. The Locomotive . se the last formula gives the highest results. 52,W0 y s- Therefore M = 4(>,h()0 inch lbs. 2 X 36 The required resistance of the two angle irons against bending,also called the section modulus, which we will designate by S, cannow be obtained from the formula: fin which / equals the fibre stress. 46,600 therefore 6 14 12,500 The cross section of the angles which has to possess this sectionmodulus is at the pin holes, where the greatest moment was foundto exist, and is shown in fig. 3. The section modulus of a cross section of a beam is obtainedby dividing the moment of in
. The Locomotive . se the last formula gives the highest results. 52,W0 y s- Therefore M = 4(>,h()0 inch lbs. 2 X 36 The required resistance of the two angle irons against bending,also called the section modulus, which we will designate by S, cannow be obtained from the formula: fin which / equals the fibre stress. 46,600 therefore 6 14 12,500 The cross section of the angles which has to possess this sectionmodulus is at the pin holes, where the greatest moment was foundto exist, and is shown in fig. 3. The section modulus of a cross section of a beam is obtainedby dividing the moment of inertia of that section by the distanceof the outermost fibre to the neutral axis. This distance is markeda in fig. 3. The neutral axis is a line coinciding with an imaginaryplane called the neutral surface which is supposed to pass throughthe center of gravity of the cross section. To find the distance of the center of gravity of the section tothe bottom line marked b in fig. 3, we can make use of thefollowing rule:. Fig. 3. I9I9] THE LOCOMOTIVE. 233 Taking the sum of the areas ./j, . I., and .i;, each multiplied by therespective distances of their centers to the bottom line and dividethis by the total area. The result is distance b. We therefore have: ^ J 2 • / 4 ■ -^^s + ? X H X / + J X 1/2 X H 1/2 X JVi + 2 X Vz + 3 X 1/2I = /..?/5 /;n-//, and X = 5 — /..?/5 = .?.6,V5 /;/r//The moment of inertia (/) of the section can now be calculatedin the following manner: Moment of inertia of area Ay with respect to its own center is A = ^j, = ^-jj-^ = .OS/ Moment of inertia of area A., with respect to its own center is = .OS/ The two moments /, and /:• will now have to be reduced tomoments of inertia about the neutral axis of the whole sectionwhich can be done by the following rule: Add to each of the moments /i and h the product of theirrespective areas times the square of the distance of their centersto the neutral axis. The results are their moments of inertia aboutthe neutral
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