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# Aerospace Structures EAE 135

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This 69 page Class Notes was uploaded by Heloise Bernhard DVM on Tuesday September 8, 2015. The Class Notes belongs to EAE 135 at University of California - Davis taught by Valeria Saponara in Fall. Since its upload, it has received 49 views. For similar materials see /class/191750/eae-135-university-of-california-davis in Engineering Aerospace Sci at University of California - Davis.

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EAE 135 Aerospace Structures Fundamental of Elasticity V La Saponara Class notes 1 Introduction To analyze a structure quantities like displacement strain and stress are used How much is the structure moving How much is it deforming Will it break How much more load can it withstand before it breaks These notes aim to de ne these quantities and the relationships among them within the general framework of the theory of elasticity 2 Displacements It is easier to have a physical sense of how much a structure is moving and deforming by looking at it It is more difficult to look at a structure and immediately guess its state of stress and know how much load it can carry before breaking down and why it will break down tension bending torsion fatigue corrosion buckling creep dynamic instability etc Lot of structural analysis techniques e g Finite Elements Methods use displacements as their variables and use well defined relationships to connect displacements to other important quantities such as strains and stresses Define a CaItesian coordinate system X y z with origin 0 Figure 1 When something moves we need to know how much is it moving and in which direction We need to use vectors to best describe displacements Vector s coordinates u v w ie X u i v j w k where L j k are the unit vectors oriented along axes X y and 2 Point P moves into point P P 9 P z 13 quotG 39lt X Figure 1 CaItesian coordinate system with displacement vector X PP Consider a generic body typically shaped as a potato Look at the distance between two internal points say P and Q Figure 2 If this distance PQ remains constant after the body has displaced we say that the body had a rigid body motion a translation andor a rotation that did not change the internal distance between any couple of internal points Rigid body motions do not cause the body to deform If the distance changes the body has deformed and we need to look at strains as well as stresses Figure 2 Examples of a rigid body motion PQ P Q b motion with deformation PQ i P Q 3 Strains Consider a potatoshaped body We will focus on a parallelepiped formed by points of this body as they displace and deform A body can deform because of a 1 change of volume 2 change of shape 3 a combination of both See Figures 3 4 V dX Figure 3 Deformation with change of volume xy View section PQSR changes into P Q S R Figure 4 Deformation with change of shape Xy view section PQSR changes into PQ S R Change of volumes are associated to the socalled longitudinal strain also called normal strain Change of shapes are associated to the socalled engineering shearing strain 31 Longitudinal strains They are given in terms of the change of length deformed length 7 original length normalized with respect to the original length projected in each direction X y z dxd2f dx g dyd2f dyg dzd2f dz 8X Z dx y dy dz Consider 8x Using Taylor s series it can be shown that 2 2 2 12 dxd2f dx6 udx dx 3 de 6x 6x 6x 2 2 2 12 9 gx126 u6 u 6 w 1 1 6x 6x 6x 6x In most materials strains are very small When measured with strain gauges for example the units used are microstrains 10396 lengthlength Therefore by using the fact that strains are small and the relation 1Xa N l a X for X ltltl it is possible to neglect different terms in Equation 1 those with higher order of magnitude and obtain Bu 9 g m 2a x ax Analogously 8 m g Z m w 2b y z 32 Engineering shearing strain Let us consider Figure 5 for the simpler case of deformation in the xy plane Bu MSS dy R 53 Figure 5 Engineering shearing strain xy View The engineering strain component in the xy plane yXy is de ned as the sum of the two deformation angles in Figure 5 Ky x l Using trigonometry and Figure 5 it is possible to show that for small deformations the following equation holds Bu 6V m 3339 7W 6y ax Analogously 7y E 7 m 3b Another definition of shearing strains is l Bu 6V 1 6w 6V 1 Bu 6w 8 m 8 gzm 8 gxzm 8 y 2 6y 6x y y 2 By 62 y 2 62 6x These are called tensorial shearing strains or elasticity shearing strains 33 Kinematics Equations 2 and 3 are the socalled kinematics relationships which connect the 6 strains 3 normal strains associated to the change of volume 3 engineering shearing strains associated to the change of shape to the 3 displacements in the X y z coordinate system 4 Stress 41 De nition and nomenclature One should have by now been exposed to the concept that a bar with area A subject to a tensile force F has a uniform stress distributed along it equal to F A Figure 6 In general stress may not be uniform or directed normally to the area it is applied to F F lt gt Figure 6 A bar subjected to a tensile force F Consider a potatoshaped body Cut a piece of the potato Figure 7a and remove it Figure 7b The effect of the nowremoved piece of potato is represented by forces acting on the surface with normal 11 a Figure 7 Sketch for definition of stress Consider a force F acting on a small area element with normal 3 dAn Figure 7c A stress vector 3 is defined as 3 lim dAn7gt0 dAn The existence of this limit is assumed by the theory of elasticity Is the stress a perpendicular vector to area dAn In general this is not the case Normal as well as shearing stresses are present To express this a volume element of the potato with dimensions dx dy dz is shown around point P Figure 8 2 dz gt y O y dy X x Figure 8 Sketch of stresses note that some components are missing for clarity sake The nomenclature adopted for the stresses is the following in cab the first index a indicates the normal to the face where the stress is applied the second index b indicates the direction of the stress For example cxx is the stress acting on the face with normal x and is in the xdirection 62x is the stress acting on the face with normal z and is in the xdirection Figure 8 If one counts these components atotal of 9 stresses will be found cxx SW 622 cxy cyx 6x2 62x cyz 62y Consider also that in the volume