method of section
-
Upload
muhd-nadzri -
Category
Documents
-
view
225 -
download
0
Transcript of method of section
-
8/12/2019 method of section
1/28
-
8/12/2019 method of section
2/28
Method of Joints Example
Using the method of joints, determine the
force in each member ofthe truss.
-
8/12/2019 method of section
3/28
Method of Joints Example
Draw the free bodydiagram of the truss and
solve for the equations
x x
x
y y
y
0
0 lb
0 2000 lb 1000 lb
3000 lb
F C
C
F E C
E C
-
8/12/2019 method of section
4/28
Method of Joints
Example
Solve the moment about C
C
y
0 2000 lb 24 ft 1000 lb 12 ft 6 ft
10000 lbC 3000 lb 10000 lb 7000 lb
M E
E
-
8/12/2019 method of section
5/28
Method of Joints
Example
Look at joint A
y AD
AD AD
x AD AB AB
AB AB
40 2000 lb
52500 lb 2500 lb C
3 30 2500 lb
5 51500 lb 1500 lb T
F F
F F
F F F F
F F
-
8/12/2019 method of section
6/28
Method of Joints
Example
Look at joint D
y AD DB DB
DB DB
x AD DB DE
DE
DE DE
4 4 4 40 2500 lb5 5 5 5
2500 lb 2500 lb T
3 30 5 5
3 32500 lb 2500 lb
5 5
3000 lb 3000 lb C
F F F F
F F
F F F F
F
F F
-
8/12/2019 method of section
7/28
Method of Joints
Example
Look at joint B
y BD BE
DE
DE DE
x BD BA BE BC
BC
BC DE
4 4
0 1000 lb5 54 4
2500 lb 1000 lb5 53750 lb 3750 lb C
3 305 5
3 32500 lb 1500 lb 3750 lb
5 5
5250 lb 5250 lb T
F F F
F
F F
F F F F F
F
F F
-
8/12/2019 method of section
8/28
Method of Joints
Example
Look at joint E
y EB EC
DE
EC EC
x EB ED EC
EC
EC EC
4 4
0 10000 lb5 54 4
3750 lb 10000 lb5 58750 lb 8750 lb C
3 305 5
3 33750 lb 3000 lb
5 5
8750 lb 8750 lb C
F F F
F
F F
F F F F
F
F F
-
8/12/2019 method of section
9/28
Method of Joints
Example
Look at joint C to check the
solution
y CE
x CE CB x
40 7000 lb
54
8750 lb 7000 lb 0 OK!5
305
38750 lb 5250 lb 0 0
5
F F
F F F C
-
8/12/2019 method of section
10/28
Method of Joints Class Problem
Determine the forces BC,
DF and GE. Using themethod of Joints.
-
8/12/2019 method of section
11/28
Method of Sections -Truss
The method of joints is most effective whenthe forces in all the members of a truss are tobe determined. If however, the force is onlyone or a few members are needed, then themethod of sections is more efficient.
-
8/12/2019 method of section
12/28
Few simple guidelines of section truss analysis :
Pass a section through a maximum of 3 members ofthe truss, 1 of which is the desired member where itis dividing the truss into 2 completely separate parts,
At 1 part of the truss, take moments about the point(at a joint) where the 2 members intersect and solvefor the member force, using M = 0,
Solve the other 2 unknowns by using the equilibriumequation for forces, using Fx = 0 and Fy = 0 .
-
8/12/2019 method of section
13/28
Method of Sections -Truss
If we were interested inthe force of member CE.We can use a cutting lineor section to breakup thetruss and solve by takingthe moment about B.
-
8/12/2019 method of section
14/28
Method of Sections Example
Determine the forces in members FH, GH and GIof the roof truss.
-
8/12/2019 method of section
15/28
Method of Sections Example
Draw a free body diagram and solve for thereactions.
RAx
RAy
L
x Ax
Ax
y
Ay
0
0 kN
0
20 kN
F R
R
F
L R
-
8/12/2019 method of section
16/28
Method ofSections Example
Solve for themoment at A.
RAx
RAy
L
A
Ay
6 kN 5 m 6 kN 10 m 6 kN 15 m
1 kN 20 m 1 kN 25 m 30 m
7.5 kN
12.5 kN
M
L
L
R
-
8/12/2019 method of section
17/28
Method of Sections Example
Solve for the member GI. Take a cut between thethird and fourth section and draw the free-bodydiagram.
HI HI
HI
1 o
8 m 10 m8 m
15 m 10 m 15 m
5.333 m
8 mtan 28.1
15 m
l l
l
-
8/12/2019 method of section
18/28
Method of Sections
Example
The free-body diagram of
the cut on the right side.
H GI
GI GI
1 kN 5 m 7.5 kN 10 m 5.333 m
13.13 kN 13.13 kN T
M F
F F
-
8/12/2019 method of section
19/28
Method of Sections Example
Use the line of action of the forces and take the momentabout G it will remove the F GI and F GH and shift F FH to the
perpendicular of G.
-
8/12/2019 method of section
20/28
Method ofSections Example
Take the moment at G
G
oFH
FH FH
1 kN 5 m 1 kN 10 m 7.5 kN 15 m
cos 28.1 8 m
13.82 kN 13.82 kN C
M
F
F F
-
8/12/2019 method of section
21/28
Method of Sections Example
Use the line of action of the forces and take the momentabout L it will remove the F GI and F FH and shift F GH to point
G.
1 o5 mtan 133.25.333 m
-
8/12/2019 method of section
22/28
Method ofSections Example
Take the moment at L
oL GH
GH GH
1 kN 5 m 1 kN 10 m cos 43.2 15 m
1.372 kN 1.372 kN C
M F
F F
-
8/12/2019 method of section
23/28
Method of Sections Class Problem
Determine the forces in members CD and CE using method of sections.
-
8/12/2019 method of section
24/28
Homework (Due 2/24/03)
Problems:
6-34, 6-37, 6-38, 6-40, 6-45, 6-63
-
8/12/2019 method of section
25/28
Truss Bonus Problem
Determine whether themembers are unstable,
determinate orindeterminate.
-
8/12/2019 method of section
26/28
Truss Bonus Problem
Determine the loads ineach of the members.
-
8/12/2019 method of section
27/28
Truss Bonus Problem
Determine the loads ineach of the members.
-
8/12/2019 method of section
28/28
Truss Bonus Problem
Determine the loads ineach of the members.