Elementary applied mechanics, by T. Alexander (and A.W. Thomson). |
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Side 5
... foot of = the substance . augmentation per foot of length , expressed in feet . Hence the total augmentation or amount of strain in feet equals the length in feet multiplied by the strain . If the augmentation equal the length , that is ...
... foot of = the substance . augmentation per foot of length , expressed in feet . Hence the total augmentation or amount of strain in feet equals the length in feet multiplied by the strain . If the augmentation equal the length , that is ...
Side 7
... foot long . Find the strain . strain = 1 . 11. A cast - iron pillar bears a strain of 001 ; its original length was 10 feet . Find its altered length . augmentation = 12 inches , altered length = 119.88 دو 12. A pillar 40 feet high ...
... foot long . Find the strain . strain = 1 . 11. A cast - iron pillar bears a strain of 001 ; its original length was 10 feet . Find its altered length . augmentation = 12 inches , altered length = 119.88 دو 12. A pillar 40 feet high ...
Side 10
... lbs . 2 sq . in . = 16,000 lbs . per sq . in . .75 in . strain = ' 000625 . length 1200 in . stress 16000 E = = 25,600,000 lbs . per sq . in . strain • 000625 15. A cast - iron pillar one square foot in 10 APPLIED MECHANICS .
... lbs . 2 sq . in . = 16,000 lbs . per sq . in . .75 in . strain = ' 000625 . length 1200 in . stress 16000 E = = 25,600,000 lbs . per sq . in . strain • 000625 15. A cast - iron pillar one square foot in 10 APPLIED MECHANICS .
Side 11
... foot in sectional area bears a weight of 2000 tons , what strain will this produce , E for cast iron being 17,000,000 lbs . ? total stress = 2000 tons per sq . foot . = 2000 × 2240 lbs . 2000 × 2240 per sq . foot . 31111 - i lbs . per ...
... foot in sectional area bears a weight of 2000 tons , what strain will this produce , E for cast iron being 17,000,000 lbs . ? total stress = 2000 tons per sq . foot . = 2000 × 2240 lbs . 2000 × 2240 per sq . foot . 31111 - i lbs . per ...
Side 17
... 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 0005 × 30,000,000 62 sq . in . B The weight of the rod ...
... 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 0005 × 30,000,000 62 sq . in . B The weight of the rod ...
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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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