Question

24 M9.4 Shear stress in a standard steel shape scenes 246 For the wide flange beam subjected to the indicated shear force V, determine the transverse shear stress acting at locations H and K. 712 84 664 216 V=930 KN TH (MPa) dimensions: mm TK (MPa) 24 Try one 1st 2nd 3rd attempt enter

Answer 1

0 Given that 246 mm - 18 mm - summ mm 712 930 216 mm 356 mm 24 mm - Since the given plane is symmetrical about horizontal & veatical anes, the Centroid of the plane lies at the mid point of the plane. ic. on the y-anis at a distam T = 712 - 356 mm from g- anis. - Ao Shear Stness at any point on the Cross. Section is. 2= FAG when - Innb. F manimom shear fource = 930 KN. Ian b-> width of the beam at that point

= 06100 INA = *moment of ineabia of plane aboot Cenkooidal anis (INA) z Iit It Is Plane Agiea me. I of individual Plane lamin Sum of the Centroidal anis. about 246 mm ---- - - - - 2uinn Ii 0 —s ī 9,= 3 uumm 3,- (INA), = Inat A. (9.)? II = (216) (24)3 + (246) (24) (344) ? 2 12 (INA), = I = 6.989x108 mmy a both the Franges of I- section are at having same dimensions and at same distance from the Centeroidal anis. Therefore INA) = INA), . (INA)2 = I2 = 6.989 x 108 mmy

DUTE 664MIT (Inn) = Ich= bd3/12 (INA) - Is = (18) (664) 3 = 4.391 x 108mmy . Area moment of inentia of the plane about the Centeroid anis / Neatonal anis is INCA = I,+ Iz+ 13. =(6.9898108 ) + (6.989x108) + (1.391x108) INA = 18.369 x108 mm" Shear stress at point k : 246 mm mm b= 1 Ty=20 cm 1 18 mm 248 mm k Ek = F (Any + A2Y2) mm (INA) (6) zu (930 x103) C 246x 24 x 260)+(248x 18 x 124) 18.369 x 108 x 18

=> /?k= 58.74 Mipal transvease shear stress at . the point k o Sheas skoless at point H 18 16mm 2x = (F) ((Amy, IA242) (INA) (b) o TH= (930x103) | (246x 24 x 128) + (116x 18 858) 246 mm 18.369 X108 x 18 Izumm Z H = 24.66 mpa) - teansvegse sheae storess at the point H. shear stress at NA! Eman= (930x103)+((246x24X344) +(332 x 18 x 166)] 18.369x108x18 man = 85.02 mpa I at the NIA