Simplified Mechanics and Strength of MaterialsWiley, 1951 - 275 páginas |
Dentro del libro
Resultados 1-3 de 28
Página 91
... width bd3 6 X 103 6 " and depth 10 " is I = = 12 = 500in . For the rec- 12 tangle whose width and depth are 4 " and 8 " respectively , the mo- bd3 4 X 83 ment of inertia is I = = = 170.6in1 . Since both of 12 12 these moments of inertia ...
... width bd3 6 X 103 6 " and depth 10 " is I = = 12 = 500in . For the rec- 12 tangle whose width and depth are 4 " and 8 " respectively , the mo- bd3 4 X 83 ment of inertia is I = = = 170.6in1 . Since both of 12 12 these moments of inertia ...
Página 121
... width of the beam at the point at which q is to be computed . The maximum horizontal shearing unit stress of a rectangular cross section occurs at the neutral surface , and its magnitude is 3 V q = X that is , 11⁄2 times the average . 2 ...
... width of the beam at the point at which q is to be computed . The maximum horizontal shearing unit stress of a rectangular cross section occurs at the neutral surface , and its magnitude is 3 V q = X that is , 11⁄2 times the average . 2 ...
Página 244
... width , is 8 " and using formula ( 6 ) , = M Kb d = = 200,000 V 131 X 8 = 13.9 " , the effective depth Formula ( 8 ) is A , = pbd ; hence A 、 0.0075 X 8 X 13.9 = = 0.835in2 , the required area of the longitudinal tensile reinforce ...
... width , is 8 " and using formula ( 6 ) , = M Kb d = = 200,000 V 131 X 8 = 13.9 " , the effective depth Formula ( 8 ) is A , = pbd ; hence A 、 0.0075 X 8 X 13.9 = = 0.835in2 , the required area of the longitudinal tensile reinforce ...
Contenido
CHAPTER | 1 |
Elements of a Force | 9 |
Equilibrant | 15 |
Derechos de autor | |
Otras 64 secciones no mostradas
Otras ediciones - Ver todas
Términos y frases comunes
allowable axial load allowable load allowable unit stress angle bars center of moments centroid column compressive stresses compressive unit stress Compute the maximum concentrated load cross section cross-sectional area deflection deformation determine diameter distance double bearing elastic limit EXAMPLE EXAMPLE factor of safety fillet weld flexure formula force polygon free body diagram funicular polygon hence indicated in Fig inertia intersection length line of action linear foot magnitude material maximum bending maximum shear modulus of elasticity moment of inertia neutral surface parallel parallelogram of forces pier plate pounds per linear pounds per square pressure PROBLEMS R₁ radius of gyration reactions reinforced concrete resisting respect resultant rivet rods section modulus shaft shear diagram shearing stress shearing unit stress shown in Fig simple beam single bearing slenderness ratio SOLUTION span square inch stirrups Table tensile stresses thickness three forces truss uniformly distributed load weight width zero