... total deflectionsat points 1, 2 and 3 due to P 1 ,P2and P3can be written asD 1 = P 1 f 11 + P2f 12 + P3f 13 D2= P 1 f 21 + P2f22 + P3f23D3= P 1 f 31 + P2f32 + P3f33Written ... becomes(9 .14 )DDD 1 23 11 12 13 21 22 23 31 32 33 1 23ÏÌÔÓÔ¸˝Ô˛Ô=ÈÎÍÍ͢˚˙˙˙ÏÌÔÓÔ¸˝Ô˛ÔfffffffffPPPFig. 9 .19 Flexibility coefficients for a loaded beam Steel ... [][] 11 ,KtC BDBACAeTTd[]=[] [][][]ÈÎ͢˚˙[] Ú 11 ,UCBDBC={}[] [][][]()[][]{} ÚÚ 1 2 11 ddeTTTed vol,\{}=[]{}=[][]{}-edBA BC 1 e Steel Designers' Manual...