Validation Problem on Torsion
| Applies To | |||
| Product(s): | STAAD.Pro | ||
| Version(s): | All | ||
| Environment: | N/A | ||
| Area: | Wiki | ||
| Subarea: | Torsion Validation Problem as per EN 1993-1-1:2005 | ||
| Original Author: | Shayan Roy,Technical Support Engineer(Structural), Bentley Kolkata | ||
1. Introduction:
The following is a validation problem of a detailed torsion check including the warping stresses on a UB 203x203x60 which has been subjected to a uniformly distributed load across the full span and a torsional moment at the centre of the span.
The detailed torsion check in the EN 1993-1-1:2005 implementation in STAAD.Pro has been invoked by using the TOR 2 parameter.
The combination of the uniformly distributed load and the torsional moment resembles the load acting at a distance from the shear center.
The implementation calculates the torsional stresses and capacities as per P057. Please refer to STAAD.Pro EN 1993-1-1:2005 implementation: Calculation of Torsional Capacities of Cross Sections as per SCI P057 for details on this.
2 . Validation Problem:
.
A British I section beam (UB 203x203x60, S275 grade steel) , 4 m long fixed at both the ends has been subjected to a uniformly distributed load of 1 KN/m , a concentrated point load 1000 KN at the midpoint and a concentrated torque of 7.5 KNm at the midpoint.
The maximum Bending Moment (Mz) at the end is 51.33 KNm and at mid point is 50.7 KNm.
Maximum Shear Force (VY) is 52KN is at the supports.
3.1 . Text Input Information.
The following is the text input information of the Staad model:
STAAD SPACE
START JOB INFORMATION
ENGINEER DATE 07-Oct-13
END JOB INFORMATION
INPUT WIDTH 79
UNIT METER KN
JOINT COORDINATES
1 0 0 0; 2 4 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 BRITISH
1 TABLE ST UC203X203X60
CONSTANTS
MATERIAL STEEL ALL
SUPPORTS
1 2 FIXED
LOAD 1 LOADTYPE None TITLE LOAD CASE 1
MEMBER LOAD
1 CMOM GX 7.5 2
1 CON GY -100 2
1 UNI GY -1
PERFORM ANALYSIS
PRINT MEMBER PROPERTIES
PARAMETER 1
CODE EN 1993-1-1:2005
SGR 1 ALL
CMT 3 ALL
TORSION 2 ALL
TRACK 2 ALL
CHECK CODE ALL
FINISH
3.2 . Manual Calculation:
Section: UB 203x203x60
Sectional Properties as per Bluebook
A = 76.4 cm²; D = 209.6 mm ; b = 205.8 mm ; t f = 14.2 mm; t w = 9.2 mm; r = 10.2 mm
Zz = 584 cm2 ; Zy = 201 cm2 ; Iz = 6120 cm4 ; Iy = 2060 cm4
J = 47.2 cm4; H = 1.97x10 11 mm6 ; G = 78846.15
Hence, torsional bending constant, a = sqrt (EH/GJ) =sqrt (205000*1.97*1011 / 78846.15*47.2*104 ) = 1041.72 mm.
A. Calculation of Torsional Properties of the section:
Normalized Warping Function
ByAppendix A, P075, pg 112 ,
Wno = hB/4 = (209.6-14.2)*205.8 / 4 = 10053.3 mm2 .
Warping Statical Moment
Sw1 = hB2T/16 = (195.4*205.8^2*14.2)/6 = 7344862.36 mm4
(In Staad, a member is designed at 13 intermediate sections along the length of the member and the most critical section is reported. For the calculation purpose, we will only design only at z =1m and z =2m.)
B. Validation of the ratio reported for Clause 6.2.7(1):
The critical ratio for this clause is evaluated at 1m from start of the member as per the STAAD Output File.
As the ends are fixed for torsion and warping and the member is subjected to a concentrated torque, we will use Case 5 of Appendix B, P057 (as directed in Table 6.1, P057). This condition is specified by specifying the CMT parameter as 3.
z =1000, a = 1041.72, α= 0.5, αL/a = 1.9199, z/a = 0.9599, L/a = 3.8398
K1 = {(1/0.999*(1-3.483)+1/23.297*(3.483-1)+3.3368-1.9199)} / {1/23.247*(23.269+3.483*23.269-3.483-1)-1.9199-3.3368}
= 0.9989
K2 = 1/0.9989*1/23.247*(3.483-1)+(3.483-23.269+3.8398*23.247)/23.247
= 3.0955
K3 = 0.043+3.3368- 3.483/0.999
= -0.1067
K4 = 3.3368-3.4867+1.009
= 0.851
K5 = 1/0.99891*1/0.999*(1-3.483)+(1-3.483*23.269)/23.247
= -5.931
Now,
Ф’ = (7.5*106) / (78846.15*47.2*104 )*1/1.9989*(0.744*1.1143-1.497+1)
= 3.3476*10-5
Ф’’’ = (7.5*106)/(78846.15*47.2*1041.722 *104 )*1/1.9989*(0.744*1.1143-1.497)
= - 6*10-11
As per equation 2.2 of P057,
Pure torsion component: Tp = GJ Ф’ = 78846.15 * 47.2*104 * 3.3476*10-5 / 106 = 1.25KNm (as reported by Staad)
Warping Torsion component: Tw = -EH Ф’’’ = 205000*1.97*1011 * -6*10-11 / 106 = -2.42KNm (against the value reported as 2.5 by STAAD)
Pure Torsion Capacity:
Tt, Rd= fy/√3 *J/t = 275/√3*(47.4*104)/14.2*1/106
= 5.3 KNm (as reported by Staad)
Warping Torsion Capacity:
Tw, Rd = ( t*b^2*fy)/6 = ( 14.2*205.82*275)/6 1/106
= 27.6KNm (as reported by Staad).
