Torsion Properties for Line Segments and Computational Scheme
for Piecewise Straight Section Calculations
Closed Thin walled Sections
the new material consists of the "corrections for ω and Qω
A = enclosed area
definition of
s

dΩc = hc −

J
1
⋅ ⋅ds = dωc
2⋅ A t 
b
( ∆Xi) 2 + ( ∆Yi) 2
⌠ 1
circ_integral = 
ds =
 t
⌡0
i
∑
s
J=
⌠
⌠ 1
J ⌠
1
2⋅ A
ω c =  hc ds −
⋅
ds =
⋅
ds

b
 t
2⋅ A  t
⌡
⌠
⌡0
⌡
 1 ds 0
 t
⌡0
ti
=
∑
i
∆li
ti
2
b
⌠ 1

ds
 t
⌡0
as
if we define
segment
length:
4⋅ A
( ∆Xi) 2 + ( ∆Yi) 2
∆li =
we now need calculation of the enclosed area A in this expression
area of triangle determined by two points and the origin:
area = area1 + area2 − area3
y
y
x1,y1
area
x0,y0
0,0
area1 :=
x
1
2
⋅y1⋅ x1
area := area1 + area2 − area3
y
x1,y1
area2 :=
1
2
x
area1
0,0
⋅(y0 + y1) ⋅(x0 − x1)
x1
area3 :=
area simplify →
x1,y1
x0,y0
area2 x0,y0
x0,y0
0,0
y
x1,y1
−1
2
⋅y0⋅ x1 +
1
2
⋅y1⋅x0
area3
0,0
x
1
2
x
⋅y0⋅ ( x0)
area_2_pts_origin :=
1
2
⋅(y1⋅x0 − y0⋅ x1
x1)
area between three points:
x1,y1
area
x2,y2
area :=
x0,y0
1
2
area :=
etc.........
1
⋅(y1⋅x0 − y0⋅x1) −
1
2
1
2
⋅(y1⋅x0 − y0⋅x1) +
area_enclosed :=
1
2
⋅
⋅(y1⋅x2 − y2⋅x1)
y2
1
2
⋅(y2⋅x1 − y1⋅x2)
x2
∑ ( X ⋅YY
i
i+ 1 −
X i+ 1⋅Yi)
i
notes_15a_tor_prop_clsd_calc.mcd
for a straight line segment ρ c = constant
∆ω c = ρ c⋅ L −
2area_enclosed ∆li
⋅
ti
circ_integral
and is linear along line
see torsion properties (open) for derivation of ρc part:
∆ω c =
x 1 + x0
2
⋅ ( y1 − y0) −
∆ω c = xm⋅( ∆y) − ym⋅∆x −
y1 + y0
2
⋅ ( x1 − x0) −
2area_enclosed ∆li
⋅
ti
circ_integral
2area_enclosed ∆li
⋅
ti
circ_integral
xm = mid-point
∆x = x1 - x0
∆y = y1 - y0
s
dΩD = dωD = hD ⋅ ds
s
⌠
⌠
Ω D( s) =  hD ds =  hC − xD⋅sin( α ) + yD ⋅cos( α ) ds
⌡
⌡0
0
=>
(
)
(
)
∆Ω D( s) = ∆ωc − xD⋅ y1 − y0 + yD ⋅ x1 − x0
calculation of ΩD and ωD aka ω identical to open
if we set Ω D0 = 0 at the start of a line segment, then Ω D1 = Ω D0 + ∆Ω D
calculate "centroid" of warping wrt shear center:
∆Q ΩD =
i
Iω =
(
)
t⋅ s1 − s0
3
ai
2
⋅ Ω D + Ω D

i

i+ 1
Ω Dcg =
⋅  ( ωD1) + ωD 0⋅ωD1 + ( ωD 0)

2
∑ ∆QΩD
A
IyωD =
2

IxωD =
(
)
t⋅ s1 − s0
6
(
)
t⋅ s1 − s0
6
ω D = Ω D − Ω Dcg
j
j
(
)
(
)
⋅ 2⋅ x1⋅ωD1 + x0⋅ωD0 + x0⋅ωD 1 + x1⋅ωD 0


⋅ 2⋅ y1⋅ω D1 + y0⋅ωD 0 + y0⋅ωD1 + y1⋅ωD0


first moment of ω needs "correction" also
q( s , x)




 Tω 
=−
 I ⋅ Qω −
 ωω 






⌠
 Q ds
ω

⌡
⌠



⌡
1
t
ds








thus a "correction"
⌠
 Q ds
ω

⌡
⌠



⌡
1
t
is applied to Qω for
ds
the closed section. the ω is for the closed section
(with it's correction applied)
⌠
first calculate ω as open which we did above, now calculate  Qω ds

