1. Shear resistance of web problem: Determine PG Web thickness for given load without stiffener.
Aim : Checking of adequacy of web thickness of a PG for a given factored load.
Given: PG size 1200x300x25x10, Factored shear force = 1000 KN, fy</math>">fy= 250 Nmm2</math>">mm^2
Solution:
1. d/tw= (1200 -2x50)/10 =111 > 67 ε</math>">ε (ε=1)</math>">(ε=1)
check for shear buckling required
2. Minimum thickness requirment of web
dtw</math>">d/tw< 200 xss=removed tabindex="0" role="presentation">ε</math>">ε
11100/10 = 111 < 200 xss=removed tabindex="0" role="presentation">ε</math>">ε ( 0K)
3. Shear capacity Check
Vd=Av×fyw/3×γmo</math>">Vd=Av×fyw/√3×γmo = 1100x10x250/1.732x1.1 = 1443418 N = 1443 KN
0.6Vd</math>">Vd = 865 KN < V>
Not safe in shear
4. Check for shear buckling as per IS 800: 2007 , Page-59
Vcr=</math>">Vcr=A_v xxtau_B`
shear stress correspondind to web buckling
τB=fyw3×λw2</math>">τB=fyw/√3×λ^2w
λw2=fyw3×τcr</math>">λ^2w=√fyw/√3×τcr
Kv</math>">Kv= 5.35 when transverse stiiffners are provided only at support
E = 2x105N/mm2</math>">10^5N/mm^2,
μispoisson′sratio=0.3</math>">μispoisson'sratio=0.3
τcr</math>">τcr = = 79.84
λw</math>">λ^2w = fyw3×τcr</math>">√(fyw/√3×τcr) =2501.732×79.84</math>">√2501.732×79.842501.732×79.84 = 1.3445
τB</math>">τB = 250/1.732x 1.34452</math>">1.344521.34452 = 79.849
Vcr</math>">Vcr = 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
2. Check whether intermediate stiffener required for given load.
Aim: 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
Given: Size of PG is 1500x500x30x12, Factored shear force = 2000 KN, fy=250N/mm2</math>">fy=250N/mm2
Solution:
1. d/tw = (1500-2x30)/12 = 120 > 67ε</math>">ε
shear buckling check requred
2. Minimum web thickness requirment
dtw</math>">d/tw 120 < 200 xss=removed tabindex="0" role="presentation">ε</math>">ε (ε</math>">ε=1) ( OK)
3. Shear capacity Check
Vd=Av×fyw/3×γmo</math>">Vd=Av×fyw/√3×γmo= 1440x12x250/1.732x1.1 = 2267478.47 N = 2267. 47 KN
0.6Vd</math>">Vd= 1360.48 KN < V>
4. Check for shear buckling as per IS 800: 2007 , Page-59
Vcr=</math>">Vcr=A_v xxtau_B`
shear stress correspondind to web buckling
τb=fyw3×λw2</math>">τb=fyw/(√3×λ2w)
λw2=fyw3×τcr</math>">λ^2w=√(fyw/√3×τcr)
τcr=Kv×π2×E/12(1-μ2)(dtw)2</math>">τcr=Kv×π2×E/12(1−μ2)(d/tw)^2
Kv</math>">Kv = 5.35 when transverse stiiffners are provided only at support
E = 2x105N/mm2</math>">105N/mm^2, μispoisson′sratio=0.3</math>">μispoisson'sratio=0.3
τcr</math>">τcr = 5.35x π2</math>">π^2 x 2x105</math>">105105/ 12 ( 1- 0.32</math>">0.320.32) (144012)2</math>">(144012)^2(144012)^2 = 67.09
λw</math>">λw = fyw3×τcr</math>">√(fyw/√3×τcr) =2501.732×67.09</math>">√2501.732×67.092501.732×67.09 = 1.466
τb</math>">τb = 250/1.732x 1.4662</math>">1.46621.4662 = 67.16
Vcr</math>">Vcr = 1440x12 x 67.16 = 1160524.8 N = 1160.52 < 2000 KN
5. Check for the requirment of siffener
If dtw</math>">d/tw> 85 then vertical stiffener is required
If dtw</math>">d/tw> 200 then longitudinal stiffener is required
If dtw</math>">d/tw> 250 then additional vertical stiffener is required
λ</math>">λ =dtw</math>">d/tw = (1500-2x30)/12 = 120 > 85ε</math>">ε
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/γmo</math>">Fw=(b1+n2)tw×fyw/γmo
b1</math>">b1 = 2(2.5x 30) = 150 mm
Fw = (0.0 + 150)x 12 x250/1.1 = 409090.9 = 409 KN < 2000>
Web is not capable to resist the factored shear force so stiffeners are required
Considering the thickness tq</math>">tq of stiffener as 12 mm ( page 65, IS 800)
outstand L0=14tw×ε</math>">L0=14tw×ε=14= 14x12 = 168 mm
