8. analysis of truss part ii, method of section, by-ghumare s m

Analysis of Truss-Method of Section

1. Analysis of Truss - Part-II
(Method of Section )
Sub- Engg. Mechanics
By: Mr. Ghumare S. M.

Truss

2. Stability of Truss
Depending on no. of members, stability of the truss is
checked by using following equation.
m = 2j-r r = No. of Reactions = 3
m = 2j-3
Where, m = Number of members,
j = Number of Joints

1. m = 2j-3 -----Satisfied Eq. ----Stable Truss
1. m 2j-3 -----Not Satisfied Eq.----Unstable



3. Methods of Section
Use: When forces in few members are required, method of
section is used.
Assumptions in the Analysis of Truss
1. Truss should be perfect truss
2. Self wt. of the truss members are neglected
3. External load acts on the joints only
4. All joints are assumed as hinge or pin joints
5. Members carrying axial loads only, Tension or Compression
in nature.
Sign Convention:
Assume Tensile force +ve (T) and Compression force – ve (C)

4. Analysis of Truss using Method of Section
Steps:
1.Draw F.B.D. of the given Truss, check stability if
reqd.
2.Apply conditions of equilibrium to entire truss and
find external reactions at supports
3.Take section or pass section through the members
where forces to be determined.
4.At a time, section should not pass through more than
three members.
5.Consider any one side of section i.e. L.H.S. or R.H.S.
for analysis.
6.Apply conditions of Equilibrium to selected section.
7.Find the forces in the corresponding members.

5. Example 1. Determine the forces in the members BC, HC
and HG for the truss shown in Fig. State weather the
forces are in tension or compression.
To check stability of the truss:
m = 2j-r , r = 3, m =13 , j =8
13 = 2x 8-3 , 13=16-3 , 13=13 Given truss is stable

6. Draw F.B.D. of the given Truss
Solution:
Draw F.B.D.
Find angles
E
 -1
H
H
θ= tan(3/2)
θ= 56.30
Apply conditions of equim. and find reactions at supports 0A= 0,
xx
F 0
YE
YE
A+ R-6 - 10 - 6 = 0,
A+ R= 22 KN = 0,-(1)
y
F

 E
A
E
E
E
-6x3 - 10x5 - 6x7 + Rx10 =0,
-18 - 50 - 42 + 10xR=0
10R= 110,
R= 11
M=0Anticlockwise
KN(), A= 11 N
-V
(
e
K)
y



7. L. H. S.

R. H. S.

8. Consider L.H.S. Assume all are tensile
Apply conditions of equim. 01.
Y HC
HC
HC
A- 6 - Fsin56.3= 0,
11- 6 - Fsin56.3 = 0,
F=6 KN (T)
y
F 2.
,
3
@
HGHC
BC
BC
C
H
B
Fx3 - 11x= 0,
Fx3 =
Mome
33,
F
ntofF,F6H
= 11
=
KN
M=
0
0
(T)
 03.
HG BC HC
HG
HG
HG
F+ F+ Fcos56.3= 0,
F+11 + 6 x 0.554= 0,
F= -14.32 KN
F= 14.32 KN (C)
x
F HC BC HG
F=6 KN (T),F= 11 KN (T)F= 14.32 KNAns: (C)

9. Example 2. Determine the forces in the members GF and GB
for the truss shown in Fig. State nature of the forces.
To check stability of the truss:
m = 2j-r , r = 3, m =11 , j =7
11 = 2x 7-3 , 11=14-3 , 11=11 Given truss is stable

10. Draw F.B.D. of the given Truss and find support reactions
Apply conditions of equillibrium. 0A= 0,
xx
F 0
YD
YD
A+ R-6 - 8 = 0,
A+ R= 14 KN = 0, (1)
y
F

 D
A
D
D
D
-6x3.05 - 8x5.49+ R x 8.54 =0,
-62.22 + 8.54R= 0
8.54R= 62.22,
R= 7.285 KN(
M=0Anticlockwise
), A= 6.715
-V
(
e
KN)
y



11. Take section 1-1
Consider L.H.S.

12. Consider L.H.S. Assume all are tensile
Apply conditions of equim. 01.
Y GB
GB
GB
A- F= 0,
6.715 - F= 0,
F=6.715 KN (T)
y
F 2.
AB
G
AB
AB
GBGF
Fx3.05 - Ayx3.05= 0,
Fx3.05 - 6.715x3.05= 0
MomentofF,
,
F = 6
F@
.71
G =
5 KN
M=0
0
(T)
 6.715
. 03
AB GF x
GF
GF
HG
F+ F+ A= 0,
+F+ 0= 0,
F= -6.715 KN
F= 6.715 KN(C)
x
F GB AB HG
F=6.715 KN (T),F= 6.An 715 KN (T)F= 6.715 KNs: (C)

13. Example 3. Determine the forces in the members DF, DG and
EG for the truss shown in Fig. State nature of the forces.
To check stability of the truss:
m = 2j-r , r = 3, m =21 , j =12
21 = 2x 12-3 , 21= 24-3 , 21=21 Given truss is stable

14. Draw F.B.D and Find support reactions yL
yL
yL
Given loa
A+ R-2-
d is Sy
2-2-2-2 =0
A+
mmetric
R= 10--------(1)
A= 5KN(), R=
a
5 KN)
0,
(
l
=
y
F

 0
0
,x
x
A
F



15. To find forces in members pass section through DF,
DG and EG
Considering the L.H.S to find the forces in the
members DF, DG and EG -1
D
0
D
θ= tan(3/4)
θ= 36.86

16. To find forces in members DF, DG and EG DG
DG
DG
DG
DG
5 - 2 - 2 - Fsin36.86 =0,
1 - Fsin36.86 =0,
1 - Fsin36.86 =0
F= 1
F
=0,
/
= 1.67KN(T
sin36.86,
)
y
F EG
EG
EG
EG
EG
D
-Ax 7 + 2 x 4 + Fx 3 =0,
-5 x 7 + 2 x 4 + Fx 3 =0,
-35 + 8 + Fx 3=0,
-27 + Fx 3=0
F= 9K
M=0
)
,
N (T
y


17. To find force in member DF X EG DF DG
DF
DF
D
F
F
F
D
D
A+ F+ F+ Fcos36.86 = 0,
0 + 9 + F+1.67cos36.86 = 0,
9 + F+ 1.67cos36.86 = 0,
F+ 1.67cos36.86 = -9,
F= -10.33
F= 10.3
=0,
3 K
KN
N(C)
X
F

18. Thank You