CE 579 Homework 6
a) Compute the critical buckling load and mode shape for the frame.
The MATLAB code idealizing the eigenvalue approach to find Pcr was used. The input file is attached as well as the outputs.
Input File
clear;
close all;
%------
%Problem 9.2 from book
%First will define the connectivity between the local and global
%coordinates
%
id = [0 1 0 4 7;...
0 2 0 5 8;...
0 3 0 6 9;...
1 4 4 7 0;...
2 5 5 8 0;...
3 6 6 9 0];
%Form element stifness matrices - 1st order
[k1] = frame(0,0,0,288,29000,248,13.3);
[k2] = frame(0,288,240,288,29000,2850,24.8);
[k3] = frame(240,0,240,288,29000,248,13.3);
[k4] = frame(240,288,720,288,29000,2850,24.8);
[k5] = frame(720,288,720,0,29000,248,13.3);
%
%Assemble global stifness matrix - 1st order
KO = zeros(9,9);
[KO] = addk(KO,k1,id,1);
[KO] = addk(KO,k2,id,2);
[KO] = addk(KO,k3,id,3);
[KO] = addk(KO,k4,id,4);
[KO] = addk(KO,k5,id,5);
%
% Form element geometric stiffness matrices - linearized
[k1g] = frameg(0,0,0,288,29000,248,13.3,0.5);
[k2g] = frameg(0,288,240,288,29000,2850,24.8,0.1);
[k3g] = frameg(240,0,240,288,29000,248,13.3,1);
[k4g] = frameg(240,288,720,288,29000,2850,24.8,0.1);
[k5g] = frameg(720,288,720,0,29000,248,13.3,0.5);
%
%Assemble global geometric stifness matrix - linearized
KG = zeros(9,9);
[KG] = addk(KG,k1g,id,1);
[KG] = addk(KG,k2g,id,2);
[KG] = addk(KG,k3g,id,3);
[KG] = addk(KG,k4g,id,4);
[KG] = addk(KG,k5g,id,5)
%
%Solving the eigenvalue problem
[V,D]=eig(KO,KG);
buckle=V
Pcr=D
buckle =
-0.0020 -1.0000 0.5622 -0.0128 0.8948 -0.9997 -1.0000 -0.7075 -0.0000
-0.7368 -0.0284 -0.0125 -0.7110 0.6401 -0.0019 0.1638 -0.6649 -1.0000
-0.0002 -0.0022 0.0014 -0.0003 0.0759 -0.0003 0.1065 0.2434 -0.0000
-0.0101 0.6351 0.2227 -0.0074 0.9337 -1.0000 -0.9201 -0.6521 -0.0000
-0.2632 0.0174 0.0274 1.0000 -0.5866 0.0010 -0.4307 1.0000 -1.0000
-0.0004 0.0013 0.0005 0.0001 0.0388 0.0000 -0.0760 -0.3616 -0.0000
-0.0130 -0.2738 -1.0000 0.0385 1.0000 -0.9995 -0.8373 -0.9000 0.0000
1.0000 0.0110 -0.0149 -0.2890 -0.0534 0.0010 0.2669 -0.3351 -1.0000
-0.0005 -0.0004 -0.0019 0.0005 0.0181 -0.0006 -0.1279 0.5958 -0.0000
Pcr =
4.3957e+00600000000
02.4864e+0060000000
009.2378e+005000000
0001.2572e+00600000
000 0793010000
000001249000
0000004030700
0000000216210
00000000Inf
Therefore, the critical buckling load (Pcr) = 1249 kips corresponding to the following mode shape
= [-0.99971 -0.001944 -0.00033866 -1, 0.0009877 1.1568e-005 -0.99951 0.0009563 -0.00060994]T
The elastic critical buckling load as well as the modal shape was validated with the following MASTAN 2 model.
(b) Perform a linear elastic analysis using any software for P=0.5 Pcre obtained in (a).
I have chosen to stick with MASTAN 2.
P = 0.5Pcre = 0.5*1249 = 624.5 kips.
Therefore: 0.1P = 62.45 kips0.5P = 312.25 kips
The results are shown below. Scale factor = 5.
