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AERSP 301Shear 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 (t2, t3, …)
• Closed Section Beams– Consider both shear and torsion loading
Force equilibrium: General stress, Strain, and Displacement Relationships
• S – the distance measured around the c/s from some convenient origin
• σz – Direct stress (due to bending moments or bending action of shear loads)
– Shear stresses due to shear loads or torsion loads (for closed section)
• σs – 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
• vt and vn: 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
(center of twist [for loading such as pure torsion])
• 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 [for loading such as pure torsion])
Center of twist cont…
• Also from
• Comparing Coefficients with:
Position of Center of Twist
Shear of Open Section Beams
• The open section beam supports shear loads Sx and Sy such that there is no twisting of the c/s (i.e. no torsion loads)
• 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, Ixy = 0. Then eq. for qs simplifies to:
Shear sample problem
• Determine the shear flow distribution in the thin-walled z-section shown due to shear load Sy applied through its shear center (no torsion).
• Where is the shear center?
• And the centroid?
t
y
xh
h/21 2
43
Shear Flow Distribution (Sx = 0):
s
xyyyxx
yyys
xyyyxx
xyys ydst
III
ISxdst
III
ISq
0202
Shear sample problem
8
12
3
333 thI
thI
thI xyyyxx
dsyx
h
SydstIxdstI
III
Sq
sys
yy
s
xyxyyyxx
ys )48.632.10(
03002
Bottom Flange: 1-2, y = -h/2, x =-h/2 + S1
0 ≤ S1 ≤ h/2
dshShh
Sq
dsyxh
Sq
sy
sys
))2/(48.6)2/(32.10(
)48.632.10(
1
0 1312
03
Show this: EXAM TYPE PROBLEM
Shear sample problem
h
Sq)h(S
qS
hSSh
S
dshShh
Sq
y
y
sy
42.02/ 2@
0)0( 1@
)74.116.5(
))2/(48.6)2/(32.10(
21
11
12
13
0 1312
1
t
y
xh
h/21 2
43
Shear sample problem
222
23
2220323
42.342.342.0
)(48.6)(42.3(2
shshh
S
qdsshh
Sq
y
sy
In web 2-3:
y =-h/2 + S2
x = 0 for 0 ≤ S2 ≤ h
Symmetric distribution about Cx with max value at S2 = h/2 (y = 0) and positive shear flow along the web
Shear Flow S2 = 0h
Sq y42.0
2
Shear sample problemIn web 3-4:
y =h/2 x = S3 for 0 ≤ S3 ≤ h
330 3334 )(24.3)(32.10(3
qdshsh
Sq
sy
Shear Flow Distribution in z-section
Calculation of Shear Center
• If a shear load passes 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 (Cx) axisat some distance ξs from the web. If a shear load Sy passes through the shear center it will produce no twist.
Let’s look at the shear flow distribution due to Sy:Since Ixy = 0 and Sx = 0
S
xx
ys dsty
I
Sq
0
)6
1(12
3
h
bthI xx Further:
Sy
s dsy
hb
h
Sq
03 61
12Then:
Sample problem shear center
Along the bottom flange 1-2, y = -h/2
12
0 13
12
61
62/
61
12 1
s
hb
h
Sdsh
hb
h
Sq ySy
At point 2: S1 = b
b
hb
h
Sq y
61
6
22
Along the web 2-3 , y = -h/2 + S2
Sample problem shear center
b
hb
h
Sssh
hb
h
S
qdssh
hb
h
Sq
yy
Sy
61
6
2261
12
2/61
12
2
22
23
20 223
23
2
At point 3: S1 = h
22
3
61
6qb
hb
h
Sq y
Along the top flange 3 - 4 , y = h/2
Sample problem shear center
b
hb
h
Ssh
hb
h
S
qdsh
hb
h
Sq
yy
Sy
61
6
261
12
2/61
12
23
3
30 33
34
3
At point 4: S3 = b
04 q Good Check!
Sample problem shear center
Shear Flow Distribution due to Sy
The moments due to this shear flow distribution should be equal to zero about the shear center
20 0 23112 22 dsqds
hq s
b h
S ξs
Sy
Solve for ξs to find the shear center location:
s
hyb h
syy dsb
hb
h
Sds
ssh
hb
h
Sdss
h
hb
h
S
0 22
0 0 2
22
23
112 61
6
2261
12
261
62
hb
h
bs
1
3 2