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pickupchik [31]
3 years ago
15

Two parts are to be assembled in a way that if one part fails, the entire assembly fails. Each of the parts have undergone exten

sive individual analysis and you are provided with statistics on the stresses that are expected. Specifically, you are provided with the mean and standard deviation of the stress, and the yield stress of the material for each part. Failure of a part will be defined as a part experiencing stress exceeding the yield stress. What is the probability that the assembly fails? (Hint: Use Table A-10 in the textbook to find the probability of failure for each part.)
Part #1: σnom = 260 MP a, U (σ) = 25 MP a, σyield = 310 MPa

Part #2: σnom = 475 MP a, U (σ) = 5 MP a, σyield = 490 MPa

Engineering
1 answer:
e-lub [12.9K]3 years ago
5 0

Answer:

attached below

Explanation:

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Write short notes on: (any four) a) Suspended ground floor b) Soil exploration c) Baulking of sand d) Bearing capacity of soil e
vredina [299]

Answer:

a) A suspended floor is a ground floor with a void underneath the structure. The floor can be formed in various ways, using timber joists, precast concrete panels, block and beam system or cast in-situ with reinforced concrete. However, the floor structure is supported by external and internal walls.

b) Soil exploration consists of determining the profile of the natural soil deposits at the site, taking the soil samples and determining the engineering properties of soils using laboratory tests as well as in-situ testing methods

c) Bulking in sand Occurs When dry sand interacts with the atmospheric moisture. Presence of moisture content forms a thin layer around sand particles. This layer generates the force which makes particles to move aside to each other. This results in the increase of the volume of sand.

d) In a nutshell, bearing capacity is the capacity of soil to support the loads that are applied to the ground above. It depends primarily on the type of soil, its shear strength and its density. It also depends on the depth of embedment of the load – the deeper it is founded, the greater the bearing capacity.

Explanation:

<h2>please follow me</h2>
6 0
3 years ago
Calculate the osmotic pressure of seawater containing 3.5 wt % NaCl at 25 °C . If reverse osmosis is applied to treat seawater,
AlladinOne [14]

Answer:

Highest osmotic pressure that membrane may experience is

' =58.638 atm

Explanation:

Suppose sea-water taken is M= 1 kg

Density of water = 1000 kg/m3

Therefore Volume of water= Mass,M/Density of water

V= 1 kg/(1000 kg/m3)

V= 10-3 m3= 1 Litre

Since mass of Nacl is 3.5 wt%,Therefore in 1 kg of water

Mass present of NaCl= m= 0.035*1000 g

m= 35 g

Since molecular weight of NaCl= 58.44 g/mol =M.W.

Thus its Number of moles of Nacl= m/M.W

nNaCl= 35g/58.44 gmol-1

= 0.5989 mol

ans since volume of solution is 1 L thus concentration of NaCl is ,C= number of moles/Volume of solution in Litres

C= 0.5989mol/ 1L

=0.5989 M

Since 1 mol NaCL disssociates to form 2 moles of ions of Na+ andCl- Thus van't hoff factor i=2

And osmotic pressure  = iCRT ------------------------------(1)( Where R= 0.0821 L.atm/mol.K and T= 25oC= 298.15 K)

Putting in equation 1 ,we get  = 2*(0.5989 mol/L)*(0.0821 L.atm/mol.K)*298.15 K

=29.319 atm

Now as the water gets filtered out of the membrane,the water's volume decreases and concentration C of NacL increases, thus osmotic pressure also increases.Thus, at 50% water been already filtered out, the osmotic pressure at the membrane will be maximum

Thus Volume of water left after 50% is filtered out as fresh water= 0.5 L (assuming no salt passes through semi permeable membrane)

Thus New concentration of NaCl C'= 2*C

C'=2*0.5989 M

=1.1978 M

and Since Osmotic pressure is directly proportional to concentration, Thus As concentration C doubles to C', Osmotic Pressure  ' also doubles from  ,

Thus,Highest osmotic pressure that membrane may experience is,  '=2*  

=2*29.319 atm

' =58.638 atm

3 0
3 years ago
Air enters the compressor of a simple gas turbine at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s. The compressor press
Zepler [3.9K]

Answer:

a) 3581.15067 kw

b) 95.4%

Explanation:

<u>Given data:</u>

compressor efficiency = 85%

compressor pressure ratio = 10

Air enters at:    flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K

At turbine inlet : pressure = 950 kPa, temperature = 1400k

Turbine efficiency = 88% , exit pressure of turbine = 100 kPa

A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW

attached below is a detailed solution to the given question

6 0
3 years ago
Any change in the system from one equilibrium state to another is called: A) Path B) Process C) Cycle D) None of the above
dexar [7]

Answer:

B) Process

Explanation:

In thermodynamics a process is a passage of a thermodynamic system from an initial to a final state of thermodynamic equilibrium.

A thermodynamic process path is the series of states through which a system passes from an initial to a final state.

Cycle is a process in which initial and final state are identical.

7 0
3 years ago
(40 points) Program the following sorting algorithms: InsertionSort, MergeSort, and QuickSort. There are 9 test les uploaded for
babunello [35]

Answer:

Explanation:

MERGE SORT

#include<stdlib.h>

#include<stdio.h>

#include<string.h>

void merge(int arr[], int l, int m, int r)

{

int i, j, k;

int n1 = m - l + 1;

int n2 = r - m;

 

int L[n1], R[n2];

for (i = 0; i < n1; i++)

L[i] = arr[l + i];

for (j = 0; j < n2; j++)

R[j] = arr[m + 1+ j];

i = 0;

j = 0;

k = l;

while (i < n1 && j < n2)

{

if (L[i] <= R[j])

{

arr[k] = L[i];

i++;

}

else

{

arr[k] = R[j];

j++;

}

k++;

}

while (i < n1)

{

arr[k] = L[i];

i++;

k++;

}

while (j < n2)

{

arr[k] = R[j];

j++;

k++;

}

}

void mergeSort(int arr[], int l, int r)

{

if (l < r)

{

int m = l+(r-l)/2;

mergeSort(arr, l, m);

mergeSort(arr, m+1, r);

merge(arr, l, m, r);

}

}

void printArray(int A[], int size)

{

int i;

for (i=0; i < size; i++)

printf("%d ", A[i]);

printf("\n");

}

int main()

{

int arr[1000] = {0};

int arr_size =0;

int data;

char file1[20];

strcpy(file1,"data.txt");

FILE *fp;

fp = fopen(file1,"r+");

if (fp == NULL) // if file not opened return error

{

perror("Unable to open file");

return -1;

}

else

{

fscanf (fp, "%d", &data);    

arr[arr_size]=data;

arr_size++;

while (!feof (fp))

{  

fscanf (fp, "%d", &data);  

arr[arr_size]=data;

arr_size++;    

}

}

printf("Given array is \n");

printArray(arr, arr_size);

mergeSort(arr, 0, arr_size - 1);

printf("\nSorted array Using MERGE SORT is \n");

printArray(arr, arr_size);

return 0;

}

3 0
3 years ago
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