Essentials in the theory of framed structures . moment at C for the beam in Fig. 158. Equation (12) is applicable to the uniform loads, whereWi = 1,000, Wi = 2,000, h = 12 and h = 18; hence M - (1,000 X 12^) + (2,000 X 18^) _ ^- 8(12 + 18) ~55^°° which agrees with the bending moment at C for the beam inFig. 160. The total bending moment at C for the combined uni-form and concentrated loads is M = —6,300 — SS,8oo = —62,100 as previously determined. 166. The continuous beam in Fig. 163 supports a uniformload of 1,000 lb. per foot over a part of the span AC. The areaPQSTW is the M-diagram, when ^
Essentials in the theory of framed structures . moment at C for the beam in Fig. 158. Equation (12) is applicable to the uniform loads, whereWi = 1,000, Wi = 2,000, h = 12 and h = 18; hence M - (1,000 X 12^) + (2,000 X 18^) _ ^- 8(12 + 18) ~55^°° which agrees with the bending moment at C for the beam inFig. 160. The total bending moment at C for the combined uni-form and concentrated loads is M = —6,300 — SS,8oo = —62,100 as previously determined. 166. The continuous beam in Fig. 163 supports a uniformload of 1,000 lb. per foot over a part of the span AC. The areaPQSTW is the M-diagram, when ^C is considered as a simple 262 THEORY OF FRAMED STRUCTURES Chap. VI span. Let the tangent to the elastic curve be drawn throughC, and let h and tz represent the tangential deviations at Aand B respectively. In finding the area-moment of PQSTWabout P, the parabolic area QST is encountered. This areahas all the properties of the area QST, which is the M-diagramfor a simple beam 12 ft. long supporting a uniform load of i ,000 tOOO* Pig. 163. lb. per foot over its entire length; hence the area-moment ofPQSTW about P may be found as follows: area PQNarea QBOarea QTOarea TOWarea QST 43,200 X 3 X 4 43,200 X 6 X 10 57,600 X 6 X 14 57,600 X 6 X 22 = 518,400= 2,592,000= 4,838,400= 7,603,200 18,000 Xi2XMXi2= 1,728,000 17,280,000 Eltx = 17,280,000 -f lf(i5 X 20)Elh = M(3o X 40) 2t\ = —ti M = —19,200 The reactions are statically determinate when M is value of M may also be determined by the use of Eq. (5) Sec. II RESTRAINED AND CONTINUOUS BEAMS 263 in which P is a concentrated load at the distance kh from the present case let P represent the weight of an element, oflength dkh at the distance kh from A, thenP = 1,000 lidk i,oooZi(^ — k^)dk 2(/l + h) The value of M may be found by integrating between thelimits k = and k = , hence ,000 X 30 whence dM = M = _ rh 2( (k - k^)dk (30 + 60) = — I SOjOOO = —19,200 167. The beam in Fig. 164 is contin
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