Modern Machine-Shop Practice Part 226

1.0689 = width of plate between rivets.

.5625 = thickness of plate.

------ 53445 21378 64134 53445 --------- Area of plate = .60125625 between rivets

Here then we find the area of plate left between the rivet holes to be 6.01 square inches, and as the area of the rivet is 6.01 square inches, the two are shown to be equal.

We may now place the various rivet diameters and the pitches that will make the rivet area and plate area in a single riveted joint equal in a table as follows:

TABLE OF RIVET DIAMETERS AND PITCHES FOR SINGLE RIVETED LAP JOINTS.

-------------------+------------------+-------- Thickness of Plate.|Diameter of Rivet.| Pitch.

-------------------+------------------+-------- 1/4 | 1/2 | 1-1/4 5/16 | 5/8 | 1-5/8 3/8 | 11/16 | 1-11/16 7/16 | 3/4 | 1-3/4 1/2 | 3/4 | 1-5/8 9/16 | 7/8 | 2 5/8 | 7/8 | 1-7/8 11/16 | 7/8 | 1-3/4 3/4 | 1 | ..

13/16 | 1 | 2 7/8 | 1 | 1-1/8 15/16 | 1-1/8 | 2-1/8 1 | 1-1/8 | 2-1/8 1-1/16 | 1-1/8 | 2-1/8 1-1/8 | 1-3/16 | 2-1/4 1-1/4 | 1-3/16 | 2-1/8 -------------------+------------------+--------

The rivets in double riveted lap joints, and in b.u.t.t strap joints having a single cover, are s.p.a.ced alike, because in both cases there are two rivets in one pitch, and the rivets are in single shear.

As there are two rivets in one pitch (instead of only one as in a single riveted joint), therefore the percentage of rivet section is doubled, and the plate section must therefore be doubled if the plate and rivet sections are to be made equal, and the rule for finding the required pitch is as follows:

_Rule._--To the amount of rivet area in one pitch, divided by the thickness of the plate, add the diameter of the rivet.

_Example._--Let the plate thickness be as in the last example 9/16, decimal equivalent = .5625, and the rivet diameter be 7/8 inch = decimal equivalent .875, the area of one rivet being .6013 square inch, and the pitch is calculated as follows:

.6013 = area of one rivet.

2 = the rivets in one pitch.

------ Plate thickness = .5625 ) 1.2026 ( 2.1377 1.1250 ------ 7760 2.137 5625 .875 = rivet diameter.

---- ----- 21350 3.012 = pitch.

16875 ----- 43750 39375 ----- 43750 39375 ----- 4375

We find, therefore, that the pitch is 3.012, or 3 inches (which is near enough for practical purposes), and we may now make it clear that this is correct.

[Ill.u.s.tration: Fig. 3257.]

In Fig. 3257 the joint is shown drawn one-half full size, and the length a of plate left between the rivet holes measures (as nearly as it is necessary to measure it) 2-5/32 inches, or 2.156, and if we multiply this by the thickness of the plate = .5625 inch, we get 1.2 square inches as the area of the plate left between the rivet holes.

Now there are two rivets in a pitch (as one-half of B, one-half of C, and the whole of F), and as the area of each rivet is .6, therefore the area of the two will be 1.2, and the plate section and rivet section are shown to be equal.

The area at _a_ is obviously the same as that at A, because the pitches of both rows of rivets are equal, this being an ordinary zigzag riveted joint.

We may now consider the diagonal pitch of the rivets, using the rule below.

The pitch 6, + 4 times the rivet diameter ------------------------------------------- = the diagonal divided by 10 pitch _p__{D}.

In this example the pitch has been found to be 3 inches, hence we have

.875 = diameter of rivet.

4 = constant.

----- 3.500

3 = pitch of the rivets.

6 = constant.

-- 18 3.5 = rivet diameter multiplied by 4.

---- 10 ) 21.5 (2.15 = the diagonal pitch.

20 -- 15 10 -- 50

The diagonal pitch, that is, the distance _p__{D}, Fig. 3257, is therefore found to be 2.15, or 2-1/8 inch full.

The amount of metal left between the rivets, measured on the diagonal pitch, is twice the dimension H multiplied by the thickness of the plate, and as this (with the diagonal pitch determined as above) always exceeds the pitch A or _a_, therefore if the plate fails, it will be along the line _a_, and not through the diagonal pitch.

We may now consider the total amount that the plates overlap in a double riveted lap joint zigzag riveted, this amount being twice the distance E, added to the distance V between the rows of rivets.

The distance E, Fig. 3257, is usually made one and a half times the diameter of the rivet, this being found to give sufficient strength to prevent the edge of the plate from tearing out and to prevent the rivet from shearing the plate out to the edge, rupture not being found to occur in either of these directions.

The rule for finding the distance V, when the diagonal pitch has been determined by the rules already explained, is as follows:

_Rule._--To the pitch multiplied by 11, add 4 times the rivet diameter, then multiply by the pitch, plus 4 times the rivet diameter. Then extract the square root and divide by 10.

Placed in formula, the rule appears as follows, _d_ representing the rivet diameter, and _p_ the pitch.

__________________________ /(11_p_ + 4_d_)(_p_ + 4_d_) ---------------------------- = distance V between the rows of rivets.

10

As this rule involves the extraction of the square root of the sum of quant.i.ties above the line, and as in determining the diagonal pitch, we have already determined the distance V, it is unnecessary to our purpose to carry out this latter calculation, as it is easier to find the diagonal pitch, and then, after drawing the joint, the distance between the rows of rivets can be measured if it is required, as it might be in finding the length of plate required to roll into a strake for a boiler of a given diameter and having a double riveted lap joint.

We may now consider chain riveted joints in comparison with zigzag riveted joints, which is especially necessary, because it has been a.s.sumed by some that the second row of rivets in a chain riveted joint added nothing to the strength of the joint.

[Ill.u.s.tration: Fig. 3258.]

Fig. 3258 represents a chain riveted joint, having the same thickness of plate, rivet diameter and pitch as the zigzag riveted joint in Fig.

3257, and it will be seen that the plate sections at a and at _a_ are the same in the two figures, and as there are four half rivets, which are equal to two rivets, in one pitch, therefore the strength of the two joints is equal.

Each joint can be as efficiently caulked as the other, as the rivet s.p.a.cing is the same and the edge of the plate is the same distance from the rivets in both cases.

The pitch of the rivets is obtained by the same rule as for zigzag riveted joints, and all we have now to consider is the distance apart of the two rows of rivets or distance V in the Fig. 3258, and for this there are two rules, the first being that it shall not be less than twice the diameter of the rivet, which would leave a dimension at H in the figure equal to the diameter of the rivet. The second rule is that a better proportion than the above is to multiply the diameter of the rivet by 3. This makes the dimension at H equal to twice the rivet diameter.

When the joints have double b.u.t.tstraps, the rivets may be s.p.a.ced as wide as the necessity for tight caulking will admit, because, on account of the rivets being in double shear, the rivet percentage exceeds the plate percentage.

[Ill.u.s.tration: Fig. 3259.]

The allowance for the rivets being in double shear is 75 per cent., or in other words, a rivet in double shear is allowed 1.75 times the area of the same size rivet in single shear.

STATIONARY ENGINE BOILERS.

The simplest form of horizontal boiler is the plain cylinder boiler, an example of which is given in Fig. 3259, and which is largely used in iron works and coal mines.

Boilers of this cla.s.s are easily cleaned, because the whole interior can be readily got at to clean.

As the bottom of this boiler gets thinned from wear, the boiler is turned upside down, thus prolonging its life.