![]() ![]() spans, and it also gives the values of the shears on each side of the supports. 2.12, Vx ϭ R1 ϩ R2 ϩ R3 Ϫ P1 Ϫ P2 Ϫ P3 Mx ϭ R1 (l1 ϩ l2 ϩ x) ϩ R2 (l2 ϩ x) ϩ R3x Ϫ P1 (l2 ϩ c ϩ x) Ϫ P2 (b ϩ x) Ϫ P3a Mx ϭ M3 ϩ V3x Ϫ P3a Table 2.1 gives the value of the moment at the various supports of a uniformly loaded continuous beam over equal TLFeBOOKīEAM FORMULAS 59 FIGURE 2.12 Continuous beam. The ultimate load is wL ϭ 4MP (1 ϩ k) ϭ 6.1 MP (2.8) L L In any continuous beam, the bending moment at any section is equal to the bending moment at any other section, plus the shear at that section times its arm, plus the product of all the intervening external forces times their respective arms. TLFeBOOKĥ8 CHAPTER TWO kMP ϭ w x (L Ϫ x) Ϫ x kMP 2 L w L Ϫ kMP L ϩ KMP Ϫ 1 Ϫ kMP kMP ϭ 2 wL 2 wL 2 wL2 2 leading to k2M 2 Ϫ 3kMP ϩ wL2 ϭ0 P 4 wL2 (2.7) When the value of MP previously computed is substituted, 7k2 ϩ 4k ϭ 4 or k (k ϩ 4͞7) ϭ 4͞7 from which k ϭ 0.523. TLFeBOOKĥ7 FIGURE 2.11 Ultimate-load possibilities for a rigid fraĪme of constant section with fixed bases. The two moment diagrams are combined in (c), producing peaks at which plastic hinges are assumed to form. In (b) are shown the moment diagrams for this loading condition with redun- dants removed and for the redundants. For equilibrium, TLFeBOOKĥ6 CHAPTER TWO FIGURE 2.10 Continuous beam shown in (a) carries twice as much uniform load in the center span as in the side span. Maximum moment occurs in the interior spans AB and CD when xϭ L Ϫ M (2.5) 2 wL or if M ϭ kMP when xϭ L Ϫ kMP (2.6) 2 wL A plastic hinge forms at this point when the moment equals kMP. Using this method with the vertical load at midspan equal to 1.5 times the lateral load, the ultimate load for the frame is 4.8MP /L laterally and 7.2MP /L vertically at midspan. 2.10c, equilib- rium is maintained when MP ϭ wL2 Ϫ 1 MB Ϫ 1 MC 4 2 2 wL2 ϭ 4 Ϫ kMP wL2 (2.4) ϭ 4(1 ϩ k) The mechanism method can be used to analyze rigid frames of constant section with fixed bases, as in Fig. Figure 2.10b shows the moment diagram for the beam made determinate by ignoring the moments at B and C and the moment diagram for end moments MB and MC applied to the determinate beam. ![]() 2.10, the ratio of the plastic moment for the end spans is k times that for the cen- ter span (k Ͼ 1). TLFeBOOKīEAM FORMULAS 55 For the continuous beam in Fig. The equilibrium method, based on the lower bound theorem, is usually easier for simple cases. The load corresponding to an equilibrium condition with arbitrarily assumed values for the redundants is smaller than, or at best equal to, the ultimate loading-provided that everywhere moments do not exceed MP. TLFeBOOKĥ4 CHAPTER TWO FIGURE 2.9 Moments due to deflection of a fixed-end beam. A load computed on the basis of an assumed link mechanism is always greater than, or at best equal to, the ultimate load. ULTIMATE STRENGTH OF CONTINUOUS BEAMS Methods for computing the ultimate strength of continuous beams and frames may be based on two theorems that fix upper and lower limits for load-carrying capacity: 1. TLFeBOOKīEAM FORMULAS 53 FIGURE 2.8 (Continued) Characteristics of loadings. TLFeBOOKįIGURE 2.7 Chart for fixed-end moments due to any type of loading. In a similar manner the fixed-end moment for a beam with one end hinged and the supports at different levels can be found from MF ϭ K d (2.3) L where K is the actual stiffness for the end of the beam that is fixed for beams of variable moment of inertia K equals the fixed-end stiffness times (1 Ϫ C FLC RF). ![]() The sum and product of three distinct positive integers are 15 and 45, respectively.BEAM FORMULAS 51 FIGURE 2.6 Fixed-end moments for a prismatic beam. Value of t for a germ population to double its original value Ratio of force of water to force of oil acting on submerged plateįind the approximate height of a mountain by using mercury barometerĮquivalent head, in meters of water, of 150 kPa pressureĬompute for the discharge on the sewer pipeĬoefficient of discharge of circular orifice in a wall tank under constant headĪbsolute pressure of oil tank at 760 mm of mercury barometerĪbsolute pressure at 2.5 m below the oil surfaceįind $x$ from $xy = 12$, $yz = 20$, and $zx = 15$ Which curve has a constant first derivative?ĭepth and vertex angle of triangular channel for minimum perimeterĬalculation for the location of support of vertical circular gate Spacing of Rivets or Bolts in Built-Up Beams.Load and moment diagrams for a given shear diagram. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |