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A theory of the plastic yielding due to bending of cantilevers and fixed-ended beams. Part I.
Journal of the Mechanics and Phys/cs of Solids, 1954, Vol. 8, pp. I to 15.
A
THEORY
BENDING
OF OF
THE
PLASTIC
CANTILEVERS BEAMS.
Pergamon Pr¢~ Ltd., London.
YIELDING AND
DUE
TO
FIXED-ENDED
PART I.
By A. P. GREEN British Iron a n d Steel Research Association Metal Working L a b o r a t o r i e s , Sheffield
(Received 14th May, 1954)
SUMMARY PLANE strain a n d plane stress solutions are obtained for the plastic yielding of horizontal uniformly tapering cantilevers of rectangular section carrying a vertical e n d load. An isotropic plastic-rigid material is assumed, and, for the plane stress solutions, a Miszs yield criterion. T h e effects of weak end support, a n d of axial loading in addition to vertical loading, are analysed for uniform cantilevers in plane strain.
1.
INTRODUCTION
THE simple theory of plastic bending of a uniform beam by equal and opposite couples with or without axial load is well known (see for example NADAI 1981, RODERICK 1948, HILL 1950, p. 81), and is well supported by experiment (RoDERICK and PHILLIPS 1949, BAKER et al. 1949, NEAL 1950). The same theory is often applied to the calculation of bending moments in beams bent under shearing loads. I t is in fact assumed, at present, in the plastic limit design of structures (BAKER 1949, SYMONDS and NEAL 1951) that the effect of shearing loads on the bending moment is negligibl6 for most practical purposes. The assumption is known to be incorrect, however, if tile mean shear stress across a transverse section is high (HENDRY 1950). The present work is an a t t e m p t to assess theoretically the effect of shearing loads. In P a r t I, plain strain and plain stress solutions are described for the problem of a cantilever of rectangular transverse section bent under end loading. Only those solutions are considered for which yielding does not extend to the loaded end of the cantilever. The only published work on this problem appears to be a paper by HORNE ( 1 9 5 1 ) to which more detailed reference is made below. An isotropic plastic-rigid material is assumed, i.e. the rigidity modulus is assigned an indefinitely large value. The yield-point for a given loading system is precisely defined, therefore, as the moment when local distortion first begins. The yieldpoint load has the practical significance that the load for an actual material with a finite rigidity may be raised very near to it and yet cause a permanent distortion of elastic order of magnitude only (HILL and SIEBEL 1958). In P a r t II, the solutions for uniform cantilevers are applied to the problems 1
2
A . P . GREEN
of yielding of a centrally loaded simply supported beam, and of centrally or uniformly loaded fixed-ended beams. All the plane stress solutions for uniformly thick cantilevers and fixed-ended beams are then applied to corresponding problems for I-beams bent about axes perpendicular to their webs. Finally, some of the theoretical results, both yield-point loads and modes of deformation, are compared with experiment. 2.
FORMULATION
OF T H E
PROBLEM
Consider a uniformly tapering horizontal cantilever of length 1 which is symmetrical about a horizontal plane through Oy (Fig. 1). Its horizontal width w is constant, and any transverse section perpendicular to Oy is rectangular. Its thick-
Fig. 1. Uniformly tapering cantilever of rectangular cross-section. ness is a minimum t at the unsupported end O. It is supported in position and direction at one end A B and the other end X O X carries a uniform shearing stress S and a uniform axial pressure P, just sufficient to cause yielding. The weight of the cantilever itself is neglected since it is small in comparison with the yieldpoint loads. Solutions are obtained under the two extreme conditions of plane strain and plane stress,* corresponding to a wide and a narrow beam respectively. The approximation to one or other of these conditions should be very close if w is larger or smaller than the thickness of the beam at the regions of deformation by a factor of at least six. The plane strain solutions are valid for any yield criterion; in the plane stress solutions a Mises yield criterion is assumed. The possibility of buckling in a narrow beam is not considered here. Most of the solutions obtained are incomplete in that no stress distribution is constructed in the non-deforming regions. Hence, according to HILL'S (1951) extremum principles for a plastic-rigid body,the yield-point loads m a y be overestimated, though this is unlikely, especially in view of the experimental confirmation obtained for deformation modes. • For a detailed a c c o u n t o f the theories of plane p l ~ t i e strain a n d plane plastic stress see R . HXLL, Th e Mathematical T / w o ~ of P/asfici~, Chs. 6 and I I (Clm~ndon Press, Oxford). T h e m i n e n o t a t i o n is used in this imper.
Plastic yielding due to bending of cantilevers and fixed ended beams
8
HORXE (1951) considered a uniformly thick cantilever carrying a vertical load at one end. He assumed plane stress conditions and a Tresea yield criterion. His solutions are incomplete, however, because he did not construct a mode of deformation compatible with his stress distribution. They are equilibrium states of stress which obey the stress boundary conditions and which do not violate the yield criterion except in two very small areas. Such solutions tend to underestimate the yield-point loads (HILL 1951). They are not compared here with the present solutions because a different yield criterion was assumed. 3.
PLANE STRAIN SOLUTIONS
(a) Shearing without axial load First let us suppose that the cantilever is rigidly fixed at .4B and yields under a vertical load St only at tile other end (Fig. 2). S depends on O and the ratio l/t i.e. on the geometry of the cantilever. D
I
''Y'~
0
t
..........
io
! st
(°) Z
A
/ ,/ / / / / / / /
X
(,)
I
Fig. 2.
y-
~f
/ / / / / / / /
FOR LARGE VALUES OF t
(d) EOUILIBRIUH STRESS FIELD FOR O • 7Se
Plate strain SoIutiovL~ for strongly supported tapering cantilevers under shear loading. (a) I ; (b) II ; (c) I for large values of l/t ; (d) Equilibrium stress field for 0 = 75 °.
If l/t and ~9 are the formation of a greatest, while the II shown in Figs. bending of a wide
sufficiently large the cantilever m a y be expected to yield by plastic ' hinge ' at its fixed end where the bending m o m e n t is rest of it remains rigid. This suggests the slip-line fields I and 2a and 2b. These fields are similar to those devised for the bar notched symmetrically on opposite sides with deep wedge
4
A.P.
GnEE~
shaped notches ( G R E E N 1958a). In the present problem, solution I applies to the higher values of l/t and solution I I to the lower values. In ADS and B E T the field, which is determined by the straight stress-free edge, consists of straight lines inclined at angles ~t/4 to the edge. It is extended round the singularities at A and B to form the regions A Y S and B Y T (on A P S and BQT), consisting of straight lines and circular acrs. These regions meet at a point Y in I, but in II they are connected by the single curved slip-line PYQ. Relative to its rigid support the cantilever yields by the rotation of its rigid end, in I about Y, and in II by sliding over the arc PYQ. Therefore, PYQ must be a circular are, and A P Y Q B a continuous slip-line across which there is a velocity discontinuity. The material in A Y S D (or APSD) is extended and in B Y T E (or BQTE) is compressed. Hence, the mean compressive stress p is equal to -- k in ADS and to k in BET, where k is the yield stress in pure shear of the material. In I there is a stress discontinuity across the point Y. TABLE 1.
PLANE STRAIN.
Strongly ,upported cantilevers with uniform taper yielding under shear load. 0
l/t
S/k
d/t
R/t
8
Go
13-78 5"00 2"00 1"00 0.73
0 0.038 O.IOS 0-281 0.518 0.635
0"707 0.625 0.560 0"401 0'183 0.076
0 0 0"086 0.298 0.590 0.733
0 8 ° 10' 8 0 10'
85 °
6.62 5.82 3.00 2.£0 0.88
0-347 0.351 0.406 0"485 0.715
1 "409 1'169 0.788 0.564 0"164
0 0 0-199 0.371 0"782
0 5 ° 40" 5 ~ 40' 5° 40' 5 ° 40'
80 °
3"63 8"41 2"50 1"50 1"17
0.684 0"684 0.702 0.775 0'820
1 "398 1 "298 0 "962 0"532 0.362
0 0 0'288 0.598 0.754
0 3 ~ 10' 30 1 0 '
90 °
8 ° 10' 8 ° 10' 8 ° 10'
3 Q10' 3 ° 10'
Let us denote the angle 8 B Y ( = SAP) by 8, the length AS ( = B T ) by d, the radius of curvature of the arc P Q by R and its angular span by 2A. These dimensions are uniquely determined by the stress boundary conditions and the shape of the cantilever. The fields must be symmetrical about Oy because the cantilever is symmetrical and P = 0. Therefore, in I Y lies on Oy, and in II OYC, the line which joins O to C the centre of curvature of the are PQ, coincides with Oy. I t follows that the initial direction of motion of 0 is vertically downwards. In I I p varies from - - k to + k along S P Q T , and by applying the Heneky relations ( H I L L 1950, p. 185) we find that 28 = 0 - - ~Ir - - ½ and 2A = 0 -- ~ r + ~ . I n I 8 is o n l y limited by the condition t h a t the yield criterion m u s t n o t be violated
Plastic yielding due to bending of eantflevers and fixed ended beams
5
in the corner of the rigid region between S Y and Y T , i.e. the change in p in turning through the angle of this corner must not exceed that prescribed by the H e n c k y relations along a slip line turning through such an angle (HILL 1954). Hence, 2~ ~< 0 - ~ r - ½, and solution I changes over to I I when the equality sign holds and R = 0. The shapes of the fields are finally determined by taking moments a b o u t Y or C for the portion of the cantilever to the right of .4 Y B . Their dimensions and the correspoflding values of S/1¢ are given in Table 1 for values of 0 = 90 °, 85 °, 80 ° and 75 °. The relations between S / k and lit are shown in Fig. 8. When
, .