element of Figure 8 stresses are applied also to the back faces 42 Equilibrium equations The stress components are related to each other by the equilibrium equations in the form of Newton s second law 2 E mg Q 2 M Q Consider a volume element subject to its weight and to the stress components The inertia force mg is a body force ie a force proportional to the volume for example gravity or a magnetic force Set elemental body forces equal to p E dx dy dz where p density E body force per unit mass dx dy dz volume 9 p E dx dy dz has units of force For simplicity consider the xy view and equilibrium in the xdirection Figure 9 The element will have sides dx and dy while dimension dz is out of the plane 2 FX 0 9 amalde Gwalde 039DC agicbcjdydz 0 x 6039yy 039 d W k y n 60 y 039 60 dy 02y a W dx gt 6y x 039xy Bx A 6 039 a B 039 xx dx 7 dX y 6x 4oyx 39 X Uyy V V Figure 9 Sketch for equilibrium along XaXis Note that zcomponents of the stress are not shown for sake of clarity of the picture 6039 x 1 y dy dxd2 6y azxdxdy 0392X 60 d2 62 dxdy prdxdydz 0 6039 a 6039 W 6 VZ pr 0 4a 6x 6 62 Analogously equilibrium along the other two directions is given by 6039 6039 6039 y yy ZypB 0 4b 6x 6y 62 y 40 6 amp p320 6y 6 6x 2 It can be shown that the equilibrium of the moments gives the following in case of absence of body moments e g moments from magnetic eld xy ny 6x2 62x 62y Gyz Therefore equilibrium equations relate the stress components to each other in a very speci c 43 Boundary conditions To solve the partial differential equations 4a4c some boundary conditions are needed Boundary conditions have to do with forces applied to the boundaries of the body traction boundary conditions or to speci ed displacement elds displacement boundary conditions If the problem gives both types of boundary conditions it is called a mixed boundary conditions problem and is typically more complicated to solve We still use equilibrium for the boundary conditions in the form of Newton s second law However this law is de ned in terms of a volume and we need to work with the surface where the external force EexLis applied as well as with the internal forces related to stresses To do so consider the volume immediately next to the entire surface of the body to be divided in many tetrahedra pyramids with one face on the surface and the rest of each tetrahedron inside the body Figure 10 Figure 10 Sketches of a body subject to external force b tetrahedron used for the calculations c view in xy plane Note that axes x y and z are still assumed perpendicular to each other Call dA face ADB of the tetrahedron under analysis with normal 3 and external force Em Call dAx face PCB perpendicular to the xaxis dAy face APC perpendicular to the yaxis and dAZ face APB perpendicular to the zaxis The areas of these faces are related to each other by cosines of the angles between the normal 3 positive outward and the axes perpendicular to the faces eg dAZ cosnz dA 1 Introduce the tractions If you are looking at internal forcesarea you use stresses If you are looking at external forcesarea you use tractions In particular as the external force Em is projected along the three reference axes and is divided by the area dA you get traction components TX Ty TZ units forcearea Apply force equilibrium along xaxis using traction Tx all the stresses in that direction and the body force Bx 1 From geometry dAZdA area of PBAarea of ABC PDAB2 CDAB2 PDCD From Figure 10c PD CD cosy 2 FX 0 9 Tdi aridl aydiy Judl prdz dAZ Tdi 039DC cosn xdA 0y cosn ydA 039 cosnzdA prdz cosnzdA Note that the last term on the right hand side is smaller than the others and can be neglected 9 T oquot cosnx 0y cosn y 0 cosn 2 5a Analogously Ty oquoty cosn x 039yy cosn y O39Zy cosnz 5b T2 O39XZ cosn x O39yz cosn y 039Z cosnz 5c These equations relate the internal stress to an applied load distributionarea acting on the surface of a body Example Consider a piece of sheet metal with rectangular shape as in Figure 11 Assume that its stress is characterized by the following matrix 039 OX Ox 5 2 0 039 ayx ayy ayz 3 0 02x Ozy 022 0 Calculate the tractions on 2 faces one located at a y 0 and one located at X b 0 y a Figure 11 The tractions are the external forcesarea applied to this structure which are in equilibrium with the internal forcesareas the stresses To use Equations 5 we need to calculate the direction cosines ofthe normal to the two faces ie cosnX cosny cosnz Face a y 0 has normal parallel to the XaXis ie characterized by COSIIX l cosny 0 cosnz 0 Therefore Equations 5 becomes TX 039 cosn x ayx cosn y ozx cosn z gt TX 5 1 2 0 0 0 Ty oquoty cosn x 039yy cosn y O39Zy cosn2 gt Ty 2 l 3 0 00 T2 039 cosnx O39yz cosny 039Z cosn2 gt T2 00 00 40 Consequently the traction applied to face a y 0 in equilibrium with that state of stress is l 5 1 21 Analogously face X b 0 has normal parallel to the yaXis ie COSIIX 0 cosny l cosn2 0 The traction is calculated as TX 039 cosn x ayx cosn y 02x cosn 2 gt TX 5 0 2 l 00 Ty oquoty cosn x 039yy cosn y O39Zy cosn2 gt Ty 20 3 l 00 T2 039 cosn x O39yz cosn y 039Z cosn2 gt T2 0 Consequently the traction applied to face X b 0 in equilibrium with that state of stress is l 217 3 j 5 Compatibility equations The last relationships under considerations are the socalled compatibility equations They are written in terms of strain components They ensure that the solution of siX strain components 8 8y 82 yxy yy ya ya yyz yzy leads to a unique and continuous displacement field with three components u V and w Failure to satisfy the compatibility equations would lead to a displacement field solutions that has gaps and overlaps This solution would therefore not be continuous and is not considered acceptable in the theory of elasticity as well as in the numerical techniques based on it e g Finite Elements Methods 2 62 62 6 8f 8y 7V 6a 6y 6x 6x6y 628 628 62y DC 22 Z 622 6x2 6x62 2 62 62 6 82 82 7yz 6c 6y 62 6y62 6 6 2 gr yyzayxz KW 6822 62K 6x 6y 62 6x6y 6 6 62 2r 6 6yk 6x 6y 62 6x62 6 6 2 3 2m 6f 6xk 6x 6y 62 6y62 6 Stress transformation 61 Tensors It seems that stresses may be very dependent on the reference system used to analyze the body Is it possible to calculate the stress eld for any other reference system once we know the values in xyz Can we find out a condition for maximum or minimum stresses and the planes in which these stresses exist It can be shown that stress transformation laws from one reference xyz to a new reference xyz are given