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:

a uniform gravity load along the beam, w = 2 kips/ft (dead + live)
a lateral wind load, F = 15 kips
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} = \sqrt{\frac{2 π^{2} E}{F_{y}}} = \sqrt{\frac{2 π^{2} 29, 000}{36}} = 126.1

Member 1

Size W 12X26, L = 10 ft., a = 7.65 in², S_z = 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 · F_Y = 19.8 ksi

Bending Major Axis

F_{c z} = \frac{50 {(10)}^{6} C_{b}}{S_{x c}} (\frac{I_{y c}}{l}) \sqrt{0.772 \frac{J}{I_{y c}} + 9.87 {(\frac{d}{l})}^{2}} \leq 0.55 F_{y}

Where:

C_b = 1.75 + 1.05(M1/M2)+0.3x(M1/M2)²

M1 = 0, so C_b = 1.75
S_zc= Section modulus with respect to the compression flange =204/(0.5 · 12.22) = 33.38789 in³

I_YC= t_b 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} = \frac{50 {(10)}^{6} 1.75}{33.38789} (\frac{8.6564}{120}) \sqrt{0.772 \frac{0.28389}{8.6564} + 9.87 {(\frac{11.46}{120})}^{2}} 10^{- 3} = 64.375 k s i

which is larger than 0.55 · F_Y = 19.8 ksi, so F_CZ = 19.8 ksi

Axial Compression

Critical (kL/r) = 1.0 · 120/1.5038 = 79.7978

As (kL/r) < C_c , the allow axial stress in compression is given by:

F_{a} = \frac{F_{y}}{F . S .} [1 - \frac{{(k L / r)}^{2} F_{y}}{4 π^{2} E}] = \frac{36}{2.12} [1 - \frac{{(79.79)}^{2} 36}{4 π^{2} 29, 000}] = 13.58 k s i

Actual Stress

Actual axial stress, f_a= 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 = f_bz = 56.5 · 12/33.4 = 1.692 · 12 = 20.3 ksi

F_{e z} = \frac{π^{2} E}{F . S . {(k L / r)}^{2}} = \frac{π^{2} 29, 000}{2.12 {(120 / 5.17)}^{2}} = 250.6 k s i

From Table 10-36A, C_mz = 0.85

Equation 10-42

\frac{f_{a}}{F_{a}} + \frac{C_{m z} f_{b z}}{(1 - \frac{f_{a}}{{F'}_{e z}}) F_{b z}} + \frac{C_{m y} f_{b y}}{(1 - \frac{f_{a}}{{F'}_{e y}}) F_{b y}} = \frac{3.26}{13.58} + \frac{0.58 \cdot 20.3}{(1 - \frac{3.26}{250.6}) 19.8} + 0 = 1.122

For the end section, use Equation 10.43:

\frac{f_{a}}{0.472 F_{y}} + \frac{f_{b z}}{F_{b z}} + \frac{f_{b y}}{F_{b y}} = \frac{3.26}{0.472 (36)} + \frac{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 · F_Y = 19.8 ksi

Bending Major Axis

F_{c z} = \frac{50 {(10)}^{6} C_{b}}{S_{x c}} (\frac{I_{y c}}{l}) \sqrt{0.772 \frac{J}{I_{y c}} + 9.87 {(\frac{d}{l})}^{2}} \leq 0.55 F_{y}

Where:

C_b = 1.75 + 1.05(M1/M2)+0.3x(M1/M2)²

M1 = 39.44 and M2 = 677.96, so C_b = 1.69
S_zc= Section modulus with respect to the compression flange =204/(0.5 · 12.22) = 33.38789 in³

I_YC= t_b 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} = \frac{50 {(10)}^{6} 1.69}{33.38789} (\frac{8.6564}{60}) \sqrt{0.772 \frac{0.28389}{8.6564} + 9.87 {(\frac{11.46}{60})}^{2}} 10^{- 3} = 227.34 k s i

which is larger than 0.55 · F_Y = 19.8 ksi, so F_CZ = 19.8 ksi

Axial Compression

Critical (kL/r) = 1.0 · 60/1.504 = 39.92

As (kL/r) < C_c , the allow axial stress in compression is given by:

