Determine the force in each member of the truss. state of the members are in tension or compression

6–3.

Determine the force in each member of the truss.

State if the members are in tension or compression.Determine the force in each member of the truss. S

Answer

General guidance

Concepts and reason
When slender members are attached together at their ends, then it forms a truss. Members that are used to form a truss are metal bars and wood. Planar trusses are the trusses which lie on a single plane. Each member of truss is considered as two force member. Either the member is subjected to tensile force or member is subjected to compressive force. Usually members that are subjected to compression are very thick than members that are in tension.

To solve this problem, make a cut across the three forces that have to be determined. Use force equilibrium or moment equilibrium to determine the forces.

Fundamentals

There are three equilibrium conditions of forces:

• Horizontal equilibrium

Summation of Forces in the direction of x will be equal to zero. It is mathematically represented as ΣF = 0.

• Vertical equilibrium:

Summation of Forces in the direction of y will be equal to zero. It is mathematically represented as ΣΕ, = 0.

• Moment equilibrium:

Summation of Forces about a point will be zero. It is mathematically represented as o=w3.

Step-by-step

Step 1 of 2

Consider the joint A.

АВ
FAC
1301b

Write the force equilibrium along y direction.

ΣΕ, = 0
(Ρ.)-Bas0) -0
(Fac) = 150 lb(compression)

Here, is the force in member .

Write the force equilibrium along x direction.

ΣΕ = 0
F. Εκ (130) = 0
F.-(150)-(130) = 0
Ελμ = 140 lb(tension)

Here, is the force in member .

Consider the joint B.

В
FARK
► FBD
AB

Write the force equilibrium along x direction.

ΣF = 0
F - F = 0
Fan -(140) = 0
FBD = 140 lb (tension)

Here, is the force in member.

Write the force equilibrium in vertical direction.

ΣΕ, = 0
Frc = 0 lb

Force in member AC,AB, BD, BC is 150 lb (compression), 140 lb (tension), 140 lb (tension), respectively.


Using the equilibrium of force, forces in each members are determined.

Step 2 of 2

Consider the following diagram at joint C.

Write the force equilibrium along y direction.

ΣΕ, = 0
σω) (0)=0
()-(150) = 0
Fn =150 Ib(tension)

Here, force in member CD is .

Write the force equilibrium along x direction.

ΣΕ = 0
-F. + 3(ru) • ξω) = 0
-Fi + 3(150) + (150) = 0
Fς = 180 Ib(compression)

Here, force in member CE is .

Draw the forces at joint D.

FBD
D
> FDF
F CD
FDE

Write the force equilibrium along y direction.

ΣF, = 0
(F») - (+) = 0
(Fe)-(150) = 0
FpE = 120 lb (compression)

Here, force in member DE is .

Write the force equilibrium along x direction.

ΣF = 0
F, (ω)- F = 0
Fpx = 230 lb (tension)

Here, force in member DF is .

Consider the forces at joint D.

FDE
FCE

Write the force equilibrium along x direction.

ΣF = 0
F-()=0
(180)-(F)=0
F = 300 lb(compression)

Here, force in member EF is .

The force in member CD, CE,DE, DF,EF is 150 lb (tension), 180 lb (compression), 120 lb (compression), 230 lb (tension), 300 lb (compression)respectively.


Using force equilibrium conditions, the force in each member is calculated.

Answer

Force in member AC,AB, BD, BC is 150 lb (compression), 140 lb (tension), 140 lb (tension), respectively.

The force in member CD, CE,DE, DF,EF is 150 lb (tension), 180 lb (compression), 120 lb (compression), 230 lb (tension), 300 lb (compression)respectively.

Hottest videos

Leave a Reply

Your email address will not be published. Required fields are marked *

Related Posts