Week 5 Challenge
Design of Plate Girder
Q1. Shear resistance of web problem
Determine PG Web thickness for given load without stiffener
For a PG of size 1200 x 300 x 25x 10, Check if the web is adequate to carry a factored shear force of 1000 kN. Yield strength of plate fy is 250 Mpa.
Aim & Introduction- Checking of adequacy of web thickness of a PG for a given factored load.
- PG size 1200x300x25x10
- Factored shear force = 1000 KN
- fyfy= 250 Nmm2mm2
Procedure-
1) dtwdtw = (1200 -2x50)/10 =111 > 67 εε (ε=1)(ε=1)
check for shear buckling required
2) Minimum thickness requirment of web-
dtwdtw< 200εε
11100/10 = 111 < 200εε ( 0K)
3) Shear capacity Check -
Vd=Av×fyw/√3×γmoVd=Av×fyw/3×γmo
= 1100x10x250/1.732x1.1 = 1443418 N = 1443 KN
0.6VdVd = 865 KN < V = 1000 KN
Not safe in shear
4) Check for shear buckling as per IS 800: 2007 , Page-59 -
Vcr=Vcr=A_v xxtau_B`
shear stress correspondind to web buckling
τB=fyw√3×λ2wτB=fyw3×λw2
λ2w=√fyw√3×τcrλw2=fyw3×τcr
τcr=Kv×π2×E/12(1−μ2)(dtw)2τcr=Kv×π2×E/12(1-μ2)(dtw)2
KvKv = 5.35 when transverse stiiffners are provided only at support
E = 2x105N/mm2105N/mm2 , μispoisson'sratio=0.3μispoisson′sratio=0.3
τcrτcr = 5.35x π2π2 x 2x105105/ 12 ( 1- 0.320.32) (110010)2(110010)2
= 79.84
λwλw = √fyw√3×τcrfyw3×τcr
=√2501.732×79.842501.732×79.84 = 1.3445
τBτB = 250/1.732x 1.344521.34452 = 79.849
VcrVcr = 1100x10 x 79.849 = 878339 N = 878.339 < 1000 KN
The web shear force corresponding to the web buckling is less than maximum Factored shear force , hence not safe
Result- Thus it is concluded that web is not adequate to carry a factored shear force of 1000 KN
Q-2. Check whether intermediate stiffener required for given load
For a PG of size 1500 x 500 x 30 x 12,check if the intermediate web stiffener is required to carry a factored shear force of 2000 kN, if required, find the spacing of stiffener. Yield strength of plate fy is 250 Mpa.
Aim & Introduction - To check whether intermediate web stiffener is required for a PG of size 1500x500x30x12 carrying a factored shear force of 2000 KN. Spacing of stiffener is also to be found out in case it is required
- Size of PG is 1500x500x30x12
- Factored shear force = 2000 KN
- fy=250N/mm2fy=250N/mm2
Procedure-
1) dtwdtw = (1500-2x30)/12 = 120 > 67εε
shear buckling check requred
2) Minimum web thickness requirment-
dtwdtw 120 < 200 εε (ε=1)(ε=1) ( OK)
3) Shear capacity Check -
Vd=Av×fyw/√3×γmoVd=Av×fyw/3×γmo
= 1440x12x250/1.732x1.1 = 2267478.47 N = 2267. 47 KN
0.6VdVd = 1360.48 KN < V = 2000 KN ( Not safe in shear)
