚𝑖𝑛 = 4
3 + 𝑣 𝑅12 + (1 − 𝑣)𝑅22
• The maximum radial stress is found by consideration of the equation (4.13) is
𝜎𝑟 = 3 + 𝑣
𝜌𝜔2
8
2 2
𝑅 𝑅
𝑅12 + 𝑅22 − 1 2 2 − 𝑟 2
𝑟
𝑑𝜎𝑟
• This will be a maximum when
=0
𝑑𝑟
𝑑
𝑅12 𝑅22
2
2
i.e.
when 0 =
𝑅1 + 𝑅2 − 2 − 𝑟 2
𝑑𝑟
𝑟
2 2 2
0 = 𝑅1 𝑅2 3 − 2𝑟 ∴ 𝑟 =
𝑟
𝑅1 𝑅2
(4.16)
• Substituting for r in eqn. (4.13)
• .; 𝜎𝑟𝑚𝑎𝑥 = 3 + 𝑣
𝜌𝜔2
8
𝑅1 − 𝑅2
2
(4.17
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Rotating Disc of Uniform Strength
• Consider, therefore, an element of a disc subjected to
equal hoop and radial stresses, i.e 𝜎𝐻 = 𝜎𝑟 = 𝜎
• The condition of equal stress can only be achieved, as in the case of uniform strength
cantilevers, by varying the thickness.
• Let the thickness be t at radius r and (t + δt) at radius (r + δr).
• Then centrifugal force on the element
• = 𝑚𝑎𝑠𝑠 × 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 𝜌𝑡𝑟𝛿𝜃𝛿𝑟 𝜔2 𝑟 = 𝜌𝑡𝜔2 𝑟 2 𝛿𝜃𝛿𝑟
• The equilibrium equation is then
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1
2 2
𝜌𝑡𝜔 𝑟 𝛿𝜃𝛿𝑟 + 𝜎 𝑟 + 𝛿𝑟 𝛿𝜃 𝑡 + 𝛿𝑡 = 2𝜎𝑡𝛿𝑟 sin 𝛿𝜃 + 𝜎𝑟 𝑡𝛿𝜃
2
Rotating Disc of Uniform Strength
• i.e. in the limit
𝜎𝑡𝑑𝑟 = 𝜌𝑡𝜔2 𝑟 2 𝑡𝑑𝑟 + 𝜎𝑡𝑑𝑟 + 𝜎𝑟𝑑𝑡
∴
𝜎𝑡𝑑𝑟 = −𝜌𝜔2 𝑟 2 𝑡𝑑𝑟
𝑑𝑡
𝜌𝜔2 𝑟𝑡
∴
=−
𝑑𝑟
𝜎
• Integrating,
• i.e. for uniform strength the thickness of
the disc must vary according to the
following equation,
𝑡 = 𝑡0 𝑒
−𝜌𝜔2 𝑟 2 Τ2𝜎
(4.18)
𝜌𝜔2 𝑟 2
log 𝑒 𝑡 = −
+ log 𝑒 𝐴
2𝜎
• Where log 𝑒 𝐴 is a convenient constant
𝑡 = 𝐴𝑒
−𝜌𝜔2 𝑟 2 Τ2𝜎
• Where r = o ; t = A = t0
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Thin Rotating Ring or Cylinder
• Consider a thin ring or cylinder as shown in Fig. 3-5
subjected to a radial pressure p caused by the centrifugal
effect of its own mass when rotating.
• The centrifugal effect on a unit length of the circumference is
𝑝 = 𝑚𝜔2 𝑟
• Thus, considering the equilibrium of half the ring shown in the figure, 2F = p x 2r
(assuming unit length)
F = pr
• Where F is the hoop tension set up owing to rotation.
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Thin Rotating Ring or Cylinder
• The cylinder wall is assumed to be so thin that the centrifugal effect can be assumed
constant across the wall thickness.
• ∴ 𝐹 = 𝑚𝑎𝑠𝑠 × 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 𝑚𝑤 2 𝑟 2 × 𝑟
• This tension is transmitted through the complete circumference and therefore is resisted by
the complete cross-sectional area.
