Barber J.R. Intermediate Mechanics of Materials

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5.7 Unloading, springback and residual stress 265

This reduction in curvature is known as ‘springback’. The ﬁnal radius of curvature

on complete unloading can be determined as





max

−

∗

−

max

, (5.36)

where (1/R)

max

is the curvature at the maximum bending moment M

max

Example 5. 9

The T-beam of Figure 5.26 is made of an aluminium alloy with E = 80 GPa and

= 200 MPa. It is bent to a radius of 10 m and then released. Find the ﬁnal radius

of curvature and the residual stress distribution.

170

100

all dimensions in mm

Figure 5.26

We must ﬁrst determine the applied moment M

max

and the corresponding stress

distribution. From equation (5.15), we have

d =

200 ×10

×10

80 ×10

= 0.025 m = 25 mm.

The location of the neutral axis is determined by the condition that the axial force

F =0. However, since d is considerably less than the total depth of the section, it is

likely that there is an upper plastic zone and that it extends down into the web as

shown in Figure 5.27 (a). In this case, the stress distribution will be as shown in

Figure 5.27 (b), where c is an unknown dimension to be determined. The condition

F = 0 then requires that

(0.1 ×0.01 + c ×0.01)S

−([0.17 −0.05 −c] ×0.01)S

= 0 ,

since the elastic region makes self-cancelling contributions to the axial force. Solving

for c, we have

c = 0.01 m.

266 5 Non-linear and Elastic-Plastic Bending

170

neutral

axis

-S

(a) (b)

Figure 5.27

The corresponding bending moment is

max

= S

(0.1 ×0.01 ×0.04 + 0.01 ×0.01 ×0.03)

−S

(0.11 ×0.01 ×(−0.08)) + 2S



0.01 ×0.025



×0.025



= 200 ×10

×13.517 ×10

−3

= 27,033 Nm.

For the unloading problem, we require the location of the centroid and the second

moment of area I

. Using the method of §4.3 and treating the web and the ﬂange as

areas A

respectively, we can construct the ﬁrst three rows of Table 5.1 (in mm).

Distances are measured from the bottom of the section.

i 1 2

1700 1000

¯y

85 175

4.09 ×10

8.3 ×10

( ¯y

− ¯y) −33.3 56.7

Table 5.1

Thus, the total area is

A = 1700 + 1000 = 2700 mm

and the location of the centroid is deﬁned by

5.7 Unloading, springback and residual stress 267

A ¯y = 1700 ×85 + 1000 ×175 ,

giving

¯y = 118.3 mm.

This permits us to complete the last row of the table, after which we obtain

= 4.09 ×10

+ 8.3 ×10

+ 1700 ×33.3

+ 1000 ×56.7

= 9.202 ×10

−6

The residual curvature is therefore given by (5.36) as

−

27033

80 ×10

×9.202 ×10

−6

−1

from which

= 15.8 m.

Notice that this represents a substantial springback from the radius of 10 m at M

max

The ﬁnal stress distribution is obtained by subtracting the distribution

max

27033y

9.202 ×10

−6

= 2938y MPa

(y in mm) from the distribution of Figure 5.27 (b), noting however that y is mea-

sured from the elastic neutral axis. This gives the piecewise linear stress distribution

sketched in Figure 5.28, which is completely determined by the stresses at the points

A,B,C,D, in Figure 5.27 (a), which are

= 200 −2938(0.18 −0.1183) = 18.7 MPa

= 200 −2938(0.16 −0.1183) = 77.5 MPa

= −200 −2938(0.11 −0.1183) = −175.6 MPa

= −200 −2938(−0.1183) = 148 MPa.

