I am Sulochana. I am a Mechanical Design Engineer—M.tech in design Engineering, B.tech in Mechanical Engineering. I have worked as an intern at Hindustan Aeronautics limited in the design of the armament department. I have experience working in R&D and design. I am skilled in CAD/CAM/CAE: CATIA | CREO | ANSYS Apdl | ANSYS Workbench | HYPER MESH | Nastran Patran as well as in Programming languages Python, MATLAB and SQL.
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Primary and secondary stresses in pressure vessels | Pressure vessel stress analysis | Pressure vessel design procedure:
The first step in designing a vessel is the purpose of application and specifications that function the container characteristics. The environment and nature of liquid and gases are another important factors. The parameters involves in the designing:
The temperature and pressure (maximum safety).
Factor of safety.
The capacity to contain the volume.
Corrosion allowance
Design temperature.
Spherical vessel:
where, M = mass, (kg) P = pressure difference (the gauge pressure), (Pa) V = volume, = The density of the vessel material, (kg/m3) = The maximum working stress that material can tolerate. (Pa)
p is internal gauge pressure, r is the sphere’s inner radius, The thickness of the sphere wall is denoted by t.
Pressure vessel equations for stress | Pressure vessel equations | Pressure vessel formula | longitudinal stress pressure vessel:
sigma = stress in the longitudinal direction, p is internal gauge pressure, and sigma = stress in the longitudinal direction r is the sphere’s inner radius, The thickness of the sphere wall is denoted by t.
Mechanical design of pressure vessel | Pressure vessel design | pressure vessel calculations | how to design pressure vessel | pressure vessel dimensions
create outline of the design: Design and create the requirements of the vessel using the dimensions. Include dimensions such as shape, diameter, length, pressure, temperature, and construction material. Find out mechanical strength: Find out mechanical calculations using the software. Software gives both 2D or 3D drawings: pressure vessel design drawing:
Design standards: The purpose of application of vessel. Operating pressure and temperature Materials for fabrication Vessel head type Orientation: Horizontal or vertical Dimensions Openings and connections Requirements for heating and cooling Surface finish External factors Design stresses are adjusted using safety factors applied to material properties, including: Yield strength (design temp) Ultimate tensile strength (room temperature) Creep strength (design temp)
Gasket design for pressure vessel:
A gasket is designed in such way that the flanges should be able to create the specific amount of compressive load on the surface of the vessel. It created an seal with no pressure. The gasket should be attached to the flange surfaces and be compressed to reduce the internal voids and spaces.
Design of a noncircular pressure vessel:
Because of the geometry of the cylindrical shape, most pressure vessel and pipe flanges have a circular cross section. However, there are some pressure vessels or pressure ducts where a rectangular or other non-circular shape is required, whether for space or process reasons.
Water pressure vessel design:
Hydrostatic testing uses water for the test. It is a method that comprises pipeline systems, gas cylinders, boilers, and pressure vessel. These components are tested to check the strength and any kind of leakage from the system. Hydro tests are quite required for the repair and replacements of the equipment that will operate under the desired conditions. Hydrostatic test is the type of pressure test that can work by using the water and filling water in the components that removes the air contained within the system. and it pressurizes system with up to 1.5 times the design pressure.
How to calculate static head in pressure vessel:
Pressure vessel end cap design(heads): Design pressure of vessel includes: Static head= Pressure resulting from weight of liquid Acting on internal of the pressure. Higher liquid height results in higher pressure. Static fluid pressure is independent of the liquid’s form, total mass, or surface area. pressure= weight/area=mg/A
Pressure vessel skirt design:
Generally skirt support is provided to the tall columns. The vertical orientation of the container is supported by the skirt support in pressure vessels. The benefit of using skirt supports is that it reduces the amount of stress at the supports. Skirt is a cylindrical shell column with a diameter equal to or greater than the vessel’s outside diameter. The skirt is welded to the vessel’s bottom and sits over the bearing plate. The bearing plate is located on top of the concrete foundation system.
Pressure vessel skirt support design:
The vessel’s dead weight.
The vessel’s operating weight.
Lateral loads
Wind load
Seismic load
Skirts are the supports that are used in the vertical pressure vessels. They don’t take load from the pressure of the fluid inside the container. Weight of the vessel and fluid inside and the environmental loads altogether are considered for the designing of the skirt support. Skirts gives the less expensive design for the support of the taller pressure vessels. W+Fw+Ew= Total load.
Design of a jacketed pressure vessel:
A jacketed vessel is a container designed to control the temperature of its contents by encircling the vessel with a cooling or heating “jacket” through which a cooling or heating fluid is circulated. A jacket is an exterior cavity that provides for a consistent heat exchange between the fluid moving inside it and the vessel’s walls. Liner-less composite pressure vessels (CPVs), also known as type 5 (type V) tanks in some sectors, are the most efficient composite pressure vessels (burst pressure x volume/weight).
Vacuum pressure vessel design:
vacuum pressure vessel design uses a design pressure which is in accordance with the full vacuum of the vessel state that the internal pressure is vacuum and external pressure becomes 100kpa that is atmospheric pressure.
Pressure vessel fatigue calculations:
Fatigue life of material is determined first. The fatigue of material is determined by testing many samples to check the failure of the material. At each stress level, the number of cycles should be able to be calculated. Test samples are highly polished round bars that are as close to identical as manufacturing can make them. A test bar is rotated with a load applied so that a fiber at the bar’s surface is in tension and then in compression as the bar rotates, resulting in a complete stress reversal as shown.
There are several stress cycles, each with a different stress magnitude and number of cycles. Fatigue damage from each stress cycle adds up, so the total effect of all stress cycles must be calculated. The rule of Miner:
Pressure vessel shapes:
Although pressure vessels can potentially be any form, the majority are made up of portions of spheres, cylinders, and cones. A popular design is a cylinder with end caps called as heads. The most frequent head forms are hemispherical or dished.
Design of a vertical pressure vessel support:
They have a better pressure distribution, making them more secure. They use less energy because gravity allows their contents to flow easily and effortlessly. They require less ground space for their inhabitation.
Area compensation method in pressure vessel:
Nozzle reinforcement is the method of the area compensation. This method is used when there is opening in the cut section of the pressure vessel.
An area is removed from the shell and the head. The removed area should be equal to area added and it should reinforce by an equal amount of area near the opening.
composite pressure vessel analysis:
The objective of the analysis of the composite pressure vessel system is that It should increase the storage capacity of the system to the specific level. Hence, using the steel vessel, detailed analysis of the vessel design should be performed according to the multi axial stresses those are resulted from the tank design system in the transition region of cylinder and head.
Minimum wall thickness for pressure vessel:
1/16 inches is the minimum wall thickness is used for the pressure vessels. pressure vessel volume formula:
where, V= volume, r= radius of the internal surface a= area of the vessel I= moment of inertia.
Pressure vessel principal stresses:
There are two principal stresses in the pressure vessel. Hoop stress longitudinal stress This shows that the stress along the surface of the vessel should have resultant that balances the internal pressure.
FAQ/short notes:
What is the purpose of a pressure vessel:
Gases and liquids are held at high pressures within pressure vessels. Pressure vessels are used in boilers, reservoirs, highly pressurized pneumatic cylinders, and industrial uses, among other things.
How do pressure vessels work:
It works at higher pressure or increasing pressures. It reaches the pressure that makes the application function work such that it holds the gases or the liquids in the storage tanks. It provides the pressure through the valves or through the heat transfer.
What are the types of pressure vessels:
Pressure vessel types depends on the design of the vessels for the functionality of the applications in the industries. Mainly pressure vessels can be divided into the types according to their purpose for the applications. According to above factors mainly pressure vessels have three types:
Storage vessels:
These tanks are mainly useful for the industrial applications. These typically used in horizontal or vertical manner. It stores liquids and gases such as oil, chlorine and natural gases. It can be available in any size ranges. It is available in variable shapes like cylindrical or spherical for their vertical or horizontal manners. The material used in for the manufacture of the the type of product is carbon steel considering the external environment. Such vessels need careful construction as the internal substances can be bad without proper maintenance. Process vessels:
Process vessels are designed as per the requirements of the application while construction to reach the required specifications. Various processes can be performed in pressure vessels. Pressure vessels can be used in conjunction with other products depending on the application. So the manufacturing material required for such vessel components can be of unique material or multiple different materials. These pressure requires following important factors: Proper designing Proper material selection depending on the properties that reaches the applications requirements. Careful and proper construction as per the specification.
What is the distinction between an autoclave and a pressure vessel:
An autoclave is a type of pressure vessel. The main difference between the both is that the autoclaves are the type of pressure vessels that uses high pressure and high temperatures, the body should be capable of sustaining such high temperatures and pressures.
Pressure Vessel Design is vast topic, we will continue publishing article on Pressure Vessel. For more articles, click here.
Pressure vessel definition | what is pressure vessel | high pressure vessel | large pressure vessel
A pressure vessel is a container that holds a lot of pressure. It is a container that is designed to hold gases or liquids at a pressure higher than atmospheric pressure. It is a closed vessel with the capacity to store high pressure liquids or gases at internal or external pressures, regardless of the pressure vessel’s size, shape, or dimensions.
The liquids/gases are contained in these leak-proof vessels. These containers are designed based on the application’s purpose. Depending on the pressures, the operating temperatures of the containers change. The vessel works on the internal preconditioning pressures that are lower or higher than the air pressure.
Due to external tensile forces acting on the container’s internal surfaces, the container was able to resist the gas pressure. The thickness of a pressure vessel is proportional to the radius of the tank and inversely related to the maximum allowable normal stress of the material for the container’s internal surface. The normal tensile stress is related to the vessel’s pressure and radius, but inversely proportional to the vessel’s thickness.
Pressure vessel fabrication is an complicated process. For the fabrication and assembling of the parts the steps required are follows: Select the material for the fabrication.
cutting and burning of the material as per the requirement machining of parts cooling of weld and sand blasting Assembling and welding of parts Fabrication processes basic conditions: Design conditions. Procedures for welding to be used Welding Specifications Procedures for heat treatment will be used.
Construction of the container is tested to check the cracks, defects or any other existing failures. Hydrostatic test:
Hydrostatic test use water for the test. This test is safer method as it releases small amount of energy whenever fracture occurs. Pneumatic test:
Pneumatic test use air or gas for the test. mass production often represent samples testing for the destruction in controlled environment. Testing on pressure vessel is done to make sure the vessel is free from the defects, cracks or any other failures . Visual Tests (VT): Visual test is a type of test that provide information and overview regarding the pressure vessel by the observation of the internal and external substances of the tanks.
Liquid Penetrant Testing (LPT) is a form of pressure vessel test that uses thin liquids as a penetrant on the pressure vessel’s surface. The fissures in the vessel’s surface are readily visible. Using a chemical and a penetrant, proper visualization can be observed under UV light.
Magnetic particle testing is performed in conjunction with magnetic current to detect flaws. Whenever there is a defect ,there will be disturbance in the magnetic current.
Radiographic Test (RT): This type of test is tested using the X-rays to find out the defects on the external or internal surfaces of the vessel.
Ultrasonic Testing (UT): Ultrasonic testing is the testing that detect the defects using the sound waves. Whenever there are cracks on the external and internal surfaces of the vessel, the ultrasonic waves experience disturbances.
Reactor pressure vessel:
A reactor pressure vessel is a nuclear power plant that contains nuclear reactor coolant, a shroud, and the reactor core.
The classifications are as follows: Reactor for light-water – Reactor with graphite as a medium –
Thermal reactor cooled by gas –
Heavy-water pressurised reactor –
Reactor cooled by liquid metal –
Reactor for molten salt –
Components of the reactor vessel:
Body of the reactor vessel:
The large component containing the fuel assembly, coolant, and fittings to support coolant structures is the reactor body. A reactor head is attached at the top of vessel.
Assembling the fuel:
The fuel assembly of nuclear fuel, which is typically composed of uranium or uranium–plutonium mixtures. Typically, it is a rectangular block of gridded fuel rods.Reactor vessel body
Ammonia pressure vessel:
It is a Low pressure vessel. In this container Ammonia is force fed for the storage by the circulation at low pressure in the vessel.
Pressure vessel material | High temperature pressure vessel material:
Use of pressure vessel | purpose of pressure vessel
Pressure vessels are primarily used to store gases and liquids at high pressures. Pressure vessel applications are based on the requirements:
Industry of Oil and Gas: A container is used as a receiver at high temperatures and pressures. Chemical Industry: It is a pressure vessel in which a process (chemical reaction) needs to take place, culminating in a fundamental change in the container’s content.
Energy (Power Generation) Industry: The energy (power generation) industry emits polluted gases.Therefore pressure vessels are used to store such gases. nuclear power plant uses reactor pressure vessels.
Pressure vessel head types | different types of pressure vessel heads:
There are different types of tank heads and they vary according to the shape according to the advantage to the application: Ellipsoidal Head: Most economical. H=1/4D (Height =H, Diameter =D) has a radius ratio of 2:1 on the major and minor axes, allowing it to withstand greater pressure. Head with a Hemispherical Shape
This is a more spherical head, with a radius equal to the tank’s cylindrical section.
It aids in the even distribution of pressure across its surface.
The dish and cylinder share a toroidal-shaped transition known as the knuckle.
type 4 pressure vessel:
Type 4 pressure vessel is all carbon fibre pressure vessel containing polyamide or polyethylene plastic.It has low weight and high strength.carbon fibre gives more strength to the vessel that it can sustain high loads.It also increases the corrosion resistance and fatigue resistance of pressure vessels. This type of vessel has maximum volume , Hence it has capacity to store hydrogen at high pressure.
type V pressure vessel: high-volume pressure container of type v Type V approaches depend on advancements in three major technological fields: materials, design, and tooling. It uses single material to manufacture a laminate system that gives structural strength at high pressures. It also form barrier layers to persist fluids and gases substances.
