Week 2 Challenge
1. Net Area for tension member calculation problem
Calculate the net section area for the tension member as shown in Figure below. The plate size is 260 x 20 thick and Bolt size is 20 mm.
Plate – Drilled Hole
Ans) To calculate the net section are the formulat is An=(b-nDh)t
Here An= Net section area
b= width of the plate= 260mm
n= number of bolts= 12 nos
Dh= Dia of bolt = 20 mm
t= thinckness of plate = 20 mm
Hence An = (260- 12*20) 20= 400 mm2
Plate – Zigzag bolting
Ans) To calculate the net section area the formulat is An=(b-sum(Dh)+Sum(p^2/4g))t
Here An= Net section area
b= width of the plate
Dh= Dia of bolt
t= thinckness of plate
p= Pitch distance
g= gauge distance
Hence An = (260- (7*20)+(((60^2/4*60)*3)+((50^2/4*60)*3)))* 20= 3925 mm2
Angle (200 x 100 x 12 with M20 bolt)
Ans) To calculate the net section area the formulat is An=(b-sum(Dh)+Sum(p^2/4g))t
Since the section is non uniform the net effective area of the section is Anet= Alpha*An
Here An= Net section area
b= width of the plate
Dh= Dia of bolt
t= thinckness of plate
p= Pitch distance
g= gauge distance
alpha= factors affecting section
Hence An = (300- (6*20)+(((50^2/4*90)*3)+((50^2/4*130)*3)))* 12= 2583 mm2
Here the total no of bolt is 6 hence alpha=0.8
Hence Anet = 0.8*2583=2066.4 mm2.
2. Sag rod design problem
Design a sag rod as shown in Figure below. Purlins are supported at 1/3rd points by sag rods.
Assume centre to centre distance of Rafter spaced at 8 m , spacing of purlin = 1.2 m, Self weight of roofing (including thermal insulation)= 60 kg/m2 , Wind pressure intensity = 150 kg/m2,Slope of rafter = 15 degrees, Live Load =75 kg/m2. Assume self weight of purlin as 100 kg/m.
Answer
To design sag rod we need to find the dia of the rod
Formulat to calculate the dia of the rod is Areq= R/0.6fy
That is (Pi*d^2/4)=R/0.6fy
F=Rcos (teta)
=> R=F/Cos(teta)
F = 5/8 x w x sin(teta) x s x 4
F = 5/8 x (100 x 9.81) x sin15 x 8 x 4
F = 5078.03 N
R=5078.03/cos(15)=5257.16N
=>Pi*d^2/4 = 5257.16 / 0.6 x 250
d = 6.68 mm
Let's consider 8mm dia sag rod.
3. Determine section size for given tensile load
Design a single angle member of length 5m to carry axial tensile force of 500 kN. Yield strength fy of the material to be considered is 350 N/mm2.
Answer
fy = 350MPa, L=5 m and tensile force =500Kn
Step 01: Calculate gross area
Tensile strength = 0.909fyAg
=>Ag=Tdg/0.909fy= 500000/0.909*350= 1571.59 mm^2
Step 02: From IS 808 a required angle sections will be selected provided that the Gross area is 25-40% more thant the Ag calculated in step 01
From Table 5.1 we get ISA 90X90X12 mm having Ag 2020 mm^2 which is more than 25% of Ag calculated. Hence Satisfactory
design strength due to yielding of gross section ,
Tdg = 0.909fyAg= 0.909*350*2020 = 642.66kn
It is greater than Tensile force hence safe
design strength due to rupture at net section,
Tdn = 0.72An fu + Beta
Tdn = 0.72x2020x410
Tdn= 596.30 Kn
It is greater than Tensile force hence safe
design strength due to block shear,
for calculation of block shear, length of weld is required
P1 x 90 = 350 x (90 -31)= 229.44Kn
P2 = 350 - 229.44= 120.56 Kn
Assuming 4mm thick welding
strength of 4mm weld = 0.7 x 4 x 0.462 x 410 = 530.37 kn/mm
effective length of weld on head side = 229.44x 1000/530.37 = 432.60mm
effective length of weld on toe side = 120.56 x 1000/530.37 = 227.31mm
Total welded length = 432.60 + 227.31 = 659.91mm
Avg = Avn = 659.91 x 12 = 7918.92 mm^2
Atg = Atn = 90 x 12 = 1080 mm^2
Tdb1 = 0.525Avg x fy + 0.72 x Atn x fy
= 0.525 x 7918.92 x 350/1000 + 0.72 x 1080 x 410/1000 = 1773.92kn
Tdb2 = 0.416 Avn x fu + 0.91Atg fy
= 0.416 x 7918.92 x 410/1000 + 0.91 x 1080 x350/1000 = 1694.63Kn
tensile strength provided = 642.66kn, 596.30KN and 1694.63KN > 500KN
hence safe.