Shear Stress from Shear Load in closed non symmetric section
Q
t1
t2
Shear load is applied such that (pure) bending occurs
can't use symmetry to determine where to start s = 0 arc length parameter
approach: divide into two problems and superpose:
Q
t1
=
t2
q*
+
q* is the shear flow we have developed to date opening the section and
q1 is a constant shear flow in the closed section
superposition => we add the two flows for the actual shear flow
how do we calculate each. i.e q = q* + q1
We have one condition; the net has to match the applied load
the second comes from the physical situation; the slip at the cut must be 0
1/7
q1
⌠
slip =  γ ds
⌡
⌠
⌠
 γ ds = 

⌡
⌡
⌠


⌡
G
q_star
t
⌠
−

⌡
q1 :=
τ
b
where integral is circular and γ is the shear strain
ds =
⌠
1  q
⋅
ds = 0
G  t
⌡
⌠

ds + 

⌡
q_star( s)
t( s) q1
t
G the shear modulus = constant =>
ds = 0
ds
where the numerator is the integral around the cross
section and the denominator is as well (circular)
0
⌠


⌡
is constant =>
q1
1
t( s)
ds
⌠
q_star( s) := ⋅m_star(
m_star s) and 

I
⌡
Q
b
q_star( s)
t( s)
0


Q 
q( s) = ⋅  m_star( s) −
I




⌠


⌡
Q ⌠
ds = ⋅ 
I 
⌡
b
0
b
m_star( s)
t( s)
0
⌠


⌡
1
t( s)
ds
let's do an example
2/7

ds 





m_star( s)
t( s)
ds
in this case the example is
symmetric wrt z axis Iyz = 0 would
need to account for assymmetry
as above if necessary
Example of closed rectangular cross section non-symmetric
subject to shear force
b := 10
dd := 8
t1 = 1
t2 =
Q
Q := 100
3
2
dd
g
t2
t1
N
td := t1
ts := t1
tb := t1
tls := t2
h
n_elements := 4
A_total :=
index 0 mcd
∑ An
reference baseline, h is distance between NA and
lower segment; g = dd - h is distance to top
segment
 dd 
 dd
 
2 
d := 
 0 
 dd 

 2 
A_total = 40
n

1 ⋅t 3⋅b 
d
 12

 1 ⋅dd3⋅t 
s
 12
i0 := 

 1 ⋅tb3⋅b 
 12

 1 3 

12 ⋅dd ⋅tls


I :=
t1
b
n := 0 .. n_elements − 1
 td⋅b 

 ts⋅dd 
A := 

 tb⋅b 
 t ⋅dd
 ls 
A
∑ (dn⋅An)
h :=
n
A_total
h= 4
 10 
8
A=  
 10 
 12
 
 8

4
d =  
 0

 4
g := dd − h
g= 4
dg := h
  i + A ⋅ d 2 − A_total⋅d 2
g
n ( n) 
  0n
n

∑
I=
1285
3
removed 2 as this is not half section
deck (top)
my_top( s) := g⋅td⋅s⋅(0 < s ≤ b)
0<s≤b
3/7
right side
1 2
1 2
my_side( s) :=  g⋅t d⋅b + ts⋅  g⋅ s − g⋅ b − ⋅s + s⋅ b − ⋅b  ⋅[ [b < s ≤ (b + g + h)] ]
2
2 


[b < s ≤ (b + g + h)]
1
1 2
2
mybgh := g⋅ td⋅b + ts⋅ g⋅( b + g + h) − g⋅ b − ⋅(b + g + h) + (b + g + h)⋅b − ⋅b 
2
2 

bottom
my_bottom( s) :=  mybgh − h⋅ tb⋅(s − b − g − h)  ⋅[b + g + h < s ≤ (2⋅ b + g + h)]


[b + g + h < s ≤ (2⋅ b + g + h)]
left side
same for as right side with s starting at 2*b +dd, with initial value my(2*b + dd)
s
⌠
my ( s) =  y ⋅ t ds
⌡
0
⌠
my ( s) = 
⌡
2⋅ b+dd
0
⌠
y ⋅ t ds + 
⌡
s
2⋅ b+dd
 ⌠
y ⋅ t ds = my (2⋅ b + dd) + t s 
⌡

s



−h + [s − (2⋅b + dd)] ds
2⋅ b+dd
[2⋅ b + dd < s ≤ (2⋅ b + 2dd)]
using ll for 2*b + dd
s
s
⌠
 −
−h
h + ( σ − ll) ddσ
σ→
→
⌡
ll
ll
my (2⋅ b + dd) = my_bottom(2⋅ b + dd)
my_bottom(2⋅ b + dd) = 0
ll := 2⋅b + dd
1 2
1 2
my_left_side( s) :=  my_bottom(2⋅ b + dd) + t s⋅  −4⋅s + ⋅s − ll⋅s + 4⋅ ll + ⋅ll   ⋅[ [2⋅ b + dd < s ≤ (2⋅ b + 2dd)] ]
2
2 


4/7
s := 0, 0.1 .. b + dd + b + dd
my ( s) := my_top( s) + my_side( s) + my_bottom( s) + my_left_side( s)
need also t(s)
t_top( s) := td⋅(0 ≤ s ≤ b)
t_side( s) := ts⋅[b < s ≤ (b + g + h)]
t_bottom( s) := tb⋅(b + g + h < s < 2⋅ b + g + h)
t_left_side( s) := tls⋅[2⋅ b + dd ≤ s ≤ (2⋅ b + 2dd)]
t( s) := t_top(s) + t_side( s) + t_bottom( s) + t_left_side( s)
Q = 100
q( s) :=
Q⋅ my ( s)
Q⋅ my ( s)
τ ( s) :=
I
I⋅ t( s)
60
15
40
10
my( s )
q( s )
20
τ ( s)
0
0
20
5
5
0
10
20
30
40
0
10
20
30
40
s
s
1.6
1.4
t( s )
1.2
1
0
10
20
s
5/7
30
40
plot of stress distribution around cross section for q_star
magnitudes shown positive is out of section
this is shear stress
15
stress superposed on geometry
10
5
0
5
10
15
0
5
10
15
stress superposed on geometry
⌠


⌡
m_star( s) := my ( s)
q1 :=
2⋅ b+2⋅ dd
m_star( s)
t( s)
⌠


⌡
2⋅ b+2⋅ dd
1
t( s)
25
ds
55613827
q1 =
2524920
0
0
6/7
20
ds
q( s) :=
Q
(
)
⋅ m_star( s) − q1
I
τ ( s) :=
q( s)
t( s)
10
10
5
q( s )
0
τ ( s)
0
5
10
0
20
40
s
10
0
20
40
s
t1 ≡ 1.0
plot of stress distribution around cross section
magnitudes shown positive outward
this is shear stress
10
8
6
stress superposed on geometry
4
2
0
2
4
6
8
10
0
2
4
6
8
10
12
stress superposed on geometry
14
16
18
now that we have considered on such cross section, what if there are one or more adjacent?
7/7
t2 ≡ 1.5