12.1.

Introduction
Thus far, we have considered loadings that cause the material of a member to behave
elastic ally. We are now concern with the behavior of mac hine and struc tural c omponents
when stresses exc eed the proportional limit. In such cases it is nec essary to make use of
the stressstrain diagrams obtained from an ac tual test of the material, or an idealized
stressstrain diagram. For purposes of analysis, a single mathematical expression will often
be employed for the entire stressstrain diagram. Deformations and stresses in members
made of elastic plastic and rigid-plastic materials having various forms will be determined
under single and combined loadings. Applic ations inc lude c ollapse load of structures, limit
design, membrane analogy, rotating disks, and pressure vessels.

The plastic ity desc ribes the inelastic behavior of a material that retains permanent yielding
on complete unloading. The subject of plastic ity is perhaps best introduced by recalling the
principal c harac teristic s of elastic behavior. First, a material subjected to stressing within
the elastic regime will return to its original state upon removal of those external influenc es
causing application of load or displacement. Second, the deformation c orresponding to a
given stress depends solely on that stress and not on the history of strain or load. In plastic
behavior, opposite c harac teristic s are observed. The permanent distortion that takes place
in the plastic range of a material can assume c onsiderable proportions. This distortion
depends not only on the final state of stress but on the stress states existing from the start
of the loading proc ess as well.

The equations of equilibrium (1.14), the c onditions of compatibility (2.12), and the strain
displacement relationships (2.4) are all valid in plastic theory. New relationships must,
however, be derived to connect stress and strain. The various yield criteria (disc ussed in
Chap. 4), which strictly speaking are not required in solving a problem in elasticity, play a
direct and important role in plasticity. This chapter can provide only an introduction to what
is an ac tive area of contemporary design and researc h in the mec hanics of solids.[*] The
basic s presented c an, however, indicate the potential of the field, as well as its
complexities.
[*] For a de tailed discussion of problem s in plasticity, se e Re fs. 12.1 through 12.3.
12.2. Plastic Deformation
We shall here deal with the permanent alteration in the shape of a polycrystalline solid
subjec t to external loading. The crystals are assumed to be randomly oriented. As has been
demonstrated in Section 2.15, the stresses acting on an elemental c ube c an be resolved
into those assoc iated with change of volume (dilatation) and those c ausing distortion or
change of shape. The distortional stresses are usually referred to as the deviator stresses.

The dilatational stresses, such as hydrostatic pressures, c an clearly decrease the volume
while they are applied. The volume change is rec overable, however, upon removal of
external load. This is bec ause the material c annot be compelled to assume, in the absence
of external loading, interatomic distances different from the initial values. When the
dilatational stresses are removed, therefore, the atoms revert to their original position.
Under the conditions described, no plastic behavior is noted, and the volume is essentially
unc hanged upon removal of load.

In contrast with the situation described, during change of shape, the atoms within a c rystal
of a polycrystalline solid slide over one another. This slip action, referred to as disloc ation, is
a c omplex phenomenon. Disloc ation c an oc cur only by the shearing of atomic layers, and
consequently it is primarily the shear c omponent of the deviator stresses that controls
plastic deformation. Experimental evidence supports the assumption that assoc iated with
plastic deformation essentially no volume change occ urs; that is, the material is
incompressible (Sec. 2.10):

Equation 12.1

Therefore, from Eq. (2.39), Poissons ratio for a plastic material.

Slip begins at an imperfec tion in the lattice, for example, along a plane separating two
regions, one having one more atom per row than the other. Because slip does not occ ur
simultaneously along every atomic plane, the deformation appears discontinuous on the
microscopic level of the crystal grains. The overall effect, however, is plastic shear along
certain slip planes, and the behavior described is approximately that of the ideal plastic
solid. As the deformation c ontinues, a loc king of the disloc ations takes place, resulting
in strain hardening or cold working.

In performing engineering analyses of stresses in the plastic range, we do not usually need
to c onsider disloc ation theory, and the explanation offered previously, while overly simple,
will suffic e. What is of great importanc e to the analyst, however, is the experimentally
determined c urves of stress and strain.
12.3. Idealized StressStrain Diagrams
The bulk of present-day analysis in plasticity is predic ated on materials displaying idealized
stressstrain c urves as in Fig. 12.1aand c. Suc h materials are referred to as rigid, perfec tly
plastic and elastic -perfec tly plastic, respec tively. Examples include mild steel, c lay, and
nylon, which exhibit negligible elastic strains in c omparison with large plastic deformations at
practic ally constant stress. A more realistic portrayal inc luding strain hardening is given
in Fig. 12.1b for what is called a rigid-plastic solid. In the c urves, a and b designate the
tensile yield and ultimate stresses, yp and u, respec tively. The curves of Fig.
12.1c and d do not ignore the elastic strain, which must be inc luded in a more general
stressstrain depic tion. The latter figures thus represent idealized elasticplastic diagrams
for the perfec tly plastic and plastic materials, respec tively. The material in Figs.
12.c and d is also called a strain-hardening or work-hardening material. For a linearly
strain-hardening material, the regions ab of the diagrams become a sloped straight line.

Figure 12.1. Idealized stressstrain diagrams: (a) rigid, perfectly plastic material;
(b) rigid-plastic material; (c) elastic, perfectly plastic material; and (d) elastic, rigid-
plastic material.

[View full size im age ]

True StressTrue Strain Relationships

Consider the idealized true stressstrain diagram given in Fig. 12.2. The frac ture stress and
fracture strain are denoted by f and f, respectively. The fundamental design utility of the
plot ends at the ultimate stress u. Note that a c omparison of a true and nominal plot
is shown in Fig. 2.12. In addition to the preceding idealized cases, various theoretical
analyses have been advanc ed to predic t plastic behavior. Equations relating stress and
strain beyond the proportional limit range from the rather empirical to those leading to
complex mathematic al approaches applic able to materials of specific type and structure. For
purposes of an ac curate analysis, a single mathematical expression is frequently employed
for the entire stressstrain curve. An equation representing the range of diagrams,
especially useful for aluminum and magnesium, has been developed by Ramberg and Osgood
[Ref. 12.4]. We c onfine our discussion to that of a perfectly plastic material displaying a
horizontal straight-line relationship and to the parabolic relationship desc ribed next.

Figure 12.2. Idealized true stressstrain diagram.

For many materials, the entire true stresstrue strain c urve may be represented by the
parabolic form

Equation 12.2

where n and K are the strain-hardening index and the strength c oeffic ient, respectively.
The definitions of true stress and true strain are given in Section 2.7. The curves of Eq.
(12.2) are shown in Fig. 12.3a. We observe from the figure that the sloped/d grows
without limit as approac hes zero for n 1. Thus, Eq. (12.2) should not be used for small
strains when n 1. The stressstrain diagram of a perfec tly plastic material is represented
by this equation when n = 0 (and hence K = yp). Clearly, for elastic materials (n = 1 and
hence K = E), Eq. (12.2) represents Hookes law, wherein E is the modulus of elasticity.

Figure 12.3. (a) Graphical representation of = K n; (b) true stress versus true
strain on loglog coordinates.

For a partic ular material, with true stresstrue strain data available, K and n are readily
evaluated inasmuc h as Eq. (12.2) plots as a straight line on logarithmic coordinates. We
can thus rewrite Eq. (12.2) in the form

Equation a

Here n is the slope of the line and K the true stress associated with the true strain at 1.0
on the loglog plot (Fig. 12.3b). The strain-hardening c oeffic ient n for commerc ially used
materials falls between 0.2 and 0.5. Fortunately, with the use of computers, much refined
modeling of the stressstrain relations for real material is possible.
12.4. Instability in Simple Tension
We now desc ribe an instability phenomenon in uniaxial tension of practic al importance in
predic ting the maximum allowable plastic stress in a rigid-plastic material. At the ultimate
stress u in a tensile test (Fig. 12.2), an unstable flow results from the effec ts of strain
hardening and the decreasing cross-sectional area of the spec imen. These tend to weaken
the material. When the rate of the former effec t is less than the latter,
an instability occ urs. This point c orresponds to the maximum tensile load and is defined by

Equation a

Sinc e axial load P is a function of both the true stress and the area (P = A), Eq. (a) is
rewritten

Equation b

The condition of inc ompressibility, Ao Lo = AL, also yields

Equation c

as the original volume Ao Lo is c onstant. Expressions (b) and (c) result in

Equation d

From Eqs. (2.25) and (d), we thus obtain the relationships

Equation 12.3

for the instability of a tensile member. Here the subscript o denotes the engineering strain
and stress (Sec. 2.7).

