Question

Coordinates (x, y, z): A (0, +7mm, 0) B (0, 0, -7mm) C (0, -7mm, 0) 0 (20cm, 0, 12cm) Fixed support 2 (y direction) 35 KN (x direction) 22 kN (z direction) 16 kN-m

Answer 1

Forces that are forces ar going to act at Section containing pointe due to all 1 Due to 22KN-loadi- (22x12) Ki-cm • Anomal force 22 KN! Toode in ge-direction. - A Bending moment of (22x12)KM-Cm Counter clerkwise 22km 235 kW 235x12) KN-cm 12) Due to 35 KN-load :- A shear force of 35 kn lood in y direction. A Bending moment of (35x 20) KN-cm Counterclockwise A torsionar rommt op (35x12) KNl-cms . Clockwise. (35x20) KHCM. <33 Due to 16 KN-m moment A Bending momont of 16 ki-m. Ibichom (!) Mormal force and Bonding moment produces Normal Stresses and, Shear force and Torsional moment produces Shear Stresses. on combination of all the above forces and moments, we have in $(22x12)km-em 22.KN. 35 KN 35 KM (35x12) Ki-em OKN. 22 KN Okn. Mx = (35X0:12) KNM = 4.2 KN m (35420)KN-cm + (16)KM_03 My = (22 x 8-12) Klum 2:64 KNM Mz = (25*0-2 4 16)k.m: 20.2 km-om. 'Now, lets collabte the effect of Each forces and moment one by one, Effect of For 142.92x10 KN B = 22 T (Ixio 3)² m? 22 KN A

(B) Effect of Fy 35 from the shear stress distribution we are seeing that To=o. [- VAY INAD C) Effect of fz AS Fz=0 .: To CD) Effect of Ma 4.2Knig Do 8 Ic= (4.2) x x 672163 In (2x7x10 37 41 32 T= M 8 J : 7795.34 x103KN Hote. - This figure is to find the direction of Shear Stress by torsional moment. seeing that , At Point C, the stress is directed towards positive Z-axis- Z As we are [zy: Tyz = the Hote! * (28) J. Polar moment of inertia = 7 de 32 32 8= distance from Neutral axis. I= Moment of inertia - Tat 64 *(28)* 64

LEX Effect of My 2.64 KN-m 32.64 KM-M A B Ecco. (Point a c is at Neutral z axis) LAS Effect of Mz :.= 20.2. X (4x103) = 74983.79 x103 KM me ^(2x7x10-3) Olio 64 20.2 kN-M (3) 50, by and 67 at point (142-92x103) e rayon 142.92 MPa 142.92 x 103 KN by Due to fa 74983 -79 MPa - 62 = (74983.79x103) kn/m² 2 74983.79 x 10 km 4 Due to MZ L43 Try at point C, Try=0 453 Taz at point C Taz=0 L6> łyz at point C 7795.34 MPa. Tyz = 5795.34X10% km = 7795:37x10en by (Due to Me)