# Shear of Open Cross-Section Beams - Comcast.net

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15 Νοε 2013 (πριν από 4 χρόνια και 5 μήνες)

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AERSP 301

Shear of beams

(Open Cross
-
section)

Jose Palacios

Shear of Open and Closed Section Beams

Megson

Ch. 17

Open Section Beams

Consider only shear loads applied through shear center (no twisting)

Torsion loads must be considered separately

Assumptions

Axial constraints are negligible

Shear stresses normal to beam surface are negligible

Near surface shear stress = 0

Walls are thin

Direct and shear stresses on planes normal to the beam surface are const through
the thickness

Beam is of uniform section

Thickness may vary around c/s but not along the beam

Thin
-
Walled

Neglect higher order terms of t (t
2
, t
3
, …)

Closed Section Beams

Force equilibrium: General stress, Strain, and Displacement
Relationships

S

the distance measured
around the c/s from some
convenient origin

σ
z

Direct stress
(due to
bending moments or bending

Shear stresses due to
(for closed section)

σ
s

Hoop stress, usually zero
(non
-
zero due to internal
pressure in closed section
beams)

zs

=

sz

=

shear flow; shear

force per unit length

q =

t

(positive in the direction of s)

Force equilibrium (cont’d)

From force equilibrium considerations in z
-
direction:

Force equilibrium in s
-
direction gives

Stress Strain Relationships

Direct stress:

z

and

s

strains

z

and

s

Shear stress:

strains

(=

zs

=

zs
)

Express strains in terms of displacements of a
point on the c/s wall

v
t

and v
n
: tangential and normal
displacements in xy plane

Not used

(1/r: curvature of wall in x
-
y plane)

Stress Strain Relationships

To obtain the shear strain, consider the element below:

Shear strain:

Center of Twist

Equivalent to pure rotation about some pt. R

For the point N

Origin O of axes
chosen arbitrarily,
and axes undergo
disp.
u, v,

Center of Twist (cont’d)

But

Equivalent to pure rotation about some pt. R

Center of twist cont…

Also from

Comparing Coefficients with:

Position of Center of Twist

Shear of Open Section Beams

The open section beam supports
x

and S
y

such that there
is no twisting of the c/s (i.e. no torsion

For this, shear loads must pass
through a point in the c/s called the
SHEAR CENTER

Not necessarily on a c/s member

Use the equilibrium eqn.

And obtaining

z

from basic bending theory

(no hoop stresses,

s

= 0)

Shear of Open Section Beams cont…

From:

Shear of Open Section Beams cont…

Integrating with respect to s starting from an origin at an
open edge (q = 0 at s = 0) gives:

For a c/s having an axis of symmetry, I
xy

= 0. Then eq.
for q
s

simplifies to:

Shear sample problem

Determine the shear flow distribution
in the thin
-
walled z
-
section shown
S
y

applied through
its shear center (no torsion).

Where is the shear center?

And the centroid?

t

y

x

h

h/2

1

2

4

3

Shear Flow Distribution (S
x

= 0):

s
xy
yy
xx
yy
y
s
xy
yy
xx
xy
y
s
yds
t
I
I
I
I
S
xds
t
I
I
I
I
S
q
0
2
0
2
Shear sample problem

8

12

3
3
3
3
th
I
th
I
th
I
xy
yy
xx

ds
y
x
h
S
yds
t
I
xds
t
I
I
I
I
S
q
s
y
s
yy
s
xy
xy
yy
xx
y
s
)
48
.
6
32
.
10
(
0
3
0
0
2
Bottom Flange: 1
-
2,
y =
-
h/2, x =
-
h/2 + S
1

0
≤ S
1
≤ h/2

ds
h
S
h
h
S
q
ds
y
x
h
S
q
s
y
s
y
s
))
2
/
(
48
.
6
)
2
/
(
32
.
10
(
)
48
.
6
32
.
10
(
1
0
1
3
12
0
3
Show this: EXAM TYPE PROBLEM

