V. AASHTO 17th Ed ASD - Design Frame
The following compares the solution of a design performed using STAAD.Pro against a hand calculation.
Reference
Following step by step hand calculation of Allowable Stress Steel Design per the AASHTO Standard Specifications for Highway Bridges, 17th Edition (2002).
Problem
Determine the allowable stresses (AASHTO code) for the members of the structure as shown in figure. Also, perform a code check for these members based on the results of the analysis.
Members 1, 2 = W12X26, Members 3, 4 = W14X43
Members 5, 6, 7 = W16X36, Memb8= L40404,
Member 9 = L50506
The frame is subject to the following load cases:
Only the AASHTO steel design elements are check here. No structural analysis calculations are included in these hand verifications.
Though the program does check shear per the AASHTO specifications, those calculations are not reflected here. Only the controlling stress ratios are presented.
As all members are grade 36 steel, the following critical slenderness parameter applies to each:
Member 1
Size W 12X26, L = 10 ft., a = 7.65 in2, Sz = 33.39 in3
From observation, Load case 1 will govern
Calculate the allowable stress as per Table 10.32.1A.
Bending Minor Axis
Allowable minor axis bending stress:
FTY = FTZ = 0.55 · FY = 19.8 ksi
Bending Major Axis
Where:
Cb = 1.75 + 1.05(M1/M2)+0.3x(M1/M2)2
- M1 = 0, so Cb = 1.75
- Szc = Section modulus with respect to the compression flange =204/(0.5 · 12.22) = 33.38789 in3
IYC = tb 3/12 = 0.38 · 6.493/12 = 8.6564 in4 |
J = (2 x 6.49 · 0.383+ (12.22 – 2 · 0.38) · 0.233)/3 = 0.28389 in4 |
which is larger than 0.55 · FY = 19.8 ksi, so FCZ = 19.8 ksi
Axial Compression
Critical (kL/r) = 1.0 · 120/1.5038 = 79.7978
As (kL/r) < Cc , the allow axial stress in compression is given by:
Actual Stress
Actual axial stress, fa= 25/7.65 = 3.26 ksi
The critical moment occurs at the end node of the beam. So we use the AASHTO equation 10.42 in section 10-36 to calculate the design ratio.
Actual bending stress = fbz = 56.5 · 12/33.4 = 1.692 · 12 = 20.3 ksi
From Table 10-36A, Cmz = 0.85
Equation 10-42
For the end section, use Equation 10.43:
The critical stress ratio is thus 1.217. The value calculated by STAAD is 1.218
Member 2
Size W 12X26, L = 5 ft., a = 7.65 in2, Sz = 33.4 in3
From observation Load case 1 will govern, Forces at the midspan are
Calculate the allowable stress as per Table 10.32.1A.
Bending Minor Axis
Allowable minor axis bending stress:
FTY = FTZ = 0.55 · FY = 19.8 ksi
Bending Major Axis
Where:
Cb = 1.75 + 1.05(M1/M2)+0.3x(M1/M2)2
- M1 = 39.44 and M2 = 677.96, so Cb = 1.69
- Szc = Section modulus with respect to the compression flange =204/(0.5 · 12.22) = 33.38789 in3
IYC = tb 3/12 = 0.38 · 6.493/12 = 8.6564 in4 |
J = (2 x 6.49 · 0.383+ (12.22 – 2 · 0.38) · 0.233)/3 = 0.28389 in4 |
which is larger than 0.55 · FY = 19.8 ksi, so FCZ = 19.8 ksi
Axial Compression
Critical (kL/r) = 1.0 · 60/1.504 = 39.92
As (kL/r) < Cc , the allow axial stress in compression is given by:
Actual Stress
Actual axial stress, fa= 8.71 / 7.65 = 1.138 ksi
The critical moment occurs at the end node of the beam. So we use the AASHTO equation 10.42 in section 10-36 to calculate the design ratio.
Actual bending stress = fbz = 56.5 · 12/33.4 = 1.691 · 12 = 20.3 ksi
(KL/r)z = 1 · 60/5.16 = 11.618 |
From Table 10-36A, Cmz = 0.85
Equation 10-42
For the end section, use Equation 10.43:
The critical stress ratio is thus 1.092. The value calculated by STAAD is 1.093.
Member 3
Size W 14X43, L = 11 ft., a = 12.6 in2, Sz = 62.7 in3
From observation Load case 3 will govern, Forces at the end are
Calculate the allowable stress as per Table 10.32.1A.
