Challenge

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

do=20-2=18mm

no. of bolts=((b-2xe)/p)+1

P=2.5d=2.5x20=50

e=1.5d=1.5x20=30

no. of bolts=((260-2x60)/50)+1

no. of bolts=6

Net section area=(b-nxdo)xt

Net section area=(260-6x18)x20

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=2.5x20=50

e=1.5d=1.5x20=30

No. of bolts=((260-2x30)/50)+1

No. of bolts=6

Net section area=(b-nd0)xt

Net section area=((260-6x18)x20

Net section area=3000 mm^2

Angle (200 x 100 x 12 with M20 bolt)

Ans:

b=260mm

d=20mm

d0=20-2=18mm

t=12 mm

No. of bolts=((b-2xe)/p)+1

P=2.5d=2.5x20=50

e=1.5d=1.5x20=30

No. of bolts=((200-2x30)/50)+1

No. of bolts=6

Net section area=(b-nd0)xt

Net section area=((200-6x18)x20

Net section area=1840 mm^2

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=5/8xwxsinoxsx4

F=5/8x100x9.81xsin15x8x4

F=5061.96 N

Area=F/0.6fy

pixd^22/4=5016.96/0.6x250

d=6.5mm

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.

Ans:

Assume 4mm weld size,fy=250 Mpa,fu=410 Mpa

Gross area=1.1Tdg/fy=1.1x250x1000/250=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)x8=768mm^2

area of outstanding leg=(75-8/2)x8=568 mm^2

Ag=768+568=1336mm^2,hence ok

Design strength due to yielding of gross section

Tdg=Agxfy/rmo=0.91x1336x250/1000=303.94 KN

Design strength due to rupture at net section

Tdn=aphaxAnfu/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

P1x100=250x(100-31)=172.50 KN

P2=250-172.50=77.5 KN

Strength of 4mm weld=0.7x4x0.462x410=530.37 KN/mm

Effective length of weld on tow side=77.5x1000/530.37=146mm

Total welded length=325+146=471mm

Avg=Avn=471x8=3768 mm^2

Atg=Atn=100x8=800mm^2

Tdb1=0.525Avg x fy+0.72x Atn x fy

=0.525x3768x250/1000+0.72x800x410/1000=730.31 KN

Tdb2=0.416x3768x410/1000+0.91x800x250/1000=824.67 KN

Tensile strength provided=438.21 KN,303.94 KNand 730.71 KN>250 KN.

Hence safe.

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