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a difference of something approaching 100 per cent.: Tredgold the highest, Clark th lowest, and Barlow about midway. New sets of experiments are desirable, especially fo large scantlings. From 1866 D. Kirkaldy has made a number of experiments on full size timbers, and of late years Mr. Lanza has made some at the Massachusetts Institute His results show that the ordinary formulæ require revision. For instance, a spruce bean 12 inches by 2 inches and 15 feet long broke with a central load of 5894 lbs. According to Tredgold’s formula, it ought to have carried a load of 8928 lbs. before breaking. Thi result was corroborated by other tests, and the general conclusion arrived at is that, wherea we have been accustomed to use as a constant in the familiar formula W=

cbd2

4 cwt. fo fir or pine beams (in fact, in one of his examples 530 lbs, is used), we ought really to use constant of not more than 2} cwt. The more thoroughly large size specimens, whether o wood or iron, are tested, the more will our knowledge of their strength be increased, an we shall be less dependent upon theories. (J. Slater, 1887). As an instance, put forwarı fo long since as :871-72 by Captain (now Colonel) Seddon from experiments by D. Kir kaldy on (1) white Riga fir, and (2) red Dantzic fir, where each piece was 20 feet long

White Riga fir was 18 inches square=169 square inches area.
Ultimate stress 331,260 lbs. = 1960 lbs. per square inch.

or 147-88 tons=126.04 tons per square foot.
Gave way at knots 2 feet 9 inches from centre. De flection 642.
Red Dantzic fir was 13.5 by 13-2 =178 square inches area.
Ultimate stress 309,120 lbs.= 1742 lbs. per square inch.

or 138.00 tons=112-02 tons per square foot.
Gave way at knot 0.9 off centre. Deflection 548 only.

Transactions, Royal Institute of British Architects, 1871-72, pp 156-164 Strictly speaking, the law that the breaking weight depended on the variations of stres was known before 1849, when it was shown that cast iron bars bruke with halt' tiie statica breaking weight when subjected to continued repetitions of load. Subsequently Sir W Fairbairn carried out an experiment on a riveted girder subjected to continual loading and unloading for a period of two or three years. It brok9 with two-fifths of the statica breaking load, and after repairing, with one-third the statical breaking weight, after over one million changes of load in each case. The statical breaking stress was not, as com monly assumed, the exact measure of the structural value of a material. (Prof. Unwin in Builder, 1887, p. 741).

16299. PROBLEM I.-To find the breaking weight of a beam, the load being in the middle and all the dimensions known. The ends loose or supported.

For A, Timber beams :-
b d’S lbs.
bd? S cwt.
bd2 s tons

a 4d S
W lbs. 1
=Wcwt. 1

W tons.

1 W lbs.
I feet
I feet
I feet

lius.
Here S, in the first formula, represents the value of the breaking weight in pounds in the
middle, taken from the preceding table : in the other two it would be deduced from it;
359 lbs.

3:21 thus, taking Riga fir =3.21 for cwt., and

•160 for tons: b breadth in

20 cwt. inches; i depth in inches ; 1 distance between the points of support, in feet; a area of section. These letters will be continued in these problems, until other values are attached to them. W will always represent the breaking weigh

112 lbs.

bd2 F
67

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1629h. For B, Wrought or cast iron rectangular beams :-
b d2 Slbs.

bd? Scwt.
= W lbs.
=Wcwts.

W lbs.
Here F represents the weight in pounds borne by a rod 1 inch square, when the strain is
as great as the rod will bear without destroying part of its elastic force, = 1.5,300 for cast
iron. From a mean of 265 experiments hy Hodgkinson and Fairbairn, it appears i by
Gregory) that a weight of 454:4 pounds in the middle of a bar of cast-iron, 1 inch square
and 4:5 feet bearing, produced fracture. Therefore, for a bar of any other dimensions,
we have :
bd2C=2045
bd2 C=18.25

bd?C='912
W lbs. 1
W cwts

W tons.
1629i. For C, Cast iron girder (Tredguld's section).

*(1-9 *pa)= W Ibs. Here q, difference between the breadth in the iniddle and the extreme'breadth= .625 as found to answer in practice; and p, depth of the narrow part in the middle = '7 as found to answer. When the middl, part of the beam is omitted, except susli sient uprights to connect the top and bottom bars, then *(1-p')= W lbs. Here

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bdF
61

bd?

