Question

al — al - a2 + a3 — Determine the shear force and bending moment in the beam. Use al = 2.4 m, a2 = 2.1 m, a3 = 2.5 m, F1 = 32 N, F2 = 21 N and M = 250 Nm for your calculation.

Shear force in the section AB in N, Shear force in the section BC in N, Shear force in the section CD in N, Shear force in the section DE in N, Bending moment at B in Nm, Bending moment at D in Nm, Maximum bending moment in the beam in Nm, Minimum bending moment in the beam in Nm

Please answer ALL components of question in the way asked. Thank you

Answer 1

321 RIN 200N.m B 2.4m c) 20lm o 2.5m TE EMASO REX 904 =) (32 x 2 4) + 250 + (21 X 809) 2 RE =) 50-180N {fyzo RA +50-180 = 32 +21 RA= 2.82N Shear Fonce in section AB= +282 N Shear Foue in section B(2) 2-82-322 - 29-18N - 29-180 MM Shear fola in Section (M) - 29-18 N Shear Force in section DE - 50.18 ON Baselina po mad at Ba) (2-8 2x24) 3) +6.708 N-m Az

Berling Moment at Da) + (50-130x15) a) 105-40NM AU Minimum bending Moment - O'Nm at A and E Man lending Moment a) (2-8284-sa) +250 - (32 / 24) Iman. Berding Moment ) +186-736 N-m at a

For Section AB x { Fyzo AB LY + 2.82 KN A M =) 2082 X =) 2082 X 24 =) Mp=76-765 2-82120 For Section BC 32kN 204 AT 2282 / 204 l {fy=0 32+ VBC = 2.82 VBL 2) – 29:18 KN m Ma 3.821 - 3268_204) Il McA (2.82 84:8) – (32x204) +250-/186-736 extra moment at c. Same For CD FOR DE VO VOET E {Fyzo 500181 VDE + 50 18 = 0 I UDE = -J0-18 KN Ary M = 50.18X (la 205) Mp = 5018&225 = 125-45 KN-M

AT 204 B M = 2082x - 32(3-2-4) 2002 D 012) 48 ( Just Left of ( ) Main) (2: 12 x 408) - B2X24) Mmin = -63-36 IN-m