Strength of Materials |
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Page 7
... applied loads passes through the centroid of that surface . It does not follow , however , that positioning the load through the cen- troid of the section always results in a uniform stress distribution . For example , in Fig . 1-7 is ...
... applied loads passes through the centroid of that surface . It does not follow , however , that positioning the load through the cen- troid of the section always results in a uniform stress distribution . For example , in Fig . 1-7 is ...
Page 70
... applied at B. 315. A 16 - ft shaft rotating at 126 rpm has 100 hp applied at a gear that is 6 ft from the left end where 30 hp are removed . At the right end , 40 hp are removed and another 30 hp leaves the shaft at 5 ft from the right ...
... applied at B. 315. A 16 - ft shaft rotating at 126 rpm has 100 hp applied at a gear that is 6 ft from the left end where 30 hp are removed . At the right end , 40 hp are removed and another 30 hp leaves the shaft at 5 ft from the right ...
Page 523
... applied and then removed , determine the residual force in each bar . Ans . Pa = -8100 lb ; P. = 16,200 lb 1416. The bar shown in Fig . P - 1416 is firmly attached to rigid supports . The yield strengths for steel and aluminum alloy are ...
... applied and then removed , determine the residual force in each bar . Ans . Pa = -8100 lb ; P. = 16,200 lb 1416. The bar shown in Fig . P - 1416 is firmly attached to rigid supports . The yield strengths for steel and aluminum alloy are ...
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allowable stresses aluminum angle applied assumed axial load beam in Fig beam loaded beam shown bending bending moment bolt bronze cantilever beam caused centroid column compressive stress Compute the maximum concentrated load concrete continuous beam cross section deformation Determine the maximum diameter elastic curve end moments equal equivalent Euler's formula factor of safety fibers flange flexure formula free-body diagram ft long ft-lb Hence Hooke's law horizontal ILLUSTRATIVE PROBLEMS inertia lb/ft length loaded as shown main plate maximum shearing stress maximum stress midspan deflection modulus Mohr's circle moment of area moment of inertia neutral axis obtain plane positive proportional limit R₂ radius reaction Repeat Prob resisting restrained beam resultant segment shaft shear diagram shearing force shown in Fig Solution Solve Prob span statically indeterminate steel strain tensile stress three-moment equation torque torsional uniformly distributed load vertical shear weld zero ΕΙ