University:
Angelo State University
Course:
ENGR 2301 | Engineering Mechanics - Statics
Academic year:

2024
Views:

210

Pages:

9
Author:

iwanankgfd

Problem 6.172 3 kN 3 kN 3 kN F 3 kN D H 3 kN 1.5 kN 6.75 m 1.5 kN B J A L C 3m E G 3m 3m I 3m K 3m 3m A Pratt roof truss is loaded as shown. Using the method of sections, determine the force in members FH , FI, and GI. 3 kN 3 kN Problem 6.172 Solving Problems on Your Own 3 kN F 3 kN D H 3 kN 1.5 kN B J 6.75 m 1.5 kN A L C 3m E G 3m 3m I 3m K 3m 3m A Pratt roof truss is loaded as shown. Using the method of sections, determine the force in members FH , FI, and GI. 1. Draw the free-body diagram of the entire truss, and use this diagram to determine the reactions at the supports. 2. Pass a section through three members of the truss, one of which is the desired member. Problem 6.172 3 kN Solving Problems on Your Own 3 kN F 3 kN A Pratt roof truss D H 6.75 m is loaded as 1.5 kN 1.5 kN shown. Using the B J method of A L sections, C I E G K determine the force in members 3m 3m 3m 3m FH , FI, and GI. 3m 3m 3. Select one of the two portions of the truss you have obtained, and draw its free-body diagram. This diagram should include the external forces applied to the selected portion as well as the forces exerted on it by the intersected members before these were removed. 4. Now write three equilibrium equations which can be solved for the forces in the three intersected members. 3 kN 3 kN 3 kN 3 kN Problem 6.172 Solution 3 kN F 3 kN D H 3 kN 1.5 kN B J L A Ax 1.5 kN C E I G K Ay Draw the free-body diagram of the entire truss, and use it to determine reactions at the supports. + Σ Fx = 0: Ax = 0 L 3m 3m 3m 3m 3m 3m Total load = 5(3 kN) + 2(1.5 kN) = 18 kN By symmetry Ay = L = 1 2 (18kN) = 9 kN 3 kN 3 kN Problem 6.172 Solution 3 kN F 3 kN D H 3 kN 1.5 kN B J 1.5 kN L A C E I G K 9 kN 9 kN 3m 3m 3m 3m 3m 3m Pass a section through three members of the truss, one of which is the desired member. 3 5 Free Body: Portion HIL 4 F 3 kN FFH H 3 kN FFI 6.75 m Problem 6.172 Solution J a G FGI 1.5 kN L I 3m Slope of FHJL K 9 kN 3m Select one of the two portions of the truss you have obtained, and draw its free-body diagram. 3m 6.75 = 3 4 9.00 FG 6.75 tan α = GI = 3.00 3 5 4 α = 66.04 o Now write three equilibrium equations which can be solved for the forces in the three intersected members. 3 5 Free Body: Portion HIL 4 F 3 kN Write the three equilibrium equations. FFH 3 kN H FFI 6.75 m J G Problem 6.172 Solution 66.04o FGI 1.5 kN L I K 9 kN 3m 3m Force in FH: 3m 4 2 + Σ MI = 0: FFH ( 5 3 x 6.75 m) + (9 kN)(6 m) - (1.5 kN)(6m) - (3 kN)(3m) = 0 4 5 FFH (4.5 m) + 36 kN-m = 0 FFH = -10.0 kN FFH = 10.0 kN C 3 5 Free Body: Portion HIL 4 F 3 kN Write the three equilibrium equations. FFH H 3 kN FFI 6.75 m J G Problem 6.172 Solution 66.04o FGI 1.5 kN L I K 9 kN 3m FFH = 10.0 kN C 3m Force in FI: 3m + Σ ML = 0: -FFI sin 66.04 o (6 m) + (3 kN)(6 m) + (3 kN)(3 m) = 0 FFI sin 66.04o(6 m) = 27 kN-m FFI = + 4.92 kN FFI = 4.92 kN T 3 5 Free Body: Portion HIL 4 F 3 kN Write the three equilibrium equations. FFH H 3 kN FFI 6.75 m J G Problem 6.172 Solution 1.5 kN 66.04o FGI FFH = 10.0 kN C FFI = 4.92 kN T L I K 9 kN 3m 3m Force in GI: 3m + Σ MF = 0: -FGI (6.75 m) - (3 kN)(3 m) - (3 kN)(6 m) - (1.5 kN)(9 m) + (9 kN)(9 m) = 0 FGI (6.75 m) = +40.5 kN-m FGI = + 6.00 kN FGI = 6.00 kN T