Week 2 Challenge
1.Net Area for tension member calculation problem
Aim: To Calculate the net section area for the tension member for below,
- Plate – Drilled Hole
- Plate – Zigzag bolting
- Angle (200 x 100 x 12 with M20 bolt)
Observation:
1. Plate – Drilled Hole
Formula: Net section area = (b - nd)t
Given, b = 260mm, t = 20mm, d = 20 mm, do = 20+2 = 22mm, n = 4
Solution: An = (260 - 4(22))20 = 3440mm2
Result: An=3440mm2
2. Plate – Zigzag bolting
Formula: Net section area = (b - nd + sum(P2/4g))t
Given:
b = 260mm, t = 20mm, d = 20 mm, do = 20+2 = 22mm
Solution:
Case 1: Along section AD or BF : An = (260 - 2(22))20 = 4320mm2
Case 2 : Along section CEG : An = (260 - 3(22))20 = 3880mm4
Case 3 : Along section ACEG : An = (260 - 4(22) + 60^2/4(60) )20 = 3740mm2
Case 4 : Along section BCEG : An = (260 - 4(22) + 50^2/4(60) )20 = 3648.33mm2
Case 5 : Along section ACF : An = (260 - 3(22) + 60^2/4(60) + 50^2/4(60))20 = 3948mm2
Result:
The minimum value is the net section area,
An = 3648.33 mm2
3. Angle (200 x 100 x 12 with M20 bolt)
Formula: Net section area = (b - nd + sum(P2/4g))t
Given:
angle section: 200mmX100mm, t = 12mm, d = 20 mm, do = 20+2 = 22mm
Solution:
An = (200+100 - 3(22) + 50^2/4(90) + 50^2/4(130) ) 12 = 2949.02mm2
Result:
An = 2949.02mm2
2. Sag rod design problem
Aim: To design a sag rod.
Given: centre to centre distance of Rafter = 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.
Solution:
DL + LL = (60+75)*9.8*1.2*8 = 12700
Load on each sag rod = (12700/3 + 100*9.8*8/3)Sin(15) = 6846.66*Sin(15) = 1772 N
Therfore,
Load on threaded rod (Tdn)= 0.9*An*fu/1.25
1772 = 0.9*An*fu/1.25
An = 1772*1.25/(0.9*250) (fu = 250 Mpa)
An = 9.844
d = (4*An/pi)^0.5 = (4*9.844/pi)^0.5
d = 3.54 mm
Result: d = 3.54 mm
3. Determine section size for given tensile load
Aim: To 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.
Solution:
Assume 4mm weld size, fy = 250MPa, fu = 410 Mpa
Approximate gross area = 1.1Tdg/fy = 1.1 x 250 x 1000/250 = 1100 mmA2
from steel table, select ISA 100x75x8mm connected to gusset plate by longer length Cxx
area of connected leg = ( 100 - 8/2) x 8 = 768mm2
area of outstanding leg = ( 75 - 8/2) x 8 = 568mm2
Ag = 768+568 — 1336mm2 > 1100mm2 ,hence Safe.
Design strength due to yielding of gross section ,
Tdg = Agfy/rmo = 0.91 x 1336 x 250 11000 = 303.94kN
Design strength due to rupture at net section,
Tdn = alpha*An*fu/rml
fu = o.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 mm2
Atg = Atn = 100 x 8 = 800mm2
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 + 0.91 x Atg x fy = 0.416 x 3768 x 410/1000 + 0.91 x 800 x 250/1000 = 824.67KN
Tensile strength provided = 438.21kN, 303.94KN and 730.71KN > 250KN
Hence safe.