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3 years ago

5

A steel sphere with radius 1.0010 cm at 41.0°C must slip through a brass ring that has an internal radius of 1.0000 cm at the sa

me temperature. To what temperature must the brass ring be heated so that the sphere, still at 41.0°C, can just slip through? Coefficient of linear expansion α for brass is 19.0 × 10−6 K−1.

1 answer:

tatiyna3 years ago

4 0

Answer:

$\Delta T = 52.6 ^o C$

Explanation:

As we know that radius of the brass ring is given as

$R_{brass} = 1.0000 cm$

radius of the sphere is given as

$R_{sphere} = 1.0010 cm$

now by thermal expansion formula we know that

$L = L_o(1 + \alpha \Delta T)$

so we will have

$1.0010 = 1.0000(1 + (19\times 10^{-6})\Delta T)$

so we have

$\Delta T = 52.6 ^o C$

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Given $\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$ at a point. What is the force per unit area at this point acting normal to the surface with $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$ ? Are there any shear stresses acting on this surface?

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$\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau = \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z$

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Since $\tau \neq 0$ , there are shear stresses acting on the surface.

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3 years ago