F_{a} = \frac{F_{y}}{F . S .} [1 - \frac{{(k L / r)}^{2} F_{y}}{4 π^{2} E}] = \frac{36}{2.12} [1 - \frac{{(39.92)}^{2} 36}{4 π^{2} 29, 000}] = 16.13 k s i

Actual Stress

Actual axial stress, f_a= 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 = f_bz = 56.5 · 12/33.4 = 1.691 · 12 = 20.3 ksi

(KL/r)_z= 1 · 60/5.16 = 11.618

F_{e z} = \frac{π^{2} E}{F . S . {(k L / r)}^{2}} = \frac{π^{2} 29, 000}{2.12 {(11.618)}^{2}} = 998.5 k s i

From Table 10-36A, C_mz = 0.85

Equation 10-42

\frac{f_{a}}{F_{a}} + \frac{C_{m z} f_{b z}}{(1 - \frac{f_{a}}{{F'}_{e z}}) F_{b z}} + \frac{C_{m y} f_{b y}}{(1 - \frac{f_{a}}{{F'}_{e y}}) F_{b y}} = \frac{1.138}{16.13} + \frac{0.58 \cdot 20.3}{(1 - \frac{1.138}{998.5}) 19.8} + 0 = 0.942

For the end section, use Equation 10.43:

\frac{f_{a}}{0.472 F_{y}} + \frac{f_{b z}}{F_{b z}} + \frac{f_{b y}}{F_{b y}} = \frac{1.138}{0.472 (36)} + \frac{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

F_x = 25.5 kip (compression)
M_z = 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 · F_Y = 19.8 ksi

Bending Major Axis

F_{c z} = \frac{50 {(10)}^{6} C_{b}}{S_{x c}} (\frac{I_{y c}}{l}) \sqrt{0.772 \frac{J}{I_{y c}} + 9.87 {(\frac{d}{l})}^{2}} \leq 0.55 F_{y}

Where:

C_b = 1.75 + 1.05(M1/M2)+0.3x(M1/M2)²

M1 = 0, so C_b = 1.75
S_zc= Section modulus with respect to the compression flange = 428/(0.5 · 13.66) = 62.7 in³

I_YC= 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} = \frac{50 {(10)}^{6} 1.75}{62.7} (\frac{22.61}{132}) \sqrt{0.772 \frac{0.917}{22.61} + 9.87 {(\frac{12.6}{132})}^{2}} 10^{- 3} = 83.19 k s i

which is larger than 0.55 · F_Y = 19.8 ksi, so F_CZ = 19.8 ksi

Axial Compression

Critical (kL/r) = 1.0 · 132/1.894 = 69.69

As (kL/r) < C_c , the allow axial stress in compression is given by:

F_{a} = \frac{F_{y}}{F . S .} [1 - \frac{{(k L / r)}^{2} F_{y}}{4 π^{2} E}] = \frac{36}{2.12} [1 - \frac{{(69.69)}^{2} 36}{4 π^{2} 29, 000}] = 14.39 k s i

Actual Stress

Actual axial stress, f_a= 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 = f_bz = 112.17 · 12/62.7 = 1.789 · 12 = 21.467 ksi

(KL/r)_z= 1 · 132/5.828 = 22.649

F_{e z} = \frac{π^{2} E}{F . S . {(k L / r)}^{2}} = \frac{π^{2} 29, 000}{2.12 {(22.648)}^{2}} = 263.18 k s i

From Table 10-36A, C_mz = 0.85

Equation 10-42

\frac{f_{a}}{F_{a}} + \frac{C_{m z} f_{b z}}{(1 - \frac{f_{a}}{{F'}_{e z}}) F_{b z}} + \frac{C_{m y} f_{b y}}{(1 - \frac{f_{a}}{{F'}_{e y}}) F_{b y}} = \frac{2.024}{14.39} + \frac{0.58 \cdot 21.467}{(1 - \frac{2.024}{263.21}) 19.8} + 0 = 1.069

For the end section, use Equation 10.43:

\frac{f_{a}}{0.472 F_{y}} + \frac{f_{b z}}{F_{b z}} + \frac{f_{b y}}{F_{b y}} = \frac{2.024}{0.472 (36)} + \frac{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

F_x = 8.75 kip (tension)
M_z = 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 · F_Y = 19.8 ksi