in the form from old reference to new reference 9 039 c 039 or 7 0106 Gary oxz where 039 039 039yy 039yz is the matrix with the stress components and the matrix of 039 039 039Z Zr zy cDC cy cm direction cosines is c CyX cyy Cy where cab cosab cosine between the new axis 62 czy 622 a and the old axis b eg cxy is the cosine between x and y The T in Equation 7 stands for transpose The stress is a tensor of the second order A tensor is a mathematical quantity that can be represented by a multidimensional array and that satisfies a very specific transformation law from one reference system to another reference system This law is indicated by a matrix equation Equation 7 The tensor is called of the second order when it can be represented by a 2D array a matrix with rows and columns ie it is characterized by values given to two parameters or two indices In this case the two indices are those of the stress components ie cab A tensor of the second order also requires multiplication with two matrices of direction cosines in this case c and its transpose cT for the transformation A tensor of the first order would have one parameter or index that changes and have one matrix for the transformation An example of a tensor of the rst order is a vector a 1D array in a cartesian reference system A a1 1 a j a3 l The transformation law is a c A ie a l cxx cxy cxz al v a 2 cyx cyy cyz 02 v a 3 czx czy 622 a3 62 Principal stresses Going back to the transformation of stresses and to simplify our discussion let us assume a plane stress case such that 6x2 0 cyz 622 A plate with thickness much smaller than the other two dimensions and subject to inplane loading is a typical case of plane stress because the components of the stress along the axis perpendicular to the plate are very small We will get back to this de nition later on in the course Assume a counterclockwise rotation about 2 2 Figure 12 ie axes x and y are each rotated counterclockwise to reach the new positions x and y A zz Figure 12 Plane rotation around zz axes For the case in Figure 12 cxx cosxx cos9 cxy cosxy cos909 sin9 cyx cosyx cos 90 9 sin9 cyy cosyy cos9 Substituting into Equation 7 leads to a a 0 039 OX 0 cos 6 sin 6 0 0 039 CW 0 c ayx 0W 0 cT where c sin6 cos6 0 8 0 0 0 0 0 0 0 0 l L J L J We use now trigonometric identities to express the angle 9 in terms of 29 which will turn out to be convenient for the derivative process2 that will follow Equation 8 becomes 039 039DC 039yy039m 039yycos 26039Xy sin 26 9a l l 039 3039DC 039 5O39m 039 cos 26 03y s1n 26 9b l 6 6 oXy E039DC 039yys1n2 0y cos2 90 We want now to calculate at what angle of rotation 29 and consequently 9 each of the stress components 039 039 03y 1s a max1mum or a m1n1mum 29 corresponds to a part1cular plane 2 Namely sin2 6 l cos 26 cos2 6 l cos 26 611039DC 039 0 at99x9 tan2t9x 10a 6129 1 E Oquot10C O39yy do 039 W 0 9at99y9 tan2t9 y 10b d26 y 1 Ogtoc oyy 2 1 do Eogtoc oyy v 0 9at9959 tan2t95 10c d26 039 Note that the angles 9x and By are the same It turns out that when one of the normal stresses is maximum say 039 the other 039 is minimum It can be shown that you cannot have both stresses being maximum or both being minimum at the same time Substituting 10a in 9a 9b 9c plus relating tan29x to cos29x and sin29x leads to the following 039 0 annual 12 1 l 3039m 039 0m 039 z 0 039aVg 039s max1mum normal stress lla OJO 12 1 l 2 mm 3039 039 039 2 0W 7031Vg 0395 m1n1mum normal stress 1 lb 0W 0 zero shear stress 1 1c 7121501 Recall that tan29 is a periodic function If one chooses to be in the rst quadrant 0329lt112 the stress in Equation lla will be larger than the stress in Equation llb If one chooses to be in the third quadrant 11 29lt3n2 signs change and one obtains a0 2 1 that becomes the m 1m minimum normal stress while 039 t 1 becomes maximum Because of this it is more M 1m appropriate to label the stresses as l 039 039 039 039 max1mum normal stress 12a 1 2 xx yy 5 l 0392 3039DC 039 0395 m1n1mum normal stress 12b 1 where as 10 039yy 2 6y These stresses are called principal stresses and the plane 039 characterized by 26 tan391 W is called principal plane Remember that there is no 1 5O39xx 03y shear acting on this plane because of Equation llc These stresses are used M in structural analysis and failure criteria If one substitutes Equation 10c into 9a through 9c it is found out that on the plane where maximum shear occurs there are still normal stresses These are equal to each other and to l 039 Vg 30 039y The maX1mum shear 1tselftums out to be equal to a i 1 2 2 0395 1au 039yy 0y Note that or indicates the direction of the shear stress along the faces Another way of obtaining the principal stresses is to set up an eigenvalue problem For details see Chapter 1 from the notes of Bauchau and Craig 63 Mohr s circle Mohr s circle is a graphic tool that allows to identify the principal directions the principal stresses as well as to nd the stress components at any directions given an initial state of stress cxx cyy cxy Figure 13 Positive yface shearing stress n yface6 6 A yy Ky Compressive normal stress 1 C Tens1le normal stress A Xface cxx cxy 7an V Positive Xface shearing stress Figure 13 Mohr s circle How to build Mohr s circles 039 Given a stress eld cxx cyy cxy calculate 0390Vg This will be the center of Mohr s circle In a xy plane such as in Figure 13 draw point X and point Y by using the values of cxx cyy cxy and the convention on the horizontal and vertical axes of the Mohr s plane eg positive shear stress on the Xface will be in the lower half of the plane positive shear stress on the Y face will be in the upper half of the plane Connect the two points and thus build the diameter of Mohr s circle The angle between X and Y is 180 deg This corresponds to a physical angle of 90 deg between the two planes which is what we should expect The radius can either be found through the drawing itself or just using 1 R as 2O39m 039yy2 6y Once the radius is known with respect to the origin of the reference system it is possible to calculate points C and B which are respectively the maximum principal stress and the minimum principal stress They lie on the xaxis which is the axis of normal stresses No shear stresses indicated along y are possible on these principal planes The direction of the principal plane is to be calculated with respect to point X Remember that the angles are in terms of 29 therefore the actual physical angle is equal to half that Points A and A are