Introduction of Eq. (12.2) into Eq. (12.3) results in

or

Equation 12.4

That is, at the instant of instability of flow in tension, the true strain has the same
numerical value as the strain-hardening index. The state of true stress and the true strain
under uniaxial tension are therefore

Equation e

The problem of instability under simple compression or plastic buckling is disc ussed
in Section 11.6. The instability condition for c ases involving biaxial tension is derived
in Sections 12.12 and 12.13.

Example 12.1. Three-Bar Structure

Determine the maximum allowable plastic stress and strain in the pin-jointed
structure sustaining a vertical load P, shown in Fig. 12.4. Assume that = 45
and that each element is construc ted of an aluminum alloy with the following
properties:

yp = 350 MPa, K = 840 MPa, n = 0.2

AAD = ACD = 10 105 m2, ABD = 15 105 m2, h = 3 m

Figure 12.4. Example 12.1. Pin-connected structure in axial tension.

Solution
The structure is elastically statically indeterminate, and the solution may readily
be obtained on applying Castiglianos theorem (Sec. 10.7). Plastic yielding begins
upon loading:

P = yp ABD + 2 yp AAD cos

= 350 106[15 + 2 10 cos 45]105 = 101,990 N

On applying Eqs. (e), the maximum allowable stress

= Knn = 840 106(0.2)0.2 = 608.9 MPa

occ urs at the following axial and transverse strains:

1 = n = 0.2, 2 = 3 = 0.1

We have L = h/cos . The total elongations for instability of the bars are
thus BD = 3(0.2) = 0.6 m and AB = CD= (3/cos 45)(0.2) = 0.85 m.

Example 12.2. Tube in Axial Tension

A tube of original mean diameter do and thickness t o is subjected to axial tensile

loading. Assume a true stressengineering strain relation of the form
and derive expressions for the thic kness and diameter at the instant of
instability. Let n = 0.3.

Solution

Differentiating the given expression for stress,

This result and Eq. (12.3) yield the engineering axial strain at instability:

Equation f

The transverse strains are o/2, and hence the dec rease of wall thickness
equals nt o/2(1 n). The thickness at instability is thus
Equation g

Similarly, the diameter at instability is

Equation h

From Eqs. (g) and (h), with n = 0.3, we have t = 0.79t o and d = 0.79do. Thus,
for the tube under axial tension, the diameter and thickness dec rease
approximately 21% at the instant of instability.
12.5. Plastic Axial Deformation and Residual Stress
As pointed out earlier, if the stress in any part of the member exc eeds the yield strength of
the material, plastic deformationsoccur. The stress that remains in a structural member
upon removal of external loads is referred to as the residual stress. The presence of residual
stress may be very harmful or, if properly controlled, may result in substantial benefit.

Plastic behavior of ductile materials may be conveniently represented by considering an

idealized elastoplastic material shown inFig. 12.5, where yp and yp designate the yield
strength and yield strain, respectively. The Y corresponds to the onset of yield in the
material. The part OB is the strain c orresponding to the plastic deformation that results from
the loading and unloading of the specimen along line AB parallel to the initial portion OY of
the loading c urve.

Figure 12.5. Tensile stressstrain diagram for elastoplastic material.

The distribution of residual stresses can be found by superposition of the stresses owing to
loading and the reverse, or rebound,stresses due to unloading. (The strains c orresponding
to the latter are the reverse, or rebound, strains.) The reverse stress pattern is assumed to
be fully elastic and c onsequently can be obtained applying Hookes law. That means the
linear relationship between and remains valid, as illustrated by line AB in the figure. We
note that this superposition approac h is not valid if the residual stresses thereby found
exceed the yield strength.

The nonuniform deformations that may be c aused in a material by plastic bending and
plastic torsion are c onsidered in Sections 12.7 and 12.9. This sec tion is concerned with a
restrained or statically indeterminate structure that is axially loaded beyond the elastic
range. For such a case, some members of the structure experience different plastic
deformation, and these members retain stress following the release of load.

The magnitude of the axial load at the onset of yielding, or yield load Pyp, in a statically
determinate ductile bar of cross-sec tional area A is yp A. This also is equal to the plastic ,
limit, or ultimate load Pu of the bar. For a statically indeterminate structure, however, after
one member yields, additional load is applied until the remaining members also reach their
yield limits. At this time,unrestricted or uncontained plastic flow oc curs, and the limit
load Pu is reac hed. Therefore, the ultimate load is the load at which yielding begins
in all materials.

Examples 12.3, 12.7, and 12.12 illustrate how plastic deformations and residual stresses
are produced and how the limit load is determined in axially loaded members, beams, and
torsional members, respectively.
Example 12.3. Residual Stresses in an Assembly

Figure 12.6a shows a steel bar of 750-mm2 c ross-sectional area plac ed

between two aluminum bars, each of 500-mm2 cross-sectional area. The ends of
the bars are attac hed to a rigid support on one side and a rigid thic k plate on
the other. Given: Es = 210 GPa, ( s)yp = 240 MPa, Ea = 70 GPa, and ( a )yp =
320 MPa. Assumption: The material is elasticplastic .

Figure 12.6. Example 12.3. (a) Plastic analysis of a statically

indeterminate three-bar structure; (b) free-body diagram of end plate.

[View full size im a ge ]

Calculate the residual stress, for the c ase in whic h applied load P is increased
from zero to Pu and removed.

Solution

Material Behavior. At ultimate load Pu both materials yield. Either material

yielding by itself will not result in failure because the other material is still in the
elastic range. We therefore have

Pa = 500(320) = 160 kN, Ps = 750(240) = 180 kN

Henc e,

Pu = 2(160) + 180 = 500 kN

Applying an equal and opposite load of this amount, equivalent to a release load
(Fig. 12.6b), causes each bar to rebound elastically.

Geometry of Deformation. Condition of geometric fit, a = s gives

Equation a
Condition of Equilibrium. From the free-body diagram of Fig. 12.6b,

Equation b

Solving Eqs. (a) and (b) we obtain

Superposition of the initial forc es at ultimate load Pu and the elastic rebound
forc es owing to release of Pu results in:

(Pa )res = 79.1 160 = 80.9 kN

(Ps)res = 341.7 180 = 161.7 kN

The associated residual stresses are thus

Comment

We note that after this prestressing process, the assembly remains elastic as
long as the value of Pu = 500 kN is not exceeded.
12.6. Plastic Deflection of Beams
In this sec tion we treat the inelastic deflec tion of a beam, employing the mec hanics of
materials approach. Consider a beam of rec tangular section, as in Fig. 12.7a, wherein the
bending moment M produces a radius of curvature r. The longitudinal strain of any fiber
located a distance y from the neutral surfac e, from Eq. (5.9), is given by

Figure 12.7. (a) Inelastic bending of a rectangular beam in pure bending; (b) stress
strain diagram for an elastic-rigid plastic material.

Equation 12.5

Assume the beam material to possess equal properties in tension and c ompression. Then,
the longitudinal tensile and compressive forces cancel, and the equilibrium of axial forces is
satisfied. The following describes the equilibrium of moments about the z axis (Fig. 12.7a):

Equation a

For any spec ific distribution of stress, as, for example, that shown in Fig. 12.7b, Eq. (a)
provides M and then the deflection, as is demonstrated next.

