 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

Expert

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

Expert

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 $⇒Shearstra\in =\frac{Shearstress}{Shear\text{mod}\underset{―}{e}s}$
$⇒\gamma =\frac{\tau }{4}$
$⇒\gamma =\frac{0.05}{0.2}=0.25rad$

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