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Shear stress in beams Dr Alessandro Palmeri Senior Lecturer in Structural Engineering <A.Palmeri@lboro.ac.uk>

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Learning Outcomes   When we have completed this unit (2 lectures + 1

tutorial), you should be able to: ◦  Gain a broad understanding of shear stress in

beams ◦  Apply the Jourawski’s formula to solid rectangular

sections and web of T and I sections

  Schedule: ◦  Lecture #1: Jourawski’s formula, with application

to solid rectangular sections ◦  Lecture #2: Application to web of T and I

sections ◦  Tutorial

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JOURAWSKI’S FORMULA, WITH APPLICATION TO SOLID RECTANGULAR SECTIONS

Lecture #1

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Shear Force & Bending Moment  At a given cross section of the beam, we

have generally both Bending Moment (BM) My and Shear Force (SF) Vz

  The SF gives the slope of the BM diagram:

  That is, the larger is the rate of variation of the BM along the beam, the larger is the SF

Vz =dMy

dx

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Shear Force & Bending Moment

 

W

W

x

x

−

+

W W

54W

34W

x

x

+

−

+

My

Vz

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Shear Force & Bending Moment

  2W L

W

W

x

x−

+

W W

W

W

x

x−

+

+ +

My

Vz

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Normal stress (Sigma)   The bending moment M sets up

(longitudinal) normal stress σx, which varies linearly with the distance from the neutral axis:

where ◦  Iyy is the second moment of the cross sectional

area ◦  z is the distance from the neutral axis (positive if

downward)

σ x =My z

I yy

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Shear stress (Tau)

  Key point: As the SF depends on the variation of the BM along the beam, similarly the shear stress τxz depends on the variation of the direct stress σx over the cross section

M M

Side View

tension

compression

Cross Section

t 0σ >

c 0σ <

Bending Stress

σ x < 0

σ x > 0

My My

x

M M

Side View

tension

compression

Cross Section

t 0σ >

c 0σ <

Bending Stress

σ x < 0

σ x > 0x

z

y x

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Jourawski’s formula   An engineering solution for the shear stress

distribution over the beam’s cross section can be obtained by using simple equilibrium considerations

  Since the elasticity theory will be not involved, this is just an approximate solution

  You must recognize the intrinsic limitations of this formula:

τ xy(P) =

Vz Qy(P)I yy b(P)

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Jourawski’s formula   Without lack of

generality, let us consider two cross sections 1 and 2 very close together, i.e. the distance is infinitesimal (dx), so that under a distributed load the BM experiences an infinitesimal variation (dMy)

dx1 2

My My + dMy

Vz =dMy

dx

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Jourawski’s formula

 

MdM M+

M

dM M+

FdF F+

cσ c cdσ σ+

dx

A A

Ab

Fx + dFx

b(P)

σ xFx σ x + dσ x

My

My + dMy

My + dMy

My

dx

x

P P

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Jourawski’s formula   Looking closely to the equilibrium of the top

“slice” of the beam between sections 1 and 2, i.e. above the ideal horizontal cut P-P, one can observe that a force difference dFx exists between these two sections, which must be balanced by a shear stress of resultant dRx

Fx Fx + dFx

dRxA A

τ zx (P)1 2

P P

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Jourawski’s formula   The forces Fx and Fx+dFx applied to sections 1

and 2, respectively, can be determined from the direct stress σx, which in turn depends on the BM at these sections, along with second moment of the cross sectional area Iyy and distance z from the neutral axis

Fx Fx + dFx

dRxA A

τ zx (P)1 2

P P

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Jourawski’s formula   It can be shown that the force acting on the plane P-P

is given by:

  where Qy(P) is the first moment of the cross sectional area above (or below) P-P with respect to the neutral axis, which can be computed as:

  AP being the area above (or below) P-P, while yA is

the distance of the centroid of AP with respect to the neutral axis

dFx =dMy Qy (P)

I yy

Qy (P) = AP × zP

zP

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Jourawski’s formula   Approximation! Under the assumption that the

shear stress τxz is uniformly distributed over the area P-P, we have that the difference force dFx is given by:

where b(P) is the width of the beam’s cross section at the level of P-P

  Rearranging the above equations, and taking into account that Vz= dMy/dx, we get the Jourawski’s formula for shear stress distribution

dFx =dMy Qy (P)

I yy= τ zx (P)b(P)dx

τ zx (P)horizontal

= τ xz (P)vertical

=dM y

dx

Qy (P)

I yy b(P)=Vz

Qy (P)

I yy b(P)

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Jourawski’s formula   Note! The Jourawski’s formula gives

the values of both horizontal and vertical shear stress at the level P-P

  This expression can be used to find beam’s shear stress in solid sections, e.g. square, rectangular, circular, and etcetera

