. The elements of railroad engineering . e find the safe working value forshearing stress per square inch along the grain for pine is 200 lb., andto sustain a stress of 40,000 lb., it will require as many square inches ofsurface as 200 is contained in 40,000, which is 200 sq. in. Ans. 2. An oak tie-beam is subjected to a shearing stress across the grainof 62,000 lb. ; how many square inches of surface are necessary tosustain this stress ? Ans. sq. in. 1811. To find the safe dimensions for a rectan-gular horizontal beam of given span, supported atboth ends, and >vhich is at the same ti
. The elements of railroad engineering . e find the safe working value forshearing stress per square inch along the grain for pine is 200 lb., andto sustain a stress of 40,000 lb., it will require as many square inches ofsurface as 200 is contained in 40,000, which is 200 sq. in. Ans. 2. An oak tie-beam is subjected to a shearing stress across the grainof 62,000 lb. ; how many square inches of surface are necessary tosustain this stress ? Ans. sq. in. 1811. To find the safe dimensions for a rectan-gular horizontal beam of given span, supported atboth ends, and >vhich is at the same time subjectedboth to a transverse strain and to a longitudinaltensile or pulling one, or to a longitudinal com-pressive one : Rule.— When the longitudinal strain is tensile, find sepa-rately the safe dimensions as if for a beam alone, and as iffor a tie alone, and add the tiuo resulting areas the longitudinal strain is eompressive, find separately 1250 RAILROAD STRUCTURES. iNi Planh. 3x8 3x8 Separators. k I tr J 5* 3 -r^l. Fin. 629. RAILROAD STRUCTURES. 1257 the safe dimensions as if for a beam alone, and as if for apillar alone, and add the t7Co resulting areas together. Example.—A rectangular horizontal beam of yellow pine of 14 feetspan is to sustain a uniformly distributed transverse load of 40,000 lb.,and a pulling stress of 30,000 lb., with a factor of 6; what should be thedimensions of the beam ? Solution.—A uniformly distributed load of 40,000 lb. is equivalentto a center load of 20,000 lb. A safe center load of 20,000 lb. with afactor of 6 is equivalent to a center breaking load of 120,000 lb. Apply-ing the rule given in Art. 1804, we have 120,000x14 = 1,680, constant for center breaking loads for yellow pine (Art. 1802) is550. 1,680,000 -T- 550 = 3,054, the cube root of which is This givesin inches the side of a square beam which will safely carry the givenload. From Art. 1810 we conclude the safe working stress in tension persquare inc
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