When a girder is loaded with several different loads the strains are found graphically, as Fig. 970 (similarly to Fig. 957); and, arithmetically, they are proved by working each load out separately and adding the results together. On Fig. 970 it will be noticed that, in the small diagram, the perpendicular line of loads is divided proportionately to the reactions, and the depth line thrown out at right angles from that point The other working is precisely the same as Fig. 957.

Graphic diagrams in this class of work are also of service to the designer, in that they help him to economise judiciously by riveting plates on, intermediately, along the flanges, to withstand irregular strains without increasing the general sectional area. For instance, the diagram of strains, Fig. 971, clearly points out that an additional plate of sectional area, to resist the strain indicated by A B, and of a length equivalent to C D, can be judiciously added to the length of girder E F, designed to resist a strain amounting to G H.

Sometimes we have, in practice, to find out, by calculation or stress diagrams, the strains caused by certain loads at various intermediate points along a girder; and this is done mathematically by multiplying the reaction by the distance from the support, and dividing the result by the depth; while it is graphically treated (alternatively to Figs. 962 to 964, where the perpendiculars are scaled) as Fig. 972, which represents a girder of 12 feet span, 2 feet deep, loaded centrally with 10 tons (though this rule is applicable to any weight and position), and it is required to find the strain caused by the load at every 2 feet.

The girder is drawn to a scale by which feet and tons are made equivalent; and, perpendicularly upwards from A on the bottom flange, at the point of support, a line A B is measured to represent the reaction at that support; and. Iron, the top point B, a line is drawn .cross, parallel to the prder, and of indefinite length. The points at which it is required to find the strain are marked off along the top flange; and from A a line is drawn through each one to the indefinite line from B, and the distance from B, cut off by that intersecting line from A, gives the required answer when reduced to weight according to scale. The strains on the bottom flange are worked out conversely to the above, as shown on the lower part of the Fig.ure; though this is not necessary, as the strains on both flanges are alike in amount, only different in character.

Under Several Disttnet Loads PracticalBuildingConstruction01 775Under Several Disttnet Loads PracticalBuildingConstruction01 776

Fig. 971.

Under Several Disttnet Loads PracticalBuildingConstruction01 777

Proof of this is obtained by calculation on lines just laid down: -

(5 tons reactionx 2 feet from support)/(depth of 2 feet = 10/2=5 tons strain 2 feet away (5 tons reactionx 4 feet from support)/(depth of 2 feet = 20/2=10 tons strain 4 feet away

If the load be placed at any intermediate point between the supports the reactions of course are unequal, but when multiplied by the distance of the load from each point of support, the same result should be given.

Under Several Disttnet Loads PracticalBuildingConstruction01 778

Fig. 973.

For, supposing this girder were loaded 3 feet from one support, the reactions would then be in the proportion of 3 and 9 (to make up the 12 feet) - i.e., they would be respectively 7 1/2 tons and 2 1/2 tons, which give the same result when multiplied by the distances away as 7 1/2 tons x 3 feet, and 2 1/2 tons x 9 feet, each equal 21 1/2 tons strain under the load; and the strains, at any intermediate points between load and supports, can be ascertained in like manner by multiplying the reaction by the distance the point (at which it is required to find the strain) is away from that support. Fig. 973 represents the form of the strains on the girder in Fig. 972.

The - shearing - on girders is equal to the reaction at each support from the centre or load to that support when it is a concentrated load; while a uniform load causes a gradual shearing from nothing immediately under it to the amount of the reaction at the support, as Figs. 974, 975,

976, and 977, will explain. So that, in the latter case, the "shearing" at any point is equivalent to the amount of load distributed between that point and the centre.