Question

Problem 2 Below is shown a planar stress element for a given stress state o.. . and known stresses) Using a free body diagam ("wedge" element) shown in Fig. (c) below, static equations of equilibrium, Fx =0 and Fy=0, and trig, Identities, Show that the normal and shear stresses at an arbitrary angle ,0x, Txy are defined by the following equations: NOTE: These are parametric equations for Mohr's Circle 0x = 6 0; + 0Os cos 2€ + sin 20 1 y = - 2 sin 20 + , cos 20

Answer 1

.. unit thickness of this Pwedge Gym the wedge the equilibrium of consider that as per the principle Abe: we know face "Be of the shear, the of simple edge will be subjected to a clockwise Rny (complementary. stress equal to - shear ... shear stress) Now, Horizontal force Acting on the face AC, Pa = by the: (*) Pg= May xae (b) vertical force, force for Horizontal the face Bey PB = Ming X Be.(F)

- vertical, force on the face Be, Paz. by X, BC (W) Now, Resolving the forces, perpendicular - the face AB, P= P, CSO: 1P2 sino tB CSO + Pasino : Gn he est + Try Aesing I By Be caso + GuiBc sino Now Resolving the force tangential to: AB, P -P, sino + please 1 sino +3, caso. == Gute sing tray Ae. USE... TG4 BC sino + Gybe caso

Normal stress ( hermoss the section AB) Kny. A Cont - Gri Agasa Tiny Alsing + Zmy BC. cosa AB đ, 8 Sin 8 AB besing + Treo) Ray Be Post A Acceso Aed US by be sino + Feno) G Cas? + Ray sino Gost t Teny sind us # Gy sin? 6. C1760526) + 4 (i-Cos20) - teny sinza

To out on to come and eas20 + Riny s?h28] - shear stress across the plane AB est May Be Sint En aesino AB sy Be ess 6. AB T IAC Josy May Ae caso ...Gin Ae sina Fit (Aeloss) Mng Bc sing (Be/sino) T Gy be eso T Besina

- Gr sind caso + by sino caso in triny (6656- sin36) Su Oy sin 20 + Myy easzo ::C : En roy si n 26 trny las za