1) The documents provide examples of solving for forces in truss members by using free body diagrams and equilibrium equations. 2) The solutions involve drawing FBDs of the trusses or sections of trusses, then writing the ΣF and ΣM equations and solving the systems of equations. 3) The determined forces are then identified as either tension or compression forces in the members.

1. Examples
EX (1):
Determine the force in each member of the truss, and
state if the members are in tension or compression.
Solution
Free Body Diagram:

2. First, we should calculate reactions at A and B:
=0AM∑
) (2) + (900) (2) + (600) (4) = 0y(B
2100 N = 2100N-=y. B.
.
=0yF∑
y= By= 0 AyB-yA
2100N = 2100 N-=yA
= 0xF∑
(1)----------1500-=xB–x+ 900+ 600 = 0 AxB–xA
Node A:

3. = 0xA
= 2100 N (Tension)1F
By substituting in eqn 1, we get:
=1500 NxB
Node D:
1341.6 N-=6+ 600 = 0 F6F
𝟏
√𝟓
= 1341.6 N ( compression )6F
(1341.6) = 0
𝟐
√𝟓
-2F
(Tension)= 1200 N2F
Node c:
(compression)= 1341.6 N5= F6Fand= 03FCE is a zero force member,so
Node E:

4. 900-sin45 =4F
1272.8 N = 1272.8 (compression)-=4F
EX(2):
The truss, used to support a balcony, is subjected to the loading
shown. Approximate each joint as a pin and determine the force
in each member. State whether the members are in tension or
.2 KN=2KN, P3=1Pcompression. Set
Solution
Diagram:Free Body

5. =0CM∑
= 8 KNx(1) = 0 ExE-2(1) + 3(2)
= 0xF∑
x= Ex= 0 CxE-xC
= 8 KNxC
= 0yF∑
(1)---------= 5y+ Ey3 = 0 C–2-y+ EyC
Node E:
= 0yE
)compression= 8 KN (1F8-=1F
By substituting in eqn 1, we get:

6. = 5 KNyC
Node C:
= 7.07 KN ( Tension )√ 𝟐= 52Fsin45 = 52F
= 83cos45 + F√ 𝟐= 8 53cos45 + F2F
= 3 KN (Tension)3F
Node B:
= 3 KN (Tension)5F
= 2 KN (Tension)4F
Node A:

7. = 0yF∑
sin45 = 063 + F
= 4.24 KN (Tension)√ 𝟐= 36F
EX(3):
Determine the force in each member of the truss and state if the
and= 2 KN1Pmembers are in tension or compression. Set
= 1.5 KN.2P
Solution
m:Free Body Diagra

8. =0AM∑
) = 0√ 𝟑(2x(1.5) (6)+(2) (3) + E
4.33 KN = 4.33 KN-=xE
= 0xF∑
= 4.33 KNx4.33) = 0 A-+ (xA
= 0yF∑
= 3.5 ……………..(1)y+ EyA
Node C:
= 3 KN (Tension)2Fsin30 = 1.52F

9. 2.6 KN=2.6 KN (compression)-=√ 𝟑1.5-=1F+ 3cos30 = 01F
Node D:
= 2 KN (Tension)3F
2.6 KN =2.6 KN (compression)-=4F
Node E:
2.6 = 0-cos30 + 4.335F
2 KN = 2 KN (compression)-=5F
= 1 KNy2) sin30 = 0 E-+ (yE
By substituting in eqn(1), we get
= 2.5 KNy+ 1 = 3.5 AyA
Node A:

10. cos60 = 2.56F
= 5 KN (Tension)6F
EX(4):
Determine the force in each member of the truss and state if the
= 0.2PandKN4=1Pmembers are in tension or compression. Set
Solution
Free Body Diagram:

11. =0AM∑
= 2 KNy(4) (2) = 0 E–(4)yE
= 0yF∑
2 KN = 2 KN-=y+ 2 = 0 AyA
= 0xF∑
(1)---------x= Ax= 0 ExA-xE
FD & BG are zero force members, so
and as a result= 02Fand= 09F
AG & GC & FC & FE became zero force members,so
= 08= F10= F3= F1F
Node A:

