Statically indeterminate stresses in stiff framed structures . - (36). tfc*Af/>* _n_) (36a) From equations 29 and 29a of paragraph 24, f-lO±3&). (29). (29a) The last two equations are essentially the same as equations 36 and36a. 49. (c). Uniformly Distributed Load. Frame Symmetrical aboutthe Vertical Center Line. The uniform load is equal to W/L poundsper lineal foot. In this case F/L equals WL/12, and equations 29end 29a become £(- &OL±3£.J (37). (37e) (d). Two Point Loading. Frame and Loads Symmetrical aboutthe Vertical Center Line. See Fig. 41. Here F/L equals Pab/L. Equations
Statically indeterminate stresses in stiff framed structures . - (36). tfc*Af/>* _n_) (36a) From equations 29 and 29a of paragraph 24, f-lO±3&). (29). (29a) The last two equations are essentially the same as equations 36 and36a. 49. (c). Uniformly Distributed Load. Frame Symmetrical aboutthe Vertical Center Line. The uniform load is equal to W/L poundsper lineal foot. In this case F/L equals WL/12, and equations 29end 29a become £(- &OL±3£.J (37). (37e) (d). Two Point Loading. Frame and Loads Symmetrical aboutthe Vertical Center Line. See Fig. 41. Here F/L equals Pab/L. Equations 29 and 29a become Af*-/ie* €^Jo(- 2+**) .... (38).A/c*/y*= £r J (38a) If a « L/3 , these equations become faff*. *£(- (39). 41. Afc=A?£> = ^( -Qr ) (39a) For other symmetrical loadings, see values of f/L in Table III,page 29, to be substituted in equations 29 and 29a. 31. FBAME VtfITH HORIZONTAL LO/D ON COLUMNS. (a). Concentrated Load at any Point. Fig. 42 shows a rect-angular frame with a concentrated load applied on member AD, at a 50,. distance a from the bottom. Applying eq-uations 1 and 2, substituting Z for l/x,and transferring it to the left hand mem-ber of the equation , gives MdAZ0 = 2E(29d+9a-3R) - £f^0.^ . .(a). h h M/JDZ0 = -2B(B©A+0D-3R) - £f^,a . .(b). n h MABZ! = 2E(29A+9B) (c). MBAZl = -2E(29B+9A) (d). MBCZ2 s 2E(29B+9C-3R) (e). McBZ2 = -2E(29C+9B-3R) (f). MCDZ3 « 2E(29C+9d) (g). MdCZ3 = -2E(29^9C) (h). Let the shear in AD below P be represented by and the shear inBC by H2 . Then MB = -Hixi, = Hi(h-^) - Pb,s -H2(h-x2).MC = these four equations, gives Ma -Mb +Mq -Md = Ph - Pb s Pa. . (i).Combining equations (a) to (h), to eliminate all values of 9 end R,gives Ma(Z0+Z;l) * MB{Zi+Z2) + Mc(Z2*Z3) ? Md(Z3+Z0) = -PabZ0 MAZi + MBZi - Mc(2Z3+Z2) - M£(2Z3*Zo) = PabZ (J). .... (k). 51. MAZi + MB(2Zi+3Z2) + Mc(2Z3+3Z£) * MdZ3 = 0 (1). The last four equations are rewritten in Table V. TABLE V. Equations for the Rectangular .Framewith Co
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