## United States Standard Dimensions Of Bolts And Nuts
* Diameter at the root of the thread. The basis of the Franklin Institute or United States standard for the heads of bolts and for nuts is as follows: The short diameter or width across the flats is equal to one and one-half times the diameter plus 1/8 inch for rough or unfinished bolts and nuts, and one and one-half times the bolt diameter plus, 1/16 inch for finished heads and nuts. The thickness is, for rough heads and nuts, equal to the diameter of the bolt, and for finished heads and nuts 1/16 inch less. Fig. 155. Fig. 156. The hexagonal or hexagon (as they are termed in the shop) heads of bolts may be presented in two ways, as is shown in Figures 155 and 156. The latter is preferable, inasmuch as it shows the width across the flats, which is the dimension that is worked to, because it is where the wrench fits, and therefore of most importance; whereas the latter gives the length of a flat, which is not worked to, except incidentally, as it were. There is the objection to the view of the head, given in Figure 156, however, that unless it is accompanied by an end view it somewhat resembles a similar view of a square head for a bolt. It may be distinguished therefrom, however, in the following points: If the amount of chamfer is such as to leave the chamfer circle (as circle A, in Figure 153) of smaller diameter than the width across the flats of the bolt-head, the outline of the sides of the head will pass above the arcs at the top of the flats, and there will be two small flat places, as A and B, in Figure 156 (representing the angle of the chamfer), which will not meet the arcs at the top of the flats, but will join the sides above those arcs, as in the figure; which is also the case in a similar view of a square-headed bolt. It may be distinguished therefrom, however, in the following points: If the amount of chamfer is such as to leave the chamfer circle (A, Figure 153) of smaller diameter than the width across the flats of the bolt-head, the outline of the sides will pass above the arc on the flats, as is shown in Figure 157, in which the chamfer A meets the side of the head at B, and does not, therefore, meet the arc C. The length of side lying between B and D in the side view corresponds with the part lying between E and F in the end view. Fig. 157. If we compare this head with similar views of a square head G, both being of equal widths, and having their chamfer circles at an equal distance from the sides of the flats, and at the same angle, we perceive at once that the amount of chamfer necessary to give the same distance between the chamfer circle and the side of the bolt (that is, the distance from J to K, being equal to that from L to M), the length of the chamfer N for the square head so greatly exceeds the length A for the hexagon head that the eye detects the difference at once, and is instinctively informed that G must be square, independently of the fact that in the case of the square head, N meets the arc O, while in the hexagon head, A, which corresponds to N, does not meet the arc C, which corresponds to O. When, however, the chamfer is drawn, but just sufficient to meet the flats, as in the case of the hexagon H, and the square I, in Figure 157, the chamfer line passes from the chamfer circle to the side of the head, and the distinction is greater, as will be seen by comparing head H with head I, both being of equal width, having the same angle of chamfer, and an amount just sufficient to meet the sides of the flats. Here it will be seen that in the hexagon H, each side of the head, as P, meets the chamfer circle A. Whereas, in the square head these two lines are joined by the chamfer line Q, the figures being quite dissimilar. Fig. 158. It is obvious that whatever the degree or angle of the chamfer may be, the diameter of the chamfer circle will be the same in any view in which the head may be presented. Thus, in Figure 158, the line G in the side view is in length equal to the diameter of circle G, in the end view, and so long as the angle of the chamfer is forty-five degrees, as in all the views hitherto given, the width of the chamfer will be equal at corresponding points in the different views; thus in the figure the widths A and B in the two views are equal. Fig. 159. If the other view showing a corner of the head in front of the head be given, the same fact holds good, as is shown in Figure 159. That the two outside flats should appear in the drawing to be half the width of the middle flat is also shown in Figure 158, where D and E are each half the width of C. Let us now suppose, that the chamfer be given some other angle than that of 45 degrees, and we shall find that the effect is to alter the curves of the chamfer arcs on the flats, as is shown in Figure 160, where these arcs E, C, D are shown less curved, because the chamfer B has more angle to the flats. As a result, the width or distance between the arcs and line G is different in the two views. On this account it is better to draw the chamfer at 45 degrees, as correct results may be obtained with the least trouble. If no chamfer at all is to be given, a hexagon head may still be distinguished from a square one, providing that the view giving three sides of the head, as in Figure 158, is shown, because the two sides D and E being half the width of the middle one C, imparts the information that it is a hexagon head. If, however, the view showing but two of the sides and a corner in front is given, and no chamfer is used, it could not be known whether the head was to be hexagon or square, unless an end view be given, as in Figure 161. Fig. 160. If the view showing a full side of the head of a