Question

please answer all correctly!!

Coordinates (x, y, z): A (0, +7mm, 0) | B (0, 0, -7mm) C(0, -7mm, 0) 0 (20cm, 0, 12cm) Fixed support y direction) 135 kN (x direction) 22 kN-m (z direction) 16 kN

Answer 1

Moment at M Axial Force and Bending any cross-section will give normal stresses where as Shear Force and Torsional Moment give Direct Shear stress and Indirect Shear stress respectively. Example o=my to t > P Normal Stress = f (Tensile) = Bending stress which is Above NA it will be compressive below NA it will be Tensile in nature INA - Moment of thertia about bending axis NA "INA NA INA А type of Normal stress. & て二太v TO J indirect Shear stressi A Direct Shear stress Torsional shear stress J = Polar MOI 9 tried to visualise the forsional shear stress. here. over the cross section.

2. We will transfer the forces and moment acting over o to the cross-section where B is located. The direction of 22 kNom moment (Torsional Moment) can be obtained from right hand thumb rule Moreover any force can be transferred to another location in addition of a moment (if any) due to eccentricity upto that another location, с K X F=35kN Fy = 35KN * Please do watch carefully how the loads got transferred. Ime 160 d My = 16a My=160 m А Me=22 kN-m My= 22kN-m M = 356. Mx=356 'c =35a M2=35a DMZ LPM1/2 = 350 F z = 16 KN Fz= 16 KN =35kN b M2=35a Mu=22kN-m F = 35kn 자 * Fz= 16 KN My = 16a My = 22 KN-m Direction of moment can be obtained by right hand thumb rule Mx = Moment about X-Axis Ez = 16 KN

We notice from the given co-ordinates. and a tc- 20cm b = 12cm. Therefore forces acting over B can be shown as: Fx = Axial force in X-direction = 0 fr Shear Force in Y-direction = 35kN Fz Shear Force in Z-direction = 16KN ( Torsional Moment about X-axis). Mae 22-356 = 17.8 KN-m ( Bending Moment about y-axis). My 16(a+c) = 3.2 kN-m Mz 35(a+c) 7 KN-m (Bending moment about Z-axis). 3. Calculations for of, of, zat Bi ß along x-direction ya Normal stress over 7 7 A • Axial stress Ex My XZ MzY A Tyy Izz a Bending stress Here Fx=0, Z=7cm, y=0, Iyy = Iz z X (14)*mm* = 1885-74 mm). and A = 7 (14) mm² = 153-94 mm? I Pa = IN/m2 . 3.2X106N-mm x 70mm 118.7868103 MPa IMPa = 1N/mm² 1885.74 mmt IGPa = 105 103MPa. = 118.786 Gpa 64 x =

Sign Convention - Take Tensile stress as -ive Take Compressive stress as time and so x = - 118.786 Gla. why? - Because Bending moment (My) will try to compress the fibres on left of NA K-Axis ( which is also Neutral Axis dere to symmetry of the cross-section) and will try to elongate right side of NA So at B there will be tensile stress due to My. on the o ] 19 Oz = 0 Here you may argue that fy and Ę forces & are acting t as well as there is moment Mx too then also why oz It is they are shear force and Tersional moment which will gine shear stress at 8 and not normal stress. over B be cause 6 Calculations for Txy at 8: shear stress at B in x-y plane. Tuy Zry Zry Zxy Zry Direct shear stress due to Shear fora Torsional moment. Indirect ) try, 1) zny of (As there is no shear stressin tat)

♡ Calculations for Zxz at Bi Zxz = Shear stress in X-Z plane Fenz (You can better see the figure and assume the blone; then it will be you will have © Calolations for Zyz at B a clear bicture of the situation). Zyz Shear stress in y-z plane at B. Tyz Tyzi + Zyzz Zmax at NA ie at the level of B. Cove to Due to D Shear forca Torsional Moment Shear stress Distribution, due to fy. Tyz, Emax = 4 Zang (For solid Circular cross- section) Tyzi x 358103N 753.94 mm2 303-148 MPa (Acting in upward direction) A at B. Zy Zz Tr J t where I = Polar Mol. = Iyy + Izz T= Torsional moment = Mx -Acting upward at B Tytz 17.8810x x 7X10 N-mm2 2 x 1885.74 mmt GMx = 17.8 KN ( Acting upward at 6) ZyZz 330.3748103 MPa 330.677X103 - 330-3740103 MPa 1330-677 330-374 Gla/ (Acting upward) Tyz Note- Why 9 did not consider Fz here which is also a shear force at the cross-section of location of B. which will gine shear Stress in Y-Z blane only. ?

please see here, Rofack Z clockwise to make it as 4-axis) chofat Fa fz Zmax A B shear stress Distribution due to Fz have noticed that due to Fz; the shear stress at from the shear stress distribution figure Here you B = Noten However use v- SF this formula to find Direct shear stress てこ VA A- Anea where we want to find a Distance of a from NA Con Mo 6 - Width at the section