Demonstration of Lamés formula for the strength of thick hollow cylinders Subjected to uniform normal pressure

329

Zanza--Demonstration of ~Lamd's Eormula.

DEMONSTRATION OF LAM]~'S F O R M U L A F O R T H E STRENGTH OF THICK HOLLOW CYLINDERS SUBJECTED TO UNIFORM NORMAL PRESSURI~.

By GAETANOLANZA, Prof. Mass. Institute of Technology. The widely differing results obtained by the use of the various formulm for the' thickness of the walls of hydrostatic presses (thick hollow cylinders) seem to render it desirable that in the case of a formula which is as reliable as Lured's*, a demonstration should be given which could be read by a person who has not previously studied the theory of elasticity. This I have endeavored to do in the following article, the steps followed being virtually those taken by Lamg, but all matter foreign to the subject in hand being eliminated. In this paper the word strain is used to indicate the elongation or shortening of a body pe~ unit of length : the force producing it being indicated by the word stress. ~ Let OA ---~ R ~ inside radius ; OB ~ .R I ~ out~ side radius; let the length of the portion of the ~ ! cylinder under consideration be unity; let • : ~D ~ intensity of internal normal pressure ; kdI /)~ ~ intensity of external normal pressure. ~ ~ssume any r i n g E F e f w h e r e O E ~ r , ~

~

O~-r

+dr;p=intensityofnormal

stresson

the inside ; p + d T ~ intensity of normal stress on the outside ; t ~ intensity of tangential stress. Suppose that after the pressure is applied the ring E E e f becomes E1 E1 elf1 where 0 E t ~-- r + u ; 0 el ~ r + u + d r + d u (u being a small quantity). I. A s to the strains in the radial and in the tangential direetwn.

In the radial direction the widtl~ .E e ~ d r has become : E 1 ea ~ d r + d u ;. hence the strain in this d i r e c t i o n -

du dr"

In the tangential direction the length 2 r r of the circumference ~7 E has become : 2r, u u 2 ~r(r + u) ; hence strain in the tangentialdirection = 2~r r ~ r

*

"

Le?ons sur Elasticitd des Corps Solides," par M. ,G. Lamd.

330

Civil and Mechanical "Engineering.

II. To determine the relation between p, t, and r eons~tent with equilibrium. Consider the forces acting on the upper half ring _E15"~ elf,; 1st, upward force due to internal pressure : ~ 2 p (r + u ) * ; 2d, downward force due to external pressure: = . o (p + dp) (r + ' ~ + d r + d u ) ; 3d, tangential stress (represented as acting upward) :

2 t (dr + du). (In the ordinary case when there is no external pressure on.th0 cylinder, d p and p have contrary signs, and 2 t ( d r + d u ) is a downward force and consequently a hoop tension); hence for equilibrium wq must have:

2 (p + dp)(r + u + dr + d u ) ~ 2p (r + u ) - - 2 t ( d r + du)=o, or2pr-+-2pu+2p d r + 2 p d u + 2 r d p + 2 u d p + 2 d p dr+ 2dp du--2pr--2pu--2t d r - - 2 t du--~-o. Cancelling 2 p r and 2 p u with - - 2p r and - - 2 p u and omitting those terms that contain infinitesimals of a higher order than the first, viz.: 2p du, 2u dp,'2dp dr, 2 d p du, 2t d u , anddividing by 2, we obtain :

p d r + r d p - - t dr---o. dp + p . T t ~---o as the desired relation. Dividing by d r, we have ~Zr III. It is evident that the stresses p and t must depend upon, and

du

u

therefore be, functions of the strains d~r and --'r' and moreover these

du

functions must be of such a nature as to vanish whenever b o t h e r r and -u-u vanish, since we know that when there are no strains there r are no stresses; hence if we express the stresses p and t in series of ascending powers and products o f ddUr and --r' u the development must not contain any term independent both of du a I l d -u- .

dr

r

The total upward force acting on the ring in oonsequence of the internal normal pressure will be the same as that acting on a section of the cylinder, made by a plane passing through its axis and the diameter El FI • The area of this section will be 2 (r + u ) X 1 ~ 2 (r-}- u), hence the total upward force will be 2 (r + u) ~ p, or 2 p ( r + u).

Lanza--Demonstration du

831

o f L a m g ' s Formula.

~t

Moreover since d-r and --r are themselves small quantities, we may omit all powers higher than the first, and all products of these strains, and use as a sufficiently close appro.ximation only those terms containing their first powers; hence we may write

where p and t are (as has been stated) the intensi~ies of a pair of stresses at right angles to each other at any point of the ring

E1/Pl el fl" Now if the properties of the metal are independent of direction, i. e., if the elasticity is the same in all directions; we ought, by du u interchanging coSrdinates and hence by interchanging drr and r to deduce either of the aboye equations from the other: hence we must havep---~l

(;) + i l l (u) d-r "

T ~ s can only be when

¢1"-~ fl and fll ~ e~; hence we have :

where ~ and ¢~, in the ease of a homogeneous body whose elasticity is constant in all d~rections, are constant quantities depending on the properties of the magerial. From all the preceding we have the following three equations : dp

p--t --

d r -} r

o

(1)

du+ ~ u P~CZ

t=

fl ~

dr

(~)

r

+ ,~

('~)

. d2u~;

Differentiate (2)andwe have

-----~-d~ 'z+ -

dup~

dr

r~

Subtract (3) from (2) and we have :

dp

Substitute these values of ~ d ~u

fl d u

~ -d-~r~ ~ r

dr

flu

and p - - t in (1) and we have : ~ d u

r~ + r

dr

~ u

fl d u

~

- - - r d--r + ~

j~ u ~

o.

