Elastic Clinders Subjected to End Loadings mi@seu.edu.cn
Outline Elastic Clinders with End Loading ( 端部受载柱体 ) Etension of Clinders ( 拉伸 ) Torsion of Clinders ( 扭转 ) Stress Function Formulation ( 应力函数体系 ) Displacement Formulation ( 位移体系 ) Membrane nalog ( 薄膜比拟 ) Boundar Equation Scheme ( 边界方程解法 ) Fourier Method Rectangular Section ( 傅立叶法 ) Multipl Connected Cross-Sections ( 多连通截面扭转 ) Hollow Sections ( 中空截面 ) Bending of Clinders ( 弯曲 ) Bending of Circular, Rectangular, and Elliptic Clinders without Twisting ( 圆形 矩形 椭圆截面 )
Elastic Clinders Subjected to End Loadings Semi-Inverse Method Zero lateral forces: 0. Let us guess the most general form for stresses F 0 F 0 F 0 0, 0, 0 3
Elastic Clinders Subjected to End Loadings Beltrami-Michell Equations 1 F F F F 1 1 1 F F F F 1 1 1 1 F F, 0 1 F F F F 1 1 0 satisfied 0 0 C C C C C C 1 3 4 5 6 1 1 1 1 F F F F C4 1 C5 1 4
Etension of Clinders ssumptions Load P is applied at centroid of crosssection so no bending effects Using Saint-Venant Principle, eact end tractions are replaced b staticall equivalent uniform loading Thus assume stress is uniform over an cross-section throughout the solid P P, 0 Using stress results into Hooke s law and combining with the straindisplacement relations gives u u P E v, 0, v v P E w, 0, w w P E u 0 R Integrating and dropping rigid-bod motion terms such that displacements vanish at origin S l P u E P v E P w E 5
Torsion of Clinders Guided b Observations from Strength of Materials: Projection of each section on,-plane rotates as rigid-bod about central ais mount of projected section rotation is linear function of aial coordinate Plane cross-sections will not remain plane after deformation thus leading to a warping displacement T R S l 6
Torsional Deformation In-plane / projected displacements u r sin, v r cos ngle of twist:. The warping displacement is assumed to be a function of onl the in-plane coordinates u, v, w w(, ). The displacement equilibrium equations require G G u v u G u G G w G u v v v w w w F F F 0 satisfied 0 satisfied w w 0 0 7
Stress Function Formulation Strain and stress field 0 0 u 1 w w v G w w(, ) 1 w w G Equilibrium equations result in Two stress components satisf F 0 G, G = (,) Stress compatibilit is Prandtl Stress Function automaticall satisfied. 8
Stress Function Formulation: BCs On lateral surface T T n n n n n n n n 0 0 0 0 0 0 T n n n n 0 d d d 0 0, i.e. set: 0. ds ds ds On ends: n n n T, T, T 0 More interested in satisfing the resultant end-loadings 1 : 0 P dd, : 0 P dd 3 : 0 P 5 : 0 M dd, 4 : 0 M dd dd, 6 : T M dd n d, ds n d ds
Stress Function Formulation: BCs On ends (with a big help from the Green s theorem) 3 5 are automaticall satisfied. 1 : dd dd d n 0 satisfied S S ds : dd dd d n 0 satisfied S S ds 6: T dd dd d d S dd T dd dd The assumed stress function ields a governing Poisson equation. The stress function vanishes on the lateral boundar. The overall torque is related to the integral of the stress Function. The remaining end conditions are automaticall satisfied. 10
Displacement Formulation Epressing the equilibrium equation with warping displacement 0 w w 0 w w G, G BCs on the lateral surface: n 0 T n n n w w For arbitrar n:, or equivalentl: w n w n dw d dn ds w w G n G n d d n n ds ds n d, ds n d ds
