Question

8. Refer to the axial loading problem 3 above. In that problem you were asked to calculate normal and shear stress components on a given plane (45° CCW as shown in that figure). Considering it as 2D plane stress

Answer 1

• 9.3) - Axial Member is considered as uniformly Stressed - Stress Tensor at a point qiven as -T/A o of where Oxx P/A To oo Lo o P= 10KN & A= 10 cm I OXIDN loxint e = 10 MPa - Now We have to determine Stress At a ploue 45' Cow in - 2 direction Let Say plane denoted as - let In - Nomal component of stress at New Plane .gt t - Tangential component of stress at new planes • Now, In 8 In can be fomulard as, • Tn - Oxx cose + Oyy sino + 2 Oxy sing we In = Oxy costo - Oxy sin²o - Coxx - Omy ) sino caso On= (10) cos 45 + (0) categy +260) sin 45 casus • On = 5Mpa • In=co) cos²5 - (0) sin 45-(10-0) sinus cosus In = -5 Mpa no In = smpa • By Figure r In = Supa • By Figures . All in . Mpa In = 5Mpa YY=0

8.8) Refering Above Problem - The Mohr's Circle can be drawon by co-ordinates as Sex = 10MPa, Suyo Try=0 . (0,5 t - - - . (20-90) (DO) Couyo -smpa (5,0) (Oxx-lompa) (0,5) - [w = 5Mpa I-Our Original plane of problem is O-A-B where Oxx = 10Mpa 1-Plane Rotared in ecw directiono=450 so in Mohr's circle it will be Rotated by2eie go in Cew direktion. ... New plane of solution is c-A-D . On = 04 Redius of Mohr's Circle = 5Mpa Tn = AD = Redius of Mohr's circle = snapa (Magnitude) Hence And on 5Mpa & In- smpa | (By Magnitude)