. Practical structural design; a text and reference work for engineers, architects, builders, draftsmen and technical schools;. EXTERNAL FORCES 15 line of the triangle being merely a closing line. There is no bend-ing moment at D. At (c) is drawn the shear diagram. The load being concen-trated at the end, the shear of course is constant from the pointof application of the Gio) Mo) Oho] load to the shear may befound at any point bymeasuring a verticalline at the point acrossthe shear diagram. r .4 ...,i —7—>i<--//—-—»j — 8 3 1_. Moment Diagram Relation between Shear and Moment
. Practical structural design; a text and reference work for engineers, architects, builders, draftsmen and technical schools;. EXTERNAL FORCES 15 line of the triangle being merely a closing line. There is no bend-ing moment at D. At (c) is drawn the shear diagram. The load being concen-trated at the end, the shear of course is constant from the pointof application of the Gio) Mo) Oho] load to the shear may befound at any point bymeasuring a verticalline at the point acrossthe shear diagram. r .4 ...,i —7—>i<--//—-—»j — 8 3 1_. Moment Diagram Relation between Shear and Moment Diagram Assume line C,C„ to :^be dropped across theshear diagram. Thenthe area of the sheardiagram to the right ofthe line, that is to thefree end, gives thebending moment at CSimilarly, the area ofthe shear diagram tothe right of line B„B,gives the bending mo-ment at B. The areais in foot pounds be-cause the vertical di-mension is expressed inpounds and the horizontal dimension in feet,in all cases and must not be for got ; In Fig. 4 is shown the case of several loads on a beam. The bend-ing moment at D = 0. The bending moment at C = 8 X 600, andthis is plotted as line CC,. The bending moment at 5 = (15 X 600)+ (7 X 400), and this is plotted as hne BB,. The bending mo-ment at A = (19 X 600) + (11 X 400) + (4 x 200), and this isplotted as line A A,. The shear diagram, at (c), shows the shear at the right end
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