12. = 24(0.75/1.25)F
= 10/3 = 3.33 KN (Tension)4F
= 8/3 = 2.67 KN4= (1/1.25) FxA
By substituting in eqn (1), we get
= 2.67 KNxE
Node B:
= 3.33 KN (Tension)5= F4F
Node F:
= 27(0.75/1.25) F
= 10/3 =3.33 KN (Tension)7F
Node D:

13. = 3.33 KN (Tension)6= F7F
EX(5):
Determine the force in each member of the truss and state if the
andN1200=1Pmembers are in tension or compression. Set
= 1500 N.2P
Solution
Free Body Diagram:
=0cM∑
500(3.6) =0–x1.5 B

14. = 1200 NxB
=0xF∑
+1200 +1200 =0xC
2400 N-=xC
=0yF∑
= 500 Ny500 = 0 C–yC
Node D:
1300 N = 1300 N (compression)-=2F500-=2(1.5/3.9)F
= 2400 N (Tension)1F= 12002+ (3.6/3.9)F1F
AB & AC are zero force members, so
= 05= F3F
Node B:
500 N = 500 N (compression)-=4F
EX(6):

15. Determine the force in each member of the truss and state if the
members are in tension or compression.
Solution
Free Body Diagram:
=0cM∑
(2.4) = 0yA-3(1.2) + 4.5(3.6)
= 8.25 KNyA
= 0x=0 CxF∑
=0yF∑
= 0.75 KNy= 0 C-4.5–3–+ 8.25yC
BD and EA are zero force members, so

16. = 08= F3F
Node C:
= 1.25 KN (Tension)4F= 0.754(0.9/1.5)F
+(1.2/1.5) (1.25) = 01F
1 KN = 1 KN (compression)-=1F
Node B:
= 1 KN (compression)2= F1F
Node D:
+ 0.75 + 3 = 05(0.9/1.5)F
6.25 KN = 6.25 KN (compression)-=5F
5 = 0–6F–1

17. 4 KN = 4KN (compression)-=6F
Node E:
= 4 KN (compression)7= F6F
Node F:
= 4.59(0.9/1.5) F
= 7.5 KN (Tension)9F
EX(7):
Determine the force in members GE, GC, and BC of the truss.
Indicate whether the members are in tension or compression.
Solution
•Choose sectiona-a since it cuts through the three members

18. •Draw FBD of the entire truss
= 400Nx= 0 AxA-= 0; 400Nx∑F+
= 900Ny(12) = 0 Dy400(3) + D-1200(8)-= 0;A∑M
= 300Ny1200 + 900 =0 A–y= 0; Ay+ ∑F
• Draw Free Body Diagram for the sectionportion
∑MG = 0; - 300(4) – 400(3) + FBC(3) = 0 FBC = 800N (Tension)
∑MC = 0; - 300(8) + FGE(3) = 0 FGE = 800N (Compression)
+ ∑Fy = 0; 300 -
3
5
FGC = 0 FGC = 500N (Tension)
EX(8):

19. Determine the force CF of the truss shown in the figure. Indicate
whether the member is in tension or compression. Assume
each memberis pin connected.
Solution
Free Body Diagram:
Section aa will be used since this
section will “expose”the internal force in member CF as “external”·
on the free-body diagram of either the right or left portion of the
truss. It is first necessary, however, to determine the supportreactions
on either the left or right side.
The free-body diagram of the right portion of the truss, which is the
easiest to analyze, is shown in the following figure:

20. Equations of Equilibrium: We will apply the moment equation
about point 0 in order to eliminate the two unknowns F FG and FcD.
The location of point 0 measured from E can be determined from
proportional triangles. i.e. 4/(4 + x) = 6/(8 + x ) . x = 4 m. Or,
stated in another manner. The slope of member GF has a drop of 2 m
to a horizontal distance of 4 m. Since FD is4 m. Fig. then from
D t0 O the distance must be 8 m.
An easy way to determine the moment of FCF about point O is to use
the principle of transmissibility and slide FCF to point C. and then
resolve FCF into its two rectangular components. We have
= 0;O∑M+
4.5(4) = 0–sin45 (12) + 3(8)CFF-
(compression)= 0.589 KNCFF
EX(9):

21. Determine the force in member EB of the roof truss shown in
the figure. Indicate whether the member is in tension or
compression.
Solution