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Snezhnost [94]
3 years ago
9

Two steel plates of 15 mm thickness each are clamped together with an M14 x 2 hexagonal head bolt, a nut, and one 14R metric pla

in washer under the nut.
a) What is a suitable length for the bolt, rounded up to the nearest 5 mm?
b) What is the bolt stiffness?
Engineering
1 answer:
Blababa [14]3 years ago
7 0

Answer:

a) 50 mm

b) 808.24 MN/m

Explanation:

Given:

Thickness of each steel plate = 15mm

a) To find suitable length of bolt, we'll use:

Length of bolt = grip length + height of nut

To find the grip length since there is a washer, we'll use:

Grip length = plate thickness + washer thickness

Since we have 2 plates of 15mm thickness,

Plate thickness = 15 + 15 = 30mm

Using the table, a metric plain washer has a thickness of 3.5mm

Grip length = 30 + 3.5 =33.5 mm

Height of nut: Using table A-31, height of hexagonal nut is 12.8 mm

Therefore,

Length of bolt = 33.5 + 12.8 = 46.3

Rounde up to the nearest 5mm, we'll get 50mm

Length of bolt = 50mm

b) Bolt stiffness:

Threaded length for L ≤ 125mm

LT = 2*d + 6

Where d = 14mm, from table(8-7)

= 2*14 + 6

= 34 mm

Area of unthreaded portion:

Ad= \frac{\pi}{4} d^2 = \frac{pi}{4} * 14^2 = 153.94 mm^2

Length of unthreaded portion in grip:

Ld = 50 - 34 = 16mm

Length of threaded portion in grip:

Lt = 33.5 - 16 = 17.5mm

Bolt stiffness = \frac{A_d A_t E}{A_d l_t + A_t l_d}

= \frac{153.94 * 115 * 207}{(153.94*17.5)+(115*16)} = 808.24

Bolt stiffness = 808.24 MN/m

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Stone has been used as a building material since ancient times because of its compressive strength, which is the __________.
Doss [256]

Complete Question:

Stone has been used as a building material since ancient times because of its compressive strength, which is the?

Group of answer choices

A. ability to support pressure without breaking.

B. ability to press down and become solidly fixed in the ground.

C. relative weight of a block of stone to its size.

D. force a block exerts on the blocks around it.

Answer:

A. ability to support pressure without breaking.

Explanation:

Stone has been used as a building material since ancient times because of its compressive strength, which is the ability to support pressure without breaking.

Compressive strength can be defined as the ability of a structural element or particular material to withstand an applied, which is aimed at reducing the size of the structural element.

Simply stated, it is the ability of a structural element or material to withstand an applied load without deflections, fracture or having any crack.

In this context, a stone possesses the ability to resist or withstand compression loads.

Some examples of other materials or structural elements having good compressive strength are steel, bones, concrete etc. The standard unit of measurement of the compressive strength of a material is Mega Pascal (MPa) or pound-force per square inch (psi) in the United States of America.

7 0
4 years ago
Assign numMatches with the number of elements in userValues that equal matchValue. userValues has NUM_VALS elements. Ex: If user
erica [24]

Answer:

import java.util.Scanner;

public class FindMatchValue {

  public static void main (String [] args) {

     Scanner scnr = new Scanner(System.in);

     final int NUM_VALS = 4;

     int[] userValues = new int[NUM_VALS];

     int i;

     int matchValue;

     int numMatches = -99; // Assign numMatches with 0 before your for loop

     matchValue = scnr.nextInt();

     for (i = 0; i < userValues.length; ++i) {

        userValues[i] = scnr.nextInt();

     }

     /* Your solution goes here */

         numMatches = 0;

     for (i = 0; i < userValues.length; ++i) {

        if(userValues[i] == matchValue) {

                       numMatches++;

                }

     }

     System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);

  }

}

6 0
3 years ago
At a construction site, there are constant arguments and conflicts amongst workers of different contractors and sub-contractors.
jeyben [28]

Interpersonal skill is required by the construction manager to resolve the arguments and conflicts among workers and between the contractors.

Answer: Option(d)

<u>Explanation:</u>

  • Interpersonal skills are a set of behavioral action that involves the person to interact freely with other peoples effectively.
  • A person before interacting with others should pay attention, listen to their words and give responses to them after they finish with their arguments.
  • In the above answer, the construction manager has to develop interpersonal skills to resolve the issues, conflicts, and arguments raised by his employees.
  • The manager must listen to their needs and demands possessed and should take necessary action in resolving the issues.
  • For that, the manager has to interact freely and effectively with their employees in resolving the issues.
6 0
4 years ago
Calculate the time taken to completely empty aswimming pool 15
user100 [1]

Answer:

Time needed to empty the pool is 401.35 seconds.

Explanation:

The exit velocity of the water from the orifice is obtained from the Torricelli's law as

V_{exit}=\sqrt{2gh}

where

'h' is the head under which the flow of water occurs

Thus the theoretical discharge through the orifice equals

Q_{th}=A_{orifice}\times \sqrt{2gh}

Now we know that

C_{d}=\frac{Q_{act}}{Q_{th}}

Thus using this relation we obtain

Q_{act}=C_{d}\times A_{orifice}\times \sqrt{2gh}

Now we know by definition of discharge

Q_{act}=\frac{d}{dt}(volume)=\frac{d(lbh)}{dt}=Lb\cdot \frac{dh}{dt}

Using the above relations we obtain

Lb\times \frac{dh}{dt}=AC_{d}\times \sqrt{2gh}\\\\\frac{dh}{\sqrt{h}}=\frac{AC_{d}}{Lb}\times \sqrt{2g}dt\\\\\int_{1.5}^{0}\frac{dh}{\sqrt{h}}=\int_{0}^{t}\frac{0.62\times 0.3}{15\times 9}\times \sqrt{2\times 9.81}\cdot dt\\\\

The limits are put that at time t = 0 height in pool = 1.5 m and at time 't' the height in pool = 0

Solving for 't' we get

\sqrt{6}=6.103\times 10^{-3}\times t\\\\\therefore t=\frac{\sqrt{6}}{6.103\times 10^{3}}=401.35seconds.

4 0
3 years ago
3 cm of water evaporated from a 200-hectare vertical walled reservoir during 24 hours. Storm water was added to the reservoir at
ycow [4]

Answer:

water released through the bottom of the reservoir in 24 hrs is 1992 ha-cm

Explanation:

Given the data in the question;

A = 200 hectare =  2 × 10⁶ m²

water evaporated Ve = 2 × 10⁶ m² × 3 × 10⁻² = 60000 m³ { in 24 hrs }

Water added by storm in 24hrs Vi = 3 × 24 × 3600 = 259200 m³

now let water released be Vr

ΔV = V_ini - V_final = 0

Vi - Ve - Vr = 0

Vr = Vi - Ve

Vr = 259200 m³ - 60000 m³

Vr = 199200 m³ = 19920000 m² - cm

Vr = 1992 ha-cm

Therefore, water released through the bottom of the reservoir in 24 hrs is 1992 ha-cm

7 0
3 years ago
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