148

-175.6

77.5

18.7

Figure 5.28

268 5 Non-linear and Elastic-Plastic Bending

5.7.2 Reloading and shakedown

The results of the last section show that a beam that is plastically deformed and then

unloaded will generally exhibit a state of residual stress opposite to that experienced

at the most extreme loading condition. This residual stress is beneﬁcial if the beam is

subsequently loaded again in the same direction. Indeed, since the unloading process

is completely elastic, it is clear that we could reload the beam to within an arbitrarily

small value below the initial maximum moment M

max

without producing any addi-

tional plastic deformation. Perhaps more important from an engineering perspective

is the fact that loading the deformed beam between zero and a moment lower than

max

will result in maximum stresses lower than those that would be experienced

by a beam without residual stresses. This means that the fatigue resistance of a beam

subjected to zero to maximum loading can be enhanced by applying an initial over-

load in the direction of the maximum loading, sufﬁcient to produce some plastic

deformation. In effect, the residual stress induced shifts the mean value of the cyclic

stresses closer to zero, without affecting the alternating component.

If the bending moment in a beam oscillates from zero to a maximum value in the

range M

< M < M

, plastic deformation will occur during the ﬁrst loading cycle,

but in subsequent cycles it will just reach the yield limit but not surpass it and no

subsequent plastic deformation will occur. This is a common phenomenon in elastic-

plastic structures subjected to cyclic loading and is known as shakedown. In effect,

the structure tends to deform plastically early in the process in such a way as to

develop a state of residual stress that discourages further plastic deformation. Indeed,

there is a theorem, known as Melan’s theorem

, which proves that if any state of

residual stress can be found that would be sufﬁcient to inhibit plastic deformation

during cyclic loading, the structure will in fact shake down until the behaviour is

purely elastic. It is almost as though the structure would do its utmost to avoid its

being subjected to repeated plastic deformation!

It is clear, however, that if the beam is plastically deformed under a bending

moment M and is then subjected to a bending moment of opposite sign, the residual

stress from the ﬁrst loading will increase the tendency for plastic deformation, so

that a smaller moment is needed to cause ﬁrst yield in the reverse direction. In most

cases, if a beam is plastically deformed by a bending moment M

, (M

< M

elastic deformation will persist on reversed loading as long as

−2M

< M < M

. (5.37)

In other words, the range of bending moments under which the defomation is purely

elastic is equal to 2M

as it would be for a beam without residual stress (−M

M < M

), but the mean value is shifted in the direction of the original moment M

Equation (5.37) may overstate the elastic range under initial unloading if the beam

is unsymmetric and M

is close to the fully plastic moment, but under cyclic loading

the system will still shake down eventually as long as (5.37) is satisﬁed.

P.S.Symonds (1951), Shakedown in continuous media, ASME Journal of Applied Mechan-

ics, Vol.18, pp.85–89.

5.8 Limit analysis in the design of beams 269

5.8 Limit analysis in the design of beams

We have seen in §§ 5.4–5.6 that the fully plastic moment M

is generally easier to

calculate than M

and exceeds M

by a fairly modest percentage. Since all design

calculations are subject to safety factors to allow for unforeseen overloads and other

uncertainties, an alternative procedure is therefore to perform a fully plastic ‘col-

lapse’ or ‘limit’ analysis of the problem and apply a somewhat larger safety factor.

5.8.1 Plastic hinges

Figure 5.29 shows a simply supported beam with a central load and the correspond-

ing bending moment diagram.

L /

Figure 5.29: A simply supported beam with a central load

Failure will occur when the maximum bending moment FL/4 reaches M

and

hence

F =

Once this happens, unlimited bending of the beam will occur at B without further

increase in F and the beam will collapse as shown in Figure 5.30 (a).

Some plastic deformation will occur at all points where M > M

and this deﬁnes

a range of points around B as shown in Figure 5.30 (b). However, for most beam sec-

tions, the greater part of the deformation is concentrated near B as in Figure 5.30 (a)

and the beam behaves as though it had a sticky hinge at B which requires a moment

to move it. We describe this process by saying that the beam develops a plastic

hinge at B. Once the hinge is developed, the beam becomes a mechanism in the sense

that unlimited further deformation can occur at the hinge without further increase in

force or additional deformation elsewhere in the beam.

