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2/15

GIVEN

Single story, single bay frame

shown in the figure. In this frame,column marked as (1) is

500500 mm while the other

column is 400400 mm. Beam is,4002000 mm.


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GIVEN

Moment-curvature diagrams of these two columns are as

shown in the figure. The M-K diagrams are drawn for N1=150

kN and N2=400 kN considering the effect of the lateral load.


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REQUIRED

Assuming that the uniformly distributed

load on the beam remaining constant,sketch the variation of the base shear inthe columns, as the lateral load increases.


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SOLUTION

Since the beam is very deep, in the analysisit can be assumed to be infinitely rigid. Thismeans that the point of inflection will be at

the midheight of the columns.


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EQUILIBRIUM

N1 + N2 = 5 100 = 500 kN

Taking moment about R and G;N2 = 250 + F

N1 = 250 F

Ignoring second order effects;

V1 5 = M1V2 5 = M2

V1 + V2 = F


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COMPATIBILITY

1 = 2 =


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ELASTIC DISPLACEMENTS

It is assumed that the columns reach their ultimatelimit states when, M=My,and after this stage plastic

hinge fully forms. In the elastic stage, however, thecolumn flexural stiffnesses were calculated byassuming

Icr= 1/2 Ic and Ec = 20,000 MPa

Thus;

For Column 1: EI= 50,000 kN/m2

,For Column 2: EI= 20,000 kN/m2

Please note, the slopes of M-K diagrams

correspond to the values given above.


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ELASTIC DISPLACEMENTS

Hence,

1

1

1 0001666053

2

2

550000

2M.

M

=

=

2

2

2 M0004166.05

3

2

2

520000

M

2

=

=


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M1

/M2

RATIO

As 1 = 2 = , we can determine theM

1/M

2ratio:

5.2M

M

M0004166.0M0001666.0

2

1

21

=

=

V1 5 = M1

V2 5 = M2

From equilibrium:

therefore; V1 / V2 = 2.5


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AT THE FIRST YIELD

As M1/M2 =2.5 Column 1 will reach the yield pointfirst.

m.kN180Mandm.kN450MThus 21 ==

N2 = 250 + F

N1 = 250 F

From equilibrium:

therefore;

kN1263690F

kN.36VandkN90V:areforcesshearingCorrespond

y

21

=+=

==

N2 = 250 + 126 = 376 kN

N1 = 250 F = 124 kN

and: mm75)mm.N10450M(0001666.053

2

2

550000

M

2

61

1

1===

=

mm150y1 ==


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AT THE SECOND YIELD

M1=450 kN.m(constant, cannot be increased)

M2

=250 kN.m

From the moment curvature diagrams

K2 =250/20,000 = 0.0125 1/mm

/2 = 0.0125 5/2 2/3 5 = 0.10417 m = 208.3 mm

V1 = 90 kN (sabit) , V2 = 250/5 = 50 kN ,

F = V1 + V2 = 140 kN

From equilibrium,

N1 = 110 kN , N2 = 390 kN


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AT THE SECOND YIELD

104150754505432

2540090

21 ...K... =+=

K1 = 0.0183 rad/m


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RESULT


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RESULT

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