Question

Complete the shear moment diagrams.

600 N 400 N/m С PA B -2 m -2 m 2 m

Answer 1

bon FBD- Given Beam i's present at point is internal hinge So brea brewning into to parts and drawing the FBD for the Reaction. it for detum determining I Geron Ro yoo N/m HA с. Р 2m B 2m Rc RB internal Breaking the hinge with equal and opposite dummy load. for whole beam: RAIRE 60ot 400x2 RA TRc = tro +800 RA+Rc 1400 N Talling moment from Right about point B. Y as at internal hinge B.M. is 2000} - Rcx 2 T U OU X 2 *1 = 0 EMB=0 = > Rc 40x2 4 UD N

Rc UVON this from equation RA 1400 - Hoo RA = 1000 N B Ffr/ Set Paling moment foom left side about point Em3=0 =) -MATRAX 4 Geox 2 >) - MA + 1000 *4- 1200 MA Ivoox 4 – 1200 그 -) MA 2800 Nm (07) ACW For SFD. Sechon A to p. (S.F)x TRA = t1ooo Sechin B p to (SF)x RA - p = +1000-600 = 400 С B to Section (SF)x & RA-P TRB - RB. 400(x-4) (SF)x 400 - 400131-4) point where (SF)x -0,

400-400 (9-4)=0 40019-4) = 400 71-4 = 1 725m + 400 (SFDB (SF) = +400-400 (6-4) -400N + 100E N +1 OOON TULON tuto N + Hiem olt im Beim - UUON For BMD. Section AP MA 2800 Nm - 2800 + Rgxx 2-2800 +1000 de Mp - 2800t 1000 x2 -800 Nm P to B. Sechon Mp 800 Nm (MX) = -2800 +R9x2 - px (2-2) MB - 0 Nm

En Sedon B to C Mc=0 MB=0 2 Mx 2800 + 1000 x x - Gero (2-2) -400 (7-4 2 { from left side Finding the point where it is zero. (Me=0) 0 =-28 + 10x - 6x +12 - 4192716– 8x) -28 + 47+12 - 2012:32 +16 24 9.222 +20% -48-0 -)-2? + 1D X -824-0 >> 3742-107+ 24 =0 § at point c} Xa = 9 point B? da {at Maximum or minimum. &Mx dx 1000 - foo -80o (9-4). a 400 - 400 (2-4) -0 a = 5 Mx 25 + 200 Nm

+ 2050 Nm 20 am Im im 3 C -800nm 2800 Nm

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