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STRESSES IN FLAT PLATES DUE TO BENDING
CHAPTER XXXI
STRESSES IN FLAT PLATES DUE TO BENDING 336. Statement of the Problem and Assumptions.-When a flat plate of material is supported at its boundary and loaded by forces applied at right angles to its surface, flexure takes place just as it does in the case of a beam. But the flexure of a flat plate presents a more difficult problem than that ofa beam since the curvature is not confined to taking place in parallel planes. The general theory of the bending of Hat plates is difficult, and we shall here consider only an approximate theory analogous to the simple theory of flexure in beams. The cases which are covered by this theory include most of those which are of practical interest, and it will appear that even this theory involves us in sufficiently laborious algebra. We assume that the plate is of uniform thickness and material, that the deflections of the middle surface are small * compared with the thickness of the plate. The plane midway between the faces of the plate is referred to as the middle-plane, or, when deformed, the middlesurface, of the plate; we assume that this surface is unextended. We also assume that elements of the plate originally straight and perpendicular to the middle-plane remain straight and become perpendicular to the middle surface when strained. We neglect normal stresses across planes parallel to the middle-surface. The simplest case is that of a circular plate loaded by forces which are uniformly disposed round the axis, so that stresses and strains are the same for all radii. We shall therefore deal with this first. SYMMETRICALLY LOADED CIRCULAR PLATE
337. General Equations.-In Fig. 373 let OA be the axis of the plate, and BO a section of the strained middle surface by any meridian plane. Take the origin 0 at the point where the axis of the plate cuts the middle plane of the unstrained plate. Let P and Q be points on the middle surface, at distances rand r + ~r from the axis. Then the normals at P and Q intersect the axis OA at points which ultimately coincide and become the centre of curvature • The flexure of plates when the deflections are not small compared with the thickness is treated by J. Prescott in Phil. Mag., Jan., 1922, or see the same writer's book, Applied Ela8ticity. For an account of some experimental work, see Enginu"""" 1927(i).
489
STRENGTH OF MATERIALS
490
+
of the middle-surface atP, when 8r+0. Let ~ and ~ 8~ be theinclination of these normals to the axis OA. ,' , Let w == the deflection of P from its initial position. ! ! Let M be a point on the normal ~rp;"-~ through P and let PM == z, z being positive in the same direction as w. :! 0 :: ---..-Let P, and Pe == the radial and circumferential direct stresses at M. Let e, and ee be the corresponding strains. We neglect shearing stresses in radial planes and direct stresses FIG. 373. across the middle surface. Before strain the distance of M from the axis was r; after strain it becomes r zqJ. Hence the circumferential strain is given by ,
I
" 1 I
,
-II
I
"
I
+
_ 2n(r
ee - -
+ zqJ)
2nr _
-
zqJ
- -
2nr
r
·
(i)
If N be the point on the normal through Q corresponding with M on the normal through P, Le. QN == z, the length of N M changes from ~r to ~r + z~qJ. Hence the radial strain is given by z~qJ
e == Lt. -
~r
r
Also, since
qJ
dqJ
== z-
. (ii)
dr
is a small angle, we can ,vrite ffJ
dw dr
== - - .
Hence the strains become eo
= -
-=-r . dw and e dr
r
= - zd2w. 2 dr
. (iii)
The stress-strain equations are P, Pe d 2w - - - ==e = - z - E mE r dr 2 Pe
Pr
_
_
E - mE - ee - -
z dw -;: dr·
Solving these for P, and Pe we get, mEz
(1 dw + mddr22W)
P r = - m 2 _ I ;: dr
mEz (m dw
Pe = - m2 - 1 -;: dr
+ ddr2W) 2
(I) .
(2)
We must now consider the stress-resultants and stress-couples, that
BENDING STRESSES IN FLAT PLATES
491
is the resultant forces and couples on an element of the plate due to the stresses. Consider the element ABGD (Fig. 374), the dimensions of which are shown in the figure. We shall first find the resultant force and couple acting on the element due to the stresses Pe. Let t'= the thickness of the plate. Let LKM be the intersection of the middle-surface with the element; let O'N be the radius bisecting KM, and NT the tangent at N to the arc MK. Let QR8 be an elementary layer of the plate, at a distance z from the middle-surface, and of thickness <5z. Let P be the middle point of QR, and U the middle point of R8. We shall take the radial and circumferential stresses separately and L·
(F+8F)(r +8r)69
" ;~'AI'~ r~l
~~ :.-:_-~:~ ~: -:. :.o>
X
~f'2
2
M;+8M
(
GD
IFrOfJ
Side Elevation
FIG.
374.
estimate their total action on the element ABO . .. H. Then, by considering the equilibrium of the element as a whole, we shall deduce an expression for the sheaIjng force over the face ADEF, and finally form a differential equation for finding the displacement. Consider first the effect of the circumferential stresses Pe. The total action over the strip QR is a force Pe • <5r. <5z acting through P at right angles to QR and parallel to the middle surface of the element. This force can be resolved into components p' and p", respectively perpendicular and parallel to the radius through U, as shown more clearly in the small plan view. The stress on the face ODEH will give equal components at P' as shown. The components p' on the two faces cancel so that there is no resultant force perpendicular to the radius. Again, we see from (2) that Pe, and therefore p", is proportional to z, so that p" changes sign with z and the integral of p" oyer each of the faces ABGF and DOBE will vanish.' Hence the hoop stresses Pe produce
STRENGTH OF MATERIALS
492
no resultant radial force on the element. a couple: we have
p"
= p~r.~z
. sin M 2
They do, however, produce
= !p~r.~z.M},
since ~O is a small angle. The moment of this force about the line NT is tPe.c5r.c5z.c50·.z, and each face AG and DE will contribute a couple of moment
t
r'l t
oZ •
~r
· dz · MJ,
-"2 Let M 1 == the resultant couple about NT due to the hoop stresses, then, using (2), ,ve have
M1
= -
M1
or
mE (rlt -.- dw m'l. - I r' dr
== _
.1nEt
12(m 2
3
1)
-
+ d-
c 2W) f2 c5r. c50
dr 2
z2dz
-2
+ d2W)~r. 2
(m. dw r dr
t
dr
~O
'
•
(3)
This is a couple tending to rotate the element about a line parallel to NT, and represents the whole action of the hoop stresses. Now consider the radial stresses on the face ADEF. The force on the strip R8 is Pr • rb() . c5z, acting along the radius through U. From (1) we see that P r changes sign with z, so that the resultant force over the ,vhole face ADEF is zero. The resultant force parallel to NT is also zero, but there is a couple about NT, the moment of which is t
·dz · z.
