A moment M of magnitude 50 kN-m is transmitted to a column flange

SOLUTION

Concept

 Force in a rivet joint due to moment is acting is given by

\(F = \frac{{M \times r}}{{{\rm{Σ }}{r^2}}}\)

 Where M is the moment 

 r is the radial distance from C.G 

Calculations

 Here all rivets are at equal distance from C.G 

The centroid of rivets will be on the symmetric lines and each rivet will be at

 \(\begin{array}{l} r = \sqrt {{{\left( {50} \right)}^2} + {{\left( {50} \right)}^2}} \\ = 50\sqrt 2 \;mm \end{array}\)

Shear force in rivet A is given by

\({F_A} = \frac{{Mr}}{{\sum {r^2}}} = \frac{{50 \times {{10}^3} \times 50\sqrt 2 }}{{4 \times {{\left( {50\sqrt 2 } \right)}^2}}}\)

 = 176.8 kN

 Similarly in Rivet B Force is 

 \(\begin{array}{l} r = \sqrt {{{\left( {50} \right)}^2} + {{\left( {50} \right)}^2}} \\ = 50\sqrt 2 \;mm \end{array}\)

\({F_B} = \frac{{Mr}}{{\sum {r^2}}} = \frac{{50 \times {{10}^3} \times 50\sqrt 2 }}{{4 \times {{\left( {50\sqrt 2 } \right)}^2}}}\) = 176. 8 kN

 ∴ Ratio of forces \(\frac{{{F_A}}}{{{F_B}}} = \frac{{176.8}}{{176.8}} = 1\)