the maximum principal shear stress possible positive or negative corresponds to the direction of the vector the value does not change When the shear stress is maximum the normal stresses cxx and SW are nonzero are equal to each other and to cavg Example 1 Draw a Mohr s circle based on this stress field cxx cyy cxy 30 MPa 10 MPa 15 MPa Show that the direction for maximum normal stress is 9 28 deg counterclockwise and that the direction of the maximum shear stress is 17 deg clockwise Draw the corresponding stress components on a representative body element y Shearing Astress MPa yface 1015 Tensile normal stress MPa DE 10 MPa EF 10 MPa DY 15 MPa 6mg 20 xface 30 15 BY 18 MPa 7 Figure 14 O39an T 20 MPa The center of the c1rcle Will be at 20 un1ts away from the origin of the reference system The Xface point X is located in the lower right quadrant at coordinates 30 MPa 15 MPa Figure 14 Recall that the ycoordinate along the lower right quadrant is by de nition associated to shear stresses along the xface The Yface point Y is located in the top right quadrant Connecting the two points gives the diameter which is equal to 18 MPa looking at the coordinates or using the formula From point X draw the angle 29 to point C That is equal to tan29 FX EF 1510 15 rad 9 9 28 deg counterclockwise from X to C C has location 3an R 20 18 MPa 38 MPa This is the value of the maximum principal stress B has location 6an R 2 MPa This is the value of the minimum principal stress at 90 deg with respect to the direction along which the maximum stress is acting 2 9 180 deg Both stresses are tensile both C and B are on the positive side of the normal stresses Finally the maximum shear stress is found on the plane 29 with angle drawn between X and A One can see that this is the complementary to 90 deg of the principal plane clockwise Therefore the direction 9 itself where the maximum shear stresses are found is equal to 9 74528 deg 17 deg where the indicates a clockwise angle The maximum shear stress has value equal to the radius of the circle 18 MPa The normal stress coordinate 6an is associated to this shear stress because these stresses are present on the element at the same time Figure 15 Figure 15 a d and assoclated normal stress Example 2 Mohrs clrcle wlll allow to explaln why for erample cracks form aMS deg m a shaft loaded ln Lorslon For erample a Lorslonal loadlng T aeurrg on a sha produces a slate of stress equal to Orr 6w 6x11 007I stresses and comment on the results our ll ar e g for a hellcoptermll rotor hug www Vtol orngdfstrurzSOgdf clrcle Shear stress xface 01 AX Tensile normal stress yface 01 a e V The maximum principal normal stress is given by 61 5 point C The minimum principal normal stress is given by 62 39c point B The maximum value of the shear stress is 139 points A A The principal plane is located 90 deg clockwise with respect to X ie at 45 deg in the physical plane 45 deg is the direction of maximum normal stress and cracks will develop along that direction 7 Strain transformation 71 Principal strains To obtain a strain field in a different plane strain transformation laws can be followed They are the same as for the stress field Section 61 g l cl 8 617 g g 8 where 8 8y 8 8 Strains are tensors of the second order like stresses 8 82 8 Note that the tensorial shearing strains as de ned on p 4 have been used The matrix c is the same as in section 61 The engineering shearing strains do not follow the transformation law because of the factor 2 between engineering shearing strains and tensorial strains getting in the way For the case in which three strain components are equal to zero plane strain such as 812 0 gyz 2 one can wr1te 8 8m 8 yy810C 8 cos 26 8W sin 26 13a 1 1 29 39 26 13b 8w 3 gyy 38 gyycos 8W sm t 1 8W E m 8yys1n2t9gxy cos2t9 130 Also analogously to what done for stresses it is possible to de ne principal strains such that tan2t9 1 y 3 8m 8 indicates the plane principal plane where the strains are maximum and 12 l l 2 m1n1mum are normal stra1ns and are glven by 8172 58 8 i gyyz yxy On this plane no shearing strain is possible Mohr s circles for strains can also be drawn 8 Strain measurements Strain gauges are used to measure strains They consist of a thin layer of material with thin electric wires attached to it A strain gauge is bonded to the structure under study As the structure deforms the wire of the gauge deforms from a nominal length and its electrical properties namely resistiVity change This change is recorded compared to the resistiVity related to the original length and related to the deformation of the structure the gauge was attached to a b Direction in which strain is measured Figure 16 a Closeup of typical strain gauge b package containing 5 strain gauges from Vishay Measurements online catalog Strains are o en measured in microstrains 10396 lengthlength To fully identify the three components of the strain vector in a plane strain case ie 8m 8W yxy it is necessary to measure the strain in three directions A socalled rosette pattern can be used Figure 17 or three strain gauges can be positioned at speci c angles with respect to each other Figure 18 Brands 4 Vtshay MeasuVEmEHtS Glnuu 4 strain Gages 4 ThreerElement Rosette Pattern NIicl39oJIeasul39ements Strain Gages ThreeElement Rosette Pattern The typical quotrectangularquot l osene panem shown here has its three independent grids oriented are 45 and 90 a at quotDeltaquot pattems with grids at 0 60 and 110 degrees are also available Planar rosettes like the nu lm m h r i an a n Stacked rosettes are also available with separate grids quotstackedquot on top of one another with three independent measm emems about a paint the principal strains and their direclians can be calculaied EliE33 Figure 17 Strain rosette from Vishay Measurements online catalog 20 Figure 18 Strain rosette Ref hnn39 www zernmer h n 1 Mn 1mm Using Figure 12 and Equation 13a one can write three equations one for each strain gauge 81 18n 8 18n 7 8W005261 8W sin 261 2 2 g 1a 8 n 7 8WCOS 262 8W sin 262 g 8n 8 n 7 8WCOS 263 8W sin 263 The left hand sides of those equations as well as 61 62 63 are known measured by the strain gauges and defined by the rosette configuration The unknowns are an 8W 7W which can be solved for Once the strain components are known one can solve for the stress by using the relationships connecting stress to strain called constitutive equations 9 Constitutive equations These are the equations that relate stress