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.

a. Plate – Drilled Hole

Answer:-

Given data:

Plate size : 260 x 20 mm

Bolt size : 20mm

No. of bolts : 4

Bolt diameter (d_h) = 22mm

As per IS 800 Clause 6.3.1

Net area (A_n) = (b - nd_h)t

Where,

b - Width of plate

t - Thickness of plate

d_h - Diameter of bolt hole (2mm in addition to the diameter of the hole)

n - Number of bolts in critical section

(A_n) = (260 - 4 x 22)20

(A_n) = 3440 mm²

Result:-

Net area for chain bolting is 3440 mm²

b. Plate – Zigzag bolting

Answer:-

As per iS 800 Clause 6.3.1

For Staggared Arrangement of bolts.

Net Area (A_n) = (b - nd_h + (P²/4g))t

Where,

b - Width of plate

t - Thickness of plate

d_h - Diameter of bolt hole (2mm in addition to the diameter of the hole)

n - Number of bolts in critical section

P - Staggered pitch length between line of bolt holes

g - Gauge length between the bolt holes

Path AD and BF have two bolts:

Net area = (260 - 2 x 22) x 20

(A_n) = 4320 mm²

Path CEG have three bolts:

Net area = (260 - 3 x 22) x 20

(A_n) = 3880 mm²

Path ACEG have four linear bolts and one stagger bolt:

Net area = (260 - 4 x 22 + (60²)/(4 x 60)) x 20

(A_n) = 3740 mm²

Path BCEG have four linear bolts and one stagger bolt:

Net area = (260 - 4 x 22 + (50²)/(4 x 60)) x 20

(A_n) = 3648 mm²

Path ACF and BCD have three linear bolts and two stagger bolts:

Net area = (260 - 3 x 22 + (60²)/(4 x 60) + (50²)/(4 x 60)) x 20

(A_n) = 4388 mm²

Path BCF and ACD have three linear bolts and two stagger bolts:

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

(A_n) = 4380 mm²

The minimum net area is (3648 mm²) for path BCEG have four linear bolts and one stagger bolt.

Result :-

Net area for zig zag bolting is 3648 mm²

c. Angle (200 x 100 x 12 with M20 bolt)

For M20 bolt, d_h = 22mm

As per IS 808 Part 1 Table 6.1

Gross Area for L angle (200 x 100 x 12) = 3480 mm²

Path AC have two bolt holes:

Net area = Gross area - nd_h x t

Net area = 3459 - 2 x 22 x 12

(A_n) = 2952 mm²

Path ABC have two linear bolts and one staggered bolt:

Net area = Gross area - nd_ht+ (P²/4g)t

P = 50

g = 70+60-(6+6) = 118

Net area = 3459 - 3 x 22 x 12 + (50²)/(4 x 118) x 12

(A_n) = 2752 mm²

Path ABE have three staggered bolts:

Net area = Gross area - nd_ht+ (P²/4g)t

P = 50

g (AB) = 90

g (BE)= 70+60-(6+6) = 118

Net area = 3459 - 3 x 22 x 12 + (50²)/(4 x 118) x 12 + (50²)/(4 x 90) x 12

(A_n) = 2835 mm²

The minimum net area is (2731 mm²) for path ABC have two linear bolts and one stagger bolt.

Result :-

Net area for L Angle bolting is 2752 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 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.

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.

Given Data:-

Dead load from roofing = 60 kg/m²

Self weight of purlin = 100kg/m

Live load = 75 kg/m²

Wind load = 150 kg/m²

Panel size = 8 x 1.2 m

θ = 15 degree

F_y = 250 N/mm²

Answer:-

Calculation of total load acting on purlin.

W_D = Load of Roofing icluding Isulation x Purlin spacing + Self weight of Purlin

W_D = (60 x 1.2 + 100) x 9.81 = 1687 N/m

W_L = Live load x Purlin spacing

W_L = 75 x 1.2 x 9.81 = 883 N/m

W = W_D + W_L

W = 2570 N/m

W_x = W Sin θ = 2570 Sin 15 = 665 N/m

W_y = W Cos θ = 2570 Cos 15 = 2482 N/m

F = 5/8 W_x x S x 4

= 5/8 x 665 x 8 x 4

F = 13300 N

R = F/Cos θ = 13300/Cos 15 = 13770 N

A_req = R/0.6F_y

𝝿/4 d²_req = 13770/(0.6 x 250)

d_req = 10.81mm

Result:-

Hence Provide 12 mm diameter steel bar as Sag rods.

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.

Given data:-

L = 5m

Load = 500kN

F_y = 350 N/mm²

Answer:-

Obtain the gross area, Ag, required to carry the design load Tu from the equation using the yield stress.
T_u = fy A_g / γ_m0
Here, T_u = 500 kN, fy = 350 N/mm², and

As per IS 800 Table 5 γ_m0 = 1.10

A_g = 500 x 1000 x 1.10 / 350

Therefore, A_g = 1572 mm²

Hence Provide L Angle 110 x 110 x 8mm

Design Strength of member due to yielding:

As per IS 800 Clause 6.2

T_dg = fy A_g / γ_m0

= 350 x 1702 / 1.1 = 541545 N = 541.545 kN

Design strength of member tearing of net section for single angles:

As per IS 800 Clause 6.3.3

For preliminary sizing of one or two bolts

γ_m1 = 1.25

Assume fu = 500 N/mm²

T_dn = 0.6 A_n f_u / γ_m1

= 0.6 x 1702 x 500 / 1.25 = 408480 N=408.480 kN

Slenderness ratio = Kl/r_min

= 5000 / 21.4

Kl/r_min = 232.55

As per IS 800-2007 Table 8a.

Checked its is safe.