Week 4 Challenge
1) AIM & INTRODUCTION: calculate section classification of beams
PROCEDURE :
assume all sections are Fy 250.
E = (250/fy)^1/2 = (250/250)^1/2=1
ISMB 500
From table ,
d = 500 mm
b = 180 mm
tr = 17.2 mm
tw = 10.2 mm
As per IS 800 : 2007 Table 2
b/tf = 180/17.2 = 10.47 < 10.5 E
d/tw = 500/10.2 = 49.02 < 84 E
Hence it is classified as a compact section (class 2)
ISMC 200
From Table ,
h = 200 mm
b = 75 mm
tf = 11.4 mm
tw = 6.1 mm
As per IS 800 : 2007 Table 2
b/tf = 75/11.4 = 6.58 < 9.4 E
d/tw = 200/6.1 = 32.79 < 84 E
Hence it is classified as a plastic section (class 1)
PG 1300 x 450 x 20 x 12
As per IS 800 : 2007 Table 2
b/tf = 450/20 = 22.5 < 29.3 E
Hence it is classified as a plastic section ( class 1)
Double angle (Back to back) - 2 x ISA 75x75x6
From table ,
b = 75 mm
tf = 6 mm
As per IS 800 : 2007 Table 2
b/tf = 75/6 = 12.5 < 15.7 E
Hence it is classified as a semi compact section (class 3)
RHS - 150 x 100 x 10
d/tw = 150/10 = 15<42 E^2
Hence it is classified as a plastic section (class 1)
Built - up box - 1000 x 450 x 20 x 12
d/tw = 1000/12 = 83.33 < 84 E
Hence it is classified as a plastic section (class 1)
RESULT: calculated section classification of beams
2)Effective length problem:
AIM: TO find the effective length of rafter
PROCEDURE :
Rafters are in compression which is a continously supported .
Length of Rafter = 5.36/sin 15 = 20.71 m
As per IS 800 : 2007 Table 15 ,
Effective length = 0.85 L
therefore , Effective length of Rafter = 0.85 x 20.71 = 17.60 m
RESULT: founde the effective length of rafter succesfully
3) Deflection of a beam problem:
- AIM: calculation of deflectiom at the centre of the beam
PROCEDURE :
GIVEN DATA :
a = 2 m
b = 3 m
L = 7 m
W = 100 N
w = 6 N/m
EI = 20 X 10^6 Nm^2
solution:
Deflection at centre = WL^3 /24EI x [3a/L - 4(A/L)^3] + 5wL^3 /384EI
=100 X 7^3 / (24 X 20 X 10^6) X [(3 x 2/7) - 4 x (2/7)^3] + 5 x 6 x 7^3 /(384 x 20 x 10 ^6 )
delta max (at centre ) = 0.056 mm
RESULT : calculated of deflectiom at the centre of the beam
AIM : calculate of deflection at the centre of the beam
PROCEDURE :
given data
a = 1.5 m
b = 1 m
c = 2.5 m
L = 5 m
W1 = 2 KN
W2 = 5 KN
w = 1.5 KN/m
EI = 25 X 10 ^6 Nm^2
solution
Deflection at centre = w1L^3/48EI x [3a /L - 4 (a/L)^3] + W2 L^3 /48EI x [3c/L-4(c/L)^3] + 5wL^3/384EI
=2000 X 5^3 / (48 x 25 x 10^6 ) x [(3 x 1.5 /5) - 4 x (1.5 /5)^3] + 5000 X 5^3 /(48 x 25 x 10^6) x [(3 x 2.5 /5) - 4 x (2.5/5)^3] + 5 x 1500 x 5^3/ (384 x 25 x 10^6)
delta max (at center) = 0.783 mm
RESULT:calculateed of deflection at the centre of the beam
AIM :Calculate the flexural stiffness
PROCEDURE :
diven data:
L=10 m
w = 10 kn/m
delta max (at center) = 5 mm
solution:
delta max (at centre) = 5wL^3/384EI
EI = 5WL^3/384 X DELTA MAX
= 5 X 10000 X 10^3/(384 X 5 X 10^-3)
EI = 26.04 x 10 ^6 N/m^2
RESULT :Calculated the flexural stiffness succesfully
4) AIM :Moment capacity determination problem
PROCEDURE:
Given data
length = 6 m
Dead load = 5 kn/m^2
Live load = 6 kn/m^2
A point load at midpoint = 100 kn
Bearing Length = 50 mm
step 1
Load calculation:
DL = 5 x 3 = 15 kN/m
LL = 6 x 3 = 18 kN/m
Total UDL = 33 kN/m
point load at the center of span = 100 kN
STEP 2:
Shear Force :
RA + RB = 33 x 6 + 100 = 298
RA = RB
Taking moment about A
RB x 6 = 33 x 6 + 100 = 298
RA=RB
Taking moment about A
RB x 6 = 100 x 3 + 33 x 6 x 3
RB = 149 kN
RA=149 kN
SF at C (center ) = 50 kN
SF at B = -149 kN
SF at C (center) = 50 kN
SF at B = -149 kN
Hence maximum shear force = 149 kN
STEP 3:
Bending Moment:
MA = 149 x 6 - 100 x 3 - 33 x 6 x 3 = 0
MC = 149 x 3 - 33 x 3 x 1.5 = 298.5 kN m
MB = 0
Factored Moment = 1.5 x 298.5 = 447.75 kN m
step 4:
section modulus:
Zp (required) = M x ym0/fy = 447.75 x 10^6 x 1.1 x /250=1970.1 cm^3
Choosing section ISWB 600 @ 133.7 kg/m, section properties from IS 808 : 1989 Table 2.2.
