Part 9 - Estimation of magnitude of the unbalanced centrifugal forces driving tectonic movement Based on the above observations, let us assume as a working hypothesis that the Earth can be modelled as a rotating body where the centre of mass is offset from the principal axis of rotation. For the purposes of this paper the author will consider the two principal approaches to determine the circumferential forces associated with an unbalanced rotating body Model 1. Rigid body dynamics As discussed in Section 5, the Pacific plate has all the appearances of being in compression while the almost diametrically opposed African plate appears to be subject to tensile forces The simplest model is to consider the Earth as an eccentrically rotating solid body such as an unbalanced flywheel. Although this model (shown in Figs 8 & 9 and enumerated in Appendix 1) accounts for the compressive and tensile stresses developed in the outer rim, it does not describe the unbalanced centripetal forces which the author believes to be linked to tectonic forces resulting in plate movement. Fig 8 below. Force/Vector diagram showing the centripetal force P and the circumferential or tensile force F in the outer rim of a solid rotating body. Rotating Machines are designed to ensure that the developed circumferential stress F, is less than the design hoop stress. Fig 9 below. Force/Vector diagram showing the differential circumferential stress induced in the outer rim of a solid rotating body, whose centre of mass is not co-incident with the principal axis of rotation. Under these conditions the tensile forces are increased on the 'heavier' side and are decreased on the 'lighter' side. Thus the hoop stress in the rim will be in tension on the 'heavier side' and in compression on the 'lighter side'. Model 2. Outer rim is able to slide relative to main body In order to determine the magnitude and direction of the forces postulated as being responsible for tectonic movement the model used in one in which the thin crust is able to slide relative to the solid body at the crust /mantle interface. By way of illustration Fig. 10 shows that if an unbalanced disc with an outer annular ring containing fluid is rotated about its principal axis, the liquid will move to the ‘lighter’ side. Fig. 11 shows an analogous situation with the sliding continental plates. If we consider the crust as being able to move relative to the mantle, albeit it over a long geological time scale, then a simple force diagram (Fig 12) can be constructed by making the following assumptions: the crust is a thin shell that is able to slide relative to the mantle, the forces due to eccentricity are superimposed on the stress caused by the general rotation and gravity, and the stress that is of interest for the purposes of tectonic movement is the differential stress due to this eccentricity. By approaching the problem in terms of a thin shell moving relative to the mantle, it is possible to consider what increments of the tensile force are responsible for putting the Pacific basin under compression (note crumpled profile) and the African Plate under tension. (note the Rift Valley). The calculations to derive the expression of the circumferential stress at the surface of the earth, which are based on the consideration of the eccentrically induced loads on the thin crust are detailed in Appendix 11. The term radius of eccentricity’ was introduced to denote the distance between the centre of mass and the major axis of rotation. From Appendix 2 the following relationship was derived: Eq.2 shows that the magnitude of the circumferential forces or in this case the derived circumferential stress is dependent on the distance between the geometric centre and the centre of mass, i.e. the ‘radius of eccentricity’. In a limiting case, if the ‘radius of eccentricity’ is zero, the rotating body will be balanced and the net force will be zero. Fig 13 shows this relationship in between F (circumferential stress) and the E (radius of eccentricity). Fig. 13 below. Relationship between the radius of eccentricity and the circumferential stress Having derived an equation which relates the circumferential stress with the rotational velocity and the centre of eccentricity it would seem appropriate to consider the possibility of tectonic activity on the planet Venus. Due to the low peripheral velocity of Venus (1 Revolution in 243 days = 6.5 kmhr -1 ) as compared with 531.5 kmhr -1 on Earth, the centrifugal forces available as compared to the similar sized planet Earth will be in the ratio of (42.5) 2 / (531.5) 2 = 1806.25/282492.25 = 0.006: 1. This would give a stress value of 3.9x10 -3 Nmm -2 . (0.059 psig). The unbalanced centripetal forces thus needed for tectonic activity are negligibly small. In order to better understand the magnitude of the calculated circumferential stress in the continental crust, it is helpful to relate the model to more familiar applications. (These are shown pictorially in Cartoons 2, 3 &4) The stress value of 7.29x10 -2 Nmm -2 if applied to a 1 tonne braked motor vehicle with a rear surface area of 1000 mm x 1300 mm=1.3x10 6 mm 2 will yield a push force 94770 N. In imperial units this equates to a push of 21305 lbf (pound force) or 9.5 tonf (ton force). Rounded up and put more simply, this equates to the vehicle being pushed by 118 people each of whom weighs 180 pounds (81.8 kg) (see Cartoon 1). As the incline between the height of the Andes (taken as 5 km) and distance between the Peru- Chile trench and the Cordillera –Real (taken as c.1000 km) is c. 1:250, the vehicle can be considered to be on a level surface for purposes of scaling. However a 3 tonne hoist will easily pull the vehicle up a 1:3 incline onto a pick-up truck It is also worth noting that an upward acting net force of 2.37 x 10 -2 N/mm 2 (3.5 psig) on a 60 metre long wing span of an aircraft is sufficient to keep a large 350 tonne aircraft flying. A puff of