Conical head pressure vessel:
Conical Head:
It is also called as tapered tank head. It is used for vessel bottom or cover plates. It has concentric cone shape. The conical shaped head contain large and small end cone.
Applications: Depending on the thickness of the material it can be equipped with around 8000 mm dia. and wall thickness of 20mm. The conical head forced on the bottom of the pressure vessel to accomodate internal materials and connect two stage vessels of different diameters.
Difference between boiler and pressure vessel: A pressure vessel is a container which contains the fluids ,gases or combination at high pressures. whereas a boiler is a container that contains the liquid that is water such that it can be boiled by the heat source at higher temperatures.
Pressure vessel dished ends dimensions | pressure vessel end caps:
Dished ends are the caps that are attached to the end of the main body by welding process. They are manufacured using diffrenet methods so as to reach the application requirements taht depends on the type of dished end.
The type of each dish end gives the characteristics of the end caps. For plate thicknesses of 25 mm / 1.0 inch or greater. Plates with a thickness of less than 25 mm / 1.0 inch. For plate thicknesses of 25 mm / 1.0 inch or greater.
Pressure vessel plumbing:
The pressure vessel is the container having switches that control the opening and closing of the container. It requires minimal amount of pressure when the tap is opened and it slacks when the tap is closed. When it reaches to the lowest pressure, the pump stops and the pressure also starts dropping. The pressure then drop in the pipes to the on switch pump and pump starts again.
Pressure vessel failure modes include ductile rupture, brittle fracture, and abrasion. abnormal deformation, insecurity (buckling),
ratcheting (progressive deformation),
fracture due to fatigue,
rupture due to creep,
ratcheting creep,
interaction between creep and fatigue,
buckling creep,
and the impact of the environment on cracking.
Heating pressure vessel | central heating pressure vessel:
An heating pressure vessel is the expansion tank. It is a small tank and protects closed water heating which are not open to ambient temperature. systems and hot water systems from high pressures.
The container contains air having compressibility cushion shocks caused by hammering and absorbing excessive water pressure caused due to the thermal expansion. Domestic applications Automotive applications
Hot water expansion vessel pressure setting | expansion vessel pressure setting:
water pressure should be -60 Psi. Thermal expansion container contains an pressurized compressed air. It expands and contracts in response to the expanded water from the water heater. Check the expansion tank’s air pressure.
Pressure vessel with lug support:
Vertical vessels with a height-to-diameter ratio of 2-3 are typically equipped with bracket supports. These are made of plates and attached to the vessel with the shortest possible weld length.
It is less expensive.
Can be easily attached to the vessel with a short weld.
It is simple to level.
If a sliding arrangement is provided, it can absorb diametrical expansions.
Because of their ability to absorb bending stresses eccentrically of loads, thick wall vessels are best suited for them.
In order to measure the liquid level in a pressure vessel, the gaseous pressure in the vessel’s head must be measured with a second transducer. To obtain the hydrostatic pressure due only to the column of liquid, subtract the head pressure from the overall pressure.
These vessels operate by reaching a specific level of pressure to meet the application’s requirements. The design is specification of the vessel is the application purpose such as storing ,containing , heat exchange and chemical reaction processing of the products. Valves, release gauges or heat transfer are used to proper delivery in the vessel.
The pressure level of the the normal atmospheric pressure is approximately 15 psi and it cam increase up to 15000 psi.
Replacement pressure vessels:
Repair of the pressure vessels is done to maintain its operating conditions. The replacement should be in order to maintain the safe operations and to maintain trouble free service. The vessel condition repair contains following considerations: mechanical problems,
Rules for construction of pressure vessels:
The pressure vessel construction requires specific prohibition and non mandatory guidance for the material slections , design of the vessel, design of the components , inspection and testing of the vessel and the parts ,markings and the reports , high pressure protection and certifications of the vessels .
The pressure applied on the internal and external surfaces of the container should be between 10-10000psi ,it may go up to 70000 psi that is the maximum limit. pressure vessels can be fired or unfired. The pressure applied can be from external sources or the application of heat transfer.
Vertical pressure vessel:
Vertical vessel is the orientation of the vessel which represents the container in the vertical direction(upright). It has different supports than the horizontal pressure vessel. It fits with different types of supports foe example skirt and lug that is able to hold the weight of the vessel. They can perfectly fit into the small spaces.
Water pressure vessel design | hydrostatic pressure vessels | hydrostatic test procedure for pressure vessel:
Hydrostatic testing uses water for the test. It includes components such as piping systems, gas cylinders, boilers, and pressure vessels. Theses components are tested to check the strength and any kind of leakage from the system.
Hydro tests are quite required for the repair and replacements pf the equipment that will operate under the desired conditions. Hydrostatic test is the type of pressure test that can work by using the water and filling water in the components that removes the air contained within the system. and it pressurizes system with up to 1.5 times the design pressure.
What is unfired pressure vessel:
This is type of the vessel that gains the heat from the source either directly or indirectly. To avoid overheating such containers should observe the caution measurement while handling the system.
Industries that utilize unfired pressure vessel: petrochemical power generation oil and gas Types: Thermal oil heaters Boilers.
Proof testing pressure vessels:
Proof pressure testing is the testing used to verify whether a component can sustain the pressure above the operating pressure without any permanent damage to the system. It is a form of stress that can demonstrate the fitness of the expansion joint under the high pressure conditions.
The test can also prove whether the component can sustain the high pressures. It is a non-destructive testing procedure, as opposed to other methods.
Different types of nozzles in pressure vessels:
Radial Nozzle Non-Radial Nozzle Hill Side Nozzle Tangential Nozzles Angular Nozzles.
Pressure vessel closures:
The pressure vessel closures provide closure guidance. These are commonly employed in medium to large pressure containers. It also has locking mechanisms and attachments for secure use. Pressure Vessel Closures Have Arrived.
Products are available.
Closures for Pressure Vessels
Aluminum pressure vessel:
Aluminum is being investigated as a replacement for stainless steel, with the major draw being its lower density and the expectation of a significantly lower tare weight.
Pressure vessel with cladding:
A cost-effective solution is to apply a layer of corrosion-resistant material of appropriate thickness to the equipment’s contact surfaces, made of a cost-effective and structurally strong material such as carbon steel. The technique of integrating two layers of different materials is known as cladding or lining.
While the word Lining is broad and can refer to a variety of materials, the term Cladding is used when the corrosion-resistant layer given is metallic and well-bonded to the surface. As a result, the word Cladding is frequently used to refer to steel-fabricated equipment such as pressure tanks and shell and tube heat exchangers.
Column pressure vessel:
Pressure vessels operate at a pressure greater than atmospheric pressure, whereas columns operate at atmospheric pressure. Furthermore, pressure vessels are subjected to pressure on all sides of their internal surfaces.
This is in contrast to columns, which only experience pressure in one direction.
Pressure vessels are built to hold liquids and gases at high pressures. A column’s principal function, on the other hand, is to separate gases from liquids using trays. In summary, you can select high performance pressure tanks using the information in this guide.
Ultrasonic testing of pressure vessels:
Ultrasonic testing is the testing that detect the defects using the sound waves. Refers to the thickness of the material’s plate. Whenever there are cracks on the external and internal surfaces of the vessel, the ultrasonic waves experience disturbances.
Difference between pressure vessel and storage tank:
The primary distinction between a pressure vessel and a storage tank is that pressure vessels contain liquids or gases at a pressure greater than atmospheric pressure. Storage tanks, on the other hand, contain liquids or gases under normal air pressure. Because pressure vessels can be highly catastrophic, they have more stringent safety requirements.
Storage tank safety design requirements are not as stringent as those of their counterparts.
Different types of pressure vessels:
Pressure vessel types depends on the design of the vessels for the functionality of the applications in the industries. Mainly pressure vessels can be divided into the types according to their purpose for the applications. According to above factors mainly pressure vessels have three types: Storage vessels:
These tanks are mainly useful for the industrial applications. These typically used in horizontal or vertical manner. It can be available in any size ranges. It is available in variable shapes like cylindrical or spherical for their vertical or horizontal manners. The material used in for the manufacture of the the type of product is carbon steel considering the external environment. Such vessels need careful construction as the internal substances can be damaged without proper maintenance.
Process vessels: Process vessels are designed as per the requirements of the application while construction to reach the required specifications. Various processes can be performed in pressure vessels. Pressure vessels can be used combined with other products as per the application requirements. So the manufacturing material required for such vessel components can be of unique material or multiple different materials.
Other types include:
High-Pressure Vessels: Autoclaves
Tanks for Expansion,
Exchangers of heat,
Tanks for high-pressure water,
Tanks for Vacuuming,
Pressure Vessels ASME,
Pressure Vessels with thin Walls,
Boilers are closed pressure vessels that heat fluids, most commonly water.
Jacketed pressure vessel | Pressure vessel jacket | Design of a jacketed pressure vessel:
A jcketed vessel is a container designed to control the temperature of its contents by encircling the vessel with a cooling or heating “jacket” through which a cooling or heating fluid is circulated. A jacket is an exterior chamber that facilitates consistent heat exchange between the fluid moving in it and the vessel’s walls.
Liner less composite pressure vessels (CPVs) have the highest pressure vessel efficiency (burst pressure x volume/weight) of any composite pressure vessel. They are also known as type 5 (type V) tanks in some sectors.
Pressure vessel for liquid nitrogen:
Cryogenic liquid cylinders are vacuum-jacketed, insulated pressure containers. To prevent the cylinders from pressure buildup, they are outfitted with safety release valves and rupture discs. These containers can withstand pressures of up to 350 psig and hold 80 to 450 litres of liquid.
Internal polishing. Internal cleaning and drying is automated. Cleaning with oxygen. Flushing with de-ionized water.
Cleaning with steam.
Shotblasting on both the inside and outside of the building.
Rinses with solvents
Baking in the oven to remove contaminants.
Coating both internally and externally
NVR (non-volatile residue) analysis
Particulate matter counts
Surface finish is measured using a profilometer gauge (Ra)
Measurements of coating thickness
Dimensions of the anchor profile
Pressure vessel relief valve:
Pressure vessel relief valve is the device that protect the container by the release of high pressures. The operation is automatic Valve can be opened and closed. the valve is opened at certain level and and it closes when the level return back to normal position.
Pressure vessel welding is the joining process that is used to connect the metal plates of vessel using the heat or the pressure. It should be good quality that should sustain loading conditions. Pressure vessel are used to store the liquids and gases at higher pressure rather than at atmospheric pressure. Welding of the container should be high quality structures and high strength materials as it should sustain the loading conditions.
If the good surface is used, then welding will be easy. There might be occurrence of the errors during the welding process.so it is required to apply some testing test to detect the errors. Porosity is one of the major factor that can occur during the welding. porosity occurs mostly in any component during the welding process.It creates gas bubbles that look like voids during the testing.To avoid such defects it is advisable to use proper welding methods.
Another important factor is the nitride which is highly adherent contaminant. That can cause edges brittle and create porosity in welding processes. Inclusions can be mixed with the weld pool and get stuck in the component during the solidification. This can be eliminated by using brush before the solidification.
Thin walled pressure is the type of vessel that has wall thickness smaller than overall size of vessel. t wall<r(vessel) The internal pressure is higher than external pressure.
Thick walled pressure | Definition of Thick Walled Pressure:
This is a vessel with the a wall thickness that is 1/10 or 1/20 more than its radius. The wall encounters more circumferential stress on the interior surface and lessens as it approaches the exterior diameter. Advantages of composite pressure vessels: Better performance results. Fibers carry the load on the composite. The load on the fibers is distributed by the resin matrix. The filament winding procedure is used to create a composite pressure vessel.
Air pressure vessel | Air receiver pressure vessel | Air pressure vessel testing:
Air pressure vessels are used to store the fluids, vapors and gases at high pressure. It is also called as air pressure tanks, tanks storage and containment units. Pressure testing is used to maintain the integrity of the vessels at high pressure levels. Non-destructive test.
FAQ/Short Notes
How do you test a pressure vessel:
Testing on pressure vessel is done to make sure the vessel is free from the defects, cracks or any other failures . Visual Tests (VT):
Visual test is a type of test that provide information and overview regarding the pressure vessel by the observation of the internal and external substances of the tanks. Liquid Penetrant Testing (LPT):
This is a technique of test in which transparent liquids are used as penetrants on the surface of a pressure vessel. It shows clearly the cracks on the vessel surface. Under the U.V. light, proper visualization can be observed using fluorescent chemical with the penetrant. Magnetic Particle Testing (MT):
Magnetic particle testing detects defects by using a magnetic current. whenever there is a defect ,there will be disturbance in the magnetic current. Radiographic Test (RT): This type of test is tested using the X-rays to find out the defects on the external or internal surfaces of the vessel. Ultrasonic Testing (UT): Ultrasonic testing is the testing that detect the defects using the sound waves. Whenever there are cracks on the external and internal surfaces of the vessel, the ultrasonic waves experience disturbances.
What is the distinction between a pressure vessel and a storage tank?
The difference between pressure vessels and storage tank is that the pressure vessels works at higher pressures and storage tanks works at normal atmospheric pressures. Storage tanks store the fluids. Pressure vessel hold the fluids at high pressures. Whenever a vessel reaches a certain pressure, it becomes a pressure vessel. When pressures reach 15 Mpa or greater. What is the frequency with which a pressure vessel should be tested: At least once in every five years.
