3 3 member-to-member redistribution
TRANSCRIPT
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Member-to-MemberRedistribution
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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