Bulletin Volume 103 PDF Download

This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1917 edition. Excerpt: ...in the following general form MAB = 2EK26A + eB-3R)-CAB (11) MBA = 2EK(2dB+0A-3R)+CBA (12) These are the general slope deflection equations which apply to any condition of loading and restraint. The sign of the constant C may be determined as follows: In a fixed beam the sign of the resisting moment at the end of a member is opposite to that of the moment of external loads. For instance, in Fig. 4 the moment of external loads about the end A is clockwise, so the resisting moment CAB is counter clockwise or negative; and since the moment of the loads is counter clockwise about B, CBA is clockwise or Figure 5 positive. If the loads were upward instead of downward, the signs of Cab and Cba would be reversed. With signs thus treated, C becomes merely a numerical, or scalar, quantity. It has been noted that equations (11) and (12) apply to any condition of restraint of the ends of a member. Fig. 5 shows a member restrained at A and hinged to the support at B, so that the resisting moment at B is zero. Equations (11) and (12) may be written: MA B = 2EK2 dA + dD-3R)-CAB O = Ml)A = 2EK(26D+ 6A-3R)+CBA Combining these two equations to eliminate 9n gives MAB=EK(36A-3R)-(CAB+) (13) If the beam is fixed at A and hinged at B, with the supports on the same level. 6A and R in equation (13) are zero, and therefore the term--(CAB---T-) represents the resisting moment at the end A, and can be readily calculated for any given loading. By similar reasoning, when a beam is restrained at the end B and hinged to support at A, it is found that MBA = EK(36B-3R)+(CBA+) (14) n For more convenient reference let the quantity (CAn+----) be Q denoted by HAn, and the quantity (CUA---5--) by HnA Equations (13) and (14) then take the general form MAB=EK3BA-3R)-HAB (15)...