shear stresses (2nd year)

Shear stress distribu-on + Shear centre

Dr Alessandro Palmeri <[email protected]>

Teaching schedule Week Lecture 1 Staff Lecture 2 Staff Tutorial Staff 1 Beam Shear Stresses 1 A P Beam Shear Stresses 2 A P --- --- 2 Shear centres A P Basic Concepts J E-R Shear Centre A P 3 Principle of Virtual

forces J E-R Indeterminate Structures J E-R Virtual Forces J E-R

4 The Compatibility Method

J E-R Examples J E-R Virtual Forces J E-R

5 Examples J E-R Moment Distribution -Basics

J E-R Comp. Method J E-R

6 The Hardy Cross Method

J E-R Fixed End Moments J E-R Comp. Method J E-R

7 Examples J E-R Non Sway Frames J E-R Mom. Dist J E-R 8 Column Stability 1 A P Sway Frames J E-R Mom. Dist J E-R 9 Column Stability 2 A P Unsymmetric Bending 1 A P Colum Stability A P 10 Unsymmetric Bending 2 A P Complex Stress/Strain A P Unsymmetric

Bending A P

11 Complex Stress/Strain A P Complex Stress/Strain A P Complex Stress/Strain

A P

Christmas Holiday

12 Revision 13 14 Exams 15 2

Mo-va-ons

Shear failure of a RC beam and (b) load–displacement curve. Engineering Structures, Volume 59, 2014, 399 - 410

Diagonal failure: short beam (a) and long beam (b); crushing of the compressive zone (c). Engineering Structures, Volume 59, 2014, 399 - 410 3

Learning Outcomes When we have completed this unit (3 lectures + 1 tutorial), you should be able to: •  Apply the Jourawski’s formula for solid, thick-‐walled and

thin-‐walled sec-ons, including rectangular and circular sec-ons, T, I and channel sec-ons

•  Iden-fy the posi-on of shear centre for beam sec-ons with one or two axes of symmetry

Schedule: •  Lecture #1: Review of Jourawski’s formula •  Lecture #2: Applica-on to thin-‐walled cross sec-ons •  Lecture #3: Shear centre •  Tutorial (week 2)

4

Further reading

•  R C Hibbeler, “Mechanics of Materials”, 8th Ed, Pren-ce Hall – Chapter 7 on “Transverse Shear”

•  T H G Megson, “Structural and Stress Analysis”, 2nd Ed, Elsevier – Chapter 10 on “Shear of Beams” (eBook)

5

Stresses in beams (1/2) When a beam is subjected to external forces or deforma-ons, different stresses may rise. Engineers need to know the level of stress in the material, and specifically they need to check: •  Normal stress σx (parallel to the

beam’s axis), which tends to stretch (or shorten) the longitudinal fibres of the beam (Greek leger sigma)

•  Shear stresses τxz and τxy (orthogonal to the x axis in the beam’s cross sec-on), which tend to produce a rela-ve slipping between adjacent fibres (Greek leger tau)

6

Stresses in beams (2/2) Integra-ng the stresses σx, τxz and τxy over the beam’s cross sec-onal area A, one obtains the internal forces, which are in equilibrium with external forces and support reac-ons

N = σ x dAA∫

!!Vz = τ xz dA

A∫ Vy = τ xy dA

A∫

My = σ xzdAA∫

!!Mz = − σ xydA

A∫

T = τ xzy−τ xyz( )dAA∫

Normal force:

Shear forces:

Bending moments:

Torque: 7

Jourawski’s formula for thick beam sec-ons with axis of symmetry (1/3)

The Jourawski’s formula gives the average shear stress on a generic chord of length b, orthogonal to the shear force V applied along the axis of symmetry

!!τ =

Vz ′Qy

bIyy

τ

8

Jourawski (1821-‐1891) was a Russian engineer and one of the pioneers of bridge construc-on and structural mechanics in Russia.


