Question

The figure on the left below shows a stress block (plane strain condition is assumed, z axis is not shown here). Its components as well as the material properties are, • 0,= 10 MPa . O, is currently unknown • Txy = 30 MPa • Young's Modulus E = 150 GPa • Poisson's Ratio V=-0.25 • Note: Poisson's Ratio is negative Rotate the x-y axes in the anticlockwise direction by 45° (as shown in the figure on the right below), resulting a new coordinate system x'-y'. In the new coordinate system xuy, the normal stress componento, = 45 MPa. y len

Answer 1

& ginen data by = ? 6x = lompa, Txy = 30MPa. = 150hpa M = -0-25. 5x = 45 MPa. exy 45 n G92 x o = 45° (+) 6x+6y 2 2 5x 5 2464 + (Grace) les 20 + 2xy bin28 45 – 10+64 + 10-6y & Losso + 0 singo 45 57 +30 sy = 20 MP / Aus in 20 MPa 2x 30 (2) fanzop - 2 ray/ -64 10-20 - 40.268 49.73 (3) 61, a Top= - 48.2680) Ans [us < op <45] principal stresses. Maxon principal stress, 62 = Min principal stress- 5 -

;tzo ܚܘ ] * ,cxrc_< ܘ 6 = 6x+64 + 2 45.U) MPa 10420 + |(10-2)* + 230) To, 2 us.umpa Ang 53 10720 - (10-20)? +B)? Tor - 15.41 MPa And 2 - 15.41 MPa ISMPa way = 15MPa a to 15 =) - 6x- by binzo] + fery los2o.] 159) -(10-20 fin20 + 30 Coszo 15 = 5 sinza + 3020120 3 - sinza + 6 Cos 20 square both sides sin?20 + 36 (os? 20 + 12 Cos 20 hinga. w s Sin²20 + 60120 1 . As all know

9 = 1+3560220 + 1260120 binad Cas20 = 11 - Sin? Sinza Si-loped. 8 = het los2o = x 35x² + 12x 51-x² by solving x = -0.63 Los 20 = -0-63 Sin2o > 0.7766 los 20 = – 0-63 0.2) Mom 64.525° 3 (5) strain in aby & shear stran in kry. Ex = 6r - uby E É 10 +0.25 x 20 150 X 103 150x103 Ex = = 1x104 = 100xico Ex = 100u / Ang saya 150 u tus w

Tages Creation " G = 2 tund . 1 od 29 E - 150 21-25) 1.25) 2(1+1) 30 1300 e 100 x 10 100 x103 30 1714 (6) fanzop: fan 20p Tayla/ Ex-ley & 724 / Ex-by Ese-Ey tanzop =) 300 100-150 op =) - 40.268° -40.268 o 49.73 Aus. Ex + Ey Muain - /377-07 ! A 225 + 152.07 225 - 152.07 Emin - ) [ 72.93 4 Aug.