to strain in a material Under uniaxial tension metallic materials behave as in Figures 19 and 20 ULTlMATF TENSlLE STRENGTH oi 5 FRACTURE POINT STRESS U 12 04 05 03 10 12 14 ELONGATIGN 7a Figure 19 Typical stressstrain curve for a steel specimen under uniaxial tension Ref v quot101 v1 591mquot StressStrain Curve for Brittle Material m Elastic Ultlmahe lelt Strength au u 7 S l t l r quot a e s m Rupture 5 Elastic Region 0 a nquot n a39 n a 2 Mung m 5min gt Figure 20 Stressstrain curve of a typical brittle material linearlyelastic in this case Ref L 39 39 39 39 39 2 91 Generalized Hooke s law Figures 19 and 20 applied to metallic specimens which were loaded in one single direction and the strain was measured in the same direction Let s apply tensile loading along X and measure strain in the y direction change of length divided by the original length in the ydirection Let s assume that the material is linearlyelastic It can be shown that there is a small shrinking in the ydirection This is the socalled Poisson s ratio effect the amount of strain is related to the strain in the Xdirection by UK EyV XVE 15 v is Poisson s ratio it varies between 0 and 05 for incompressible materials such as rubber Many metals have a Poisson s ratio equal to 03 In isotropic materials Young s modulus E the shear modulus G and Poisson s ratio v are related to each other3 so that only two properties are needed to characterize the materials Poisson s ratio identi es a multidimensional problem we only look at it when we are investigating deformation in a direction perpendicular to loading If a tensile load is applied along the ydirection and the change of length along x is measured it can be found that there will be a shrinkage in the xdirection This can be represented by the same Poisson s ratio used in Equation 15 because in isotropic materials all directions behave in the same way In linear elasticity the principle of superposition can be used Therefore let s apply loads in all directions leading to three normal stresses cxx SW and 622 and three shear stresses cxy cyz and 6x2 and measure the strains in the corresponding directions The generalized Hooke s law for isotropic materials represents such behavior and is given by the following system of equations 039 gxO vi vO ZZ 16a E E E a 010 O 8 v v ZZ 16b y E E E 039 8 0 VO V 16c 039 1039 ag 16d m G yz 2 G 039 1039 as l6e 72 G a 26 039 1039 M 7 160 These equations can be written in matrix form as 8 S 6 with S called compliance matrix lE vE vE 0 0 0 vE lE vE 0 00 8x 0 8y VE VE lE 0 0 0 ay 17 0 0 0 lG 0 0 82 02 7 0 0 0 0 lG 0 d yz yz ya 0 0 0 0 0 lG a Z 7x an 3 E For example G Thls can be shown usmg MohI s Circles 21 v 23 An anisotropic material will have a different generalized Hooke s law This is due to the fact that more material properties are needed to characterize the material This number can range from 21 independent properties fully anisotropic to 5 transversely isotropic which means one plane of the material is isotropic depending on the number of the planes of symmetry for the material under study A typical generalized Hooke s law looks as follows 5 S11 S12 s13 s14 s15 S15 7 5W 521 522 523 S24 525 526 5 22 531 S32 533 S34 s35 536 22 532 S41 S42 S43 S44 S45 S46 72 52 s51 552 553 554 555 556 an ex 561 562 563 564 565 566 my 18 Note that normal strains and shearing strains are fully coupled there are no zeros in Equation 18 Orthotropic materials are a class of anisotropic materials represented by 9 independent material properties They are of interest for aerospace applications because many composite materials belong to this category The generalized Hooke s law for this case is given by L vJ x quotZr n n 0 Ex Ey Ez 3 i 32 n n 0 an Ex E E2 3953 39 D by vxz v i 0 0 0 If 522 Ex E E2 3972 5 z 7 z y 0 0 0 1 0 0 y czr do 023 C 7 V 0 0 0 l 0 V 26323 0 0 0 0 0 where 5quot E1quot In orthotropic materials there is no coupling between normal strains and shearing strains there are zeros in the equation hence it is easier to design with them The different parameters are material properties in specific directions eg Ex is the Young s modulus in the Xdirection4 Gx is the shear modulus in the xy plane etc 4 x y and z are related respectively to the direction of the reinforcement longitudinal direction and to the directions perpendicular to it transverse directions Equation 19 in a generic xyz reference system will need to include sines and cosines of angles between the material directions and the reference system 24 Summarizing the constitutive laws in addition to equilibrium kinematics straindisplacement boundary conditions and compatibility allow to solve for displacements deformation and stresses of any structure under exam There are numerous equations that need to be satis ed and it is important to assess whether there are any assumptions that can be made to reduce the number of equations and unknowns 25 Equations for Linear Elastic Fracture Mechanics and Fatigue EAE 135 Winter 2006 Classes on 03062006 03082006 This document provides further information with respect to the presentation le fracturefatiguelecturepdf 1 Linear Elastic Fracture 39 LEFM is a methodology that allows you to predict study and measure fracture toughness Fracture toughness characterizes the resistance of a material to cracking and it depends on a variety of factors such as temperature environment loading rate etc Out of the three crack opening modes see presentation file the one that is analyzed is Mode 1 which is typically more critical from the design standpoint lower fracture toughness with respect to the Mode 11 and Mode III fracture toughness Note that mixed mode is not addressed in this document but exists KIFSJE 1 Equation 1 shows the Mode I stress intensity factor which characterizes the behavior of the material when a crack of length 2a is present in it F is a factor that depends on the geometry of the specimen and the crack itself S is the nominal stress When the stress intensity factor reaches a critical value the crack grows The document attached at the end provides some nonfocused tables that show how F called Y in the document changes with respect to the specimen s geometry and crack length divided by the width of the specimen K It 1s poss1ble to express the stress at a d1stance r ahead of a crack 1n terms of As r 9 0 the r stress should theoretically be in nite but this is not possible in nature The material will yield so there will be a plastic zone at the