Outstand of the face of the web on either side of stiffener = 20tqε</math>">tqε = 2x20x12x1 = 480 mm
Core area = 2x 168 x12+ 480x12 = 9792 mm2</math>">mm^2
I c(or)e= 480x 123</math>">123123/12+ 2{ 12x 1683</math>">16831683/12 + 168x12x902</math>">902902} = 42.211x106mm4</math>">10^6mm^4
rmin=IA</math>">rmin=√I/A (42.211×106/9792</math>">√42.211×106/979242.211×10^6/9792 )= 65.65
Slenderness ratio KL/r = 0.7x 1440/65.65 = 15.35
fcd</math>">fcdfrom table 9c of IS 800 :2007
By Interpolation fcd</math>">fcd= 225.395 N/mm2</math>">mm^2
Fc=A c(or)e×fcd= 9792x 225.395 = 2207067.84 N = 2207.067 KN > Vumax</math>">Vu max=2000 KN
Hence safe
7. Spacing of intermediate stiffener (C)
3d≥C≤d</math>">3d≥C≤d3
Let C /d= 1.4
C = 1.4x 1440 = 2016 say 2000 mm c/c
3x1440 ≥</math>">≥≥ 2000 ≤</math>">≤≤1440
4320≥2000≤1440</math>">4320≥2000≤14404320≥2000≤1440 (OK)
C/d = 2000/1440 = 1.38 < 2</math>">√2
Is=1.5d3×tw3/C2</math>">I(s)=1.5d^3×tw^3/C^2=1.5×14403×123/20002</math>">1440^3×12^3/2000^2 = 19.402×106mm4</math>">×10^6 mm^4
Iprovided</math>">I (provided)= 12×14403/12</math>">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
3. Check whether intermediate stiffener required for given load.
Aim : Web thickness is to be calculated in order to avoid stiffeners
Given: Size of PG is 1500x500x30x12 ,Factored shear force = 2000 KN, fy=250N/mm2</math>">fy=250N/mm^2
Solution:
1. d/tw = (1500-2x30)/12 = 120 > 67ε</math>">ε
shear buckling check requred
2. Minimum web thickness requirment
dtw</math>">(d/tw)120 < 200 xss=removed tabindex="0" role="presentation">ε</math>">ε (ε=1)</math>">(ε=1) ( OK)
3. Shear capacity Check
Vd=Av×fyw/3×γmo</math>">Vd=Av×fyw/√3×γmo= 1440x12x250/1.732x1.1 = 2267478.47 N = 2267. 47 KN
0.6Vd</math>">Vd = 1360.48 KN < V>
4. Check for shear buckling as per IS 800: 2007 , Page-59
Vcr=</math>">Vcr=A_v xxtau_B
shear stress correspondind to web buckling
τB=fyw3×λw2</math>">τB=fyw/√3×λ2w
λw2=fyw3×τcr</math>">λ2w=√(fyw/√3×τcr)
τcr=Kv×π2×E/12(1-μ2)(dtw)2</math>">τcr=Kv×π^2×E/12(1−μ2)(dtw)^2
Kv</math>">Kv = 5.35 when transverse stiiffners are provided only at support
E = 2x105N/mm2</math>">10^5N/mm^2 , μispoisson′sratio=0.3</math>">μispoisson'sratio=0.3
τcr</math>">τcr= 5.35x π2</math>">π^2 x 2x105</math>">105/ 12 ( 1- 0.32</math>">0.3^2) (146012)2</math>">(146012)^2= 65.27
λw</math>">λw = fyw3×τcr</math>">√fyw√3×τcr=2501.732×65.27</math>">√2501.732×65.272501.732×65.27 = 1.487
τB</math>">τB= 250/1.732x 1.4872</math>">1.48721.4872 = 65.278
Vcr</math>">Vcr = 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/γmo</math>">Fw=(b1+n2)tw×fyw/γmo
b1</math>">b1= 2(2.5×tf)</math>">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>
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=</math>">Vcr==A_v xxtau_b`
shear stress correspondind to web buckling
τb=fyw3×λw2</math>">τb=fyw/√3×λ2w
λw2=fyw3×τcr</math>">λ2w=√(fyw/√3×τcr)
τcr=Kv×π^2×E/12(1−μ^2)(d/tw)^2
Kv</math>">Kv= 5.35 when transverse stiiffners are provided only at support
E = 2x105N/mm2</math>">10^5 N/mm^2, μispoisson′sratio=0.3</math>">μispoisson'sratio=0.3
τcr</math>">τcr = 5.35x π2</math>">π^2x2x105</math>">10^5/ 12 ( 1- 0.32</math>">0.3^2) (144016)2</math>">(144016)^2 = 119.27
λw</math>">λw = fyw3×τcr</math>">√(fyw√3×τcr)=2501.732×119.27</math>">√(250/1.732×119.272501.732×119.27) = 1.10
when 0.8 < λw</math>">λw< 1>
τb=[1-0.8(λw-0.8)](fyw3)</math>">τb=[1−0.8(λw−0.8)](fyw√3)= [ 1- 0.8 ( 1.10- 0.8) 250/1.732] = 109.699
Vcr</math>">Vcr= 1440x16 x 109.7 = 2527488 N = 2527.488> Vmax</math>">Vmax= 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
