LINEAR PLATE BENDING

Transcription

1 LINEAR PLATE BENDING

2 1 Linea plate bending A plate is a body of which the mateial is located in a small egion aound a suface in the thee-dimensional space. A special suface is the mid-plane. Measued fom a point in the mid-plane, the distance to both sufaces of the plate is equal. The geomety of the plate and position of its points is descibed in an othonomal coodinate system, eithe Catesian (coodinates {x, y, }) o cylindical (coodinates {, θ, }). Fig. 1.1 : Plate with cuved mid-plane and vaiable thickness 1.1 Geomety The following assumptions about the geomety ae supposed to hold hee : the mid-plane is plana in the undefomed state, the mid-plane coincides with the global coodinate plane = 0, the thickness h is unifom. Along the edge of the plate a coodinate s is used and pependicula to this a coodinate n. y x s θ h n Fig. 1. : Plana plate with unifom thickness

3 1. Extenal loads The plate is loaded with foces and moments, which can be concentated in one point o distibuted ove a suface o along a line. The following loads ae defined and shown in the figue : foce pe unit of aea in the mid-plane : s x (x,y),s y (x,y) o s (,θ), s t (,θ) foce pe unit of aea pependicula to the mid-plane : p(x,y) o p(,θ) foce pe unit of length in the mid-plane : f n (s),f s (s) foce pe unit of length pependicula to the mid-plane : f (s) bending moment pe unit of length along the edge : m b (s) tosional moment pe unit of length along the edge : m w (s) p x θ y s f (s) m b (s) s f s (s) m w (s) n f n (s) 1.3 Catesian coodinate system Fig. 1.3 : Extenal loads on a plate In the Catesian coodinate system, thee othogonal coodinate axes with coodinates {x, y, } ae used to identify mateial and spatial points. As stated befoe, we assume that the midplane of the plate coincides with the plane = 0 in the undefomed state. This is not a estictive assumption, but allows fo simplification of the mathematics. y x s θ h n Fig. 1.4 : Plate in a Catesian coodinate system

4 1.3.1 Displacements The figue shows two points P and Q in the undefomed and in the defomed state. In the undefomed state the point Q is in the mid-plane and has coodinates (x,y,0). The out-ofplane point P has coodinates (x,y,). As a esult of defomation, the displacement of the mid-plane point Q in the (x, y, )-coodinate diections ae u, v and w, espectively. The displacement components of point P, indicated as u x, u y and u, can be elated to those of point Q. θ y θ θ x P T P S R y x u Q v w Q Fig. 1.5 : Displacement of a out-of-midplane point due to bending u x = u SR = u QRsin(θ x ) u y = v ST = v QT sin(θ y ) u = w + SQ QR = cos(θ y ) QT = cos(θ x ) SQ = cos(θ) u x = u sin(θ x )cos(θ y ) u y = v cos(θ x )sin(θ y ) u = w + cos(θ) No out-of-plane shea When it is assumed that thee is no out-of-plane shea defomation, the so-called Kichhoff hypotheses hold : staight line elements, initially pependicula to the mid-plane emain staight, staight line elements, initially pependicula to the mid-plane emain pependicula to the mid-plane. The angles θ x and θ y can be eplaced by the otation angles φ x and φ y espectively. This is illustated fo one coodinate diection in the figue.

5 φ x P Q P Q φ x x Fig. 1.6 : Displacement of out-of-midplane point due to bending with no shea θ x = φ x ; θ y = φ y ; θ = φ Small otation With the assumption that otations ae small, the cosine functions appoximately have value 1 and the sine functions can be eplaced by the otation angles. These otations can be expessed in the deivatives 1 of the -displacement w w..t. the coodinates x and y. cos(φ) = cos(φ y ) = cos(φ x ) = 1 sin(φ y ) = φ y = w,y ; sin(φ x ) = φ x = w,x u x = u w,x u y = v w,y u = w + } Constant thickness The thickness of the plate may change due to loading pependicula to the mid-plane o due to contaction due to in-plane defomation. The diffeence between and is obviously a function of. Tems of ode highe than ae neglected in this expession. It is now assumed that the thickness emains constant, which means that ζ(x,y) = 0 has to hold. With the assumption, which is coect fo small defomations and thin plates, the displacement components of the out-of-plane point P can be expessed in mid-plane displacements. (x,y) = (x,y) (x,y) = ζ(x,y) + η(x,y) + O( 3 ) h = ζ(x,y)( h ) ζ(x,y)( h ) 1 Deivatives ae denoted as e.g. h = 0 ζ(x,y) = 0 ( ) = ( ),x and ( ) = ( x x y ),xy