.
7_,.
V"
.
.
.
~ ...... 08
~e .75°
.
\\
~0.72o22, --_--
e,75o
S
.....
0-4
.
._r"
~" ~
O'6
.
..............
. --..
t. ..o° ~e=ES°
0"2, PLAN E ST R~N'~ ~ ~ ' ~ - . . . . . . . . . _ --'--PLANE STRAIN (ALTERNA'~" S O L U T I O N _ ~ - - - . ~ . - . ~ - " - " ' ~ . ~ . AFTER HARDENING) . . . . . . . . . . ~- ~~[e,90 ° ------PLANE STRESS ~--'~ I [ I I i I l hO 2"0 30 4'0 S-O bO 70
¥
Fig. 3. Relations between S/k and l/t for strongly supported tapering cantilevers under shear loading.
0 :-- 75 °, S = k for all values of I/t. For higher values of 0 S / k varies and, as we would expect, decreases as l/t increases. The solutions are discussed below in further detail.
Solution I For a given O, as 1/t increases, 8 decreases.
When 8 = 0
l/t = ½ tan 8 (sin 28 -- cos 28), and S / k = 2 sin 28.
(1)
This limiting solution is also valid for all greater values of l/t since, with S = 0, the corners A', B ' of the field do not have to be anchored at the corners of the cantilever (Fig. 2e). When O = ½~r, 8 -~ 0 only as l/t -> oo, and in the limit S = 0. As 8 is decreased the minimum value of l/t for which G = 0 also decreases, while the eorresponding value of S increases, finally becoming equal to k when 0 = 75 °. Fig. 2d shows an equilibrium stress field, involving stress discontinuities across the lines X Y , which covers the whole cantilever. This is, therefore, a complete solution. I t is clear that all cantilevers with 8 ~ 75 ° yield when S = k by shearing at the faee X O X .
6
A.P.
Gnc.E.~
Solution I I F o r a given ~, as I/r/t decreases, R / t increases, and d/t decreases (see Table 1). The solution breaks down when yielding extends to the end X O X . I t can be shown (by s y m m e t r y and other arguments) t h a t the m a x i m u m average shear stress across a plane section of the (assumed) rigid cantilever occurs across YO. This means shear stress increases as l/t decreases and becomes equal to k, with d still finite. L e t us denote the variables in this limiting solution b y the subscript 0 (their values are given in Table 1 in the last line for each value of 0). As 0 decreases, lo/t o and S O increase while Ro/d o decreases, until for 8 -~ 75 ° 0.76' lo/t o -=- 2-53, So/k ~-0.9998, and R = 0. Over the minute range 75 ° 0 . 7 6 ' > ~ > 75 ° the mean shear stress reaches the value k across Y O for solutions of type 1. Finally, for 8 = 75 ° the limit occurs when S = 0, the same slip-line field as t h a t discussed above (Fig. 2d). Solutions for which lit < loft o are discussed in another p a p e r (GRFE~" 1954). Solutions for other shapes of cantilever m a y be constructed in similar manner. Calculations have been made by the writer (GRErx~ 1958b) for cantilevers with concave circular sides instead of a uniform taper.
A A
-Y
g
iX
_y . . . . . . . . . . . . .
(
)
(b) t
Fi~. 4. Alternative modes of deformation in plane strain : (a) l/t' > 1.81 ; (b) l/t' > 1-81.
Alternative Modes of Deformation Any mode of deformation is only instantaneously valid if the shape of the b o d y a n d / o r the distribution of hardness in the deforming region is changing. I f the rate of hardening is rapid, deformation quickly spreads outwards from the original regions in which yielding commences. Some parts m a y cease to deform and m a y even unload elastically, particularly when deformation can proceed under falling loads. If, however, the applied loads m u s t be increased in order to continue yielding, an entirely new mode of deformation m a y be" initiated in regions which were previously rigidly constrained throughout. If, in the present problem, the cantilever and its vertical support are one piece of material of uniform width, an alternative mode involving deformation of the support m a y occur, as shown in Fig. 4. The slip-lifie field depends on the value of l/t' (where t' = A B)
Plastic yielding due to bending of cantilevers and fixed ended beams
7
but not of 0. I f Ill' > 1.31 it consists of a circular arc, A Y B (Fig. 4a) with p ---- 0 at its mid-point, Y. The rigid material to the right of A Y B rotates by sliding over the arc. Assuming that the change in shape of the cantilever is negligibIe up to the moment when this alternative defornmtion begins, the relations between S/k and Ill for 8 = 90 ° and 85 ° have been calculated, and are shown in Fig. 3. I t can be seen that S/k is always greater than its corresponding value for the solutions discussed previously, and the smaller 0 the greater the difference. When 8 ---- 90 °, the differences are small if l/'t is large, and consequently the alternative mode soon comes into operation after initial yielding, if the rate of hardening is high. This has been confirmed by experiment (HUNDY 1954). When l i t ' < 1.31 the slip-line fieht (Fig. 4b) includes regions of tension and compression near A and B adjacent to the horizontal faces of the blocks. No calculations have been made for this solution.
Effect of Weaker Support Naturally the load which a cantilever can carry depends on the precise manner in which it is suppported at its fixed end. The solutions so far described have been for perfectly strong rigid support. In order to find out how sensitive to the end conditions the yield-point loads are likely to be, we now consider a weaker method of support. Suppose that a uniformly thick cantilever fits into a horizontal slot A J K B in a rigid vertical support (Fig. 5). Provided that the length of cantilever in the
//!
AIXS~-dl .
_~__~x .
.
.
.
(°) Fig. 5.
.
o
.
,
,
St
5t
(b)
Plane strain solutions for weakly supported uniform cantilevers under shear loading. (a) I ; (b) II.
slot is sufficiently long, it is reasonable to assume that yielding does not extend to the end JK. The beam is supported by an upward pressure over some region to the left of B and a downward pressure over some region to the right of J . The surface of the beam in the region of .4 is assumed to be clear of tlle support. Possible slip-line field solutions of types I and I I are shown in Fig. 5. They differ from those described above in that there is no singularity on the stress-free upper surface (i.e., ~1 = 0), and they are not therefore, symmetrical about Oy.* In II, ~2 = ¼ ~ - - ½ and 2A---- ¼ ~ + ~. The change-over from I I to I occurs when l / t = 7.59. When Ift = 0.71 the mean shear stress across CO is k. Since the solutions are unsymmetrical this may not be the precise value of 1/'t at which solution I I breaks down, but it is probably a close approximation. The lack of * N o t a t i o n : lengths a nd angles whiell are equal in tlw s y n n n e t r i e a l solutions I a n d I I a r e denoted in similar but
unsym~trlcal solutions by the subscripts 1 an(i 2 in tile tensile a nd compressive regions respectively.
8
A.P.
GREEN
s y m m e t r y also m e a n s t h a t 0 no l o n g e r m o v e s v e r t i c a l l y d o w n w a r d s w h e n y i e l d i n g c o m m e n c e s , b u t i n s t e a d m o v e s s l i g h t l y o u t w a r d s in a d i r e c t i o n inclined a t t h e s m a l l angle ~ (say) to t h e v e r t i c a l . T h e c a l c u l a t e d d i m e n s i o n s of t h e fields a n d t h e c o r r e s p o n d i n g v a l u e s of S / k a n d ~ a r e given in T a b l e 2. C o m p a r i n g t h e s e results w i t h t h o s e for s t r o n g s u p p o r t , t h e r e d u c t i o n in S / k is o n l y a b o u t 4 % for t h e s m a l l e s t v a l u e s of I/t c o n s i d e r e d a n d decreases to a b o u t 1 ~ when I//t ,-~ 6 a n d to zero as I/t --> oo. TABLE 2.
PLANE STRAIN.
Weakly supported uniform cantilevers yielding under shear load. l/t
S/k
dl/t
d2/t
R/t
yc
16"69 18'81 11.64 9"97 8 "65 7"59 5"34 8.66 2"34 1.64 1 "35 0"847 0.710
0.0807 0"0372 0.0443 0.0520 0.0602 0.O689 0.0906 0-147 0-234 0.830 0"393 0"558 0.620
0"717 0"719 0 "720 0"723 0"725 0"727 0"706 0-651 0"567 0.473 0-412 0.251 0,190
0"598 0'588 0'579 0'570 0"562 0"553 0.584 0.484 0'408 0"322 0"267 0'121 0'065
0 0 0 0 0 0 0.027 0"097 0'204 0 "322 0.400 0"604 0"681
0"002 0-008 0.008 0"019 0"085 0"047 0.079 0"091
T h e s o l u t i o n s o f F i g . 5 m a y b e a p p l i e d to o t h e r c o m m o n c o n d i t i o n s of s u p p o r t such as t h o s e s h o w n in Figs. 6 a n d 7. T h e c a n t i l e v e r in F i g . 6 is w e l d e d or f i r m l y c l a m p e d t o t h e flat t o p o f a rigid pillar. Fig. 7 shows a s i m p l y s u p p o r t e d b e a m
I
T
St Fig. 6. Uniform cantilever firmly clamped to the top of a rigid vertical pillar.
t
4
[
I st
Fig. 7.
