Question

The cylindrical elements are joined by a link to produce a connecting rod-crank mechanism for machinery and are represented as shown in subsection (a). At the end they act: P1 = 2.0 kips, P2 = 1.5 kips and P3 = 1.0 kips. The cylindrical elements are made of Al 6061 - T6 profile (part b)). (A) Determine the state of stresses acting at point B of section A and represent them in an element of analysis

У. -3.0 in B 0.7 in P1 = 2.0 kips 1.5 in N by - 4,5 in Py = 1.5 kips by - 1.5 in Ps - 1.0 kips (a) (b) Area = . (@- sen ) Cos 10 r Sen 4r 3 0-sen o 0011 * II (o – sen 9+2 sen Ø. sen? 9)

Answer 1

Given Line ie Presentation: Cross-section у % 26:3 in sin dest > zo b2 = 4 sin P=2.0 kips P=15 kips 53ersin P3I Kips At Section -A Procedure: Let us discuss Stresses developed due to each at Point B? force CS Scanned with CamScanner

which develops develops compressive Stress is Force B = 1.5 Kips" acts along negative x-axis at Point 'B' of section A. Fa (compressive Strea] - Where, A is area of Cross-section A- I 05- 6-7 A 1.277 in 1500 Ps: 1-277 = 117.5 Psi (compresse) in Force 'p': otis: Due to theo thy pes of stresses are developed a) Bending stress due to 'My acting about to where, y = P x (6, +63) N, 1000x (5) Ibtain M, 1000 10-in shear stress is also developed due to Sheour eyle P2 = 2Kip Suanned wi. Cunusuarner cs

a) bending Storss due to M = 9000 ly.in Bending stress developed is unsile in nature along si axis calculated from Bending Equation Points y Iz where 1=0-35 in = 0.2286 in 12 = (15)*- (07* Tag - 9000 lbf-in x 0.35 in 0-2285 in ng = 13785.56 Psi tensile) TV2 = 13.7856 Kpsi] Csik) b) Transverse shear stress due to P, = 2 kips Transverce Shear Shress is developed along "negative y direction in x-plane as shown 'Ting is aluulated from Punfarged with CamScanner following formula X

Tay = VA INA where, v= shiarforce = P = 2 kips Teen Neutral z B A Arear above Point B Chatched area) A = ($ - Sino) A = (0.757 [ 2-17 -0.820) A = 0.318 in 0=265) 0:2 Cost 0 = 9.174 b=width where 'B' Present b=arsin (2) 2 (0.75) sin(247) 16 = 1.326 in INA Moment of Inertia about neutral axis that is z-axis . INA Iz = 0.2285 inyl calculated in partca 1) CS Scanned with CamScanner

ý = distance between neutral axis and Centroidal axis of Arear A as shown above. given, 9=43. (sin( 0 - Sint = 475) 2 2-17 - Sin(2:17) G = is (0.5144] g 0.7716 in Substituting all the above values in tay equation in Tay 2000 x 0-3780-7716 16x in in Inno 0.2285 x 1.326 try = 1925.24 Psi - 1.9252 KPS: iii) Force Pli → Due to only shear stress is developed in the direction of "Positive z-axis" on - Plane that is the as shown CS Taganner

As the cross-section is same and Point 'B' is same so tir value is also qual to ty Exy -1-9232 kpsi ... 42- Note: Due to 'p3 bending stress will not be developed at 6 because 'Point &" is on Neutral axis-y-axis as shown. Neubal-axis alt Stresses en flamesite: Finally indicating Points Tx = Tag-Ta, -(13-1856 - 1.1745) KPS ICS Scanned with CamScanner - 12-611 KPS

Please give POSITIVE rating if you understood the answer thank you...