FORCES AND MOMENTS IN CFD ANALYSIS

Transcription

1 ORCES AND OENS IN CD ANALYSIS Authors Ing. Zdeněk Říha, PhD., Institute of Geonics ASCR v.v.i., Ing. Josef oldna, CSc., Institute of Geonics ASCR v.v.i. Anotace V příspěvku je pojednáno o naleení působiště celkové síl v případě, že náme celkovou sílu a celkový moment. S daným tpem úloh se setkáváme, kdž řešíme proudění pomocí CD programu. Správná interpretace působiště síl je a daných podmínek velmi důležitá v případě dalších navaujících analý. Annotation he paper shows how to find the force point of action, when we know the total force and the total moment. It is usual case if we solve fluid flow with help of a CD code. Correct interpretation of force s act is ver important for net possible analsis. 1. Introduction Various CD codes offer possibilit of calculating of the total force and of the total moment to the given wall surface, see [3]. Sometimes, the knowledge of these values could be important for subsequent design of the given mechanical part, for eample dimensioning of a rotating shaft with vanes in a mier, etc. he total force and the total moment knowledge are important especiall in cases when we cannot use some EA code (e.g. ANSYS) in net design step. he paper describes relationships between calculated forces and moments. We will also tr to find correct interpretation of forces and moments calculated in a CD code for subsequent strength analsis. 2. Calculation of forces and moments on given surfaces In CD analsis, ever surface is divided into elementar areas. his is consequence of geometr meshing, when we divide given domain to cells where base equations of flow are solved. igure 2.1 Elementar forces description

2 In fluid flow calculations, we can decompose total elementar force E to normal and tangential direction with regards to small elementar area, where force acts, see figure 2.1. Normal force created b pressure is called pressure force EP. angential force created b viscosit is called viscous force EV. hen, the total elementar force has to be vector sum of the pressure force and of the viscous force. E = EP + EV (2.1) Described total force can be also fragmented ( E, E, E ) according to directions of eisting coordinate sstem,,. We obtain components of total elementar force and we can write following relation: E = i + j + k. (2.2) E E E Small area depicted on the picture 2.1 can be part of a wall surface, where we are going to determine total force. If given wall surface contains m small areas where forces act, we can write following equations: m = E, (2.3) E= 1 m = E, (2.4) E= 1 m = E, (2.5) E = 1 = i + j + k. (2.6) orces,, are components of total force. Vectors i, j, k in equations (2.2) and (2.6) are unit vectors, which define given directions,, of the given coordinate sstem. Now, we have calculated non-eisting total force acting on given wall surface. Values from equations (2.3) (2.4) and (2.5) represent standard output from above mentioned CD codes. We are able to calculate inclination of line, where non-eisting total force lies with respect to aes of given coordinate sstem. Unfortunatel, we do not know coordinates of an point of action, which lies on the above mentioned line. In other words, we know sie of total force and its direction, but we do not know where the total force acts. It is consequence of simplification, when we substitute m real elementar forces b one non-eisting total force. Nevertheless, sometimes it is necessar to know total forces for net analsis, e.g. dnamic analsis of whole sstem, where our solved piece represents onl a part of the assembl. herefore, correct interpretation of the non-eisting total force is ver important. Let s find the total moment given b above mentioned elementar forces. oment can be defined as consequence of the force acting. Generall speaking, the moment equals to vector product of the position vector and the force vector, see [2]. = r (2.7) he equation (2.7) can be written in matri form: r i r j k r. (2.8) hen, we can find components of moment using Sarrus s rule for the determinant calculation. = i r i r (2.9)

3 = j r j r (2.10) = k r k r (2.11) oment is vector sum of its components. = + + (2.12) Now, we come back to the wall surface with elementar areas and elementar forces. We will calculate moment of the elementar forces to given point E. We can directl use equations (2.9), (2.10) and (2.11) because components of position vector are known, see figure 2.2. Let s suppose that we can define global coordinate sstem, where the moment is calculated and local coordinate sstem with the same orientation as global coordinate sstem on the elementar area, where elementar forces act, see figure 2.2. Position vector r E defines distance between booth coordinate sstems. According to above mentioned equations, we can write equations for components of the elementar moment E : igure 2.2 Elementar moments description E E = r r (2.13) E E E E E E E = r r (2.14) E E = r r (2.15) E E E E otal components of moments can be identicall calculated as components of the total forces which act on given elementar areas, see equations (2.3), (2.4) and (2.5). m = E E= 1 (2.16)

