Elementary applied mechanics, by T. Alexander (and A.W. Thomson). |
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Side 6
... ? Which sustains the greater ? the strains . For the 100 - fathom cable Give the ratio of strain = augmentation length = 3 in . 7200 in . = ' 000417 . For the 90 fathom cable strain = augmentation length 2.75 6 APPLIED MECHANICS .
... ? Which sustains the greater ? the strains . For the 100 - fathom cable Give the ratio of strain = augmentation length = 3 in . 7200 in . = ' 000417 . For the 90 fathom cable strain = augmentation length 2.75 6 APPLIED MECHANICS .
Side 17
... gives 1433 cub . in .; reckoned at 480 lbs . per cubic foot gives 398 lbs . Hence .. gross load = 398 lbs . live load = 4480 lbs . dead load = = 9358 9358 stress E - area stress = strain 9358 or 30,000,000 = area × 0005 ° 9358 ... area ...
... gives 1433 cub . in .; reckoned at 480 lbs . per cubic foot gives 398 lbs . Hence .. gross load = 398 lbs . live load = 4480 lbs . dead load = = 9358 9358 stress E - area stress = strain 9358 or 30,000,000 = area × 0005 ° 9358 ... area ...
Side 18
... gives a dead load = 248,000 lbs . The live load must not exceed one half of this . NOTE . Other considerations limit the strength of the pillar if it be long . RESILIENCE . DEF . The Resilience of a body is 18 APPLIED MECHANICS .
... gives a dead load = 248,000 lbs . The live load must not exceed one half of this . NOTE . Other considerations limit the strength of the pillar if it be long . RESILIENCE . DEF . The Resilience of a body is 18 APPLIED MECHANICS .
Side 23
... gives instant strain elongation resilience of rod = = 97200 lbs . per sq . in . 00324 ft . per ft . of length . = ' 02268 ft . live load x elongation . 48600 lbs . × 02268 ft . = 1102 ft . - lbs . 32. The chain of a crane is 30 feet ...
... gives instant strain elongation resilience of rod = = 97200 lbs . per sq . in . 00324 ft . per ft . of length . = ' 02268 ft . live load x elongation . 48600 lbs . × 02268 ft . = 1102 ft . - lbs . 32. The chain of a crane is 30 feet ...
Side 25
... gives an additional elongation equal to that for a dead load of 1,800 lbs . elongation ( due to 1800 lbs . ) 1800 · 006 ft . = 3900 * 1800 × .006 3900 = ' 00277 ft . inst . elongation for live load = Now both the 3900 lbs . and the 900 ...
... gives an additional elongation equal to that for a dead load of 1,800 lbs . elongation ( due to 1800 lbs . ) 1800 · 006 ft . = 3900 * 1800 × .006 3900 = ' 00277 ft . inst . elongation for live load = Now both the 3900 lbs . and the 900 ...
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amount of resilience amount of stress angle of repose augmentation axis Cambridge chain component stresses cos² Crown 8vo dead load direction and intensity earth Edition ellipse elongation equal Equating equilibrium Eton College Extra fcap feet long find the intensity find the stress ft.-lbs greater greatest principal stress Hence Hooke's law horizontal inch in section instant internal stress J. P. MAHAFFY LATIN layer length live load modulus of elasticity normal component OM² Owens College pair of rectangular parallelopiped plane BB plane inclined plane of greatest planes of principal position of cc principal stresses Professor proof strain proof stress rectangular planes right angles School shearing shearing stress sin² sloping solid square inch stress on cc stretched surface tangential component tangential stress thrust total stress velocity wall weight
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