𝐹
𝑚𝜔2 𝑟 2
∴ ℎ𝑜𝑜𝑝 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝐴 = 𝐴
• where A is the cross-sectional area of the ring.
• Now with unit length assumed, m/A is the mass of the material per unit volume, i.e. the
density ρ. ∴ ℎ𝑜𝑜𝑝 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝑚𝜔2 𝑟 2
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Rotating Thick Cylinders or Solid
Shafts
• In the case of rotating thick cylinders the longitudinal stress σL must be taken into account
and the longitudinal strain is assumed to be constant.
• Thus, writing the equations for the strain in three mutually perpendicular directions ,
𝜀𝐿 =
𝑙
𝐸
𝜎𝐿 − 𝑣𝜎𝐻 − 𝑣𝜎𝑟
𝑙
(4.19)
𝑑𝑠
𝜀𝑟 = 𝐸 𝜎𝑟 − 𝑣𝜎𝐻 − 𝑣𝜎𝐿 = 𝑑𝑟
𝑙
𝜀𝐻 = 𝐸
𝜎𝐻 − 𝑣𝜎𝐿 − 𝑣𝜎𝑟
• From eqn. (4.21);
(4.20)
𝑠
=𝑟
(4.21)
𝐸𝑠 = 𝑟 𝜎𝐻 − 𝑣(𝜎𝑟 + 𝜎𝐿 )
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Rotating Thick Cylinders or Solid
Shafts
𝑑
𝑑𝜎
𝑑𝜎
𝑑𝜎
• Differentiating, 𝐸
𝑠
𝑑𝑟
=𝑟
𝐻
𝑑𝑟
−𝑣
𝑟
𝑑𝑟
−𝑣
𝐿
𝑑𝑟
+ 𝑙 𝜎𝐻 − 𝑣𝜎𝑟 − 𝑣𝜎𝐿
• Substituting for E(ds/dr) in eqn. (4.20) and simplify
𝑑𝜎𝐻
𝑑𝜎𝑟
𝑑𝜎𝐿
0 = (𝜎𝐻 − 𝜎𝑟 )(1 + 𝑣) + 𝑟
− 𝑣𝑟
− 𝑣𝑟
𝑑𝑟
𝑑𝑟
𝑑𝑟
• Now, since EL is constant, differentiating eqn. (4.19), and simplify
𝑑𝜎𝐻
𝑑𝜎𝑟
2
∴
0 = 𝜎𝐻 − 𝜎𝑟 1 + 𝑣 + 𝑟 1 − 𝑣
− 𝑣𝑟(1 + 𝑣)
𝑑𝑟
𝑑𝑟
• Dividing through by(1 + 𝑣), 0 = 𝜎𝐻 − 𝜎𝑟 + 𝑟 1 − 𝑣
𝑑𝜎𝐻
𝑑𝜎𝑟
− 𝑣𝑟
𝑑𝑟
𝑑𝑟
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Rotating Thick Cylinders or Solid
Shafts
• But the general equilibrium equation will
be the same as that obtained in 4.2, eqn. (4.1),
𝑑𝜎
𝜎𝐻 − 𝜎𝑟 − 𝑑𝑟𝑟 = 𝜌𝜔2 𝑟 2
i.e.
• Therefore substituting for 𝜎𝐻 − 𝜎𝑟 ,
2 2
0 = 𝜌𝜔 𝑟
𝑑𝜎𝑟
𝑑𝜎𝐻
𝑑𝜎𝑟
+𝑟
+𝑟 1−𝑣
− 𝑣𝑟 ;
𝑑𝑟
𝑑𝑟
𝑑𝑟
𝑑𝜎𝐻 𝑑𝜎𝑟
0 = 𝜌𝜔 𝑟 + 𝑟 1 − 𝑣
−
𝑑𝑟
𝑑𝑟
2 2
𝑑𝜎𝐻 𝑑𝜎𝑟
𝜌𝜔2 𝑟
∴
−
=−
𝑑𝑟
𝑑𝑟
(1 − 𝑣)
𝜌𝜔2 𝑟 2
• Integrating, 𝜎𝐻 + 𝜎𝑟 = − 2 1−𝑣 + 2𝐴
• where 2A is a convenient constant of integration.