Shape factor = zp/ze = 1.14
so,ze (required ) = 1970.1/1.14 = 1728.16 cm^3
zp(provided) = 4035.6 cm^3
zp (requred) = 3540 cm^3
r1 =17
h=600mm
bf=250mm
tf=21.3mm
tw=11.2mm
Ixx=106198 cm^4
Iyy=4700 cm^4
rxx=250 mm
ryy=52.5 mm
d = h - 2(r1+tf) = 600-2x(17+21.3)
=523.4 mm
step 5:
classification of section:
As per IS 800:2007 Table 2
(bf/2)/tr = (250/2)/21.3 = 5.86<9.4
d/tw = 523.4 / 11.2 = 46.73 < 84
Hence the section is plastic
step 6:
Calculation of shear capacity of section:
Vd = (Av x fy)/(`3` x ymo) = (600 x 11.2 x 250) (`3` x 1.1) =881797 N
Vd= 881.797 kN > 1.5 x 149 = 223.5 kN
Hence safe in shear
0.6 vd = 0.6 x 881.797 = 529.08 kN > V = 223.5 kN
Hence Low shear
step 7:
Calculation of design capacity of moment - by interpolation method
Design strength for laterally unsupported beam
kL/ry = 6000/52.5 = 114.28
hf/tf =(600 - 21.3 )/21.3 = 27.16
fcrb = 240 N/mm^2
delta LT = (FY/Fcr)^1/2 = (250/240)^1/2 = 1.02
chi LT = 0.5 [1 + ALPHA lt(delta LT -0.2) + 1.02^2]
=0.5[1+.21(.21-.2)+1.01^2]
=1.1063
xlt =1/[CHILT+(chi^2 - del ^2LT)^.5<-1.0
=1/[1.1063 + (1.1063^2 - 1.02^2)
=0.652<1
Hence ok.
fbd = XLT X fy /ym0
=0.652 x 250 /1.1 = 148.18 N/mm^2
Md = Bb+ Zp + fbd
=1.0 x 4035.6 x1000 x 148.18
Md = 598 kNm> M (447.75 kNm)
Hence section is afe in bending stress
step 8:
check for deflection:
del = [5wl^4 / 384EI] + [WI^3/48EI]
=[5 X 18 X 6000^4 /384] + [100 x 6000^3 /48] x 1 / (20 x 10^4 x 106198 x 10^4)
del = 1.43 + 21.18 = 22.61 mm
permissible maximum deflection asa per IS 800 Table 6 for Live load with Elastic cladding
=L/240 = 6000/240 =25 mm > 22.61 mm
Hence safe in defelction
step 9:
Check for Wen Buckling :
d/tw = 523.4 /11.2 = 46.73 < 67 E
Hence ok
so no need to check for web buckling
step 10:
check for web crippling
n1 = 2.5 x (tr + r1) = 2.5 (21.3 + 17 ) = 95.75 mm
As per IS 800 : 2007 Clause 8.7.4
Fw = [(b1 + n2) x tw x fy )]/rm0
where b1 = Length of load dispersion
=[(50 + 95.75 ) x 11.2 x 250]/1.1
=371000 N = 371 kN > V = 223.5 kN
Hence safe in bearing
Result:
Thus , it is concluded that the beam ISWB 600 @ 133.7 KG/M , is capable to resist bending moment of 548.70 kN m safely.