In the Jourawski’s formula, is the first moment of the par-al area below the chord b, that is: while Iyy is the second moment of area about the y axis (orthogonal to the shear force Vz and passing through the centroid G)

!!′Qy = zdA

′A∫ = b

′A∫ zdz

′Qy

′A

!!Iyy = z2 dA

A∫ = b

A∫ z2 dz

9


The average shear stress along the chord goes to zero at the top and bogom fibres, and take the maximum value close to the centroid G If the length of the chord b does not vary along the depth of the cross sec-on (e.g. in a rectangular cross sec-on), the the maximum of the shear stress happens at the centroidal posi-on

!τmax

10

Limita-ons of the Jourawski’s formula

As only the average value of the shear stress along is computed, the Jourawski’s formula should only be used for design purposes when ligle varia-ons are expected for the shear stress along the chord b, e.g. in the web, not for the ver-cal component of the shear stress in the flanges of an I sec-on

11

Rectangular cross sec-on (1/3) First moment of the par-al area: Second moment of area:

!!

′Qy = ′A ′z = b d2− z

⎛⎝⎜

⎞⎠⎟

′A

d4+ z2

⎛⎝⎜

⎞⎠⎟

′z

= b2

d2

4− z2

⎛

⎝⎜⎞

⎠⎟

Iyy =bd3

1212

Rectangular cross sec-on (2/3) Shear stress (quadra-c func-on of the depth z, i.e. parabolic distribu-on)

!!

τ =Vz ′Qy

bIyy=

Vz b 2

b 2 d3 12d2

4− z2

⎛

⎝⎜⎞

⎠⎟

=6Vz d

2

4bd 31− 4z2

d2

⎛

⎝⎜⎞

⎠⎟

= 32Vzbd

1− 2zd

⎛⎝⎜

⎞⎠⎟

2⎡

⎣⎢⎢

⎤

⎦⎥⎥

13

Rectangular cross sec-on (3/3) The maximum value of the shear stress then happens at the centroidal posi-on (z=0): while for z=±d/2 (bogom and top fibres) one gets

τmax =32Vz

bd=1.5

Vz

A

τ = 0

14

Circular cross sec-on (1/3) In this case we have:

b = 2Rcos(α )

!!z = Rsin(α ) dz = Rcos(α )dα

′Qy = b′A∫ zdz= 2R3 cos2(α )sin(α )dα

α

π /2

∫

= 23R3 cos3(α )

!!

Iyy = bA∫ z2 dz= 2R4 cos2(α )sin2(α )dα

−π /2

π /2

∫

= π R4

415

Circular cross sec-on (2/3) According to the Jourawski’s formula, the average shear stress along the ver-cal axis is:

τ =Vz ′Qy

bIyy=Vz

23R3 cos 3 2(α )

π R 5 2

2cos(α )

= 43

Vz

π R2cos2(α )= 4

3Vz

π R21− z

R⎛⎝⎜

⎞⎠⎟

2⎡

⎣⎢⎢

⎤

⎦⎥⎥

16

Circular cross sec-on (3/3) The maximum value of the shear stress then happens at the centroidal posi-on (z=0): while again for the extreme fibres, when z=±R, one gets Moreover, also in this case the distribu-on of the shear stress is parabolic along along the depth of the cross sec-on

τmax =43

Vz

π R2=1.33

Vz

A

τ = 0

17

Thick-‐walled I sec-on with unsymmetrical flanges (1/10)

The Jourawski’s formula can also be used to evaluate the average shear stress in the web and the flanges of an I sec-on Let’s find the area first of the thick-‐walled example sec-on: A = 8u2

At

+8u2

Aw

+6u2

Ab

= 22u2

18


To find the posi-on of the centroid G, the first moment of the cross sec-onal area about a reference axis Y must be evaluated:

!!

QY = At

t2+ Aw t + d

2⎛⎝⎜

⎞⎠⎟+ Ab d + 3

2t

⎛⎝⎜

⎞⎠⎟

= 4u3 +24u3 +33u3 = 61u3

dG =QY

A= 61u3

22u2= 2.77u

19


Iyy =Bt3

12+ At dG −

t2

⎛⎝⎜

⎞⎠⎟

2

+ 2td3

12+ Aw t + d

2−dG

⎛⎝⎜

⎞⎠⎟

2

+ bt3

12+ Ab d + 3

2t −dG

⎛⎝⎜

⎞⎠⎟

2

= 98.20u4

It is now possible to calculate the second moment of area:

20


Let’s also define the radius of gyra-on (denoted with the Greek leger rho): This allows rewri-ng the Jurawsky’s formula in the equivalent form:

!!ρy =

IyyA

= 98.20u4

22u2= 2.11u

τ =′Qy

bρy2

Vz

A

21


We can then calculate the average shear stress at the key loca-ons in the beam sec-on The minus sign means that the flow of shear stresses exits the area

τ1 =′Qy (1)

b1 ρy2

Vz

A=− ′A1 ⋅ dG −

t2( )⎡

⎣⎤⎦

t ⋅4.45t2Vz

A

=− 3t2 ⋅2.27u⎡⎣ ⎤⎦

4.45u3

Vz

A= −1.53

Vz

A

!! ′A122


!!

