1.5 basic loadings: bending and torsion external forces ... · ... bending and torsion ... table of...

IAM-DT-I : SS2005 Prof. Dr.- Ing. P. J. Mauk Universität Duisburg-Essen Institut für Angewandte Materialtechnik Umformtechnik

LECTURE-DT-1_03 Seite 1 von 20

1.5 Basic loadings: Bending and Torsion External forces and internal stresses:

a) Tension stresses → tension forces b) Compression stresses → compression forces c) Shear stresses → shear forces

Other basic loading of machine parts:

Bending of a beam: d) Mb: bending moment, e) Tt: torsion torque, f) normal and shear

stresses combined



The beam deflects by the force applied to the free (cantilever beam):

Distance l to fixed end is important for the deflection y at the free end.

If the force F is increased the deformation increases on the magnitude of the force and the length l (lever arm) of the force

bF l M⋅ = (bending moment) the bending moment causes internal forces which causes strains: extensions on upper side ε + → σ + tension stresses compressions on lower side ε − → σ − compression stresses

Strains (deformations) due to bending:

a) undeformed element in the beam, b) deformed element in the beam



Deformed element : the normal strain along a line element s∆ is:

( )( )

with and

s ssx s r s r y

r y rr

yr

ε

ϑ ϑ

ϑ ϑε

ϑ

ε

′∆ − ∆=

∆′∆ = ∆ = ∆ ∆ = + ∆

+ ∆ − ∆=

∆

=

i i

i ii

The strain varies linear with y from the neutral axis. Let c the maximum distance be to the outer fiber from neutral axis:

maxmax

yyr

C Cr

ε ε εε

= ⇒ = i

This is a linear relationship for the strain distribution:

max

yC

ε σ

σ σ=

∼

i ; according to Hooke’ s law

View on the cross section of the beam: All internal moments must be equal to external bending moment

( ) ( )b bM ex M in=∑ ∑ Strain and stress distribution in a bar (beam) ( ) ( ) max

max

y y yyC C

εε ε

ε= ⇒ = i linear relation of strain

if: ( )( )( ) ( )

max:

0 : 0 0

:

y C y C

y y

y C y C y

ε ε

ε

ε ε

= = =

= = =

< < = −



from a linear strain distribution follows: ( ) maxyyC

σ σ= i acc. to Hooke’ s law

maxσ : max. bending stress due to external bending moment

All external bending moments must be equal to all internal bending moments dM dx dy y

dA yσσ

Ι =

=

i i ii i

For equilibrium condition:

max

max

2max

I bA A

bA

A

dM y dA M

yC

yM y dAC

y dAC

σ

σ σ

σ

σ

= ⋅ ⋅ =

= ⋅

= ⋅ ⋅ ⋅

= ⋅ ⋅

∫ ∫

∫

∫

This integral is a special parameter in bending:

2

Ay dA∫ i : area moment of inertia = yΙ (section moment 2nd degree)



For a rectangular bar: b h× The area moment of inertia is: ( about y-axis ; the bending axis)

2 2

2 21 1

+ + +

− − −= ⋅ = ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅∫ ∫ ∫ ∫ ∫

a

e

h by xy h by x

AI y dA y dy dx y dy dx ;

3 32

2

22

2 23

2 233 3

−+

+

−−

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎛ ⎞⎛ ⎞⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠⎜ ⎟

⎜ ⎟⎝ ⎠

− −= ⋅ = − ⋅

hb

bh

h hb bxy

( )3 3 3 3 3

3 43 3 24 24 122 3 2 3

⎛ ⎞ ⎛ ⎞⎡ ⎤ ⎡ ⎤⎜ ⎟ ⎜ ⎟ ⎣ ⎦ ⎣ ⎦⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

−= − ⋅ = + = = Ι =⋅ ⋅

⋅ i iyh h h h b hb b mm mm mm

For the bending moment is:

3max

3 2

max max

12 22

12 6

withbb h hM C

Cb h b h

h

σ

σ σ

⋅= ⋅ =

⋅ ⋅ ⋅= ⋅ = ⋅⋅

For bending the maximum stress at the outer fiber of the section is of interest:

2

12Ι

Ι = =i yy by

b h WC

section modulus for y-axis (Widerstandsmoment)

Therefore in bending calculations the section module is used:

max

23

2 16

σ σ

σ⎡ ⎤

⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦

= =

= = =ii

bb

by

bb by

MW

M mm mmW mmb h



Stress distribution in cantilever bending: The neutral axis = bending axis gives the neutral plane where no stress are acting or this is the stress free plain in the bending section:

For many standard sections: rectangular areas, circular areas, hollow circular areas and others the values for the following data:

- area moment of inertia for x-and y- axis - section (bending) modulus are tabulated.

