Shear stress due to Shear load (pure bending) multi-cell closed
cross-section
Q
1
i-1
2
i
n
i+1
W ith resulting distribution of shear flow q(x,y) or q(s)
=
open section
with shear flow
q*(s)
1
i-1
2
q1
→
q 1↑
q 2↑
q1
1
qi-1
→
q2
→
q1 ↓
q2 ↓
←
q2
i
↑ q3
qi
→
↑ qi
↑ qi-1
qn
qi+1
→
qi ↓
→
qn ↑
↑ qi+1
qn ↓
qi+1 ↓
qi-1 ↓
2
n
i+1
←
←
i-1 qi-1
i qi
←
←
n closed loop
(constant)
shear flows in
each closed
cell
n qn
i+1 qi+1
for open section portion:
s
⌠
m_star( s) =  y ( s)⋅t( s) ds
⌡
q_star( s) =
0
otherwise:
s
⌠
m_stary ( s) =  y ( s)⋅t( s) ds
⌡
0
1
Q
I
⋅m_star( s)
IF Iyz = 0
remember y(s) is distance in y direction
from centroid
s
⌠
m_starz( s) =  z( s)⋅t( s) ds
⌡
0
notes_18_shear multicell_clsd.mcd
q_star( s) =
Vy ( x)
 −I 2 + I ⋅I 
yy zz
 yz
∑
q_cell (s) = q_star_cell ( s) +
i
i
(
⋅ Iyz m_starz( s) − Iyy m_stary ( s)
qik = qi − qk
q
ik
(i , k)
⌠
 γ ds = 0
⌡
)
1 = 1, 2, 3, ...., n
k = adjacent cell to i (i+1 and i-1 in above figure)
= qi where no adjacent cell
1 = 1, 2, 3, ...., n
integral is circular
this is condition of no slip
⌠
⌠
1 ⌠
1
 γ ds = ⋅ τ ds = ⋅ 
G ⌡
G 
⌡
⌡
q
t
as τ =
ds = 0
q
t
G≠ 0
=> for each cell i
⌠
0=

⌡
⌠
 q 1 ds −
ds =
 i t
t
celli ⌡
common_side_i_k
q
∑
∑
⌠
 ⌠
 q_stari
1
  q ds + 
ds
t
 k t

⌡
 ⌡
where integration is summed over each wall element (circular integral in q_star case)
but since qi and qk are constant over all walls it can be extracted from the sum =>
⌠
qi⋅ 

⌡
⌠
qk⋅ 
ds −

t
⌡
k
1
∑
⌠

ds = −
t

⌡
1
q_star
t
i
first and rhs integrals are circular
whereas second is over wall common to
⌠
 q_stari
i and k and 
ds is the integral
t

⌡
ds
this is a system of n linear equations:
of the open shear flow around cell i
for example cell 1 lhs
cell 1
cell 2
⌠
q1⋅ 

⌡
⌠
ds − q2⋅ 

t
⌡
1
1
t
ds
1,2 => integral along common wall of 1 and 2
1
=> circular integral around cell 1
2
notes_18_shear multicell_clsd.mcd
and entire system of equations becomes:
1
⌠
q1⋅ 

⌡
1,2
⌠
ds − q2⋅ 

t
⌡
1
1
⌠
−q1⋅ 

⌡
1
⌠

= −

⌡
ds
t
q_star
1
ds
t
1,2
⌠
ds + q2⋅ 

t
⌡
⌠
ds − q3⋅ 

t
⌡
1
1,2
1
2
⌠
−q2⋅ 

⌡
1
t
⌠

= -

⌡
ds
⌠
ds + q3⋅ 

t
⌡
1
⌠
ds − q4⋅ 

t
⌡
1
3
1
t
⌠

= −

⌡
ds
...............
⌠
q ⋅
n− 1 
⌡
⌠
let each element of the matrix 

⌡
1
t
..............
⌠
ds − qn⋅ 

t
⌡
1
n-1,n
t
ds
t
q_star
t
3
ds
3,4
..............
1
2
2,3
2,3
⌠
where η ik = 

⌡
q_star
= ........
⌠

ds = −
t

⌡
1
q_star
t
n
ds
n
ds be expressed by η
ds integral along wall separating i and k
i,k
⌠
1
and η ii = 
ds integral around cell i
 t
⌡
if wall thickness is piecewise constant walls =>
sik
η ik =
tik
4
and η ii =
∑
j= 1
sij
tij
=
si1
ti1
+
si2
ti2
+
si3
ti3
+
si4
ti4
where sij , tij is the length and thickness of wall j of cell i
we observe that the matrix is symmetric i.e. η ik = η ki for the figure we are analyzing
3
notes_18_shear multicell_clsd.mcd
the adjacency can be general as shown:
qk
←
qi-1
↓qi-1
i-1
qk↑
q
kkk
i+1
← →
←
qi
iq ↑ qi+1
i
qi-1↑ ↓q i
q ↑
i
↓qi+1 i+1
q
i
qi-1
qi+1
→
→
→
⌠

qi⋅ η ii − q ⋅ η i−1 , i − q ⋅ η ki − q ⋅ η i , i+ 1 − q ⋅ η im = −
i− 1
k
i+ 1
m

⌡
with contribution =>
⌠
if t not constant η ik = 

⌡
1
t( s)
q_star
i
t
ds along wall between i and k
we now have n equations and n unknowns; qi
solution:
first calculate q_star from m_star (it's tedious) then
sik
calculate η ik = or
tik
4
calculate η ii =
∑
j= 1
sij
tij
⌠


⌡
=
1
t( s)
si1
ti1
+
ds along wall between i and k
si2
ti2
+
si3
ti3
+
si4
ti4
or
⌠


⌡
1
t( s)
ds around cell i
check problem on page 117 Hughes with one additional cell added: new sheet
4
notes_18_shear multicell_clsd.mcd
ds