Week 2 Challenge
Design of Tension Member
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)
b = 260mm
d = 20mm
d0 = 20 - 2 = 18mm
t = 20mm
No. of bolts = ((b - 2xe)/p )+1
P = 2.5d = 2.5 x 20 = 50
e = 1.5d = 1.5 x 20 = 30
No. of bolts = ((260 - 2 x 30)/50)+1
No. of bolts = 6
Net section area = ( b - ndo )t
Net section area = ( 260 - 6 x 18) 20
Net section area = 3000 mm^2
Plate – Zigzag bolting
ans)
b = 260mm
d = 20mm
d0 = 20 - 2 = 18mm
t = 20mm
No. of bolts = ((b - 2xe)/p )+1
P = 2.5d = 2.5 x 20 = 50
e = 1.5d = 1.5 x 20 = 30
No. of bolts = ((260 - 2 x 30)/50)+1
No. of bolts = 6
Net section area = ( b - ndo )t
Net section area = ( 260 - 6 x 18) 20
Net section area = 3000 mm^2
Angle (200 x 100 x 12 with M20 bolt)
ans)
b = 200mm
d = 20mm
d0 = 20 - 2 = 18mm
t = 12mm
No. of bolts = ((b - 2xe)/p )+1
P = 2.5d = 2.5 x 20 = 50
e = 1.5d = 1.5 x 20 = 30
No. of bolts = ((200 - 2 x 30)/50)+1
No. of bolts = 6
Net section area = ( b - ndo )t
Net section area = ( 200 - 6 x 18) 20
Net section area = 1840 mm^2
Q2)
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.
ans)
F = 5/8 x w x sinO x s x 4
F = 5/8 x 100 x 9.81 x sin15 x 8 x 4
F = 5061.96 N
Areq = F / 0.6fy
pixd^2/4 = 5061.96 / 0.6 x 250
d = 6.5 mm
Q3)
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.
Tip Steps :
- Calculate required area of cross section from factored load
- Choose an angle section from required area calculated
- Calculate design strength of the member due to yielding
- Calculate design strength of member by tearing of net section
- For a given member assume bolts placed with appropriate edge/end/pitch/gauge distance and calculate block shear capacity
- Check slenderness is within limit
ans)
assume 4mm weld size , fy = 250MPa, fu = 410 Mpa
approximate gross area = 1.1Tdg/fy = 1.1 x 250 x 1000/250 = 1100 mm^2
from steel table, select ISA 100x75x8mm connected to gusset plate by longer length Cxx = 31mm
area of connected leg = ( 100 - 8/2) x 8 = 768mm^2
area of outstanding leg = ( 75 - 8/2) x 8 = 568mm^2
Ag = 768+568 = 1336mm^2>1100mm^2 ,hence ok
design strength due to yielding of gross section ,
Tdg = Agfy/rmo = 0.91 x 1336 x 250 /1000 = 303.94kn
design strength due to rupture at net section,
Tdn = a An fu/rm1
fu = 0.8x1336x410/1000
fu = 438.21 Kn
design strength due to block shear,
for calculation of block shear, length of weld is required
P1 x 100 = 250 x (100 -31)= 172.50Kn
P2 = 250 - 172.50 = 77.5 Kn
strength of 4mm weld = 0.7 x 4 x 0.462 x 410 = 530.37 kn/mm
effective length of weld on head side = 172.5 x 1000/530.37 = 325mm
effective length of weld on toe side = 77.5 x 1000/530.37 = 146mm
Total welded length = 325 + 146 = 471mm
Avg = Avn = 471 x 8 = 3768 mm^2
Atg = Atn = 100 x 8 = 800mm^2
Tdb1 = 0.525Avg x fy + 0.72 x Atn x fy
= 0.525 x 3768 x 250/1000 + 0.72 x 800 x 410/1000 = 730.31kn
Tdb2 = 0.416 Avn x fu + 0.91Atg fy
= 0.416 x 3768 x 410/1000 + 0.91 x 800 x250/1000 = 824.67Kn
tensile strength provided = 438.21kn, 303.94KN and 730.71KN > 250KN
hence safe.