6 u x (x,y,) = u(x,y) w,x u y (x,y,) = v(x,y) w,y u (x,y,) = w(x,y) + η(x,y) 1.3. Cuvatues and stains The linea stain components in a point out of the mid-plane, can be expessed in the midplane stains and cuvatues. A sign convention, as is shown in the figue, must be adopted and used consistently. It was assumed that staight line elements, initially pependicula to the mid-plane, emain pependicula to the mid-plane, so γ x and γ y should be eo. Whethe this is tue will be evaluated late. ε xx = u x,x = u,x w,xx = ε xx0 κ xx ε yy = u y,y = v,y w,yy = ε yy0 κ yy γ xy = u x,y + u y,x = u,y + v,x w,xy = γ xy0 κ xy ε = u, = η(x,y) γ x = u x, + u,x = η,x γ y = u y, + u,y = η,y ε = ε 0 κ y ε xx0 ε xy0 ε xy0 ε yy0 x κ xx κ yy κ xy κ xy Fig. 1.7 : Cuvatues and stain definitions Stesses It is assumed that a plane stess state exists in the plate : σ = σ x = σ y = 0. Because the plate may be loaded with a distibuted load pependicula to its suface, the fist assumption is only appoximately tue. The second and thid assumptions ae in consistence with the Kichhoff hypothesis, but has to be elaxed a little, due to the fact that the plate can be loaded with pependicula edge loads. Howeve, it will always be tue that the stesses σ, σ x and σ y ae much smalle than the in-plane stesses σ xx, σ yy, and σ xy.

7 x y σ yx σ xy σ xx σ yy Fig. 1.8 : In-plane stesses Isotopic elastic mateial behavio Fo linea elastic mateial behavio Hooke s law elates stains to stesses. Mateial stiffness (C) and compliance (S) matices can be deived fo the plane stess state. ε xx ε yy γ xy σ xx σ yy σ xy = 1 E = E 1 ν 1 ν 0 ν (1 + ν) 1 ν 0 ν (1 ν) σ xx σ yy σ xy ε xx ε yy γ xy ; ε = ν E (σ xx + σ yy ) shot notation : ε = Sσ σ = S 1 ε = Cε = C(ε 0 κ ) Inconsistency with plane stess The assumption of a plane stess state implies the shea stesses σ y and σ x to be eo. Fom Hooke s law it then immediately follows that the shea components γ y and γ x ae also eo. Howeve, the stain-displacement elations esult in non-eo shea. Fo thin plates, the assumption will be moe coect than fo thick plates. ε = η η(x,y) = ε = 1 γ y = ν (1 ν) (w,xxy + w,yyy ) 0 ν E (σ xx + σ yy ) = ν (1 ν) (w,xx + w,yy )

8 γ x = ν (1 ν) (w,xxx + w,xyy ) Coss-sectional foces and moments Coss-sectional foces and moments can be calculated by integation of the in-plane stess components ove the plate thickness. A sign convention, as indicated in the figue, must be adopted and used consistently. The stess components σ x and σ y ae much smalle than the elevant in-plane components. Integation ove the plate thickness will howeve lead to shea foces, which cannot be neglected. Ñ = = M D x = N xx N yy N xy M xx M yy M xy h/ h/ = = h/ h/ h/ h/ σ d = h/ σ d = σ x d ; D y = {C(ε 0 κ )}d = hcε 0 h/ h/ h/ h/ h/ {C(ε 0 κ )} d = 1 1 h3 Cκ σ y d y N xx N xy N xy D y N yy x D x M yy M xy M xx M xy Fig. 1.9 : Coss-sectional foces and moments Stiffness- and compliance matix Integation leads to the stiffness and compliance matix of the plate. [ Ñ M = [ Ch 0 0 Ch 3 /1 [ ε 0 κ [ ε 0 κ = [ S/h 0 0 1S/h 3 [ Ñ M

