Chapter 7 Thin Cylinder |  | |
| Topics Covered | The topic menu above allows you to move directly to any of the four sections for each topic. The sections are: Case Intro: To help introduce and understand the basic principles, a case study is presented. Theory: This section will review the basic principles and equations that you should know to answer the exam questions. It does not give detailed derivations of the theory. Case Solution: The case study is solved in detail in this section. Graphics, narrations, animations, and equations are used to help you understand how the problem was solved. | |
Friday, 8 July 2011
THIN CYCLINDER
PRESSURE VESSELS
| Introduction | ||
 Cylindrical Gas Pressure Vessels on Rail Car | A rail car needs to transport pressurized gas and due to shipping constraints, each cylinder can have only a diameter of 34 cm (inside dimension). The end caps are not considered in this initial design. What is known:
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 Dimensions | Question | |
| What is the minimum wall thickness in the cylindrical section of the gas pressure vessel? | ||
| Approach | ||
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| Thin-walled Pressure Vessels | |||||
 Cylindrical Pressure Vessel with Internal Pressure | Both cylinderical and spherical pressure vessels are common structures that are used ranging from large gas storage structures to small compressed air tanks in industrial equipment. In this section, only thin-walled pressure vessels will be analyzed. A pressure vessel is assumed to be thin-walled if the wall thickness is less than 10% of the radius (r/t < 10). This condition assumes that the pressure load will be transfered into the shell as pure tension (or compression) without any bending. Thin-walled pressure vessels are also known as shell structures and are efficient storage structures. If the outside pressure is greater than the inside pressure, the shell could also fail due to buckling. This is an advanced topic and is not considered in this section. | ||||
| Cylindrical Pressure Vessels | |||||
 Cylindrical Vessels will Expierence Both Hoop and Axial Stress in the Mid-section | Only the middle cylindrical section of a cylinder pressure vessel is examined in this section. The joint between the end caps and the mid-section will have complex stresses that are beyond the discussion in this chapter. In the mid-section, the pressure will cause the vessel to expand or strain in only the axial (or longitudinal) and the hoop (or circumferential) directions. There will be no twisting or shear strains. Thus, there will only be the hoop stress, σh and the axial stress, σa. as shown in the diagram at the left. | ||||
 Cross Section Cut of Cylindrical Vessel | Pressure vessels can be analyzed by cutting them into two sections, and then equating the pressure load at the cut with the stress load in the thin walls. In the axial direction, the axial pressure from the discarded sections will produce a total axial force of p(πr2) which is simply the cross section area times the internal pressure. It is generally assumed that r is the inside radius. The axial force is resisted by the axial stress in the vessel walls which have a thickness of t. The total axial load in the walls will be σa(2πrt). Since the cross section is in equilbrium, the two axial forces must be equal, giving p(πr2) = σa(2πrt) This can be simplified to
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 Hoop Section Cut from Cylindrical Vessel | In addition to the axial stress, there will be a hoop stress around the circumference. The hoop stress, σh, can be determined by taking a vertical hoop section that has a width of dx. The total horizontal pressure load pushing against the section will be p(2r dx) as shown in the diagram. The top and bottom edge section will resist the pressure and exert a load of σh(t dx) (each edge). The edge loads have to equal the pressure load, or p(2r dx) =σh(2t dx) This can be simplified to
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| Spherical Pressure Vessel | |||||
 Spherical Pressure Vessel Cut in Half | A spherical pressure vessel is really just a special case of a cylinderical vessel. No matter how the a sphere is cut in half, the pressure load perpendicular to the cut must equal the shell stress load. This is the same situation with the axial direction in a cylindrical vessel. Equating the to loads give, p(πr2) = σh(2πrt) This can be simplified to
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 Cylindrical Gas Storage Tank | A gas storage tank needs to be designed to hold pressurized gas at 10 MPa. The tank inside diameter is set at 34 cm due to tank stacking system on a rail car. For safety reasons, a factor of safety of 2.0 is required. The material is steel with a yield stress of 250 MPa. The thickness of the material needs to be determined. To account for the stress interaction between the hoop and axial directions, the maximum distortion energy theory (von Mises' Yield Criterion) will be used to predict failure. It is assumed that the end caps will not fail and only the cylinder middle section will be considered (end cap stresses are complex and not studied in this eBook). | |
| Hoop and Axial Stresses | ||
| Functions for the hoop and axial stress can be determined for a cylindrical pressure vessel. These are σh = Pr/t = (10 MPa)(0.17 m)/t = 1.7/t σh = Pr/(2t) = (10 MPa)(0.17 m)/(2t) = 0.85/t Both the stresses are functions of t. | ||
| Failure Criteria | ||
 Stress Element at Cylinder Section | The maximum distortion energy criteria takes into consideration stresses in multiple directions. The equation is or for this case,  The yield stress is given as 250 MPa. However, to account for a factor of safety of 2.0, the actual yield stress is reduced in half. Substituting into the failure equation gives,  2.890 + 0.7225 - 1.445 = 15,625 t2 t = 0.1178 m = 11.78 mm | |