Bending Major Axis

F_{c z} = \frac{50 {(10)}^{6} C_{b}}{S_{x c}} (\frac{I_{y c}}{l}) \sqrt{0.772 \frac{J}{I_{y c}} + 9.87 {(\frac{d}{l})}^{2}} \leq 0.55 F_{y}

Where:

C_b = 1.75 + 1.05(M1/M2)+0.3 · (M1/M2)²

M1 = -191.36 Kip-in , M2 = -1346.08 Kip-in so C_b= 1.606

I_YC= 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} = \frac{50 {(10)}^{6} 1.606}{62.6} (\frac{22.61}{48}) \sqrt{0.772 \frac{0.911}{22.61} + 9.87 {(\frac{12.6}{48})}^{2}} 10^{- 3} = 508.8 k s i

which is larger than 0.55 · F_Y = 19.8 ksi, so F_CZ = 19.8 ksi

Axial Tension

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

F_t = 0.55 · F_Y = 19.8 ksi

Actual Stress

Actual axial stress, f_a= 8.75 / 12.6 = 0.694 ksi

Actual bending stress = f_bz = 112.17 · 12/62.7 = 1.789 · 12 = 21.47 ksi, which exceeds F_CZ.

f_bz/F_CZ = 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.

\frac{f_{a}}{0.472 F_{y}} + \frac{f_{b z}}{F_{b z}} + \frac{f_{b y}}{F_{b y}} = \frac{0.694}{0.472 (36)} + \frac{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 · F_Y = 19.8 ksi

Bending Major Axis

F_{c z} = \frac{50 {(10)}^{6} C_{b}}{S_{x c}} (\frac{I_{y c}}{l}) \sqrt{0.772 \frac{J}{I_{y c}} + 9.87 {(\frac{d}{l})}^{2}} \leq 0.55 F_{y}

Where:

C_b = 1.75 + 1.05(M1/M2)+0.3 · (M1/M2)²

M1 = 40.14, M2 = -684.4 so C_b= 1.81

S_zc= Section modulus with respect to the compression flange = 448/(0.5 · 15.86) = 56.5 in in³

I_YC= 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} = \frac{50 {(10)}^{6} 1.81}{56.5} (\frac{12.238}{60}) \sqrt{0.772 \frac{0.5}{12.238} + 9.87 {(\frac{15}{60})}^{2}} 10^{- 3} = 263.1 k s i

which is larger than 0.55 · F_Y = 19.8 ksi, so F_CZ = 19.8 ksi

Axial Compression

Critical (kL/r) = 1.0 · 60/1.52 = 69.69

As (kL/r) < C_c , the allow axial stress in compression is given by:

F_{a} = \frac{F_{y}}{F . S .} [1 - \frac{{(k L / r)}^{2} F_{y}}{4 π^{2} E}] = \frac{36}{2.12} [1 - \frac{{(39.474)}^{2} 36}{4 π^{2} 29, 000}] = 16.15 k s i

Actual Stress

Actual axial stress, f_a= 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 = f_bz = 57.04 · 12/56.5 = 1.001 · 12 = 12.115 ksi

(KL/r)_z= 1 · 60/6.5= 9.231

F_{e z} = \frac{π^{2} E}{F . S . {(k L / r)}^{2}} = \frac{π^{2} 29, 000}{2.12 {(9.231)}^{2}} = 1, 584.4 k s i

From Table 10-36A, C_mz = 0.85

Equation 10-42

\frac{f_{a}}{F_{a}} + \frac{C_{m z} f_{b z}}{(1 - \frac{f_{a}}{{F'}_{e z}}) F_{b z}} + \frac{C_{m y} f_{b y}}{(1 - \frac{f_{a}}{{F'}_{e y}}) F_{b y}} = \frac{1.323}{16.149} + \frac{0.85 \cdot 12.115}{(1 - \frac{1.323}{1, 584.4}) 19.8} + 0 = 0.602

For the end section, use Equation 10.43:

\frac{f_{a}}{0.472 F_{y}} + \frac{f_{b z}}{F_{b z}} + \frac{f_{b y}}{F_{b y}} = \frac{1.323}{0.472 (36)} + \frac{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 · F_Y = 19.8 ksi

