Simplified Mechanics and Strength of MaterialsWiley, 1951 - 275 páginas |
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Página xvi
... solution of a problem involves some basic principle or law of equilibrium . What is the principle involved ? This is the key to the solution . Perhaps the fundamental principle can be expressed by an equation . Write it down . Now ...
... solution of a problem involves some basic principle or law of equilibrium . What is the principle involved ? This is the key to the solution . Perhaps the fundamental principle can be expressed by an equation . Write it down . Now ...
Página 80
... SOLUTION . If this rivet fails by shear , there are two planes ( at the faces of the enclosed plate ) at which the rivet must fail . This rivet is in double shear . Since , from the foregoing example , we found that the allowable ...
... SOLUTION . If this rivet fails by shear , there are two planes ( at the faces of the enclosed plate ) at which the rivet must fail . This rivet is in double shear . Since , from the foregoing example , we found that the allowable ...
Página 175
Harry Parker. SOLUTION . The key to the solution of this problem is the flexure formula , M = S. This requires that we first compute the maxi- f mum bending moment . Be certain , when using the flexure formula , that M is in units of ...
Harry Parker. SOLUTION . The key to the solution of this problem is the flexure formula , M = S. This requires that we first compute the maxi- f mum bending moment . Be certain , when using the flexure formula , that M is in units of ...
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