4. Proportioning problem.
Aim :
- Minimum depth and web thickness of a beam with given span and UDL is required to be found out.
- Flange depth and thickness of assumed section is also to be found out to carry the moment.
Given: span of the beam = 15 m, UDL =100 KN/m, section is plastic, Assume steel grade Fe 415, γmo</math>">γmo=1.1
Solution:
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 Mumax</math>">Mumax = wl2/8</math>">wl2/8= 155.625x 15x15/8 = 4376.95 KNm
Maximum Shear force Vumax</math>">Vumax= wl/2 = 155.625x15/2 = 1167.18 KN
3. Design of web
Optimum depth of web plate dw=(Mk/fy)0.33</math>">dw=(Mk/fy)0.33
k=dtw</math>">k=d/tw
If k=dtw</math>">k=d/tw=>200 then intermediate transverse stiffeners are required
To avoid the transverse stiffeners asuume k=dtw</math>">k=d/tw=180
dw=(4376.95×106×180/250)0.33</math>">dw=(4376.95×106×180/250)0.33=(4376.95×106×180/250)0.33 = 1363.032 mm
provide dw</math>">dw=1500 mm
Thickness of web tw=(Mk2fy)0.33</math>">tw= (4376.95×106/1802×250)0.33</math>">(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×dw</math>">Af=M×γmo/fy×d= 4376.95 x106</math>">106106x1.1/250x1400 = 13756.12 mm2</math>">mm^2
From table 2 page 18 0f S 800:2007
To keep the flange in plastic ategory
btf≤8.4</math>">b/tf≤8.4
b2tf≤8.4</math>">b/2tf≤8.4
b = 16.8xtf</math>">tf
Af=bf×tf</math>">Af=bf×tf=13756.12 = 16.8x tf×tf</math>">tf×tf
tf</math>">tf= 28.6 mm sa y 30 mm
Width of flange = Aftf</math>">Af/tf= 13756.12/30 = 458.53 mm say 500 mm
assume flange size 500x 30 mm
4.1. Section classification
btf</math>">b/tf= 250/30 = 8.333 < 8>
5. Check for moment capacity ( cl 8,2.1.2, page-53 of is 800:2007)
Assuming Vumax</math>">Vumax < 0.6Vd</math>">0.6Vd
Vd=Avfyw(3γmo)</math>">Vd=Avfyw/(√3γmo) ( cl. 8.4.1, page-59, IS 800: 2007)
1x 1500x10x250/1.732x1.1 = 1968297.29 N = 1968.29 KN
0.6Vd</math>">Vd = 1180.97 KN > Vumax</math>">Vumax = 1167.18 KN
Md=βbZpfyγmo</math>">Md=βbZpfy/γmo <= Zefyγmo</math>">Zefy/γmo
Md</math>">Md β=1</math>">β=1 for plastic section
Zp=2×bf×tf×D-tf2</math>">Zp=2×bf×tf×D−tf2 = 2x 500x30x (1500+30)/2 = 22.95×106mm3</math>">22.95×106mm322.95×106
Md</math>">Md= 1x 22.95x106</math>">10^6x250/1.1 = 5215.90 x106N-mm</math>">10^6N−mm10^6
Md</math>">Md = 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.