4 m = E E= 1 m = E E = 1 hen, the total moment equals: (2.17) (2.18) = i + j + k = + + (2.19) Values given b equations (2.16), (2.17) and (2.18) represent standard output from CD codes. Calculated total moment is real eisting value, because there is no simplification introduced into the calculation. he total moment is direct consequence of elementar forces acting on given elementar areas of the observed wall surface. Now, we have calculated the total force on given surfaces wall and the total moment in the given point. In other words, we have standard output of forces and moments from a CD code. 3. Location of the total force and its moment We can find point of action of the total force if we use equations (2.9)-(2.11) describing relations between forces and moments. When we know the moment and the force, then, we have three equations and three variables. If we are finding componets of the posotion vector, it is possible to show that the sstem of equations (2.9)-(2.11) is singular. he sstem of equations should describe line where total force is ling. However, there is ver important condition, which has to be fulfilled. Equations (2.9)-(2.11) describe points, which can create line of the total force if vector of the total force and vector of the total moment are perpendicular to each other. If this condition is not valid, then we can calculate more than one line where the total force can act. hus, such interpretation of the total force and total moment is incorrect. igure 3.1 2D case, the total force and the total moment

5 Let s find cases, where the total force and the total moment alwas will be perpendicular to each other. In 2D space, vectors alwas have to be perpendicular, because force and position vector r alwas lie in plane where we solve fluid flow. hus, the moment alwas is perpendicular to plane (and force ) where we solve fluid flow. Let s suppose that 2D geometr is defined in XY plane, see figure 3.1. In this case we can find point of action using equation (2.11). One coordinate has to be selected as parameter p because onl one equation (2.11) eists and second coordinate will be calculated. hen, we can write: = p (3.1) + p = (3.2) We can see that the point of action is eas to find for 2D case. Equations (3.1) and (3.2) also describe line, where total calculated force can act. eature of the force line is that moment is ero here, because position vector is alwas ero. In 3D space, it can be that vectors and are not perpendicular to each other. Nevertheless, let s suppose that vectors of the total moment and of the total force calculated b some CD code are perpendicular to each other in 3D space. It means that cosϕ of angle ϕ between force and moment vectors is ero (ϕ =90 ο ). his can be checked b equation (3.3), see [1]. We can see that cosϕ also is ero if scalar product of above listed vectors is ero. + + cosϕ = = (3.3) igure 3.2 3D case, the total force and the total moment hen, we can find point of action using equations (2.9)-(2.11). Equation sstem is singular. It means that we have to choose one coordinate as a parameter p and calculate the rest of coordinates. = p (3.4)

