V. AISC LRFD - Non Compact Wide Flange 3

To determine the design bending strength of a non-compact American wide flange section bent about its strong axis.

Reference

AISC Load Factor Resistance Design, 3rd Edition, Example 5.2, case (b), page 5-16.

Problem

W21x48 shape

F_y = 50 ksi

Length = 40 ft

L_b = 160 in.

Validation

The plastic limiting unbraced length is given by:

L_{p} = 1.76 r_{y} \sqrt{\frac{E}{F_{y}}} = 1.76 (1.66) \sqrt{\frac{29,000}{50}} = 70.4 in.

The inelastic limiting unbraced length is given by:

L_{r} = \frac{r_{y} X_{1}}{F_{L}} \sqrt{1 + \sqrt{1 + X_{2} {F_{L}}^{2}}}

(F1-6)

where

F_L	=	F_yf - F_r = 50 ksi - 10 ksi = 40 ksi
X₁	=	$\frac{π}{S_{x}} \sqrt{\frac{E G J A}{2}} = \frac{π}{93.0} \sqrt{\frac{29,000 (11,200) (0.803) (14.1)}{2}} = 1,449$
X₂	=	$4 \frac{C_{w}}{I_{y}} {(\frac{S_{x}}{G J})}^{2} = 4 (\frac{3,950}{38.7}) {[\frac{93.0}{11,200 (0.803)}]}^{2} = 0.0437$

L_{r} = \frac{1.66 \times 1,449}{40} \sqrt{1 + \sqrt{1 + 0.0437 {(40)}^{2}}} = 184.6 in.

L_p < L_b < L_r; thus, the section is inelastic.

The nominal flexural strength is:

M_{n} = C_{b} [M_{p} - (M_{p} - M_{r}) (\frac{L_{b} - L_{p}}{L_{r} - L_{p}})] < M_{p}

(F1-2)

M_r = F_LS_x = 40 × 93.0 = 3,720 in·k

(F1-7)

M_p = F_yZ_x = 50 × 107 = 5,350 in·k

Conservatively assume C_b = 1.0 (this is the default behavior in STAAD.Pro when the parameter is not directly specified).

M_{n} = 5,350 - (5,350 - 3,720) (\frac{160 - 70.4}{184.6 - 70.4}) = 4,071 in·k

The lateral-torsional buckling limit state capacity:

ϕ_bM_n = 0.9×(4,071) = 3,664 in·k

Comparison

The reported STAAD.Pro value is based on the term MNZ in the STAAD.Pro output.

Table 1. Comparison of results
Result Type	Reference	STAAD.Pro	Difference	Comments
Design flexural strength, ϕ_bM_n (in·kips)	3,664	3,658.52	negligible

STAAD Input

The file C:\Users\Public\Public Documents\STAAD.Pro CONNECT Edition\Samples\ Verification Models\09 Steel Design\US\AISC\AISC LRFD - Non Compact Wide Flange 3.STD is typically installed with the program.

STAAD SPACE BENDING CAPACITY PER AISC LRFD 3RD ED
START JOB INFORMATION
ENGINEER DATE 23-Sep-18
END JOB INFORMATION
*
*  EXAMPLE PROBLEM 5.2, CASE (B), PAGE 5-16, AISC LRFD 3RD ED.
*  CAPACITY (MNZ) SHOULD BE ABOUT 306 KIP-FT
*
UNIT FEET KIP
JOINT COORDINATES
1 0 0 0; 2 40 0 0;
*
MEMBER INCIDENCES
1 1 2;
*
MEMBER PROPERTY AMERICAN
1 TABLE ST W21X48
*
DEFINE MATERIAL START
ISOTROPIC MATERIAL1
E 4.176e+06
POISSON 0.3
END DEFINE MATERIAL
CONSTANTS
MATERIAL MATERIAL1 ALL
*
SUPPORTS
1 PINNED
2 FIXED BUT MZ
*
LOAD 1
MEMBER LOAD
1 CON GY -10 13.333
1 CON GY -10 26.667
*
PERFORM ANALYSIS
*
PARAMETER 1
CODE LRFD
MAIN 1 ALL
FYLD 7200 ALL
UNT 13.333 ALL
TRACK 1 ALL
CHECK CODE ALL
FINISH

STAAD Output

                         STAAD.Pro CODE CHECKING - (LRFD 3RD EDITION)   v1.0
                         ***********************
 ALL UNITS ARE - KIP  FEET (UNLESS OTHERWISE Noted)
 MEMBER     TABLE       RESULT/   CRITICAL COND/     RATIO/     LOADING/
                          FX            MY             MZ       LOCATION
 =======================================================================
      1  ST   W21X48                   (AISC SECTIONS)
                           PASS     LRFD-H1-1B-C       0.437         1
                        0.00 C          0.00        -133.33       13.33
 +---------------------------------------------------------------------+
 | DESIGN STRENGTHS FOR MEMBER     1       UNITS - KIP IN              |
 | PNC=   35.84  PNT=  634.50  MNZ= 3658.52  MNY=  636.75  VN=  194.67 |
 +---------------------------------------------------------------------+