9 1.3.6 Equilibium equations Conside a small column cut out of the plate pependicula to its plane. Equilibium equiements of the coss-sectional foces and moments, lead to the equilibium equations. The coss-sectional shea foces D x and D y can be eliminated by combining some of the equilibium equations. N xx,x + N xy,y + s x = 0 N yy,y + N xy,x + s y = 0 D x,x + D y,y + p = 0 M xy,x + M yy,y D y = 0 M xy,y + M xx,x D x = 0 M xx,xx + M yy,yy + M xy,xy + p = 0 The equilibium equation fo the coss-sectional bending moments can be tansfomed into a diffeential equation fo the displacement w. The equation is a fouth-ode patial diffeential equation, which is efeed to as a bi-potential equation. M xx M yy M xy = Eh 3 1(1 ν ) 1 ν 0 ν (1 ν) M xx,xx + M yy,yy + M xy,xy + p = 0 w,xx w,yy w,xy 4 w x w y w x y = ( x + y )( w x + w y ) = p B B = Eh 3 1(1 ν = plate modulus ) Othotopic elastic mateial behavio A mateial which popeties ae in each point symmetic w..t. thee othogonal planes, is called othotopic. The thee pependicula diections in the planes ae denoted as the 1-, - and 3-coodinate axes and ae efeed to as the mateial coodinate diections. Linea elastic behavio of an othotopic mateial is chaacteied by 9 independent mateial paametes. They constitute the components of the compliance (S) and stiffness (C) matix.

10 ε 11 ε ε 33 γ 1 γ 3 γ 31 = E1 1 ν 1 E 1 ν 31 E ν 1 E1 1 E 1 ν 3 E ν 13 E1 1 ν 3 E 1 E G G G 1 31 σ 11 σ σ 33 σ 1 σ 3 σ 31 ε = Sσ C = 1 σ = S 1 ε = Cε 1 ν 3 ν 3 E E 3 ν 13 ν 3 +ν 1 ν 31 ν 3 +ν 1 E E 3 1 ν 31 ν 13 ν 1 ν 3 +ν 31 ν 1 ν 31 +ν 3 E 1 E 3 ν 1 ν 3 +ν 13 E 1 E 3 ν 1 ν 13 +ν 3 1 ν 1 ν 1 E 1 E E 1 E E E E 1 E E 1 E G G G 31 with = (1 ν 1ν 1 ν 3 ν 3 ν 31 ν 13 ν 1 ν 3 ν 31 ν 1 ν 3 ν 13 ) E 1 E E 3 Othotopic plate In an othotopic plate the thid mateial coodinate axis (3) is assumed to be pependicula to the plane of the plate. The 1- and -diections ae in its plane and otated ove an angle α (counteclockwise) w..t. the global x- and y-coodinate axes. Fo the plane stess state the linea elastic othotopic mateial behavio is chaacteied by 4 independent mateial paametes : E 1, E, G 1 and ν 1. Due to symmety we have ν 1 = E E 1 ν 1. y α 1 x Fig : Othotopic plate with mateial coodinate system ε 11 ε γ 1 = E 1 1 ν 1 E 1 ν 1 E 1 1 E G 1 1 σ 11 σ σ 1