τ 2 =′Qy (2)

b2 ρy2

VzA=− At ⋅ dG −

t2( )⎡⎣ ⎤⎦

2t ⋅4.45t2VzA

=− 8u2 ⋅2.27u⎡⎣ ⎤⎦

8.90u3

VzA= −2.04

VzA

23


!!

τG =′Qy (G)

bG ρy2

VzA

=′Qy (2)− 2t ⋅ 12 dG − t( )2⎡

⎣⎢⎤⎦⎥

2t ⋅4.45t2VzA

=−18.16u3 − 2u ⋅ 12 1.77u( )2⎡

⎣⎢⎤⎦⎥

8.90u3

VzA

= −2.39VzA

24


τ 3 =′Qy (3)

b3 ρy2

Vz

A=

Ab ⋅32 t +d −dG( )⎡

⎣⎤⎦

2t ⋅4.45t2Vz

A

=6u2 ⋅2.73u⎡⎣ ⎤⎦8.90u3

Vz

A=1.84

Vz

A

!!

τ 4 =′Qy (4)

b4 ρy2

VzA=

′A4 ⋅32 t +d −dG( )⎡⎣ ⎤⎦t ⋅4.45t2

VzA

=2u2 ⋅2.73u⎡⎣ ⎤⎦4.45u3

VzA=1.23

VzA

25


For this loading condi-on, with the shear force Vz ac-ng along the web, the varia-on of the shear stresses is: •  Linear in the flanges (which

are orthogonal to Vz) •  Quadra-c in the web (which

is parallel to Vz) •  Zero at extreme fibres of

the flanges •  Maximum at the posi-on of

the centroid G in the web

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If the I beam is made of two -mber board glued to a -mber shaq, and are the values of the average shear stress in the top and bogom layers of glue Shear stresses in orthogonal planes take the same values (complementary shear stresses)

!τ 2 τ 3

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Thin-‐walled sec-ons If the thickness of the panels cons-tu-ng open or closed cross sec-ons is negligible in comparison with their width and depth, the cross sec-on is said to be thin-‐walled The shear stress can be assumed to be uniform over the thickness of the panels It is possible to work all the quan--es with respect to the centreline of each panel

τ

28

Thin-‐walled I sec-on with unsymmetrical flanges (1/5)

Let’s analyse the same cross sec-on studied before, assuming this -me that the thickness t is much less than all the other geometrical dimensions A = 8ut

At

+8utAw

+6utAb

= 22ut

QY = Aw

d2+ Abd

=16u2t +24u2t = 40u3

!!dG =

QY

A= 40u 2 t22 u t

=1.82u29

dG


!!

Iyy = At dG2

+2td3

12+ Aw

d2−dG

⎛⎝⎜

⎞⎠⎟

2

+Ab d −dG( )2= 66u3t

!!ρy =

IyyA

= 66u3t22ut

=1.73u

30


!!

τ1 =′Qy (1)

b1 ρy2

VzA=− ′A1 ⋅dG⎡⎣ ⎤⎦t ⋅3u2

VzA

=− 4ut ⋅1.82u⎡⎣ ⎤⎦

3u2tVzA= −2.43

VzA

τ 2 =′Qy (2)

b2 ρy2

Vz

A=− At ⋅dG⎡⎣ ⎤⎦2t ⋅3u2

Vz

A

=− 8ut ⋅1.82u⎡⎣ ⎤⎦

6u2tVz

A= −2.43

Vz

A

31


τG =′Qy (G)

bG ρy2

Vz

A=

′Qy (2)− 2t ⋅ 12dG2⎡⎣ ⎤⎦

2t ⋅3u2

Vz

A

=−14.56u2t − t ⋅ 1.82u( )2⎡

⎣⎢⎤⎦⎥

6u2tVz

A

= −2.99Vz

A

!!

