Aurel Vlaicu University

Student: Teacher:
Malau Lino Adriana Motica

Date, June 2023

Introduction
As subject of a practical problem, I will present the motor axis of an angle grinder (Figure 1)
which has a motor with the following specifications:
Power 900 W
Speed 11.000 rpm

Figure 1: The motor axis of an angle guinder

This motor shaft is subject to torsion and bending. In what follows you presented the torsional
stress throught theory-based calculations, and the bending stress through a simulation in
SolidWorks design program. The maximum engine torque of which the engine is capable is
determined, based on the technical characteristics of the engine, starting from the relationship,
known from phyics, between power, moment and angular speed:
P=M ∙ ω
2π ∙n
ω=
60
P – power
M – Moment maxim
ω – angular velocity
n – speed of the engine
So,
2 π ∙ n 2 π ∙11000 rad
ω= = =1151, 91( )
60 60 s
P 900
M= = =0.781 [ N . m ] =781 [ N . mm ]
ω 1151.0
Figure 2 – Technical drawing of the motor axis of the angle grinder
The shaft is made by turning S460N steel, according to the EN 10025-3 standard from 2004.
This type of steel is a high quality structural steel, which has the following technical
characteristics:
 Tear strength Rm = 540-720 [N/mm2]
Upper yield strength (at 16 mm in diameter) ReH = 460 [N/mm2]
Elongation at break An = 17
Character of sock bending (at 20˚C) -Longitudinal: KV = 55[J]
-Transversal: : KV = 31[J]

I make AB the axis of an angle grinder with the specification according to the technical drawing
(figure 2), length (=68.4 mm actuated by the moment M=7811 Nmm) to simplify the
calculations I will consider a section of 14mm in diameter. Is required:
1) Calculating the value of the reactions and creating the force diagram.
2) Determination of the tensional moment the effort capable torsion, of check the tensional
axis and the tensional deformation energy.
Solution:
1) To determinate the reactions, write the equilibrium equations.

∑ F ix =0 → H A =0
∑ F iy =0 →V A −V B=0
M 781
∑ M A =0 → V B ∙68.4−M →V B= 68.4 = 68.4 =11.42
−M −781
∑ M B=0 → V A ∙68.4 + M → V B= 68.4 = 68.4 =−11.42
After determinate the reactions, the scheme is divided into section and draw the diagram (figure
3)
(A-1)
N=H A=0

T =V A =−11.42

M A =V A ∙ X A

M A ( X =0 )=0

M A ( X =15 )=−11.42∙ 15=−171.3

(B-1)
N=0
T =V B=11.42
M A ( x=0 )=0

As check we calculate the moment at point 1 on the other side

M B ( X =53.4 ) =V A ∙68.4 +V B ∙ 53.4=−781.1+609.8=−171.3

Figure 3 – Diagram of torsional forces

2.
2.1 Determination of the torsional moment
To determinate the torque to which the engine is actually subjected, it starts from the engine
power expressed in horse power and the speed Rotation per minute.
P [ CP ]
M t =716.2∙ daN ∙ m
n [ rot /min ]
900 W =1.20691988 CP
1.20691988
→ M t =716.2 ∙ =0.0785814 daN ∙m=785,814 N ∙ mm
11000
The maximum torque that can be supported by the material at the section corresponding to 14
mm is determined with the formula M t =G∙ θ ∙ I p, were:
G - corresponding transverse modulus of elasticity, the material that is 79000N/mm2
θ – specific twist of the material from the tables – 0,28˚/m = 0,00028˚/mm
4 4
πd π ∙14 4
I p= = =3771 mm → M t max =79000 ∙ 0,00028∙ 3771=83414 , 52 N ∙ mm
32 32
Due to the fact that the twisting moment to which the shaft is subjected by motor is much smaller
than that which the material resists, it follows that the spindle successfully resists the stress.
2.2 Determination of specific twist
But M t =G∙ θ ∙ I p → that one can determinate the effective specific twist θef , with
Mt
θef =
G∙I p

Were: M t – effective torque to which the shaft is subjected

785.814
θef = =0.0000026377652236 <¿ from the table in ˚/mm.
79000 ∙3771
2.3 Determination of the torsional stress
The determination of the torsional stress is done using the formula:
M t 785,814 3 3
τ= = =1.46=τ a were W p = π d = π ∙ 14 =539 mm3−¿ resistance module polar
Wp 539 16 16
where d=14 – diameter of the section.
2.4 Determining the capable effort and checking the torsion axis
W t cap=τ a ∙ W p =1 , 46 ∙539=786.94

For a part to resist the torsional stress produced by the moment of torsion it is necessary that
Mt 785,814
τ max= = =0,998≤ τ a=1 , 46 ≤
W p cap 786 , 94

It follows that the part resists the torsional stress

2.5 Energy determination
The energy accumulated under the effect of a constant torque is determined with:
2 2
Mt 785,814
E= = =0.0010363964207124
2 Gτ ∙ I p 2 ∙79000 ∙ 3771