Displacement Formulation BCs on the ends 1 : 0 P dd, : 0 P dd 3 : 0 P 5 : 0 M dd, 4 : 0 M dd dd, 6 : T M dd 1 5 are automaticall satisfied. w w 6: T M dd G G dd w w T G ( ) dd J w w J G d d... Torsional Rigidit 1
Formulation Comparison Stress Function Formulation Displacement Formulation G 0 Lateral Surface T dd Ends Relativel Simple Governing Equation Relativel Simple Boundar Conditions R w w 0 dw 1 d dn ds Lateral Surface R T G ( ) dd Ends w w Simple Governing Equation Complicated Boundar Condition 13
Membrane nalog Consider a thin elastic membrane stretched with uniform tension over a closed frame and subjected to a uniform pressure. Deflected Membrane Nd Nd p Nd d Nd d d R Nd pdd Nd S Membrane Element Static Deflection of a Stretched Membrane Equilibrium of Membrane Element The equilibrium of a membrane element requires F Nd d Nd Nd d Nd pdd 0 p N 14
Membrane nalog The membrane is stretched over the boundar of the frame 0 on S The volume enclosed b the deflected membrane and the - plane Membrane Equations p N 0 on S V dd V dd Torsion Equations T 0 on S dd G Equations are same with: =, p/n = G, T = V 15
Membrane nalog n long an contour line, the resultant shear stress must be tangent d d, n, ds ds d d d d 0 n n n 0 ds ds ds ds d d t n n n n dn dn n nt n cos sin 0 0 0 cos sin 0 nt t t sin cos 0 0 0 sin cos 0 n t 0 0 1 0 0 0 1 0 0 cos sin 0 0 sin cos cos sin sin cos 0 16
Membrane nalog The shear stress at an point in the crosssection is given b the negative of the slope of the membrane in the direction normal to the contour line through the point. The maimum shear stress appears alwas to occur on the boundar where the largest slope of the membrane occurs. Using membrane visualiations, a useful qualitative picture of the stress function distribution can be determined and approimate solutions can be constructed. The torque is given as twice the volume under the membrane. 17
Solutions Derived from Boundar Equation If boundar is epressed b relation f(,) = 0, this suggests possible simple solution scheme of epressing stress function as = Kf(,) where K is an arbitrar constant. This form satisfies boundar condition on S, and for some simple geometric shapes it will also satisf the governing equation with appropriate choice of K. Unfortunatel this is not a general solution method and works onl for special cross-sections of simple geometr. f (, ) 0 S R Boundar-Value Problem G R T 0 S dd Ends 18
Elliptical Section a b 1 Look for Stress Function Solution: K a b 1. satisfies the BC and will satisf the governing equation if K a b G a b G a b a b a b 1. Since the governing equation and the BC are satisfied, we have found the solution. 19
Elliptical Section Relation to the carring torque a b G 1 1 T dd dd dd dd a b a b 3 3 3 3 a b G 1 a b 1 ab a b G ab a b a 4 b 4 a b ngle of twist per unit length a b T a b G ab a b Stress field (two-fold smmetr) T 1 3 3 T T T ab a b ab a b =, 3 3 4 4 The maimum shear stress (Strength of Materials vs. Membrane analog) T ma 0, b ab 0
Elliptical Section Warping displacement (two-fold smmetr) G w d G T 1 d 3 ab G T 3 ab G T b a 3 3 a b G w If a = b, the results degenerate to torsion of circular shafts. 1
Equilateral Triangular Section For stress function tr product form of each boundar line equation 3 3 K a a a satisfies boundar condition and will satisf governing equation G G, if K. 6a Since governing equation and boundar condition are satisfied, we have found the solution.