4) Check for shear buckling as per IS 800: 2007 , Page-59
Vcr=Vcr=A_v xxtau_B`
shear stress correspondind to web buckling
τb=fyw√3×λ2wτb=fyw3×λw2
λ2w=√fyw√3×τcrλw2=fyw3×τcr
τcr=Kv×π2×E/12(1−μ2)(dtw)2τcr=Kv×π2×E/12(1-μ2)(dtw)2
KvKv = 5.35 when transverse stiiffners are provided only at support
E = 2x105N/mm2105N/mm2 , μispoisson'sratio=0.3μispoisson′sratio=0.3
τcrτcr = 5.35x π2π2 x 2x105105/ 12 ( 1- 0.320.32) (144012)2(144012)2
= 67.09
λwλw = √fyw√3×τcrfyw3×τcr
=√2501.732×67.092501.732×67.09 = 1.466
τbτb = 250/1.732x 1.46621.4662 = 67.16
VcrVcr = 1440x12 x 67.16 = 1160524.8 N = 1160.52 < 2000 KN
5 ) Check for the requirment of siffener-
If dtwdtw> 85 then vertical stiffener is required
If dtwdtw> 200 then longitudinal stiffener is required
If dtwdtw> 250 then additional vertical stiffener is required
λλ = dtwdtw = (1500-2x30)/12 = 120 > 85εε
Only vertical stiffener need to be provided
6 ) Check for bearing capacity ( as per clause 8.7.4 0f IS 800: 2007 Page-67)
Fw=(b1+n2)tw×fyw/γmoFw=(b1+n2)tw×fyw/γmo
b1b1= 2(2.5×tf)2(2.5×tf) = 2(2.5x 30) = 150 mm
`F_w = (0.0 + 150)x 12 x250/1.1 = 409090.9 = 409 KN < 2000 KN
Web is not capable to resist the factored shear force so stiffeners are required
Considering the thickness tqtq of stiffener as 12 mm ( page 65, IS 800)
outstand L0=14tw×εL0=14tw×ε= 14x12 = 168 mm
Outstand of the face of the web on either side of stiffener = 20tqεtqε
= 2x20x12x1 = 480 mm
Core area = 2x 168 x12+ 480x12 = 9792 mm2mm2
IcoreIcore= 480x 123123/12+ 2{ 12x 16831683/12 + 168x12x902902}
= 42.211x106mm4106mm4
rmin=√IArmin=IA √42.211×106/979242.211×106/9792 = 65.65
Slenderness ratio KL/r = 0.7x 1440/65.65 = 15.35
fcdfcd from table 9c of IS 800 :2007
By Interpolation fcdfcd= 225.395 N/mm2mm2
Fc=Acore×fcdFc=Acore×fcd
= 9792x 225.395 = 2207067.84 N = 2207.067 KN > VumaxVumax =2000 KN
Hence safe
7) Spacing of intermediate stiffener (C) -
3d≥C≤d3d≥C≤d
Let C /d= 1.4 =
C = 1.4x 1440 = 2016 say 2000 mm c/c
3x1440 ≥≥ 2000 ≤≤1440
4320≥2000≤14404320≥2000≤1440 (OK)
C/d = 2000/1440 = 1.38 < √22
Is=1.5d3×t3w/C2Is=1.5d3×tw3/C2
= 1.5X14403×123/2000214403×123/20002 = 19.402×106mm4×106mm4
IprovidedIprovided = 12×14403/1212×14403/12 = 29.859
Result - Thus it is concluded that intermediate stiffeners are required for the given PG to resist shear force of 200 KN and c/c spacing of stiffener is 2000 mm
Q-3. Check whether intermediate stiffener required for given load
For the above problem, determine the web thickness required to avoid the stiffeners.
