Simplified Mechanics and Strength of MaterialsWiley, 1951 - 275 páginas |
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Página 29
... hence their resultant acts at the mid - length of the upper chord as shown in the diagram . The line of action of the resultant is continued until it intersects the line of action of the right reac- tion , which , we know , is vertical ...
... hence their resultant acts at the mid - length of the upper chord as shown in the diagram . The line of action of the resultant is continued until it intersects the line of action of the right reac- tion , which , we know , is vertical ...
Página 74
... hence , the cross - sectional area is 5.5 X 7.5 or 41.25in2 . P The load supported is 40,000 # ; therefore , f = — 970 # / in2 , the actual compressive unit stress . = 40,000 A 41.25 Since the timber block stands upright , the load is ...
... hence , the cross - sectional area is 5.5 X 7.5 or 41.25in2 . P The load supported is 40,000 # ; therefore , f = — 970 # / in2 , the actual compressive unit stress . = 40,000 A 41.25 Since the timber block stands upright , the load is ...
Página 234
... hence the average stress is 12fc . The area of beam in compression is bXkd . Therefore 12fcbkd = C , the sum of all the compressive stresses . The forces C and T ( Fig . 129 ) constitute a mechanical couple the lever arm of which is jd ...
... hence the average stress is 12fc . The area of beam in compression is bXkd . Therefore 12fcbkd = C , the sum of all the compressive stresses . The forces C and T ( Fig . 129 ) constitute a mechanical couple the lever arm of which is jd ...
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