Foundation course in mensuration and mechanical drawing by Poore John Bentley 1866-

Author:Poore, John Bentley, 1866- [from old catalog]
Language: eng
Format: epub, pdf
Tags: Measurement, Mechanical drawing
Publisher: Washington, N.J., J. B. Poore
Published: 1916-03-25T05:00:00+00:00

edges of the middle part of the washer. Then frorri the point d draw d — h perpendicular to c — d and representing a length of 3". From h draw h — i perpendicular to d — h and representing a length of 6". The points d and i give you the direction for drawing the line d — e so that the washer will fit the truss. You can then complete the wing d-e-f-b. Draw the other wing in the same manner.

To find the contents of the washer we must find the area of the side as shown and multiply by the width; then deduct the core hole. The area of the middle part a-b-c-d is ij4" x 4" = 5 sq. in. The two wings are alike, so we will consider b-d-e-f. It is not a parallelogram, but the sides b — d and e — f are parallel, and its area is half the sum of its parallel sides multiplied by the perpendicular distance between them; b — d is i>^" and e — f is %"; the perpendicular between them is 3". ^ of (i^" plus %") X 3" = 3 %6 sq. in., the area of side of one wing; area of two wings 1= 6}i sq. in. Adding 5 sq. in. and 6}i sq. in. we have 11^ sq. in., the area of the side of the washer; and multiplying by 6'^ the width, we have 68j4 cu. in., from which we must deduct the hole. The hole is ij^" in diam. and ij^" deep or long. In shape it is a cylinder. Its contents must be the square of i>^" x .7854 x ij4" = ij4 cu. in. (1.25 cu. in.) Deducting 1.25 cu. in from 68>4 cu. in. (68.25 cu. in.) we have 67 cu. in. of cast iron weighing .26 lb. to the cu. in., or a total weight of 17.4 lbs.

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