1. Shear stress in beams
Dr Alessandro Palmeri
Senior Lecturer in Structural Engineering
<A.Palmeri@lboro.ac.uk>
1/
30

2. 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:
2/
30
◦ Lecture #1: Jourawski’s formula, with application
to solid rectangular sections
◦ Lecture #2: Application to web of T and I
sections
◦ Tutorial

3. Lecture #1
JOURAWSKI’S FORMULA,
WITH APPLICATION TO
SOLID RECTANGULAR
SECTIONS
3/
30

4. 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:
Vz =
 That
dM y
dx
is, the larger is the rate of variation of
the BM along the beam, the larger is the SF
4/
30

5. Shear Force & Bending Moment
W W
W
−
My
Vz
5/
30
+
x
W
5
4
+
x
W
x
+
−
x
3
W
4

6. Shear Force & Bending Moment
W
My
Vz
2W L
W
+
W
W
+
x
−
W
6/
30
+
x
+
x
−
x
W

7. Normal stress (Sigma)
 The
bending moment M sets up
(longitudinal) normal stress σx, which varies
linearly with the distance from the neutral
axis:
σx =
where
My z
I yy
◦ Iyy is the second moment of the cross sectional
area
◦ z is the distance from the neutral axis (positive if
downward)
7/
30

8. Shear stress (Tau)

M
My
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
σ
σ
σ cx < 0 σ cx < 0
M
My
M
x
M
compression compression
y
tension
x
8/
30
Side View
Side View
tension
σ tx > 0 σ tx > 0
σ >0 σ >0
z
Cross Cross
Bending Bending
Section Section
Stress Stress
x

9. Jourawski’s formula


Since the elasticity theory will be not involved,
this is just an approximate solution

9/
30
An engineering solution for the shear stress
distribution over the beam’s cross section can be
obtained by using simple equilibrium
considerations
You must recognize the intrinsic limitations of this
formula:
τ xy ( P) =
Vz Qy ( P)
I yy b( P)

10. Jourawski’s formula

10/
30
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)
1 2
dx
My
Vz =
dM y
dx
M y + dM y

11. Jourawski’s formula
σ xc
σ
F + dFx
Fx + dF
σ dσ
σ cx + dσ xc
Fx
F
P
A
A
P
x
M
My
My + dMy
M + dM
My
M
11/
30
+ dM
M y + dMy
dx
bA
b(P)

12. 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
τ zx (P)
P
A
12/
30
2
1
dRx
A
P
Fx + dFx

13. 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
τ zx (P)
P
A
13/
30
2
1
dRx
A
P
Fx + dFx

14. Jourawski’s formula

It can be shown that the force acting on the plane P-P
is given by:
dFx =

dM y Q y (P)
I yy
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:
Q y (P) = AP × zP

14/
30
zP
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

15. Jourawski’s formula


15/
30
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:
dM y Qy (P)
dFx =
= τ zx (P)b(P)dx
I yy
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
dM y Qy ( P)
Qy ( P)
τ zx ( P) = τ xz ( P) =
= Vz
   


dx I yy b( P)
I yy b( P)
horizontal
vertical

16. Jourawski’s formula


τ zx ( P)
 

horizontal
y
P
x
τ xz ( P)
 

16/
30
vertical
z
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

17. Beam’s shear stress in a solid
rectangular section

17/
30
We need to apply the Jourawski’s formula to the
cross section shown below, which is subjected to a
vertical shear force Vz

18. Beam’s shear stress in a solid
rectangular section

The shear stress at a generic level P takes a parabolic
(i.e. quadratic) expression:
⎛d
⎞
A(P) = b ⎜ − zP ⎟
⎝2
⎠
⎞
1⎛ d
zP = ⎜ + zP ⎟
2⎝ 2
⎠
18/
30
⎞
1⎛ d2
2
Q y (P) = A(P) zP = ⎜ − zP ⎟
2⎝ 4
⎠
τ xz (P) =
Vz Q y (P)
I yy b
⎞
Vz ⎛ d 2
2
=
⎜ 4 − zP ⎟
2 I yy ⎝
⎠

