1 Bending of a beam with a rectangular section

x x1

3 Epaisseur b

MM 2h

x

x2

Figure 1 : Geometry of the beam and applied load

The beam in figure 1 has a rectanguar section (thickness 2h, width b. The applied load is a purebending moment. As a consequence, the internal forces are represented by a uniaxial stress tensor, wherethe only non-zero component σ11 linearly depends on x3. The exercice starts with a purely static analysis.A kinematic analysis will follow in question 6.

The behaviour of the material is elastic (E, ν) perfectly plastic (σy).

1. What are the stress and the strain distributions in the elastic domain ?

Pure bending around x2 axis of a beam whose neutral axis is x1 is characterized by aonedimensional stress tensor. The component σ11 is linear wrt x3 (let us assume σ11 = kx3). As aconsequence, the strain component ε11 is also linear wrt x3. On pose σ11 = kx3. The tensors arerepresented by : σ 0 0

0 0 00 0 0

and

σ/E 0 00 −νσ/E 00 0 −νσ/E

(1)

The stress vector in a current section of normal e1 writes σ11e1. This classically produces a zeroresulting force on the section (0 6 x2 6 b et −h 6 x3 6 h). The moment produced by the internalforces can be computed, after denoting by (0,x2,x3) the components of OM vector :

M =ZZ

(OM×T )dS = M2e2 +M3e3 (2)

with :

M2 =ZZ

x3 σ11 dx2 dx3 =ZZ

k x23 dx2 dx3 (3)

M3 =−ZZ

x2 σ11 dx2 dx3 =−ZZ

k x2 x3 dx2 dx3 (4)

(5)

The component M3 s equal to zero (integral of x3 between −h and h). The expression obtned forM2, that will be simply denoted M later, is (see Fig.2a) :

M = kbZ +h

−hx2

3 dx3 =23

kbh3 (6)

One can then express k as a function of the moment, using the notation I = 2bh3/3. The value ofσ11 is :

σ11(x3) = σ(x3) = Mx3/I (7)

This is an odd function in x3, whose maximum value, σm, reached at x3 = h, is 3M/2bh2.

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2. Which is the value Me of the aplied moment corresponding to the offset of plastic flow ?

The condition is reached when σm = σy, that is : Me = 2bh2σy/3

σ− Μ

σΜ

σ0−

0σ

σ0−

0σ

x3

Compression

Traction

−h

+h

a−a

ba

x3

c

x3

Figure 2 : Stress profile of σ11 in a beam under circular bending :(a) Elasticity, (b) During plastic flow, (c) Limit load

3. What is the stress distribution when M becomes larger than Me. Show that there is a limit value ofthe flexure moment M∞ for which the strain becomes infinite.

The value of the absolute value of σ11 is limited at σy. Consequently, three zones can befound in the beam width when M > Me. The material remains elastic in the middle of thebeam (−a ≤ x3 ≤ a), and two plastic zones appear, in tension (for x3 > a), and in compression(x3 < −a). In the elastic zone, the stress profile is still linear w.r.t. x3 (σ11 = kx3). In the plasticzones, the stress values are respectively σ11 = +σy,(x3 > a), and σ11 =−σy,(x3 <−a) (Fig.2b).The two unknowns of the problem, k and a, have to verify : – the boundary condition (1) :Z +h

−hx3 σbdx3 = M

– the continuity of the stress vector at the internal boundary between elastic and plastic zones :ka = σy, so that k = σy/a.M value is obtained by replacing σ11 by its expression in equation (1)

M/2 =Z a

0x3(σy x3/a)bdx3 +

Z h

ax3 bσy dx3

M = bσy(h2−a2/3)

Remarks– If a = h : M value is Me = (2/3)bσyh2

– If a = 0 : M = M∞ = bσyh2 = 3Me/2In both cases, the elastic and plastic solutions are in good agreement with the limit values.For M = M∞, the plastic zone covers the whole beam section. This corresponds to the maximumbending moment. (Fig.2c).

4. What will happen, while releasing external load (M = 0) ?i) after a maximum bending moment Mm ≤Me,

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ii) after a maximum bending moment Me < Mm < M∞ ?What is the shape of the so called residual stress field obtained in this last case ?

If the maximum vlaue of M during the preloading remains smaller than Me, the whole sectionremains elastic, so that the beam will recover its initial shape after unloading, without any stress.On the contrary, if the initial loading has produced plastic flow, the local permanent deformationof the material will generate stresses, namely compressive stress in the elongated zone, and tensilestress in the area previously in compression. Assuming that the unloading is fully elastic (it has tobe confirmed later) the final state is obtained by means of a superposition procedure. A stress fieldcorresponding to a −Mm moment has to be added to the state obtained at the end of the loadingphase. The elastic field is simply defined, for −h ≤ x3 ≤ h, by : σ11 = −Mmx3/I. The resulting

profile is then (see Fig.3a) :– for −a≤ x3 ≤ a σ = σyx3/a−3Mmx3/2bh3

– for x3 ≥ a σ = σy−3Mmx3/2bh3

– for x3 ≤−a σ =−σy−3Mmx3/2bh3

Remarks– The slope −3Mm/2bh3 is negative for |x3|> a.