What are the uses of pressure vessels:
To hold fluids at high pressures. High reactive chemicals , petroleum products can be stored at high pressures in pressure vessels. For the heat exchange and removal of excess heat. For the chemical reactions at certain pressures and temperatures.
Which material is used in manufacturing a pressure vessel:
steel made of carbon Steels with low alloy content Steels with a high alloy content Carbon steel, manganese steel, and so forth.
Why are semispherical end caps used on cylindrical pressure vessels rather than flat ones:
Cylinders are used as they are less expensive than spheres but the spheres are more stronger at the corners. So the spherical or rounded ends are fitted at end caps rather than flat ones . The following are some of the advantages of a spherical pressure vessel over a cylindrical pressure vessel: Spherical pressure vessel has smaller surface area per unit than any other pressure vessel shapes. As there is lesser surface area, the amount of the heat transfer from the high temp area will be less than other shapes. So the spherical pressure vessel is more efficient than any other pressure vessels.
Figure 1: Spherical pressure vessel
Figure 2: Cylindrical pressure vessel
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Invention: John Oldham (1821)-solve paddle steamer design. Flexible coupling has 3 types. Oldham coupling is one of the type of the flexible coupling.
Oldham coupling definition:
Oldham coupling transmits torque between shaft through mating slots on the center discs mounted on the hub. Oldham coupling is useful for parallel alignment applications. It can acquire axial and angular misalignments.
Basic applications of couplings: To transmit power and torque. To accommodate misalignments (angular, axial, parallel) To absorb shock loads and vibrations.
Oldham coupling parts:
Oldham Coupling has three discs. All discs are attached to each other using the grooves. The discs are mounted at the mid-sections of the shafts. The disc is mounted on the input shaft. Another the disc mounted on the output shaft. The three discs are connected to each other.
Central disc: It is the coupling part consisting of two hubs. It contains two shafts on the sides of the discs perpendicular to the hub axis and plugged into the hub axis. The center is used to reduce the backlash. And hence it is press-fitted at the center. It is designed to behave as a mechanical fuse. Mainly Oldham coupling working is based on the parallel misalignment, so the discs sliding motion acquire a large amount of parallel misalignment.
The hubs: Oldham coupling consists of two hubs that are connected at the end of the shafts. It is a round circular type of disc. Hubs are grooved into the center of the discs.
Oldham coupling: Working principle
Oldham coupling has three main parts(three discs).
One input shaft is connected to the disc and the disc is connected to the output shaft. each one is coupled to input and output shafts, and one is joined to the first two discs with the grooves. The disc at the center rotates at same speed and same axis as both the shafts. Its center rotates about the center axis orbit twice per rotation at the midpoint of input and output shafts. Springs are used in the mechanism to reduce the backlash. The discs are used to connect the driving and the driver shaft in mechanical power transmission assemblies.
Oldham coupling advantages | advantages and disadvantages of Oldham coupling
It has a low moment of inertia.
Material used for the center disc is plastic. Hence the electrical isolation between the discs is not possible. All three are aligned to each other.
If the first disc breaks, the torque limit is exceeded in the center disc so as to prevent the torque transmission.
It is economical compared to other couplings.
It has a high torque capacity.
Easy to dismantle and disassemble.
Due to the backlash, there is a need for replacement. Hence the replacement is inexpensive compared with other couplings as it is compact in size.
Accommodate large parallel and radial misalignments.
It gives protection to the driven components and support bearings.
It is compact in size, so it is easy installation.
It has high torsional stiffness.
Longer service life
Tighter machining tolerances with one shaft connected to another maintain a consistent performance rate.
Good surface finish.
Very low restorative force compared with other coupling designs:
Low maximum speed=3000 rpm.
Oldham coupling disadvantages | Advantages and Disadvantages of Oldham coupling
It can accommodate even small angular misalignments.
It can accommodate very limited angular displacement.
At the high torque, the axial load which is reactive, can be applied on the support bearings.
Oldham coupling applications:
It can accomodate slightest misalignments. The driving shafts and the driven shafts generate at a similar speed. It behaves as an electrical insulator. It can transmit power and torque. The material used in the three discs leads to the use of Oldham of coupling in many device applications.
Oldham shaft coupling:
Oldham coupling applications are mainly used to join the two parallel coaxial rotating shafts. It can transmit the torque at the same speed and same rotation mechanism. An Oldham type coupling is part of the flexible type coupling. It is a better design for the coupling shafts, which has misalignments, and it can transmit torque between the shafts at the same speeds and same direction. The coupling can acquire any slightest amount of misalignment. Oldham coupling has three discs and two hubs, and one center connected by the use of grooves that fit the fins on the midsection, each side, and perpendicular to each other.
Shaft coupling linkages system requirements:
It should be easy to disassemble It should allow the misalignment between the rotating shafts rotating along any axis. It should not be any projecting parts. It should reduce the further misalignments that can occur in the running operations that can increase the power transmission and machine runtime. It should reach the manufacturer’s target machine train to define non-zero alignment.
Oldham coupling used in:
Robotics and servo applications Printer and copy machines.
Uses of Oldham coupling:
The purpose of the couplings is to connect two rotating shaft types of equipment, allowing some misalignment. Careful selection and installation, and maintenance of the couplings at reduced costs. The coupling provides the connection between the shafts that are even manufactured separately. It also provides disconnections between the repairs and replacements.
Couplings allow misalignments along with lateral, axial, and angular directions. It provides protection to the support bearings and shaft hubs from overloading. It can change the vibration characteristics of the rotating shafts.
Oldham coupling design:
It is an old design mechanism with identical slotted elements put together to slide between them. The Oldham designing is used for the machine shafts, which have a parallel misalignment.
Bore diameter= Diameter of coupling bore mounted on the connecting shafts. Overall length is the length of the coupling end to the end face. Hub width is the end face to internal face width. Slider block material Coupling diameter Maximum rated torque Lateral offset – Lateral offset is the parallel misalignment. Angular offset – Angular misalignment Axial offset – maximum axial deviation along with the shaft Shaft coupling fastening method
Design calculation of Oldham coupling| Oldham coupling dimensions:
The length L, L=1.75d, The diameter, D2=2d, The thickness of the flange, t=0.75d, Diameter of the disc, D=3L, Centerline distance, a=D-3d, The breadth of groove W=D/6, The groove thickness, h1=W/2, The disc thickness, h=W/2, The total pressure on each coupling=F=1/4(p*D*h) The torque on each side of coupling=T Ttc=2Fh=p*D*2h/6
Oldham coupling drawing:
Oldham coupling material:
For the Oldham coupling, the material used is different for the parts of the mechanism, Mentioned as follows: Slider block is manufactured by the polymer material (nylon, acetal, or combination) to reduce the backlash. The slotted mating halves are manufactured from aluminum to reduce inertia. It can also be manufactured from brass material of smaller sizes. Materials can be used for mid disc: Delrin Materials can be used for side disc(hubs): stainless steel. Aluminum alloy.
Oldham coupling mechanism:
An Oldham coupling is the mechanism that is a combination of the rigid connections bodies(kinematic linkages) with definite relative motion. Mechanisms generally have linkages that can move relative to each other—for example, gear and gear train, belt and chain drive, cam and follower.
This also includes the friction devices such as brake and clutches, and structural components, fasteners, bearings, springs, lubricants, etc. And the different types of the machine element parts like spline, pins, and keys. To perform the mechanical work machine elements mechanism transmits power to the resistance to overcome. A coupling is a device that is a connection between the two shafts for transmission of the power.
A coupling can allow the misalignment but it does not allow disconnection of shafts during the procedure. The torque limiting couplings can disconnect or slip due to the torque limit is exceeded. Oldham coupling is the inversion of the double slider crank mechanism. It is obtained by the connecting links. It can join the elements having lateral misalignments.
Oldham coupling consists of three discs and two flanges with the slots with two tongues attached to the central floating part perpendicularly to each other. Pins are provided, which pass through the flanges and the floating disc. When the tounge1 are fitted in the slot, it allows relative motion between shafts.
The tounge2 is fitted in the other slot allowing the relative vertical motion between the shafts. The resultant motion of the two components allows motion to accommodate the misalignment of the rotating shafts. The Oldham coupling mechanism is based on the three discs with hubs. The sliding friction develops wear of slider block that creates a backlash in the misaligned systems.
Oldham coupling mechanism applications:
Oldham coupling mechanism is mostly used for the stepper motor-driven positioning stages. The elements of the coupling absorb the shock loads and the vibrations from the frequent starts and stop of the load reversals. The Oldham coupling is designed for the motion systems, which are idle half of the time. The latter is available for mounting the driving and the driven shafts without any disturbance within the shaft alignments.
Misalignments in the Oldham coupling:
Misalignment is the displaced alignment between the shafts.
Misalignment can be parallel, angular or axial, or combinations.
In the Oldham coupling, parallel misalignment is generally observed.
An angular misalignment in the machine train is that in which the shafts intersect at angles less than 180.
Tighter alignments have higher energy efficiency.
The tighter alignments have less wear on the machine parts.
Generalized Oldham Coupling:
Oldham coupling is the inversion of the double slider crank mechanism[citation] an Oldham coupling is the device used to transmit motion and power between parallel rotating coaxial shafts. Oldham coupling is the flexible coupling part. It gets its flexibility from the disc materials,
The following shows Oldham coupling types based on the shape: with circular slots, with curvilinear slots
The figure shows the generalized Oldham coupling with circular slots (link1), an input disc mounted at the midsection of the input shafts (link2) an output disc mounted at the midsection of the output shaft(link4),floating disc(link3).
The radii r1 and r2 are the radii of the centerlines of the slots. The radii may or may not be equal .It is not necessary that the radii has to be equal. The radii intersect at the centerline axis of the floating disc.
The primary Oldham coupling type transmits torque at the same speed of both the shafts. The generalized Oldham coupling has two types of slots. One is circular slots and the another is curvilinear slots. Using either of the slots. the Oldham coupling can transmit non uniform motion and also It can produce quick return motion. and It can used in many applications where devices require non-uniform transmission.
FAQ/Short Notes:
How to reduce backslash in Oldham coupling, and what is it?
Backlash is the angular movement in the mating parts of the mechanical parts. It is often possible that backlash occurs in the coupling movement. The excessive backlash can wear out the coupling parts.Backlash is the angular movement in between the mating parts. Coupling inserts inspection needs for the proper precision of the shaft alignment. A backlash has to be less than a 2-degree angular movement.
Control and reduce of the backlash methods: Replacement of the couplings inserts and defective components. Reduce the backlash by rotating shafts, and maintain torque at a consistent level.
Coupling is the device that is used for the connection of shafts to transmit power and torque.
Coupling is the connection of shaft units that are manufactured separately. Coupling failure occurs due to key and bolts. So it can be covered by flanges to avoid failure. The shafts may have all-direction axes. Couplings are used for variable-type applications. Examples: Motor and generator Motor and pump
Flexible Couplings Definition:
Flexible couplings are the coupling device used for the flexible connection. It is the type of coupling used to join shafts having misalignment in lateral or angular directions. It has flexible connection allowing the misalignments. The elements used have the capacity to absorb the shock loads and the vibrations.
Machines, servomechanisms, instrumentation, light machinery, steel industry, the petrochemical industry, utilities, off-road vehicles, and heavy machinery, etc.
Elastomeric coupling has elastic properties. It is the part of the flexible coupling which can attain their flexibility from the material’s tension and compression ability. Material Example: Rubber, plastic, etc.
Advantages:
Material uses: Rubber or plastic leading to low in cost and allows temperature rise. Resistant to fatigue failure. It provides longer life at a minimal cost. Regular maintenance is not required as it doesn’t use any lubrication. No need to use lubrication, so regular maintenance is not required.
Mechanically flexible couplings –
Mechanically flexible couplings are part of the flexible coupling, and it gets its flexibility from the loose-fitting parts and the rolling and sliding parts. It requires regular lubrication. Example: Nylon gear coupling
Gear coupling:
A gear coupling is the application part of a flexible coupling It is the type of coupling the shafts with gears mounted on the hubs. Sleeves (hollow cylinders) have internal gear teeth.
Regular lubrication is required. lubrication used: grease, oil
Advantages : It provides good torque characteristics.
Metallic membrane couplings –
Metallic membrane coupling is the application of flexible coupling. It gets its flexibility from the flexing of the metallic discs (thin).
Miscellaneous couplings –
Miscellaneous coupling is part of the flexible coupling. This type of coupling gets its flexibility from the combinations of the mechanisms like spring couplings.
Flexible coupling functions:
It Transmits power. It Transmits torque. The selection of the flexible coupling depends on the max rotational speed. Power loss is due to the friction heat from the sliding and rolling at high speeds. Loss of efficiency is due to frictional losses. The flexible coupling has the advantage that it is the coupling that can give efficiency more than 99%.
Flexible coupling drawing:
Advantages of flexible coupling:
It allows small misalignments. It absorbs shock loads and vibrations. It can transmit a high amount of torques. Simple in construction
Disadvantages of flexible coupling:
It has a high cost due to additional parts. Require more space.
Bush pin type flexible coupling:
It is the type of coupling used to connect shafts with smaller misalignments.
Transmits the torque from the high tensile material to the input shaft.
Material used: Rubber and leather(bush)
The material gives the flexibility to the coupling.
The grooved coupling having the flexible connection is Flexible grooved coupling. It allows some misalignment in the shaft connection.
