Explanation Given: The service load is 10 kip . 25 % is dead load. 75 % is live load. Concept Used: Write the expression for the factored load. P = P u 2 = 1.2 D + 1.6 L 2 ..... (I) Here, the dead load is D , the live load is L , the factored load is P u , the load on single angle member is P . Write the expression for the direct shear stress in x direction. f y = P L ..... (II) Here, the direct shear stress is f y , the total length of the weld is L . Write the expression for the centroid in x direction. x ¯ = a 1 y 1 + a 2 y 2 a 1 + a 2 ..... (III) Here, the area of the weld in the connection is a 1 , a 2 , the distance from the centroid of the areas is y 1 , y 2 respectively in the x direction. Write the expression for the centroid in y direction. y ¯ = a 1 y 1 + a 2 y 2 a 1 + a 2 ..... (IV) Here, the area of the weld in the connection is a 1 , a 2 , a 3 , the distance from the centroid of the areas is y 1 , y 2 , y 3 respectively in the y direction. Write the expression for the moment about the centroid. M = P e ..... (V) Here, the moment is M , the eccentricity is e . Write the expression for the moment of inertia about x direction. I x = ∑ ( b d 3 12 + A ( y ¯ − y ) 2 ) ..... (VI) Here, the moment of inertia about x axis is I x , the breadth of the member is b , the depth of the member is d , the area of the member is A . Write the expression for the moment of inertia about y direction. I y = ∑ ( b d 3 12 + A ( y ¯ − y ) 2 ) ..... (VII) Here, the moment of inertia about x axis is I y . Write the expression for the polar moment of inertia. J = I x + I y ..... (VIII) Here, the polar moment of inertia is J . Write the expression for the shearing stress in x direction. f 2x = M y ¯ J ..... (IX) Write the expression for the shearing stress in y direction. f 2y = M x ¯ J ..... (X) Write the expression for the resultant shearing stress. f v = ( ∑ f x ) 2 + ( ∑ f y ) 2 ..... (XI) Here, the resultant shear stress is f v . Calculation: Calculate the load on single angle member. P = ( 1.2 × 0.25 × 10 ksi ) + ( 1.6 × 0.75 × 10 ksi ) 2 = ( 3 + 12 ) ksi 2 = 7.5 ksi Calculate the direct stress. Substitute 7.5 kips for P , 15 in for L in the Equation (II). f y = 7.5 kips 15 in = 0.5 kips / in Calculate the centroid in x direction. Substitute 5 in for a 1 , 5 in for a 2 , 2.5 in for y 1 , 2.5 in for y 2 , in the Equation (III). x ¯ = ( 5 in × 2 .5 in ) + ( 5 in × 2.5 in ) ( 5 + 5 + 5 ) in = 25 in 2 15 in = 1.67 in Calculate the centroid in y direction. The section is symmetrically welded in the y direction. Hence, the centroidal distance in y direction is 2.5 in . Calculate the moment about the centroid. Substitute 7.5 kips for P , 12 + 5 − 1