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.

AASHTO ASD verification problem

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:

1. a uniform gravity load along the beam, w = 2 kips/ft (dead + live)
2. a lateral wind load, F = 15 kips
3. a combination of 75% of load 1 and 75% of load 2

Solution

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:

$C c = 2 π 2 E F y = 2 π 2 29 , 000 36 = 126.1$

Member 1

Size W 12X26, L = 10 ft., a = 7.65 in2, Sz = 33.39 in3

From observation, Load case 1 will govern

• Fx = 25.0 kip (compression)
• Mz= 56.5 k-ft

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

$F c z = 50 ( 10 ) 6 C b S x c ( I y c l ) 0.772 J I y c + 9.87 ( d l ) 2 ≤ 0.55 F y$

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

 $F c z = 50 ( 10 ) 6 1.75 33.38789 ( 8.6564 120 ) 0.772 0.28389 8.6564 + 9.87 ( 11.46 120 ) 2 10 − 3 = 64.375 ⁢ k s i$

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:

$F a = F y F . S . [ 1 − ( k L / r ) 2 F y 4 π 2 E ] = 36 2.12 [ 1 − ( 79.79 ) 2 36 4 π 2 29 , 000 ] = 13.58 k s i$

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

$F e z = π 2 E F . S . ( k L / r ) 2 = π 2 29 , 000 2.12 ( 120 / 5.17 ) 2 = 250.6 k s i$

From Table 10-36A, Cmz = 0.85

Equation 10-42

$f a F a + C m z f b z ( 1 − f a F ′ e z ) F b z + C m y f b y ( 1 − f a F ′ e y ) F b y = 3.26 13.58 + 0.58 ⋅ 20.3 ( 1 − 3.26 250.6 ) 19.8 + 0 = 1.122$

For the end section, use Equation 10.43:

$f a 0.472 F y + f b z F b z + f b y F b y = 3.26 0.472 ( 36 ) + 20.3 19.8 + 0 = 1.217$

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

• Fx = 8.71 kip (compression)
• Mz= 56.5 k-ft

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

$F c z = 50 ( 10 ) 6 C b S x c ( I y c l ) 0.772 J I y c + 9.87 ( d l ) 2 ≤ 0.55 F y$

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

 $F c z = 50 ( 10 ) 6 1.69 33.38789 ( 8.6564 60 ) 0.772 0.28389 8.6564 + 9.87 ( 11.46 60 ) 2 10 − 3 = 227.34 ⁢ k s i$

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:

$F a = F y F . S . [ 1 − ( k L / r ) 2 F y 4 π 2 E ] = 36 2.12 [ 1 − ( 39.92 ) 2 36 4 π 2 29 , 000 ] = 16.13 k s i$

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

 $F e z = π 2 E F . S . ( k L / r ) 2 = π 2 29 , 000 2.12 ( 11.618 ) 2 = 998.5 k s i$

From Table 10-36A, Cmz = 0.85

Equation 10-42

$f a F a + C m z f b z ( 1 − f a F ′ e z ) F b z + C m y f b y ( 1 − f a F ′ e y ) F b y = 1.138 16.13 + 0.58 ⋅ 20.3 ( 1 − 1.138 998.5 ) 19.8 + 0 = 0.942$

For the end section, use Equation 10.43:

$f a 0.472 F y + f b z F b z + f b y F b y = 1.138 0.472 ( 36 ) + 20.3 19.8 + 0 = 1.092$

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

• Fx = 25.5 kip (compression)
• Mz = 112.17 k-ft

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

$F c z = 50 ( 10 ) 6 C b S x c ( I y c l ) 0.772 J I y c + 9.87 ( d l ) 2 ≤ 0.55 F y$

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

 $F c z = 50 ( 10 ) 6 1.75 62.7 ( 22.61 132 ) 0.772 0.917 22.61 + 9.87 ( 12.6 132 ) 2 10 − 3 = 83.19 ⁢ k s i$

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:

$F a = F y F . S . [ 1 − ( k L / r ) 2 F y 4 π 2 E ] = 36 2.12 [ 1 − ( 69.69 ) 2 36 4 π 2 29 , 000 ] = 14.39 k s i$

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

 $F e z = π 2 E F . S . ( k L / r ) 2 = π 2 29 , 000 2.12 ( 22.648 ) 2 = 263.18 k s i$

From Table 10-36A, Cmz = 0.85

Equation 10-42

$f a F a + C m z f b z ( 1 − f a F ′ e z ) F b z + C m y f b y ( 1 − f a F ′ e y ) F b y = 2.024 14.39 + 0.58 ⋅ 21.467 ( 1 − 2.024 263.21 ) 19.8 + 0 = 1.069$

For the end section, use Equation 10.43:

$f a 0.472 F y + f b z F b z + f b y F b y = 2.024 0.472 ( 36 ) + 21.467 19.8 + 0 = 1.203$

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

• Fx = 8.75 kip (tension)
• Mz = 112.17 k-ft

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

$F c z = 50 ( 10 ) 6 C b S x c ( I y c l ) 0.772 J I y c + 9.87 ( d l ) 2 ≤ 0.55 F y$

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

 $F c z = 50 ( 10 ) 6 1.606 62.6 ( 22.61 48 ) 0.772 0.911 22.61 + 9.87 ( 12.6 48 ) 2 10 − 3 = 508.8 ⁢ k s i$

which is larger than 0.55 · FY = 19.8 ksi, so FCZ = 19.8 ksi

Axial Tension

Note: No connection information is specified and no reduction of section is assumed.