270 5 Non-linear and Elastic-Plastic Bending

plastic hinge

plastic deformation

range

(a)

(b)

Figure 5.30: (a) Mode of plastic collapse for the simply supported beam, (b) the

corresponding bending moment diagram

5.8.2 Indeterminate problems

For indeterminate problems, collapse requires that the beam or structure develop

enough plastic hinges to remove the redundancy and convert the structure to a mech-

anism. It is usually fairly easy to determine where these hinges must occur. A good

method is to sketch the probable shape of the shear force and/or bending moment

diagrams. Remember that

= V , (5.38)

where V is the shear force, so the maximum and minimum values of bending moment

correspond to points where the shear force passes through zero. In particular, no

plastic hinge can develop in the middle of a beam segment with no lateral load, since

the shear force there must be constant.

Once a probable arrangement of plastic hinges is identiﬁed, the problem is re-

duced to one which can be solved using equilibrium considerations alone. This is

almost always a great deal easier than the solution of the corresponding elastic inde-

terminate problem — for example using Castigliano’s Second Theorem (§3.10).

Example 5. 10

The built-in beam of Figure 5.31 (a) is loaded by a uniformly distributed load w per

metre. The beam is made of steel with S

=300 MPa and its cross section is as shown

in Figure 5.31 (b). Find the value of w at plastic collapse.

5.8 Limit analysis in the design of beams 271

120

all dimensions in mm

(a) (b)

Figure 5.31

We ﬁrst need to determine the fully plastic moment M

. The total area of the

section is

A = 80 ×10 + 120 ×10 = 2000 mm

so the median line cuts the section at a distance 100 mm (1000 mm

/10 mm) from

the bottom of the section as shown in Figure 5.32.

neutral

axis

100

all dimensions in mm

Figure 5.32

Taking moments about the bottom of the section, the fully plastic moment is then

determined as

= 300 ×10

(0.08 ×0.01 ×0.125 + 0.02 ×0.01 ×0.11 −0.1 ×0.01 ×0.05)

= 21600 Nm.

8 m

w per metre

272 5 Non-linear and Elastic-Plastic Bending

(a)

Figure 5.33

The bending moment diagram will be as shown in Figure 5.33 (a), so ﬁrst yield

will occur either at the supports or at the mid-point. However, the development of a

single plastic hinge is not sufﬁcient to cause the beam to collapse. The load can be

increased further until the beam is converted to a mechanism by the establishment of

plastic hinges at all three points ABC as shown in Figure 5.33 (b). Taking moments

about C in the free-body diagram

for the segment BC (Figure 5.34) then yields the

equilibrium equation

−4w ×2 = 0

and hence

w =

(4 m ×2 m )

= 5400 N/m.

w per metre

Figure 5.34

5.9 Summary

In this chapter, we have developed methods for determining the stress distribution

and the curvature for beams that are loaded into the plastic range. The same methods

can also be used for non-linear elastic materials. The strain is always a linear function

Notice that the moments M

must oppose the relative rotations in Figure 5.33 (b).

(b)

w per metre

4 m

4 w

5.9 Summary 273

of position in the section and the three constants describing this deformation can in

principle always be found from three equilibrium equations deﬁning the transmitted

force and moments.

If a beam is bent about an axis of symmetry, this will also be the neutral axis if

there is no axial force. Symmetric regions yielded in tension and compression are

separated by an elastic region. If the moment axis is not an axis of symmetry, the

neutral axis generally moves as plastic deformation progresses.

The maximum bending moment (the fully plastic moment) corresponds to the

situation where the elastic zone has shrunk to zero and the beam then develops a

plastic hinge. For a rectangular beam section, the shape factor f = 1.5, meaning that

the load can be increased by 50% beyond ﬁrst yield before collapse occurs. This

‘residual strength’ for an elastic-beam is a useful reserve in design to accommodate

unlikely extreme loading events. However, more ‘efﬁcient’ sections such as I-beams

have values of f much closer to unity, since most of the section area is concentrated

at the same distance from the neutral axis. In such cases, a good design strategy is to

calculate the fully plastic moment and use an appropriate safety factor. This method

is particularly useful in designing complex indeterminate structures for strength.

If a beam is loaded into the plastic range and then unloaded, it will be left in a

self-equilibrated state of residual stress and will have some residual curvature. This

is important in forming operations involving bending, since the elastic recovery or

‘springback’ must be allowed for in designing appropriate forms and tools. Also, the

tensile residual stresses may inﬂuence the fatigue life of formed components.