1 Bending of a beam with a rectangular section

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1 1 Bending of bem with rectngulr section x3 Episseur b M x 2 x x 1 2h M Figure 1 : Geometry of the bem nd pplied lod The bem in figure 1 hs rectngur section (thickness 2h, width b. The pplied lod is pure bending moment. As consequence, the internl forces re represented by unixil stress tensor, where the only non-zero component σ 11 linerly depends on. The exercice strts with purely sttic nlysis. A kinemtic nlysis will follow in question 6. The behviour of the mteril is elstic (E, ν) perfectly plstic (σ y ). 1. Wht re the stress nd the strin distributions in the elstic domin? Pure bending round x 2 xis of bem whose neutrl xis is x 1 is chrcterized by onedimensionl stress tensor. The component σ 11 is liner wrt (let us ssume σ 11 = k ). As consequence, the strin component ε 11 is lso liner wrt. On pose σ 11 = k. The tensors re represented by : nd σ/e νσ/e νσ/e (1) The stress vector in current section of norml e 1 writes σ 11 e 1. This clssiclly produces zero resulting force on the section (0 x 2 b et h). The moment produced by the internl forces cn be computed, fter denoting by (0,x 2, ) the components of OM vector : M = (OM T )ds = M 2 e 2 + M 3 e 3 (2) with : M 2 = σ 11 dx 2 d = k x3 2 dx 2 d (3) M 3 = x 2 σ 11 dx 2 d = k x 2 dx 2 d (4) The component M 3 s equl to zero (integrl of between nd h). The expression obtned for M 2, tht will be simply denoted M lter, is (see Fig.2) : M = kb x3 2 d = 2 3 kbh3 (6) One cn then express k s function of the moment, using the nottion I = 2bh 3 /3. The vlue of σ 11 is : σ 11 ( ) = σ( ) = M /I (7) This is n odd function in, whose mximum vlue, σ m, reched t = h, is 3M/2bh 2. 1 (5)

2 2. Which is the vlue M e of the plied moment corresponding to the offset of plstic flow? The condition is reched when σ m = σ y, tht is : M e = 2bh 2 σ y /3 +h σ Μ σ Μ Trction Compression b c Figure 2 : Stress profile of σ 11 in bem under circulr bending : () Elsticity, (b) During plstic flow, (c) Limit lod 3. Wht is the stress distribution when M becomes lrger thn M e. Show tht there is limit vlue of the flexure moment M for which the strin becomes infinite. The vlue of the bsolute vlue of σ 11 is limited t σ y. Consequently, three zones cn be found in the bem width when M > M e. The mteril remins elstic in the middle of the bem ( ), nd two plstic zones pper, in tension (for > ), nd in compression ( < ). In the elstic zone, the stress profile is still liner w.r.t. (σ 11 = k ). In the plstic zones, the stress vlues re respectively σ 11 = +σ y,( > ), nd σ 11 = σ y,( < ) (Fig.2b). The two unknowns of the problem, k nd, hve to verify : the boundry condition (1) : σbd = M the continuity of the stress vector t the internl boundry between elstic nd plstic zones : k = σ y, so tht k = σ y /. M vlue is obtined by replcing σ 11 by its expression in eqution (1) Z M/2 = (σ y /)bd + bσ y d 0 M = bσ y (h 2 2 /3) Remrks If = h : M vlue is M e = (2/3)bσ y h 2 If = 0 : M = M = bσ y h 2 = 3M e /2 In both cses, the elstic nd plstic solutions re in good greement with the limit vlues. For M = M, the plstic zone covers the whole bem section. This corresponds to the mximum bending moment. (Fig.2c). 4. Wht will hppen, while relesing externl lod (M = 0)? i) fter mximum bending moment M m M e, 2

