V.SNiP SP16 2011 - I Section with UDL

Verify the design capacity per SP16.13330.2011 of an I section subject to a uniformly distributed load.

Details

Uniformly distributed load of 100 kN/m along the entire span. Member is simply supported span 5m in length. Section used is an HD320x127.

Grade of Steel: S235 - E = 206,000 MPa, f_y = 235 MPa, γ_c1 = 1.1, γ_c2 = 1.1

Validation

Check for Flexure

\frac{M}{W_{n, m i n} R_{y} γ_{c}} \leq 1

[Eqn. 41 of SNiP SP16.13330.2011]

where

M	=	312.5
W_n,min	=	W_x = 1,926.5 cm³
R_y	=	235 MPa

\frac{M}{W_{n, m i n} R_{y} γ_{c}} = 0.628 \leq 1 (OK)

Check for Shear

\frac{Q S}{I t_{w} R_{s} γ_{c}} \leq 1

[Eqn. 42 of SNiP SP16.13330.2011]

where

Q	=	q.l/2 = 250 kN
S	=	1,070 cm³
I	=	30,820 cm⁴
R_s	=	0.58×235 MPa
t_w	=	11.5 mm

\frac{Q S}{I t_{w} R_{s} γ_{c}} = 0.5034 \leq 1 (OK)

Check for Lateral Torsional Buckling

\frac{M_{x}}{ϕ_{b} W_{c x} R_{y} γ_{c}} \leq 1

[Eqn. 43 of SNiP SP16.13330.2011]

where

α	=	$1.54 \frac{I_{t}}{I_{y}} {(\frac{l_{e f}}{h})}^{2} = 9.1603$ [From Eqn. G.4] 0.1 ≤ α ≤ 40 so from Table G.1: ψ = 1.6 + 0.08α = 2.333
ϕ₁	=	$ψ \frac{I_{y}}{I_{x}} {(\frac{h}{l_{e f}})}^{2} \frac{E}{R_{y}} = 0.00078 < 0.85$ from Table G.3: $ϕ_{b} = 0.68 + 0.21 ϕ_{1} = 1.2073 \geq 1.0$ so use $ϕ_{b} = 1.0$

\frac{M_{x}}{ϕ_{b} W_{c x} R_{y} γ_{c}} = \frac{312.5}{1.0 \times 1,926.5 (10^{- 6}) \times 235 (10^{3}) \times 1.1} = 0.628 \leq 1 (OK)

Check for Combined Shear & Flexure

\frac{0.87}{R_{y} γ_{c}} \sqrt{σ_{x}^{2} - σ_{x} σ_{y} + σ_{y}^{2} + 3 τ_{x y}^{2}} \leq 1

[Eqn. 44 of SNiP SP16.13330.2011]

where

σ_x	=	M / W_x = 312.5×10³ / 1,926.5 = 162.2 MPa
σ_y	=	0 (No weak axis bending)
τ_xy	=	QS / (I × t_w) = 0 MPa Note: Moment is maximum at mid-span, where shear is zero.

\frac{0.87}{235 \times 1.1} \sqrt{{162.2}^{2}} = 0.55 \leq 1 (OK)

From Table 11 of SP16 code:

λ_ub = 0.35 + 0.0032×b/t + (0.76-0.02×b/t)×b/h = 0.35 + 0.0032×300/20.5 + (0.76-0.02×300/20.5)×1.0016 = 0.865

[Eqn. 73]

λ_{b} = \frac{I_{e f}}{b} \sqrt{\frac{R_{y}}{E}} = 0.5629

λ_b ≤ λ_ub, OK

Check for Deflection

f_{\max} = \frac{5}{384} \frac{q l^{4}}{E I_{x}} = \frac{5}{384} \frac{100 \times 5^{4}}{206,000 \times 10^{3} \times 30,820 \times 10^{-8}} = 0.0128 m

Maximum member deflect = L/200 = 0.025 m

Ratio = 0.0128 / 0.025 = 0.512

Results

Table 1. Comparison of results
Result Type	Reference	STAAD.Pro	Difference
Cl. 8.2.1 (41)	0.628	0.63	Negligible
Cl. 8.2.1 (44)	0.55	0.55	None
F.69	0.628	0.628	None
Max deflection	0.025	0.025	None
Deflection	0.0128	0.0128	None
Deflection ratio	0.512	0.512	None

STAAD.Pro Input File

The file C:\Users\Public\Public Documents\STAAD.Pro CONNECT Edition\Samples\ Verification Models\09 Steel Design\Russia\SNiP SP16 2011 - I Section with UDL.std is typically installed with the program.