Consider the true stresstrue strain relationship of the form = K n. Introduc ing this
together with Eq. (12.5) into Eq. (a), we obtain

Equation b

where
Equation 12.6

From Eqs. (12.5), = K n, and (b) the following is derived:

Equation c

In addition, on the basis of the elementary beam theory, we have, from Eq. (5.7),

Equation d

Upon substituting Eq. (c) into Eq. (d), we obtain the following equation for a rigid plastic
beam:

Equation 12.7

It is noted that when n = 1 (and henc e K = E), this expression, as expected, reduces to
that of an elastic beam [Eq. (5.10)].

Example 12.4. Rigid-Plastic Simple Beam

Determine the deflection of a rigid-plastic simply supported beam subjec ted to a

downward concentrated forc e Pat its midlength. The beam has a rectangular
c ross sec tion of depth 2h and width b (Fig. 12.8).

Figure 12.8. Examples 12.4 and 12.8.

 Solution

The bending moment for segment AC is given by

Equation e

where the minus sign is due to the sign convention of Section 5.2.

Substituting Eq. (e) into Eq. (12.7) and integrating, we have

Equation f

where

Equation g

The constants of integration c 1 and c 2 depend on the boundary conditions v(0)

= dv/dx(L/2) = 0:

Upon introduction of c 1 and c 2 into Eq. (f), the beam deflection is found to be

Equation 12.8

Interestingly, in the case of an elastic beam, this becomes

 For x = L/2, the familiar result is

The foregoing proc edure is applicable to the determination of the deflec tion of beams
subjec t to a variety of end conditions and load c onfigurations. It is clear, however, that
owing to the nonlinearity of the stress law, = K n, the principle of superposition cannot
validly be applied.
12.7. Analysis of Perfectly Plastic Beams
By neglec ting strain hardening, that is, by assuming a perfec tly plastic material, considerable
simplific ation c an be realized. We shall, in this section, foc us our attention on the analysis of a
perfec tly plastic straight beam of rec tangular section subject topure bending (Fig. 12.7a).

The bending moment at whic h plastic deformation impends, Myp, may be found directly from the
flexure relationship:

Equation 12.9a

or

Equation 12.9b

Here yp represents the stress at which yielding begins (and at which deformation continues in a
perfec tly plastic material). The quantity S is the elastic modulus of the cross section. Clearly, for the
beam c onsidered, . The stress distribution corresponding to Myp,
assuming identic al material properties in tension and compression, is shown in Fig. 12.9a. As the
bending moment is increased, the region of the beam that has yielded progresses in toward the
neutral surface (Fig. 12.9b). The distance from the neutral surface to the point at which yielding
begins is denoted by the symbol e, as shown.

Figure 12.9. Stress distribution in a rectangular beam with increase in bending moment: (a)
elastic; (b) partially plastic; (c) fully plastic.

It is c lear, upon examining Fig. 12.9b, that the normal stress varies in ac cordance with the relations

Equation a
and

Equation b

It is useful to determine the manner in whic h the bending moment M relates to the distanc e e. To do
this, we begin with a statement of the x equilibrium of forces:

Canc elling the first and third integrals and c ombining the remaining integral with Eq. (a), we have

This expression indicates that the neutral and centroidal axes of the cross sec tion coinc ide, as in
the case of an entirely elastic distribution of stress. Note that in the case of a nonsymmetric al c ross
section, the neutral axis is generally in a loc ation different from that of the centroidal axis
[Ref. 12.5].

Next, the equilibrium of moments about the neutral axis provides the following relation:

Substituting x from Eq. (a) into this equation gives the elastic plastic moment, after integration,

Equation 12.10a

Alternatively, using Eq. (12.9a), we have

Equation 12.10b
The general stress distribution is thus defined in terms of the applied moment, inasmuch as e is
c onnec ted to x by Eq. (a). For the c ase in which e = h, Eq. (12.10) reduces to Eq. (12.9)
and M = Myp. For e = 0, whic h applies to a totally plastic beam, Eq. (12.10a) becomes

Equation 12.11

where Mu is the ultimate moment. It is also referred to as the plastic moment.

Through applic ation of the foregoing analysis, similar relationships can be derived for other c ross-
sectional shapes. In general, for any cross section, the plastic or ultimate resisting moment for a
beam is

Equation 12.12

where Z is the plastic section modulus. Clearly, for the rec tangular beam analyzed here, Z = bh2.
The Steel Construc tion Manual(Sec. 11.7) lists plastic section moduli for many common geometries.
Interestingly, the ratio of the ultimate moment to yield moment for a beam depends on the geometric
form of the cross section, called the shape fac tor:

Equation 12.13

For instance, in the c ase of a perfec tly plastic beam with (b 2h) with rectangular c ross sec tion,
we have andZ = bh2. Substitution of these area properties into the preceding
equation results in f = 3/2. Observe that the plastic modulus represents the sum of the first
moments of the areas (defined in Sec. C.1) of the cross section above and below the neutral
axisz (Fig. 12.7a): Z = bh(h/2 + h/2) = bh2.

As noted earlier, when a beam is bent inelastically, some plastic deformation is produc ed, and the
beam does not return to its initial c onfiguration after load is released. There will be some residual
stresses in the beam (see Example 12.7). The foregoing stresses are found by using the principle of
superposition in a way similar to that desc ribed in Section 12.5 for axial loading. The unloading
phase may be analyzed by assuming the beam to be fully plastic , using the flexure formula, Eq. (5.4).

Plastic Hinge
We now consider the momentc urvature relation of an elastoplastic rectangular beam in pure
bending (Fig. 12.7a). As desc ribed in Section 5.2, the applied bending moment M produces a
c urvature which is the reciprocal of the radius of curvature r. At the elastic plastic boundary of the
beam (Fig. 12.9b), the yield strain is x = yp. Then, Eq. (12.5) leads to

Equation c
The quantities 1/r and 1/r yp c orrespond to the curvatures of the beam after and at the onset of
yielding, respec tively. Through the use of Eqs. (c),

Equation 12.14

Carrying Eqs. (12.14) and (12.9) into Eq. (12.10b) give the required momentc urvature relationship
in the form:

Equation 12.15

We note that Eq. (12.15) is valid only if the bending moment becomes greater than Myp. When M <
Myp, Eq. (5.9a) applies. The variation of M with curvature is illustrated in Fig. 12.10. Observe
that M rapidly approac hes the asymptotic value , which is the plastic moment Mu. If 1/r =
2/r yp or e = h/2, eleven-twelfths of Mu has already been reac hed. Obviously, positions A, B, andE of
the curve correspond to the stress distributions depic ted in Fig. 12.9. Removing the load at C, for
example, elastic rebound occurs along the line CD, and point D is the residual c urvature.

Figure 12.10. Momentcurvature relationship for a rectangular beam.

Figure 12.10 depic ts the three stages of loading. There is an initial range of linear elastic response.
This is followed by a curved line representing the region in whic h the member is partially plastic and
partially elastic . This is the region of contained plastic flow. Finally, the member continues to yield
with no inc rease of applied bending moment. Note the rapid ascent of eac h c urve toward
its asymptote as the sec tion approac hes the fully plastic condition. At this stage, unrestric ted
plastic flow occ urs and the corresponding moment is the ultimate or plastic hinge moment Mu.
Therefore, the c ross section will abruptly continue to rotate; the beam is said to have developed
a plastic hinge. The rationale for the term hinge becomes apparent upon describing the behavior of a
beam under a c oncentrated loading, discussed next.