11 σ 11 σ σ 1 = 1 1 ν 1 ν 1 E 1 ν 1 E 1 0 ν 1 E E (1 ν 1 ν 1 )G 1 ε 11 ε γ 1 Using a tansfomation matices T ε and T σ, fo stain and stess components, espectively, the stain and stess components in the mateial coodinate system (index ) can be elated to those in the global coodinate system. c = cos(α) ; s = sin(α) ε xx c s cs ε yy = s c cs γ xy cs cs c s σ xx σ yy σ xy = c s cs s c cs cs cs c s σ 11 σ σ 1 ε 11 ε γ 1 = T ε 1 ε ε = T σ 1 σ σ ε = T 1 ε idem = T ε σ 1 σ σ ε = T 1 ε ε = T 1 ε S σ = T 1 ε S T σ = S σ σ = T σ 1 σ σ = T 1 σ C ε = T 1 σ C T ε = C ε ε 1.4 Cylindical coodinate system In the cylindical coodinate system, thee othogonal coodinate axes with coodinates {,θ,} ae used to identify mateial and spatial points. It is assumed that the mid-plane of the plate coincides with the plan = 0 in the undefomed state. We only conside axisymmetic geometies, loads and defomations. Fo such axisymmetic poblems thee is no dependency of the cylindical coodinate θ. Hee we take as an exta assumption that thee is no tangential displacement. p() y x f θ m b f s mw Fig : Plate in a cylindical coodinate system f n

12 1.4.1 Displacements All assumptions about the defomation, descibed in the fome section, ae also applied hee : no out-of-plane shea (Kichhoff hypotheses), small otation, constant thickness. The displacement components in adial (u ) and axial (u ) diection of an out-of-plane point P can then be expessed in the displacement components u and w of the coesponding same x and y coodinates point Q in the mid-plane. φ P Q P Q φ Fig. 1.1 : Displacement of a out-of-midplane point due to bending u (,θ,) = u(,θ) w, u (,θ,) = w(,θ) + η(,θ) 1.4. Cuvatues and stains The linea stain components in a point out of the mid-plane, can be expessed in the midplane stains and cuvatues. A sign convention, as indicated in the figue, must be adopted and used consistently. It was assumed that staight line elements, initially pependicula to the mid-plane, emain pependicula to the mid-plane, so γ = 0. ε = u, = u, w, = ε 0 κ ε tt = 1 u γ t = u,t + u t, = 0 = 1 u w, = ε tt0 κ tt ε = ε 0 κ ε = u, = η(,) ε = u, + u, = η, γ t = u t, + u,t = 0

13 ε 0 ε tt0 κ tt κ θ Fig : Cuvatues and stain definitions Stesses A plane stess state is assumed to exist in the plate. The elevant stesses ae σ and σ tt. Othe stess components ae much smalle o eo. σ tt σ θ Fig : In-plane stesses Isotopic elastic mateial behavio Fo linea elastic mateial behavio Hooke s law elates stains to stesses. Mateial stiffness (C) and compliance (S) matices can be deived fo the plane stess state. [ ε ε tt [ σ σ tt = 1 [ [ 1 ν σ E ν 1 σ tt = E [ [ 1 ν ε 1 ν ν 1 ε tt ; ε = ν E (σ + σ tt ) shot notation : ε = Sσ σ = S 1 ε = Cε = C (ε 0 κ )

14 1.4.5 Coss-sectional foces and moments Coss-sectional foces and moments can be calculated by integation of the in-plane stess components ove the plate thickness. A sign convention, as indicated in the figue, must be adopted and used consistently. The stess component σ is much smalle than the elevant in-plane components. Integation ove the plate thickness will howeve lead to a shea foce, which cannot be neglected. Ñ = M = D = [ N N t [ M M t h h = h = σ d [ σ h σ tt h [ σ h σ tt d = h h d = C (ε 0 κ ) d = hcε 0 h h C (ε 0 κ ) d = 1 1 h3 Cκ D N N t M t M θ Fig : Coss-sectional foces and moments Stiffness- and compliance matix Integation leads to the stiffness and compliance matix of the plate. [ Ñ M = [ Ch 0 0 Ch 3 /1 [ ε 0 κ [ ε 0 κ = [ S/h 0 0 1S/h 3 [ Ñ M Equilibium equations Conside a small column cut out of the plate pependicula to its plane. Equilibium equiements of the coss-sectional foces and moments, lead to the equilibium equations. The coss-sectional shea foce D can be eliminated by combining some of the equilibium equations.

