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 R_yn = 235 MPa. γ_c1 = 1.1, γ_c2 = 1.1

Validation

R_y = R_yn/ γ_m = 223.8 MPa

R_s = 0.58×R_y/ γ_m = 129.8 MPa

Check for Flexure

Need to satisfy the following equation:

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

(Eq. 41)

where

100 (5)² / 8 = 312.5 kN·m

Thus, the ratio is $\frac{312.5}{1,926.5 \times 235 {(10)}^{-3} \times 1.1} = 0.63 < 1$

Check for Shear

Need to satisfy the following equation:

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

(Eq. 42)

where

0 kN

Thus, the ratio is 0.0 < 1

Check for Stability

ϕ_{x} = \frac{M_{x}}{W_{c} \times γ_{c}} = \frac{312.5}{1,926.5 \times {(10)}^{-3} \times 1.1} = 147.5 MPa

{\bar{λ}}_{u b} = [0.35 + 0.0032 \frac{b}{t} + (0.76 - 0.02 \frac{b}{t}) \frac{b}{h}] \sqrt{\frac{R_{y}}{ϕ_{x}}}

{\bar{λ}}_{u b} = [0.35 + 0.0032 \frac{300}{20.5} + (0.76 - 0.02 \frac{300}{20.5}) \frac{300}{320}] \sqrt{\frac{235}{147.5}} = 1.054

{\bar{λ}}_{b} = \frac{l_{e f}}{b} \sqrt{\frac{R_{y}}{E}} \frac{10}{0.3} \sqrt{\frac{235}{206,000}} = 1.126 > {\bar{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:

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

(Eq. 69)

α = 1.54 \frac{I_{t}}{I_{y}} {(\frac{l_{e f}}{h})}^{2} = 1.54 \frac{225.1}{9,239} {(\frac{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 \frac{I_{y}}{I_{x}} {(\frac{h}{l_{e f}})}^{2} \frac{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 $\frac{312.5}{0.936 \times 1,926.5 {(10)}^{- 3} \times 235 \times 1.1} = 0.67 < 1$

Check for Combined Flexure & Shear

Need to satisfy the following equation:

\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

(Eq. 44)

where

σ_x	=	M_x / W_x = 312.5 (10)³ / 1,926.5= 162.2 MPa
σ_y	=	M_y / W_y = 0 MPa
τ_xy	=	0 MPa

Thus, the ratio is $\frac{0.87}{235 \times 1.1} \sqrt{{(162.2)}^{2}} = 0.55 < 1$

Check for Deflection

f_{\max} = \frac{5}{384} \frac{q l^{4}}{E I_{x}} = \frac{5}{384} \frac{100 \times 5^{4}}{206 {(10)}^{6} 30,820 {(10)}^{- 8}} = 0.0128 m

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

Thus, the ratio is 0.0128 / 0.025 = 0.51

Results

Result Type	Reference	STAAD.Pro	Difference
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

STAAD.Pro Input

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
LOAD 1 LOADTYPE None  TITLE LOAD CASE 1
MEMBER LOAD
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 Output

                       STAAD.PRO CODE CHECKING - (SP 16.13330.2017)   V1.0
                       ********************************************
   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      0.67         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