Aim & Introduction - Web thickness is to be calculated in order to avoid stiffeners
- Size of PG is 1500x500x30x12
- Factored shear force = 2000 KN
- fy=250N/mm2fy=250N/mm2
Procedure-
Procedure-
1) dtwdtw = (1500-2x30)/12 = 120 > 67εε
shear buckling check requred
2) Minimum web thickness requirment-
dtwdtw 120 < 200 εε (ε=1)(ε=1) ( OK)
3) Shear capacity Check -
Vd=Av×fyw/√3×γmoVd=Av×fyw/3×γmo
= 1440x12x250/1.732x1.1 = 2267478.47 N = 2267. 47 KN
0.6VdVd = 1360.48 KN < V = 2000 KN ( Not safe in shear)
4) Check for shear buckling as per IS 800: 2007 , Page-59
Vcr=Vcr=A_v xxtau_B`
shear stress correspondind to web buckling
τB=fyw√3×λ2wτB=fyw3×λw2
λ2w=√fyw√3×τcrλw2=fyw3×τcr
τcr=Kv×π2×E/12(1−μ2)(dtw)2τcr=Kv×π2×E/12(1-μ2)(dtw)2
KvKv = 5.35 when transverse stiiffners are provided only at support
E = 2x105N/mm2105N/mm2 , μispoisson'sratio=0.3μispoisson′sratio=0.3
τcrτcr = 5.35x π2π2 x 2x105105/ 12 ( 1- 0.320.32) (146012)2(146012)2
= 65.27
λwλw = √fyw√3×τcrfyw3×τcr
; =√2501.732×65.272501.732×65.27 = 1.487
τBτB = 250/1.732x 1.48721.4872 = 65.278
VcrVcr = 1440x12 x 65.278 = 1128003.84 N = 1128.003 < 2000 KN
6 ) Check for bearing capacity ( as per clause 8.7.4 0f IS 800: 2007 Page-67)
Fw=(b1+n2)tw×fyw/γmoFw=(b1+n2)tw×fyw/γmo
b1b1= 2(2.5×tf)2(2.5×tf) = 2(2.5x 30) = 150 mm
`F_w = (0.0 + 150)x 12 x250/1.1 = 409090.9 = 409 KN < 2000 KN
Web is unsafe in shear capacity
Thickness is required to be increased in order to to resist the factored shear force .
Let us assume the thickness of web be 16 mm
Check for shear buckling as per IS 800: 2007 , Page-59
Vcr=Vcr=A_v xxtau_b`
shear stress correspondind to web buckling
τb=fyw√3×λ2wτb=fyw3×λw2
λ2w=√fyw√3×τcrλw2=fyw3×τcr
τcr=Kv×π2×E/12(1−μ2)(dtw)2τcr=Kv×π2×E/12(1-μ2)(dtw)2
KvKv = 5.35 when transverse stiiffners are provided only at support
E = 2x105N/mm2105N/mm2 , μispoisson'sratio=0.3μispoisson′sratio=0.3
τcrτcr = 5.35x π2π2 x 2x105105/ 12 ( 1- 0.320.32) (144016)2(144016)2
= 119.27
λwλw = √fyw√3×τcrfyw3×τcr
; =√2501.732×119.272501.732×119.27 = 1.10
when 0.8 < λwλw< 1.2
then τb=[1−0.8(λw−0.8)](fyw√3)τb=[1-0.8(λw-0.8)](fyw3)
τbτb = [ 1- 0.8 ( 1.10- 0.8) 250/1.732] = 109.699
VcrVcr= 1440x16 x 109.7 = 2527488 N = 2527.488> VmaxVmax= 2000 KN
Now the web is safe in shear
Result- It is concluded that thickness of the web of PG is required to be increased to to 16 mm from 12 mm for making it to resist the given factored shear force of 2000 KN
Q-4. Proportioning problem
Calculate the minimum depth & web thickness requirement of a beam of span 15 m to carry a UDL of 100 kN/m. Also calculate the flange depth & thickness considering section to be plastic and check if the assumed size is adequate to carry the moment.
Aim & Introduction - (1) Minimum depth and web thickness of a beam with given span and UDL is required to be found out.
(2) Flange depth and thickness of assumed section is also to be found out to carry the moment.
