Axial, Shear and Moment Interaction of Single Plate “Shear Tab” Connections
Influence of Axial & Shear
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Transcript of Influence of Axial & Shear
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INFLUENCE OF AXIAL FORCE ON THE VALUE
OF THE PLASTIC MOMENT
If an axial load is applied to the CrossSection of a short
column, then the load will give rise to a uniform compressive
stress over the section. Addition of small bending moment will
then produce a linear variation of elastic stress across the
section. Further increase of B.M. with axial load remainingconstant will cause yield on one face followed by yield on the
other face and eventually yield on the whole Cross-Section.
During this process, the zero-stress axis shifts progressively
towards the final position in the fully plastic state.
To explain this, take a rectangular section on which an
axial load P is acting at the centre line of cross section, in
addition to a bending moment Mp sufficient to cause the whole
section to be fully plastic. Due to force P across the Cross-
Section, the zero stress axes must shift from the centre line by
plastic
Fully
tension
Yield in
in CompYield
B.M.Medium
B.M.smallNo B.M.
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an amount d. It will be seen that the fully plastic stress
distribution may be regarded as being composed of two parts,
i.e. on the original fully plastic distribution in the absence of
axial load may be superimposed a fictitious distribution
involving stress of magnitude 2 fL. This distribution must be
equivalent to a total force P and it will reduce the value of the
full plastic moment by an amount P
d2
1
P = 2bdfL = o
Mp = Mpo P
d2
1
= Mpo 2
12d. Po
Mpofbdbdfd2
1dPo2
1L
2
L
2 ===
Mp = Mpo (1 2)
2
C
2 Bddtc
Bd
2t1
tt
+=
T
CC
T
Bd
b
2d
c c
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o is the squash load of the Cross-Section in the absence
of bending moment and Mpo is the full plastic moment (bd2fL)
in the absence of axial load.
So2
2
Po
P11
Mpo
M
==
=> 1Po
P
Mpo
M2
=
+
This gives the doubly symmetric curve. This is true for
tensile load also. In a system of Cartesian coordinates with
Po
Pas abscissa and (M/Mpo) as ordinate the complete plot is
as shown.
For the
general case of a cross section having at least one axis of
symmetry, it may be supposed that the axial load P causes the
zero-stress axis to shift so that an area a is transferred from
tension to compression (or vice-versa).
If the centre of the transferred area a is at distance
y
from the original zero-stress axis.
P = 2fLa
3
oP/P
po/ MpM
+1
+1
-1
-1
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Mp = Mpo - P
y
i.e. Zp
= Zpo
- 2a
y
I Section subjected to bending with respect to its centroidal
axis of maximum moment of inertia (strong axis)
A
da2
Po
P=
If the neutral layer is in the web i.e.,
N/Np < AW/A, AW = area of web
4
=+
t ctc
2tc
y
ha
h
y
hX X
b
t
M
N
d
a
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A = total area
Zpx = Zp along X-X axis
So that Zd = ad2
pxZ
Zd1
Mp
M= 1
2
Po
P
= 1
2
L
L
fA
adf2
= 1 -2
22
A
da4
= 1 -Zp
Zd1
ah
ad42
2
=
aZp4
A
yA
yda41
Zp
ad1
Mp
M 2
22
2222
==
= 1 -aZp4
A
Po
P 22
forA
Av
Po
P 0.15 Po, the linear interaction as given
may be used.
15.0Po
Pfor)
Po
P1(18.1
Mp
M=
For bending about the weak axis, M approximately equals
Mp as long as P < 0.4 Po and
40.0Po
P
for)Po
P
1(19.1Mp
M2
=
Influence of the shear force on the value of the plastic
moment
The effect of shear force on a beam of general cross-
section is much more complex than that of the axial load. With
the latter, the resulting stresses could be superimposed
directly on the bending stresses, since they were all
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longitudinal. Shear combined with bending gives rise to a two-
dimensional stress system. Here, the special case of I
section with approximate method will be discussed. It is
commonly assumed in elastic design that the shear stresses
are uniformly distributed over the web of an I-section and the
flanges not contributing at all to the carrying of the shear
force.
If this same assumptions are made in the plastic analysis
of the problem then an empirical solution may be obtained
which gives good agreement with experimental results.
Full plastic moment of an I section is
Mp = BT (D T) fL + L2
ftT2
D
= Mf (flange) = BT (D T) fL
WW (Web) = L2
tT2
Df
If the shearing force F acts on the web causing an
uniform shear stress in the web, so
F = (D 2T) t.
