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Statics problem just need the highlighted questions which is the " Pin Radius for G and H " just one question

Part 3: Design cross-section and material for member ABC, pins G and H and select the diameter of hydraulic cylinder GH In Part 3 of this OEMP project you will be designing the material and cross-section of member ABC, the material and cross-section of pins G and H, as well as the diameter of the hydraulic cylinder based on your static equilibrium analysis from Part 2. For this project, all team’s will be using US Customary units (pounds, slugs, feet, inches). Before beginning the design process, your team will need to develop a list of at least four design criteria that will be used to guide your material selection for member ABC and pins G and H. Do not limit yourself to only criteria included on the Table of Material Properties. Include a brief justification/explanation of the significance of each design criterion placed on your list. (Note: the material itself is NOT a design criterion, rather it is the design criteria that will help you select the material.) Mechanics of Materials (also called Mechanics of Solids or Strength of Materials) is a branch of engineering that studies the behavior of solid objects subjected to stress and strain due to applied loads. The task of an engineer is to properly design a structure or machine to withstand the anticipated stress and strain that occurs under loading without failing. Materials subjected to loadings undergo stress. You will have the opportunity to learn about stress and strain in more detail in courses such as Mechanics of Solids (EAS209), Analysis of Structures (MAE315), and Finite Element Structural Analysis (CIE426). However, for the purpose of this project, you will be introduced to two general types of stress; normal stress (¢) and shear stress (7). Normal stress is the result of an axial load, a bending load, or both. Typically, the bending load produces the largest normal stress. Shear stress is the result of a shear force producing an action similar to the blades of a pair of scissors sliding over each other. The strength of a material, and subsequently its resistance to failure, is often described by its yield strength and ultimate strength A material’s yield strength (ay) is the stress beyond which plastic deformation (permanent deformation) occurs. The ultimate strength (oy), or ultimate tensile strength (UTS), is the maximum stress a material can withstand prior to fracture. (You will be referencing the Table of Material Properties posted on UBLearns for this part of the project.) Material selection and sizing are important decisions made by engineers as part of the design process. Mechanical failure can occur in a variety of ways including plastic deformation, buckling, fracture, and joint failure. The internal bending moment in a beam, or member, often causes the most amount of stress leading to possible failure. Bending stress (0) may be calculated using the formula = No (Eq. 1) where M is the bending moment acting at a given location along the member, / is the centroidal area moment of inertia of the beam’s cross section, and ¢ is the distance from the centroidal axis, or neutral axis, to the point of interest. Figure 1. Cross-sectional diagram showing a red differential element for which bending stress may be computed using oc = = Bending stress is zero along the beam’s neutral axis, increasing linearly as the point of interest moves away from this axis reaching a maximum along the top or bottom edge of the cross-section, or the farthest point away from the neutral axis. For example, consider the following circular and rectangular cross-sectional areas shown in Figure 2. Bending stress at the top edge of each cross- : . Mi : : section is given by Opa = it where c¢ represents the distance from the neutral axis to the element lx being analyzed within the cross-sectional cut and I, is the area moment of inertia about the neutral axis perpendicular to the dimension c. y 1 ir - . - [a] 1 | | Figure 2. Circular cross-sectional area (left) and rectangular cross-sectional area (right) showing an element farthest from the neutral axis having maximum stress. Consider various cross-sectional designs for member ABC and whether you want your cross- section to be solid or hollow. Discuss the relationship between the cross-section and your list of design criteria. Design and sketch the cross-section of member ABC. Use the equation for bending stress (Eq. 1) to help you design a cross-section capable of withstanding the maximum bending load experience within member ABC for your Pool Lift. Use Eq. 1 to relate the maximum bending STESS, Opqx, to the cross-sectional dimension(s). Based on your team’s list of design criteria and the bending stress equation (while referring to the Table of Material Properties), select the material and cross-sectional dimensions of vertical member ABC for your Pool Lift. Consider the impact of selecting a standard size versus a custom size for this member. Justify your material choice and cross-sectional design by showing mathematically that the maximum bending stress in member ABC does not exceed the yield strength (a) based on your design. Evaluate whether the cross-sectional size of member ABC is physically reasonable and your Pool Lift will still be considered “portable”. If not, consider revising your design. The next components of the Pool Lift your team will be designing are pins G and H. All teams will be modeling both pin supports as a Clevis Joint. A Clevis Joint, as pictured in the left image of Figure 3, is a U-shaped coupling with holes on either side through which the pin passes allowing for rotation. This type of joint produces a balanced distribution of the applied force across the U- shape. The advantage of a Clevis