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For the beam with shown loading, show that the equation of the elastic curve is Wo 7w,L3 + x3 y = EI 120L 36 360 В A L-

For the beam with shown loading, show that the equation of the elastic curve is Wo 7w,L3 + x3 y = EI 120L 36 360 В A L-

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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For the beam with shown loading, show that the equation of the elastic curve is 1 Wo 7w,L3 y = El x5 + 120L 36 360 В A L-
Transcribed Image Text:For the beam with shown loading, show that the equation of the elastic curve is 1 Wo 7w,L3 y = El x5 + 120L 36 360 В A L-
TABLE OF EQUATIONS Normal Stress Bearing Stress P A td Shear Stress (Single) Shear Stress (Double) P = 1 2A Tangential Stress (Cylindrical) PD Pr O = =T Longitudinal Stress (Cylindrical) PD Pr O = 2t 4t Stress on Spherical Vessels Hooke's Law PD Pr Os = 4t O = Ee 2t Deformation Axial Strain PL ΔΙ. E = AE 1-1-, Generalized Hooke's Law dy E Ex = Dilatation dy Ey = E e = €x + €y + €z E dy E, = -v E E Shear Strain Shear Modulus E G = 2(1+ v) y = Thermal Strain Thermal Deformation 8, = a(AT)L ET = aAT Shear Stress due to Torsion Shear Stress at any Point Tc Tmax = T=-Tmax Polar Moment of Inertia of Circle Minimum Shear Stress (Hollow) Tmin =-Tmax C2 Angle of Twist Polar Moment of Inertia of Hollow Circle TL JG Shear Stress due to Torsion (Non-Circular Tubes) Power T P = T 2tA Angle of Twist (Non-Circular Tubes) TL ( ds Bending Stress Ox = --om 4A?G Maximum Bending Stress Mc Om =T Elastic Section Modulus Deflection of a Beam d?y Edr = M Second Moment-Area Theorem tc/p = (area bet.C and D)x Area of General Spandrel bh A = n+1 Centroid of General Spandrel b n+2 Deflection in Simply Supported Beams 8p = 0Ạx - tạ/A tc/A Deflection in Cantilever Beams 8R = (area of M/El diagram)Xg Maximum and Minimum Stress in Beams with Combined Axial and Lateral Loads in Beams Combined Axial and Lateral Loads P My A I P. Mc Omax,min = t- O =-- Transformation of Plane Stress cos 20 +Txy sin 20 2 Principal Stresses O + dy Ox - 0y 2 Ox-dy sin 20 + Tyy cos 20 2 cos 20 - Tzy sin 20 Omar, min= 2 Maximum Shear Stress and Corresponding Angle Angle of Principal Plane + Tây Tmax = 21.xy_ O- dy 2Txy tan 20 tan 20, %3D Coordinates of Center of Mohr's Circle (*.0) (0x+ dy 2
Transcribed Image Text:TABLE OF EQUATIONS Normal Stress Bearing Stress P A td Shear Stress (Single) Shear Stress (Double) P = 1 2A Tangential Stress (Cylindrical) PD Pr O = =T Longitudinal Stress (Cylindrical) PD Pr O = 2t 4t Stress on Spherical Vessels Hooke's Law PD Pr Os = 4t O = Ee 2t Deformation Axial Strain PL ΔΙ. E = AE 1-1-, Generalized Hooke's Law dy E Ex = Dilatation dy Ey = E e = €x + €y + €z E dy E, = -v E E Shear Strain Shear Modulus E G = 2(1+ v) y = Thermal Strain Thermal Deformation 8, = a(AT)L ET = aAT Shear Stress due to Torsion Shear Stress at any Point Tc Tmax = T=-Tmax Polar Moment of Inertia of Circle Minimum Shear Stress (Hollow) Tmin =-Tmax C2 Angle of Twist Polar Moment of Inertia of Hollow Circle TL JG Shear Stress due to Torsion (Non-Circular Tubes) Power T P = T 2tA Angle of Twist (Non-Circular Tubes) TL ( ds Bending Stress Ox = --om 4A?G Maximum Bending Stress Mc Om =T Elastic Section Modulus Deflection of a Beam d?y Edr = M Second Moment-Area Theorem tc/p = (area bet.C and D)x Area of General Spandrel bh A = n+1 Centroid of General Spandrel b n+2 Deflection in Simply Supported Beams 8p = 0Ạx - tạ/A tc/A Deflection in Cantilever Beams 8R = (area of M/El diagram)Xg Maximum and Minimum Stress in Beams with Combined Axial and Lateral Loads in Beams Combined Axial and Lateral Loads P My A I P. Mc Omax,min = t- O =-- Transformation of Plane Stress cos 20 +Txy sin 20 2 Principal Stresses O + dy Ox - 0y 2 Ox-dy sin 20 + Tyy cos 20 2 cos 20 - Tzy sin 20 Omar, min= 2 Maximum Shear Stress and Corresponding Angle Angle of Principal Plane + Tây Tmax = 21.xy_ O- dy 2Txy tan 20 tan 20, %3D Coordinates of Center of Mohr's Circle (*.0) (0x+ dy 2
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