crack tip The size of this plastic zone depends on whether for example the specimen is subject to a plane strain or a plane stress case If the specimen has dimensions x y z plane strain corresponds to the case 82 0 for example a beam with width z where there is no deformation in the width direction and plane stress to the case 62 0 for example a xy plate where there is no perpendicular loading In a plane stress problem the plastic region is found to be larger However the plastic region could be even larger than in the case of plane stress if the material has yielded in a great part of the specimen Linear elastic fracture mechanics LEFM is used when the plastic zone ahead of the crack is much smaller than the dimensions of the crack or of the specimen When this is no longer the case for example in a cracked specimen with 80 yielding present one has to use Elastic Jemtegal covered m an advanced Fracture Mecnamcs course IC VHMdT an LAMA a matenal Constant SEE PRESENTATION FOR FURTHER DETAILS 2 Faun a ran a amn There are other analysrs tools Lhathave to do wrtn SVN curveseffec1 ofresrdual stress Goodman equauonsM1ners rule etc wnrcn are based on predrctrng faugue hfe based on the load hrstory f e a a o the structur based on acture mechamcs Tensile 3 a 93 A max 39 a m S I 0 E 1 Time 3 an u V mm Compressive Ref http www roymecn co ukUsefuliTablesFangueStxessilevels htrnl ans rs avery common sketch c c u A stress or R same ratro between mrnrmum load and maxrmum load R Introducrng the stress rntensrty factor range AK Kmax Kmm F Smelm e F o39me cltmltgtm dN dadN means how crack growth a changes with number of cycles N C and m are constants found typically by curve tting of experimental data When expressing this equation in a loglog plot m is the slope of the curve Fatigue test data is affected by much more scatter than quasi static test data so basic statistics has to be taken into consideration Walker s Equation allows to take into account also the stress ratio in the crack propagation da m1 WC1AK where A is a function of the stress ratio R AK and a material constant y and m1 is a di erent constant from the one used in Paris Equation Using either equation say Paris equation it is possible to make an assessment of the number of cycles to failure from a crack that grows from an initial value ai to the final value af 7 da m in 1 da AK7FAS39g WCltAKgt 9Nfallmeigmw9 17m2 pm2 a a Nfailure f I m 2 chASJZl 1 m2 recall that m is a constant obtained from experimental data on a material A typical crack growth fatigue problem involves the calculation of the number of cycles to failure if a structure is subject to a given fatigue load given Smax Smin The final length af can Z l K be calculated know1ng the fracture toughness and through iterationsl 61 S m J and the 7239 max initial length could be the minimum crack length detectable with a given inspection method Another type of problem has to do with how many load cycles will pass before an inspection is able to detect a crack of a given length Overall fracture mechanics allows to predict crack propagation Damage tolerance and fail safe design are based on the assumption that cracks exist in a structure due to manufacturing process material defects residual stresses etc and will be allowed to grow within limits but will be repaired during inspections able to detect them 1 F is a function of the specimen s geometry and of the crack length so the procedure to find the crack length corresponding to a given fracture toughness will involve iterations There are plenty of engineering cases in which inspections failed to do so for a reason or the other Inspections are expensive and sometimes the area where the crack is growing is not easily 39 quot the J for missing cracks are catastrophic see examples in the presentation le MORE INFO ON STRESS INTENSITY FACTOR FOLLOWING Stress xmehsty ht www 5v w eauetessesMsaznuy meE ankWClasstJanalkxm Stress Intensity The Ltbetty Bell Phtladelphta PA Jefferson Ktm MSE 2094 Term Project EduedR D an 35410 t t stress thtehstty hear L stresses extends A 10 to theoretical strength valueswheretmymtema1 and extremal surface cracks ereate htghet stresses hearthese r k The ongthal as fabrteated duhhg the W 701 asoluuon Unhke tr Tm quott tt K is eth t t t load type 11 or en chls hm dw W r H or w h H P h km W twont stress state predteted by the Stress Intehstty Factor Stress Analysis of Cracks laf 3141EIEI6 7 37 PM 1 www 5v w senexassssMsaznuy ma ankWClasstJanalkxm Stress Imenaty MmeH Nbdrhl Frame 3 a Base modes a mm mm mm mack imam dswacnmsms Gener F Hertzberg p321 to the encounteredm engmeenng desxgn and Wm be explamed here m more data m K a a a crack as well as the geometry ofthe solxdpxece where the cracks are detected Fxg 8 5 Hertxberg p323 The Vud W m w de mm Kand posmon can be wnuen as Mm leI c 139 I m i v Van quotI2 25 u y z nsunea snssmbmmu arsmssssm mamlvm am up 2 HM gamma 7 37 PM Stress Intenmy m www 5v w eauelnssesMseznuy meE ankWClasstJanalklm Fracture Toughness K1 terms othe stress intensity factor K but at a enneal stress state as chevu ME y r nl slgma 15 cracklength 1 Fl H W Generally for llnee dlfferent Y s where h V VW m y mils g an l However Yfaetor ls 1 0 for the plate oflnflmte wldth and 1 1 for aplate of semlelnflmte wldth When the yleld strength 2 ane am P P K2 3quot Yi whereEZZS i we 47y fracture toughness from plane stress to plane saaln see g 8 12 Calllster p194 Me l 1ln 1 mm mm Inlunllvr Slclumur HelmN u Mylme sannnnmpm nandvm vgmmd uufphur euwnnr nn mu m 3 HM zlAznn 7 37 PM ht www 5v w eauslassesMsaznuy meE ankWclasstJamlktm Stress lmehaty twmtur Designing and Preventing Fracture with K1E ch stress and Y machmery mammum allowable stze ofthe crack ean be wntten as 1K 2 120 i I 47 Frwm VM m we aeetdehts For example m alrcra eomponehts there are a lot of nvetholes and small cracks whlch bnng Y staess cracks whlch bhhgs down y caltbrauon factor and the also the the staess concentration Addmonally damages m h quot Vanous cleanmg methods are descnbedm the table below swam ml mam numm hahum 1M mmm mums 39Suggrurdckmmxwltnwl39 Mttmmu 34mLIam wttaetumtmws hat estam tumult rm u m 30 M mt m a D am awe tmme mlae the th m s new u w rum Amlrml w r u I39wwuhml nmmlb Rd t 4 s Table A l Hertzberg p752 Conclusion V durablllty and safe operation mmnnt 7 37 PM Stress Intenmy 1 www 5m awaysmama meEankWclasstJamlkm AppendAX B Hertzberg p7 K Cahbrau ons 57 forTypxcal Test Specimen Geomemes m 9 mm umw n Juwl