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Kipish [7]
2 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]2 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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Explanation:

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2 years ago
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QUESTÃO 13. Explique o uso das aspas no trecho "Darei a cada uma de vocês
lesya [120]

Answer: speaks Portuguese

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6 0
3 years ago
A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.
Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = \frac{PL}{AE}    ................1

here δ is change in length and P is applied load  and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = \frac{PL}{AE}  

δ = \frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}  

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm

7 0
3 years ago
Use phasor techniques to determine the impedance seen by the source given that R = 4 Ω, C = 12 μF, L = 6 mH and ω = 2000 rad/sec
Zielflug [23.3K]

Answer:

Z = 29.938Ω ∠22.04°

I = 2.494A

Explanation:

Impedance Z is defined as the total opposition to the flow of current in an AC circuit. In an R-L-C AC circuit, Impedance is expressed as shown:

Z² = R²+(Xl-Xc)²

Z = √R²+(Xl-Xc)²

R is the resistance = 4Ω

Xl is the inductive reactance = ωL

Xc is the capacitive reactance =

1/ωc

Given C = 12 μF, L = 6 mH and ω = 2000 rad/sec

Xl = 2000×6×10^-3

Xl = 12Ω

Xc = 1/2000×12×10^-6

Xc = 1/24000×10^-6

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I =V/Z

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7 0
3 years ago
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