wind with dynamic pressure as low as 0.135 x 10 -2 N/mm2 (0.2 psig) acting on the large surface area of a ship’s sail will cause a boat to move across water. Thus the unbalanced centrifugal forces created by placing the centre of mass of the Earth 1 km off centre are large and cannot be ignored. The calculated circumferential forces if applied to the cross sectional area of the South American plate are more than sufficient to push it over the Nazca plate. What is important to remember is longitudinal stresses are half as much as the circumferential stresses. Therefore, we can say that longitudinal strength is twice as strong as circumferential strength. This is only true for illustration purposes LECTURE 15 Members Subjected to Axisymmetric Loads Pressurized thin walled cylinder: Preamble : Pressure vessels are exceedingly important in industry. Normally two types of pressure vessel are used in common practice such as cylindrical pressure vessel and spherical pressure vessel. In the analysis of this walled cylinders subjected to internal pressures it is assumed that the radial plans remains radial and the wall thickness dose not change due to internal pressure. Although the internal pressure acting on the wall causes a local compressive stresses (equal to pressure) but its value is neglibly small as compared to other stresses & hence the sate of stress of an element of a thin walled pressure is considered a biaxial one. Further in the analysis of them walled cylinders, the weight of the fluid is considered neglible. Let us consider a long cylinder of circular cross - section with an internal radius of R 2 and a constant wall thickness‘t' as showing fig. This cylinder is subjected to a difference of hydrostatic pressure of ‘p' between its inner and outer surfaces. In many cases, ‘p' between gage pressure within the cylinder, taking outside pressure to be ambient. By thin walled cylinder we mean that the thickness‘t' is very much smaller than the radius R i and we may quantify this by stating than the ratio t / R i of thickness of radius should be less than 0.1. An appropriate co-ordinate system to be used to describe such a system is the cylindrical polar one r, q , z shown, where z axis lies along the axis of the cylinder, r is radial to it and q is the angular co-ordinate about the axis. The small piece of the cylinder wall is shown in isolation, and stresses in respective direction have also been shown. Type of failure: Such a component fails in since when subjected to an excessively high internal pressure. While it might fail by bursting along a path following the circumference of the cylinder. Under normal circumstance it fails by circumstances it fails by bursting along a path parallel to the axis. This suggests that the hoop stress is significantly higher than the axial stress. In order to derive the expressions for various stresses we make following Applications : Liquid storage tanks and containers, water pipes, boilers, submarine hulls, and certain air plane components are common examples of thin walled cylinders and spheres, roof domes. ANALYSIS : In order to analyse the thin walled cylinders, let us make the following assumptions : • There are no shear stresses acting in the wall. • The longitudinal and hoop stresses do not vary through the wall. • Radial stresses s r which acts normal to the curved plane of the isolated element are neglibly small as compared to other two stresses especially when The state of tress for an element of a thin walled pressure vessel is considered to be biaxial, although the internal pressure acting normal to the wall causes a local compressive stress equal to the internal pressure, Actually a state of tri-axial stress exists on the inside of the vessel. However, for then walled pressure vessel the third stress is much smaller than the other two stresses and for this reason in can be neglected. Thin Cylinders Subjected to Internal Pressure: When a thin – walled cylinder is subjected to internal pressure, three mutually perpendicular principal stresses will be set up in the cylinder materials, namely • Circumferential or hoop stress • The radial stress • Longitudinal stress now let us define these stresses and determine the expressions for them Hoop or circumferential stress: This is the stress which is set up in resisting the bursting effect of the applied pressure and can be most conveniently treated by considering the equilibrium of the cylinder. In the figure we have shown a one half of the cylinder. This cylinder is subjected to an internal pressure p. i.e. p = internal pressure d = inside diametre L = Length of the cylinder t = thickness of the wall Total force on one half of the cylinder owing to the internal pressure 'p' = p x Projected Area = p x d x L = p .d. L (1) The total resisting force owing to hoop stresses s H set up in the cylinder walls = 2 .s H .L.t (2) Because s H .L.t. is the force in the one wall of the half cylinder. the equations (1) & (2) we get 2 . s H . L . t = p . d . L s H = (p . d) / 2t Circumferential or hoop Stress (s H ) = (p .d)/ 2t Longitudinal Stress: Consider now again the same figure and the vessel could be considered to have closed ends and contains a fluid under a gage pressure p.Then the walls of the cylinder will have a longitudinal stress as well as a ciccumferential stress. Total force on the end of the cylinder owing to internal pressure = pressure x area = p x p d 2 /4 Area of metal resisting this force = pd.t. (approximately) because pd is the circumference and this is multiplied by the wall thickness Assessing the Biomechanical Impact of Medical Devices Using MSC.Software at Texas A&M University Contest Winner By Lucas H. Timmins, PhD Student, Texas A&M University, College Station, TX The Vascular Biomechanics Group in the Department of Biomedical Engineering at Texas A&M University investigates the role of biomechanics in the treatment of vascular disease. In particular, we