Simply supported beam concentrated load.
under
c a r r y i n g a v e r t i c a l l o a d o v e r a finite l e n g t h B~ B2 = a. More d e t a i l e d e x a m i n a t i o n is r e q u i r e d of t h e stresses in t h e ( a s s u m e d ) rigid region b e l o w B 1 B 2 to d e t e r m i n e h o w l a r g e t h e r a t i o a / d 1 m u s t be t o a v o i d v i o l a t i n g t h e y i e l d c r i t e r i o n there. T h e s o l u t i o n is c l e a r l y i n v a l i d if a / d is v e r y s m a l l b u t is p r o b a b l y v a l i d if a :> ~/2 d I. Since t h e b e n d i n g m o m e n t s a r e e q u a l across t h e b e a m a t B 1 a n d B2, y i e l d i n g t a k e s p l a c e on t h e side for w h i c h t h e y i e l d - p o i n t b e n d i n g m o m e n t is least. T h i s is n o t n e c e s s a r i l y t h e side for w h i c h S is l e a s t (i.e. l g r e a t e s t ) as will be seen l a t e r in t h e discussion o f results ( P a r t I I ) .
Plastic yielding due to b e n d i n B o f cantilevers a n d fixed e n d e d b e a m s
9
(b) Uniform cantilever sheared under axial load The material is assumed to be isotropic and hence solutions with tensile axial loads are identical to those with compressive loads of equal magnitude and S reversed, except t h a t the stresses and strain-rates are reversed at every point in the body. I t is sufficient therefore to consider only compressive loading. Solutions are obtained for a cantilever of uniform thickness t firmly supported in position and direction at the end `4B. The application of pressure upsets the s y m m e t r y of loading, and consequently, the slip-line fields, are also unsymmetrical. The solutions I and I I with P = 0 are modified b y expanding the compressive and contracting the tensile regions* (Fig. 8), st and m a y be conveniently (o) (b) Ib subdivided into two different kinds, which are labelled (a) and (b) according to whether 81 is respectively greater t h a n or equal to zero. As the pressure is increased on a given cantilever Sx decreases. I f P is sufficiently large, ~ = 0, the c o m e r of the tensile region .4 (d) rTb need not be anchored at IIo the corner of the cantiFig. 8. P l a n e strain solutions for strongly s u p p o r t e d lever, and its distance u n i f o r m cantilevers u n d e r c o m b i n e d shear a n d axial increases as the pressure loading : (a) I a ; (b) I b ; (e) I I a ; (d) I I b . is increased. I n the slip-line fields of solution I, (8 x + 82) m u s t be less t h a n (¼w -- ½) to avoid violating the yield criterion in the rigid corner to the right of Y. In solution I I , the H e n c k y relations determine t h a t S1 + 82 = ~w - - ½ and the angular span of the circular arc PQ is 2A = ~ + ½. The ranges of validity of the different types of solution are shown in Fig. 9 in terms of I/t and P. The transition from I to I I occurs when R = 0, and the greater P the smaller is the value of l/t at which it occurs. R decreases with increasing lit for a given pressure, or with increasing pressure for a given lit. B o t h R and 81 are zero when I/t = 5.3 and P / k : 0.66. In solution I (b) as P is increased on a given cantilever both 81 and d 1 decrease, becoming zero simultaneously when S ---- 0, P ---- 2k, and ~ = tan -l [t/2 (l -- t)] = ~ o say. The limiting slip-line field then consists only of the compressive region B E Y and l/t m u s t be greater than 2 for E to lic inside the cantilever. The mode of deformation (rotation a b o u t Y) is unique when d I :> 0 but in the limit d~ = S ~ 0,
Io
c.],=
• See f o o t n o t e
on
p, 7 for
notation.
ofx]~Z C~_ Ap_D _ == _-_--x4
.
.
.
.
.
.
.
10
A, P. GItEEN
there are an infinite n u m b e r of possible modes. P r o v i d e d t h a t there is no constraint on the motion, ~, ean take a n y value between Yo and -- 70 (the condition t h a t the plastic work m u s t be positive precludes tile possibility of y < 70). I n practice the situation would be unstable and the slightest deviation f r o m u n s y m metrical loading would lead to bending of the cantilever with ~, --- q: 7o. Below certain critical values of l/t, which depend on P, yielding extends to the end XOX, and solutions 1 and I I are not valid. A close a p p r o x i n m t i o n to 20
x (0)
I f{~
l(b)
~1,-,
8
C)
\
\
~'-
rr¢o) " % ~
4
...-..~_.
,1 0"73 .............. 0
"/- ........
-
0'5
{'0
_P
-
_-
tit --
{.5
2.0
k
Fig. 9. Range of validity of the different plane strain solutions for a strongly supported uniform
cantilever. this limit is p r o b a b l y provided b y calculating the values of I/t and P at which the m e a n shear stress across OY is equal to k. Thc dashed line in Fig. 9 is the result of such calculations ; below this line solutions I and I I are certainly n o t valid. -k,l t 0.4
O2
t=4
0
0.5
i'O
t.5
2-0
k
Fig. 10. Relations between 8/k and P/k (i.e. yield loci) for strongly supported uniform cantilevers in plane strain.
All results, including the dimensions of the slip-line fields are t a b u l a t e d in Table 8 (no results have been calculated for solution I (b)). The relations between S/k and P/k for various values of I/t are shown in Fig. 10. These curves m a y be
Plastic yielding due to bending of cantilevers and fixed e n d e d b e a m s
]l
r e g a r d e d a s p a r t o f t h e y i e l d loci f o r t h e c a n t i l e v e r s , r e p r e s e n t i n g p o s s i b l e y i e l d point states.
A complete locus would be symmetrical about both axes since the
c a n t i l e v e r is o f s y m m e t r i c a l s h a p e a n d t h e m a t e r i a l is a s s u m e d t o b e i s o t r o p i c .
TABLE 3.
PLANE STRAIN.
Strongly supported uniform cantilevers yielding under combined shear and axial pressure. l/t 8.0
S/k
P/k
81 c
~2 c
dl/t
d2/t
0.583 0.557 0.508 0.223 0.050
0"616 0"657 0.754 1.047 1 "278
0.038 0.021 0 0 o
0.003 0.008 0.018 0.046 0.065
0.0656 0.0646 0.0602 ! 0.0848 i 0.0093
0.098 0.279 _ 0.014 1.401 1-867
0.128 0.083 0 0
0.168 0.208 0.283 0.15') 0"07,1
4"0
0.186 0.184, 0.181 0.124 0"105 0.068 ; 0.034 0.O09
0-092 0-275 0.459 0.658 1.089 1.481 1 "769 1.942
0"123 0-083 0"043 0 0 0 0 0
0"168 0"203 0'243 0'285 0.285 0.210 0'14 ) 0"07 ,)
0.51$ 0.498 0.477 0.469 0.870 0.204, 0.093 0.024
0"552 0.593 0.643 0.706 0.841 1.020 1.170 1-303
O.120 0'107 0-082 0.041 o 0 0 I 0
0"005 0"014 0.025 0.088 0-096 0"108 0.181 0.152
2.0
0.280 0.278 0.273 0.264 0.245 0.222 0.184 0.125 0.068 0.019
0.089
0.123 0.083 0.043 O
0"163 0"203 0.243 0"285 0"285 0"285 0'285 0"285 0.2 O
0.419 0.461 0-512 0.577 0.672 0.758 0.864, 1.001 1.128 1.272
0.284 0.257 ] 0.214 0.170 0.143 O.085 0 0
0.097 0-021 0-086 0.057 0-095 0.129 0.178 0.244 0.809
o.1 o
0.385 0.361 0.345 0.340 0"280 0.240 0.200 0.171 0.085 0.021
0
0.886
0.518 0-517 0"513 0.507 0.499 0"497 0"483
0.084 0-252 0.421 0.608 0.736 0.769 0"980
0-203 0-243 0.285 0.285 0.285 0.285
0"167 0"141 0"123 0"114 0"080 0 "072 0 "020
0.200 0.241 0.290 0.351 0'399 0.411 0"488
0.589 0.577 0.554 0.515 0.500 0"496 0"470
0.005 0"014 0.026 0.043 0-072 0"079 0"129
1.0
0"267 0"445 0"638 0"891 1 "090 1"340 1'618 1.820 1.957
0
0
0 o
0 0 0
0.083 0.043 0 0 0 0
4.
PLANE STRESS SOLUTIONS
Solutions have been obtained only for shearing without axial load, though is i n t e n d e d t o e x t e n d
it
the result to include combined shear and pressure on a
u n i f o r m l y t h i c k c a n t i l e v e r . T h e s y m m e t r i c a l c h a r a c t e r i s t i c fields i n p l a n e s t r e s s f o r t a p e r i n g c a n t i l e v e r s a r e a n a l o g o u s t o t h e s l i p - l i n e fields d e s c r i b e d i n § 3a, a n d r e f e r e n c e will b e m a d e t o t h e s a m e F i g . 2, t h o u g h t h e p r e c i s e s h a p e s o f t h e fields a r e s l i g h t l y d i f f e r e n t ( F i g . 11).
