Question

(1) Determine the resultant internal loadings on the cross sections through point D and E. Points E is just to the right of the 20 kN load. 20 KN 15 kN/m TITUT E - 2m - 2m--- 1.5 m -.--1.5m - (2) The pipe has a mass of 10 kg/m. If it is fixed to the wall at A, determine the resultant internal loadings acting on the cross section at B. SOON Im 2 m BON

Answer 1

1. Given. beam. w/ RokNet homtoam matsman ndeterminate by one dejove. So Choosing Redundant Ra for Calculation DB an= pzm 1.5 mills Ve IT the - At for a t A deflection will be zero. - SAI=&AR for uniformely variang wad Choosing a strip of dx width from end al at distance x.. is a kN/m - 2DEN Ar Hims + 2001:59 Fodples) ZEL - 1744.25 EL dady(tas y la xar) taa) = 934 + e n + + 123, Som = 1749.25 this will be equal to deflection due to RA. SA₂ = PA (7) = 343RA Equating SA, = San 174.25 - 3438 Pa = 15.26 V

- Now all the Internal Reaction can be - 7.5kN/m determined. for points 2ma R = 15-26KW for shean force at D & Ef=0 VE RA - 1. (7.5) (2) V- 15.26-7,5 Vo= 7,76 kN] for moment at D x {Mo=o. -Mo- 2 (7.5) (2) 2 / 3 + PA(2)=0 TM25.52 kN-m ISKYM 2010 for pointe, k zml.sm zm PA=15.26100 for shear force & Efy=-0. Br(4) 15) – Ve–2020 15.26-30-VE-2020 Ne =-34.74 kN for moment at E. EME=0 -4.-€ (4) (5)(1577)+(5-26) (5:5) = 0 Me=01-07 kNam

2 2. 22 given. man of pipe. to kg/m - So It will act as uniformily distributed loud. taking to kg/m 7 Loon/m. r (g=10m/sec). YOON 22 5oon 22 at D, resolving Boon. into. y &z direction. Tale -- y component 7 (300)) + 2400 Shirestin 2 component 7 goo () 7 160N (1) Eve) z load due to weight of pipe co, & TooN m) x2m 4 2000 (4) Yoon at B, shear in a direction. 4 . Soo V272 loom 2uops. V = summation of louds in a direction = 200 +1801 - 581 10/21 x = summation of loadly in x direction --> 2400 = 5OON GON - So Resultant Shecy = Jų ²+U² Shear = 680 765.76 N.1 Axial tension at B 7 Summation of loads in y direction Axial tension = 400+240 = 6YON? torsion at B = (summation of momenents about y axis) Horsion atB = 550002 = 1000N-m) Bending moment at B about x Aris 7 (200).(2) +(180)2) + 400(2) +[)2) () | Mxx = 1860 Nm = 17600-m about z Axos a (500 x2) = 1000 Nom | Mzz = 1000 nm