shear stress in bending - vsb.cz

Department of Structural Mechanics

Faculty of Civil Engineering, VSB - Technical University Ostrava

Elasticity and plasticity

Theme 6Shearing stress in bending

• Basic relationships and conditions of solutions• Shearing stress in chosen cross-sections

• Shear flux and shear centre• Dimension of members in shear

• Composite beams

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Shearing stress in beams

Basic relations and conditions of solution

ba

+

VV

RbzRaz

F

-

VV

In bending there come up usually together with bending moments also shear forces, which causes the shearing stress.

0, ≠yz MV0==== zxy MMVNIn the plane xz:

Plane bending: external and internal forces are in planes xy or xz – the main

planes.

0, ≠zy MV0==== yxz MMVNIn the plane xy :

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Basic examples of shearing stress


Cracks in bridges after flood in 2002, South Czech, photo: Prof. Ing. Vladimír Tomica, CSc.

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Cracks in bridges after flood in 2002, South Czech, photo: Prof. Ing. Vladimír Tomica, CSc.


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Crack in the support of the concrete beam

photo: Prof. Ing. Radim Čajka, CSc.


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Základní vztahy a předpoklady řešení

Detail of the screw connection

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The formula of reciprocity of the shearing stresses

y

z

x

[ ]

=

z

yzy

xzxyx

σ

τσ

ττσ

σ

sym.

Tenzor of stresses:

{ } { }T

xyzxyzzyx τττσσσσ =

Vector of stresses:

more topic n.8

Just 6 stresses components

S

zd

yd

xd

xyτyxτ

yxτ

xyτ

yxxy ττ = zyyz ττ = xzzx ττ =similarly

Basic relations and conditions of solution

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Basic relations for derivation of shearing stress

Shearing stress in chosen cross-sections

+y

+z

A B

T

z( )zb

ySA,

( )zy

yz

zxxzbI

SV

.

.==ττ

statical moment of separate part of cross section...yS

zV ... Shear force in section

yI ... moment of inertia of the whole cross section

( )zb ... Width of cross-section in the assessing place

Grashof formula

Cross sectional characteristics for shear stress

in bending are:

- Sy

- Iy

[ ]3

Tparty mzAS ⋅=

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Shear stress in a rectangular cross section


z

y

b

h

Cross section

maxτ

o2

Distribution of xzτ

z3.

12

1hbI y =

( ) bb z =

Tz

( ) A

V

bh

V

bbh

hbhV

bI

SVZZ

z

zy

yz

axm2

3

2

3

12

142

.

.

32/1 ==

⋅

⋅⋅===ττ

[ ]3

Tprůrčástiy mzAS ⋅=

statical moment of separate part of cross section...yS

zV ... Shear force in section

yI ... moment of inertia of the whole cross section

( )zb ... Width of cross-section in the assessing place

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Design and assessment of rectangular cross section in shear

Dimensioning

Assessment after Ultimate Limit

State

Design

Realisation

Dimensioning

dreqEd fAV ,,

RdEd VV ≤

RdVTo make bigger section

M

kd

ff

γ=

3.

2

3max

dz f

A

V≤=τ

d

zreq

f

VA

.2

.33=

1≤Rd

Ed

V

V

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Shearing stress in thin-walled members


httt wf <<,,

t

h

Symmetric I

hwh

ft

ft

fb

wt

w (web)

f (flange)

Thin-walled beam

Open cross sections: I, U, T, C, Z Hollow box-beam sections (pipe):

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Shear stress in I profile


Section

Wall Distribution of

z

y

Flange Distribution of

xzτ

xyτ

max,xzτ

o2

hwh

ft

ft

fb

wt

o1

Condition of solutions:• shearing stress is constant in the vertical cut to the fragment of the wall (see Detail)• is parallel to circumference of the section

xyτDetail

Detail

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Shear stress in the wall of the I-profile


Distribution

z

y

xzτ

max,xzτ

o2

hwh

ft

ft

fb

wt

z

tI

SV

y

yz

x.

.=τBasic formula:

t ... Thickness in the assessing

fragment

yS ...statical moment of a

separate partn of the section

xτ ... Shearing stress in the plane perpendicular to the axis x

xzτxyτ

vertical part

horizontal part

Shearing stress in the wall xzτ

(quadratic function)

[ ]3

Tparty mzAS ⋅=

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Shear stress in the flange of I-profile


z

y hwh

ft

ft

fb

wt

Distribution of xyτ

o1

tI

SV

y

yz

x.

.=τBasic formula

Shearing stress in the flange xyτ

(Linear function)

[ ]3

Tparty mzAS ⋅=

Tz

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Maximum stress in the I-profile


Distribution of

z

y

Distribution of

xzτ

xyτ

max,xzτ

o2

hwh

ft

ft

fb

wt

o1

( )( )[ ]2

max ....4...8

htthtbttI

Vwfwf

wy

z +−−=τ

0=z

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Shear centre

Shear flux and shear centre

At double symmetric cross sections the resulting force goes throught the centre of

gravity, at non symmetric sections it is different – if a plane of the load is not a plane

of symmetry, than the load must go throught the shear centre A, otherwise the

member would be twisted (stress from torsion).We obtain resulting shear forces Qf in the profile by the integration of shearing

stresses along the single walls of the open profile. They are equivalent to shear force Vz.

+y

+z

TA

0h

0h

t

t

zV

o45

fQ

fQ

xτ

xτ

zV

fQ

fQ

2

zf

VQ =

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Shear centre of the U-profile


z

y T0h

A

zV

fQ

fQ

0ba

wQ

xyτ

xyτ

xzτ

shth

stS ffy ...2

1

2.. 0

0 ==

s

y

z

fy

yz

xyI

shV

tI

SV

.2

..

.

.0==τ

Flange:

Profiles U, UE, UPE - size a in tables

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Shear centre of U-profile


z

y T0h

A

zV

fQ

fQ

0ba

wQ

xyτ

xyτ

xzτ

s

M

( )[ ]22

000 4....4...8

zhthbttI

Vwf

wy

zxz −+=τ

Web

... likewise I profile

zw VQ =

0.. hQaV fz =Statical moments to point M → y

f

z

f

I

hbt

V

hQa

.4

... 2

0

2

00 ==

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Composite members

Composite members

a b

b

a b

0≠V

welds, screws, bolts

a

a

( ) ( )( ) y

yz

zy

yz

zzxzxI

SV

bI

SVbbQ

.

.

...* === τ

[kN/m]

aI

SVaQQ

y

yz

xx ..

.* ==[kN]

Shear force for one connecting member

↓

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Questions for the exam

1. Shearing stress in bending of the rectangular cross section

2. Shearing stress in bending of thin-walled members, shear centre

3. Assessment of the members under shear stress in bending

4. Composite members

Questions for the exam

Documents

shear stress in bending - vsb.cz