Question

For the beam shown, a) Draw the bending moment diagram, b) Determine the maximum normal stress due to bending. 300 N 400 N/m + D А B 200 N.m Hinge 2.00 m 2.00 m 2.00 m 300 mm 750 mm x = 292 mm 560 mm NA: Neutral Axis

Answer 1

given BOON 400 N/m RAH (200Nom Rp RAI am am km Ha ik At equilibrium Eforizontal farce no Ranco Moment at hinge (right) Roxa. 400x2x1 =0 - Rp=400N Evertical force o RA-300 - 400x2 + Rp to RA=700N Moment at hinge C Cigt RAX U-300x2 - MA +200 -0. MA=2400

Coneider a section ara at a dietance a from chear force at x x = RA (x-ojo - 300 (+2) - 400 (4) Moment at u My = RA (x-o) - 300 (x-2) - uoo (x-4)² MA (2-0) 0 + 200 (&-2) 2 (S.FARA = PA = TooN. (right) MA= - – MA – 2400 Nom (right) 8.FB : 700- 300 - 400N (rights MB= "BF – 2400 + 700x27 - 1000 Nom Clepty MB = -2400 + 200 x 2 +200 = -800 Nom (aight) (sf) c = 700-300 = coon. Mc 2 – 2400 X tooxy + 200 - 300x2 - 400 N (left (Sf) = 700 - 500 - 800 = 700x 6 - 2400 - 300x47 200-400X2X) = 0 Mp =

Shear fare Diaghram TOON yoon YOON A IB - - 400) k am am am x 3 (s.fl=0-700-300 900 (x-4) To 400 - 400 (7-4) co x-q-1 x= cm M X-5m - tooxs - 2400-300x3 + 200 - 400xIxor 200 N.m. 2000 m 0 800 1000 NIM 24oonim Mmant Zubonom.

250 mm I 292mm 300 130 INA INAO + LN'A INAO ar Jooxaso + (300X700) (378-294; + 130x750? + (130x70°) ( 450 - 2921" 3 36 aro_294) + 130x703 t (150x2°) (Aso 36 10 9 - 1.20969x10^° + 1609452100 +1609432roo 10 mm I= 1.5315765x10 Mmon - - 2400 x103 Nimm

AL point a Mmon it ye 80 at top of the over-pection tengile bending abues auto 3 ofenelle Ma 2400XIO bending a bin a y = X 458 top 1.531576rxido St.be - 0.071769 MPa At point A at the bottom of the vore- fection - compressive bending etrece acts oc.be o M 2 y bortom a 2400 X 10?x292 1.5315709 x 100 o cibt = 0.045756 M Ra C comp)