Calculate the areas under the stress-strain curve (toughness) for the materials shown in Fig. below, (a) plot them as a

. The flexure strength test was performed on a concrete beam having a cross section of 0.15m by 0.15m and a span of 0.45m. If th

ivann1987 [24]

Answer:

σ =5.39Mpa

Explanation:

step one:

The flexure strength is defined as the tendency with which unreinforced concrete yield to bending forces

Flexural strength test Flexural strength is calculated using the equation:

σ = FL/ (bd^2 )----------1

Where

σ = Flexural strength of concrete in Mpa

F= Failure load (in N).

L= Effective span of the beam

b= Breadth of the beam

step two:

Given data

F=40.45 kN= 40450N

b=0.15m

d=0.15m

L=0.45m

step three:

substituting into the expression we have

σ = 40450*0.45/ (0.15*0.15^2 )

σ =18202.5/ (0.15*0.15^2 )

σ =18202.5/ (0.15*0.0225 )

σ =18202.5/0.003375

σ =5393333.3

σ =5393333.3/1000000

σ =5.39Mpa

Therefore the flexure strength of the concrete is 5.39Mpa

5 0

3 years ago

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 M P a m. If the plate is expos

bekas [8.4K]

Answer:

Answer for the question is given in the attachment.

Explanation:

3 0

3 years ago

Read 2 more answers

A completely reversible heat pump produces heat at a rate of 300 kW to warm a house maintained at 24 °C. The exterior air, which

trapecia [35]

Answer:

No.

Explanation:

The Coefficient of Performance of the reversible heat pump is determined by the Carnot's cycle:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

$COP_{HP} = \frac{297.15\,K}{297.15\,K-280.15\,K}$

$COP_{HP} = 3.339$

The power required to make the heat pump working is:

$\dot W = \frac{300\,kW}{3.339}$

$\dot W = 89.847\,kW$

The heat absorbed from the exterior air is:

$\dot Q_{L} = 300\,kW - 89.847\,kW$

$\dot Q_{L} = 210.153\,kW$

According to the Second Law of Thermodynamics, the entropy generation rate in a reversible cycle must be zero. The formula for the heat pump is:

$\frac{\dot Q_{L}}{T_{L}} - \frac{\dot Q_{H}}{T_{H}} + \dot S_{gen} = 0$

$\dot S_{gen} = \frac{\dot Q_{H}}{T_{H}} - \frac{\dot Q_{L}}{T_{L}}$

$\dot S_{gen} = \frac{300\,kW}{297.15\,K}-\frac{210.153\,kW}{280.15\,K}$

$\dot S_{gen} = 0.259\,\frac{kW}{K}$

Which contradicts the reversibility criterion according to the Second Law of Thermodynamics.

5 0

3 years ago

A fluid with a relative density of 0.9 flows in a pipe which is 12 m long and lies at an angle of 60° to the horizontal At the t

Minchanka [31]

Answer:

$Q=7.3\times 10^{-3} m^3/s$

Explanation:

Given that

At top $d_2=30 mm,P_2=860 KPa ,P_1=1000 KPa,d_1=85 mm$

$\rho =900\dfrac{Kg}{m^3}$

We know that

$\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2$

$A_1V_1=A_2V_2$

$\frac{V_1}{V_2}=\left(\dfrac{d_2}{d_1}\right)^2$

$\frac{V_1}{V_2}=\left(\dfrac{30}{85}\right)^2$

$V_2=8.02V_1$

$Z_2=12 sin60^{\circ}$

$\dfrac{1000\times 1000}{900\times 9.81}+\dfrac{V_1^2}{2\times 9.81}+0=\dfrac{860\times 1000}{900\times 9.81 }+\dfrac{V_2^2}{2\times 9.81}+12 sin60^{\circ}$

So $V_1=1.30$ m/s

We know that flow rate Q=AV

$Q=A_1V_1$

By putting the values

$A_1=\dfrac{\pi}{4}d^2$

$Q=7.3\times 10^{-3} m^3/s$

To find the flow rate we do not need the direction of flow,because we are just doing balancing of energy at inlet and at the exits of pipe.

4 0

3 years ago

Describe the basic mechanism by which positive photolithography transfers a pattern, and comment on the chemical interactions of

ankoles [38]

Photo-lithography is the method of giving geometric shapes on a mask to the surface of a silicon wafer.

<u>Explanation:</u>

The fabrication of an integrated circuit (IC) requires a variety of physical and chemical processes conducted on a semiconductor (e.g., silicon) substrate. In common, the numerous methods used to create an IC fall into three divisions: film deposition, patterning, and semiconductor doping.

Films from both conductors (such as polysilicon, aluminum, and extended recently copper) and nonconductors (various forms of silicon dioxide, silicon nitride, and others) are utilized to combine and separate transistors and their parts.

Selective doping of different regions of silicon permits the conductivity of the silicon to be altered with the application of voltage. By building structures of these various parts millions of transistors can be assembled and wired together to form the complex circuitry of a modern microelectronic device.

Fundamental to all of these methods is lithography, i.e., the development of three-dimensional relief images on the substrate for subsequent transfer of the model to the substrate.

6 0

3 years ago