Consider a simply supported beam of rec tangular c ross section, subjected to a load P at its midspan
(Fig. 12.11a). The corresponding bending moment diagram is shown in Fig. 12.11b. Clearly,
when Myp < |PL/4| < Mu, a region of plastic deformation occ urs, as indic ated in the figure by the
shaded areas. The depth of penetration of these zones can be found from h e, where eis
determined using Eq. (12.10a), because M at midspan is known. The length of the middle portion of
the beam where plastic deformation occurs c an be readily determined with reference to the figure.
The magnitude of bending moment at the edge of the plastic zone is Myp = (P/2) (L Lp)/2, from
whic h

Figure 12.11. Bending of a rectangular beam: (a) plastic region; (b) moment diagram; (c)
plastic hinge.

Equation d

With the increase of P, Mmax Mu, and the plastic region extends farther inward. When the
magnitude of the maximum momentPL/4 is equal to Mu, the c ross sec tion at the midspan becomes
fully plastic (Fig. 12.11c). Then, as in the case of pure bending, the curvature at the center of the
beam grows without limit, and the beam fails. The beam halves on either side of the midspan
experienc e rotation in the manner of a rigid body about the neutral axis, as about a plastic
hinge, under the influenc e of the constant ultimate moment Mu. For a plastic hinge, P = 4Mu/L is
substituted into Eq. (d), leading to Lp = L(1 Myp/Mu).

The capac ity of a beam to resist collapse is revealed by c omparing Eqs. (12.9a) and (12.11). Note
that the Mu is 1.5 times as large as Myp. Elastic design is thus c onservative. Considerations such as
this lead to c oncepts of limit design in struc tures, disc ussed in the next section.

Example 12.5. ElasticPlastic Analysis of a Link: Interaction Curves

A link of rectangular c ross sec tion is subjected to a load N (Fig. 12.12a). Derive general
relationships involving Nand M that govern, first, the case of initial yielding and, then,
fully plastic deformation for the straight part of the link of length L.

Figure 12.12. Example 12.5. (a) A link of rectangular cross section carries load P
at its ends; (b) elastic stress distribution; (c) fully plastic stress distribution.

[View full size im a ge ]

Solution

Suppose N and M are such that the state of stress is as shown in Fig. 12.12b at any
straight beam section. The maximum stress in the beam is then, by superposition of the
axial and bending stresses,

Equation e

The upper limits on N (M = 0) and M (N = 0), c orresponding to the condition of yielding,

are

Equation f

Substituting 2hb and I/h from Eq. (f) into Eq. (e) and rearranging terms, we have

Equation 12.16
If N1 is zero, then M1 must achieve its maximum value Myp for yielding to impend.
Similarly, for M1 = 0, it is nec essary for N1 to equal Nyp to initiate yielding. Between
these extremes, Eq. (12.16) provides the infinity of combinations of N1 and M1 that will
result in yp.

For the fully plastic case (Fig. 12.12c), we shall denote the state of loading
by N2 and M2. It is apparent that the stresses ac ting within the range
e < y < e contribute pure axial load only. The stresses within the range e < y <h and
e > y > h form a couple, however. For the total load system described, we may write

Equation g

Equation h

Introduc ing Eqs. (12.11) and (g) into Eq. (h), we have

Finally, dividing by Mu and noting that and Nyp = 2bh yp, we

obtain

Equation 12.17

Figure 12.13 is a plot of Eqs. (12.16) and (12.17). By employing these interac tion
c urves, any combination of limiting values of bending moment and axial force is easily
arrived at.

Figure 12.13. Example 12.5 (continued). Interaction curves for N and M for a
rectangular cross-sectional member.
Let, for instanc e, d = h, h = 2b = 24 mm (Fig. 12.12a), and yp = 280 MPa. Then the
value of and, from Eq. (e), Myp/Nyp = h/3 = 8. The radial line representing

is indicated by the dashed line in the figure. This line intersects

the interaction curves at A(0.75, 0.25) and B(1.24, 0.41). Thus, yielding impends
for N1 = 0.25Nyp = 0.25 (2bh yp) = 40.32 kN, and for fully plastic deformation, N2 =
0.41Nyp = 66.125 kN.

Note that the distance d is assumed constant and the values of N found are
c onservative. If the link deflection were taken into acc ount, d would be smaller and the
c alculations would yield larger N.

Example 12.6. Shape Factor of an I-Beam

An I-beam (Fig. 12.14a) is subjected to pure bending resulting from end c ouples.
Determine the moment causing initial yielding and that results in complete plastic
deformation.

Figure 12.14. Example 12.6. (a) Cross section; (b) fully plastic stress distribution.
 Solution

The moment corresponding to yp is, from Eq. (12.9a),

Refer now to the completely plastic stress distribution of Fig. 12.14b. The moments of
forc e owing to yp, taken about the neutral axis, provide

Combining the prec eding equations, we have

Equation i

From this expression, it is seen that , while it is for a beam of rectangular

sec tion (h1 = 0). We conclude, therefore, that if a rec tangular beam and an I-beam are
designed plastic ally, the former will be more resistant to c omplete plastic failure.

Example 12.7. Residual Stresses in a Rectangular Beam

Figure 12.15 shows an elastoplastic beam of rec tangular c ross sec tion 40 mm by 100
mm c arrying a bending moment of M. Determine (a) the thickness of the elastic c ore; (b)
the residual stresses following removal of the bending moment. Given: b = 40 mm, h = 50
mm, M = 21 kN m, yp = 240 MPa, and E = 200 GPa.

Figure 12.15. Example 12.7. Finding the thickness (2e) of the elastic core.

Solution

a. Through the use of Eq. (12.9a), we have

 Then Eq. (12.10b) leads to

Solving,

e = 31 103 m = 31 mm and 2e = 62 mm

Elastic c ore depicted as shaded in Fig. 12.15.

b. The stress distribution c orresponding to moment M = 21 kN m is illustrated in Fig.

12.16a. The release of moment M produces elastic stresses, and the flexure
formula applies (Fig. 12.16b). Equation (5.5) is therefore

Figure 12.16. Example 12.7. Stress distribution in a rectangular beam:

(a) elasticplastic; (b) elastic rebound; (c) residual.

[View full size image]

By the superimposition of the two stress distributions, we can find the residual
stresses (Fig. 12.16c). Observe that both tensile and compressive residual
stresses remain in the beam.

Example 12.8. Deflection of a Perfectly Plastic Simple Beam

Determine the maximum deflection due to an applied forc e P acting on the perfectly
plastic simply supported rec tangular beam (Fig. 12.8).

Solution

The c enter deflec tion in the elastic range is given by

Equation j

At the start of yielding,

Equation k

Expression (j), together with Eqs. (k) and (12.9a), leads to

Equation l

In a like manner, we obtain

Equation m

for the center deflection at the instant of plastic c ollapse.

12.8. Collapse Load of Structures: Limit Design
On the basis of the simple examples in the previous section, it may be deduced that
struc tures may withstand loads in excess of those that lead to initial yielding. We recognize
that, while such loads need not cause structural collapse, they will result in some amount of
permanent deformation. If no permanent deformation is to be permitted, the load
configuration must be suc h that the stress does not attain the yield point anywhere in the
struc ture. This is, of course, the basis of elastic design.

When a limited amount of permanent deformation may be tolerated in a structure, the

design c an be predicated on higher loads than correspond to initiation of yielding. On the
basis of the ultimate or plastic load determination, safe dimensions can be determined in
what is termed limit design. Clearly, such design requires higher than usual factors of
safety. Examples of ultimate load determination are presented next.

Consider first a built-in beam subjec ted to a c oncentrated load at midspan (Fig. 12.17a).
The general bending moment variation is sketc hed in Fig. 12.17b. As the load is
progressively inc reased, we may anticipate plastic hinges at points 1, 2, and 3, because
these are the points at which maximum bending moments are found. The c onfiguration
indic ating the assumed location of the plastic hinges (Fig. 12.17c) is the mechanism of
collapse. At every hinge, the hinge moment must clearly be the same.