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Rotating Thick Cylinders or Solid
Shafts
• Thus hoop and radial stresses in rotating thick cylinders can be
obtained from the equations for rotating discs provided that Poisson's
ratio u is replaced by 𝑣 /(l - 𝑣), e.g. the stress at the centre of a
rotating solid shaft will be given by eqn. (4.11) for a solid disc
modified as stated above,
i.e.
𝜎𝐻 =
𝑣
𝜌𝑅2 𝜔2
3 + (1−𝑣) 8
(4.22)
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Combined Rotational and Thermal Stresses in Uniform
Discs and Thick Cylinders
• Consider, therefore, a disc initially unstressed and subjected to a temperature rise T.
• Then, for a radial movement s of any element, eqns. (4.2) and (4.3) may be modified to
account for the strains due to temperature thus:
𝑑𝑠
1
=𝐸
𝑑𝑟
𝑠
1
𝜎𝑟 − 𝑣𝜎𝐻 + 𝐸𝛼𝑇 ; 𝑟 = 𝐸
𝜎𝐻 − 𝑣𝜎𝑟 + 𝐸𝛼𝑇
(4.23)
• where α is the coefficient of expansion of the disc material.
• From eqn. (4.23),
•
𝑑𝑠
1
=
𝑑𝑟
𝐸
𝜎𝐻 − 𝑣𝜎𝑟 + 𝐸𝛼𝑇 + 𝑟
𝑑 𝜎𝐻
𝑑 𝜎𝑟
𝑑𝑇
−𝑣
+ 𝐸𝛼
𝑑𝑟
𝑑𝑟
𝑑𝑟
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Combined Rotational and Thermal Stresses in
Uniform Discs and Thick Cylinders
• Therefore from eqn. (4.23),
1
𝐸
∴
𝜎𝑟 − 𝑣𝜎𝐻 + 𝐸𝛼𝑇
1
=𝐸
𝜎𝐻 − 𝜎𝑟 1 + 𝑣
𝜎𝐻 − 𝑣𝜎𝑟 + 𝐸𝛼𝑇 + 𝑟
𝑑𝜎𝐻
𝑑𝜎𝐻
𝑑𝜎𝑟
𝑑𝑇
− 𝑣 𝑑𝑟 + 𝐸𝛼 𝑑𝑟
𝑑𝑟
𝑑𝜎𝑟
𝑑𝑇
+ 𝑟 𝑑𝑟 − 𝑣𝑟 𝑑𝑟 + 𝐸𝛼𝑟 𝑑𝑟 = 0
(4.24)
𝑑𝜎𝑟
• but, from the equilibrium eqn. (4.1), 𝜎𝐻 −𝜎𝑟 − 𝑟 𝑑𝑟 = 𝜌𝑟 2 𝜔2
• Therefore substituting for 𝜎𝐻 − 𝜎𝑟 in eqn. (4.23),
𝑑𝜎𝐻 𝑑𝜎𝑟
𝑑𝑇
2
+
= − 1 + 𝑣 𝜌𝑟𝜔 − 𝐸𝛼
=0
𝑑𝑟
𝑑𝑟
𝑑𝑟
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Combined Rotational and Thermal
Stresses in Uniform Discs and Thick
Cylinders
𝜌𝑟 2 𝜔2
• Integrating, 𝜎𝐻 − 𝜎𝑟 = − 1 + 𝑣
− 𝐸𝛼𝑇 + 2𝐴
2
• where, again, 2A is a convenient constant.