3 ii) fter mximum bending moment M e < M m < M? Wht is the shpe of the so clled residul stress field obtined in this lst cse? If the mximum vlue of M during the preloding remins smller thn M e, the whole section remins elstic, so tht the bem will recover its initil shpe fter unloding, without ny stress. On the contrry, if the initil loding hs produced plstic flow, the locl permnent deformtion of the mteril will generte stresses, nmely compressive stress in the elongted zone, nd tensile stress in the re previously in compression. Assuming tht the unloding is fully elstic (it hs to be confirmed lter) the finl stte is obtined by mens of superposition procedure. A stress field corresponding to M m moment hs to be dded to the stte obtined t the end of the loding phse. The elstic field is simply defined, for h, by : σ 11 = M m /I. The resulting for σ = σ y / 3M m /2bh 3 profile is then (see Fig.3) : for σ = σ y 3M m /2bh 3 for σ = σ y 3M m /2bh 3 Remrks The slope 3M m /2bh 3 is negtive for >. The residul stresses re blnced : σ d = 0. One cn check tht threre is no reverse plstic flow, since, for the mximum bending moment, M m = M = bσ y h 2, the compressive stress σ c obtined by superposition t = h remins lrger thn σ y σ c = σ y (3bσ y h 2 /2bh 3 )h = σ y /2 σ c /2 C σ T C T T σ0 /2 b Figure 3 : σ 11 stress profile fter unloding : () fter n elstoplstic preloding, (b) when the preloding reches the limit lod 5. Restrting the resolution of the problem with n horizontl force P in ddition to the bending moment. the question is to define in the P M plne the elstic domin of the system, chrcterizing the vlues of the couples (P, M) t the onset of plsticity, nd the limit lod, corresponding to the filure of the structure. If n horizontl force is pplied in ddition to the bending moment, the shpe of the stress tensor is the sme (component σ 11 only), but the neutrl line moves. The vlue is just : σ 11 = M /I + P/2bh The boundry of the elstic domin is then segment in ech qudrnt of the pln P M. The limit lod is lredy defined for pure bending. Without ny bending moment, the limit lod in tension is nothing but the lod tht initite plstic flow : P = P e = 2hbσ y 3

4 The combined vlues of the set (P r, M r ) leding to the filure of the bem cn be found directly, since they do not depend on the loding pth. The expressions of the moment nd of the horizontl force write in this cse (Fig.4) : if < : σ 11 = σ y It comes then : Z P = if > : σ 11 = σ y. σ y bd + Z M = σ y bd + σ y bd = 2σ y b σ y bd = bσ y (h 2 2 ) After dividing respectively by P e nd M e, P r = P e /h ; M r = 3M e (1 2 /h 2 )/2, the resulting digrm is shown in Fig.4b : M r /M e = (3/2)(1 (P r /P e ) 2 ) () Figure 4 : () σ 11 stress profile t limit stte, for xil tension nd pure bending, (b) illustrtion of the elstic nd plstic domin in thep M plne (b) 6. Clculte the deflection during the loding, nd the residul deflection. Assuming tht plne section remins plne, the displcement field in the bem cn be deduced from three vribles : u 1 = U(s) + θ U(s) is the horizontl displcement of the section, θ is the rottion ngle round x 2, u 3 = V (s) V (s) is the verticl displcement. The ngle is the opposite of the derivtive of the deflection with respect to x 1 (this is needed to hve zero 13 sher strin) : θ +V,1 = 0 The xil strin cn be deduced from the rottion, since ε 11 = u 1,1 = θ,1 For the cse of n elstic behviour, the term θ,1 cn be expressed s function of the pplied lod M = σ 11 bd = EIθ,1 For the cse of perfectly plstic behviour, the rottion kinemtics is imposed by the elstic zone, so tht the orienttio is prescribed by the slope between et. In this re : ε 11 = σ 11 /E = (σ y /E)( /) 4

5 θ,1 = σ y /E Consequently, the curvture V,11 is : for the elstic regime : V,11 = M/EI for the elstoplstic regime : V,11 = σ y /E Integrting these two equtions for bem of length 2L ( L x 1 L) tht is simply supported t its two ends leds to the following expression for the verticl displcement : for the elstic regime : V = (M/2EI)(L 2 x 2 1 ) for the elstoplstic regime : V = (σ y /2E)(L 2 x 2 1 ) The mximum vlue of the deflectionis obtined for x 1 = 0. After replcing by its expression ( is function of M in the elstoplstic regime), the globl response is obtined : σ y L 2 V = 2Eh 3(1 M/bσ y h 2 ) Remrks This expression is consistent with the expression found for the elstic cse when M = M e = (2/3)bσ y h 2. The deflection tends to infinity when M tends towrd M = bσ y h 2. In this lst cse, the smll perturbtion ssumption no longer works, fr from filure, so tht the computtion should introduce terms coming from the non liner geometricl evolution. The residul displcement cn be clculted by superposing to the previous expression (obtined for the moment M m ) the displcement field obtined by mens of n elstic clultion with M m pplied moment, so tht : V = σ y L 2 2Eh 3(1 M m /bσ y h 2 ) M ml 2 2EI 5