Bending Minor Axis
Allowable minor axis bending stress:
FTY = FTZ = 0.55 · FY = 19.8 ksi
Bending Major Axis
Where:
Cb = 1.75 + 1.05(M1/M2)+0.3x(M1/M2)2
- M1 = 0, so Cb = 1.75
- Szc = Section modulus with respect to the compression flange = 428/(0.5 · 13.66) = 62.7 in3
IYC = tb3/12 = 0.53 · 8.03/12 = 22.61in in4 |
J = (2 · 8.0 · 0.533+ (13.66 – 2 · 0.53) · 0.3053)/3 = 0.913 in4 |
which is larger than 0.55 · FY = 19.8 ksi, so FCZ = 19.8 ksi
Axial Compression
Critical (kL/r) = 1.0 · 132/1.894 = 69.69
As (kL/r) < Cc , the allow axial stress in compression is given by:
Actual Stress
Actual axial stress, fa= 25.5 / 12.6 = 2.024 ksi
The critical moment occurs at the end node of the beam. So we use the AASHTO equation 10.42 in section 10-36 to calculate the design ratio.
Actual bending stress = fbz = 112.17 · 12/62.7 = 1.789 · 12 = 21.467 ksi
(KL/r)z = 1 · 132/5.828 = 22.649 |
From Table 10-36A, Cmz = 0.85
Equation 10-42
For the end section, use Equation 10.43:
The critical stress ratio is thus 1.203. The value calculated by STAAD is 1.204.
Member 4
Size W 14X43, L = 4 ft., a = 12.6 in2, Sz = 62.6 in3
From observation Load case 3 will govern, Forces at the end are
Calculate the allowable stress as per Table 10.32.1A.
Bending Minor Axis
Allowable minor axis bending stress:
FTY = FTZ = 0.55 · FY = 19.8 ksi
Bending Major Axis
Where:
Cb = 1.75 + 1.05(M1/M2)+0.3 · (M1/M2)2 |
M1 = -191.36 Kip-in , M2 = -1346.08 Kip-in so Cb = 1.606 |
IYC = tb3/12 = 0.53 · 8.03/12 = 22.61 in4 |
J = (2 · 8.0 · 0.533+ (13.66 – 2 · 0.53) · 0.3053)/3 = 0.913 in4 |
which is larger than 0.55 · FY = 19.8 ksi, so FCZ = 19.8 ksi
Axial Tension
Ft = 0.55 · FY = 19.8 ksi
Actual Stress
Actual axial stress, fa= 8.75 / 12.6 = 0.694 ksi
Actual bending stress = fbz = 112.17 · 12/62.7 = 1.789 · 12 = 21.47 ksi, which exceeds FCZ.
fbz/FCZ = 21.47/19.8 = 1.084
The critical moment occurs at the end node of the beam. So we use the AASHTO equation 10.43 in section 10-36 to calculate the design ratio for the end section.
The critical stress ratio is thus 1.125. The value calculated by STAAD is 1.126.
Member 5
Size W 16X36, L = 5 ft., a = 10.6 in2, Sz = 56.5 in3
From observation Load case 3 will govern, Forces at the end are
Calculate the allowable stress as per Table 10.32.1A.
Bending Minor Axis
Allowable minor axis bending stress:
FTY = FTZ = 0.55 · FY = 19.8 ksi
Bending Major Axis
Where:
Cb = 1.75 + 1.05(M1/M2)+0.3 · (M1/M2)2 |
M1 = 40.14, M2 = -684.4 so Cb = 1.81 |
- Szc = Section modulus with respect to the compression flange = 448/(0.5 · 15.86) = 56.5 in in3
IYC = tb3/12 = 0.43 · 6.993/12 = 12.238 in4 |
J = (2 · 6.99 · 0.433+ (15.86 – 2 · 0.43) · 0.293)/3 = 0.5 in4 |
which is larger than 0.55 · FY = 19.8 ksi, so FCZ = 19.8 ksi
Axial Compression
Critical (kL/r) = 1.0 · 60/1.52 = 69.69
As (kL/r) < Cc , the allow axial stress in compression is given by:
Actual Stress
Actual axial stress, fa= 14.02 / 10.6 = 1.323 ksi
The critical moment occurs at the end node of the beam. So we use the AASHTO equation 10.42 in section 10-36 to calculate the design ratio.