f~lrr~O
Denoting this by M 2 we have from (1) M2
or
d2W dw + m 2 )rc50. 1 r dr dr
== - ~(~ 2 m
M2
=-
-
3
mEt (~~W 12(m 2 - 1) r dr
t
J2 tz2 . dz -"2
+ md W)rlJ() 2
(4)
dr 2
This represents the whole action of the radial stresses on the face ADEF. Similarly the radial stresses on the face BORG produce a couple M2
2c5r , + lJM == M + dM dr 2
about a line parallel to NT. 3
2
From (4) we have
mEt (d- 2W lJM 2 == - -----2 12(m - 1) dr 2
2
d w + mrd3w) ~(). + mdr drS 2
lJr
.
(4A)
BENDING STRESSES IN FLAT PLATES
493
Now let us consider the equilibrium of the whole element. Let F == the shearing force on the face ADEF, per unit length of arc, as shown in Fig. 374. Then the total shearing force on ADEF is F . r~O, and that on the face BORG is (F + ~F)(r + ~r)~O. Let p == the pressure per unit area on the face ABOD, where p may be a function of r but not of O. Then taking moments about the line NT we have
(M 2
+ dM + (F + dF)(r + dr)dO . dr 2)
This reduces to
+ FrdO . dr -
~M2
Substituting for Fr~O
. dr == -
mEt 12(m 2
-
(m - -dw I) r dr
dr. dO
3
(d 2W.
2 -
F
=
M 1 == O.
M 1 == O.
+ d~W) dr~
+ 12(mmEt
Hence
-
from (4A) and M 1 from (3), we get
~M2 3
M2
m 2Et 3
I)
3
2
d w + rnrd W) dr. dO -dr + 1n-dr dr 3
2
2
(d 3W
1 d 2w 1 dW) 12(m 2 _ 1) dr +;: dr 2 -;:2 dr
(5)
Again, resolving vert.ically, we have p . r . ~O . dr - (F
+ ~F)(r
+ ~r)~O + FrdO == 0
which reduces to
pr . ~O . ~r - rdF . ~O -
. ~O == O.
F~r
Hence, in the limit, after dividing by rdr. dO, we get
dF dr
F
+;:- == p.
Substituting for F from (5) ,ve get
m 2Et 3 12(m 2 -
I)
(d~W
dr 4
2 d 3w
+;: dr3
1 d 2w
-;:2 dr 2
1 dW)
+;:a. dr = P
·
(6)
Let us, for brevity, write m 2Et 3 - .. -------. 12(m 2 - 1)
==
D.
(7 )
Then we can put (6) in the form
1;: dT rdr {I;: dr rdrdW)} ] -_ d[
d
d(
p
jj
(8).
This is the general differential equation for the deflection of a circular plate when the load is distributed symmetrically about the axis. When w has been found the radial and hoop stresses are given by (1) and (2), and the shearing force by (5).
STRENGTH OF MATERIALS
494
338. General Solution when the Load is Uniform.-When the
load is uniformly distributed, p is constant and we have from (8)
. • · · .
·
{I
d d ( dU')} rdr ;: dr rdr
!..~(rdW) ~(rdW)
dr
dr
=
2
+A
I
r
2D
== L- r 2 + Al log r + A 2
dr
4D
= L- r 3 + AIr log r + A 2 r
dr
dw r dr
pr
== 2D
D
== P- r + Al
~{~ ~(rdW)}
dr r dr r dr
=?-.r
~[r ~{~ ~(r~~)}J
4D
4
pr 16D
2
A Ir ( 1 +'2 og r -
] '~d
+"21 A
= 1:Dr3 + tA1r(1og r - !)
:~
=
:. W
pr 4 64D
+
A Ir2 4 (log r - I)
2
r2
+
4 .4
a
+ !A r +:t; .
+ lA
(9)
2
2
r2
+ Aalog r + A
4
(10)
There are four constants of integration to be determined, AI' A 2 , A aJ and A 4 • If the plate is continuous to the centre Al and A a must be zero since W does not become infinite at the centre. It will be convenient here to obtain general expressions for the stresscouple M 2 and the shearing force F. From (9) we have
d 2w dr 2
=
3pr 2 16D
d 3w _ 3pr dr 3 - 8D
+ !A
I
(log r
+ i) + i A 2 -
Aa
-;:2
+ A + 2A a I
7
2r
3
2
dw.In (4 ) an d (-0), we fi n, d a fter sIm. utIng f or dw - , d- w an d SU bS t 1Ot' dr dr 2 dr 3 plification,
M~ = _ (3m + 1) . .p r
r . ~()
m
2
16
_
!![A1{(m m
2
+ 1) log r + t(m -
A + 2"(m + I) 2
F ==pr
2
+ AID r
where D is given by (7).
In
A a - I) ] ;:2(m
(11) .
(12)
BENDING STRESSES IN FLAT PLATES Similarly from (I) and (2), we have when z p,
== _ 3{3m
+
l)pr 8mt 2
2
mEt
_
2{m 2
1)
-
== ~
2
+ 1) 10' g r + "fl(m -
[A 1 {(m 2
+ A~m + 1) -
3 A ;:2(m - 1)
2
Pe
= -
+
3(m 3)pr 8mt 2
2
mEt 2{m 2 - I)
_
[A2
1{(m
495
+ 1) 10
r -
J
l.(m -
g·2
+ A2 (m + 1) + rA (m 3
2
2
I)}
I)J
(13)
In (14)
These expressions give the maximum values of the radial and hoop tensile stresses at any radUUB. The constants of integration can be obtained in various cases from the following condUtions : If an edge is freely supported, wand M 2 vanish at that edge.
H an edge is encastre or clamped, wand dw vanish at that edge.
dr At a free edge, M 2 and F must vanish. If the load has dUfferent uniform values PI' P2 ... , over different regions of the plate,M2 and w must have the same values for the PI region and the P2 region where they meet. If only one edge is supported the value of F round that edge is determined by the total load on the plate, and thus Al is given by (12). We shall now consider some particular cases. 339. Solid Circular Plate, Uniformly Loaded over the Whole Area: Ed~e Freely Supported.-Let a be the radius of the plate. Since it is continuous up to the centre we must have Al == A a == o. When r == a we must have M 2 == 0 and w using (II), gives _ (3m
+ 1)pa
m
whence
A2
2
_ DA 2 (m 162m
== o.
