Question

2. (35pts) For a 3D combined loading, please determine the 3D stress state at point A and B in the circular cross-section of a-a. The radius of the cross-section is R. The end of the structure is clamped. We are interested in the stress state on the cross-section that is cut by a-a plane. Please go through the following steps: 2.1) (5pts) List the contributions (axial force, bending moment, or twisting torque) from F2. Clearly indicate what kind of contribution (bending, torsion, uniaxial loading, etc) with respect to (or along) which axis. 2.2) (10pts) Calculate the stresses at point A and point B due to F2. 2.3) (15pts) Repeat the above two questions for F3. 2.4) (5pts) Insert all these stress components into a 3 x 3 stress matrix. a-a Az Figure 2. Problem 2. For a circular cross-section, the centroid of the bottom half circle is 4R/(37) away from the center.

Answer 1

a-a F3 Different kind of loading at cross-section due to force feu AS Mz = FzL, C Radius of cross-section - R. Mr = F2 H → No transverse shear load and Twisting moment :-- Compressive load along -ve 4-axis = - F2 q-az → Bending moment about 2-azis [cow from top] = F2L → Bending moment about X-atis [ccm from right side ] = F2H --Scanned with CamScanner

Shessej at point A due to this loading ve F2 Compressive Stress - - F2 Gy - F2 TTR2 TPR2. Bending Shes by = F2 H XC-R) IT XR4 366 = M 2 Tx1. | Compression on bottom Fibre je.at A] 4 F2H TTR3 Shear Shell TS=0 want me to Sheses at point B due to this loading i.e. fa Compressive Shess - Gya Bonding stress. 6039 Gy: MzQ Tension on right side | fibre le. at B - IZZ by - F2 LIR I R“ 60= 4F241 69 TR3 TTR3 shear stress Is = 0 → Different kind of loading at cls. a-a due to force f3. Max-f312 1714 = F30 CS Scanned with CamScanner

fror Direct transverse Shear force along tvc Z-azis - F3 ->Bending moment about X-axis [ clockwise a Right-side vier] - Fala → Twisting moment about y-aris [clockraise from frent): f3 LI -7 Stresses at point À due to loading f3 . Direct Shear Ezy =0 try. The Ri Twisting Shear t z T (R) Јуу - Bending stress by Tay= 2 Fall M 2 s stress is opposite to ? I applied torque και το Μα 7 laa Tension in l at A bottom fibre! by = f3 L2 (R) IRG 7 Stresses at point B due to loading F3 Direct Shear tzy. (SF) Ağ be width at point A Ixıb try: F3 TR (48) try some CS Scanned with TR4 TT R4 X 2R CamScanner

Twisting sheer tegen met R4 - Tayo - 2 - 2 F3 L1 TTR3 Bending Stress -Net Stress at point A and stress matrix for point a 67 = 636=0 Tay: - Tya = 2 Fall The = tar = Tyz = czy=0 3D State Stress matrix [632A = 273 Li TTR3 2 fati fan - 4 fez ! tafada o | - Pfali -2F3L, TTR3 H 4f512 TR3 TR3 OR? 636 = 62=0 the Teyze se - 2010 Taz = TzI = Tay = Tya -O Scanned with CamScanner

| [63] =|| - oo- 3 LI 張繼襲 TTR3 Se: TR3 23/ TTR3 Scanned with CamScanner