Question

The bar is made of mild steel (Sy =245 MPa) and is loaded by the forces Fx = 300 N, F, = 450 N, F2 = 300 N, and Mx = 55 N m. - 100 mm (A) Find the principal stresses and the max shear stress at A of the member. (B) Compute the von Mises stress at A. TINTI (C) Compute the factors of safety of the member using the distortion-energy (DE) and maximum-shear-stress (MSS) theories. 15-mm D. ME

Answer 1

- Fu Mx FZ 1. Due to tension Fx entire steel is unden rt= 300 xy n (0.015)? = 1.698 MPa

Due to Fy, Bending monent of magnitudii Fyx l. will act at fined end. =D 4500 ort=45mm Part A will be under compression due to bending as Fy is upwards. -

ob = -32 My - 32 (45) ndi - 260.015)} = -135 mea

- Due to Fz, 8 will be under compression & D under tersien as it is in 2-der?. . Bf to berding stress varies linearly 1. across cross section. - At A (6=0) due to Fz t ve -veli

a constant gives sheer Due to torque stress on Torsioned My there is os sheft . It outer fibres.. sheer stress = - = 16 I

t = 16 Mx = 16X55 - - 83.038 MPa 1nc0.0153) n d3 - similar to tersile stress or the magnitud of transverse shear stress is negligible due to Fy, Fg Fz & hence it is ignored. Tensile stress is also ignored - OHARD state of stress Of A z Ts, 2= ontoy Ion-oy) + truy 12 = 2

7,2 = 5 * 56²+ zo 1,2 = -135 25+ (49)* + 22.018 ( 55,2)= -67.5 I 107.012 =39.52 MPa =-174.52 MPa 0 - Imam = Veggy + 83.01e* = 107 ve os outro Enare - (on-oy) + tay = 107,012 n Cmar - L Tronco + Try = cy)2 102.012 Mpa ® von Mises stress at a For berding tersile stress & sheer stress, or when T2,2 are known Tum=50? Tirette?

btw of tol =wno (25.hti-) + (256t+) X7568-728.68 =

To calculate Fos critical section needs to be identified. 1 Assumptions - Tersile stress due to fx is neglected as its magnitude is less compared to bending stress. Transverse shear stress is neglected. for same reason. Note - fy will produce berding stress at A &c. Fz will produce bending stress at B & D. (Maninun] → since magnitude of Fy> Fz both - A & C are critical section. stress shear is same ie equel at A&c on all outer fibers. @ Tum= 197.28mpa [ Distorsiona ererary fos- sy =245 - 1.242 - Iner = 107.012= Syt (mex shear 2(n) stress theory) Fos= 245 h = 1.145 2x107012 2 n=
Note that shaft subjected to combined load is always designed based on bending stress and and torsional stress. The magnitude of transverse shear stress and axial stress are very less when compared with the above two.

It is important to understand the distribution of bending and torsional shear stress across the cross section.

Due to Fy point A is under compression stress. But due to Fz there is no stress as seen from stress distribution.

Further calculations are show in the figure.

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