Figure 12.17. (a) Beam with built-in ends; (b) elastic bending moment diagram; (c)
mechanism of collapse with plastic hinges at 1 through 3.

The equilibrium and the energy approaches are available for determination of the ultimate
loading. Elec ting the latter, we refer toFig. 12.17c and note that the change in energy
associated with rotation at points 1 and 2 is Mu , while at point 3, it isMu(2). The
work done by the c onc entrated force is P v. Ac cording to the principle of virtual work, we
may write

Pu(v) = Mu() + Mu() + Mu(2) = 4Mu()

where Pu represents the ultimate load. Because the deformations are limited to small values,
it may be stated that and . Substituting in the prec eding expression
for v, it is found that

Equation a

where Mu is calc ulated for a given beam using Eq. (12.12). It is interesting that, by
introduction of the plastic hinges, the originally static ally indeterminate beam is rendered
determinate. The determination of Pu is thus simpler than that of Pyp, on whic h elastic
analysis is based. An advantage of limit design may also be found in noting that a small
rotation at either end of the beam or a slight lowering of a support will not influenc e the
value of Pu. Moderate departures from the ideal case, such as these, will, however, have a
pronounc ed effec t on the value of Pyp in a statically indeterminate system.

While the positioning of the plastic hinges in the preceding problem is limited to the single
possibility shown in Fig. 12.17c, more than one possibility will exist for situations in which
several forces act. Correspondingly, a number of collapse mechanisms may exist, and it is
incumbent on the designer to select from among them the one associated with the lowest
load.

Example 12.9. Collapse Analysis of a Continuous Beam

Determine the c ollapse load of the c ontinuous beam shown in Fig. 12.18a.

Figure 12.18. Example 12.9. (a) A beam is subjected to loads P and 2P;
(bd) mechanisms of collapse with plastic hinges at 2 through 4.
 Solution

The four possibilities of collapse are indic ated in Figs. 12.18b through d. We
first c onsider the mechanism of Fig. 12.18b. In this system, motion occurs
bec ause of rotations at hinges 1, 2, and 3. The remainder of the beam remains
rigid. Applying the principle of virtual work, noting that the moment at point 1 is
zero, we have

P(v) = Mu(2) + Mu() = 3Mu()

Bec ause

this equation yields Pu = 6Mu/L.

For the collapse mode of Fig. 12.18c,

and thus Pu = 2Mu/L.

The collapse mec hanisms indicated by the solid and dashed lines of Fig.
12.18d are unac ceptable because they imply a zero bending moment at section
3. We conc lude that c ollapse will occur as in Fig. 12.17c, when P 2Mu/L.
Example 12.10. Collapse Load of a Continuous Beam

Determine the c ollapse load of the beam shown in Fig. 12.19a.

Figure 12.19. Example 12.10. (a) Variously loaded beam; (b) mechanism
of collapse with plastic hinges at 2 and 3.

Solution

There are a number of c ollapse possibilities, of which one is indicated in Fig.

12.19b. Let us suppose that there exists a hinge at point 2, a distance e from
the left support. Then examination of the geometry leads to 1 = 2(L
e)/e or 1 + 2 = L2/e. Applying the principle of virtual work,

or

from which

Equation b

The minimization c ondition for p in Eq. (b), dp/de = 0, results in

Equation c
 Thus, Eqs. (b) and (c) provide a possible collapse configuration. The remaining
possibilities are similar to those discussed in the previous example and should be
c hecked to ascertain the minimum collapse load.

Determination of the collapse load of frames involves muc h the same analysis. For c omplex
frames, however, the approaches used in the foregoing examples would lead to extremely
cumbersome c alculations. For these, spec ial-purpose methods are available to provide
approximate solutions [Ref. 12.6].

Example 12.11. Collapse Analysis of a Frame

Apply the method of virtual work to determine the c ollapse load of the structure
shown in Fig. 12.20a. Assume that the rigidity of member BC is 1.2 times
greater than that of the vertic al members AB and CD.

Figure 12.20. Example 12.11. (a) A frame with concentrated loads; (b and
c) mechanism of collapse with plastic hinges at A, B, C, and D.
 Solution

Of the several collapse modes, we c onsider only the two given in Figs.
12.20b and c. On the basis of Fig. 12.20b, plastic hinges will be formed at the
ends of the vertic al members. Thus, from the principle of virtual work,

Substituting u = L, this expression leads to Pu = 8Mu/L.

Referring to Fig. 12.20c, we have MuE = 1.2Mu, where Mu is the c ollapse

moment of the vertic al elements. Applying the principle of virtual work,

Noting that u = L and , this equation provides the following expression

for the collapse load: Pu = 2.56Mu/L.
12.9. ElasticPlastic Torsion of Circular Shafts
We now consider the torsion of c ircular bars of ductile materials, which are idealized as
elastoplastic , stressed into the plastic range. In this case, the first two basic assumptions
associated with small deformations of circular bars in torsion (see Sec. 6.2) are still valid.
This means that the c ircular cross sections remain plane and their radii remain straight.
Consequently, strains vary linearly from the shaft axis. The shearing stressstrain curve of
plastic materials is shown in Fig. 12.21. Referring to this diagram, we can proceed as
discussed before and determine the stress distribution across a section of the shaft for any
given value of the torque T.

Figure 12.21. Idealized shear stressshear strain diagrams for (a) perfectly plastic
materials; (b) elastoplastic materials.

[View full size im age ]

The basic relationships given in Section 6.2 are applic able as long as the shear strain in the
bar does not exceed the yield strain yp. It is recalled that the condition of torque
equilibrium for the entire shaft (Fig. 6.2) requires

Equation a

Here , are any arbitrary distance and shearing stress from the c enter O, respec tively,
and A the entire area of a cross section of the shaft. Inc reasing in the applied torque,
yielding impends on the boundary and moves progressively toward the interior. The cross-
sectional stress distribution will be as shown in Fig. 12.22.

Figure 12.22. Stress distribution in a shaft as torque is increased: (a) onset of yield;
(b) partially plastic; and (c) fully plastic.

[View full size im age ]

At the start of yielding (Fig. 12.22a), the torque T yp, through the use of Eq. (6.1), may be
written in the form:

Equation 12.18

The quantity J = c 4/2 is the polar moment of inertia for a solid shaft with radius r = c .
Equation (12.18) is called the maximum elastic torque, or yield torque. It represents the
largest torque for which the deformation remains fully elastic .

If the twist is increased further, an inelastic or plastic portion develops in the bar around an
elastic core of radius 0 (Fig. 12.22b). Using Eq. (a), we obtain that the torque resisted by
the elastic c ore equals

Equation b

The outer portion is subjected to c onstant yield stress yp and resists the torque,

Equation c

The elastic plastic or total torque T, the sum of T 1 and T 2, may now be expressed as
follows

Equation 12.19
When twisting bec omes very large, the region of yielding will approac h the middle of the
shaft and will approach zero (Fig. 12.22c). The corresponding torque T u is the plastic, or
ultimate, shaft torque, and its value from the foregoing equation is

Equation 12.20

It is thus seen that only one-third of the torque-carrying capacity remains after yp is
reached at the outermost fibers of a shaft.

The radius of elastic core (Fig. 12.22b) is found, referring to Fig. 6.2, by
setting = yp and = 0. It follows that

Equation 12.21a

in whic h L is the length of the shaft. The angle of twist at the onset of
yielding yp (when 0 = c ) is therefore

Equation 12.21b

Equations (12.21) lead to the relation,

Equation 12.22

Using Eq. (12.19), the ultimate torque may then be expressed in the form:

Equation 12.23

This is valid for > yp. When < yp, linear relation (6.3) applies.

A sketch of Eqs. (6.3) and (12.23) is illustrated in Fig. 12.23. Observe that after yielding
torque T yp is reac hed, T and are related nonlinearly. As T approaches T u, the angle of
twist grows without limit. A final point to be noted, however, is that the value of T u is
approached very rapidly (for instance, T u = 1.32 T yp when = 3 yp).