12
A.P. Gmmm
In ASD the field determined b y the straight stress-free edge, consist~of straight lines inclined to the edge at the angle fl ~ tan -1 ~/~ ---- 54o44 , ; the state of stress is uniaxial tension Y = x/8 k parallel to the edge. HILL (1952) has shown t h a t the stresses in the extension of this field round the singularity A, are % = k cos ~,
~ = 2k cos ~b,
~r¢ ----k sin ~b,
(2)
where the polar angle ~b is measured from the base-line indicated in Fig. 11, which is inclined at the angle 2fl to the edge A B. The curved characteristics such as S P have the equation r # sin ~b = const while the other family are straight lines through A. As in the plane strain solutions, 0 moves vertically downwards at yielding, and deformation takes place b y rotation of the rigid cantilever either about the point of stress discontinuity Y (solution I. Fig. 2a) or about a centre of rotation C lying on Oy (solution II, Figs. 2b and 11). The latter solution involves local necking along P Y and local bulging along YQ, with pure shear at Y the mid-point of this are. Local necking along a narrow band through the thickness of the
A /~ ~//'~ ~'~/ ~/ // / f / // / / _ ~ _ : - ~.\, . . ~ / \\ @
.
.
~, o L
the cantilever in this problem) can only occur along a characteristic (HILL 1952). The vector representing the relative velocity / / j of the material on either side of the neck is perpendicular to the ~other characterstic direction and is inclined to the neck at an angle ~b which varies with the stress Fig. 11. Characteristic field for a strongly supported state along the neck (e.g. ~b : 0 uniform cantilever in plane stress. is pure shear and ~b = sin -1 (1//3) in pure tension). The same applies to local bulging (i.e., thickening instead of thinning) though this can only occur with a strain-softening material or with a metal like annealed mild steel* t h a t has a sharp drop in yield stress after the initial yield-point. Whether or not local bulging is in fact possible, solution II provides at least an upper bound to yield-point loads (HILL 1951). The diffuse mode of deformation for a strain hardening material is not known.
Y
The equation linking the state of stress to the state of strain-rate along PYQ is found to be tan A ~ 2 tan ~b, (8) • Local bulging has been obmerved in a mild steel preming by B. P. GREEN and B. B. HUNDY ( u n p u b ~ h e d ) .
P l a s t i c yielding d u e to b e n d i n g of c a n t i l e v e r s a n d fixed e n d e d b e a m s
13
and hence the polar equation of the arc referred to C as origin is r 2 cos A : constant, where the polar angle ~ is measured from C Y O . This is, of course, the same t y p e of curve as t h a t round the singularity at A. F o r the purpose of calculating the resultant forces acting across the arc P Y Q it is convenient to regard C P Y Q as a TABLE 4.
PLANE STRESS.
Stro~gly supported cantilevers with uniform taper yiddi~ under shear load. l/t 90 °
~o
75 °
0"122
0.0489
15 ° 5 4 '
0"178 0 "240
0"439 0"375
0"327
0"290
0.110 0.187 0.290
15 ° 5 4 ' 15 ° 5 4 ' 15 ° 5 4 '
0"420 0 "499 0"555 0"595 0"618
0"200
0.400 0-492
15 ° 5 4 ' 15 ° 5 4 ' 15 ° 5 4 '
2.67 1"94 1 "41 1 "07 0"865
3-01 2.84 1"81
~°
0 15 ° 5 4 ' 15 ° 54"
3.77
6"88 8"88 8.87
R/t 0 0 0.0254
0 0"0799 0-102
~O
d/t 0"612 0"530 0"509 0"489
5'65 4"48
0"750 0-675 0"636 85 °
S/k
0"301 0-319 0-331 0.344
1-36
0"380 0"431 0.501
1.04 0.889
0'579 0.646
0'724
0.691
3"35 2"83 2.41 2-02
1.65 1 "29
0"123 0"070 0"030 0'008
0.557 0.605 0-682
15 ° 5 4 '
0 0-0404
11 ° 3 0 ' 11 ° 8 0 '
0"0749 0"156
11 ° 8 0 '
15 ° 5 4 '
1 "220 0"887 0.809 0-749 0.624 0-502
11 ° 11 ° 11 ° 11 °
30' 80' 30' 30'
0-370 0.246
0"251 0"370 0.4,92
0"148
0-594
0"085
0"662
0.592
1 '206
0
0.595 0-605 0.623
l "050 0.907 0.764 0.613 0.446
1.00
0.653 0.698 0-751
0 0-0907 0'191 0.306 0.446
0"293
0"587
0 7° 7° 7° 7° 7° 7°
0"875
0"780
0"219
0"659
7~ 2V
2"28 2-16 2"05 1.72 1 "47
0-866 0.866
1"183 1"126
0
0-868 0"874 0"885 0.901
1.061
0 0 0"0580 0"216 0"352 0"490
1.24
0"863 0"704
0.550
11 ° 80' 11 ° 8 0 '
21' 11' 11' 21' 21' 2V
3 ° 24" 3 ° 24' 8 ° 24' 3 ° 24' 3 ° 24'
fully plastic region in stress equilibrium, the stresses being given b y equation (2) with ~ = ½~r - - A. The resultant forces are then evaluated over the boundaries C P and CQ instead of integrating t h e m over the arc P Y Q . The angular span
14
A.P. GREEN
of the arc is given by the ~'ondition that the stress-state should be continuous along the continuous characteristic A P Y Q B . Hence it is related to the angle of taper of the cantilever bv the equation 2A - tan -I (½ tan A) ~ 19 + ~Ir -- 2ft.
(4)
The dimensions of the characteristic fields and values of S/k which have been calculated are given in Table 4. The length of A P : BQ is denoted by d, and of CP = CQ by R. The relations between S k and I t are compared in Fig. 3 with those for plane strain. For a given cantilever the plane stress value of S / k is about 14% less t h a n the plane strain value over the whole range of l,t for > 75 °. This was to be expected for long thin cantilevers with a MIsEs yield criterion because there is the same difference between the corresponding yield stresses in compression or tension. I t is interesting, however, that the percentage difference remains almost the same even for the smallest value of lit considered ; in plane stress the stress state in the deforming region then varies between pure tension at the edge and pure shear at the centre, whereas in plane strain it is pure shear throughout. Since the general outlines of the plane stress and plane strain solutions are very similar only the points of difference are mentioned in the following discussion of the two types of solution.
Solution I When 3 ---- 0.
l/t ---- tan 0 (sin 20 -- ~/2 cos 20)/2~/~,
S / k ---- ~/8 sill 20,
(5)
and, as with (3) in plane strain, this limiting solution is valid for all greater values of 1/t. When 0 = 7 2 ° 2 2 ', S----k for all values of l/t and a stress field can be constructed over the whole cantilever similar to t h a t shown in Fig. 2d, the magnitude of the stress parallel to the surfaces being Y = V'Sk instead of 2k. The relation between l/t and S / k for a uniformly thick cantilever (O = ½~r) in the range of solution I (i.e. l/t > 5.65) is given by the equations
[ I/t= ½cot$
1 1 + ~/SsinSsin(B-t- $
)]
sin2~ '
and
S / k - - sln . (~ + ~ "
(6)
Solution I I The variables in the limiting solution when the mean shear stress across Oy is equal to k are again denoted b y the subscript 0. When 8 = 72°24 ', R o = o, lo/t o = 1.89 and So/k = 0.9985. Only over the minute range 72o24 ' > 8 > 72022 ' is this limit reached for solutions of type I. An alternative mode of deformation similar to t h a t shown in Fig. 4a can be constructed b u t no calculations have been made.
ACKNOWLEDGMENTS I am grateful to Mr. D. SPENCELEYof the General Plasticity Section of B.I.S.R.A. for his assistance in calculating the results and drawing the diagrams for this paper which is based on B.I.S.R.A. Reports MW/B/14/58 and MW/B/9/54.
Plastic yielding due to bending of cantilevers and fixed ended beams REFFRENCES
BAKER, J. F. BAKER, J. F., HORNS, M. R. and RODERICK, J. W. GREEN, A. P.
HENDRY, A. W. HILL, R.
15
1949
J. Inst. Cir. Eng., 31, 188.
1949 1953a 1953b 1954 1950 1950
Proc. Roy. Soc. A, 198, 493. Quart. J. Mech. App. Maths., 6, 228. B . I . S . R . A . Report MW/B/14/53. J. Mech. Phys. Solids, 2, 197. The Structural Engineer, 28, 811. The Mathematical Theory of Plasticity (Clarendon
1951 1952 1954 1953 1951 1~o54 1931 1950 1948
Phil. Mag., 42, 868. J Mech. Phys. Solids, 1, 19. J. Mech. Phys. Solids, 2, 000. J. Mech. Phys. Solids, 1,207. Proc. Roy. Soc. A, 207, 216. MetaUurgia, 109. Plasticity, Ch. 22 (McGraw-Hill). Phil. Trans. A, 242, 197. Phil. Mag., 39, 529.
Press, Oxford).
HILL, R. and SIEBEL, M . . P . L . HORNS, M. R. I~[UNDY, B. B. NADAI, A. NF.aL, B. G. RODKRICK, J. W. RODmaICK, J. W. and 1949 PHILLIPS, I. H. SYMONDS, P. S. and NEaL, B. G. 1951
Research Engineering Structures Supplement, p. 9. J. Franklin Inst., 252, 388 and 469.