7
D
TB
t
fc
fL
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If the web remains fully plastic then the longitudinal
stress in the web available to resist the B.M. will be reduced
below the value fL. There are several yield criterion for
combined stress system, but the Von-Mises yield condition fits
best in this case.
L
22 fZ3f =+
The stress distribution over the cross section under
combined shear and bending sufficient to cause full plasticityis as shown.
The contribution from the flanges remains unaltered, but
the contribution from the web is reduced.
Experiments indicate that the influence of the shear force
is negligible for a beam with rectangular cross section and a
ratio of span l to depth h of at least 10. The influence of the
shear force V is also negligible for an I beam with a similar
ratio as long as V does not exceed fL Aw/ 3 which
corresponds to full plastification of the web.
Example
8
Z
shear stressBending stress
f
Lf
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Suppose that the ISHB 300 is subjected to a shearing
force of 300 kN. fL = 250 N/mm2. Find the moment carrying
capacity.
=
2mm/N5.114
4.9)26.10300(
000,300=
2
L
L
)f/(31f
f =
=2
250
5.11431
= 0.608
Zpo = Zf + Zw Zf + ZW
Zp = Zf + Zw.
Lf
fZf + Zw
Lf
f
= Zpo Zw
Lf
f1
So the original plastic modules Zpo reduces by
Zw (1 f/fL) = Zw (0.392)
Zw = 4.9x)4.139(tT2
D 2w
2
=
Effect of
axial load on
I section.
9
9.4
250
10.6
300
D
TB
t
fL
BD
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Mp = Mpo 2D2 t.fL
P = 2tDfL
If the mean axial stress is denoted by p so that P = p.A
where A is the total cross-sectional area and if n = p/fL then
equation
Zp = Zpo 2D2t
2tD2 = 2L
2
L
222
f
f.
t4
Dt4
=t4
1.
f
Ap
tf4
P2
L
22
2
L
2
=
= 22
n.t4
A
So Zp = Zpo 22
n.t4
A
Combined effects of shear and axial load. This can be
found by considering the figure as given.
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Zw = Zwo (1 2) .L
f
f
P = Po
Lf
f
Where Po is the squash load of the web alone. From yield
equation
Zw = Zwo)f/(31
)Po/P()f/(31
2
L
22
L
In this equation, the mean web shearing stress is
computed, as before, assuming that none of the shear force is
carried by the flanges of the I-section.
Biaxial Bending
11
Z
shear stressBending stress
f
Lf
DBD
consined with
axial stress
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When a section is subjected to bending about both axes
such that full plasticity is produced, the neutral axis changes
its inclination. However, if no overall axial thrust or tension is
present, the neutral axis must continue to divide the cross-
section into two equal are as
A section of dimension b x d is subjected simultaneously
to bending moments Mx about XX and My about yy, causing
tensile and compressive plastic zones as shown on wither side
of neutral axis NN. For purposes of calculation, this stress
distribution may be considered as the sum of the distribution
plastic bending about XX only and zones OAC with stress + 2 y
and OBD with -2 y. If Y is used to define the inclination of the
neutral axis.
Mx = 3/Y2y2
b
2
12.
4
bdyy
2
My = 22
b
3
22y
2
b
2
1y
12
stress
Zero
A
C
y
d/2
d/2
b/2b/2
+=
N
N
XX
Y
Y
y-2- y
y+2y+
B
D
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These expressions are valid provided Y< d/2. The
dimension Y may be eliminated by using expressions for the
plastic module of the section about XX and YY
==
4
dbZpy&
4
bdZpx
22
1Zpy
My
4
3
Zpx
Mx2
yy
=
+
0 < 1Zpx
Mx
3
2
Zpy
My
yy
If the N.A. is assumed to intersect the edges of dimension
b, it is found that
1Zpy
My
Zpx
Mx
4
3
y
2
y
=
+
1Zpy
My
3
2
Zpx
Mx0
yy
Empirically, it is found that the relationship.
1Zpy
My
Zpx
Mx71.1
y
71.1
y
=
+
and gives a very close approximation to the accurate
result.
Hollow, square and rectangular sections gives the
identical with those for all solid rectangular sections.
Circular sections (both solid and hollow) give the
relationship.
1Zpy
My
Zpx
Mx2
y
2
y
=
+
*****
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