square-headed bolt is given, then either an end view must be given, as in Figure 162, or else a single view with a cross on its head, as in Figure 163, may be given. It is the better plan, both in square and hexagon heads, to give the view in which the full face of a flat is presented, that is, as in Figures 155 and 163; because, in the case of the square, the length of a side and the width across the head are both given in that view; whereas if two sides are shown, as in Figure 161, the width across flats is not given, and this is the dimension that is wanted to work to, and not the width across corners. In the case of a hexagon the middle of the three flats is equal in width to the diameter of the bolt, and the other two are one-half its width; all three, therefore, being marked with the same set of compasses as gives the diameter of the body of the bolt, were as shown in Figure 152. For the width across flats there is an accepted standard; hence there is no need to mark it upon the drawing, unless in cases where the standard is to be departed from, in which event an end view may be added, or the view showing two sides may be given. Fig 161. Fig. 162. Fig. 163. Fig. 164. To draw a square-headed bolt, the pencil lines are marked in the order shown by figures in Figure 164. The inking in is done in the order of the letters Fig. 165 To draw the end view of a hexagon head, first draw a circle of the diameter across the flats, and then rest the triangle of 60 degrees on the blade Fig. 166. If the other view of the head is to be drawn, then first draw the lines Fig. 167. If the diameter across corners of a square head is given, and it be required to draw the head, the process is as follows: For a view showing one corner in front, as in Figure 168, a circle of the given diameter across corners is pencilled, and the horizontal centre-line Fig. 168. Fig. 169. If the flats are to lie in the other direction, the pencilling will be done as in Figure 169. The circle is marked as before, and with the triangle placed as shown at T 1, line Fig. 170. For a hexagon head we have the processes, Figures 170 and 171. The circle is struck, and across it line Fig. 171. The chamfer circles are left out of these figures to reduce the number of lines and so keep the engraving clear. Figure 171 shows the method of drawing a hexagon head when the diameter across corners is given, the lines being drawn in the alphabetical order marked, and the triangle used as will now be understood. Fig. 172. Fig. 173. It may now be pointed out that the triangle may be used to divide circles much more quickly than they could be divided by stepping around them with compasses. Suppose, for example, that we require to divide a circle into eight equal parts, and we may do so as in Figure 172, line Let it be required to draw a polygon having twelve equal sides, and the triangle of sixty is used, marking all the lines within the circle in Figure 173, except Fig. 174. Fig. 175. In both of these examples we have assumed that the diameter across corners of the polygon was given, but suppose the diameter across the flats were given, and the construction is a little more complicated. Circle Fig. 176. It is obvious, also, that the triangle may be used to draw slots radiating from a centre, as in Figure 176, where it is desired to draw a chuck-plate having 6 slots. The triangle of 60 is used to draw the centre lines, If the slots are not to radiate from the centre of the circle the process is as follows: The outer circle Fig. 177. For a five-sided figure it is best to step around the circumference of the circle with the compasses, but for a three-sided one, or trigon, the construction is as follows: It will be found that the compasses set to the radius of a circle will accurately divide it into six equal divisions, as is shown in Figure 178; hence every other one of these divisions will be the location for a corner of a trigon. The circle being drawn, a line A, 179, is drawn through its centre, and from its intersection with the circle as at Fig. 178. Fig. 179. Fig. 180. Draw the line O P, Figure 181, and at a right angle to it the line O B; divide these two lines into parts of one inch, as shown in the cut, which is divided into inches and quarter inches, and from these points of division draw lines crossing each other as shown. Fig. 181. From the point O, draw diagonal lines, at suitable angles to the line O P. As shown in the cut, these diagonal lines are marked:
But still others could be added for figures having a greater number of sides. 1. Now it will be found as follows: Half the diameter, or the radius of a piece of cylindrical work being given, and the number of sides it is to have being stated, the length of one side will be the distance measured horizontally from the line O B to the diagonal line for that particular number of sides. Example.—A piece of work is 2-1/2 inches in diameter, and is required to have 9 sides: what will be the length of the sides or flats? Now the half diameter or radius of 2-1/2 inches is 1-1/4 inches. Then look along the line O B for 1-1/4, which is denoted in the cut by figures and the arrow A; set one point of the compasses at A, and the other at the point of crossing of the diagonal line with the 1-1/4 horizontal line, as shown in the figure at Again: A piece of work, 4 inches in diameter, is to have 9 sides: how long will each side be? Now half of 4 is 2, hence from B to But suppose that from the length of each side, and the number of sides, it is required to find the diameter to which to turn the piece; that is, its diameter across corners, and we simply reverse the process thus: A body has 9 sides, each side measures 27/32: what is its diameter across corners? Take a rule, apply it horizontally on the figure, and pass it along till the distance from the line O B to the diagonal line marked 9 sides measures 27/32, which is from 1-1/4 on O B to For any other number of sides the process is just the same. Thus: A body is 3-1/2 inches in diameter, and is to have 5 sides: what will be the length of each side? Now half of 3-1/2 is 1-3/4; hence from 1-3/4 on the line O B to the point C, where the diagonal line crosses the 1-3/4 line, is the length of each of the sides. 