Cyclopedia of mechanical engineering; a general reference work Editor-in-chief Howard Monroe Raymond Assisted by a corps of mechanical engineers, technical experts, and designers of the highest professional standing . ting the sec-tion of the arm for strength is to consider the base of the Tonly, of rectangular section, breadth I, and depth A, for which the internal moment of resistance is ^ ? -O Also, it is simplest to assume one dimension, say the breadth,and the allowable fibre stress, and figure for the depth. Takingthe breadth at IJ inches, which looks about right, and the fibrestress at
Cyclopedia of mechanical engineering; a general reference work Editor-in-chief Howard Monroe Raymond Assisted by a corps of mechanical engineers, technical experts, and designers of the highest professional standing . ting the sec-tion of the arm for strength is to consider the base of the Tonly, of rectangular section, breadth I, and depth A, for which the internal moment of resistance is ^ ? -O Also, it is simplest to assume one dimension, say the breadth,and the allowable fibre stress, and figure for the depth. Takingthe breadth at IJ inches, which looks about right, and the fibrestress at 2,500, and equating the external moment to the internal,we have ^^ggg^2,500x|125><A^ 6X7,666^ 2,^ •*h = i/TO = (say 41)Drawing in this size, and tapering the arm to the rim as inthe case of the pulleys, making the depth of the rim according tothe suggested proportions given in Part II, Gears, giving thecenter leg of the T a thickness of J inch tapering to 1 inch, andheavily filleting the arms to the rim and center flange, we have afairly well proportioned gear. The. next thing to determine is the size of the driving circle upon which their centers lie may be 11 inches in diam-. Fig. la 244 MAC HIKE DESIGX 47 eter, and there will naturally be six bolts, one between each arm. These bolts are in pure shear, and the material of which they are to be made ought to be good for at least 8,0Q0 pounds per square inch fibre stress. The force acting at the circumference of an circle would be -^———-=13,091 pounds. Equating the load on each bolt to the resisting shear gives „ ,y^^ . 8, Let A=area resisting »,UUUXA- - Let (/=dia. of bolt. IT di , Then A- ,4 d= v/.3o (say .6) Jg-inch bolts would do. But |-inch bolts are pretty small to use in connection with such heavy They look out of proportion to the adjacent parts. Hence ^-inch bolts have been substituted as being better sui
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