Bending stress –

strain of bar exposed

to bending moment

• Basic principles and conditions of solution

• Calculation of bending (direct) stress

• Design of bar exposed to bending moment

• Combined stress of bar

Department of Structural Mechanics

Faculty of Civil Engineering, VSB - Technical University Ostrava

Elasticity and Plasticity

2 / 27

Bars under bending

Basic principles and conditions of solution

The bending moments and shear forces become in the bar in the course of

bending.

ba

M

l

M

MM

M

V

Simple bending

0, yz MV0 zxy MMVNIn xz plane hold true:

Plane bending: inner and external forces are situated in xy plane or xz plane

– principal plains.

0, zy MV0 yxz MMVNIn xy plane hold true:

x

a

Raz

b

Rbzl

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Simple bending

Laboratory test

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Simple bending

Testing of structures

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Basic conditions

b) axial fibres are not mutually in compression

0 zy

a) deformated cross-sections stay on plane

figure and perpendicular to deformated axis

(Bernoulli hypothesis)

Character of condition is deformation-geometrical.

a

M

b

M

z

x

z

y

y

z

Daniel Bernoulli(1700 - 1782)

Basic principles and conditions of solution

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Relations between inner forces and stress in cross-section

+z

+x

+y

N

zV

yV

Cross-section

Centre of gravity

Central line

y

z

AN x d.d ANA

xd

xxy

xz AzzNMA

xy d..

AyyNMA

xz d..

likewise

Calculation of bending (direct) stress

Placement of

inner forces

resultant

The resultant force can be

replaced by the normal force

on the normal and the two

shear forces in y, z

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Normal stress in bending

z

x

max

max

Neutral axes is the same as the

central line only at simple

loadind by the bending moment.zAB

C ED

d

xdxdxd

r

x = n

y

y

xI

z.M

e

IW

y

y

Extrem of stress is on outer fibres where z = e.

y

y

y

y

W

Me

I

Mmax

e

Distribution of normal stress x in

bending is linear over the hight of

beam and extreme values are in

outer fibres.

Zerro value of x is on neutral axes.

Wy - section modulus

for outer fibres [m3]

Iy - moment of inertia

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Normal ( bending) stress at simple bending

ANA

xd

Simple bending: suma N = 0

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Extrem of normal stress in bending - symmetrical cross section

z

y

x

lowery

y

lowerxW

M

,

,

Upper fibres:

uppery

y

upperW

M

,

Lower fibres:

y

y

lowerxupperxW

M max,,max,,

upperx,

lowerx,

Minus stress

Positive stress

We can determine signe of stress according to bending moment, after

deformation in bending there are clear tensile or compressed fibres.

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x1e

z

y 2e

Neutral axes in centre of gravity of section

fibrestensileW

Me

I

M

ey

y

y

y

ex

1,

11, .

fibrescompressedW

Me

I

M

ey

y

y

y

ex

2,

22, .

1

1e

IW

y

e,y 2

2e

IW

y

e,y

… Section modullus for outer fibres [m3]

0x

upper

upperW

M

lower

lowerW

M

lowerxupperx max,,max,,

Distance of outer fibres from axes of center of

gravity e1,2 (or c1,2)

In farther fibres from neutral axes there are with higher

stress ( je Wy,min )

Extrem of normal stress in bending - asymmetrical cross section

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Combination of stresses

a

Raz

b

Rbzl

Rax

F

σx

A

N xN

zI

M

y

yM

N

M

V

N MM

0N

c

In section c stress is

calculated by

superposition and it is

possible to gain:

x = n

Movement of neutral axes

-

+

- -

+

- - -

+ +

-

-

+

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Limited validitation of derived relation

Limited validation

y

y

xI

zM .

• Relation is valid for case of simple bending, constant cross-section and

the height of beam h << l (span).

xz

x

max

ha

Raz

b

Rbzl

(compression)

(tension)

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Limited validation of derived relation

x

a

Raz

b

Rbzl

Relation is not valid in abrupt changes of cross-section.

hy

y

xI

zM .

Limited validation

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Limited validation of derived relation

Limited validation

x

a

Raz

b

Rbzl

Relation is not valid in

case of bearing walls,

where l < 3h .

h

z (tension)

(compression)

y

y

xI

zM .

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Cross sectional characteristics

Cross sectional characteristics

x1c

2,cx

z

y

1,cx

2c

Neutral axis in center of gravity

1,

11, .cy

y

y

y

cxW

Mc

I

M

2,

22, .cy

y

y

y

cxW

Mc

I

M

Cross-section modulus calculation in case of simple shapes

b

h

1

1,c

IW

y

cy 2

2,c

IW

y

cy

3..12

1hbI y

hbI z ..12

1 3

2..

6

1

2

hbh

IW

y

y

hbb

IW

y

z ..6

1

2

2

d64

. 4dI

32

.

2

3d

d

IW

… Cross-section modulus to outer fibres [m3]

0x

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Design and reliability assessment of bar exposed to bending moments

Design of bar exposed to bending moment

Reliability assessment of design

Limit state of carrying capacity

Design of carrying

structure

Realization

Dimensioning

dEd fWM ,, min

dRdEd fWMM .min

RdMAdjusted design

d

Ed

f

MW min

M

kd

ff

dEd MM max

Assumption in design:

The same strength of

material in case of tension

and compression (steel), no

shear stresses influence

1Rd

Ed

M

M

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Vertical, horizontal and unsymmetrical bending

z

y

y

y

xI

zM .

z

zx

I

yM .

a b

Vertical bending Horizontal bending

z

z

y

y

xI

yM

I

zM .. My and Mz – combined stress of bar

(unsymmetrical bending)

Combined stress of bar

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Eccentric tension and compression

+z

+x

+y

N

Centre of gravity

Central line of beam

ze

ye

Neutral axis

nz

ny

Tension

zy eNM . yz eNM .

- acting N +My + Mz or another expression is when the N force whose position is placed

against the centrer of gravity on eccentrities ey and ey. Positive N force on positive

eccentrities causes moments:

Bending stress is the sume of individual stresses:

yI

Mz

I

M

A

N

z

z

y

y

x ..

(the expres. in brackets =0, y intersection is

obtained by substituting zero for z-coordinate)

A

Ii

y

y 2

A

Ii zz 2

22

..1.

z

y

y

zx

i

ye

i

ze

A

N

Segments of neutral axis:

0z 0.

1.2

z

y

i

ye

A

N y

zn

e

iy

2

0yz

y

ne

iz

2

Combined stress of bar

into:

is possible to modify by substitution:

0x