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Kipish [7]
3 years ago
12

How many sets of equations (V and M equations) would you need to describe shear and moment as functions of x for this beam? In o

ther words, how my segments of the beam need to be analysed separately? There is a uniform distributed load (w )on the left and a concentrated force P applied to the beam
Engineering
1 answer:
den301095 [7]3 years ago
5 0

Shear and moment as functions of x is described below .

Explanation:

1. Beam is the slender bar that carries transverse

loading; that is, the applied force are perpendicular to the bar.

2. In a beam, the internal force system consist of a shear force and

a bending moment acting on the cross section of the bar.

3. The  shear force and the bending moment usually vary continuously

along the length of the beam.

4. The internal forces give rise to two kinds of stresses on a

transverse section of a beam:

(1) normal stress that is caused by

bending moment and

(2) shear stress due to the shear force.

Knowing the distribution of the shear force and the bending

moment in a beam is essential for the computation of stresses

and deformations.

Shear- Moment Equations

The determination of the internal force system acting at a given

section of a beam : draw a free-body diagram that expose these

forces and then compute the forces using equilibrium equations.

The goal of the beam analysis -determine the shear force  V and  the bending moment  M at every cross section of the beam.

To derive the expressions for  V and M in terms of the distance x

measured along the beam. By plotting these expressions to scale,

obtain the shear force and bending moment diagrams for the

beam.

The shear force and bending moment diagrams are convenient

visual references to the internal forces in a beam; in particular,  they identify the maximum values of  V and  M

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Answer:

The power developed by engine is 167.55 KW

Explanation:

Given that

V_d=2.5\times 10^{-3} m^3

Mean effective pressure = 6.4 bar

Speed = 2000 rpm

We know that power is the work done per second.

So

P=6.4\times 100\times 2.5\times 10^{-3}\times \dfrac{2\pi \times2000}{120}

We have to notice one point that we divide by 120 instead of 60, because it is a 4 cylinder engine.

P=167.55 KW

So the power developed by engine is 167.55 KW

4 0
3 years ago
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A type of shoot in which continuous lighting used is: 1) studio.

<h3>What is a photoshoot?</h3>

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ANSWER:

Detail drawing
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Air flows from a large reservoir in which the pressure and temperature are 1 MPa and 30°C, respectively, through a convergent–di
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A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the othe
vodomira [7]

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V_f = V_g = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v_f = 0.001057 m³/kg

v_g = 1.0037 m³/kg

u_f = 486.82 kJ/kg

u_g 2524.5 kJ/kg

h_g = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V_f/v_f + V_g/v_g

we substitute

m₁ = 0.002/0.001057  + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v_g

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m_{out = m₁ - m₂

m_{out = 1.89414  - 0.003985

m_{out = 1.890155 kg

so, Initial internal energy will be;

U₁ = m_fu_f + m_gu_g

U₁ = (V_f/v_f)u_f  + (V_g/v_g)u_g

we substitute

U₁ = (0.002/0.001057)(486.82)  + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E_{in -  E_{out = ΔE_{sys

QΔt - m_{outh_{out = m₂u₂ - U₁

QΔt - m_{outh_g = m₂u_g - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

5 0
3 years ago
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