⌡
we know ∆ω is piecewise linear over s ( ∆ω = hD*∆s as hD constant over segment) thus over segment:
2
notes_15a_tor_prop_clsd_calc.mcd
 ω1 − ω0 ⋅s + ω0 − s0⋅  ω1 − ω0 
ω ( s) := 


 s1 − s0
 s1 − s0 
y1 − y0
y=
x1 − x0
⋅x + y0 − x0
( y1 − y0)
( x1 − x0)
s
⌠
Qω ( s ) :=  ω ( σ )⋅t dσ + Q ω0
⌡
s0
s
⌠
1 ω1 − ω0 2 1 2⋅ω0⋅s1 − 2⋅ s0⋅ω1
1
−s0⋅ω1 − ω0⋅ s0 + 2⋅ω0⋅s1
⋅t⋅s + ⋅
⋅t⋅ s − ⋅s0⋅
⋅t
 ω ( σ ) ⋅t dσ collect, s → ⋅
⌡
2 s1 − s0
2
s1 − s0
2
s1 − s0
s0
s1
⌠
s
 ⌠
ω ( σ )⋅t dσ + Qω0
 
⌡s0

∆Q ω ( s) :=
ds

t
⌡s0
s1
⌠




⌡
1
(s0⋅ω1 + ω0⋅ s0 − 2⋅ω0⋅s1)
−1 (ω1 − ω0)
2 1 (2⋅ω0⋅s1 − 2⋅ s0⋅ω1)
⋅
⋅t⋅s − ⋅
⋅t⋅s − ⋅s0⋅
⋅t + Q ω0
2 ( −s1 + s0)
2
(−s1 + s0)
2
(−s1 + s0)
t
simplify
ds
s0
 1 ⋅ω
→
collect, t, Q ω0 , s1 , s0 , t  6
copied from off the page:
=>
=
 1 ⋅ω1 +

6
(s1 − s0)⋅
1
3


 −1 ⋅ω1 −
3
2
⋅ω0 ⋅s1 + 
Q ω0
t
Qω0
1
1

 2
⋅ω0 ⋅s0⋅ s1 +  ⋅ω1 + ⋅ω0 ⋅s0 + (s1 − s0)⋅
3
3
t

6

2
1
1
2

+  ⋅ω1 + ⋅ω0 (s1 − s0)
3
6

a reference Heins calculates this increment as
∆Q ω ( s) :=
( Q + Qω0) ⋅ 
2 ω1
1
s1 − s0 
t

+
1
12
⋅ (ω0 − ω1)⋅(s1
ω1
− s0)
2
it can be shown that these are equivalent, but it is not obvious what motivated the second form
from open development:
1
Qω1 := Qω0 + ⋅(ω1 + ω0) ⋅ t⋅ (s1 − s0
s0)
2
 1 Q + Q ⋅ (s1 − s0)
 ( ω1 ω0)
t
2
+
1
12
linear => area is half end-point *t* distance, t constant
⋅ (ω0 − ω1)⋅(s1 − s0)
2


simplify
 1 ⋅ω1 + 1 ⋅ω0 ⋅s12 +  − 2 ⋅ω0 − 1 ⋅ω1 ⋅s0⋅ s1 +  1
→


collect, t, Q ω0 , s1 , s0  6
3

3
3

 6
copied from off the page: (it is the same as above)
 1 ⋅ω1 +

6
1
3


2
 −1 ⋅ω1 −
3
⋅ω0 ⋅s1 + 
Qω0
1
1

 2
⋅ω0 ⋅s0⋅ s1 +  ⋅ω1 + ⋅ω0 ⋅s0 + (s1 − s0)⋅
3
3
t

6

2
3
notes_15a_tor_prop_clsd_calc.mcd
⌠
 Q ds
ω

⌡
⌠



⌡
1
t
Qω_corr =
∑
i
  si+ 1 − si 
1
⋅Qω +  ⋅ω i+ 1 +

i 6
 ti 
correction is then:
∑
or ....
i
1+ 1
+ Qω  ⋅ 
i
∆li = si+ 1 − si
∑
Qω := Qω − Qω_corr
i
i
i
Qω_corr =
∆li

 Qω ⋅
 i ti
ti
∑
∑
i
∑
si+ 1 − si
ti
∆li
ti
∆li
1
 Qω + Q ω  ⋅
i ti
 2  i+ 1
∑
i
4

1 
1
2
+  ⋅ω i+ 1 + ⋅ω i ( ∆li) 
3 
6

i
or ....


+ ( ω i − ω i+ 1)( si+ 1 − s1)

i
and .
2
ti
 si+ 1 − s1 
1
∑  2 Qω
3


⋅ω i ( s i+ 1 − s 1)
s i+ 1 − s i
i
ds
1
+
1
12

( ω i − ω i+1) ( ∆li) 2

∆li
ti
notes_15a_tor_prop_clsd_calc.mcd
2

