This section is from the book "The American House Carpenter", by R. G. Hatfield. Also available from Amazon: The American House Carpenter.

In this last polygon, a peculiarity seems to indicate an error: the line FG has no length; it begins and ends at the same point; or, rather, the polygon is complete without it. This is easily understood when it is considered that the two lines FG and G H do not contribute any strength towards sustaining the loads PQ and QR, and in so far as these weights are concerned they might be dispensed with, and the space occupied by the three triangles F, G, and H left free, and be designated by only one letter instead of three. Thus it appears that there are only four instead of five forces at the point PEFG Q, and that the four are represented by the lines of the polygon QPEFQ.

The peculiarity above explained arises from considering loads only on the top chord: the analysis of the case is correct as worked from the premises given; but in practice there is always more or less load on the bottom chord at the middle, which should be considered. This will be included in a case proposed in the next article. One half of the diagram of forces is now complete. The other half being exactly the same, except that it is in reversed order, need not here be drawn.

198. - Framed Girders: Load on Both Chords. - Let

Fig. 54 represent the axial lines of a girder carrying an equally distributed load on each chord, represented by the arrows and balls shown in the figure. Let each bay measure 10 feet, or the length of the girder be 50 feet, and its height be 4 1/2 feet. The diagram of forces (Fig. 55) for this girder is obtained thus:

Fig. 54.

Fig. 55.

The plan of the girder, Fig. 54, requires to be lettered as shown; having one letter within each panel and outside the frame, and one between every two weights or strains. Then, in Fig. 55, mark the vertical line K V at L, M, N, O, and P, dividing it by scale into equal parts, corresponding with the weights on the top chord represented by the arrows. For example, if the load at each arrow equals 6 1/2 tons, make K L, L M, M N, etc., each equal to 6 1/2 parts of the scale. Then KP will equal the total load on the top flange. Make the distance P V equal to the sum of the loads on the bottom chord. Then K V equals the total load on the girder. Bisect K V in U; then K U or U V equals half the total load; consequently, equals the reaction of the bearing at K or P of Fig. 54.

Now, to obtain the polygon of forces converging at K A U, Fig. 54, we have one of these forces, K U, or the reaction of the bearing at K A U, equal to K U, Fig. 55. From Udraw U A parallel with U A of Fig. 54, and from K draw K A parallel with the strut KA, Fig. 54, and intersecting the line U A at A, a point which marks the limit of K A and U A, and closes the polygon K A U K, the sides of which are in proportion respectively to the three strains which converge at the point A U K, Fig. 54. For example, since the line K U by scale measures the vertical reaction, K U, of the bearing at A U K, Fig. 54, therefore the line K A of the diagram of forces by the same scale measures the strain in the strut K A, Fig. 54, and the line A U of the diagram by the same scale measures the strain in the bottom chord at A U, Fig. 54. For the strains converging at K A B L, Fig. 54, of which two, K A and K L, are already known, we draw from A the line A B parallel with the line A B, Fig. 54, and from L draw LB parallel with L B, Fig. 54, meeting A B at B, a point which limits the two lines and closes the polygon K A B L K, the lines of which are in proportion respectively to the strains converging at the point K A B L,Fig. 54, as before explained. Of the five strains converging at U A B C T, we already have three - T U, U A,and A B; to obtain the other two, make U Q equal to P V, equal to the total load upon the lower flange; divide U Q into four equal parts, Q R, R S, S T, and T U, corresponding with the four weights on the lower chord, and represented, by the four balls, Fig. 54. Now, from T, the point marking the first of these divisions, draw T C parallel with T C, Fig. 54, and from B draw B C parallel with the strut B C, Fig. 54, meeting T C in C, a point which limits the lines B C and T C and closes the polygon T U A B C T, the sides of which are in proportion respectively to the strains converging in the point T U A B C T, Fig. 54. Of the five forces converging at M L B C D, we already have three - M L, L B, and B C; to obtain the other two, from M draw M D parallel with M D, Fig. 54, and from C draw CD parallel with C D, Fig. 54, meeting M D at D, a point limiting the lines M D and CD and closing the polygon M L B C D M, the sides of which are in proportion to the strains converging at the point M L B C D, Fig. 54. Of the five forces converging at the point S T C D F, three - S T, T C, and C D - are known; to obtain the other two, from S draw S F parallel with S F, Fig. 54, and from D draw D F, parallel with the strut D F, Fig. 54, meeting the line S F in F, a point limiting the two lines S F and D F and closing the polygon S T C D F S, the sides of which are in proportion to the strains converging at S T C D F, Fig. 54. One half of the strains in Fig. 54 are now shown in its diagram of forces, Fig. 55; and since the two halves of the girder are symmetrical, the forces in one half corresponding to those in the other, hence the lines of the diagram for one half of the forces may be used for the corresponding forces of the other half.

199. - Framed Girders: Dimensions of Parts. - The parts of a framed girder are the two horizontal chords (top and bottom) and the diagonals - the struts and ties. The top chord is in a state of compression, while the bottom chord experiences a tensile strain. Those of the diagonal pieces which have a direction from the top to the bottom chord, and from the middle towards one of the bearings of the girder, as K A, B C, or D F, Fig. 54, are struts, and are subjected to compression. The diagonal pieces which have a direction from the bottom to the top chord, and from the middle towards one of the supports, as A B or C D, Fig. 54, are ties, and are subjected to extension, (Art. 83). The amount of strain in each piece in a framed girder having been ascertained in a diagram of forces, as shown in Arts. 197 and 198, the dimensions of each piece may be obtained by rules already given. The dimensions of the pieces in a state of compression are to be ascertained by the rules for posts in Arts. 107 to 114, and those in a state of tension by Arts. 117 to 119 (see Arts. 226 to 229). Care is required, in obtaining the size of the lower chord, to allow for the joints which necessarily occur in long ties, for the reason that timber is not readily obtained sufficiently long without splicing. Usually, in cases where the length of the girder is too great to obtain a bottom chord in one piece, the chord is made up of vertical lamina, and in as long lengths as practicable, and secured with bolts. A chord thus made will usually require about twice the material; or, its sectional area of cross-section will require to be twice the size of a chord which is in one whole piece; and in this chord it is usual to put the factor of safety at from 8 to 10.

The diagonal ties are usually made of wrought iron, and it is well to secure the struts, especially the end ones, with iron stirrups and bolts. And, to prevent the evil effects of shrinkage, it is well to provide iron bearings extending through the depth of each chord, so shaped that the struts and rods may have their bearings upon it, instead of upon the wood.

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