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Lilit [14]
3 years ago
7

Problem 1. A. Using singularity functions method, find out the equations for q, v, and M. B. Using appropriate boundary conditio

ns, find out the constants. C. Draw shear force and bending moment diagrams.

Engineering
1 answer:
PIT_PIT [208]3 years ago
7 0

Answer:

Hello the required diagram for the question  is missing attached is the diagram AND A FREE BODY DIAGRAM

B)Rav = 3.375 KN  Rbv = 4.625 KN

Explanation:

Attached is the detailed diagram of the bending moment diagrams and shear force

A) using singularity functions

y(x) = Displacement shape of Elastic curve

dy/dx = ∅ (x) ( slope )

d^2x/dy^2 = m(x) / E I  ( moment function )

d^3x / dy^3 = I / EI V(x) ( shear function )

d^4x / dy^4 = I / EI q(x) ( loading function )

B) using the appropriate boundary conditions find out the constants

summation of F(x)  = 0

Rah = 0

Rav - 8 + Rbv = 0

Rav + Rbv = 8 ----- equation 1

summation of M(a) = 0

= -8(4) + Rbv (8) - 5 = 0

therefore Rbv = 37 / 8 = 4.625 KN

insert the value of Rbv back into equation 1

Rav = 8 - 4.635 = 3.375 KN

C) ATTACHED ARE THE DIAGRAMS

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nirvana33 [79]

Answer:

a)  the temperature to which the pin must be cooled for assembly is T_2 = -101.89^ \ ^0}C

b) the radial pressure at room temperature after assembly is P_f = 62.8 \ MPa

c) the  safety factor in the resulting assembly = 6.4

Explanation:

Coefficient of thermal expansion \alpha = 12.3*10^{-6} \  ^0 C

Yield strength \sigma_y = 400 MPa

Modulus of elasticity (E) = 209 GPa

Room Temperature T_1 = 20°C

outer diameter of the collar D_o = 95 \ mm

inner diameter of the collarD_i = 60 \ mm

pin diameter D_p = 60.03 \ mm

Clearance c = 0.06 mm

a)

The temperature to which the pin must be cooled for assembly can be calculated by using the formula:

(D_i - c )-D_p = \alpha * D_p(T_2-T_1)

(60-0.06)-60.03=12.3*10^{-6}*60.03(T_{2}-20^{0}C)

-0.09 = 7.38369*10^{-4}(T_{2}-20^{0}C)

-0.09 = 7.38369*10^{-4}T_2  \ \ - \ \ 0.01476738

-0.09 +  0.01476738 = 7.38369*10^{-4}T_2

−0.07523262 =7.38369*10^{-4}T_2

T_2 = \frac{-0.07523262}{7.38369*10^{-4}}

T_2 = -101.89^ \ ^0}C

b)

To determine the radial pressure at room temperature after assembly ;we have:

P_f = \frac{E * (D_p-D_i)(D_o^2-D_1^2)}{D_i*D_o} \\ \\ \\  P_f = \frac{209*10^9* 0.03(95^2-60^2)}{60*95^2}  \\ \\ P_f = 62815789.47 \ Pa \\ \\ P_f = 62.8 \ MPa

c)  the safety factor of the resulting assembly is calculated as:

safety factor =  \frac{Yield \ strength }{walking \ stress}

safety factor =  \frac{400}{62.8}

safety factor = 6.4

Thus, the  safety factor in the resulting assembly = 6.4

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3 years ago
The net potential energy EN between two adjacent ions, is sometimes represented by the expression
Anastaziya [24]

Answer:

as answered in the attached file.

Explanation:

The detailed steps, derivation and appropriate differentiation is as shown in the attachment

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3 years ago
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Determine the maximum volume in gallons​ [gal] of olive oil that can be stored in a closed cylindrical silo with a diameter of 3
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Answer:

V=1601gal

Explanation:

Hello! This problem is solved as follows,

First we must raise the equation that defines the pressure at the bottom of the tank with the purpose of finding the height that olive oil reaches.

This is given as the sum due to the atmospheric pressure (1atm = 101.325kPa), and the pressure due to the weight of the olive oil, taking into account the above, the following equation is inferred.

P=Poil+Patm

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α=density of water=1000kg/m^3

g=gravity=9.81m/s^2

h= height that olive oil reaches

solving

P=Poil+Patm

P=Patm+0.86αgh

h=\frac{P-Patm}{0.86\alpha g } =\frac{179214.28-101325}{(0.86)(1000)(9.81)} \\h=9.23m[/tex]

Now we can use the equation that defines the volume of a cylinder.

V=V=\frac{\pi }{4} D^{2} h

D=3ft=0.9144m

h=9.23m

solving

V=\frac{\pi }{4} (0.9144)^{2} 9.23=6.06m^3

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Investigating how slime molds reproduce is an example of applied research.<br> True<br> False
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Answer:

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Explanation:

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