Timber Bridge. The type of truss shown is more suitable for somewhat smaller spans than 100 ft., but the stresses and scantlings are worked out on the assumption that the type must not be altered. The main objection is that the top boom, where it slopes to the abutment at either end, is 37 ft. 6 in. long, and has to resist in compression the maximum boom load owing to the shallow depth of the truss at the ends. To avoid using such a heavy stick of timber as this condition would require it will be necessary to use portal bracing to support it at mid length; this, again, will only leave a headroom on to the bridge of half the depth of the truss, say 8 ft. If the waggon will clear this, it will be all right; but if not, either the heavy section of timber must be used or the type of truss altered. It will be necessary to use the same scantling of timber right through the top boom, so that this strut practically fixes the section of the boom. The wide bays also cause a large bending moment on the bottom boom, while shorter bays would reduce it. The quickest way to arrive at the dead-load stresses will be by a stress diagram, while to obtain the live-load stress it will be necessary to draw a separate diagram for each position of the live load. As, however, we can tell by inspection the position to place the live load in to obtain the maximum stresses, it will be quicker to place the load in the required position and obtain the stresses by calculation. Draw first a frame diagram, assuming a central depth of 16 ft., and making all bays equal and 17 ft. long; this gives a total length of 17 x 6 = 102 ft. as the length of truss, centre to centre of supports, which allows 2 ft. landing at each end of truss. It is only possible to assume a dead load, and then, when all the sizes have been got out, a correct estimate can be made of the weight of the whole bridge, not forgetting the deck, rail bolts, connecting plates, etc. If this actual weight be much greater than that assumed, a fresh design must be got out, assuming a new dead load greater than the calculated weight, as the increased size of the numbers under the new conditions will still further add to the weight. We will assume the dead load at 1 ton per bay. The vertical members omitted from the frame diagram simply act as suspenders for the load on the bottom chord, and transfer the load direct to the top boom. After lettering the diagram as shown-i.e. between each external force in each space-the stress diagrain may be drawn. Draw the vertical line AD, and beginning at the top, set off the loads in the direction in which they act. Thus AB, the first load, is set off downwards, also BC and CD, while DE is set off upwards, as it represents the reaction, then EF and FG downwards, and finally GA upwards. Lines are now drawn parallel to the respective lines in the frame. The lines in the force diagram measured by the force scale, to which the loads were set off on AD, give the stresses in the corresponding lines in the frame diagram. Thus, AH in force diagram scales 5-87 tons, and equals the stress in AH on the frame diagram. The omitted suspenders are each subject to a tension of 1 ton-rather less, really, as about of the dead load is really due to the top half of the girder; but it is not necessary to allow for this in a small girder. The live load of 6 ft. by 6 36 ft. long 6 x 2 12 tons 6 tons full girder. It so nearly covers two bays that we may take the panel load as 3 tons. The maximum stress in HJ will occur when the head of the train has advanced halfway along the second panel; the panel point at G has then a maximum load of 3 tons. The reaction at R2=3=0·5 ton, and this is the shear on HJ. Then 0.5 × cosec = = = = = = The stress, or 0·5 × 2·294·58 tons, where = angle HJ makes with horizontal =25° 48'. By loading panel points 2 and 3 the maximum shear is obtained on KL, and also with the load coming on the reverse way on LM. reaction R, here +3x=2.5 tons. The stress KL and LM = 2.5 x cosec 43° 16' 25 x 1·463.65 tons. The maximum stress in JK occurs when panels 2+3 are loaded; it equals the vertical component of the stress in BJ, so that we must know this before we can calculate JK, as it is impossible to take a section anywhere through the truss, including JK, where there are not four unknown forces. Pass a section through the line a . . . . b, cutting three members. By taking moments about the point of intersection of two of the members we find the stress in the third. Thus R1 =3+x=3.5 tous. Then stress on BJ = SBJ × 14·4 = (3·5 × 34), SBJ 3.5 x 34 8.2 tons. = = The stress on JK is therefore 8.2 × sin 25° 48′ = 8·2 × 0·435 = 3.5 tons. The maximum stress in the top boom occurs in the member AH, and is a maximum when panel points 1+ 2 are loaded. The reaction R1 = 3 + 1 × 3 = 4 tons... SAH = 4.5 × cosec 25° 48′ = 4·5 × 2.29 103 tons. The maximum live-load stress in the bottom boom occurs in the part GH and is a maximum when apices 1 and 2 are loaded. It equals the reaction R1 × cotan = 4·5 × 2.069.27 tons. The live-load diagram is drawn the same way as the dead. That illustrated shows the live-load stresses when apices 2 and 3 are loaded. The other diagrams may be constructed in the same manner. We now have live Allowing for the ultimate strength of the timber in tension 8000 lbs. and in compression 5000 lbs., the safe loads with a factor of safety of 8 are 1000 tension and 750 compression. Let 7= length of member in inches; r = radius of gyration = minimum width x 0.289; c = constant = 3000; S = safe load on member in pounds per sq. in. Assume 12 in. by 12 in. timber, r= 0.289 × 12=34. Then 0-42 ton = 940 lbs. per sq. in. The size assumed is sufficient to meet the stresses, but 12 in. by 12 in. is better, as an allowance must be made for For the sake of the joints the width should be the same as the top and bottom members. Try a section 12 in. x 8 in., T=0·289 × 8. Then The same size timber may be used as for HJ, or it may be slightly reduced if desired. 5.5 tons x 2240 lbs. = 1 in. wide. 1000 lbs. per sq. in. x 12 in. 12 in. by 6 in. would be a suitable size, and the same for the other verticals. Diagonal braces should be used in the top bracing and also in the deck to resist the wind stresses. The wind will not affect the boom section very much, as the maximum wind stress is in the middle of the bridge, and the boom stress is less there than at the ends. The vertical posts are subject to bending from the wind, but they are ample in size. The end member AH will be most affected; but to enter fully into the wind stresses would take up too much space. A plate containing several examples of stress diagrams appears opposite, which should be carefully studied and worked out by the reader. |