Actual bending stress = fbz = 57.04 · 12/56.5 = 1.001 · 12 = 12.115 ksi
(KL/r)z = 1 · 60/6.5= 9.231 |
From Table 10-36A, Cmz = 0.85
Equation 10-42
For the end section, use Equation 10.43:
The critical stress ratio is thus 0.689. The value calculated by STAAD is 0.690.
Member 6
Size W 16X36, L = 16 ft., a = 10.6 in2, Sz = 56.5 in3
From observation Load case 3 will govern, Forces at the end are
Calculate the allowable stress as per Table 10.32.1A.
Bending Minor Axis
Allowable minor axis bending stress:
FTY = FTZ = 0.55 · FY = 19.8 ksi
Bending Major Axis
Where:
Cb = 1.75 + 1.05(M1/M2)+0.3 · (M1/M2)2 |
M1 = 8.947 M2 = 183.05 so Cb = 1.69 |
- Szc = Section modulus with respect to the compression flange = 448/(0.5 · 15.86) = 56.5 in in3
IYC = tb3/12 = 0.43 · 6.993/12 = 12.238 in4 |
J = (2 · 6.99 · 0.433+ (15.86 – 2 · 0.43) · 0.293)/3 = 0.5 in4 |
which is larger than 0.55 · FY = 19.8 ksi, so FCZ = 19.8 ksi
Axial Compression
Critical (kL/r) = 1.0 · 192/1.52 = 126.3
As (kL/r) > Cc , the allow axial stress in compression is given by:
Actual Stress
Actual axial stress, fa= 10.2 /10.6 = 0.962 ksi
The critical moment occurs at the end node of the beam. So we use the AASHTO equation 10.42 in section 10-36 to calculate the design ratio.
Actual bending stress = fbz = 62.96 · 12/56.5 = 1.114 · 12 = 13.37 ksi
(KL/r)z = 1 · 192/6.51= 29.49 |
From Table 10-36A, Cmz = 0.85
Equation 10-42
For the end section, use Equation 10.43:
The critical stress ratio is thus 0.732. The value calculated by STAAD is 0.732
Member 7
Size W 16X36, L = 4 ft., a = 10.6 in2, Sz = 56.5 in3
From observation Load case 3 will govern, Forces at the midspan are
Calculate the allowable stress as per Table 10.32.1A
Bending Minor Axis
Allowable minor axis bending stress:
FTY = FTZ = 0.55 · FY = 19.8 ksi
Bending Major Axis
FCZ = 0.55 · FY = 19.8 ksi
Axial Tension
Fa = 0.55 · FY = 19.8 ksi
Actual Stress
Actual axial stress, fa= 24.05 /10.6 = 2.268 ksi, hence, ok.
Actual bending stress = fbz = 62.96 · 12/56.5 = 1.114 · 12 = 13.37 ksi
So the combined ratio is
fa/Fa + fbz/FTZ + fby/FTY = 2.268/19.8 + 13.37/19.8 + 0 = 0.790
The critical moment occurs at the end node of the beam. So we use the AASHTO equation 10.43 in section 10-36 to calculate the design ratio for the end section.
The critical stress ratio is thus 0.809. The value calculated by STAAD is 0.809.
Member 8
Size L4x4x1/4, L = 7.07 ft., a = 1.938 in2
From observation Load case 1 will govern, Forces
Fx = 23.04 kip (compression)
Calculate the allowable stress as per Table 10.32.1A
Axial Compression
Critical (KL/r)y = 1.0 · 7.07 · 12/0.795 = 106.7
As (kL/r) < Cc , the allow axial stress in compression is given by:
Actual Stress
Actual axial stress, fa= 23.04 /1.938 = 11.88 ksi
fa/Fa = 11.88/10.89 = 1.091
The value calculated by STAAD is 1.091.
Member 9
Size L5x5x3/8, L = 5.657 ft, A = 3.61 in2
From observation Load case 1 will govern, Forces
Fx = 48.44 kip (compression)
Calculate the allowable stress as per Table 10.32.1A
Axial Compression
Critical (KL/r)y = 1.0 · 5.567 · 12/0.99 = 68.57
As (kL/r) < Cc , the allow axial stress in compression is given by:
Actual Stress
Actual axial stress, fa= 48.44 /3.61 = 13.42 ksi
fa/Fa = 13.42/14.47 = 0.927
The value calculated by STAAD is 0.928.