A
THEORY
BENDING
OF OF
THE
PLASTIC
CANTILEVERS BEAMS.
Pergamon Pr¢~ Ltd., London.
YIELDING AND
DUE
TO
FIXED-ENDED
PART I.
By A. P. GREEN British Iron a n d Steel Research Association Metal Working L a b o r a t o r i e s , Sheffield
(Received 14th May, 1954)
SUMMARY PLANE strain a n d plane stress solutions are obtained for the plastic yielding of horizontal uniformly tapering cantilevers of rectangular section carrying a vertical e n d load. An isotropic plastic-rigid material is assumed, and, for the plane stress solutions, a Miszs yield criterion. T h e effects of weak end support, a n d of axial loading in addition to vertical loading, are analysed for uniform cantilevers in plane strain.
1.
INTRODUCTION
THE simple theory of plastic bending of a uniform beam by equal and opposite couples with or without axial load is well known (see for example NADAI 1981, RODERICK 1948, HILL 1950, p. 81), and is well supported by experiment (RoDERICK and PHILLIPS 1949, BAKER et al. 1949, NEAL 1950). The same theory is often applied to the calculation of bending moments in beams bent under shearing loads. I t is in fact assumed, at present, in the plastic limit design of structures (BAKER 1949, SYMONDS and NEAL 1951) that the effect of shearing loads on the bending moment is negligibl6 for most practical purposes. The assumption is known to be incorrect, however, if tile mean shear stress across a transverse section is high (HENDRY 1950). The present work is an a t t e m p t to assess theoretically the effect of shearing loads. In P a r t I, plain strain and plain stress solutions are described for the problem of a cantilever of rectangular transverse section bent under end loading. Only those solutions are considered for which yielding does not extend to the loaded end of the cantilever. The only published work on this problem appears to be a paper by HORNE ( 1 9 5 1 ) to which more detailed reference is made below. An isotropic plastic-rigid material is assumed, i.e. the rigidity modulus is assigned an indefinitely large value. The yield-point for a given loading system is precisely defined, therefore, as the moment when local distortion first begins. The yieldpoint load has the practical significance that the load for an actual material with a finite rigidity may be raised very near to it and yet cause a permanent distortion of elastic order of magnitude only (HILL and SIEBEL 1958). In P a r t II, the solutions for uniform cantilevers are applied to the problems 1
2
A . P . GREEN
of yielding of a centrally loaded simply supported beam, and of centrally or uniformly loaded fixed-ended beams. All the plane stress solutions for uniformly thick cantilevers and fixed-ended beams are then applied to corresponding problems for I-beams bent about axes perpendicular to their webs. Finally, some of the theoretical results, both yield-point loads and modes of deformation, are compared with experiment. 2.
FORMULATION
OF T H E
PROBLEM
Consider a uniformly tapering horizontal cantilever of length 1 which is symmetrical about a horizontal plane through Oy (Fig. 1). Its horizontal width w is constant, and any transverse section perpendicular to Oy is rectangular. Its thick-
Fig. 1. Uniformly tapering cantilever of rectangular cross-section. ness is a minimum t at the unsupported end O. It is supported in position and direction at one end A B and the other end X O X carries a uniform shearing stress S and a uniform axial pressure P, just sufficient to cause yielding. The weight of the cantilever itself is neglected since it is small in comparison with the yieldpoint loads. Solutions are obtained under the two extreme conditions of plane strain and plane stress,* corresponding to a wide and a narrow beam respectively. The approximation to one or other of these conditions should be very close if w is larger or smaller than the thickness of the beam at the regions of deformation by a factor of at least six. The plane strain solutions are valid for any yield criterion; in the plane stress solutions a Mises yield criterion is assumed. The possibility of buckling in a narrow beam is not considered here. Most of the solutions obtained are incomplete in that no stress distribution is constructed in the non-deforming regions. Hence, according to HILL'S (1951) extremum principles for a plastic-rigid body,the yield-point loads m a y be overestimated, though this is unlikely, especially in view of the experimental confirmation obtained for deformation modes. • For a detailed a c c o u n t o f the theories of plane p l ~ t i e strain a n d plane plastic stress see R . HXLL, Th e Mathematical T / w o ~ of P/asfici~, Chs. 6 and I I (Clm~ndon Press, Oxford). T h e m i n e n o t a t i o n is used in this imper.
Plastic yielding due to bending of cantilevers and fixed ended beams
8
HORXE (1951) considered a uniformly thick cantilever carrying a vertical load at one end. He assumed plane stress conditions and a Tresea yield criterion. His solutions are incomplete, however, because he did not construct a mode of deformation compatible with his stress distribution. They are equilibrium states of stress which obey the stress boundary conditions and which do not violate the yield criterion except in two very small areas. Such solutions tend to underestimate the yield-point loads (HILL 1951). They are not compared here with the present solutions because a different yield criterion was assumed. 3.
PLANE STRAIN SOLUTIONS
(a) Shearing without axial load First let us suppose that the cantilever is rigidly fixed at .4B and yields under a vertical load St only at tile other end (Fig. 2). S depends on O and the ratio l/t i.e. on the geometry of the cantilever. D
I
''Y'~
0
t
..........
io
! st
(°) Z
A
/ ,/ / / / / / / /
X
(,)
I
Fig. 2.
y-
~f
/ / / / / / / /
FOR LARGE VALUES OF t
(d) EOUILIBRIUH STRESS FIELD FOR O • 7Se
Plate strain SoIutiovL~ for strongly supported tapering cantilevers under shear loading. (a) I ; (b) II ; (c) I for large values of l/t ; (d) Equilibrium stress field for 0 = 75 °.
If l/t and ~9 are the formation of a greatest, while the II shown in Figs. bending of a wide
sufficiently large the cantilever m a y be expected to yield by plastic ' hinge ' at its fixed end where the bending m o m e n t is rest of it remains rigid. This suggests the slip-line fields I and 2a and 2b. These fields are similar to those devised for the bar notched symmetrically on opposite sides with deep wedge
4
A.P.
GnEE~
shaped notches ( G R E E N 1958a). In the present problem, solution I applies to the higher values of l/t and solution I I to the lower values. In ADS and B E T the field, which is determined by the straight stress-free edge, consists of straight lines inclined at angles ~t/4 to the edge. It is extended round the singularities at A and B to form the regions A Y S and B Y T (on A P S and BQT), consisting of straight lines and circular acrs. These regions meet at a point Y in I, but in II they are connected by the single curved slip-line PYQ. Relative to its rigid support the cantilever yields by the rotation of its rigid end, in I about Y, and in II by sliding over the arc PYQ. Therefore, PYQ must be a circular are, and A P Y Q B a continuous slip-line across which there is a velocity discontinuity. The material in A Y S D (or APSD) is extended and in B Y T E (or BQTE) is compressed. Hence, the mean compressive stress p is equal to -- k in ADS and to k in BET, where k is the yield stress in pure shear of the material. In I there is a stress discontinuity across the point Y. TABLE 1.
PLANE STRAIN.
Strongly ,upported cantilevers with uniform taper yielding under shear load. 0
l/t
S/k
d/t
R/t
8
Go
13-78 5"00 2"00 1"00 0.73
0 0.038 O.IOS 0-281 0.518 0.635
0"707 0.625 0.560 0"401 0'183 0.076
0 0 0"086 0.298 0.590 0.733
0 8 ° 10' 8 0 10'
85 °
6.62 5.82 3.00 2.£0 0.88
0-347 0.351 0.406 0"485 0.715
1 "409 1'169 0.788 0.564 0"164
0 0 0-199 0.371 0"782
0 5 ° 40" 5 ~ 40' 5° 40' 5 ° 40'
80 °
3"63 8"41 2"50 1"50 1"17
0.684 0"684 0.702 0.775 0'820
1 "398 1 "298 0 "962 0"532 0.362
0 0 0'288 0.598 0.754
0 3 ~ 10' 30 1 0 '
90 °
8 ° 10' 8 ° 10' 8 ° 10'
3 Q10' 3 ° 10'
Let us denote the angle 8 B Y ( = SAP) by 8, the length AS ( = B T ) by d, the radius of curvature of the arc P Q by R and its angular span by 2A. These dimensions are uniquely determined by the stress boundary conditions and the shape of the cantilever. The fields must be symmetrical about Oy because the cantilever is symmetrical and P = 0. Therefore, in I Y lies on Oy, and in II OYC, the line which joins O to C the centre of curvature of the are PQ, coincides with Oy. I t follows that the initial direction of motion of 0 is vertically downwards. In I I p varies from - - k to + k along S P Q T , and by applying the Heneky relations ( H I L L 1950, p. 185) we find that 28 = 0 - - ~Ir - - ½ and 2A = 0 -- ~ r + ~ . I n I 8 is o n l y limited by the condition t h a t the yield criterion m u s t n o t be violated
Plastic yielding due to bending of eantflevers and fixed ended beams
5
in the corner of the rigid region between S Y and Y T , i.e. the change in p in turning through the angle of this corner must not exceed that prescribed by the H e n c k y relations along a slip line turning through such an angle (HILL 1954). Hence, 2~ ~< 0 - ~ r - ½, and solution I changes over to I I when the equality sign holds and R = 0. The shapes of the fields are finally determined by taking moments a b o u t Y or C for the portion of the cantilever to the right of .4 Y B . Their dimensions and the correspoflding values of S/1¢ are given in Table 1 for values of 0 = 90 °, 85 °, 80 ° and 75 °. The relations between S / k and lit are shown in Fig. 8. When
, .
