Explanation Given information: Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below. For summation of forces along x -direction is equal to zero ( ∑ F x = 0 ) , consider the forces acting towards right side as positive ( → + ) and the forces acting towards left side as negative ( ← − ) . For summation of forces along y -direction is equal to zero ( ∑ F y = 0 ) , consider the upward force as positive ( ↑ + ) and the downward force as negative ( ↓ − ) . For summation of moment about a point is equal to zero ( ∑ M at a point = 0 ) , consider the clockwise moment as negative and the counter clockwise moment as positive. Method of joints: The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates Tension (T). Calculation: Show the angle of the truss member as shown in Figure 1. Refer Figure 1. Consider the horizontal and vertical reactions at A are A x and A y . Consider the vertical reaction at D is D y . Calculate the value of the angle α , β , γ as follows: tan α = 24 32 α = tan − 1 ( 24 32 ) α = 36.869 ° β = 180 ° − 90 ° − α = 180 ° − 90 ° − 36.869 ° = 53.131 ° tan γ = 24 7 γ = tan − 1 ( 24 7 ) γ = 73.739 ° Find the length of the top chord members of truss as follows: ( A G ) 2 = 24 2 + 32 2 A G = 24 2 + 32 2 A G = 40 ft By symmetry of the truss. A G = G D = 40 ft A E = E G = G F = F D = 20 ft Consider the point load acting on the truss at A, E, G, F, and D are denoted by P A , P E , P G , P F , P D . Find the load acting on truss at A using the relation: P A = 300 lb/ft × A E 2 = 300 lb/ft × 20 2 ft = 3 , 000 lb × ( 1 k 1 , 000 lb ) = 3 k Find the load acting on the truss at E using the relation: P E = 300 lb/ft × A G 2 = 300 lb/ft × 40 2 ft = 6 , 000 lb × ( 1 k 1 , 000 lb ) = 6 k Find the load acting on the truss at G in span AG using the relation: P G = 300 lb/ft × E G 2 = 300 lb/ft × 20 2 ft = 3 , 000 lb × ( 1 k 1 , 000 lb ) = 3 k Find the load acting on the truss at G in span AD using the relation: P G = 150 lb/ft × G F 2 = 150 lb/ft × 20 2 ft = 1 , 500 lb × ( 1 k 1 , 000 lb ) = 1.5 k Find the load acting on the truss at F using the relation: P F = 150 lb/ft × G D 2 = 150 lb/ft × 40 2 ft = 3 , 000 lb × ( 1 k 1 , 000 lb ) = 3 k Find the load acting on the truss at D using the relation: P D = 150 lb/ft × D F 2 = 150 lb/ft × 20 2 ft = 1 , 500 lb × ( 1 k 1 , 000 lb ) = 1.5 k Show the load acting on the truss as shown in the Figure 2. Refer Figure 2. Find the distance AH , DJ using the relation: A H A E = D J D F = A I A G A H 20 = D J 20 = 32 40 A H = D J = 16 ft Find the distance EH , FJ using the relation: E H A E = F J D F = G I A G E H 20 = F J 20 = 24 40 E H = F J = 12 ft Calculate the value of the reactions at A and D as follows: Take the sum of the forces in the vertical direction as zero. ( − 3 − 6 − 3 + 1.5 + 3 + 1.5 ) cos α + A y + D y = 0 ( − 3 − 6 − 3 + 1.5 + 3 + 1.5 ) cos 36.869 ° + A y + D y = 0 − 4.8 + A y + D y = 0 A y + D y = 4.8 (1) Take the sum of the moment about A as zero. ∑ M A = 0 [ − ( 6 × 12 12 2 + 16 2 × 12 ) − ( 6 × 16 12 2 + 16 2 × 12 ) − ( 3 × 24 24 2 + 32 2 × 32 ) − ( 3 × 32 24 2 + 32 2 × 24 ) + ( 1