RESISTENCIA DE MATERIALES
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1. The document contains two engineering problems involving the calculation of stresses in mechanical parts. 2. The first problem calculates the increase in a circular tube's torque capacity if its semicircular cross section is inverted. The capacity increases by a factor of 1.66. 3. The second problem determines the stress in a simply supported floorboard experiencing bending from a moment applied 3 degrees off the z-axis. It finds the stress is 925 psi, higher than the 800 psi if applied along z, and calculates the neutral axis angle as -25.3 degrees.
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About:
A helium boosting and decanting system is typically used in various industrial applications, particularly in the production and handling of gases, including helium including leak test of reciprocating cylinder. Heres a brief overview of its components and functions:
Components
1. Helium Storage Tanks: High-pressure tanks that store helium@ 150 bars.
2. Boosting Pumps: Designed to boost helium pressure up to 150 bar, ensuring efficient flow throughout the system.
3. Decanting Unit: Separates liquid helium from gas, facilitating decanting at pressures of up to 2 bars.
4. Pressure Regulators: Maintain and control the pressure of helium during transport.
5. Control Valves: automatic control valve is provided for the flow and direction of helium through the system.
6. Piping and Fittings: High-quality, corrosion-resistant materials for safe transport.
Functions
Boosting Pressure: The system boosts helium pressure up to 150 bar for various applications.
Decanting: Safely decants helium, separating liquid from gas at pressures of up to 2 bar.
Safety Measures: Equipped with relief valves and emergency shut-off systems to handle high pressures safely.
Monitoring and Control: Sensors and automated controls monitor pressure and flow rates.
Application:
Cryogenics: Cooling superconducting magnets in MRI machines and particle accelerators.
Welding: Used as a shielding gas in welding processes.
Research: Crucial for various scientific applications, including laboratories and space exploration.
Key Features:
Helium Storage & Boosting System
Decanting System
Pressure Regulation & Monitoring
Valves & Flow Control
Filtration & Safety Components
Structural & Material Specifications
Automation & Electrical Components
RESISTENCIA DE MATERIALES
- 1. Escuela Profesional de Mec叩nica, Mec叩nica El辿ctrica y Mecatr坦nica R.C. Hibbeler, 2017, Mec叩nica de Materiales II, 9na.Ed. Problemas 5-109 6-113 2014203101 PINTO VALDIVIA ROBIN MARCOS SECCION B
- 2. 1. Problema 5-109 Para un esfuerzo promedio m叩ximo dado, determine el factor por el cual se incrementa la capacidad de carga de par si la secci坦n semicircular se invierte de la l鱈nea discontinua posici坦n a la secci坦n que se muestra. El tubo tiene 0.1 pulg. de espesor. Enunciado
- 3. 1. Problema 5-109 Resoluci坦n Para un esfuerzo promedio m叩ximo dado, determine el factor por el cual se incrementa la capacidad de carga de par si la secci坦n semicircular se invierte de la l鱈nea discontinua posici坦n a la secci坦n que se muestra. El tubo tiene 0.1 pulg. de espesor. Am=(1.10)(1.75) 0.552 2 Am=1. 44982 A卒 =(1.10)(1.75) + 0.552 2 Am=2.4002 2 = 2＄基 = 2 基
- 4. 1. Problema 5-109 Para un esfuerzo promedio m叩ximo dado, determine el factor por el cual se incrementa la capacidad de carga de par si la secci坦n semicircular se invierte de la l鱈nea discontinua posici坦n a la secci坦n que se muestra. El tubo tiene 0.1 pulg. de espesor. ＄ = 2 基卒 2 基 = 基卒 基 = 2.4002 1.4498 = 1.66 Resoluci坦n
- 5. 2. Problema 6-113 El tablero se utiliza como viga de piso simplemente apoyada. Si se aplica un momento flector de M = 800 lb 揃 pie a 3 属 de la eje z, determine la tensi坦n desarrollada en el tablero en el esquina A. Compare este estr辿s con el desarrollado por el mismo momento aplicado a lo largo del eje z (u = 0 属). 多Cu叩l es el 叩ngulo a para el eje neutro cuando u = 3 属? Comentario: normalmente, piso Las tablas se clavar鱈an en la parte superior de las vigas para que u 0 属 y no se producir鱈a la alta tensi坦n debida a la desalineaci坦n. Enunciado
- 6. 2. Problema 6-113 El tablero se utiliza como viga de piso simplemente apoyada. Si se aplica un momento flector de M = 800 lb 揃 pie a 3 属 de la eje z, determine la tensi坦n desarrollada en el tablero en el esquina A. Compare este estr辿s con el desarrollado por el mismo momento aplicado a lo largo del eje z (u = 0 属). 多Cu叩l es el 叩ngulo a para el eje neutro cuando u = 3 属? Comentario: normalmente, piso Las tablas se clavar鱈an en la parte superior de las vigas para que u 0 属 y no se producir鱈a la alta tensi坦n debida a la desalineaci坦n. Mz=800cos3属 Mz=798.904 lb.ft My=800sen3属 Mz=41.869lb.ft Iz= 1 12 (2)(63) Iz=36 4 Iy= 1 12 (6)(23 ) Iy=4 4 Resoluci坦n
- 7. El tablero se utiliza como viga de piso simplemente apoyada. Si se aplica un momento flector de M = 800 lb 揃 pie a 3 属 de la eje z, determine la tensi坦n desarrollada en el tablero en el esquina A. Compare este estr辿s con el desarrollado por el mismo momento aplicado a lo largo del eje z (u = 0 属). 多Cu叩l es el 叩ngulo a para el eje neutro cuando u = 3 属? Comentario: normalmente, piso Las tablas se clavar鱈an en la parte superior de las vigas para que u 0 属 y no se producir鱈a la alta tensi坦n debida a la desalineaci坦n. = ю 腫 + 腫 = 798.904(12)(3) 腫 + 41.869(12)(1) 腫 =925psi tan 留 = 腫 腫 tan牢 tan 留 = 36 4 tan 3 属 留 = 25.3属 = = 800(12)(3) 36 =800 psi Resoluci坦n