.
7_,.
V"
.
.
.
~ ...... 08
~e .75°
.
\\
~0.72o22, --_--
e,75o
S
.....
0-4
.
._r"
~" ~
O'6
.
..............
. --..
t. ..o° ~e=ES°
0"2, PLAN E ST R~N'~ ~ ~ ' ~ - . . . . . . . . . _ --'--PLANE STRAIN (ALTERNA'~" S O L U T I O N _ ~ - - - . ~ . - . ~ - " - " ' ~ . ~ . AFTER HARDENING) . . . . . . . . . . ~- ~~[e,90 ° ------PLANE STRESS ~--'~ I [ I I i I l hO 2"0 30 4'0 S-O bO 70
¥
Fig. 3. Relations between S/k and l/t for strongly supported tapering cantilevers under shear loading.
0 :-- 75 °, S = k for all values of I/t. For higher values of 0 S / k varies and, as we would expect, decreases as l/t increases. The solutions are discussed below in further detail.
Solution I For a given O, as 1/t increases, 8 decreases.
When 8 = 0
l/t = ½ tan 8 (sin 28 -- cos 28), and S / k = 2 sin 28.
(1)
This limiting solution is also valid for all greater values of l/t since, with S = 0, the corners A', B ' of the field do not have to be anchored at the corners of the cantilever (Fig. 2e). When O = ½~r, 8 -~ 0 only as l/t -> oo, and in the limit S = 0. As 8 is decreased the minimum value of l/t for which G = 0 also decreases, while the eorresponding value of S increases, finally becoming equal to k when 0 = 75 °. Fig. 2d shows an equilibrium stress field, involving stress discontinuities across the lines X Y , which covers the whole cantilever. This is, therefore, a complete solution. I t is clear that all cantilevers with 8 ~ 75 ° yield when S = k by shearing at the faee X O X .
6
A.P.
Gnc.E.~
Solution I I F o r a given ~, as I/r/t decreases, R / t increases, and d/t decreases (see Table 1). The solution breaks down when yielding extends to the end X O X . I t can be shown (by s y m m e t r y and other arguments) t h a t the m a x i m u m average shear stress across a plane section of the (assumed) rigid cantilever occurs across YO. This means shear stress increases as l/t decreases and becomes equal to k, with d still finite. L e t us denote the variables in this limiting solution b y the subscript 0 (their values are given in Table 1 in the last line for each value of 0). As 0 decreases, lo/t o and S O increase while Ro/d o decreases, until for 8 -~ 75 ° 0.76' lo/t o -=- 2-53, So/k ~-0.9998, and R = 0. Over the minute range 75 ° 0 . 7 6 ' > ~ > 75 ° the mean shear stress reaches the value k across Y O for solutions of type 1. Finally, for 8 = 75 ° the limit occurs when S = 0, the same slip-line field as t h a t discussed above (Fig. 2d). Solutions for which lit < loft o are discussed in another p a p e r (GRFE~" 1954). Solutions for other shapes of cantilever m a y be constructed in similar manner. Calculations have been made by the writer (GRErx~ 1958b) for cantilevers with concave circular sides instead of a uniform taper.
A A
-Y
g
iX
_y . . . . . . . . . . . . .
(
)
(b) t
Fi~. 4. Alternative modes of deformation in plane strain : (a) l/t' > 1.81 ; (b) l/t' > 1-81.
Alternative Modes of Deformation Any mode of deformation is only instantaneously valid if the shape of the b o d y a n d / o r the distribution of hardness in the deforming region is changing. I f the rate of hardening is rapid, deformation quickly spreads outwards from the original regions in which yielding commences. Some parts m a y cease to deform and m a y even unload elastically, particularly when deformation can proceed under falling loads. If, however, the applied loads m u s t be increased in order to continue yielding, an entirely new mode of deformation m a y be" initiated in regions which were previously rigidly constrained throughout. If, in the present problem, the cantilever and its vertical support are one piece of material of uniform width, an alternative mode involving deformation of the support m a y occur, as shown in Fig. 4. The slip-lifie field depends on the value of l/t' (where t' = A B)
Plastic yielding due to bending of cantilevers and fixed ended beams
7
but not of 0. I f Ill' > 1.31 it consists of a circular arc, A Y B (Fig. 4a) with p ---- 0 at its mid-point, Y. The rigid material to the right of A Y B rotates by sliding over the arc. Assuming that the change in shape of the cantilever is negligibIe up to the moment when this alternative defornmtion begins, the relations between S/k and Ill for 8 = 90 ° and 85 ° have been calculated, and are shown in Fig. 3. I t can be seen that S/k is always greater than its corresponding value for the solutions discussed previously, and the smaller 0 the greater the difference. When 8 ---- 90 °, the differences are small if l/'t is large, and consequently the alternative mode soon comes into operation after initial yielding, if the rate of hardening is high. This has been confirmed by experiment (HUNDY 1954). When l i t ' < 1.31 the slip-line fieht (Fig. 4b) includes regions of tension and compression near A and B adjacent to the horizontal faces of the blocks. No calculations have been made for this solution.
Effect of Weaker Support Naturally the load which a cantilever can carry depends on the precise manner in which it is suppported at its fixed end. The solutions so far described have been for perfectly strong rigid support. In order to find out how sensitive to the end conditions the yield-point loads are likely to be, we now consider a weaker method of support. Suppose that a uniformly thick cantilever fits into a horizontal slot A J K B in a rigid vertical support (Fig. 5). Provided that the length of cantilever in the
//!
AIXS~-dl .
_~__~x .
.
.
.
(°) Fig. 5.
.
o
.
,
,
St
5t
(b)
Plane strain solutions for weakly supported uniform cantilevers under shear loading. (a) I ; (b) II.
slot is sufficiently long, it is reasonable to assume that yielding does not extend to the end JK. The beam is supported by an upward pressure over some region to the left of B and a downward pressure over some region to the right of J . The surface of the beam in the region of .4 is assumed to be clear of tlle support. Possible slip-line field solutions of types I and I I are shown in Fig. 5. They differ from those described above in that there is no singularity on the stress-free upper surface (i.e., ~1 = 0), and they are not therefore, symmetrical about Oy.* In II, ~2 = ¼ ~ - - ½ and 2A---- ¼ ~ + ~. The change-over from I I to I occurs when l / t = 7.59. When Ift = 0.71 the mean shear stress across CO is k. Since the solutions are unsymmetrical this may not be the precise value of 1/'t at which solution I I breaks down, but it is probably a close approximation. The lack of * N o t a t i o n : lengths a nd angles whiell are equal in tlw s y n n n e t r i e a l solutions I a n d I I a r e denoted in similar but
unsym~trlcal solutions by the subscripts 1 an(i 2 in tile tensile a nd compressive regions respectively.
8
A.P.
GREEN
s y m m e t r y also m e a n s t h a t 0 no l o n g e r m o v e s v e r t i c a l l y d o w n w a r d s w h e n y i e l d i n g c o m m e n c e s , b u t i n s t e a d m o v e s s l i g h t l y o u t w a r d s in a d i r e c t i o n inclined a t t h e s m a l l angle ~ (say) to t h e v e r t i c a l . T h e c a l c u l a t e d d i m e n s i o n s of t h e fields a n d t h e c o r r e s p o n d i n g v a l u e s of S / k a n d ~ a r e given in T a b l e 2. C o m p a r i n g t h e s e results w i t h t h o s e for s t r o n g s u p p o r t , t h e r e d u c t i o n in S / k is o n l y a b o u t 4 % for t h e s m a l l e s t v a l u e s of I/t c o n s i d e r e d a n d decreases to a b o u t 1 ~ when I//t ,-~ 6 a n d to zero as I/t --> oo. TABLE 2.
PLANE STRAIN.
Weakly supported uniform cantilevers yielding under shear load. l/t
S/k
dl/t
d2/t
R/t
yc
16"69 18'81 11.64 9"97 8 "65 7"59 5"34 8.66 2"34 1.64 1 "35 0"847 0.710
0.0807 0"0372 0.0443 0.0520 0.0602 0.O689 0.0906 0-147 0-234 0.830 0"393 0"558 0.620
0"717 0"719 0 "720 0"723 0"725 0"727 0"706 0-651 0"567 0.473 0-412 0.251 0,190
0"598 0'588 0'579 0'570 0"562 0"553 0.584 0.484 0'408 0"322 0"267 0'121 0'065
0 0 0 0 0 0 0.027 0"097 0'204 0 "322 0.400 0"604 0"681
0"002 0-008 0.008 0"019 0"085 0"047 0.079 0"091
T h e s o l u t i o n s o f F i g . 5 m a y b e a p p l i e d to o t h e r c o m m o n c o n d i t i o n s of s u p p o r t such as t h o s e s h o w n in Figs. 6 a n d 7. T h e c a n t i l e v e r in F i g . 6 is w e l d e d or f i r m l y c l a m p e d t o t h e flat t o p o f a rigid pillar. Fig. 7 shows a s i m p l y s u p p o r t e d b e a m
I
T
St Fig. 6. Uniform cantilever firmly clamped to the top of a rigid vertical pillar.
t
4
[
I st
Fig. 7.