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.

$f a 0.472 F y + f b z F b z + f b y F b y = 0.694 0.472 ( 36 ) + 21.467 19.8 + 0 = 1.125$

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

• Fx = 14.02 kip (compression)
• Mz= 57.04 k-ft

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

$F c z = 50 ( 10 ) 6 C b S x c ( I y c l ) 0.772 J I y c + 9.87 ( d l ) 2 ≤ 0.55 F y$

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

 $F c z = 50 ( 10 ) 6 1.81 56.5 ( 12.238 60 ) 0.772 0.5 12.238 + 9.87 ( 15 60 ) 2 10 − 3 = 263.1 ⁢ k s i$

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:

$F a = F y F . S . [ 1 − ( k L / r ) 2 F y 4 π 2 E ] = 36 2.12 [ 1 − ( 39.474 ) 2 36 4 π 2 29 , 000 ] = 16.15 k s i$

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

 $F e z = π 2 E F . S . ( k L / r ) 2 = π 2 29 , 000 2.12 ( 9.231 ) 2 = 1 , 584.4 k s i$

From Table 10-36A, Cmz = 0.85

Equation 10-42

$f a F a + C m z f b z ( 1 − f a F ′ e z ) F b z + C m y f b y ( 1 − f a F ′ e y ) F b y = 1.323 16.149 + 0.85 ⋅ 12.115 ( 1 − 1.323 1 , 584.4 ) 19.8 + 0 = 0.602$

For the end section, use Equation 10.43:

$f a 0.472 F y + f b z F b z + f b y F b y = 1.323 0.472 ( 36 ) + 12.115 19.8 + 0 = 0.689$

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

• Fx = 10.2 kip (compression)
• Mz= 62.96 k-ft

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

$F c z = 50 ( 10 ) 6 C b S x c ( I y c l ) 0.772 J I y c + 9.87 ( d l ) 2 ≤ 0.55 F y$

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

 $F c z = 50 ( 10 ) 6 1.81 56.5 ( 12.238 192 ) 0.772 0.5 12.238 + 9.87 ( 15 192 ) 2 10 − 3 = 287.9 ⁢ k s i$

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:

$F a = π 2 E F . S . ( k L / r ) 2 = π 2 29 , 000 2.12 ( 126.3 ) 2 = 8.46 k s i$

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

 $F e z = π 2 E F . S . ( k L / r ) 2 = π 2 29 , 000 2.12 ( 29.49 ) 2 = 155.2 k s i$

From Table 10-36A, Cmz = 0.85

Equation 10-42

$f a F a + C m z f b z ( 1 − f a F ′ e z ) F b z + C m y f b y ( 1 − f a F ′ e y ) F b y = 0.962 8.46 + 0.85 ⋅ 13.37 ( 1 − 0.926 155.2 ) 19.8 + 0 = 0.691$

For the end section, use Equation 10.43:

$f a 0.472 F y + f b z F b z + f b y F b y = 0.962 0.472 ( 36 ) + 13.37 19.8 + 0 = 0.732$

The critical stress ratio is thus 0.732. The value calculated by STAAD is 0.732

Note: The program gives this value when the slenderness check is suppressed (MAIN 1.0 for member 6); otherwise the member fails as a compression member with a slenderness parameter greater than 120).

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

• Fx = 24.05 kip (tension)
• Mz= 62.96 k-ft

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

Note: No connection information is specified and no reduction of section is assumed.

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.

$f a 0.472 F y + f b z F b z + f b y F b y = 2.268 0.472 ( 36 ) + 13.37 19.8 + 0 = 0.809$

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:

$F a = F y F . S . [ 1 − ( k L / r ) 2 F y 4 π 2 E ] = 36 2.12 [ 1 − ( 106.7 ) 2 36 4 π 2 29 , 000 ] = 10.89 k s i$

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:

$F a = F y F . S . [ 1 − ( k L / r ) 2 F y 4 π 2 E ] = 36 2.12 [ 1 − ( 68.57 ) 2 36 4 π 2 29 , 000 ] = 14.47 k s i$

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

Table 1. Comparison of results
Member Number STAAD.Pro Results Hand Calculation Difference
1 1.216 1.217 none
2 1.091 1.092 none
3 1.207 1.203 none
4 1.130 1.126 none
5 0.691 0.689 none
6 1.052 0.732 43.7% 1
7 0.812 0.808 <1%
8 1.114 1.091 2.1%
9 0.920 0.927 <1%
Note: (1) The ratio is 0.832 when the slenderness check is suppressed, which results in a 13.7% difference.

The file C:\Users\Public\Public Documents\STAAD.Pro CONNECT Edition\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
5 TO 7 UNI Y -2
5 FX 15
1 0.75 2 0.75
PERFORM ANALYSIS
PRINT MEMBER FORCES
PARAMETER 1
CODE AASHTO
TRACK 0 ALL
CHECK CODE ALL
FINISH


                         STAAD.Pro CODE CHECKING - ( AASHTO - ASD)   v1.0
***********************
ALL UNITS ARE - KIP  FEET (UNLESS OTHERWISE Noted)
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.