joint, also referred to as a double-shear joint, is that the force at the pin is applied over two cross-sectional areas rather than one, thereby reducing the amount of shear stress at either point of contact in half (right image in Figure 3). « > Double shear Figure 3. Clevis joint (left) and double-shear diagram (right). Next, you will be selecting the material and diameter of the pins at joints G and H. Calculating shear stress can be considerably more complicated than the equations for computing bending stress. However, for this analysis, a simplified equation will be used where the average shear stress, caused by a shearing force parallel to the pin cross-section, is given by _ Finear Tavg = A where Fypeqr represents the shear force acting across the surface area and 4 is the cross-sectional area transmitting the shear stress. Each pin will be modeled as a cylinder having a solid circular cross-sectional area, A = mr?. Joints G and H will be modeled as clevis joints. Assuming that the average shear stress in the pins must be less than half the material’s ultimate strength (0.50), select the material and diameter of pins G and H based on the maximum force experienced at these joints. mathematically that the pins will not break under the anticipated shear force. Consider whether your selected pin radius is reasonable. Be sure your units are consistent throughout your analysis. (Note: the unit ksi represents a kip per square inch or 1000 pounds per square inch) Finally, you will need to select the diameter of your hydraulic cylinder GH. The hydraulic force is related to its operating pressure by F = pA, where F is the force experienced by the hydraulic, p is the operating pressure, and A is the hydraulic piston’s circular cross-sectional area. Recall, the hydraulic cylinder is constrained by a maximum operating pressure of 1000 psi. Justify your choice of diameter by showing mathematically that your hydraulic cylinder can support the maximum load experienced by the hydraulic in your Pool Lift. Consider whether your design is reasonable. Susk need the Pin Radius (Pin & wal) with =f deps ard EBD. Need “he yoalue of Pin fFodiss. Pool Lift Loading Values from Part 2 Max Bending Moment in Member Max force in hydraulic GH (Ib) ABC (lbrin) 3(3(24 Thin| 2%63 Ib Cross-sectional sketch of member ABC (Include dimensions and label the neutral axis on your sketch using a dashed line) - Member ABC Cross-section shape Cross-section dimensions y Material Material’s yield x strength (ay) (ksi) Y Pins G and H Hydraulic Cylinder GH Jus > ce | Pin radius (inches) Vi Diameter of hydraulic n — ff (inches) hist Material Material’s ultimate strength (ay) (ksi) lo=36im , lpe=' in, kyz 9m, haz1\§ in 8 = (6. X= G6 ah hse Pd = -51.0 we Ea nSe Leglees 63 / mes ‘ egy Wem Le ney hock = ¥ Mn J Didnnce Lom wSeY Figure 1: Side View (y-z plane) of Pool Lift with Lifting Assembly in a raised position where the a measrires the angle of member EB with respect to the horizontal.

Part 3: Design cross-section and material for member ABC, pins G and H and select the diameter of hydraulic cylinder GH In Part 3 of this OEMP project you will be designing the material and cross-section of member ABC, the material and cross-section of pins G and H, as well as the diameter of the hydraulic cylinder based on your static equilibrium analysis from Part 2. For this project, all team’s will be using US Customary units (pounds, slugs, feet, inches). Before beginning the design process, your team will need to develop a list of at least four design criteria that will be used to guide your material selection for member ABC and pins G and H. Do not limit yourself to only criteria included on the Table of Material Properties. Include a brief justification/explanation of the significance of each design criterion placed on your list. (Note: the material itself is NOT a design criterion, rather it is the design criteria that will help you select the material.) Mechanics of Materials (also called Mechanics of Solids or Strength of Materials) is a branch of engineering that studies the behavior of solid objects subjected to stress and strain due to applied loads. The task of an engineer is to properly design a structure or machine to withstand the anticipated stress and strain that occurs under loading without failing. Materials subjected to loadings undergo stress. You will have the opportunity to learn about stress and strain in more detail in courses such as Mechanics of Solids (EAS209), Analysis of Structures (MAE315), and Finite Element Structural Analysis (CIE426). However, for the purpose of this project, you will be introduced to two general types of stress; normal stress (¢) and shear stress (7). Normal stress is the result of an axial load, a bending load, or both. Typically, the bending load produces the largest normal stress. Shear stress is the result of a shear force producing an action similar to the blades of a pair of scissors sliding over each other. The strength of a material, and subsequently its resistance to failure, is often described by its yield strength and ultimate strength A material’s yield strength (ay) is the stress beyond which plastic deformation (permanent deformation) occurs. The ultimate strength (oy), or ultimate tensile strength (UTS), is the maximum stress a material can withstand prior to fracture. (You will be referencing the Table of Material Properties posted on UBLearns for this part of the project.) Material selection and sizing are important decisions made by engineers as part of the design process. Mechanical failure can occur in a variety of ways including plastic deformation, buckling, fracture, and joint failure. The internal bending moment in a beam, or member, often causes the most amount of stress leading to possible failure. Bending stress (0) may be calculated using the formula = No (Eq. 