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Given the statically indeterminate system, determine the reactions (magnitude and direction) at A and E. (Esteel = 200 GPa; Eprass = 105 GPa) Lengths are in millimeters 50 50 90 60 A DI E Steel B. Brass

Given the statically indeterminate system, determine the reactions (magnitude and direction) at A and E. (Esteel = 200 GPa; Eprass = 105 GPa) Lengths are in millimeters 50 50 90 60 A DI E Steel B. Brass

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Refer to the table of equations upon answering the problem.
TABLE OF EQUATIONS Normal Stress Bearing Stress On = td A Shear Stress (Single) Shear Stress (Double) P P A = 1 2A Tangential Stress (Cylindrical) PD O, = 2t Longitudinal Stress (Cylindrical) PD 4t Stress on Spherical Vessels PD の、= Hooke's Law o = E€ 4t Deformation Axial Strain PL 8 = AE E = Generalized Hooke's Law Oy E Ox Ex = E Dilatation Oy Ox -- v E e = €x + €y + €z Ey = E Oz Ez = E E -- v- Shear Strain Shear Modulus E G = 2(1+ v) Thermal Deformation Thermal Strain δ, - α(ΔΤ)L ET = aAT Shear Stress due to Torsion Shear Stress at any Point Tc max = T=-Tmax Polar Moment of Inertia of Circle Minimum Shear Stress (Hollow) C1 Tmax C2 J = Tmin Polar Moment of Inertia of Hollow Circle Angle of Twist TL J = JG Shear Stress due to Torsion (Non-Circular Tubes) Power T P = Tw 2tA Angle of Twist (Non-Circular Tubes) ds Bending Stress TL O, = Om 4A²G, Maximum Bending Stress Mc Elastic Section Modulus Om =T
Transcribed Image Text:TABLE OF EQUATIONS Normal Stress Bearing Stress On = td A Shear Stress (Single) Shear Stress (Double) P P A = 1 2A Tangential Stress (Cylindrical) PD O, = 2t Longitudinal Stress (Cylindrical) PD 4t Stress on Spherical Vessels PD の、= Hooke's Law o = E€ 4t Deformation Axial Strain PL 8 = AE E = Generalized Hooke's Law Oy E Ox Ex = E Dilatation Oy Ox -- v E e = €x + €y + €z Ey = E Oz Ez = E E -- v- Shear Strain Shear Modulus E G = 2(1+ v) Thermal Deformation Thermal Strain δ, - α(ΔΤ)L ET = aAT Shear Stress due to Torsion Shear Stress at any Point Tc max = T=-Tmax Polar Moment of Inertia of Circle Minimum Shear Stress (Hollow) C1 Tmax C2 J = Tmin Polar Moment of Inertia of Hollow Circle Angle of Twist TL J = JG Shear Stress due to Torsion (Non-Circular Tubes) Power T P = Tw 2tA Angle of Twist (Non-Circular Tubes) ds Bending Stress TL O, = Om 4A²G, Maximum Bending Stress Mc Elastic Section Modulus Om =T
Given the statically indeterminate system, determine the reactions (magnitude and direction) at A and E. (Esteel = 200 GPa; Eprass = 105 GPa) %3D Lengths are in millimeters 50 50 90 60 A DI E Steel B ● Brass 40 kN 30 kN d 3 20 тm d = 15 mm
Transcribed Image Text:Given the statically indeterminate system, determine the reactions (magnitude and direction) at A and E. (Esteel = 200 GPa; Eprass = 105 GPa) %3D Lengths are in millimeters 50 50 90 60 A DI E Steel B ● Brass 40 kN 30 kN d 3 20 тm d = 15 mm
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