Thomas flexible disc coupling:
Some couplings don’t require regular maintenance. It can work properly on its own. The coupling is all flexible metal parts coupling. The coupling is known as Thomas flexible disc coupling.
Advantages of flexible coupling over rigid coupling:
Flexible coupling can be used with low levels of torque transmission and small misalignments. It allows slight misalignment and still can transmit the same amount of torque as rigid coupling.
Difference between rigid and flexible grooved coupling:
As you know that Flexible coupling creates flexible connections between equipment and the components used to assemble the coupling fixed with some amount of loose, that why some amount of misalignment it can absorb. But metallic flexible type has greater torque capability than other flexible couplings, and some torque will be lost during complete operation of equipment.
Flexible shaft coupling:
Flexible shaft coupling is the flexible connection of the coupling shafts It prevents coupling failure. It reduces noise, vibrations and protects coupling components.
Flexible spacer type coupling:
It is the type of coupling which has an extra length shaft installed in the shafts. It provides space to remove the mechanical seal during maintenance.
Flexible coupling alignment tolerance:
Up to 400 mils.
Flexible coupling design:
The flexible coupling is designed to calculate the torque and power transmission, considering the misalignment in any direction. Due to less misalignment, There is less movement in the parts. This leading to less axial and bending stresses development in the shafts. Torque=P/rotational speed, If the speed of the shaft increases, power increases, and torque decrease. There is the possibility of a loss of torque.
Parallel offset: This type of offset occurs in the flexible couplings shaft connection where both the shaft axes are parallel to each other and not in the same line. Angular offset: This type of misalignment occurs in the flexible coupling shafts where the axes of the shafts meet at the center points of the coupling. Combined parallel and angular offset: This type of misalignment is the flexible coupling offset of the shafts where the axes of the shafts do not intersect and are not parallel to each other.
Flexible coupling material:
Brass. Aluminum. Cast Iron. Stainless Steel. Carbon Steel. Rubber etc.
Flexible coupling at high temperature:
At the high temperature, the couplings will transmit thrust between the machines.
A the temperature increases, the couplings will be more flexible and it will give more flexible connection.
At increasing temperatures, The coupling losses it’s torsional stiffness, increasing more pressure.
Flexible chain coupling:
Roller chain coupling, in which sprockets are attached at adjacent ends of the two abutting shafts, then wrapped together by a common roller chain segment that spans both sprockets. Clearance between the chain and the sprockets allows up to degrees of angular shaft-centerline misalignment and up to about 0.010 inches parallel shaft-centerline misalignment. Roller chain couplings are low-cost, high-torque devices but may be noisy. Wear or fretting wear is a potential failure mode.
Flexible coupling encoder:
It is a device used to provide maximum mechanical protection.
It is the protected type flexible coupling device which protects the coupling.
Flexible disc type coupling:
Flexible disk couplings are the disc type coupling that allows smaller angular and parallel misalignment. Misalignment of about one degree of angular misalignment and an inch of parallel shaft misalignment can be observed.
Flexible grid coupling:
A grid coupling is a system consisting of two shafts, a grid spring(metallic), and a split cover. This type of coupling is the coupling device that transmits torque between the coupling shafts using a metallic grid spring. Advantages: High torque density. The grid coupling spring elements have the ability to absorb shock loads and peak loads. It also has the capacity to dampening the vibrations. flexible type of coupling tend to have ability to allow the misalignment.
Flexible coupling Problems and solutions:
1)The motor power is 50 KW, and the speed given in rpm is 300rpm. The bearing pressure on bush is equal to 0.5MPa. Allowable shear stress is 25MPa, and the bearing stress is 50MPa. The shear yield strength given is 60MPa. Given data: shaft dia. = 50mm,pins diameter(PCD)=140mm. Determine the dimensions of the rubber bush for flexible coupling.
Solution:
Torque transmitted,
T=159N-m.
shaft diameter
d=23.8mm Let, d=25mm,
n= no. of pins,
n=4, dneck= 8mm shear stress,
yield stress of the pin material. d=Dpin+2*t(sleeve) d=20mm t=6mm Bush length, T= 159Nm, p = 0.5MPa, dbush =0.02m and dc = 0.14m ,L = 56.78 mm.
2)Design a bushed-pin type flexible coupling shaft transmitting 50 Kw at 1000rpm. The bearing pressure in the rubber bush is 0.5MPa and allowable shear stress in the pins is 25Mpa. The Diameter of shaft is 60 mm. Given: P = 50KW; N = 1000 rpm, d = 50 mm,
What are the three types of flexible compression couplings:
Jaw type coupling Donut type coupling Pin and bushing type coupling.
Flexible coupling vs. solid coupling:
solid coupling is rigid coupling. a Rigid coupling is the coupling device that is rigid in connection, whereas flexible coupling is the coupling device that is flexible in connection. Flexible coupling elements have the ability to absorb vibrations and shock loads, whereas rigid coupling is free of vibrations and shock loads.
Shielded coupling vs. flexible coupling:
Shielded coupling is a unique type of coupling. Shielded coupling is encased in a metal case. It is used for underground applications.
Flexible coupling types- According to their uses:
General-purpose Flexible coupling Gear type Chain type Grid type coupling special purpose Flexible coupling mechanically flexible type, etc.
A Coupling is a connecting device, which connects two rotating shafts. A coupling is used for power transmission and torque transmission. Connected at the ends of the shafts, there is the possibility of failure or slippage depending on the torque limit of the shafts.
Applications:
⦁ The purpose of coupling is to transmit power and torque.
⦁ Transportation of shafts becomes easier by dismantle and assembling the shafts by the use of a coupling.
⦁ Connect the driving part to the driven element.
⦁ To reduce transmission shocks.
⦁ Protects the system.
There are some shafts that are manufactured separately and still can be joined together by the use of coupling:
A rigid coupling is a coupling device used to connect shafts that are perfectly aligned, or there is no misalignment in the shafts in all directions.
These type of couplings are mostly used in vertical actions in the system. Rigid couplings transmit rotational as well as an axial motion to the two connected shafts rotating at certain rpm. Rigid couplings transmit power and torque between the shafts and between the two systems only if the shafts are appropriately aligned. Example: Vertical pump.Example: electric motor.
A rigid coupling is connected from the equipment shaft to the motor shaft.
The coupling shafts transmit axial thrust.
Configurations: Split configuration: Split along the centerline. Flanged configuration: The two couplings and the flanges are bolted together. Flanged rigid couplings use adjusting plates that are utilized to set up proper vertical position. .
Rigid couplings types:
The flanged type rigid coupling The clamp-type rigid coupler The sleeve type rigid coupling
Flange rigid coupling:
Flange coupling is the device that is used to connect to shafts if both the machine shafts are properly aligned to each other. Flange coupling is used where free access is available for both the shafts.
It is mostly used coupling, and the couplings flanges and the shafts are bolted together.
Advantages of flange coupling:
Flange coupling is less expensive as compared to the other type. Less space is required for installation. Interchangeable.
Material used :
Flanged couplings are constructed using various materials, including grey cast iron, malleable iron, carbon steels and carbon steels series ranging from 1035 to 1050.
Rigid couplings can be manufactured from most metal materials. This allows Rigid couplings to be used in many applications and variable conditions. It can transmit more power.
MATERIAL AND ITS PROPERTIES:
The manufacturing process used to manufacture flanged coupling is the casting process, as it contains recess and projection. The flange coupling is commonly made from grey cast iron those which are characterized by graphite microstructure, causing a fracture to the material to appear as grey.
Cast iron is the most commonly used material due to its casting properties having less tensile strength than compressive strength. Alloys of iron contain carbon and silicon 2.5-4% and 1.3%, respectively.
Cast iron experiences less solidification shrinkage.
Silicon is corrosion resistant, and in the casting process, it leads to an increase in fluidity and offers good weldability.
Advantages of rigid coupling:
⦁ Rigid couplings can be used in complex motion systems.
⦁ Rigid coupling provides more torque between the shafts.
⦁ It is also useful for better positioning as it has High torsional stiffness.
⦁ Easy availability.
⦁ Cost-efficient.
⦁ It has precision with zero backlashes.
⦁ Rigid couplings are used for the proper alignment and rigid connection.
⦁ Easy to assemble and disassemble.
⦁ Easy for the maintenance operations
Specifications: Rigid flange couplings
⦁ Rigid couplings are stiff connected, and it do not absorb vibrations leading to the possibility of replacement of the coupling due to wear on the parts are not properly aligned. ⦁ It requires routine check-up for wear and alignment check. ⦁ Apply lubrication regularly.
Difference between rigid and flexible couplings:
Rigid coupling is the coupling device used to connect shafts, and the connection between the shafts is the rigid connection where the two shafts are closely connected, whereas, in the case of flexible coupling, the connection between the two shafts is the flexible connection.
Flexible coupling provides the connection between the shaft components, which assemble the fixed coupling with some amount of loose connection. This gives some misalignment between the shafts.
Rigid coupling gives the smooth transmission of torque between both the shafts and the components, whereas in the flexible coupling, only the metallic type flexible coupling has a large capacity than other flexible couplings, and there is the possibility of some torque loss during the operation.
Flange coupling adapter:
A flange coupling adapter connects the end of the ductile iron pipe to the flanged pipe, valve or fitting.
Design procedure for flange coupling:
Assembly of muff coupling:
A hollow cylinder is attached at the ends of both the shafts using the sunk key. The hollow cylinder is called the sleeve. Torque and power are transmitted through shafts using these hollow cylinders.
First, it is transmitted from the first shaft to the sleeve; From the sleeve, it is transmitted to the key. Then from the key, it is again transmitted to the hollow cylinder(sleeve).
It is easy to manufacture and design and difficult to assemble and disassemble. The material used: Cast iron The factor of safety =6-8 (on the ultimate strength) It is required to have more axial space and less radial space dimensions.
Sleeve standards :
Sleeve outer dia. D = 2d + 13 Length of the sleeve, L = 3.5d d= diameter of the shaft.
Design of Shafts:
Shaft design is based on the torsional shear stress.
For torque transmission, shear stress T is given by,
Where, T = Torque acting on shafts, J = shaft polar moment of inertia, r = d/2
Allowable shear stress=[τ] determines the dimensions of the shaft.
Sleeve Design:
D = 2d + 13 L = 3.5d,
Consider a hollow shaft, The torsional shear stress in the sleeve is calculated,
Design of Key:
Cross-section of the key selected corresponding to the shaft dia and key dimensions. cross-sections of the keys: Square and rectangular Length of the key in each shaft,
Shear and crushing stresses,
shear stress,
where, w= Width of the key. h= height of the key.
Clamp coupling:
Clamp coupling is compression coupling or also called split muff coupling. Split coupling is coupling in whose sleeves are split into two halves along the plane passing through the shaft axis.
These split sleeves are attached using bolts and placed in the recesses. Assembling and disassembling is easy for the clamp coupling. Clamp coupling balancing is difficult for high speeds and shock loads.
Design of Bolts:
Bolts design is based on torque transmission.
Let [σt] = permissible tensile stress,
dc = diameters of bolts, n = number of bolts
Clamping force of each bolt,
clamp force is applied equally on each shaft.
Frictional Torque,
Flange Coupling:
Flange coupling is the coupling device consisting of two flanges that are keyed to the shafts. The flanges are joined together using the bolts on a circle concentric to the shaft.
Power transmission is from the driving shaft to the flange on the driving shaft with the help of the key and from the flange on the driving shaft to the flange on the driven shaft using the key again.
For the proper alignment, projection and recess is used with the flanges Inner hub, flanges and protective circumferential flanges – Protected type flanges Flange coupling design dimensional proportions:
Flange coupling: The outer diameter of hub, D = 2 Bolts diameter, D1 = 3 d Flange diameter, D2 = 4 d Hub length L = 1.5 d tf = 0.5 d tp = 0.25 d
Design of Hub:
A hollow shaft is considered, with the inner diameter = diameter of the shafts, Outer diameter= 2* inner diameter. For torsional shear stress.
Where, T = In designing Hub required Twisting moment (or torque) J = shaft’s Polar moment of inertia ( axis of rotation) r = D/2
Design of Flange:
The hub is for the toque transmission through the bolts, The flange is subjected to the shear.
Tangential force,
Shear stress,
Design of Bolts:
Let n be the total number of bolts. Force acting on each bolt, where D1 is the pitch circle diameter of bolts. Area resisting shear,
where, dc = core diameter of bolts Shear stress,
Area under crushing Crushing stress,
Bolts are subjected to both shear and crushing stress, Due to the transmission of torque, the force acts perpendicular to the bolt axes.
Types of flange coupling as follows: ⦁ Protected type flange coupling ⦁ Marine flange coupling. ⦁ Unprotected type flange coupling.
Unprotected type flange coupling:
In unprotected type flange coupling, No of bolts used= 3-6 The keys are attached at the right angle along the circumference of the shafts dividing the keyways.
Unprotected flange coupling and cast iron flange coupling dimensions:
d= diameter of the shaft, then D = 2 d Hub length, L = 1.5 d, flange, D2 = D1 + (D1 – D) = 2D The thickness of flange, tf = 0.5 d Number of bolts = 3,
Flanges are attached using the bolts.
Protected type flange coupling:
A protective circumferential rim is used. The rim covers the nut and the bolt.
It consists of the the following protective procedures:
Perform visual inspections, Check signs of wear or fatigue, Clean couplings regularly and change the lubricants regularly. Maintenance is required in operating conditions and adverse situations.
Advantages of the protective type flange coupling:
It can transmit high torque.
It is simple to construct.