Figure 12.23. Torqueangle of twist relationship for a circular shaft.

When a shaft is strained beyond the elastic limit (point A in Fig. 12.23) and the applied
torque is then removed, rebound is assumed to follow Hookes law. Thus, once a portion of a
shaft has yielded, residual stresses and residual rotations ( B ) will develop. This proc ess
and the application of the preceding relationships are demonstrated in Example 12.12.
Static ally indeterminate, inelastic torsion problems are dealt with similarly to those of axial
load, as was discussed in Section 12.5.

Example 12.12. Residual Stress in a Shaft

Figure 12.24 shows a solid circular steel shaft of diameter d and

length L c arrying a torque T. Determine (a) the radius of the elastic core; (b)
the angle of twist of the shaft; (c) the residual stresses and the residual
rotation when the shaft is unloaded. Assumption: The steel is taken to be an
elastoplastic material. Given: d = 60 mm, L = 1.4 m, T = 7.75 kN m, yp = 145
MPa, and G = 80 GPa.

Figure 12.24. Example 12.13. Torsion of a circular bar of elastoplastic

material.
 Solution

We have c = 30 mm and J = (0.03)4/2 = 1272 109 m4.

a. Radius of Elastic Core. The yield torque, applying Eq. (12.18), equals

Equation (12.19), substituting the values of T and yp, gives

Solving, 0 = 0.604 (30) = 18.1 mm. The elasticplastic stress distribution

in the loaded shaft is illustrated inFig. 12.25a.

Figure 12.25. Example 5.13 (continued). (a) Partial plastic

stresses; (b) elastic rebound stresses; (c) residual stresses.

[View full size image]

b. Yield Twist Angle. Through the use of Eq. (6.3), the angle of twist at the
onset of yielding,

Introducing the value found for yp into Eq. (12.22), we have

c. Residual Stresses and Rotation. The removal of the torque produces elastic
stresses as depic ted in Fig. 12.25b, and the torsion formula, Eq. (6.1),
leads to reversed stress as
 Superposition of the two distributions of stress results in the residual
stresses (Fig. 12.25c).

Permanent Twist. The elastic rebound rotation, using Eq. (6.3), equals

The preceding results indicate that residual rotation of the shaft is

res = 8.03 6.11 = 1.92

Comment

We see that even though the reversed stresses max exceed the yield
strength yp, the assumption of linear distribution of these stresses is valid,
inasmuch as they do not exc eed 2 yp.
12.10. Plastic Torsion: Membrane Analogy
Rec all from Chapter 6 that the maximum shearing stress in a slender bar of arbitrary section
subjec t to pure torsion is always found on the boundary. As the applied torque is inc reased,
we expect yielding to occ ur on the boundary and to move progressively toward the interior,
as sketched in Fig. 12.26a for a bar of rec tangular section. We now determine the ultimate
torque T u that c an be carried. This torque corresponds to the totally plastic state of the
bar, as was the case of the beams previously disc ussed. Our analysis treats only perfec tly
plastic materials.

Figure 12.26. (a) Partially yielded rectangular section; (b) membraneroof analogy
applied to elasticplastic torsion of a rectangular bar; (c) sand hill analogy applied to
plastic torsion of a circular bar.

The stress distribution within the elastic region of the bar is governed by Eq. (6.9),

Equation 12.24

where represents the stress function ( = 0 at the boundary) and is the angle of twist.
The shearing stresses, in terms of , are

Equation a

Inasmuc h as the bar is in a state of pure shear, the stress field in the plastic region is,
acc ording to the Mises yield c riterion,expressed by

Equation 12.25
where yp is the yield stress in shear. This expression indicates that the slope of the
surface remains constant throughout the plastic region and is equal to yp.

MembraneRoof Analogy
Bearing in mind the condition imposed on by Eq. (12.25), the membrane analogy (Sec.
6.6) may be extended from the purely elastic to the elasticplastic c ase. As shown in Fig.
12.26b, a roof abc of c onstant slope is erected with the membrane as its base. Figure
12.26c shows suc h a roof for a c ircular section. As the pressure acting beneath the
membrane increases, more and more contact is made between the membrane and the roof.
In the fully plastic state, the membrane is in total contact with the roof, membrane and roof
being of identical slope. Whether the membrane makes partial or complete contac t with the
roof clearly depends on the pressure. The membraneroof analogy thus permits solution of
elastic plastic torsion problems.

Sand Hill Analogy

For the c ase of a totally yielded bar, the membraneroof analogy leads quite naturally to
the sand hill analogy. We need not construc t a roof at all, using this method. Instead, sand
is heaped on a plate whose outline is cut into the shape of the c ross sec tion of the torsion
member. The torque is, ac cording to the membrane analogy, proportional to twic e the
volume of the sand figure so formed. The ultimate torque corresponding to the fully plastic
state is thus found.

Referring to Fig. 12.26c, let us apply the sand hill analogy to determine the ultimate torque
for a circular bar of radius r. The volume of the corresponding cone is ,
where h is the height of the sand hill. The slope h/r represents the yield point stress yp.
The ultimate torque is therefore

Equation 12.26

Note that the maximum elastic torque is T yp = (r 3/2) yp. We may thus form the ratio

Equation 12.27

Other solid sections may be treated similarly [Ref. 12.7]. Table 12.1 lists the ultimate
torques for bars of various cross-sectional geometry.

Table 12.1. Torque Capacity for Various Common Sections

Cross section Radius or sides Torque Tu for full plasticity
Circ ular r

Equilateral triangle a

Rec tangle a, b (b > a)

Square a

Thic k-walled tube b: outer a: inner

The procedure may also be applied to members having a symmetrically located hole. In this
situation, the plate representing the c ross sec tion must contain the same hole as the actual
cross section.
12.11. ElasticPlastic Stresses in Rotating Disks
This sec tion treats the stresses in a flat disk fabric ated of a perfectly plastic material,
rotating at constant angular velocity. The maximum elastic stresses for this geometry are,
from Eqs. (8.30) and (8.28) as follows:

For the solid disk at r = 0,

Equation a

For the annular disk at r = a,

Equation b

Here a and b represent the inner and outer radii, respectively, the mass density,
and the angular speed. The following discussion relates to initial, partial, and c omplete
yielding of an annular disk. Analysis of the solid disk is treated in a very similar manner.

Initial Yielding
According to the Tresc a yield condition, yielding impends when the maximum stress is equal
to the yield stress. Denoting the critic al speed as 0 and using , we have, from Eq.
(b),

Equation 12.28

Partial Yielding
For angular speeds in exc ess of 0 but lower than speeds resulting in total plastic ity, the
disk contains both an elastic and a plastic region, as shown in Fig. 12.27a. In the plastic
range, the equation of radial equilibrium, Eq. (8.26), with yp replacing the maximum
stress , bec omes

Figure 12.27. (a) Partially yielded rotating annular disk; (b) stress distribution in
complete yielding.
Equation 12.29

or

The solution is given by

Equation c

By satisfying the boundary c ondition r = 0 at r = a, Eq. (c) provides an expression for the
constant c 1, whic h when introduced here results in

Equation d

The stress within the plastic region is now determined by letting r = c in Eq. (d):

Equation 12.30

Referring to the elastic region, the distribution of stress is determined from Eq. (8.27)
with r = c at r = c , and r = 0 at r = b. Applying these c onditions, we obtain

Equation e

The stresses in the outer region are then obtained by substituting Eqs. (e) into Eq. (8.27):

Equation 12.31

To determine the value of that causes yielding up to radius c , we need only

substitute for yp in Eq. (12.31) and introduce c as given by Eq. (12.30).

Complete Yielding
We turn finally to a determination of the speed 1 at whic h the disk becomes fully plastic.
First, Eq. (c) is rewritten

Equation f

Applying the boundary conditions, r = 0 at r = a and r = b in Eq. (f), we have

Equation g

and the c ritic al speed ( = 1) is given by

Equation 12.32
Substitution of Eqs. (g) and (12.32) into Eq. (f) provides the radial stress in a fully plastic
disk:

Equation 12.33

The distributions of radial and tangential stress are plotted in Fig. 12.27b.
12.12. Plastic StressStrain Relations
Consider an element subject to true stresses 1, 2, and 3 with c orresponding true
straining. The true strain, whic h is plastic , is denoted 1, 2, and 3. A simple way to derive
expressions relating true stress and strain is to replace the elastic

constants Eand v by Es and , respectively, in Eqs. (2.34). In so doing, we obtain

equations of the total strain theory or the deformational theory, also known as Henckys
plastic stressstrain relations:

Equation 12.34a

The foregoing may be restated as

Equation 12.34b

Here Es, a func tion of the state of plastic stress, is termed the modulus of
plastic ity or secant modulus. It is defined by Fig. 12.28

Equation 12.35

Figure 12.28. True stresstrue strain diagram for a rigid-plastic material.

in whic h the quantities e and e are the effec tive stress and the effective
strain, respec tively.

Although other yield theories may be employed to determine e, the maximum energy of
distortion or Mises theory (Sec. 4.7) is most suitable. According to the Mises theory, the
following relationship c onnec ts the uniaxial yield stress to the general state of stress at a
point:

Equation 12.36

where the effective stress e is also referred to as the von Mises stress. It is assumed that
expression (12.36) applies not only to yielding or the beginning of inelastic ac tion ( e = yp)
but to any stage of plastic behavior. That is, e has the value yp at yielding, and as
inelastic deformation progresses, e inc reases in ac cordance with the right side of Eq.
(12.36). Equation (12.36) then represents the logical extension of the yield condition to
desc ribe plastic deformation after the yield stress is exceeded.

Collecting terms of Eqs. (12.34), we have

Equation a

The foregoing, together with Eqs. (12.35) and (12.36), leads to the definition

Equation 12.37a

or (on the basis of 1 + 2 + 3 = 0), in different form,

Equation 12.37b

relating the effective plastic strain and the true strain components. Note that, for simple
tension, 2 = 3 = 0, 2 = 3 = 1/2, and Eqs. (12.36) and (12.37) result in

Equation b
Therefore, for a given e , e can be read direc tly from true stressstrain diagram (Fig.
12.28).

Henc kys equations as they appear in Eqs. (12.34) have little utility. To give these
expressions generality and convert them to a more convenient form, it is useful to employ
the empiric al relationship (12.2)

e = K( e)n

from whic h

Equation c

The true stressstrain relations, upon substitution of Eqs. (12.36) and (c) into (12.34),
then assume the following more useful form:

Equation 12.38a

Equation 12.38b

Equation 12.38c

where = 2/ 1 and = 3/ 1.

In the case of an elastic material (K = E and n = 1), it is observed that Eqs. (12.38) reduce
to the familiar generalized Hookes law.

Example 12.13. Analysis of Cylindrical Tube by Henckys Relations

A thin-walled c ylindric al tube of initial radius r o is subjected to internal

pressure p. Assume that the values of r o,p, and the material properties (K and n)
are given. Apply Henckys relations to determine (a) the maximum allowable
stress and (b) the initial thickness t o for the cylinder to become unstable at
internal pressure p.

Solution

The current radius, thickness, and the length are denoted by r, t, and L,
respectively. In the plastic range, the hoop, axial, and radial stresses are

Equation d

We thus have = 2/ 1 and = 0 in Eqs. (12.38).

Corresponding to these stresses, the components of true strain are, from Eq.
(2.24),

Equation e

Based on the c onstancy of volume, Eq. (12.1), we then have

or

Equation f

The first of Eqs. (e) gives

Equation g
The tangential stress, the first of Eqs. (d), is therefore

from which

Equation h

Simultaneous solution of Eqs. (12.38) leads readily to

Equation i

Equation (h) then appears as

Equation j

For material instability,

whic h, upon substitution of p/ 1 and p/ 1 derived from Eq. (j), becomes

or

Equation k

In Eqs. (12.38) it is observed that 1 depends on and . That is,

Equation l
Differentiating, we have

Equation m

Expressions (k), (l), and (m) lead to the instability c ondition:

Equation 12.39

a. Equating expressions (12.39) and (12.38a), we obtain

and the true tangential stress is thus

Equation 12.40

b. On the other hand, Eqs. (d), (f), (g), and (12.38) yield

From this expression, the required original thic kness is found to be

Equation 12.41

wherein 1 is given by Eq. (12.40).

In the case under c onsideration, = 1/2 and Eqs. (12.39), (12.40), and
(12.41) thus become
Equation 12.42

For a thin-walled spheric al shell under internal pressure, the two principal
stresses are equal and hence = 1. Equations (12.39), (12.40), and
(12.41) then reduce to

Equation 12.43

Based on the relations derived in this example, the effective stress and the
effective strain are determined readily. Table 12.2 furnishes the maximum
true and effective stresses and strains in a thin-walled cylinder and a thin-
walled sphere. For purposes of c omparison, the table also lists the results
(Sec. 12.4) pertaining to simple tension. We observe that at instability,
the maximum true strains in a sphere and c ylinder are muc h lower than the
c orresponding longitudinal strain in uniaxial tension.

Table 12.2. Stress and Strain in Pressurized Plastic Tubes

Member Tensile bar Cylindrical tube Spherical tube

1 n

2 0

1/K (n)n

2/K 0
It is signific ant that, for loading situations in which the components of stress do not
increase continuously, Henckys equations provide results that are somewhat in error, and
the inc remental theory (Sec. 12.13) must be used. Under these c irc umstances, Eqs.
(12.34) or (12.38) cannot describe the complete plastic behavior of the material. The
latter is made c learer by c onsidering the following. Suppose that subsequent to a given
plastic deformation the material is unloaded, either partially or completely, and then
reloaded to a new state of stress that does not result in yielding. We expec t no c hange to
occ ur in the plastic strains; but Henc kys equations indic ate different values of the plastic
strain, because the stress c omponents have c hanged. The latter cannot be valid because
during the unloadingloading process, the plastic strains have, in reality, not been affec ted.
12.13. Plastic StressStrain Increment Relations
We have already discussed the limitations of the deformational theory in c onnection with a
situation in which the loading does not c ontinuously increase. The incremental theory offers
another approach, treating not the total strain associated with a state of stress but rather
the inc rement of strain.

Suppose now that the true stresses at a point experience very small changes in
magnitude d 1, d 2, and d 3. As a consequence of these increments, the effec tive
stress e will be altered by d e and the effec tive strain e by d e. The plastic strains thus
suffer increments d 1, d 2, and d 3.

The following modific ation of Henc kys equations, due to Lvy and Mises, desc ribe the
foregoing and give good results in metals:

Equation 12.44a

An alternative form is

Equation 12.44b

The plastic strain is, as before, to occur at c onstant volume; that is,

d 1 + d 2 + d 3 = 0

The effec tive strain inc rement, referring to Eq. (12.37b), may be written

Equation 12.45

The effec tive stress e is given by Eq. (12.36). Alternatively, to asc ertain d e from a
uniaxial true stressstrain curve such asFig. 12.28, it is necessary to know the inc rease of
equivalent stress d e. Given e and d e , it then follows that at any point in the loading
process, application of Eqs. (12.44) provides the inc rement of strain as a unique func tion of
the state of stress and the inc rement of the stress. Ac cording to the LvyMises theory,
therefore, the deformation suffered by an element varies in acc ordance with the specific
loading path taken.