Torsional Resistance TRd = Tt, Rd+ Tw, Rd
Now, Tp/(Tt,Rd) = 1.25/5 = 0.236
And, (Tw )/(Tw,Rd) = 2.42/27.6 = 0.08.
Hence, the critical ratio of 0.236 is reported by Staad for this clause. An extract from code check details of the STAAD Output file is as below:
C. Validation of the ratio calculated for clause 6.2.7(5):
Clause 6.2.7(5) refers back to the yield criterion of clause 6.2.1(5). The critical ratio for this clause is reported at a distance of 2 metres from the start of the member.
Torsion at section z=2m
z =2000, a = 1041.72, α= 0.5, αL/a = 1.9199, z/a = 1.9199, L/a = 3.8398
ф = (7.5*106*1041.72)/(78846.15*47.2*104 )*1/(1+0.9989)*[(0.9989*(-.1067)+0.851)*(2.483)-3.3368+1.9199
= 0.045
Ф’ = (7.5*106)/(78846.15*47.2*104 )*1/1.9989*(0.744*3.3368-3.483+1)
= -4.243*10-8
Ф’’ = (7.5*106)/(78846.15*47.2*1041.72*104 )*1/1.9989*(0.744*3.483-3.3368)
= - 7.2146*10-8
Ф’’’ = (7.5*106)/(78846.15*47.2*1041.722 *104 )*1/1.9989*(0.744*3.3368-3.483)
= -9.295*10-11
As per equation 2.2 of P057,
Pure torsion component: Tp = GJ Ф’ = 78846.15 * 47.2*104 * 0.045 / 106 = 0.0016
Warping Torsion component: Tw = -EH Ф’’’ = 205000*1.97*1011 * -9.295*10-11 / 106 = -3.75KNm
Stress Check at 2m, Check for Elastic Verification as per CL 6.2.1.(5)
The program calculates the stress at 4 points of any cross sections as illustrated:
Let us calculate the stress at mid span, z = 2m from the start of the member.
Point 1:
Bending Stress:
σbz = Mz/Zz = (50.7*106)/(584*103 ) (Zz= Iz/(y/2)
= 86.82 N/mm²
σbyt = Myt/Zy = Ф* Mz/Zy ( CL 2.3, P057, pg 14)
= (50.7*0.0045)/(201*103
= 11.34 N/mm² (Zy= Iz/(y/2)
Warping Stress:
Warping Normal Stress: σwf =- E*Wns*φ” ( CL 2.1.5 of P057, pg 10)
=-205000*10053.3*-7.2146*10^-8 = 148.7 N/mm²
Warping Shear Stress: σtw = 0. (Warping Shear is not produced at end point of flange)
Pure Torsion Stress:
Shear Stress: = Gt Ф’ ( CL 2.1.4 of P057, pg 9)
= 78846.15 * 14.2 * 4.23*10^-8 = 0.0475
Direct Shear Stress
Shear Stress due to Direct Shear V : τb = 0. (Shear Stress due to bending is not produced at the flange end)
The stress check will be performed using equation 6.1 of EN 1993-1-1:2005 as given below:
σxEd = Axial stress + (Bending Stress)z +(Bending Stress)y + Normal Stress due to warping.
τEd =Shear stress due to Direct Shear + Shear stress due to pure torsion + Shear stress due to warping torsion.
σxEd = 86.82 + 11.34 + 148.7 = 246.86 N/mm²
τEd = 0.0475
Hence, for point 1, the ratio is calculated as (246.86/275)2 + 3(0.0475/275)2 = 0.805.
Point 2:
Bending Stress:
σbz = 86.82 N/mm²
σbyt = 11.34 N/mm²
Warping Stress:
Warping Normal Stress: σwf = 0 (Normal warping Stress is 0 at mid of flange)
Warping Shear Stress: σtw = - (E*Wns*φ’’)/t = (-205000*734.486.104*-9.295*10 11 / 14.2 )
= 9.86 N/mm²
Pure Torsion Stress:
Shear Stress: = 0.0475 (Same as point 1)
Direct Shear Stress
Shear Stress due to Direct Shear V : τb =(Vy*Qf) /(Iz*t)
Where Vy is the shear at the section and Q f is the statical moment at the flange.
