1. Find: FCD [kN] 25.6 (tension) 25.6 (compression) 26.3 (tension)B D
2 [m]
20 [kN]
7[m] A H G F E
5 [m]
5 [m]
5 [m]
5 [m]
25.6 (tension)
25.6 (compression)
26.3 (tension)
26.3 (compression)
Find the force in member CD, in kiloNewtons. [pause] In this problem, ---
60 [kN]

2. Find: FCD [kN] 25.6 (tension) 25.6 (compression) 26.3 (tension)B D
2 [m]
20 [kN]
7[m] A H G F E
5 [m]
5 [m]
5 [m]
5 [m]
25.6 (tension)
25.6 (compression)
26.3 (tension)
26.3 (compression)
A truss is subjected to a 60 kiloNewton force, downward, at joint F, and also subjected a ----
60 [kN]

3. Find: FCD [kN] 25.6 (tension) 25.6 (compression) 26.3 (tension)B D
2 [m]
20 [kN]
7[m] A H G F E
5 [m]
5 [m]
5 [m]
5 [m]
25.6 (tension)
25.6 (compression)
26.3 (tension)
26.3 (compression)
20 kiloNewton force, to the right, at joint B. Externally, the truss is connected ---
60 [kN]

4. Find: FCD [kN] 25.6 (tension) 25.6 (compression) 26.3 (tension)B D
2 [m]
20 [kN]
7[m] A H G F E
5 [m]
5 [m]
5 [m]
5 [m]
25.6 (tension)
25.6 (compression)
26.3 (tension)
26.3 (compression)
with a pin connection, at joint A, and a roller connection, at joint E.
60 [kN]

5. Find: FCD [kN] 25.6 (tension) 25.6 (compression) 26.3 (tension)B D
2 [m]
20 [kN]
7[m] A H G F E
5 [m]
5 [m]
5 [m]
5 [m]
25.6 (tension)
25.6 (compression)
26.3 (tension)
26.3 (compression)
The dimensions of the truss, are also provided. [pause] The force in member CD can be determined, ----
60 [kN]

6. Find: FCD [kN] C B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m] 5 [m] 5 [m]
by cutting an imaginary section through members, CD, DG and FG, ---
60 [kN]

7. Find: FCD [kN] ΣMG = 0 θ =-60 [kN] * dFG+FE,y * dEGo
+FCD * dDG * cos(90-θ)
7[m]
FDG
FFG G F E
summing the moments at point G, setting the value to zero, and solving for the for the force ---
5 [m]
5 [m]
60 [kN]
FE,y

8. Find: FCD [kN] ΣMG = 0 θ =-60 [kN] * dFG+FE,y * dEGo
+FCD * dDG * cos(90-θ)
7[m]
FDG
60 [kN] * dFG-FE,y*dEG
FCD =
FFG F E
dDG * cos(90-θ) o
in member CD. [pause] This equation assumes, ----
5 [m]
60 [kN]
FE,y

9. Find: FCD [kN] ΣMG = 0 θ =-60 [kN] * dFG+FE,y * dEGo
+FCD * dDG * cos(90-θ)
7[m]
FDG
60 [kN] * dFG-FE,y*dEG
FCD =
FFG F E
dDG * cos(90-θ) o
member CD experiences a tensile force. Currently, the unknown variables in this equation include ---
5 [m]
tensile
60 [kN]
FE,y

10. Find: FCD [kN] ΣMG = 0 θ =-60 [kN] * dFG+FE,y * dEGo
+FCD * dDG * cos(90-θ)
7[m]
FDG
60 [kN] * dFG-FE,y*dEG
FCD =
FFG F E
dDG * cos(90-θ) o
the length of member DG, the vertical force at joint E, and the angle theta, made by members CD and DG.
5 [m]
tensile
60 [kN]
FE,y

11. Find: FCD [kN] ΣMG = 0 θ =-60 [kN] * dFG+FE,y * dEGo
+FCD * dDG * cos(90-θ)
7[m]
FDG
60 [kN] * dFG-FE,y*dEG
FCD =
FFG F E
dDG * cos(90-θ) o
To solve for the vertical force at joint E, ---
5 [m]
tensile
60 [kN]
FE,y

12. Find: FCD [kN] ΣMA = 0 C B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m] 5 [m]
We’ll sum the moments about joint A, which includes ---
FE,y
60 [kN]

13. Find: FCD [kN] ΣMA = 0 φ C B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m]
the 20 kiloNewton force at joint B, ---
FE,y
60 [kN]
0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE

14. Find: FCD [kN] ΣMA = 0 φ C B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m]
the 60 kiloNewton force at joint F, ---
FE,y
60 [kN]
0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE

15. Find: FCD [kN] ΣMA = 0 φ C B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m]
and the vertical reaction force, at the roller connection, at point E. Solving for the vertical force at joint E, ---
FE,y
60 [kN]
0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE

16. Find: FCD [kN] ΣMA = 0 φ C B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m]
we’re left without knowing the angle phi, ----
FE,y
60 [kN]
0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE
20 [kN]*dAB*cos(φ) + 60[kN]*dAF
FE,y=
dAE

17. Find: FCD [kN] ΣMA = 0 φ C B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m]
or the distances between the joints, A and B, A and E, and A and F.
FE,y
60 [kN]
0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE
20 [kN]*dAB*cos(φ) + 60[kN]*dAF
FE,y=
dAE

18. Find: FCD [kN] ΣMA = 0 φ C B D 2 [m] dAB= (5[m])2+(7[m])2 7[m] 7[m] AH G F E
5 [m]
5 [m]
5 [m]
5 [m]
ΣMA = 0
The length of member AB equals the square root of
FE,y
60 [kN]
0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE
20 [kN]*dAB*cos(φ) + 60[kN]*dAF
FE,y=
dAE

19. Find: FCD [kN] ΣMA = 0 φ C B D 2 [m] dAB= (5[m])2+(7[m])2 7[m] 7[m]H G F E
5 [m]
5 [m]
5 [m]
5 [m]
ΣMA = 0
meters. The length from joint A to joint E, equals, ---
FE,y
60 [kN]
0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE
20 [kN]*dAB*cos(φ) + 60[kN]*dAF
FE,y=
dAE

20. Find: FCD [kN] ΣMA = 0 φ C B D 2 [m] dAB= (5[m])2+(7[m])2 7[m]H G F E
dAE = 20 [m]
5 [m]
5 [m]
5 [m]
5 [m]
ΣMA = 0
20 meters, and the length from joint A to joint F, equals, ----
FE,y
60 [kN]
0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE
20 [kN]*dAB*cos(φ) + 60[kN]*dAF
FE,y=
dAE

21. Find: FCD [kN] ΣMA = 0 φ C B D 2 [m] dAB= (5[m])2+(7[m])2 7[m]H G F E
dAE = 20 [m]
dAF = 15 [m]
5 [m]
5 [m]
5 [m]
5 [m]
ΣMA = 0
15 meters. [pause] The angle phi is calculated ---
FE,y
60 [kN]
0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE
20 [kN]*dAB*cos(φ) + 60[kN]*dAF
FE,y=
dAE

22. Find: FCD [kN] ΣMA = 0 φ=tan-1 φ C 5 [m] B D 2 [m] 7 [m] 7[m] 7[m] A HG F
dAB= 74 [m] E
dAE = 20 [m]
dAF = 15 [m]
5 [m]
5 [m]
5 [m]
5 [m]
ΣMA = 0
by taking the arctangent of 5 meters, divided by 7 meters, which equals, ---
FE,y
60 [kN]
0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE
20 [kN]*dAB*cos(φ) + 60[kN]*dAF
FE,y=
dAE

23. Find: FCD [kN] ΣMA = 0 φ=tan-1 φ=35.5 φ C 5 [m] B D 2 [m] 7 [m] 7[m]o φ
7[m]
7[m] A H G F
dAB= 74 [m] E
dAE = 20 [m]
dAF = 15 [m]
5 [m]
5 [m]
5 [m]
5 [m]
ΣMA = 0
35.5 degrees. [pause] Plugging in the angle, the lengths, and the forces, ---
FE,y
60 [kN]
0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE
20 [kN]*dAB*cos(φ) + 60[kN]*dAF
FE,y=
dAE

24. Find: FCD [kN] ΣMA = 0 φ=35.5 φ C B D 2 [m] 20 [kN] 7[m] A H G Fo φ
7[m] A H G F
dAB= 74 [m] E
dAE = 20 [m]
dAF = 15 [m]
5 [m]
5 [m]
5 [m]
5 [m]
ΣMA = 0
the vertical reaction force at joint E equals ---
FE,y
60 [kN]
0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE
20 [kN]*dAB*cos(φ) + 60[kN]*dAF
FE,y=
dAE