5) Purlin design problem:
design channel purlin , considering sag rod at mid sag rod at mid span for the roof truss having following data
span of truss = 20 m
spacing of Truss = 4 mc/c
Roof slope : 20 degrees spacing of purlin : 2m
wind suction : 1 kN/m^2
Live Load : 0.75 kN/m^2
Dead Load : 1.4 kN/m^2
Given data:
span of truss = 20 m
spacing of Truss = 4 m c/c
Roof slope = 20 degrees
spacing of purlin = 2 m
Wind suction = 1 kN/m^2
Live Load : 0.75 kN/m^2
Dead Load = 1.4 kN/m^2
solution:
step 1:
calculation of Load :
Dead Load perpendicular to the rafter = 1.4 Cos 20 degree
=1.4 x .939 = 1.3146 kN/m^2
Dead Load per meter (load /m^2 x spacing of purlin) = 1.3146 x 2 = 2.6292 kN/m
Live Load perpendicular to the rafter = 0.75 cos 20 degree
0.939 = 0.7042 kN/m^2
Dead Load per meter (Load /m^2 x spacing of purlin) = 0.7042 x 2 = 1.4084 kN/m
Wind load perpendicular to the rafter = 1.0 cos 0 degree
= 1.0 kN/m^2
Wind load per meter (load /m^2 x spacing of purlin ) = 1.0 x 2 = 2.0 kN/m^2
Load combination (1) = 1.5 x (DL + LL)
= 1.5 X (2.6292 + 1.408)
=6.0558 kN/m
Load combination (2) = 1.5 x (DL + LL - WL)
=1.5 x (2.6292 + 1.408 - 2)
worst load combination = 6.0558 kn/m
dead load parallel to the rafter = 1.4 sin 20 degree
=1.4 x .342 =0.4788 kN/m^2
Dead load per meter (load /m^2 x spacing of purlin ) = 0.4788 x 2 = 0.9576 kN/m
live load parallel to the rafter = 0.75 sin 20
=.75 x 0.342
= 0.2565 kN/m^2
live load per meter (load /m^2 x spacing of purlin ) = 0.2565 x 2 = 0.513 kN/m
wind load parallel to mrafter = 0.00
1.5 x (DL +LL ) = 1.5 x (0.9576 + .513 ) = 2.2059 kN/m
step 2:
bending moment calculation:
BM about Z-Z Mz = wl^2 /10 = 6.0558 x 4 ^2 /10 = 9.689 kN m
(purlin is continous over PR of truss)
BM about y-y My = wl^2 /10 = 2.2059 x 4^2 /10 = 3.529 kN m
step 3 :
calculation of shear force :
Fz=wl/2 = 6.0558 x 4 / 2 = 12.117 kN
Fy = wl/2 = 2.2059 x 4 / 2 = 4.419 kN
step 4 :
selection of trial channel section for purlin:
ISMC 200 SELECTED for purlin
properties taken from steel tbles
h= 200 mm
bf = 75 mm
tf = 11.4 mm
tw = 6.1 mm
h1 = 150.2 mm
Izz = 1819.3 mm^4
Zpz = Mz x ymo /fy + (2.5 D / bf ) My x ym0 /fy
= 9.689 x 10 ^6 x 1.1 /250 + (2.5 x 200 /75 ) x 3.529 x 10^6 x 1.1 /250
= 42631.6 + 6.6666 x 15527.6 = 146147 mm^3
x - coordinate C.G of the channel section
200 x 6.1 + 2 x 11.4 x = 2 x (68.9 - x ) x 11.4
x = 7.69 mm
Zpy =(600 x 6.1 ) (6.1/2 + 7.69) + 2[(68.9 x 11.4 ) (68.9 /2 -7.69)]
=39308.4 + 42037.819
=81346.2 mm^3
step 5:
section classification:
d/tw = 150.2/6.1 = 24.62 < 42E
Hence the section is plastic
step 6:
calculation of shear capacity of section:
About Z AXIS
Vd = (Av x fy )/(3^1/2 x ymo) = (200 x 6.1 x 250 ) (3^1/2 x 1.1 ) = 160.083 kN
0.6 Vd = 0.6 x 160.08 = 96.05 kN > Fz = 12.117 kN
Hence safe
About Y axis
Vd = (Av x fy )/(3^1/2 x ymo) = (2 x 75 x 11.4 x 250 ) (3^1/2 x 1.1 ) = 224.38 kN
0.6 Vd = 0.6 x 224.38 = 134.628 kN > Fy = 4.419 kN
Hence safe
design moment in z direction
Mdz = Bb x Zpz x fy /rm0
= 1.0 x 146147 x 250 /1.1 = 33215227.3 N mm
Mdz = 33.215 kN m > Mz (9.689 kN m)
Hence section is afe in moment in Z direction
Design moment in y direction
Mdy = Bb x Zpy x fy /rm0
=1.0 x 81346.2 x 250 /1.1 = 18487772 N mm
Mdy = 18.48 kN m > My (3.529 kN m)
Hence section is afe in moment in y direction
step 7 :
overall strength condition :
MZ x Mdz + My /Mdy <-1
9.689 /33.215 +3.529/18.48 =0.292+0.191= 0.483<1
Hence safe.
step 8 :
Check for deflection:
Actual deflection = 5wl^2 /384EI = (5 x 1.4084 x 4000^4) / (384 x 20 x 10^4 x 1819.3 x 10 ^3) = 12.89 mm
Permissible maximum deflection as per IS 800 : 2007 Table 6 for Live Load with Elastic cladding = L/240 = 4000/240 = 16.66 mm
Actual deflection is lesser than permissible deflection
Hence safe in deflection
result:
It is concluded that for purlin of truss ISMC 200 section is afe in shear moment , strength and deflection.