. Theory of structures and strength of materials. ction,/is also constant, and eq. (2)becomes (3) Also, since DE is the difference between DF^.wA EF, J{DF ~ EF)EF. ds = o =fDF. EF. ds~ f EFds (4) Remark.—Eq. i expresses the fact that the span remainsinvariable when a series of bending moments, , act atpoints along the rib. These, however, are accompanied by athrust along the arch, and the axis of the rib varies in lengthwith the variation of thrust. Let //„ be the horizontal thrust for that symmetrical loadingwhich makes the linear arch coincide with the axis of the rib. Let T^
. Theory of structures and strength of materials. ction,/is also constant, and eq. (2)becomes (3) Also, since DE is the difference between DF^.wA EF, J{DF ~ EF)EF. ds = o =fDF. EF. ds~ f EFds (4) Remark.—Eq. i expresses the fact that the span remainsinvariable when a series of bending moments, , act atpoints along the rib. These, however, are accompanied by athrust along the arch, and the axis of the rib varies in lengthwith the variation of thrust. Let //„ be the horizontal thrust for that symmetrical loadingwhich makes the linear arch coincide with the axis of the rib. Let T^ be the corresponding thrust along the rib at E. The shortening of the element ds at E of unit section T — T= ^ ds. Example i. Let the axis of a rib of uniform section andhinged at both ends be a semicircle of radius r. Let a single weight W be placed at a point upon the ribwhose horizontal distance from O, the centre of the span, is a. jee THEORY OF STRUCTURES. The linear arch (or bending-moment curve) consists oftwo straight Hnes DA, Draw any vertical line intersecting the axis, the lineararch, and the springing line AC in E\ D, F, respectively. Let OF = X, and let dx be the horizontal projection uponAC oi the element ds at E. Then -r- = cosec E OF = t^vv »ax E F or EFds = rdx (i) Applying condition (4), f^ DFrdx + f DFrdx = f EFrdx, or f^ DFdx 4- f DFdx = f EFdx, or area of triangle ADC = area of semicircle. And if z be the vertical distance of D from AC, zr = —, ARCHED RIB WITH HIXGED ENDS. 767 or 7ir 2 = one-half of length of rib. (2) .-. DE = DF- EF Ttr -V? a . (3) Hence, if /i be the horizontal thrust on the arch due to IV, a = M= W- 2r (4) Similarly, if there are a number of weights W^, W,, W^, . .upon the rib, and if /^,, h^, /i^, . . are the corresponding hori-zontal thrusts, the total horizontal thrust //^will be the sum ofthese separate thrusts, ^=/^ + /^,+ (5) It will be observed that the apices (/?,, D^, D^, . .) of the several linear ar
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