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Lecture 3 Plate Bending

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Lecture 3 Plate Bending Topics in Ship Structural Design (Hull Buckling and Ultimate Strength) Lecture 3 Plate Bending Reference : Ship Structural Design Ch.09 NAOE Jang, Beom Seon OPen INteractive Structural Lab 9.1 Small Deflection Theory "Long" Plates(Cylindrical Bending) . An equation relating the deflections to the loading can be developed as for the beam. Saddle shape . Cylindrical Bending: A plate which is bent about one axis only(a>>b) Transverse deformation does not occur, since such a deformation would required a saddle shape deformation. y x y = x y x EE E εy -εx y x y x y 0 ε E E x -εy (1-2 ) = x x E E = =E x1- 2 x x Plate in cylindrical bending 2 OPen INteractive Structural Lab Reference Poisson Ratio trans axial Strain Δεy induced by εy y x y E 1 ( ) x E x y z 1 ( ) y E y z x 1 ( ) z E z x y OPen INteractive Structural Lab 9.1 Small Deflection Theory Formulas between a beam and a long plate . E΄ : effective modulus of elastictiy, always > E. E E (1 2 ) → A plate is always “stiffer” than a row of beams . For a long prismatically loaded plate, the extra stiffness may be fully taken into account for by using E΄ in place of E. For simply supported plates: 5pb4 5 pb 4 (1- 2 ) == max 384E I 32 Et 3 . For clamped plates: pb4 pb 4(1- 2 ) == max 384E I 32 Et3 4 OPen INteractive Structural Lab 9.1 Small Deflection Theory Formulas between a beam and a long plate . Moment curvature relation may be obtained from External bending moment = moment of stress force 23 tt/2 /2 E z Et1 M zdz= ( ) dz = tt/2x /2 22 1rrxx 12(1- ) E where =/zr x (1 2 ) x xx EDt13 3 M ==Et 2 D 2 12(1- ) rrxx 12(1- ) . D : the flexural rigidity of the plate the constant of proportionality between moment and curavature analogous with EI in beam theory EI M 5 OPen INteractive Structural Lab 9.1 Small Deflection Theory Formulas between a beam and a long plate . The radius of curvature of the plate can be expressed in terms of the deflection w of the plate 2w 1 2w M D M v 2 x2 x rx EI . Also for a unit strip in a long plate the max. stress is max=/M max c I where the section modulus I / c=(t32 /12)/(t/2)=t /6 . For a uniform pressure p, the maximum bending moment is proportional to pb2. The stress is expressed in 2 b max =kp t . The coefficient k depends on the boundary conditions: k=3/4 for simply supported edges k=1/2 for clamped edges 6 OPen INteractive Structural Lab Reference Beam Theory stress – strain - curvature - Moment 1) strain-curvature 3) moment-curvature relation Geometric relation 2)strain-stress relation M x ydA A y 2 Ey 1 E 1 3 Ey 1 M 4 EI 4) Stress-moment relation My x I OPen INteractive Structural Lab Reference Beam theory 1) Strain-Curvature Relation Initial length of Line e-f = dx Length after bending = L1 y L ( y)d dx dx 1 dx d compression Natural axis No change in length L1 dx y x y Tension dx OPen INteractive Structural Lab Reference Beam theory 3) moment-curvature relation dM ydA Mx x ydA A 1 M Ey2dA E y2dA EI EI Ey A A 1 M : EI = bending rigidity EI I y 2dA : Moment of Inertia A OPen INteractive Structural Lab Reference Homework #1 Derive in detail Beam Theory : Relation between load, shear force, Moment . Force equilibrium → Relation between shear force (V) and load (q) dV F 0 : V qdx (V dV ) 0 q vert dx . Moment equilibrium → Relation between Moment (M) and shear force (V) dx dM M 0 : M qdx (V dV )dx (M dM ) 0 V 2 dx . Combining two relations d 2M q dx2 . Relation between deflection and distributed load 2 d M d 4v q dx2 EI dx4 EI OPen INteractive Structural Lab 9.1 Small Deflection Theory Forces and moments in a plate element . Force equilibrium → Relation between shear force (V) and load (q) qx qy dV + +p =0 q q :shear force per unit length xy dx m : moment per unit length . Moment equilibrium mm m m dM xy- y +q =0 yx +x -q =0 V xyy yxx dx . Combining two relations 2 22 m mm 2 x -2xy + y +p=0 d M 22 q x x y y d 2 x . Relation between deflection and distributed load 2w 2w 2w 2w m D v x 2 2 my D 2 v 2 4 x y y x 4w 4w 4w p d v q 2 4 2 2 4 2w 4 x x y y D m D(1 v) dx EI xy xy OPen INteractive Structural Lab 9.1 Small Deflection Theory Derivation of the Plate Bending Equation The previous theory is only applicable if : . Plane cross sections remain plane. The deflections of the plate are small(wmax not exceeding 3/4 t) . The maximum stress nowhere exceeds the plate yield stress(i.e., the material remains elastic) A panel of plating will have curvature in two directions = x y = y x x EE y EE v v2 v y x y E E x（1-2） = + E xy Differential element of plating OPen INteractive Structural Lab 9.1 Small Deflection Theory Derivation of the Plate Bending Equation . Assumptions stated previously give rise to the strain-curvature relations 2w 2w (z) (z) x 2 y 2 x y E 2w 2w (z) v x 2 2 2 1 v x y Et3 D 2 2 2 12(1- ) t / 2 t / 2 E w w 2 mx x zdz v z dz t / 2 t / 2 2 2 2 1 v x y 2w 2w m D v x 2 2 x y x 2 （1-） = xy + E 2w 2w m D v y 2 2 y x OPen INteractive Structural Lab Reference – Mechanics of Material Beam Theory : Deflection curve . Relation between deflection and distributed load From Geometric relation ds dx d ds 1 d dx 1 d 2 1 d dv 2 tan dx ds dx d 1 M tan dx EI 2 d M d 4v q 2 2 4 d M dx EI dx EI q dx2 OPen INteractive Structural Lab 9.1 Small Deflection Theory Forces and moments in a plate element . The lateral load pdxdy is carried by the distributed shear forces q acting on the four edges of the element. In the general case of bending, twisting moments will also be generated on all the four faces Without the lateral load pdxdy With the lateral load pdxdy Forces and moments in a plate element OPen INteractive Structural Lab 9.1 Small Deflection Theory Forces and moments in a plate element . Sine convention in beam theory – Determined how deform the material – Shear force : rotate an element clockwise (+) – Bending moment : shorten the upper skin (+) Sine convention in beam theory Sine convention in plate bending OPen INteractive Structural Lab 9.1 Small Deflection Theory Forces and moments in a plate element . From equilibrium of vertical forces : qx (qx dx)dy qxdy q x q x +y +p =0 xyqy (q dy)dx q dx pdxdy 0 y y y . Taking moments parallel to the x-axis, at Point A mm xy y myx mx -q +q:sheary =0 force per unit+ length -qx =0 m xy yx (m xy dx)dy m dy m dx m : moment per unit length xy x xy y m (m y dy)dx q dxdy 0 y y y Point A . Taking moments parallel to the y-axis OPen INteractive Structural Lab 9.1 Small Deflection Theory Forces and moments in a plate element . Because of the principle of complementary shear stress, it follows that myx=-mxy, so that qx qy + +p =0 mxy mx xy - + -q =0 yxx . Substitutingmm for qx andm qy m xy- y +q =0 yx +x -q =0 xyy yxx 2m 22mm x -2xy + y +p=0 x22 x y y m m m m ( xy x ) ( y xy ) p 0 x y x y y x OPen INteractive Structural Lab 9.1 Small Deflection Theory Forces and moments in a plate element . If a point A in the plate a distance z from the neutral surface is displaced a distance v in the y direction, . The change in slop of the line AB will be v v dx v v x dx x . The change in slop of the line AD u x . The shear strain is Shear strain and plate deflection u =+ xy Shear strain is induced by . The shear stress deflection. That is, the vertical deflection generates shear strain. u =G + xy OPen INteractive Structural Lab 9.1 Small Deflection Theory Forces and moments in a plate element . from the geometry of the slope of w in each direction (x and y): w w u z v z x y . Making these substitutions gives w u=-z u(), z(), (), x x 2w 2Gz xy (+) convention τ . The twisting moment 부호 정의 필요 2 3 2 t / 2 w 2 Gt w E mxy 2G z dz t / 2 G xy 6 xy 2(1 v) τ Et 3 2w Et 3 (1 v) 2w 2w m D(1 v) xy 12(1 v) xy 12(1 v)(1 v) xy xy -τ +τ OPen INteractive Structural Lab 9.1 Small Deflection Theory Forces and moments in a plate element .
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