Question

A rectangular plate of dimensions X = 265 mm and Y = 100 mm is formed by welding two triangular plates (see figure). The plate is subjected to a compressive stress ox of 4 MPa in the long direction and a tensile stress Oy of 14.5 MPa in the short direction. Using Mohr's Circle (2 points for the correct drawing); determine the normal stress ow acting perpendicular to the line of the weld and the shear stress Tw acting parallel to the weld. (Assume that the normal stress Ow is positive when it acts in tension against the weld and the shear stress tw is positive when it acts counterclockwise against the weld.) (Mpa) Weld (Mpa) Y(mm) X (mm) Answer: Ow (MPa) TW (MPa) =

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Answer 1

Solution : by G =-ampa 190 1 = 265m goce -62 by =+14.5mpa, y = loomm x will be anis » And as there is no shear stress acting on the plate (combined) these 6x4 by Primiple > centre of mohn's cincle (co) -> (2,0) = [bnt by o d radius of mohn's cincle 101 62 - by -45 = 9.25 " this formula is valid only for principle plane → So monr's circle {-4+149,0] = [5.25,0] =) ra 4 - 14.5 | 47day 17 9.25 with respect to major Tw 6-40) 120 114.5,0) 6 5,03 > Angle of plane of weld NIC principle plane o tam " ( 26 ) hence on Mona's cingle represent planemo at angle'20' from magen boinciple plane. - 20.67° = 5.95 20 = 41.350

-Counterclockwise e from mons's cincle Tw = 8 simeo = 9.25 Sim 41.35 = 6.14 MPa 3 6w im perpendicular direction: & Gw 5.25 -8cus zo = 5.25 9.25 COS 41.35 -1.88 MPa