19. Beam’s shear stress in a solid
rectangular section

At the bottom edge:
d
zP =
2
Vz
τ xz (P) =
×0 = 0
2 I yy

At the neutral axis:
z P = zG = 0
Vz
d2
τ xz (G) =
×
= max τ xz (G)
2 I yy 4
{
19/
30
}
Vz
bd3
12 Vz
I yy =
⇒ τ max =
= 1.5
12
8 bd
A
50% more than the
nominal average
shear stress V/A

20. Lecture #2
APPLICATION TO WEB
OF T AND I SECTIONS
20/
30

21. Wrong results for flanges of T and I
sections

The Jourawsky’s formula:
τ xz (P) =
Vz Qy (P)
I yy b(P)
cannot be applied to the flanges of T and I
sections, since the results so obtained are wrong
The parabolic distribution in
the web, although
approximate, is acceptable
from an engineering point of
view
21/
30
This is not correct, and
underestimates the actual
shear stress in the flange

22. 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
22/
30
τ web

23. Wrong results for flanges of T and I
sections

23/
30
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

24. Shear stress in the web of T
sections
1.0 m
20 kN
20 kN
+
Vz
V
20 kNm
24/
30
My
M
−

25. Shear stress in the web of T
sections
150
150
106.6
43.4
B
X neutral
B
X
axis
35.4
39.6
τ xz ( P) =
16
2
Vz Qy ( P)
I yy b( P)
⎛
150 × 16
16 ⎞
+ (150 × 16 ) × ⎜ 43.4 − ⎟
12
2⎠
⎝

35.4
134

 
2
⎛
16 × (150 − 16)3
134 ⎞
+
+ (134 × 16 ) × ⎜ 106.6 −
= 9.63× 106 mm 4
12
2 ⎟
⎝
⎠
 


39.6
I yy =
25/
30
16
3

26. xz
150
150
106.6
43.4
Shear stress in the web of T
sections
τ (P)
B
X neutral
16
B
X
axis
35.4
39.6
Shear stress at the
top of the web
We consider here the area above B-B
16
A(B) = 150 × 16 = 2,400 mm
2
⎛
16 ⎞
z B = − ⎜ 43.4 − ⎟ = −35.4 mm
2⎠
⎝
Q(B) = 2,400 × (−35.4) = −85,000 mm 3
26/
30
The negative sign
means that the flow of
Vz Qy (B) 20,000 × (−85,000)
shear stress exits the
τ xz (B) =
=
= −11.0 MPa considered area, i.e. is
I yy b(B)
9.63× 106 × 16
downward

27. Shear stress in the web of T
sections
150
150
106.6
43.4
τ XX
B
B
X neutral
G
16
axis
X
35.4
Shear stress at the
neutral axis
39.6
We consider here the area below G
16
A(G) = 106.6 ×16 = 1,706 mm 2
zG =
106.6
= 53.3 mm
2
Q y (G) = 1,706 × 53.3 = 90,900 mm 2
27/
30
τ xz (G) =
Vz Q y (G)
I yy b(G)
=
20,000 × 90,900
= 11.8 MPa
6
9.63 ×10 ×16
The positive sign means
that the flow of shear
stress enters the the
considered area, i.e. is
downward

28. Shear stress in the web of T
sections
150
150
106.6
43.4
B
G
16
B
X neutral
τ XX
axis
X
35.4
39.6
We obtain the same result
when considering in the
calculations the area
above the neutral axis
16
⎛ 43.4 − 16 ⎞
Qy (G) = (150 × 16 ) × (−35.4) + ⎡( 43.4 − 16 ) × 16 ⎤ × ⎜ −
= −91,000 mm 2
⎟
⎦ ⎝
  ⎣


2
⎠
Qy (B)
The only difference is the
20,000 × (−91,000)
τ xz (G) =
=
= −11.8 MPa negative sign, because this
6
I yy b(G)
9.63 ×10 ×16
time the flow of shear stress
exits area above the neutral
axis, i.e. is downward
Vz Q y (G)
28/
30

29. Shear stress in the web of T
sections
150
150
106.6
43.4
B
B
X neutral
G
16
axis
X
τ BB = 11.0 MPa
τ XX = 11.8 MPa
16

29/
30
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

30. Shear stress in the web of other
sections
30/
30