-



Table of area moments of inertia and section modulus for solid, hollow or combined

sections



1.6 Basic loadings: Torsion of a circular shaft A further basic loading is the torsion of a circular shaft:

Deformation at the end of the shaft:

ϑϑ γ=i

rr

l the strain comes from internal shear stresses

ϑ ϑτ γ∼r r as like to Hooke’s law for torsion, there is a linear relation for shear stress

ϑ ϑτ γ= ir rG G: shear module 2

N

mm⎡ ⎤⎢ ⎥⎣ ⎦

The maximum shear deformation is at the outer diameter, there is a linear relationship for the shear stresses:

( ) ϑτ γ= =i iir G rl

G

The external torque Mt = T is equal to the internal torques in the twisted section:



Torsion of a circular section: a) distortion of two sections with stress element; b) distorted grid on the surface

Calculation of internal torques in a circular section:

( ) ( )2 2

0 0 0 0

2 2 30 0 0 0

R Rt

R R

r

polararea momentof inertia

M T r r dA r r r dr d

G Gr r r dr d r dr dl l

π π

π π

α

τ τ α

ϑ ϑα α= =

= = ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅

⋅ ⋅= ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅

∫ ∫ ∫ ∫

∫ ∫ ∫ ∫ ;

4 4 42 2300 0

04 4

24 4 2 2

2 2 32

RR

P r

P

r R R DI r dr d with R

D DI

π π

α

πα α π

π π

= =

⎛ ⎞⎜ ⎟⎝ ⎠

⋅= ⋅ ⋅ = ⋅ = ⋅ = =

⋅= ⋅ =

∫ ∫

According to the bending of beams, the maximum shear stress is at the surface of the circular section:

2= P

tIW D

: section torsion modulus

4 32

32 16

τπ π

= = =i i it

t

T T TW D D

D

: maximum torsion shear stress

ϑ = ii P

T lG I

: maximum twist angle at the end of the circular section



The G – Module (Shear Module) is a elastic material constant:

( ) ( ) 2 ;807692.62 1 2 1 0.3

Poisson' s ratioNmm

E E EG νν

= == ≈ ≈⋅ + ⋅ +

281000 for 210000≈ ≈ NG Emm

Torsion and bending are the most applied technical loadings to machine elements: Example: Torsion of shaft: a motor connected to a machine = rotational power of a torque:

2 1 min60 30 sec min secπ πω

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎣ ⎦

⎣ ⎦⎣ ⎦= = = = =i i i i i i ii i i i

n n N m N mP T T T W

n = rpm of the motor = revolutions per minute

2 withp

AI r dA dA r d drα= ⋅ = ⋅ ⋅∫ ;

( )

4 4 4

4 222 3 30 0 0

04 4

2 2 2 16 32

4

24 2

RR

PA A

D D D

rI r r d dr r d dr r dr d

R R

ππ

π π π

α α α α

ππ

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

⋅⎛ ⎞= ⋅ = ⋅ =⎜ ⎟⎝ ⎠

= ⋅ ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ = ⋅

⋅= ⋅ =

∫ ∫ ∫ ∫

( ) ( )

( ) max

2

3max

with

A A

A

rrR

T r r dr d r r r dr d

T r dr dR

τ τ

τ α τ α

τ α

=

= ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅

= ⋅ ⋅

⋅

∫ ∫

∫



In the case of torsion the torque rotates around the central axis of the section: a polar area moment of inertia is defined. For the polar area moment of inertia is:

