Question

A shaft with a diameter of 43 mm, is shown below On the right hand side at location D a wheel has a force F of 4824N applied. The diameter of this wheel is 150 mm. The torque produced by F is transmitted through the entire shaft to location A where the torque is reacted. There are no other constraints at location A. Bearings, are located at B and C, and provide radial constraint. The bearing at B also provides axial constraint to the shaft. An axial force of 1000 N is applied at the center of the wheel at D. It is in the direction of location A.

a. Create a torsion and axial force diagram for the shaft.

b.Find the reactions at B and C.

c. Create a shear force, V, and bending moment, M diagram for the shaft.

d. Consider where the highest state of stress might occur, both in terms of distance along the beam and where on the cross-section.

e. Draw a stress element at the highest point of stress and find the value of the stresses that act onthe stress element.

f. What are the principle stresses at this point?

g. If the material is AISI 4140 steel and has a yield strength of 655 MPa, what is the factor of safety according to the MNST, and MSST theories?

Combined Loading Problem 1: A shaft, with a diameter of 43 mm, is shown below. On the right-hand side at location D a wheel has a force F of 4824N applied. The diameter of this wheel is 150 mm. The torque produced by F is transmitted through the entire shaft to location A where the torque is reacted. There are no other constraints at location A. Bearings, are located at B and C, and provide radial constraint. The bearing at B also provides axial constraint to the shaft. An axial force of 1000 N is applied at the center of the wheel at D. It is in the direction of location A. 250 mm 100 mm a. Create a torsion and axial force diagram for the shaft. b. Find the reactions at B and C. c. Create a shear force, V, and bending moment, M diagram for the shaft. d. Consider where the highest state of stress might occur, both in terms of distance along the beam and where on the cross-section. e. Draw a stress element at the highest point of stress and find the value of the stresses that act on the stress element. f. What are the principle stresses at this point? g. If the material is AISI 4140 steel and has a yield strength of 655 MPa, what is the factor of safety according to the MNST, and MSST theories?

Answer 1

Solution (a) Ecos 20° comp Fcos 20° component of the force will be responsible for torsion by the torque :: Torque, T = F Cos 20° ( 0.150) = 4824 COS 20° (0.075) = 339.98 N-m Also, axial force only restrained by bearing B 339.99N-m 1000N 235938 M - m. afacere fis1000N Torsion , (N + m - 339.99 N-on TOOON a D . Axial force - (N) Axial force diagram

ETTE there is no torque that is acting between pointe A and D (Except at AfD). the torque would not change in he would not change in t torque diagram. Similarly think for axial force. (6) (RB)y (Rcly lesinzoo (RB) x → SM0.25 m 0.1 m 2 K looon KZ - (RB), Rc) z KEcos20 EFx = 0 = (Re)x = 1000 N Efy = 0 = (RB), + (Re)y = Fsin 20º = 1649.90 N { Fz = 0 = (RB), + (Re), = F Cos 20º = 4533.07 N - Esin 20 = 1649.90 N = {(Mc ) yani's = 0 -> (RB) (0.25) + Fcos 200 10.10) = 0 - 1813.2.3 N (RC)2 = 6846 :30 N Now, (Mc) z asis = 0 (Re)y (0.25) + Fsim 209(0-10) = 0 WEITERE HERE De

ESSAISERS INTRO TESSA = - 659.962 N v = 2309.867 N. .. Reaction at B is given by Ro - (1000 î - 659.96 ſ + 1813.23 Ŕ) N - Reaction at c is given by - Å )N. RR = (2309.86 Î -6346.30 (0) Shear force dinge Shear force diagram for Horizontal plane and Bending moment diagram for horizontal plane -1-913.23 Shear Force diagram c. -4533-07 453-30 M(Nm) -- Bending moment diagram - CARTOON

Now, Bending a bending moment at C = 1813.23 (0.25) = 4 5 3 30 N - Bending moment ng moment at DE1813.23 0.35) - L6346.30 (0010)) - O Shear for force and Bending the vertical plane - and Bending moment diagram 1649.5 Shear force diagram -659-96 M(N-m Bending moment diagram :- 164.9905 Now, Bending moment at c =-659.96 (0.25) - - 16409 N-7 Bending moment at D :659.96 (0.35) + 2309.8 (0.10) - ON-m.

(d) from Bending Clear that highe bending moment diagram , it is Occur at 'c'i highest state of stress will oint c, Bending moment Bending moment in Y-direction = 4 5 3 3 0 '- m. Bending moment in oz-direction = 164.9 Nom 0.25 m Olom Bo cho Ao D : At the point Ć (cross section), maximum stress due to Bending moment in will take place Y-direction . Due to bending moment = 453-30 it will experience tension in x-direction. This part will experience compression As far as 'same for section't torsion every is point concerned, it will be on periphery at point Ć is orlom from point D. TESTO STEIFELLEL

em.SK exy 2 A :. = 32 M Aq3 32 (45330) = 53.0743 MPa A 10.043) Zxy = 16 T 21.778 MPa. 16 (339.98) (0.043) 3 Td 3 If Principal stresses : 5701 - 77004116)+ziy 58.07+ 0 .07,2 + 1077) i o = 65.33 MPa to: -7.25 MPa . 65.33-1-7-2 Zmane - - 725) =36-29 MPa

(ģ) Sy = 659 - 655 MP a factor of safety ito of safety, Fos. According to stress to Maximum Normal theory 'MNST! . Sy = o (Fos) → FOs = sy. 55.5% = 10.02 655 - 10.02 S = 65.33 According 'MSST 10 ig to to maximum shear stress theory. maxi 0:5 Sy = fos (Zmar) => fos= 0.5 Sy 0.5 (655) 36.29 - 9o 2 . Iman Hence, According to MNST, FOs = 10:02 According to MSST, FOS = 9.02