6 p = (3.5) + p = (3.6) If we choose coordinate as a parameter p, see (3.4), we can derive rest of coordinates from equations of the moment (2.10) and (2.11), see equations (3.5) and (3.6). Equation 2.9 is not used in this case. Equations (3.4), (3.5) and (3.6) describe points of the spatial line, where the total force lies and can act. Again, feature of the force line is that moment of the is ero here, because position vector r is alwas ero, see figure 3.2. here is also another feature of the depicted line, if we move the total force along this line to an point ling on this line, we alwas produce the same total moment to given point in the space (for eample point [0,0,0] in the global coordinate sstem, see figure 3.1 or 3.2). We have solved determination of point of action for simple 2D and 3D cases. Now, let s suppose that the total moment and the total force vectors are not perpendicular to each other. 4. General relation between the total force and the total moment. Generall speaking, calculated total force and calculated total moment from CD code does not have to be perpendicular to each other. he total force is created b m elementar forces E. Ever elementar force has line where it can act. Influence of m elementar force s lines is removed and substituted b one line of the total force. he substitution of m lines b one line interrupts mathematical coupling between total force and total moment. herefore, we alwas should check perpendicularit between and using the relation (3.3) for 3D case. If we will ignore above mentioned conditions of perpendicularit, then we can easil obtain three parallel lines, where the total force could act. hree parallel lines can be obtained, because we have three pairs of independent equations (one coordinate is constant elective parameter), which describe line in space, where the total force can act. his can be derived from relations (2.9)-(2.11). Of course, this result is not acceptable, because given total force can have one line of action onl. Let s solve case, when cosϕ is not equal to ero. It means that the vector of the total force and vector of the total moment are not perpendicular to each other. Remember, that we would like find point of action for above mentioned conditions. In other words, we would like to find parametric equations for onl one line, where the total force lies and acts. We can define plane ρ using the total force vector and the total moment vector if we move down the total force vector along position vector r, see figure 3.2. We know that the force is cause and the moment crates consequence of the force acting. We can perform decomposition of the total moment to the total force direction and direction perpendicular to the direction of the total force. We obtain moment and moment 0, see figure 4.1. ollowing equations are valid: = + 0 (4.1) 2 2 = (4.2) 2 = (4.3) = sinϕ (4.4)

7 0 cosϕ (4.5) = igure 4.1 3D case, the total force and the total moment in plane ρ he moment is directl produced b calculated total force and the moment 0 is rest given b non perpendicularit between the total force and the total moment. Now, we know perpendicular moment and we are able to find parametric equations of onl one line, where the total force lies. We can calculate sie of the total moment using the equation (4.2) and we can also calculate sie of components and O using equations (4.4) and (4.5). If we want to find point of action, we have to find components of the moment in global coordinate sstem. hen, it will be possible to use equations (2.9)-(2.11), resp. (3.4)-(3.6) and calculate point of action. One wa how to find components of is to calculate sie and components of position vector r s (see figure 3.1) in global coordinate sstem. hen, we can use equations (2.9)-(2.11) for components of calculation in global coordinate sstem. he position vector r s is special vector, which allows to calculate the moment using relation (2.7) resp. (4.6). his vector is perpendicular to the plane ρ. Sie of position vector r s can be calculated b following equation which is valid for vector product, see [1]: = r s sinϕ. (4.6) As can be seen in igure 4-1, ϕ equals 90 o and sinϕ equals 1. herefore, we can write: r s =. (4.7) Now, we need calculate components of the position vector r s in global coordinate sstem. It is possible to use direction cosines, see [1]. We can find direction cosines from vector product of the total force and the total moment, because position vector r s is perpendicular to plane ρ created b moments,, 0 and force. Let s define vector and its components.

8 = (4.8) 2 2 = (4.9) = (4.10) = (4.11) = (4.12) We can calculate searched direction cosines. cos α =, (4.13) cos β =, (4.14) cos γ =. (4.15) hen components of position vector r s can be calculated using following equations: r = s r s cosα (4.16) r = s r s cos β (4.17) r = s r s cosγ (4.18) We have full defined position vector r s. he calculated position vector r s is shortest distance between lines, where the total force and the total moments act. Coordinates of the position vector r s also define one point of action on line where the total force acts. Now, we can calculate searched components of the moment, if we install calculated components of r S to equations (2.9) (2.11), see equations (4.19)-(4.21). = r r (4.19) s s s s = r r (4.20) = r r (4.21) s s If we substitute components of the position vector r s b general coordinates,, we obtain three parametric equations, which describe line where the total force lies. Sstem of equations is singular. In other words, we can find a lot of (infinitel) points on line, where the total force can act. It means that we have to choose one coordinate as constant parameter p and to calculate rest of coordinates: = p (4.22) p = (4.23) + p = (4.24)