τ 3 =′Qy (3)

b3 ρy2

VzA=

Ab ⋅ d −dG( )⎡⎣

⎤⎦

2t ⋅3u2

VzA

=6ut ⋅2.18u⎡⎣ ⎤⎦

6u2tVzA= 2.18

VzA= τ 4

32


The distribu-on of the shear stresses is similar to what we have seem for the thick-‐walled case, with the important difference that this -me we consider the centrelines

33

Shear-‐stress resultants in thin-‐walled cross sec-ons (1/3)

If the thin-‐walled panel is orthogonal to the shear force, the shear stress varies linearly along the panel, meaning that the resultant shear force in the panel can be computed knowing the value of the shear stress at the extreme fibres

For instance, let’s consider the resultant forces Rt and Rb at the top and bogom flanges in the previous example, respec-vely

34

Rt = τ t

At

2=0+ τ1

2τ t

4

2ut

At

= 2.43Vz

2211ut

2 ut = 0.221Vz

!!

Rb = τb

Ab2=0+ τ 4

2τb

3utAb

=1.82Vz

44 ut3 ut = 0.149Vz


If the thin-‐walled panel is parallel to the shear force, the varia-on of the shear stress is quadra?c, meaning that evaliua-ng the resultant shear force in the panel requires also the value of the shear stress at midpoint (this scheme of integra-on is known as the Simpson’s rule)

This is the case of the resultant force Rw in the web of the previous example

35

Rw = τw Aw =τ 2 + 4 τ 5 + τ 3

6τw

Aw


We can consider the top flange and the top half of the web to calculate the shear stress at the mid-‐web posi-on (point 5 in the figure)

!!

τ 5 =′Qy (5)

b5 ρy2

VzA=− Bt( )dG − 2t d

2( ) dG −d4( )

2t 1.73u( )2VzA

= −8ut( ) 1.82u( )+ 2t 2u( ) 0.82u( )

2t3u2

VzA

= −17.846

VzA= −2.77

VzA

36

Rw = 2.43+ 4×2.97+2.186

Vz

AAw = 2.75

Vz

22 ut8 ut =Vz

Shear force not passing through an axis of symmetry (1/4)

The Jourawski’s formula can also be applied if the shear force does not pass through an axis of symmetry, e.g. in the case in which a horizontal shear force Vy is applied orthogonally to the web of the I sec-on considered before

37

Izz =t 8u( )312

+t 6u( )312

= 60.67u3t

ρz =IzzA= 60.67u3t

22ut=1.66u


This -me the flanges are parallel to the shear force, and therefore the varia-on of the shear stress is quadra-c, with the maxima happening at the centroidal posi-ons (points 6 and 7 in the figure) and zero values of the extreme fibres

38

!!

τ 6 =′Qz(6)

b6 ρz2

Vy

A=

4 u t( )2 ut 2.76 u2( )

Vy

A= 2.90

Vy

A

τ 7 =′Qz(7)

b7 ρz2

Vy

A=

3 u t( )1.50 ut 2.76 u2( )

Vy

A=1.63

Vy

A


If the shear force Vz is parallel to the web, then the web takes the whole shear force

39 Rw =Vy


If the shear force Vz is parallel to the flanges, then the flanges takes the whole shear force

40

!!

Rt = τ t At =0+ 4

2τ 6 +0

63

Vy

AAt

= 232.90

Vy

22 ut8 ut = 0.70Vy

Rb = τb Ab =0+ 4

2τ 7 + 0

63

Vy

AAt

= 231.63

Vy

22 ut6 ut = 0.30Vy!!Rt +Rb = 0.70Vy +0.30Vy =Vy

Shear centre (1/2) The distribu-on of the shear stress in the cross sec-on must give the same resultant ac?ons in the horizontal (y) direc-on and ver-cal (z) direc-on, as well as the same moment about any given point In the case when the shear force is horizontal, let’s use the resultant forces in the two flanges to calculate the moment Mτ of the shear stresses about the centroid G (taken as posi-ve if an-clockwise)

41

Mτ = Rt dG −Rb d −dG( )= 0.70Vy( ) 1.82u( )− 0.30Vy( ) 2.18u( )= 1.274−0.654( )Vy u = 0.62Vy u

Shear centre (2/2)

This allows calcula-ng the posi-on where the shear force needs to applied in order to have a distribu-on of the shear stresses consistent with the Jourawski formula (of pure shear)

This is the posi-on of the so-‐called “shear centre”

