Question

2) A beam supports the loading shown below. Determine the internal normal force, shear force and beading moment acting just to the left of point B and to the right of point C of the 8 kN force. 8 KN loknom A D oo 6 ) B C 2m 5m 1

Answer 1

solution - 8KN 70 10KN-m of * x am ² Sm → support reactims SKN JOKN-m A am 5m 'RAY 'Roy (1) Fy=0; RAY + Roy = 8 KN. et ŽMo=0; Ray C 7) – 8 (5) – 10 = RAY 50/7 ~ 7.142 KN. RAY = 7.142 KN & Roy= 0.858 KN. By passing C5086-sections throug points B and Considering the free body diagrams of the left segments AB and Ac of beam. ca B MB A NB The points B and c are just left & just right to 8KN face am VB RAY SKN Mc Nc Hence consider 2m Vc AB ² AC (approximately) (ie am) RAY

Here, M= N= Normal force, v= shear force 8 Bending moment Equations of equilibrium For past AB (t) {Fx=0; No=0 Using NB=0 41)=Fy=0; RAY - VB=0 VB= RAY = 7.142 kN. [VB ~ 7.142KN G+ Mg=0; MB = Ray (2) + MB=0 7.142 x 2 = 14.284 KN-m. MB = 14.284 KN-m For section AC : 6) FxO; Nc=0 (-ve ve mennys opposite direction) 41) Fy=0) RAY-8KN -Ve=0 Vc = 7.142-82-0.858 KN (U) Vc = -0.858KN 0.858 KN EMC=0; - Rayx 2) + Mc =0 Mc = 7.142 X2 = 14.284 KN-m. Mc ~ 14.284 KN-m Therefore, Requised Internal loadings, & MB = 14.284 KN-m Ve= 0.858 KN (1) Mc *14.284 kN-m No=0, VB= 7.142 KN Nc=0 5