Question

A cantilever beam supports the applied loads and moments as shown. (a) Calculate the support reactions. (b) Use the graphical method to construct the shear-force and bending moment diagrams for the beam. Also label the values of shear-force and bending=moment at all key points. 30 kN/m 25 kN 12 kN/m 80 kN.m х A В C D E F 1 m 3 m 1 m 1 m 1 m

Answer 1

Efy=0 30kN/m 25kN Tehn 12kan 80kN-m Go A 1 Re= (5x30x2)+($942 TH Im Im lm 3m im - + 12x2x2 – 25x2 BOB Bell 40kN MF = 80+ ($x30x3 % (3x30x3x (3+3 ME=258kN-m Span AB (x from AS Sx=0 My = 80 MA=MB=-80 kN-m SA=SB=0 1 Span B C (x from B) Sx = 1x sex xxx Mx = -800 (1 x 30 xx xx xx) SB = =0 (r=o) Se = 45kN MB = -80 KN-m (x=0) Me (x= 3 =- 125kN-m (x=3)

-12R (r=o) = 45kN Mc (x=0 (x=1) Span CD (u from ) Sx = 0+ (2x30x3) + (124) My = -80 - (6x30x3x{1+k) Se like So=57kN - 125kN-m (x=0) Mo = -176 kN-m • Span DE (x from ) Sx = 0+/x30x3) + (12x) -25 - (2 x30x3*(1+k) 32 KN (x-1) SE = 44 KN (x-2) =-176 kon-m (=1) ME = -214 KN-m (n=2) So = My = - 80 -12x2 +25(x-1) Mp Span EF (x frome) My = ME - 80- Sp=0+(4x3013)+(1242) – 95 (6x30x3x/3+x) - 12x2X (1+k) + 25 (1+k) SE= Sf = 4URN =-214 KN-m (x=0) ME = -258 KN-m (X= 1)

30 kN/m 25kN The 12 kN/m Goknom im im im 3m In 57 ti 44 45 u u SEO 1 ( ) 32 80 - 258 3 BMD 195 176 2 214 '