Comparison
STAAD Input
The file C:\Users\Public\Public Documents\STAAD.Pro 2023\Samples \Verification Models\09 Steel Design\US\AASHTO\AASHTO 17th Ed ASD - Design Frame.STD is typically installed with the program.
STAAD PLANE VERIFICATION PROBLEM FOR AASHTO CODE
START JOB INFORMATION
ENGINEER DATE 22-Sep-18
END JOB INFORMATION
*
* THIS DESIGN EXAMPLE IS VERIFIED BY HAND CALCULATION
* FOLLOWING AASHTO ASD 97 CODE.
*
UNIT FEET KIP
JOINT COORDINATES
1 0 0 0; 2 25 0 0; 3 0 10 0; 4 25 11 0; 5 0 15 0; 6 25 15 0; 7 5 15 0;
8 21 15 0;
MEMBER INCIDENCES
1 1 3; 2 3 5; 3 2 4; 4 4 6; 5 5 7; 6 7 8; 7 8 6; 8 3 7; 9 4 8;
MEMBER PROPERTY AMERICAN
1 2 TABLE ST W12X26
3 4 TABLE ST W14X43
5 TO 7 TABLE ST W16X36
8 TABLE ST L40404
9 TABLE ST L50506
MEMBER TRUSS
8 9
DEFINE MATERIAL START
ISOTROPIC MATERIAL1
E 4.176e+06
POISSON 0.3
END DEFINE MATERIAL
CONSTANTS
MATERIAL MATERIAL1 ALL
SUPPORTS
1 2 PINNED
LOAD 1 DL + LL
MEMBER LOAD
5 TO 7 UNI Y -2
LOAD 2 WIND FROM LEFT
JOINT LOAD
5 FX 15
LOAD COMBINATION 3
1 0.75 2 0.75
PERFORM ANALYSIS
LOAD LIST 1 3
PRINT MEMBER FORCES
PARAMETER 1
CODE AASHTO
TRACK 0 ALL
CHECK CODE ALL
FINISH
STAAD Output
STAAD.Pro CODE CHECKING - ( AASHTO - ASD) v1.0 *********************** ALL UNITS ARE - KIP FEET (UNLESS OTHERWISE Noted) MEMBER TABLE RESULT/ CRITICAL COND/ RATIO/ LOADING/ FX MY MZ LOCATION ======================================================================= * 1 ST W12X26 (AISC SECTIONS) FAIL AASHTO 10-43 1.216 1 25.00 C -0.00 56.50 10.00 * 2 ST W12X26 (AISC SECTIONS) FAIL AASHTO 10-43 1.091 1 8.72 C 0.00 56.50 0.00 * 3 ST W14X43 (AISC SECTIONS) FAIL AASHTO 10-43 1.207 3 25.50 C -0.00 -112.20 11.00 * 4 ST W14X43 (AISC SECTIONS) FAIL AASHTO 10-43 1.130 3 8.83 T 0.00 -112.20 0.00 5 ST W16X36 (AISC SECTIONS) PASS AASHTO 10-43 0.691 3 14.02 C -0.00 -57.00 5.00 * 6 ST W16X36 (AISC SECTIONS) FAIL KL/R ratio 1.052 1 5.65 C 0.00 -15.25 0.00 7 ST W16X36 (AISC SECTIONS) PASS AASHTO 10-43 0.812 3 24.13 T 0.00 63.00 0.00 * 8 ST L40404 (AISC SECTIONS) FAIL AASHTO 10-42 1.114 1 23.03 C 0.00 0.00 0.00 **WARNING: For Memb # 8 SECTIONS WHICH ARE NOT I-SHAPED CANNOT BE DESIGNED FOR BENDING. AS PER THE AASHTO CODE 17th EDITION. PLEASE USE THE AISC 9th EDITION CODE FOR DESIGNING THE SECTION. 9 ST L50506 (AISC SECTIONS) PASS AASHTO 10-42 0.920 3 48.55 C 0.00 0.00 0.00 **WARNING: For Memb # 9 SECTIONS WHICH ARE NOT I-SHAPED CANNOT BE DESIGNED FOR BENDING. AS PER THE AASHTO CODE 17th EDITION. PLEASE USE THE AISC 9th EDITION CODE FOR DESIGNING THE SECTION.