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SashulF [63]
4 years ago
10

The pads are 200mm long, 150 mm wide and thickness equal to 12mm. 1- Determine the average shear strain in the rubber if the for

ce P = 15 kN and the shear modulus for the rubber is G = 830 kPa. 2- What is the relative horizontal displacement between the interior plate and the outer plates Esteel = 200 GPa Gste
Engineering
1 answer:
lord [1]4 years ago
5 0

Answer:

a) 0.3

b) 3.6 mm

Explanation:

Given

Length of the pads, l = 200 mm = 0.2 m

Width of the pads, b = 150 mm = 0.15 m

Thickness of the pads, t = 12 mm = 0.012 m

Force on the rubber, P = 15 kN

Shear modulus on the rubber, G = 830 GPa

The average shear strain can be gotten by

τ(average) = (P/2) / bl

τ(average) = (15/2) / (0.15 * 0.2)

τ(average) = 7.5 / 0.03

τ(average) = 250 kPa

γ(average) = τ(average) / G

γ(average) = 250 kPa / 830 kPa

γ(average) = 0.3

horizontal displacement,

δ = γ(average) * t

δ = 0.3 * 12

δ = 3.6 mm

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Is a steam power plant a heat engine?
Novosadov [1.4K]

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Yes

Explanation:

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A five legged intersection has safety issues. in 2018, the number of crashes reported was 48 and the average 24 hour volume ente
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Crash rate is calculated using average crash frequency.

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4 0
4 years ago
Do all websites use the same coding to create?
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3 years ago
An open glass tube is inserted into a pan of fresh water at 20 °C. What tube diameter is needed to make the height of capillary
kiruha [24]

Answer:

The tube diameter is 2.71 mm.

Explanation:

Given:

Open glass tube is inserted into a pan of fresh water at 20°C.

Height of capillary raise is four times tube diameter.

h = 4d

Assumption:

Take water as pure water as the water is fresh enough. So, the angle of contact is 0 degree.

Take surface tension of water at 20°C as 72.53\times 10^{-3} N/m.

Take density of water as 100 kg/m3.

Calculation:

Step1

Expression for height of capillary rise is gives as follows:

h=\frac{4\sigma\cos\theta}{dg\rho}

Step2

Substitute the value of height h, surface tension, density of water, acceleration due to gravity and contact angle in the above equation as follows:

4d=\frac{4\times72.53\times10^{-3}\cos0^{\circ}}{d\times9.81\times1000}

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d=2.719\times10^{-3} m.

Or

d=(2.719\times10^{-3}m)(\frac{1000mm}{1m})

d=2.719 mm

Thus, the tube diameter is 2.719 mm.

 

5 0
4 years ago
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