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Aviation History
1931
1931 - 0290.PDF
18 SUPPLEMENT TO FLIGHT MARCH 27, 1931 THE AIRCRAFT ENGINEER and equal to MH (if Mi, had been negative £2m]t would have been in the opposite direction) and similarly k,mk in the direction A'A and equal to MA. At mB and mA erect perpendiculars to BJB and A!A respectively, which cut each other at X. With OX as diameter draw a, circular arc jnuXmA. By shading in the space between the two arcs the bending moment diagram is obtained. Suppose the moment at x ins. from A is required. The linear measurement a; is converted into an angular measure- ment by multiplying by 57-3 [A, giving 57-3 \ix.° Draw the vector OR, so that D ROA1 = 57-3 [xx degrees : the required moment is given by the intercept r1J"2 between the two arcs. Where the arc k^2 is below the arc mAw,B, the moment is -f w ; where the reverse is the case it is — ve. It is seen that the two arcs intersect at two points, Cx and C8. Here obviously the bending moment is zero, and, therefore, these points represent points of contraflexure on the beam. The variation of the bending moment as the beam is traversed from A to B is obtained from the diagram by swinging the vector OR from OA1 to OB1. At OA1 (A on the beam) the moment = kt wA = MA and + ve. It diminishes to zero at c,, reaches a maximum —ve value at OX, becomes zero again at C2, and finally reaches a + ve value equal to k% mI( = MB at OB1 (B on the beam). The end load P affects the shear, aa well as the bending moment in the beam, and the true value of the shear can also be read off the diagram. For example, the shear at x in. from A, i.e., at OR in the diagram, is obtained by measuring the length of the line joining r2, the point of intersection of OR and the arc mAmB, to X, on the bending moment scale, and then multiplying the result by y.. The ease of the single bay with more complicated loading will not be dealt with at this juncture, as the correct pro- cedure is fully covered by the following treatment of the continuous beam. In this treatment it will be found that for any particular bay two diagrams are drawn ; the first enables the bending moments at the supports to be found, and the second is the actual bending moment diagram for the bay based on the end moments and other quantities found by means of the first diagram. When, therefore, it is desired to draw the bending moment diagram for a single bay in which the end moments are known, the procedure is to draw the first diagram as if the bay formed part of a continuous beam, and as if the end moments were unknown ; then to draw the actual bending moment diagram based on the known end moments and the data obtained by means of the first diagram. c Let C, B, A, be three consecutive supports of a continuous beam. As in the ordinary theorem of three moments, the method depends on finding an expression for the slope at B, firstly in terms of the bending moments at A and B and known constants, and, secondly, of those at C and B and known constants for that bay. Equating the two expres- sions so found gives a relation between MA, MB and M» When dealing with bays AB and BC, it is convenient to reckon the directions from A to B and from C to B positive, i.e., directions towards the common point B are + re. Before equating the two expressions for the slope *'B it will, therefore, be necessary to change the sign of one of them. Once the values of the bending moments at the supports are obtained, the actual drawing of the bending moment diagram is a simple matter, as nothing but straight lines and arcs of circles enter into the construction. Probably the clearest description of the procedure can be given by dividing it into three parts dealing with I. Changes of spar loading and concentrated loads. II. Changes of moments of inertia of spar section. III. Combination of I and II. The table of data below is first prepared. Each sub-bay is marked with a number, and the numerical suffixes to the various tabulated quantities denote the values of those quantities for the corresponding bays, e.g., a3 is the value of <x for sub-bay 3. When \it appears under sub-bay 3 it means that the value of \i for sub-bay 3 is the same as that TABLE I. Changes of Spar Loading and concentrated Loads. c I P/GI7 h oc us H-x ex., U-' //, 2 U/l A NOTE : P and I are constant over each main bay for this case. 272$
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