Answered: Part A - Reaction force What is the…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Learning Goal: To calculate the elastic deflection in an axially loaded member. For a bar subject to axial loading, the change in length, or deflection, between two points A and B is 8 = fr N(x) dx A(x)E(x)' internal normal force, A is the cross-sectional area, E is the modulus of elasticity of the material, L is the original length of the bar, and is the position along the bar. This equation applies as long as the response is linear elastic and the cross section does not change too suddenly. In the simpler case of a constant cross section, homogenous material, and constant axial load, the This NL integral can be evaluated to give d = AE shows that the deflection is linear with respect to the internal normal force and the length of the bar. In some situations, the bar can be divided into multiple segments where each one has uniform. internal loading and properties. Then the total deflection can be written as a sum of the NL deflections for each part, 8 = AE Figure , where N is the Li C F₁ ↑ d₂ D < 1 of 1 F₂ Part A Reaction force The circular rod shown (Figure 1) has dimensions d₁ = 6.2 cm, L₁ = 6 m, d₂ = 4.8 cm, and L2 = 5 m with applied loads F₁ = 130 kN and F2 = 60 kN. The modulus of elasticity is E = 95 GPa. Use the following steps to find the deflection at point D. Point B is halfway between points A and C.

<U4-Ch4 Tutorial
Elastic Deformation of an Axially Loaded Member
Learning Goal:
To calculate the elastic deflection in an axially
loaded member
For a bar subject to axial loading, the change in
length, or deflection, between two points A and B
N(a) da
A(z)E(x)'
is 8 = fo
where N is the
internal normal force, A is the cross-sectional area,
E is the modulus of elasticity of the material, L is
the original length of the bar, and at is the position
along the bar. This equation applies as long as the
response is linear elastic and the cross section
does not change too suddenly.
In the simpler case of a constant cross section,
homogenous material, and constant axial load, the
NL This
This
AE
integral can be evaluated to give
shows that the deflection is linear with respect to
the internal normal force and the length of the ban
In some situations, the bar can be divided into
multiple segments where each one has uniform
internal loading and properties. Then the total
deflection can be written as a sum of the
NL
deflections for each part, 6-ΣAE
Figure
AIB с
d₁
L
F₁
↑
d₂
L₂
< 1 of 1 >
D
F₂
▾ Part A - Reaction force
What is the reaction force at A? Let a positive reaction force be to the right.
Express your answer with appropriate units to three significant figures.
▸View Available Hint(s)
The circular rod shown (Figure 1) has dimensions d₁ = 6.2 cm, L₁ = 6 m, d₂ = 4.8 cm, and
L₂=5 m with applied loads F₁ = 130 kN and F₂ = 60 kN. The modulus of elasticity is E951
GPa. Use the following steps to find the deflection at point D. Point B is halfway between
points A and C.
FA=
Cº
Submit
Part B - Segment the rod
8-
=Σ ΑΕ
Check all that apply.
▸ View Available Hint(s)
AB
CD
BC
Value
AC
AD
BD
For the given rod, which segments must, at a minimum, be considered in order to use.
NL
Submit
Part C - Calculate the deflection
Submit
to calculate the deflection at D?
őd= Value
Units
Provide Feedback
What is the deflection of the end of the rod, D? Let a positive deflection be to the right.
Express your answer with appropriate units to three significant figures.
▸ View Available Hint(s)
4
?
R
1 of 5
Units
Review
?
>
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