N = 3; if the far end is fixed, N = 4; and if the angular rotation at the two ends is equal in magnitude but opposite in sense or direction, N = 2. In general, N depends upon the restraints at C, D, and E, and varies between 2 and 6. 12. Solutions of Problems for Partially Restrained Beams by the Slope-Deflection and Area-Moment Method. Having derived an expression for the restraining moment at the face of a column the equations can be written for all the moments about any single joint, such as the intersection, at the point B, of the continuous beam with the column illustrated in the diagrams of Fig. 26. Example 6. It is assumed that the ends of the beam at A and C are I Center-Line of Beam FIXED and that R = 0. The = (b) เ In these equations the values of CBA and CBC are taken from Table (II), page 130. Since N = 4 we have from the general Formula (42) on page (137), and as the algebraic sum of the moments about the point B must equal zero, MBA+MBC + MBD + MBE = 0 Substituting the values of the moments in this equation: Substituting this value of 0 in the moment-equations, 66 666 SEK1+8EK2 For the columns, Fig. 26(b), the moment of inertia of the section of the concrete is in which b is the width and d the thickness of the column, the dimension, d, being parallel to the beam. For the steel I = (6 X 1)(9.5)2(14) 7 560 in1 = in which 6 is the number of bars; 1 the sectional area of each bar, in square inches: 9.5 the distance in inches of the centroid of the reinforcement from the median axis; and 14 the value of (n − 1). The total value of the moment of inertia for the columns is K2 = = = 293 in3 in which 120 is the clear story-height, in inches. For the beam, Fig. 26(c), for the section of the concrete, In this equation 82 is the width, in inches, of the flange of the T beam determined from the equation in which b' is the width of the web (10 in) and t the thickness of the slab (6 in); and d is the depth of the section, assuming the neutral axis at the bottom of the slab. in which kd, the distance of the neutral axis below the top of the slab, is the slab-thickness. The total value of the moment of inertia for the beam-section is in which 240 is one span of the beam, in inches. Substituting these values in the moment-equations, This example illustrates the method of determining the moments at the supports of a continuous beam when the restraint due to the columns is considered. If K2 = 0, that is, if the restraint at the columns is neglected, making the conditions similar to those of simple continuity at B, then, from the preceding equations, and M BA D Example 7. As another example, with conditions similar to those often encountered in practice, the frame illustrated in Fig. 27, representing a beam. AB which frames into an exterior column CAD, is considered. It is assumed that the beam is loaded uniformly with a total load W, and that it is continuous beyond the point B where the continuity is such that the tangent to the elastic curve is horizontal, making B 0. This is the condition, which in this discussion designates a beam as FIXED at that support. R is again made equal to zero. the purpose of simplifying the numerical work, the values of K for the columns are considered equal to those for the beam. = K W = 30000 lb A K B -20 K For C Fig. 27. Continuous Beam, Uni formly Loaded, and Framing into a Two-Story Outside Wall Column Following the same general procedure of writing the moment-equations for any particular joint and solving simultaneously, with an equation of equilibrium for the value of the slope, we have: For ends of columns fixed at C and D (N = 4). Substituting this value of the slope in the moment-equations, For the ends of the columns hinged at C and D (N = 3), and This example illustrates the difference in the values of the negative moment in the beam at the support, resulting from the two opposite assumptions in regard to the condition of restraint existing at the far ends of the column. Obviously the true value of this moment in this problem lies somewhere between the two, and as these values are only about 10% apart the design-moment that should be used is determined with sufficient accuracy. 13. Solutions of Problems for Building-Frames by the Slope-Deflection and Area-Moment Method. In a monolithic building frame such as is shown in Fig. 28 each member exerts a certain degree of restraint upon those members connected with it. It has been shown that where the beams or girders can be assumed to be either simply supported or FIXED in relation to the supports, the determination of the moments is relatively simple. A method, also, has been derived for solving conditions of partial restraint by determining the value of the angle 0, which is defined as the slope of the elastic curve at the joint under consideration. If the frame shown in Fig. 28 were uniformly loaded on all bays, the joints A and B would be practically FIXED. If, on the other hand, the full design-load were placed on spans ag and dh while the spans aA and Bd were subject to dead loads only, the conditions at A and B would be considerably |