Question

Problem 4. (65 points). For the beam shown, calculate the Reactions at A and B, and then draw the Shear and Bending-Moment diagrams. Consider the loading P= 29 kN/m. Р C D А B 40 KN 40 KN toom 0.6 m 1.8 m

Answer 1

40KN 40KN 29 KN/m RA R * 0.6m 1.8m 0.6 m Support reactions: using equilibrium conditions for vertical forces, RAT R 40+ 40+29*log 132.2 → RATRB moment about B Taking Rax (0.6+1.8+0.6) – 900(1.8+0.6) – 29x1•8464,2 +0.6) 40x0.6 = 0. 66.1 KN RA Ans. :. from equation RB 2 132.2 -66.1 RB = 66.1 kN. Ass.

Shear force force (SF) calculation SF at A = RA KN. 66.1 c (just left) 66.1 KN SF at at a ( just right) 6601 - 40 Z 26.1 KN SF at D (just left) Z 26.1-29x108 2 -26.1 KN. SF at D ( just right) - 26.140 66.1 Kna SF at B N. .66.1 KN. SF diagram 66-1 561 26.1 + B A С. - 2611 - 6611 -6601

Bending moment calculations: BM at Azo. BM at C RAXO.6 66.1 x0.6 = 39.66 kNam. BM at D Rax (0.6 +158) - 40x118 – 2981-8x 2 2 6601x2.4 - 40x108 29x 1.82 2 239.66 kN-m. BM at B=0. om at mid span RAX (0.6+ 18 ) - 40x108 !! 2 1.8 29 1.Q 2 2*2 =66.1 x 1.5-40*0.9-29x 1.82 8 2 51.405 Kno-m. BM diagram 510405 39.66 39.66 С. B A Mid span