l x WinnWI u A W m 4 mmm u WI r umwr amwrl x W Wr w 391 mIW quot iwwwlm mm mm unrwwv nomm39 7 1 Ianw mmwrl gamma 7 37 PM m www 5v w awaysMammy ma ankWclasstJanalkxm Stress Intenmy 5 HM um PW Md Imam K nw quotWI mm in m MM w 5 mm w 32 q m r m ulnarWI IAWAV IH A uWI x17vl r VaM umw IJMVWI39I m W W Wm Wk mu k k WI ul WWI MNquot W W mm MW 1 1 kanlw K unEuwnllww m WW 7 MW mm mm m 4 2W min A Sunands and In I nut 114mm 12 m Am mm mm 9 77 5 mLSuwlryrW r Bvranmnnm mum sm Au m References calmer w Mammals Sums and Engxeenng John Way and Sons New York1994 n n h r w T M May and 5man York 1996 Table of Contents Submt edby fe non K Kim Vthma Tech Mmemz Science and EMgneevmg I SSexMSEZ0947NoteBooW97UasstjanaIlam memtv hm h p www w v mm a Last updated 5500 gamma 7 37 PM Introduction to Fracture Mechanics and Fatigue EAE 135 Winter 2007 Valeria La Saponara PhD March 14th 2007 MAE UCD Histo cal Prospective I Mechanical failures due to fatigue investigated in more than 150 years I Mysterious failure in 1919 of tank with 2 million gallons molasses in Boston I 1940s T2 tankers Liberty Ships 2700 ships prefabricated allwelded construction built in as little as 4 days Some ships broke in two 1500 brittle failures Schenectady 1943 MAE ucn Historical Prospective Cont d 19505 British De Havilland Comet the world s first commercial jet airliner Four crashes in 19531955 First example of metal fatigue due to high altitude ights Thousands of pressurized climbsdescents 9 thin metal around rectangular windows racked 9 catastrophic failure Redesigned round windows Mmucn Historical Prospective Con d 1969 F111 crash with 100 hrs ight Lost Ie wing in lowlevel training ight Failure due to fatigue crack from sharpedged forging defect in the wingpivot tting Material had low fracture toughness 9 Efforts in development of fracture mechanics damaqe tolerant d9 Inn may I Historical and Economical Prospective Cont d The word faz igue first used in 1839 book on mechanics by J V Poncelet A Wohler started studying railway axle failures in Germany in 1850s I The annual cost of fatigue to the US economy is 3 of the gross national product MAE UCD Fracture Mechanics Modes Mode I opening Mode ll sliding Mode lll tearing Mode l is typically the most critical Ref Fracture Mechanics from Theory to Practice byV Z Parton MAE UCD I Fracture Toughness in Materials m men Vmuuans mu nullsvumlwn Lamas run Hanna MW MEALB M men Wm M meme 1 0mm Mum u m VMERS MD EEumu v H i x mp mpwn mswsssarsuutwm WWW 7 VEMPEMWJHE tom 5 mm 7 7 mm mm Am 5 39m x m quotW 7 quot quotquot quotquot MW WNW k 7 m Im mm W 7 7 7 r 5 mm Mm mm m m an Mammy m mu mu r H 331quot Hf mfg mm m W W M 7 Hr w mg m Izw nu n 1 1 1 hummus um um m5 Mm u m u 52 mm mm m w W Dnlnmlmlmwww w W M u m 1qu u um m u 29mm 5m mm mm m r mm M quotL s m Ib IUWNHIIWMilTvmmmNylllmxHllinmimumi l unmumumM thmeth quotmwnmm m L755 4 m s 3 r W mumMUYM M quotW m um nun V Mihuu mquot m P W W kquot NH m mlsnms K m m Ref Mechanical Behavior of Materials by N E Dowling Prentice Hall 1999 r nquot law Wlw 5 quotquot quot i sin 5 ml mka MAE UcD Fracture Toughness Cont d Tradeoff strengthfracture toughness a mth n 5 6am 5 5 m 7 g rmmvedam a 39039 2m KrS iss Mum n D 2A 40 m 393 Crack Lengm mm mam alum Figm 85 Failure mm for mam plain orzumo Al mum m A c g m mu mm Omngc m 2 E vi a crack Lsngln 0 539 Crack mam for a awgtsunglh nigmaughum m emnn a m mnmmn crack Icngm a smngn Inwlunghnuss mmquot Flgurl as and or my 39al b H m comams inmmnl n M mug m mnsmn a is annulled by mule mmm 8 Ref Mechanical Beha vior of Materials by N E Dowiing Prentice Hall 1999 MAEUCD Fracture Toughness Cont d KIC decreases with temperature Temperame 5 m 4 no 5 a 5mm u 03 53945355 x chluve Yougmass MPa so a mo mmer c Ra MechanicsBehaviorofMareHas byN E Duwhng Prermce Hat 1999 Mmucn rdea mack lea max merar I Fracture Toughness Cont d nmnm H aslin svress a 39 J yielded Iedr slnbmed sues ceram r 5 A plasliczane E 1 r 239s Plashczane ame Bil Fmiw urn5m and Human null m 125 11 mm m um um I n Home 12mm 39 a u n WWW um lurrm JUL m pmumy mm y Validity of linear elastic fracture mechanics LEFM Ref Mechanical Behavior of Materials by N E Dowiing Prentice Hall 1999 MAE UCD x Figure tslimm lpproxi 1 mm l Fracture Toughness KIC HH Slael nquot 25nku a an m was at Spoclmun mva mum my ml be 39r 39um um mum m an all 7 m L summm Ianquot 33 En or hwkn sg l72U MPAL lrLLIptcd mm to ihc high sircngm m a gt quotI ll copyrighl AS I M reprinted mm gamma Ref Mechanical Behavior of Materials by N E Dowling Prentice Hall 1999 MAE UCD Profiles of fractures for toughness tests on compact specimen of 7075T651 Aluminum LEFM vs Elasto Plastic Fracture Mechanics i r 7 Ref Mechanical Behavior of Net a n a n z 2 5 K 3 Kiet f shg li g Dowllng Yes No 41 7 7 magmaVniszEvga and nf n plane I quot 0 3931 on an 7 means applicame i 7 7 m 1 7 m planat duugnsmns 1 i Vi W v Is he oad below a 0 or he tuny pmsuc va ueV r Yes Adjusi K values Sing a um Mm me mm M m mm A d mum sz mule mummy hm i npccml if m tsfoev rg bie cay K Ic chE Imnv mughw mung MAE UCD plume I Fatigue Flynn 91a Fnclul mm mm mull nal Ivmm 1mm mbcr Mum I of A mdzysu chnghauw s deechnology Cm mm m gumle Fzmgmrcmcknrigmm Illnnms hedznmllcgtpcmm arms L Lu A m r mu Mm mm um 9310 gm quotg a 7m MP 50 0 mm um crack n he m uhc m lughcr mast gag Fhm V by A mmy Wcsnnghuuse em m Tummvlogx cu Ref Mechanical Behavior of n sec a n arm c an 39 39 x mum D I g u r lal ddl Materials by N E Dowilng Prentice Hall 1999 MAE UCD 13 I Safe life vs Fail safe Fatigue design philosophies Failsafe structure has defects Needed redundant structural members load transfer inspection routines Examples stiffened wing skins stiffened fuselage skins Safelife structure is resistant to defects Needed knowledge of fatigue environmental effects Examples landing gear wingfuselage joints hinges on variable geometry wings MAE UCD 14 Fatigue Safety Factors L Ref Mechanical Behaw39orofMaterias by N E Dowling Prentice Hall 1999 MAE UCD 15 Inspections 5 Stress N Cycles 39gure 112 VariaLion of worstcase crack length a and swam 3 be 9m nns are required 7 Ref Mechanical Behaw39orofMaterias by N E Dowiing Prentice Hall 1999 MAE UCD 16 I Paris and Walker s Equations Ref Mechanical quot Behavior of Materials b N E Dowiing Preniic Hall 19 9 mm n m a w MD 5 n A many n7 5m Ms nu WW m m mu 11m Jw m ins n m gmmh mo 10 an m 4 m umlnlvmh m man I u s y a mum mum m mu F1 AK Hag mm in mm Elmquot FluIIamongmullmlmmcrumdcmugu mss my Wudul l cpms vcjsxlalml Mlcgmmufkh