– The residual stresses are balanced :Z +h

−hσ dx3 = 0.

– One can check that threre is no reverse plastic flow, since, for the maximum bending moment,Mm = M∞ = bσyh2 , the compressive stress σc obtained by superposition at x3 = h remainslarger than −σy

σc = σy− (3bσyh2/2bh3)h =−σy/2

σc

σT σ0−

0σ− /2

0 /2

−aa

a b

σ0

σ

TC

C

T

x3 x3

Figure 3 : σ11 stress profile after unloading : (a) after an elastoplastic preloading, (b) when the preloadingreaches the limit load

5. Restarting the resolution of the problem with an horizontal force P in addition to the bendingmoment. the question is to define in the P–M plane the “elastic domain” of the system,characterizing the values of the couples (P, M) at the onset of plasticity, and the “limit load”,corresponding to the failure of the structure.

If an horizontal force is applied in addition to the bending moment, the shape of the stress tensoris the same (component σ11 only), but the neutral line moves. The value is just :

σ11 = Mx3/I +P/2bh

The boundary of the elastic domain is then a segment in each quadrant of the plan P–M. The limitload is already defined for pure bending. Without any bending moment, the limit load in tension isnothing but the load that initiate plastic flow :

P∞ = Pe = 2hbσy

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The combined values of the set (Pr, Mr) leading to the failure of the beam can be found directly,since they do not depend on the loading path. The expressions of the moment and of the horizontalforce write in this case (Fig.4a) :

if x3 < a : σ11 =−σy

if x3 > a : σ11 = σy.

It comes then :

P =Z a

−h−σybdx3 +

Z h

aσybdx3 =−2σyab

M =Z a

−h−σyx3bdx3 +

Z h

aσyx3bdx3 = bσy(h2−a2)

After dividing respectively by Pe and Me, Pr = −Pea/h ; Mr = 3Me(1− a2/h2)/2, the resultingdiagram is shown in Fig.4b :

Mr/Me = (3/2)(1− (Pr/Pe)2)

(a) (b)Figure 4 : (a) σ11 stress profile at limit state, for axial tension and pure bending, (b) illustration of the

elastic and plastic domain in theP–M plane

6. Calculate the deflection during the loading, and the residual deflection.

Assuming that a plane section remains plane, the displacement field in the beam can be deducedfrom three variables :

u1 = U(s)+θx3 U(s) is the horizontal displacement of the section,θ is the rotation angle around x2,

u3 = V (s) V (s) is the vertical displacement.The angle is the opposite of the derivative of the deflection with respect to x1 (this is needed tohave a zero 13 shear strain) :

θ+V,1 = 0

The axial strain can be deduced from the rotation, since

ε11 = u1,1 = θ,1x3

For the case of an elastic behaviour, the term θ,1 can be expressed as a function of the applied load

M =Z +h

−hσ11x3bdx3 = EIθ,1

For the case of a perfectly plastic behaviour, the rotation kinematics is imposed by the elastic zone,so that the orientatio is prescribed by the slope between −a et a. In this area :

ε11 = σ11/E = (σy/E)(x3/a)

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θ,1 = σy/Ea

Consequently, the curvature V,11 is :– for the elastic regime : V,11 =−M/EI– for the elastoplastic regime : V,11 =−σy/EaIntegrating these two equations for a beam of length 2L (−L≤ x1 ≤ L) that is simply supported atits two ends leads to the following expression for the vertical displacement :– for the elastic regime : V = (M/2EI)(L2− x2

1)– for the elastoplastic regime : V = (σy/2Ea)(L2− x2

1)The maximum value of the deflectionis obtained for x1 = 0. After replacing a by its expression (ais a function of M in the elastoplastic regime), the global response is obtained :

V =σyL2

2Eh√

3(1−M/bσyh2)

Remarks– This expression is consistent with the expression found for the elastic case when M = Me =(2/3)bσyh2.– The deflection tends to infinity when M tends toward M∞ = bσyh2. In this last case, thesmall perturbation assumption no longer works, far from failure, so that the computation shouldintroduce terms coming from the non linear geometrical evolution.The residual displacement can be calculated by superposing to the previous expression (obtainedfor the moment Mm) the displacement field obtained by means of an elastic calulation with a−Mm

applied moment, so that :

V =σyL2

2Eh√

3(1−Mm/bσyh2)−MmL2

2EI

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