Question

*Part 1 For the simply supported beam subjected to the loading shown, derive equations for the shear force V and the bending moment M for any location in the beam. (Place the origin at point A.) Let a-3.00 m, b=4.00 m, Pg - 35kN, and Pc = 90kN. Construct the shear-force and bending-moment diagrams on paper and use the results to answer the questions in the subsequent parts of this GO exercise. Calculate the reaction forces Ay and Dy acting on the beam. Positive values for the reactions are indicated by the directions of the red arrows shown on the free-bady diagram below. (Note: Since Ay = 0, it has been omitted from the free-body diagram.) Answers: Ay = 0 KN

Answer 1

95k 9 0 BUL More To sm ß 3m a um to Eify = 0 RałRB. = 125- O IMA=0 (85x 3 )+(90x6) - Rox 10 Ro= 64.5kn egn -RA = GO-SK 685kN 760-5 RN 125.5 im 25-5 KM tve cum 30 -ve sbeces forces SEⓇA left = 0 SF a night = 60.5 KN(I) SEO = 60:5-35 = 25-SKN -- | SEQC = 25:5-90 = -64.SIN SER SEO -64.5KN Sf@brighto porght 64-S N SC

- Part 2 La] V@x=20 = GO-SEN b]_V@ x= cm = 25-5KIN. c] V@ x=8m =-64.SITY pont B Bending momento BmQx-o 0 Bm@x+20 = (60-5x2) = 121 kv.m BM@x-gps = (60.5*3) = 181.5 Inn BM@x=40 = (60-5x4) - (35x1) = 207 KN:M_ → BM@x-60 = (60.5x6.) - (85x3)=958 kiyim Bm@xgm = (60-5x8)- (35x5)- (90x2) = 129 kNem # Part 4 = Xmax- 6m, BMmax= 258 kNem