+
The first condition,
1) =0
+ l)pa -' == - (3m ----_.._.2
8(m
+ I)D
== 0 when r == a gives, from (IO), A == _ pa4 + (3m + l)pa4 == (5m + I)pat .
The condition w &
64D
32{m
+ l)D
64{m
+ l)D
Inserting the values of the constants in (IO), we get after some simplification,
w = 3p(m
2
1)(a2 16m 2Et 3
_
r 2 )(5m
m
+ la +1
2._
r 2)
•
. (-I 5)
STRENGTH OF MATERIALS
496
The stresses are then given by (13) and (14) : -
+ l)p(a 2 _
== 3(3n~
P
r 2)
81nt 2
r
Po == -3p 2 {(3m 8mt
+ l)a
•
(m
2 -
+ 3)r
2}
These both have their greatest value when r == 0, giving 3(3m + 1)pa 2 P r (max.) == Po (max.) == - - - - - 8mt 2 If we take the strain energy as the criterion of failure and elastic limit of the material in pure tension we must have 2 p r 2 + Po 2 - m -P:Po ==12 r where P r and Po have the values given by (18). 9(3m
+ 1)2(m -
1) 2 4 -12 P a -
32m 3t4
f
.
(16)
.
(17)
.
(18)
be the
This gives
From this we find that the pressure required to cause elastic failure is given by
1
=
3(3~m+T)
-vi
2m
== 0
and -
m
1·
~:.
(19)
340. Solid Circular Plate, Uniformly Loaded over the Whole Area: Edge Clamped.-As before we must have Al == A a == O. When r == a we must also have w
dw dr
dition gives, from (9),
== o. The latter con-
pa 2
The condition
1.0
A 2 === - --. 8D == 0 when r == a gives, from (10), pa 4 pa 4 pa4 A. == - 64D + 32D == 64D· pr 4 pa 2 r 2 w == 64D - 32D
Hence we find
== _~(a2
64D or, inserting the value of D, w
pa1:
+ 64D
r 2)2
-
__1)p (a 2 16m 2Et 3
== ~(11L-=-_-=
_
r2) 2.
From (13) and (14) the maximum stresses at radius r are found to be Pr = Po
3p {(m 8mt 2
= . 3p 2 {(m 8mt
+ 1)a 2 -
(3m
+ 1)a
(m
2
-
+ 1)r 2}
+ 3)r
2}
•
•
(20)
•
(21)
BENDING STRESSES IN FLAT PLATES
497
As in the case of the plate with the edge freely supported, these stresses are greatest at the centre, and there 3(m + 1)pa 2 Pr (max.) = Ps (max.) = 8mt 2 (22) • • Comparing (18) with (22) it will be seen that the maximum stress when the edge is clamped is only about 0·4 times the maximum stress when the edge is free, if m be 10/3. 341. Annular Ring Freely Supported at the Outer Edge and Loaded Uniformly Round the Inner Edge (Fig. 375 shows a diametral section).-Let the radii of the outer and inner boundaries be a and b. Let there be a total load W distributed uniformly round the inner edge of the plate, the surface of the plate being free from load.
375.
FIG.
We must have M s putting p = 0,
A
and
=0
when r
=
a and r = b, hence, from (11),
A + ~(m + I) 2
-.2{(m
+ I) log a + !(m -
I)}
A1
. + 1) log b + l(m -
In + A2~(m
2
2 {(m
.
A == ~(m2
+ 1) =
a
- 1)
A
ii
Hence, by subtraction, (m -
1)(-!. - b~)A3 == A2 (m + I) log~ b a A m +1 a b A I a m _ 1 " 2(a2 _ b2) og r;" 1
2
2
2 2
.
••
3 -
-
1
(
"a)
Also, multiplying the first of the above equations by a 2 , the second by b2, and subtracting, we get (a 2
-
b 2 )(m
+ I)A s =
+bA 2
Hence
_ [b
1
{(m
-
a 2 A 1 {(m
+ 1) log a + !(m -
+ 1) log b + l(m -
b-
I)}
I)}
log a 2 log a m - 1 ] L4 1 • • (P) a 2 -b 2 2(m+l) The shearing force per unit length of outer edge is W/2'Jla, hence from (12) we have, since p = 0, W A2
-
2
A 1 =-2nD
STRENGTH OF MATERIALS
498
Thus the constants A H A a and A a are determined; A 4 is found from the condition w = 0 when r = a. Omitting the details of the algebra, we finally get * w
W
[1 {
3m
+ 1+ a2 a_2 b2 1og ab + 1og7Jr} r
== 2nD 4" - 2(m + 1)
+~
+
2 2
m 1. a b log~ . log ~+~~ 2 m - 1 a 2 - b2 b r 4 a 2 - b2
log~
b
2
+ (3m + l)o,2J(23) 8(m + 1)
From this we can calculate the stresses by means of (13) and (14).
342. Solid Plate Uniformly Loaded Round a Circle: Edge Freely Supported (see Fig. 376).-The conditions of this problem are
~a~
~ ~b~ FIG.
376.
the same as the last except that the plate is continuous to the centre. A total load W is uniformly distributed round the circumference of a circle of radius b, the radius of the plate being a. We must divide the plate into two regions: a> r> {) and b> r> O. For the first region we must have M 2 =0, W =0, when r =a. F == W /2na, when r = a. For the region r b the constants Al and A a in (10) disappear, dw and F = 0 for all values of r. When r = b, M a and wand - must dr have the same values whether calculated for the first or second region. Let us write r b, w == :i-A2r2 + A 4 (i) A' r> b, w' == _ l r 2 (log r - 1) + !A'2r2 + A'alog r + A ' 4 (ii) 4 Then: Wi == 0, r == a gives A' . a2 (log a-I) + !A'2a2 A'alog a + A ' 4 == 0 . (iii)
<
<
-i
+
M 2 == 0, r == a, gives A' -t{(m 1) log a !(m - I)}
+
F
+
= ~ when 2na
r
=
+ A' -t(m + 1) -
a gives
A'I=~ *
A' a 23(m - 1) = 0 (iv)
2nD
Kelvin and Tait, Natural Philo8ophy, Pt. II, p. 198.
