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)

意味

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.

表 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 2023\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 | 
 +---------------------------------------------------------------------+