use MSC.Software to investigate the solid mechanical implications of implanting vascular stents and aortic stent grafts (Figure 1). Figure 1: A. Vascular stents on balloon catheters. B. Aortic stent grafts Cardiovascular diseases are the principal cause of death in the United States, with total cost associated with the disease in upwards of $450 billion USD. As a result, an extensive effort has gone into developing biomedical devices to treat these diseases. As the placement of these devices alters the biomechanical environment inside an artery, proper computational mechanical analysis must be carried out to ensure their treatment efficacy, but also to examine the mechanical loads that they place on the arterial wall. MSC.Software allows for proper analysis of these inherently difficult (e.g. non-homogeneous, non-linear) contact mechanics problem. MSC.Software allows for construction and simulation of vascular stent implantation to examine the mechanical implications of varying specific design parameters on the arterial wall mechanics (Figure 2). Through the modeling of biological tissues and the application of physiologic boundary conditions, appropriate finite element models can be developed in MSC.Patran and solved with MSC.Marc. [...]... cylinder walls is hemisphere the same and the thickness of the It is assumed that the internal diameter of both are If the internal pressure of the shell is The Stresses in the cylinder walls are :- then :- (7) (8) Then (9) (10) The Stresses in the hemispherical Ends are :- (11) (12) (13) If there is to be distortion under pressure, at the junction between the cylinder and the ends (14) (15) (16) Taking... the cylinder walls is and the thickness of the hemisphere It is assumed that the internal diameter of both are the same If the internal pressure of the shell is then :The Stresses in the cylinder walls are :- (7) (8) Then (9) (10) The Stresses in the hemispherical Ends are :- (11) (12) (13) If there is to be distortion under pressure, at the junction between the cylinder and the ends (14) (15) (16) Taking... same for both For the drum material (25) For the Cylinder :- (26) (27) (28) ( 29) (30) (31) (32) For the hemispherical ends :The question states that the hoop Strain in the ends is the same as in the cylinder thus :- (33) (34) (35) As the pressure on the water originally inside the cylinder increases, there is a loss in volume which is made up with additional water Decrease in the Volume of the original water... (27) (28) ( 29) (30) (31) (32) For the hemispherical ends :The question states that the hoop Strain in the ends is the same as in the cylinder thus :- (33) (34) (35) As the pressure on the water originally inside the cylinder increases, there is a loss in volume which is made up with additional water Decrease in the Volume of the original water is given by :- (36) (37) The Additional Volume of Water required... Stress in the cylinder walls The equilibrium equation is therefore :- (40) (41) After th tank has been filled with water so that the pressure raised to 200 lb./sq.in Let bars and Cylinder be the final tensile stress in the tie the final Tensile Stress in the The equilibrium equation is now :- (42) (43) The Hoop Stress in the cylinder is no affected by the Tie bars so :- (44) (45) The Strain equation Thew increase... Poisson's ratio of 0.3 (17) Note that the maximum Stress will then occur in the hemispherical ends i.e (18) Which is greater than the hoop stress in the cylinder For equal maximum stress should equal 0.5 Volumetric Strain On The Capacity Of A Cylinder The capacity of a cylinder is the dimensions increase by so if there will be an increase in volume and :- ( 19) (20) Neglecting the products of small quantities... Poisson's ratio of 0.3 (17) Note that the maximum Stress will then occur in the hemispherical ends i.e (18) Which is greater than the hoop stress equal maximum stress in the cylinder For should equal 0.5Volumetric Strain On The Capacity Of A Cylinder The capacity of a cylinder is so if the dimensions increase by there will be an increase in volume and :- ( 19) (20) Neglecting the products of small quantities... to 3 tons/sq.in and the tank filled with water, find the increase in capacity when the pressure is raised to 200 lb./sq.in and find the final stress in the tie bars ( 39) Initially As the tie bars are in tension they are compressing the walls of the Tank It is therefore possible to write an equation which equates the compressive force on the cylinder walls to the tensile force in the tie bars In this... find the increase in capacity when the pressure is raised to 200 lb./sq.in and find the final stress in the tie bars Initially As the tie bars are in tension they are compressing the walls of the Tank It is therefore possible to write an equation which equates the compressive force on the cylinder walls to the tensile force in the tie bars In this initial condition there is no Hoop Stress Let be the. .. Cylinder of 6 in internal diameter and 0.1 in thick has ends closed by rigid plates and then filled with water When an external axial pull of 4000 lb is applied to the ends, the water pressure is observed to drop by 7 lb./sq.in Determine the value of Poisson's Ratio.(U.L.) (54) Assuming that the cylinder remains full then the increase in the volume of water must equal the increase in the capacity of the . Part 9 - Estimation of magnitude of the unbalanced centrifugal forces driving tectonic movement Based on the above observations, let us assume as a working hypothesis that the Earth. and the thickness of the hemisphere . It is assumed that the internal diameter of both are the same. If the internal pressure of the shell is then :- The Stresses in the cylinder walls are :-. stress value of 3.9x10 -3 Nmm -2 . (0.0 59 psig). The unbalanced centripetal forces thus needed for tectonic activity are negligibly small. In order to better understand the magnitude of the calculated