  Adjustments are needed when dealing with ◦  open sections, e.g. I sections, T

sections, C sections, angles and etcetera; ◦  hollow sections, e.g. pipes or hollow

rectangular sections

x y

z

τ zx (P)horizontal

τ xz (P)

vertical

P

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Beam’s shear stress in a solid rectangular section   We need to apply the Jourawski’s formula to the

cross section shown below, which is subjected to a vertical shear force Vz

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Beam’s shear stress in a solid rectangular section

  The shear stress at a generic level P takes a parabolic (i.e. quadratic) expression:

A(P) = b d2− zP

⎛⎝⎜

⎞⎠⎟

zP =12d2+ zP

⎛⎝⎜

⎞⎠⎟

Qy (P) = A(P) zP =12d 2

4− zP

2⎛⎝⎜

⎞⎠⎟

τ xz (P) =Vz Qy (P)I yy b

=Vz2 I yy

d 2

4− zP

2⎛⎝⎜

⎞⎠⎟

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Beam’s shear stress in a solid rectangular section

  At the bottom edge:

  At the neutral axis:

zP =d2

τ xz (P) =Vz2 I yy

× 0 = 0

zP = zG = 0

τ xz (G) =Vz2 I yy

× d2

4= max τ xz (G){ }

I yy =bd 3

12⇒ τmax =

128Vzbd

=1.5VzA

50% more than the nominal average shear stress V/A

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APPLICATION TO WEB OF T AND I SECTIONS

Lecture #2

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Wrong results for flanges of T and I sections   The Jourawsky’s formula:

cannot be applied to the flanges of T and I sections, since the results so obtained are wrong

τ xz (P) =Vz Qy (P)I yy b(P)

This is not correct, and underestimates the actual shear stress in the flange

The parabolic distribution in the web, although approximate, is acceptable from an engineering point of view

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Wrong results for flanges of T and I sections

  Experimental and numerical analyses confirm that the “flow” of the shear stress in the flange of an I section is similar to what depicted below

τ flange

τweb

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Wrong results for flanges of T and I sections

  At the free edge of a member, there cannot be a finite component of the stress acting at right angle to the edge, simply because there is nothing to provide a reaction

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Shear stress in the web of T sections   1.0m

20 kN

20 kN

20 kNm

V

M

+

−My

Vz

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Shear stress in the web of T sections

  150

XB

16

16

  150

  43.4

  106.6

XB

I yy =150×163

12+ 150×16( )× 43.4− 16

2⎛⎝⎜

⎞⎠⎟

35.4

2

+ 16× (150−16134

)3

12+ 134×16( )× 106.6− 134

2⎛⎝⎜

⎞⎠⎟

39.6

2

= 9.63×106 mm4

neutral axis

35.4

39.6

τ xz (P) =

Vz Qy (P)

I yy b(P)

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Shear stress in the web of T sections

  150

XB

16

16

  150

  43.4

  106.6

XB

A(B) = 150×16 = 2,400 mm2

neutral axis

35.4

39.6

zB = − 43.4− 16

2⎛⎝⎜

⎞⎠⎟= −35.4 mm

Q(B) = 2,400× (−35.4) = −85,000 mm3

τ xz (B) =

VzQy (B)

I yy b(B)= 20,000× (−85,000)

9.63×106 ×16= −11.0 MPa

τ xz (P)

Shear stress at the top of the web

We consider here the area above B-B

The negative sign means that the flow of shear stress exits the considered area, i.e. is downward

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Shear stress in the web of T sections

  150

XB

16

16

  150

  43.4

  106.6

XB

A(G) =106.6×16 =1,706 mm2

neutral axis

35.4

39.6

zG = 106.62

= 53.3mm

Qy (G) =1,706×53.3 = 90,900 mm2

τ xz (G) =Vz Qy (G)I yy b(G)

= 20,000× 90,9009.63×106 ×16

=11.8 MPa

XXτ

Shear stress at the neutral axis

We consider here the area below G

G

The positive sign means that the flow of shear stress enters the the considered area, i.e. is downward

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Shear stress in the web of T sections

  150

XB

16

16

  150

  43.4

  106.6

XB

neutral axis

35.4

39.6

Qy (G) = 150×16( )× (−35.4)

Qy (B)

+ 43.4−16( )×16⎡⎣ ⎤⎦ × − 43.4−162

⎛⎝⎜

⎞⎠⎟= −91,000 mm2

τ xz (G) =Vz Qy (G)I yy b(G)

= 20,000× (−91,000)9.63×106 ×16

= −11.8 MPa

XXτ

We obtain the same result when considering in the calculations the area above the neutral axis

G

The only difference is the negative sign, because this time the flow of shear stress exits area above the neutral axis, i.e. is downward

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Shear stress in the web of T sections

  Please note that, if the width b is constant along the web, then: ◦  the maximum shear stress τxz happens at the level of the

neutral axis ◦  the distribution of the shear stress is parabolic

  150

XB

16

16

  150

  43.4

  106.6

XB

neutral axis

BB 11.0 MPaτ =XX 11.8MPaτ =G

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Shear stress in the web of other sections