2. It will be found that the length of a side of a square being given, the size of the square, measured across corners, will be the length of the diagonal line marked 45 degrees, from the point O to the figures indicating, on the line O B or on the line O P, the length of one side. Example.—A square body measures 1 inch on each side: what does it measure across the corners? Answer: From the point O, along diagonal line marked 45 degrees, to the point where it crosses the lines 1 (as denoted in the figure by a dot). Again: A cylindrical piece of wood requires to be squared, and each side of the square must measure an inch: what diameter must the piece be turned to? Now the diagonal line marked 45 degrees passes through the 1-inch line on O B, and the inch line on O P, at the point where these lines meet; hence all we have to do is to run the eye along either of the lines marked inch, and from its point of meeting the 45 degrees line, to the point O, is the diameter to turn the piece to. There is another way, however, of getting this same measurement, which is to set a pair of compasses from the line 1 on O B, to line 1 on O P, as shown by the line D, which is the full diameter across corners. This is apparent, because from point O, along line O B, to 1, thence to the dot, thence down to line 1 on O P, and along that to O, encloses a square, of which either from O to the dot, or the length of the line D, is the measurement across corners, while the length of each side, or diameter across the flats, is from point O to either of the points 1, or from either of the points 1 to the dot. Fig. 182. After graphically demonstrating the correctness of the scale we may simplify it considerably. In Figure 182, therefore, we have applications shown. A is a hexagon, and if one of its sides be measured, it will be found that it measures the same as along line 1 from O B to the diagonal line 45 degrees, which distance is shown by a thickened line. At 1-1/2 is shown a seven-sided figure, whose diameter is 3 inches, and radius 1-1/2 inches, and if from the point at 1-1/2 (along the thickened horizontal line), to the diagonal marked 49 degrees, be measured, it will be found exactly equal to the length of a side on the polygon. At C is shown part of a nine-sided polygon, of 2-inch radius, and the length of one of its sides will be found to equal the distance from the diagonal line marked 52-1/2 degrees, and the line O B at 2. Let it now be noted that if from the point O, as a centre, we describe arcs of circles from the points of division on O B to O P, one end of each arc will meet the same figure on O P as it started from at O B, as is shown in Figure 181, and it becomes apparent that in the length of diagonal line between O and the required arc we have the radius of the polygon. Example.—What is the radius across corners of a hexagon or six-sided figure, the length of a side being an inch? Turning to our scale we find that the place where there is a horizontal distance of an inch between the diagonal 45 degrees, answering to six-sided figures, is along line 1 (Figure 182), and the radius of the circle enclosing the six-sided body is, therefore, an inch, as will be seen on referring to circle A. But it will be noted that the length of diagonal line 45 degrees, enclosed between the point O and the arc of circle from 1 on O B to one on O P, measures also an inch. Hence we may measure the radius along the diagonal lines if we choose. This, however, simply serves to demonstrate the correctness of the scale, which, being understood, we may dispense with most of the lines, arriving at a scale such as shown in Figure 183, in which the length of the side of the polygon is the distance from the line O B, measured horizontally to the diagonal, corresponding to the number of sides of the polygon. The radius across corners of the polygon is that of the distance from O along O B to the horizontal line, giving the length of the side of the polygon, and the width across corners for a given length of one side of the square, is measured by the length of the lines A, B, C, etc. Thus, dotted line 2 shows the length of the side of a nine-sided figure, of 2-inch radius, the radius across corners of the figure being 2 inches. Fig. 183. The dotted line 2-1/2 shows the length of the side of a nine-sided polygon, having a radius across corners of 2-1/2 inches. The dotted line 1 shows the diameter, across corners, of a square whose sides measure an inch, and so on. Fig. 184. This scale lacks, however, one element, in that the diameter across the flats of a regular polygon being given, it will not give the diameter across the corners. This, however, we may obtain by a somewhat similar construction. Thus, in Figure 184, draw the line O B, and divide it into inches and parts of an inch. From these points of division draw horizontal lines; from the point O draw the following lines and at the following angles from the horizontal line O P. Fig. 185. A line at 75° for polygons having 12 sides. From the point O to the numerals denoting the radius of the polygon is the radius across the flats, while from point O to the horizontal line drawn from those numerals is the radius across