MECHANICS  CASE STUDY SOLUTION

Basic Canopy Truss
Joint Labels and Forces


The canopy currently exists and a new hoist needs to be installed under the
canopy to move packages and equipment around on the dock. The logical location
for the hoist is to attach it to the center truss node as shown in the diagram.
The actual load W is not known and needs to be determined. The roof load is given,
60 lb/ft^{2}, and the truss configuration is known.
The question is what is the maximum load W before one of the members buckle?
The canopy requires a factor of safety of 2 which needs to be included in the
calculations. There is a canopy truss every 5 feet along the length of the dock,
so each truss needs to support at least 5 feet of the roof load.
The first step will be to solve for the loads in the truss members using both
the
joint method and the
section method from statics.






Tension and Compression Members

Tension (blue) and
Compression (red) Members


Before calculating the force in every member, some members can be eliminated
from consideration. Recall, the problem is only concerned about buckling so just the
compression members are of interest.
The top two members, AB and BC, will be in tension as the canopy is pushed downward
by the roof and the hoist loads. Next, the full hoist load will be carried
in tension by member BE since they are collinear at joint E.
Likewise, from inspection members DB, DE, and EC will be in compression resisting
the roof and hoist loads. If DB is in compression then AD will be in tension.
This leaves only members DB, DE, and EC.






Roof Load Distribution

Roof Load Distribution on Truss 

Since the trusses are spaced every 5 feet along the length of the dock, each
truss will need to carry the roof load over 5 feet. The total load will be
Truss Total Load = (60
lb/ft^{2})(8
ft)(5 ft) = 2,400 lb
The roof is connected to the truss through the joint points and so the roof load
is distributed to each of the joints as concentrated forces. However, the distribution
is not equal. The outside two joints (A and C) will only carry half of what the
center joint will carry.
F_{A} = F_{C} = (60
lb/ft^{2})(2 ft)(5 ft) = 600 lb
F_{B} = (60 lb/ft^{2})(4
ft)(5 ft) = 1,200 lb
The angle θ can be determined from the geometry,
tanθ = 3/8 => θ =
20.56^{o}






Member Loads in EC and DE

Joint E Free Body Diagram 

Notice that all members at joint E are perpendicular. This requires the force
in member DE equal the force in member EC. Thus,
F_{DE} = F_{EC}
There are many ways to find the load in truss members, but one useful method
is the
method of joints. This method requires one to analyze a single joint with
only two unknown member forces. This can be done at joint C.




Joint C Free Body Diagram 

The equilibrium equations can be applied to joint C, giving
ΣF_{y} = 0
600 + F_{BC} sin20.56 = 0
F_{BC} = 1,709 lb
ΣF_{x} = 0
1,709 cos20.56  F_{EC} = 0
F_{EC} = 1,600 lb
The negative value means it is under compression, as originally assumed. Notice,
that both members are not effected by the hoist load. In other words, none of
the hoist load will be supported by members BC or EC. Therefore, members BC or
EC are not critical when the hoist is used.






Member Load in DB

Truss Section


The other compression member is DB. The load in DB can best be determined using
the
method of sections. Making a section cut through members DE, DB, and AB allows
solving for all three members. Taking moments about joint C
will give one equation and one unknown,
ΣM_{C} = 0
W (4) +1,200 (4) + F_{DB}cosθ (1.5)
+ F_{DB}sinθ (4) = 0
F_{DB} = 1.424W  1,709 lb
The compression force in F_{DB} is a function of the hoist weight W.






Buckling load of Member DB



The buckling load of member DB can be determined using Euler's Formula for simply
supported (pinnedpinned ends) columns.
The moment of inertia for the 1.5 inch pipe with a wall thickness of 1/16 in
is
= 0.07304 in^{4}
The length of member DB is
L = (4^{2} + 1.5^{2})^{1/2} = 4.272 ft = 51.26 in
The objective is to find the largest hoist load, W, that can be applied without
buckling with a a factor of safety of 2.0. This gives,
W = 1,593 lb
Thus, the hoist and its load can be a maximum of 1,593 lb. This should be sufficient
to assist in moving most packages and equipment around on the dock.