25. Find: FCD [kN] ΣMA = 0 φ=35.5 φ C B D 2 [m] 20 [kN] 7[m] A H G Fo φ
7[m] A H G F
dAB= 74 [m] E
dAE = 20 [m]
dAF = 15 [m]
5 [m]
5 [m]
5 [m]
5 [m]
ΣMA = 0
52 kiloNewtons. And we can add this value to the figure.
FE,y
60 [kN]
0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE
20 [kN]*dAB*cos(φ) + 60[kN]*dAF
FE,y=
=52 [kN]
dAE

26. Find: FCD [kN] ΣMA = 0 φ=35.5 φ C B D 2 [m] 20 [kN] 7[m] A H G Fo φ
7[m] A H G F
dAB= 74 [m] E
dAE = 20 [m]
dAF = 15 [m]
5 [m]
5 [m]
5 [m]
5 [m]
ΣMA = 0
[pause] Returning to the section we cut earlier, ---
52 [kN]
60 [kN]
0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE
20 [kN]*dAB*cos(φ) + 60[kN]*dAF
FE,y=
=52 [kN]
dAE

27. Find: FCD [kN] ΣMG = 0 θ =-60 [kN] * dFG+FE,y * dEGo
+FCD * dDG * cos(90-θ)
7[m]
FDG
60 [kN] * dFG-FE,y*dEG
FCD =
FFG F E
dDG * cos(90-θ) o
we can plug in our value of 52 kiloNewtons for the vertical reaction at joint E.
5 [m]
60 [kN]
FE,y

28. Find: FCD [kN] ΣMG = 0 θ =-60 [kN] * dFG+FE,y * dEGo
+FCD * dDG * cos(90-θ)
7[m]
FDG
60 [kN] * dFG-FE,y*dEG
FCD =
FFG F E
dDG * cos(90-θ) o
The other two unknown variables needed to solve for the force ---
5 [m]
60 [kN]
52 [kN]

29. Find: FCD [kN] ΣMG = 0 θ =-60 [kN] * dFG+FE,y * dEGo
+FCD * dDG * cos(90-θ)
7[m]
FDG
60 [kN] * dFG-FE,y*dEG
FCD =
FFG F E
dDG * cos(90-θ) o
in member CD are, the length of member DG, and the angle, theta. [pause]
5 [m]
60 [kN]
52 [kN]

30. Find: FCD [kN] dDG θ 60 [kN] * dFG-FE,y*dEG FCD = dDG * cos(90-θ) C
2 [m]
dDG= (5[m])2+(7[m])2
FCD D θ
dDG
7[m]
FDG
FFG G F E
The length of member DG, is the same hypotenuse we previously calculated, ---
5 [m]
5 [m]
60 [kN]
52 [kN]
60 [kN] * dFG-FE,y*dEG
FCD =
dDG * cos(90-θ)

31. Find: FCD [kN] dDG θ 60 [kN] * dFG-FE,y*dEG FCD = dDG * cos(90-θ) C
2 [m]
dDG= (5[m])2+(7[m])2
FCD D θ
dDG= 74 [m]
dDG
7[m]
FDG
FFG G F E
the square root of 74, meters. [pause] To find the angle theta, ---
5 [m]
5 [m]
60 [kN]
52 [kN]
60 [kN] * dFG-FE,y*dEG
FCD =
dDG * cos(90-θ)

32. Find: FCD [kN] θ 60 [kN] * dFG-FE,y*dEG FCD = dDG * cos(90-θ) C C
2 [m]
FCD D θ D
7[m]
FDG
FFG G F E
we’ll extract the triangle CDG from the truss, and define the angles, ----
5 [m]
5 [m]
60 [kN]
52 [kN] G
60 [kN] * dFG-FE,y*dEG
FCD =
dDG= 74 [m]
dDG * cos(90-θ)

33. Find: FCD [kN] θ1 θ θ2 60 [kN] * dFG-FE,y*dEG FCD = dDG * cos(90-θ) C
2 [m]
FCD D θ1 θ D
7[m] θ2
FDG
FFG G F E
theta 1 and theta 2. Which makes theta equal to ---
5 [m]
5 [m]
60 [kN]
52 [kN] G
60 [kN] * dFG-FE,y*dEG
FCD =
dDG= 74 [m]
dDG * cos(90-θ)

