a)Consider the problem sketch. Ignoring the supports, the sketch contains how many two-force members. b)The magnitude of the reaction moment about Support-A is _____ kN*m.
Plane Trusses
It is defined as, two or more elements like beams or any two or more force members, which when assembled together, behaves like a complete structure or as a single structure. They generally consist of two force member which means any component structure where the force is applied only at two points. The point of contact of joints of truss are known as nodes. They are generally made up of triangular patterns. Nodes are the points where all the external forces and the reactionary forces due to them act and shows whether the force is tensile or compressive. There are various characteristics of trusses and are characterized as Simple truss, planar truss or the Space Frame truss.
Equilibrium Equations
If a body is said to be at rest or moving with a uniform velocity, the body is in equilibrium condition. This means that all the forces are balanced in the body. It can be understood with the help of Newton's first law of motion which states that the resultant force on a system is null, where the system remains to be at rest or moves at uniform motion. It is when the rate of the forward reaction is equal to the rate of the backward reaction.
Force Systems
When a body comes in interaction with other bodies, they exert various forces on each other. Any system is under the influence of some kind of force. For example, laptop kept on table exerts force on the table and table exerts equal force on it, hence the system is in balance or equilibrium. When two or more materials interact then more than one force act at a time, hence it is called as force systems.
a)Consider the problem sketch. Ignoring the supports, the sketch contains how many two-force members.
b)The magnitude of the reaction moment about Support-A is _____ kN*m.
![Article 4/5 Space Trusses 199
SAMPLE PROBLEM 4/5
The space truss consists of the rigid tetrahedron ABCD anchored by a ball-
dsocket connection at A and prevented from any rotation about the x-, y-, or
as by the respective links 1, 2, and 3. The load L is applied to joint E, which
rigidly fixed to the tetrahedron by the three additional links. Solve for the
rres in the members at joint E and indicate the procedure for the determina-
dion of the forces in the remaining memberg of the truss.
E
1
4 m
4 m
A.
Solution. We note first that the truss is supported with six properly placed
anstraints, which are the three at A and the links 1, 2. and 3. Also, with m = 9
members and j = 5 joints, the condition m + 6 = 3j for a sufficiency of members
to provide a noncollapsible structure is satisfied.
The
external reactions at A, B, and D can be calculated easily as a first step,
А,
although their values will be determined from the solution of all forces on each
O of the joints in succession.
We start
three unknown forces act, which in this case is joint E. The free-body diagram of
joint E is shown with all force vectors arbitrarily assumed in their positive ten-
a sion directions (away from the joint). The vector expressions for the three un-
known forces are
Helpful Hints
O Suggestion: Draw a free-body dia-
gram of the truss as a whole and ver-
ify that the external forces aeting on
the truss are A, = Li, A, = Lj, A, =
(4L/3)k, B - 0, D,
-(4L/3)k.
with a
joint on which at least one known force and not more than
-Lj, D =
FEB
(-i - j),
FEC
FED ="D (-3j - 4k)
FED
FER =
FEC =
(-3i - 4k),
2 With this assumption, a negative
numerical value for a force indicates
/2
Equilibrium of joint E requires
compression.
[EF = 0]
L + FER + Frc + FEn = 0
or
L.
BI
FEB
-Li +
FED
(-3j - 4k) = 0
E
(-i - j) +
(-3i + 4k) +
FER
FEc/FRD
Rearranging terms gives
AF Bc ED
4 m
4 m
FEB 3FEC
FEB 3F
k = 0
Equating the coefficients of the i-, j-, and k-unit vectors to zero gives the three
equations
3 m
3F EC = -L
FEB +
FEB 3FED = 0
FEc + FED = 0
Solving the equations gives us
FED = 5L/6
Ans.
FEB = -L//2
FEC = -5L/6
Thus, we conclude that FER and FEc are compressive forces and Fgp is tension.
Unless we have computed the external reactions first, we must next analyze
Joint C with the known value of Fge and the three unknowns FCB, FCA, and Fen.
The procedure is identical with that used for joint E. Joints B, D, and A are then
nalyzed in the same way and in that order, which limits the scalar unknowns to
three for each joint. The external reactions computed from these analyses must,
course, agree with the values which can be determined initially from an analy-
tis of the truss as a whole.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6cf78908-a1f9-4330-9ece-f4021df10613%2Fceacf426-a276-4c37-9fab-36b8bc87732d%2Fbff99o_processed.jpeg&w=3840&q=75)
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