Equilateral Triangular Section Stress field (three-fold smmetr) G ( a ), a G a a ( ) 3 5 3T ma ( a,0) G a 3 18a Warping displacement (three-fold smmetr) w d w 3 G 6a Relation to the loading torque 7 3 5 3 5 4 T dd Ga G I p 5 3T 5T, I 4 p dd 3 3a 7Ga 3GI p 4 3
dditional Eamples Using Boundar Equation Scheme a c r = acos a a c r. a r = b a c a c Section with Higher Order Polnomial Boundar K a c a c ma ( )( ) G K, c 3 8 4 a (1 ) ( a,0) (0, a) G a Circular Shaft with Circular Kewa G b r 1 acos r ( ) G a ( ) Ga ma kewa s b/ a 0 : ma shaft Stress Concentration 4
When Boundar Equation Scheme Does Not Work? = m 1 = a = -m General Triangular Section Rectangular Section Tring the previous scheme of products of the boundar lines does not create a stress function that can satisf the governing equation. 5
Rectangular Section Fourier Method Solution Previous boundar equation scheme will not create a stress function that satisfies the governing equation. Thus we must use a more fundamental solution technique - Fourier method. Thus look for stress function solution of the standard form: G a h p p with (, ) ( ) homogeneous solution must then satisf 0, ( a, ) 0, (, b) G ( a ) h h h Separation of Variables Method X Y h(, ) X ( ) Y( ) : h 0 X Y XY 0 X Y X X 0 X sin Bcos Y Y 0 Y C sinh Dcosh 1 h( a, ) 0 a n, n 1,3,5, 6
Rectangular Section Fourier Method Solution Homogeneous solution n n h(, ) BD cos cosh Bncos cosh, a a n1 nb n h(, b) G( a ) Bncosh cos, n1 a a B Fourier cosine series n 1,3,5, n 1,3,5, 1 n l n f a0 ancos, an f cos d, l l 0 l n1 n b a n 3 G a ( 1) Bncosh G ( a ) cos d B 0 n a a a 3 3 n b n cosh a G a h p ( n1)/ ( n1)/ 3 G a ( 1) n n ( ) cos cosh 3 n 1,3,5 3 n b a a n cosh a n 0,1,, 7
Rectangular Section Fourier Method Solution Stress field T ( n1)/ 16 G a ( 1) n n cos sinh n 1,3,5 n b n cosh a a a n cosh a 16G a 1 ma ( a,0) G a n 1,3,5 n b n cosh a Relation to the loading torque ( n1)/ 16 G a ( 1) n n G sin cosh n 1,3,5 n b a a 3 4 16G a b 104Ga 1 nb dd tanh 3 n a Warping displacement 5 5 n1,3,5 w d w n cosh a ( n1)/ 3 a ( 1) n n sin sinh 3 G n 1,3,5 3 n b a a 8
Rectangular Section Fourier Method Solution 9
Thin Rectangles (a<<b): Open Thin-Walled Tubes Investigate results for special case of a ver thin rectangle with a << b. Under conditions of b/a >> 1 nb nb 16 1 3T a a 3 3 Gt 3 cosh, tanh 1, a t T G a b G t, t 3T t d 3T G, ma Gt 4 t 4 d t t Torsion of sections composed of thin rectangles: Neglecting local regions where rectangles are joined, we can use thin rectangular solution over each section. Stress function contours shown justif these assumptions. Thus load carring torque for such composite N section will be given b: 3T G t The maimum shear stress can be estimated for the narrowest rectangle. i1 i i b a 30
Multipl Connected Cross-Sections On all lateral boundaries: d 0 i Constan t. ds Value of i ma be arbitraril chosen onl on one boundar, commonl taken as ero on S o. The single-valuedness of warping displacement requires: w G, G w w w 1 0 S S S G 1 sds i sds G, G S S dw d d d d d d where is the area enclosed b an arbitrar closed-path S that is encircling an inner boundar S i. The value of i on the inner boundar S i must be chosen so that the above relation is satisfied. S 31
Multipl Connected Cross-Sections The BCs on clinder ends require n n 1 : 0 P T dd, : 0 P T dd 3 : 0 P 5 : 0 M 1 T n T For multiple holes: dd, 4 : 0 M T n 1 5 are automaticall satisfied. n dd n n dd, 6 : T M ( T T ) dd 1 1 1 1 1 1 n n 6: T T T dd dd dd d d dd S o, C, S 1 d d dd dd S T dd T dd ii i 3
