Oberlaudacu

2022-01-11

If a frictional force of 50 N is applied to each side of the tires, determine the average shear strain in the rubber. Each pad has cross-sectional dimensions of 20 mm and 50 mm.
${G}_{r}=0.20MPa$

Jenny Sheppard

The base is 20 mm and 50 mm in size. The strain in the tire must be ascertained.Friction force
${F}_{t}=50N$
Using the expressed shear stress, we will determine T.
Surface cross-sectional area:
$A=50\cdot 20$
$=1000m{m}^{2}$
$\tau =\frac{{F}_{t}}{A}$
$=\frac{50}{1000}$
$\tau =0.05MPa$
We can also calculate the shear stress in the following way.
$\tau =\gamma \cdot G$
Where:
$\gamma$- shear strain
$\tau$- shear strain
$⇒\gamma =\frac{\tau }{G}$
$=\frac{0.05}{0.20}$
$\gamma =0.25rad$
The solution is:
$\gamma =0.25rad$

Shannon Hodgkinson

Given that $⇒$ Area of cross section $\left(Acs\right)=50×20$
$⇒Acs=1000m{m}^{2}$
Forse $\left(F\right)=50N$, $G=0\cdot 2MPa$
$\because ShearstressPSK\left(\tau \right)=\frac{F}{Acs}=\frac{50}{1000}=0.05MPa$
As we know $⇒Shearstra\in =\frac{Shearstress}{Shear\text{mod}\underset{―}{e}s}$
$⇒\gamma =\frac{\tau }{4}$
$⇒\gamma =\frac{0.05}{0.2}=0.25rad$

nick1337

$\gamma =0.00025$
Explanation:
$\gamma =\frac{\Delta x}{L}$
where:
$\gamma$ is the shear strain,
$\Delta x$ is the change in length,
and $L$ is the original length.
Given that the pads have cross-sectional dimensions of 20 mm and 50 mm, we can assume that the original length $L$ is 20 mm. Since the pads are made of rubber, we can assume that the shear modulus of rubber (${G}_{r}$) is 0.20 MPa.
The shear stress ($\tau$) is given by:
$\tau =\frac{F}{A}$
where:
$\tau$ is the shear stress,
$F$ is the applied force, and
$A$ is the cross-sectional area.
In this case, the applied force ($F$) is 50 N, and the cross-sectional area ($A$) can be calculated as follows:
$A=\text{{width}\right\}×\text{{height}\right\}$
Plugging in the values, we have:
$A=20\phantom{\rule{0.167em}{0ex}}\text{mm}×50\phantom{\rule{0.167em}{0ex}}\text{mm}$
Now, we can substitute the values into the formula for shear stress to find the shear strain:
$\tau ={G}_{r}·\gamma$
Solving for $\gamma$, we get:
$\gamma =\frac{\tau }{{G}_{r}}$
Plugging in the values for $\tau$ and ${G}_{r}$, we have:
$\gamma =\frac{50\phantom{\rule{0.167em}{0ex}}\text{N}}{20\phantom{\rule{0.167em}{0ex}}\text{mm}×50\phantom{\rule{0.167em}{0ex}}\text{mm}}·\frac{1\phantom{\rule{0.167em}{0ex}}\text{MPa}}{0.20\phantom{\rule{0.167em}{0ex}}\text{MPa}}$
Now, we can simplify the equation:
$\gamma =\frac{50\phantom{\rule{0.167em}{0ex}}\text{N}}{1000\phantom{\rule{0.167em}{0ex}}{\text{mm}}^{2}}·5$
Converting the units to meters, we have:
$\gamma =\frac{50\phantom{\rule{0.167em}{0ex}}\text{N}}{1000000\phantom{\rule{0.167em}{0ex}}{\text{mm}}^{2}}·5$
Simplifying further:
$\gamma =\frac{50×5}{1000000}$
Thus, the average shear strain in the rubber is $0.00025$.
Therefore, the average shear strain in the rubber is $\gamma =0.00025$.

Vasquez

To solve the problem, we can use the equation for shear strain, which is given by:

Given that the shear stress is equal to the frictional force applied on each side of the tire (50 N), and the shear modulus is given as G_r = 0.20 MPa, we can substitute these values into the equation to find the shear strain.
First, let's convert the shear modulus from MPa to N/m²:
${G}_{r}=0.20×{10}^{6}\phantom{\rule{0.167em}{0ex}}\text{Pa}$
The cross-sectional area of each pad can be calculated as follows:
$A=\text{{width}\right\}×\text{{height}\right\}=\left(20\phantom{\rule{0.167em}{0ex}}\text{mm}\right)×\left(50\phantom{\rule{0.167em}{0ex}}\text{mm}\right)$
Next, we can calculate the shear strain using the formula mentioned earlier:

Finally, we can simplify the units by noting that 1 mm = 0.001 m:
$\gamma =\frac{50}{0.20×{10}^{6}}×\frac{1}{0.001}=\frac{50}{0.20×{10}^{6}×0.001}\phantom{\rule{0.167em}{0ex}}\text{m/m}$
Therefore, the average shear strain in the rubber is $\frac{50}{0.20×{10}^{6}×0.001}\phantom{\rule{0.167em}{0ex}}\text{m/m}$.

RizerMix

To determine the average shear strain in the rubber brake pads of a bicycle tire, we can use the formula:

Given:
- Shear stress, $\tau =50\phantom{\rule{0.167em}{0ex}}\text{N}$
- Cross-sectional dimensions: ${d}_{1}=20\phantom{\rule{0.167em}{0ex}}\text{mm}$ (width) and ${d}_{2}=50\phantom{\rule{0.167em}{0ex}}\text{mm}$ (height)
- Shear modulus, ${G}_{r}=0.20\phantom{\rule{0.167em}{0ex}}\text{MPa}=0.20×{10}^{6}\phantom{\rule{0.167em}{0ex}}{\text{N/m}}^{2}$
First, let's convert the dimensions to meters:
${d}_{1}=20\phantom{\rule{0.167em}{0ex}}\text{mm}=20×{10}^{-3}\phantom{\rule{0.167em}{0ex}}\text{m}$
${d}_{2}=50\phantom{\rule{0.167em}{0ex}}\text{mm}=50×{10}^{-3}\phantom{\rule{0.167em}{0ex}}\text{m}$
Now, we can calculate the average shear strain:

Substituting the given values:
$\text{{Shearstrain}\right\}=\frac{50\phantom{\rule{0.167em}{0ex}}\text{N}}{0.20×{10}^{6}\phantom{\rule{0.167em}{0ex}}{\text{N/m}}^{2}}$
Simplifying the expression:
$\text{{Shearstrain}\right\}=\frac{50}{0.20×{10}^{6}}=0.25×{10}^{-3}$
Therefore, the average shear strain in the rubber brake pads is $0.25×{10}^{-3}$ rad.

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