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By H. Werner Rosenthal Dipl. Ing. FRIBA (auth.)

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65 (continued). , whenever a floor or a wall is to be supported. The numerical relationships for a centrally supported point load, and for a distributed load of equal magnitude is very simple. Because they are so common and also so useful we include them here: In Fig. 6 7, the diagram shows a beam Lm long loaded centrally by a load of W kN. The reaction (support) at each end equals W/2, so that maximum central bending moment equals M= W/2 xL/2 = WL kNm* 4 ~o o f'l o oo{_ b Fig. 66. a *Note that live loads are measured in "kN" (kilonewtons) and bending moments in kilonewtons x metres.

31 p p p Fig. 57. p Fig. 58. These moments can be plotted and the resulting graph indicates the distribution of moments along the beam, with a maximum, in this case, occurring at the centre. Already, it is apparent that a beam must be designed to withstand this maximum, at least in the centre. The concept of maximum bending moment is therefore of vital importance. When there are two symmetrical loads the moments are distributed differently (Fig. 59). In this case, the maximum is spread over a considerable distance along the span.

60a, b, c. Let the load be 50 kg, acting 3 units from the support which also provides a reaction of 50 kg. At the load point, as before, the moment has reached a magnitude of 50 x 3. At a point 1 unit further along, the load (if left to itself) would tum the beam back at the rate of- 50 x 1, so here we have + 50 X 4 - 50 X 1 = 50 X 3 At a point 2 units along, we have + 50 X 5 - 50 X 2 = 50 X 3 and so on. The bending moment along the balcony shown in Fig. 48 grows linearwise, from zero at the "free" end to a maximum at the supporting end (Fig.

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