Easy to assemble and disassemble
Marine flange coupling:
This is a type of coupling where the flanges are attached to the shafts using the tapered headless bolts. thickness, t=d/3, bolts, D1=1.6d, D2=2.2d,
Advantages:
It is cheap.
It is simple in structure.
More efficient.
Maintainance is not required.
Disadvantages:
1.It cannot be de-engaged in motion. 2.This type of coupling cannot transmit the torque between the shafts that are not linear.
Checking the coupling balance:
Balancing requires cost and it is difficult to balance. The amount of the coupling unbalance can be tolerated by the system. The analysis gives the detailed functions and the characteristics of the system and the connected machines.
Rigid flange couplings are less expensive than the flexible couplings. Rigid type couplings have rigid connections so they are torsional stiff and does not give access to any misalignment between the shafts. Due to the thermal effect, parts have misalignment during the operation, and both the shafts are physically aligned.
Rid couplings are couplings having rigid connections. It does not absorb vibrations leading to the replacement of the parts. Due to wear on the parts, misalignment occurs. The operators require routine maintenance and checking of the parts for wear and alignment.
Flanged pin bush couplings:
Flanged pin bush coupling is also called as bush pin type coupling.
This coupling works as protective type flange coupling with better modifications.
This coupling device has pins, and it is used to work with coupling bolts.
The material used: Rubber The rubber bushing can absorb vibrations and shocks during its operations.
Flange compression coupling:
The flange compression coupling of the coupling device.
Flange compression couplings have two cones which is used to place over the shafts.
The shafts should be coupling shafts.
The hollow cylinder is a sleeve that is used to fit over the cones.
Sleeve coupling flange:
Sleeves are attached to the shafts.
To lock the coupling in position, two threaded holes are provided.
Split flange coupling:
Split flange coupling is the coupling device the sleeves are split into two halves made up of the cast iron. These split parts are connected using mild steel bolts.
Advantages of the split flange coupling:
Easy assembling and dismantling without changing the position of the shafts. It can be used to connect two shafts of heavy transmission at moderate speed.
FAQS:
Flange coupling is what type of coupling:
Rigid type coupling.
Flange coupling specifications. Explain.
⦁ It should be easy to assemble or disassemble. ⦁ Flange coupling should transmit torque and power. ⦁ Maintain proper alignment. ⦁ Minimize the shock loads transmission.
Requirements to ensuring of the shaft alignment before attaching the fixing bolts:
⦁ If it is easy to connect or disconnect the coupling. ⦁ No projecting parts There should be less misalignment in running operation, leading to maximum power transmission.
Why is the key used in protective type flange coupling?
Keys are used to preventing rotational motion. The surfaces of the shaft and hubs parts provide cut to mount the keys, joints.
Why did recess provide in flange coupling?
To provide the clearance in the flanges, recess is provided. The flanges are tightly fitted with the use of bolts using the torque to be transmitted.
The minimum number of bolts required in flange coupling:
Four, six, or up to 12 bolt assemblies.
What is the grade of cast iron used to make rigid type flange coupling?
Bulk modulus is the ability of the material to be resistant to compression. It is the volumetric elasticity and is inversely proportional to Compressibility. The object having incompressibility deform in all directions when the load is applied from all directions.
Bulk modulus is the volumetric stress over volumetric strain.
The ratio of the increase in pressure relative to the decrease in volume.
P=pressure V=initial volume dP/dV = derivative of pressure with respect to volume.
Hence,
Bulk modulus unit:
SI unit of Bulk modulus of elasticity: N/m^2(Pa)
Dimension of Bulk modulus:
Bulk modulus pressure:
The effect of pressure on incompressibility explained from the below graph:
The value of Bulk modulus is required to determine:
The value of the Bulk modulus is required to determine the Mach number. Mach number is a dimensionless quantity.
Bulk modulus measurement:
How incompressible a solid is measured by the Bulk modulus. Hence bulk modulus also referred as incompressibility.
The incompressibilityof fluid:
The fluid volume modulus is the measure of resistance to compression. It is the ratio of fluid volume stress to volumetric strain.
Bulk modulus of various materials:
Materials : Bulk modulus values
The Bulk modulus of water:2.2Gpa
The Bulk modulusof water at high pressure:2.1Gpa
The Bulk modulus of air:142Kpa isentropic, 101Kpa isothermal
Bulk modulusfor steel:160Gpa
Bulk modulusof mineral oil:1.8Gpa
Bulk modulusof mercury:28.5Gpa
The adiabatic Bulk modulus of air:142Kpa
The Bulk modulus of diesel:1.477Gpa (at 6.89Mpa and 37.8°C)
The Bulk modulus of ice:11-8.4Gpa(0K-273K)
The Bulk modulus of hydraulic oil:
The Bulk modulus of concrete:30-50Gpa
The Bulk modulus of diamond:443Gpa
The Bulk modulusof rubber:1.5-2Gpa
The Bulk modulus of water at high pressure:2-5Gpa
Hydraulic fluid incompressibility:
Hydraulic fluid incompressibility is the Compressibility resistant property of the material. Hydraulic fluid gets affected by the applied pressure. As the applied pressure increases, the volume of the body decreases.
The Bulk modulus of elasticity:
The modulus of elasticity of liquid varies depending on the specific gravity and the temperature of the liquid. K is always constant within elastic limit of the material. This is the Bulk modulus of elasticity.
K=Volumetric stress/volumetric strain
The sign indicates the decrease in volume.
Volume modulus is associated with a change in volume.
Compressibility is calculated as reciprocal of incompressibility. Compressibility represented as, Compressibility=1/K SI unit: m^2/N or Pa^-1. Dimensions of Compressibility: [M^-1L^-1T^2]
Derivation of Bulk modulus of elasticity:
Bulk fluid modulus is the ratio of the change in pressure to change in volumetric strain. \\frac {-\\delta V} {V}=\\frac {\\delta P} {K} δV: change in volume δp: change in pressure V: actual volume K: volume modulus δp tends to zero
V=1/density
The Bulk modulus of incompressible liquid:
The volume of incompressible fluid does not change. As the force is applied, the change in volume is zero due to the volumetric strain of the incompressible fluid is zero.
Temperature dependence:
The modulus of incompressibility evolves due to the volumetric stress evolves periodically.
It is coupled to shear modulus, Assume constant Poisson’s ratio.
The time-dependent modulus is represented as,
Elastic constants relationships:
Relationships between Poisson’s ratio, Young’s modulus and shear modulus with bulk modulus:
For an incompressible fluid, the maximum limit of poisson’s ratio be 0.5. For K to be positive μ should be always than 0.5. n = 0.5. 3G = E. K = ∞. E= 3K(1-2 μ) E= 2G(1+μ) 2G(1+μ)=3K(1-2 μ)
Distinguish between young’s modulus and bulk modulus:
Young’s modulus is related with longitudinal stress and longitudinal strain of the body.
Incompressibility is the form of volumetric stress and volumetric strain. Bulk modulus exists in solid, liquid and gas, whereas Young’s modulus exists in only solids. Young’s modulus gives the change in length of the body, whereas Bulk modulus gives the change in volume of the body.
Distinguish between Shear modulus and Bulk modulus :
Bulk modulus is the form of volumetric stress and volumetric strain. It involves the effect of the applied pressure. as the pressure increase , the volume of the body decreases. This gives the negative sign to the ratio of the stress to strain. The ratio is associated with the volume of the body. In case of shear modulus, shear modulus is the form of shearing stress and shearing strain. It involves the effect of shear stress on the body. It is the response to the deformation of the body. The ratio is associated with the shape of the body.
where, T=shear stress gamma=shear strain Incompressibility exists in solid, liquid and gas, whereas shear modulus exists in only solids.
Isentropic Bulk modulus:
Incompressibility of the body at the constant entropy is called isentropic bulk modulus. The ratio of change in applied pressure to fractional volume change in the body due to the pressure change is a form of the isentropic incompressibility.
Isothermal Bulk modulus:
When the temperature is constant throughout the incompressibility is called isothermal bulk modulus. The ratio of change in applied pressure to fractional volume change in the body due to the pressure change is a form of the isentropic incompressibility
Negative Bulk modulus:
Why negative:
Bulk modulus has a negative sign because of the decrease in volume due to an increase in pressure.
Adiabatic Bulk modulus:
Adiabatic Bulk modulus is the ratio of the pressure to change in fractional volume in the adiabatic process when there is no heat exchange with the surrounding.
Dimensional analysis is the process of solving a physical problem by reducing no relevant variables and appealing it to the dimensional homogeneity. Processing: Experimental data interpretation Solve physical problems Presentation of equations Establish relative importance Physical modelling
Bulk modulus,
P= pressure = [M L-1 T-2] V=volume= L3 dP=change in pressure= [M L-1 T-2] dV=change in volume= L3
1) A solid ball has initial volume v; it is reduced by 20% when subjected to volumetric stress of 200N/m^2.Find the Bulk modulus of the ball.
Solution: V1=v, Volumetric strain= Final volume to initial volume *100 Volumetric stress related to volumetric strain=200N/m^2 K= (volumetric stress/Volumetric strain) = (200/0.02) =10^4N/m^2
2) The initial pressure of the system is 1.0110^5Pa. The system undergoes a change in pressure to 1.16510^5Pa. Find out the incompressibility of the system.
Solution: P1=1.0110^5Pa, P2=1.16510^5Pa, At 20°c change in volume=20% Bulk modulus=-dP/(dV/V) =- (1.01×10^5−1.165×10^5)/0.1 =1.55*10^5Pa.
3)5 litres of water is compressed at 20atm.Calculate the volume change in water.
Given:
K of water =20*10^8 N/m^2
Density of mercury=13600 kg/m^3 g=9.81m/s^2
Normal atm.=75cm of mercury
Original volume=5L=510^-3 m^3 Pressure dP=20atm=207510^-2136009.8 Solution: Volumetric stress= pressure intensity=dp K = dp/(dv/v)
Change in volume=dpV/K =5*10^-6 m^3 =5 cc.
Frequently asked questions:
What is the Bulk modulus of granite? 50Gpa.
Can incompressibility be negative: No.
Bulk modulus formula speed: The speed of sound depends on the Bulk modulus and density,
Bulk modulus of air at 20 c: Density of air at 20°C =1.21kg/m^3 Speed of sound=344m/s So, K can be calculated from the above formula,
K=143186.56N/m^2 Hence, K=0.14Mpa
Flexural modulus and incompressibility: Bulk modulus is the volumetric elasticity and is inversely proportional to the Compressibility. The object having incompressibility deform in all directions when the load is applied from all directions. Flexural modulus is the ability of the material to resist bending. Flexural modulus is the ratio of stress to the strain in flexural deformation.
Modulus of elasticity and incompressibility:
Modulus of elasticity is the ability of the material to resist deformation elastically when applied to external forces. Modulus of elasticity occurs under the elastic deformation region in stress-strain curve. Incompressibility is the volumetric elasticity and is inversely proportional to the Compressibility. The object having volume modulus deform in all directions when the load is applied from all directions
What material has the highest bulk modulus values? Diamond.
Why is the value of K maximum for a solid but a minimum for gases? Incompressibility is the resistance to compression of the substance. High pressure is required to compress the solid rather than compressing a gas. Hence the modulus of solid is maximum, and that of gas is low.
If Young’s modulus E is equal to the incompressibility K then what is the value of Poisson’s ratio:
K=E/3(1-2u) K=E 3(1-2u)=1 1-2u=1/3 u=1/3 So the value of Poisson’s ratio=1/3.
With increase in pressure, Does compressibility decrease or increase ?
As the pressure increases, volume of the body decreases. Decrease in volume gives rise to increase in incompressibility. Incompressibility is the ability to resist the compression of the body. So as it increases the compression of the body decreases. Hence the compressibility of the decreases.
What is the effect of the temperature increase?
As the temperature increases, The resistance to compression decreases. As the ability of compression of the body decreases, the bulk modulus decreases, leading to an increase in compressibility.
When the incompressibility of a material becomes equal to the shear modulus what would be the Poisson’s ratio:
2G(1+u)=3K(1-2u) as G=K, 2(1+u)=3(1-2u) 8u=1 u=1/8 Hence the value of Poisson’s ratio=1/8.
What will the velocity of sound in water m/s be if the volume modulus of water is 0.2*10^10 N/m 2:
c=2*10^6m/s.
To compress a liquid by 10% of its original volume, the pressure required is 2*10^5 N /m^2. What is the K (modulus of the liquid)?
“Thermal stress is the stress in the material due to the temperature change and this stress will lead to plastic deformation in the material.”
Thermal stress equation | Thermal stress formula:
The stress induced due to temperature change: σ=Eα∆T It is documented that changes in temp will cause elements to enlarge or contract and if increment in the length of a uniform bar of length L and ∆L is the change f length because of its temp has been changed from T0 to T then ∆L could be represented as ∆L = αL (T – T0) where α the coefficient of thermal expansion.
Thermal stress unit:
SI unit: N/m^2
Thermal hoop stress:
Stress generated for thermal change. Let us assume a thin tire having diameter ‘d’ has fitted on to the wheel of diameter ‘D’. If the temp of the tire has been changed in such a way that the diameter of the tire is increased and it has become equal to the diameter of the wheel and if temp of the tire is decreased to original, the tire diameter tries to return to its original dimension and because of this process a stress has been generated in the tire material. This stress is an example of Thermal hoop stress. so, the temperature difference=t degree. thermal strain=D-d/d Hoop stress= e. E Hence, Hoop stress=(D-d).E/d
The objective of the thermal analysis is to study the behavior of material after applying thermal loading and thermal stress. To study heat transfer within an object or between objects and thermal analysis is utilized for temp measurement, thermal gradient, and warmth flux distributions of the body.