15 N t d N dθ Ddθ M t d pddθ (M + M, d)( + d)dθ (N + N, d)( + d)dθ dθ M dθ (D + D, d)( + d)dθ M t d N t d Fig : Foces on infinitesimal pat of the plate N, + 1 (N N t ) + s = 0 D, + D + p = 0 M, + M M t D = 0 M, + M, M t, + p = 0 The equilibium equation fo the coss-sectional bending moments can be tansfomed into a diffeential equation fo the displacement w. The equation is a fouth-ode patial diffeential equation, which is efeed to as a bi-potential equation. [ M M t = Eh 3 1(1 ν ) [ 1 ν ν 1 M, + M, M t, + p = 0 d 4 w d 4 + d 3 w d 3 1 d w d + 1 dw 3 d = p B [ w, 1 w, ( d d + 1 ) ( d d w d d + 1 ) dw = p d B B = Eh 3 1(1 ν = plate modulus ) Solution The diffeential equation has a geneal solution compising fou integation constants. Depending on the extenal load, a paticula solution w p also emains to be specified. The integation constants have to be detemined fom the available bounday conditions, i.e. fom the pescibed displacements and loads.

16 The coss-sectional foces and moments can be expessed in the displacement w and/o the stain components. ( d w M = B M t = B w = a 1 + a + a 3 ln() + a 4 ln() + w p ( 1 d + ν ) dw d dw d + ν d w d ) D = dm d + 1 (M M t ) = B ( d 3 w d d w d 1 ) dw d Examples This section contains some examples of axisymmetic plate bending poblems. Plate with cental hole Solid plate with distibuted load Solid plate with vetical point load Plate with cental hole A cicula plate with adius R has a cental hole with adius R. The plate is clamped at its oute edge and is loaded by a distibuted load q pependicula to its plane in negative -diection. The equilibium equation can be solved with pope bounday conditions, which ae listed below. q R R

17 bounday conditions ( ) dw w( = R) = 0 ( = R) = 0 d m b ( = R) = M ( = R) = 0 f ( = R) = D( = R) = 0 diffeential equation geneal solution d 4 w d 4 + d 3 w d 3 1 d w d + 1 dw 3 d = q B w = a 1 + a + a 3 ln() + a 4 ln() + w p paticulate solution w p = α 4 substitution 4α + 48α 1α + 4α = q B α = q 64B solution w = a 1 + a + a 3 ln() + a 4 ln() q 64B 4 The 4 constants in the geneal solution can be solved using the bounday conditions. Fo this pupose these have to be elaboated to fomulate the pope equations. dw d = a + a 3 + a 4( ln() + ) ( d w M = B d + ν ) dw d = B ( M t = B ν d w = B q 16B 3 { (1 + ν)a (1 ν) 1 a 3 + (1 + ν)ln()a 4 + (3 + ν)a 4 (3 + ν) q ) dw d d + 1 { (1 + ν)a + (1 ν) 1 a 3 + (1 + ν)ln()a 4 + (3ν + 1)a 4 (3ν + 1) D = dm d + 1 (M M t ) ( = B 4 a 4 + q ) B 16B } } q 16B The equations ae summaied below. The 4 constants can be solved. Afte solving the constants, the vetical displacement at the inne edge of the hole can be calculated. w( = R) = a 1 + 4a R + a 3 ln(r) + 4a 4 R ln(r) q 4B R4 = 0 dw d ( = R) = 4a R + 1 R a 3 + 4a 4 R ln(r) + a 4 R q B R3 = 0 { M ( = R) = B (1 + ν)a (1 ν) 1 R a 3+

18 q ((1 + ν)ln(r) + (3 + ν))a 4 (3 + ν) 16B R} = 0 ( D( = R) = B 4 R a 4 + q } B R = 0 w( = R) = qr4 64B { 9(5 + ν) + 48(3 + ν)ln() 64(1 + ν)(ln()) } 5 3ν Fo the values listed in the table, calculated values ae shown as a function of the adial distance. inne adius R 0.1 m oute adius R 0. m thickness h 0.01 m Young s modulus E 100 GPa Poisson s atio ν global load q -100 Pa w x dw/d 1.4 x D M t M Fig : Calculated values as a function of adial distance. Solid plate with distibuted load A solid cicula plate has adius R and unifom thickness. It is clamped at its oute edge and loaded with a distibuted foce pe unit of aea q in negative -diection. The equilibium equation can be solved with pope bounday conditions, which ae listed below. Besides the obvious bounday conditions, thee ae also two symmety conditions, one fo the defomation and the othe fo the load.

