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40 kips/ft 40 kip A 1.5 + 1.5 ft

40 kips/ft 40 kip A 1.5 + 1.5 ft

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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For the simply supported beam with shown loading, determine the deflection at midspan. (EI = 750 kips · ft²) 40 kips/ft 40 kips/ft В A 1.5 ft_, 1.5 ft -4 ft-
Transcribed Image Text:For the simply supported beam with shown loading, determine the deflection at midspan. (EI = 750 kips · ft²) 40 kips/ft 40 kips/ft В A 1.5 ft_, 1.5 ft -4 ft-
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 Hooke's Law PD O, = 4t o = Ee 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 G = 2(1+ v) Thermal Deformation Thermal Strain ốr = a(AT)L ET = aAT Shear Stress due to Torsion T'c TmaxT Shear Stress at any Point 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 Bending Stress Maximum Bending Stress Mc Om =T Elastic Section Modulus Deflection of a Beam Second Moment-Area Theorem d'y dr El = M tC/p = (area bet. C and D)X, Area of General Spandrel bh A = Centroid of General Spandrel b c=+2 n+1 Deflection in Simply Supported Beams Deflection in Cantilever Beams 8g = (area of M/El diagram)Xg tCIA Maximum and Minimum Stress in Beams with Combined Axial and Lateral Loads in Beams Combined Axial and Lateral Loads P My AT P. Mc Gmazmin =7+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 + Tầy Omax.min D 2 6-dy sin 20 + ry cos 20 sin 20 + Txy Cos 20 2 Maximum Shear Stress and Corresponding Angle Angle of Principal Plane Tmax = + Ty tan 20, = tan 28, = --0, 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 Hooke's Law PD O, = 4t o = Ee 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 G = 2(1+ v) Thermal Deformation Thermal Strain ốr = a(AT)L ET = aAT Shear Stress due to Torsion T'c TmaxT Shear Stress at any Point 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 Bending Stress Maximum Bending Stress Mc Om =T Elastic Section Modulus Deflection of a Beam Second Moment-Area Theorem d'y dr El = M tC/p = (area bet. C and D)X, Area of General Spandrel bh A = Centroid of General Spandrel b c=+2 n+1 Deflection in Simply Supported Beams Deflection in Cantilever Beams 8g = (area of M/El diagram)Xg tCIA Maximum and Minimum Stress in Beams with Combined Axial and Lateral Loads in Beams Combined Axial and Lateral Loads P My AT P. Mc Gmazmin =7+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 + Tầy Omax.min D 2 6-dy sin 20 + ry cos 20 sin 20 + Txy Cos 20 2 Maximum Shear Stress and Corresponding Angle Angle of Principal Plane Tmax = + Ty tan 20, = tan 28, = --0, Coordinates of Center of Mohr's Circle
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