Simply supported beam concentrated load.
under
c a r r y i n g a v e r t i c a l l o a d o v e r a finite l e n g t h B~ B2 = a. More d e t a i l e d e x a m i n a t i o n is r e q u i r e d of t h e stresses in t h e ( a s s u m e d ) rigid region b e l o w B 1 B 2 to d e t e r m i n e h o w l a r g e t h e r a t i o a / d 1 m u s t be t o a v o i d v i o l a t i n g t h e y i e l d c r i t e r i o n there. T h e s o l u t i o n is c l e a r l y i n v a l i d if a / d is v e r y s m a l l b u t is p r o b a b l y v a l i d if a :> ~/2 d I. Since t h e b e n d i n g m o m e n t s a r e e q u a l across t h e b e a m a t B 1 a n d B2, y i e l d i n g t a k e s p l a c e on t h e side for w h i c h t h e y i e l d - p o i n t b e n d i n g m o m e n t is least. T h i s is n o t n e c e s s a r i l y t h e side for w h i c h S is l e a s t (i.e. l g r e a t e s t ) as will be seen l a t e r in t h e discussion o f results ( P a r t I I ) .
Plastic yielding due to b e n d i n B o f cantilevers a n d fixed e n d e d b e a m s
9
(b) Uniform cantilever sheared under axial load The material is assumed to be isotropic and hence solutions with tensile axial loads are identical to those with compressive loads of equal magnitude and S reversed, except t h a t the stresses and strain-rates are reversed at every point in the body. I t is sufficient therefore to consider only compressive loading. Solutions are obtained for a cantilever of uniform thickness t firmly supported in position and direction at the end `4B. The application of pressure upsets the s y m m e t r y of loading, and consequently, the slip-line fields, are also unsymmetrical. The solutions I and I I with P = 0 are modified b y expanding the compressive and contracting the tensile regions* (Fig. 8), st and m a y be conveniently (o) (b) Ib subdivided into two different kinds, which are labelled (a) and (b) according to whether 81 is respectively greater t h a n or equal to zero. As the pressure is increased on a given cantilever Sx decreases. I f P is sufficiently large, ~ = 0, the c o m e r of the tensile region .4 (d) rTb need not be anchored at IIo the corner of the cantiFig. 8. P l a n e strain solutions for strongly s u p p o r t e d lever, and its distance u n i f o r m cantilevers u n d e r c o m b i n e d shear a n d axial increases as the pressure loading : (a) I a ; (b) I b ; (e) I I a ; (d) I I b . is increased. I n the slip-line fields of solution I, (8 x + 82) m u s t be less t h a n (¼w -- ½) to avoid violating the yield criterion in the rigid corner to the right of Y. In solution I I , the H e n c k y relations determine t h a t S1 + 82 = ~w - - ½ and the angular span of the circular arc PQ is 2A = ~ + ½. The ranges of validity of the different types of solution are shown in Fig. 9 in terms of I/t and P. The transition from I to I I occurs when R = 0, and the greater P the smaller is the value of l/t at which it occurs. R decreases with increasing lit for a given pressure, or with increasing pressure for a given lit. B o t h R and 81 are zero when I/t = 5.3 and P / k : 0.66. In solution I (b) as P is increased on a given cantilever both 81 and d 1 decrease, becoming zero simultaneously when S ---- 0, P ---- 2k, and ~ = tan -l [t/2 (l -- t)] = ~ o say. The limiting slip-line field then consists only of the compressive region B E Y and l/t m u s t be greater than 2 for E to lic inside the cantilever. The mode of deformation (rotation a b o u t Y) is unique when d I :> 0 but in the limit d~ = S ~ 0,
Io
c.],=
• See f o o t n o t e
on
p, 7 for
notation.
ofx]~Z C~_ Ap_D _ == _-_--x4
.
.
.
.
.
.
.
10
A, P. GItEEN
there are an infinite n u m b e r of possible modes. P r o v i d e d t h a t there is no constraint on the motion, ~, ean take a n y value between Yo and -- 70 (the condition t h a t the plastic work m u s t be positive precludes tile possibility of y < 70). I n practice the situation would be unstable and the slightest deviation f r o m u n s y m metrical loading would lead to bending of the cantilever with ~, --- q: 7o. Below certain critical values of l/t, which depend on P, yielding extends to the end XOX, and solutions 1 and I I are not valid. A close a p p r o x i n m t i o n to 20
x (0)
I f{~
l(b)
~1,-,
8
C)
\
\
~'-
rr¢o) " % ~
4
...-..~_.
,1 0"73 .............. 0
"/- ........
-
0'5
{'0
_P
-
_-
tit --
{.5
2.0
k
Fig. 9. Range of validity of the different plane strain solutions for a strongly supported uniform
cantilever. this limit is p r o b a b l y provided b y calculating the values of I/t and P at which the m e a n shear stress across OY is equal to k. Thc dashed line in Fig. 9 is the result of such calculations ; below this line solutions I and I I are certainly n o t valid. -k,l t 0.4
O2
t=4
0
0.5
i'O
t.5
2-0
k
Fig. 10. Relations between 8/k and P/k (i.e. yield loci) for strongly supported uniform cantilevers in plane strain.
All results, including the dimensions of the slip-line fields are t a b u l a t e d in Table 8 (no results have been calculated for solution I (b)). The relations between S/k and P/k for various values of I/t are shown in Fig. 10. These curves m a y be
Plastic yielding due to bending of cantilevers and fixed e n d e d b e a m s
]l
r e g a r d e d a s p a r t o f t h e y i e l d loci f o r t h e c a n t i l e v e r s , r e p r e s e n t i n g p o s s i b l e y i e l d point states.
A complete locus would be symmetrical about both axes since the
c a n t i l e v e r is o f s y m m e t r i c a l s h a p e a n d t h e m a t e r i a l is a s s u m e d t o b e i s o t r o p i c .
TABLE 3.
PLANE STRAIN.
Strongly supported uniform cantilevers yielding under combined shear and axial pressure. l/t 8.0
S/k
P/k
81 c
~2 c
dl/t
d2/t
0.583 0.557 0.508 0.223 0.050
0"616 0"657 0.754 1.047 1 "278
0.038 0.021 0 0 o
0.003 0.008 0.018 0.046 0.065
0.0656 0.0646 0.0602 ! 0.0848 i 0.0093
0.098 0.279 _ 0.014 1.401 1-867
0.128 0.083 0 0
0.168 0.208 0.283 0.15') 0"07,1
4"0
0.186 0.184, 0.181 0.124 0"105 0.068 ; 0.034 0.O09
0-092 0-275 0.459 0.658 1.089 1.481 1 "769 1.942
0"123 0-083 0"043 0 0 0 0 0
0"168 0"203 0'243 0'285 0.285 0.210 0'14 ) 0"07 ,)
0.51$ 0.498 0.477 0.469 0.870 0.204, 0.093 0.024
0"552 0.593 0.643 0.706 0.841 1.020 1.170 1-303
O.120 0'107 0-082 0.041 o 0 0 I 0
0"005 0"014 0.025 0.088 0-096 0"108 0.181 0.152
2.0
0.280 0.278 0.273 0.264 0.245 0.222 0.184 0.125 0.068 0.019
0.089
0.123 0.083 0.043 O
0"163 0"203 0.243 0"285 0"285 0"285 0'285 0"285 0.2 O
0.419 0.461 0-512 0.577 0.672 0.758 0.864, 1.001 1.128 1.272
0.284 0.257 ] 0.214 0.170 0.143 O.085 0 0
0.097 0-021 0-086 0.057 0-095 0.129 0.178 0.244 0.809
o.1 o
0.385 0.361 0.345 0.340 0"280 0.240 0.200 0.171 0.085 0.021
0
0.886
0.518 0-517 0"513 0.507 0.499 0"497 0"483
0.084 0-252 0.421 0.608 0.736 0.769 0"980
0-203 0-243 0.285 0.285 0.285 0.285
0"167 0"141 0"123 0"114 0"080 0 "072 0 "020
0.200 0.241 0.290 0.351 0'399 0.411 0"488
0.589 0.577 0.554 0.515 0.500 0"496 0"470
0.005 0"014 0.026 0.043 0-072 0"079 0"129
1.0
0"267 0"445 0"638 0"891 1 "090 1"340 1'618 1.820 1.957
0
0
0 o
0 0 0
0.083 0.043 0 0 0 0
4.
PLANE STRESS SOLUTIONS
Solutions have been obtained only for shearing without axial load, though is i n t e n d e d t o e x t e n d
it
the result to include combined shear and pressure on a
u n i f o r m l y t h i c k c a n t i l e v e r . T h e s y m m e t r i c a l c h a r a c t e r i s t i c fields i n p l a n e s t r e s s f o r t a p e r i n g c a n t i l e v e r s a r e a n a l o g o u s t o t h e s l i p - l i n e fields d e s c r i b e d i n § 3a, a n d r e f e r e n c e will b e m a d e t o t h e s a m e F i g . 2, t h o u g h t h e p r e c i s e s h a p e s o f t h e fields a r e s l i g h t l y d i f f e r e n t ( F i g . 11).