Q f = (205.8*14.2*97.7)/2 = = 142757 mm3
Hence,
τb= (50.7*103*142757) / (6120*104*14.2) = 8.32 N/mm²
σxEd = 86.82 + 11.34 = 98.16 N/mm²
τEd = 8.32+9.86+0.0475 = 18.23 N/mm²
Hence, for point 2, the ratio is calculated as (98.16/275)2 + 3(18.23/275)2 = 0.14.
Point 3:
Stress at point 3 will be same as in point 2.
Point 4:
Bending Stress:
σbz = 0 ( At mid web, bending stress is always 0)
σbyt = 11.34 N/mm²
Warping Stress:
Warping Normal Stress: σwf = 0 (Normal warping Stress is 0 at web, warping is always at flange)
Warping Shear Stress: σtw = 0 (Warping Shear Stress is 0 at web, warping is always at flange)
Pure Torsion Stress:
Shear Stress: τtw = 0 ( Shear stress at mid web is 0 due to pure torsion)
Direct Shear Stress
Shear Stress due to Direct Shear V: τb =(Vy*Qw) /(Iz*t)
Where Vy is the shear at the section and Q w is the statical moment at the web.
Q w = (205.8*14.2*97.7) + (90.6*9.2*90.6)/2
= 323273 mm3
Hence, τb = (50.7*103*323273)/(6120*104*9.2) = 29.11 N/mm²
σxEd = 11.34 N/mm²
τEd = 29.11 N/mm²
Hence, the ratio for point 4 is calculated as (11.34/275)2 + 3(29.11/275)2 = 0.035.
Hence critical ratio is 0.805 at location 1 of the cross section. The same ratio is reported by Staad. An extract from the STAAD Output File is as below:
D. Validation of the ratio calculated for ratio 6.2.7(9):
As per clause 6.2.7(9) of the code, for combined shear and Torsional moment the plastic shear resistance accounting for Torsional effects should be reduced from Vpl,Rd to V pl,T,Rd and the design shear force should be less than this reduced shear strength. The critical ratio for this clause is calculated at a distance of 1 metre from the start of the section.
Shear Area: Av = A-2btf + (tw+2r) +tf
= 7640-2*(205.8*14.2)+(9.2+2*10.2)*14.2
= 2215.6 mm2
Hence, V pl,Rd = (2215.6*275)/√3
= 351.77 KN
And,
τ t, Ed = Gt Ф’ = 78846.15*14.2*3.347*10-5
= 37.48 Nmm (Critical Section is 1 m from start of member, hence Ф’ corresponding to z =1m is used)
Hence, V pl, T,Rd = √(1-37.48/(1.25 275/√3)) )*351.7 = 316.6 KN = 316.6 KN
V Ed at 1m is 51KN (from Staad)
So,
( V Ed ) / (V pl,T,Rd ) = 51/316.6 = 0.16(against the value reported by Staad as 0.155)
An extract from the output file shows the ratio reported as below:
E. Torsion Check as per CL A.2 of EN 1993-6:2007
This clause is essentially for crane run-away beams and found in EN 1993-6:2007, Annex A. However, this is checked for I sections subjected to bending and torsion in order to assess the interaction between torsion and lateral torsional buckling.
Clause A.1 states that members that are subjected to combined bending and torsion should satisfy:
The critical ratio is reported at the start of the member.
Thus, z = 0.
At z =0, we have Ф = 0.
Where My,ed = Major Axis Moment = 51.3 KNm
My,Rd =Resistance Moment about the major axis = 180.4 KNm
Mz,ED = Minor Axis Moment = Ф x My,ed = 0. (CL 2.3, P057, pg 14)
Tw, ED = Design value of warping torsion = 3.75 KNm
Tw, RK = Warping Torsional resistance moment = 27.5 KNm
kw = 0.7- (0.2*τ w,Ed) /( τ w,Rk /γM1 ) = 0.7- (0.2*3.75)/(27.5/1.0) = 0.6727
kzw = 1- (Mz,Ed/(Mz,Rk/γM1 ) = 1-0 = 1
kα = 1/ (1- (Mz,Ed)/Mz,Cr )) = 1/(1-51.3/288.9) = 1.216 (The value of Mcr =288.9 KNm is taken from Staad result)
And, Mb,RD = χLT Wy fy / γM1
Hence, χLT = 148.4/180.4 = 0.8226 (The Mb,RD , Wy , fy values has been taken from Staad)
So equation A.1 stands to:
(51.3)/(0.8226*180.4) + 0 + (0.6727*1*1.216*3.75)/(27.5)
= 0.46(STAAD reports the ratio as 0.45)
An extract from the output file as below shows the ratio reported:
A pdf version of this article can be downloaded from the following link: