# Shear Stress Yielding Analysis

4 Pages197 Words171 Views
|
|
|
Answer (2): Given data:d=0.2mL=1mT=20πStress at point (A) Fx=0Vx20π=0Fy=0Vy=20πMy=0σyb=NA=200.1π=200π=63.66σy=MyII=πR44=π(0.1)44=0.25πx104m4y=4R3πσyb=20πx1x4x0.13πx0.25πx104¿3.395305x104σy=63.66+33953.05σy=34016.71Find, shear stressτBy=VyQItQ=A1y1¿π(0.1)2(4R3π)¿π(0.1)2(4x0.13π)
Q=13.33x104I=πR44=π(0.1)44=0.25πx104m4τA=20πx13.33x1040.25πx104x0.2τA=5332kN/m2Answer (B)Yielding not occur at point “A”.τmax=±(σx+σy2)2+τxy2¿(80000+02)2+53322¿40353.81kNm2¿40.353MpaAs per given condition if maximum shear stress greater then yield stress then yield occur, but value of maximum shear stress is lower than yield stress therefore yielding not occur at point “B” Principal stress at point A, σ1,2=(σx+σy2)±(σx+σy2)2+(τxy)2¿34016.71+02±(34016.71+02)2+(5332)2¿17008.355±34432.060σA=51440.4153kNm2(Tensile)σA=17423.705kNm2(Compressive)Answer (c)Yielding not occur at point “A”.τmax=±(σx+σy2)2+τxy2

## End of preview

Want to access all the pages? Upload your documents or become a member.

Related Documents
|8
|428
|190

|6
|319
|83

|10
|467
|172

|8
|408
|191

|7
|1071
|15

|12
|1550
|200

### Support

#### +1 306 205-2269

Chat with our experts. we are online and ready to help.