64

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d shole depth; and p depth of part omitted. If the thickness of the web be about sth
or both of the depth of the beam, then--
ad C=514

ad C=26
W cwts. | W tons.

ad C=2.166

W tons.
I inches
7 inches

I feet Here 514 ewts. may be used for side castings, or 536 cwts. for erect castings. The other 514 cwts.

536 quantities are obtained thus : = 257, called 26 tons; 26-8, called 27 tons.

20 for tons When l is used in feet, 29 =2-166: a represents area of bottom flange in inches.

1629k. For D, Cast iron girder (Hodgkinson's pattern):adC=514

ad
=W tons.

= W tons. itd

. = P tons. 7 inches

I feet

1 ft. Here a and d as before; P permanent load distributed, or about one-fourth of the breaking weight distributed ; and multiplied by 2 when the ends are fixed = one-half BW. From the experiments above quoted from Gregory, we obtaina d 4852

ad 43-33
W lbs. 1 Wcwts.

ad 2-166

W tons. 16291. Gregory's work also states an arbitrary formula given by Mr. Dines, which he found to be tolerably correct in all cases where the length of the girder did not exceed 25 fuet; its depth in the centre not greater than 20 inches; the breadth of the bottom flange

not less than one-third, or more than half, the depth; the thickness of the metal not less than Jath of the depth. Then1792 [o da-(6–6.)?,?] =W lbs. | [bd? (6–6,)d,?] = Wcwts.

018 [b d?? (0–69)d,?] =W tons. Here b entire breadth of bottom flange; b, thickness of the vertical part; d depth of whole girder ; d, depth without the lower flange, all in inches ; l length in feet.

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ad C=6.25

I feet

7

WI

W=150 xI= 100

88d

1629p. For F, Wrouyht-iron p'ate girder :ad (= 1500 7 inches - W cwts.: area of the top Aange being | greater than the bottom flange, and thi

adc thickness of the web about or be of the depth of the beam.

1 inches

- W tons, in which case C = 75 for deep plates, as 22 inches; cr= 60 for less depth, as 7 inches. tons, when d is more than and the area of the top flange is 1.75 of the bottom fange Here a area of bottom flange in inches'; some calculators deduct the rivet holes from it total width. For depths under 12 inches, the width of the top flange should be half the depth; and when over 12 inches, one-third. With the latter proportion, feathers of stiffening pieces should be used to supply the deficiency in lateral stiffness occasioned by the reduced width of flange (par. 1629c.). The usual thickness of web for all depth: under 3 feet is inch. Fairbairn ( Tubular Bridges, p. 247) discovered that the top flang should have an area double that of the lower one to give the strongest form of wrought irou beam, a contrary principle to that obtained in cast iron. 16299.

To find the area of either of the Aanges at the centre of a girder supported a both ends, the formula is

8sd

Here W represents live and dead loads uniformly distributed in tons; 1 span in feet; d depth in inches; and s sale strain per square inch of metal on the flange, in tons.

Therefore, say 8x8=5x8-33 sect. area

45 squar inches, excluding rivet holes in the bottom flange. This formula is tre equivalent o W=weight per foot run x 12

When the sectional area of the top fange is to be greate than that of the bottom fange, multiply the latter area by 1.2. The average sectiona area of a theoretically proportioned fanged girder may be taken at frds of the central sectional area. To find the sectional areas of either fange at any point along the whold length of the girder, the formula is

2sd

(1-2)=a, excluding rivet holes in bottom flange. Here W weight per foot run in tons; x lesser segment into which the span is divided 1-r greater segment; s and d as before. To find the sectional area at any length of the web, the formula is -= a square inches. Here x is the distance from the centre; Wand s as before. The vertical strain, at the centre of the beam, when one-half of the girder is fully loaded, is equal to of the fully loaded beam, that is Wxl. At the ends or pillars, the vertical strain is greatest, and is equal to } Wxl. The strain at the centre, when the load is uniformly distributed, is obtained from the formula = 8. Here W distributed load in tons; I length in feet; d depth in feet; s strain in tons of compression in the top flange and tension in the bottom flange. Half the load collected at the centre of a girder being equal to the load distributed, the above formula becomes At any other point, ratios of strain will be as the square of half the span to the square of the segments into which a given point divides the span. The approximate strain at the centre, per square inch, on any beam, may be obtained from the formula

Here W distributed load in tons; l length in terms of the depth; a sectional area in square inches; and s strain in tons per square inch. To find the sectional area of a fange for a plate girder lixed at one end and free at the other the formula is exclusive of rivet holes in the top flange; W weight in tons at the end of the girder; x length in feet froin loaded end to the point where the seetional areas are required; d depth in feet; s safe strain per square inch in tons. When the load is uniformly distributed, using the same notations as before, except that W in tons is

W r2 the load per foot run, the fremula is

2 sd

=a. (Engineer's, Architect's, fc., Pucket-book, 1865.)