STAAD PLANE
START JOB INFORMATION
ENGINEER DATE 18-Mar-19
END JOB INFORMATION
INPUT WIDTH 79
UNIT METER KN
JOINT COORDINATES
1 0 0 0; 2 5 0 0;
MEMBER INCIDENCES
1 1 2;
DEFINE MATERIAL START
ISOTROPIC STEEL
E 2.05e+008
POISSON 0.3
DENSITY 76.8195
ALPHA 1.2e-005
DAMP 0.03
TYPE STEEL
STRENGTH FY 253200 FU 407800 RY 1.5 RT 1.2
END DEFINE MATERIAL
MEMBER PROPERTY EUROPEAN
1 TABLE ST HD320X127
CONSTANTS
MATERIAL STEEL ALL
SUPPORTS
1 PINNED
2 FIXED BUT FX MZ
LOAD 1 LOADTYPE None  TITLE LOAD CASE 1
MEMBER LOAD
1 UNI GY -100
PERFORM ANALYSIS
PRINT FORCE ENVELOPE NSECTION 2 ALL
PARAMETER 1
CODE RUSSIAN 2011
BEAM 1 ALL
ENSGR 1 ALL
ENMAIN 1 ALL
GAMC2 1.1 ALL
GAMC1 1.1 ALL
TB 1 ALL
DFF 200 ALL
TRACK 2 ALL
CHECK CODE ALL
FINISH

STAAD.Pro Output

                       STAAD.PRO CODE CHECKING - (SP 16.13330.2011)   V2.0
                       ********************************************
   ALL UNITS ARE - KN METRE
   ========================================================================
   MEMBER     CROSS          RESULT/   CRITICAL COND/    RATIO/    LOADING/
              SECTION NO.      N             Mx            My      LOCATION
   ========================================================================
      1  I      HD320X127    PASS     SP cl.8.2.1(41)    0.63         1
                          0.000E+00      3.125E+02    0.000E+00   2.500E+00
      1  I      HD320X127    PASS     SP cl.8.2.1(42)    0.00         1
                          0.000E+00      3.125E+02    0.000E+00   2.500E+00
      1  I      HD320X127    PASS     SP cl.8.2.1(44)    0.55         1
                          0.000E+00      3.125E+02    0.000E+00   2.500E+00
      1  I      HD320X127     PASS      SP cl.8.4.1      0.63         1
                          0.000E+00      3.125E+02    0.000E+00   2.500E+00
      1  I      HD320X127     PASS      SP cl.DISPL      0.51         1
                          0.000E+00      3.125E+02    0.000E+00   2.500E+00
   MATERIAL DATA
      Steel                         = S235        EN10025-2
      Modulus of elasticity         = 206.E+06 kPa
      Design Strength (Ry)          = 235.E+03 kPa
   SECTION PROPERTIES (units - m, m^2, m^3, m^4)
      Member Length                 = 5.00E+00
      Gross Area                    = 1.61E-02
      Net Area                      = 1.61E-02
                                         x-axis      y-axis
      Moment of inertia (I)         :   308.E-06    924.E-07
      Section modulus (W)           :   193.E-05    616.E-06
      First moment of area (S)      :   107.E-05    470.E-06
      Radius of gyration (i)        :   138.E-03    757.E-04
      Effective Length              :   5.00E+00    5.00E+00
      Slenderness                   :   0.00E+00    0.00E+00
   DESIGN DATA (units -kN,m) SP16.13330.2011
      Axial force                   :   0.000E+00
                                         x-axis      y-axis
      Moments                       :   312.5E+00    0.000E+00
      Shear force                   :   0.000E+00    0.000E+00
      Bi-moment                     :   0.000E+00 Value of Bi-moment not being entered!!!
      Class of beam cross-section   :   1 class
   CRITICAL CONDITIONS FOR EACH CLAUSE CHECK
      F.(41)  M/(Wn,min*Ry*GammaC)= 312.5E+00/( 1.93E-03* 235.0E+03* 1.10E+00= 6.28E-01=&lt;1
      F.(44)  0.87/(Ry*GammaC)*SQRT(SIGMx^2+3*TAUxy^2)=
              0.87/( 235.0E+03* 1.10E+00)*SQRT(-162.2E+03^2+3* 0.000E+00)=
              5.46E-01=&lt;1
              TAUxy/(Rs*GammaC)= 0.000E+00/( 136.3E+03* 1.10E+00)= 0.00E+00=&lt;1
      F.(69)  M/(FIb*Wcx*Ry*GammaC)=
              312.5E+00/( 1.00E+00* 1.93E-03* 235.0E+03* 1.10E+00)= 6.28E-01=&lt;1
          LIMIT SPAN/DEFLECTION (DFF) =    200.00   (DEFLECTION LIMIT=      0.025 M)
          SPAN/DEFLECTION = 390.4E+00 (DEFLECTION=  1.281E-02M)
          LOAD=    1     RATIO=    0.512