The displacements at the nodes as well as the member loads are shown below:
Node / Location / x-disp (in) / y-disp (in) / RotationRad / Deg
2 / Left / 6.115 / -0.221 / -0.00329 / -0.1883
3 / Middle / 6.101 / -0.4709 / -0.00069 / -0.0398
5 / Right / 6.087 / -0.2408 / -0.00308 / -0.1767
c) Perform a second-order elastic analysis using the same software and loads.
The MASTAN2 results are shown below. Scale factor = 5.
The displacements at the nodes as well as the member loads are shown below:
Node / Location / x-disp (in) / y-disp (in) / RotationRad / Deg
2 / Left / 10.96 / -0.4074 / -0.004969 / -0.2847
3 / Middle / 10.95 / -0.6729 / -0.007351 / -0.4212
5 / Right / 10.93 / -0.4414 / -0.006073 / -0.348
c) Add a notional load of Ni = 0.003Yi, and perform the second order analysis again.
Total gravity load applied in the vertical direction = 624.5 + 312.25 + 312.25 = 1249 kips
Ni = 0.003*1249 = 3.75 kips
Total lateral load at joint = 64.25 + 3.75 = 68 kips
The MASTAN2 results are shown below. Scale factor = 5.
The displacements at the nodes are shown below:
Node / Location / x-disp (in) / y-disp (in) / RotationRad / Deg
2 / Left / 11.93 / -0.4414 / -0.00531 / -0.3042
3 / Middle / 11.91 / -0.7102 / -0.0007586 / -0.0435
5 / Right / 11.9 / -0.4784 / -0.006682 / -0.3829
i) What is the value of B1 and B2 factors?
The results of the first order analysis are shown below:
Element / Description / V (kips) / Mend1 (kip-in) / Mend2 (kip-in) / P (kips) / Cm1 / Left Column / 20.38 / -3017 / 2853 / 295.9 / 0.98
2 / Left Beam / 16.34 / 2853 / -1069 / 42.07 / 0.75
3 / Center Column / 21.68 / -3139 / 3105 / 630.6 / 1.00
4 / Right Beam / 10.2 / 2036 / -2859 / 20.39 / 0.88
5 / Right Column / 20.39 / 2859 / -3013 / 322.4 / 0.98
Where Cm = 1 – (Mlow/Mhigh)
Mlow corresponds to the lower absolute value of moment and Mhigh corresponds to the higher absolute moment value.
B1:
For the left column.
From alignment chart, K1 = 0.52
kips
Pr = 295.9 kips (See Part iv)
For the center column.
From alignment chart, K1 = 0.515
kips
Pr = 630.6 kips (See Part iv)
For the right column.
From alignment chart, K1 = 0.53
kips
Pr = 322.4 kips (See Part iv)
B2:
H = 64.25 kips
H = 6.1 in (From Part b)
L = 288 in
Rm = 0.85 for Moment Frames
Pe2 = Rm*HL/H = 0.85*64.25*288 / 6.1 = 2578.4 kips
Pu = 295.9 + 630.6 + 322.4 = 1249 kips
ii) What is the ratio of second-order to first-order drift? Is it ≥ 1.5
Node / Location / Ratio of Second Order / First Order2 / Left / 1.792
3 / Middle / 1.795
5 / Right / 1.796
Therefore, the second order drift to the first order drift ratio is about 1.795. Yes, this is larger than 1.5. Therefore, the direct analysis approach given in Appendix 7 must be used.
iii) Summarize the required member strengths (Pr, Vr, Mr) obtained from (b, c, and d).
Required member strengths from Part a)
Element / Description / V (kips) / Mmax (kip-in) / P (kips)1 / Left Column / 20.38 / 3017 / 295.9
2 / Left Beam / 16.34 / 2853 / 42.07
3 / Center Column / 21.68 / 3139 / 630.6
4 / Right Beam / 10.2 / 2859 / 20.39
5 / Right Column / 20.39 / 3013 / 322.4
Required member strengths from Part b)
Element / Description / V (kips) / Mmax (kip-in) / P (kips)1 / Left Column / 36.43 / 5273 / 281
2 / Left Beam / 30.05 / 5062 / 36.78
3 / Center Column / 37.78 / 5246 / 636.9
4 / Right Beam / 16.72 / 4929 / 23.24
5 / Right Column / 35.73 / 5177 / 327.8
Required member strengths from Part c)
Element / Description / V (kips) / Mmax (kip-in) / P (kips)1 / Left Column / 39.69 / 5745 / 277.8
2 / Left Beam / 32.95 / 5519 / 39.89
3 / Center Column / 41.12 / 5739 / 638.3
4 / Right Beam / 18.08 / 5356 / 25.2
5 / Right Column / 38.84 / 5629 / -329
iv) Compare the required member strengths (Pr, Vr, Mr) from (b) with those obtained from (c) and (d). What are the percent differences? What causes them?
