Question

What is the reaction Ax to the nearest 0.01 kN for the following frame? El are constant and the same for all members. Consider only bending energy. W = 67 kN/m w -5 m-+ 10 m ---5 m 5 m А B

Answer 1

STEP I: GIVEN Q7kN/m с D Tom 5m EIE CONSTANT 5m 5m B STEP I: TO FIND REACTION AT 'H' (Ax) = ? -1 STEP I: DEGREE OF INDETERMINACY (DS) Ds = 3m +2-3j-h ma 3, 8-4 sj=4, h=0 Ds = 3x3+4-3x4-0 = STEP TV: PRIMARY STRUCTURE CONSIDERING THE REACTION AT À (Ax) TO BE REDUNDANT AND THUS REMOVING IT FROM GIVEN STRUCTURE, WE OBTAIN: 67k NA im D C Tom 5m 5m xz B x A LEBY 4-Bx THEN, WE WOULD HAVE TO SATISFY; AxVA + DA=0 VA: DISPLACEMENT DUE TO UNIT LDAD STEPT: SUPPORT REACTIONS 0 <fx=O Bx=0 g@fy =0>) Ay+By –67X10503) Ay+By = 670 GEMA=D=> Byx20 –67X10X 10=0= By = 335kN, SO, Ay = 335kN, STEP VI: METHOD OF VIRTUAL WORK FOR PRIMARY STRUCTURE, DUE TO SYMMETRY 5 Mm dx = 250x la doc, + OEI +25°(1675 +335 23–33.523)(-145) dez 15 dec, +SE352)+(-AS) day ДА EI EI m: MOMENT DUE TO UNIT LOAD (IN THE SAME DDRECTTON AS REDUNDANT) DA= 0 +(-41875) + 111666.667) =-153541.667 EI EI EI AND, UA=S mm dx = 25%!2;}e () dx, +25° -1x$)y {\x5)doz +25E1x5); 6–125) das EI +

I UA = 250 250 350 + 250 STEP VIL: REACTION AT 'A' 503.333 EI EI Ax 583.333 + (-153541.667) = 0. EI EI Ax= 263. 214 kN. ZANS.