# V. AASHTO 2nd Ed LRFD - Design Beam

The following compares the solution of a design performed using STAAD.Pro against a hand calculation of steel design per the AASHTO (LRFD) code.

## Reference

The following step by step hand calculation as per AASHTO LRFD(1998) code.

## Problem

Determine the allowable resistances (AASHTO LRFD, 1998 code) for the member of the structure as shown in figure. Also, perform a code check for the member based on the results of the analysis.

A 48 foot long, non-composite girder is assumed to be simply supported. The section is a plate girder with 16" x 1" plate flanges and a 34" x 3.6" plate web (36 in. total depth). All plates are grade 36 steel.

The beam is subject to a 9.111 kip/ft uniform load.

## Solution

 A = 154.4 in2, Iz = 21,594 in4, Iy = 814.9 in4

 Sz = 21,594(2)/36 = 1,200 in3

 $r y = I y A = 814.9 154.4 = 2.30 i n$

 $r z = I z A = 21 , 594 154.4 = 11.83 i n$

From observation Load case 1 will govern,

Mz = 2,624 kip-ft

## Axial Compression Capacity

Refer Clause 6.9.4 of the code.

 (kL/r)y = 0.333 × 576/ 2.297 = 83.50

 (kL/r)z = 1 × 576/ 11.826 = 48.71

 (kL/r)crit = 83.50 < 120, ok.

Calculation of Width/Thickness ratio for axial compression

Plate buckling coefficients taken from Table 6.9.4.2-1:

kw = 1.49, kf = 0.56

Slenderness ratio for the web:

(d - 2·tf)/tw = (36 - 2 · 1) / 3.6 = 9.444

Slenderness ratio for the 1/2 flange:

bf/(2·tf) = 16/(2 · 1) = 8.0

Critical ratio for web:

$k w E F y = 1.49 29 , 000 36 = 42.29 ⁢ > 9.444$

Critical ratio for flange:

$k f E F y = 0.56 29 , 000 36 = 15.89 > 8.0$

Thus, OK [AASHTO LRFD Cl. 6.4.9.2]

Slenderness ratio about major and minor axis:

 $λ z = ( K l r z π ) 2 F y E = ( 1.0 ⋅ 576 11.83 π ) 2 36 29 , 000 = 0.298$

 $λ y = ( K l r y π ) 2 F y E = ( 0.333 ⋅ 576 2.30 π ) 2 36 29 , 000 = 0.877$

λy governs, thus λ = 0.877

λ < 2.25, so Equation 6.9.4.1-1 is used to determine the nominal compressive resistance

Pn = 0.66λFyAs = 0.660.877·36·154.4 = 3,861 kips

The factored compressive resistance, Pr = φcPn = 0.9 · 3,861 = 3,475 kips

## Major Axis Bending Capacity

The compression flange moment of inertia:

 Iyc = 1(16)3/12 = 341.3 in4

 Iyc/Iy = 341.3/814.9 = 0.419, > 0.1 and < 0.9

Thus, OK [AASHTO LRFD Cl. 6.10.2.1]

$2 D o t w = 2 ( 17 ) 3.6 = 9.445 < 6.77 E F y = 6.77 29 , 000 36 = 192.1$

Thus, OK [AASHTO LRFD Cl. 6.10.2.2]

Calculation of depth of the web in compression at the plastic moment, Dcp(Clause 6.10.3.3.2)

Area of Web, Aw = (36 – 2 × 1) × 3.6 = 122.4 in2

Area of flange area in tension, Aft = Area of flange in compression, Afc = 16 × 1 = 16 in2

$D c p = ( D w 2 A w F y ) F y ( A f t + A w − A f c ) = ( 34 2 ⋅ 122.4 ⋅ 36 ) 36 ( 16 + 122.4 − 16 ) = 17 i n$
$2 D c p t w = 2 ( 17 ) 3.6 = 9.445 < 3.76 E F y = 3.76 29 , 000 36 = 108.1$

Thus, OK [AASHTO LRFD Cl. 6.10.4.1.2]

$b f t f = 16.0 2 ( 1.0 = 8.0 < 0.382 E F y = 0.382 29 , 000 36 = 10.98$

Thus, OK [AASHTO LRFD Cl. 6.10.4.3]

Check clause 6.10.4.1.6a

 (B/t)flange  < 0.75 × (B/t)flange_limit = 0.75 ×10.978

 (D/T)web < 0.75 × (D/T)web_limit =0.75×108.057

Check clause 6.10.4.1.7

 Mpz   = Pz × Fy= 1,600 in3(36 ksi) = 57,614 in-k

 $L b = [ 0.124 − 0.0759 ( M l / M p z ) ] × [ r y E F y ] = [ 0.124 − 0.0759 ( 0 / 57 , 614 ) ] × [ 2.30 ( 29 , 000 ) 36 ] = 229.5 i n$

Unsupported length, Lu = 576 in > Lb.; section is non-compact.

Check clause 6.10.4.1.9

A notional section comprised of the compression flange and one-third of the depth of the web in compression, taken about the vertical axis.

