Page 428. Sample Problem 6.1.Solve this problem completely;but replace theapplied point-forceon the rightfrom 1.5 kN to 2 kN.Let the one on the left remain unchanged.At the same time, instead of computing the averageshearing stress in each joint at the provided sectionn-n, please find the average shearing stress in each jointfortheworst section of the beamwherever it is(and state/show where it is), not necessarily at the providedlocation ofsectionn-n.Keep in mind that this change will result in many other changes across the board.Discuss all practicalissues in your findings.For example, where is the system likelyto fail in terms of the glued parts, and why?
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.
Page 428. Sample Problem 6.1.Solve this problem completely;but replace theapplied point-forceon the rightfrom 1.5 kN to 2 kN.Let the one on the left remain unchanged.At the same time, instead of computing the averageshearing stress in each joint at the provided sectionn-n, please find the average shearing stress in each jointfortheworst section of the beamwherever it is(and state/show where it is), not necessarily at the providedlocation ofsectionn-n.Keep in mind that this change will result in many other changes across the board.Discuss all practicalissues in your findings.For example, where is the system likelyto fail in terms of the glued parts, and why?
![0.020 m
Neutral axis
Neutral axis
0.100 m-
Fig. 3 Using area above section a-a to
find Q.
0.020 m
= 0.0417 m
₂= 0.0583 m
0.060 m
Fig. 4 Using area below section b-b to
find Q.
ANALYSIS:
Shearing Stress in Joint a. Using Fig. 3, pass the section a-a through
the glued joint and separate the cross-sectional area into two parts. We
choose to determine Q by computing the first moment with respect to the
neutral axis of the area above section a-a.
Q=Ay₁ = [(0.100 m) (0.020 m)] (0.0417 m) = 83.4 x 10m³
Recalling that the width of the glued joint is t = 0.020 m, we use Eq. (6.7)
to determine the average shearing stress in the joint.
Tave =
Tave =
VQ
It
VO
It
=
Shearing Stress in Joint b. Using Fig. 4, now pass section b-b and
compute Q by using the area below the section.
=
(1500 N) (83.4 x 10 m³)
(8.63 x 10
m²) (0.020 m)
Q = Ay₂ = [(0.060 m)(0.020 m)] (0.0583 m) = 70.0 x 10-6; m³
(1500 N) (70.0 x 10 m²)
(8.63 x 10 m²) (0.020 m)
Tave = 725 kPa
Tave = 608 kPa](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb4ef3dc5-68e2-4116-aeaf-14e594acc0f6%2F70eab36e-5ccf-446f-a92c-8a52bb9504a6%2Fhm4d2r4_processed.jpeg&w=3840&q=75)
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