𝜌𝑟 2 𝜔2
• Subtracting eqn. (4.1), 2𝜎𝑟 + 𝑟 𝑑𝑟 = − 2
𝑑 𝜎𝑟
• But
𝑑𝜎𝑟
𝑑
2𝜎𝑟 + 𝑟 𝑑𝑟 = 𝑑𝑟
∴
2
(4.25)
3 + 𝑣 − 𝐸𝛼𝑇 + 2𝐴
1
𝑟 𝜎𝑟 × 𝑟
2
𝑑 2
𝜌𝑟 2 𝜔2
𝑟 𝜎𝑟 = 𝑟 −
3 + 𝑣 − 𝐸𝛼𝑇 + 2𝐴
𝑑𝑟
2
𝜌𝑟 4 𝜔2
𝜎𝑟 =- 8
2𝐴𝑟 2
− 𝐸𝛼 ‫ 𝑟𝑑𝑟𝑇 ׬‬+ 2 − 𝐵
• Integrating, 𝑟
3+𝑣
where, as in eqn. (4.7), -B is a second convenient constant of integration.
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Combined Rotational and Thermal Stresses in
Uniform Discs and Thick Cylinders
𝐵
𝜌𝑟 4 𝜔2
∴ 𝜎𝑟 = 𝐴 − 𝑟 2 − 8
𝐸𝑎
3 + 𝑣 − 𝑟 2 ‫𝑟𝑑 𝑟𝑇 ׬‬
(4.26)
• Then, from eqn. (4.26),
𝐵
𝜎𝐻 = 𝐴 + 𝑟 2 −
1 + 3𝑣
𝜌𝑟 2 𝜔2
𝐸𝑎
− 𝐸𝛼𝑇 + 𝑟 2 ‫𝑟𝑑 𝑟𝑇 ׬‬
8
(4.27)
• For thick cylinders with an axial length several times the outside diameter the above plane
stress equations may be modified to the equivalent plane strain equations by replacing 𝑣 by
𝑣 / (1 − 𝑣), E by E/(1 - 𝑣 2 ), E by E/(1 + u)a. i.e. 𝐸𝛼 becomes 𝐸𝛼 1 − 𝑣
• In the absence of rotation the equations simplify to
𝐵
𝐸𝑎
𝜎𝑟 = 𝐴 − 𝑟 2 − 𝑟 2 ‫𝑟𝑑 𝑟𝑇 ׬‬
𝐵
𝐸𝑎
𝜎𝐻 = 𝐴 + 2 + 2 ‫𝑟𝑑 𝑟𝑇 ׬‬
𝑟
𝑟
(4.28)
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(4.29)
Combined Rotational and Thermal
Stresses in Uniform Discs and Thick
Cylinders
• With a linear variation of temperature from
T = 0 at r = 0, i.e. with
T = Kr
𝐵
𝜎𝑟 = 𝐴 − 𝑟 2 −
𝐸𝑎𝐾𝑟
3
𝐵
𝐸𝑎𝐾𝑟
𝜎𝐻 = 𝐴 + 𝑟 2 + 2 3
(4.30)
(4.31)
• With a steady heat flow, for example, in the
case of thick cylinders when Ea becomes
𝑟𝑑𝑇
∴
𝑑𝑇 𝑏
= 𝑎𝑛𝑑 𝑇 = 𝑎 + 𝑏 log 𝑒 𝑟
𝑑𝑟 𝑟
Eα/(1 - 𝑣 2 ). 𝑑𝑟 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 𝑏;
• The equations become
𝐵
𝐸𝛼𝑇
𝐵
𝐸𝛼𝑇
𝜎𝑟 = 𝐴 − 𝑟 2 − 2(1−𝑣)
(4.32)
𝐸𝑎𝑏
𝜎𝐻 = 𝐴 + 𝑟 2 − 2 1−𝑣 − 2(1−𝑣) (4.33)
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Example 2-4
A steel ring of outer diameter 300 mm and internal diameter 200 mm is shrunk onto a solid
steel shaft. The interference is arranged such that the radial pressure between the mating
surfaces will not fall below 30 MN/m2 whilst the assembly rotates in service. If the maximum
circumferential stress on the inside surface of the ring is limited to 240 MN/m2, determine the
maximum speed at which the assembly can be rotated. It may be assumed that no relative slip
occurs between the shaft and the ring.