 $A r t = A c f + ( c − t f 3 ) t w = 16 + ( 18 − 1.0 3 ) 3.6 = 36.4 i n 2$

 $I r t = t f [ ( c − t f 3 ) t w 3 12 ] = 1 [ ( 18 − 1 3 ) 3.6 3 12 ] = 363.4 i n 4$

 $r r t = 363.4 36.4 = 3.149 i n$

 $L p = 1.76 ( 3.159 ) 29 , 000 36 = 159.8 i n$

 Lb > Lp

Clause 6.10.4.2.6

$L r = 4.44 I y c d S x c E F y = 4.44 341.3 ( 36 ) 1 , 200 29 , 000 36 = 403.3 i n$

Minimum radius of gyration of the compression flange taken about the vertical axis:

$r t = I y c A c f = 341.3 16 = 4.619 i n$

Hybrid factor, Rh = 1.0 (For Homogeneous sections, Hybrid factors shall be taken as 1.0 per clause 6.10.4.3.1)

As per clause 6.10.4.3.2, Load-shedding factor, Rb

If area of the compression flange, Acf ≥ the area of the tension flange, Atf

 lb= 5.76

 Dc = 17.00

 $2 D c t w = 2 ( 17 ) 3.6 = 9.445 < L b E F y = 5.76 29 , 000 36 = 163.5$

Thus:

Rb,comp = Rb,ten = 1.0

Lr < Lb

Cb  = 1.75 + 1.05(M1/M2)+0.3x(M1/M2)2

Here M1 = 0, M2 = 0 so Cb = 1.75

 $M n z , c o m p = C b z R b , c o m p R h M y 2 ( L r L b ) 2 = 1.75 ( 1.0 ) ( 1.0 ) [ 36 ( 1 , 200 ) 2 ] ( 408.4 576 ) 2 = 19 , 000 k i p ⋅ i n$

 $M n z , t e n = R b , t e n R h F y Z = 1.0 ( 1.0 ) ( 36 ) ( 1 , 200 ) = 43 , 188 k i p ⋅ i n$

 Mnz = 19,000 kip·in

Resisting Moment

Mr = Qf · Mn= 1.0 (19,000) = 19,000 kip·in

Actual Moment = 31,488 kip·in

Interaction ratio = 31,488 / 19,000 = 1.657

## Comparison

Table 1. Comparison of results
Value Hand Calculations STAAD.Pro Results Difference
Critical Interaction Ratio 1.657 1.654 none

The following input is used in this verification example.

STAAD SPACE
START JOB INFORMATION
ENGINEER DATE 23-May-11
END JOB INFORMATION
INPUT WIDTH 79
UNIT FEET KIP
JOINT COORDINATES
1 0 0 0; 2 38 0 0;
MEMBER INCIDENCES
1 1 2;
START USER TABLE
TABLE 1
UNIT INCHES KIP
WIDE FLANGE
AASHTOGIRDER
154.4 36 3.6 16 1 21593.8 814.86 539.4 129.6 32
END
UNIT FEET KIP
DEFINE MATERIAL START
ISOTROPIC STEEL
E 29000
POISSON 0.3
DENSITY 0.000283
ALPHA 6e-06
DAMP 0.03
TYPE STEEL
STRENGTH FY 36 FU 58 RY 1.5 RT 1.2
END DEFINE MATERIAL
MEMBER PROPERTY AMERICAN
1 UPTABLE 1 AASHTOGIRDER
CONSTANTS
MATERIAL STEEL ALL
SUPPORTS
1 PINNED
2 FIXED BUT FX MY MZ
UNIT INCHES KIP
1 UNI GY -0.76
PERFORM ANALYSIS
PARAMETER 1
CODE AASHTO LRFD
TRACK 2 ALL
CHECK CODE ALL
FINISH


                       STAAD.PRO CODE CHECKING - (   AASHTO - LRFD)   v1.0
********************************************
*    1  ST  AASHTOGIRDER             (UPT)
FAIL      Slenderness        1.654         1
0.00            0.00           0.00        0.00
Section Properties (in)
-----------------------
Ax =     154.40   Ay =     129.60   Az =      32.00
Iz =   21593.80   Iy =     814.86
Rz =      11.83   Ry =       2.30
Sz =    1199.66   Sy =     101.86
Input Parameters (Kip-in)
--------------------------
Fyld  =      36.00   Fu    =      58.00
Lz    =     456.00   Ly    =     456.00
Kz    =       1.00   Ky    =       1.00
UNL   =       0.00   Ratio =       1.00
Design Results (Kip-in)
-----------------------
Klrz =      38.56   Klry =     198.49
CBy  =       0.00   CBz  =       0.00
CAPACITIES (Kip-in)
--------------------
Compressive Capacity =       0.00
Tensile Capacity     =       0.00
Moment Capacity_y    =       0.00
Moment Capacity_z    =       0.00
Shear Capacity       =       0.00
DESIGN FORCES (Kip-in)
-----------------------
Compressive Force    =       0.00
Tensile Force        =       0.00
Moment Y             =       0.00
Moment Z             =       0.00
Shear Force          =     173.28