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For the given beam with shown loàding, determine the reacCtion at the roller support. 20 kips/ft B A

For the given beam with shown loàding, determine the reacCtion at the roller support. 20 kips/ft B A

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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Show detailed solution refer to the table of equations.
TABLES OF EQUATIONS Normal Stress Bearing Stress td Shear Stress (Single) Shear Stress (Double) P 2A Tangential Stress (Cylindrical) Longitudinal Stress (Cylindrical) PD PD 2t 4t Stress on Spherical Vessels PD O, = Hooke's Law o = Ee 4t Deformation Axial Strain PL 8 = AE Generalized Hooke's Law Oy V下 Ex E Dilatation dy Ey = E e = e, +€y +6, E E, = Shear Strain Shear Modulus E y == G = 2(1+v) Thermal Deformation Thermal Strain ốr = a(AT)L Er = aAT Shear Stress due to Torsion Shear Stress at any Point T'c Tmax =T =Tmar Polar Moment of Inertia of Circle Minimum Shear Stress (Hollow) Tmin= -Tmax Polar Moment of Inertia of Hollow Circle Angle of Twist TL JG Shear Stress due to Torsion (Non-Circular Tubes) T Power P = Ta 2tA Angle of Twist (Non-Circular Tubes) TL ( ds 4A2GT Bending Stress a. = --0,m Maximum Bending Stress Mc Om =T Elastic Section Modulus Deflection of a Beam Second Moment-Area Theorem d'y = M El Tr tc/p = (area bet. C and D)x, Area of General Spandrel bh A = Centroid of General Spandrel b n+1 n+2 Deflection in Simply Supported Beams Deflection in Cantilever Beams 8g = (area of M/El diagram)Xg Maximum and Minimum Stress in Beams with Combined Axial and Lateral Loads in Beams P My AT Combined Axial and Lateral Loads P. Mc Gmaxmin =+T Transformation of Plane Stress Ox + dy, O - dy 2 cos 20 + Tyy Sin 20 Principal Stresses 2 cos 20 -Ty sin 20 2 d, + dy + r, Omax.min = 2 Tutu =- y sin 20 + Try Cos 20 sin 20 + Txy Cos 20 2 Maximum Shear Stress and Corresponding Angle Angle of Principal Plane Tmax = + Ty tan 20, = tan 28, = --0y tan 20, = Coordinates of Center of Mohr's Circle
Transcribed Image Text:TABLES OF EQUATIONS Normal Stress Bearing Stress td Shear Stress (Single) Shear Stress (Double) P 2A Tangential Stress (Cylindrical) Longitudinal Stress (Cylindrical) PD PD 2t 4t Stress on Spherical Vessels PD O, = Hooke's Law o = Ee 4t Deformation Axial Strain PL 8 = AE Generalized Hooke's Law Oy V下 Ex E Dilatation dy Ey = E e = e, +€y +6, E E, = Shear Strain Shear Modulus E y == G = 2(1+v) Thermal Deformation Thermal Strain ốr = a(AT)L Er = aAT Shear Stress due to Torsion Shear Stress at any Point T'c Tmax =T =Tmar Polar Moment of Inertia of Circle Minimum Shear Stress (Hollow) Tmin= -Tmax Polar Moment of Inertia of Hollow Circle Angle of Twist TL JG Shear Stress due to Torsion (Non-Circular Tubes) T Power P = Ta 2tA Angle of Twist (Non-Circular Tubes) TL ( ds 4A2GT Bending Stress a. = --0,m Maximum Bending Stress Mc Om =T Elastic Section Modulus Deflection of a Beam Second Moment-Area Theorem d'y = M El Tr tc/p = (area bet. C and D)x, Area of General Spandrel bh A = Centroid of General Spandrel b n+1 n+2 Deflection in Simply Supported Beams Deflection in Cantilever Beams 8g = (area of M/El diagram)Xg Maximum and Minimum Stress in Beams with Combined Axial and Lateral Loads in Beams P My AT Combined Axial and Lateral Loads P. Mc Gmaxmin =+T Transformation of Plane Stress Ox + dy, O - dy 2 cos 20 + Tyy Sin 20 Principal Stresses 2 cos 20 -Ty sin 20 2 d, + dy + r, Omax.min = 2 Tutu =- y sin 20 + Try Cos 20 sin 20 + Txy Cos 20 2 Maximum Shear Stress and Corresponding Angle Angle of Principal Plane Tmax = + Ty tan 20, = tan 28, = --0y tan 20, = Coordinates of Center of Mohr's Circle
For the given beam with shown loading, determine the reaction at the roller support. 20 kips/ft C B A - 4 ft - - 4 ft -
Transcribed Image Text:For the given beam with shown loading, determine the reaction at the roller support. 20 kips/ft C B A - 4 ft - - 4 ft -
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