Question

Shown in the figure below is a solid cylinder. The x-axis is along the longitudinal direction. The angle o is measured with respect to the x-axis, positive if counterclockwise, i.e. towards the y-axis as shown below; and negative if clockwise. 01 and 02, with 01202 , are the principal stresses, and Imax is maximum shear stress. Oz Y Fracture surface Normal to Fracture surface Suppose that the cylinder is subjected to a uniform tensile stress of 5000 psi in the longitudinal direction. Suppose that the bar is made of A36 steel. The angle between the normal to the fracture surface and x axis is:

Answer 1

VO fractw.) -O 2 2 So shear Value =0 Solution : Güven: nauis along longitud ml Anie O meiswed from meanis. 5,- Major Primnipol strand on =50oolsi T2 = Minor Principal struy i q bar is made of A36 steel Matrual will fail si Sheer mode inice maliziad ieduchile fructwu orcurs at Imeu plame which is 45° to powincipal anis. So 0-45°C kvm os Cup Icom (11) Iman - 01-02 olo =Ty=0.55n So None of there is correct l'iy smee Only longitudnal tensile stress acts it will cause no shesin B) 6 (0) 1,2 [contg): Von wy) ++42kg 1->$1,24][onfront] Toon Gazon-on=o So option is None of the above Orientation of te will be 90°+ Orientation of ori tom 20 = 0=0° foro, 190° for 52 Pennipal plane imentation wil on=5ooopsi (vii) Orientatwn of co' already discussed above (ving r1=on already discused matericel is Cast Iron (which is brittle in nature) (1) Material will fail due to normal stress 8=90° (11) By co Already discussed Clio ima plame at45° to honinepolplame (Alreadly discussed) ܐ 20 onog ziny zo Now