Simplified Mechanics and Strength of MaterialsWiley, 1951 - 275 páginas |
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Página 122
... maximum unit stress occurs at the neutral surface , and this is the stress with which we are prin- cipally concerned . The lengths of the arrows shown in Fig . 67 ( d ) indicate the distribution of horizontal shearing stresses in a rec ...
... maximum unit stress occurs at the neutral surface , and this is the stress with which we are prin- cipally concerned . The lengths of the arrows shown in Fig . 67 ( d ) indicate the distribution of horizontal shearing stresses in a rec ...
Página 134
... maximum value . 9,750 X 2 = 19,500 ' # M ( x = 2 ) = = M ( x = 6 ) = ( 9,750 × 6 ) – ( 6,000 × 4 ) = 34,500 ' # M ( x = 12 ) = ( 9,750 × 12 ) — [ ( 6,000 × 10 ) + ( 4,000 × 6 ) ] = 33,000 ' # The bending moment diagram is shown in Fig ...
... maximum value . 9,750 X 2 = 19,500 ' # M ( x = 2 ) = = M ( x = 6 ) = ( 9,750 × 6 ) – ( 6,000 × 4 ) = 34,500 ' # M ( x = 12 ) = ( 9,750 × 12 ) — [ ( 6,000 × 10 ) + ( 4,000 × 6 ) ] = 33,000 ' # The bending moment diagram is shown in Fig ...
Página 143
... maximum bending moment ? SOLUTION . As the maximum bending moment is - Pl , — MPI = − ( 2,000 × 5 ) = -10,000 ' # or 120,000 " # = — PROBLEMS 79 - A . A simple beam has a span of 21 ′ 0 ′′ with concentrated loads of 4,500 # each at the ...
... maximum bending moment ? SOLUTION . As the maximum bending moment is - Pl , — MPI = − ( 2,000 × 5 ) = -10,000 ' # or 120,000 " # = — PROBLEMS 79 - A . A simple beam has a span of 21 ′ 0 ′′ with concentrated loads of 4,500 # each at the ...
Contenido
CHAPTER | 1 |
Elements of a Force | 9 |
Equilibrant | 15 |
Derechos de autor | |
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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