For steel, 𝜌 = 7470 𝑘𝑔/𝑚3 , 𝑣 = 0.3, E = 208 GN/m2
Solution
𝐵
From eqn. (3.7); 𝜎𝑟 = 𝐴 − 𝑟 2 −
3+𝑣
2 2
𝜌𝑟
𝜔
8
𝐵
3.3
Now when r = 0.15, 𝜎𝑟 , = 0; 0 = 𝐴 − 0.152 − 8 𝜌𝑟 2 0.15 2
(1)
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(2)
Example 2-4 Continues
Also, when r = 0.1, 𝜎𝑟
𝐵
6
2
= -30 MN/m ; −30 × 10 = 𝐴 −
(2) – (3)
𝐵 = 0.54 × 106 + 0.693 𝜔2
From (3),
𝐴 = 24 × 106 + 100.1𝜔2
0.12
−
3.3
2
2
𝜌𝜔
0.1
8
(3)
But since the maximum hoop stress at the inside radius is limited to 240 MN/m2, from eqn.
𝐵
1+3𝑣
(4.8), 𝜎𝐻 = 𝐴 + 𝑟 2 − 8 𝜌𝑟 2 𝜔2
i.e.
6
2
0.54
×
10
+
0.693𝜔
1.9
6
6
2
2
240 × 10 = 24 × 10 + 100.1𝜔 +
−
×
7470
×
0.01𝜔
0.12
8
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Example 2-4 Continues
240 × 106 = 78 × 106 + 169.3𝜔2 − 17.7𝜔2
∴ 151.7𝜔2 = 162 × 106 ;
𝜔
2
162×106
=
= 1.067 × 106 ;
151.7
𝜔 = 1033 𝑟𝑎𝑑Τ𝑠 = 9860𝑟𝑒𝑣/𝑚𝑖𝑛
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Example 2-5
A steel rotor disc which is part of a turbine assembly has a uniform thickness of 40 mm. The
disc has an outer diameter of 600 mm and a central hole of 100 mm diameter. If there are 200
blades each of mass 0.153 kg pitched evenly around the periphery of the disc at an effective
radius of 320 mm, determine the rotational speed at which yielding of the disc first occurs
according to the maximum shear stress criterion of elastic failure. For steel, E = 200 GN/m2, υ
= 0.3, ρ = 7470 kg/m3 and the yield stress σy in simple tension = 500 MN/m2.
Solution
Total mass of blades = 200 x 0.153 = 30.6 kg
Effective radius = 320 mm
Therefore centrifugal force on the blades = 𝑚𝜔2 𝑟 = 30.6 × 𝜔2 × 0.32
2
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Now the area of the disc rim = πdt = π × 0.6 × 0.004 = 0.024π𝑚
Example 2-5 Continues
The centrifugal force acting on this
area thus produces an effective radial
stress acting on the outside surface of
the disc since the blades can be
assumed to produce a uniform
loading around the periphery.
Now eqns. (4.7) and (4.8) give the general form of the
expressions for hoop and radial stresses set up owing
to rotation,
Therefore radial stress at outside
surface
When r = 0.05,
2
30.6 × 𝜔 × 0.32
=
0.024𝜋
= 130𝜔2 𝑁Τ𝑚2 (𝑡𝑒𝑛𝑠𝑖𝑙𝑒)
i.e.
𝐵
𝜎𝑟 = 𝐴 − 𝑟 2 −
3+𝑣
2 2
𝜌𝑟
𝜔
8
(1)
𝐵
1+3𝑣
𝜎𝐻 = 𝐴 + 2 −
𝜌𝑟 2 𝜔2
𝑟
8
𝜎𝑟 = 0
3.3
∴ 0 = 𝐴 − 400𝐵 − 8 𝜌𝜔2 0.05 2
When r=0.3
(2)
(3)
𝜎𝑟 = +130𝜔2
3.3
2
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∴ 130𝜔2 = 𝐴 − 11.1𝐵 − 8 𝜌𝜔2 0.3
(4)
Example 2-5 Continues
3.3
∴ 130𝜔2 = 388.9𝐵 − 8 𝜌𝜔2 9 − 0.25 10−2
(4) - (3),
130 + 270 2
𝐵=
𝜔 = 1.03𝜔2
388.9
Substituting in (3),
𝐴 = 412𝜔
2
3.3
+ 8 × 7470 0.05 2 𝜔2 = 419.7𝜔2 = 420𝜔2
Therefore substituting in (2) and (l),
The stress conditions at the inside surface are
𝜎𝐻 = 420𝜔2 + 412𝜔2 − 4.43𝜔2 = 827𝜔2
with 𝜎𝑟 = 0
The stress condition at the outside surface are
𝜎𝐻 = 420𝜔2 + 11.42𝜔2 − 159𝜔2 = 272𝜔2 with
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𝜎𝑟 = 130𝜔2
Example 2-5 Continues
The most severe stress conditions therefore
occur at the inside radius where the maximum
shear tress is greatest
i.e.
