Week 2 Challenge

Design of Tension Member

Question: 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.

Tip Steps:

• Refer Slide 26 for Net area calculation formula
• For angle section Flatten the member and calculate the gauge distance

Aim:

To be determine the net section area for the tension member as shown in figures below.

Plate – Drilled Hole

Given Data:

b = width of the plate = 260 mm

t = Thickness of the plate = 20 mm

d_h = Diameter of bolt holes = Diameter of Bolts + 2 mm = 20 + 2 = 22 mm

n = Number of bolts = 4 nos.

Procedure:

Net Section Area (A_n) = (b – n d_h) t

(A_n) = (260 – (4) x (22)) x 20

(A_n) = 3440 mm²

Result:

The Net Section area for the tension member (A_n) = 3440 mm².

Plate – Zigzag bolting

Given Data:

b = width of the plate = 260 mm

t = Thickness of the plate = 20 mm

d_h = Diameter of bolt holes = Diameter of Bolts + 2 mm = 20 + 2 = 22 mm

Procedure:

For AD & BF

Net Area (A_n) = (b – n d_h) t

Number of bolts (n) = 2 Nos.

(A_n) = (260 – 2 x 22) x 20

(A_n) = 4320 mm².

For CEG

Net Area (A_n) = (b – n d_h) t

Number of bolts (n) = 3 Nos.

(A_n) = (260 – 3 x 22) x 20

(A_n) = 3880 mm².

For ACEG

(A_n) = (b – n d_h + (P² / 4g)) x t

Number of bolts (n) = 4 Nos.

(A_n) = (260 – 4 x 22 + (60)²/ (4 x 60)) x 20

(A_n) = 3740 mm².

For BCEG

(A_n) = (b – n d_h + (P² / 4g)) x t

Number of bolts (n) = 4 Nos.

(A_n) = (260 – 4 x 22 + (50)²/ (4 x 60)) x 20

(A_n) = 3648 mm².

For ACF and BCD

(A_n) = (b – ∑d_h + ∑ (P² / 4g)) x t

Number of bolts (n) = 3 Nos.

(A_n) = (260 – 3 x 22 + (60)²/ (4 x 60)) + ((50)² / (4 x 60)) x 20

(A_n) = 4388 mm².

For BCF and ACD

Number of bolts (n) = 3 Nos.

(A_n) = (260 - 3 x 22 + 2(50²)/ (4 x 50) x 20

(A_n) = 4380 mm²

Result:

The BCEG having minimum value of Net Section area (A_n) = 3648 mm² with 4 nos. of bolts

So Net Section Area for Zig Zag bolting (A_n) = 3648 mm²

Angle (200 x 100 x 12 with M20 bolt)

Given Data:

(l x h) = 200 x 100 mm, b = 200 + 100 = 300 mm

Thickness (t) = 12 mm

Diameter of the holes (d_h) = Diameter of the bolt + 2 mm = 20 + 2 = 22 mm.

Procedure:

Net Section Area (A_n) = (b – n d_h + ∑ (P² / 4g)) x t

(A_n) = (300 – 3 (22) + (50)²/ 4 x (90) + (50)²/ 4 x (130)) x 12

(A_n) = 2949.02 mm².

Result:

The Net Section area for the angle (A_n) = 2949.02 mm².

Question: 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 center to center 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.

Tip Steps:

• Calculate total gravity load (DL & LL) on the purlin
• Calculate Wind load acting on the purlin. Note that wind load acts perpendicular to the member.
• Resolve and calculate the load acting parallel to the roof
• Force contribution to each sag rod is calculated
• Calculate required net area and chose a desired sag rod size.

Aim:

To be determine the load on sag rod and purlin.

Given Data:

Rafter space = 8m

Purlin spacing = 1.2m

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

Assumed self-weight of purlin = 100 kg/m.

Procedure:

Total Gravity Load (T_g) = m x g x h

= (60 + 75) x 9.8 x 1.2 x 8

(T_g) = 12700 N

Load on a Sag rod or Threaded rod

Td_n = (T_g/ 3 + w x g x s/3) Sin Ө

= (12700/3 + 100 x 9.8 x (8/3)) x Sin (15 ̊)

Td_n = 1772 N

Desired size of Sag rod.

Td_n = 0.9 x A_n x F_u / 1.25

F_u = 250 MPa

1772 = 0.9 x A_n x F_u / 1.25

A_n = (1772 x 1.2) / (0.9 x 250)

(A_n) = 9.844 mm².

D = (4 x A_n / Π) ^0.5

D = (4 x 9.844 / Π) ^0.5

D = 3.54 mm

Force Contribution to each sag rod

F = (5/8) x w x Sin Ө x s x 4

F = (5/8) x 100 x 9.81 x Sin 15 ̊ x 8 x 4

F = 5062.00 N

Net Area required

An = F/ 0.6 x f_y

An = 5062 / (0.6 x 250)

An = 6.5 mm².

Result:

• Total gravity load (T_g) = 12700 N
• Load on sag rod or threaded rod (Td_n) = 1772 N
• Size of the sag rod (D) = 3.54 mm
• Force contribution to sag rod (F) = 5062.00 N
• Required Net Area (An) = 6.5 mm².

Question: 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.

Tip Steps :

• Calculate required area of cross section from factored load
• Choose an angle section from required area calculated
• Calculate design strength of the member due to yielding
• Calculate design strength of member by tearing of net section
• For a given member assume bolts placed with appropriate edge/end/pitch/gauge distance and calculate block shear capacity
• Check slenderness is within limit

Aim:

To be determine the calculations to design the strength of member.

Given Data:

Single angle member of length = 5m

Axial tensile force = 500 kN

Yield strength F_y = 350 N/mm².

Procedure:

Here to be consider the data as follows

Weld size (t) = 4 m,

F_y = 250 MPa,

F_u = 410 MPa

Approximate gross area = 1.1 x Td_g/f_y = 1.1 x 250 x 1000/250 = 1100 mm²

From steel table, select ISA 100 x75 x 8mm connected to gusset plate by longer length C_xx

Area of connected leg = (100 - 8/2) x 8 = 768mm²

Area of outstanding leg = (75 - 8/2) x 8 = 568mm²

A_g = 768+568 = 1336mm2 > 1100mm2

Therefore it’s Safe.

Design strength due to yielding of gross section

Td_g = A_g F_y/ rmo

Td_g = (0.91 x 1336 x 250)/ 11000

Td_g = 304.00 KN

Design strength due to rupture at net section

Td_n = α x A_n x F_u /rml

F_u = 0.8 x 1336 x 410/1000

F_u = 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

S = 530.37 KN/mm

Effective length of weld on head side (E_L1)

(E_L1) = (172.5 x 1000) /530.37

(E_L1) = 325 mm

Effective length of weld on toe side (E_L2)

(E_L2) = (77.5 x 1000) /530.37

(E_L2) = 146 mm

Total welded length (L) = (E_L1) + (E_L2)

(L) = 325 + 146

(L) = 471 mm

Av_g = Av_n = 471 x 8 = 3768 mm²

At_g = At_n = 100 x 8 = 800mm²

Td_b1 = 0.525 Av_g x F_y + 0.72 x At_n x F_y

Td_b1 = 0.525 x 3768 x 250/1000 + 0.72 x 800 x 410/1000

Td_b1 = 730.31 KN

Td_b2 = 0.416 Av_n x F_u + 0.91 x At_g x F_y

Td_b2 = (0.416 x 3768 x 410) /1000 + (0.91 x 800 x 250) /1000

Td_b2 = 824.67 KN

Result:

Tensile strength provided = 438.21 KN, 304.00 KN and 730.71 KN

All Provided strengths are > 250KN

Therefore the Design of the angle member is in safe condition.

Conclusion:

Here for all 3 questions are solved or designed for the given tension members as per their data and tips in the questions.