Homework Help is Here – Start Your Trial Now!
Engineering
Mechanical Engineering
Question 3 - Torsion ( A composite shaft is shown in the figure below which is fixed against rotations at ends A and E. Section AC is a solid cylindrical shaft made from aluminium and Section CE is a solid cylindrical shaft made from steel Diameters of sections ACand CE are dencted as dac and dog respectively. Both sections are perfecty joined at C such that there is strain compatibility at the interface. There is a torque Ti applied at the mid-span of Section AC, and a torque T applied at the mid-span of Section CE. Directions of the applied torques are as shown in the Figure. The shear modulus of aluminium is Gat = 50 GPa and the shear modulus of steel is G = 110 GPa The dimensions of the shafts and the magnitude of applied torques are given by: T = 0.5 kNm T= 4.5 kNm dAc = 120 mm dce = 70 mm LAB = LBC = 4 m Lcp = LDe = 2 m Assume the shaft to remain linear elastic during loading. T, T T, A. T, В D E Part One: Equilibrium Write an equation of equilibrium for the structure by filling in the coefficients: TA+ Te+ Note that any expression that can be simplified to a correct equilibrium equation will be occepted. You may or may not need both unknowns. If you do not need one of the unknowns, then please enter 0 as the coefficient. DO NOT LEAVE ANY BOX EMPTY Part Two: Compatibility Write an equation of compatibility for the structure in terms of thne unknown reactions by filling in the coefficients: ]Te+[ Note that any expression that can be simpiified to a correct compatibility equation will be accepted. You may or may not need both unknowns. If you do not need one of the unknowns, then please enter O as the coefficient. DO NOT LEAVE ANY BOX EMPTY 2.

Question 3 - Torsion ( A composite shaft is shown in the figure below which is fixed against rotations at ends A and E. Section AC is a solid cylindrical shaft made from aluminium and Section CE is a solid cylindrical shaft made from steel Diameters of sections ACand CE are dencted as dac and dog respectively. Both sections are perfecty joined at C such that there is strain compatibility at the interface. There is a torque Ti applied at the mid-span of Section AC, and a torque T applied at the mid-span of Section CE. Directions of the applied torques are as shown in the Figure. The shear modulus of aluminium is Gat = 50 GPa and the shear modulus of steel is G = 110 GPa The dimensions of the shafts and the magnitude of applied torques are given by: T = 0.5 kNm T= 4.5 kNm dAc = 120 mm dce = 70 mm LAB = LBC = 4 m Lcp = LDe = 2 m Assume the shaft to remain linear elastic during loading. T, T T, A. T, В D E Part One: Equilibrium Write an equation of equilibrium for the structure by filling in the coefficients: TA+ Te+ Note that any expression that can be simplified to a correct equilibrium equation will be occepted. You may or may not need both unknowns. If you do not need one of the unknowns, then please enter 0 as the coefficient. DO NOT LEAVE ANY BOX EMPTY Part Two: Compatibility Write an equation of compatibility for the structure in terms of thne unknown reactions by filling in the coefficients: ]Te+[ Note that any expression that can be simpiified to a correct compatibility equation will be accepted. You may or may not need both unknowns. If you do not need one of the unknowns, then please enter O as the coefficient. DO NOT LEAVE ANY BOX EMPTY 2.

Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter3: Torsion
Section: Chapter Questions
Problem 3.4.21P: A nonprismatic bar ABC with a solid circular cross section is loaded by distributed torques (sec...
