Week 5 Challenge
1 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.
GIVEN DATA
shear capacity v = 1000 kN
fy = 250 n/mm2
vd = Av*fy/ `sqrt3` ymo
= d*tw *fy / `sqrt3`* 1.1
= (1200-2*50)*10 250/`sqrt3` * 1.1 = 1443 kN
since Vd > V , so web section is adequate.
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.
given data
shear force V = 2000 kN
fy =250 n/mm2
h=1500 mm
bf = 500m
tf = 30 mm
tw= 12 mm
d= (1500-2*30) = 1440 mm
Vd = Av*fy/`sqrt3`* ymo
= d*tw *fy / `sqrt3` ymo
= 1440*12 *250/ `sqrt3` *1.1 = 2268 kN
since Vd > V provided ,
so web section is in adequate .no need to provide stiffner.
d/tw = 1440/ 12
=120 <200
3. Check whether intermediate stiffener required for given load
For the above problem, determine the web thickness required to avoid the stiffeners.
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.
GVEN DATA =
L= 15M
LOAD = 100KN/M
SELF WT OF GIRDR = 100 *15/200 = 7.5KN =0.5KN/M
TOTAL LOAD = 100+0.5=100.5 KN/M
TOTAL FACTORED LOAD = 1.5*100.5 = 150.75 KN/M
BENDING MMENT M = WI2/8 = 150.75*15*15/8=4239.84 KNM
SHEAR FORCE V= WL/4 = 150.75*15/4 =562.5 KN
CALCULATION OF DEPTH OF WEB =
d = ( M *K/FY)^1/3
WHERE M = MOMENT , K = 67 , IF STIFFNER NOT REQUIRED,
d= 4239.8 *67 *10^6/250 = 975 mm
ASSUME d =1000 mm
k= d/tw
tw = d/k
= 1000/67 =14.92
assueme tw = 15 mm
M = Af*fy*d/ymo
4239.8 *10^6 = Af *250 *1000/1.1
Af= 18655mm2
Af= 44239.8*1.1/250*1000
since section is plastic
so bf /2tf = 9.4 to 10.5
bf = 2*10.5 tf
bf= 21 tf
A = bf*tf
= 21*tf^2 = 18655 mm2
tf = 30 mm
thickness of flange tf = 30 mm
width of flange =
18655 = bf*tf 
bf= 18655/30 = 621
bf (flange) = 630 mm
check for the moent
Md = Ze*fy/ymo
ze= Ixx/ ymax
Ymax= d/2+tf = 1000/2+ 30 = 530mm
h = d/2+tf/2 = 515mm
Ixx = 2( bd3/12 + A*h2) + bd3/12
=2( 630*30^3 +630*30 *515^2) + 15 *1000^3/12
Md = 4849 knm > M , SO section is safe