The STD code below is the example of a singly symmetric (top flange is greater than the bottom flange) and simply supported "I" beam subject to uniform distributed load acting downwards and also subject to axial compression. For this particular case, equation H2-1 (i.e.: tension fiber) of AISC 335-89 is applicable.
The actual axial compression stress is fa = 3 ksi
The allowable axial compression stress is Fa = 24.63 ksi (Eq. E2-1)
The calculated actual stresses at the Tension and Compression Fibers are 25 ksi and 14.04 ksi respectively
The allowable bending stresses for the Tension and Compression Fibers are 30 ksi (Eq. F1-5) and 24.69 ksi (Eq. F1-6) respectively
With the stresses obtained above, we can proceed to calculate the stress ratios for each fiber:
TENSION FIBER (Equation H2-1)
In this Equation the term fa/Ft is meant to be for Axial tension, however in this case we have Axial Compression, so then from the last paragraph of page 5-55 this term will not be included, and we have:
fb/Fb = 25/30 = 0.83
COMPRESSION FIBER (Equation H1-3)
fa/Fa + fb/Fb = 3/24.63 + 14.04/24.693 = 0.69
CONTROLLING FIBER
From the stress ratios calculated above, it is noted that the tension fiber is controlling, therefore the Stress Ratio = 0.83 and the Controlling Equation is H2-1.
However STAAD is reporting a stress ratio of 1.251 and Critical Condition H1-3
ADDITIONAL RESULTS FROM CALCULATIONS, FOR YOUR REFERENCE:
y_bar_comp = 21.58197 in
1/3 Compression Web Height = (y_bar_comp - tf_top) / 3 = 6.53 in
I_rT = 2,304.23 in4
A_rT = 52.90 in2
rT = 6.60 in
REMARKS:
1.- It appears that STAAD is calculating the bending ratio by dividing the actual bending tensile stress by the allowable bending compression stress and in in my opinion that is not correct. I understand that would be "CONSERVATIVE" but it is not "CORRECT". Please explain why STAAD does that?
2.- Please inform for what other codes STAAD is NOT calculating stress ratios "BY FIBER"
3.- User Provided Tables were created and the values mentioned above were obtained when using the “WIDE FLANGE” option to input the geometric properties (i.e.: widths and thicknesses), however when the option of “ISECTION” is used, STAAD reports a different value for FA because it reports 20.70 ksi which is different than FA of 24.63 ksi reported when “WIDE FLANGE” is used. The KL/R_max and the FYLD are the same in both cases, so STAAD should calculate exactly the same FA for both cases. In this case, the value of FA = 24.63 ksi is correct. Please explain how STAAD calculated the value of FA = 20.70 ksi?
4.- There is also a discrepancy in the values reported for FCZ. They are 22.13 ksi vs 22.20 ksi for "WIDE FLANGE" and "ISECTION" respectively. In this case, both values are wrong in my opinion. My calculation shows that the correct value is 24.69 ksi from Equation F1-6. Please explain why the discrepancy?
Please try to provide answers for each item above by separate
Best Regards,
David G
STAAD SPACE
START JOB INFORMATION
ENGINEER DATE 22-Jul-16
END JOB INFORMATION
INPUT WIDTH 79
UNIT FEET KIP
JOINT COORDINATES
1 0 0 0; 2 40 0 0;
MEMBER INCIDENCES
1 1 2;
START USER TABLE
TABLE 1
UNIT FEET KIP
WIDE FLANGE
W-FLG
0.741319 5 0.0625 2 0.166667 2.81297 0.127669 0.00373019 0.3125 0.296296 -
1.33333 0.083333
TABLE 2
UNIT FEET KIP
ISECTION
I-SEC
5 0.0625 5 2 0.166667 1.33333 0.083333 0.3125 0.308519 0.00389995
END
DEFINE MATERIAL START
ISOTROPIC STEEL
E 4.176e+006
POISSON 0.3
DENSITY 0.489024
ALPHA 6e-006
DAMP 0.03
TYPE STEEL
STRENGTH FY 5184 FU 8352 RY 1.5 RT 1.2
END DEFINE MATERIAL
MEMBER PROPERTY
1 UPTABLE 1 W-FLG
CONSTANTS
MATERIAL STEEL ALL
SUPPORTS
1 PINNED
2 FIXED BUT FX MY MZ
LOAD 1 LOADTYPE None TITLE LOAD CASE 1
MEMBER LOAD
1 UNI GY -15.81287
JOINT LOAD
2 FX -320.6873
PERFORM ANALYSIS
PARAMETER 1
CODE AISC
BEAM 1 ALL
FYLD 7200 ALL
LY 20 ALL
TRACK 2 ALL
CHECK CODE ALL
FINISH