Comparisons of the values for Part a and Part b are shown below.
% Differences [(Values in table (Case b)-Values in table(Case a))/Values in table (Case a)
Element / Description / V ratio % / Mmax ratio % / P ratio %1 / Left Column / 78.75 / 74.78 / -5.04
2 / Left Beam / 83.90 / 77.43 / -12.57
3 / Center Column / 74.26 / 67.12 / 1.00
4 / Right Beam / 63.92 / 72.40 / 13.98
5 / Right Column / 75.23 / 71.82 / 1.67
Comparisons of the values for Part a and Part b are shown below.
% Differences [(Values in table (Case b)-Values in table(Case a))/Values in table (Case a)
Element / Description / V ratio % / Mmax ratio % / P ratio %1 / Left Column / 94.75 / 90.42 / -6.12
2 / Left Beam / 101.65 / 93.45 / -5.18
3 / Center Column / 89.67 / 82.83 / 1.22
4 / Right Beam / 77.25 / 87.34 / 23.59
5 / Right Column / 90.49 / 86.82 / 2.05
There are significant differences when comparing the results from the first order analysis to the second order analysis. Second order effects cause the shears and moment demands in each member to increase by 75% (At least by 67 %). The lateral load causes the beam to sway prior to buckling causing a P- effect. Therefore, the axial loads have an eccentricity in relation to the supports. This eccentricity subjects the member to curvature. Moment equilibrium takes into account the offset of the axial loads, which therefore increases the moment demand. Note that if the lateral load is removed from the structure, there are insignificant differences between the first order and the second order analysis.
A second order analysis marginally changes the axial load demand in the columns. However, the total axial load in all columns from Case a and Case b were very similar.
v) What is the impact of adding the notional load to the frame?
Although the increase in lateral load appeared to be small with respect to the initial applied axial load, the shear and moment demand still increased significantly (about 9% higher than without the notional load). The notional load increases the lateral displacement of the frame which therefore further increases the P- effects. The % differences between the member forces before to after the addition of the notional load are shown below.
Element / Description / V ratio % / Mmax ratio % / P ratio %1 / Left Column / 8.95 / 8.95 / -1.14
2 / Left Beam / 9.65 / 9.03 / 8.46
3 / Center Column / 8.84 / 9.40 / 0.22
4 / Right Beam / 8.13 / 8.66 / 8.43
5 / Right Column / 8.70 / 8.73 / 0.37
vi) Would it make sense to reduce the member flexural stiffness to EI*? Why?
EI* = 0.8EI(1)(App. 7, obtained from class handout)
Based on the results obtained in Part ii, the Direct Analysis Method in Appendix 7 must be used. Therefore, we have to consider the above equation according to AISC-2005 (as interpreted from the class handout). This parameter takes into account the initial residual stresses in the cross section. With this parameter, all of the initial imperfections in the structure and members that have been discussed in this course will be taken into consideration for this homework problem. Initial residual stresses have a significant impact on the flexural stiffness of the member if the demand causes early yielding. However, an elastic analysis was conducted although the selection of a member may include an inelastic capacity. Therefore, the accuracy of just stating that EI* is equal to the reduction shown above is not as accurate as performing a second-order inelastic analysis of the structure which includes the initial residual stresses in each member. The problem identified is that EI* should equal EI in some cases and should be altered as the flanges undergo yielding. I still feel that EI* should be considered as it increases the deflections significantly and therefore increases the moment and shear demands as well. As a trial, E was reduced to 0.8E in the MASTAN2 model. The results drifts and moment demands in the members are shown below. More rigorous analysis will need to be conducted to investigate the accuracy of Equation 1.
2 / Left / 18.73
3 / Middle / 18.72
5 / Right / 18.69