The load on a bolt consists of an axial pull of 10 KN together with a transverse shear force of 5 KN. Find the diameter of bolt required according to I. Maximum principal stress theory; 2. Maximum shear stress theory; 3. Maximum principal strain theory, 4. Maximum strain energy theory, and 5 Maximum distortion energy theory. Take permissible tensile stress at elastic limit = 100 MPa and poisson's ratio = 0.3​

Answers

Answer 1

Answer:

hey. its a big question. solved from *c hegg

Explanation:

The Load On A Bolt Consists Of An Axial Pull Of 10 KN Together With A Transverse Shear Force Of 5 KN.
The Load On A Bolt Consists Of An Axial Pull Of 10 KN Together With A Transverse Shear Force Of 5 KN.
The Load On A Bolt Consists Of An Axial Pull Of 10 KN Together With A Transverse Shear Force Of 5 KN.
The Load On A Bolt Consists Of An Axial Pull Of 10 KN Together With A Transverse Shear Force Of 5 KN.
The Load On A Bolt Consists Of An Axial Pull Of 10 KN Together With A Transverse Shear Force Of 5 KN.

Related Questions

Your organization recently purchased 20 Android tablets for use by the organization's management team. To increase the security of these devices, you want to ensure that only specific apps can be installed. Which of the following would you implement?
A. Credential Manager.
B. App whitelisting.
C. App blacklisting.
D. Application Control.

Answers

c! hope this helps i got it right
Answer : Application whitelisting

Explanation : an application whitelisting is the security approach used by organisations and administrators to secure the organisation devices and system. The application whitelisting works in such a way that, case administrator powers to restrict users or employees from using any malicious or any other application which is not allowed to be used in the organisation . the administrator will use the application whitelisting to only allowed those applications for the employees to access which the administrator wants to . Therefore the apps not listed in the The whitelist of applications are not allowed to be used in the devices provided by the organisation . Hence , to increase the security of the 20 Android tablets purchase by the organisation "Application whitelisting" will be the best approach to do so , by allowing only those apps which are allowed by the organisation to be installed and worked upon in those tablets . Above provided question is answered and explained feel free to ask any questions in the comments section below.

A structure is designed using 4 circular columns. Due to a quirky design, the four columns will all carry different loads of 1800 N, 2100 N, 2275 N, and 2200 N. A factor of safety of 5 is used to design the columns. The diameter of each of the columns is supposed to be 50 cm, at most. Determine the maximum height of the structure (i.e. the column height) so that the structure will not fail. Assume that all columns may be modeled as Euler columns for your analysis. Assume a pinned-pinned boundary condition for your analysis, and assume the elastic modulus of the column material is 10 MPa.

Answers

Answer:

5.16 M

Explanation:

Loads ; 1800N, 2100N, 2275N, 2200N

safety factor = 5

diameter of each column = 50 cm = 0.5 m

Elastic modulus = 10 MPa

Calculate the max height of structure

moment of inertia for a circular section ( I ) = πd^4 / 64

lets represent the required maximum height of the column as L

Applying Euler column theory

The bucker load of the column =  ( attached below )

attached below is the remaining solution

Carbon dioxide at a temperature of 0oC and a pressure of 600 kPa (abs) flows through a horizontal 40-mm- diameter pipe with an average velocity of 2 m/s. Determine the friction factor if the pressure drop is 235 N/m2 per 10-m length of pipe.

Answers

Answer:

f = 0.04042

Explanation:

temperature = 0°C = 273k

p = 600 Kpa

d = 40 millemeter

e = 10 m

change in  P = 235 N/m²

μ = 2m/s

R = 188.9 Nm/kgk

we solve this using this formula;

P = ρcos*R*T

we put in the values into this equation

600x10³ = ρcos * 188.9 * 273

600000 = ρcos51569.7

ρcos = 600000/51569.7

=11.63

from here we find the head loss due to friction

Δp/pg = feμ²/2D

235/11.63 = f*10*4/2*40x10⁻³

20.21 = 40f/0.08

20.21*0.08 = 40f

1.6168 = 40f

divide through by 40

f = 0.04042

Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping station is 140 m and that of the exit point is 150 m. The required terminal head is 10 m. Estimate the pipe diameter and pumping head using the explicit design procedure g

Answers

Answer:

[tex]D=0.41m[/tex]

Explanation:

