Question

I need help with shear moment diagram on my solids class.

30 kN/m 100 KN 100 B M 4,- -30mm 9 m 3 m 3 m

Answer 1

For a cantelever beam we can assume that its center is at a distance of 2l/m from the fixed end and (l-2l/3) from the second end by acting it as a simple force of magnitude Wl/2. Since Node A and Node C are fixed and thus they do not have any degree of freedom. But node B can rotate. The rest of solution of this que.tion is given as in these figure,
elassuste Pag 07 100KN B A С. - Romm 3m 3m * Go 8 Velocity -diag VERN) 252 a C A 74.2 74.8 bending Moment ding $7.5 A B-18.2 - 1669

Date Page 06 But from equ 0 MBA + Mec to 1 + 2 EI 202 +0.0100) - 121.5 2E I (202-001500 175 solving Value (General) V 1.111 do (EI) = 0.00278 E It 46.5 EI in KNm² is EI = 25400 kNm² from @ 4 Qe = to.oouis rad Step 4: calculate bending Moment LEI ZE, 0.004 isto. 0100) +81 + 160 kNm Now MAR Similarly MGA MBC -18.2kNm +18.3kNm - 166.9 KNM MCB

so Again mode A is fired REI (2de + 0.01o0)-121-5 MBA 9 end near member member B is Moving moment at Be at point B. MBC = REI ( 20 et de - 3 Pac) + FEMEL = LEI (2de + de 310.00500)) + : Node L is also fixed 4 75 . 2 2de- 0.015oo MBL +75 2 СІ 6 Now similarly - 75 similsly for CB MCB = 2 E1 (29e + P2 - 3 4 e) A FEMER EZEI ( 20ct de - 3(0-00560) Again C is fixed so Meg = 2 EI (de - 8 81500) - 15 node

Step 3 slope deflection Equation The general slope deflection equation having neither of beam element I have pintor 2 E1 (2 dn't Qf -34]+ FEMME hinge at and is Manf L og point A. for moment at end member Ab at A is neer side A & B is for siche MAR 2&I (2 dat de - 3 WAL) + PEMAS. (20A + de-3-0.00333) + 81 L - ZEI But since node QA to A is fixed. e Made 2 EI 9 Qe + 0.0100 + 8 for moment at end B point B, foy member AB side fa sicle MBA = 2 EI (2 PBA QA - 3 Was) + FEMEA 2E I (2007 09-36-0.00332)) - 121:5 - ZEI 9

Pure 03 mb m 100 N ch A 2 -- C +81 kNm ze am SFTSKNM - 1215 Km 6m - KNM 2/3 KNM FEMAB= wl x 20 2 we 3 304 SL- 30 SIKNM 30 um FEMBA od we x lx 1 im kNm 2 10 30x 81 - 121.5KNM 20 20 100x 6 +75kNm [EMBC = PL 8 00 x 6 75 kNm FEM CB = -PL 8

Po 02 ay point b equilibrium to sum محور کا فلی whese is moment at end of memb AB at node b. Member BU at node B. Then only moments that be maintained sum of moments around mode B Step Chord rotation for member and BL YAB = rotation (chord YAB = LAB YAB = -30mm 9000 mm Lfor clockwise sotation and 64 AR=-0.00333 rad for AB -ve sign + DG 30 LBL 6000 4BL = toooosorod tue sign for anti-clockwicy (countor-clockwice rotation. 2: Calculation fined End Moments: (FEM) Stup ܣܛܘ

classmate Page 1 9 need with help P sheer moment, diag solids 30kN/m TON Og = -3ommor >$ 3m 3m am c ماه solution: Nodes А fined and not have freedom ( Dors) o de any degree of move nonizontally since it is retrained by mootion member AB and BC, which both fined horizontally Node & in alio the roller support that location. however node can rotate. The total rotation of node B is Overall this stru tuse DOF. only In addition to external distributed load member AB if point load at centre member BC je support og 20 мм downwards at B QB one a oy point For Cole rotational DO F at node the equilibrium condition will MBL = 0 +