Chapter 12
Trusses
Truss bridges are one of the oldest types of modern
bridges. The Pratt truss bridge was one of the most
popular designs of truss bridges.
The Clear Creek Bridge is a pin connected Warren
truss bridge near Bellwood Nebraska. It was
constructed in 1891 and is still in service for light
traffic today.
1.
Marian Physics
Homework
Trusses
Using the method of joints, calculate all the support reaction forces and tension/compression forces in all members.
1.As shown in the diagram below, we have a truss supported by a pinned joint at Point A and supported by a
roller at point D. A vertical load of 500 lb. acts at point F, and a horizontal load of 800 lb. acts at Point C. For
this structure we wish to determine the values of the support reactions, and the force (tension/compression) in
members AB, AE, BE, BC, EF , EC, CD and FD.
1. Find external forces at contact points “A” and “D”.
“A” is pinned and has both “X” contact forces and “Y” contact forces.
“D” is on rollers and has only “Y” contact forces.
2.
Apply Equilibrium conditions:
a. 12,000lbs = FED sin66.4° ⇒ FED = 13,100lbs(C)
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b. ∑ Fy = 0 ⇒ Ay + Dy − 500lbs = 0
c. ∑t = 0
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Select “A” as pivot point, counter clockwise is positive.
Tourques
about “A”
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ported by a roller at point D. For this structure we wish to determine the value of all
the support forces acting on the structure, and to determine the force in all members
−(500lbs)(8 ft.) + Dy (12 ft.) − (800lbs)(3 ft.) = 0
Dy = 533lbs
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Now we know Dy so we can go to equation “b” above and solve for Ay
Ay = −33lbs
Now to do the method of joints for the internal forces.
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Joint “E”
(no external forces)
B
Start with the joint at “A”.
AB
800 lbs
756lbs
37°
A
AE
F
EF
33 lbs
33lbs
Up=Down
FEC sin 37° = 33lbs ⇒ FEC = 55lbs(tension)
Since the 33 lb force is down AB has to be up (tension)
Up= Down
FAB sin 37° = 33lbs ⇒ FAB = 55lbs
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FEC
Left=Right
756lbs = FEC cos 37° + FEF ⇒ FEF = 712lbs(tension)
Left=Right
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800lbs = FAB cos 37° + FAE ⇒ FAE = 756lbs
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About “C”
On to Joint “B”
FCD
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F
BE
55lbs
F
37°
800lbs
44lbs
BC
FCF
55lbs
Left=Right
44lbs + 55cos 37° + FCD cos 37° = 800lbs ⇒ FCD = 891lbs(compression)
Up=Down
(891lbs)sin 37° = (55lbs)sin 37° + FCF ⇒ FCF = 500lbs(tension)
E
Since AB is under tension the forces are to the left and down. So BE must be up
(compression) and BC must be to the right (tension).
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About “F”
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712lbs
Up=Down
FBE = 55sin 37° ⇒ FBE = 33lbs(compression)
Left=Right
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55cos 37° = FBC ⇒ FBC = 44lbs(tension)
we are only interested in left/right (we already know CF from above)
D
Left=Right
FFD= 712 lbs tension
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About “E”
Up=Down
13,100lb(sin66.4°) = FCE sin66.4° ⇒ FCE = 13,100lbs(T)
13,100lbs
Left=Right
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F
FFE
F
CE
= 0 ⇒ −12,000lbs − 20,000lbs + FAY + FDY = 0
∑F
= 0 ⇒ FAX = 0
y
x
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F
1
∑t = 0
B
Choose “A” as pivot, counterclockwise as positive
(−12,000lbs)(4 ft) − (20,000lbs)(12 ft) + FDY (24 ft) = 0 ⇒ FDY = 12,000lbs
Substitute 12,000lbs into first equation to get:
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12,000lbs
CD
FED
Since 12.000lb force is up, the only possible down force is from ED. So ED mus
be down and to the right (Compression).
Since ED is to the right, CD must be to
the left (Tension)
C
lbs
E
5,240lbs
13,100sin66.4° = FFC sin66.4° ⇒ FFC = 13,100lbs(C)
FAY = 20,000lbs
About “D”
F
00
1
3,
E
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One more joint “C”
Since FCE is up and to the right, only FFC can be down. Since FCE, FCD and FCF
are to the right, FBC must be to the left.
2. External forces:
∑F
13,100cos66.4° + FCE cos66.4° = FFE ⇒ FFE = 10,500lbs(C)
13,100cos66.4° +13,100cos66.4° + 5240 = FBC ⇒ FBC = 15,950lbs(T)
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Up=Down
12,000lbs = FED sin66.4° ⇒ FED = 13,100lbs(C)
Left=Right
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FCD = FED cos66.4° ⇒ FCD = 5,240lbs(T)
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