Hollow Elliptical Section For this case lines of constant shear stress coincide with both inner and outer boundaries, and so no stress will act on these lateral surfaces. Therefore, hollow section solution is found b simpl removing inner core from solid solution. This gives same stress function and stress distribution in remaining material. a b G a b a b ( ka) ( kb ) 1 Constant value of stress function on inner boundar is: i 1 1 b a a b G k 1. a b Load carring capacit is determined b subtracting load carried b the removed inner clinder from the torque relation for solid section 3 3 3 3 a b G ( ka) ( kb) G G 3 3 4 T a b (1 k ) a b ( ka) ( kb) a b T 1. ab 1 k Maimum stress still occurs at = 0 and = b: ma 4 33
Closed Thin-Walled Tubes Thin wall thickness implies that the membrane slope BC can be approimated b a straight line. Since the membrane slope equals resultant shear stress: s 1 t s. Determine the value of 1 : S c G c sds G c 1, 1 ds Sc t s Load carring relation: 1 T dd 11 1 1 1 c. where = section area, 1 = area enclosed b S 1. Combining these relations: T T T 1 s 4 Sc c ct s G c t s τ ma occurs across the narrowest wall. where S c = length of tube centerline, c = area enclosed b tube centerline. 1 ds o 1 34
Open vs. Closed Thin-Walled Tubes Cut creates an open tube and produces significant changes to stress function, stress field and load carring capacit. Open tube solution can be approimatel determined using results from thin rectangular solution. Stresses for open and closed tubes can be compared and for identical applied torques, the following relation can be established Open Tube Closed Tube 3T t T t c c Open Tube 6, but since c 1 Closed Tube Open Tube Closed Tube Stresses are higher in open tube and thus closed tube is stronger. Cut 35
Bending of Clinders From the general formulation The other three nonero stresses will be determined to satisf the equilibrium and compatibilit relations and all associated boundar conditions. 0 Motivated from the general formulation and Strength of Materials F F F 0 0 0 C1 C C3 C4 C5 C6 B C l B equilibrium equations: 0 0 B C 0 36
Bending of Clinders Stress function 1 1 B C 0 B C 0 1 1 B B 1 1 C C Beltrami-Michell Equations 1 F F 1 B 1 B 0 1 1 F F 1 C C 0 1 1 B 0 1 C B k C 1 0 1 37
Bending of Clinders Determination of the integral constant k 1 v u 1 v u 1 v w u w 1 1 1 C B k G G G 1 Selecting the section origin ( = = 0) at the center of twist 1 G k k G C B G 1 The governing equation: BCs on the lateral surface 1 d 1 d ds ds d 1 d d C B ds ds ds n n 0 B C 0 38
Bending of Clinders Fleural portion Torsional portion C B R 1 1 d C d B d S 61 3 3 p h, p, h 0 BCs on = l C B G d ds 0 S 1 : P dd, : P dd, 3 : ( ) dd P P o 0 1 3 1 1 P B dd B C d1d B dd 1 1 1 3 d C d B d d d d C d B d 1 1 3 d d C d B d S S 3 dd dd B C d d Green's Theorem: u d u d u u d d. S 1 1,1 1, 1 1 R 39
Bending of Clinders Determination of the coefficients 3 1 P BI CI P I P I P I P I B, C P BI CI Relationship among torsion and fleure I I I I I I 1 1 op 0P C B dd 1 dd dd C B dd 1 op 0P J C B d ngle of twist can be inferred from the fleural stress function. Can be used to determine the shear center / center of fleure. Without torsion, i.e. α = 0, this equation can be independentl used for two cases to generate two equations for locations o and o. d 40