34. Find: FCD [kN] θ = θ1 + θ2 θ1 θ θ2 60 [kN] * dFG-FE,y*dEG FCD =
2 [m]
θ = θ1 + θ2
FCD D θ1 θ D
7[m] θ2
FDG
FFG G F E
theta 1 plus theta 2. From the given dimensions of the truss, ---
5 [m]
5 [m]
60 [kN]
52 [kN] G
60 [kN] * dFG-FE,y*dEG
FCD =
dDG= 74 [m]
dDG * cos(90-θ)

35. Find: FCD [kN] θ = θ1 + θ2 θ1 θ θ1=tan-1 60 [kN] * dFG-FE,y*dEG FCD =
2 [m]
θ = θ1 + θ2
FCD D
2 [m] θ1 θ
5 [m]
7[m]
FDG
2 [m]
θ1=tan-1
FFG G F E
5 [m]
Theta 1 equals the arctangent of 2 meters divided by 5 meters, or ---
5 [m]
5 [m]
60 [kN]
52 [kN]
60 [kN] * dFG-FE,y*dEG
FCD =
dDG= 74 [m]
dDG * cos(90-θ)

36. Find: FCD [kN] θ = θ1 + θ2 θ1 θ θ1=tan-1 θ1=21.8
2 [m]
θ = θ1 + θ2
FCD D
2 [m] θ1 θ
5 [m]
7[m]
FDG
2 [m]
θ1=tan-1
FFG G F E
5 [m] o
θ1=21.8
21.8 degrees. [pause] Likewise, theta 2 equals, ---
5 [m]
5 [m]
60 [kN]
52 [kN]
60 [kN] * dFG-FE,y*dEG
FCD =
dDG= 74 [m]
dDG * cos(90-θ)

37. Find: FCD [kN] θ = θ1 + θ2 θ θ2 θ2=tan-1 60 [kN] * dFG-FE,y*dEG FCD =
2 [m]
θ = θ1 + θ2
FCD D θ
5 [m]
7[m] θ2
FDG
7 [m]
7 [m]
FFG
θ2=tan-1 F E
5 [m]
the arctangent of 7 meters divided by 5 meters, which equals, ---
5 [m]
5 [m]
60 [kN]
52 [kN]
60 [kN] * dFG-FE,y*dEG
FCD =
dDG= 74 [m]
dDG * cos(90-θ)

38. Find: FCD [kN] θ = θ1 + θ2 θ θ2 θ2=tan-1 θ2=54.5
2 [m]
θ = θ1 + θ2
FCD D θ
5 [m]
7[m] θ2
FDG
7 [m]
7 [m]
FFG
θ2=tan-1 F E
5 [m]
54.5 degrees. [pause] Adding the values of theta 1, ---
5 [m]
5 [m] o
θ2=54.5
60 [kN]
52 [kN]
60 [kN] * dFG-FE,y*dEG
FCD =
dDG= 74 [m]
dDG * cos(90-θ)

39. Find: FCD [kN] θ = θ1 + θ2 θ θ1=21.8 θ2=54.5 60 [kN] * dFG-FE,y*dEG
2 [m]
θ = θ1 + θ2
FCD D θ o
θ1=21.8
7[m] o
θ2=54.5
FDG
FFG G F E
and theta 2, theta equals, ---
5 [m]
5 [m]
60 [kN]
52 [kN]
60 [kN] * dFG-FE,y*dEG
FCD =
dDG= 74 [m]
dDG * cos(90-θ)

40. Find: FCD [kN] θ = θ1 + θ2 θ θ1=21.8 θ2=54.5 θ=76.3
2 [m]
θ = θ1 + θ2
FCD D θ o
θ1=21.8
7[m] o
θ2=54.5 o
FDG
θ=76.3
FFG G F E
76.3 degrees. [pause] From observation, the length of ---
5 [m]
5 [m]
60 [kN]
52 [kN]
60 [kN] * dFG-FE,y*dEG
FCD =
dDG= 74 [m]
dDG * cos(90-θ)

41. Find: FCD [kN] θ = θ1 + θ2 θ θ1=21.8 θ2=54.5 θ=76.3
2 [m]
θ = θ1 + θ2
FCD D θ o
θ1=21.8
7[m] o
θ2=54.5 o
FDG
θ=76.3
FFG
dFG=5 [m] G F E
member FG equals 5 meters, and the length between joints E and G equals ---
5 [m]
5 [m]
60 [kN]
52 [kN]
60 [kN] * dFG-FE,y*dEG
FCD =
dDG= 74 [m]
dDG * cos(90-θ)

42. Find: FCD [kN] θ = θ1 + θ2 θ θ1=21.8 θ2=54.5 θ=76.3
2 [m]
θ = θ1 + θ2
FCD D θ o
θ1=21.8
7[m] o
θ2=54.5 o
FDG
θ=76.3
FFG
dFG=5 [m] G F E
dEG=10 [m]
10 meters. Plugging these values into our equation,
5 [m]
5 [m]
60 [kN]
52 [kN]
60 [kN] * dFG-FE,y*dEG
FCD =
dDG= 74 [m]
dDG * cos(90-θ)

43. Find: FCD [kN] θ = θ1 + θ2 θ θ1=21.8 θ2=54.5 θ=76.3
2 [m]
θ = θ1 + θ2
FCD D θ o
θ1=21.8
7[m] o
θ2=54.5 o
FDG
θ=76.3
FFG
dFG=5 [m] G F E
dEG=10 [m]
the tensile force in member CD equals, ---
5 [m]
5 [m]
60 [kN]
52 [kN]
60 [kN] * dFG-FE,y*dEG
FCD =
dDG= 74 [m]
dDG * cos(90-θ)

44. Find: FCD [kN] θ = θ1 + θ2 θ θ1=21.8 θ2=54.5 θ=76.3 FCD =-26.3 [kN]
2 [m]
θ = θ1 + θ2
FCD D θ o
θ1=21.8
7[m] o
θ2=54.5 o
FDG
θ=76.3
FFG G F E
FCD =-26.3 [kN]
negative 26.3 kiloNewtons, in tension, which is equivalent a compressive force of, --
5 [m]
5 [m]
(tension)
60 [kN]
52 [kN]
60 [kN] * dFG-FE,y*dEG
FCD =
dDG= 74 [m]
dDG * cos(90-θ)

45. Find: FCD [kN] θ = θ1 + θ2 θ θ1=21.8 θ2=54.5 θ=76.3 FCD =26.3 [kN]
2 [m]
θ = θ1 + θ2
FCD D θ o
θ1=21.8
7[m] o
θ2=54.5 o
FDG
θ=76.3
FFG G F E
FCD =26.3 [kN]
26.3 kiloNewtons.
5 [m]
5 [m]
(compression)
60 [kN]
52 [kN]
60 [kN] * dFG-FE,y*dEG
FCD =
dDG= 74 [m]
dDG * cos(90-θ)

46. Find: FCD [kN] θ = θ1 + θ2 θ θ1=21.8 θ2=54.5 θ=76.3 25.6 (tension)
2 [m]
θ = θ1 + θ2
FCD D θ o
θ1=21.8
7[m] o
θ2=54.5 o
FDG
θ=76.3
25.6 (tension)
25.6 (compression)
26.3 (tension)
26.3 (compression)
FFG G F E
FCD =26.3 [kN]
When reviewing the possible solutions, ---
5 [m]
5 [m]
(compression)
60 [kN]
52 [kN]
60 [kN] * dFG-FE,y*dEG
FCD =
dDG= 74 [m]
dDG * cos(90-θ)

47. Find: FCD [kN] θ = θ1 + θ2 θ θ1=21.8 θ2=54.5 θ=76.3 AnswerD
2 [m]
θ = θ1 + θ2
FCD D θ o
θ1=21.8
7[m] o
θ2=54.5 o
FDG
θ=76.3
25.6 (tension)
25.6 (compression)
26.3 (tension)
26.3 (compression)
FFG
AnswerD G F E
FCD =26.3 [kN]
the answer is D.
5 [m]
5 [m]
(compression)
60 [kN]
52 [kN]
60 [kN] * dFG-FE,y*dEG
FCD =
dDG= 74 [m]
dDG * cos(90-θ)

48. ( ) ? τ [lb/ft2] γclay=53.1[lb/ft3] Index σ’v = Σ φ γ Δ d ˚ H*C σfinal
Find: σ’v ρc d = 30 feet
(1+wc)*γw
wc+(1/SG)
σ’v = Σ φ γ Δ d ˚ d
Sand
10 ft
γT=100 [lb/ft3]
100 [lb/ft3]
φ=α1-α2
10 [ft]
20 ft
Clay
= γsand dsand
+γclay dclay A W S
V [ft3]
W [lb]
40 ft
text
wc = 37% ? Δh
20 [ft]
τ [lb/ft2]
(5 [cm])2 * π/4
( )
H*C σfinal
ρcn=
1+e σinitial
log ‘ φ
c=0
400
1,400 σ3
Sand σ1