9 Equations (4.22)-(4.24) describe line of the total force in case that the calculated total force and the calculated total moment are not perpendicular to each other. Now, we are able to interpreted acting of total force for net possible analsis. We have found place where the total force acts and moment has to be ero. Nevertheless, we must not forget that moment 0 acts in point, where the total moment is calculated. or further possible analsis, it is necessar to present the total force in given point of action C and the moment 0 in point D, where the total moment was calculated, see figure 4.2. igure 4.2 possible interpretation of calculated results, eample of beam 5. Eample Let s go to take an eample. We have solved some fluid flow using ANSYS LUEN CD software. Geometr consist of rotating shaft with vanes, bo, input pipe and output pipe. luid flows from input into the output and it is also mied b rotating vanes, see figure 5.1. We would like to find point of action for correct interpretation of loading for net analsis. igure 5.1 ier It is eas to print out the total force and the total moment (in other words loading of the rotating shaft with vanes) calculated b ANSYS LUEN. hen, components are following: = - 598,6 N; = -469,4 N; = -1120,5 N; = 1,17 Nm; = 1,85 Nm; = -75,2 Nm; otal moment is calculated to point [0,0,0] (middle of the solved geometr on ais of rotation). irstl, we should check if the total force and the total moment are perpendicular to each other. When we use equation (3.3), then, cos(ϕ) equals 0,83. his implies that ϕ = 35,7

10 and tested vectors are not perpendicular to each other. We have to use laout listed in chapter 4. Now it is necessar decompose total moment to direction of total force and direction perpendicular to direction of total force. We can sie of desired components using equations (4.4) and (4.5). hen, = 43,93 Nm; 0 = 61,03 Nm. is important moment which is perpendicular to total force line and is direct result of action of the force. Components of this moment in absolute coordinate sstem fill equations (4.19) (4.21) which allow to find an point of action. Let s calculate sie of position vector r s which is perpendicular to both moment and force. he result from equation (4.7) is following: r s = 0,032 m. Now, we will determine direction cosines of r s vector using vector product and then we will calculate r s vector components. Using of equations (4.8) (4.15) we obtain following values: cos α = 0,63; cos β = -0,78; cos γ = -0,0039. hen, equations (4.16) - (4.18) define components of r s vector. r s = 0,020 m ; r s = -0,025 m; r s = -0,0003m. We can calculate components of moment using equation (4.19)-(4.21) = 28,15 Nm; = 23 Nm; = -24,67 Nm; Now, we have collected all necessar variables and we are able to build equations (4.22)- (4.24). In other words we are able to find point of action. or eample, if we take = 0,03m then we obtain rest of coordinates according to equations (4.23) and (4.24) = 0,091m; = 0,132m; It is also necessar to calculate component of O moments. irstl, we will calculate direction cosines of according to equations (4.13)-(4.15), where we substitute sie of vector and its component b sie of the total force vector and its component. hen, we can calculate components of 0 with help of equations (4.16)-(4.18), where we substitute sie of r s b 0 and we use calculated direction cosines of. Results are following: 0 = -26,97 Nm; 0 = -21,15 Nm; 0 = -50,49 Nm; Components of vectors,, 0 and the point of action coordinates full define loading for net possible analsis. 6. Conclusion he paper shows how to calculate point of action, if we know the total force and the total moment to the given wall surface from a CD code. We can use equations presented in section 3, which come from equations in section 2, see (2.9)-(2.11). If we solve 2D case of fluid flow, we can alwas use components of the calculated total force and components of the total moment in equations (2.9)-(2.11) and derive equations like (3.1) and (3.2), because the total force vector and the total moment vector alwas are perpendicular to each other. When we solve 3D fluid flow, firstl, we have to check if above mentioned vectors are perpendicular to each other using the relationship (3.3). When condition of perpendicularit is fulfilled for 3D case (cosϕ =0 resp. ϕ =90 ο ) we can do the same as for 2D case. If condition of vector s perpendicularit between the total force and the total moment is not fulfilled, we should proceed according to the procedure presented in the section 4. It means to find the real moment produced b the total force. inall, the total force and its position have to be completed about the moment 0 and its position in presentation of loading.

11 5. References [1] Rektors, K., a kol.: Přehled užité matematik I, publishng house Prometheus, Prague 2000, ISBN , [2] lorian, Z., Pellant, K., Schánek.: echnická mechanika I Statika, [3] Anss-luent User s guide,