. Railway mechanical engineer . ram Fig. 9).The shearing unit stress in the bolt = 11,000 V = = 11,080 !b. per sq. in. Since the bolt (1) is subjected to a shearing stress aswell as a tensile stress it is necessary to find the maximumcombined stresses. This can readily be done by using theformulae T = a p + V V + ?i P (0. = V v= + ^4 p= (2) Where T = maximum tensile unit stress S := maximum shearing ur p = applic V = apphe Substituting the values already given for p and j we find: 2,990 of the ijolt it is necessary to resolve it into its component inthe direction of Qf and of its n
. Railway mechanical engineer . ram Fig. 9).The shearing unit stress in the bolt = 11,000 V = = 11,080 !b. per sq. in. Since the bolt (1) is subjected to a shearing stress aswell as a tensile stress it is necessary to find the maximumcombined stresses. This can readily be done by using theformulae T = a p + V V + ?i P (0. = V v= + ^4 p= (2) Where T = maximum tensile unit stress S := maximum shearing ur p = applic V = apphe Substituting the values already given for p and j we find: 2,990 of the ijolt it is necessary to resolve it into its component inthe direction of Qf and of its normal. These are shown inthe adjacent .sketch Fig. 10, where p represents an addi-tional pull of 1,000 lb. and p a shear of 6,700 lb. Then,tensile unit-stress, 7,400 4 p = = 8,460 It., per sq. in.; shearing unit-stress = 6,700 lit stressunit stressshearing unit stress + * (U,080:i = lb. per sq. Let us now investigate the nature and intensity of thestresses in one of the bolts in the horizontal flange 0-N. The. := lb. per sq. in. combining this witli the shearing stress of 2,430 lb. per sq. to the moment Qd, we find the combined shear to lb. per sq. in. The solution of the problem would beincomplete if we did not ascertain whether bolt (5) is worst case is obviously that of bolt (5). Here we have—Secondary resistance =^ 50,000 X X 12 Qf = =7,400 lb. 526 which is in the nature of a pull. The moments of the component of P about KK also causesa shearing stress in this bolt. 48,750 X X 13 Qf = ■ = 2,420 lb. 656 Shearing unit stress= 2,420 = 2,430 lb. per sq. in. There is also the direct resistance p which was alreadyfound to be 7,140 lb. As this force p is inclined to the axis
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Keywords: ., bookcentury1900, bookdecade1910, booksubjectrailroadengineering