In a particular case of straining of sheet metal under biaxial tension, the LvyMises
equations (12.44) become

Equation 12.46

where = 2/ 1 and 3 taken as zero. The effective stress and strain inc rement, Eqs.
(12.36) and (12.45), is now written

Equation 12.47

Equation 12.48

Combining Eqs. (12.46) and (12.48) and integrating yields

Equation 12.49

To generalize this result, it is useful to employ Eq. (12.2), e = K( e )n, to include strain-
hardening c haracteristics. Differentiating this expression, we have

Equation 12.50

Note that, for simple tension, n = e = and Eq. (12.50) reduces to Eq. (12.3).

The utility of the foregoing development is illustrated in Examples 12.14 and 12.15. In
closing, we note that the total (elasticplastic ) strains are determined by adding the elastic
strains to the plastic strains. The elastic plastic strain relations, together with the
equations of equilibrium and compatibility and appropriate boundary c onditions, c ompletely
desc ribe a given situation. The general form of the LvyMises relationships, inc luding the
elastic inc remental components of strain, are referred to as thePrandtl and Reuss equations.

Example 12.14. Analysis of Tube by LvyMises Equations

Redo Example 12.13 employing the LvyMises stressstrain increment

relations.

Solution

For the thin-walled cylinder under internal pressure, the plastic stresses are

Equation a

where r and t are the current radius and the thic kness. At instability,

Equation b

Bec ause , introduc tion of Eqs. (a) into Eq. (b) provides

Equation 12.51

Clearly, dr/r is the hoop strain increment d 2, and dt/t is the incremental
thickness strain or radial strain incrementd 3. Equation (12.51) is the c ondition
of instability for the cylinder material.

Upon applic ation of Eqs. (12.47) and (12.48), the effec tive stress and the
effective strains are found to be

Equation c
It is observed that axial strain does not oc cur and the situation is one of plane
strain. The first of Eqs. (c) leads to , and c ondition (12.51)
gives

from which

A comparison of this result with Eq. (12.50) shows that

Equation 12.52

The true stresses and true strains are then obtained from Eqs. (c)
and e = K( e )n and the results found to be identic al with that obtained using
Henc kys relations (Table 12.2).

For a spherical shell subjec ted to internal pressure 1 = 2 = pr/2t, = 1

and d 1 = d 2 = d 3/2. At stability dp= 0, and we now have

Equation d

Equations (12.47) and (12.48) result in

Equation e

Equations (d) and (e) are combined to yield

Equation f
From Eqs. (f) and (12.50), it is concluded that

Equation 12.53

True stress and true strain are easily obtained, and are the same as the values
determined by a different method (Table 12.2).
12.14. Stresses in Perfectly Plastic Thick-Walled Cylinders
The case of a thic k-walled c ylinder under internal pressure alone was considered in Section
8.2. Equation (8.11) was derived for the onset of yielding at the inner surface of the
cylinder owing to maximum shear. This was followed by a discussion of the strengthening of
a c ylinder by shrinking a jacket on it (Sec. 8.5). The same goal c an be ac hieved by applying
sufficient pressure to cause some or all of the material to deform plastically and then
releasing the pressure. This is briefly desc ribed next [Ref.12.8].

A continuing increase in internal pressure will result in yielding at the inner surfac e. As the
pressure inc reases, the plastic zone will spread toward the outer surface, and an elastic
plastic state will prevail in the c ylinder with a limiting radius c beyond whic h the cross
section remains elastic. As the pressure increases further, the radius c also increases until,
eventually, the entire c ross sec tion becomes fully plastic at the ultimate pressure. When
the pressure is reduced, the material unloads elastically. Thus, elastic and plastic stresses
are superimposed to produc e a residual stress pattern (see Example 12.15). The
generation of suc h stresses by plastic action is called autofrettage. Upon reloading, the
pressure needed to produc e renewed yielding is greater than the pressure that produced
initial yielding; the cylinder is thus strengthened by autofrettage.

This sec tion concerns the fully plastic and elasticplastic ac tions in a thick-walled cylinder
fabric ated of a perfectly plastic material of yield strength yp under internal pressure p as
shown in Fig. 12.29a.

Figure 12.29. (a) Thick-walled cylinder of perfectly plastic material under internal
pressure; (b) fully plastic stress distribution in the cylinder at the ultimate pressure
for b = 4a; (c) partially yielded cylinder.

[View full size im age ]

Complete Yielding
The fully plastic or ultimate pressure as well as the stress distribution c orresponding to this
pressure is determined by application of the LvyMises relations with e = yp. In polar
coordinates, the axial strain increment is

Equation 12.54
If the ends of the cylinder are restrained so that the axial displac ement w = 0, the problem
may be regarded as a case of plane strain, for which z = 0. It follows that d z = 0 and Eq.
(12.54) gives

Equation a

The equation of equilibrium is, from Eq. (8.2),

Equation b

subjec t to the following boundary c onditions:

Equation c

Based on the maximum energy of distortion theory of failure, Eqs. (4.4b) or (12.36),
setting 1 = , 3 = r, and results in . From this,
we obtain the yield condition:

Equation 12.55

Alternatively, according to the maximum shearing stress theory of failure (Sec. 4.6), the
yield c ondition is

Equation 12.56

Introducing Eq. (12.55) or (12.56) into Eq. (b), we obtain d r/dr = k yp/r, whic h has the
solution

Equation d
The constant of integration is determined by applying the sec ond of Eqs. (c):

c 1 = k yp ln b

Equation (d) is thus

Equation e

The first of conditions (c ) now leads to the ultimate pressure:

Equation 12.57

An expression for can now be obtained by substituting Eq. (e) into Eqs. (12.55) or
(12.56). Consequently, Eq. (a) provides z. The complete plastic stress distribution for a
spec ified yp is thus found to be

Equation 12.58

The stresses given by Eqs. (12.58) are plotted in Fig. 12.29, whereas the distribution of
elastic tangential and radial stresses is shown in Fig. 8.4a.

Example 12.15. Residual Stresses in a Pressurized Cylinder

A perfectly plastic c losed-ended cylinder ( yp = 400 MPa) with 50- and 100-mm
internal and external diameters is subjected to an internal pressure p (Fig.
12.29a). On the basis of the maximum distortion energy theory of failure,
determine (a) the complete plastic stresses at the inner surfac e and (b) the
residual tangential and axial stresses at the inner surfac e when the cylinder is
unloaded from the ultimate pressure pu.
 Solution

The magnitude of the ultimate pressure is, using Eq. (12.57),

a. From Eqs. (12.58), with r = a,

Note, as a c hec k, that z = ( r + )/2 yields the same result.

b. Unloading is assumed to be linearly elastic. Thus, we have the following

elastic stresses at r = a, using Eqs. (8.13) and (8.20):

The residual stresses at the inner surfac e are therefore

( )res = 141.7 533.7 = 392 MPa ( z)res = 89.2 106.7 = 195.9

MPa

The stresses at any other location may be obtained in a like manner.

Partial Yielding
For the elastic segment for whic h c r b (Fig. 12.29c), the tangential and radial
stresses are determined using Eqs. (8.12) and (8.13) with a = c. In so doing, we obtain

Equation 12.59a

Equation 12.59b
Here pc is the magnitude of the (compressive) radial stress at the elasticplastic
boundary r = c where yielding impends. Acc ordingly, by substituting these expressions into
the yield c ondition r = k yp, we have

Equation 12.60

This pressure represents the boundary condition for a fully plastic segment with inner
radius a and outer radius c . That is, constant c 1 in Eq. (d) is obtained by applying ( r )r =
c = pc:

Substituting this value of c 1 into Eq. (d), the radial stress in the plastic zone bec omes

Equation 12.61a

Then, using the yield condition, the tangential stress in the plastic zone is obtained in the
following form:

Equation 12.61b

Equations (12.59) and (12.61), for a given elasticplastic boundary radius c , provide the
relationships necessary for c alculation of the elastic plastic stress distribution in the
cylinder wall (Fig. 12.29c).