Truss examples

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:
=0 A M∑
) (2) + (900) (2) + (600) (4) = 0 y (B
2100 N = 2100N -= y . B
.
.
=0 yF∑
y= By = 0 Ay B -y A
2100N = 2100 N -= y A
= 0 x F∑
(1) ----------1500 -= x B –x + 900+ 600 = 0 Ax B –x A
Node A:

3. = 0 x A
= 2100 N (Tension) 1F
By substituting in eqn 1, we get:
=1500 N xB
Node D:

1341.6 N-= 6 + 600 = 0 F 6F
�
√??????

= 1341.6 N ( compression )6 F
(1341.6) = 0
�
√??????
-2 F
(Tension) = 1200 N 2 F
Node c:
(compression) = 1341.6 N 5= F 6Fand = 0 3FCE is a zero force member,so
Node E:

4. 900 -sin45 = 4 F
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. =0C M∑
= 8 KN x (1) = 0 E xE -2(1) + 3(2)
= 0 x F∑
x= E x= 0 C xE - xC
= 8 KN xC
= 0 y F∑
(1) ---------= 5y + Ey 3 = 0 C –2 - y + Ey C
Node E:

= 0 y E
)compression= 8 KN ( 1F8 -= 1 F
By substituting in eqn 1, we get:

6. = 5 KN yC
Node C:

= 7.07 KN ( Tension ) √�= 5 2Fsin45 = 5 2F
= 8 3cos45 + F√� = 8 53 cos45 + F 2F
= 3 KN (Tension) 3F
Node B:

= 3 KN (Tension) 5F
= 2 KN (Tension) 4F
Node A:

7. = 0 yF∑
sin45 = 0 63 + F
= 4.24 KN (Tension) √�= 3 6F

EX(3):
Determine the force in each member of the truss and state if the
and = 2 KN 1Pmembers are in tension or compression. Set
= 1.5 KN. 2P

Solution
m:Free Body Diagra

8. =0 A M∑
) = 0 √� (2 x(1.5) (6)+(2) (3) + E
4.33 KN = 4.33 KN -= xE
= 0 xF∑
= 4.33 KN x4.33) = 0 A -+ ( xA
= 0 yF∑
= 3.5 ……………..(1) y+ E yA
Node C:

= 3 KN (Tension) 2Fsin30 = 1.5 2F

9. 2.6 KN=2.6 KN (compression)-= √�1.5 -= 1F+ 3cos30 = 0 1F
Node D:

= 2 KN (Tension) 3F
2.6 KN =2.6 KN (compression) -= 4F
Node E:

2.6 = 0 -cos30 + 4.33 5F
2 KN = 2 KN (compression) -= 5F
= 1 KN y2) sin30 = 0 E-+ ( yE
By substituting in eqn(1), we get
= 2.5 KN y+ 1 = 3.5 A yA
Node A:

10. cos60 = 2.5 6F
= 5 KN (Tension) 6F

EX(4):
Determine the force in each member of the truss and state if the
= 0. 2Pand KN 4= 1Pmembers are in tension or compression. Set

Solution
Free Body Diagram:

11. =0 A M∑
= 2 KN y(4) (2) = 0 E –(4) yE
= 0 yF∑
2 KN = 2 KN -= y+ 2 = 0 A yA
= 0 xF∑
(1) --------- x= A x= 0 E xA - xE
FD & BG are zero force members, so
and as a result = 0 2Fand = 0 9F
AG & GC & FC & FE became zero force members,so
= 0 8= F 10= F 3= F 1F
Node A:

12. = 2 4(0.75/1.25)F
= 10/3 = 3.33 KN (Tension) 4F
= 8/3 = 2.67 KN 4= (1/1.25) F xA
By substituting in eqn (1), we get
= 2.67 KN xE
Node B:
= 3.33 KN (Tension) 5= F 4F
Node F:

= 2 7(0.75/1.25) F
= 10/3 =3.33 KN (Tension) 7F
Node D:

13. = 3.33 KN (Tension) 6 = F 7F

EX(5):
Determine the force in each member of the truss and state if the
and N1200 = 1Pmembers are in tension or compression. Set
= 1500 N. 2P

Solution
Free Body Diagram:


=0 c M∑
500(3.6) =0 – x1.5 B

14. = 1200 N xB
=0 x F∑
+1200 +1200 =0 xC
2400 N -= xC
=0y F∑
= 500 N y500 = 0 C – yC
Node D:

1300 N = 1300 N (compression) -= 2F500 -= 2(1.5/3.9)F
= 2400 N (Tension) 1F= 1200 2+ (3.6/3.9)F 1F
AB & AC are zero force members, so
= 0 5= F 3F
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:

=0 c M∑
(2.4) = 0 yA -3(1.2) + 4.5(3.6)
= 8.25 KN yA
= 0 x=0 Cx F∑
=0 y F∑
= 0.75 KN y= 0 C -4.5 –3 –+ 8.25 yC
BD and EA are zero force members, so

16. = 0 8= F 3F
Node C:

= 1.25 KN (Tension) 4F= 0.75 4(0.9/1.5)F
+(1.2/1.5) (1.25) = 0 1F
1 KN = 1 KN (compression) -= 1F
Node B:
= 1 KN (compression) 2= F 1F
Node D:

+ 0.75 + 3 = 0 5(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= F 6F
Node F:

= 4.5 9(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 section a-a since it cuts through the three members

18. •Draw FBD of the entire truss

= 400N x= 0 A xA -= 0; 400N x∑F+
= 900N y(12) = 0 Dy400(3) + D -1200(8) -= 0; A∑M
= 300N y1200 + 900 =0 A – y= 0; A y+ ∑F

• Draw Free Body Diagram for the section portion


∑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 member is 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 support reactions
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 KN CFF

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