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

Girders of solid timber are useful for the support of floors only where posts are admissible as supports, at intervals of from 8 to 15 feet. For unobstructed long spans it becomes requisite to construct a frame to serve as a girder (Arts. 163, 182). A frame of this kind requires two horizontal pieces, a top and a bottom chord, and a system of struts and suspension-pieces by which the top and bottom chords are held in position, and the strains from the load are transmitted to the bearings at the ends of the girders. Various methods of arranging these struts and ties have been proposed. One of the most simple and effective is shown in Fig. 52, forming a series of isosceles triangles. The proportion between the length and height of a girder is important as an element of economy both of space and cost. When circumstances do not control in limiting the height, it may be determined by this equation from Transverse Strains, Art. 624 -

d = (175+l)l/2400; (90)

in which d is the depth or height between the axes of the top and bottom chords, and / is the length between the centres of bearings at the supports (d and l both in feet). This equation in words is as follows:

Rule LXVI. - To the length add 175; multiply the sum by the length; divide the product by 2400, and the quotient will be the required height between the axes of the top and bottom chords.

Example. - What should be the depth of a girder which is 40 feet long between the centres of action at the supports? For this the rule gives -

d = (175+40) x 40/2400 = 3.5 8 1/3;

or, the proper depth for economy of material is 3 feet and 7 inches.

The number of bays, panels, or triangles into which the bottom chord may be divided is a matter of some consideration. Usually girders from -

20 to 59 feet long should have 5 bays. | ||||||||

59 | " | 85 | " | " | " | " | 6 | " |

85 | " | 107 | " | " | " | " | 7 | " |

107 | " | 127 | " | " | " | " | 8 | " |

127 | " | 146 | " | " | " | " | 9 | " |

197.-Framed Girder and Diagram of Forces. - Let Fig. 52 represent a framed girder of six bays of, say, 11 feet each, or of a total length of 66 feet.

The lines shown are the axial lines, or the imaginary lines passing through the axes of the several pieces composing the frame. The six arrows indicate the six pressures into which the equally distributed load is supposed to be divided. Each of these is at the apex of a triangle, the base of which lies along the lower chord.

The spaces between the arrows are lettered; so, also, the space between the last arrow at either end and the point of support has a letter, and so has each triangle, and there is one for the space beneath the lower chord. These letters are to be used in describing the diagram of forces, as was explained in Art. 195. The diagram of forces (Fig. 53) for this girder-frame is drawn as follows, namely: Upon a verti-

cal line A N mark the points A, O, P, Q, R, S, and N, at equal distances, to represent the six equal vertical pressures indicated by the arrows in Fig. 52. The equal distances A O, OP, etc., may be made of any convenient size; but it will serve to facilitate the measurement of the forces in the diagram if they are made by a scale of equal parts, and the number of parts given to each division be made equal to the number of tons of 2000 pounds each which is contained in the pressure indicated by each arrow. On this vertical line the distance A 0 represents the load at the apex of the triangle B, or the points OCB (Art. 195); the distance OP represents the weight at the second arrow, or at the point O PE D C, and so of the rest. If the weights upon the points in the upper chord had been unequal, then the division of the vertical line A N would have had to be correspondingly unequal, each division being laid off by the scale, to accord with the weight represented by each. The line of loads, A N, being adjusted, the other lines are drawn from it (Art. 195), so as to make a closed polygon for the forces converging at each point of the frame, Fig. 52 - commencing with the point A B T, Fig. 52, where there are three forces, namely, the force acting through the inclined strut A B, the horizontal force in B T, and the vertical reaction A T at the point of support. This last is equal to half the entire load, or equal to the pressure indicated by the three arrows, A O, O P, and P Q, and is represented in Fig. 53 by A Q or A T. From the point Q draw a horizontal line Q B; this is parallel with the force B T of Fig. 52, in the lower chord. From the point A draw A B parallel with the strut A B of Fig 52. This line intersects the line B T in i? and closes the polygon A B T A; the point B defines the length of the lines A B and B T, and these lines measured by the scale by which the line of loads was constructed give the required pressures in the corresponding lines, A B and B T, of Fig. 52.

Fig. 53.

Taking next the point A B C O, where four forces meet, of which we already have two, namely, the force in the strut A B and the load A O - from the point O draw the horizontal line O C; this is parallel to the horizontal force O C of Fig. 52. Now from B draw B C parallel with the suspension-piece B C of Fig. 52. This line intersects O C in C, and the point C limits the lines O C and B C and closes the polygon A B C O A, the four sides of which are respectively in proportion to the four forces converging at the point A B C O of Fig. 52, and when measured by the scale by which the line of loads was constructed give the required strains respectively in each. Taking next the point B C D T, where four forces converge, of which we already have two, B C and B T - from B extend the horizontal line T B to D; from C draw C D parallel with C D of Fig. 52, and extend it to intersect T D in D, and thus close the polygon T B C D T.

The lines in a part of this polygon coincide - those from B to T; this is because the two strains B T and D T, Fig. 52, lie in the same horizontal line. Again, taking the point OCDE P, where five forces meet, three of which, O P, O C, and CD, we already have - draw from D the line D E parallel with D E of Fig. 52, and from P the line PE horizontally or parallel with PE of Fig. 52. These two lines intersect at E and close the polygon PO CDEP, the sides of which measure the forces converging in the point PO CD E, Fig. 52. Next in order is the point DEFT, Fig. 52, where four forces meet, two of which, TD and D E, are known. From E draw EF parallel with E F in Fig. 52; and from T, TF parallel with TFin Fig. 52; these two lines meet in F and close the polygon TD E F T, the sides of which measure the required strains in the lines converging at the point DEFT, Fig. 52. Taking next the point PEFG Q, Fig. 52, where five forces meet, of which we already have three, QP, PE, and EF - from F draw a line parallel with F G of Fig. 52, and from Q a line parallel with Q G of Fig. 52. These two intersect at G and complete the polygon QPEFG Q, the lines of which measure the forces converging at PEFG Q in Fig. 52.

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