Given data:
Allowable tensile stress = 150 N/mm2
Size of ISA 100 100 10 mm
Number of rivets = 6
Diameter of rivets = 21.5 mm gross

To find:
Capacity of single ISA 100 100 10 mm as tension member connected by one leg using 6 number of 21.5 mm gross rivets

Solution:
1. Area of ISA 100 100 10 mm
Given ISA size is 100 mm × 100 mm which means it has a cross-sectional area of 100 × 100 = 10,000 mm2.
Cross-sectional area of the web = thickness × depth = 10 × 100 = 1000 mm2
Effective area = 10,000 – 2 × 1000 = 8000 mm2

2. Shear capacity of one rivet
The shear capacity of one rivet can be calculated as follows:
Shear capacity of one rivet = 0.577 × π × (d/2)2 × τ
where d is the diameter of the rivet and τ is the allowable shear stress of the rivet material.
Given diameter of rivet = 21.5 mm gross
Net diameter of the rivet = 21.5 – 1.5 = 20 mm
Allowable shear stress of rivet material = 100 N/mm2 (taken from IS 1929)
Shear capacity of one rivet = 0.577 × π × (20/2)2 × 100
= 157.1 kN

3. Tensile capacity of one rivet
The tensile capacity of one rivet can be calculated as follows:
Tensile capacity of one rivet = π × (d/2)2 × σ
where σ is the allowable tensile stress of the rivet material.
Given allowable tensile stress = 150 N/mm2
Tensile capacity of one rivet = π × (20/2)2 × 150
= 117.8 kN

4. Total capacity of 6 rivets
Total shear capacity of 6 rivets = 6 × 157.1 = 942.6 kN
Total tensile capacity of 6 rivets = 6 × 117.8 = 706.8 kN

5. Capacity of the tension member
The minimum of the shear and tensile capacities of the tension member and rivets should be considered as the capacity of the tension member.
Capacity of the tension member = min (8000, 942.6, 706.8) = 706.8 kN

Therefore, the capacity of single ISA 100 100 10 mm as a tension member connected by one leg using 6 number of 21.5 mm gross rivets is 706.8 kN.