This section is from the book "Safe Building", by Louis De Coppet Berg. Also available from Amazon: Code Check: An Illustrated Guide to Building a Safe House.

(g/f) for steel as = 15000 pounds, instead of 10000 pounds, can take the curve marked "15000 pounds - tension steel," in any of the three tables.

Or, if he wishes to Figure his iron at only 10000 pounds safe stress for tension, that is ( t/f) = 10000 pounds, he will, instead of using the curve marked "12000 pounds - tension wrought-iron," take the curve marked " 10000 pounds - single shear steel."

The writer, however, always sticks to the one set of values for tension, and for pins and rivets to those given in tables (also after Formula 115), as they are certainly safe values, and yet not low enough to make the work excessively costly. Iron contractors will frequently quote off-hand opinions of celebrated engineers saying these values are much too low and thus backing up their economical tendencies in trying to add lightness (and beauty) to roof trusses and plate girders, but as a rule the opinions are either not authentic, or else it is found that the celebrated engineer, when delivering the opinion, had a similar axe to grind, as had the contractor. Good engineers as well as good architects, will attempt to save their clients all they can, but hardly at the risk of taking chances in their most important constructive works. The Figures along the tops of Tables XXXVIII to XL give the values of either iron or steel, according, of course, to which curve is used.

The dotted curved lines in the Tables XXXVIII to XL give the safe bending-moments for iron and steel pins and rivets, and for these the lower horizontal lines of Figures are used.

The use of the tables is simple, similar to the other tables with curves. Thus if we have a 5/8" steel rivet bearing against a 7/16" steel plate, we use Table XXXV and pass horizontally to the right from the vertical Figure (or diameter marked) 5/8" till we strike the (fourth) slanting full bearing line, marked 7/16". This is one-third way between the vertical lines marked (below for steel) 4000 and the next vertical unmarked line to the right; as each vertical space at the bottom (steel) evidently is one-fourth of a thousand or 250 pounds, a third space will, of course, be 83 pounds, or a 5/8" steel rivet bearing against a 7/16" steel plate will resist safely 4083 pounds. Had we calculated arithmetically we should have had

For different values.

Curve of Bend-ing-moment.

How to use Tables.

5/8 7/16. 15000 =4101 pounds.

Had the rivet and plate been of wrought-iron we should have used the upper line of Figures; here the intersection is one-third way between the second and third unmarked lines after 3000; as there are five spaces above between each thousand, each space evidently represents one-fifth of a 1000 or 200 pounds for iron, so that the bearing value for a 5/8" iron rivet against a 7/16" iron plate would be 3000+ 200 1/3. 200 = 326 7 pounds. By calculation we should have had 5/8. 7/16. 12000 = 3281 pounds.

The single shearing value of a 5/8" steel rivet at (g/f) = 10000 pounds would be = 306 7 pounds ; for, the horizontal line 5/8" in Table XXXVIII strikes the single shearing curve for steel about one-third way between vertical line marked at the top 3000 and the next unmarked line to the right. Each space evidently represents 200 pounds, so that we should have for the steel 5/8" rivet in single shear 3000 +1/3. 200 = 306 7 pounds. By arithmetical calculation we should have had the area of a 4" circle multiplied by 10000 pounds or

0,3068. 10000 = 3068 pounds.

The double shearing value of a 5/8" rivet would be double this, or

= 6136 pounds, and this is confirmed by the Table (XXXVIII), as the horizontal line 5/8" strikes the double shearing curve for steel

20000 pounds about two-thirds way between vertical line marked above 6000 and its next unmarked line to the right, or

6000 +2/3.200=6134 pounds.

For the safe bending-momenl we find the horizontal line 5/8" strikes dotted curved line marked "safe bending-momenl on steel at 18000 pound" on the second vertical unmarked line to the left of the one marked at the bottom 500; each space below is evidently 20 pounds, and as they increase to the left we must add the two spaces, or

Bearing Values.

Shearing

Values.

500+40 = 540 pounds, which would be the safe bending-moment on a 5/8" steel rivet or pin.

Or to illustrate further take the last Example (III). We had a 1" steel plate and 1" steel rivet. The actual bending-moment we found was u.l/8 = 135000.1/8 = 16875 From Table XXXVIII we see that the safe bending-moment on a 1" steel rivet is the intersection of horizontal line 1" being one-third way between third and fourth vertical unmarked lines to the left of 1700 = 1700 + 3 1/2. 20 = 1767 pounds-inch.

Dividing the actual bending-moment at the joint 16875 pounds-inch, by the safe bending-moment on each rivet, will, of course, give the number of rivets required, or

16875/1767 = 9,55 being the same result as before, namely, 10 rivets. Take Example II, we had the same rivet, plate and strain, but there was but one cover-plate and the rivet was practically a cantilever. The bending-moment was u.l/2 =135000.1/2 = 62500 pounds-inch.

This divided by the safe bending-moment on each rivet (1767 pounds-inch) as found in Table XXXVIII gives the number of rivets required, or

62500/1767 = 35,4 or same as before.

The use of Tables for pins or tie-rods, is, of course, exactly similar to use for rivets, as already described.

For instance, we have a 2" pin bearing against a 1 1/8 inch thick eye-bar or head of a tie-rod. We use Table XXXVI, the 2 inch horizontal line, strikes the full slanting line marked 1 1/8 inch exactly on one of the vertical lines. If both pin and eye-bar are wrought-iron - (or if either were wrought-iron we should use the smaller value) - we use the top line of Figures, or we find our vertical line the third to the right of 24000 and as each space evidently represents 1000 pounds, the safe bearing value of our 2 inch pin against a 1 1/8 inch thick eye-bar would be 27000 pounds, if one or both are of wrought-iron. Had both pin and eye-bar been of steel, we should have used the bottom line of Figures, where each space evidently represents 1250 pounds, and we should have had 33750 pounds.

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