Determination of the location of the shear center through use of the
sectorial area function
Vz
r⊥
ds
P
P
Vy
q
If the shear forces Vy and Vz act through the shear center, P, and hence
produce only bending of the beam, then the shear flow, q, that produces
those shear forces must not produce any moment about P, i.e.
L
∫ q r ds = 0
⊥
0
However, since q(0) =0, we have
q (s)
VI
(
=
z yz
− Vy I yy ) Qz + (Vy I yz − Vz I zz ) Qy
I yy I zz − I yz2
= Fz Qz + Fy Qy
A′ ( s )
s
where
Qz =
Qy =
∫
ydA = y ( s ) A′ ( s )
∫
zdA = z ( s ) A′ ( s )
A′( s )
A′( s )
q(s)
z (s)
C
so
L
L
0
0
Fz ∫ r⊥Qz ds + Fy ∫ r⊥Qy ds = 0
y (s)
C is the centroid
of the entire
cross-section
y and z are measured
from C to the centroid of A'
L
L
0
0
Fz ∫ r⊥Qz ds + Fy ∫ r⊥Qy ds = 0
Since this must be true for arbitrary Fz , Fy we must have
L
∫ r Q ds = 0
⊥
z
0
L
∫ r Q ds = 0
⊥
y
0
Consider
L
L
∫ r Q ds = ∫ r y ( s ) A′(s)ds = 0
⊥
0
Let
Then
⊥
z
0
d ω = r⊥ ds
r⊥ y ( s ) A′( s )ds = y ( s ) A′( s )dω = d ⎡⎣ y ( s ) A′( s )ω ⎤⎦ − ω d ⎡⎣ y ( s ) A′( s ) ⎤⎦
So we have
L
∫ r Q ds = yA′ω
⊥
z
0
y ( L ) = 0,
But
s=L
s =0
L
− ∫ ω d [ yA′] = 0
0
A′(0) = 0
so the first term on the right hand side of
the above equation vanishes and we find
L
∫ ω d [ y A′] = 0
0
Since y A′ =
∫ y dA
it follows that d [ y A′] = y dA and we obtain , finally
A′
L
∫ ω y dA = 0
0
L
In an entirely similar fashion from
L
∫ ω z dA = 0
0
∫ r Q ds = 0
⊥
0
y
we can show
Summary:
We have shown that in order for the shear forces to produce only bending,
we must have satisfied
L
∫ ω y dA = 0
0
L
∫ ω z dA = 0
0
ω is called the sectorial area function. We see it is given by
s
ω ( s ) = ∫ r⊥ ds + ω0
0
constant
To specify a value for the constant (which does not affect the above two
conditions), we can also require
L
∫ ω dA = 0
0
( this gives what is called the principal sectorial area function)
The principal sectorial area function satisfies
L
L
L
∫ ω dA = 0
∫ ω z dA = 0
∫ ω y dA = 0
0
0
0
Geometrical
interpretation of ω:
r⊥
ds
P
area d Ω =
1
r⊥ ds
2
so
ω = 2Ω + ω0
Thus, the sectorial area function is just twice the area swept out by the
radius from P to the centerline of the cross-section (to within a constant).
Sign convention:
ω = + 2 |Ω|
Ω
P
ω = - 2 |Ω|
Ω
P
sectorial area function coming from area swept out counterclockwise is
positive, coming from area swept out clockwise is negative