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bars, or the spaces which each link of the chain occupies in the span of the bridge. If therefore the sum of these cosines, multiplied by the radius b, be deducted from the span of the bridge, the difference will be the length of the horizontal space occupied by the two upper links; and half of this space, multiplied by the secant of a, will be the length of one of those links. The sum of all the links will be the length of the chain. The sum of the differences of the drop-bars, added to the deflection of the upper link, will be the total deflection of the chain. The roadway may be made to rise with a fair curve, by making the rise bear a certain proportion to the fall or deflexion of the chain.

The sum of the deflexion of the chain, the length of the centre dropbar, and the rise of the road, will be the height of the point of suspension at the standard.

Example.

11 = 15° = angle of suspension.

b = 5 feet = length of each link.

41 = 17 = number of drop-bars.

98.625 : distance between the points of suspension.

3.5 feet = length of centre drop-bars.

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9.5250 : ditference.

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2.5418 = deflection of upper link. ft. in. ft. in. . 5 X 16 -I- 9.8602 X 2 99.7204 length of chain. The sum of column No. 5 7.5068 deflection of ditto. Ditto No. 6 1.5014 rise of roadway. 7.5068 -|- 1.5214 + 3.5 12.5082 height of the point Of suspension at standard. N. B. Column 5 is found by multiplying column 4 by 5 feet. Column 6 is one-fifth of column No. 5. Column 7 is equal to columns 5th --|- 6th -+ 3 .5 feet.

The geometrical construction of this problem will answer as. a proof to the foregoing rule, and will be of assistance in making plans of suspension bridges.

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In the right-angled triangle ABC make the angle A : 15°‘: angle of suspension, and the side AB 2 5 feet = length of one link of the chain. Divide the side CB into as many spaces, commencing at C, as there are drop-bars in -§— the space I 832- spaces, and join An A 2 n, &c. From the centre A with the radius AB describe‘ the arc BD, and complete the lines shewing the sines and cosines of the angles formed by the line AB and the radii An, A 2 n, A 3 n, &c. Then as these radii are parallel to the links of the chain, the sines of the angles E 1, E 2, E 3, &c. are the differences between thelengths of the drop-bars l, '2, 3, 4, &c. and the cosines of these angles are. the spaces which the links of the chain occupy in the space of the bridge. ‘ Supposing n I: length of the centre drop-bar, the other drop-bars will

be as follows :

Centre bar "8th, n + E 8. 7th, n + E 8 + E 7. . 6th, n + E 8 + E 7 + E 6, and so on. This does not in- clude the rise of the road, however, which is an arbitrary quantity.

X.-——Table shewing the W'eight or Pressure which a cylihdrical wroughtiron Bolt will sustain when supported at the ends, and bonded in the middleof its Length. By Captain J. THOMSON, Engineers.‘

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Observations on the foregoing Table.
There are two ways in which the bolt may be broken, either by a

cross strain, or by detrusion, which is the pulling out the part of the

bolt from between the points of support: besides these two ways in
which the fastening may be broken, the bolt may crush and cut away
the eye of the link which presses upon it.
1- If wg-—-weight or pressure in tons,
l=lengt_h of the bolt between the points of support in inches,

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support a cross strain; but when lbecomes less than ( 267

_ 4 . bolt will be liable to detrusion, to avoid which, d=-(.08 w) . But

detrusion can never take place when both the bolt and the link are formed of iron, or the same metal, because when 1 becomes less than

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( w )’l the link may be cut by the bolt; to obviate which, the"

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value of d should be = w

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3 These rules are taken from Tmcnoonn, the arbitrary quantities assumed by him being corrected by a comparison made, and 0 mean, taken from the best anthorlties.

This last equation supersedes the first

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when w=7 1.5 I’. This place is marked * in the table.
' Remarks on keys, hold-fasts, 8;c.
Put b=the breadth in inches,

d=the depth in inches,

10;-weight in tons,

l=l|ength of bearing in inches ; then the breadth should never

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As an example, suppose a bar 1 inch square to support 8 tons was fastened by a key -, required the breadth and depth?

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To support the accuracy of this table, a set of experiments was commenced. but the results from them were so unsatisfactory, that they were not continued. But during the proof of three bridges in which bolts of from 1% in. to 9} in. were used, with various lengths of bearing, and pressures of from 20 to 15 tons, the dimensions marked in the table were found suficiently strong i-n every instance; but the diameter of the bolt thus given could not be reduced much, or what was the same thing, the length of bearing could not be decreased with out a risk of failure.

- The best Swedish iron bolts did not sustain a greater pressure W than the ordinary Englishbolt iron, (rolled,not hammered.) The Swedish iron when strained in excess bent, and became dented as in the marginal figure: the side a was bulged or rose half as much as b was indented or bent, on the other side ; when the bolts were formed of English bolt iron (unhammered), numerous cracks opened on the convex surface of the bolts at a and 00, when the indentation at b amounted to Q; ofthe diameter of the bolt ; the bolt failed by these cracks meeting each other, and the centre part of the bolt was drawn out.

The bars, which these bolts connected, were calculated to sustain 9tons per square inch of section, and the eyes 7 tons, but when the whole were proved by a tension §rd greater than the calculated strength, the eyes broke more frequently than either the bars or bolts.

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The following table, for which we are also indebted to Captain

J. THOMSON, Engineers, will serve as a practical continuation of tho observations on roofing, in the last number of the J ournal,

'XI.—-A Table of the Scantlings of Beams of Teak or Saul Wood, to sus

tain a Terrace Roof not exceeding seven inches in thickness ,- the de. flea-ion not to exceed one-fortieth of an inch for each foot of length.

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Distance 12part ofBeams, LENGTH OF HEARING IN FEET. one foot from centre to cen

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Burgahs require to be made six times stiffer than beams, in order to prevent cracks in the terrace roof; and as they are invariably placed one foot apart, and have a breadth of three inches, they should be as many inches in depth as they are feet in length of bearing between the beams.

Explanation of the Table, with Ea-amples of its use.

I The table shews on inspection the scantlings of beams to support roofs not exceeding 80 lbs. per square foot, including the weight of the timber. lt has been calculated, according to the rule in Tnnnoomfs Carpentry, Section II. par. 90, the value of the constant quantity, a being taken at '01. The scantlings given in the table are measured in the middle of the beam ; the lower side is supposed to be cut straight, and the upper side with a curve of one or two inches, versed sine, for

each 10 feetin length of the beam. .

As the stiffest beam that can be cut out of a round timber has its breadth to its depth in the proportion of '6 to 1 nearly, the proportion of the breadth to the diameter will be as '5 to 1, or the breadth will be % the depth.

As the cost of timber is partly proportioned to its contents, the deeper the beams are made, the cheaper the roof will be within certain limits; and as the cutting of timbers through the heart or centre of the wood is supposed to render the beams more durable, all the timbers should be cut into two beams, particularly as the strength of the timber is not at all reduced by this measure.

There is, however, a proportion between the depth and breadth which cannot be exceeded without the risk of the beam breaking side. ways. T11nnoo1.n’s rule is, (Sec. 11. par. 82,) “ the breadth in inches should not be less than six-tenths of the length in feet, divided by the square root of the depth in inches." ' '

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