 # V. SNiP SP16 2017 - I section with UDL

Design an I section subjected to a uniform distributed load per the SP 16.13330.2017 code.

## Details

A 5m long, simply supported beam has a European HD320X127 section. The beam is subjected to a uniform distributed load of 100 kN/m in the Y direction. The steel used has a modulus of elasticity of 206,000 MPa and a Ryn = 235 MPa. γc1 = 1.1, γc2 = 1.1

## Validation

Ry = Ryn/ γm = 223.8 MPa

Rs = 0.58×Ry/ γm = 129.8 MPa

Check for Flexure

Need to satisfy the following equation:

 $M W n , min R y γ c ≤ 1$ (Eq. 41)
where
 M = 100 (5)2 / 8 = 312.5 kN·m

Thus, the ratio is $312.5 1,926.5 × 235 ( 10 ) -3 × 1.1 = 0.63 < 1$

Check for Shear

Need to satisfy the following equation:

 $Q S I t w R s γ c ≤ 1$ (Eq. 42)
where
 Q = 0 kN

Thus, the ratio is 0.0 < 1

Check for Stability

$λ ¯ u b = [ 0.35 + 0.0032 b t + ( 0.76 − 0.02 b t ) b h ] R y ϕ x$
$λ ¯ u b = [ 0.35 + 0.0032 300 20.5 + ( 0.76 − 0.02 300 20.5 ) 300 320 ] 235 147.5 = 1.054$
$λ ¯ b = l e f b R y E 10 0.3 235 206,000 = 1.126 > λ ¯ u b$

So, the stability of the beam is not ensured per Cl. 8.4.4.b. Check per Cl 8.4.1 needs to be performed.

Check for Lateral-Torsional Buckling

Use an effective length of 10 m.

Need to satisfy the following equation:

 $M x ϕ b W c x R y γ c ≤ 1$ (Eq. 69)
 $α = 1.54 I t I y ( l e f h ) 2 = 1.54 225.1 9,239 ( 10 0.32 ) 2 = 36.64$ (Eq. G.4)

Since 0.1 < α < 40, from Table G.1:

$ψ = 1.6 + 0.08 α = 4.531$
 $ϕ 1 = ψ I y I x ( h l e f ) 2 E R y = 1.219$ (Eq. G.4)
$ϕ b = 0.68 + 0.21 ϕ 1 = 0.68 + 0.21 ( 1.219 ) = 0.936$

Thus, the ratio is $312.5 0.936 × 1,926.5 ( 10 ) − 3 × 235 × 1.1 = 0.67 < 1$

Check for Combined Flexure & Shear

Need to satisfy the following equation:

 $0.87 R y γ c σ x 2 − σ x σ y + σ y 2 + 3 τ x y 2 ≤ 1$ (Eq. 44)
where
 σx = Mx / Wx = 312.5 (10)3 / 1,926.5= 162.2 MPa σy = My / Wy = 0 MPa τxy = 0 MPa

Thus, the ratio is $0.87 235 × 1.1 ( 162.2 ) 2 = 0.55 < 1$

Check for Deflection

The maximum member deflection is limited to l / 200 = 0.025 m

Thus, the ratio is 0.0128 / 0.025 = 0.51

## Results

Ratio of Flexure (Eq. 41) 0.63 0.63 none
Ratio of Shear (Eq. 42) 0 0 none
Ratio of LTB (Eq. 69) 0.67 0.67 none
Ratio of Combined Shear & Flexure (Eq. 44) 0.55 0.55 none
Deflection (m) 0.0128 0.01281 negligible
Deflection Ratio 0.51 0.51 none

The file C:\Users\Public\Public Documents\STAAD.Pro CONNECT Edition\Samples\ Verification Models\09 Steel Design\Russia\SNiP SP16 2017 - I section with UDL.std is typically installed with the program.

STAAD SPACE
START JOB INFORMATION
ENGINEER DATE 1-Sep-20
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+08
POISSON 0.3
DENSITY 76.8195
ALPHA 1.2e-05
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
1 UNI GY -100
PERFORM ANALYSIS
PARAMETER 1
CODE RUSSIAN
CMN 8 ALL
CMM 6 ALL
GAMM 2 ALL
LY 10 ALL
LZ 10 ALL
PY 235000 ALL
GAMC2 1.1 ALL
GAMC1 1.1 ALL
TB 1 ALL
DFF 200 ALL
TRACK 2 ALL
CHECK CODE ALL
FINISH


                       STAAD.PRO CODE CHECKING - (SP 16.13330.2017)   V1.0
********************************************
ALL UNITS ARE - KN METRE
========================================================================
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      0.67         1
0.000E+00      3.125E+02    0.000E+00   2.500E+00
1  I      HD320X127     PASS         DISPL         0.51         1
0.000E+00      3.125E+02    0.000E+00   2.500E+00
MATERIAL DATA
Steel                         = User
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              :   1.00E+01    1.00E+01
Slenderness                   :   0.00E+00    0.00E+00
DESIGN DATA (units -kN,m) SP16.13330.2017
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!!!
Stress-strain state checked as:   Class    1
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+00^2)=
5.46E-01=&lt;1
TAUxy/(Rs*GammaC)= 0.000E+00/( 136.3E+03* 1.10E+00)= 0.00E+00=&lt;1
LAMBDA_b=(Lef/b)*SQRT(Ry/E)=
( 100.0E-01/( 3.000E-01))*SQRT( 235.0E+03/ 206.0E+06)= 1.126E+00
SIGMA_x=Mx/(Wc*GammaC)= 312.5E+00/( 192.6E-05* 110.0E-02)= 1.475E+05 kPa
LAMBDA_ub=(0.35+0.0032*b/t+(0.76-0.02*b/t)*b/h)*delta*SQRT(Ry/SIGMA_x)=
=(0.35+0.0032* 1.463E+01+(0.76-0.02* 1.463E+01)* 1.002E+00)* 1.000E+00* 1.262E+00
= 1.092E+00&lt; LAMBDA_b= 1.126E+00
**Warning- Stability of the beam is not ensured according to cl. 8.4.4 b)
F.(69)  Mx/(FIb*Wcx*Ry*GammaC)=
312.5E+00/( 9.36E-01* 1.93E-03* 235.0E+03* 1.10E+00)= 6.70E-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     LOCATION=    2.500