. Practical structural design; a text and reference work for engineers, architects, builders, draftsmen and technical schools;. nails donot split the pieces the reinforcement is not effective. For effec-tive arch action there must be substantial abutments there are no substantial abutments a tie rod is necessary to tiethe ends together and take the thrust. There being no tie rod,it is evident that the lower part of the joist will have to act asa tie. We know that when a beam is loaded the lower fibers arestressed in tension and the upper fibers are stressed in compres-sion. To incr
. Practical structural design; a text and reference work for engineers, architects, builders, draftsmen and technical schools;. nails donot split the pieces the reinforcement is not effective. For effec-tive arch action there must be substantial abutments there are no substantial abutments a tie rod is necessary to tiethe ends together and take the thrust. There being no tie rod,it is evident that the lower part of the joist will have to act asa tie. We know that when a beam is loaded the lower fibers arestressed in tension and the upper fibers are stressed in compres-sion. To increase the tension in the bottom by adding to it theamount required to take care of the thrustin the diagonal reinforcing strips is nothelpful. This old-time method is, therefore,based on a fallacy and should be abandoned. In Fig. 65 is shown another method, thereinforcing being spiked along the top edgeto make a beam of T-section. This raises theneutral surface so the increased area in compression is supposedto be offset by an increased area in tension. Assume a joist 3 ins. x 14 ins. of wood in which a fiber stress of 102. Fig. 65 — T-beam ofWood GIRDERS AND TRUSSES 103 1200 lbs. per square inch is used. A strip 1 in. x 4 ins. is spiked oneach side along the top. How much is the strength increased? The original strength, Mr =^-^ = ^^^^ X^^ X ^^ = 117,600 in. lbs., the neutral plane being in the middle of the joist. To find theposition of the neutral plane in the T-section use the method ofmoments, taking rnoments about the lower edge: The original piece, 3 X 14 x 7 = 294 One added piece Ix 4x12= 48 Second added piece 1 x 4 x 12 = 48 = ms. Area = (3 x 14) + (2 x 4) = 50 First the area of the beam was multiplied by the distance ofthe center of gravity above the bottom, the result being the area of each added piece was multiplied by the distanceof its center of gravity above the bottom of the beam. This was12 ins., being half the depth of the piece added to the dif
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