Types of thermal analysis:
There are two sorts of thermal analysis:
Steady-state thermal analysis:
Steady-state thermal analysis aims to seek out the temperature or heat flux distribution in structures when an equilibrium is reached.
Transient thermal analysis:
Transient thermal analysis sets bent determine the time history of how the temperature profile and other thermal quantities change with time Also, thermal expansion or contraction of engineering materials often results in thermal stress in structures, which may be examined by conducting thermal-stress analysis.
Thermal stress importance:
Thermal Stress Analysis is essential to determine the thermal stresses due to temperature changes in structures. We can proceed to
Solve Equation K. T = q ⦁ To obtain the temperature change fields initially apply the temperature change ΔT as initial strain ⦁ The stress-strain relations due to temperature change were determined by first using 1D case materials. The thermal strain (or initial strain): εo = αΔT
Case study with ANSYS Workbench:
Material: Aluminum k = 170 W/(m · K) ρ = 2800 kg/m3; c = 870 J/(kg · K) E = 70GPa; v = 0.3 α = 22 × 10–6/°C Boundary conditions: Air temperature of 28°C; h = 30 W/(m2 · °C). Steady state: q′ = 1000 W/m2 on the base. Initial conditions: Steady-state: Uniform temperature of 28°C.
Start ANSYS workbench
Create a steady-state thermal analysis system:
Add new material: provided with all given data.
Launch design modeler program.
Create body
Launch the steady-state thermal program
Generate mesh
Apply boundary conditions.
Solve and retrieve results.
Thermal analysis of water-cooled engine:
The following steps are followed after finalizing the engine specification.
Design of water-core and head-core system.
Design of liner system. (Based on its parameters like bore, stoke and thickness etc.)
Design of water pump and installation.
Cooling system design and it’s subsystems such as radiators, fans, oil-cooler design.
Aspects of thermal analysis of engine block:
Cylinder head valve bridge water velocities (design of cross section in headwater core).
Piston and valve cooling aspect analysis.
Liner cavitation analysis.
Cylinder head gasket design analysis.
Thermal stress weathering:
Thermal stress weathering is the thermal fracture is a mechanical breakdown of rock due to thermal expansion or contraction caused by the change in temperature.
Effects of thermal stresses in weld joints: Thermal stress in welding and in bonded joints:
The temperature of the body is uniformly1 increased, The normal strain of the body is, x = y = z = α(T) Here, α is the co-efficient of thermal expansion. T is the temp variation. The stress is represented as σ1 =− E =−α(T)E in a similar means, if a consistent flat plate is restrained at the sides and also subjected to a constant temp rise. σ2 =− α(T)E(1−ν) The stresses σ1, σ2 are called thermal stresses. They arise due to a natural process during a clamped or restrained member.
Thermal stress equation for cylinder| Thermal stress in thick walled cylinder:
Thin-walled Cylinder:
Thick-walled cylinder:
Thermal stress relief process:
The process of heat treatment is used to decrease the residual thermal stresses in the materials. First, the part needs to be heated at 1100-1200degree F, leading to relief of the stresses and hold it there for an hour per inch of thickness, and then left to chill in tranquil air at temperature.
Thermal Expansion:
When a solid material experiences an increase in temp or temperature difference, the volume of the structure of solid material increases, this phenomenon is acknowledged as thermal expansion and this volume increment will lead to an increase in stress of the structure.
Coefficients of Thermal Expansion:
(Linear Mean Coefficients for the Temperature Range 0–100°C):
Aluminum: 23.9(10)−6 Brass, cast: 18.7(10)−6
Carbon steel: 10.8(10)−6 Cast iron: 10.6(10)−6
Magnesium :25.2(10)−6 Nickel steel: 13.1(10)−6
Stainless steel: 17.3(10)−6 Tungsten: 4.3(10)−6
Thermal stresses in composite bars formula: Thermal stress in compound bars:
Compound bars and composite bars, when undergoing temperature change, tend to contract or expand. Generally thermal-strain is a reversible process so material will return to its actual shape when the temp also decreased to its actual value, though there are some materials that does not behave according to thermal expansion and contraction.
Bars in series:
Thermal stress and strain: Thermal stress and strain definition:
The stress produced due to change in temperature is known as thermal-stress. Thermal stress=α(t2-t1).E The strain corresponding to thermal stress is known as thermal strain. Thermal strain=α(t2-t1)
Engine, radiator, exhaust, heat exchangers, power plants, satellite design, etc.
Thermal residual stress:
Differences within the temperatures during the manufacturing and dealing environment are the most explanation for thermal (residual) stresses.
Thermally induced stress
σ=E ∆L/L
Thermal stress calculation in pipe:
Pipes expand and contract due to variable temperatures. The coefficient of thermal expansion shows the rate of thermal expansion and contraction.
Factors affecting thermal stress:
Temperature gradient.
Thermal expansion contraction.
Thermal shocks.
Thermal stress is dependent on the thermal expansion coefficient of the material and if change of temp is more, then the stress will be more too.
Modulus of Elasticity in thermal expansion:
If the bar is prevented from completely expanding within the axial direction, then the typical compressive stress-induced is σ=E ∆L/L where E is the modulus of elasticity. So the thermal stress needed is, α = –αE (T – T0) In general, in an elastic continuum, the natural process is non-uniform throughout and this is a function of time and space usually. therefore the space coordinates (x, y, z), i.e. T = T(t, x, y, z).
Limitations of thermal stress analysis:
The body into account could also be restrained from expansion or movement in some regions, and external tractions could also be applied to other regions and stress calculation under such circumstances may be quite complex and difficult to compute. This also having following case is constrained.
Thin circular disks with equal temp difference.
Long circular cylinder. (This could be hollow and solid)
Sphere having radial temperature variation. (This could be hollow and solid)
Straight beam of arbitrary cross section.
Curved beam case.
Thermal stress problems and solutions:
1) A steel rod of length 20m having temperature 10-degree Celsius. Temperature is raised to 50 degrees Celsius. Find the thermal stress produced. Given: T1=10, T2=50, l=20, α=1210^-6, E=20010^9
Thermal stress=α(t2-t1).E
=1210^-6(50-10)20010^9
=9610^6 N/m^2.
FAQ/Short Notes:
What is the effect of thermal stresses ?
This has a significant effect on the materials and can lead to fracturing, and plastic deformation depends on the temperature and material type.
Which material can be used as thermal insulator and why?
Cellulose. Because it blocks air better than fiberglass and has low thermal conductivity.
What are the three most common types of heat stress ?
Types of heat stress commonly used:
Tangential
radial
axial.
How to calculate thermal stresses in glass?
Thermal stress in glass is varied at different temperatures.
Thermal stress and deformation:
Thermal deformation is the property of a substance to expand with heating and contract with cooling, normally kind of deformation because of temp change and this is stated by linear expansion coefficient α. α=ΔL/L×Δt Here, ⦁ α is that the linear expansion coefficient of a substance (1/K). ⦁ ΔL is that the expansion or contraction value of a specimen(mm). ⦁ L is the actual length. ⦁ Δt is the temperature difference measured in Kelvin or degree Celsius. The higher the thermal expansion coefficient, gives higher the value of thermal deformation.
Thermal stress weathering:
Thermal stress weathering is the thermal fracture a, mechanical breakdown of rock due to thermal expansion or contraction caused by the change in temperature.
What is the formula for thermal expansion stress and strain?
Thermal stress formula:
α(t2-t1). E
Thermal strain formula:
α(t2-t1).
What is the relationship between thermal stress and thermal strain?
Thermal stress and Thermal strain in 2D-3D cases: Temperature changes do not yield shear strains. In both 2-D and 3-D cases, the entire strain is often given by the following vector equation: ε = εe + εo And the stress-strain relation is given by σ = Eεe = E(ε − εo).
Which parameters have to be defined for isotropic materials for structural and thermal analysis in ANSYS?
Isotropic Thermal Conductivity
Material
Heat transfer coefficient
If strain causes stress, then in free thermal expansion why is stress absent even though there is thermal strain:
Stress is the internal resistance when applied to an external load. When the material is undergone any load or force, the material tries to resist the force leading to stress generation. If the material is undergoing free thermal expansion, the material won’t experience any internal stress leading to no stress generation.
What are some examples of thermal expansion in everyday life?
What is the application of thermal diffusivity in real world?
⦁ Insulation.
Does Hooke’s law fails in case of thermal expansion?
Hook’s law applies to a thermal expansion only when there is a restriction to the object undergoing thermal stress. If there is no applied stress, there won’t be any expansion and Hook’s law states that stress is directly proportionate to strain.
Why does copper have such a low thermal expansion ?
If the coefficient of thermal expansion is nearly equal for both steel and concrete, then why is a concrete structure considered a better firefighter If the coefficient of thermal expansion is almost equal for both steel and concrete, then why is a concrete structure considered a better firefighter: A concrete structure has low thermal conductivity and does not heat up quickly. Hence If the coefficient of thermal expansion is nearly equal for both steel and concrete, then why is a concrete structure considered a better firefighter.
Why we do static structure buckling modal thermal nonlinear fatigue based on stress and strain in Ansys?
It is a finite element method. To predict the exact and accurate strength of the structures, nonlinear analysis is performed. It takes into the changes in the parameters as the load is applied.
What does thermal capacity mean?
The thermal capacity of the material is the amount of heat required to change the material temp by unit mass of material.
What is the difference between the thermal expansion coefficients of steel and copper?
Thermal expansion coefficients 20 °C (x10−6 K−1) copper=17 steel=11-13.
What is the use of thermal conductivity?
Thermal conductivity is that the ability of an object to conduct heat. It measures the amount of heat that transfers through the material.
Do any materials have a zero thermal expansion coefficient?
There exist few materials which have zero thermal expansion coefficient. Mesopores.
Hooke’s law| Hooke’s law for thermal stress:
σth = Eϵth If the material is undergoing free thermal expansion, the fabric won’t experience any internal stress leading to no stress generation.
What is thermal shrinkage in concrete:
When the hot concrete cooled down at ambient temperature, the volume of the concrete reduces; this process is called thermal contraction or thermal shrinkage in concrete.
What is the best simulation and analysis software for mechanical engineering mainly structural analysis and dynamic analysis thermal not required?
Ansys, Nasttan, Abaqus, 1-deas NX, etc.
Thermal stress-strain: Why the bar does not bend when it is heated from the bottom with one end alone fixed:
Thermal stresses in cantilever beams:
Case1: Fixed free bar: If a rod is heated through temp rise, the rod will tend to expand by amount εo=αLΔT, if the rod is free at other ends, undergoes thermal expansion ε=αΔT, ε = εo, εe = 0, σ =E(ε- εo)=E(αΔT- αΔT)= 0 That is, there is no thermal stress in this case.
Case2: Fixed-fixed bar If there’s a constraint on the right-hand side, that is, the bar cannot expand to the proper, then we have: ε = 0, εe =−εo σ=E(ε-εo)=E(0- αΔT)= = −αΔT, σ = −EαΔT Thus, thermal stress exists.
Shear strains do not change only normal strains change.
If the temperature changes, the size of the body changes, though it will not change the shape of the body. So, considering this fact, the shear-strain of the body does not change.
Principal Stress is the maximum and minimum stresses derived from normal stress at an angle on a plane where shear stress is zero.
How to calculate principal stress ?
Principal stress equation | Principal stress formula: Maximum and Minimum principal stress equations:
Principal stress derivation | Determine the principal planes and the principal stresses
Normal stresses:
Differentiate,
“p” represents the principal plane.
There are two principal stresses, one at angle and other at Maximum and Minimum principal stresses:
R=
substitute in equation 1:
substitute value of R
Maximum and minimum normal stresses are the principal stresses:
The state of Stress:
The principal stress is the reference co-ordinate axes to the representation of the stress matrix and this stress components are the significance of the state of stress could be represented as,
Stress tensor:
Principal stresses from stress tensor and stress invariants |principal stress invariants
There are three principal planes at any stressed body, with normal vectors n, called principal directions where the stress vector is in the same direction as normal vector n with no shear stresses and these components depend on the alignment of the co-ordinate system.
A Stress vector parallel to the normal unit vector n is specified as,
Where, \\lambda represents constant of proportionality.
The principal stress vectors represented as,
The magnitude of three principal stresses gives three linear equations. The determinant of the coefficient matrix is equal to zero and represented as,
Principal stresses are the form of normal stresses, and the stress vector in the coordinate system is represented in the matrix form as follows:
I1, I2, I3 are the stress invariants of the principal stresses, The stress invariants are dependent on the principal stresses and are calculated as follows,
The principal stresses equation for stress invariants:
Principal stress trajectories | Principal directions of stress
Stress trajectories show the principal stress directions and their varying magnitude of the principal stresses.
Von mises stress vs principal stress
Von mises principal stress equation
Von Mises is the theoretical measure of the stress yield failure criterion in ductile materials. The positive or negative sign depends on the principal stresses. Principal stresses Boundary conditions:
Theories of failure give the yield stresses of the components subjected to multiaxial loading. Further, when it is compared with the yield point of the components shows the margin of the safety of the component.
Maximum principal stress is considered for brittle elements such as casting components (i.e., clutch housing, gearbox, etc.) Von-mises stress theory is based on shear strain energy theory is suggested for ductile materials like aluminum, steel components.
Why von mises stress is recommended for ductile and Principal Stress for brittle materials?
