Question

For the shaft shown in figure 1, made of AISI steel 1030 CD with Sut = 520 MPa and Sy = 440 MPa., determine the factor of safety based on failure theories for ductile of Maximum Shear Stress and Distortion Energy (Von Misses). The point selected for analysis is determined form the following table. The loads are F= 2.13 kN, P= 5.13 kN, and T = 31.3 N. m. 100 mm B 15-mm D. Р T

Answer 1

Given Information :

• 
  F= 2.13 kN
  
• 
  P= 5.13 kN
  
• 
  T = 31.3 N-m
  
• Length of the shaft :
  1 = 100 mm
  
• Diameter of the shaft :
  d= 15 mm
  

Solution :

The shaft is loaded with :

• Axial Tensile load
• Transverse shear load
• Torsional moment

The stresses induced in the shaft at point A are :

• Axial Tensile stress
• Normal Bending stress
• Torsional Shear stress

Axial Tensile Stress

The normal stress due to axial load is given by the formula

Р ба

Where

DO
Axial Stress

F
Axial Load
$A=$ Cross sectional area

Substituting the formula of A in the above equation

4* P 0 0 12

The torsional shear stress

From the equation of pure torsion we can write

= Tm RO G* L J

where
$T=$ Applied Torque
$J=$ Polar Moment of Inertia

Tm
Maximum Torsional Shear Stress

RO
Outer Radius of the beam
$G=$ Modulus of Rigidity
$\theta=$ Angle of Twist
$L=$ Length of beam

Rearranging the equation to obtain the relationship between Maximum Torsional Shear stress and the applied torque :

$\frac{T*R_0}{J}=\tau_m$

16 * Т Tm

The Normal Bending stress :

From the equation of pure bending

0b E INA y R

where
$M=$ Maximum bending moment
$I_N_A=$ Area Moment of Inertia about Neutral Axis

40
Maximum bending stress
$y=$ Distance between the neutral axis and the fiber at which bending stress is being calculated
$E=$ Young's Modulus of elasticity

R=
Radius of curvature of elastic curve

Rearranging the equation to obtain the relationship between Maximum Bending stress and the applied Transverse Shear load :

$\frac{M*y}{I_N_A}=\sigma_b$

32 * М оъ П

POINT A

Now that we have established the stresses acting at point A, we can now calculate the magnitudes :

Axial Tensile Stress

4* P 0 0 12

4 * 2.13 * 1000 σα Π* 152

oa 12.05 MPa

Torsional Shear stress

16 * Т Tm

16 * 31.3 * 1000 Tm * 153

Tm = 47.23 MPa

Normal Bending stress

M *у оъ INA

Since the point A is at the Neutral axis of the cross-section

y 0

op = 0 MPa-

Transverse Shear stress

Since there point A is at the circumference, there will not be any Transverse shear stress

Calculating the Principal Stresses using the following formula :

Or 01,2 Ost Oy 2 -2 ту 2

Here :

от – да

0

ry m

Substitute the values :

01 E ੪॥ + + 2 2

12.05 01 + (12.052 + 47.232 4 2. -

01 53.63 M Pa

da 02 = + 2 2

12.05 12.052 +47.232 4 2

ση –41.58 MPα

The Factor of safety according to :

(a) Maximum Shear stress theory :

Ans:

The maximum shear stress is given by the radius of the Mohr's Circle at the point :

0 R= = 504)2 + Tiny 2

R= -2 m 4

12.052 R= + 47.232 4

R= 47.61 MPa

Therefore the maximum shear stress is :

Тат = 47.61 MPа

According to Maximum shear stress theory :

Oyt Tmax 2* N

Where

Oyt
Yield strength of the material
$N=$ Factor of safety

Substitute the values

47.61 440 2 * N

N = 4.62

Therefore the factor of safety based on Maximum Shear Stress Theory is :
N = 4.62

(b) Von-Mises Theory

Ans :

The Von Mises stress is given by :

de = (01 - 02)2 + (02 - 03)2 – (03 - 01)

For the given loading :

-Oe 201 + 20 202 – 201 * 02

Accroding to Von Mises Theory :

N :) 20

20 + 202 – 201* 02 = 24

yt + 01 02 =

Substitute the values

440 53.632 + (-41.58) - (53.63 * -41.58) =(-

440 6835.0087 = ( N

440 82.67 =( N

N = 5.32

Therefore the Factor of safety according to Von-Mises Theory for point A is :  
N = 5.32

POINT B

Now that we have established the stresses acting at point A, we can now calculate the magnitudes :

Axial Tensile Stress

4* P 0 0 12

4 * 2.13 * 1000 σα Π* 152

oa 12.05 MPa

Torsional Shear stress

16 * Т Tm

16 * 31.3 * 1000 Tm * 153

Tm = 47.23 MPa

Normal Bending stress

32 * М оъ П

32 * F* 1 Oь #

$\sigma_b=\frac{32*2.13*1000*100}{\pi*15^3}$

$\sigma_b=642.84\ MPa$

The Normal Bending stress at point B is tensile in nature.

NOTE : Since the Normal Bending stress at point B is exceeding the Ultimate Tensile strength of the material, The shaft will undergo fracture, after extensive permanent deformation.