See similar textbooks
icon
Related questions
Question
Question 3 - Torsion ( A composite shaft is shown in the figure below which is fixed against rotations at ends A and E. Section AC is a solid cylindrical shaft made from aluminium and Section CE is a solid cylindrical shaft made from steel Diameters of sections ACand CE are denoted as dac and dog respectively. Both sections are perfecty joined at C such that there is strain compatibility at the interface. There is a torque Ti applied at the mid-span of Section AC, and a torque T applied at the mid-span of Section CE. Directions of the applied torques are as shown in the Figure. The shear modulus of aluminium is Gat = 50 GPa and the shear modulus of steel is G = 110 GPa The dimensions of the shafts and the magnitude of applied torques are given by: T = 0.5 kNm T = 4.5 kNm dAc = 120 mm dce = 70 mm LAB = LBC = 4 m Lcp = LDe = 2 m Assume the shaft to remain linear elastic during loading. T, T T, A. T, B D E Part One: Equilibrium Write an equation of equilibrium for the structure by filling in the coefficients: TA+ Te+ Note that any expression that can be simplified to a correct equilibrium equation will be accepted. You may or may not need both unknowns. If you do not need one of the unknowns, then please enter 0 as the coefficient. DO NOT LEAVE ANY BOX EMPTY Part Two: Compatibility Write an equation of compatibility for the structure in terms of thne unknown reactions by filling in the coefficients: Note that any expression that can be simpiified to a correct compatibility equation will be accepted. You may or may not need both unknowns. If you do not need one of the unknowns, then please enter O as the coefficient. DO NOT LEAVE ANY BOX EMPTY 2.
Transcribed Image Text:Question 3 - Torsion ( A composite shaft is shown in the figure below which is fixed against rotations at ends A and E. Section AC is a solid cylindrical shaft made from aluminium and Section CE is a solid cylindrical shaft made from steel Diameters of sections ACand CE are denoted as dac and dog respectively. Both sections are perfecty joined at C such that there is strain compatibility at the interface. There is a torque Ti applied at the mid-span of Section AC, and a torque T applied at the mid-span of Section CE. Directions of the applied torques are as shown in the Figure. The shear modulus of aluminium is Gat = 50 GPa and the shear modulus of steel is G = 110 GPa The dimensions of the shafts and the magnitude of applied torques are given by: T = 0.5 kNm T = 4.5 kNm dAc = 120 mm dce = 70 mm LAB = LBC = 4 m Lcp = LDe = 2 m Assume the shaft to remain linear elastic during loading. T, T T, A. T, B D E Part One: Equilibrium Write an equation of equilibrium for the structure by filling in the coefficients: TA+ Te+ Note that any expression that can be simplified to a correct equilibrium equation will be accepted. You may or may not need both unknowns. If you do not need one of the unknowns, then please enter 0 as the coefficient. DO NOT LEAVE ANY BOX EMPTY Part Two: Compatibility Write an equation of compatibility for the structure in terms of thne unknown reactions by filling in the coefficients: Note that any expression that can be simpiified to a correct compatibility equation will be accepted. You may or may not need both unknowns. If you do not need one of the unknowns, then please enter O as the coefficient. DO NOT LEAVE ANY BOX EMPTY 2.
Part Three: Reactions Calculate the unknown reactions (use positive if their direction matches the assumed direction in the figure above, or use negative if acts in opposing direction) TA kNm TE kNm Part Four: Internal Torgues Calculate the internal torque in each section of the shaft (use the right hand thumb rule from class as the positive sign convention) TAB kNm THC = kNm Tcp= kNm TDE kNm Part Five: Stress and Twist Calculate the magnitude of twist at point C radians Calculate the magnitude of maximum shear stress across both materials composing the shaft MPa Tmaz
Transcribed Image Text:Part Three: Reactions Calculate the unknown reactions (use positive if their direction matches the assumed direction in the figure above, or use negative if acts in opposing direction) TA kNm TE kNm Part Four: Internal Torgues Calculate the internal torque in each section of the shaft (use the right hand thumb rule from class as the positive sign convention) TAB kNm THC = kNm Tcp= kNm TDE kNm Part Five: Stress and Twist Calculate the magnitude of twist at point C radians Calculate the magnitude of maximum shear stress across both materials composing the shaft MPa Tmaz
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 3 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Strain Energy
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:
9781337093347
Author:
Barry J. Goodno, James M. Gere
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS