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Svetradugi [14.3K]
4 years ago
6

A load P is applied horizontally while the other end is fixed to a structure. A load P is applied horizontally while the other e

nd is fixed to a structure. How many bolts are required to make the assembly safe? Use 8 mm bolts (M8-1.25) that yield at 0.5 MPa. (You must provide an answer before moving to the next part.)
Engineering
1 answer:
sveta [45]4 years ago
8 0

Answer:

The number of bolt is 4.

Explanation:

Given that,

\tau = 0.5\ MPa

distance = 8 mm

Suppose, Load P = 205 N

We need to calculate the shear area of bolt

Using formula of shear stress

\tau=\dfrac{P}{A}

Here is double shear.

So, \tau will be double.

A=\dfrac{P}{2\tau}

Put the value into the formula

A=\dfrac{205}{2\times0.5\times10^{6}}

A=205\times10^{-6}\ m^2

We need to calculate the area of one bolt

Using formula of area

A'=\dfrac{\pi}{4}d^2

Put the value into the formula

A'=\dfrac{\pi}{4}\times(8\times10^{-3})^2

A'=50.26\times10^{-6}\ m^2

We need to calculate the number of bolt

Using formula of number of bolt

n=\dfrac{A}{A'}

Put the value into the formula

n=\dfrac{205\times10^{-6}}{50.26\times10^{-6}}

n=4.0

Hence, The number of bolt is 4.

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A charge of 11.748 nC is uniformly distributed along the x-axis from −2 m to 2 m . What is the electric potential (relative to z
svp [43]

Answer:

electric potential  =  22.36 volt

Explanation:

given data

charge Q =  11.748 nC

distance d = 5 - 2 = 3 m

length = 2 + 2 = 4 m

Coulomb constant = 8.98755 × 109 N·m²/C ²

solution

electric potential is express as

electric potential  = \frac{Q}{4\pi \epsilon  _o L} ln(1+\frac{L}{d})   ..............1

electric potential  = \frac{KQ}{L} ln(1+\frac{L}{d})  

put here value

electric potential  = \frac{8.98755\times 10^9\times 11.748\times 10{-9}}{4} ln(1+\frac{4}{3})  

electric potential  =  22.36 volt

6 0
3 years ago
A piston–cylinder device contains 15 kg of saturated refrigerant-134a vapor at 280kPa. A resistance heater inside the cylinder w
attashe74 [19]

Answer:

18.77 A

Explanation:

To solve this we use the energy balance equation, that is:

E_{in}-E_{out}=\Delta E_{sys}\\\\Q_{in}+W_{in}+W_{out}=\Delta U\\\\Q_{in}+W_{in}=\Delta U\\\\But\ W_{in}=Voltage(V)*Current(I)*change\ in\ time(\Delta t)=VI\Delta t,\Delta U=m(h_2-h_1)\\\\Given\ that\ m=15kg,V=110\ V,\Delta t=6\ min = (6*60\ s)=360\ s,Q_{in}=20.67\ kJ/kg*15\ kg=310.05\ kJ=310050\ J\\\\From\ table: At\ P_1=280kPa, h_1=249.71\ kJ/kg=249710\ J/kg;At\ P_2=280kPa \ and\  T_1=75^oC,P_2=319.95\ kJ/kg=319950\ J/kg\\\\Substituting:\\\\310500+(110*360*I)=15(319950-249710)\\\\

39600I=1053600-310500\\\\39600I=743100\\\\I=18.77\ A

3 0
3 years ago
What are the four basic parts of process plan
Rus_ich [418]

Answer:

1. Goal setting

2. Developing the planning premises

3. Reviewing Limitations

4. Deciding the planning period

5. Formulation of policies and strategies

6. Preparing operating plans

7. Integration of plans

Explanation:

I don't have an explanation but i thru in a few extra answers for like extra credit or something if your teacher does that

8 0
3 years ago
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
Strike441 [17]

Answer:

the elongation of the metal alloy is 21.998 mm

Explanation:

Given the data in the question;

K = σT/ (εT)ⁿ

given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,

strain-hardening exponent n = 0.22

we substitute

K = 345 / 0.02^{0.22

K = 815.8165 Mpa

next, we determine the true strain

(εT) = (σT/ K)^1/n

given that σT = 412 MPa

we substitute

(εT) = (412 / 815.8165 )^(1/0.22)

(εT) = 0.04481 mm

Now, we calculate the instantaneous length

l_i = l_0e^{ET

given that l_0 = 480 mm

we substitute

l_i =480mm × e^{0.04481

l_i =  501.998 mm

Now we find the elongation;

Elongation = l_i - l_0

we substitute

Elongation = 501.998 mm - 480 mm

Elongation = 21.998 mm

Therefore, the elongation of the metal alloy is 21.998 mm

6 0
3 years ago
Write a program that randomly chooses between three different colors for displaying text on the screen. Use a loop to display tw
11Alexandr11 [23.1K]

Answer:

INCLUDE Irvine32.inc

.data

msgIntro  byte "This is Your Name's fourth assembly extra credit program. Will randomly",0dh,0ah

         byte "choose between three different colors for displaying twenty lines of text,",0dh,0ah

         byte "each with a randomly chosen color. The color probabilities are as follows:",0dh,0ah

         byte "White=30%,Blue=10%,Green=60%.",0dh,0ah,0

msgOutput byte "Text printed with one of 3 randomly chosen colors",0

.code

main PROC

;

//Intro Message

       mov edx,OFFSET msgIntro  ;intro message into edx

       call WriteString         ;display msgIntro

       call Crlf                ;endl

       call WaitMsg             ;pause message

       call Clrscr              ;clear screen

       call Randomize           ;seed the random number generator

       mov edx, OFFSET msgOutput;line of text

       mov ecx, 20              ;counter (lines of text)

       L1:;//(Loop - Display Text 20 Times)

       call setRanColor         ;calls random color procedure

       call SetTextColor        ;calls the SetTextColor from library

       call WriteString         ;display line of text

       call Crlf                ;endl

       loop L1

exit

main ENDP

;--

setRanColor PROC

;

; Selects a color with the following probabilities:

; White = 30%, Blue = 10%, Green = 60%.

; Receives: nothing

; Returns: EAX = color chosen

;--

       mov eax, 10              ;range of random numbers (0-9)

       call RandomRange         ;EAX = Random Number

       .IF eax >= 4          ;if number is 4-9 (60%)

       mov eax, green           ;set text green

       .ELSEIF eax == 3         ;if number is 3 (10%)

       mov eax, blue            ;set text blue

       .ELSE                    ;number is 0-2 (30%)

       mov eax, white           ;set text white

       .ENDIF                   ;end statement

       ret

setRanColor ENDP

6 0
3 years ago
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