42

!!MV =Vy dG −dC( )MV =Mτ ⇒ Vy 1.82u−dC( ) = 0.62Vy u

1.82u−dC = 0.62 u⇒ dC =1.20 u

The moment of the shear stresses must be equal to the moment of the shear force Vy about the same point (in this example, the centroid G)

Centroid and shear centre (1/2) If the axial force N is not applied at the centroid G (or centre of mass), then the cross sec-on is subjected to a bending moment M=N d

If the shear force V does not pass through the shear centre C, then the cross sec-on is subjected to a torque (twis?ng moment) T=V d

43

Centroid and shear centre (2/2) Some key points to remember 1.  To avoid the addi-onal shear stresses due to the torque T, the shear

force must pass through the shear centre 2.  Addi-onally, if the line of ac-on of the shear force does not pass

through the shear centre, then the cross sec-on is subjected to a rotatory (in-‐plane) movement

3.  If a cross sec-on has an axis of symmetry, then both the centroid G and the shear centre C belong to it, but in general they do not coincide

4.  The posi-on of the centroid along the axis of symmetry can be obtained by dividing the first moment Q about an orthogonal axis by the area A

5.  The posi-on of the shear shear centre along the axis of symmetry requires: i) considering a dummy shear force V, which is fic--ously applied orthogonally it; and ii) equa-ng the twis-ng moment due to the shear force, MV, to the twis-ng moment due to the distribu-on of the shear stresses, Mτ

44

Channel sec-on (1/6)

In the case of a thin-‐walled channel sec-on with symmetrical flanges, one can show that the shear centre C is ‘outside’ the cross sec-on, on the axis of symmetry, opposite to the centroid G with respect to the web A shear force Vz parallel to the web is necessary to find the posi-on dC of the shear centre

45


Let’s find first the posi-on of the centroid G

46

!!

A= 2Af + Aw = 2btf +dtw= 2⋅120⋅3+200⋅5=1,720mm2

!!QZ = 2 Af

b2= b 2 tw = 43,200mm3

dG =QZ

A= 43,200

1,720= 25.1mm


The second moment of area about the horizontal axis y can be computed assumed that web and flanges are thin

47

Iyy =tw d

3

12+2Af

d2

⎛⎝⎜

⎞⎠⎟

2

= 5⋅2003

12+2⋅360 ⋅1002

=10.53×106


A dummy shear force Vz= 1 kN is used to find the posi-on of the shear centre C

48

!!

τ1 =′Qy (1)Vzb1 Iyy

= −Af

d2Vz

tf Iyy

= −360 ⋅ 100 ⋅ 1,000

3⋅10.53× 106= −1.14MPa

τ1 tf = τ 2 tw ⇒ τ 2 =τ1 tftw

= −0.68MPa

τG =′Qy (G)Vz

bG Iyy== −1.16MPa


Once the distribu-on of the shear stresses is known, one can compute the resultant shear forces

49

Rf = τ f Af =τ1

2Af = 0.57 ⋅360 = 205N

!!

Rw = τw Aw =τ 2 + 4 τG + τ 2

6Aw

= 0.68+ 4⋅1.16+0.686

1,000

=1,000N =Vz


The posi-on of the shear centre is then obtained by enforcing the equivalence between the moment Mτ due to the shear stresses and the moment due to the shear force MV

50

Mτ (1)= Rf d

!!MV (1)=VzdC

!!

MV (1)=Mτ (1) ⇒ VzdC = Rf d

dC =Rf dVz

= 205⋅ 200

1,0005

= 41mm

Shear centre of typical cross sec-ons

In many cases the posi-on of the shear centre can be immediately determined, based on simple geometrical considera-ons

51

Key learning points 1.  The Jourawski’s formula can be used to calculate the distribu-on of

the shear stresses when they can be considered almost constant along a chord which splits the cross sec-ons in two parts

2.  In thin-‐walled cross sec?ons, one can neglect the thickness of the cons-tu-ng panels in comparison to the other geometrical dimensions, and the shear stresses are uniform along the thickness

3.  The distribu-on of the shear stresses must be equivalent to the applied shear force, and the equivalence of moments allow calcula-ng the posi-on of the shear centr

4.  There is no twis-ng moment (and therefore no in-‐plane rota-on of the cross sec-on) if and only if the shear force passes through the shear centre

5.  Similarly to the centroid, the shear centre always belongs to an axis of symmetry

52

shear stresses (2nd year)

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