Influence of Axial & Shear

8/3/2019 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.


2/13

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

6


7/13

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


10/13

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.

10


11/13

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

=

+

*****

13

Influence of Axial & Shear

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