332

Civil and Mechanical Engineering.

fl du

flu

Cancelling - d-r a n d -

~- w i t h - - r

f du

flu

dr-- and - ~

respectively,

and dividing by e~ we obtain : d2 zt I d u dr 2 + r dr

u r2 - - o .

(r

d~u',~

By transposition we have : 1 du r dr

but

Jdr2

u

dr

r

d r ; hence ~rr~ . . . . .

dr~--

du

u

have d r r - ~ - - - r -

d~

+

)

r2 ;

u u r2 is the differential coefficient o f -

to r, or to

.'.

1 du

dr 2=-

with regard

dr

d r ; orperformingtheintegrationwe

2 a (2 a being a constant whose value is

not yet determined). Multiplying by r and transposing we obtain : du r~r +u=2ar; du

but r -dr ~- u is the differential coefficient of r u with regard to r or d(ru)

to - - ~

; hence we have :

d(-ru--)--2ar; dr -dr

multiplying by d r and integrating

dr=2a

r dr, orru=ar

2~-b

(b being an-

b u = a r ~- - - , r and from this equation we obtain directly the two strains, viz.: other undetermined c o n s t a n t ) . . ' , du dr =a--

b

~

u

ands- ~a

b

+r2.

Substituting these in the values of p and t we have: 5

,:~(a--~) -~,~(a -{-b):a(~-]-f)--~i-(~--P) b

Lanza--JDemonstration o)~ Zamd's Formula.

838

:Next we proceed to determine the values of a and b, and to do this we must observe that when r ~ / ~ , or when the thin ring is taken as the ring whose radius is 0 A, we have p = - - P (the minus sign being used because it is a compressive force), and when r = RI, we have

p=--P1. Making these substitutions we obtain: b two equations of the first degree from which to determine a and b; any method of elimination will give us:

1

PR2---__PI.Rle~

a = ~-_{_ ~..~(

/i~12__ /~ 2

1

((P--P,)R2R,2~

,], and/3 = t0~ _ _ ~ \

./1~12-- ~ $ -

,].

Substituting these values of a and b in the last value of t we have. PR2--PI"R12 1 ( . P - - -P1) R2 R12 t---- R~~ _ R 2 -+- ~ R1~ _ R 2 , which is the intensity of the hoop tension in the ring ~'~ $'1 e~f~. The greatest value of t, or the greatea~hoop tension evidently occurs in the inside ring, or when r ~ R, when the maximum value of t becomes: /)1~2-TP1R, 2 ( P - - P1) R, 2 P R~-- P~ R~Z + P R,Z-- P~ R,~ --

R12_~11~

2

-~-

/~12_R

2

--

R12_R

2

,

and the greatest intensity of hoop tension :

P (R ~ + R~~)- - 2 P1 R~~ R l ~ - - R2 If the intensity o f ' t h e working strength of the material be k. we must therefore have, ~n order to ensure safety : --

k=

P (R ~ + B1~ ) - 2 P1 R~~. .R12 - - R z

or, clearing fractions, we have:

k R1~ - k R

2=PR 2 + PR, 2-2P1Rl~;

transposing and factoring: R, ~ (It - - P "+ 2 -Pl) = R ~ (7c + . P ) ; dividing by k ~ P + 2 -P1 we have : k+P Rl~ - - k - - P + 2 P~ R2"

/

k+p

Extracting the square root R ~ = R ~/]c Z I 9 + ~ - p ' which determines the external radius when the internal radius, the outside and inside pressures, and the working strength of the material are known. In the more common ease when there is no external pressure, as in the

Civil and Mechanical .Engineering.

834

case of the hydrostatic press, when the pressure of the external air is left out of account, P1 is to be made equal to zero, and the above formulm become : Intensity of maximum hoop tension:

+ Pp "

= P (R2 + R12), and RI =

Rl ~ - - R 2 Hence, if in the case of a thick hollo~'~ylinder, we have P ~ intensity of internal normal pressure ; -/)1~ intensity of external normal pressure ; R = inside radius of cylinder ; R1 ~ outside radius of cylinder, and/c intensity of working strength of the material ; or, which is the same thing, ~--- intensity of maximum hoop tension consist.ent Wi~h safety, we shall have: 1st, Suppose P, P , k and R are known, and R~ required : R~ ~--R ~

k-]-P

k - - P + 2 P~

/ ~ = / ~ ~/ k + P

when there is an external pressure,

when tLere is no external pressure.

2d, Suppose with the same data that the thickness of the walls be required, we shall have : Thickness when there is an externa,1 pressure :

---~R~--R~-R{,/

le+P

1 ~

k - - 1 ) + 2 P, Thickness when there is no external pressure:

J

= R , -- R = II{ ~,~k~pik + P - - 1 } 3d, Suppose R, R~ and h' to be known~ and it be required to determine the greatest internal normal pressure that can be safely used, and we shall have : R12~ R 2 Intensity of greatest pressure ~ }9 ~ / c R, 2 + R:" 4th, I f R, R . P and PI be given, a,nd it be required to determine the actual intensity of the greatest hoop tension, we shall have : Intensity of greates~ hoop tension, when there is an external pressure:

1) (R,2-',-li z) -- 2 P, R, = -

n ; , - : n

-

,

Intensity of greatest hoop tension when there is no external pressure : 2 R,-i A) 2

RI;--R.[