Failure of brittle materials used to uni-axial test is along a plane vertical to the axis of loading. So, the failure is because of normal stress generally. Out of all theories of failure, principal stress theory is based on normal Stress. Hence for brittle materials, principal stress theory is recommended,
Ductile materials fail at 45 degrees inclined at the plane of loading. So, the failure is due to shear stress. Out of all theories of failure shear strain energy or von-mises theory and maximum shear stress theory is based on shear stress. By comparison, von mises gives better results. Hence for ductile materials, von mises theory is recommended.
Different types of stress
Absolute principal Stress | Effective principal Stress:
The principal stresses are based on maximum Stress and minimum Stress. So, the range of the Stress is between the maximum and minimum Stress, (stress range is limited and less) and might lead to higher fatigue life. So, it is important to find out the effective principal Stress that gives the maximum value out of the two over the given period of time.
What is Maximum Normal Stress theory?
This states that brittle failure occurs when the maximum principal Stress exceeds the compression or the tensile strength of the material. Suppose that a factor of safety n is considered in the design. The safe design conditions require that.
Maximum principal stress equation
Where σ1, σ2, σ3 are three principal stresses, maximum, minimum, and intermediate, in the three directions, Sut and Suc are the ultimate tensile strength and the ultimate compressive strength, respectively.
To avoid brittle failure, the principal stresses at any point in a structure should lie within the square failure envelope based on the maximum normal stress theory.
Maximum principal stress theory |Maximum principal stress definition
consider two-dimensional stress state and the corresponding principal stresses such as σ1 >σ2 >σ3 Where σ3=0, σ2 may be compressive or tensile depending on the loading conditions where σ2 may be less or greater than σ3.
According to maximum principal stress theory, failure will occur when σ1 or σ2 =σy or σt The conditions are represented graphically with coordinates σ1,σ2. If the state of Stress with coordinates (σ1,σ2 ) falls outside of the rectangular region, failure will occur as per the maximum principal stress theory.
Mohr’s circle principal stresses
Explain the Mohr’s circles for the three-dimensional state of stress:
Consider a plane with a reference point as P. Sigma is represented as normal Stress and tau by the shearing stress on the same plane.
Take another plane with reference point Q representing sigma and tau as normal stress and shear stress, respectively. Different planes are passing through point p, different values of principal and shear stress.
For each plane n, a point Q with coordinates as shear stress and principal Stress can be located.
Determine the normal and shear stresses for point Q in all possible directions of n.
Obtain three principal stresses as maximum principal Stress, minimum principal Stress, and intermediate principal Stress and represent them in ascending order of the values of the stresses.
Draw three circles with diameters as the difference between the principal stresses.
The shaded area region is the Mohr’s circle plane region.
The circles represent the Mohr’s circles.
(σ1-σ3) and the associated normal stress is (σ1+σ3)
There are three normal stresses, so are three shear stresses.
The principal shear planes are the planes where shear stresses act and principal normal stress acts at a plane where shear stress is ‘0’ and shear stress act at a plane where normal principal stress is zero. The principal shear stress act at 45° to the normal planes.
The shear stresses are denoted by And the principal stresses are denoted by
Third principal stress
3rd principal Stress is relative to the maximum compressive stress due to the loading conditions.
3D principal stress examples:
For three-dimensional case, all three planes have zero shear stresses, and these planes are mutually perpendicular, and normal stresses have maximum and minimum stress values and these are the normal stresses that represent the principal maximal and minimal stress.
These principal stresses are denoted by, σ1,σ2, σ3. Example: 3D Stress in hub-a a steel shaft is force-fitted into the hub. 3D Stress in machine component.
Principal deviatoric stress:
Principal deviatoric stresses are obtained by subtracting mean Stress from each principal Stress.
Intermediate principal Stress:
The principal Stress, which is neither maximum nor minimum, is called intermediate Stress.
Principal stress angle | Orientation of principal stress: θP
Orientation of the principal Stress is computed by equating shear stress to zero in x-y direction at the principle plane rotated through an angle theta. Solve θ to get θP, the principal stress angle.
Important Frequently Asked Questions (FAQs):
Maximum principal stress theory is applicable for which material?
Answer: Brittle materials.
What are the 3 principal stresses?| What is maximum and minimum principal stress ?
Maximum principal Stress | Major principal stress: Most tensile (σ1) Minimum principal Stress | Minor Principal Stress: Most compressive (σ3) Intermediate principal Stress (σ2)
Principal Stress vs normal Stress:
Normal Stress is the force applied to the body per unit area. Principal Stress is the stress applied to the body having zero shear stress principal Stress is in the form of normal Stress giving maximum and minimum stresses on the principal plane.
Principal Stress vs Bending Stress:
Bending Stress is the Stress that occurs in the body due to the application of a large amount of load that causes the object to bend.
Principal Stress vs axial Stress:
Axial Stress and principal stress are the parts of normal stress.
What is the significance of principal Stress?
Principal Stress shows the maximum and minimum normal stress. Maximum normal Stress shows the component’s ability to sustain the maximum amount of force.
What are the principal stresses in a shaft with torque applied ?
The normal Stress is maximum or minimum, and shear stress is zero.
when shear stress=0,
Important Principal stress problems:
1) A rectangular stress vector having shear stress in XY direction of 60Mpa and normal tensile Stresses of 40Mpa.How to find principal stresses ?
Solution: Given: Principal stresses are calculated as,
σ1=100Mpa
σ2=-20Mpa
2)What are the coordinates of the center of Mohr’s circle for an element subjected to two mutually perpendicular stresses, one tensile of magnitude 80MPa and other compressive of magnitude 50MPa?
σx = 80 MPa, σy = -50 MPa Co-ordinates of center of Mohr’s circle =[ ½( σx + σy),0] = [(30/2),0] = (15,0)
3)A body was subjected to two mutually perpendicular stresses of -4MPa and 20MPa, respectively. Calculate the shear stress on the plane of the shear.
σx+σy /2= -4+20/2 = 8Mpa Radius= σ1-σ2/2 = 20-(-4)/2 = 12 where σx,σy are principal stresses at pure shear stress,σn=0 shear stress= squareroot12^2-8^2= 8.94Mpa.
4) Application of principal stress | Find the principal stresses for the following cases.
i)σx=30 Mpa, σy=0, \\tau=15Mpa.
solution:
σ1=36.21Mpa
σ2=-6.21Mpa ii)σx=0,σy=80MMpa, \\tau=60Mpa.
σ1=97Mpa
σ2=12.92Mpa
iii)\\tau=10Mpa, σx=50Mpa,σy=50Mpa.
σ1=60Mpa
σ2=40Mpa
5) The maximum principal Stress is given 100 Mpa, and the Minimum principal Stress is 50 MPa. Calculate the maximum shear stress and the orientation of the principal plane using Mohr’s circle.
Given: Maximum principal stress=100Mpa(compressive) Minimum principal stress=50 Mpa(compressive) Solution: Maximum shear stress is the radius of the Mohr’s circle, then we can write as follows.
R=
2θ = 90, from the maximum principal stress direction. So, the orientation at that point is θ = 45 from the maximum principal stress direction.
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Shear modulus is the ratio of the shear stress to the shear strain.
Shear modulus is defined as the measure of the elastic shear stiffness of the material and it is also acknowledged as ‘modulus of rigidity’. So, this parameter answers the question of how rigid a body is? Shear modulus is the material response to a deformation of body because of the shearing stress and this work as ‘the resistant of the material to shearing deformation’.
In the above figure, Side lengths of this element will not change, though the element experiences a distortion and shape of element is changing from the rectangle to a parallelogram.
Why do we calculate the modulus of rigidity of the material ? Shear modulus equation | Modulus of Rigidity equation
Shear modulus is the ratio of the shear stress to the shear strain, which is measures the amount of distortion, is the angle (lower case Greek gamma), always ex-pressed in radians and shear stress measured in force acting on an area. Shear modulus represented as, G= Where, G= shear modulus τ=shear stress = F/A ϒ = shear strain=
modulus of rigidity symbol
G or S or μ
What is the SI unit of rigidity modulus ?
Shear modulus units | Unit of modulus of rigidity
Pascal or usually denoted by Giga-pascal. Shear modulus is always positive.
What is the dimensional formula of modulus of rigidity ?
Shear modulus dimensions:
Shear modulus of materials:
Shear modulus of steel | Modulus of rigidity of steel
Structural steel:79.3Gpa Modulus of rigidity of stainless steel:77.2Gpa Modulus of rigidity of carbon steel: 77Gpa Nickel steel: 76Gpa
Modulus of rigidity of mild steel: 77 Gpa
What is the Rigidity modulus of copper in N/m2 ? Modulus of rigidity of copper wire:45Gpa Shear modulus of Aluminum alloy: 27Gpa A992 Steel: 200Gpa Shear modulus of concrete | Modulus of rigidity of concrete: 21Gpa Silicon shear modulus: 60Gpa Poly ether ether ketone (PEEK):1.425Gpa Fiberglass shear modulus: 30Gpa Polypropylene shear modulus: 400Mpa Polycarbonate shear modulus: 5.03Gpa Polystyrene shear modulus:750Mpa
Shear modulus derivation | Modulus of rigidity derivation
If the co-ordinate axes (x, y, z) coincides with principle axes and intended for an isotropic element, the principal strain axes at (0x,0y,0z ) point, and considering alternative frame of reference directed at (nx1, ny1, nz1) (nx2, ny2, nz2) point and in the meantime, Ox and Oy are at 90 degree to each other. So we can write that, nx1nx2 + ny1ny2 + nz1nz2 = 0 Here Normal stress (σx’) and the shear-stress ( τx’y’) has been computed utilizing Cauchy’s formulation. The resultant stress vector on the plane will have components in (x-y-z) as τx=nx1σ1. τy=nx2 σ2. τz=nx3 σ3.
The normal stress on this x-y plane has been computed as the summation of the component’s projections along the normal directions and we can elaborate as σn= σx=nx^2 σ1+nx^2 σ2+nx^2 σ3.
Similarly, the shear stress component in x and y plane nx2, ny2, nz2. Thus τxy=nx1nx2σ1+ny1ny2σ2+nz1nz3σ3 Considering as ε1, ε2, ε3 are the principal strains and the normal-strain is in x-direction, then we can write as εx’x’=nx1^2ε1+ny^2ε2+nz^2ε3. The shear strain is obtained as,
εx’=εy’
Substituting the values of σ1, σ 2 and σ 3,
τx’y’=μϒx’y’ Here, μ= shear modulus usually represented by term G. By taking other axis as Oz¢ with direction cosines (nx3, ny3, nz3) and at right-angle with the Ox¢ and Oy¢. This Ox¢y¢z¢ will create conventional forms an orthogonal set of axes, therefore we can write as,
strain components,
Elastic constants and their relations:
Young’s modulus E:
The young’s modulus is the measure of the stiffness of the body and acts as resistance of the material when the stress is functional. The young’s modulus is considered only for linear stress-strain behavior in the direction of stress.
E=
Poisson’s ratio (μ):
The Poisson’s ratio is the measure of the deformation of the material in the directions perpendicular to the loading. Poisson’s ratio ranges between -1 to 0.5 to maintain young’s modulus, shear modulus (G), bulk modulus positive. μ=-
Bulk Modulus:
Bulk modulus K is the ratio of the hydrostatic pressure to the volumetric strain and better represented as K=-v
E and n are generally taken as the independent constants and G and K could be stated as follows:
G=
K=
for an isotropic material, Hooke’s law is reduced to two independent elastic constants named as Lame’s co-efficient denoted as l and m. In terms of these, the other elastic constants can be stated as follows.
If bulk modulus considered to be +ve the Poisson’s ratio never be more than 0.5 (maximum limit for incompressible material). For this case assumptions are n = 0.5. 3G = E. K = ∞. ⦁ In terms of principal stresses and principal strains:
⦁ In terms of rectangular stress and strain components referred to an orthogonal coordinate system XYZ:
Young’s modulus vs shear modulus | relation between young’s modulus and modulus of rigidity
Hook’s law for shear stress: τxy=G.ϒxy where, τxy is represented as Shear-stress, Shear-modulus is G and Shear strain is ϒxy respectively. Shear-Modulus is resistant to the deformation of the material in response to shear stress.
Dynamic shear modulus of soil:
Dynamic shear modulus gives information about dynamic one. Static shear-modulus gives information about static one. These are determined using shear wave velocity and density of the soil.
Shear Modulus Formula soil
Gmax=pVs2
Where, Vs=300 m/s, ρ=2000 kg/m3, μ=0.4.
Effective shear modulus:
The ratio of the average stresses to average strains is the effective shear-modulus.
Modulus of rigidity of spring:
The modulus of rigidity of the spring is the measurement of the stiffness of the spring. It varies with the material and processing of the material.
For Closed Coil Spring:
For Open Coil Spring:
Where, R= mean radius of the spring. n = number of coils. d= diameter of the wire. N= shear modulas. W= load. δ=deflection. α= Helical angle of the spring.
Modulus of Rigidity- Torsion | Modulus of Rigidity Torsion test
The rate change of strain undergoing shear stress and is a function of stress subjected to torsion loading.
The main objective of the torsion experiment is to determine the shear-modulus. The shear stress limit is also determined using the torsion test. In this test, one end of the metallic rod is subjected to torsion, and the other end is fixed. The shear strain is calculated by using the relative angle of twist and gauge length. γ = c * φG / LG. Here c – cross-sectional radius. Unit of φG measured in radian. τ = 2T/(πc3),
shear-stress is linearly proportionate to shear-strain, if we measured at the surface.