Bending of Circular Beams without Twisting Governing equation P 0, P P, B Cl B 0 P I C I I P I I P I 0 C B P r cos 1 1 I P 3 3 P 3 f r cos f r 3cos cos 3, I 61 I 41 f 1 f 1 f r r r r f 0 R n BCs on the lateral surface f r sin n cosn d 1 d d C B r a 3 f ds ds ds 1r cos 3r cos 3 1 d 1 P 3 d 1 P 3 a sin a 3sin sin 3 a d I d 8 I n1 f 0 41
Bending of Circular Beams without Twisting Fleural Stress Function 3 P r 3 1r cos 3 r cos 3 I 81 41 3 3 3 d P a 3 P 3a a 1a sin 33 a sin 3 sin sin 3 d I 81 81 I 8 8 a 3a a 1 81 8 1 33 81 8 3 3 1 3 3 81 3 P 3 r 1 3 a r cos cos r cos 3 I 81 81 4 P 3 1 1 a I 81 81 41 3 1 4 1 a 4
Bending of Circular Beams without Twisting Stress field P 3 1 1 a I 8 1 8 1 4 1 3 1 P1 B 4I 1 1 P 3 3 1 C a I 81 81 81 P B Cl l I P 3 P 3 P 3 ma 0,0 a a 4 I 8 1 a 4 8 1 a 1 The maimum shear stress occurs at the origin and is the same as that of Strength of Materials theor when ν = ½. The displacement field can determined b the usual procedure. 43
Bending of Rectangular Beams without Twisting Governing equation P 1 I P 3 f a I 6 1 BCs on the lateral surface d 1 P d ds I ds R S b d d 1 P d For a : ds d 0 f 0 ds d I d d d 1 P 3 For b : ds d b f a ds d I 6 1 Proposition of the homogeneous solution, f 0 n n,,, f n sin cosh n1 a a,,, 44
Bending of Rectangular Beams without Twisting Fleural Stress Function For a : f 0, satisfied 3 n n n b 3 a 1 For b : f nsin cosh a n n 1 a a 61 3 3 n b 1 n cosh a n n 3 n sin cosh P a 1 a a 3 b a 3 3 I 1 n b n 1 n cosh 6 1 a Stress field n n n sin sinh a P 1 a a P, 1 l I n 1 n n b cosh I a n n n 1 cos cosh P P P 1a b 3 a a a I I 61 I n b n 1 n cosh a 45
Bending of Rectangular Beams without Twisting Comparison to the Strength of Materials solution 3P b 3 3 3 a 8ab 8ab 1 n1 n Strength of Materials n n n cos cosh P 1a 1 a a n b cosh a ma 3 Strength of Materials Correction Term 3 P ap 6 1 n b a,0 1 sech 4ab 4b 1 n1 n a Strength of Materials Correction Term n 3 P ap 1 1 nb 0 0, 0 1 sech 3 4ab 8b 1 n1 n a Correction Term The maimum shear stress occurs at ( = ± a, = 0) and is the same as that of Strength of Materials theor when ν = 0. For the case of a thin rectangular section with b >> a, the elasticit solution reduces to that of the Strength of Materials. 46
Bending of Elliptic Beams without Twisting Governing equation P 1 I P f 3, f 0 I 61 BCs on the lateral surface 1 d P P b S d I I a d P df P df 1 b b 1 b d I d 1 I a d a 1 Epect up to quadratic stress distribution m n (, ) mn f m0 n0 R b a 3 3 01 0 03 10 11 1 0 1 30 0 0 0 0 f 0 603 1 0 1 630 1 330 0 303 1 0 47
Bending of Elliptic Beams without Twisting Epect up to quadratic stress distribution On S : df 10 11 d 1 b 0 a b 1 1 330 ab a a 1 10 1 b 3 b a 1 a b 1 11 0 1 1 0 0 03 0 a b a 1 a b 10 1b b 30 3a b a 1 b b a b 1 f(, ) 01 30 3 a b 3 b b 10 1 30 a b 6a 61 a 1 1 Fleural stress function P a b 3 b b I a b 6a 1 a 1 1 Stress field... 48
Outline Elastic Clinders with End Loading ( 端部受载柱体 ) Etension of Clinders ( 拉伸 ) Torsion of Clinders ( 扭转 ) Stress Function Formulation ( 应力函数体系 ) Displacement Formulation ( 位移体系 ) Membrane nalog ( 薄膜比拟 ) Boundar Equation Scheme ( 边界方程解法 ) Fourier Method Rectangular Section ( 傅立叶法 ) Multipl Connected Cross-Sections ( 多连通截面扭转 ) Hollow Sections ( 中空截面 ) Bending of Clinders ( 弯曲 ) Bending of Circular, Rectangular, and Elliptic Clinders without Twisting ( 圆形 矩形 椭圆截面 ) 49