Question

A strain gauge rosette attached to an aircraft component fabricated from thin sheet material recorded the following strains: E = 550 JE E45 = -450 ME €90 = -150 ue The thin sheet material properties are: E = 70 GPS v = 0.3 (a) Determine the state of stress in the 0°-90° axis system at the strain gauge location. [11 marks) (b) Determine the principal stresses at the strain gauge location and the maximum shear stress. [6 marks) (c) With the aid of the previously calculated stress states, illustrate the 'invariance of stress'. [2 marks] (d) Determine the orientation of the principal stress axis to the direction of the E, strain gauge, and the maximum-minimum shear stress axis to the direction of the E, strain gauge. [6 marks]

Answer 1

. Giren : So - 550 к. 550 у.е. £и, -450 дk &qo - 15 и E= 70 apa у : 0.3 Ел & ? Ем -4 (20) Say sin (20) 2 (uso rosse fre) ot kaь 6-4° os (2~~s) • Во-р sin(s» ѕ) tus" 2 2. 2 • Eus - (o +40) Vorao Vay 2» (-450) - ( 550 – 50) Vry Bay — 130 и 4 shear strain -

СА state of stress in o-qu axis systery би €о- , : - убу 550 x 56 Prot - о. 3. Су » У D Ego - Ey . Бу E E -1 50 х бу towie 0.3 - Б. 70x1 mocy 21 Ф- 32. ? Ги = MP4 Бу 1.2 Mpa 42 E 2 (112) t- GVry 2(140.3) - 24.42 3. -11 - 4. 20.42 Кра exy Trey = -35 мга бу - 12 +38. ? = 6. '3.7 new gf

a Eminor to calculate Mow, calculare Emejor principal stresses Snoy u 5min Emay Emin min E 4 Smay E 2 Emojor/minor Bryty 1993. () to the oferta)'. More 1/1624095. Changes 550 - 150 Emoj/min 200 + 738.24) 938.241 x 106 Emog'or = 200 + 728. zu Eminor -538:241 ob 938.241 xob Emejor =

E 5 minor mayor = o mayor E E E (9.332 4/64) major 0.3% 5 minor similarity Eminor * E - Ominor 0.3 mayor E -5.38 x104 5min may' x 0.3 solving these ③ - equations major - 59.738 MPa -19.73% 15 minor : Mpa sum of stresses on stress invarience at perpendicular plane will remain any of rotation about mutually perpendicular axis) 0 o maior Ominor 38.8 +1.2 their a 59.738 - 19.738 կ է Чо hence correes

orientation of principal planes +2 Tuy 2-35 tan (20p) = Би-б 38.8-1.2- - 30.88 OP2 90-30.88 59.120° check which plane is for mayor 38.8+1.2 38.8-1.2 - los (2*430.81)) –350 sin(2- (-20.42) 2 2 51.738 MPa i - 30.88 So the plane for major 59.120° minor 49 1 14.718 MPa B ! 59.120° 30.88 59.718 Mpa

Mohr's Circle 60.00 T Max shear = 39.73 50.00 (38.80, 35.00) (38.80, 35.00) 40.00 х 30.00 20.00 10.00 Min prin = -19.73 20.00, 0.0) 20.0 40.0 0.00 010 -60.0 -40.0 -20.0 60.0 o 80.0 100.0 Max prin = 59.73 - 10.00 -20.00 -30.00 (1.200,-35.00) -40.00 (1.200,-35.00) -50.00 Min shear = 39.73 -60.00 Original Stress Principal Stress Transformed Stress - Extreme Shear Stress

Element Rotated to Max Shear Stress (14.12 deg) o.y = 20.00 T=-39.73 x = 20.00 х T = 39.73 0.X = 20.00 T= 39.73 TE-39.73 O.y = 20.00
please give some positive rating.. i have less cf score..

Thank You So Much..