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quester [9]
3 years ago
11

An axial compressive load of 708 kN is applied to a cylindrical component, 81 mm in diameter and 418 mm long, made of aluminium.

Calculate the compressive stress, fc
Engineering
1 answer:
dalvyx [7]3 years ago
4 0

Answer:

The compressive stress of aplying a force of 708 kN in a 81 mm diamter cylindrical component is 0.137 kN/mm^2 or 137465051 Pa (= 137.5 MPa)

Explanation:

The compressive stress in a cylindrical  component can be calculated aby dividing the compressive force F to the cross sectional area A:

fc= F/A

If the stress is wanted in Pascals (Pa), F and A must be in Newtons and square meters respectively.

For acylindrical component the cross sectional area A is:

A=πR^

If the diameter of the component is 81 mm, the radius is the half:

R=81mm /2 = 40.5 mm

Then A result:

A= 3.14 * (40.5 mm)^2  = 5150.4 mm^2

In square meters:

A= 3.14 * (0.0405 m)^2  = 0.005150 m^2

Replacing 708 kN to the force:

fc= 708 kN / 5150.4 mm^2 = 0.137 kN/mm^2

Using the force in Newtons:

F= 70800 N

Finally the compressive stress in Pa is:

fc= 708000 / 0.005150 m^2 = 137465051 Pa = 137 MPa

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Consider a steam turbine, with inflow at 500oC and 7.9 MPa. The machine has a total-to-static efficiency ofηts=0.91, and the pre
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Answer: \dot m_{in} = 23.942 \frac{kg}{s}, \dot H_{out} = 39632.62 kW

Explanation:

Since there is no information related to volume flow to and from turbine, let is assume that volume flow at inlet equals to \dot V = 1 \frac{m^{3}}{s}. Turbine is a steady-flow system modelled by using Principle of Mass Conservation and First Law of Thermodynamics:

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- \dot W_{out} + \eta\cdot (\dot m_{in} \dot h_{in} - \dot m_{out} \dot h_{out}) = 0

This 2 x 2 System can be reduced into one equation as follows:

-\dot W_{out} + \eta \cdot \dot m \cdot ( h_{in}- h_{out})=0

The water goes to the turbine as Superheated steam and goes out as saturated vapor or a liquid-vapor mix. Specific volume and specific enthalpy at inflow are required to determine specific enthalpy at outflow and mass flow rate, respectively. Property tables are a practical form to get information:

Inflow (Superheated Steam)

\nu_{in} = 0.041767 \frac{m^{3}}{kg} \\h_{in} = 3399.5 \frac{kJ}{kg}

The mass flow rate can be calculated by using this expression:

\dot m_{in} =\frac{\dot V_{in}}{\nu_{in}}

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The enthalpy rate at outflow is:

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A glass plate is subjected to a tensile stress of 40 MPa. If the specific surface energy is 0.3 J/m2 and the modulus of elastici
Alika [10]

Answer:

The maximum length of a surface flaw is 8.24 μm

8.24 μm

Explanation:

Given that:

The modulus of elasticity E = 69 GPa

The specific surface energy \delta_s = 0.3 J/m²

The length of the surface flaw "a" = ??

From the theory of the brittle fracture;

\sigma _c = \bigg (  \dfrac{2E \delta_s}{\pi a}  \bigg )^{1/2}

Making a the subject of the formula; we have:

a = \bigg (  \dfrac{2 \times E \times \delta_s}{\pi \sigma _c ^2}  \bigg )

a= \bigg (  \dfrac{2 \times 69*10^9 \times 0.3}{\pi (40*10^6)^2}  \bigg )

a = 8.24 × 10⁻⁶ m

a = 8.24 μm

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