Question

Homework - State Loading Advanced (Part 2) A solid round 20 mm diameter hur is based as shown below, where the lead iP XOAN, the downward tr adise.SSAN, the lig h the is M-20 Nm, and the true lead is T-30 Nm. This results in certain stresses in the bar the base al Points located at the top of the har d ew from (located ISO w from 4-y il Decated 270cw from located in the center of the Cross 100 mm from the applied loads as shown. The bar is of AISI 1006 cold dramaselli, 20 MPa and 10 MP We pesme that this w here value afer yielding freising D i SHOW ALL WORK) a) Fill in the wheelow with EQUATIONS for the resulting stress NOTE 16T 321 4F by The state factor of safety based upon the dist ) Theatefactor of the pot The state face of safety and enthed ) The face of pe e n they (DET A h whey MSTA e hoy (DE) r shey (MST) D DPT DFDM "NOTE: These are ONLY valid of the member is und and DOP Dato Detto F D M
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Answer 1

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torsional Shear stress due to turque *Norsmal stress due to pl Bending Stress due to M Side viero equal at every point © Mauimum tve) and Minimum Eve) at Extreme fibre o zero at Neutral plane © Maximum at outer radit o Zeleo at le=o CCentre Now Shear Stress due to lond F Side viewo Normal She Stress does not induce sheay Stsex o Bending stsex does not induce shear stress tousional Shear Stress does not induce parabolic vasciation • Zero at Extreme fibre - Maximuem Somewheree in Middle Not Necessary at Neutral asis Note: Normal loading does not duce shear stress Bending does not induce shease Stress tousioral loading does not create Normal strey direct shear loading does not induce direct © O Stress

Normal loading e Stress at 8x103 - 25. 464 Mpa Toxy = 0 (20) Torsional loading T at Soetseme fibres. A 7. BY Txy = 167 = 16 x 30x10 = 19.09 MPa Tx=0 I ay ad² X (20)3 DJ Note at E point Try = o C from stress distribution) Moment (M) at top and bottom fibre A + x = 32 M (tensile) ad3 C ) Te * -32M (compresion) (12), - 32 x 20X10 – 25.964 Mpa (Tx), - -32 x 20 x 10° -25.464 Mpa XX(2018 ł x (20) Not at point B,E,D T O C poleks lies on Neutral plane) C Stress distribution) Try = 0 C does not produce shear stress) direct force (transverse div"). F. at point B,D,E } at point B, D, E t = 8 4F = 4x0.55x 10° 2 2.3342 T 3A 3X A (20) Note: When F is transfere to observing point A,B, C, D, E it will Mesult in Bending stresses at A and C using principle of Mechanics 4 e F F

Tx = 32Fxd at polnt A and C C tve, -Ve Yespectively) ad3 Tx 32 FXL. 32 * 0.55 x 10 x 100 XX (20) 3 1d3 = 70.02817 Mpa table Stress at a I due to p I due to I 25.464 Mpa 0 due to F due to M 70.028 Mpal 25.464 19.09 MPa Stress at B due top due to I due tor due to M 25.464 MPa Txy 19.09 MPa 2.33 42 MPa Stress at C due top due to I due to F due tom 25.464 MPa o -70.028 MPa 25,464 Try o 19.09 MPa C o stress at D due to F due to M due to p due to T 25.464 Mpal 19.09 MPa 2.3342 MPa Stress at E due to P due to T due to F due to M 25.464 Mpal 2,3342 Mpa

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