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8

WI
d

WI
4d

WI

=S. 8 a

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=a, ds

1629r. For G, Rolled irons or bars :ad C=6

= W tons, Here a area of bottom flange includes to above the upper part of I feet

the swelling for flange, as b c (fig. 6139.); or to the whole of the angle in plate girder F. A railway bar is often useful in country places; the intricate formulæ for the strength of the various parts will be found in

Barlow; and in the Engineer's gr., Pocket-book for Fig. 61.39.

1861.

I'III

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16291. For I, Reversed Tee irons: When z is one-fourth of the table or flange b c, and the form as 5: 12 of a rectangle, then ba 27 #x2:41

=SW. It was stated in the Oldham Mil Report that this form of beam, which might be considered to support a weight of say 1000 lbs., may be broken if reversed, that

is the flange placed uppermost, as T, with a weight of say 340 lbs. Hodgkinson experimented on two bars 4 feet inches long, the flange 4 inches wide, rib 11 inch deep, with

a thickness of metal of about inch. One bar was tried with the flange uppermost, the other bar with the flange downwards. The former broke with 2 cwt., the latter with

9 est. Experiments on three girders of this shape, the web being 2 inches high and inch thick, the flange 2 inches wide by 1 inch thick, and 24 inches long, were made by Cooper of Drury Lane. He stated that the gain in strength over a fliteh 0 2 inches by inch was 25 per cent.; the loss in stiffness being 30 per cent. The strength arising from the accumulation of the quantity submitted to tensile action bears out an adequate result, or 580 times its own weight, instead of 400, as 0 2 inches by 1 inch, and 0 0 2 inches by inch each, placed inch apart, showing over them an increase of strength of nearly 50 per cent. In using this form of section, it makes no difference whether the load be placed wholly on the top of the vertical web, or on the lower flange; the result obtained in either case was the same.-Builder, 1845, vol. iii. p. 593. The results of some other experiments on this useful form of iron are given in the Engineer's Pocket-book for 1861, p. 205. The formula W tons is also applicable to the trough-shaped section, as N ( fig. 613r.), as to the inverted Tee or

1 shape M, taking the two vertical ribs to be equivalent to one rib of the same depth and double the thickness. The thickness of the horizontal and vertical parts of these girders should be equal, or nearly equal, so as to obtain as near an equality in cooling as possible.

1629u For E. Mixed beams; Flitch beams and double fliich beams :- These beams are composed of an iron plate (cast or wrought) placed betwen two pieces of fir timber, fig. 615s., or of a plate placed on each side of the solid timber beam, fig. 613t. These plates again may have a table or flange, as in the case of the single plate; or of a half flange, as

in the case of the plates on each side of the beam. All these should be bolted, or otherwise secured together, to render them as homogeneous as possible. Hurst gives the formula

Tier (C6 +30t) = W in cwts. Here t breadth of the one, or. two, wrought iron flitches; b breadth of wood, both in inches ; C, coefficient = 4 teak, 3 oak, 2.5 fir, and 2-0 elm. Fairbairn considers that “the addition of the timber on each side of the plate gives increased stiffness, and renders it less liable to warp under strain. It is called a sandwich beam.” He states this beam “to be weak, comparing the results with those of the simple plate girder ; and its elasticity, although considerable, is nevertheless so imper-feet as to render it inadmissible for the support of great loads, whether proceeding from a dead weight, or one in motion over its surface. With riveted angles or flanges, the timbers on each side might have been useful in preventing lateral flexure, but they would not have contributed, in any great degree, to the vertical bearing powers of the beam.” (Application, &c., p. 284-5.) Rolled flat irons can now be obtained about 13 to 14 inches deep, from inch to an inch in width, up to 30 feet in lengih, and for special cases somewhat longer.

