Math is needed here. First some assumptions need to be made. For the purpose of this analysis, let the exterior column be HSS14x14x5/16 tubes at 25%
of the maximum axial load prior to any damage. The column has the following properties:
Similar to HSS 14x14x5/16
A = 15.7in^2
I = 739 in^4
S = 92.3 in ^3
Pn = 557k (from AISC LRFD 3rd, table 4-6 with an unbraced length, KL = 12’-4”)
Pu = ¼* 557k = 139k
Mn = 92.3in^3*46ksi = 4645 kip*in (Mn = maximum bending capacity)
The column itself will bend inwardly until it snaps if the column itself ever becomes inelastic. This can be defined by the ratio (I’ve simplified
this a bit): Mu/Mn + Pu/Pn < 1. The other limit state is P-delta. When the exterior column is pulled inwards, it deflects. This deflection(Δ)
generates a moment, specifically P2*Δ. This moment, creates more deflection, which further magnifies the moment, creating more moment, and so forth.
P-delta has two outcomes: the moment reaches equilibrium at some point, or becomes unstable and continues to grow. This phenomenon can be easily shown
with a simple experiment. Take a straw and try to compress it between your fingers. It has a surprising amount of strength. Now push the middle in
slightly. This is p-delta.
We know that the floor diaphragm along the wall were greatly damaged due to fire and the impact of the aircraft. Let us assume that there are two
floor diaphragms that are damaged to the point that they no longer provide bracing against buckling. There is thus a pull-in force of 6kips at two
places along the length of the column.
CALCULATION 1: DIAPHRAGM DAMAGE, NO FIRE EFFECTS
Unbraced length = 37’-0”
Pu = 139k
Pn = 465k (from AISC LRFD 3rd, table 4-6 with an unbraced length, KL = 37’-0”)
Mn = 4645 kip*in
Mu = P*a
Mu = 6kip*1/3*37ft
Mu = 74kip*ft or 888 kip*in
Deflection = P*a *(3L^2 – 4*a^2)/(24*E*I) (Formula from AISC LRFD 3rd)
a = 1/3*L
= P * 1/3L *(3L^2 – 4/9*L^2)/(24*E*I)
= 23*P*L^3/(1296*E*I)
= 23*6k*(37ft*12in/ft)^3/(1296*29000ksi*739in)
= 0.435in
Additional moment due to P-delta
Mu+ = 0.435in*139k = 61.02 kip*in
Additional Deflection = Mu+*L^2 / (4*EI)
= 61.02kip*in*(37 * 12ft\in)^2 / (4*29000ksi*739in^3)
= 0.140in
Additional moment due to P-delta2
Mu++ = (0.435+0.140)in*139k = 79.93 kip*in
Additional Deflection = Mu++*L^2 / (4*EI)
= 79.93kip*in*(37 * 12ft\in)^2 / (4*29000ksi*739in^3)
= 0.184in
As seen, the first p-delta iteration results in an increased deflection of 0.140in. The second results in a deflection of only 0.184in. We can thus
conclude that p-delta will eventually converge and that no further iterations are necessary. The 6kip pull-in force with no effect of fire will not
result in the column becoming unstable.
In a 600C fire, the Modulus of Elasticity will have reduced to approximately 0.3 of its original value, and the yield strength to 0.5 of its original
value. The effect of the Modulus of Elasticity being so greatly lowered is of far greater important than the yield strength, however.
CALCULATION 2: DIAPHRAGM DAMAGE, 500C FIRE
Unbraced length = 37’-0”
E = 0.3*29000ksi = 8700ksi
Pu = 139k
Pn = 465k*0.5 = 233k
Mn = 4645 kip*in *0.5 = 2323kip*in
Mu = P*a
Mu = 6kip*1/3*37ft
Mu = 74kip*ft or 888 kip*in
Deflection = P*a *(3L^2 – 4*a^2)/(24*E*I) (Formula from AISC LRFD 3rd)
a = 1/3*L
= P * 1/3L *(3L^2 – 4/9*L^2)/(24*E*I)
= 23*P*L^3/(1296*E*I)
= 23*6k*(37ft*12in/ft)^3/(1296*8700ksi*739in)
= 1.45in
Additional moment due to P-delta
Mu+ = 1.45in*139k = 201.6 kip*in
Additional Deflection = Mu+*L^2 / (4*EI)
= 201.6kip*in*(37 * 12ft\in)^2 / (4*8700ksi*739in^3)
= 1.55in
Additional moment due to P-delta2
Mu++ = (1.45+1.55)in*139k = 417 kip*in
Additional Deflection = Mu++*L^2 / (4*EI)
= 417kip*in*(37 * 12ft\in)^2 / (4*8700ksi*739in^3)
= 3.20in
This results in the column becoming unstable due to p-delta. This can easily be seen in that the deflection due to P-delta2 is double that of
P-delta1. It can therefore be concluded that it was necessary for both fire and damage to result in the collapse of the towers.
[edit on 5-7-2008 by ThroatYogurt]
