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Today: Return Exam 3 Questions on Computer Problem Trusses: Two force members Method of Joints New Homework
FAB FAC
FAD
FAC
Homework 23: Prob 6.4 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. compression. A
375 lb
500 lb C
B
6 ft
8 ft
D
8.4 ft
Prob 6.12: Determine the force in each member of the Fink truss shown. State whether each member is in tension or compression.
4.2 kN 2.8 kN
D
2.8 kN E
B C
3m
F G
A
2m
2m
2m
2m
FAB FAD
Prob 6.23: For the roof truss shown, determine the force in each of the members located to the left of member GH. State whether each member is in tension or compression. 1.5 kN 1.5 kN
1.5 kN
1.5 kN 1.5 kN
G
1 kN
D
1 kN
I
E
B 1m A
F
C
2.4 m
H
1.2 m
2.4 m
2.5 m
K M
J
L
2.4 m
2.4 m
1.2 m
Prob 6.30: For the given loading, determine the zero force members in the truss.
B
A D
C F
E
G
H I
J
K
L
Structural Analysis: Truss-- A structure composed of slender two force members, joined at their ends.
Members of a truss may only be subjected to tensile or compressive forces.
Two-force member: A two-force member is any body which has forces applied to it at only two points of contact. In order for the body to remain in equilibrium, the line of the force acting at each contact, must act along a line that passes through each point.
If the body is a straight member, then it is said to be in either tension or compression
Examples of Two Force Members:
Trusses are built out of straight, two-force members. Forces are only applied at the connections between individual members.
A Method of Joints: In the method of joints, point equilibrium is set up for forces that act on the pin at each joint of the body.
64 in
Process: Assume all members are in tension Identify zero-force members by inspection Find the support reactions of the entire truss or Find a joint which has less than three members and find member forces.
Example:
720 lb
B
50 in
C
48 in D
A
A
720 lb
720 lb B 64 in
B
50 in
48 in
C
D
FCy
FD
FCx
C
Start by finding the reactions of the entire body:
D
FCx
∑ M C = 0
FCy
0 = −720lb (50in ) + FD (48in )
FD =
∑F
x
(720lb )(50in )
=0
(48in )
=750 lb
∑ Fy = 0
0 = FCx + 720lb
0 = FCy + F D
FCx = −720 lb
FCy = − FD = −750 lb
FD
Next set up the equilibrium at point B because there are only two members coming together there.
∑F
x
= 0:
0=−
48
50 FBA = 750 lb
∑ Fy = 0 :
0=
FBD =
14 50
14
FBA =
14 50
FBA
14
P=720 lb 48
FBA − F BD
50
Joint B
50
FBA + 720lb
FBD
(750) =210 lb
Next move to Joint A and apply point equilibrium:
∑ Fx = 0 :
0=
48 50
FAD
FAB +
F AD 80 80 =− F AB 50 80 = −
∑ Fy = 0 :
0= −
48
14
50
(750)
FAB −
64
= −
FAB FAC 1200lb
FAD −F AC
FAD
48
80
48 14
50 64
50 80 14 64 FAC = − FAB − F AD 50 80 14 64 FAC = − (750) − ( −1200) =750lb 50 80 Joint C
FCA
Finally, look at Joint C.
∑ Fx = 0 :
FCD
FCx
0 = FCx + F CD
FCD = − FCx = −( −720 lb) =720 lb FCy
Summary: FBA = FAB = 750 lb ( T )
FCD = FDC = 720 lb ( T )
FBD = FDB = 210 lb ( T )
FAC = FCA = 750 lb ( T )
FAD = FDA = −1200 lb =1200 lb ( C)
Example 2:
Use method of joints to find the forces in each of the truss members. 4 kN
6 kN
3m
4 kN
3m
C
E 45o
45o
D
3m
3m F
G
45o
45o B
A
Solution: Start by finding the reactions of the entire body:
∑ MA = 0 0 = −6kN (3m ) − 4kN (6m ) + FB (6m )
FB =
∑ Fx = 0
6(3) + 4(6) 6
=7 kN
∑ Fy = 0
0 = F Ax
0 = FAy + FB − 6kN − 4kN −4kN
F Ax = 0
FAy = 14 − FB = 14 − 7 = 7 k
Next inspect for zero force members:
4kN
Identify EF and GC as ZFM Start at Joint E:
FED
FEA Joint E
∑F
x
= 0:
0 = F ED
∑ Fy = 0 :
0 = −4 − F EA
FEA = −4 kN
Next note that this truss is symmetric, so
FCB = −4 kN and F CD = 0
Next Look at Joint D: 0
6 kN
FDE
∑F
x
= 0:
0 = − FDF cos 45o + F DG cos 45o
FDF = F DG
∑ Fy = 0 :
45o FDF
0 FDC
45o FDG
0 = − 6 − FDF sin 45o −F DG sin 45o 2 F DF sin 45o = − 6 FDF = F DG =
−6 o
2sin 45
= −4.24
Note that the force in members FA and GB will be the same as in DF and DG, so FFA = −4.24 kN and FGB = −4.24 kN Finally, look at Joint A to find force in AB.
∑F
x
= 0:
0 = FAx + FAB + F AF cos 45o
Joint A
FAE FAF FAB
FAx
FAB = − FAF cos 45o = −( −4.24) cos 45o =3 k
Summary: FAB = 3 kN (T )
FAy
FAF = FFD = FDG = FGB = −4.24 =4.24 kN ( C)
FAE = FBC = −4 = 4 kN ( C)
FED = FEF = FDC = FGC =0
Example 3: Use method of joints to find the forces in each of the truss members.
6 kN
B
A
0.9 m 3 kN
C E 1.2 m
Start at Joint E.
D
1.2 m
Discussion Problem: For the given loading, how many and where are the zero-force members in the truss shown.
