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For the beam with shown loading, determine the deflection at C when E = 200 GPa,I = 26.9 × 106 mm*.

For the beam with shown loading, determine the deflection at C when E = 200 GPa,I = 26.9 × 106 mm*.

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
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For the beam with shown loading, determine the deflection at C when E = 200 GPa,I = 26.9 × 106 mm*. 60 kN C В A 1 т 1 m
Transcribed Image Text:For the beam with shown loading, determine the deflection at C when E = 200 GPa,I = 26.9 × 106 mm*. 60 kN C В A 1 т 1 m
TABLE OF EQUATIONS Normal Stress Bearing Stress P A td Shear Stress (Single) Shear Stress (Double) P P A = 1 2A Tangential Stress (Cylindrical) PD Pr O = =- Longitudinal Stress (Cylindrical) PD Pr 2t t 4t 2t Stress on Spherical Vessels Hooke's Law PD Pr O = Ee 4t 2t Deformation Axial Strain PL ΔΙ. 8 = AE E = Generalized Hooke's Law dy E Ex = Dilatation dy Ey = E e = €x + €y + €z E E, = -v E E Shear Strain Shear Modulus E G = 2(1+ v) y = Thermal Strain Thermal Deformation 8, = a(AT)L Er = 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 IG 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 Second Moment-Area Theorem d?y ElT = M tc/p = (area bet.C and D) dx Area of General Spandrel bh A = n+1 Centroid of General Spandrel b n+2 Deflection in Simply Supported Beams 8 = 0,x - 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 Р Му A I Combined Axial and Lateral Loads P. Mc Omax,min = t- O =-- Transformation of Plane Stress cos 20 +Txy sin 20 Principal Stresses 2 Oz + dy Ox - 0y 2 dy = cos 20 - Tzy sin 20 Omax, min= 2 7. O- dy cin 20 + Tyy Cos 20 y =- 2 Maximum Shear Stress and Corresponding Angle Angle of Principal Plane 2Txy + rảy Tmax = tan 20 tan 20, = Coordinates of Center of Mohr's Circle (0x+ dy 2.0) 2
Transcribed Image Text:TABLE OF EQUATIONS Normal Stress Bearing Stress P A td Shear Stress (Single) Shear Stress (Double) P P A = 1 2A Tangential Stress (Cylindrical) PD Pr O = =- Longitudinal Stress (Cylindrical) PD Pr 2t t 4t 2t Stress on Spherical Vessels Hooke's Law PD Pr O = Ee 4t 2t Deformation Axial Strain PL ΔΙ. 8 = AE E = Generalized Hooke's Law dy E Ex = Dilatation dy Ey = E e = €x + €y + €z E E, = -v E E Shear Strain Shear Modulus E G = 2(1+ v) y = Thermal Strain Thermal Deformation 8, = a(AT)L Er = 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 IG 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 Second Moment-Area Theorem d?y ElT = M tc/p = (area bet.C and D) dx Area of General Spandrel bh A = n+1 Centroid of General Spandrel b n+2 Deflection in Simply Supported Beams 8 = 0,x - 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 Р Му A I Combined Axial and Lateral Loads P. Mc Omax,min = t- O =-- Transformation of Plane Stress cos 20 +Txy sin 20 Principal Stresses 2 Oz + dy Ox - 0y 2 dy = cos 20 - Tzy sin 20 Omax, min= 2 7. O- dy cin 20 + Tyy Cos 20 y =- 2 Maximum Shear Stress and Corresponding Angle Angle of Principal Plane 2Txy + rảy Tmax = tan 20 tan 20, = Coordinates of Center of Mohr's Circle (0x+ dy 2.0) 2
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