| Example | ||
 A Partial Section of a Penstock | A penstock for a hydraulic power plant has an inside diameter of 1.5 m and is composed of wooden staves bound together by steel hoops. The cross-sectional area for each steel hoop is 300 mm2. If the allowable tensile stress for the steel is 130 MPa, what is the maximum space, L, between the hoop bands under a head of water of 30 m? The mass density of water is 1,000 kg/m3. The water pressure can be assumed to be the same at all interior locations of the penstock. | |
| Solution | ||
 Spacing of the Steel Hoops in the Penstock | The pressure corresponding to a head of 30 m water is given by p = ρgh = (1,000) (9.81) (30) = 294 kPa Circumferential stress in the steel bands is considered as the failure criteria for a safe design. If the maximum spacing between hoops is denoted as L, then each hoop must resist the water pressure over a length L of the penstock. | |
 Direct Evaluation of Bursting Force F | The bursting force F, acting over the flat surface of the fluid equals the pressure intensity p multiplied by the area, DL, over which it acts. F = pDL = (294) (1.5) L = 441 L This bursting force will be resisted by the equal forces P acting on each cut surface of the cylindrical wall. Assuming the whole resisting force will be given by the steel hoops, P = Aσ = (300×10-6) (130×103) kN = 39 kN Applying the summation of forces, ΣF = 0 F - 2P = 0 F = 2P 441 L = 2 (39) L = 0.1769 m = 177 mm | |
Thursday, 7 July 2011
INTRODUCTION
Chapter 1 Introduction |  | |
| Topics Covered | The topic menu above allows you to move directly to any of the four sections for each topic. The sections are: Case Intro: To help introduce and understand the basic principles, a case study is presented. Theory: This section will review the basic principles and equations that you should know to answer the exam questions. It does not give detailed derivations of the theory. Case Solution: The case study is solved in detail in this section. Graphics, narrations, animations, and equations are used to help you understand how the problem was solved. | |
HOOKE'S LAW
| Introduction | ||
 Cross Section of Test System  Specimen Testing Set Up Problem Animation Click to view movie (101k) | Susan Placer, a new engineer at AAA Materials Testing Lab, is asked to find the Poisson's ratio for a new epoxy adhesive. The test method used at the lab compresses a 1 in cube specimen under compression. The test frame consists of a 1 in cube hole to place the sample. Then a load is place on the top surface (see figure). The total deflection of the top surface is monitored. The test sample is a 1 in cube which is placed into a perfectly fitting frame. The frame is substantially more stiff than the sample and will not deflect during the test. What is known:
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| Question | ||
| What is the Poisson's ratio of the new epoxy adhesive? Also, what stresses are generated on the sides of the test frame? | ||
| Approach | ||
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| One Dimensional Hooke's Law | ||||||||||||||
 1-D Hooke's Law | Recall, Hooke's Law in one dimension (uniaxial loading), relates the normal stress and normal strain as
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| Poisson's Ratio | ||||||||||||||
 Poisson's Effect Click to view movie (21k)  Contraction in y- and z-direction when Stressed in x-direction | When analyzing more than one dimension, interaction between all directions needs to be considered. This is done through Poisson's ratio. Basically, Poisson's ratio is the amount of transverse contraction, or negative strain, when strained in a given direction. For a basic object pulled or strained in the x-direction, the Poisson's Ratio is defined as
For a three dimensional object, Poisson's ratio will occur in equally in both perpendicular directions. If the load is in the x-direction, then strain in the y- and z-direction will be εy = εz = -νεx | |||||||||||||
| Two Dimensional Stress-Strain | ||||||||||||||
 Stress in Two Directions | If a material is isotropic (homogenous in all directions, such as a solid metal) and is pulled in two directions, then due to Poisson's ratio, the overall normal strain will be the total of the two strains. For example, if there are normal stresses in both the x- and y-directions, then the total normal strain in the x-direction is εx total = εx due to σx + εx due to σy = σx /E - νσy/E εx = (σx - νσy) / E Similarly, the normal strain in the y-direction would be εy = (σy - νσx) / E | |||||||||||||
 Pure Shear Stress in a 2D plane Click to view movie (29k)  Shear Angle due to Shear Stress | However, Hooke's Law also relates shear strain and shear stress. If the shear stress and strain occurs in a plane then the stress and strain are related as
γ = τ/G = [2(1 + ν)/E] τ where G is the shear modulus (a material property) and γ is the shear strain. The shear strain is defined as the angle (radians) caused by the shear stress as shown in the diagram at the left. The shear modulus is related to Young modulus and Poisson's ratio,
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| Three Dimensional Stress-Strain | ||||||||||||||
 Stress Directions in 3D (τxy = τyx, τyz = τzy, τxz = τzx) | Just like 1D or 2D, Hooke's Law can also be applied to material undergoing three dimensional stress (triaxial loading). The development of 3D equations is similar to 2D, sum the total normal strain in one direction due to loads in all three directions. For the x-direction, this gives, εx total = εx due to σx + εx due to σy + εx due to σz = σx /E - νσy/E - νσz/E εx = (σx - νσy - νσy) / E Similarly, the other directions can also be determined. The final equations are summarized in the table below. | |||||||||||||