𝜏𝑚𝑎𝑥 =
𝜎1 −𝜎3
=
2
827𝜔2 −0
Thus, for failure according to this theory, 2
827𝜔2
=
2
i.e.
827𝜔2 = 𝜎𝑦 = 500 × 106
2
Now the maximum shear stress theory of
elastic failure states that failure is assumed to
occur when this stress equals the value of
𝜏𝑚𝑎𝑥 at the yield point in simple tension,
i.e.
𝜎𝑦
𝜎𝑦 −0
𝜎𝑦
𝜎1 −𝜎3
𝜏𝑚𝑎𝑥 = 2 = 2 = 2
∴ 𝜔
2
× 106 = 0.604 × 106
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𝜔
= 780 𝑟𝑎𝑑Τ𝑠 = 7450 𝑟𝑒𝑣Τ𝑚𝑖𝑛
500
= 827
Example 2-6
The cross-section of a turbine rotor disc is designed for uniform strength under rotational
conditions. The disc is keyed to a 60 mm diameter shaft at which point its thickness is a
maximum. It then tapers to a minimum thickness of 10 mm at the outer radius of 250 mm
where the blades are attached. If the design stress of the shaft is 250 MN/m2 at the design
speed of 12000 rev/min, what is the required maximum thickness? For steel ρ = 7470 kg/m3.
Solution
From eqn. (3.18) the thickness of a uniform strength disc is given by 𝑡 = 𝑡0 𝑒
−𝜌𝜔2 𝑟 2 Τ2𝜎
where t0 is the thickness at r = 0.
𝜌𝑟 2 𝜔2
7470
Now at r = 0.25, 2𝜎 = 2×250×106
2𝜋 2
12000 × 60 × 0.252 = 1.47
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(1)
Example 2-6 Continues
𝜌𝑟 2 𝜔2
7470
and at r = 0.03, 2𝜎 = 2×250×106
But at r = 0.25,
2𝜋 2
9×10−4
2
12000 × 60 × 0.03 =1.47 × 625×10−4 = 0.0212
t = 10 mm
Therefore substituting in (l),
0.01 = 𝑡0 𝑒 −1.47 = 0.2299 𝑡0
0.01
𝑡0 =
= 0.0435𝑚 = 43.5 𝑚𝑚
0.2299
Therefore at r = 0.03
𝑡 = 0.0435𝑒 −0.0212 = 0.0435 × 0.98
= 0.0426𝑚 = 42.6𝑚𝑚
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Example 2-7
(a) Derive expressions for the hoop and radial stresses developed in a solid disc of radius R
when subjected to a thermal gradient of the form T = Kr. Hence determine the position Rings,
Discs and Cylinders Subjected to Rotation and Thermal Gradients 133 and magnitude of the
maximum stresses set up in a steel disc of 150 mm diameter when the temperature rise is
150°C. For steel, α = 12 X 10-6 per °C and E = 206.8 GN/m2.
(b) How would the values be changed if the temperature at the centre of the disc was
increased to 30"C, the temperature rise across the disc maintained at 150°C and the thermal
gradient now taking the form T = a + br?