Frequently Asked Questions:
What are the 3 Modulus of elasticity?
Young’s modulus:
This is the ratio of longitudinal stress to longitudinal strain and could be better represented as
The ratio of hydrostatic pressure to volume strain is called the Bulk modulus denoted as
Bulk Modulus(K)=volume stress/volume strain.
Modulus of Rigidity:
The ratio of shear stress to the shear strain of the material may well characterized as
Shear Modulus(η)=shear stress/shear strain.
What does a Poisson ratio of 0.5 mean?
Passion’s ratio ranges between 0-0.5.at small strains, an incompressible isotropic elastic material deformation gives Poisson’s ratio of 0.5. Rubber has a higher bulk modulus than the shear-modulus and Poisson’s ratio nearly 0.5.
What is a high modulus of elasticity?
The modulus of elasticity measures the resistance of the material to the deformation of the body and if modulus increse then material required additional force for the deformation.
What does a high shear modulus mean?
A high shear-modulus means the material has more rigidity. a large amount of force is required for the deformation.
Why is shear modulus important?
The shear-modulus is the degree of the stiffness of the material and this analyze how much force is required for the deformation of the material.
Where is shear modulus used ?| What are the applications of rigidity modulus?
The Information’s of shear-modulus is used any mechanical characteristics analysis. For calculation of shear or torsion loading test etc.
Why is shear modulus always smaller than young modulus?
Young’s modulus is the function of longitudinal strain and shear modulus is a function of transverse strain. So, this gives the twisting in the body whereas young’s modulus gives the stretching of the body and Less force is required for twisting than stretching. Hence shear modulus is always smaller than the young’s modulus.
For an ideal liquid, what would be the shear modulus?
In ideal liquids shear strain is infinite, the shear modulus is the ratio of shear stress to the shear strain. So the shear modulus of ideal liquids is zero.
When the bulk modulus of a material becomes equal to the shear modulus what would be the Poisson’s ratio ?
As per the relation between bulk modulus, shear modulus and poissons ratio, 2G(1+μ)=3K(1-2 μ) When, G=K 2(1+ μ)=3(1-2 μ) 2+2 μ=3-6 μ 8 μ=1 μ =1/8
Why the required shear stress to initiate dislocation movement is higher in BCC than FCC?
BCC structure has more shear stress values critical resolved than FCC structure.
What is the ratio of shear modulus to Young’s modulus if poissons ratio is 0.4, Calculate by considering related assumptions.
Which has a higher modulus of rigidity a hallow circular rod or a solid circular rod ?
Modulus of rigidity is the ratio of shear stress to the shear strain and shear stress is the Force per unit area. Hence shear stress is inversely proportional to the area of the body. solid circular rod is stiffer and stronger than the hollow circular rod.
Modulus of Rigidity vs Modulus of Rupture:
The modulus of rupture is the fracture strength. It is the tensile strength of the beams, slabs, concrete, etc. Modulus of rigidity is the strength of material to be rigid. It is the stiffness measurement of the body.
If the radius of the wire is doubled how will the rigidity modulus vary? Explain your answer.
Modulus of rigidity does not vary by change of the dimensions and hence modulus of rigidity remains the same when the radius of the wire is doubled.
Coefficient of viscosity and modulus of rigidity:
The coefficient of viscosity is the ratio of the shear stress to the rate of shear strain which varies by the velocity change and displacement change and the modulus of rigidity is the ratio of shear stress to the shear strain where shear strain is due to transverse displacement. The ratio of shear-modulus to the modulus of elasticity for a Poisson’s ratio of 0.25 would be For this case we may consider that. 2G(1+μ)=3K(1-2 μ) 2G(1+0.25) =3K(1-0.5) 2G(1.25)=3K(0.5) G/K=0.6 Answer = 0.6
What material has modulus of rigidity equal to about 0.71Gpa ?
Answer: Nylon(0.76Gpa) Polymers range between such low values.
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Spring constant is the measure of the stiffness of the spring. Springs having higher stiffness are more likely difficult to stretch. springs are elastic materials. when applied by external forces spring deform and after removal of the force, regains its original position. The deformation of the spring is a linear elastic deformation. Linear is the relationship curve between the force and the displacement.
Spring constant formula:
F= -Kx
Where,
F= applied force,
K= spring constant
x = displacement due to applied load from normal position.
Spring constant units:
the spring-constant represented as K, and it’s unit is N/m.
How to find spring constant?
Spring constant equation:
The Spring-constant is determined according to the Hooke’s law stated as below:
The applied force on the springs is directly proportional to the displacement of the spring from the equilibrium.
The proportionality constant is the spring’s constant. The spring force is in opposite direction of force. So, there is a negative sign between the relation of the force and the displacement.
F= -Kx
Therefore,
K= -F/x(N/m)
Dimension of spring constant:
K=-[MT^-2]
Constant force spring:
Constant force spring is the spring that does not obey Hooke’s law. The spring has the force it exerts over its motion range is constant and does not vary by any means. Generally, these springs are constructed as springs rolled up such that the spring is relaxed when fully rolled up and after unrolling the restoring force takes place as the geometry remains constant as the spring unrolls. The constant force spring exerts the constant force for unrolling due to the change in radius of curvature is constant.
The constant spring force does not give constant force at all the time. Initially, it has a finite value and after the spring is deflected 1.25 times its diameter it reaches full load and maintains the constant force in the spring despite the deformation. These springs are made with metal strips and not with wires The springs are made up of materials like stainless steel, High carbon steel, etc. springs give tension in the linear direction.
The performance, corrosion elements, temperature affects the fatigue of such springs. They are more likely to have a lifespan of 2500 cycles to more than one million depending on the size and load applied.
Spring constant examples
Spring constant of a rubber band:
Rubber band acts like spring within certain limitations. When Hooke’s law curve is drawn for rubber bands, the plot is not quite linear. But if we stretch the band slowly it might follow Hooke’s law and have spring-constant value. Rubber band can stretch only its elastic limit that
also depends on the size, length, and quality.
Spring constant values:
Spring constant value is determined using the Hooke’s law. As per the Hooke’s law, when spring is stretched, the force applied is directly proportional to the increase in length from the original position.
How to determine spring constant?
F=-Kx
K=-F/x
Spring constants of materials :
Spring constant for Steel=21000 kg/m3
Spring constant for Copper = 12000 kg/m3
How to find spring constant from graph ?
Spring constant graph:
Can the spring constant be negative?
This can not be negative.
Spring constant formula with mass:
T=
where,
T= period of spring
m=mass
k=spring constant
Effective spring constant:
Parallel: When two massless springs which obey Hooke’s law and connected through the thin vertical rods at the ends of the springs, connecting two ends of springs are said to be parallel connection.
The constant force direction is perpendicular to the force direction.
Spring constant K written as,
K=K1+K2
Series:
When springs are connected to each other in a series manner such that the total extension combination is the sum of total extension and spring’s constant combination all the springs.
The Force is applied at the end of the end spring. The force direction is in the reverse direction as the springs compressed.
Hooke’s law,
F1=k1x1
F2=k2x2
x 1+ x 2 =
Equivalent spring constant:
K =
Torsional spring constant:
A torsion spring is twisted along the axis of the spring.When it is twisted it exerts torque in opposite direction and is proportional to the angle of the twist.
A torsional bar is a straight bar that is subjected to twisting gives shear stress along the axis torque applied at its end.
Examples:
Helical torsional spring, torsion bar, torsion fiber
Applications:
clocks-clocks has spring coiled up together in a spiral, It is a form of helical torsional spring.
torsional spring constant formula | Torsion coefficient
Within elastic limit torsional springs obey Hook’s law as it twisted within elastic limit,
Torque represented as,
τ = -Kθ
τ = − κ θ
K is displacement called the torsional spring coefficient.
The -ve sign specifies that torque is acting in reverse to twist direction.
The energy U, in Joules
U= ½*Kθ^2
Torsional balance:
Torsion balance is torsional pendulum. It works as a simple pendulum.
To measure the force, first, need to find out the spring’s constant. If the force is low, it’s difficult to measure the sparing constant. One needs to Measure the resonant vibration period of the balance.
The frequency depends on the Moment of Inertia and the elasticity of the material. So, the frequency is chosen accordingly.
Once the Inertia is calculated, springs constant is determined,
F=Kδ/L
Harmonic Oscillator:
Harmonic oscillator is a simple harmonic oscillator when undergoes deformation from the original equilibrium position experiences restoring force F is directly proportional to the displacement x.
Mathematically written as follows,
F= -Kx
Torsional Spring rate:
Torsional spring rate is the force of spring travelled around 360 degrees. This can be further calculated by the amount of force is divided by 360 degrees.
Factors affecting spring constant:
Wire diameter: The diameter of the wire of the spring
Coil diameter: The diameters of the coils, depending on the stiffness of the spring.
Free length: Length of the spring from equilibrium at rest
The number of active coils: The number of coils that compress or stretch.
Material: Material of the spring used to manufacture.
Constant torque spring:
Constant torque spring is a type of spring that is a stressed constant force spring traveling between 2 spools. After the release of the compressed spring torque is calculated from the output spool as the spring returns back to its original equilibrium position in the storage spool
Spring constant range:
k = k’ δ’/δ,
K Varies from
Minimum= 0.9N/m
Maximum=4.8N/m
Spring’s constant depends on the number of turns n.
Ideal spring constant:
The spring constant is the measure of the stiffness of the springs. The larger the value of k, the stiffer is the spring and it is difficult to stretch the spring. Any spring that obeys Hooke’s law equation is said to be an ideal spring.
Constant force spring assembly:
A Constant Force Spring is mounted on a drum by wrapping it around the drum. The spring has to be tightly wrapped. Then the free end of the spring is attached to the loading force such as in a counterbalance uses or vice-versa.
The drum diameter should be larger than the inside diameter.
Range: 10-20% drum diameter> Inside diameter.
One and a half-wrapped spring should be on the drum at extreme extension.
The strip will be unstable at the larger extensions so it is advisable to keep it smaller.
Pulley diameter must be greater than the original diameter.
FAQs:
Why is spring constant important?
The spring-constant is important as it shows the basic material property. This gives exactly how much force is required to deform any spring of any material. The higher spring’s constant shows the material is stiffer and the lower spring’s constant shows the material is less stiff.
Can spring constant change?
Yes. spring-constant can change as per the force applied and the extension of the material.
Can spring constant be 0 ?
No. The spring-constant cannot be zero. If it is zero, the stiffness is zero.
Can spring constant has negative value?
No. the Spring-constant always has a positive value.
When are Young’s modulus and Hooke’s spring constant equal?
When the ratio of the length to that area of the spring is unity, then the young’s modulus and the spring’s constant value will be equal.
Spring constant is represented as, K=-F/x,
The above mentioned equation shows the relationship between springs constant and the extension of the spring for the same applied force
Why a spring is cut in half, its spring constant changes?
This is inversely proportional to the extension of the spring. when the spring is cut into half, the length of the spring reduces hence the spring’s constant will be doubled.
Does Newton’s third law fails with a spring ?
Answer : No
Spring constant problems:
Q1) A spring is stretched by 20cm and a 5kg load is added to it. Find the spring constant.
Given:
Mass m = 5kg.
Displacement x=20cm.
Solution:
1.Find out the force applied on the spring
F= m*x
= 5*20*10^-2
= 1N.
The load applied on the spring is 1N. So, the spring will apply an equal and opposite load of -1N.
2. Find out the spring constant
K= -F/x
=-(-1/20*10^-2)
= 5N/m
The constant of the spring is 5N/m.
Q2)A force of 25 KN is applied on the spring of spring constant of 15KN/m.Find out the displacement of the spring.
Given:
Applied force= 2.5KN
Spring-constant=15KN/m
Solution:
1.Find out the displacement of the spring
The spring will apply equal and opposite force of -2.5KN
F=-Kx
X=-F/K
= – 2.5/15
= 0.167m
Hence the spring is displaced by 16.67cm.
Q3)A spring with a force constant of 5.2 N/m has a relaxed length of 2.45m and spring’s perpendicular length 3.57m. When a mass is attached to the end of the spring and permitted to rest. What is elastic potential energy stored in the spring?
Solution:
Given:
Force constant= 2.45m
x = 2.45m
L= 3.57m
Force constant spring:
F= -Kx
The work was done due to stretching of the spring= Elastic potential energy of the spring.
W=Kx^2/2
Extension x = 3.57-2.45
=1.12
W=5.2*1.12^2/2
=3.2614 J.
Q4) A massless spring with force constant k 400N/m hangs vertically from the ceiling. A 0.2 kg block is attached to the end of the spring and released. The highest elastic strain energy kept in the spring is (g= 10m/s^2).
Given:
Force constant= 400N/m
m = 0.2kg
g= 10m/s^2
Solution:
Maximum elastic strain energy=1/2*K*x^2
=
=0.02J
Spring constant with multiple springs
A spring is cut into 4 equal parts and 2 are parallel What is the new effective spring constant of these parts?
The spring’s constants of the four springs is k1, k2, k3, k4
respectively,
Parallel:
Equivalent spring’s constant (k5) = k1 + k2
Series;
Total equivalent springs constant of the system:
K=
If a spring constant of 20N /m and it is stretched by 5cm what is the force acting on the spring:
Given:
K=2 N/m.
x = 5cm.
According to Hooke’s law,
F= -Kx
= – 20*5*10^-2
=-1N
Spring force is in opposite direction
Hence spring force = 1N.
An object with a weight of 5.13 kg placed on top of a spring compresses it by 25m What is the force constant of the spring How high will this object go when the spring releases its energy.