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 1.0 inch Specimen Cube Inserted into Test Frame | A simple cube of material is inserted into a snug fitting test frame and then a 3,000 lb load is applied on the top surface. The sides of the test frame are stiff and will not deflect at the test load level. Also, the sides are smooth so there is no friction forces as the specimen is loaded. The objective in testing the specimen is to determine the Poisson's ratio of the material. The Young's modulus, E, is known, 2,000 ksi. Also, the total deflection after loading is 0.001 inches. | |
| Determining Poisson's ratio | ||
 Cube Under Loading | A good place to start solving this problem is the full 3D Hooke's law equations.  While these look complex, many terms are easily determined. The vertical stress can be calculated as σz = F/A = 3000 lb/(1 in)2 = 3,000 psi = 3.0 ksi The other two horizontal stresses are unknown, but must be the same due to symmetric geometry and loading. Thus, σx = σy = p | |
 Problem Diagram | Next, the three strains can be determined. The vertical strain can be found since the deflection is known, giving εz = δ/L = (0.001 in) / (1.0 in) = 0.001 The other two strains are also known since the outside walls do not deflect (due to their high stiffness). Thus, εx = εy = 0.0 Also, Young's modulus is known, E = 2,000 ksi. | |
 Graph of Function f(ν) = (1-ν) / [(1+ν)(1-2ν)] | Simplifying the 3D equations to two equations give,  Next, substituting specific values reduce these equations to  The negative signs represent compression for both strain and stress. There are now two unknowns, p and ν. The second equation can be rearranged to give,  This can be manipulated to give 3ν2 + 0.5ν - 0.5 = 0 Using the quadratic equation, the positive root gives ν = 0.3333 This equation can also be solved numerically by graphical methods or by using numerical programs like MathCad, Mathematica, MathLab, or even Excel. The equation is graphed at the left which also gives ν = 0.3333 | |
| Stresses | ||
| | The stresses in the x and y directions can now be determined,  p = σx = σy = -1.5 ksi = 1,500 psi | |
| Example | ||
 Plate with Applied Loads and Displacement | A 0.5 cm thick rectangular plate is pulled in tension by two loads in the x and y directions as shown. The total deflection in the x and y direction is 0.021 cm and 0.009 cm, respectively. What is the Poisson's ratio of the material? The z direction has no load and the deflection is not known. | |
| Solution | ||
 Stresses and Strains on Plate | This problem involves loading from two directions, and thus requires at least the 2-D Hooke's Law. The 3-D Hooke's Law could be used, but since σz is zero, those equations will reduce to the 2-D equations. The equations are, εx = (σx - ν σy)/E εy = (σy - ν σx)/E The strains and stresses in the x and y direction need to be calculated. σx = Px/Ax = 5/[(0.05)(0.005)] = 20 MPa σy = Py/Ay = 9/[(0.10)(0.005)] = 18 MPa εx = 0.021/10 = 0.0021 cm/cm εy = 0.009/5 = 0.0018 cm/cm Substituting the stresses and strains into the 2-D Hooke's Law equations, gives 0.0021 E = 20 - ν 18 0.0018 E = 18 - ν 20 Solving for ν gives, ν = 0.1876 | |
DEFINITION
Mechanical engineering is a discipline of engineering that applies the principles of physics and materials science for analysis, design, manufacturing, and maintenance of mechanical systems. It is the branch of engineering that involves the production and usage of heat and mechanical power for the design, production, and operation of machines and tools. It is one of the oldest and broadest engineering disciplines.
What is Dynamic?
In the field of physics, the study of the causes of motion and changes in motion is dynamics. In other words the study of forces and why objects are in motion. Dynamics includes the study of the effect of torques on motion. These are in contrast to Kinematics, the branch of classical mechanics that describes the motion of objects without consideration of the causes leading to the motion.
What is static?
Statics is the branch of mechanics concerned with the analysis of loads (force, torque/moment) on physical systems in static equilibrium, that is, in a state where the relative positions of subsystems do not vary over time, or where components and structures are at a constant velocity. When in static equilibrium, the system is either at rest, or its center of mass moves at constant velocity.
What is Strength of Material?
In materials science, the strength of a material is its ability to withstand an applied stress without failure. The applied stress may be tensile, compressive, or shear. Strength of materials is a subject which deals with loads, deformations and the forces acting on the material. A load applied to a mechanical member will induce internal forces within the member called stresses. Those stresses acting on the material cause deformations of the material. Deformation of the material is called strain, while the intensity of the internal forces are called stress. The strength of any material relies on three different type of analytical method: strength, stiffness and stability, where strength refers to the load carrying capacity, stiffness refers to the deformation or elongation, and stability means refers to the ability to maintain its initial configuration. Material yield strength refers to the point on the engineering stress-strain curve (as opposed to true stress-strain curve) beyond which the material experiences deformations that will not be completely reversed upon removal of the loading. The ultimate strength refers to the point on the engineering stress-strain curve corresponding to the stress that produces fracture.
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