- span of the beam = 15 m
- UDL =100 KN/m
- section is plastic
- Assume steel grade Fe 415
- γmoγmo =1.1
Procedure -
1) Load calculation-
Factored Load ( UDL) = 1.5x100 =150 KN/m
Self weight of plate girder = wl/200 = 100x15x15 /400 = 56.25
self wt of girder/m length = 56.25/15 = 3.75 KN/m
Factored self weight = 1.5x3.75 =5.625 KN/m
Total factored load = 150 +5.625 = 155.625 KN/m
2) Calculation of maximum BM and SF
Maximum Bending moment MumaxMumax = wl2/8wl2/8= 155.625x 15x15/8
= 4376.95 KNm
Maximum Shear force VumaxVumax= wl/2 = 155.625x15/2 = 1167.18 KN
3) Design of web-
Optimum depth of web plate dw=(Mk/fy)0.33dw=(Mk/fy)0.33
k=dtwk=dtw
If k=dtwk=dtw >200 then intermediate transverse stiffeners are required
To avoid the transverse stiffeners asuume k=dtwk=dtw=180
dw=(4376.95×106×180/250)0.33dw=(4376.95×106×180/250)0.33 = 1363.032 mm
provide dwdw =1500 mm
Thickness of web tw=(Mk2fy)0.33tw=(Mk2fy)0.33
= (4376.95×106/1802×250)0.33(4376.95×106/1802×250)0.33
= 7.976 mm say 10 mm
Assume web plate 1500x 10 mm
4) Design of flange--
Area of the flange Af=M×γmo/fy×dwAf=M×γmo/fy×dw
= 4376.95 x106106x1.1/250x1400 = 13756.12 mm2mm2
From table 2 page 18 0f S 800:2007
To keep the flange in plastic ategory
btf≤8.4btf≤8.4
b2tf≤8.4b2tf≤8.4
b = 16.8xtftf
Af=bf×tfAf=bf×tf
13756.12 = 16.8x tf×tftf×tf
tftf = 28.6 mm sa y 30 mm
Width of flange = AftfAftf = 13756.12/30 = 458.53 mm say 500 mm
Assume flange size 500x 30 mm
5) SECTION CLASSIFICATION-
btfbtf= 250/30 = 8.333 < 8.4 ( flange is Plastic)
5) Check for moment capacity ( cl 8,2.1.2, page-53 of is 800:2007)
Assuming VumaxVumax < 0.6Vd0.6Vd
Vd=Avfyw(√3γmo)Vd=Avfyw(3γmo) ( cl. 8.4.1, page-59, IS 800: 2007)
1x 1500x10x250/1.732x1.1 = 1968297.29 N = 1968.29 KN
0.6VdVd = 1180.97 KN > VumaxVumax = 1167.18 KN
Md=βbZpfyγmoMd=βbZpfyγmo≤≤ ZefyγmoZefyγmo
HereMdMd β=1β=1 for plastic section
Zp=2×bf×tf×D−tf2Zp=2×bf×tf×D-tf2
= 2x 500x30x (1500+30)/2 = 22.95×106mm322.95×106mm3
MdMd= 1x 22.95x106106x250/1.1 = 5215.90 x106N−mm106N-mm
MdMd = 5215.90 KNm > Max Bending moment =4376.95 KNm
Result - It is concluded that our Plate girder is capable to bear the maximum Bending moment
Q-5. Design of Load carrying stiffener
Design a intermediate stiffener in ISMB 500 beam to carry a load of 1000 kN to resist exceeded force to its shear capacity.
Aim & Introduction - Designing of Intermediate stiffener in ISMB 500 beam for a load of 1000KN to resist exceeded force to its shear capacity.
- Beam is ISMB 500
- Load =100 KN
- Assume steel grade Fe 415
- γmoγmo =1.1
Procedure - Shear force = 1000KN
Design/Factored Shear force = 1.5x 1000 = 1500 KN
1) Section properties of ISMB 500-
bfbf = 180 mm , tftf= 17.2 mm , twtw= 10.2 mm
A = 110.74cm2cm2 , h = 500 mm
2) Shear capacity Check -
Vd=Av×fyw/√3×γmoVd=Av×fyw/3×γmo
= (500- 2x17.2)x 10.2 x250/1.732x1.1 = 623178.66 N = 623.17 KN
0.6VdVd = 373.90 KN < V = 15000 KN ( Not safe in shear)
Transverse transverse web stiffener are required
3) Design of web stiffener-
Shear force to be resisted by Web stiffener = 1500- 623 .17 = 876.83 KN
Let us assume the size of stiffener as 300x 10 mm
AvAv= 2x350x10 = 7000 mm2mm2
Vd=Av×fyw/√3×γmoVd=Av×fyw/3×γmo
= 7000x 250/1.732x 1.1 = 918.53 KN > 876.83 ( safe)
Result- It is cincluded that for exceeded shear force over and above the given factored force of 1500 KN , Intermediate stiffener of size 350x 10 on both side of web is required for resisting it safely.