.
(v)
BENDING STRESSES IN FLAT PLATES
499
w' = w when r = b gives A'lb 2 (log b - 1)
+ A'2b 2 + A'alog b + A', = A 2b2 + A,.
4 4 4
. (vi)
The condition t.hat M 2 calculated from (ii) == M 2 calculated from (i), when r == b, gives A' A' A' A2 y{(m+l) log b+!(m-l)}+-i(m+l)---- b23(m-l)== 2 (m+l) (vii) dw'
dw
== b
-d == ---, when r r dr
.
gIves
-A'lb (1 og b _ 2·1) 2
+ A'2b + A'a _ A - - -2b b
2
(viii)
2
The equations (iii)-(viii) determine the six constants of integration. Considerations of space prevent us from giving the details of the algebra; the steps for finding the constants are as follows: from (vii) and (viii) find A 'a in terms of A' l' then (iv), (iii), (viii), (vi), give A '2' A'«, A 2 and A«, in that order, in terms of A'lJ which is known from (v). Finally we get 2 2 (3m+l)a -(m-l)b (a 2_r 2)] (24)
w= W [-(r2+b2) 10 ~+(r2-b2)+ 8nD g b w':;:= W [_ (r2
8nD
2(1111
+ b2) 10 !!.. + (3m + l)a 2 g r
2(m
+ l)a 2
)J
(m - l)b\a 2 - r 2
+ l)a 2
(25)
343. Solid Plate with Load Concentrated at the Centre.*-If the load could be truly concentrated at a point at the centre of the plate the stresses at that point would be infinitely great. In practice the load will either be distributed round a very small circle or over the areal of a very small circle. In the former case the above analysis will apply, and if we take the expression for w', rna.king b infinitesimal we shall get an expression for the deflection except near the point of application of the load. Doing this we find, with the edge freely supported, w
== -W [ - r 2 log -a + J. 3m 8nD
t:
+ 1(a
m+l
2
2 -
r 2)
J
. (26)
Froln this we have, by substitution in (1) and (2),
3JV( 1+ -1) log-a
p =T
Po ===
2nt2
m
. (27)
r
3lV [(1 +!) log!!... + (1 _-!.)J
2nt 2
m
r
m
(28)
at any point on the surface outside the region of application of the load. • For circular plates of varying thickness see an article by G. D. Birkhoff, Phil. Mag., May, 1922. See also Prescott's Applied Elasticity. For effect of shear stress OD deflection, see Engineering, Vol. 123 (1927), p.. 343 and Vol. 125 (1923), p. 31.
STRENGTH OF MATERIALS
500
344. Rectangular Plate. Supported at the Edges.*-B. C. Laws
(loc. cit. infra) gives the following formulre for the maximum str~8_and deflection·: the greatest principal stress occurs at the middle points of the, longer sides and is given by a2 pb 2 . (29) 2 2 2a + 6b • t2
whilst the maximum deflection is given by . a'
a'
pb'
+ 2b' · 32Et
3
. (30)
•
where t is the thickness of the plate.
EXAMPLES XXXI 1. In the case considered in § 342, if the edge of the plate be clamped, show that W [ a (at + bt)(a ' - r l ) ] r < b, w = 81CD - (r 2 + b 2 ) log b + r 2 - b 2 + 2a 2 .
r > b, w
=
W [
81CD -
(r t
a
+ b 2 ) log r +
(at
+ b 2 )(a t 2a t
-.r l ) ]
Hence show that the deflection due to a concentrated load at the centre is given by
w =
S:D[-
r1log;
+ !(a l -
rl)J.
2. Show that the displacement of a circular plate of radius a, clamped at the edge, and carrying a load W at a point 0 distant c from the centre is given by t
w
=-.!V [_R21ogCRI + 1(~R'as.s-R2)] 81CD aR
where R denotes the distance of any point in the plate from the point 0, and R' is the distance of the same point from 0' the inverse of 0 with respect to the circle. 3. A circular plate of radius a is freely supported at the edge and carries a total load W uniformly distributed over the area of a concentric circle of radius b. • The formu1~ given here are not exact, although the complete solution has been worked out; the solution for a rectangular plate with clamped edges has not been worked out, but an approximate method of solution has been given by Ritz, J. J. Math. (GreUe), Bd. 135, 1909. See also Edinburgh Proc. Royal Soc., 1912, for the application to the case of a square by C. G. Knott. See also papers in Trans. [nst. Naval Architects: A. M. Robb, 1921 (or Engineering, April, 1921); G. H. Bryan in Vol. XXXV; I. G. Boobnoft in Vol. XLIV: J. Montgomerie in Vol. LXI; T. B. Abell, 1923; aJao E1Ifineering, June 7, 1929. Also paper by Hankey: Der Spannung8ZU8tand in rechteclcigen platten, R. Oldenburg, Berlin, 1913; and B. C. Laws, Proc. [nat. G.E., 1921-2, I. t J. H. Michell, Proc. London Math. Soc., 1902. See also Love's Theory oj Ela8ticity, 3rd Ed., p. 495.
BENDING STRESSES IN FLAT PLATES
501
Show that 'r
> b,
_
W [(
w - 8nD r< b, W
=
r
I
W [( l 8nD r
+
!!:..) I ..!:... 2(3m + 1)01 -- (m -I)b ' _ 2 og a + 4(m + I) .
bl)
+ 2"
+
log
ab + r'-b' ~ -
2(3m
+
l)a
l
4(m
-
4rna ' -(m-l)bl 4(m
+
I)a l
(3m + l)r ' ] 2(m + I)
(r
t
-
bl)
+ l)b ' ]
(7m
+ I)
Hence show that, at the centre, the maximum stresses are given by
pr
= P6 =
3(m+I)W[ m 2nmt t m +1
a m - I bl] m + l' 4a I •
+ log b -
4. If the plate in question (3) be clamped at the edge show that the stresses ou the surface, at the centre, are given by'
pr
= P6 =
whilst at the circumference p,
3(1n + I)W( a 2nmtl 10gb
=
3W (
2nt t 1 -
+
hi) 4a 1
;
l
b ) 2a 2
•
5. The end of a gas-engine piston may be regarded as a diaphragm 12* diameter and iN thick clamped at its edge. If the piston is subjecte d to a pressure of 400 Ibs./in. 1 determine the intensity of greatest radial and hoop stress. (Mech. Sc. Trip., B., 1915.)
STRESSES IN FLAT PLATES DUE TO BENDING 336. Statement of the Problem and Assumptions.-When a flat plate of material is supported at its boundary and loaded by forces applied at right angles to its surface, flexure takes place just as it does in the case of a beam. But the flexure of a flat plate presents a more difficult problem than that ofa beam since the curvature is not confined to taking place in parallel planes. The general theory of the bending of Hat plates is difficult, and we shall here consider only an approximate theory analogous to the simple theory of flexure in beams. The cases which are covered by this theory include most of those which are of practical interest, and it will appear that even this theory involves us in sufficiently laborious algebra. We assume that the plate is of uniform thickness and material, that the deflections of the middle surface are small * compared with the thickness of the plate. The plane midway between the faces of the plate is referred to as the middle-plane, or, when deformed, the middlesurface, of the plate; we assume that this surface is unextended. We also assume that elements of the plate originally straight and perpendicular to the middle-plane remain straight and become perpendicular to the middle surface when strained. We neglect normal stresses across planes parallel to the middle-surface. The simplest case is that of a circular plate loaded by forces which are uniformly disposed round the axis, so that stresses and strains are the same for all radii. We shall therefore deal with this first. SYMMETRICALLY LOADED CIRCULAR PLATE
337. General Equations.-In Fig. 373 let OA be the axis of the plate, and BO a section of the strained middle surface by any meridian plane. Take the origin 0 at the point where the axis of the plate cuts the middle plane of the unstrained plate. Let P and Q be points on the middle surface, at distances rand r + ~r from the axis. Then the normals at P and Q intersect the axis OA at points which ultimately coincide and become the centre of curvature • The flexure of plates when the deflections are not small compared with the thickness is treated by J. Prescott in Phil. Mag., Jan., 1922, or see the same writer's book, Applied Ela8ticity. For an account of some experimental work, see Enginu"""" 1927(i).
489
STRENGTH OF MATERIALS
490
+
of the middle-surface atP, when 8r+0. Let ~ and ~ 8~ be theinclination of these normals to the axis OA. ,' , Let w == the deflection of P from its initial position. ! ! Let M be a point on the normal ~rp;"-~ through P and let PM == z, z being positive in the same direction as w. :! 0 :: ---..-Let P, and Pe == the radial and circumferential direct stresses at M. Let e, and ee be the corresponding strains. We neglect shearing stresses in radial planes and direct stresses FIG. 373. across the middle surface. Before strain the distance of M from the axis was r; after strain it becomes r zqJ. Hence the circumferential strain is given by ,
I
" 1 I
,
-II
I
"
I
+
_ 2n(r
ee - -
+ zqJ)
2nr _
-
zqJ
- -
2nr
r
·
(i)
If N be the point on the normal through Q corresponding with M on the normal through P, Le. QN == z, the length of N M changes from ~r to ~r + z~qJ. Hence the radial strain is given by z~qJ
e == Lt. -
~r
r
Also, since
qJ
dqJ
== z-
. (ii)
dr
is a small angle, we can ,vrite ffJ
dw dr
== - - .
Hence the strains become eo
= -
-=-r . dw and e dr
r
= - zd2w. 2 dr
. (iii)
The stress-strain equations are P, Pe d 2w - - - ==e = - z - E mE r dr 2 Pe
Pr
_
_
E - mE - ee - -
z dw -;: dr·
Solving these for P, and Pe we get, mEz
(1 dw + mddr22W)
P r = - m 2 _ I ;: dr
mEz (m dw
Pe = - m2 - 1 -;: dr
+ ddr2W) 2
(I) .
(2)
We must now consider the stress-resultants and stress-couples, that
BENDING STRESSES IN FLAT PLATES
491
is the resultant forces and couples on an element of the plate due to the stresses. Consider the element ABGD (Fig. 374), the dimensions of which are shown in the figure. We shall first find the resultant force and couple acting on the element due to the stresses Pe. Let t'= the thickness of the plate. Let LKM be the intersection of the middle-surface with the element; let O'N be the radius bisecting KM, and NT the tangent at N to the arc MK. Let QR8 be an elementary layer of the plate, at a distance z from the middle-surface, and of thickness <5z. Let P be the middle point of QR, and U the middle point of R8. We shall take the radial and circumferential stresses separately and L·
(F+8F)(r +8r)69
" ;~'AI'~ r~l
~~ :.-:_-~:~ ~: -:. :.o>
X
~f'2
2
M;+8M
(
GD
IFrOfJ
Side Elevation
FIG.
374.
estimate their total action on the element ABO . .. H. Then, by considering the equilibrium of the element as a whole, we shall deduce an expression for the sheaIjng force over the face ADEF, and finally form a differential equation for finding the displacement. Consider first the effect of the circumferential stresses Pe. The total action over the strip QR is a force Pe • <5r. <5z acting through P at right angles to QR and parallel to the middle surface of the element. This force can be resolved into components p' and p", respectively perpendicular and parallel to the radius through U, as shown more clearly in the small plan view. The stress on the face ODEH will give equal components at P' as shown. The components p' on the two faces cancel so that there is no resultant force perpendicular to the radius. Again, we see from (2) that Pe, and therefore p", is proportional to z, so that p" changes sign with z and the integral of p" oyer each of the faces ABGF and DOBE will vanish.' Hence the hoop stresses Pe produce
STRENGTH OF MATERIALS
492
no resultant radial force on the element. a couple: we have
p"
= p~r.~z
. sin M 2
They do, however, produce
= !p~r.~z.M},
since ~O is a small angle. The moment of this force about the line NT is tPe.c5r.c5z.c50·.z, and each face AG and DE will contribute a couple of moment
t
r'l t
oZ •
~r
· dz · MJ,
-"2 Let M 1 == the resultant couple about NT due to the hoop stresses, then, using (2), ,ve have
M1
= -
M1
or
mE (rlt -.- dw m'l. - I r' dr
== _
.1nEt
12(m 2
3
1)
-
+ d-
c 2W) f2 c5r. c50
dr 2
z2dz
-2
+ d2W)~r. 2
(m. dw r dr
t
dr
~O
'
•
(3)
This is a couple tending to rotate the element about a line parallel to NT, and represents the whole action of the hoop stresses. Now consider the radial stresses on the face ADEF. The force on the strip R8 is Pr • rb() . c5z, acting along the radius through U. From (1) we see that P r changes sign with z, so that the resultant force over the ,vhole face ADEF is zero. The resultant force parallel to NT is also zero, but there is a couple about NT, the moment of which is t
·dz · z.
f~lrr~O
Denoting this by M 2 we have from (1) M2
or
d2W dw + m 2 )rc50. 1 r dr dr
== - ~(~ 2 m
M2
=-
-
3
mEt (~~W 12(m 2 - 1) r dr
t
J2 tz2 . dz -"2
+ md W)rlJ() 2
(4)
dr 2
This represents the whole action of the radial stresses on the face ADEF. Similarly the radial stresses on the face BORG produce a couple M2
2c5r , + lJM == M + dM dr 2
about a line parallel to NT. 3
2
From (4) we have
mEt (d- 2W lJM 2 == - -----2 12(m - 1) dr 2
2
d w + mrd3w) ~(). + mdr drS 2
lJr
.
