Week 2 Challenge

CHALLENGE WEEK 2

Question: -1

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

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

Question: - 2

2. Sag rod design problem

Design a sag rod as shown in Figure below. Purlins are supported at 1/3^rd 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

Question: - 3

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.