Simplified Mechanics and Strength of MaterialsWiley, 1951 - 275 páginas |
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Página 49
... zero moment . Again , let us write an equation of moments with point C as the center of moments . Then ( 1,240 X 10 ) ... zero , the sum of all the vertical components of all the forces equals zero , and the sum of the moments of all the ...
... zero moment . Again , let us write an equation of moments with point C as the center of moments . Then ( 1,240 X 10 ) ... zero , the sum of all the vertical components of all the forces equals zero , and the sum of the moments of all the ...
Página 50
... zero , ZH = 0 , or the sum of the horizontal forces to the right equals the sum of the horizontal forces to the left ... zero and hence their moments are also zero . For problems of this type in which certain forces are 50 MOMENTS OF FORCES.
... zero , ZH = 0 , or the sum of the horizontal forces to the right equals the sum of the horizontal forces to the left ... zero and hence their moments are also zero . For problems of this type in which certain forces are 50 MOMENTS OF FORCES.
Página 118
... zero , we shall equate the expression to zero . Then , solving for x , the unknown , we shall have found the distance from R1 at which the shear passes through zero . Thus 1,800 - ( 200 X x ) = 0 200 X x = 1,800 x = 9'0 " EXAMPLE ...
... zero , we shall equate the expression to zero . Then , solving for x , the unknown , we shall have found the distance from R1 at which the shear passes through zero . Thus 1,800 - ( 200 X x ) = 0 200 X x = 1,800 x = 9'0 " EXAMPLE ...
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