16290. The method of trussing a beam is explained in Carpentry (par. 2021, et seq.). 1629w. The formulæ for finding the strength for examples IV.and V., fig. 675., are and 12+ W2=s. Here l length in feet; d depth in feet—both measured from the points of intersection of the stay, tension rod, and top beam ; W load in tons uniformly distributed; h horizontal thrust on beam in tons; and s strain on inclined part of tension rod in tons. When the truss has more than one stay, h, i, and d will represent the same ; and 4 tensile strain on the horizontal portion of the rod. The strain on the inclined tie rod will be 14+nd+ =s; n the number of times that the horizontal distance between the pier and the nearest stay is contained in l. If any load be placed on the middle, the strain h will be doubled. If any load be placed on each of the stays, then I will represent the distance of each loaded stay from the nearest pier; d depth as before; h horizontal thrust on the part next the pier ; s tension on each of the inclined ties. Then

=h; and V + W5=s. To resist the strains of the inclined tie rods with safety, allow an inch of sectional area in the tie rod for every 5 tons of strain. The stay, being in compression, should be calculated as a column capable of supporting the load if in the middle, or one half if distributed. The beam, though in compression, should be capable of supporting the

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Load between the stays, as a beam exposed to transverse strain, according to the rules before given. Tie rods, when exposed to great strains, are not generally of much value, because the iron stretches.

16291. Mr. Cubitt experimented on an equal flanged cast iron girder, 27 feet long, I inches deep, and 4 inches broad across the flanges; the rods were 1 inch diameter. When the ends of the rods were placed above the beam, it was found to be weaker than having no rud at all. When the fastening was made at the upper end of the girder, and giving a distance to the rod of 6 inches below the gird:r instead of an inch, an increas:d stiffnes: was obtained of above a ton (Warr, p. 259). Some experiments are recorded in the Builder, 1857, on iwo beams of Dantzic timber, each 28 feet long, 14 inches square, with and without a tie rod. Barlow records an experiment (p. 158) on four beams, two being trussed similarly to the figure on plate xxxix. of Nicholson's Carpenters' New Guide. Mr. Cooper's experiments on trussed beams are given in the Builder, 1845, p. 612. Fou Trelis girders, another mode of trussing a beam, Fairbairn, p. 129, uses the same formu'a as for the plate girder F, but with the constant 60. For this, the student is referred to the Application &c. of Iron, enlarged edition, 1864.

Other conditions than that of the Weight in the Viddle. 1629y. To find the ultimate strength of a beam (section A or B), when a weighi

is placed somewhere between the middle and the end. RuleMultiply twice the length of the longer end, A, fig. 613u, by twice the length of the shorter end B, and divide the product by the whole length C, which will give the effec: tive length to be used as the divisor for the calculatior of strength under the conditions of the beam :- Thus say

a beam is 2x10)*(x5)

= 13.33 effi ctive length; and (S=2548 cast iron)x2x62 15 long

= 13,762.64 lbs, weight Hurst puts it as,

(37)** W

product of tuo lengths from each end. 1629z. Barlow (p. 39-40) has stated a case where a beam has to support tvo equa weights between the points of support, FF', as at D and E, Example I. fig. 6130 then since IC=iC= fli, and W=W', the general expression becomes (ID+E, C, W I DHE

W=f. And if we suppose further ID=;E, then it becomes simply ID.W=1

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Ilg. 613

13:33

li

-2

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W

W

Fig. 613v.

Now, if both weights act at the centre, it appears from the preceding investigation, that fli. (2W) = { Ii. W=IC.W=f. Whence the strain in the two cases will be to each other as ID to IC; and hence the following practical deduction:When a beam is loaded with a weight, and that weight is appended to an inflexible bar or bearing, as DE, in Ex. 2, the strain upon the beam will vary as the distance ID, or a the difference between the length of the beam and the length of the bearing; for tl bearing DE being inflexible, the strains will be exerted in the points D and F, exactly in the same manner as if the bearing was removed, and half the weight hung on at each o these paints. This remark may be worth the consideration of practical men in variou architectural constructions. He also puts the case of a beam, which, instead of being fixed a cach end, merely rests on two props, and extends beyond them on each side equal to half thuis distance, as Ex. 3: if the weights W W' were suspended from these latter points, each equal to one-fourth of the weight W, then this would be double of that which would be necessary to produce the fracture in the common case ; for, dividing the weight W into four equal parts, we may conceive two of these parts employed in producing the strain oi fracture at E, and one of each of the other parts as acting in opposition to W and W iind by these means tending to produce the fractures at F and F. This is the casi which has been erroneously confounded with a former one (fixed at each end), but the dis:inction between them is sufficiently obvious; because here the tension of the fibres, is the places where the strains are excited, are all equal; whereas in the former the middle one was double of each of the other two.

1630. Experiments are recorded in the Civil Engineer Journal, 1849, xi, page 44, or para'lel bars of cast iron, 4 feet 8f inches long and 4 inches square, placed on two

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