Shear sample problem

h
S
q
)
h
(S
q
S
hS
S
h
S
ds
h
S
h
h
S
q
y
y
s
y
42
.
0
2
/

2
@
0
)
0
(

1
@
)
74
.
1
16
.
5
(
))
2
/
(
48
.
6
)
2
/
(
32
.
10
(
2
1
1
1
1
2
1
3
0
1
3
12
1

t

y

x

h

h/2

1

2

4

3

Shear sample problem

2
2
2
2
3
2
2
2
0
3
23
42
.
3
42
.
3
42
.
0
)
(
48
.
6
)
(
42
.
3
(
2
s
hs
h
h
S
q
ds
s
h
h
S
q
y
s
y

In web 2
-
3:

y =
-
h/2 + S
2

x = 0 for 0

S
2

h

C
x

with max value at
S
2

= h/2 (y = 0)

and positive shear flow along the web

Shear Flow
S
2

= 0

h
S
q
y
42
.
0
2

Shear sample problem

In web 3
-
4:

y =h/2

x = S
3

for 0

S
3

h

3
3
0
3
3
34
)
(
24
.
3
)
(
32
.
10
(
3
q
ds
h
s
h
S
q
s
y

Shear Flow Distribution in z
-
section

Calculation of Shear Center

through the shear center,
it will produce NO TWIST

M = 0

If c/s has an axis of
symmetry, the shear
center lies on this axis

For cruciform or angle
sections, the shear center
is located at the
intersection of the sides

Sample Problem

Calculate the shear
center of the thin
-
walled
channel shown here:

Sample problem shear center

The shear center (point S) lies on the horizontal
(C
x
)

axis

at some distance
ξ
s

from the web
S
y

passes through

the shear center it will produce
no twist.

Let’s look at the shear flow distribution due to S
y
:

Since
I
xy

= 0
and
S
x

= 0

S
xx
y
s
ds
ty
I
S
q
0
)
6
1
(
12
3
h
b
th
I
xx

Further:

S
y
s
ds
y
h
b
h
S
q
0
3
6
1
12
Then:

Sample problem shear center

Along the bottom flange 1
-
2,
y =
-
h/2

1
2
0
1
3
12
6
1
6
2
/
6
1
12
1
s
h
b
h
S
ds
h
h
b
h
S
q
y
S
y

At point 2:
S
1

= b

b
h
b
h
S
q
y

6
1
6
2
2
Along the web 2
-
3 ,
y =
-
h/2 + S
2

Sample problem shear center

b
h
b
h
S
s
s
h
h
b
h
S
q
ds
s
h
h
b
h
S
q
y
y
S
y

6
1
6
2
2
6
1
12
2
/
6
1
12
2
2
2
2
3
2
0
2
2
3
23
2
At point 3:
S
1

= h

2
2
3
6
1
6
q
b
h
b
h
S
q
y

Along the top flange 3
-

4 ,
y = h/2

Sample problem shear center

b
h
b
h
S
s
h
h
b
h
S
q
ds
h
h
b
h
S
q
y
y
S
y

6
1
6
2
6
1
12
2
/
6
1
12
2
3
3
3
0
3
3
34
3
At point 4:
S
3

= b

0
4

q
Good Check!

Sample problem shear center

Shear Flow Distribution due to
S
y

The moments due to this shear flow

distribution should be equal to zero about the shear center

2
0
0
23
1
12
2
2
ds
q
ds
h
q
s
b
h

S

ξ
s

S
y

Solve for
ξ
s

to find the shear center location:

s
h
y
b
h
s
y
y
ds
b
h
b
h
S
ds
s
s
h
h
b
h
S
ds
s
h
h
b
h
S

0
2
2
0
0
2
2
2
2
3
1
1
2
6
1
6
2
2
6
1
12
2
6
1
6
2

h
b
h
b
s
1
3
2