(4A)
BENDING STRESSES IN FLAT PLATES
493
Now let us consider the equilibrium of the whole element. Let F == the shearing force on the face ADEF, per unit length of arc, as shown in Fig. 374. Then the total shearing force on ADEF is F . r~O, and that on the face BORG is (F + ~F)(r + ~r)~O. Let p == the pressure per unit area on the face ABOD, where p may be a function of r but not of O. Then taking moments about the line NT we have
(M 2
+ dM + (F + dF)(r + dr)dO . dr 2)
This reduces to
+ FrdO . dr -
~M2
Substituting for Fr~O
. dr == -
mEt 12(m 2
-
(m - -dw I) r dr
dr. dO
3
(d 2W.
2 -
F
=
M 1 == O.
M 1 == O.
+ d~W) dr~
+ 12(mmEt
Hence
-
from (4A) and M 1 from (3), we get
~M2 3
M2
m 2Et 3
I)
3
2
d w + rnrd W) dr. dO -dr + 1n-dr dr 3
2
2
(d 3W
1 d 2w 1 dW) 12(m 2 _ 1) dr +;: dr 2 -;:2 dr
(5)
Again, resolving vert.ically, we have p . r . ~O . dr - (F
+ ~F)(r
+ ~r)~O + FrdO == 0
which reduces to
pr . ~O . ~r - rdF . ~O -
. ~O == O.
F~r
Hence, in the limit, after dividing by rdr. dO, we get
dF dr
F
+;:- == p.
Substituting for F from (5) ,ve get
m 2Et 3 12(m 2 -
I)
(d~W
dr 4
2 d 3w
+;: dr3
1 d 2w
-;:2 dr 2
1 dW)
+;:a. dr = P
·
(6)
Let us, for brevity, write m 2Et 3 - .. -------. 12(m 2 - 1)
==
D.
(7 )
Then we can put (6) in the form
1;: dT rdr {I;: dr rdrdW)} ] -_ d[
d
d(
p
jj
(8).
This is the general differential equation for the deflection of a circular plate when the load is distributed symmetrically about the axis. When w has been found the radial and hoop stresses are given by (1) and (2), and the shearing force by (5).
STRENGTH OF MATERIALS
494
338. General Solution when the Load is Uniform.-When the
load is uniformly distributed, p is constant and we have from (8)
. • · · .
·
{I
d d ( dU')} rdr ;: dr rdr
!..~(rdW) ~(rdW)
dr
dr
=
2
+A
I
r
2D
== L- r 2 + Al log r + A 2
dr
4D
= L- r 3 + AIr log r + A 2 r
dr
dw r dr
pr
== 2D
D
== P- r + Al
~{~ ~(rdW)}
dr r dr r dr
=?-.r
~[r ~{~ ~(r~~)}J
4D
4
pr 16D
2
A Ir ( 1 +'2 og r -
] '~d
+"21 A
= 1:Dr3 + tA1r(1og r - !)
:~
=
:. W
pr 4 64D
+
A Ir2 4 (log r - I)
2
r2
+
4 .4
a
+ !A r +:t; .
+ lA
(9)
2
2
r2
+ Aalog r + A
4
(10)
There are four constants of integration to be determined, AI' A 2 , A aJ and A 4 • If the plate is continuous to the centre Al and A a must be zero since W does not become infinite at the centre. It will be convenient here to obtain general expressions for the stresscouple M 2 and the shearing force F. From (9) we have
d 2w dr 2
=
3pr 2 16D
d 3w _ 3pr dr 3 - 8D
+ !A
I
(log r
+ i) + i A 2 -
Aa
-;:2
+ A + 2A a I
7
2r
3
2
dw.In (4 ) an d (-0), we fi n, d a fter sIm. utIng f or dw - , d- w an d SU bS t 1Ot' dr dr 2 dr 3 plification,
M~ = _ (3m + 1) . .p r
r . ~()
m
2
16
_
!![A1{(m m
2
+ 1) log r + t(m -
A + 2"(m + I) 2
F ==pr
2
+ AID r
where D is given by (7).
In
A a - I) ] ;:2(m
(11) .
(12)
BENDING STRESSES IN FLAT PLATES Similarly from (I) and (2), we have when z p,
== _ 3{3m
+
l)pr 8mt 2
2
mEt
_
2{m 2
1)
-
== ~
2
+ 1) 10' g r + "fl(m -
[A 1 {(m 2
+ A~m + 1) -
3 A ;:2(m - 1)
2
Pe
= -
+
3(m 3)pr 8mt 2
2
mEt 2{m 2 - I)
_
[A2
1{(m
495
+ 1) 10
r -
J
l.(m -
g·2
+ A2 (m + 1) + rA (m 3
2
2
I)}
I)J
(13)
In (14)
These expressions give the maximum values of the radial and hoop tensile stresses at any radUUB. The constants of integration can be obtained in various cases from the following condUtions : If an edge is freely supported, wand M 2 vanish at that edge.
H an edge is encastre or clamped, wand dw vanish at that edge.
dr At a free edge, M 2 and F must vanish. If the load has dUfferent uniform values PI' P2 ... , over different regions of the plate,M2 and w must have the same values for the PI region and the P2 region where they meet. If only one edge is supported the value of F round that edge is determined by the total load on the plate, and thus Al is given by (12). We shall now consider some particular cases. 339. Solid Circular Plate, Uniformly Loaded over the Whole Area: Ed~e Freely Supported.-Let a be the radius of the plate. Since it is continuous up to the centre we must have Al == A a == o. When r == a we must have M 2 == 0 and w using (II), gives _ (3m
+ 1)pa
m
whence
A2
2
_ DA 2 (m 162m
== o.
+
The first condition,
1) =0
+ l)pa -' == - (3m ----_.._.2
8(m
+ I)D
== 0 when r == a gives, from (IO), A == _ pa4 + (3m + l)pa4 == (5m + I)pat .
The condition w &
64D
32{m
+ l)D
64{m
+ l)D
Inserting the values of the constants in (IO), we get after some simplification,
w = 3p(m
2
1)(a2 16m 2Et 3
_
r 2 )(5m
m
+ la +1
2._
r 2)
•
. (-I 5)
STRENGTH OF MATERIALS
496
The stresses are then given by (13) and (14) : -
+ l)p(a 2 _
== 3(3n~
P
r 2)
81nt 2
r
Po == -3p 2 {(3m 8mt
+ l)a
•
(m
2 -
+ 3)r
2}
These both have their greatest value when r == 0, giving 3(3m + 1)pa 2 P r (max.) == Po (max.) == - - - - - 8mt 2 If we take the strain energy as the criterion of failure and elastic limit of the material in pure tension we must have 2 p r 2 + Po 2 - m -P:Po ==12 r where P r and Po have the values given by (18). 9(3m
+ 1)2(m -
1) 2 4 -12 P a -
32m 3t4
f
.
(16)
.
(17)
.
(18)
be the
This gives
From this we find that the pressure required to cause elastic failure is given by
1
=
3(3~m+T)
-vi
2m
== 0
and -
m
1·
~:.
(19)
340. Solid Circular Plate, Uniformly Loaded over the Whole Area: Edge Clamped.-As before we must have Al == A a == O. When r == a we must also have w
dw dr
dition gives, from (9),
== o. The latter con-
pa 2
The condition
1.0
A 2 === - --. 8D == 0 when r == a gives, from (10), pa 4 pa 4 pa4 A. == - 64D + 32D == 64D· pr 4 pa 2 r 2 w == 64D - 32D
Hence we find
== _~(a2
64D or, inserting the value of D, w
pa1:
+ 64D
r 2)2
-
__1)p (a 2 16m 2Et 3
== ~(11L-=-_-=
_
r2) 2.
From (13) and (14) the maximum stresses at radius r are found to be Pr = Po
3p {(m 8mt 2
= . 3p 2 {(m 8mt
+ 1)a 2 -
(3m
+ 1)a
(m
2
-
+ 1)r 2}
+ 3)r
2}
•
•
(20)
•
(21)
BENDING STRESSES IN FLAT PLATES
497
As in the case of the plate with the edge freely supported, these stresses are greatest at the centre, and there 3(m + 1)pa 2 Pr (max.) = Ps (max.) = 8mt 2 (22) • • Comparing (18) with (22) it will be seen that the maximum stress when the edge is clamped is only about 0·4 times the maximum stress when the edge is free, if m be 10/3. 341. Annular Ring Freely Supported at the Outer Edge and Loaded Uniformly Round the Inner Edge (Fig. 375 shows a diametral section).-Let the radii of the outer and inner boundaries be a and b. Let there be a total load W distributed uniformly round the inner edge of the plate, the surface of the plate being free from load.
375.
FIG.
We must have M s putting p = 0,
A
and
=0
when r
=
a and r = b, hence, from (11),
A + ~(m + I) 2
-.2{(m
+ I) log a + !(m -
I)}
A1
. + 1) log b + l(m -
In + A2~(m
2
2 {(m
.
A == ~(m2
+ 1) =
a
- 1)
A
ii
Hence, by subtraction, (m -
1)(-!. - b~)A3 == A2 (m + I) log~ b a A m +1 a b A I a m _ 1 " 2(a2 _ b2) og r;" 1
2
2
2 2
.
••
3 -
-
1
(
"a)
Also, multiplying the first of the above equations by a 2 , the second by b2, and subtracting, we get (a 2
-
b 2 )(m
+ I)A s =
+bA 2
Hence
_ [b
1
{(m
-
a 2 A 1 {(m
+ 1) log a + !(m -
+ 1) log b + l(m -
b-
I)}
I)}
log a 2 log a m - 1 ] L4 1 • • (P) a 2 -b 2 2(m+l) The shearing force per unit length of outer edge is W/2'Jla, hence from (12) we have, since p = 0, W A2
-
2
A 1 =-2nD
STRENGTH OF MATERIALS
498
Thus the constants A H A a and A a are determined; A 4 is found from the condition w = 0 when r = a. Omitting the details of the algebra, we finally get * w
W
[1 {
3m
+ 1+ a2 a_2 b2 1og ab + 1og7Jr} r
== 2nD 4" - 2(m + 1)
+~
+
2 2
m 1. a b log~ . log ~+~~ 2 m - 1 a 2 - b2 b r 4 a 2 - b2
log~
b
2
+ (3m + l)o,2J(23) 8(m + 1)
From this we can calculate the stresses by means of (13) and (14).
342. Solid Plate Uniformly Loaded Round a Circle: Edge Freely Supported (see Fig. 376).-The conditions of this problem are
~a~
~ ~b~ FIG.
376.
the same as the last except that the plate is continuous to the centre. A total load W is uniformly distributed round the circumference of a circle of radius b, the radius of the plate being a. We must divide the plate into two regions: a> r> {) and b> r> O. For the first region we must have M 2 =0, W =0, when r =a. F == W /2na, when r = a. For the region r b the constants Al and A a in (10) disappear, dw and F = 0 for all values of r. When r = b, M a and wand - must dr have the same values whether calculated for the first or second region. Let us write r b, w == :i-A2r2 + A 4 (i) A' r> b, w' == _ l r 2 (log r - 1) + !A'2r2 + A'alog r + A ' 4 (ii) 4 Then: Wi == 0, r == a gives A' . a2 (log a-I) + !A'2a2 A'alog a + A ' 4 == 0 . (iii)
<
<
-i
+
M 2 == 0, r == a, gives A' -t{(m 1) log a !(m - I)}
+
F
+
= ~ when 2na
r
=
+ A' -t(m + 1) -
a gives
A'I=~ *
A' a 23(m - 1) = 0 (iv)
2nD
Kelvin and Tait, Natural Philo8ophy, Pt. II, p. 198.
.
