Articoli correlati a Principles of mechanics, an abstract of lectures
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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1874 Excerpt: ...3, 4 (Fig. 25), then we have a load of 1 ton at 2, and a load of 1 ton at 3. Let F1 be the supporting force, equal and opposite to the downward force at point of support; S1 = stress on the bar 1, 2, or 3, 4; H, = horizontal thrust upon the bar 1, 4. Since the angle at 1 is an angle of 60, we have the following relations existing between the forces, F,: S1: H::-86: 1:-5, and therefore-= Si and-577 = E, and if-ob-oo Fi = 1 ton, then we have Sl5 or stress on bar 1,2, or 3, 4 =-= 1 16 ton; and H, or hori Via. 27. We have now to calculate the stresses on the bars of the second frame 5, 2, 3, 6 (Fig. 26); and since we have And by multiplying these values by the same ratios as in case 1, p. 63, we have for the inclined stresses, 2 90 tons = stress along 1, 2, or 3, 4. 1-74 ton = „ 2,5, or 3,6. 0-58 ton = „ 7, 5, or 7, 6. The horizontal stresses are found by finding the bending moments at the points 2, 5, 7, 6, and 3 (as explained at p. 54), and dividing the moments so found by the depth of the girder, thus:--Tons. Supporting force at 1 = 2 5 2 = 1-5 5 = 0-5 7 = O'O 'Bending Moments. Bound 2 = 2-5 x-5 =1-25 „ 5 = (1-25 + 1-5 x-5).. =2-00 „ 7 = (1-25 +-75 +-5 x-5) =2-25 And the horizontal stresses are, 1-25 At 2 or 3, or their opposites = = 1-45. 'OO At 5 or 6, or their opposites = = 2-32. 2-25 At 7, or its opposite =--53-= 2-61. '00 On comparison of the two methods it will be seen that the results are similar. FORMS OF GIRDERS. Plate Girders. Fig. 28 shows in section a wrought-iron plate'girder; Fig. 29 shows part of same girder in elevation. A and B are the flanges of wrought-iron plates, suitably joined at their extremities by covering plates; C is the web, generally of thin plate. Angle-irons D are riveted to the web, the flanges being a...
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