12
A.P. Gmmm
In ASD the field determined b y the straight stress-free edge, consist~of straight lines inclined to the edge at the angle fl ~ tan -1 ~/~ ---- 54o44 , ; the state of stress is uniaxial tension Y = x/8 k parallel to the edge. HILL (1952) has shown t h a t the stresses in the extension of this field round the singularity A, are % = k cos ~,
~ = 2k cos ~b,
~r¢ ----k sin ~b,
(2)
where the polar angle ~b is measured from the base-line indicated in Fig. 11, which is inclined at the angle 2fl to the edge A B. The curved characteristics such as S P have the equation r # sin ~b = const while the other family are straight lines through A. As in the plane strain solutions, 0 moves vertically downwards at yielding, and deformation takes place b y rotation of the rigid cantilever either about the point of stress discontinuity Y (solution I. Fig. 2a) or about a centre of rotation C lying on Oy (solution II, Figs. 2b and 11). The latter solution involves local necking along P Y and local bulging along YQ, with pure shear at Y the mid-point of this are. Local necking along a narrow band through the thickness of the
A /~ ~//'~ ~'~/ ~/ // / f / // / / _ ~ _ : - ~.\, . . ~ / \\ @
.
.
~, o L
the cantilever in this problem) can only occur along a characteristic (HILL 1952). The vector representing the relative velocity / / j of the material on either side of the neck is perpendicular to the ~other characterstic direction and is inclined to the neck at an angle ~b which varies with the stress Fig. 11. Characteristic field for a strongly supported state along the neck (e.g. ~b : 0 uniform cantilever in plane stress. is pure shear and ~b = sin -1 (1//3) in pure tension). The same applies to local bulging (i.e., thickening instead of thinning) though this can only occur with a strain-softening material or with a metal like annealed mild steel* t h a t has a sharp drop in yield stress after the initial yield-point. Whether or not local bulging is in fact possible, solution II provides at least an upper bound to yield-point loads (HILL 1951). The diffuse mode of deformation for a strain hardening material is not known.
Y
The equation linking the state of stress to the state of strain-rate along PYQ is found to be tan A ~ 2 tan ~b, (8) • Local bulging has been obmerved in a mild steel preming by B. P. GREEN and B. B. HUNDY ( u n p u b ~ h e d ) .
P l a s t i c yielding d u e to b e n d i n g of c a n t i l e v e r s a n d fixed e n d e d b e a m s
13
and hence the polar equation of the arc referred to C as origin is r 2 cos A : constant, where the polar angle ~ is measured from C Y O . This is, of course, the same t y p e of curve as t h a t round the singularity at A. F o r the purpose of calculating the resultant forces acting across the arc P Y Q it is convenient to regard C P Y Q as a TABLE 4.
PLANE STRESS.
Stro~gly supported cantilevers with uniform taper yiddi~ under shear load. l/t 90 °
~o
75 °
0"122
0.0489
15 ° 5 4 '
0"178 0 "240
0"439 0"375
0"327
0"290
0.110 0.187 0.290
15 ° 5 4 ' 15 ° 5 4 ' 15 ° 5 4 '
0"420 0 "499 0"555 0"595 0"618
0"200
0.400 0-492
15 ° 5 4 ' 15 ° 5 4 ' 15 ° 5 4 '
2.67 1"94 1 "41 1 "07 0"865
3-01 2.84 1"81
~°
0 15 ° 5 4 ' 15 ° 54"
3.77
6"88 8"88 8.87
R/t 0 0 0.0254
0 0"0799 0-102
~O
d/t 0"612 0"530 0"509 0"489
5'65 4"48
0"750 0-675 0"636 85 °
S/k
0"301 0-319 0-331 0.344
1-36
0"380 0"431 0.501
1.04 0.889
0'579 0.646
0'724
0.691
3"35 2"83 2.41 2-02
1.65 1 "29
0"123 0"070 0"030 0'008
0.557 0.605 0-682
15 ° 5 4 '
0 0-0404
11 ° 3 0 ' 11 ° 8 0 '
0"0749 0"156
11 ° 8 0 '
15 ° 5 4 '
1 "220 0"887 0.809 0-749 0.624 0-502
11 ° 11 ° 11 ° 11 °
30' 80' 30' 30'
0-370 0.246
0"251 0"370 0.4,92
0"148
0-594
0"085
0"662
0.592
1 '206
0
0.595 0-605 0.623
l "050 0.907 0.764 0.613 0.446
1.00
0.653 0.698 0-751
0 0-0907 0'191 0.306 0.446
0"293
0"587
0 7° 7° 7° 7° 7° 7°
0"875
0"780
0"219
0"659
7~ 2V
2"28 2-16 2"05 1.72 1 "47
0-866 0.866
1"183 1"126
0
0-868 0"874 0"885 0.901
1.061
0 0 0"0580 0"216 0"352 0"490
1.24
0"863 0"704
0.550
11 ° 80' 11 ° 8 0 '
21' 11' 11' 21' 21' 2V
3 ° 24" 3 ° 24' 8 ° 24' 3 ° 24' 3 ° 24'
fully plastic region in stress equilibrium, the stresses being given b y equation (2) with ~ = ½~r - - A. The resultant forces are then evaluated over the boundaries C P and CQ instead of integrating t h e m over the arc P Y Q . The angular span
14
A.P. GREEN
of the arc is given by the ~'ondition that the stress-state should be continuous along the continuous characteristic A P Y Q B . Hence it is related to the angle of taper of the cantilever bv the equation 2A - tan -I (½ tan A) ~ 19 + ~Ir -- 2ft.
(4)
The dimensions of the characteristic fields and values of S/k which have been calculated are given in Table 4. The length of A P : BQ is denoted by d, and of CP = CQ by R. The relations between S k and I t are compared in Fig. 3 with those for plane strain. For a given cantilever the plane stress value of S / k is about 14% less t h a n the plane strain value over the whole range of l,t for > 75 °. This was to be expected for long thin cantilevers with a MIsEs yield criterion because there is the same difference between the corresponding yield stresses in compression or tension. I t is interesting, however, that the percentage difference remains almost the same even for the smallest value of lit considered ; in plane stress the stress state in the deforming region then varies between pure tension at the edge and pure shear at the centre, whereas in plane strain it is pure shear throughout. Since the general outlines of the plane stress and plane strain solutions are very similar only the points of difference are mentioned in the following discussion of the two types of solution.
Solution I When 3 ---- 0.
l/t ---- tan 0 (sin 20 -- ~/2 cos 20)/2~/~,
S / k ---- ~/8 sill 20,
(5)
and, as with (3) in plane strain, this limiting solution is valid for all greater values of 1/t. When 0 = 7 2 ° 2 2 ', S----k for all values of l/t and a stress field can be constructed over the whole cantilever similar to t h a t shown in Fig. 2d, the magnitude of the stress parallel to the surfaces being Y = V'Sk instead of 2k. The relation between l/t and S / k for a uniformly thick cantilever (O = ½~r) in the range of solution I (i.e. l/t > 5.65) is given by the equations
[ I/t= ½cot$
1 1 + ~/SsinSsin(B-t- $
)]
sin2~ '
and
S / k - - sln . (~ + ~ "
(6)
Solution I I The variables in the limiting solution when the mean shear stress across Oy is equal to k are again denoted b y the subscript 0. When 8 = 72°24 ', R o = o, lo/t o = 1.89 and So/k = 0.9985. Only over the minute range 72o24 ' > 8 > 72022 ' is this limit reached for solutions of type I. An alternative mode of deformation similar to t h a t shown in Fig. 4a can be constructed b u t no calculations have been made.
ACKNOWLEDGMENTS I am grateful to Mr. D. SPENCELEYof the General Plasticity Section of B.I.S.R.A. for his assistance in calculating the results and drawing the diagrams for this paper which is based on B.I.S.R.A. Reports MW/B/14/58 and MW/B/9/54.
Plastic yielding due to bending of cantilevers and fixed ended beams REFFRENCES
BAKER, J. F. BAKER, J. F., HORNS, M. R. and RODERICK, J. W. GREEN, A. P.
HENDRY, A. W. HILL, R.
15
1949
J. Inst. Cir. Eng., 31, 188.
1949 1953a 1953b 1954 1950 1950
Proc. Roy. Soc. A, 198, 493. Quart. J. Mech. App. Maths., 6, 228. B . I . S . R . A . Report MW/B/14/53. J. Mech. Phys. Solids, 2, 197. The Structural Engineer, 28, 811. The Mathematical Theory of Plasticity (Clarendon
1951 1952 1954 1953 1951 1~o54 1931 1950 1948
Phil. Mag., 42, 868. J Mech. Phys. Solids, 1, 19. J. Mech. Phys. Solids, 2, 000. J. Mech. Phys. Solids, 1,207. Proc. Roy. Soc. A, 207, 216. MetaUurgia, 109. Plasticity, Ch. 22 (McGraw-Hill). Phil. Trans. A, 242, 197. Phil. Mag., 39, 529.
Press, Oxford).
HILL, R. and SIEBEL, M . . P . L . HORNS, M. R. I~[UNDY, B. B. NADAI, A. NF.aL, B. G. RODKRICK, J. W. RODmaICK, J. W. and 1949 PHILLIPS, I. H. SYMONDS, P. S. and NEaL, B. G. 1951
Research Engineering Structures Supplement, p. 9. J. Franklin Inst., 252, 388 and 469.