Essentials in the theory of framed structures . ,B and C necessaryfor equilibrium. {a) Algebraic Method.—We first resolve each inclined forceinto horizontal and vertical components, and indicate them inthe sketch (Fig. 46). The magnitude of the force D is 100 lb. It acts to the leftand upward, sloping at the bevel of 3 in. horizontally to 4 ; as indicated by the arrow-head at D and the sides;^and J ^ of the triangle jO^. Hence, fj ?.4:5Let D = 100 lb. represent the magnitude of the force, Dh represent the horizontal component,and Dy represent the vertical
Essentials in the theory of framed structures . ,B and C necessaryfor equilibrium. {a) Algebraic Method.—We first resolve each inclined forceinto horizontal and vertical components, and indicate them inthe sketch (Fig. 46). The magnitude of the force D is 100 lb. It acts to the leftand upward, sloping at the bevel of 3 in. horizontally to 4 ; as indicated by the arrow-head at D and the sides;^and J ^ of the triangle jO^. Hence, fj ?.4:5Let D = 100 lb. represent the magnitude of the force, Dh represent the horizontal component,and Dy represent the vertical , iifjk is considered as a force triangle,Dh:D,:D Sec. IV EQUILIBRIUM OF COPLANAR FORCES IS or Dh : Dv : loo :: 3 in. : 4 in. : 5 = 60 lb., acting to the leftand Dv = 80 lb., acting upward. The magnitude and sense of the force A are unknown. Forthe present we shall assume (or guess) that the sense is awayfrom the point O, and let Ah = 8a., acting to the right,Av = 15a., acting upward; thenand A = aVS^ + 15^ = FiGAe Assume that the force B acts away from O, and letBh = 126, acting to the right,5„ = 56, acting downward, and B = h\/i2^-\- s^ = 13b. l6 THEORY OF FRAMED STRUCTURES Chap. I The forces £ and C (C is assumed to act upward) and thecomponents of the inclined forces D, A and B are shown in thesketch (Fig. 46). Three independent equations are requiredfor a solution of the three unknown quantities C, a and h. IH = oor forces acting to the left = forces acting to the right. 60 = 64 + Sd + 12b (i) IF = o or forces acting upward = forces acting downward. C + 80 + 15a = 56 (2) IM = oor clockwise moments = counter-clockwise momentsabout any point. Let us choose the point F. The arm of thehorizontal forces is p = 2 in. and the arm of the vertical forcesis 5 = 3 in. IM 1 •P Then 2 X 64 = 128 2 X 60 = 120 2 X 8a = i6a 3 X C = 3C 2 X 126 = 24& 3 X 80 = 240 3 X 56 = 15& 3 X 15a = 45^ 128 + 16a + 39& = 360 + sC + 45*^ (3)Equations (i), (2) and (3)
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Keywords: ., bookcentury1900, bookdecade1920, booksubjectstructu, bookyear1922