corners of the polygon. Fig. 186. A hexagon measures two inches across the flats: what is its diameter measured across the corners? Now from point O to the horizontal line marked 1 inch, measured along the line of 60 degrees, is 1 5-32nds inches: hence the hexagon measures twice that, or 2 5-16ths inches across corners. The proof of the construction is shown in the figure, the hexagon and other polygons being marked simply for clearness of illustration. Fig. 187. Fig. 188. Let it be required to draw the stud shown in Figure 185, and the construction would be, for the pencil lines, as shown in Figure 186; line 1 is the centre line, arcs, 2 and 3 give the large, and arcs 4 and 5 the small diameter, to touch which lines 6, 7, 8, and 9 may be drawn. Lines 10, 11, and 12 are then drawn for the lengths, and it remains to draw the curves in. In drawing these curves great exactitude is required to properly find their centres; nothing looks worse in a drawing than an unfair or uneven junction between curves and straight lines. To find the location for these centres, set the compasses to the required radius for the curve, and from the point or corner A draw the arcs Fig. 189. Similarly for the curve Fig. 190. Fig. 191. To draw the piece shown in Figure 188, the lines are drawn in the order indicated by the letters in Figure 189, the example being given for practice. It is well for the beginner to draw examples of common objects, such as the hand hammer in Figure 190, or the chuck plate in Figure 191, which afford good examples in the drawing of arcs and circles. In Figure 191 Fig. 191 a. In Figure 192 is shown the pencilling for a link having the hubs on one side only, so that a centre line is unnecessary on the edge view, as all the lengths are derived from the top view, while the thickness of the stem and height of the hubs may be measured from the line A. In Figure 193 there are hubs (on both sides of the link) of unequal height, hence a centre line is necessary in both views, and from this line all measurements should be marked. Fig. 192. Fig. 193. In Figure 194 are represented the pencil lines for a double eye or knuckle joint, as it is sometimes termed, an example that it is desirable for the student to draw in various sizes, as it is representative of a large class of work. These eyes often have an offset, and an example of this is given in Figure 195, in which A is the centre line for the stem distant from the centre line B of the eyes to the amount of offset required. Fig. 194. Fig. 195. Fig. 196. Fig. 197. In Figure 196 is an example of a connecting rod end. From a point, as A, we draw arcs, as B C for the width, and E D for the length of the block, and through A we draw the centre line. It is obvious, however, that we may draw the centre line first, and apply the measuring rule direct to the paper, and mark lines in place of the arcs B, C, D, E, and F, G, which are for the stem. As the block joins the stem in a straight line, the latter is evidently rectangular, as will be seen by referring to Figure 197, which represents a rod end with a round stem, the fact that the stem is round being clearly shown by the curves A B. The radius of these curves is obtained as follows: It is obvious that they will join the rod stem at the same point as the shoulder curves do, as denoted by the dotted vertical line. So likewise they join the curves E F at the same point in the rod length as the shoulder curves, both curves in fact being formed by the same round corner or shoulder. The centre of the radius of A or B must therefore be the same distance from the centre of the rod as is the centre from which the shoulder curve is struck, and at the same time at such a distance from the corner (as E or F) that the curve will meet the centre line of the rod at the same point in its length as the shoulder curves do. Fig. 198. Figure 198 gives an example, in which the similar curved lines show that a part is square. The figure represents a bolt with a square under the head. As but one view is given, that fact alone tells us that it must be round or square. Now we might mark a cross on the square part, to denote that it is square; but this is unnecessary, because the curves F G show such to be the case. These curves are marked as follows: With the compasses set to the radius E, one point is rested at A, and arc B is drawn; then one point of the compass is rested at C, and arc D is drawn; giving the centre for the curve F by a similar process on the other side of the figure, curve G is drawn. Point C is obtained by drawing the dotted line across where the outline curve meets the stem. Suppose that the corner where the round stem meets the square under the head was a sharp one instead of a curve, then the traditional cross would require to be put on the square, as in Figure 199; or the cross will be necessary if the corner be a round one, if the stem is reduced in diameter, as in Figure 200. Fig. 199. Fig. 200. Fig. 201. Figure 201 represents a centre punch, giving an example, in which the flat sides gradually run out upon a circle, the edges forming curves, as at A, B, etc. The length of these curves is determined as follows: They must begin where the taper of the punch joins the parallel, or at C, C, and they must end on that part of the taper stem where the diameter is equal to the diameter across the flats of the octagon. All that is to be done then is to find the diameter across the flats on the end view, and mark it on the taper stem, as at D, D, which will show where the flats terminate on the taper stem. And the curved lines, as A, B, may be drawn in by a curve that must meet at the line C, and also in a rounded point at line D. Triangles Curves Feedback |