Bending Major Axis

F_{c z} = \frac{50 {(10)}^{6} C_{b}}{S_{x c}} (\frac{I_{y c}}{l}) \sqrt{0.772 \frac{J}{I_{y c}} + 9.87 {(\frac{d}{l})}^{2}} \leq 0.55 F_{y}

Where:

C_b = 1.75 + 1.05(M1/M2)+0.3 · (M1/M2)²

M1 = 8.947 M2 = 183.05 so C_b= 1.69

S_zc= Section modulus with respect to the compression flange = 448/(0.5 · 15.86) = 56.5 in in³

I_YC= 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} = \frac{50 {(10)}^{6} 1.81}{56.5} (\frac{12.238}{192}) \sqrt{0.772 \frac{0.5}{12.238} + 9.87 {(\frac{15}{192})}^{2}} 10^{- 3} = 287.9 k s i

which is larger than 0.55 · F_Y = 19.8 ksi, so F_CZ = 19.8 ksi

Axial Compression

Critical (kL/r) = 1.0 · 192/1.52 = 126.3

As (kL/r) > C_c , the allow axial stress in compression is given by:

F_{a} = \frac{π^{2} E}{F . S . {(k L / r)}^{2}} = \frac{π^{2} 29, 000}{2.12 {(126.3)}^{2}} = 8.46 k s i

Actual Stress

Actual axial stress, f_a= 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 = f_bz = 62.96 · 12/56.5 = 1.114 · 12 = 13.37 ksi

(KL/r)_z= 1 · 192/6.51= 29.49

F_{e z} = \frac{π^{2} E}{F . S . {(k L / r)}^{2}} = \frac{π^{2} 29, 000}{2.12 {(29.49)}^{2}} = 155.2 k s i

From Table 10-36A, C_mz = 0.85

Equation 10-42

\frac{f_{a}}{F_{a}} + \frac{C_{m z} f_{b z}}{(1 - \frac{f_{a}}{{F'}_{e z}}) F_{b z}} + \frac{C_{m y} f_{b y}}{(1 - \frac{f_{a}}{{F'}_{e y}}) F_{b y}} = \frac{0.962}{8.46} + \frac{0.85 \cdot 13.37}{(1 - \frac{0.926}{155.2}) 19.8} + 0 = 0.691

For the end section, use Equation 10.43:

\frac{f_{a}}{0.472 F_{y}} + \frac{f_{b z}}{F_{b z}} + \frac{f_{b y}}{F_{b y}} = \frac{0.962}{0.472 (36)} + \frac{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 · F_Y = 19.8 ksi

Bending Major Axis

FCZ = 0.55 · F_Y = 19.8 ksi

Axial Tension

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

F_a = 0.55 · F_Y = 19.8 ksi

Actual Stress

Actual axial stress, f_a= 24.05 /10.6 = 2.268 ksi, hence, ok.

Actual bending stress = f_bz = 62.96 · 12/56.5 = 1.114 · 12 = 13.37 ksi

So the combined ratio is

f_a/F_a + f_bz/F_TZ + f_by/F_TY = 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.

\frac{f_{a}}{0.472 F_{y}} + \frac{f_{b z}}{F_{b z}} + \frac{f_{b y}}{F_{b y}} = \frac{2.268}{0.472 (36)} + \frac{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

F_x = 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) < C_c , the allow axial stress in compression is given by:

F_{a} = \frac{F_{y}}{F . S .} [1 - \frac{{(k L / r)}^{2} F_{y}}{4 π^{2} E}] = \frac{36}{2.12} [1 - \frac{{(106.7)}^{2} 36}{4 π^{2} 29, 000}] = 10.89 k s i

Actual Stress

Actual axial stress, f_a= 23.04 /1.938 = 11.88 ksi

f_a/F_a = 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

F_x = 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) < C_c , the allow axial stress in compression is given by:

F_{a} = \frac{F_{y}}{F . S .} [1 - \frac{{(k L / r)}^{2} F_{y}}{4 π^{2} E}] = \frac{36}{2.12} [1 - \frac{{(68.57)}^{2} 36}{4 π^{2} 29, 000}] = 14.47 k s i

Actual Stress

Actual axial stress, f_a= 48.44 /3.61 = 13.42 ksi

f_a/F_a = 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.

STAAD Input

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
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.