Simplified Mechanics and Strength of MaterialsWiley, 1951 - 275 páginas |
Dentro del libro
Resultados 1-3 de 21
Página 242
... reinforcement . No other sizes should be employed . = For protection to the steel reinforcement we have 12 " or 2 " of concrete below the rods . Since 15.6 " is the depth to the center of the row of reinforcing rods , 15.6 + 0.37 +1.5 ...
... reinforcement . No other sizes should be employed . = For protection to the steel reinforcement we have 12 " or 2 " of concrete below the rods . Since 15.6 " is the depth to the center of the row of reinforcing rods , 15.6 + 0.37 +1.5 ...
Página 252
... reinforcement , = = 40 # / in2 and f = 20,000 # / in2 . using the following specification data : ve 136 - B . A simple beam has a span of 14 ' 0 " with a distributed load , including its own weight , of 2,000 pounds per linear foot . If ...
... reinforcement , = = 40 # / in2 and f = 20,000 # / in2 . using the following specification data : ve 136 - B . A simple beam has a span of 14 ' 0 " with a distributed load , including its own weight , of 2,000 pounds per linear foot . If ...
Página 254
... reinforcement being four rods 3/4 " in diameter . Investigate the bond stress . 137 - C . A total uniformly distributed load of 7,200 # is placed on a beam whose effective depth is 11 " . The longitudinal tensile reinforcement consists ...
... reinforcement being four rods 3/4 " in diameter . Investigate the bond stress . 137 - C . A total uniformly distributed load of 7,200 # is placed on a beam whose effective depth is 11 " . The longitudinal tensile reinforcement consists ...
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