Civil Engineering Reference
In-Depth Information
The product of horizontal flange stress and flange thickness at the extremities of the
web is, from Eq. (10.14)
S
y
I
z
B
8
(
D
τ
f(h)
t
f
=
+
d
)
t
f
(10.16)
Comparing Eqs (10.15) and (10.16) we see that
τ
w
t
w
=
2
τ
f(h)
t
f
(10.17)
The product
stress
thickness
gives the
shear force per unit length
in the walls of the
section and is known as the
shear flow
, a particularly useful parameter when considering
thin-walled sections. In the above example we note that
τ
f(h)
t
f
is the shear flow at the
extremities of the web produced by considering one half of the complete flange. From
symmetry there is an equal shear flow at the extremities of the web from the other half
of the flange. Equation (10.17) therefore expresses the equilibrium of the shear flows
at the web/flange junctions. We shall return to a more detailed consideration of shear
flow when investigating the shear of thin-walled sections.
×
In 'thick' I-section beams the horizontal flange shear stress is not of great importance
since, as can be seen from Eq. (10.17), it is of the order of half the magnitude of the
vertical shear stress at the extremities of the web if
t
w
t
f
. In thin-walled I-sections
(and other sections too) this horizontal shear stress can produce shear distortions
of sufficient magnitude to redistribute the direct stresses due to bending, thereby
seriously affecting the accuracy of the basic bending theory described in Chapter 9.
This phenomenon is known as
shear lag.
E
XAMPLE
10.3
Determine the distribution of vertical shear stress in a beam of
circular cross section when it is subjected to a shear force
S
y
(Fig. 10.8).
The area
A
of the slice in this problem is a segment of a circle and therefore does not
lend itself to the simple treatment of the previous two examples. We shall therefore
use Eq. (10.5) to determine the distribution of vertical shear stress. Thus
D
/
2
S
y
b
0
I
z
τ
=−
by
1
d
y
1
(10.18)
y
where
π
D
4
64
I
z
=
(Eq. (9.40))
Integration of Eq. (10.18) is simplified if angular variables are used; thus, fromFig. 10.8
D
2
cos
θ
D
2
cos
φ
D
2
sin
φ
D
2
cos
φ
d
φ
b
0
=
2
×
b
=
2
×
y
1
=
d
y
1
=
Equation (10.18) then becomes
π/
2
16
S
y
π
D
2
cos
θ
cos
2
φ
sin
φ
d
φ
=−
τ
θ