19 q R bounday conditions w( = R) = 0 ( dw d ( ) dw ( = R) = 0 d ) ( = 0) = 0 D( = 0) = 0 diffeential equation geneal solution d 4 w d 4 + d 3 w d 3 1 d w d + 1 dw 3 d = q B w = a 1 + a + a 3 ln() + a 4 ln() + w p paticulate solution w p = α 4 substitution 4α + 48α 1α + 4α = p B α = q 64B solution w = a 1 + a + a 3 ln() + a 4 ln() q 64B 4 The 4 constants in the geneal solution can be solved using the bounday and symmety conditions. Fo this pupose these have to be elaboated to fomulate the pope equations. The two symmety conditions esult in two constants to be eo. The othe two constants can be detemined fom the bounday conditions. Afte solving the constants, the vetical displacement at = 0 can be calculated. dw d = a + a 3 + a 4( ln() + ) q 16B 3 D = dm d + 1 (M M t ) ( = B 4 a 4 + q ) B bounday conditions ( ) dw ( = 0) = 0 d a 3 = 0 D( = 0) = 0 a 4 = 0

20 displacement in the cente ( ) dw ( = R) = a R q d 16B R3 = 0 a = q 3B R w( = R) = a 1 + a R q 64B R4 = 0 a 1 = q 64B R4 w( = 0) = q 64B R4 Fo the values listed in the table, calculated values ae shown as a function of the adial distance. inne adius R 0 m oute adius R 0. m thickness h 0.01 m Young s modulus E 100 GPa Poisson s atio ν global load q -100 Pa w 0.5 x dw/d.5 x D M t M Fig : Calculated values as a function of adial distance. Solid plate with vetical point load A solid cicula plate has adius R and unifom thickness d. The plate is clamped at its edge. It is loaded in its cente ( = 0) by a point load F pependicula to its plane in the positive -diection. The equilibium equation can be solved using appopiate bounday conditions, listed below. Besides the obvious bounday conditions, thee is also the symmety condition that in the

21 cente of the plate the bending angle must be eo. In the same point the total coss-sectional foce K = πd must equilibate the extenal foce F. F R bounday conditions w( = R) = 0 ( dw d ( ) dw ( = R) = 0 d ) ( = 0) = 0 K( = 0) = F diffeential equation geneal solution d 4 w d 4 + d 3 w d 3 1 d w d + 1 dw 3 d = 0 w = a 1 + a + a 3 ln() + a 4 ln() The deivative of w and the coss-sectional load D can be witten as a function of. The symmety condition esults in one integation constant to be eo. Because the deivative of w is not defined fo = 0, a limit has to be taken. The coss-sectional equilibium esults in a known value fo anothe constant. The emaining constants can be detemined fom the bounday conditions at = R. Afte solving the constants, the vetical displacement at = 0 can be calculated. dw d = a + a 3 + a 4( ln() + ) D = dm d + 1 (M M t ) = B ( 4 a 4 bounday conditions ( ) dw { lim = lim a + a } 3 0 d 0 + a 4( ln() + ) = 0 a 3 = 0 K() = πd() = 8πBa 4 = F a 4 = F 8πB w( = R) = a 1 + a R + F 8πB R ln(r) = 0 ( ) dw ( = R) = Ra + F (R ln(r) + R) = 0 a = d 8πB ) a 1 = FR 16πB F (ln R + 1) 16πB

22 displacement at the cente w( = 0) = lim 0 FR { ( ) ( ) ( 1 + ln 16πB R R R) } = FR 16πB The stesses can also be calculated. They ae a function of both coodinates and. We can conclude that at the cente, whee the foce is applied, the stesses become infinite, due to the singulaity, povoked by the point load. σ = 3F { ( πd (1 + ν)ln R)} σ tt = 3F { ( πd 3 ν + (1 + ν)ln R)}