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Example 2-7 Continues
Solution
(a) The hoop and radial stresses are given by eqns. (4.28) and (4.29) as follows:
𝐵
𝐸𝑎
𝜎𝑟 = 𝐴 − 𝑟 2 − 𝑟 2 ‫𝑟𝑑 𝑟𝑇 ׬‬
In this case
(1)
2
‫𝑟𝑑 𝑟 ׬ 𝐾 = 𝑟𝑑 𝑟𝑇 ׬‬
𝐵
𝐸𝑎
𝜎𝐻 = 𝐴 + 𝑟 2 + 𝑟 2 ‫𝑟𝑑 𝑟𝑇 ׬‬
(2)
𝐾𝑟 3
=
, the constant of integration being incorporated
3
into the general constant A.
∴
𝐵
𝑎𝐸𝐾𝑟
𝜎𝑟 = 𝐴 − 𝑟 2 − 3
𝐵
𝑎𝐸𝐾𝑟
𝜎𝐻 = 𝐴 + 𝑟 2 + 3 − 𝑎𝐸𝐾𝑟
(3)
(4)
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Now in order that the stresses at the centre of the disc, where r = 0, shall not be infinite, B
must be zero and hence B/r2 is zero.
Also 𝜎𝑟 = 0 at r = R
Example 2-7 Continues
Therefore substituting in (3), 0 = 𝐴 −
𝑎𝐸𝐾𝑟
𝑎𝐸𝐾𝑟
𝑎𝑛𝑑
𝐴
=
3
3
Substituting in (3) and (4) and rearranging,
𝑎𝐸𝐾
𝜎𝑟 =
𝑅−𝑟
3
𝑎𝐸𝐾
𝜎𝐻 =
𝑅 − 2𝑟
3
The variation of both stresses with radius is linear and they will both have maximum values at
the center where r = 0.
.
𝑎𝐸𝐾𝑅
12×10−6 ×206.8×109 ×𝐾×0.075
𝜎𝑟𝑚𝑎𝑥 = 𝜎𝐻𝑚𝑎𝑥 = 3 =
3
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Now T = Kr and T must therefore be zero at the centre of the disc where r is zero.
Example 2-7 Continues
Thus, with a known temperature rise of 150°C, it follows that the temperature at the outside
radius must be 150°C. 150 = K × 0.075, therefore K = 2000°/m
12×10−6 ×206.8×109 ×𝐾×0.075
2
i.e. 𝜎𝑟𝑚𝑎𝑥 = 𝜎𝐻𝑚𝑎𝑥 =
=
124
𝑀𝑁/𝑚
3
(b) With the modified form of temperature gradient,
‫ 𝑎 ׬ = 𝑟𝑑𝑟𝑇 ׬‬+ 𝑏𝑟 𝑟𝑑𝑟 = ‫ 𝑟𝑎 ׬‬+ 𝑏𝑟
2
𝑎𝑟 2
𝑏𝑟 3
𝑑𝑟 = 2 + 3
Substituting in (1) and (2),
𝐵
𝐸𝑎 𝑎𝑟 2
𝑏𝑟 3
𝜎𝑟 = 𝐴 − 2 − 2
+
𝑟
𝑟
2
3
𝐵
𝐸𝑎 𝑎𝑟 2
𝑏𝑟 3
𝜎𝐻 = 𝐴 + 𝑟 2 + 𝑟 2 2 + 3
(5)
− 𝛼𝐸𝑇
(6)
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Now
T = a+br
Example 2-7 Continues
Therefore at the inside of the disc where r = 0 and T = 30°C,
30 = a + b(0) and
a = 30
(7)
At the outside of the disc where T = 180°C,
180 = a + b(0.075)
(8)
(8) - (7) 150 = 0.075b
:. b = 2000
Substituting in (5) and (6) and simplifying,
𝐵
𝜎𝑟 = 𝐴 − 2 − 𝐸𝛼 15 + 667𝑟
𝑟
𝐵
𝜎𝐻 = 𝐴 + 𝑟 2 − 𝐸𝛼 15 + 667𝑟
(9)
− 𝐸𝑇𝛼
Now for finite stresses at the centre, B=0
(10)
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Example 2-7 Continues
Also, at r = 0.075,
𝜎𝑟 = 0, and T = 180°C
Therefore substituting in (9),
0 = 𝐴 − 12 × 10−6 × 206.8 × 109 15 + 667 × 0.075
0 = 𝐴 − 12 × 206.8 × 103 × 65
𝐴 = 161.5 × 106
From (9) and (I0) the maximum stresses will again be at the centre where r = 0,
i.e. 𝜎𝑟𝑚𝑎𝑥 = 𝜎𝐻𝑚𝑎𝑥 = 𝐴 − 𝛼𝐸𝑇 = 124 𝑀𝑁/𝑚2 , as before
N.B.
The same answers would be obtained for any linear gradient with a temperature difference of
150°C.
Thus a solution could be obtained with the procedure of part (a) using the form of distribution
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T = Kr with the value of T at the outside taken to be 150°C (the value
at r = 0 being
automatically zero).
Example 2-8
An initially unstressed short steel cylinder, internal radius 0.2 m and external radius 0.3 m, is
subjected to a temperature distribution of the form T = u + b log, r to ensure constant heat flow
through the cylinder walls. With this form of distribution the radial and circumferential
stresses at any radius r , where the temperature is T. are given by
𝐵
𝛼𝐸𝑇
𝜎𝑟 = 𝐴 − 2 −
𝑟
2 1−𝑣
𝐵
𝛼𝐸𝑇
𝐸𝛼𝑏
𝜎𝐻 = 𝐴 − 2 −
−
𝑟
2 1−𝑣
2 1−𝑣
If the temperatures at the inside and outside surfaces are maintained at 200°C and 100°C
respectively, determine the maximum circumferential stress set up in the cylinder walls. For
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steel, E = 207 GN/m2, u = 0.3 and α = 11 x 10-6 per °C.
Example 2-8 Continues
Solution
𝑇 = 𝑎 + 𝑏 log 𝑒 𝑟
200 = 𝑎 + 𝑏 log 𝑒 0.2 = 𝑎 + 𝑏 0.6931 − 2.3026
200 = 𝑎 − 1.6095𝑏
Also
(1)
100 = 𝑎 + 𝑏 log 𝑒 0.3 = 𝑎 + 𝑏 1.0986 − 2.3026
100 = 𝑎 − 1.204𝑏
(2)
(2) – (1)
100=-0.4055b;
Also
𝐸𝛼
207×109 ×11×10−6
6
=
=1.6
x
10
2(1−𝑣)
2(1−0.29)
b = -246.5 = -247
Therefore substituting in the given expression for radial stress, 𝜎𝑟 =
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Example 2-8 Continues
𝐵
0−𝐴 − 𝑟 2 − 1.6 × 106 𝑇
At r = 0.3,
𝜎𝑟 = 0 and T = 100
𝐵
0 = 𝐴 − 0.09 − 1.6 × 106 × 100
(3)
At r = 0.2, 𝜎𝑟 = 0 and T = 200
𝐵
0 = 𝐴 − 0.04 − 1.6 × 106 × 200
(4) – (3)
(4)
0 = 𝐵 11.1 − 25 − 1.6 × 108
B = -11.5 x 106
and from (4), A = 25B + 3.2 x 108 = (-2.88 + 3.2)108 = 0.32 x 108
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Example 2-8 Continues
substituting in the given expression for hoop stress,
6
11.5
×
10
6
6
𝜎𝐻 = 0.32 × 108 −
−
1.6
×
10
𝑇
+
1.6
×
10
× 247
2
𝑟
𝐴𝑡 𝑟 = 0.2,
𝜎𝐻 = 0.32 − 2.88 − 3.2 + 3.96 108 = −180 𝑀𝑁Τ𝑚2
𝐴𝑡 𝑟 = 0.3,
𝜎𝐻 = 0.32 − 1.28 − 1.6 + 3.96 108 = +140 𝑀𝑁Τ𝑚2
The maximum tensile circumferential stress therefore occurs at the outside radius and has a
value of 140 MN/m2. The maximum compressive stress is 180 MN/m2 at the inside radius.
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THANK YOU
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