Week 2 Challenge

1. Given plate size 260*20 (b*d)

bolt size = 20 mm

net sectional area (An) = (b-nd)t

= (260-4*22)*20

= 3440 mm2

section will fail in tension.

Given plate Zigzag bolt

Net sectional area (An) = (b-ndh)t

failure paths = AD, BC

AED, BEC

AEFG, BEFG

AEC, BED

EFG

Path AD, BF (2 holes)

An = (b-nd)t

= (260-2*22)20

= 4320 mm2

path AED, BEC

An = (260-3*22)*20

= 3880 mm2

path AEFG, BEFG

An = (260-4*22)*20

= 3440 mm2

path AEC, BED

An = (260-3*22)20

= 3880 mm2

path EFG

An = (260-3*22)*20

= 3880 mm2

For angle section

given g1 = 90mm , g2 = 118mm

p = 50mm

ABC (An) = (b-ndh+∑(p2/4g))t

= (3480-(3*22)+(50^2/4*90)+(50^2/4*118))*12

= 2540mm2

2. Gravity load = 60*1.2+75*1.2+100

= 262kg/m

= 262*9.81

= 2.57 kn/m

resolve to roof component= 2.57*sin15

= 0.67 kn/m

load on sag road = 0.67*8/3

=1.79Kn

Factored load = 1.5*1.79 = 2.69 Kn

net area = Tdn*γm1/0.9*Fu

= 2.69*10^3*1.25/0.9*410

=9.112 mm2

provide 12mm sag road (standard)

π/4*12^2 = 84.3mm^2

3. Given

tensile load = 500kn

span = 5m

Yield strength = 350 N/mm^2

Stress = Load / area

area = load/ stress

= 1.1*500*1000/350

= 1571 mm^2

choose the section ISA 125*95*8 of area = 1698mm^2

Net sectional rupture = Tdn = α*An*fu/γm1

An = ((125-8/2)-(3*22))*8+(95-8/2)*8

An = 1340 mm^2

Tdn = 0.8*1340*410/1.25 =
Block shear failure
Tdb1= [Avgfy/(√3γmo )+0.9Atn fu/γm1 ]
Tdb2= [0.9Avnfu/(√3 γm1)+Atgfy/γm0]
in above formulae
Avg= (35+(3*50))*10 = 1850mm2
Avn = 35+(3*50)-(3.5*22))*10 = 1080mm2
Atg = (60+60+40)*10 = 1600mm2
Atn = (60+60+40)*10 = 1600 mm2
Tdb1 = 552 Kn
Tdb2 = 547 kn
Slenderness ratio = L/r = L/r =5000/28.0
= 170 ≤ 400

section is safe