Question

A 100-mm-diametre shaft is subjected to a 10-kN.m steady bending moment, an 8-kN.m steady torque, and a 150-kN axial force. The yield strength of the shaft material is 600 MPa. Use the Tresca and the Von Mises theories to determine the safety factors for the given loadings.

Answer 1

Answer:

Tresca FOS ,N=4.49

Von mises theory FOS N=4.27

Explanation:

Cross sectional area of shaft A=$\dfrac{\pi }{4}d^2$

                    $A=7.8\times 10^{-3}$

Bending stress $\sigma _b=\frac{32M}{\pi d^3}$ =101.91 MPa                 (M=10 KN-m)

Shear stress $\sigma _s=\frac{16T}{\pi d^3}$ =40.76 MPa                    (T=8 KN-m)

Axial stress$\sigma _a=\dfrac{4P}{\pi d^2}$=19.23 MPa

Now find the principle stress

$\sigma _1,\sigma_2=\dfrac{\sigma_a+\sigma_b}{2}\pm\sqrt {\left (\dfrac{\sigma_a+\sigma_b}{2}\right )^2+\tau ^2}$

Now put the values

$\sigma _1,\sigma_2=\dfrac{19.23+101.91}{2}\pm\sqrt {\left (\dfrac{19.23+101.91}{2}\right )^2+40.76^2}$

$\sigma _1=133.62 ,\sigma _2= -12.42 MPa$

From tresca theory

$\tau _{max}=\dfrac{\sigma _y}{2N}$

$\dfrac{\sigma _1-\sigma_2}{2}=\dfrac{600}{2N}$

N=4.49

From Von mises theory

$\sigma _1^2+\sigma _2^2-\sigma _1\sigma _2=\left (\dfrac{\sigma_y}{N}\right )^2$

$133.62^2+\12.42^2+133.62\times 12.42=\left(\dfrac{600}{N}\right )^2$

N=4.27