vkwnmxmllf x n3 w n hhnlulnzdlnlnwg nnfnwng pm I quotMum AA r m equal quot122 MAW quotW my 1mm m u mum w 39 2n fnrmamdynrll mis 7 a mi 2 i 6 A a i wmmm 5 2 chx39 g a 117 WNW WWW WNW n 4mm x M MW m mm mum m MAE UCD Independence on Geometry of Fatigue B ehav1or g g B u Fi re 1110 Fatigue mumquot ratedatafraO Snc be 3 demonstrating geomeu39y mm Adapted from Kiesml 80H AK MPaJ r used with pen39nission Ref Mechanical Behavior of Materials by N E Dowiing Prentice Hall 1999 MAE UCD 18 I Aircraft Constructions inleriar axlenm Ref Mechanical Behavior of p Materials by N E Dowiing Prentice Hall 1999 9quot 11 Wm M in Iim mum will a crack dclm Nun WWW Hwyum mam spacing aimmama is 15 m u rm Immay J P Eu zrin Wood 70 p 41 MAE UCD 19 I Corrosion Under atmospheric conditions corrosion leads to the decomposition of materials into their natural state The natural decomposition products of metals are minerals 6 copperbearing minerals found in the earth s crust Ref notes of Dr M L Free University of Utah metal extraction gt recycling end of useful product life MAE UCD product utilization discontinued use or corrosionfailure l Impact of corrosion Corrosion leads to loss of productivity product contamination part over design loss of life I It is conservatively estimated that 30 billion could be saved through proper use of corrosion minimization technology each year in the U S M G Fontana Corrosion Engineering 3rd ed McGrawHill NY p 15 1986 Ref notes of Dr M L Free University of Utah MAE UCD 21 21 I Corrosion Types biocorrosion bacteria assisted corrosion high temperature hydrogen induced cavitation cracking corrosion fatigue intergranular cracking pitting l CreVICe COFFOSIOn I Specianzed I dealloying stress corrosion erosion cracking fretting uniform corrosion I galvanic Ref notes of Dr M L Free University of Utah MAE UCD 22 I Cavitation type of erosion corrosion Rotary vacuum pump blade with heavy erosion corrosion near the bottom at the water line Schematic diagram of a typical magnified corrosion fatigue crack crosssection Exa m e S b ma ri n es Ref notes of Dr M L Free University of Utah MAE UCD I Corrosion Fatigue su rface metal Schematic diagram of a typical magnified corrosion fatigue crack Shaft With corr05ion fatigue fracture crosssection Ref notes of Dr M L Free University of Utah MAE UCD 24 l Fretting wear assisted corrosion mildly corrosive environment moving object 0 IIIIO Macroscopic view of bolts from a submersible pump The top bolt shows fretting nearthe middle and top in contact with metal surface metal Ref notes of Dr M L Free University of Utah MAE UCD 25 I Galvanic Corrosion corrosive environ me nt more less reactive reactive metal metal Macroscopic View of galvanic corrosion which preferentially corrodes the more reactive of two connected metals Macroscopic view of galvanic corrosion on a pipe flange Ref notes ofDr M L Free University of Utah MAE UCD 26 I High Temperature Corrosion high temperature corrosive environment Macroscopic view ofthe effect of fretting on a metal surface Ref notes ofDr M L FreeUniversityofUta1 Example gas turbines Mme Puma 39Vbimzis if 5 l mm mm Cross section ofa typical modern gas turbine engine by GE 1989 MAEUCD Stress Corrosion Cracking SCC Cracking induced from the combined influence of tensile stress and a corrosive environment Stress on aircraft parts may be residual Within the part as a result of the production process or externally applied cyclic loading Pressfit bushings tapered bolts and severe metal forming are examples of high residual tensile stresses which can lead to stress cracking httgwwwcorrosiondoctorscrg Formssccmm MAE UCD 28 l Corrosion Mechanisms Metals and some ceramic matrix composites for example CC have a unique tendency of losing electrons in an environment 9 oxidation corrosion Noble metals do not corrode easily eg Au Pt Active metals corrode easily eg Al Mg Ti MAE UCD 29 29 I Advanced Materials Boeing 777 Weight saved x Ver Uncol Nd and braces 7 z lcggvl fg fmwnm 108 mlsueverseleowl Z nglmlg 3 sehle ll Inlelcnwllnnerlme Alm 7 Mcmmsinninhibilin m und v Immemwm g P CFRPmmpJ axth a l m5nmeemylggles AMPS u i 1117273 2xxxm3 T42 T36 3 Upper skin 9 F quot quotd 5 4quot 1 29 7 5 Sulbilizer mad mugs Colnpusiles a Tullgwned CFRP I Pile core I Perfamled CFRPNnmellgr 4 Keel beam 4 Belly smug 39 2 Fuselage skin 1Tl1lck beam 139 e a K Algllierl uglleld39 39 gm macs duelmg El n cone fulel39 sleeve 39 A nmcm chine MAE UCD 30 Safety in Aircraft Fatigue Aircraft decommissioned from the Army purchased by US Forest Service USFS Public use does not need to comply to FAA regulations required inspections Examples In 2002 Lockheed C130 firetanker inflight separation of right wing 3 casualties ln service at USAF 1957 1986 bought by USFS in 1988 National Transportation Safety Board NTSB reported 12inch long fatigue crack on the lower surface of the right wing with two separate fatigue crack initiation sites at stringer attachment rivet holes MAE UCD 31 31 I Safety in Aircraft Fatigue Cont d Chalk s Ocean Airways flight 101 crash 2005 20 casualties Grumman G73T Turbine Mallard 1947 Wing separated in flight Preliminary analysis shows fatigue cracks at the wingfuselage junction MAE UCD 32 32 Flight UA 232 crash 1989 113 casualties DC 10 Engine fan rotor disintegrated due to undetected fatigue crack in titanium disk Manufacturing defect was missed by nondestructive inspections MAE UCD 33 E 135 Flow Chart Types of external loading applied on a general structure axial bending about two axes torsion due to axial loading affected by axial stiffness due to 3 D bending affected by bending stiffness dosed sections Shear easier Studied center hear ow in f 1 due to 3D bendmg ezui 1i brium formu as usmg affected by or Open open Sh t 39 39 39 bending stiffness Wlth anal Seamus sections flow Stress shear closed gt 0 en due to torsion gt noncircular gt uniform gt function gt H gt p affected by sections torsion membrane 0W seetlons secnons torsional analogy Stlffness nonuniform c1rcular sections torsion umform tors10n Solve a general 3D loading problem by uncoupling axial loading two bending problems and torsion need to use shear center 9 calculate stresses use superp0s1t10n How to model different parts ofa wing eg purpose of spar stringers panels l Design considerations materials statically indeterminate structures allow for redundancy load transfer buckling fatigue fracture mechanics aeroelasticity

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