(v)
BENDING STRESSES IN FLAT PLATES
499
w' = w when r = b gives A'lb 2 (log b - 1)
+ A'2b 2 + A'alog b + A', = A 2b2 + A,.
4 4 4
. (vi)
The condition t.hat M 2 calculated from (ii) == M 2 calculated from (i), when r == b, gives A' A' A' A2 y{(m+l) log b+!(m-l)}+-i(m+l)---- b23(m-l)== 2 (m+l) (vii) dw'
dw
== b
-d == ---, when r r dr
.
gIves
-A'lb (1 og b _ 2·1) 2
+ A'2b + A'a _ A - - -2b b
2
(viii)
2
The equations (iii)-(viii) determine the six constants of integration. Considerations of space prevent us from giving the details of the algebra; the steps for finding the constants are as follows: from (vii) and (viii) find A 'a in terms of A' l' then (iv), (iii), (viii), (vi), give A '2' A'«, A 2 and A«, in that order, in terms of A'lJ which is known from (v). Finally we get 2 2 (3m+l)a -(m-l)b (a 2_r 2)] (24)
w= W [-(r2+b2) 10 ~+(r2-b2)+ 8nD g b w':;:= W [_ (r2
8nD
2(1111
+ b2) 10 !!.. + (3m + l)a 2 g r
2(m
+ l)a 2
)J
(m - l)b\a 2 - r 2
+ l)a 2
(25)
343. Solid Plate with Load Concentrated at the Centre.*-If the load could be truly concentrated at a point at the centre of the plate the stresses at that point would be infinitely great. In practice the load will either be distributed round a very small circle or over the areal of a very small circle. In the former case the above analysis will apply, and if we take the expression for w', rna.king b infinitesimal we shall get an expression for the deflection except near the point of application of the load. Doing this we find, with the edge freely supported, w
== -W [ - r 2 log -a + J. 3m 8nD
t:
+ 1(a
m+l
2
2 -
r 2)
J
. (26)
Froln this we have, by substitution in (1) and (2),
3JV( 1+ -1) log-a
p =T
Po ===
2nt2
m
. (27)
r
3lV [(1 +!) log!!... + (1 _-!.)J
2nt 2
m
r
m
(28)
at any point on the surface outside the region of application of the load. • For circular plates of varying thickness see an article by G. D. Birkhoff, Phil. Mag., May, 1922. See also Prescott's Applied Elasticity. For effect of shear stress OD deflection, see Engineering, Vol. 123 (1927), p.. 343 and Vol. 125 (1923), p. 31.
STRENGTH OF MATERIALS
500
344. Rectangular Plate. Supported at the Edges.*-B. C. Laws
(loc. cit. infra) gives the following formulre for the maximum str~8_and deflection·: the greatest principal stress occurs at the middle points of the, longer sides and is given by a2 pb 2 . (29) 2 2 2a + 6b • t2
whilst the maximum deflection is given by . a'
a'
pb'
+ 2b' · 32Et
3
. (30)
•
where t is the thickness of the plate.
EXAMPLES XXXI 1. In the case considered in § 342, if the edge of the plate be clamped, show that W [ a (at + bt)(a ' - r l ) ] r < b, w = 81CD - (r 2 + b 2 ) log b + r 2 - b 2 + 2a 2 .
r > b, w
=
W [
81CD -
(r t
a
+ b 2 ) log r +
(at
+ b 2 )(a t 2a t
-.r l ) ]
Hence show that the deflection due to a concentrated load at the centre is given by
w =
S:D[-
r1log;
+ !(a l -
rl)J.
2. Show that the displacement of a circular plate of radius a, clamped at the edge, and carrying a load W at a point 0 distant c from the centre is given by t
w
=-.!V [_R21ogCRI + 1(~R'as.s-R2)] 81CD aR
where R denotes the distance of any point in the plate from the point 0, and R' is the distance of the same point from 0' the inverse of 0 with respect to the circle. 3. A circular plate of radius a is freely supported at the edge and carries a total load W uniformly distributed over the area of a concentric circle of radius b. • The formu1~ given here are not exact, although the complete solution has been worked out; the solution for a rectangular plate with clamped edges has not been worked out, but an approximate method of solution has been given by Ritz, J. J. Math. (GreUe), Bd. 135, 1909. See also Edinburgh Proc. Royal Soc., 1912, for the application to the case of a square by C. G. Knott. See also papers in Trans. [nst. Naval Architects: A. M. Robb, 1921 (or Engineering, April, 1921); G. H. Bryan in Vol. XXXV; I. G. Boobnoft in Vol. XLIV: J. Montgomerie in Vol. LXI; T. B. Abell, 1923; aJao E1Ifineering, June 7, 1929. Also paper by Hankey: Der Spannung8ZU8tand in rechteclcigen platten, R. Oldenburg, Berlin, 1913; and B. C. Laws, Proc. [nat. G.E., 1921-2, I. t J. H. Michell, Proc. London Math. Soc., 1902. See also Love's Theory oj Ela8ticity, 3rd Ed., p. 495.
BENDING STRESSES IN FLAT PLATES
501
Show that 'r
> b,
_
W [(
w - 8nD r< b, W
=
r
I
W [( l 8nD r
+
!!:..) I ..!:... 2(3m + 1)01 -- (m -I)b ' _ 2 og a + 4(m + I) .
bl)
+ 2"
+
log
ab + r'-b' ~ -
2(3m
+
l)a
l
4(m
-
4rna ' -(m-l)bl 4(m
+
I)a l
(3m + l)r ' ] 2(m + I)
(r
t
-
bl)
+ l)b ' ]
(7m
+ I)
Hence show that, at the centre, the maximum stresses are given by
pr
= P6 =
3(m+I)W[ m 2nmt t m +1
a m - I bl] m + l' 4a I •
+ log b -
4. If the plate in question (3) be clamped at the edge show that the stresses ou the surface, at the centre, are given by'
pr
= P6 =
whilst at the circumference p,
3(1n + I)W( a 2nmtl 10gb
=
3W (
2nt t 1 -
+
hi) 4a 1
;
l
b ) 2a 2
•
5. The end of a gas-engine piston may be regarded as a diaphragm 12* diameter and iN thick clamped at its edge. If the piston is subjecte d to a pressure of 400 Ibs./in. 1 determine the intensity of greatest radial and hoop stress. (Mech. Sc. Trip., B., 1915.)