The circular rod shown $($Figure $1)$ has dimensions $d_1=7cm,L_1=6m,d_2=3\mathrm{.}2cm,$ and $L_2=5m$ with applied loads $F_1=140kN$ and $F_2=55kN.$ 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 $C.$

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.

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Correct

The negative answer means the reaction force actually points to the left and is a tension force.

Part B $-$ Segment the rod

For the given rod, which segments must, at a minimum, be considered in order to use $={\displaystyle \underset{?}{\overset{?}{}}}\frac{NL}{AE}$ to calculate the deflection at $D?$

Check all that apply.

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$BD$

$AD$

$CD$

$AB$

$BC$

$AC$

Correct Part C $-$ Calculate the deflection

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.

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${}_D=$

Incorrect; Try Again 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 $={\displaystyle \underset{0}{\overset{L}{}}}\frac{N(x)dx}{A(x)E(x)},$ 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 $x$ 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 integral can be evaluated to give $=\frac{NL}{AE}.$ This 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 deflections for each part, $={\displaystyle \underset{?}{\overset{?}{}}}\frac{NL}{AE}.$