Strength of Materials - Lecture Notes / Mehmet Zor
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e
S
G
P
5.5.
in Simple Shear Bending of
Thin-Walled, Symmetrical Section Beams
Shear Center
tvid- 5.5
Strength of Materials - Lecture Notes / Mehmet Zor
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Shear Center
5.5.1 Reminder and warning: In order to understand this subject, "shear stress (τ) and shear flow (q) distributions in shear bending" explained in chapter 5.3 must be well understood. Our aim in this section is to determine the location of the shear center in thin-walled, symmetrical cross-section beams subjected to shear simple bending.
5.5.2 Concept of Shear Center: Let's apply a singular force P to a beam concreted against a wall from end A, passing through the center of gravity of free section B. We learned in chapter 5.3 that the shear internal force and shear flow distribution (q) in any cross-section C will be as below.
P
S
C
A
B
D
G
q: shear flow. (Internal shear force per unit length N/mm)
P
S
q
q
qmax
G
B
C
D
more…>>
(5.3.3.b)
Figure 5.5.1
Figure 5.5.2
Strength of Materials - Lecture Notes / Mehmet Zor
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Shear Center
P
S
q
b
d
f
k
q
qmax
G
B
C
D
i
n
m
q: shear flow:
V
F
F
P
G
S
z
y
C
D
e
b
h
h
V
F
F
P
G
S
z
y
C
D
b
h
h
Q: Static moment of the relevant part.
(The Q value was explained in section 5.3. It will be understood more clearly in future numerical examples.)
qa=qk=0 ; qd > qi
qd =qf , qi =qm (from symmetry)
qn =qmax (at the center of gravity.)
Shear flow direction: It is like the water filling from b flowing towards k.
Figure 5.5.3
Figure 5.5.4
Figure 5.5.5
Strength of Materials - Lecture Notes / Mehmet Zor
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Since the load P is applied from G, both bending and torsion occur.
Since the load P was applied from the shear center S, only bending occurred; there is no torsion.
5.5.3 An example of the importance of the Shear Center:
Shear Center
Figure 5.5.6
Figure 5.5.7
Strength of Materials - Lecture Notes / Mehmet Zor
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P
z
x
y
G
5.5.4 Shear center in symmetrical sections according to the V direction: In this type of sections, both the shear center (S) and the center of gravity (G) points are on the symmetry axis (y or z axis) in the V direction, but they can be in the same or different positions. The V direction is coincident with either the y or z axes. As can be seen in the examples below, since the direction of the external force P passes through S, the static equilbrium equations are satisfied, and since the moment (torsional moment) with respect to G will be zero, there will be no rotational deformation (distortion, twisting) in the beam around x.
y
z
y
z
P
V
V/2
y
z
G
G,S
G,
V
F
F
F
F
P
P
V/2
S
S
V (resultant)
Shear Center
Note that in this example, P and V are horizontal and the section is symmetrical with respect to the V direction.
F
F
In order to understand why the flow directions of shear flows in the sections are in these directions, topic 5.3 needs to be examined.
F
F
S
Figure 5.5.9
Figure 5.5.10
Figure 5.5.11
Figure 5.5.8
Strength of Materials - Lecture Notes / Mehmet Zor
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The cross-sectional dimensions of the thin-walled cantilever beam in the figure are b = 100 mm, h = 150 mm and t = 3 mm.
A vertical force of P = 20 kN will be applied to the plate attached to the free end of the beam.
According to this;
a-) Determine the shear flow distribution in any section.
b-) At what distance e should force P be applied so that there is no rotation (twisting) on the beam around its own axis (x)?
c-) Find the location and intensity of the maximum shear stress by drawing the distribution of shear stresses.
b
h
t
G
A
B
D
E
O
t
Example 5.5.1
y
z
e
P
Shear Center
Figure 5.5.12
Figure 5.5.13
Strength of Materials - Lecture Notes / Mehmet Zor
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Solution:
≈
b=100mm
h=150mm
t
A
B
D
E
z
y
G
t=3mm
To use it in our calculations, we must first calculate the moment of inertia with respect to the horizontal axis z passing through the center of gravity:
(Small differences due to t thickness are neglected.)
Shear Center
A
z
web
flange
flange
b
G
y
t
t
Figure 5.5.14
Figure 5.5.15
Strength of Materials - Lecture Notes / Mehmet Zor
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For AB and DE flanges:
Total internal shear forces in AB and DE flanges:
a-) In any cross section:
Shear force: V = P = 20kN (From static balance)
A
B
D
E
Ff
Ff
V
Shear Center
(Shear flow for a point at a horizontal distance 𝑧′ from A)
Static moment of shaded area A1 :
(N/mm)
qf
qf
qf
qw
A
B
D
E
b=100mm
t=3mm
h=150mm
G
y
z
A1=tz
Figure 5.5.16
Figure 5.5.17
Strength of Materials - Lecture Notes / Mehmet Zor
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Shear flow at distance y:
,
,
≈
Shear Center
(static moment of total shaded area):
A’
y
z
G
t
t
b
qw
Ff
Ff
For Web BD
y
A
B
D
E
qf
qw
h=150mm
b=100mm
qf
t=3mm
G
z
B
D
E
A
Fw=V=P
Figure 5.5.18
Figure 5.5.20
Figure 5.5.19
Strength of Materials - Lecture Notes / Mehmet Zor
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b-)
e
t
s
G
A
B
D
E
z
y
P=2kN
V
Ff
Ff
The absence of any rotation in the beam around its own axis (x) is possible because the net torsion moment is zero, and this is possible by applying the external force P from the shear center. Then, in this case, we are asked about the distance (e) of the shear center S.
Shear Center
h=150mm
b=100mm
Figure 5.5.21
Strength of Materials - Lecture Notes / Mehmet Zor
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Shear stress distribution in top and bottom flanges (AB, DE) :
Shear stress distribution in the Web BD
c-)
t=3mm
A
B
D
E
z
y
G
t
Shear Center
h=150mm
b=100mm
Since shear stress 𝝉 (N/mm2) is the internal force per unit area, 𝝉 stress is equal to the internal force per unit length, that is, 𝑞 (N/mm), shear flow divided by the thickness. Horizontal shear stresses in the flanges (AB, DE) are 𝜏𝑧𝑥,
and shear stresses in the web BD are 𝜏𝑦𝑥.
Stress direction
Section normal
Stress direction
Stress intensity
Stress intensity
The "(red) arrows" on the outside are to show the intensity and distribution of stresses. The actual direction of the stresses is the direction of the (blue) arrows on the cross section.
(It was calculated in answer a.)
Figure 5.5.22