From the question we are told that:

Discharge rate [tex]V_r=0.35 m3/s[/tex]

Distance [tex]d=4km[/tex]

Elevation of the pumping station [tex]h_p= 140 m[/tex]

Elevation of the Exit point [tex]h_e= 150 m[/tex]

Generally the Steady Flow Energy Equation SFEE is mathematically given by

[tex]h_p=h_e+h[/tex]

With

[tex]P_1-P_2[/tex]

And

[tex]V_1=V-2[/tex]

Therefore

[tex]h=140-150[/tex]

[tex]h=10[/tex]

Generally h is give as

[tex]h=\frac{0.5LV^2}{2gD}[/tex]

[tex]h=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]

Therefore

[tex]10=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]

[tex]D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}[/tex]

[tex]D=0.41m[/tex]

how to solve circuit theory using mesh analysis

Answers

Explanation:

Find a minimal set of cycles that covers all vertices and edges of the circuit graph. For each cycle, define a "mesh" current, and write the Kirchhoff's Voltage Law (KVL) equation with respect to each of the edges in the cycle. Where an edge is part of more than one cycle, all current(s) defined for the edge will contribute to the voltage there.

This will give as many equations as there are mesh currents. Solve the resulting system of equations. The (signed) sum of the mesh currents through any edge is the current in that circuit branch.

__

Example

Consider the attached circuit. It shows mesh currents I1, I2, and I3 in graph cycles with those numbers. The KVL equations are ...

  mesh 1: I1(R3 +R2 +R1) -I2·R1 -I3·R2 = Vi (the voltage across the current source)

  mesh 2: -I1·R1 +I2(r1 +1/(sC)) -I3(1/(sC)) = Vs

  mesh 3: -I1·R2 -I2(1/(sC)) +I3(R2 +sL +1/(sC)) = 0

You will note that the matrix of equation coefficients is symmetric.

__

In this example, you will end with I1 as a function of Vi. If I1 is a given source value, that relation can be used to find Vi.

what is geo technical

Answers

Geotechnical engineering and engineering geology are a branch of civil engineering

Question 1. If a fiber weight 3.0 g and composite specimen weighing 4.g. The composite specimen weighs 2.0 g in water. If the specific gravity of the fiber and matrix is 2.4 and 1.3, respectively, find the 1. Theoretical density of composite 2. Experimental density 3. Void fraction

Answers

Answer:

Explanation:

From the given information:

weight of fiber [tex]w_f[/tex] = 3.0 g

weight of composite specimen [tex]w_c[/tex] = 4.0 g

specimen composite weight in water [tex]C_{wm}[/tex] = 2.0 g

specific gravity of fiber [tex]S_f[/tex] = 2.4

specific gravity of matrix [tex]S_m[/tex] = 1.3

The weight of the matrix = weight of the composite - the weight of fiber

⇒ (4.0 - 3.0) g

= 1.0 g

The theoretical density of the composite [tex]\rho_{ct}[/tex] can be determined by using the formula:

[tex]\dfrac{1}{\rho_{ct}} = \dfrac{w_f}{w_cS_f}+ \dfrac{w_m}{w_cS_m}[/tex]

[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{(4.0 \times 2.4)}+ \dfrac{1.0}{(4.0\times 1.3)}[/tex]

[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{9.6}+ \dfrac{1.0}{5.2}[/tex]

[tex]\dfrac{1}{\rho_{ct}} =0.505\\[/tex]

[tex]\rho_{ct} =\dfrac{1}{0.505}[/tex]

[tex]\mathbf{\rho_{ct} = 1.980 \ g/cm^3}[/tex]

The experimental density [tex]\rho _{ce}[/tex] is determined  by using the equation:

[tex]\rho _{ce} = \dfrac{w_f + w_c}{\dfrac{w_f }{S_f} + \dfrac{w_c }{S_m} }[/tex]

[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{\dfrac{3.0 }{2.4} + \dfrac{4.0 }{1.3} }[/tex]

[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{1.250 +3.077 }[/tex]

[tex]\mathbf{\rho _{ce} = 1.620 \ g/cm^3}[/tex]

The void fraction is: [tex]= \dfrac{\rho_{ct}-\rho_{ce}}{\rho_{ct}}[/tex]

[tex]= \dfrac{1.980-1.620}{1.980}[/tex]

= 0.1818

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