( )2 2 2

2 2

4 4 4

4

4 3

; ² ² ²

2 4 64 24 2

264 32

acc. to Pythagoras

for circular sections

with

PA A

x y x yA A

Px y

bx

r x yI r dA x y dA

x dx y dy I I I I

I R d d dI I R

d dWd

π π π

π π

= += ⋅ = + ⋅

= ⋅ + ⋅ = + =

⋅ ⋅ ⋅= = = = = =⋅

⋅ ⋅ ⋅= =⋅

∫ ∫

∫ ∫

From the sum of all internal torques the external torque is calculated:

( )4 223

0 0 00

4

RRG G rT r dr d

l lππ

α α α⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

= ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅∫ ∫ ;

4

2

4 4

4 4

24 2 2

32 32

with

with p

pPp

p

mm Nmm mm

G R G R dT Rl l

d G dT Il

G II G G T lT Il l l G I

ϑ

ππ

π π

ϑϑ⎡ ⎤⋅

⎢ ⎥⎣ ⎦

⋅= ⋅ ⋅ = =

⋅= ⋅ =

⋅ ⋅⋅ ⋅ ⋅= = ⋅ = → =⋅

Or the twist angle due to the torque “T” is :

432ϑ

π= i i

i iT lG d

In mechanical engineering the max. value for twisted bars is max 0,25 /ϑ ≈ ° m length or 0.0044 rad / m is allowed from this equation. The necessary diameter for given torque T with a limit value maxϑ for the twist angle is:

44

max max

32 32π ϑ π ϑ

= ⇒ =i i i ii i i i

T l T ld dG G



Other solution of the problem:

( )

( )

max

42max max

max max2P

PP

AI

rrR

IRT r r dA r dA WR R R

τ τ

τ τ πτ τ τ

= ⋅

= ⋅ ⋅ = ⋅ ⋅ = ⋅ = ⋅ = ⋅∫ ∫

polar section module

4 3

2 2π π= =i ii

PI R RR R

for circular sections 2

= dR there for:

3 3

max3 316

162 2p

pP

I d d TWR W d

π π πτπ

⋅ ⋅ ⋅= = = → = =⋅ ⋅

for circular sections is valid:

3 3 3

max 3

:2 2

16 2 2 16 3232( )

and

therefor

x y x yP

P Px y bx by

PP bx

b b

b

I I I I II WI I W W

Wd d dW W

M MbendingW d

π π π

σπ

= + =

= = = =

⋅ ⋅= → = = =⋅

⋅= =

⋅



Strength of materials

Area moment of inertia and section modulus of sections:Bending and buckling

Cross section Area moment of inertia Axial section modulus W Polar section modulus Wp

Torsion



1.7 Basic loadings: Combined loading (bending and tension):

σ = TT

FA

( tension stress: normal stress)

at the section point A

σ = bb

b

MW

at point A bσ tension bending stress

total stress at point A is: σ σ σ= +A A A

tot T b addition of normal stresses at point B: σ σ σ−=B B B

tot T b addition of normal stresses in the same section area Same condition for shear stresses is valid: Equal types of stresses can be superimposed with respect to their sign (+) for tension stresses (-) and compression stresses. But: a linear combination of shear and normal stresses is not allowed. Combination of a bending moment and a torque: The common loading of a circular shaft of a bending moment and a torque is very often:

;= i tbM F l T σb and τ t can not added in a simple way, therefore we need an equivalent σ eq stress:

( ),σ σ τ σ= ≤eq tb allf

The equivalent stress σeq is compared with an allowable stress for the material:



There are different theories for equivalent stresses, for ductile materials the maximum-distortion-energy theory has been proved as the best one for combined normal and shear stresses: (DET): “the Mises criterion”

( )2203σ α τσ = + i tbeq

σb : max. bending stress τ t : max. shear stress due to torsion

0α : stress intensity factor

( ),lim0

,lim

0,7 0,6 0,83σ

ατ

= ≈ ≈ib

t

,limσb : limit bending stress

,limτ t : limit torsion stress if 0,2σ ≥eq PR or σby

then yielding starts

1.5 basic loadings: bending and torsion external forces ... · ... bending and torsion ... table of...

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