1. Problem 6.172
3 kN
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 F
3 kN
3 kN D H
6.75 m
1.5 kN
1.5 kN B J A L C E G I K
3 m
3 m
3 m
3 m
3 m
3 m

2. Solving Problems on Your Own
3 kN
Problem 6.172
Solving Problems on Your Own
3 kN
3 kN F
3 kN
A Pratt roof truss is loaded as shown. Using the method of sections, determine the force in members FH , FI, and GI.
3 kN D H
6.75 m
1.5 kN
1.5 kN B J A L C E G I K
3 m
3 m
3 m
3 m
3 m
3 m
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.

3. Solving Problems on Your Own
3 kN
Solving Problems on Your Own
3 kN
3 kN F
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 D H
6.75 m
1.5 kN
1.5 kN B J A L C E G I K
3 m
3 m
3 m
3 m
3 m
3 m
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.

4. S Fx = 0: Ax = 0 Ay = L = (18kN) = 9 kN
Problem Solution
3 kN
3 kN
Draw the free-body diagram of the entire truss, and use it to
determine reactions at the supports. F
3 kN
3 kN D H
1.5 kN
1.5 kN B J A L
S Fx = 0: Ax = 0 C E G I K + Ax Ay L
3 m
3 m
3 m
3 m
3 m
3 m
Total load = 5(3 kN) + 2(1.5 kN) = 18 kN
By symmetry 1 2
Ay = L = (18kN) = 9 kN

5. Pass a section through three members of the truss, one of
3 kN
Problem Solution
3 kN
3 kN F
Pass a section through three members of the truss, one of
which is the desired member.
3 kN
3 kN D H
1.5 kN
1.5 kN B J A L C E G I K
9 kN
9 kN
3 m
3 m
3 m
3 m
3 m
3 m

6. tan a = = a = 66.04o Free Body: Portion HIL
Problem Solution 5 3
Free Body: Portion HIL 4
3 kN F
FFH
Select one of the two portions of the truss you have obtained,
and draw its free-body diagram.
3 kN H
FFI
6.75 m
1.5 kN J a L
Slope of FHJL G I
FGI K
6.75
9.00 3 4 5
9 kN 3 = 4
3 m
3 m
3 m FG GI
6.75
3.00
tan a = = a = 66.04o
Now write three equilibrium equations which can be solved for
the forces in the three intersected members.

7. S MI = 0: FFH ( x 6.75 m) + (9 kN)(6 m)
Problem Solution 5 3
Free Body: Portion HIL 4
3 kN F
FFH
Write the three equilibrium equations.
3 kN H
FFI
6.75 m
1.5 kN J
66.04o L G I
FGI K
9 kN
Force in FH:
3 m
3 m
3 m 4 5 2 3
S MI = 0: +
FFH ( 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 = kN FFH = 10.0 kN C

8. FFH = 10.0 kN C S ML = 0: -FFI sin 66.04o(6 m) + (3 kN)(6 m)
Problem Solution 5 3
Free Body: Portion HIL 4
3 kN F
FFH
Write the three equilibrium equations.
3 kN H
FFH = 10.0 kN C
FFI
6.75 m
1.5 kN J
66.04o L G I
FGI K
9 kN
Force in FI:
3 m
3 m
3 m
S ML = 0: +
-FFI sin 66.04o(6 m) + (3 kN)(6 m)
+ (3 kN)(3 m) = 0
FFI sin 66.04o(6 m) = 27 kN-m
FFI = kN FFI = 4.92 kN T

9. -FGI (6.75 m) - (3 kN)(3 m) - (3 kN)(6 m)
Problem Solution 5 3
Free Body: Portion HIL 4
3 kN F
FFH
Write the three equilibrium equations.
3 kN H
FFH = 10.0 kN C
FFI = 4.92 kN T
FFI
6.75 m
1.5 kN J
66.04o L G I
FGI K
9 kN
Force in GI:
3 m
3 m
3 m
S 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) = kN-m
FGI = kN FGI = 6.00 kN T