1) where M is the bending moment acting at a given location along the member, / is the centroidal area moment of inertia of the beam’s cross section, and ¢ is the distance from the centroidal axis, or neutral axis, to the point of interest. Figure 1. Cross-sectional diagram showing a red differential element for which bending stress may be computed using oc = = Bending stress is zero along the beam’s neutral axis, increasing linearly as the point of interest moves away from this axis reaching a maximum along the top or bottom edge of the cross-section, or the farthest point away from the neutral axis. For example, consider the following circular and rectangular cross-sectional areas shown in Figure 2. Bending stress at the top edge of each cross- : . Mi : : section is given by Opa = it where c¢ represents the distance from the neutral axis to the element lx being analyzed within the cross-sectional cut and I, is the area moment of inertia about the neutral axis perpendicular to the dimension c. y 1 ir - . - [a] 1 | | Figure 2. Circular cross-sectional area (left) and rectangular cross-sectional area (right) showing an element farthest from the neutral axis having maximum stress. Consider various cross-sectional designs for member ABC and whether you want your cross- section to be solid or hollow. Discuss the relationship between the cross-section and your list of design criteria. Design and sketch the cross-section of member ABC. Use the equation for bending stress (Eq. 1) to help you design a cross-section capable of withstanding the maximum bending load experience within member ABC for your Pool Lift. Use Eq. 1 to relate the maximum bending STESS, Opqx, to the cross-sectional dimension(s). Based on your team’s list of design criteria and the bending stress equation (while referring to the Table of Material Properties), select the material and cross-sectional dimensions of vertical member ABC for your Pool Lift. Consider the impact of selecting a standard size versus a custom size for this member. Justify your material choice and cross-sectional design by showing mathematically that the maximum bending stress in member ABC does not exceed the yield strength (a) based on your design. Evaluate whether the cross-sectional size of member ABC is physically reasonable and your Pool Lift will still be considered “portable”. If not, consider revising your design. The next components of the Pool Lift your team will be designing are pins G and H. All teams will be modeling both pin supports as a Clevis Joint. A Clevis Joint, as pictured in the left image of Figure 3, is a U-shaped coupling with holes on either side through which the pin passes allowing for rotation. This type of joint produces a balanced distribution of the applied force across the U- shape. The advantage of a Clevis joint, also referred to as a double-shear joint, is that the force at the pin is applied over two cross-sectional areas rather than one, thereby reducing the amount of shear stress at either point of contact in half (right image in Figure 3). « > Double shear Figure 3. Clevis joint (left) and double-shear diagram (right). Next, you will be selecting the material and diameter of the pins at joints G and H. Calculating shear stress can be considerably more complicated than the equations for computing bending stress. However, for this analysis, a simplified equation will be used where the average shear stress, caused by a shearing force parallel to the pin cross-section, is given by _ Finear Tavg = A where Fypeqr represents the shear force acting across the surface area and 4 is the cross-sectional area transmitting the shear stress. Each pin will be modeled as a cylinder having a solid circular cross-sectional area, A = mr?. Joints G and H will be modeled as clevis joints. Assuming that the average shear stress in the pins must be less than half the material’s ultimate strength (0.50), select the material and diameter of pins G and H based on the maximum force experienced at these joints. mathematically that the pins will not break under the anticipated shear force. Consider whether your selected pin radius is reasonable. Be sure your units are consistent throughout your analysis. (Note: the unit ksi represents a kip per square inch or 1000 pounds per square inch) Finally, you will need to select the diameter of your hydraulic cylinder GH. The hydraulic force is related to its operating pressure by F = pA, where F is the force experienced by the hydraulic, p is the operating pressure, and A is the hydraulic piston’s circular cross-sectional area. Recall, the hydraulic cylinder is constrained by a maximum operating pressure of 1000 psi. Justify your choice of diameter by showing mathematically that your hydraulic cylinder can support the maximum load experienced by the hydraulic in your Pool Lift. Consider whether your design is reasonable. Susk need the Pin Radius (Pin & wal) with =f deps ard EBD. Need “he yoalue of Pin fFodiss. Pool Lift Loading Values from Part 2 Max Bending Moment in Member Max force in hydraulic GH (Ib) ABC (lbrin) 3(3(24 Thin| 2%63 Ib Cross-sectional sketch of member ABC (Include dimensions and label the neutral axis on your sketch using a dashed line) - Member ABC Cross-section shape Cross-section dimensions y Material Material’s yield x strength (ay) (ksi) Y Pins G and H Hydraulic Cylinder GH Jus > ce | Pin radius (inches) Vi Diameter of hydraulic n — ff (inches) hist Material Material’s ultimate strength (ay) (ksi) lo=36im , lpe=' in, kyz 9m, haz1\§ in 8 = (6. X= G6 ah hse Pd = -51.0 we Ea nSe Leglees 63 / mes ‘ egy Wem Le ney hock = ¥ Mn J Didnnce Lom wSeY Figure 1: Side View (y-z plane) of Pool Lift with Lifting Assembly in a raised position where the a measrires the angle of member EB with respect to the horizontal.

Apr 28, 2023

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