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

Solution:

We know, Net section area = (b - nd)t

Given, b = 260mm : t = 20mm

d = 20 mm but, do = 20+2 = 22mm

n = 4

therefore, An = (260 - 4(22))20 = 3440mm2

Plate – Zigzag bolting

Solution:

We know, Net section area = (b - nd + sum(P2/4g))t

Given, b = 260mm : t = 20mm

d = 20 mm but, do = 20+2 = 22mm

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

The minimum value is the net section area, Hence An = 3648.33 mm2

Angle (200 x 100 x 12 with M20 bolt)

Solution:

We know, Net section area = (b - nd + sum(P2/4g))t

Given, angle section: 200mmX100mm : t = 12mm

d = 20 mm but, do = 20+2 = 22mm

An = (200+100 - 3(22) + 50^2/4(90) + 50^2/4(130) ) 12 = 2949.02mm2

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.

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, we know

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

d = 3.54 mm

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.

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.